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1700	https://chatlyai.app/blog/understanding-interval-notation	Understanding Interval Notation, with Definition, Examples, and Types Use CasesPricingBlogNews Contact Support Launch App Blog / Writing Understanding Interval Notation, with Definition, Examples, and Types Written by Muhammad Bin Habib Mon Sep 08 2025 Confused by brackets and parentheses? Practice interval notation with AI Chat. Ask AI Chat by Chatly Understanding Interval Notation, with Definition, Examples, and Types What is Interval Notation?Why Use Interval Notation?Symbols in Interval Notation Types of Intervals Open Interval Closed Interval Half-Open or Half-Closed Interval Unbounded Intervals Singleton Intervals How to Write Interval Notation Step by Step Step 1: Identify the Boundaries Step 2: Decide on Inclusion or Exclusion Step 3: Write the Interval in Order Step 4: Handle Infinite Ranges Step 5: Combine Intervals if Needed Examples of Interval Notation Example 1: Simple Inequality Example 2: Mixed Boundaries Example 3: Negative Infinity Example 4: Union of Two Intervals Example 5: Intersection of Overlapping Intervals Example 6: Real-World Application Common Mistakes to Avoid in Interval Notation Using Brackets with Infinity Reversing the Order of Numbers Forgetting Boundaries Overcomplicating Simple Ranges Ignoring Consistency with Inequalities Practice Problems with Interval Notation Interval Notation Problem 1 Interval Notation Problem 2 Interval Notation Problem 3 Interval Notation Problem 4 Interval Notation Problem 5 Interval Notation Problem 6 How Chatly Can Help With Interval Notation Conclusion Suggested Reads Understanding Interval Notation, with Definition, Examples, and Types Inequalities often feel clumsy when written out in full sentences in specific cases. Interval notation fixes that problem by compressing a range of numbers into a short and precise form. Mathematicians, teachers, and students rely on it because it looks clean and communicates boundaries without confusion. Everyday problems in algebra or calculus turn simpler once interval notations are understood. Instead of writing “x is greater than 3 but less than or equal to 7,” the same idea appears as (3, 7]. One glance, and the reader knows exactly where the values start, where they end, and which boundaries are included. This guide covers everything you need: the definition of interval notation, the symbols used, different types of intervals, step-by-step writing, and examples of interval notation. By the end, you will not only recognize it but also use it confidently in practice. What is Interval Notation? Interval notation is a mathematical way of describing a set of numbers that lie between two endpoints on the real number line. Instead of writing long inequalities, this method condenses the information into brackets or parentheses. It is a standardized shorthand that saves space and avoids misinterpretation. For example, the inequality 2 ≤ x < 6 is written in interval notation as [2, 6). The square bracket shows that 2 is included in the set, and the parenthesis shows that 6 is excluded. This mix of brackets and parentheses signals inclusion or exclusion clearly, which makes interval notations much easier to scan than words. Intervals can represent finite ranges or extend endlessly toward positive or negative infinity. A common interval notation example is (−∞, 0), which represents all negative numbers. Another is [5, ∞), showing numbers greater than or equal to 5. These examples of interval notation show how a simple system can describe both limited and unlimited ranges with precision. In mathematics, this system becomes essential in algebra, precalculus, and calculus. It helps describe solution sets, domains, and ranges in a way that can be quickly understood by anyone familiar with the notation. Why Use Interval Notation? Mathematics thrives on clarity. Interval notation delivers that clarity by turning wordy inequalities into sharp, compact expressions. Teachers prefer it in algebra because it saves space and makes grading easier. Students appreciate it because long descriptions shrink into a line that fits neatly onto a number line. Intervals written this way allow quick communication of ideas. Instead of stumbling through “x is greater than or equal to 1 and less than 4,” the interval appears simply as [1, 4). This example of interval notation strips away extra words while keeping every detail intact. Another reason interval notations are used so widely is their role in advanced mathematics. Domains of functions, solution sets in calculus, and probability ranges often stretch to infinity. Writing those without interval notation would feel messy. With it, the concept is reduced to a compact format such as (−∞, ∞), representing all real numbers in one clean symbol. In practice, interval notation makes problem solving faster, reduces the chance of misinterpretation, and aligns with how math is taught globally. Symbols in Interval Notation Every interval notation relies on a handful of symbols. Once mastered, they act like a universal language for describing sets of numbers. Parentheses ( ) indicate that an endpoint is excluded. For example, (2, 8) represents all real numbers between 2 and 8, but not including 2 or 8. Brackets [ ] show inclusion. Writing [2, 8] means both 2 and 8 belong to the set, along with every value in between. Infinity Symbols (∞, −∞) mark unbounded ranges. They always appear with parentheses, never brackets. An interval notation example is (−∞, 5], which includes all numbers less than or equal to 5. Union Symbol ∪ combines intervals. A set like (−∞, 0) ∪ (5, ∞) expresses values less than 0 or greater than 5. Intersection Symbol ∩ identifies overlap between intervals. For instance, [1, 7] ∩ (4, 9) results in (4, 7], the shared portion. These symbols form the foundation of interval notations. Once you know how to read them, the structure of any example becomes clear. Even complex expressions feel manageable when broken into their symbols. Types of Intervals Interval notation is not one-size-fits-all. Different situations require different expressions, and each variation has its own purpose. Understanding these types makes the difference between writing an accurate solution and producing something misleading. Open Interval An open interval excludes both boundaries. It uses parentheses around the endpoints. Writing (2, 7) means every number greater than 2 and less than 7 is included, but not 2 or 7 themselves. This type of interval notation is common when strict inequalities appear in problems. Closed Interval A closed interval includes both endpoints. Square brackets make that clear. [2, 7] represents all numbers between 2 and 7, including both 2 and 7. This example of interval notation is used whenever the values at the edges matter as much as the values inside. Half-Open or Half-Closed Interval Some ranges include one boundary but not the other. Writing [2, 7) means 2 is included, but 7 is excluded. Writing (2, 7] flips it around. These interval notations often show up when solutions require a starting point but not an ending value, or the other way around. Unbounded Intervals Not all intervals stay within neat limits. Some stretch toward infinity. (−∞, 4) captures all numbers less than 4. [5, ∞) represents every number greater than or equal to 5. Infinity is never included, so parentheses always surround it. This type of interval notation is a favorite in calculus, where limits and infinite sets dominate. Singleton Intervals Occasionally, an interval represents just one number. Writing [3, 3] might look strange, but it indicates a set with only the number 3. This example of interval notation is rare, yet useful in precise definitions. How to Write Interval Notation Step by Step Mastering interval notation takes practice, but the process itself is straightforward once broken into clear stages. Every inequality can be translated into an interval, and every interval can be translated back into an inequality. Step 1: Identify the Boundaries Look at the inequality or range given. If the problem states −2 ≤ x < 5, the two boundaries are −2 and 5. Recognizing the limits is the first move. Step 2: Decide on Inclusion or Exclusion Brackets mean the boundary is included. Parentheses mean the boundary is excluded. In the example above, −2 carries a bracket because the symbol is ≤, and 5 carries a parenthesis because the symbol is <. Step 3: Write the Interval in Order Always write intervals from the smaller number to the larger. So [−2, 5) is the correct form. Writing (5, −2] would be meaningless. This is one of the most common mistakes beginners make with interval notations. Step 4: Handle Infinite Ranges Numbers can extend forever, but infinity is never part of the set. That means ∞ and −∞ always carry parentheses. An interval notation example: x ≥ 3 becomes [3, ∞). Similarly, x < −1 becomes (−∞, −1). Step 5: Combine Intervals if Needed When a solution has more than one range, use the union symbol ∪. For instance, x ≤ −1 or x > 4 is expressed as (−∞, −1] ∪ (4, ∞). These examples of interval notation show how complex ranges get simplified into one clean line. Examples of Interval Notation Concepts stick best when applied. Seeing multiple cases of inequalities turned into interval notations makes the method easier to internalize. Each interval notation example below shows how boundaries and symbols work together. Example 1: Simple Inequality x > 3 This means all numbers greater than 3. The boundary at 3 is excluded, so the interval is written as (3, ∞). Example 2: Mixed Boundaries −2 ≤ x < 6 The left side includes −2 because of the ≤ sign. The right side excludes 6 because of the < sign. The interval notation is [−2, 6). Example 3: Negative Infinity x ≤ −1 Numbers less than or equal to −1 stretch to negative infinity. That becomes (−∞, −1]. Notice infinity always gets parentheses. Example 4: Union of Two Intervals x < −2 or x ≥ 4 Two separate ranges are present. Together, they become (−∞, −2) ∪ [4, ∞). These types of examples of interval notation often appear in quadratic inequalities. Example 5: Intersection of Overlapping Intervals Numbers in [1, 7] and (4, 9) The overlap between these two sets is (4, 7]. This shows how intersection reduces the range to only what both sets share. Example 6: Real-World Application A temperature range between 20°C and 30°C, including 20 but not 30, would be [20, 30). Writing it in interval notation communicates the exact limits in a clean, mathematical way. Common Mistakes to Avoid in Interval Notation Even with a clear set of rules, errors happen often when writing interval notations. Small slips can change the meaning completely, which is why knowing what to avoid matters as much as knowing how to write correctly. Using Brackets with Infinity Infinity is not a number, so it cannot be included in any set. Writing [−∞, 5] or [3, ∞] is incorrect. The correct form is (−∞, 5] or [3, ∞). This mistake shows up regularly among beginners still getting comfortable with interval notation. Reversing the Order of Numbers Intervals must always progress from smaller to larger values. Writing (8, 2) does not make sense. The correct interval notation example would be (2, 8). Misordering destroys clarity and confuses the reader. Forgetting Boundaries Leaving out whether a boundary is open or closed weakens accuracy. For example, writing (2, 7) when the condition is 2 ≤ x ≤ 7 is wrong. The correct interval notation is [2, 7]. These examples of interval notation show how a small symbol carries weight. Overcomplicating Simple Ranges Sometimes writers add unions or unnecessary symbols when a single interval works fine. Instead of writing (−∞, 2) ∪ (2, 5), a simpler and correct interval notation example would just be (−∞, 5). Ignoring Consistency with Inequalities Intervals should always reflect the inequality signs used. A mismatch between ≤ or < and the chosen brackets or parentheses changes the meaning. Careful alignment prevents miscommunication. Practice Problems with Interval Notation Theory alone rarely sticks. Working through problems helps train the eye and makes interval notations second nature. Try these exercises, then check the solutions to see how each range converts into the correct interval notation example. Interval Notation Problem 1 Write the solution set for x ≥ 4 in interval notation. Answer: [4, ∞) Interval Notation Problem 2 Convert the inequality −3 < x ≤ 5 into interval notation. Answer: (−3, 5] Interval Notation Problem 3 Describe the set of all real numbers less than −2 or greater than 6 using interval notation. Answer: (−∞, −2) ∪ (6, ∞) Interval Notation Problem 4 Find the intersection of [0, 10] and (5, 15). Answer: (5, 10] Interval Notation Problem 5 Express the set of numbers between −1 and 7, not including either boundary, in interval notation. Answer: (−1, 7) Interval Notation Problem 6 Show how to represent exactly one number, say 9, in interval notation. Answer: [9, 9] How Chatly Can Help With Interval Notation Students often freeze when asked to convert inequalities into interval notations. The rules feel simple in theory, but one missed bracket or a misused infinity symbol can change everything. Chatly turns that confusion into clarity. AI Chat inside Chatly can walk step by step through any interval notation example. Ask it to explain why (2, 7] is different from [2, 7], and you’ll get a clear breakdown without extra noise. Complex problems, such as unions and intersections, become easier because the explanation comes in plain words with worked examples of interval notation built into the response. Researching background concepts takes time too. AI Search inside Chatly pulls supporting examples of interval notation, definitions, and related math explanations directly into the writing space. Instead of juggling between textbooks and web searches, everything sits in one place. Practice strengthens skill, and Chatly helps generate endless problems with solutions. Whether it’s open intervals, closed intervals, or infinite ranges, you can test yourself until the process feels automatic. For anyone learning algebra or calculus, Chatly becomes more than a helper. It becomes a practice partner for mastering interval notations. Conclusion Interval notation gives mathematics a compact voice. Instead of long sentences filled with inequality symbols, the same range appears in one clear expression. A small set of brackets and parentheses can show everything from finite ranges to infinite stretches. Precision separates good work from errors. Using [ ] instead of ( ) changes which values belong in the set. Examples of interval notation remind us that every symbol carries weight. Once practiced often enough, these notations become second nature in algebra, calculus, and beyond. Suggested Reads How to Write Exponents in Google Docs How to Write a Lease Agreement How to Write a Research Paper How to Write a UI UX Case Study How to Write a Powerful Monologue How to Write a Case Study Learn Symbols, Types, and Examples in One Place Use AI Chat to break down complex inequalities into clear interval notations. Explore examples of interval notation, test yourself with practice problems, and refine skills faster. Start Practicing with AI Chat Experience Chatly Now Frequently Asked Questions Explore the most asked questions related to interval notations. How do you write interval notations? What do () and [] mean in interval notation? What are the types of intervals? How to convert inequality to interval notation? What is an example of an interval number? What is an example of an interval notation number line? What does the ∪ symbol mean in interval notation? 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1701	https://www.sciencedirect.com/topics/veterinary-science-and-veterinary-medicine/pyramidal-tract	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Clinical Evaluation of the Nervous System 2012, Principles of Neurological Surgery (Third Edition)Gerald A. Grant, Richard G. Ellenbogen Pyramidal Tract The pyramidal tract begins in the motor strip of the cortex and courses downward through the brain and into the spinal cord. In the hemispheres it is called the coronal radiata and then becomes the internal capsule, cerebral peduncle, and pyramidal tract, which crosses at the medulla–spinal cord junction, and finally in the spinal cord becomes the corticospinal tract. Functionally, a lesion anywhere along this tract can produce the same long tract signs. Signs of pyramidal tract dysfunction include spasticity, weakness, slowing of rapid alternating movements, hyperreflexia, and a Babinski sign.8 Muscle tone is examined by manipulating the major joints and determining the degree of resistance. Spasticity is one type of increased tone (resistance of a relaxed limb to flexion and extension). Muscle strength is commonly graded from 0 to 5 using the grading system shown in Table 2.2. Acute lesions anywhere along the pyramidal tract may also produce flaccid hemiparesis, at least initially, with spasticity developing later. If the whole area of cortex supplying a limb is damaged, the extrapyramidal pathways may be unable to take over and an acute global flaccid weakness of the limb can occur. Intraoperative monitoring has been used to mitigate injury to the corticospinal tract.3 Pyramidal tract lesions typically produce weakness of an arm and leg, or face and arm, or all three together.9 Facial weakness may manifest with a slight flattening of the nasolabial fold; however, the forehead will not be weak (frontalis muscle) because the muscles on each side of the forehead have dual innervation by both cerebral hemispheres (corticopontine fibers). The less affected muscles are the antigravity muscles (wrist flexors, biceps, gluteus maximus, quadriceps, and gastrocnemius). Specific tests of grouped muscle strength can also be quite useful (Table 2.3): pronator drift (arms outstretched with the palms up), standing on each foot, hopping on one foot, walking on toes (gastrocnemius), walking on heels (tibialis anterior), and deep knee bend (proximal hip muscles). Typically, pyramidal lesions often cause rapid alternating movements to become slowed but accuracy is preserved. This is in contrast to cerebellar lesions (see later discussion), which can result in fast but inaccurate, sloppy movements. Reflexes can also be quite important in detecting subtle pyramidal tract lesions, especially if asymmetrical. Reflexes are graded by a numerical system: 0 indicates an absent reflex, trace describes a reflex that is palpable but not visible, 1+ is hypoactive but present, 2+ is normal, 3+ is hyperactive, 4+ implies unsustained clonus, and 5+ is sustained clonus. Clonus is a series of rhythmic involuntary muscle contractions induced by sudden stretching of a spastic muscle such as at the ankle. The cutaneous reflex (abdominal twitch obtained when you gently stroke someone’s abdomen) and the cremasteric reflex (L1, L2 innervation; retraction of the testicle upward with a brush along the inner thigh) may also be lost in pyramidal tract lesions. The abdominal cutaneous reflexes in the upper quadrant of the abdomen are mediated by segments T8 and T9; the lower by T10 to T12. If, for example, the lower abdominal reflexes are absent but the upper are preserved, the lesion may be between T9 and L1. The Hoffmann reflex is reflective of hyperreflexia and spasticity on that side and suggests pyramidal tract involvement. It is elicited by snapping the distal phalanx of the middle finger; a pathological response consists of thumb flexion. The Babinski reflex is the best-known sign of disturbed pyramidal tract function. The Babinski reflex is an important sign of upper motor neuron disease, but should not be confused with a more delayed voluntary knee and toe withdrawal due to oversensitive soles of the feet.10 The Babinski reflex is sought by stroking the lateral border of the sole of the foot, beginning at the heel and moving toward the toes. The stimulus should be firm but not painful. The abnormal response, referred to as the Babinski sign, consists of immediate dorsiflexion of the big toe and subsequent separation (fanning) of the other toes. The Babinski sign is present in infancy but usually disappears at about 10 months of age (range 6-12 months). When planar responses produce equivocal results, a related reflex may be tested by stroking the lateral aspect of the dorsum of the foot, and is known as the Chaddock sign. In general, the more spasticity is present, the more likely the pyramidal tract lesion is in the spinal cord, especially if the spasticity is bilateral.11 Conversely, it is unusual for a pyramidal tract lesion in the spinal cord to produce a hemiparesis or monoparesis. A hemiparesis that involves the face places the lesion somewhere above the facial nucleus, although if the hemiparesis spares the face, the lesion need not be below the facial nucleus. Mild or more chronic hydrocephalus may also cause impressive pyramidal tract dysfunction in the legs more than in the arm fibers. Bladder axons also become stretched by the dilated ventricles associated with hydrocephalus and cause urinary urgency and incontinence. Finally, it should be remembered that the spinal cord terminates normally at the level of the L1-L2 vertebral body, and therefore, neurologically L5 is anatomically in the lower thoracic region. View chapterExplore book Read full chapter URL: Book2012, Principles of Neurological Surgery (Third Edition)Gerald A. Grant, Richard G. Ellenbogen Chapter Applied Anatomy of the Brain Arteries 2009, Stroke in Children and Young Adults (Second Edition)William DeMyer Location of the Pyramidal Tract in Relation to the Genu The pyramidal tract appears as a rectangular area of hypointensity in horizontal T1-weighted magnetic resonance images of the posterior limb of the internal capsule by 4 years of age. It is hyperintense in T2-weighted images of children older than 9 years of age and thereafter for life.100 The pyramidal tract also is visualized by tensor diffusion-weighted imaging and during wallerian degeneration.101 Because of the restricted location of the pyramidal tract in the internal capsule, a small lacune can cause a pure contralateral hemiplegia, sparing the somatosensory pathways. Older texts depict the corticobulbar fibers of the pyramidal tract at the genu and the corticospinal fibers just behind. This anterior position holds only for the superior part of the capsule. As the pyramidal tract descends through the internal capsule, the fibers migrate toward the posterior part of the posterior limb before entering the midbrain basis (Fig. 2-30).102,103 A lesion high in the internal capsule at the level of the genu causes a more severe faciolingual hemiparesis than an inferior level lesion. View chapterExplore book Read full chapter URL: Book2009, Stroke in Children and Young Adults (Second Edition)William DeMyer Chapter Neurophysiological structures 2012, Fundamentals of Motor ControlMark L. Latash The pyramidal tract. The corticospinal tract One of the best-known output tracts of the cortex is the pyramidal tract with the long axons of the cortical cells projecting onto different structures participating in the control of movements (Figure 10.15). The human pyramidal tract consists of about one million fibers that originate primarily from cortical neurons in the frontal motor areas (Brodmann areas 4 and 6) with a contribution from neurons in the parietal somatosensory areas. Most of these fibers are myelinated (more than 90%); however, only a fraction of those fibers conduct action potentials at very high velocities (over 50 m/s). The pyramidal tract consists of two major groups of axons. The first group includes the axons of cortical neurons that form the corticospinal tract that goes down to the spinal cord. The second group includes fibers innervating the motor nuclei of the cranial nerves (the corticobulbar tract) that control the muscles of the face and the neck. Most of the fibers (over 80%) of the corticospinal tract cross the midline of the body at the brainstem level (decussation of pyramids) and travel in the contralateral dorsolateral column of the spinal cord to the spinal segments. A smaller number of fibers do not cross the midline and descend within the ventromedial part of the spinal cord. The uncrossed fibers innervate mostly proximal muscles, including muscles that control trunk movements. As a result, an injury to a cortical area related to motor function leads to most pronounced motor impairments in the contralateral extremities. View chapterExplore book Read full chapter URL: Book2012, Fundamentals of Motor ControlMark L. Latash Chapter Projections from the Brain to the Spinal Cord 2009, The Spinal CordCharles Watson, Alan R Harvey Pyramidal tract or corticospinal tract? Because the corticospinal fibers form the medullary pyramids in mammals, it has traditionally been called the ‘pyramidal tract’. The use of this term has the potential to cause some confusion because, by coincidence, the cells of origin are pyramidal neurons in the cerebral cortex. Moreover, it seems that the corticospinal fibers in monotreme mammals do not assemble in a paramedian pyramid as they do in other mammals (Goldby, 1939). A further reason to avoid the term ‘pyramidal tract’ is the unfortunate and now discredited ‘pyramidal/extrapyramidal’ dichotomy that was created in an attempt to explain the organization, and presumed voluntary versus involuntary components, of the different parts of the motor systems (see discussion in Brodal, 1981). For all of these reasons, we prefer to use the term ‘corticospinal tract’. View chapterExplore book Read full chapter URL: Book2009, The Spinal CordCharles Watson, Alan R Harvey Chapter Spinal Tracts – Descending/Motor Pathways 2015, Essential Clinical Anatomy of the Nervous SystemPaul Rea 9.1 Pyramidal tracts The pyramidal tracts are comprised of the corticospinal and corticobulbar tracts. These are called as pyramidal tracts as they crossover at the level of the pyramids in the medulla. They are collections of upper motor neuron fibers which go to the spinal cord (corticospinal) or the brainstem (corticobulbar) and control the motor function of the body. The corticospinal tract is comprised of a ventral and lateral tract. Many of the fibers crossover at the level of the medulla controlling muscles (and therefore movement) of the opposite side of the body. 9.1.1 Ventral and Lateral Corticospinal Tract The following summary points will highlight the key facts in relation to the ventral and lateral corticospinal tracts. This will allow quick and easy access to the key information relevant to these pathways. • : The origin of the corticospinal tracts is within the primary motor, premotor and supplementary motor cortices, as well as the cingulate motor region within the frontal lobe • : Integration of information occurs with relays between the association fibers from other motor areas. These receive information from the prefrontal, temporal and parietal lobes • : The corticospinal tracts then pass into the internal capsule at its posterior limb • : It then passes into the basis pedunculi of the midbrain • : The corticospinal tracts then pass to the pons • : Following breaking up into fasciculi and reassembling, they then create the pyramidal tract at a prominence at the medulla • : Crossover occurs at the caudal end of the medulla • : Also called decussation of the pyramids • : The majority of fibers (approximately 85%) then pass to the dorsal half of the lateral funiculus. These fibers form the lateral corticospinal tract • : The remaining 15% of fibers will then form the ventral corticospinal tract. These fibers pass in the medial portion of the ventral funiculus • : In the spinal cord, most axons of the corticospinal tract terminate in the intermediate gray matter and the ventral horn • : The corticospinal tract influences motor neurons only through interneurons within the spinal gray matter View chapterExplore book Read full chapter URL: Book2015, Essential Clinical Anatomy of the Nervous SystemPaul Rea Chapter Motor System Disorders, Part II: Spinal Cord, Neurodegenerative, and Cerebral Disorders and Treatment 2023, Handbook of Clinical NeurologyIsrael Franco Pyramidal tracts Bilateral pathologic involvement of the pyramidal tract, usually in the spinal cord, results in impaired volitional control of the urethral sphincter; therefore, the urethral sphincter may undergo inappropriate relaxation, or increased sphincter activity may occur during detrusor contraction. This imbalance is termed detrusor-urethral sphincter dyssynergia. View chapterExplore book Read full chapter URL: Handbook2023, Handbook of Clinical NeurologyIsrael Franco Review article State-of-the-Art Diagnostic Methods to Diagnose Equine Spinal Disorders, With Special Reference to Transcranial Magnetic Stimulation and Transcranial Electrical Stimulation 2019, Journal of Equine Veterinary ScienceSanne L. Journée, ... Cathérine Delesalle 2 Neuroanatomy and Function of Spinal Motor Tracts and Neural Network The pyramidal or corticospinal tract is the neurological tract that controls voluntary muscle contractions and modulates proprioception. The cell bodies of the upper motor neurons (UMNs) of the corticospinal tract are localized in the motor cortex of the brain. Their axons transfer efferent messages to the lower motor neurons (LMNs), of which the cell bodies are localized in the gray matter of the spinal cord . The LMNs innervate muscular motor units. In humans, the descending premotor and motor pathways have monosynaptic connections to the LMNs and also produce monosynaptic excitation of motor neurons. Other messages from the cortex are mediated via extrapyramidal pathways like the rubrospinal, reticulospinal, and vestibulospinal tracts and are connected to LMNs in a multisynaptic configuration. Collaterals of the corticospinal tract communicate with extrapyramidal routes at brainstem level like connections with reticular neurons and spinal neurons at different segmental levels. In humans and many primates, the corticospinal tract has a principal role in the control and regulation of motor activity and facilitates the specific capacity to perform skilled movements . By contrast, in phylogenetic older species such as horses that split off earlier from the phylogenetic pedigree than humans, it is believed that the motor activity is likely mainly regulated via the extrapyramidal system with a subordinate contribution of the corticospinal tract. According to some anatomists, the equine pyramidal system is less well developed than the extrapyramidal system [21,22]. In fact, it seems that the pyramidal tracts in horses end most likely at the level of the midcervical region of the spinal cord from where motor neurons are activated via intersynaptic connections in which propriospinal neurons are an important intermediate station. These also receive motor input from the extrapyramidal motor tracts. The brain cortex–hind limb connections appear anatomically more complex. It is unknown to what extent experimental data of smaller animals, such as rodents, cats, rats, apply to horses. It seems obvious that these animals share large neuroanatomical and neurophysiological pathways with humans and primates. The authors have deduced from various studies that TMS and TES stimuli likely are conducted via corticospinal in combination with extrapyramidal tracts [11,19]. However, the pathways that TES impulses follow are due to the interconnection via propriospinal interneurons more difficult to deduce in the thoracic and lumbar parts of the spinal cord than in the front legs and remain speculative. Nevertheless, the response measured in the hind legs (m. tibialis cranialis) appears not more delayed than can be explained by the putative length of the neural pathway. Of vital importance for the motor control of posture, position, and motor function of the body are connections with the proprioceptive and vestibular system including vestibular-spinal tracts, inhibiting connections from the cerebellum and sensory-proprioceptive afferents from the skin, tendons, and muscles. Whether this could result in observable modulation of TES- or TMS-evoked potentials is not known in horses. View article Read full article URL: Journal2019, Journal of Equine Veterinary ScienceSanne L. Journée, ... Cathérine Delesalle Chapter Hindbrain 2012, The Mouse Nervous SystemCharles Watson Group 1 – Tracts Associated with the Motor System As noted above, the hindbrain serves as a corridor used by pathways that connect rostral parts of the brain with the spinal cord. The major descending tracts that are visible in myelin stained sections are the corticospinal (pyramidal) tract and the rubrospinal tract. The corticospinal tract commences in the neocortex and travels in the internal capsule and cerebral peduncle, closely accompanied by the corticopontine tracts. After passing through the basilar pons (where they are called the longitudinal fibers of the pons), the corticospinal tract forms the pyramid of the hindbrain, which is an elongated bulge next to the ventral midline (Figs. 12.14, 12.16–12.18). In the caudal hindbrain, the corticospinal tract turns dorsally and crosses the midline to form the dorsal corticospinal tract of the spinal cord (Fig. 12.19). The rubrospinal tract arises from the red nucleus of the midbrain. The tract crosses the midline in the ventral tegmental decussation of the midbrain (Figs. 12.14, 12.16) and travels caudally just ventral to the spinal trigeminal nucleus (Figs. 12.15–12.19). A tract arising in the superior colliculus, the tectospinal tract, crosses the midline in the dorsal tegmental decussation (Figs. 12.14, 12.16) and travels caudally just ventral to the medial longitudinal fasciculus (Figs. 12.16–12.19). The tectospinal tract is very small compared to the rubrospinal tract, and its size is often exaggerated in textbook drawings. View chapterExplore book Read full chapter URL: Book2012, The Mouse Nervous SystemCharles Watson Chapter Presurgical Tractography Applications 2014, Diffusion MRI (Second Edition)Andreas J. Bartsch, ... György A. Homola Presurgical Tractography 553 23.4.1 : Motor System: The Pyramidal Tract 554 23.4.2 : Dorsal, Ventral, and Vertical Pathways of Speech, Language, and Reading 556 23.4.3 : Sensory, Brainstem, and Spinal Tracts 558 View chapterExplore book Read full chapter URL: Book2014, Diffusion MRI (Second Edition)Andreas J. Bartsch, ... György A. Homola Chapter Clinical signs 2022, Clinical Signs in Humans and Animals Associated with Minerals, Trace Elements, and Rare Earth ElementsMike Davies BVetMed CertVR CertSAO FRCVS Pyramidal signs Pyramidal tract dysfunction signs including a positive Babinski sign (big toe goes up when sole of foot is stimulated), hyperreflexia, slowing of rapid alternating movements, spasticity, weakness. \| Element \| Species \| --- \| \| Bismuth toxicity \| Humans \| View chapterExplore book Read full chapter URL:
1702	https://e-century.us/files/ijcem/10/7/ijcem0049387.pdf	Int J Clin Exp Med 2017;10(7):10918-10927 www.ijcem.com /ISSN:1940-5901/IJCEM0049387 Original Article Laparotomy versus laparoscopy for the elective cystectomy of a benign ovarian tumour during pregnancy: a retrospective cohort study Yi-Xuan Liu1,4, Yang Zhang1,2, Li Wang1,2,3 1Obstetrics and Gynecology Hospital, Fudan University, Shanghai, People’s Republic of China; 2Department of Obstetrics and Gynecology, Shanghai Medical College, Fudan University, Shanghai, People’s Republic of China; 3Shanghai Key Laboratory of Female Reproductive Endocrine Related Diseases, Shanghai, People’s Republic of China; 4Changhai Hospital, Second Military Medical University, Shanghai 200433, People’s Republic of China. Co-first authors. Received December 3, 2016; Accepted March 26, 2017; Epub July 15, 2017; Published July 30, 2017 Abstract: The aim of this manuscript is to compare the pregnancy and surgical outcomes and costs of an elective cystectomy of a benign ovarian tumour during pregnancy treated by laparotomy and laparoscopy. We completed a retrospective cohort study, and the related data was collected and analysed between September 2009 and Septem-ber 2014 in the Obstetric & Gynecology Hospital of Fudan University. A total of 65 pregnant women were included, 30 cases for laparoscopy and 35 cases for laparotomy. The laparoscopy group had higher costs (P < 0.001) and lower postoperative febrile rates (P = 0.005), a shorter postoperative hospital length of stay (P < 0.001) and a shorter antibiotic prophylaxis (P < 0.001). The gestational age at delivery, caesarean delivery rate, preterm delivery rate, length of prophylactic tocolytic therapy, rate of postoperative uterine contractions, newborn weight and rate of neonatal disorders were not significantly different between the 2 groups. Compared with laparotomic surgery for the elective cystectomy of a benign ovarian tumour during pregnancy, laparoscopy had an equal effect on the mothers and foetuses. The costs of laparoscopic surgery were higher, but it is minimally invasive. Further studies are needed to compare the long-term outcomes between the 2 groups. Keywords: benign ovarian tumour, cystectomy, laparoscopy, laparotomy, pregnancy Introduction Ovarian tumour is a common complication dur-ing pregnancy, with a morbidity of 0.19-8.8% [1-3] among pregnant women. The detection rate and the morbidity of ovarian tumours dur-ing pregnancy is growing [4-6] because of the widespread use of high-definition ultrasonogra-phy and the increasing application of assisted reproductive technology. The proportion of ovarian tumours that are benign is 97-99%, and only 1-3% of ovarian tumours are malignant . Surgery for malig-nant ovarian tumour during pregnancy is asso-ciated with serious surgical trauma, with a high-er rate of complications and a higher rate of pregnancy loss, and the radiotherapy and che-motherapy after surgery has a great influence on the growth of the foetus. A majority of women with malignant ovarian tumours who are preg-nant choose to undergo induced abortions for their own safety and then undergo comprehen-sive treatment, including chemotherapy and surgical treatment . Meanwhile, based on the meta-analysis we performed, the rates of malignant cysts were 7.9%, 8.6%, 11%, and 0% in the four studies we included [7-10]. All of our patients accepted surgery when they were sus-pected of having a malignant tumour. Benign ovarian tumours can enlarge into the broader abdominal cavity along with the uterus during pregnancy, which can induce a higher morbidity of ovarian torsion, especially during the first and second trimester, compared to benign tumours in non-pregnant women . A retrospective case-control study by Ginath Surgical treatment of benign ovarian tumour during pregnancy 10919 Int J Clin Exp Med 2017;10(7):10918-10927 reported that the rate of torsion was 56% and 7% in pregnant women and non-preg-nant women, respectively. Multi-cyst ovarian tumours had a higher torsion rate than single-cyst ovarian tumours with a rate of 86% versus 31%. The ischemic necrosis caused by ovarian torsion can release inflammatory cytokines and cause pain, and the uterine contractions caused by the inflammatory cytokines may be one of the reasons that spontaneous abortion occurs during early pregnancy. Once ovarian torsion during pregnancy is detected, emergen-cy operation is needed. Ultrasonography is one of the tools that is used to monitor the changes in ovarian tumours, and surgery is suggested for those pregnant women who have com-plaints and persistent or even enlarged ovarian tumours. Surgery is needed in cases of sus-pected complications or malignancy . Laparotomy is a traditional surgical procedure applied during pregnancy. At the same time, laparoscopy is reported to be widely used among pregnant women with benign ovarian tumour since the 1990s due to its advantages of less surgical trauma, faster recovery after the operation, and a lower rate of complica-tions, including poor wound healing and post-operative intraperitoneal adhesion. However, laparoscopic surgery has its own risks that are different from those of laparotomy and can-not be ignored. First, the lithotomy position (Trendelenburg position) can influence the blood perfusion to the uterus and placenta, which may have a negative influence on the pregnant women and their foetuses. Second, the consistent CO2 pneumoperitoneum pres-sure may increase the pressure on the dia-phragmatic muscle of pregnant women and may increase the likelihood of cardiovascular accident in pregnant women , and some animal studies have confirmed foetal acidos- is with associated tachycardia, hypertension, and hypercapnia during CO2 pneumoperitone-um [14, 15]. However, at the same time, the Trendelenburg position during an operation can increase the venous return. Thus, cardiorespi-ratory circulation can be affected to a certain degree . Case reports and case series provide a majori-ty of the previous research on the safety of laparoscopy, and only few studies have com-pared the effectiveness of laparoscopy and laparotomy; these have included different oper-ation options, such as oophorocystectomy, sal-pingo-ovariectomy, and ovarian torsion reset. Our study aimed to compare the surgical out-comes, pregnancy outcomes and costs be- tween laparotomy and laparoscopy for an elec-tive cystectomy of a benign ovarian tumour dur-ing pregnancy. Materials and methods Subjects All of the patients who underwent laparoscopy or laparotomy for benign ovarian tumours at the Obstetrics and Gynecology Hospital of Fudan University between September 2009 and September 2014 were included. All these patients had complete medical files, and their ovarian tumours were diagnosed by pathology. The data were collected from screening the electronic medical record system and the hospital registration book based on the key-words for the admitting diagnosis. The preg-nant women were split into 2 research co- horts, depending on whether they had laparos-copy or a laparotomy. We excluded cases with emergency operations, non-oophorocystecto-my operations, malignant or borderline ovarian tumours diagnosed by postoperative pathology and elective pregnancy termination in the 2 groups to exclude the effects of acute abdomi-nal pain, ovariectomy and malignant tumour on surgical and pregnancy outcomes. All of the patients signed an informed consent for the surgery, and for the patients in the laparoscopy group, they were informed that laparoscopic surgery might convert to laparotomic surgery according to the specific operation situation. Surgical procedures The preoperative preparation included a com-plete physical examination, blood and biochem-ical examination, pelvic ultrasonography, and electrocardiogram. Specific anamnesis, surgi-cal history, allergy history and marital history were recorded. The pregnant women accept- ed antibiotic prophylaxis of cefotiam (Harbin Pharmaceutical Group Holding Company, Pro- duct batch number: A1606918-2) i.v. during the 24 hours before and after the surgery if they were evaluated as having a high risk of infection. They were all forbidden to eat or drink starting at 22:00 the day before the oper-ation. The laparotomy and laparoscopy surger-Surgical treatment of benign ovarian tumour during pregnancy 10920 Int J Clin Exp Med 2017;10(7):10918-10927 ies were performed by at least 3 skilled sur-geons and 1 experienced anaesthesiologist. A combined spinal-epidural anaesthesia was applied in the laparotomic surgery, and the patients were in the supine position. The inci-sion was made 2 cm above the pubic symphy-sis or at a previous abdominal scar. The length of the incision was 6 cm to 10 cm, depending on the tumour size. General anaesthesia was applied for the laparoscopy surgery, and the patients were placed in lithotomy. A 2.5 cm inci-sion for the first puncture hole was set at the umbilicus or at a position between the umbili-cus and cartilagoensiformis, according to the size of the uterus and the size of the tumour. A 10 mm Trocar was introduced in the abdomi-nal cavity. Then, a celiac lens was placed into the abdominal area through the 10 mm Trocar and CO2 pneumoperitoneum was established, maintaining the abdominal pressure at 10-12 mmHg. Then, 3 other Trocars were introduced into the abdominal cavity under a constant visualization of the tips. The ovary was carefully exposed, avoiding the uterus, and the ovarian tumour was completely separated and taken out. An ovary suture was performed instead of electric coagulation for haemostasis. Postoperative observation and follow-up The pregnant women were observed in the resuscitation room for 1-2 hours, after which they were sent back to the wards and the Bp, P, and R was observed every 0.5 hour by ECG monitors until 8:00 the day after the operation. The foetal heartbeat was observed every 8 hours after the surgery. The first ultrasonogra-phy after the surgery was arranged 3 days after the operation, and some patients received a second ultrasonography the day before leaving the hospital. All of the pregnant women were followed up by telephone after their expected due date. Specific observation index Demographic characteristics: Age, BMI, the gestational week when the surgery occurred (calculated by last menstrual period for those who had regular periods), the average diameter of the tumour (defined by ultrasonography with-in one week before surgery), the rate of the bilateral ovarian tumours and the pathological type of the tumour were all recorded. Surgical outcomes Total surgical time, estimated blood loss, post-operative exhaust time, postoperative urinary retention, postoperative maximum tempera-ture, postoperative length of fever, postopera-tive rate of fever, rate and length of antibiotic prophylaxis, incision length, occurrence of poor wound healing, postoperative length of hospi-talization and costs were recorded. Pregnancy outcomes Foetal death, postoperative uterine contrac-tions, length of prophylactic tocolytic therapy, gestational age at delivery, caesarean delivery rate, newborn weight, preterm delivery rate and neonatal disorders were recorded. Statistical analysis SPSS 18.0 (Chicago, IL, USA) was used for the statistical analyses. A t test or a rank sum test was used for the continuous variables, and a chi square test or Fisher test was used for the frequency data. Two multivariate logistic regres-sion analyses were used to evaluate the asso-ciation between the surgical approach and postoperative fever and the association be- tween the surgical approach and uterine con-traction, controlling for potential confounding factors. P < .05 was considered statistically significant. Results Basic observation We retrospectively searched 77 cases in the first step, and 12 cases were excluded accord-ing to the indicated criteria. A total of 65 preg-nant women were included. They were divided into laparoscopy and laparotomy groups ac- cording to the surgical procedure used, and there were 30 laparoscopy cases (43.15%) and 35 laparotomy cases (53.85%). We obtained 56 cases with live delivery information; 26 cases in the laparoscopy group and 30 cases in the laparotomy group. One case of a deceased foetus of idiopathic aetiology was found in the laparoscopy group. Eight pregnant women were lost to follow-up; 3 in the laparoscopy group and 5 in the laparotomy group. Surgical treatment of benign ovarian tumour during pregnancy 10921 Int J Clin Exp Med 2017;10(7):10918-10927 Main results Demographic characteristics of the research subjects: We found no significant difference in the average age, BMI, gestational weeks at the time of surgery, tumour size before operation (tumour size under ultrasound), or the rate of bilateral ovarian tumours between the groups (Table 1). Pathology of the laparotomy and laparoscopy groups: The most common pathological type of ovarian tumour during pregnancy was mature cystic teratoma (36.9%), followed by endometri-oid cyst, mucinous cystadenoma, luteinizing cyst, and serous cystadenoma. There were no significant differences in the distribution of the pathological types between the 2 groups (P = 0.426) (Table 2). In total, the baseline variables of the research objects of these 2 groups were consistent. Distribution of the pathological types: There was no significant difference in the distribution of the pathological types of the ovarian tumours between the 2 groups, but we noticed that there was a higher proportion of mature cystic teratoma in the laparotomy group, and the lap-aroscopy group had a higher proportion of endometrial cyst (Table 2). The distribution of the pathological types of the ovarian tumours tact capsule and clear boundaries, so it is easi-er to operate on than an endometrial cyst, which is better suited to laparoscopic surgery. With improvements in the pathological diag-nostic rate of ultrasonography in our hospital, we propose a general surgical procedure ac- cording to the type and size of the ovarian tumour to reduce the surgical damage as much as possible. Surgical outcomes and costs The overall costs included anaesthesia, sur-gery, healthcare, medicine, hospital stay and material costs. Compared to laparotomy, the costs associated with laparoscopy were higher (p < 0.001). However, laparoscopy had a shorter length of hospitalization (p < 0.001) and was better accepted by the patients. The rate of antibiotic prophylaxis was lower (p < 0.001), and the length of antibiotic prophylaxis (p = 0.000) was shorter. The postoperative rate of fever (p = 0.005), the postoperative maximum tempera-ture of the patients (p = 0.001) and the postop-erative length of fever (p = 0.004) were all decreased. The estimated blood loss, total sur-gical time, postoperative exhaust time, and postoperative urine retention were not signifi-cantly different between the 2 groups (p = 0.213). There were 4 patients in the laparotomy Table 1. Baseline characteristics Characteristic Laparotomy Laparoscopy P Value Age (years; Mean ± SD) 27.1 ± 3.4 27.2 ± 4.8 .912 BMI (kg/m2; Mean ± SD) 22.4 ± 2.9 21.4 ± 2.8 .173 Gestational week when accepted surgery (weeks) 16.4 ± 3.3 17.7 ± 4.1 .153 Average tumour diameter (m; range) 0.101 (0.06-0.28) 0.0945 (0.04-0.30) .218& Rate of bilateral ovarian tumour (n, %) 9 (25.7%) 3 (10.0%) .104# T-test; &Rank sum test; #Chi square test. Table 2. Pathology of the ovarian tumours Pathology type Laparotomy Laparoscopy Total number P Value Mature cystic teratoma (n, %) 17 (48.6) 7 (23.3) 24 (36.9) .426 Endometrioid cyst (n, %) 6 (17.1) 9 (30.0) 15 (23.1) -Mucinous cystadenoma (n, %) 7 (20.0) 6 (20.0) 13 (20.0) -Luteinizing cysts (n, %) 2 (5.8) 5 (16.7) 7 (10.8) -Serous cystadenoma (n, %) 2 (5.7) 2 (6.7) 4 (6.2) -Other type of benign tumour (n, %) 1 (2.9)$ 1 (3.3)^ 2 (3.1) -$Simple cyst; ^Leiomyoma with cystic degeneration; The pathology of the two groups was not significantly different. was similar to that reported by Koo FH in 2013 . The different pathologi-cal types of ovarian tumours influence the choice of surgi-cal procedure, oper-ation time and esti-mated blood loss. The matured cystic teratoma has an in- Surgical treatment of benign ovarian tumour during pregnancy 10922 Int J Clin Exp Med 2017;10(7):10918-10927 group that showed poor wound healing (Table 3). In the multivariate analysis, there was no sig-nificant difference in the postoperative risk of fever after adjusting for confounding factors, including postoperative hospitalization, usage and length of antibiotic prophylaxis, total surgi-cal time, average tumour diameter and bilater-al ovarian tumour (Table 4). Pregnancy outcomes All cases were followed up after the expected due date. Eight pregnant women (5 laparotomy an delivery rate, newborn weight, length of prophylactic tocolytic therapy or occurrence of postoperative uterine contractions (p > 0.05) (Table 5). In the multivariate analysis, after adjusting for confounding factors, including bilateral ovarian tumour, average tumour diameter, the gesta-tional week when the surgery occurred, occur-rence of postoperative fever and length of pro-phylactic tocolytic therapy, there was also no significant difference in the risk of uterine con-tractions between the laparoscopy and lapa-rotomy groups (Table 6). Table 3. Surgical outcomes and costs between the laparotomy and laparoscopy groups Surgical outcomes and costs Laparotomy Laparoscopy P Value Total surgical time (min; range) 60 (35-140) 53.5 (25-201) .051& Estimated blood loss (ml; range) 50 (10-200) 30 (5-400) .052& Incision length (m) 0.08 (0.05-0.10) 0.025 (0.025-0.03) .000&,@ Postoperative Exhaust time (hours; range) 42.0 (6-67.5) 24.8 (7.5-47) .060& Urine retention (hours; range) 29 (17.5-91.5) 25.8 (17-75.5) .213& Maximum temperature (°C; range) 37.6 (36.8-38.6) 37.4 (36.8-38.3) .001&,@ Length of fever (hours; range) 12 (0-68) 1 (0-39) .004&,@ Rate of fever (n, %) 29 (82.9) 15 (50.0) .005#,@ Rate of antibiotic prophylaxis (n, %) 24 (68.6) 4 (13.3) < .001#,@ Length of antibiotic prophylaxis (days) 3 (0-3) 0 (0-3) .000&,@ Hospitalization (days) 6.14 ± 1.648 3.97 ± 1.326 < .001,@ Poor wound healing (n, %) 4 (13.3) 0 (0.0) .114¥ Costs (yuan) 6971.82 ± 3095.468 10034.98 ± 3382 < .001,@ T-test; &Rank sum test; #Chi square test; ¥Fisher test; @Statistical significance. Number of follow-up used as the overall number. Table 4. Multivariate analysis of the factors associated with the risk of fever Multivariate analysis 95% confidence Variable AdjOR Interval P-value Surgical approach Laparoscopy Versus Laparotomy 0.224 0.037-1.362 0.104 Bilateral ovarian tumour Yes Versus No 3.537 0.342-36.587 0.289 Average tumour diameter 0.935 0.811-1.078 0.354 Total surgical time 1.011 0.986-1.037 0.389 Usage of antibiotic prophylaxis Yes Versus No 0.098 0.002-3.728 0.207 Length of antibiotic prophylaxis 1.872 0.494-7.099 0.356 Postoperative hospitalization 1.277 0.828-1.969 0.269 and 3 laparoscopy; 12.3%) were lost to follow-up because of loss of con-tact. We could not evaluate the Apgar scores of the 2 groups be- cause 31 of the 65 pregnant women delivered in other hospitals, and therefore we could not obtain all of the Apgar data we needed. There was not a statistically significant dif-ference between these 2 groups in the rate of loss to follow-up (P = 0.716). There was only one case of foetal death that occurred in the 40 days after the surgery; this was caused by oligohydramnios. There was no statistically significant differ-ence in the gestational age at deliv-ery, preterm delivery rate, caesare-Surgical treatment of benign ovarian tumour during pregnancy 10923 Int J Clin Exp Med 2017;10(7):10918-10927 Discussion Surgical results Our study showed no significant difference in the total surgical time between these 2 study groups (p = 0.051, 60 (35-140) for laparotomy versus 53.5 (25-201) for laparoscopy). Previous comparative studies [7, 8, 17] found that com-pared to laparotomy during pregnancy, laparos-copy had outstanding surgical outcomes, including a shorter hospitalization, less blood loss during the operation, and less pain after surgery. In our research, the difference in the postoperative length of hospitalization between these 2 groups was statistically significant (p < 0.001, 6.14 ± 1.648 for laparotomy versus 3.97 ± 1.326 for laparoscopy). As for the esti-mated blood loss, the difference was not large, but the amount of blood loss was slightly lower in the laparoscopy group (p = 0.052, 50 (10-infection treatment. One of these patients had no complaints of discomfort but found a sinus in the original caesarean scar that was full of purulent fluid. Laparoscopic surgery might reduce the risk of poor wound healing from a clinical point of view, but further and larger studies are needed. Both the rate and length of antibiotic prophy-laxis in the laparoscopy group were significantly lower than those in the laparotomy group, and at the same time, the postoperative maximum temperature and the postoperative rate and length of fever were also significantly lower in the laparoscopy group in the univariate analy-sis. The postoperative infection risk was some-what lower in the laparoscopy group. Moreover, the minimally invasive surgery and the shorter postoperative hospitalization in the laparosco-py group might also reduce the risk of nosoco-mial infection. Laparoscopy might have more Table 5. Pregnancy outcomes between the laparotomy and laparoscopy groups Pregnancy outcomes Laparotomy Laparoscopy P Value Foetal death (n, %) 0 (0.0) 1 (3.7) .474¥ Postoperative Uterine contractions (n, %) 3 (8.6) 0 (0.0) .243¥ Length of prophylactic tocolytic therapy (days) 7.11 ± 4.739 7.93 ± 5.362 .516 Gestational ageat delivery (weeks) 39.25 ± 1.233 39.22 ± 1.064 .923 Caesarean delivery rate (n, %) 12 (40.0) 8 (30.8) .472& Newborn weight (g) 3393.43 ± 369.281 3202.12 ± 430.442 .079 Premature delivery (< 37 weeks) 1 (3.3) 1 (3.8) > .050¥ Neonatal disorders 0 (0.0) 0 (0.0) -Number of follow-up used as the overall number; Number of live birth deliveries used as the overall number; T-test; &Rank sum test; #Chi square test; ¥Fisher test. Table 6. Multivariate analysis of the factors associated with the risk of uterine contractions Multivariate analysis 95% confidence Variable AdjOR Interval P-value Surgical approach Laparoscopy Versus Laparotomy 0.000 0.000-+∞ 0.998 Bilateral ovarian tumour Yes Versus No 2.374 0.131-43.118 0.559 Average tumour diameter 1.014 0.807-1.274 0.908 Gestational week when accepted surgery 1.084 0.702-1.672 0.717 Postoperative Fever Yes Versus No 0.314 0.011-8.661 0.494 Length of prophylactic tocolytic therapy 1.028 0.746-1.417 0.868 200) for laparotomy versus 30 (5-400) for laparoscopy). The rate of poor wound healing between the 2 study groups was not significantly different, although the incision length was significantly shorter in the laparoscopy group (p < 0.05, 0.08 (0.05-0.10) for laparoto-my versus 0.025 (0.025-0.03) for laparoscopy). A total of 4 patients had poor wound he- aling in our study, and all belonged to the laparotomy group. Three of them had a cracked superficial layer of the wound after the operation, which healed well after anti-Surgical treatment of benign ovarian tumour during pregnancy 10924 Int J Clin Exp Med 2017;10(7):10918-10927 advantages than laparotomy based on the reduced utilization of antibiotics and the lower risk of postoperative infection for pregnant women, which are beneficial for both the moth-ers and foetuses. In the multivariate analysis, we found that surgical approach was not an independent risk factor for postoperative fever after adjusting for several confounding factors. An analysis of the overall costs of these 2 groups showed that the cost of laparoscopy was ¥3000-¥4000 higher than for laparotomy, and the difference was statistically significant (p < 0.001, 6971.82 ± 3095.468 of laparoto-my versus 10034.98 ± 3382 of laparoscopy). Despite the shorter hospitalization duration and less utilization of analgesic drugs, the high-er cost of surgical instruments and procedures made the laparoscopic surgery costs signifi-cantly higher than those of laparotomy. Most of our patients come from economically devel-oped areas, and they prefer the cosmetic advantages and reduced pain after laparoscop-ic surgery to saving money. In summary, laparoscopic surgery during preg-nancy might have a better surgical outcome than laparotomy, according to the present research. Pregnancy outcomes Compared with surgery in non-pregnant wo- men, surgery during pregnancy places more value on the pregnant women and foetal out-comes. Both laparotomy and laparoscopy had unique advantages and disadvantages. Com- pared to laparotomy, laparoscopic surgery avoids the direct stimulation of the uterus and theoretically reduces the risk of uterine con-tractions caused by the operation. However, the influence of carbon dioxide pneumope- ritoneum and abdominal pressure on the pla-cental blood supply can cause foetal acidosis. Previous research has shown that the rate of spontaneous abortion during surgery was not significantly different between laparotomy and laparoscopy. In three studies [7, 9, 10] only 2 of 160 pregnant women (84 receiving laparosco-py, 76 receiving laparotomy) in the laparotomy group had pregnancy loss, and another RCT article reported no pregnancy losses in 69 pregnant women (33 receiving laparoscopy), with no statistically significant difference be- tween the groups. These findings were consis-tent with our research. The gestational age at delivery, caesarean delivery rate, premature delivery, newborn weight, postoperative uterine contractions, and length of the prophylactic tocolytic therapy were not significantly different between the laparotomy and laparoscopy groups, which are consistent with previous multicentre retrospec-tive cohort studies . Based on a previous study, surgical approach, emergency surgery and gestational age at sur-gery might be risk factors for preterm labour . In our study, we ruled out cases of emer-gency surgeries. In the univariate analysis, there was no significant difference in the uter-ine contractions between the laparoscopy and laparotomy groups. In the multivariate analysis, we found that the surgical approach was not a risk factor for uterine contractions. There was one case of an unexplained foetal death in the laparoscopy group in our study, and there was no pregnancy loss in the lapa-rotomy group. We tracked this case as followed: the patient found a left ovarian cyst in the ultra-sound test at the 8th week without any recorded complaints and accepted left-oophorocystecto-my via laparoscopy at week 18.2. The surgery went well and lasted 40 minutes, and the pathology after surgery showed a goiter type of ovarian tumour. No abdominal pain or unusual vaginal bleeding after surgery was mentioned, and the obstetrics examination, thyroid func-tion and ultrasonography test 2 weeks and 4 weeks after the surgery showed no abnormal outcomes. The mother became aware of a disappearance of foetal movement 41 days after the surgery, and oligohydramnios and the absence of the foetal heartbeat were observed through an ultrasonography test 42 days after the surgery. The patient accepted labour induc-tion, and the pathology of the foetus and placenta were normal. The reason for the preg-nancy loss was foetal asphyxia caused by oligo-hydramnios, and there was no relevant proof that the surgery was a cause of the pregnancy loss, because the foetal heartbeat and move-ment 2 weeks after the surgery were normal. Overall, laparoscopic surgery had the same pregnancy outcomes as laparotomic surgery. It remains unclear whether laparoscopy has long-term adverse effects on the foetus or if there was a relationship between the one case of pregnancy loss in the laparoscopy group and the laparoscopic surgery itself. Larger and lon-ger studies are needed. Surgical treatment of benign ovarian tumour during pregnancy 10925 Int J Clin Exp Med 2017;10(7):10918-10927 Laparoscopy management during pregnancy and advice When women find benign ovarian tumours dur-ing pregnancy with complications, should they accept surgery? Benign ovarian tumours during pregnancy without complications are advised to accept the elective operation. The procedure for the surgery is same as that for non-pregnant women. As for the choice of operating time, Wang thought that an operation beyond 7 weeks of pregnancy was safe, but considering the side-effects discussed above, most experts thought that the second trimester was the safest tri-mester for the operation . During the first trimester, 71% of the ovary mass can disappear , and operation during this trimester is associated with a higher risk of abortion. The uterus in the third trimester is too large to perform a pelvic operation, which can cause adverse effects on the pregnancy, such as a higher rate of preterm birth . It is safer to perform the operation in the second trimester [24, 25] because the sensitivity of the uterus is lower, the uterine size leaves adequate space for the surgery, and the progesterone secreted by the well-developed placenta can adequately maintain the pregnancy. An operation on the ovary has less influence on the level of proges-terone, which creates a lower risk of abortion and fewer complications . As for the safety of the anaesthesia, in previous studies, Steinbrook considered that the use of anaesthesia during pregnancy was safe and feasible with the appropriate precautions, including close monitoring and pneumoperito-neum pressure control . The use of local anaesthesia could avoid the failure of tracheal intubation and the risk of inhalation anaesthet-ics, which could theoretically induce foetal mal-formation. However, one of Hong’s studies showed a higher rate of preterm birth associ-ated with laparotomic surgery with local anaes-thesia than with general anaesthesia when treating adnexal tumours during pregnancy. There is no evidence to indicate which of the 2 types of anaesthesia methods is better for surgery during pregnancy. A study of ovarian surgery during pregnancy showed that none of the anaesthesia methods, for example general anaesthesia or spinal anaesthesia, showed a negative impact on the pregnancy outcomes . Thus, the influence of anaesthesia during pregnancy remains controversial. In our study, the anaesthesia for laparoscopy was carried out and monitored by experienced anaesthe-tists, and none of the 65 women mentioned any complications. There was one case report of laparoendoscopic single-site surgery (LESS) for cystectomy of an ovarian tumour during pregnancy. No perio- perative or postoperative complications devel-oped, and the patient was discharged on post-operative day 2 . We believe that more LESS might be applied in the future with the development of this surgical technique and increases in aesthetic requirements. More research should be done to confirm the safety of LESS. Limitations First, our research was a retrospective study with no random assignment. Although the baseline characteristics of the 2 groups were not significantly different, we could only control for some of the common confounding factors, and those that could not be controlled might affect the results. Second, because of the short follow-up time, we used relevant short-term indicators and were unable to analyse and compare the long-term effects. Finally, new-born Apgar scores were missing because only a small proportion of research subjects delivered their babies in our hospital. Conclusion Compared with laparotomic surgery for the elective cystectomy of benign ovarian tumours during pregnancy, laparoscopy had an equal effect on the mothers and the foetuses. The costs of laparoscopic surgery were higher, but it is minimally invasive. Further studies are needed to compare the long-term outcomes between the 2 groups. Acknowledgements The present study was supported by the Shang- hai Advanced Integrated Traditional Chinese and Western Medicine Personnel Training Pro- gram (ZY3-RCPY-4-2004). Disclosure of conflict of interest None. Surgical treatment of benign ovarian tumour during pregnancy 10926 Int J Clin Exp Med 2017;10(7):10918-10927 Authors’ contribution Y-XL contributed to the conception and design of the study, performed the literature search, and extracted the study data, YZ was responsi-ble for the data analysis, contributed to the interpretation of the analyses, and wrote the manuscript. LW was responsible for the overall study planning, contributed to the interpreta-tion of the analyses, and participated in the revision of the manuscript. All of the authors approved the final version of the article. Address correspondence to: Li Wang, Obstetrics and Gynecology Hospital, Fudan University, Shang- hai 200090, People’s Republic of China. 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Peng P, Zhu L, Lang JH, Liu ZF, Sun DW and Leng JH. Clinical analysis of laparoscopic sur-gery for ovarian masses under different condi-Surgical treatment of benign ovarian tumour during pregnancy 10927 Int J Clin Exp Med 2017;10(7):10918-10927 tions during the second trimester. Chin Med J (Engl) 2013; 126: 3325-3328. Condous G, Khalid A, Okaro E and Bourne T. Should we be examining the ovaries in preg-nancy? Prevalence and natural history of ad-nexal pathology detected at first-trimester so-nography. Ultrasound Obstet Gynecol 2004; 24: 62-66. Agarwal N, Parul, Kriplani A, Bhatla N and Gup-ta A. Management and outcome of pregnan-cies complicated with adnexal masses. Arch Gynecol Obstet 2003; 267: 148-152. Schwartz N, Timor-Tritsch IE and Wang E. Ad-nexal masses in pregnancy. Clin Obstet Gyne-col 2009; 52: 570-585. Machado F, Vegas C, Leon J, Perez A, Sanchez R, Parrilla JJ and Abad L. Ovarian cancer during pregnancy: analysis of 15 cases. Gynecol On-col 2007; 105: 446-450. 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1703	https://www.amjtransplant.org/article/S1600-6135(22)14572-7/fulltext	Epstein-Barr Virus: Evasive Maneuvers in the Development of PTLD - American Journal of Transplantation Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Minireview ArticleVolume 7, Issue 2p271-277 February 2007 Open Archive Download Full Issue Download started Ok Epstein-Barr Virus: Evasive Maneuvers in the Development of PTLD Andrew L Snow Andrew L Snow Affiliations Program in Immunology Department of Surgery/Division of Transplantation, Stanford University School of Medicine, Stanford, California, USA Search for articles by this author a,b ∙ Olivia M Martinez Olivia M Martinez Correspondence Corresponding author: Dr. Olivia M. Martinez, omm@stanford.edu omm@stanford.edu Affiliations Program in Immunology Department of Surgery/Division of Transplantation, Stanford University School of Medicine, Stanford, California, USA Search for articles by this author a,bomm@stanford.edu Affiliations & Notes Article Info a Program in Immunology b Department of Surgery/Division of Transplantation, Stanford University School of Medicine, Stanford, California, USA Publication History: Received August 17, 2006; Revised October 9, 2006; Accepted October 23, 2006 DOI: 10.1111/j.1600-6143.2006.01650.x External LinkAlso available on ScienceDirect External Link Copyright: © 2006 American Society of Transplantation & American Society of Transplant Surgeons. Published by Elsevier Inc. All rights reserved. Published by Elsevier Inc. User License: Creative Commons Attribution – NonCommercial – NoDerivs (CC BY-NC-ND 4.0) \| Elsevier's open access license policy Download PDF Download PDF Outline Outline Abstract Key words: 1 Immunological Detente: the Normal EBV Life Cycle 2 Unintended Consequences: EBV in Disease and Malignancy 3 The Balancing Act: EBV and Immune Evasion 4 Applying our Knowledge: Implications for Treatment of PTLD References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract Key words: 1 Immunological Detente: the Normal EBV Life Cycle 2 Unintended Consequences: EBV in Disease and Malignancy 3 The Balancing Act: EBV and Immune Evasion 4 Applying our Knowledge: Implications for Treatment of PTLD References Article metrics Related Articles Abstract Epstein-Barr virus (EBV) infection is linked to ∼90% of B-cell lymphomas associated with posttransplant lymphoproliferative disease (PTLD), a serious complication for immunosuppressed transplant recipients. In this paper, we review the myriad ways by which EBV can inadvertently drive the genesis and persistence of B-cell lymphomas, particularly when the antiviral immune response is compromised. Probing the basic mechanisms by which EBV infection proceeds and contributes to malignancy in such cases will hopefully improve our understanding and treatment of PTLD and other EBV-associated malignancies. Key words: B-cell EBV PTLD B-cell lymphomas associated with posttransplant lymphoproliferative disease (PTLD) continue to pose a serious problem for immunosuppressed transplant recipients. Depending on the organ transplanted, ∼1–20% of transplant recipients will develop such tumors (1 1. Gottschalk, S ∙ Rooney, CM ∙ Heslop, HE Post-transplant lymphoproliferative disorders Annu Rev Med. 2005; 56:29-44 Crossref Scopus (373) PubMed Google Scholar ). Over 90% of all PTLD cases are linked to Epstein-Barr virus (EBV), which plays a prominent role in the development of several types of B-cell malignancy. EBV was first discovered in 1964 by Epstein and colleagues, who identified a unique herpesvirus present in cultured tumor biopsies derived from Burkitt's lymphoma (BL) patients (2 2. Epstein, MA ∙ Henle, G ∙ Achong, BG ... Morphological and biological studies on a virus in cultured lymphoblasts from Burkitt's lymphoma J Exp Med. 1965; 121:761-770 Crossref Scopus (174) PubMed Google Scholar ). As such, EBV became the first tumor virus candidate ever described in humans. Since its initial discovery, we now know that EBV is a B-lymphotropic, gammaherpesvirus that benignly infects over 95% of the human population for life, making it arguably one of the most successful viruses known to man (3 3. Cohen, JI Epstein-Barr virus infection N Engl J Med. 2000; 343:481-492 Crossref Scopus (1411) PubMed Google Scholar ). In recent years, numerous studies have shed light on how EBV persists in the face of a dedicated and robust host cellular immune response. EBV has evolved several immune evasion strategies designed to ensure that the virus coexists benignly with the immune system, but may inadvertently contribute to B-cell oncogenesis, including Burkitt's, Hodgkin's, PTLD and AIDS-related lymphomas. By investigating the nature of benign versus pathogenic EBV infection, basic scientists and clinicians can apply this knowledge to correctly describe the mechanisms underlying PTLD development, and thereby, identify meaningful therapeutic targets. 1 Immunological Detente: the Normal EBV Life Cycle The life cycle of EBV in human B cells reveals how a successful herpesvirus has coevolved with the host immune system to exploit normal processes of B-cell development while maintaining a balance with the T-cell response directed against the virus (4 4. Thorley-Lawson, DA Epstein-Barr virus: Exploiting the immune system Nat Rev Immunol. 2001; 1:75-82 Crossref Scopus (772) PubMed Google Scholar ). EBV establishes a lytic infection of naïve, tonsillar B cells in the follicular mantle, which also serves as the site for the production of infectious viral particles. Although a small number of infected B cells remain permissive for viral replication, EBV exists primarily as a latent infection within the host B cell. Following viral entry via CD21 on the B-cell surface, the EBV genome circularizes within 12–16 h, and the first of three possible latent gene transcription programs is initiated, known as type III latency or the “growth program” (Table 1) (5 5. Kieff, E ∙ Rickinson, AB Epstein-Barr virus and its replication Knipe, DM ∙ Howley, PM (Editors) Fields Virolgy Lippincott Williams and Wilkins, Philadelphia, PA, 2001; 2575-2627 Google Scholar ). Type III latency is characterized by the expression of the entire set of EBV latent genes, including Epstein Barr nuclear antigens (EBNAs) 1, 2, 3A, 3B, 3C, LP, latent membrane protein (LMP1, LMP2A, LMP2B), and the polyadenylated viral RNAs (EBERs 1 and 2). Expression of these genes promotes survival and expansion of infected naïve B-cell hosts that resemble activated B lymphoblasts. The unrestricted growth program also facilitates transformation of resting B cells in vitro and generation of immortalized lymphoblastoid cell lines (LCL). However, the outgrowth of transformed EBV+ B cells is normally controlled in vivo by EBV-specific cytotoxic T lymphocytes (CTL). \| Program \| Infected B cell \| Viral genes expressed \| --- \| I (Latency) \| Memory \| None (EBNA1) \| \| II (Default) \| Germinal center \| EBNA1, LMP1, LMP2A \| \| III (Growth) \| Naive \| EBNA1-EBNA6 LMP1, LMP2A, LMP2B \| Table 1: EBV latent gene transcription programs. EBV utilizes three distinct latent gene transcription programs over the course of its life cycle. Following initial lytic infection, naïve B cells in the tonsil exhibit type III latency (growth program), during which the expression of all nine latent genes promotes proliferation and survival without further differentiation. Once these cells exit the cell cycle, EBNA2 is downregulated and EBV switches to type II latency (default program) as cells migrate into follicles and differentiate into a germinal center cell phenotype. LMP2A and LMP1 are thought to functionally mimic the B-cell receptor and CD40, respectively, providing survival signals typically associated with antigen binding and T-cell help. Cells that survive this stage pass into the memory B-cell compartment, where EBV lies dormant by shutting down all latent genes except EBNA1, which is expressed during division to ensure maintenance of the EBV episome. This type I latent phenotype (latency program) allows EBV to escape CTL surveillance and persist in resting memory B cells for the lifetime of the host. Open table in a new tab From this stage, latently infected B lymphoblasts can undergo differentiation normally associated with a germinal center (GC) reaction. Infection switches to type II latency (‘default program’) by turning off expression of EBNA2, a master transactivator for latent gene transcription that mimics Notch signaling in its ability to block B-cell differentiation (6 6. Gordadze, AV ∙ Peng, R ∙ Tan, J ... Notch1IC partially replaces EBNA2 function in B cells immortalized by Epstein-Barr virus J Virol. 2001; 75:5899-5912 Crossref Scopus (51) Google Scholar , 7 7. Hofelmayr, H ∙ Strobl, LJ ∙ Marschall, G ... Activated Notch1 can transiently substitute for EBNA2 in the maintenance of proliferation of LMP1-expressing immortalized B cells J Virol. 2001; 75:2033-2040 Crossref Scopus (61) PubMed Google Scholar ). During type II latency, LMP1 and LMP2A expression provides the necessary signals required for B-cell differentiation independent of antigen-driven interactions with follicular dendritic cells (FDCs) or T helper (T H) lymphocytes. LMP2A actively suppresses signaling through the B-cell receptor to curtail growth, while simultaneously providing the tonic signals required for B-cell survival by co-opting Syk and Src-family kinases (8 8. Dykstra, ML ∙ Longnecker, R ∙ Pierce, SK Epstein-Barr virus coopts lipid rafts to block the signaling and antigen transport functions of the BCR Immunity. 2001; 14:57-67 Full Text Full Text (PDF) Scopus (141) PubMed Google Scholar ). LMP1 is a functional mimic of CD40, delivering potent survival signals to B cells normally provided through T cell help (CD154 on T H cells) (9 9. Kilger, E ∙ Kieser, A ∙ Baumann, M ... Epstein-Barr virus-mediated B-cell proliferation is dependent upon latent membrane protein 1, which simulates an activated CD40 receptor Embo J. 1998; 17:1700-1709 Crossref Scopus (385) PubMed Google Scholar ). Both viral proteins are constitutively active and signal independently of any ligands, utilizing several transmembrane domains to aggregate in lipid rafts. Hence EBV encodes two mediators for the survival of infected B cells through differentiation into a pool of long-lived memory B cells, where EBV persists for the lifetime of the host. In peripheral memory B cells, no latent genes are expressed (type I latency or ‘latency program’) allowing the host cells to escape immunosurveillance by EBV-specific CTL, explaining the nonpathogenic nature of EBV infection in most healthy carriers. Only the poorly immunogenic EBNA1 is expressed periodically in memory B cells to ensure passage of the viral episome during cell division. Expression of LMP1 and LMP2A has been detected in recirculating tonsillar memory B cells, however, which may assist in ensuring long-term survival of this latently infected memory B-cell pool (10 10. Babcock, GJ ∙ Thorley-Lawson, DA Tonsillar memory B cells, latently infected with Epstein-Barr virus, express the restricted pattern of latent genes previously found only in Epstein-Barr virus-associated tumors Proc Natl Acad Sci USA. 2000; 97:12250-12255 Crossref Scopus (121) PubMed Google Scholar ). The exact mechanism for how EBV switches between different latent programs, however, remains unknown. 2 Unintended Consequences: EBV in Disease and Malignancy Despite the balance struck between EBV persistence and the immune response in healthy individuals, EBV is associated with several diseases in which this detente is disrupted (Table 2). EBV has only been proven to play a definitive causative role in the etiology of two diseases: X-linked lymphoproliferative (XLP) syndrome and infectious mononucleosis (IM). In XLP, affected males are unable to combat proliferation of EBV-infected B cells due to a mutation in the gene encoding SAP (signaling lymphocyte activation molecule [SLAM]-associated molecule), a T-cell surface protein thought to be required for normal T–B cell interactions (11 11. Sayos, J ∙ Wu, C ∙ Morra, M ... The X-linked lymphoproliferative-disease gene product SAP regulates signals induced through the co-receptor SLAM Nature. 1998; 395:462-469 Crossref Scopus (827) PubMed Google Scholar ). IM is a short-lived lymphoproliferative disease associated with primary EBV infection during adolescence. A huge proportion of memory B cells become infected with EBV (∼50%), countered later by an overwhelming proliferation of EBV-specific T cells that can also approach 50% of the patient's peripheral T-cell pool (12 12. Callan, MF ∙ Tan, L ∙ Annels, N ... Direct visualization of antigen-specific CD8+ T cells during the primary immune response to Epstein-Barr virus In vivo J Exp Med. 1998; 187:1395-1402 Crossref Scopus (777) PubMed Google Scholar ). Our understanding of both diseases underscores the importance of a dedicated EBV-specific T-cell response in keeping the virus in check throughout the life of healthy carriers. \| Disease \| EBV latency \| Cellular origin \| --- \| Infectious mononucleosis (IM) \| III, lytic \| Naïve, memory B cells (tonsil) \| \| XLP \| III \| Naïve \| \| Burkitt's lymphoma \| I \| GC or memory B cells \| \| Hodgkin's lymphoma \| II \| GC B cells \| \| PTLD \| III \| Variable \| \| AIDS-associated B cell lymphoma \| III \| Variable \| \| Nasopharyngeal carcinoma (NPC) \| I or II \| Oropharyngeal epithelium \| Table 2: EBV-associated diseases. EBV is associated with several diseases characterized by lymphoproliferation or overt malignancy. Immunosuppression or immunodeficiency can drive XLP-, PTLD- and AIDS-associated B-cell lymphomas, which debilitates the CTL response and allows infected cells to constitutively express all latent genes. Other malignancies may be representative of inappropriate infection (NPC) or infection at the wrong stage of B-cell development (IM), when the target cell cannot exit the cell cycle. Open table in a new tab EBV is also strongly associated with several malignancies. Indeed, EBV was first unveiled in tumor cells derived from patients with BL, a malignancy endemic to equatorial Africa where EBV is associated with 95–100% of all cases (13 13. Kuppers, R B cells under influence: Transformation of B cells by Epstein-Barr virus Nat Rev Immunol. 2003; 3:801-812 Crossref Scopus (432) PubMed Google Scholar ). The hallmark of BL is a reciprocal chromosomal translocation for which the c-myc oncogene is positioned near one of the immunoglobulin (Ig) loci, resulting in dysregulated c-myc expression. Most EBV+ BLs resemble a ‘centroblast’ phenotype (based on Ig hypermutation) and exhibit type I viral latency, where only EBNA1 is expressed. The role of EBV in the development of BL remains unresolved. EBV infection has also been connected to classical Hodgkin's lymphoma, where EBV DNA can be detected in 40–60% of Hodgkin and Reed–Sternberg tumor cells (14 14. Weiss, LM ∙ Movahed, LA ∙ Warnke, RA ... Detection of Epstein-Barr viral genomes in Reed-Sternberg cells of Hodgkin's disease N Engl J Med. 1989; 320:502-506 Crossref Scopus (840) PubMed Google Scholar ). Such tumor cells are thought to arise from ‘crippled’ EBV-infected GC B cells that fail somatic hypermutation, consistent with the expression of type II latency genes (EBNA1, LMP1, LMP2A) in this compartment (15 15. Mancao, C ∙ Altmann, M ∙ Jungnickel, B ... Rescue of “crippled” germinal center B cells from apoptosis by Epstein-Barr virus Blood. 2005; 106:4339-4344 Crossref Scopus (147) PubMed Google Scholar ). EBV infection is also implicated in the pathogenesis of nasopharyngeal carcinomas (NPCs) prevalent in Southeast Asia. Most NPCs present as undifferentiated epithelial carcinomas that usually exhibit a type II latent EBV infection (16 16. Raab-Traub, N Epstein-Barr virus in the pathogenesis of NPC Semin Cancer Biol. 2002; 12:431-441 Crossref Scopus (482) PubMed Google Scholar ). The exact role of EBV in the development of NPC, some gastric carcinomas, and rare cases of EBV+ natural killer (NK) and T cell lymphomas remains elusive. PTLD is characterized by the development of EBV-infected B-cell lymphomas in immunosuppressed transplant recipients (17 17. Klein, G Post-transplantation lymphoproliferative disease Introduction. Springer Semin Immunopathol. 1998; 20:319-323 Crossref Scopus (3) Google Scholar ). PTLD represents a spectrum of disorders ranging from B-cell hyperplasia to monoclonal B-cell lymphoma, over 90% of which are EBV+. PTLD-associated lymphoma cells may derive from B cells other than naïve tonsillar cells that become infected but cannot exit the cell cycle after the growth program is expressed. The immunosuppressive drugs used to prevent graft rejection debilitate EBV-specific T cells and permit the proliferation and lymphomagenesis of these type III infected B cells. Therefore, PTLD-associated B-cell lymphomas display an activated B-cell phenotype and express the entire set of EBV latent genes and closely resemble EBV+ LCL derived in vitro. EBV infection is of particular concern in pediatric transplant recipients, since primary infection of seronegative transplant patients is more likely to lead to PTLD. Other risk factors include age after transplant, severity of immunosuppression and concurrent CMV infection (18 18. Razonable, RR ∙ Paya, CV Herpesvirus infections in transplant recipients: Current challenges in the clinical management of cytomegalovirus and Epstein-Barr virus infections Herpes. 2003; 10:60-65 PubMed Google Scholar ). Currently, there are no laboratory assays that can predict development of PTLD in at-risk patients. Measurement of EBV load is the most commonly utilized laboratory test for diagnostic purposes. Emerging studies indicate that direct quantitation of EBV-specific T-cell responses via ELISPOT or tetramer staining may prove beneficial in monitoring for the onset of EBV disease (1 1. Gottschalk, S ∙ Rooney, CM ∙ Heslop, HE Post-transplant lymphoproliferative disorders Annu Rev Med. 2005; 56:29-44 Crossref Scopus (373) PubMed Google Scholar ). Treatment of PTLD primarily involves the reduction of immunosuppressive therapy to restore the CTL response, with the unfortunate consequence of increasing the risk of graft rejection. Conversely, potent immunosuppression may select for highly proliferative B-cell lymphomas that cannot be brought to regression later by competent EBV-specific CTL. Thus PTLD represents another dangerous consequence of disturbing the balance between EBV infection in B cells and the T-cell response. Not surprisingly, phenotypically identical EBV-associated B lymphoblastoid tumors have been described in AIDS patients as well. It is in the context of studying EBV-associated cancers, particularly those exhibiting type III latency, that we begin to understand how such lymphomas develop and persist when evasive strategies exercised by EBV are left unchecked. 3 The Balancing Act: EBV and Immune Evasion Herpesviruses are particularly adept at evading immune responses, explained in part by their large genomes, which accommodate genes encoding functional homologs of cellular factors involved in cell-cycle regulation, inhibition of apoptosis and signal transduction. These proteins presumably help the virus to survive and replicate in the midst of a vigorous immune response, and may inadvertently contribute to host cell transformation in the process (Figure 1). Figure viewer Figure 1:EBV and immune evasion. Lytic cycle genes BCRF1 (viral IL-10) and BARF1 (sCSF-R) aid in blunting T-cell responses by suppressing antiviral cytokine production. BHRF1, a homolog of Bcl-2, preserves mitochondrial membrane potential and contributes to apoptosis resistance. Latent genes (EBNA1, EBNA2, LMP2A, LMP1) also protect the host B cell from multiple apoptotic stimuli, mediated by p53, Nur77, BCR and DR signals. For example, LMP1-mediated NF-κ B activation upregulates several antiapoptotic genes capable of blocking intrinsic and extrinsic cell death pathways. 3-1 Cytokine modulation One such factor expressed during the lytic phase of EBV infection is BCRF1, a viral homolog of IL-10 (vIL-10) that is highly homologous to its human counterpart despite binding to the IL-10 receptor with reduced affinity. Significant amounts of vIL-10 can be detected in serum of PTLD and IM patients, although the consequences of systemic vIL-10 levels are not known (19 19. Martinez, OM ∙ Villanueva, JC ∙ Lawrence-Miyasaki, L ... Viral and immunologic aspects of Epstein-Barr virus infection in pediatric liver transplant recipients Transplantation. 1995; 59:519-524 Crossref Scopus (50) PubMed Google Scholar , 20 20. Taga, H ∙ Taga, K ∙ Wang, F ... Human and viral interleukin-10 in acute Epstein-Barr virus-induced infectious mononucleosis J Infect Dis. 1995; 171:1347-1350 Crossref Scopus (39) PubMed Google Scholar ). Like human IL-10, vIL-10 can potently suppress Th1 immune responses through the inhibition of IL-12 and IFN-γ in vitro (21 21. Moore, KW ∙ De Waal Malefyt, R ∙ Coffman, RL ... Interleukin-10 and the interleukin-10 receptor Annu Rev Immunol. 2001; 19:683-765 Crossref Scopus (5607) PubMed Google Scholar ). On the other hand, a single amino acid change in the vIL-10 sequence removes its capacity to stimulate certain immune cells, suggesting this tailored cytokine serves as an effective immunosuppressant against the cellular immune response (22 22. Ding, Y ∙ Qin, L ∙ Kotenko, SV ... A single amino acid determines the immunostimulatory activity of interleukin 10 J Exp Med. 2000; 191:213-224 Crossref Scopus (110) PubMed Google Scholar ). It remains unclear whether vIL-10 is necessary for potentiating B-cell transformation. Whether or not vIL-10 directly participates in potentiating B-cell transformation, we propose that local vIL-10 secretion suppresses T-cell responses during the transition from lytic to latent infection until cellular IL-10 is induced following B-cell transformation. Several studies have established that cellular IL-10 is induced in EBV-transformed cells through LMP1 signaling and/or the function of EBV-encoded RNAs. We have previously shown cellular IL-10 is a critical autocrine growth factor for EBV-transformed B-cell lymphomas in vitro (19 19. Martinez, OM ∙ Villanueva, JC ∙ Lawrence-Miyasaki, L ... Viral and immunologic aspects of Epstein-Barr virus infection in pediatric liver transplant recipients Transplantation. 1995; 59:519-524 Crossref Scopus (50) PubMed Google Scholar ), and the abundant presence of IL-10 in serum from PTLD patients also implicates it in EBV-mediated PTLD pathogenesis in vivo. During lytic infection, EBV also expresses a soluble receptor for colony stimulating factor 1 (CSF-1) called BARF1. Although BARF1 appears to act as an oncogene when expressed ectopically in certain cell types, it is not clear if BARF1 contributes directly to EBV tumorigenicity in B cells, considering a BARF1-deletion mutant of EBV maintains the same transformation potential as wild-type virus (23 23. Cohen, JI ∙ Lekstrom, K Epstein-Barr virus BARF1 protein is dispensable for B-cell transformation and inhibits alpha interferon secretion from mononuclear cells J Virol. 1999; 73:7627-7632 Crossref PubMed Google Scholar ). However, BARF1 represents another weapon for immune evasion in its ability to block the function of CSF-1 in stimulating monocyte proliferation and cytokine release. Thus BARF1 may blunt cellular immune responses by interfering with secretion of antiviral factors like IFN-α. 3-2 Apoptosis resistance EBV has also evolved mechanisms for combating host-cell apoptosis following infection. For example, the lytic cycle gene BHRF1 encodes a homolog of Bcl-2 capable of inhibiting apoptosis induced by multiple stimuli, including Fas ligand (FasL) and TNF-related apoptosis-inducing ligand (TRAIL) (24 24. Kawanishi, M ∙ Tada-Oikawa, S ∙ Kawanishi, S Epstein-Barr virus BHRF1 functions downstream of bid cleavage and upstream of mitochondrial dysfunction to inhibit TRAIL-induced apoptosis in BJAB cells Biochem Biophys Res Commun. 2002; 297:682-687 Crossref Scopus (31) PubMed Google Scholar ). Earlier studies have established that EBV latent genes can also protect B cells from apoptosis in vitro in response to serum withdrawal (25 25. Gregory, CD ∙ Dive, C ∙ Henderson, S ... Activation of Epstein-Barr virus latent genes protects human B cells from death by apoptosis Nature. 1991; 349:612-614 Crossref Scopus (504) PubMed Google Scholar ). Much work on the transforming capability of EBV has focused on LMP1, an oncogene that is indispensable for EBV-mediated transformation of B cells (26 26. Kaye, KM ∙ Izumi, KM ∙ Kieff, E Epstein-Barr virus latent membrane protein 1 is essential for B-lymphocyte growth transformation Proc Natl Acad Sci USA. 1993; 90:9150-9154 Crossref Scopus (667) PubMed Google Scholar ). LMP1 upregulates the expression of several antiapoptotic genes implicated in protection from intrinsic apoptosis pathways, including bcl-2, bfl-1, mcl-1, A20, and cIAP2 (27 27. Young, LS ∙ Dawson, CW ∙ Eliopoulos, AG Epstein-Barr virus and apoptosis: Viral mimicry of cellular pathways Biochem Soc Trans. 1999; 27:807-812 Crossref Scopus (21) PubMed Google Scholar ). All of these genes represent downstream targets of NF-κB: indeed, interfering with LMP1-mediated NF-κB activation promotes spontaneous apoptosis in EBV-immortalized B cells (28 28. Cahir-McFarland, ED ∙ Davidson, DM ∙ Schauer, SL ... NF-kappa B inhibition causes spontaneous apoptosis in Epstein-Barr virus-transformed lymphoblastoid cells Proc Natl Acad Sci U S A. 2000; 97:6055-6060 Crossref Scopus (241) PubMed Google Scholar , 29 29. Kenney, JL ∙ Guinness, ME ∙ Curiel, T ... Antisense to the Epstein-Barr virus (EBV)-encoded latent membrane protein 1 (LMP-1) suppresses LMP-1 and bcl-2 expression and promotes apoptosis in EBV-immortalized B cells Blood. 1998; 92:1721-1727 Crossref PubMed Google Scholar ). Other EBV latent proteins can also interfere with certain apoptotic stimuli. For example, EBNA2 can specifically block apoptosis triggered by Nur77, a nuclear hormone receptor that induces cytochrome C release from mitochondria upon translocation to the cytoplasm (30 30. Lee, JM ∙ Lee, KH ∙ Weidner, M ... Epstein-Barr virus EBNA2 blocks Nur77-mediated apoptosis Proc Natl Acad Sci USA. 2002; 99:11878-11883 Crossref Scopus (82) PubMed Google Scholar ). Recent work indicates LMP2A signaling also promotes B-cell survival through constitutive activation of the Ras/PI3K/Akt signaling axis, which represents a potent prosurvival signal in B cells (31 31. Portis, T ∙ Longnecker, R Epstein-Barr virus (EBV) LMP2A mediates B-lymphocyte survival through constitutive activation of the Ras(PI3K(Akt pathway Oncogene. 2004; 23:8619-8628 Crossref Scopus (178) PubMed Google Scholar ). Young and colleagues have demonstrated that LMP1 also activates the Akt pathway, and that LMP2A can enhance the function of LMP1 by prolonging its half-life, suggesting that coexpression of these molecules during type II/III latency provides a powerful antiapoptotic signal to the host B cell (32 32. Dawson, CW ∙ George, JH ∙ Blake, SM ... The Epstein-Barr virus encoded latent membrane protein 2A augments signaling from latent membrane protein 1 Virology. 2001; 289:192-207 Crossref Scopus (37) PubMed Google Scholar , 33 33. Dawson, CW ∙ Tramountanis, G ∙ Eliopoulos, AG ... Epstein-Barr virus latent membrane protein 1 (LMP1) activates the phosphatidylinositol 3-kinase(Akt pathway to promote cell survival and induce actin filament remodeling J Biol Chem. 2003; 278:3694-3704 Full Text Full Text (PDF) Scopus (254) PubMed Google Scholar ). Even EBNA1, known primarily for its role in EBV episome maintenance, contributes to cell survival by binding ubiquitin-specific protease 7 (USP7) and preventing p53 stabilization by deubiquitination (34 34. Saridakis, V ∙ Sheng, Y ∙ Sarkari, F ... Structure of the p53 binding domain of HAUSP(USP7 bound to Epstein-Barr nuclear antigen 1 implications for EBV-mediated immortalization Mol Cell. 2005; 18:25-36 Full Text Full Text (PDF) Scopus (296) PubMed Google Scholar ). This unique mode of interference with p53 function explains why EBNA1 inhibition induces apoptosis in BL (35 35. Kennedy, G ∙ Komano, J ∙ Sugden, B Epstein-Barr virus provides a survival factor to Burkitt's lymphomas Proc Natl Acad Sci USA. 2003; 100:14269-14274 Crossref Scopus (216) PubMed Google Scholar ). EBV may also perturb extrinsic apoptosis stimuli triggered through death receptors (DRs). We recently discovered that EBV itself can protect latently infected BJAB B lymphoma cells from apoptosis induced through Fas and TRAIL receptors, mediated in part by LMP1-driven upregulation of cellular FLICE inhibitory protein (cFLIP), another NF-κB-responsive gene (36 36. Snow, AL ∙ Lambert, SL ∙ Natkunam, Y ... EBV can protect latently infected B cell lymphomas from death receptor-induced apoptosis J Immunol. 2006; 177:3283-3293 Crossref Scopus (41) PubMed Google Scholar ). This phenomenon may be particularly relevant to Hodgkin's lymphomagenesis, where LMP1/LMP2A may cooperate to rescue the crippled GC cells that are otherwise destined to die. Hitherto, work from our laboratory and others indicated that EBV-infected LCL show differential sensitivity to Fas-induced apoptosis (37–39 37. Durandy, A ∙ Le Deist, F ∙ Emile, JF ... Sensitivity of Epstein-Barr virus-induced B cell tumor to apoptosis mediated by anti-CD95(Apo-1(fas antibody Eur J Immunol. 1997; 27:538-543 Crossref Scopus (19) PubMed Google Scholar 38. Tepper, CG ∙ Seldin, MF Modulation of caspase-8 and FLICE-inhibitory protein expression as a potential mechanism of Epstein-Barr virus tumorigenesis in Burkitt's lymphoma Blood. 1999; 94:1727-1737 Crossref PubMed Google Scholar 39. Snow, AL ∙ Chen, LJ ∙ Nepomuceno, RR ... Resistance to Fas-mediated apoptosis in EBV-infected B cell lymphomas is due to defects in the proximal Fas signaling pathway J Immunol. 2001; 167:5404-5411 Crossref Scopus (29) PubMed Google Scholar ). We have also determined that EBV+ LCLs derived from PTLD patients show universal resistance to TRAIL-induced apoptosis (40 40. Snow, AL ∙ Vaysberg, M ∙ Krams, SM ... EBV B lymphoma cell lines from patients with post-transplant lymphoproliferative disease are resistant to TRAIL-induced apoptosis Am J Transplant. 2006; 6:976-985 Crossref Scopus (9) PubMed Google Scholar ). Variable DR apoptosis sensitivity among EBV+ B cell lines may be explained by differential latent gene expression, natural sequence variants with differing intrinsic signaling properties (e.g. LMP1) or derivation from myriad sources of host B cells; parameters that remain to be addressed experimentally. Taken together, an expanding body of work indicates that latent EBV infection, and especially LMP1 expression, can subvert important immune effector pathways of target cell elimination by inhibiting apoptosis induced by numerous stimuli. Using MHC/peptide tetramers we previously established that solid organ transplant recipients have a significant proportion of EBV-specific CD8+ T cells comparable to healthy seropositive individuals (41 41. Falco, DA ∙ Nepomuceno, RR ∙ Krams, SM ... Identification of Epstein-Barr virus-specific CD8+ T lymphocytes in the circulation of pediatric transplant recipients Transplantation. 2002; 74:501-510 Crossref Scopus (46) PubMed Google Scholar ). The fact that some PTLD lymphomas persist or relapse even in the presence of EBV-specific CTL may be due in part to apoptosis resistance, particularly if more potent cytotoxic effector molecules (perforin, granzymes) are also rendered ineffective (42 42. Lewinsohn, DM ∙ Bement, TT ∙ Xu, J ... Human purified protein derivative-specific CD4+ T cells use both CD95-dependent and CD95-independent cytolytic mechanisms J Immunol. 1998; 160:2374-2379 Crossref PubMed Google Scholar , 43 43. Katano, H ∙ Ali, MA ∙ Patera, AC ... Chronic active Epstein-Barr virus infection associated with mutations in perforin that impair its maturation Blood. 2004; 103:1244-1252 Crossref Scopus (132) PubMed Google Scholar ). 4 Applying our Knowledge: Implications for Treatment of PTLD Given the evasive tactics employed by EBV as part of its interaction with the immune system, it is imperative that therapeutic options for PTLD be evaluated with these aspects of basic virology in mind. At present, several strategies for the treatment of PTLD are being implemented in the clinic. As mentioned earlier, removal of immunosuppression is often the first approach utilized to combat PTLD lymphomas, which allows the CTL response to rebound. However, risk of graft rejection and selection for highly proliferative monoclonal lymphoma cells leave this approach far from ideal. An alternative may be found in using immunosuppressive drugs that protect the graft without promoting lymphoma outgrowth, such as rapamycin or everolimus. Both of these drugs effectively retard the growth of PTLD-associated lymphomas in vitro and in xenografted SCID mice (44–46 44. Majewski, M ∙ Korecka, M ∙ Joergensen, J ... Immunosuppressive TOR kinase inhibitor everolimus (RAD) suppresses growth of cells derived from posttransplant lymphoproliferative disorder at allograft-protecting doses Transplantation. 2003; 75:1710-1717 Crossref Scopus (144) PubMed Google Scholar 45. Nepomuceno, RR ∙ Balatoni, CE ∙ Natkunam, Y ... Rapamycin inhibits the interleukin 10 signal transduction pathway and the growth of Epstein Barr virus B-cell lymphomas Cancer Res. 2003; 63:4472-4480 PubMed Google Scholar 46. Tanner, JE ∙ Alfieri, C The Epstein-Barr virus and post-transplant lymphoproliferative disease: Interplay of immunosuppression, EBV, and the immune system in disease pathogenesis Transpl Infect Dis. 2001; 3:60-69 Crossref Scopus (115) PubMed Google Scholar ). It remains to be determined if the potential antitumor benefits of mTOR inhibitors are manifest in clinical transplantation. 4-1 Cytokine- and antibody-based immunotherapy Immunotherapy directly targeting B cells and associated cytokines must also be considered. Regimens involving the addition of cytokines like IFN-α or the neutralization of B-cell growth factors like IL-6 have shown some promise, although caution must be exercised in disturbing cytokine balances systemically. Notwithstanding the potential salutary effect on graft survival, the importance of IL-10 to proliferation of EBV+ B cell lymphomas and suppression of antiviral CTL responses suggests neutralization or inhibition of IL-10 may also prove beneficial. Our own work indicating rapamycin inhibits the growth of PTLD-associated lymphomas in part by decreasing IL-10 secretion underscores IL-10 as a pertinent target (45 45. Nepomuceno, RR ∙ Balatoni, CE ∙ Natkunam, Y ... Rapamycin inhibits the interleukin 10 signal transduction pathway and the growth of Epstein Barr virus B-cell lymphomas Cancer Res. 2003; 63:4472-4480 PubMed Google Scholar ). Clinicians have also turned to the humanized anti-CD20 mAb rituximab, which nonspecifically depletes virtually all B cells and can elicit strong anti-B-cell tumor responses. Although a thorough description of long-term effects arising from systemic B-cell depletion using this reagent awaits further study, rituximab represents an important tool for eliminating CD20+ lymphoma cells in PTLD patients. 4-2 Cellular immunotherapy Recently, much attention has focused on cellular immunotherapy for combating PTLD-associated tumors by infusing EBV-specific CTL. Although posthuman stem cell transplant patients receiving polyclonal EBV-specific CTL infusions prophylactically or in response to overt PTLD have fared well, efficacy is mixed in solid organ transplant patients, perhaps due to the presence of debilitating immunosuppressive drugs and the limited survival and expansion of the infused CTL in lymphoid-replete recipients (47 47. Gottschalk, S ∙ Heslop, HE ∙ Rooney, CM Adoptive immunotherapy for EBV-associated malignancies Leuk Lymphoma. 2005; 46:1-10 Crossref Scopus (105) PubMed Google Scholar ). In the context of EBV latency, infused CTL may also fail to control EBV+ lymphomas due to local cytokine-based suppression, resistance to CTL-induced apoptotic stimuli and/or downregulation of relevant EBV protein epitopes. Hence we suggest that subversion of T-cell immunity by EBV not only contributes to lymphomagenesis, but can also foil attempts to regain control of EBV+ lymphoma growth using autologous or allogeneic CTL therapy. 4-3 The multipronged approach: Targeting EBV itself Indeed, effective treatment of PTLD-associated tumors may not be achieved with monotherapies. For instance, our own findings prompt careful evaluation of the utility of recombinant TRAIL a promising anticancer reagent, as a potential single therapeutic for EBV-associated B-cell lymphomas (40 40. Snow, AL ∙ Vaysberg, M ∙ Krams, SM ... EBV B lymphoma cell lines from patients with post-transplant lymphoproliferative disease are resistant to TRAIL-induced apoptosis Am J Transplant. 2006; 6:976-985 Crossref Scopus (9) PubMed Google Scholar ). As with other malignancies, PTLD therapy may require a multipronged approach involving a combination of surgery, radiation, chemotherapy (e.g. CHOP), and reagents that target host B cells or boost T-cell immunity. More importantly, basic research informs us that successful approaches to eradicating PTLD-associated lymphomas must also focus on EBV infection itself. Antiviral agents such as acyclovir only target the lytic phase of viral replication, but recent work reminds us that the importance of lytic gene expression to the establishment of PTLD should not be ignored (48 48. Hong, GK ∙ Gulley, ML ∙ Feng, WH ... Epstein-Barr virus lytic infection contributes to lymphoproliferative disease in a SCID mouse model J Virol. 2005; 79:13993-14003 Crossref Scopus (191) PubMed Google Scholar ). Specifically disrupting the function of latent genes like LMP1 and LMP2A will likely prove most efficacious in combating PTLD and associated EBV+ type II/III latent cancers. The propensity of reagents that silence latent gene expression or block relevant signaling pathways (i.e. NF-κB and proteosome inhibitors) to successfully kill such lymphomas in vitro has already been established (28 28. Cahir-McFarland, ED ∙ Davidson, DM ∙ Schauer, SL ... NF-kappa B inhibition causes spontaneous apoptosis in Epstein-Barr virus-transformed lymphoblastoid cells Proc Natl Acad Sci U S A. 2000; 97:6055-6060 Crossref Scopus (241) PubMed Google Scholar , 29 29. Kenney, JL ∙ Guinness, ME ∙ Curiel, T ... Antisense to the Epstein-Barr virus (EBV)-encoded latent membrane protein 1 (LMP-1) suppresses LMP-1 and bcl-2 expression and promotes apoptosis in EBV-immortalized B cells Blood. 1998; 92:1721-1727 Crossref PubMed Google Scholar , 49 49. Keller, SA ∙ Hernandez-Hopkins, D ∙ Vider, J ... NF-kappaB is essential for the progression of KSHV- and EBV-infected lymphomas in vivo Blood. 2006; 107:3295-3302 Epub 2005 Dec 3227. Crossref Scopus (146) PubMed Google Scholar ). In conclusion, such studies demonstrate that we must redouble efforts to introduce such EBV-specific therapies into the clinic and regain the upper hand against such a ‘clever’ virus in eradicating PTLD. References 1. Gottschalk, S ∙ Rooney, CM ∙ Heslop, HE Post-transplant lymphoproliferative disorders Annu Rev Med. 2005; 56:29-44 Crossref Scopus (373) PubMed Google Scholar 2. Epstein, MA ∙ Henle, G ∙ Achong, BG ... Morphological and biological studies on a virus in cultured lymphoblasts from Burkitt's lymphoma J Exp Med. 1965; 121:761-770 Crossref Scopus (174) PubMed Google Scholar 3. Cohen, JI Epstein-Barr virus infection N Engl J Med. 2000; 343:481-492 Crossref Scopus (1411) PubMed Google Scholar 4. Thorley-Lawson, DA Epstein-Barr virus: Exploiting the immune system Nat Rev Immunol. 2001; 1:75-82 Crossref Scopus (772) PubMed Google Scholar 5. 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A single amino acid determines the immunostimulatory activity of interleukin 10 J Exp Med. 2000; 191:213-224 Crossref Scopus (110) PubMed Google Scholar 23. Cohen, JI ∙ Lekstrom, K Epstein-Barr virus BARF1 protein is dispensable for B-cell transformation and inhibits alpha interferon secretion from mononuclear cells J Virol. 1999; 73:7627-7632 Crossref PubMed Google Scholar 24. Kawanishi, M ∙ Tada-Oikawa, S ∙ Kawanishi, S Epstein-Barr virus BHRF1 functions downstream of bid cleavage and upstream of mitochondrial dysfunction to inhibit TRAIL-induced apoptosis in BJAB cells Biochem Biophys Res Commun. 2002; 297:682-687 Crossref Scopus (31) PubMed Google Scholar 25. Gregory, CD ∙ Dive, C ∙ Henderson, S ... Activation of Epstein-Barr virus latent genes protects human B cells from death by apoptosis Nature. 1991; 349:612-614 Crossref Scopus (504) PubMed Google Scholar 26. Kaye, KM ∙ Izumi, KM ∙ Kieff, E Epstein-Barr virus latent membrane protein 1 is essential for B-lymphocyte growth transformation Proc Natl Acad Sci USA. 1993; 90:9150-9154 Crossref Scopus (667) PubMed Google Scholar 27. Young, LS ∙ Dawson, CW ∙ Eliopoulos, AG Epstein-Barr virus and apoptosis: Viral mimicry of cellular pathways Biochem Soc Trans. 1999; 27:807-812 Crossref Scopus (21) PubMed Google Scholar 28. Cahir-McFarland, ED ∙ Davidson, DM ∙ Schauer, SL ... NF-kappa B inhibition causes spontaneous apoptosis in Epstein-Barr virus-transformed lymphoblastoid cells Proc Natl Acad Sci U S A. 2000; 97:6055-6060 Crossref Scopus (241) PubMed Google Scholar 29. Kenney, JL ∙ Guinness, ME ∙ Curiel, T ... Antisense to the Epstein-Barr virus (EBV)-encoded latent membrane protein 1 (LMP-1) suppresses LMP-1 and bcl-2 expression and promotes apoptosis in EBV-immortalized B cells Blood. 1998; 92:1721-1727 Crossref PubMed Google Scholar 30. Lee, JM ∙ Lee, KH ∙ Weidner, M ... Epstein-Barr virus EBNA2 blocks Nur77-mediated apoptosis Proc Natl Acad Sci USA. 2002; 99:11878-11883 Crossref Scopus (82) PubMed Google Scholar 31. Portis, T ∙ Longnecker, R Epstein-Barr virus (EBV) LMP2A mediates B-lymphocyte survival through constitutive activation of the Ras(PI3K(Akt pathway Oncogene. 2004; 23:8619-8628 Crossref Scopus (178) PubMed Google Scholar 32. Dawson, CW ∙ George, JH ∙ Blake, SM ... The Epstein-Barr virus encoded latent membrane protein 2A augments signaling from latent membrane protein 1 Virology. 2001; 289:192-207 Crossref Scopus (37) PubMed Google Scholar 33. Dawson, CW ∙ Tramountanis, G ∙ Eliopoulos, AG ... Epstein-Barr virus latent membrane protein 1 (LMP1) activates the phosphatidylinositol 3-kinase(Akt pathway to promote cell survival and induce actin filament remodeling J Biol Chem. 2003; 278:3694-3704 Full Text Full Text (PDF) Scopus (254) PubMed Google Scholar 34. Saridakis, V ∙ Sheng, Y ∙ Sarkari, F ... Structure of the p53 binding domain of HAUSP(USP7 bound to Epstein-Barr nuclear antigen 1 implications for EBV-mediated immortalization Mol Cell. 2005; 18:25-36 Full Text Full Text (PDF) Scopus (296) PubMed Google Scholar 35. Kennedy, G ∙ Komano, J ∙ Sugden, B Epstein-Barr virus provides a survival factor to Burkitt's lymphomas Proc Natl Acad Sci USA. 2003; 100:14269-14274 Crossref Scopus (216) PubMed Google Scholar 36. Snow, AL ∙ Lambert, SL ∙ Natkunam, Y ... EBV can protect latently infected B cell lymphomas from death receptor-induced apoptosis J Immunol. 2006; 177:3283-3293 Crossref Scopus (41) PubMed Google Scholar 37. Durandy, A ∙ Le Deist, F ∙ Emile, JF ... Sensitivity of Epstein-Barr virus-induced B cell tumor to apoptosis mediated by anti-CD95(Apo-1(fas antibody Eur J Immunol. 1997; 27:538-543 Crossref Scopus (19) PubMed Google Scholar 38. Tepper, CG ∙ Seldin, MF Modulation of caspase-8 and FLICE-inhibitory protein expression as a potential mechanism of Epstein-Barr virus tumorigenesis in Burkitt's lymphoma Blood. 1999; 94:1727-1737 Crossref PubMed Google Scholar 39. Snow, AL ∙ Chen, LJ ∙ Nepomuceno, RR ... Resistance to Fas-mediated apoptosis in EBV-infected B cell lymphomas is due to defects in the proximal Fas signaling pathway J Immunol. 2001; 167:5404-5411 Crossref Scopus (29) PubMed Google Scholar 40. Snow, AL ∙ Vaysberg, M ∙ Krams, SM ... EBV B lymphoma cell lines from patients with post-transplant lymphoproliferative disease are resistant to TRAIL-induced apoptosis Am J Transplant. 2006; 6:976-985 Crossref Scopus (9) PubMed Google Scholar 41. Falco, DA ∙ Nepomuceno, RR ∙ Krams, SM ... Identification of Epstein-Barr virus-specific CD8+ T lymphocytes in the circulation of pediatric transplant recipients Transplantation. 2002; 74:501-510 Crossref Scopus (46) PubMed Google Scholar 42. Lewinsohn, DM ∙ Bement, TT ∙ Xu, J ... Human purified protein derivative-specific CD4+ T cells use both CD95-dependent and CD95-independent cytolytic mechanisms J Immunol. 1998; 160:2374-2379 Crossref PubMed Google Scholar 43. Katano, H ∙ Ali, MA ∙ Patera, AC ... Chronic active Epstein-Barr virus infection associated with mutations in perforin that impair its maturation Blood. 2004; 103:1244-1252 Crossref Scopus (132) PubMed Google Scholar 44. Majewski, M ∙ Korecka, M ∙ Joergensen, J ... Immunosuppressive TOR kinase inhibitor everolimus (RAD) suppresses growth of cells derived from posttransplant lymphoproliferative disorder at allograft-protecting doses Transplantation. 2003; 75:1710-1717 Crossref Scopus (144) PubMed Google Scholar 45. Nepomuceno, RR ∙ Balatoni, CE ∙ Natkunam, Y ... Rapamycin inhibits the interleukin 10 signal transduction pathway and the growth of Epstein Barr virus B-cell lymphomas Cancer Res. 2003; 63:4472-4480 PubMed Google Scholar 46. Tanner, JE ∙ Alfieri, C The Epstein-Barr virus and post-transplant lymphoproliferative disease: Interplay of immunosuppression, EBV, and the immune system in disease pathogenesis Transpl Infect Dis. 2001; 3:60-69 Crossref Scopus (115) PubMed Google Scholar 47. Gottschalk, S ∙ Heslop, HE ∙ Rooney, CM Adoptive immunotherapy for EBV-associated malignancies Leuk Lymphoma. 2005; 46:1-10 Crossref Scopus (105) PubMed Google Scholar 48. Hong, GK ∙ Gulley, ML ∙ Feng, WH ... Epstein-Barr virus lytic infection contributes to lymphoproliferative disease in a SCID mouse model J Virol. 2005; 79:13993-14003 Crossref Scopus (191) PubMed Google Scholar 49. Keller, SA ∙ Hernandez-Hopkins, D ∙ Vider, J ... NF-kappaB is essential for the progression of KSHV- and EBV-infected lymphomas in vivo Blood. 2006; 107:3295-3302 Epub 2005 Dec 3227. Crossref Scopus (146) PubMed Google Scholar Figures (1)Figure Viewer Article metrics Related Articles Open in viewer Epstein-Barr Virus: Evasive Maneuvers in the Development of PTLD Hide CaptionDownloadSee figure in Article Toggle Thumbstrip Figure 1: Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries For Authors Guide for Authors Permissions Research Academy Submit Article Journal Information About Activate Online Access Contact Information Editorial Board Editorial Board Bios Society Information American Society of Transplantation American Society of Transplant Surgeons The content on this site is intended for healthcare professionals. We use cookies to help provide and enhance our service and tailor content. To update your cookie settings, please visit the Cookie Settings for this site. All content on this site: Copyright © 2025 Elsevier Inc., its licensors, and contributors. 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1704	https://tajimasolis.weebly.com/uploads/3/7/3/4/37346235/calcii_seq_series_solutions.pdf	Solutions to Practice Problems Series & Sequences Paul Dawkins Calculus II i © 2018 Paul Dawkins Table of Contents Chapter 4 : Series & Sequences .............................................................................................................................. 2 Section 4-1 : Sequences ............................................................................................................................................ 4 Section 4-2 : More on Sequences .............................................................................................................................. 8 Section 4-3 : Series - The Basics .............................................................................................................................. 16 Section 4-4 : Convergence/Divergence of Series .................................................................................................... 19 Section 4-5 : Special Series ...................................................................................................................................... 23 Section 4-6 : Integral Test........................................................................................................................................ 31 Section 4-7 : Comparison Test/Limit Comparison Test ........................................................................................... 37 Section 4-8 : Alternating Series Test ....................................................................................................................... 57 Section 4-9 : Absolute Convergence ....................................................................................................................... 63 Section 4-10 : Ratio Test .......................................................................................................................................... 66 Section 4-11 : Root Test .......................................................................................................................................... 70 Section 4-12 : Strategy for Series ............................................................................................................................ 72 Section 4-13 : Estimating the Value of a Series ....................................................................................................... 73 Section 4-14 : Power Series ..................................................................................................................................... 78 Section 4-15 : Power Series and Functions ............................................................................................................. 83 Section 4-16 : Taylor Series ..................................................................................................................................... 88 Section 4-17 : Applications of Series ....................................................................................................................... 95 Section 4-18 : Binomial Series ................................................................................................................................. 98 Calculus II 2 © 2018 Paul Dawkins Chapter 4 : Series & Sequences Here is a listing of sections for which practice problems have been written as well as a brief description of the material covered in the notes for that particular section. Sequences – In this section we define just what we mean by sequence in a math class and give the basic notation we will use with them. We will focus on the basic terminology, limits of sequences and convergence of sequences in this section. We will also give many of the basic facts and properties we’ll need as we work with sequences. More on Sequences – In this section we will continued examining sequences. We will determine if a sequence in an increasing sequence or a decreasing sequence and hence if it is a monotonic sequence. We will also determine a sequence is bounded below, bounded above and/or bounded. Series – The Basics – In this section we will formally define an infinite series. We will also give many of the basic facts, properties and ways we can use to manipulate a series. We will also briefly discuss how to determine if an infinite series will converge or diverge (a more in depth discussion of this topic will occur in the next section). Convergence/Divergence of Series – In this section we will discuss in greater detail the convergence and divergence of infinite series. We will illustrate how partial sums are used to determine if an infinite series converges or diverges. We will also give the Divergence Test for series in this section. Special Series – In this section we will look at three series that either show up regularly or have some nice properties that we wish to discuss. We will examine Geometric Series, Telescoping Series, and Harmonic Series. Integral Test – In this section we will discuss using the Integral Test to determine if an infinite series converges or diverges. The Integral Test can be used on a infinite series provided the terms of the series are positive and decreasing. A proof of the Integral Test is also given. Comparison Test/Limit Comparison Test – In this section we will discuss using the Comparison Test and Limit Comparison Tests to determine if an infinite series converges or diverges. In order to use either test the terms of the infinite series must be positive. Proofs for both tests are also given. Alternating Series Test – In this section we will discuss using the Alternating Series Test to determine if an infinite series converges or diverges. The Alternating Series Test can be used only if the terms of the series alternate in sign. A proof of the Alternating Series Test is also given. Absolute Convergence – In this section we will have a brief discussion on absolute convergence and conditionally convergent and how they relate to convergence of infinite series. Ratio Test – In this section we will discuss using the Ratio Test to determine if an infinite series converges absolutely or diverges. The Ratio Test can be used on any series, but unfortunately will not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of the Ratio Test is also given. Calculus II 3 © 2018 Paul Dawkins Root Test – In this section we will discuss using the Root Test to determine if an infinite series converges absolutely or diverges. The Root Test can be used on any series, but unfortunately will not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of the Root Test is also given. Strategy for Series – In this section we give a general set of guidelines for determining which test to use in determining if an infinite series will converge or diverge. Note as well that there really isn’t one set of guidelines that will always work and so you always need to be flexible in following this set of guidelines. A summary of all the various tests, as well as conditions that must be met to use them, we discussed in this chapter are also given in this section. Estimating the Value of a Series – In this section we will discuss how the Integral Test, Comparison Test, Alternating Series Test and the Ratio Test can, on occasion, be used to estimating the value of an infinite series. Power Series – In this section we will give the definition of the power series as well as the definition of the radius of convergence and interval of convergence for a power series. We will also illustrate how the Ratio Test and Root Test can be used to determine the radius and interval of convergence for a power series. Power Series and Functions – In this section we discuss how the formula for a convergent Geometric Series can be used to represent some functions as power series. To use the Geometric Series formula, the function must be able to be put into a specific form, which is often impossible. However, use of this formula does quickly illustrate how functions can be represented as a power series. We also discuss differentiation and integration of power series. Taylor Series – In this section we will discuss how to find the Taylor/Maclaurin Series for a function. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. We also derive some well known formulas for Taylor series of x e , ( ) cos x and ( ) sin x around 0 x = . Applications of Series – In this section we will take a quick look at a couple of applications of series. We will illustrate how we can find a series representation for indefinite integrals that cannot be evaluated by any other method. We will also see how we can use the first few terms of a power series to approximate a function. Binomial Series – In this section we will give the Binomial Theorem and illustrate how it can be used to quickly expand terms in the form ( ) n a b + when n is an integer. In addition, when n is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the term. Calculus II 4 © 2018 Paul Dawkins Section 4-1 : Sequences 1. List the first 5 terms of the following sequence. 2 0 4 7 n n n ∞ =     −   Solution There really isn’t all that much to this problem. All we need to do is, starting at 0 n = , plug in the first five values of n into the formula for the sequence terms. Doing that gives, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 4 0 0: 0 0 7 4 1 4 2 1: 6 3 1 7 4 2 8 8 2: 3 3 2 7 4 3 12 3: 6 2 3 7 4 4 16 4: 9 4 7 n n n n n = = − = = = − − − = = = − − − = = = − = = − So, the first five terms of the sequence are, 2 8 16 0, , , 6, , 3 3 9   − −      Note that we put the formal answer inside the braces to make sure that we don’t forget that we are dealing with a sequence and we made sure and included the “…” at the end to reminder ourselves that there are more terms to this sequence that just the five that we listed out here. 2. List the first 5 terms of the following sequence. ( ) ( ) 1 2 1 2 3 n n n n ∞ + =   −     + −     Solution Calculus II 5 © 2018 Paul Dawkins There really isn’t all that much to this problem. All we need to do is, starting at 2 n = , plug in the first five values of n into the formula for the sequence terms. Doing that gives, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 1 1 1 2: 13 13 2 2 3 1 1 1 3: 21 21 2 3 3 1 1 1 4: 89 89 2 4 3 1 1 1 5: 233 233 2 5 3 1 1 1 6: 741 741 2 6 3 n n n n n + + + + + − − = = = − + − − = = = − − + − − − = = = − + − − = = = − − + − − − = = = − + − So, the first five terms of the sequence are, 1 1 1 1 1 , , , , , 13 21 89 233 741   − − − − −      Note that we put the formal answer inside the braces to make sure that we don’t forget that we are dealing with a sequence and we made sure and included the “…” at the end to reminder ourselves that there are more terms to this sequence that just the five that we listed out here. 3. Determine if the given sequence converges or diverges. If it converges what is its limit? 2 2 3 7 3 1 10 4 n n n n n ∞ =   − +   + −   Step 1 To answer this all we need is the following limit of the sequence terms. 2 2 7 3 1 lim1 10 4 4 n n n n n →∞ − + = − + − You do recall how to take limits at infinity right? If not you should go back into the Calculus I material do some refreshing on limits at infinity as well at L’Hospital’s rule. Calculus II 6 © 2018 Paul Dawkins Step 2 We can see that the limit of the terms existed and was a finite number and so we know that the sequence converges and its limit is 1 4 − . 4. Determine if the given sequence converges or diverges. If it converges what is its limit? ( ) 2 2 3 0 1 4 n n n n ∞ − =   −     +     Step 1 To answer this all we need is the following limit of the sequence terms. ( ) 2 2 3 1 lim 4 n n n n − →∞ − + However, because of the ( ) 2 1 n− − we can’t compute this limit using our knowledge of computing limits from Calculus I. Step 2 Recall however, that we had a nice Fact in the notes from this section that had us computing not the limit above but instead computing the limit of the absolute value of the sequence terms. ( ) 2 2 2 3 3 1 lim lim 0 4 4 n n n n n n n − →∞ →∞ − = = + + This is a limit that we can compute because the absolute value got rid of the alternating sign, i.e. the ( ) 2 1 n+ − . Step 3 Now, by the Fact from class we know that because the limit of the absolute value of the sequence terms was zero (and recall that to use that fact the limit MUST be zero!) we also know the following limit. ( ) 2 2 3 1 lim 0 4 n n n n − →∞ − = + Step 4 We can see that the limit of the terms existed and was a finite number and so we know that the sequence converges and its limit is zero. Calculus II 7 © 2018 Paul Dawkins 5. Determine if the given sequence converges or diverges. If it converges what is its limit? 5 2 1 3 n n n ∞ =     −   e e Step 1 To answer this all we need is the following limit of the sequence terms. 5 5 3 2 2 5 5 lim lim lim 3 2 2 n n n n n n n n →∞ →∞ →∞ = = = −∞ − − − e e e e e You do recall how to use L’Hospital’s rule to compute limits at infinity right? If not you should go back into the Calculus I material do some refreshing. Step 2 We can see that the limit of the terms existed and but was infinite and so we know that the sequence diverges. 6. Determine if the given sequence converges or diverges. If it converges what is its limit? ( ) ( ) 1 ln 2 ln 1 4 n n n ∞ =   +     +     Step 1 To answer this all we need is the following limit of the sequence terms. ( ) ( ) ( ) 1 ln 2 1 4 2 lim lim lim 1 4 ln 1 4 4 2 1 4 n n n n n n n n n →∞ →∞ →∞ + + + = = = + + + You do recall how to use L’Hospital’s rule to compute limits at infinity right? If not, you should go back into the Calculus I material do some refreshing. Step 2 We can see that the limit of the terms existed and was a finite number and so we know that the sequence converges and its limit is one. Calculus II 8 © 2018 Paul Dawkins Section 4-2 : More on Sequences 1. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. 1 1 4 n n ∞ =       Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. This is one of those sequences that it doesn’t matter which set of information you find first and both sets should be fairly easy to determine the answers without a lot of work. Step 1 For this problem let’s get the bounded information first as that seems to be pretty simple. First note that because both the numerator and denominator are positive then the quotient is also positive and so we can see that the sequence must be bounded below by zero. Next let’s note that because we are starting with 1 n = the denominator will always be 4 4 1 n ≥ > and so we can also see that the sequence must be bounded above by one. Note that, in this case, this not the “best” upper bound for the sequence but the problem didn’t ask for that. For this sequence we’ll be able to get a better one once we have the increasing/decreasing information. Because the sequence is bounded above and bounded below the sequence is also bounded. Step 2 For the increasing/decreasing information we can see that, for our range of 1 n ≥, we have, ( ) 4 4 1 n n < + and so, ( ) 1 1 4 4 1 n n > + If we define 1 4 n a n = this in turn tells us that 1 n n a a + > for all 1 n ≥ and so the sequence is decreasing and hence monotonic. Note that because we have now determined that the sequence is decreasing we can see that the “best” upper bound would be the first term of the sequence or, 1 4 . Calculus II 9 © 2018 Paul Dawkins 2. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. ( ) { } 2 0 1 n n n ∞ + = − Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. This is one of those sequences that it doesn’t matter which set of information you find first and both sets should be fairly easy to determine the answers without a lot of work. Step 1 For this problem let’s get the increasing/decreasing information first as that seems to be pretty simple and will help at least a little bit with the bounded information. In this case let’s just write out the first few terms of the sequence. ( ) { } { } 2 0 1 0, 1, 2, 3, 4, 5, 6, 7, n n n ∞ + = − = − − − −  Just from the first three terms we can see that this sequence is not an increasing sequence and it is not a decreasing sequence and therefore is not monotonic. Step 2 Now let’s see what bounded information we can get. From the first few terms of the sequence we listed out above we can see that each successive term will get larger and change signs. Therefore, there cannot be an upper or a lower bound for the sequence. No matter what value we would try to use for an upper or a lower bound all we would need to do is take n large enough and we would eventually get a sequence term that would go past the proposed bound. Therefore, this sequence is not bounded. 3. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. { } 0 3 n n ∞ − = Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. For this sequence it might be a little easier to find the bounds (if any exist) if you first have the increasing/decreasing information. Step 1 Calculus II 10 © 2018 Paul Dawkins For this problem let’s get the increasing/decreasing information first as that seems to be pretty simple and will help at least a little bit with the bounded information. We’ all agree that, for our range of 0 n ≥ , we have, 1 n n < + This in turn gives, ( ) 1 1 1 1 3 3 3 3 n n n n − + − + = > = So, if we define 3 n n a − = we have 1 n n a a + > for all 0 n ≥ and so the sequence is decreasing and hence is also monotonic. Step 2 Now let’s see what bounded information we can get. First, it is hopefully obvious that all the terms are positive and so the sequence is bounded below by zero. Next, we saw in the first step that the sequence was decreasing and so the first term will be the largest term and so the sequence is bounded above by ( ) 0 3 1 − = (i.e. the 0 n = sequence term). Therefore, because this sequence is bounded below and bounded above the sequence is bounded. 4. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. 2 2 2 1 n n n ∞ =   −     Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. For this sequence having the increasing/decreasing information will probably make the determining the bounds (if any exist) somewhat easier. Step 1 For this problem let’s get the increasing/decreasing information first. For Problems 1 – 3 in this section it was easy enough to just ask what happens if we increase n to determine the increasing/decreasing information for this problem. However, in this case, increasing n will increase both the numerator and denominator and so it would be somewhat more difficult to do that analysis here. Calculus II 11 © 2018 Paul Dawkins Therefore, we will resort to some quick Calculus I to determine increasing/decreasing information. We can define the following function and take its derivative. ( ) ( ) 2 2 2 2 1 2 1 x x f x f x x x − + ′ = ⇒ = We can clearly see that the derivative will always be positive for 0 x ≠ and so the function is increasing for 0 x ≠ . Therefore, because the function values are the same as the sequence values when x is an integer we can see that the sequence, which starts at 2 n = , must also be increasing and hence it is also monotonic. Step 2 Now let’s see what bounded information we can get. First, it is hopefully obvious that all the terms are positive for our range of 2 n ≥ and so the sequence is bounded below by zero. We could also use the fact that the sequence is increasing the first term would have to be the smallest term in the sequence and so a better lower bound would be the first sequence term which is 7 2 . Either would work for this problem. Now let’s see what we can determine about an upper bound (provided it has one of course…). We know that the function is increasing but that doesn’t mean there is no upper bound. Take a look at Problems 1 and 3 above. Each of those were decreasing sequences and yet they had a lower bound. Do not make the mistake of assuming that an increasing sequence will not have an upper bound or a decreasing sequence will not have a lower bound. Sometimes they will and sometimes they won’t. For this sequence we’ll need to approach any potential upper bound a little differently than the previous problems. Let’s first compute the following limit of the terms, 2 2 1 1 lim lim 2 n n n n n n →∞ →∞ −   = − = ∞     Since the limit of the terms is infinity we can see that the terms will increase without bound. Therefore, in this case, there really is no upper bound for this sequence. Please remember the warning above however! Just because this increasing sequence had no upper bound does not mean that every increasing sequence will have an upper bound. Finally, because this sequence is bounded below but not bounded above the sequence is not bounded. 5. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. Calculus II 12 © 2018 Paul Dawkins 1 4 2 3 n n n ∞ = −     +   Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. For this sequence having the increasing/decreasing information will probably make the determining the bounds (if any exist) somewhat easier. Step 1 For this problem let’s get the increasing/decreasing information first. For Problems 1 – 3 in this section it was easy enough to just ask what happens if we increase n to determine the increasing/decreasing information for this problem. However, in this case, increasing n will increase both the numerator and denominator and so it would be somewhat more difficult to do that analysis here. Therefore, we will resort to some quick Calculus I to determine increasing/decreasing information. We can define the following function and take its derivative. ( ) ( ) ( ) 2 4 11 2 3 2 3 x f x f x x x − − ′ = ⇒ = + + We can clearly see that the derivative will always be negative for 3 2 x ≠− and so the function is decreasing for 3 2 x ≠− . Therefore, because the function values are the same as the sequence values when x is an integer we can see that the sequence, which starts at 1 n = , must also be decreasing and hence it is also monotonic. Step 2 Now let’s see what bounded information we can get. First, because the sequence is decreasing we can see that the first term of the sequence will be the largest and hence will also be an upper bound for the sequence. So, the sequence is bounded above by 3 5 (i.e. the 1 n = sequence term). Next let’s look for the lower bound (if it exists). For this problem let’s first take a quick look at the limit of the sequence terms. In this case the limit of the sequence terms is, 4 1 lim 2 3 2 n n n →∞ − = − + Recall what this limit tells us about the behavior of our sequence terms. The limit means that as n →∞ the sequence terms must be getting closer and closer to 1 2 − . Calculus II 13 © 2018 Paul Dawkins Now, for a second, let’s suppose that that 1 2 − is not a lower bound for the sequence terms and let’s also keep in mind that we’ve already determined that the sequence is decreasing (means that each successive term must be smaller than (i.e. below) the previous one…). So, if 1 2 − is not a lower bound then we know that somewhere there must be sequence terms below (or smaller than) 1 2 − . However, because we also know that terms must be getting closer and closer to 1 2 − and we’ve now assumed there are terms below 1 2 − the only way for that to happen at this point is for at least a few sequence terms to increase up towards 1 2 − (remember we’ve assumed there are terms below this!). That can’t happen however because we know the sequence is a decreasing sequence. Okay, what was the point of all this? Well recall that we got to this apparent contradiction to the decreasing nature of the sequence by first assuming that 1 2 − was not a lower bound. Since making this assumption led us to something that can’t possibly be true that in turn means that 1 2 − must in fact be a lower bound since we’ve shown that sequence terms simply cannot go below this value! Therefore, the sequence is bounded below by 1 2 − . Finally, because this sequence is both bounded above and bounded below the sequence is bounded. Before leaving this problem a quick word of caution. The limit of a sequence is not guaranteed to be a bound (upper or lower) for a sequence. It will only be a bound under certain circumstances and so we can’t simply compute the limit and assume it will be a bound for every sequence! Can you see a condition that will allow the limit to be a bound? 6. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded. 2 2 25 n n n ∞ = −     +   Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. For this sequence having the increasing/decreasing information will probably make the determining the bounds (if any exist) somewhat easier. Step 1 For this problem let’s get the increasing/decreasing information first. Calculus II 14 © 2018 Paul Dawkins For Problems 1 – 3 in this section it was easy enough to just ask what happens if we increase n to determine the increasing/decreasing information for this problem. However, in this case, increasing n will increase both the numerator and denominator and so it would be somewhat more difficult to do that analysis here. Therefore, we will resort to some quick Calculus I to determine increasing/decreasing information. We can define the following function and take its derivative. ( ) ( ) ( ) 2 2 2 2 25 25 25 x x f x f x x x − − ′ = ⇒ = + + Hopefully, it’s fairly clear that the critical points of the function are 5 x = ± . We’ll leave it to you to draw a quick number line or sign chart to verify that the function will be decreasing in the range 2 5 x ≤ < and increasing in the range 5 x > . Note that we just looked at the ranges of x that correspond to the ranges of n for our sequence here. Now, because the function values are the same as the sequence values when x is an integer we can see that the sequence, which starts at 2 n = , has terms that increase and terms that decrease and hence the sequence is not an increasing sequence and the sequence is not a decreasing sequence. That also means that the sequence is not monotonic. Step 2 Now let’s see what bounded information we can get. In this case, unlike many of the previous problems in this section, we don’t have a monotonic sequence. However, we can still use the increasing/decreasing information above to help us out with the bounds. First, we know that the sequence is decreasing in the range 2 5 n ≤ < and increasing in the range 5 n > . From our Calculus I knowledge we know that this means 5 n = must be a minimum of the sequence terms and hence the sequence is bounded below by 5 1 50 10 −= − (i.e. the 5 n = sequence term). Next let’s look for the upper bound (if it exists). For this problem let’s first take a quick look at the limit of the sequence terms. In this case the limit of the sequence terms is, 2 lim 0 25 n n n →∞ − = + Recall what this limit tells us about the behavior of our sequence terms. The limit means that as n →∞ the sequence terms must be getting closer and closer to zero. Now, for a second, let’s look at just the portion of the sequence with 5 n > and let’s further suppose that zero is not an upper bound for the sequence terms with 5 n > . Let’s also keep in mind that we’ve already determined that the sequence is increasing for 5 n > (means that each successive term must be larger than (i.e. above) the previous one…). Calculus II 15 © 2018 Paul Dawkins So, if zero is not an upper bound (for 5 n > ) then we know that somewhere there must be sequence terms with 5 n > above (or larger than) zero. So, we know that terms must be getting closer and closer to zero and we’ve now assumed there are terms above zero. Therefore the only way for the terms to approach the limit of zero is for at least a few sequence terms with 5 n > to decrease down towards zero (remember we’ve assumed there are terms above this!). That can’t happen however because we know that for 5 n > the sequence is increasing. Okay, what was the point of all this? Well recall that we got to this apparent contradiction to the increasing nature of the sequence for 5 n > by first assuming that zero was not an upper bound for the portion of the sequence with 5 n > . Since making this assumption led us to something that can’t possibly be true that in turn means that zero must in fact be an upper bound for the portion of the sequence with 5 n > since we’ve shown that sequence terms simply cannot go above this value! Note that we’ve not yet actually shown that zero in an upper bound for the sequence and in fact it might not actually be an upper bound. However, what we have shown is that it is an upper bound for the vast majority of the sequence, i.e. for the portion of the sequence with 5 n > . All we need to do to finish the upper bound portion of this problem off is check what the first few terms of the sequence are doing. There are several ways to do this. One is to just compute the remaining initial terms of the sequence to see if they are above or below zero. For this sequence that isn’t too bad as there are only 4 terms ( 2,3,4,5 n = ). However, if there’d been several hundred terms that wouldn’t be so easy so let’s take a look at another approach that will always be easy to do in this case because we have the increasing/decreasing information for this initial portion of the sequence. Let’s simply note that for the first part of this sequence we’ve already shown that the sequence is decreasing. Therefore, the very first sequence term of 2 29 − (i.e. the 2 n = sequence term) will be the largest term for this initial bit of the sequence that is decreasing. This term is clearly less than zero and so zero will also be larger than all the remaining terms in the initial decreasing portion of the sequence and hence the sequence is bounded above by zero. Finally, because this sequence is both bounded above and bounded below the sequence is bounded. Before leaving this problem a couple of quick words of caution. First, the limit of a sequence is not guaranteed to be a bound (upper or lower) for a sequence so be careful to not just always assume that the limit is an upper/lower bound for a sequence. Second, as this problem has shown determining the bounds of a sequence can sometimes be a fairly involved process that involves quite a bit of work and lots of various pieces of knowledge about the other behavior of the sequence. Calculus II 16 © 2018 Paul Dawkins Section 4-3 : Series - The Basics 1. Perform an index shift so that the following series starts at 3 n = . ( ) 1 1 2 3 n n n n ∞ − = − ∑ Solution There really isn’t all that much to this problem. Just remember that, in this case, we’ll need to increase the initial value of the index by two so it will start at 3 n = and this means all the n’s in the series terms will need to decrease by the same amount (two in this case…). Doing this gives the following series. ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 2 3 1 3 3 2 3 2 2 3 2 2 3 n n n n n n n n n n n n ∞ ∞ ∞ − − − − − − = = = − = − − = − − ∑ ∑ ∑ Be careful with parenthesis, exponents, coefficients and negative signs when “shifting” the n’s in the series terms. When replacing n with 2 n − make sure to add in parenthesis where needed to preserve coefficients and minus signs. 2. Perform an index shift so that the following series starts at 3 n = . 2 7 4 1 n n n ∞ = − + ∑ Solution There really isn’t all that much to this problem. Just remember that, in this case, we’ll need to decrease the initial value of the index by four so it will start at 3 n = and this means all the n’s in the series terms will need to increase by the same amount (four in this case…). Doing this gives the following series. ( ) ( ) ( ) 2 2 2 7 3 3 4 4 4 1 4 1 4 1 n n n n n n n n n ∞ ∞ ∞ = = = − + − − = = + + + + + ∑ ∑ ∑ Be careful with parenthesis, exponents, coefficients and negative signs when “shifting” the n’s in the series terms. When replacing n with 4 n + make sure to add in parenthesis where needed to preserve coefficients and minus signs. Calculus II 17 © 2018 Paul Dawkins 3. Perform an index shift so that the following series starts at 3 n = . ( ) ( ) 3 1 2 2 1 2 5 n n n n − ∞ + = − + ∑ Solution There really isn’t all that much to this problem. Just remember that, in this case, we’ll need to increase the initial value of the index by one so it will start at 3 n = and this means all the n’s in the series terms will need to decrease by the same amount (one in this case…). Doing this gives the following series. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 1 3 4 1 2 2 1 1 2 1 2 3 3 1 2 1 1 2 1 1 5 5 5 n n n n n n n n n n n n − −− − ∞ ∞ ∞ + − + − = = = − + − −+ − + = = ∑ ∑ ∑ Be careful with parenthesis, exponents, coefficients and negative signs when “shifting” the n’s in the series terms. When replacing n with 1 n − make sure to add in parenthesis where needed to preserve coefficients and minus signs. 4. Strip out the first 3 terms from the series 2 1 2 1 n n n − ∞ = + ∑ . Solution Remember that when we say we are going to “strip out” terms from a series we aren’t really getting rid of them. All we are doing is writing the first few terms of the series as a summation in front of the series. So, for this series stripping out the first three terms gives, 1 2 3 2 2 2 2 2 1 4 2 4 2 4 2 2 2 2 2 1 1 1 2 1 3 1 1 1 1 1 2 4 20 80 1 5 2 16 1 n n n n n n n n n n n n − − − − − ∞ ∞ = = − ∞ = − ∞ = = + + + + + + + + = + + + + = + + ∑ ∑ ∑ ∑ This first step isn’t really all that necessary but was included here to make it clear that we were plugging in 1 n = , 2 n = and 3 n = (i.e. the first three values of n) into the general series term. Also, don’t Calculus II 18 © 2018 Paul Dawkins forget to change the starting value of n to reflect the fact that we’ve “stripped out” the first three values of n or terms. 5. Given that 3 0 1 1.6865 1 n n ∞ = = + ∑ determine the value of 3 2 1 1 n n ∞ = + ∑ . Step 1 First notice that if we strip out the first two terms from the series that starts at 0 n = the result will involve a series that starts at 2 n = . Doing this gives, 3 3 3 3 3 0 2 2 1 1 1 1 3 1 1 0 1 1 1 1 2 1 n n n n n n ∞ ∞ ∞ = = = = + + = + + + + + + ∑ ∑ ∑ Step 2 Now, for this situation we are given the value of the series that starts at 0 n = and are asked to determine the value of the series that starts at 2 n = . To do this all we need to do is plug in the known value of the series that starts at 0 n = into the “equation” above and “solve” for the value of the series that starts at 2 n = . This gives, 3 3 2 2 3 1 1 3 1.6865 1.6865 0.1865 2 1 1 2 n n n n ∞ ∞ = = = + ⇒ = − = + + ∑ ∑ Calculus II 19 © 2018 Paul Dawkins Section 4-4 : Convergence/Divergence of Series 1. Compute the first 3 terms in the sequence of partial sums for the following series. 1 2n n n ∞ = ∑ Solution Remember that nth term in the sequence of partial sums is just the sum of the first n terms of the series. So, computing the first three terms in the sequence of partial sums is pretty simple to do. Here is the work for this problem. ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 1 2 3 3 1 2 2 1 2 2 2 10 1 2 2 2 3 2 34 s s s = = = + = = + + = 2. Compute the first 3 terms in the sequence of partial sums for the following series. 3 2 2 n n n ∞ = + ∑ Solution Remember that nth term in the sequence of partial sums is just the sum of the first n terms of the series. So, computing the first three terms in the sequence of partial sums is pretty simple to do. Here is the work for this problem. ( ) ( ) ( ) ( ) ( ) ( ) 3 4 5 2 3 6 3 2 5 2 3 2 4 38 3 2 4 2 15 2 3 2 4 2 5 416 3 2 4 2 5 2 105 s s s = = + = + = + + = + + = + + + Calculus II 20 © 2018 Paul Dawkins 3. Assume that the nth term in the sequence of partial sums for the series 0 n n a ∞ = ∑ is given below. Determine if the series 0 n n a ∞ = ∑ is convergent or divergent. If the series is convergent determine the value of the series. 2 2 5 8 2 7 n n s n + = − Solution There really isn’t all that much that we need to do here other than to recall, 0 lim n n n n a s ∞ →∞ = = ∑ So, to determine if the series converges or diverges, all we need to do is compute the limit of the sequence of the partial sums. The limit of the sequence of partial sums is, 2 2 5 8 8 lim lim 2 7 7 n n n n s n →∞ →∞ + = = − − Now, we can see that this limit exists and is finite (i.e. is not plus/minus infinity). Therefore, we now know that the series, 0 n n a ∞ = ∑ , converges and its value is, 0 8 7 n n a ∞ = = − ∑ If you are unfamiliar with limits at infinity then you really need to go back to the Calculus I material and do some review of limits at infinity and L’Hospital’s Rule as we will be doing quite a bit of these kinds of limits off and on over the next few sections. 4. Assume that the nth term in the sequence of partial sums for the series 0 n n a ∞ = ∑ is given below. Determine if the series 0 n n a ∞ = ∑ is convergent or divergent. If the series is convergent determine the value of the series. 2 5 2 n n s n = + Calculus II 21 © 2018 Paul Dawkins Solution There really isn’t all that much that we need to do here other than to recall, 0 lim n n n n a s ∞ →∞ = = ∑ So, to determine if the series converges or diverges, all we need to do is compute the limit of the sequence of the partial sums. The limit of the sequence of partial sums is, 2 lim lim 5 2 n n n n s n →∞ →∞ = = ∞ + Now, we can see that this limit exists and but is infinite. Therefore, we now know that the series, 0 n n a ∞ = ∑ , diverges. If you are unfamiliar with limits at infinity then you really need to go back to the Calculus I material and do some review of limits at infinity and L’Hospital’s Rule as we will be doing quite a bit of these kinds of limits off and on over the next few sections. 5. Show that the following series is divergent. 2 0 3 1 n n n n ∞ = + ∑ e Solution First let’s note that we’re being asked to show that the series is divergent. We are not being asked to determine if the series is divergent. At this point we really only know of two ways to actually show this. The first option is to show that the limit of the sequence of partial sums either doesn’t exist or is infinite. The problem with this approach is that for many series determining the general formula for the nth term of the sequence of partial sums is very difficult if not outright impossible to do. That is true for this series and so that is not really a viable option for this problem. Luckily enough for us there is actually an easier option to simply show that a series is divergent. All we need to do is use the Divergence Test. The limit of the series terms is, 2 3 lim lim 0 1 n n n n n s n →∞ →∞ = = ∞≠ + e Calculus II 22 © 2018 Paul Dawkins The limit of the series terms is not zero and so by the Divergence Test we know that the series in this problem is divergence. 6. Show that the following series is divergent. 2 2 5 6 8 9 3 2 n n n n n ∞ = + + + + ∑ Solution First let’s note that we’re being asked to show that the series is divergent. We are not being asked to determine if the series is divergent. At this point we really only know of two ways to actually show this. The first option is to show that the limit of the sequence of partial sums either doesn’t exist or is infinite. The problem with this approach is that for many series determining the general formula for the nth term of the sequence of partial sums is very difficult if not outright impossible to do. That is true for this series and so that is not really a viable option for this problem. Luckily enough for us there is actually an easier option to simply show that a series is divergent. All we need to do is use the Divergence Test. The limit of the series terms is, 2 2 6 8 9 lim lim 9 0 3 2 n n n n n s n n →∞ →∞ + + = = ≠ + + The limit of the series terms is not zero and so by the Divergence Test we know that the series in this problem is divergence. Calculus II 23 © 2018 Paul Dawkins Section 4-5 : Special Series 1. Determine if the series converges or diverges. If the series converges give its value. 2 1 3 0 3 2 n n n ∞ + − = ∑ Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a geometric series. Step 2 Let’s also notice that the initial value of the index is 0 n = and so we can put this into the form, 0 n n a r ∞ = ∑ At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge. In this case it’s pretty simple to put the series into the form above so here is that work. ( )( ) 2 1 3 2 1 3 3 0 0 0 0 0 3 3 3 3 2 3 3 2 2 9 2 18 18 2 8 8 n n n n n n n n n n n n n n ∞ ∞ ∞ ∞ ∞ + − − = = = = =   = = = =     ∑ ∑ ∑ ∑ ∑ Make sure you properly deal with any negative exponents that might happen to be in the terms! Also recall that all the exponents must be simply n and can’t be 3n or anything else. So, for this problem, we’ll need to use basic exponent rules to write ( ) 3 3 2 2 8 n n n = = . Step 3 With the series in “proper” form we can see that 18 a = and 3 8 r = . Therefore, because we can clearly see that 3 8 1 r = < , the series will converge and its value is, 2 1 3 3 0 0 8 3 18 144 3 2 18 8 1 5 n n n n n ∞ ∞ + − = =   = = =   −   ∑ ∑ 2. Determine if the series converges or diverges. If the series converges give its value. 1 5 6 n n ∞ = ∑ Calculus II 24 © 2018 Paul Dawkins Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a harmonic series. Step 2 So because this is a harmonic series we know that it will diverge. 3. Determine if the series converges or diverges. If the series converges give its value. ( ) 3 2 1 6 8 n n n − ∞ − = − ∑ Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a geometric series. Step 2 Let’s also notice that the initial value of the index is 1 n = and so we can put this into the form, 1 1 n n a r ∞ − = ∑ At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge. So, let’s get started on the work to put the series into the form above. First, let’s get take care of the fact that both the n’s in the exponents are negative and they should be positive. Converting to positive n’s gives, ( ) ( ) 3 2 3 2 1 1 6 8 8 6 n n n n n n − − ∞ ∞ − − = = − = − ∑ ∑ Note that how you chose to deal with the 3 and the 2 in the respective exponents is up to you. You can either do it the way we did here or strip them out and then move the terms to the numerator or denominator. As noted above we need the two exponents to be 1 n −. This is an easy “fix” if we note that using basic exponent properties we can write each term as follows, ( ) ( ) ( ) 3 1 2 2 1 1 8 8 8 6 6 6 n n n n − − − − − − = − = − − With these two rewrites the series becomes, Calculus II 25 © 2018 Paul Dawkins ( ) ( ) ( ) ( ) 3 2 1 1 1 1 1 2 2 1 1 1 1 1 6 6 8 8 8 9 4 8 8 6 2 3 6 6 n n n n n n n n n n − − − − − ∞ ∞ ∞ ∞ − − − = = = = − −     = = = −     −     − − ∑ ∑ ∑ ∑ Step 3 With the series in “proper” form we can see that 9 2 a = and 4 3 r = − . Therefore, because we can clearly see that 4 4 3 3 1 r = − = > , the series will diverge. 4. Determine if the series converges or diverges. If the series converges give its value. 2 1 3 7 12 n n n ∞ = + + ∑ Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is not a geometric or harmonic series. That only leaves telescoping as a possibility. Step 2 Now, we need to be careful here. There is no way to actually identify the series as a telescoping series at this point. We are only hoping that it is a telescoping series. Therefore, the first real step here is to perform partial fractions on the series term to see what we get. Here is the partial fraction work for the series term. ( )( ) ( ) ( ) 2 3 3 3 4 3 7 12 3 4 3 4 A B A n B n n n n n n n = = + → = + + + + + + + + + 3 3 3 4 3 3 n A A n B B = − = = → = − = − = − The series term in partial fraction form is then, 2 3 3 3 7 12 3 4 n n n n = − + + + + Step 3 The partial sums for this series are then, 1 3 3 3 4 n n i s i i =   = −   + +   ∑ Calculus II 26 © 2018 Paul Dawkins Step 4 Expanding the partial sums from the previous step give, 1 3 3 3 3 3 4 4 5 n n i s i i =   = − = −   + +   ∑ 3 5  +     3 6 − 3 6  +     3 7 − 3 1 n  +     + +  3 2 n − + 3 2 n  +   +   3 3 n − + 3 3 n  +   +   3 4 3 3 4 4 n n   −   +   = − + It is important when doing this expanding to expand out from both the initial and final values of i and to expand out until all the parts of a series term cancel. Once that has been done it is safe to assume that the cancelling will continue until we get near the end of the expansion. Note that at this point we now know that the series was a telescoping series since we got all the “interior” terms to cancel out. Step 5 At this point all we need to do is look at the limit of the partial sums to get, 3 3 3 lim lim 4 4 4 n n n s n →∞ →∞   = − =   +   Step 6 The limit of the partial sums exists and is a finite number (i.e. not infinity) and so we can see that the series converges and its value is, 2 1 3 3 7 12 4 n n n ∞ = = + + ∑ 5. Determine if the series converges or diverges. If the series converges give its value. 1 2 1 5 7 n n n + ∞ − = ∑ Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a geometric series. Step 2 Let’s also notice that the initial value of the index is 1 n = and so we can put this into the form, Calculus II 27 © 2018 Paul Dawkins 1 1 n n a r ∞ − = ∑ At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge. As noted above we need the two exponents to be 1 n −. This is an easy “fix” if we note that using basic exponent properties we can write each term as follows, 1 1 2 2 1 1 5 5 5 7 7 7 n n n n + − − − − = = With these two rewrites the series becomes, ( )( ) 1 1 1 2 1 2 1 1 1 1 1 1 1 5 5 5 5 5 25 7 175 7 7 7 7 7 n n n n n n n n n n n − + − − ∞ ∞ ∞ ∞ − − − − = = = =   = = =     ∑ ∑ ∑ ∑ Step 3 With the series in “proper” form we can see that 175 a = and 5 7 r = . Therefore, because we can clearly see that 5 7 1 r = < , the series will converge and its value is, 1 1 2 5 1 1 7 5 5 175 1225 175 7 7 1 2 n n n n n − + ∞ ∞ − = =   = = =   −   ∑ ∑ 6. Determine if the series converges or diverges. If the series converges give its value. 1 2 2 5 7 n n n + ∞ − = ∑ Step 1 Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is a geometric series. Step 2 Now, while we have correctly identified this as a geometric series it doesn’t start at either of the two standard starting values of n, i.e. 0 n = or 1 n = . This won’t stop us from determining if the series converges or diverges because that only depends on the value of r which we can determine regardless of the starting value of n with enough work. However, if the series does converge we won’t be able to use the formula for determining the value of the series as that also needs the value of a and that does require the series to start at one of the two standard starting values. Calculus II 28 © 2018 Paul Dawkins We have two options for taking care of this problem. One is to use an index shift to convert this into a series that starts at one of the standard starting values of n. In most cases this is probably the only real option. However, in this case let’s notice that this series is almost identical to the series from the previous problem. The only difference is that this series starts at 2 n = while the series in the previous problem starts at 1 n = . This means that we can use the results of the previous problem to greatly reduce the amount of work needed here. Step 3 We know that the series in the previous problem converged and since we’re only changing the starting value of n that will not affect the convergence of the series. Therefore, the series in this problem will also converge. Since we also know that the value of the series in the previous series is 1225 2 we can find the value of the series in this problem. All we need to do is strip out one term from the series in the previous problem to get, 1 2 1 2 1 2 1 2 5 5 5 7 7 7 n n n n n n + + ∞ ∞ − − − = = = + ∑ ∑ Then using the value we found in the previous problem can get the value of the series from this problem as follows, 1 1 2 2 2 2 1225 5 5 1225 875 175 175 2 7 7 2 2 n n n n n n + + ∞ ∞ − − = = = + ⇒ = − = ∑ ∑ On a quick side note if you did chose to do an index shift here are the two series (for each possible starting value of n) that you should have gotten. 1 2 3 2 1 2 1 0 5 5 5 7 7 7 n n n n n n n n n + + + ∞ ∞ ∞ − − = = = = = ∑ ∑ ∑ Both of the last two are in the “standard” form and can be used to arrive at the same result as above. 7. Determine if the series converges or diverges. If the series converges give its value. 2 4 10 4 3 n n n ∞ = − + ∑ Step 1 Calculus II 29 © 2018 Paul Dawkins Given that all three of the special series we looked at in this section are all pretty distinct it is hopefully clear that this is not a geometric or harmonic series. That only leaves telescoping as a possibility. Step 2 Now, we need to be careful here. There is no way to actually identify the series as a telescoping series at this point. We are only hoping that it is a telescoping series. Therefore, the first real step here is to perform partial fractions on the series term to see what we get. Here is the partial fraction work for the series term. ( )( ) ( ) ( ) 2 10 10 10 3 1 4 3 1 3 1 3 A B A n B n n n n n n n = = + → = − + − − + − − − − 1 10 2 5 3 10 2 5 n A A n B B = = − = − → = = = The series term in partial fraction form is then, 2 10 5 5 4 3 3 1 n n n n = − − + − − Step 3 The partial sums for this series are then, 4 5 5 3 1 n n i s i i =   = −   − −   ∑ Step 4 Expanding the partial sums from the previous step give, 4 5 5 3 1 5 5 1 3 n n i s i i =   = −   − −   = − ∑ 5 5 2 4  + −     5 3  +     5 5 − 5 4  +     5 6 − 5 5  +     5 7 − 5 7 n  +     + −  5 5 n − − 5 6 n  +   −   5 4 n − − 5 5 n  +   −   5 3 n − − 5 4 n  +     − 5 5 2 3 n n   − +   − −   5 1 5 5 5 5 2 2 1 n n n   −   −   = + − − − − Calculus II 30 © 2018 Paul Dawkins It is important when doing this expanding to expand out from both the initial and final values of i and to expand out until all the parts of a series term cancel. Once that has been done it is safe to assume that the cancelling will continue until we get near the end of the expansion. Also, as seen above these can be quite messy to expand out until everything starts to cancel out so don’t get too excited about it when it does get messy like this. It just happens sometimes and we have to be careful with all the expansion. Note that at this point we now know that the series was a telescoping series since we got almost all the “interior” terms to cancel out. Step 5 At this point all we need to do is look at the limit of the partial sums to get, 15 5 5 15 lim lim 2 2 1 2 n n n s n n →∞ →∞   = − − =   − −   Step 6 The limit of the partial sums exists and is a finite number (i.e. not infinity) and so we can see that the series converges and its value is, 2 4 10 15 4 3 2 n n n ∞ = = − + ∑ Calculus II 31 © 2018 Paul Dawkins Section 4-6 : Integral Test 1. Determine if the following series converges or diverges 1 1 n nπ ∞ = ∑ Solution There really isn’t all that much to this problem. We could use the Integral Test on this series or we could just use the p-series Test we discussed in the notes for this section. We can clearly see that 1 p π = > and so by the p-series Test this series must converge. 2. Determine if the following series converges or diverges. 0 2 3 5 n n ∞ = + ∑ Step 1 Okay, prior to using the Integral Test on this series we first need to verify that we can in fact use the Integral Test! Step 2 The series terms are, 2 3 5 n a n = + We can clearly see that for the range of n in the series the terms are positive and so that condition is met. Step 3 In this case because there is only one n in the denominator and because all the terms in the denominator are positive it is (hopefully) clear that, ( ) 1 2 2 3 5 3 5 1 n n a a n n + = > = + + + and so the series terms are decreasing. Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test. Calculus II 32 © 2018 Paul Dawkins It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 4 Now, let’s compute the integral for the test. 0 0 0 2 2 2 2 2 lim lim ln 3 5 lim ln 3 5 ln 3 3 5 3 5 5 5 5 t t t t t dx dx x t x x ∞ →∞ →∞ →∞     = = + = + − = ∞     + +     ⌠ ⌠   ⌡ ⌡ Step 5 Okay, the integral from the last step is a divergent integral and so by the Integral Test the series must also be a divergent series. 3. Determine if the following series converges or diverges. ( ) 3 2 1 2 7 n n ∞ = + ∑ Step 1 Okay, prior to using the Integral Test on this series we first need to verify that we can in fact use the Integral Test! Step 2 The series terms are, ( ) 3 1 2 7 n a n = + We can clearly see that for the range of n in the series the terms are positive and so that condition is met. Step 3 In this case because there is only one n in the denominator and because all the terms in the denominator are positive it is (hopefully) clear that, ( ) ( ) ( ) 1 3 3 1 1 2 7 2 1 7 n n a a n n + = > = + + + and so the series terms are decreasing. Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test. Calculus II 33 © 2018 Paul Dawkins It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 4 Now, let’s compute the integral for the test. ( ) ( ) ( ) ( ) ( ) 3 3 2 2 2 2 2 2 1 1 1 1 lim lim 4 2 7 2 7 2 7 1 1 1 1 1 lim 4 4 484 2 7 11 t t t t t dx dx x x x t ∞ →∞ →∞ →∞   = = −     + + +     = − + =     +   ⌠ ⌠   ⌡ ⌡ Step 5 Okay, the integral from the last step is a convergent integral and so by the Integral Test the series must also be a convergent series. 4. Determine if the following series converges or diverges. 2 3 0 1 n n n ∞ = + ∑ Step 1 Okay, prior to using the Integral Test on this series we first need to verify that we can in fact use the Integral Test! Step 2 The series terms are, 2 3 1 n n a n = + We can clearly see that for the range of n in the series the terms are positive and so that condition is met. Step 3 In this case we need to be a little more careful with checking the decreasing condition. We can’t just plug in n + 1 into the series term as we’ve done in the first couple of problems in this section. Doing that would suggest that both the numerator and denominator will increase and so it’s not all that clear cut of a case that the terms will be decreasing. Therefore, we’ll need to do a quick Calculus I increasing/decreasing analysis. Here the function for the series terms and its derivative. Calculus II 34 © 2018 Paul Dawkins ( ) ( ) ( ) ( ) ( ) 3 2 4 2 2 3 3 3 2 2 1 1 1 x x x x x f x f x x x x − − ′ = = = + + + With a quick number line or sign chart we can see that the function will increase for 3 0 2 1.2599 x < < = and will decrease for 3 2 1.2599 x = < < ∞. Because the function and series terms are the same we know that the series terms will have the same increasing/decreasing behavior. So, from this analysis we can see that the series terms are not always decreasing but will be decreasing for 3 2 n > which is sufficient for us to use to say that this condition is also met. Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 4 Now, let’s compute the integral for the test. ( ) 2 2 3 3 3 3 0 0 0 1 1 lim lim ln 1 lim ln 1 ln 1 1 1 3 3 t t t t t x x dx dx x t x x ∞ →∞ →∞ →∞     = = + = + − = ∞     + +     ⌠ ⌠   ⌡ ⌡ Step 5 Okay, the integral from the last step is a divergent integral and so by the Integral Test the series must also be a divergent series. 5. Determine if the following series converges or diverges. 2 3 3 3 2 n n n ∞ = − + ∑ Step 1 Okay, prior to using the Integral Test on this series we first need to verify that we can in fact use the Integral Test! Step 2 The series terms are, 2 3 3 2 n a n n = − + Calculus II 35 © 2018 Paul Dawkins We can clearly see that for 3 n ≥ (which matches our range of n for the series) we will have, 2 2 2 2 3 3 0 3 2 3 0 n n n n n n n n ≥ ⇒ − ≥ ⇒ − + ≥ − ≥ Therefore, the series terms are positive and so that condition is met. Note that on occasion we’ll need to do more than just state that the series terms are positive by inspection and do a little work to show that the terms really are positive! Step 3 In this case we need to be a little more careful with checking the decreasing condition. We can’t just plug in n + 1 into the series term as we’ve done in the first couple of problems in this section. Doing that the first term in the denominator would be getting larger which would suggest the series term is decreasing. However, because the second term in the denominator is subtracted off if we increase n that would suggest the denominator is getting larger and hence the series term is increasing. Because we have these “competing” interests we’ll need to do a quick Calculus I increasing/decreasing analysis. Here the function for the series terms and its derivative. ( ) ( ) ( ) 2 2 2 3 9 6 3 2 3 2 x f x f x x x x x − ′ = = − + − + With a quick number line or sign chart we can see that the function will increase for 3 2 x < and will decrease for 3 2 x > . Because the function and series terms are the same we know that the series terms will have the same increasing/decreasing behavior. So, from this analysis we can see that the series terms are always decreasing for the range n in our series and so this condition is also met. Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 4 Now, let’s compute the integral for the test. The integral we’ll need to compute is, 2 3 3 3 2 dx x x ∞ − + ⌠  ⌡ This integral will however require us to do some quick partial fractions in order to do the evaluation. Here is that quick work. Calculus II 36 © 2018 Paul Dawkins ( )( ) ( ) ( ) 3 3 2 1 1 2 1 2 A B A x B x x x x x = + → = − + − − − − − 1 3 3 2 3 3 x A A x B B = = − = − ⇒ = = = The integral is then, ( ) ( ) ( ) ( ) 3 3 3 3 3 3 3 lim lim 3ln 2 3ln 1 2 1 2 1 lim 3ln 2 3ln 1 3ln 1 3ln 2 2 1 lim 3ln 3ln 2 3ln 3ln 2 3ln 2 1 1 t t t t t t dx dx x x x x x x t t t t ∞ →∞ →∞ →∞ →∞ − = − = − − − − − − −   = − − −− −    −   = + = + =    −    ⌠ ⌠   ⌡ ⌡ Be careful with the limit of the first two terms! To correctly compute the limit they need to be combined using logarithm properties as shown and we can then do a L’Hospital’s Rule on the argument of the log to compute the limit. Step 5 Okay, the integral from the last step is a convergent integral and so by the Integral Test the series must also be a convergent series. Calculus II 37 © 2018 Paul Dawkins Section 4-7 : Comparison Test/Limit Comparison Test 1. Determine if the following series converges or diverges. 2 2 1 1 1 n n ∞ =   +     ∑ Step 1 First, the series terms are, 2 2 1 1 n a n   = +     and it should pretty obvious in this case that they are positive and so we know that we can use the Comparison Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 For most of the Comparison Test problems we usually guess the convergence and proceed from there. However, in this case it is hopefully clear that for any n, ( ) 2 2 2 1 1 1 1 n   + > =     Now, let’s take a look at the following series, 1 1 n ∞ = ∑ Because lim1 1 0 n→∞= ≠ we can see from the Divergence Test that this series will be divergent. So we’ve found a divergent series with terms that are smaller than the original series terms. Therefore, by the Comparison Test the series in the problem statement must also be divergent. Calculus II 38 © 2018 Paul Dawkins As a final note for this problem notice that we didn’t actually need to do a Comparison Test to arrive at this answer. We could have just used the Divergence Test from the beginning since, 2 2 1 lim 1 1 0 n n →∞   + = ≠     This is something that you should always keep in mind with series convergence problems. The Divergence Test is a quick test that can, on occasion, be used to quickly determine that a series diverges and hence avoid a lot of the hassles of some of the other tests. 2. Determine if the following series converges or diverges. 2 3 4 3 n n n ∞ = − ∑ Step 1 First, the series terms are, 2 3 3 n n a n = − and it should pretty obvious that as long as 3 n > (which we’ll always have for this series) that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. The “-3” in the denominator won’t really affect the size of the denominator for large enough n and so it seems like for large n that the term will probably behave like, 2 3 1 n n b n n = = We also know that the series, 4 1 n n ∞ = ∑ Calculus II 39 © 2018 Paul Dawkins will diverge because it is a harmonic series or by the p-series Test. Therefore, it makes some sense that we can guess the series in the problem statement will probably diverge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series diverges we’ll need to find a series with smaller terms that we know, or can prove, diverges. Note as well that we’ll also need to prove that the new series terms really are smaller than the terms from the series in the problem statement. We can’t just “hope” that the will be smaller. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms smaller by either making the numerator smaller or the denominator larger. In this case it should be pretty clear that, 3 3 3 n n > − Therefore, we’ll have the following relationship. 2 2 3 3 3 n n n n < − You do agree with this right? The numerator in each is the same while the denominator in the left term is larger than the denominator in the right term. Therefore, the rational expression on the left must be smaller than the rational expression on the right. Step 4 Now, the series, 2 3 4 4 1 n n n n n ∞ ∞ = = = ∑ ∑ is a divergent series (as discussed above) and we’ve also shown that the series terms in this series are smaller than the series terms from the series in the problem statement. Therefore, by the Comparison Test the series given in the problem statement must also diverge. Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different. Calculus II 40 © 2018 Paul Dawkins 3. Determine if the following series converges or diverges. ( ) 2 7 1 n n n ∞ = + ∑ Step 1 First, the series terms are, ( ) 7 1 n a n n = + and it should pretty obvious that for the range of n we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. The “+1” in the denominator won’t really affect the size of the denominator for large enough n and so it seems like for large n that the term will probably behave like, ( ) 2 7 7 n b n n n = = We also know that the series, 2 2 7 n n ∞ = ∑ will converge by the p-series Test ( 2 1 p = > ). Therefore, it makes some sense that we can guess the series in the problem statement will probably converge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 Calculus II 41 © 2018 Paul Dawkins So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. In this case it should be pretty clear that, ( ) ( ) 1 1 n n n n n n < + ⇒ < + Therefore, we’ll have the following relationship. ( ) ( ) 7 7 1 n n n n > + You do agree with this right? The numerator in each is the same while the denominator in the left term is smaller than the denominator in the right term. Therefore, the rational expression on the left must be larger than the rational expression on the right. Step 4 Now, the series, ( ) 2 2 2 7 7 n n n n n ∞ ∞ = = = ∑ ∑ is a convergent series (as discussed above) and we’ve also shown that the series terms in this series are larger than the series terms from the series in the problem statement. Therefore, by the Comparison Test the series given in the problem statement must also converge. Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different. 4. Determine if the following series converges or diverges. 2 7 4 2 3 n n n ∞ = − − ∑ Step 1 First, the series terms are, Calculus II 42 © 2018 Paul Dawkins 2 4 2 3 n a n n = − − You can verify that for 7 n ≥ we have 2 2 3 n n > + and so ( ) 2 2 2 3 2 3 0 n n n n − − = − + > . Therefore, the series terms are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. For large enough n we know that the 2 n (a quadratic term) in the denominator will increase at a much faster rate than the 2 3 n − − (a linear term) portion of the denominator. Therefore the 2 n portion of the denominator will, in all likelihood, define the behavior of the denominator and so the terms should behave like, 2 4 n b n = We also know that the series, 2 4 4 n n ∞ = ∑ will converge by the p-series Test ( 2 1 p = > ). Therefore, it makes some sense that we can guess the series in the problem statement will probably converge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. Calculus II 43 © 2018 Paul Dawkins We now have a problem however. The obvious thing to try is to drop the last two terms on the denominator. Doing that however gives the following inequality, 2 2 2 3 n n n > − − This in turn gives the following relationship. 2 2 4 4 2 3 n n n < − − The denominator on the left is larger and so the rational expression on the left must be smaller. This leads to the problem. While the series, 2 4 4 n n ∞ = ∑ will definitely converge (as discussed above) it’s terms are smaller than the series terms in the problem statement. Just because a series with smaller terms converges does not, in any way, imply a series with larger terms will also converge! There are other manipulations we might try but they are all liable to run into similar issues or end up with new terms that we wouldn’t be able to quickly prove convergence on. Hint : So, if the Comparison Test won’t easily work what else is there to do? Step 4 So, the Comparison Test won’t easily work in this case. That pretty much leaves the Limit Comparison Test to try. This test only requires positive terms (which we have) and a second series that we’re pretty sure behaves like the series we want to know the convergence for. Note as well that, for the Limit Comparison Test, we don’t care if the terms for the second series are larger or smaller than problem statement series terms. If you think about it we already have exactly what we need. In Step 2 we used a second series to guess at the convergence of the problem statement series. The terms in the new series are positive (which we need) and we’re pretty sure it behaves in the same manner as the problem statement series. So, let’s compute the limit we need for the Limit Comparison Test. 2 2 2 2 1 4 lim lim lim lim 1 2 3 4 2 3 n n n n n n n n a n n c a b b n n n n →∞ →∞ →∞ →∞       = = = = =       − − − −       Step 5 Okay. We now have 0 1 c < = < ∞, i.e. c is not zero or infinity and so by the Limit Comparison Test the two series must have the same convergence. We determined in Step 2 that the second series converges and so the series given in the problem statement must also converge. Calculus II 44 © 2018 Paul Dawkins Be careful with the Comparison Test. Too often students just try to “force” larger or smaller by just hoping that the second series terms has the correct relationship (i.e. larger or smaller as needed) to the problem series terms. The problem is that this often leads to an incorrect answer. Be careful to always prove the larger/smaller nature of the series terms and if you can’t get a series term of the correct larger/smaller nature then you may need to resort to the Limit Comparison Test. 5. Determine if the following series converges or diverges. 6 2 1 1 n n n ∞ = − + ∑ Step 1 First, the series terms are, 6 1 1 n n a n − = + and it should pretty obvious that for the range of n we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. The “-1” in the numerator and the “+1” in the denominator won’t really affect the size of the numerator and denominator respectively for large enough n and so it seems like for large n that the term will probably behave like, 3 2 6 1 n n n b n n n = = = We also know that the series, 2 2 1 n n ∞ = ∑ will converge by the p-series Test ( 2 1 p = > ). Calculus II 45 © 2018 Paul Dawkins Therefore, it makes some sense that we can guess the series in the problem statement will probably converge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. In this case we can work with both the numerator and the denominator. Let’s start with the numerator. It should be pretty clear that, 1 n n > − Using this we can make the numerator larger to get the following relationship, 6 6 1 1 1 n n n n − < + + Now, in the denominator it again is hopefully clear that, 6 6 1 n n < + Using this we can make the denominator smaller (and hence make the rational expression larger) to get, 2 6 6 6 1 1 1 1 n n n n n n n − < < = + + Step 4 Now, the series, 2 2 1 n n ∞ = ∑ is a convergent series (as discussed above) and we’ve also shown that the series terms in this series are larger than the series terms from the series in the problem statement. Therefore, by the Comparison Test the series given in the problem statement must also converge. Calculus II 46 © 2018 Paul Dawkins Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different. 6. Determine if the following series converges or diverges. ( ) 3 4 2 1 2 7 sin n n n n ∞ = + ∑ Step 1 First, the series terms are, ( ) 3 4 2 2 7 sin n n a n n + = and it should pretty obvious that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. The “+7” in the numerator and the “ ( ) 2 sin n ” in the denominator won’t really affect the size of the numerator and denominator respectively for large enough n and so it seems like for large n that the term will probably behave like, 3 4 2 2 n n b n n = = We also know that the series, 1 2 n n ∞ = ∑ will diverge because it is a harmonic series or by the p-series Test. Therefore, it makes some sense that we can guess the series in the problem statement will probably diverge as well. Calculus II 47 © 2018 Paul Dawkins Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series diverges we’ll need to find a series with smaller terms that we know, or can prove, diverges. Note as well that we’ll also need to prove that the new series terms really are smaller than the terms from the series in the problem statement. We can’t just “hope” that the will be smaller. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms smaller by either making the numerator smaller or the denominator larger. In this case we can work with both the numerator and the denominator. Let’s start with the numerator. It should be pretty clear that, 3 3 2 2 7 n n < + Using this we can make the numerator smaller to get the following relationship, ( ) ( ) 3 3 4 2 4 2 2 7 2 sin sin n n n n n n + > Now, we know that ( ) 2 0 sin 1 n ≤ ≤ and so in the denominator we can see that if we replace the ( ) 2 sin n with its largest possible value we have, ( ) ( ) 4 2 4 4 sin 1 n n n n < = Using this we can make the denominator larger (and hence make the rational expression smaller) to get, ( ) ( ) 3 3 3 4 2 4 2 4 2 7 2 2 2 sin sin n n n n n n n n n + > > = Step 4 Now, the series, 1 2 n n ∞ = ∑ is a divergent series (as discussed above) and we’ve also shown that the series terms in this series are smaller than the series terms from the series in the problem statement. Therefore, by the Comparison Test the series given in the problem statement must also diverge. Calculus II 48 © 2018 Paul Dawkins Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different. 7. Determine if the following series converges or diverges. ( ) ( ) 2 2 0 2 sin 5 4 cos n n n n n ∞ = + ∑ Step 1 First, the series terms are, ( ) ( ) 2 2 2 sin 5 4 cos n n n n a n = + and it should pretty obvious that for the range of n we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. The trig functions in the numerator and in the denominator won’t really affect the size of the numerator and denominator for large enough n and so it seems like for large n that the term will probably behave like, 2 2 1 4 4 2 n n n n n b     = = =         We also know that the series, 0 1 2 n n ∞ =       ∑ will converge because it is a geometric series with 1 2 1 r = < . Therefore, it makes some sense that we can guess the series in the problem statement will probably converge as well. Calculus II 49 © 2018 Paul Dawkins Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. In this case we can work with both the numerator and the denominator. Let’s start with the numerator. We know that ( ) 2 0 sin 5 1 n ≤ ≤ and so replacing the ( ) 2 sin 5n in the numerator with the largest possible value we get, ( ) ( ) 2 2 sin 5 2 1 2 n n n n < = Using this we can make the numerator larger to get the following relationship, ( ) ( ) ( ) 2 2 2 2 sin 5 2 4 cos 4 cos n n n n n n n < + + Now, in the denominator we know that ( ) 2 0 cos 1 n ≤ ≤ and so replacing the ( ) 2 cos n with the smallest possible value we get, ( ) 2 4 cos 4 0 4 n n n n + > + = Using this we can make the denominator smaller (and hence make the rational expression larger) to get, ( ) ( ) ( ) 2 2 2 2 sin 5 2 2 1 4 cos 4 cos 4 2 n n n n n n n n n n   < < =   + +   Step 4 Now, the series, 0 1 2 n n ∞ =       ∑ is a convergent series (as discussed above) and we’ve also shown that the series terms in this series are larger than the series terms from the series in the problem statement. Calculus II 50 © 2018 Paul Dawkins Therefore, by the Comparison Test the series given in the problem statement must also converge. Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different. 8. Determine if the following series converges or diverges. 2 3 2 n n n n − ∞ = + ∑ e Step 1 First, the series terms are, 2 2 n n a n n − = + e and it should pretty obvious that for the range of n we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 In this case let’s first notice the exponential in the numerator will go to zero as n goes to infinity. Let’s also notice that the denominator is just a polynomial. In cases like this the exponential is going to go to zero so fast that behavior of the denominator will not matter at all and in all probability the series in the problem statement will probably converge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. Calculus II 51 © 2018 Paul Dawkins In this case we can work with both the numerator and the denominator. Let’s start with the numerator. We can use some quick Calculus I to prove that n − e is a decreasing function and so, 3 1 n − − < < e e Using this we can make the numerator larger to get the following relationship, 2 2 1 2 2 n n n n n − < + + e Now, in the denominator it should be fairly clear that, 2 2 2 n n n + > Using this we can make the denominator smaller (and hence make the rational expression larger) to get, 2 2 2 1 1 2 2 n n n n n n − < < + + e Step 4 Now, the series, 2 3 1 n n ∞ = ∑ is a convergent series (p-series Test with 2 1 p = > ) and we’ve also shown that the series terms in this series are larger than the series terms from the series in the problem statement. Therefore, by the Comparison Test the series given in the problem statement must also converge. 9. Determine if the following series converges or diverges. 2 3 1 4 9 n n n n ∞ = − + ∑ Step 1 First, the series terms are, 2 3 4 9 n n n a n − = + Calculus II 52 © 2018 Paul Dawkins You can verify that for 1 n ≥ we have 2 4n n > and so 2 4 0 n n − > . Therefore, the series terms are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. For large enough n we know that the 2 n (a quadratic term) in the numerator will increase at a much faster rate than the n − (a linear term) portion of the numerator. Therefore the 2 n portion of the numerator will, in all likelihood, define the behavior of the numerator. Likewise, the “+9” in the denominator will not affect the size of the denominator for large n and so the terms should behave like, 2 3 4 4 n n b n n = = We also know that the series, 1 4 n n ∞ = ∑ will diverge because it is a harmonic series or by the p-series Test. Therefore, it makes some sense that we can guess the series in the problem statement will probably diverge as well. Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series diverge we’ll need to find a series with smaller terms that we know, or can prove, diverge. Note as well that we’ll also need to prove that the new series terms really are smaller than the terms from the series in the problem statement. We can’t just “hope” that the will be smaller. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms smaller by either making the numerator smaller or the denominator larger. We now have a problem however. The obvious thing to try is to drop the last term in both the numerator and the denominator. Doing that however gives the following inequalities, Calculus II 53 © 2018 Paul Dawkins 2 2 3 3 4 4 9 n n n n n − < + > Using these two in the series terms gives the following relationship, 2 2 2 3 3 3 4 4 4 4 9 9 n n n n n n n n − < < = + + Now the series, 0 4 n n ∞ = ∑ will definitely diverge (as discussed above) it’s terms are larger than the series terms in the problem statement. Just because a series with larger terms diverges does not, in any way, imply a series with smaller terms will also diverge! There are other manipulations we might try but they are all liable to run into similar issues or end up with new terms that we wouldn’t be able to quickly prove convergence on. Hint : So, if the Comparison Test won’t easily work what else is there to do? Step 4 So, the Comparison Test won’t easily work in this case. That pretty much leaves the Limit Comparison Test to try. This test only requires positive terms (which we have) and a second series that we’re pretty sure behaves like the series we want to know the convergence for. Note as well that, for the Limit Comparison Test, we don’t care if the terms for the second series are larger or smaller than problem statement series terms. If you think about it we already have exactly what we need. In Step 2 we used a second series to guess at the convergence of the problem statement series. The terms in the new series are positive (which we need) and we’re pretty sure it behaves in the same manner as the problem statement series. So, let’s compute the limit we need for the Limit Comparison Test. 2 3 2 3 3 1 4 4 lim lim lim lim 1 9 4 4 36 n n n n n n n n a n n n n n c a b b n n →∞ →∞ →∞ →∞       − − = = = = =       + +       Step 5 Okay. We now have 0 1 c < = < ∞, i.e. c is not zero or infinity and so by the Limit Comparison Test the two series must have the same convergence. We determined in Step 2 that the second series diverges and so the series given in the problem statement must also diverge. Be careful with the Comparison Test. Too often students just try to “force” larger or smaller by just hoping that the second series terms has the correct relationship (i.e. larger or smaller as needed) to the problem series terms. The problem is that this often leads to an incorrect answer. Be careful to always Calculus II 54 © 2018 Paul Dawkins prove the larger/smaller nature of the series terms and if you can’t get a series term of the correct larger/smaller nature then you may need to resort to the Limit Comparison Test. 10. Determine if the following series converges or diverges. 2 3 1 2 4 1 9 n n n n ∞ = + + + ∑ Step 1 First, the series terms are, 2 3 2 4 1 9 n n n a n + + = + and it should pretty obvious that for the range of n we have in this series that they are positive and so we know that we can attempt the Comparison Test for this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Hint : Can you make a guess as to whether or not the series should converge or diverge? Step 2 Let’s first see if we can make a reasonable guess as to whether this series converges or diverges. For large enough n we know that the 2 2n (a quadratic term) in the numerator will increase at a much faster rate than the 4 1 n + (a linear term) portion of the numerator. Therefore the 2 2n portion of the numerator will, in all likelihood, define the behavior of the numerator. Likewise, the “+9” in the denominator will not affect the size of the denominator for large n and so the terms should behave like, 2 3 2 2 2 n n b n n = = We also know that the series, 2 1 2 n n ∞ = ∑ will converge by the p-series Test ( 2 1 p = > ). Therefore, it makes some sense that we can guess the series in the problem statement will probably converge as well. Calculus II 55 © 2018 Paul Dawkins Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence? Step 3 So, because we’re guessing that the series converge we’ll need to find a series with larger terms that we know, or can prove, converge. Note as well that we’ll also need to prove that the new series terms really are larger than the terms from the series in the problem statement. We can’t just “hope” that the will be larger. In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms larger by either making the numerator larger or the denominator smaller. We now have a problem however. The obvious thing to try is to drop the last two terms in the numerator and the last term in the denominator. Doing that however gives the following inequalities, 2 2 3 3 2 2 4 1 9 n n n n n < + + + > This leads to a real problem! If we use the inequality for the numerator we’re going to get a smaller rational expression and if we use the inequality for the denominator we’re going to get a larger rational expression. Because these two can’t both be used at the same time it will make it fairly difficult to use the Comparison Test since neither one individually give a series we can quickly deal with. Hint : So, if the Comparison Test won’t easily work what else is there to do? Step 4 So, the Comparison Test won’t easily work in this case. That pretty much leaves the Limit Comparison Test to try. This test only requires positive terms (which we have) and a second series that we’re pretty sure behaves like the series we want to know the convergence for. Note as well that, for the Limit Comparison Test, we don’t care if the terms for the second series are larger or smaller than problem statement series terms. If you think about it we already have exactly what we need. In Step 2 we used a second series to guess at the convergence of the problem statement series. The terms in the new series are positive (which we need) and we’re pretty sure it behaves in the same manner as the problem statement series. So, let’s compute the limit we need for the Limit Comparison Test. ( ) ( ) ( ) ( ) 2 2 3 3 2 2 3 2 2 2 4 1 4 1 3 3 9 9 1 2 4 1 lim lim lim 9 2 2 2 2 lim lim 1 2 2 1 2 1 n n n n n n n n n n n n n n n a n n n c a b b n n n n n n n →∞ →∞ →∞ →∞ →∞     + + = = =     +           + + + +     = = = =   + +         Calculus II 56 © 2018 Paul Dawkins Step 5 Okay. We now have 0 1 c < = < ∞, i.e. c is not zero or infinity and so by the Limit Comparison Test the two series must have the same convergence. We determined in Step 2 that the second series converges and so the series given in the problem statement must also converge. Be careful with the Comparison Test. Too often students just try to “force” larger or smaller by just hoping that the second series terms has the correct relationship (i.e. larger or smaller as needed) to the problem series terms. The problem is that this often leads to an incorrect answer. Be careful to always prove the larger/smaller nature of the series terms and if you can’t get a series term of the correct larger/smaller nature then you may need to resort to the Limit Comparison Test. Calculus II 57 © 2018 Paul Dawkins Section 4-8 : Alternating Series Test 1. Determine if the following series converges or diverges. ( ) 1 1 1 7 2 n n n − ∞ = − + ∑ Step 1 First, this is (hopefully) clearly an alternating series with, 1 7 2 n b n = + and it should pretty obvious the n b are positive and so we know that we can use the Alternating Series Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 Let’s first take a look at the limit, 1 lim lim 0 7 2 n n n b n →∞ →∞ = = + So, the limit is zero and so the first condition is met. Step 3 Now let’s take care of the decreasing check. In this case it should be pretty clear that, ( ) 1 1 7 2 7 2 1 n n > + + + since increasing n will only increase the denominator and hence force the rational expression to be smaller. Therefore the n b form a decreasing sequence. Step 4 So, both of the conditions in the Alternating Series Test are met and so the series is convergent. 2. Determine if the following series converges or diverges. Calculus II 58 © 2018 Paul Dawkins ( ) 3 3 0 1 4 1 n n n n + ∞ = − + + ∑ Step 1 First, this is (hopefully) clearly an alternating series with, 3 1 4 1 n b n n = + + and it should pretty obvious the n b are positive and so we know that we can use the Alternating Series Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 Let’s first take a look at the limit, 3 1 lim lim 0 4 1 n n n b n n →∞ →∞ = = + + So, the limit is zero and so the first condition is met. Step 3 Now let’s take care of the decreasing check. In this case it should be pretty clear that, ( ) ( ) 3 3 1 1 4 1 1 4 1 1 n n n n > + + + + + + since increasing n will only increase the denominator and hence force the rational expression to be smaller. Therefore the n b form a decreasing sequence. Step 4 So, both of the conditions in the Alternating Series Test are met and so the series is convergent. 3. Determine if the following series converges or diverges. ( ) ( ) 0 1 1 2 3 n n n n ∞ = − + ∑ Calculus II 59 © 2018 Paul Dawkins Step 1 Do not get excited about the ( ) 1 n − is in the denominator! This is still an alternating series! All the ( ) 1 n − does is change the sign regardless of whether or not it is in the numerator. Also note that we could just as easily rewrite the terms as, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 1 1 1 2 3 1 2 3 1 1 2 3 1 2 3 n n n n n n n n n n n n n n n − − − = = = + − + − − + − + Note that ( ) 2 1 1 n − = because the exponent is always even! So, we now know that this is an alternating series with, 1 2 3 n n n b = + and it should pretty obvious the n b are positive and so we know that we can use the Alternating Series Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 Let’s first take a look at the limit, 1 lim lim 0 2 3 n n n n n b →∞ →∞ = = + So, the limit is zero and so the first condition is met. Step 3 Now let’s take care of the decreasing check. In this case it should be pretty clear that, 1 1 1 1 2 3 2 3 n n n n + + > + + since increasing n will only increase the denominator and hence force the rational expression to be smaller. Therefore the n b form a decreasing sequence. Step 4 Calculus II 60 © 2018 Paul Dawkins So, both of the conditions in the Alternating Series Test are met and so the series is convergent. 4. Determine if the following series converges or diverges. ( ) 6 2 0 1 9 n n n n + ∞ = − + ∑ Step 1 First, this is (hopefully) clearly an alternating series with, 2 9 n n b n = + and it should pretty obvious the n b are positive and so we know that we can use the Alternating Series Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 Let’s first take a look at the limit, 2 lim lim 0 9 n n n n b n →∞ →∞ = = + So, the limit is zero and so the first condition is met. Step 3 Now let’s take care of the decreasing check. In this case increasing n will increase both the numerator and denominator and so we can’t just say that clearly the terms are decreasing as we did in the first few problems. We will have no choice but to do a little Calculus I work for this problem. Here is the function and derivative for that work. ( ) ( ) ( ) 2 2 2 2 9 9 9 x x f x f x x x − ′ = = + + It should be pretty clear that the function will be increasing in 0 3 x ≤ < and decreasing in 3 x > (the range of x that corresponds to our range of n). Calculus II 61 © 2018 Paul Dawkins So, the n b do not actually form a decreasing sequence but they are decreasing for 3 n > and so we can say that they are eventually decreasing and as discussed in the notes that will be sufficient for us. Step 4 So, both of the conditions in the Alternating Series Test are met and so the series is convergent. 5. Determine if the following series converges or diverges. ( ) ( ) 2 2 4 1 1 3 n n n n n + ∞ = − − − ∑ Step 1 First, this is (hopefully) clearly an alternating series with, 2 1 3 n n b n n − = − and n b are positive for 4 n ≥ and so we know that we can use the Alternating Series Test on this series. It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that! Step 2 Let’s first take a look at the limit, 2 1 lim lim 0 3 n n n n b n n →∞ →∞ − = = − So, the limit is zero and so the first condition is met. Step 3 Now let’s take care of the decreasing check. In this case increasing n will increase both the numerator and denominator and so we can’t just say that clearly the terms are decreasing as we did in the first few problems. We will have no choice but to do a little Calculus I work for this problem. Here is the function and derivative for that work. ( ) ( ) ( ) 2 2 2 2 1 2 3 3 3 x x x f x f x x x x x − − + − ′ = = − − Calculus II 62 © 2018 Paul Dawkins The numerator of the derivative is never zero for any real number (we’ll leave that to you to verify) and since it is clearly negative at 0 x = we know that the function will always be decreasing for 4 x ≥ . Therefore the n b form a decreasing sequence. Step 4 So, both of the conditions in the Alternating Series Test are met and so the series is convergent. Calculus II 63 © 2018 Paul Dawkins Section 4-9 : Absolute Convergence 1. Determine if the following series is absolutely convergent, conditionally convergent or divergent. ( ) 1 3 2 1 1 n n n + ∞ = − + ∑ Step 1 Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms. ( ) 1 3 3 2 2 1 1 1 1 n n n n n + ∞ ∞ = = − = + + ∑ ∑ Now, notice that, 3 3 1 1 1 n n < + and we know by the p-series test that 3 2 1 n n ∞ = ∑ converges. Therefore, by the Comparison Test we know that the series from the problem statement, 3 2 1 1 n n ∞ = + ∑ will also converge. Step 2 So, because the series with the absolute value converges we know that the series in the problem statement is absolutely convergent. 2. Determine if the following series is absolutely convergent, conditionally convergent or divergent. ( ) 3 1 1 n n n − ∞ = − ∑ Step 1 Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms. Calculus II 64 © 2018 Paul Dawkins ( ) 1 2 3 1 1 1 1 1 1 n n n n n n n − ∞ ∞ ∞ = = = − = = ∑ ∑ ∑ Now, by the by the p-series test we can see that this series will diverge. Step 2 So, at this point we know that the series in the problem statement is not absolutely convergent so all we need to do is check to see if it’s conditionally convergent or divergent. To do this all we need to do is check the convergence of the series in the problem statement. The series in the problem statement is an alternating series with, 1 n b n = Clearly the n b are positive so we can use the Alternating Series Test on this series. It is hopefully clear that the n b are a decreasing sequence and lim 0 n n b →∞ = . Therefore, by the Alternating Series Test the series from the problem statement is convergent. Step 3 So, because the series with the absolute value diverges and the series from the problem statement converges we know that the series in the problem statement is conditionally convergent. 3. Determine if the following series is absolutely convergent, conditionally convergent or divergent. ( ) ( ) 1 3 3 1 1 1 n n n n + ∞ = − + + ∑ Step 1 Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms. ( ) ( ) 1 3 3 3 3 1 1 1 1 1 n n n n n n n + ∞ ∞ = = − + + = + + ∑ ∑ We know by the p-series test that the following series converges. 2 3 1 n n ∞ = ∑ If we now compute the following limit, Calculus II 65 © 2018 Paul Dawkins 2 3 2 3 3 1 lim lim 1 1 1 1 n n n n n n c n n →∞ →∞     + + = = =     + +     we know by the Limit Comparison Test that the two series in this problem have the same convergence because c is neither zero or infinity and because 2 3 1 n n ∞ = ∑ converges we know that the series from the problem statement must also converge. Step 2 So, because the series with the absolute value converges we know that the series in the problem statement is absolutely convergent. Calculus II 66 © 2018 Paul Dawkins Section 4-10 : Ratio Test 1. Determine if the following series converges or diverges. 1 2 2 1 3 1 n n n − ∞ = + ∑ Step 1 We’ll need to compute L. ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 1 2 1 2 1 2 2 2 2 2 2 1 2 2 2 1 3 1 lim lim lim 3 1 1 3 1 1 1 1 1 lim lim lim 3 3 9 1 1 1 1 9 1 1 n n n n n n n n n n n n n n a n L a a a n n n n n n n − + + + − →∞ →∞ →∞ −− − →∞ →∞ →∞ + = = = + + + + + = = = =   + + + + + +   When computing 1 n a + be careful to pay attention to any coefficients of n and powers of n. Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer! Step 2 Okay, we can see that 1 9 1 L = < and so by the Ratio Test the series converges. 2. Determine if the following series converges or diverges. ( ) 0 2 ! 5 1 n n n ∞ = + ∑ Step 1 We’ll need to compute L. ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )( )( ) 1 1 2 1 ! 1 5 1 lim lim lim 5 1 1 2 ! 2 2 ! 2 2 2 1 2 ! 2 2 2 1 5 1 5 1 5 1 lim lim lim 5 6 2 ! 5 6 2 ! 5 6 n n n n n n n n n n n a n L a a a n n n n n n n n n n n n n n n n + + →∞ →∞ →∞ →∞ →∞ →∞ + + = = = + + + + + + + + + + = = = = ∞ + + + Calculus II 67 © 2018 Paul Dawkins When computing 1 n a + be careful to pay attention to any coefficients of n and powers of n. Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer! Step 2 Okay, we can see that 1 L = ∞> and so by the Ratio Test the series diverges. 3. Determine if the following series converges or diverges. ( ) ( ) 1 3 2 1 2 2 1 5 n n n n n + ∞ + = − + ∑ Step 1 We’ll need to compute L. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 3 1 2 1 1 1 2 1 3 1 1 4 3 3 2 1 2 2 1 3 2 2 2 2 2 1 1 1 5 lim lim lim 1 5 2 1 2 2 2 2 5 lim lim 1 1 5 2 1 1 5 8 2 8 lim 5 5 1 1 n n n n n n n n n n n n n n n n n n n a n L a a a n n n n n n n n n n n n n n + + + + + + + + →∞ →∞ →∞ + + + + →∞ →∞ →∞ − + + = = = + − + − + − + = = + + − + + − + = = + + When computing 1 n a + be careful to pay attention to any coefficients of n and powers of n. Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer! Step 2 Okay, we can see that 8 5 1 L = > and so by the Ratio Test the series diverges. 4. Determine if the following series converges or diverges. ( ) 4 3 2 ! n n n ∞ = − ∑ e Step 1 We’ll need to compute L. Calculus II 68 © 2018 Paul Dawkins ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 4 1 1 1 4 4 4 4 4 4 4 4 2 ! 1 lim lim lim 1 2 ! 2 ! 2 ! lim lim lim 0 1 ! 1 2 ! 1 n n n n n n n n n n n n n n n n n a L a a a n n n n n n n + + + →∞ →∞ →∞ + + →∞ →∞ →∞ − = = = + − − − = = = = − − − − e e e e e e e When computing 1 n a + be careful to pay attention to any coefficients of n and powers of n. Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer! Step 2 Okay, we can see that 0 1 L = < and so by the Ratio Test the series converges. 5. Determine if the following series converges or diverges. ( ) 1 1 1 6 7 n n n + ∞ = − + ∑ Step 1 We’ll need to compute L. ( ) ( ) ( ) ( ) ( ) ( )( ) 1 1 1 1 1 2 1 1 1 6 7 lim lim lim 6 1 7 1 1 1 6 7 6 7 lim lim 1 6 13 6 13 1 n n n n n n n n n n n n n a n L a a a n n n n n + + + + + →∞ →∞ →∞ + + →∞ →∞ − + = = = + + − − − + + = = = + + − When computing 1 n a + be careful to pay attention to any coefficients of n and powers of n. Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer! Step 2 Okay, we can see that 1 L = and so by the Ratio Test tells us nothing about this series. Step 3 Just because the Ratio Test doesn’t tell us anything doesn’t mean we can’t determine if this series converges or diverges. In fact, it’s actually quite simple to do in this case. This is an Alternating Series with, Calculus II 69 © 2018 Paul Dawkins 1 6 7 n b n = + The n b are clearly positive and it should be pretty obvious (hopefully) that they also form a decreasing sequence. Finally, we also can see that lim 0 n n b →∞ = and so by the Alternating Series Test this series will converge. Note, that if this series were not in this section doing this as an Alternating Series from the start would probably have been the best way of approaching this problem. Calculus II 70 © 2018 Paul Dawkins Section 4-11 : Root Test 1. Determine if the following series converges or diverges. 2 1 3 1 4 2 n n n n ∞ = +     −   ∑ Step 1 We’ll need to compute L. 1 2 2 2 3 1 3 1 3 9 lim lim lim 4 2 4 2 2 4 n n n n n n n n n L a n n →∞ →∞ →∞ + +       = = = = − =       − −       Step 2 Okay, we can see that 9 4 1 L = > and so by the Root Test the series diverges. 2. Determine if the following series converges or diverges. 1 3 2 0 4 n n n n − ∞ = ∑ Step 1 We’ll need to compute L. ( )( ) 1 1 1 3 1 3 3 2 2 2 1 0 lim lim lim 0 4 4 4 16 n n n n n n n n n n n n n n L a − − − →∞ →∞ →∞ = = = = = = Step 2 Okay, we can see that 0 1 L = < and so by the Root Test the series converges. 3. Determine if the following series converges or diverges. ( ) 1 2 5 3 4 5 2 n n n + ∞ − = − ∑ Calculus II 71 © 2018 Paul Dawkins Step 1 We’ll need to compute L. ( ) ( ) ( ) 1 1 1 2 2 2 3 5 3 5 5 5 5 5 25 lim lim lim 2 2 32 2 n n n n n n n n n n L a + + − →∞ →∞ →∞ − − − − = = = = = Step 2 Okay, we can see that 25 32 1 L = < and so by the Root Test the series converges. Calculus II 72 © 2018 Paul Dawkins Section 4-12 : Strategy for Series Problems have not yet been written for this section. I was finding it very difficult to come up with a good mix of “new” problems and decided my time was better spent writing problems for later sections rather than trying to come up with a sufficient number of problems for what is essentially a review section. I intend to come back at a later date when I have more time to devote to this section and add problems then. Calculus II 73 © 2018 Paul Dawkins Section 4-13 : Estimating the Value of a Series 1. Use the Integral Test and 10 n = to estimate the value of ( ) 2 2 1 1 n n n ∞ = + ∑ . Step 1 Since we are being asked to use the Integral Test to estimate the value of the series we should first make sure that the Integral Test can actually be used on this series. First, the series terms are clearly positive so that condition is met. Now, let’s do a little Calculus I on the following function. ( ) ( ) ( ) ( ) 2 2 3 2 2 1 3 1 1 x x f x f x x x − ′ = = + + The derivative of the function will be negative for 1 3 0.5774 x > = and so the function will be decreasing in this range. Because the function and the series terms are the same we can also see that the series terms are decreasing for the range of n in our series. Therefore, the conditions required to use the Integral Test are met! Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We shouldn’t just assume that because we are being asked to use a test here that the test can actually be used! Step 2 Let’s start off with the partial sum using 10 n = . This is, ( ) 10 10 2 2 1 0.392632317 1 n n s n = = = + ∑ Step 3 Now, to increase the accuracy of the partial sum from the previous step we know we can use each of the following two integrals. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 10 10 10 2 2 2 2 2 2 11 11 11 1 1 1 1 lim lim lim 202 202 2 1 2 1 1 1 1 1 1 1 lim lim lim 244 244 2 1 2 1 1 1 t t t t t t t t t t x x dx dx x t x x x x dx dx x t x x ∞ →∞ →∞ →∞ ∞ →∞ →∞ →∞         = = − = − = + +     + +             = = − = − = + +     + +     ⌠ ⌠   ⌡ ⌡ ⌠ ⌠   ⌡ ⌡ Calculus II 74 © 2018 Paul Dawkins Step 4 Okay, we know from the notes in this section that if s represents that actual value of the series that it must be in the following range. 1 1 0.392632317 0.392632317 202 244 0.397582813 0.396730678 s s + < < + < < Step 5 Finally, if we average the two numbers above we can get a better estimate of, 0.397156745 s ≈ 2. Use the Comparison Test and 20 n = to estimate the value of ( ) 3 3 1 ln n n n ∞ = ∑ . Step 1 Since we are being asked to use the Comparison Test to estimate the value of the series we should first make sure that the Comparison Test can actually be used on this series. In this case that is easy enough because, for our range of n, the series terms are clearly positive and so we can use the Comparison Test. Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We shouldn’t just assume that because we are being asked to use a test here that the test can actually be used! Step 2 Let’s start off with the partial sum using 20 n = . This is, ( ) 20 20 3 3 1 0.057315878 ln n s n n = = = ∑ Step 3 Now, let’s see if we can get can get an error estimate on this approximation of the series value. To do that we’ll first need to do the Comparison Test on this series. That is easy enough for this series once we notice that ( ) ln n is an increasing function and so ( ) ( ) ln ln 3 n ≥ . Therefore, we get, Calculus II 75 © 2018 Paul Dawkins ( ) ( ) ( ) 3 3 3 1 1 1 1 ln ln 3 ln 3 n n n n ≤ = Step 4 We now know, from the discussion in the notes, that an upper bound on the value of the remainder (i.e. the error between the approximation and exact value) is, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 20 20 3 3 21 20 3 2 20 20 2 1 1 ln 3 ln 3 1 1 lim lim ln 3 2 ln 3 1 1 1 lim 800ln 3 2 ln 3 800ln 3 n t t t t t R T dx n x dx x x t ∞ ∞ = →∞ →∞ →∞ ≤ = <   = = −         = − =       ⌠  ⌡ ⌠  ⌡ ∑ Step 5 So, we can estimate that the value of the series is, 0.057315878 s ≈ and the error on this estimate will be no more than ( ) 1 800ln 3 0.001137799 = . 3. Use the Alternating Series Test and 16 n = to estimate the value of ( ) 2 2 1 1 n n n n ∞ = − + ∑ . Step 1 Since we are being asked to use the Alternating Series Test to estimate the value of the series we should first make sure that the Alternating Series Test can actually be used on this series. First, note that the n b for this series are, 2 1 n n b n = + and they are positive and with a quick derivative we can see they are decreasing and so the Alternating Series Test can be used here. Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We Calculus II 76 © 2018 Paul Dawkins shouldn’t just assume that because we are being asked to use a test here that the test can actually be used! Step 2 Let’s start off with the partial sum using 16 n = . This is, ( ) 16 16 2 2 1 0.260554530 1 n n n s n = − = = + ∑ Step 3 Now, we know, from the discussion in the notes, that an upper bound on the absolute value of the remainder (i.e. the error between the approximation and exact value) is nothing more than, 17 17 0.058620690 290 b = = Step 4 So, we can estimate that the value of the series is, 0.260554530 s ≈ and the error on this estimate will be no more than 0.058620690. 4. Use the Ratio Test and 8 n = to estimate the value of 1 3 2 1 3 2 n n n n + ∞ + = ∑ . Step 1 First notice that the terms are positive and so we can use the Ratio Test to do the estimate. Remember that this is a requirement only to use the Ratio Test to get an estimate of the series value and is not an actual requirement to use the Ratio Test to determine if the series converges or diverges. So, let’s start off with the partial sum using 8 n = . This is, 1 8 8 3 2 1 3 0.509881435 2 n n n s n + + = = = ∑ Step 2 Now, to get an upper bound on the value of the remainder (i.e. the error between the approximation and exact value) we need the following ratio, Calculus II 77 © 2018 Paul Dawkins ( ) ( ) 2 3 2 1 5 2 1 3 2 3 1 2 3 4 1 n n n n n n n a n n r a n n + + + + + = = = + + We’ll also potentially need the limit, ( ) 1 3 3 lim lim 4 1 4 n n n n a n L a n + →∞ →∞ = = = + Step 3 Next, we need to know if the n r form an increasing or decreasing sequence. A quick application of Calculus I will answer this. ( ) ( ) ( ) ( ) 2 3 3 0 4 1 4 1 x f x f x x x ′ = = > + + As noted above the derivative is always positive and so the function, and hence the n r are increasing. Step 4 The upper bound on the remainder is then, 6561 2,097,152 9 8 3 4 0.012514114 1 1 a R L ≤ = = − − Step 5 So, we can estimate that the value of the series is, 0.509881435 s ≈ and the error on this estimate will be no more than 0.012514114. Calculus II 78 © 2018 Paul Dawkins Section 4-14 : Power Series 1. For the following power series determine the interval and radius of convergence. ( ) ( ) ( ) 2 2 0 1 4 12 3 1 n n n x n ∞ + = − − + ∑ Step 1 Okay, let’s start off with the Ratio Test to get our hands on L. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 3 2 2 2 2 3 1 1 4 12 4 12 lim lim 1 4 12 3 1 1 3 1 1 1 1 4 12 lim 4 12 3 3 1 1 n n n n n n n n n x x L x n n n x x n + + + →∞ →∞ →∞ − + + − − = = − − + + − + + + = − = − + + Step 2 So, we know that the series will converge if, 1 4 3 4 12 1 3 1 3 3 3 4 x x x − < → − < → − < Step 3 So, from the previous step we see that the radius of convergence is 3 4 R = . Step 4 Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2. 3 3 9 15 3 4 4 4 4 x x − < − < → < < Step 5 To finalize the interval of convergence we need to check the end points of the inequality from Step 4. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 0 0 0 9 1 1 1 : 3 4 9 1 3 1 3 1 n n n n n x n n n ∞ ∞ ∞ + = = = = − = = + − + − + ∑ ∑ ∑ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 0 0 0 1 15 1 1 : 3 4 9 1 1 3 1 1 3 1 n n n n n n n n x n n n + ∞ ∞ ∞ + + + = = = − = = = + − + − + ∑ ∑ ∑ Calculus II 79 © 2018 Paul Dawkins Now, we can do a quick Comparison Test on the first series to see that it converges and we can do a quick Alternating Series Test on the second series to see that is also converges. We’ll leave it to you to verify both of these statements. Step 6 The interval of convergence is below and for summary purposes the radius of convergence is also shown. 9 15 3 Interval : 4 4 4 x R ≤ ≤ = 2. For the following power series determine the interval and radius of convergence. ( ) 2 1 3 0 2 17 4 n n n n n x + ∞ = + ∑ Step 1 Okay, let’s start off with the Root Test to get our hands on L. ( ) ( ) 1 1 1 2 2 2 1 3 3 3 lim 2 17 lim 2 17 2 17 lim 4 4 4 n n n n n n n n n n n n L x x x + + + →∞ →∞ →∞ = + = + = + Okay, we can see that , in this case, L will be infinite provided 17 2 x ≠− and so the series will diverge for 17 2 x ≠− . We also know that the power series will converge for 17 2 x = (this is the value of a for this series!). Step 2 Therefore, we know that the interval of convergence is 17 2 x = − and the radius of convergence is 0 R = . 3. For the following power series determine the interval and radius of convergence. ( ) ( ) 0 1 2 2 1 ! n n n x n ∞ = + − + ∑ Step 1 Okay, let’s start off with the Ratio Test to get our hands on L. Calculus II 80 © 2018 Paul Dawkins ( )( ) ( ) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )( )( ) 1 2 2 2 1 ! 2 2 2 1 ! lim lim 2 3 ! 2 3 2 2 2 1 ! 1 1 2 2 2 lim 0 2 3 2 2 1 n n n n n n x n n x n L n n n n n n x n x n n n + →∞ →∞ →∞ + − + + − + = = + + + + + + − + = − = + + + Okay, we can see that , in this case, 0 L = for every x. Step 2 Therefore, we know that the interval of convergence is x −∞< < ∞ and the radius of convergence is R = ∞ . 4. For the following power series determine the interval and radius of convergence. ( ) 1 2 1 0 4 3 5 n n n n x + ∞ + = + ∑ Step 1 Okay, let’s start off with the Ratio Test to get our hands on L. ( ) ( ) ( ) 1 3 2 2 1 2 1 2 4 3 4 3 5 16 16 lim lim 3 lim 3 5 5 5 5 4 3 n n n n n n n n n x x L x x x + + + + + →∞ →∞ →∞ + + = = = + = + + Step 2 So, we know that the series will converge if, 16 5 3 1 3 5 16 x x + < → + < Step 3 So, from the previous step we see that the radius of convergence is 5 16 R = . Step 4 Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2. 5 5 53 43 3 16 16 16 16 x x − < + < → − < < − Step 5 To finalize the interval of convergence we need to check the end points of the inequality from Step 4. Calculus II 81 © 2018 Paul Dawkins ( ) ( ) ( ) ( ) 1 2 1 0 0 0 4 16 1 5 4 1 53 4 4 5 : 16 5 5 16 5 5 16 5 n n n n n n n n n n n n x ∞ ∞ ∞ = = = − −   = − − = =     ∑ ∑ ∑ ( ) ( ) 1 2 1 0 0 0 4 16 43 4 4 5 5 4 : 16 5 5 16 5 5 16 5 n n n n n n n n n n n x ∞ ∞ ∞ = = =   = − = =     ∑ ∑ ∑ Now, ( ) 4 1 4 4 lim Does not exist lim 5 5 5 n n n →∞ →∞ − − = Therefore, each of these two series diverge by the Divergence Test. Step 6 The interval of convergence is below and for summary purposes the radius of convergence is also shown. 53 43 5 Interval : 16 16 16 x R − < < − = 5. For the following power series determine the interval and radius of convergence. ( ) 1 1 6 4 1 n n n x n ∞ − = − ∑ Step 1 Okay, let’s start off with the Ratio Test to get our hands on L. ( ) ( ) ( ) 1 1 6 4 1 6 4 1 6 lim lim 4 1 lim 6 4 1 1 1 1 6 4 1 n n n n n n n x n x n n L x x n n n x + − →∞ →∞ →∞ − − = = = − = − + + + − Step 2 So, we know that the series will converge if, 1 1 4 4 1 6 4 1 1 24 1 24 x x x − < → − < → − < Step 3 So, from the previous step we see that the radius of convergence is 1 24 R = . Calculus II 82 © 2018 Paul Dawkins Step 4 Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2. 1 1 1 5 7 24 4 24 24 24 x x − < − < → < < Step 5 To finalize the interval of convergence we need to check the end points of the inequality from Step 4. ( ) ( ) 1 1 1 1 1 1 1 1 6 1 5 6 1 6 : 24 6 6 n n n n n n n n n x n n n − − − ∞ ∞ ∞ − = = = − −   = − = =     ∑ ∑ ∑ 1 1 1 1 1 7 6 1 6 1 6 : 24 6 6 n n n n n n n x n n n − ∞ ∞ ∞ − = = =   = = =     ∑ ∑ ∑ Now, the first series is an alternating harmonic series which we know converges (or you could just do a quick Alternating Series Test to verify this) and the second series diverges by the p-series test. Step 6 The interval of convergence is below and for summary purposes the radius of convergence is also shown. 5 7 1 Interval : 24 24 24 x R ≤ < = Calculus II 83 © 2018 Paul Dawkins Section 4-15 : Power Series and Functions 1. Write the following function as a power series and give the interval of convergence. ( ) 4 6 1 7 f x x = + Step 1 First, in order to use the formula from this section we know that we need the numerator to be a one. That is easy enough to “fix” up as follows, ( ) 4 1 61 7 f x x = + Step 2 Next, we know we need the denominator to be in the form 1 p − and again that is easy enough, in this case, to rewrite the denominator to get the following form of the function, ( ) ( ) 4 1 6 1 7 f x x = −− Step 3 At this point we can use the formula from the notes to write this as a power series. Doing this gives, ( ) ( ) ( ) 4 4 4 0 1 6 6 7 provided 7 1 1 7 n n f x x x x ∞ = = = − − < −− ∑ Step 4 Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single x with a single exponent on it. These are easy enough rules to take care of. All we need to do is move whatever is in front of the series to the inside of the series and use basic exponent rules to take care of the x “rule”. Doing all this gives, ( ) ( ) ( ) ( ) 4 4 4 0 0 6 7 6 7 provided 7 1 n n n n n n f x x x x ∞ ∞ = = = − = − − < ∑ ∑ Step 5 To get the interval of convergence all we need to do is do a little work on the “provided” portion of the result from the last step to get, 1 1 1 4 4 4 4 4 4 1 1 1 1 7 1 7 1 7 7 7 7 x x x x x − < → < → < → < → − < < Calculus II 84 © 2018 Paul Dawkins Note that we don’t need to check the endpoints of this interval since we already know that we only get convergence with the strict inequalities and we will get divergence for everything else. Step 6 The answers for this problem are then, ( ) 1 1 4 4 4 4 0 6 1 1 Power Series : 6 7 Interval : 1 7 7 7 n n n x x x ∞ = = − − < < + ∑ 2. Write the following function as a power series and give the interval of convergence. ( ) 3 2 3 x f x x = − Step 1 First, in order to use the formula from this section we know that we need the numerator to be a one. That is easy enough to “fix” up as follows, ( ) 3 2 1 3 f x x x = − Step 2 Next, we know we need the denominator to be in the form 1 p − and again that is easy enough, in this case, to rewrite the denominator by factoring a 3 out of the denominator as follows, ( ) 3 2 1 3 1 3 1 x f x x = − Step 3 At this point we can use the formula from the notes to write this as a power series. Doing this gives, ( ) ( ) 3 3 2 2 1 1 3 3 2 1 0 3 1 provided 1 3 1 3 n n x x f x x x x ∞ = = = < − ∑ Step 4 Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single x with a single exponent on it. These are easy enough rules to take care of. All we need to do is move whatever is in front of the series to the inside of the series and use basic exponent rules to take care of the x “rule”. Doing all this gives, Calculus II 85 © 2018 Paul Dawkins ( ) ( ) ( ) ( ) ( ) 3 1 2 3 2 2 3 2 1 1 1 1 1 3 3 3 3 3 0 0 0 provided 1 3 n n n n n n n n x f x x x x x x ∞ ∞ ∞ + + = = = = = = < ∑ ∑ ∑ Step 5 To get the interval of convergence all we need to do is do a little work on the “provided” portion of the result from the last step to get, 2 2 2 1 1 3 3 1 1 3 3 3 3 x x x x x < → < → < → < → − < < Note that we don’t need to check the endpoints of this interval since we already know that we only get convergence with the strict inequalities and we will get divergence for everything else. Step 6 The answers for this problem are then, ( ) 3 1 2 3 1 3 2 0 Power Series : Interval : 3 3 3 n n n x x x x ∞ + + = = − < < − ∑ 3. Write the following function as a power series and give the interval of convergence. ( ) 2 3 3 5 2 x f x x = − Step 1 First, in order to use the formula from this section we know that we need the numerator to be a one. That is easy enough to “fix” up as follows, ( ) 2 3 1 3 5 2 f x x x = − Step 2 Next, we know we need the denominator to be in the form 1 p − and again that is easy enough, in this case, to rewrite the denominator by factoring a 5 out of the denominator as follows, ( ) 2 3 2 5 3 1 5 1 x f x x = − Step 3 At this point we can use the formula from the notes to write this as a power series. Doing this gives, Calculus II 86 © 2018 Paul Dawkins ( ) ( ) 2 2 3 3 2 2 5 5 3 2 0 5 3 1 3 provided 1 5 5 1 n n x x f x x x x ∞ = = = < − ∑ Step 4 Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single x with a single exponent on it. These are easy enough rules to take care of. All we need to do is move whatever is in front of the series to the inside of the series and use basic exponent rules to take care of the x “rule”. Doing all this gives, ( ) ( ) ( ) ( ) ( ) 1 1 3 3 2 2 2 3 3 3 3 2 2 2 2 5 5 5 5 5 5 0 0 0 3 provided 1 5 n n n n n n n n x f x x x x x x ∞ ∞ ∞ + = = = = = = < ∑ ∑ ∑ Step 5 To get the interval of convergence all we need to do is do a little work on the “provided” portion of the result from the last step to get, 1 1 3 3 3 5 125 125 125 2 2 5 5 2 8 8 8 1 1 x x x x x < → < → < → < → − < < Note that we don’t need to check the endpoints of this interval since we already know that we only get convergence with the strict inequalities and we will get divergence for everything else. Step 6 The answers for this problem are then, ( ) 1 3 2 2 3 125 125 2 5 5 8 8 3 0 3 Power Series : Interval : 5 2 n n n x x x x ∞ + = = − < < − ∑ 4. Give a power series representation for the derivative of the following function. ( ) 5 5 1 3 x g x x = − Hint : While we could differentiate the function and then attempt to find a power series representation that seems like a lot of work. It’s a good think that we know how to differentiate power series. Step 1 First let’s notice that we can quickly find a power series representation for this function. Here is that work. ( ) ( ) ( ) ( ) 5 5 5 1 5 0 0 0 1 5 5 3 5 3 5 3 1 3 n n n n n n n n g x x x x x x x x ∞ ∞ ∞ + = = = = = = = − ∑ ∑ ∑ Calculus II 87 © 2018 Paul Dawkins Step 2 Now, we know how to differentiate power series and we know that the derivative of the power series representation of a function is the power series representation of the derivative of the function. Therefore, ( ) ( ) ( )( ) 5 1 5 0 0 5 3 5 5 1 3 n n n n n n d g x x n x dx ∞ ∞ + = =   ′ = = +     ∑ ∑ Remember that to differentiate a power series all we need to do is differentiate the term of the power series with respect to x. 5. Give a power series representation for the integral of the following function. ( ) 4 2 9 x h x x = + Hint : Integrating this function seems like (potentially) a lot of work, not to mention determining a power series representation of the result. It’s a good think that we know how to integrate power series. Step 1 First let’s notice that we can quickly find a power series representation for this function. Here is that work. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 4 1 2 4 2 2 4 1 1 1 1 9 9 9 9 2 1 0 0 0 9 1 1 1 9 9 1 n n n n n n n n n n x x h x x x x x x ∞ ∞ ∞ + + = = = = = − = − = − −− ∑ ∑ ∑ Step 2 Now, we know how to integrate power series and we know that the integral of the power series representation of a function is the power series representation of the integral of the function. Therefore, ( ) ( ) ( ) ( ) ( ) 1 1 2 4 2 5 1 1 1 9 2 5 9 0 0 1 1 n n n n n n n n n h x dx x dx c x ∞ ∞ + + + + + = = = − = + − ⌠  ⌡∑ ∑ ∫ Remember that to integrate a power series all we need to do is integrate the term of the power series and we can’t forget to add on the “+c” since we’re doing an indefinite integral. Calculus II 88 © 2018 Paul Dawkins Section 4-16 : Taylor Series 1. Use one of the Taylor Series derived in the notes to determine the Taylor Series for ( ) ( ) cos 4 f x x = about 0 x = . Step 1 There really isn’t all that much to do here for this problem. We are working with cosine and want the Taylor series about 0 x = and so we can use the Taylor series for cosine derived in the notes to get, ( ) ( ) ( ) ( ) 2 0 1 4 cos 4 2 ! n n n x x n ∞ = − = ∑ Step 2 Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single x with a single exponent on it. In this case we don’t have anything out in front of the series to worry about so all we need to do is use the basic exponent rules on the 2x term to get, ( ) ( ) ( ) ( ) ( ) 2 2 2 0 0 1 4 1 16 cos 4 2 ! 2 ! n n n n n n n n x x x n n ∞ ∞ = = − − = = ∑ ∑ 2. Use one of the Taylor Series derived in the notes to determine the Taylor Series for ( ) 3 6 2x f x x = e about 0 x = . Step 1 There really isn’t all that much to do here for this problem. We are working with the exponential function and want the Taylor series about 0 x = and so we can use the Taylor series for the exponential function derived in the notes to get, ( ) 3 3 6 6 0 2 2 ! n n x x x x n ∞ = = ∑ e Note that we only convert the exponential using the Taylor series derived in the notes and, at this point, we just leave the 6 x alone in front of the series. Step 2 Calculus II 89 © 2018 Paul Dawkins Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single x with a single exponent on it. These are easy enough rules to take care of. All we need to do is move whatever is in front of the series to the inside of the series and use basic exponent rules to take care of the x “rule”. Doing all this gives, ( ) ( ) 3 3 3 3 6 6 6 6 0 0 0 2 2 2 2 ! ! ! n n n n n n n n x x x x x x x n n n + ∞ ∞ ∞ = = = = = = ∑ ∑ ∑ e 3. Find the Taylor Series for ( ) 6x f x − = e about 4 x = −. Step 1 Because we are working about 4 x = − in this problem we are not able to just use the formula derived in class for the exponential function because that requires us to be working about 0 x = . Step 2 So, we’ll need to start over from the beginning and start taking some derivatives of the function. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 6 2 6 3 3 6 4 4 6 0: 1: 6 2: 6 3: 6 4: 6 x x x x x n f x n f x n f x n f x n f x − − − − − = = ′ = = − ′′ = = − = = − = = − e e e e e Remember that, in general, we’re going to need to go out to at least 4 n = for most of these problems to make sure that we can get the formula for the general term in the Taylor Series. Also, remember to NOT multiply things out when taking derivatives! Doing that will make your life much harder when it comes time to find the general formula. Step 3 It is now time to see if we can get a formula for the general term in the Taylor Series. In this case, it is (hopefully) pretty simple to catch the pattern in the derivatives above. The general term is given by, ( ) ( ) ( ) 6 6 0,1,2,3, n n x f x n − = − = e  Calculus II 90 © 2018 Paul Dawkins As noted this formula works all the way back to 0 n = . It is important to make sure that you check this formula to determine just how far back it will work. We will, on occasion, get formulas that will not work for the first couple of n’s and we need to know that before we start writing down the Taylor Series. Step 4 Now, recall that we don’t really want the general term at any x. We want the general term at 4 x = −. This is, ( ) ( ) ( ) 24 4 6 0,1,2,3, n n f n − = − = e  Step 5 Okay, at this point we can formally write down the Taylor Series for this problem. ( ) ( ) ( ) ( ) ( ) 24 6 0 0 4 6 4 4 ! ! n n n n n n x f x x n n ∞ ∞ − = = − − = + = + ∑ ∑ e e 4. Find the Taylor Series for ( ) ( ) ln 3 4 f x x = + about 0 x = . Step 1 Okay, we’ll need to start off this problem by taking a few derivatives of the function. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( )( ) 1 2 2 3 3 3 4 4 4 5 5 5 0: ln 3 4 4 1: 4 3 4 3 4 2: 4 3 4 3: 4 2 3 4 4: 4 2 3 3 4 5: 4 2 3 4 3 4 n f x x n f x x x n f x x n f x x n f x x n f x x − − − − − = = + ′ = = = + + ′′ = = − + = = + = = − + = = + Remember that, in general, we’re going to need to go out to at least 4 n = for most of these problems to make sure that we can get the formula for the general term in the Taylor Series. Also, remember to NOT multiply things out when taking derivatives! Doing that will make your life much harder when it comes time to find the general formula. In this case we “merged” all the 4’s that came from the chain rule into a single term but left it as an exponent rather than get an actual value. This is not uncommon with these kinds of problems. The exponents we dropped down for the derivatives we left alone with the exception of dealing with the signs. Step 2 Calculus II 91 © 2018 Paul Dawkins It is now time to see if we can get a formula for the general term in the Taylor Series. Hopefully you can see the pattern in the derivatives above. The general term is given by, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 ln 3 4 0 1 4 1 ! 3 4 1,2,3, n n n n f x x n f x n x n + − = + = = − − + =  As noted this formula works all the way back to 1 n = but clearly does not work for 0 n = . It is problems like this one that make it clear why we always need to check our proposed formula for the general solution to see just how far back it works. Step 3 Now, recall that we don’t really want the general term at any x. We want the general term at 0 x = . This is, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 1 0 ln 3 0 0 1 4 1 ! 3 1 1 4 1 !3 4 1 1 ! 1,2,3, 3 n n n n n n n n n f n f n n n n + − + + = = = − − = − −   = − − =      We did a little simplification for the second one just to make it a little simpler. Step 4 Okay, at this point we can formally write down the Taylor Series for this problem. However, before we actually do that recall that our general term formula did not work for 0 n = and so we’ll need to first strip that out of the series before we put the general formula in. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 4 3 1 1 4 3 1 0 ln 3 4 ! 0 0 ! 1 1 ! ln 3 ! 1 ln 3 n n n n n n n n n n n n n n f x x n f f x n n x n x n ∞ = ∞ = + ∞ = + ∞ = + = = + − − = + − = + ∑ ∑ ∑ ∑ Don’t forget to simplify/cancel where we can in the final answer. In this case we could do some simplifying with the factorials. Calculus II 92 © 2018 Paul Dawkins 5. Find the Taylor Series for ( ) 4 7 f x x = about 3 x = −. Step 1 Okay, we’ll need to start off this problem by taking a few derivatives of the function. ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( )( ) 4 4 5 6 3 7 4 8 7 0: 7 1: 7 4 2: 7 4 5 3: 7 4 5 6 4: 7 4 5 6 7 n f x x x n f x x n f x x n f x x n f x x − − − − − = = = ′ = = − ′′ = = = = − = = Remember that, in general, we’re going to need to go out to at least 4 n = for most of these problems to make sure that we can get the formula for the general term in the Taylor Series. Also, remember to NOT multiply things out when taking derivatives! Doing that will make your life much harder when it comes time to find the general formula. In this case the only “simplification” we did was to multiply out the minus signs that came from the exponents upon doing the derivatives. That is a fairly common thing to do with these kinds of problems. Step 2 It is now time to see if we can get a formula for the general term in the Taylor Series. Hopefully you can see the pattern in the derivatives above. The general term is given by, ( ) ( ) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )( )( )( )( ) ( ) ( ) ( ) ( ) 8 8 4 2 3 7 1 4 5 6 3 2 3 2 3 4 5 6 3 7 1 6 7 1 3 ! 0,1,2,3, 6 n n n n n f x n x n x n x n − − − + = − + + = − = − + =    This formula may have been a little trickier to get. We almost had a factorial in the derivatives but each one was missing the ( )( ) 2 3 part that would be needed to get the factorial to show up. Because that was all that was missing and it was missing in each of the derivatives we multiplied each derivative by Calculus II 93 © 2018 Paul Dawkins ( )( ) ( )( ) 2 3 2 3 (i.e. a really fancy way of writing one…). We could then use the numerator of this to complete the factorial and the denominator was just left alone. Also, as noted this formula works all the way back to 0 n = . It is important to make sure that you check this formula to determine just how far back it will work. We will, on occasion, get formulas that will not work for the first couple of n’s and we need to know that before we start writing down the Taylor Series. Step 3 Now, recall that we don’t really want the general term at any x. We want the general term at 3 x = −. This is, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 4 4 4 4 4 4 7 3 1 3 ! 3 6 7 1 3 ! 6 3 7 1 3 ! 6 1 3 7 3 ! 6 1 3 7 3 ! 1,2,3, 6 3 n n n n n n n n n n f n n n n n n − + + + + + + − = − + − − + = − − + = − + = − + = =  We did a little simplification here so we could cancel out all the alternating signs that were present in the term. Step 4 Okay, at this point we can formally write down the Taylor Series for this problem. ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) 4 4 4 0 0 0 3 7 3 ! 7 3 2 1 7 3 3 3 ! 6 3 ! 6 3 n n n n n n n n n f n n n n x x x x n n ∞ ∞ ∞ + + = = = − + + + + = + = + = + ∑ ∑ ∑ Don’t forget to simplify/cancel where we can in the final answer. In this case we could do some simplifying with the factorials. 6. Find the Taylor Series for ( ) 2 7 6 1 f x x x = − + about 2 x = . Step 1 Calculus II 94 © 2018 Paul Dawkins First, let’s not get too excited about the fact that we have a polynomial here for this problem. It works exactly the same way with a few small differences. We’ll start off by taking a few derivatives of the function and evaluating them at 2 x = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0: 7 6 1 2 17 1: 14 6 2 22 2: 14 2 14 3: 0 2 0 n n n f x x x f n f x x f n f x f n f x f = = − + = ′ ′ = = − = ′′ ′′ = = = ≥ = = Okay, this is where one of the differences between a polynomial and the other types of functions we typically see with Taylor Series problems. After some point all the derivatives will be zero. That is not something to get excited about. In fact, it actually makes the problem a little easier! Because all the derivatives are zero after some point we don’t need a formula for the general term. All we need are the values of the non-zero derivative terms. Step 2 Once we have the values from the previous step all we need to do is write down the Taylor Series. To do that all we need to do is strip all the non-zero terms from the series and then acknowledge that the remainder will just be zero (all the remaining terms are zero after all!). Doing this gives, ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 1 2 3 2 2 7 6 1 2 ! 2 2 2 2 2 2 2 ! 17 22 2 7 2 n n n n n n f x x x n f f f x f x x n x x ∞ = ∞ = − + = − ′ ′′ = + − + − + − = + − + − ∑ ∑ It looks a little strange but there it is. Do not multiply/simplify this out. This really is the answer we are looking for. Also, don’t think that this is a problem that is just done to make you work another problem. There are applications of series (beyond the scope of this course however…) that really do require this kind of thing to be done as strange as that might sound! Calculus II 95 © 2018 Paul Dawkins Section 4-17 : Applications of Series 1. Determine a Taylor Series about 0 x = for the following integral. 1 x dx x − ⌠  ⌡ e Step 1 This problem isn’t quite as hard as it might first appear. We know how to integrate a series so all we really need to do here is find a Taylor series for the integrand and then integrate that. Step 2 Okay, let’s start out by noting that we are working about 0 x = and that means we can use the formula for the Taylor Series of the exponential function. For reference purposes this is, 0 ! n n x x n ∞ = =∑ e Next, let’s strip out the 0 n = term from this and then subtract one. Doing this gives, 1 1 1 1 1 ! ! n n n n x x x n n ∞ ∞ = =   −= + −=     ∑ ∑ e Of course, in doing the above step all we really managed to do was eliminate the 0 n = term from the series. In fact, that was not a bad thing to have happened as well see shortly. Finally, let’s divide the whole thing by x. This gives, 1 1 1 1 1 ! ! n n n n x x x x x n n − ∞ ∞ = = −= = ∑ ∑ e We moved the x that was outside the series into the series. This is required in order to do the integral of the series. We only want a single x in the problem and we now have that. Also note that while the function on the left has a division by zero issue the series on the right does not have this problem. All of the x’s in the series have positive or zero exponents! This is a really good thing. Of course, the other good thing that we have at this point is that we’ve managed to find a series representation for the integrand! Step 3 All we need to do now is compute the integral of the series to get a series representation of the integral. Calculus II 96 © 2018 Paul Dawkins ( )( ) 1 1 1 1 ! ! n n n n x x x dx dx c x n n n − ∞ ∞ = = − = = + ⌠ ⌠   ⌡ ⌡∑ ∑ e 2. Write down ( ) 2 T x , ( ) 3 T x and ( ) 4 T x for the Taylor Series of ( ) 6x f x − = e about 4 x = −. Graph all three of the Taylor polynomials and ( ) f x on the same graph for the interval [ ] 8, 2 − − . Step 1 The first thing we need to do here is get the Taylor Series for ( ) 6x f x − = e about 4 x = −. Luckily enough for us we did that in Problem 3 of the previous section. Here is the Taylor Series we derived in that problem. ( ) ( ) 24 6 0 6 4 ! n n n x x n ∞ − = − = + ∑ e e Step 2 Here are the three Taylor polynomials needed for this problem. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 24 24 24 2 2 3 24 24 24 24 3 2 3 4 24 24 24 24 24 4 6 4 18 4 6 4 18 4 36 4 6 4 18 4 36 4 54 4 T x x x T x x x x T x x x x x = − + + + = − + + + − + = − + + + − + + + e e e e e e e e e e e e Step 3 Here is the graph for this problem. We can see that as long as we stay “near” 4 x = − the graphs of the polynomial are pretty close to the graph of the exponential function. However, if we get too far away the graphs really do start to diverge from the graph of the exponential function. Calculus II 97 © 2018 Paul Dawkins 3. Write down ( ) 3 T x , ( ) 4 T x and ( ) 5 T x for the Taylor Series of ( ) ( ) ln 3 4 f x x = + about 0 x = . Graph all three of the Taylor polynomials and ( ) f x on the same graph for the interval [ ] 1 2 ,2 − . Step 1 The first thing we need to do here is get the Taylor Series for ( ) ( ) ln 3 4 f x x = + about 0 x = . Luckily enough for us we did that in Problem 4 of the previous section. Here is the Taylor Series we derived in that problem. ( ) ( ) ( ) ( ) 1 4 3 1 1 ln 3 4 ln 3 n n n n x x n + ∞ = − + = +∑ Step 2 Here are the three Taylor polynomials needed for this problem. ( ) ( ) ( ) ( ) ( ) ( ) 2 3 8 64 4 3 3 9 81 2 3 4 8 64 64 4 4 3 9 81 81 2 3 4 5 8 64 64 1024 4 5 3 9 81 81 1215 ln 3 ln 3 ln 3 T x x x x T x x x x x T x x x x x x = + − + = + − + − = + − + − + Step 3 Here is the graph for this problem. We can see that as long as we stay “near” 0 x = the graphs of the polynomial are pretty close to the graph of the exponential function. However, if we get too far away the graphs really do start to diverge from the graph of the exponential function. Calculus II 98 © 2018 Paul Dawkins Section 4-18 : Binomial Series 1. Use the Binomial Theorem to expand ( ) 5 4 3x + . Solution Not really a lot to do with this problem. All we need to do is use the formula from the Binomial Theorem to do the expansion. Here is that work. ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )( ) ( )( )( )( ) ( )( )( ) ( ) 5 5 5 0 1 2 3 4 5 4 3 2 1 5 2 3 4 5 5 4 3 2 2 3 4 5 5 4 3 4 3 5 5 5 5 5 4 4 3 4 3 4 3 4 3 0 1 2 3 4 5 3 5 5 4 5 4 3 4 5 4 3 4 3 4 3 5 4 3 3 2! 3! 1024 3840 5760 4320 1620 243 i i i x x i x x x x x x x x x x x x x x x − =  + =        = + + + +            +  = + + + + + = + + + + + ∑ 2. Use the Binomial Theorem to expand ( ) 4 9 x − . Solution Not really a lot to do with this problem. All we need to do is use the formula from the Binomial Theorem to do the expansion. Here is that work. ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )( ) ( )( )( ) ( ) 4 4 4 0 1 2 3 4 4 3 2 1 2 3 4 4 3 2 1 2 3 4 4 9 9 4 4 4 4 4 9 9 9 9 0 1 2 3 4 4 3 9 4 9 9 4 9 2! 6561 2916 486 36 i i i x x i x x x x x x x x x x x x − =  − = −        = + − + − + − + −           = + − + − + − + − = − + − + ∑ Calculus II 99 © 2018 Paul Dawkins 3. Write down the first four terms in the binomial series for ( ) 6 1 3x − + . Step 1 First, we need to make sure it is in the proper form to use the Binomial Series from the notes which in this case we are already in the proper form with 6 k = −. Step 2 Now all we need to do is plug into the formula from the notes and write down the first four terms. ( ) ( ) ( )( ) ( )( ) ( ) ( )( )( ) ( ) 6 0 1 2 3 2 3 6 1 3 3 6 7 6 7 8 1 6 3 3 3 2! 3! 1 18 189 1512 i i x x i x x x x x x ∞ − = −   + =     − − − − − = + − + + + = − + − + ∑   4. Write down the first four terms in the binomial series for 3 8 2x − . Step 1 First, we need to make sure it is in the proper form to use the Binomial Series. Here is the proper form for this function, ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 3 1 1 1 3 3 3 3 4 4 4 8 2 8 1 8 1 2 1 x x x x − = − = − = + − Recall that for proper from we need it to be in the form “1+” and so we needed to factor the 8 out of the root and “move” the minus sign into the second term. Also, as we can see we will have 1 3 k = Step 2 Now all we need to do is plug into the formula from the notes and write down the first four terms. ( ) ( ) ( ) ( )( ) ( )( )( ) ( )( )( )( ) 1 3 1 3 4 1 3 1 4 0 5 1 2 1 2 1 2 3 3 3 3 3 3 1 1 1 1 3 4 4 4 2 3 5 1 1 6 72 2592 8 2 2 1 2 2 1 2! 3! 2 i i x x x i x x x x x x ∞ = − = + −  = −     − − − = + − + − + − +     = − − − + ∑   Calculus II 100 © 2018 Paul Dawkins
1705	https://www.doubtnut.com/qna/10964745	If the elastic limit of copper is 1.5×108N/m2,determine the minimum diameter a copper wire can have under a load of10.0kgfind , if its elastic limit is not to be exceeded. More from this Exercise The correct Answer is:A, C, D To solve the problem of determining the minimum diameter of a copper wire that can support a load of 10 kg without exceeding its elastic limit, we can follow these steps: Step 1: Write down the given data - Elastic limit of copper (σm): 1.5×108N/m2 - Load (F): 10kg Step 2: Convert the load into force The force due to the load can be calculated using the formula: F=m⋅g where g (acceleration due to gravity) is approximately 9.8m/s2. Calculating the force: F=10kg×9.8m/s2=98N Step 3: Relate stress to force and area The stress (σ) in the wire can be defined as: σ=FA where A is the cross-sectional area of the wire. Step 4: Express the area in terms of diameter For a circular cross-section, the area A can be expressed in terms of the diameter d: A=π4d2 Step 5: Set up the equation using the elastic limit Since we want to ensure that the stress does not exceed the elastic limit, we can set up the equation: σm=FA Substituting the expression for area: σm=Fπ4d2 Step 6: Rearrange the equation to solve for diameter Rearranging the equation gives: d2=4Fπσm Taking the square root to find d: d=√4Fπσm Step 7: Substitute the values into the equation Substituting F=98N and σm=1.5×108N/m2: d=√4×98π×1.5×108 Step 8: Calculate the diameter Calculating the above expression: d=√392π×1.5×108 d≈√3924.712×108 d≈√8.32×10−7 d≈0.00091m Converting to millimeters: d≈0.91mm Final Answer The minimum diameter of the copper wire that can support a load of 10 kg without exceeding the elastic limit is approximately 0.91 mm. --- To solve the problem of determining the minimum diameter of a copper wire that can support a load of 10 kg without exceeding its elastic limit, we can follow these steps: Step 1: Write down the given data - Elastic limit of copper (σm): 1.5×108N/m2 - Load (F): 10kg Step 2: Convert the load into force The force due to the load can be calculated using the formula: F=m⋅g where g (acceleration due to gravity) is approximately 9.8m/s2. Calculating the force: F=10kg×9.8m/s2=98N Step 3: Relate stress to force and area The stress (σ) in the wire can be defined as: σ=FA where A is the cross-sectional area of the wire. Step 4: Express the area in terms of diameter For a circular cross-section, the area A can be expressed in terms of the diameter d: A=π4d2 Step 5: Set up the equation using the elastic limit Since we want to ensure that the stress does not exceed the elastic limit, we can set up the equation: σm=FA Substituting the expression for area: σm=Fπ4d2 Step 6: Rearrange the equation to solve for diameter Rearranging the equation gives: d2=4Fπσm Taking the square root to find d: d=√4Fπσm Step 7: Substitute the values into the equation Substituting F=98N and σm=1.5×108N/m2: d=√4×98π×1.5×108 Step 8: Calculate the diameter Calculating the above expression: d=√392π×1.5×108 d≈√3924.712×108 d≈√8.32×10−7 d≈0.00091m Converting to millimeters: d≈0.91mm Final Answer The minimum diameter of the copper wire that can support a load of 10 kg without exceeding the elastic limit is approximately 0.91 mm. Topper's Solved these Questions Explore 4 Videos Explore 4 Videos Explore 434 Videos Explore 17 Videos Similar Questions If the density of copper is 8.9×103kg/m3, find its relative density. The stress for elastic limit of a material is 3.5×108Nm2 .The minimum diameter of a rod made of this material which can support 500N load without exceeding elastic limit will be Knowledge Check The elastic limit of brass is 379 MPa . What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit ? The elastic limit of an elevator cable is 2×109N/m2. The maximum upward acceleration that an elevator of amss 2×103kg can have been supported by a cable would not exceed half of the elastic limit would be A steel wire of cross-sectional area 4 cm2 has elastic limit of 2.2×108N/m2. The maximum upward acceleration that can be given to a 1000 kg elevator supported by this steel wire if the stress is to exceed one-fourth of the elastic limit is [Take, g = 10 m/s2 ] The resistivity of copper is 1.7×10−8Ωm. What length of copper wire of diameter 0.1 mm will have a resistance of 34Ω ? [" The stress for elastic limit of a "],[" material is "3.5times10^(8)N/m^(2)" .The "],[" minimum diameter of a rod made of "],[" this material which can support "500N],[" load without exceeding elastic limit "],[" will be "] The elastic limit of a steel cable is 3.0×108N/m2 and the cross-section area is 4cm2. Find the maximum upward acceleration that can be given to a 900kg elevator supported by the cable if the stress is not to exceed one - third of the elastic limit. A mass of 5 kg is hung from a copper wire of 5 mm diameter and 2 m in length. Calculate the extension produced. What should be the minimum diameter of the wire so that its elastic limit is not exceeded ? Elastic limit for copper =1.5×109 dyne cm−2 and Y for copper =1.1×1012 dyne cm−2 The resistivity of copper is 1.7×10−8Ωm. Another copper wire of the same length but of half the diameter as the first is taken. What is the ratio of its resistance to that of the first wire ? DC PANDEY-ELASTICITY-Medical entrances s gallery If the elastic limit of copper is 1.5xx10^(8) N//m^(2),determine the m... The length of a metal wire is l(1) when the tension in it is T(1) and ... A load of 4 kg is suspended from a ceiling through a steel wire of len... Two wires of the same length and same material but radii in the ratio ... The elastic potential energy of a stretched wire is given by A uniform cylindrical rod of length L and cross-sectional area by forc... The ratio of hydraulic stress to the corresponding strain is known as The pressure on an object of bulk modulus B undergoing hydraulic compr... For most of the material , Young's modulus (Y) and rigidity modulus (G... The following four wires are made of the same material which of these ... The Young's modulus of a rope of 10 m length and having diameter of 2 ... A steel wire of length l and cross sectional area A is stretched by 1 ... Theoretically the value of Poisson's ratio sigma lies between A steel ring of radius r and cross section area A is fitted on to a wo... An Indian rubber cord L metre long and area of cross-section A "metre"... The density of a metal at normal pressure is rho. Its density when it ... To break a wire of one meter length, minimum 40 kg wt. is required. Th... If Poission's ratio sigma is -(1)/(2) for a material, then the materia... Two blocks of masses 1 kg and 2 kg are connected by a metal wire going... Which one the following is not a unit of Young's modulus ? There is same change in length when a 33000 N tensile force is appli... 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1706	https://www.teachhub.com/teaching-strategies/2020/07/strategies-to-teach-math-vocabulary/	Strategies to Teach Math Vocabulary - TeachHUB Skip to content K-12 Resources By Teachers, For Teachers Provided by the K-12 Teachers Alliance Search for: Teacher Resources View All Teaching Strategies Classroom Activities Classroom Management Technology in the Classroom Professional Development In the Classroom View All Lesson Plans Writing Prompts Graduate Programs Strategies to Teach Math Vocabulary Lora McKillop Elementary school principal; M.A. in Executive Leadership, Gardner-Webb University, NC July 30, 2020 Teaching Strategies HomeTeacher ResourcesTeaching Strategies Math vocabulary is an integral piece to understanding math concepts and developing math skills. Often times if a student does not understand a vocabulary term, they are unable to process and make sense of what they are doing, and this will hinder their success. If a student does not fully develop a math concept then that can, and often does, lead to a misconception or lack of understanding of the next skills since math is very much a building-block process. Students must have those foundational building blocks to advance to higher levels. Strategies to Teach Math Vocabulary In the elementary world there are several examples of vocabulary terms that students confuse as well as multiple terms for the same concept. This is something that teachers should anticipate and therefore strategically have vocabulary instruction before and as they teach the concepts so students develop a clear understanding while they master those skills. Addition and Subtraction One example begins in kindergarten when students are learning how to add and subtract. Students may see or hear the words add, sum, one more, join, and total. All of these terms mean that students are putting objects or numbers together to make more. At this early level it may be confusing to show students all of the terms up front; however, at some point while students are tackling the concept of addition teachers should introduce all of these terms and help students make the connections between them all meaning the same thing. A strategy to do that is to begin with a number line and discuss with students the concept of ‘one more’ as they move up the number line. This gives them a concrete manipulative that they can actually use as they move their finger to count up. This also gives the teacher the opportunity to discuss ‘one more’ because the students can physically see that they are moving up one space at a time. A great and easy way to make number lines is to draw them on sentence strips and laminate them. This also allows students to use a dry erase marker so that they can write on it if needed. You can use these same strategies for subtraction with the term ‘one less’ as students count down. Greater Than and Less Than As students progress in math, a concept that they struggle with is greater than and less than and the symbols that go with them. I often used to hear teachers use the words ‘alligator mouth’ when teaching this concept, but we have learned that is not best practice. Students need to comprehend what greater than and less than really mean, not that the alligator is eating the bigger number. If students rely on that way of thinking, then as they progress to higher levels of math their thinking is going to be dependent upon that way of thinking without true understanding. A strategy for teaching this is to use place value blocks or place value discs and have students manipulate them on a place value chart so that they see and can compare two values. The teacher can then discuss using the greater than and less than symbols to represent the model when writing the numbers. It is important to note that teachers should specifically discuss and help students understand which symbol to use and why. Multiplication When students graduate to multiplication, they often have a difficult time with the word ‘times’ (i.e. 4 x 6). It is essential that teachers explain what this means to students. The best strategy for teaching this vocabulary term is to use an array and discuss with students that the reason we say four times six is that you are counting four six times. Then demonstrate this for them and have them practice, practice, practice! If they do not understand that, they do not understand the concept of multiplication, which is critical for any higher math. It is important to note that students should also be taught to draw the array the correct way so that it matches how the groups are being counted. General Strategies General strategies for teaching math vocabulary throughout the year are also a great idea so that students can utilize them if they forget a term or just need a quick reminder. It also provides a record of all the vocabulary they learned throughout the year. These include: Having students keep a math vocabulary notebook Displaying a math word wall with visuals Having students use a graphic organizer so they can come up with multiple representations which helps solidify their understanding Continuing to review math vocabulary during whole-group and small-group lessons Having students turn in quick video recordings explaining a vocabulary term—this is the best way to gage understanding. A great platform for this is Seesaw because it houses all the videos in one spot, and students and teachers can give feedback to each other. I hope you have learned a few new tips to help you teach math vocabulary. Most importantly, please remember to check students’ work in their notebooks or on their graphic organizers and videos and provide them with feedback so you can correct any misconceptions. Math vocabulary is essential to student understanding, and we must be diligent as we help them build a strong foundation for their future math learning. #MathVocabulary, #TeachingStrategies More in Teaching Strategies How Gamification Can Transform Your Classroom Learning is like a game; it has rules, levels, and sometimes rewards. Just… Read More Summative Assessments: Measuring What Matters in the Classroom Summative assessments are typically the last step in the teaching and learning process…. Read More Creative Ways to Celebrate Cultural Diversity in the Classroom Multicultural education is much more than celebrating Black History Month and learning about… Read More Teacher Strategies for Student Deadline Success Deadlines are a part of life. 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1707	https://boldlyinspiredcurriculum.com/desmos-activities-for-algebra/	Skip to content CLICK HERE TO GET 10 FREE MATH WRITING PROMPTS About Teacher Resources Critical Thinking Toolkit Math Resources Teacher Planners Blog Desmos activities have become an essential tool in my high school math classroom. They are no-prep and can keep the students engaged for a full period. I love to use them as introductions, test review, and even sub plans! This blog post covers my students’ favorite Desmos activities for Algebra 2. Want to know more about Desmos and how to use it in your classroom? Check out this blog post. Table of Contents Toggle #5 Polygraph: Parabolas Polygraphs on Desmos are a math version of Guess Who? Your students will ask each other yes or no questions using math vocabulary to narrow down their options until they figure out the graph their partner chose. This activity can last from 15 to 45 minutes depending on the number of rounds you want your students to play. #4 Absolute Value Inequalities on the Number Line This Desmos activity is great for introducing what one variable absolute value inequalities look like on a number line. It can also be used as a segue from compound inequalities to absolute value inequalities. The discovery based questions will challenge your students to think critically about the graphs. #3 Polygraph: Absolute Value Similar to the Polygraph: Parabolas activity, your students will use math vocabulary to describe vertices and slopes of absolute value functions. This activity can be used as an introduction, review, or sub plans! #2 Polynomial Equations Challenges This exploratory activity challenges students to think deeper about quadratic functions and their roots. One of my favorite parts of this activity is how open-ended it is that allows for different solutions and interpretations. #1 Marbleslides: Polynomials Marbleslides are the perfect game to reinforce different parts of a function and its equation. The students will adjust the vertex, domain, and slope to catch all of the stars. This is truly the perfect mix of engaging and challenging for your Algebra 2 students. Final Thoughts about These Desmos Activities for Algebra 2 There are so many ways to use Desmos and other free math websites to enhance your lessons. This blog post just barely scratches the surface of the resources you can use (I haven’t even mentioned the graphing calculator feature yet!). The fact that all of these activities are completely no-prep and can connect directly to your Google Classroom makes them an absolute game-changer for your Algebra 2 curriculum. Looking for Desmos activities for Algebra 1? Check out my top ten favorite activities are in this blog post!
1708	https://www.rdocumentation.org/packages/edgeR/versions/3.14.0/topics/topTags	topTags function - RDocumentation ##### One week. Exclusive content. Lasting impact. Oct 6–10: How leaders are closing AI skills gaps. Unlock Access Rdocumentation -------------- powered by Learn R Programming edgeR (version 3.14.0) topTags: Table of the Top Differentially Expressed Tags Description Extracts the top DE tags in a data frame for a given pair of groups, ranked by p-value or absolute log-fold change. Usage topTags(object, n=10, adjust.method="BH", sort.by="PValue", p.value=1) Arguments object a DGEExact object (output from exactTest) or a DGELRT object (output from glmLRT), containing the (at least) the elements table: a data frame containing the log-concentration (i.e. expression level), the log-fold change in expression between the two groups/conditions and the p-value for differential expression, for each tag. If it is a DGEExact object, then topTags will also use the comparison element, which is a vector giving the two experimental groups/conditions being compared. The object may contain other elements that are not used by topTags. n scalar, number of tags to display/return adjust.method character string stating the method used to adjust p-values for multiple testing, passed on to p.adjust sort.by character string, should the top tags be sorted by p-value ("PValue"), by absolute log-fold change ("logFC"), or not sorted ("none"). p.value cutoff value for adjusted p-values. Only tags with lower p-values are listed. Value an object of class TopTags containing the following elements for the top n most differentially expressed tags as determined by sort.by:table a data frame containing the elements logFC, the log-abundance ratio, i.e. fold change, for each tag in the two groups being compared, logCPM, the log-average concentration/abundance for each tag in the two groups being compared, PValue, exact p-value for differential expression using the NB model, FDR, the p-value adjusted for multiple testing as found using p.adjust using the method specified.adjust.method character string stating the method used to adjust p-values for multiple testing.comparison a vector giving the names of the two groups being compared.test character string stating the name of the test.The dimensions, row names and column names of a TopTags object are defined by those of table, see dim.TopTags or dimnames.TopTags.TopTags objects also have a show method so that printing produces a compact summary of their contents.Note that the terms tag' andgene' are synonymous here. The function is only named as `Tags' for historical reasons. References Robinson MD, Smyth GK (2008). Small-sample estimation of negative binomial dispersion, with applications to SAGE data. Biostatistics 9, 321-332. Robinson MD, Smyth GK (2007). Moderated statistical tests for assessing differences in tag abundance. Bioinformatics 23, 2881-2887. See Also exactTest, glmLRT, p.adjust. Analogous to topTable in the limma package. Examples Run this code ```r generate raw counts from NB, create list object y <- matrix(rnbinom(80,size=1,mu=10),nrow=20) d <- DGEList(counts=y,group=rep(1:2,each=2),lib.size=rep(c(1000:1001),2)) rownames(d$counts) <- paste("gene",1:nrow(d$counts),sep=".") estimate common dispersion and find differences in expression here we demonstrate the 'exact' methods, but the use of topTags is the same for a GLM analysis d <- estimateCommonDisp(d) de <- exactTest(d) look at top 10 topTags(de) Can specify how many genes to view tp <- topTags(de, n=15) Here we view top 15 tp Or order by fold change instead topTags(de,sort.by="logFC") ``` Run the code above in your browser usingDataLab
1709	https://apps.dtic.mil/sti/tr/pdf/AD0740723.pdf	'r•, :,r' ,T4-. •t Pt hu "K tA fir" Pb no~~" t 4¾1 V4 Vf4 warn Uni No. 165tb itewi mr~ St% Av' le C p Be \|'F TWE Mi"ss' 4.. "" -F , . Best Available Copy !, hI.NATIONA/\L TECHNICAL IJIFORTIAATION• SEPV!CE A "tI1 C.TtT ~ 4 '1' A OCl~ 'xt4 E ,.-K y: rrC=CA~ Best Avai~lable copy Bos A.;'ail Cop Unclassified Seciutity Classification DOCUMENT CONTROL DATA- R & D (Security classification or title, body ,f, abstract and indexer, annotataon mut, be entered when the overall report is classilied) I. ORIGINATING AC TIVI TY (Corpora•t aulthor) 20. REPORT SECURIVy CLASSIFICATION Goodyear Aerospace Corporation Unclassified 1210 Massillon Road 2b. GROUP Akron, Ohio 44315 3 REPORT TITLE INVESTIGATION OF DYNAMIC BEHAVIOR OF TETHERED BALLOON SYSTEMS 4. DESCRIPTIVE NOTES (Type of report and inclusive dotes) Approved Scientific. Final. 1 January 1971 through 31 January 1972 29 Feb. 1972 5 AUTHORtSI (First name, a iddle Initial, lest name) Jerome J. Vorachek James W. Burbick George R. Doyle Jr. 6 REPORT DATE 7a. TOTAL NO. OF PAGES 7b. NO. OF REFPG 31 Januar7 1972 I(4 4 f 6. CONTRACT OR GRANT NO a& ORIGICI .TORS REPORT NUMBER(S) F-+9ft8-71-C-0091 GrR- ,rL ' --7659-06-01 AFCRL-72-0113 c. ~ 9b. OTMCR -RcFr- r NO(SI (Any other numbers Uuht may ba &Sslgned "DoD Element: 62101F O.this report) d. DOD Subelement: 687659 10 DISTRIBUTION STATEMENT1 A -Approved for public release; distribution unlimited I SUPPLEMENTARY NOTES 12 SPONSORING MILITARY ACTIVITY TECHW OTHER Air Force Cambridge Reseach Laboratories(LC) fL. G. Hanscom Field Bedford, Massachusetts 01730 t3 ABSTRACTN An analytical investigation of the dynamic behavior of tethered balloons was conducted. This report covers definition of tethered balloon systems and a study of stability characteristics and dynamic response to wind gusts of tethered balloon systems. Balloon systems which are investiga-ted use the British BJ Barrage Balloon,;tie Vee Balloon and a Goodyear Aerospace Model No. 1649 Single-Hull Balloon. The major tether construction -'is Columbian Rope Company's NOLARO utilizing prestretched polyester filaments. Three design altitudes, 5,000, 10,000 and 20,000 feet, are considered. The model for the tethered balloon system consists of the streamlined balloon and a tether made tip of three discrete links. The derivation of non-line;:r equations of motion for this system in three dimensions is presented. The equations are linearized for stability analysis and treated as uncoupled in :he longitudinal and lateral degrees of freedom. Characteristic equations which incorporate thr rhysical, aerodynamic, and mass characteristics of the 3ystem are developed and solved for the roots which represent the frequency and damping quali.ies. The nonlinear equations of motion are programmed for solution on a digital computer. An exploratory investigation to establish the trends of dynamic characteristics for various design parameters is reported. Design parameters considered include balloon shape and altitude, trim angle of attack, vertical location of the bridle confluence point, net static lift, tail size, reduced wind profiles, varying altitude as encountered in launch and retrieval, pay-load location, ani wind location above mean sea level. DFORM1 _ __ _ _ _ DD , NOV° 1473 Unclassified SecuitN ClassIfIcatIon Unclassified Security Classification KEY WOROS LI1IK A LINK 8 LINK C ROLE WT ROLE WT ROLE WT Aerostats Balloons Balloon Aerodynamics Balloon Dynamics Balloon Behavior Balloon Flying Qualities Balloon Mass Characteristics Balloon Stability Balloon Systems Cables Cable Dynamics Captive Balloons Dynamic Simulation Analysis Kite Balloons Stability Analysis Tethered Balloons Tethered LTA Vehicles Tethers Security C: ,ssificatior AfCKL -72 -0113 INVESTIGATION• OF DYNAMIC BEHAVIOR OF TETHERED BALLOON SYSTEMS Jerome J. Vorachek james W. Burbick George R. Doyle, Jr. Goodyear Aerospace Corporation Akron, Ohio 44315 Contract No. F19628-71-C-0091 Project No. 7659 Task No. 765906 Work Unit No. 76590601 Final Report January 1971 through 31 January 1972 Contract Monitor: Don E. Jackson, Capt., USAF Aerospace Instrumentation Laboratory Approved for public release; distribution unlimited Prepared for AIR FORCE CAhRIDGE RESEARCH LABORATORIES AIR FORCE SYSTEMS COMMAND UNITED STATES AIR FORCE BEDFORD, NASSACHUSETTS 01730 it S~ABSTRACT An analytical investigation of the dynamic behavior of tethered balloons was conducted. This report covers definition of tethered balloon systems and a study of stability characteristics and dynamic response to winrI gusts of tethered balloon systems. Balloon systems which are investiga-ted use the British BJ Barrage Balloon, the Vee Balloon and a Goodyear Aerospace Model No. 1649 Single-Hull Balloon. The major tether construction is Columbian Rope Company's NOLAPO utilizing prestretched polyester filaments. Three design altitudes, 5,000, 10,000 and 20,000 feet, are considered. The model for the tethered balloon system consists of the streamlined balloon and a tether made up of three discrete links. The derivation of non-itnear equations of motion for this system in three dimensions is presented. Tho. equations are linearized for stability analysis and treated as uncoupled in the longitudinal and lateral degrees of freedom. Characteristic equations which incorporate the physical, aerodynamic, and mass characteristics of the system are developed and solved for the roots which represent the frequency and damping qualities. The nonlinear equations of motion are programmed for solution on a digital computer. An exploratory investigation to establish the trends of dynamic characteristics for various design parameters is reported. Design parameters considered include balloon shape and altitude, trim angle of attack, vertical location of the bridle confluence point, net static lift, tail size, reduced wind profiles, varying altitude as encountered in launch and retrieval, pay-load location, and wind location above mean sea level. I ii FOREWORD This research was supported by the Air Force Systems Command, -,SAF, DOD, and was under the technical cognizance of the Air Force Cambridge Research Laboratories under Contract No. F 19628-71-C-0091. The project is being carried out under the direction of Captain Don Jackson as Contract Monitor for the Air Force Cambridge Research Laboratories. Mr. Jerome Vorachek is the Goodyear Aerospace Corporation Project Engineer. Technical effort has been provided by Mr. George Doyle for derivation of the equations of motion and characteristic equations of the tethered balloon systems, programming of equations for solution of stability and dynamic response. Mr. James BurLick generated dynamic response data and evaluation of dynamic response results with the IBM 360 computer. Mr. William Ebert developed the aerodynamic characteristics for the balloon system, and mass characteristics were generated by Mr. Walt Stricker. Technical assistance was also provided by Mr. Philip Myers and Mr. Louis Handler. The contractor's number for this report is GER-15497. i i iii TABLE OF CONTENTS Section Page I INTRODUCTION .......... ..... ..................... 1 IT. BALLOON DESCRIPTION ......... ... .................. 3 III TECHNIQUES FOR STUDY OF DYNAMIC BEHAVIOR OF TErHERED BALLOONS ........... .................... 7 1. Mathematical Models ......... ............... 7 2. General Stability Theory . . .. ........ ........ 9 3. Stability Analysis ..... ............... .... 11 4. Dynamic Response Analysis .... ............ .... 12 IV DYNAMIC RESPONSE OF T'hTHERED BALLOONS ..... ......... 15 I. Design Conditions Investigated ..... ......... 15 2. General Discussion of Dynamic Response of Tethered Balloon Systems ... ........... . . 17 3. Effects of Design Parameters on Tethered Balloon Design . . . . . . . . . . . . . . ... 29 V CONCLUSIONS AND RECOMNENDATIONS ON TETHERED BALLOON DYNAMIC BEHAVIOR ..... .... ....... 44 1. Gust Effects on Nominal BJ Balloon ....... ... 44 2. Comparison of Ballooa Types ........... 45 3. Influence of Trim. Angle on Longitudinal and Lateral Dynamic Behavior .... ............ ... 45 4. Effect of Tail Size on Longitudinal and Lateral Dynamic Behavior ...... ................ .... 46 5. Effect of Altitudes Intermediate to the Design Altitude on Behavior of the BJ Tethered Balloon System ...... ................. .... 46 6. Effects of Reduced Wind on the Behavior of the BJ Tethered Balloon System ... ........... .... 47 7. Nolaro vs Amgai Tethers ............... .... 47 8. Effect. of Payload Location ......... ..... 47 9. Effect of Winch Altitude Location ........ ... 47 10. BJ Tethered Balloon Systems Designed for Various Altitudes .... ........... . .. . 48 REFERENCES. . . . . ..... ..... .. .. . ........... 49 APPENJICES A Eouations of Motion for Dynamic Simulation A-1 B Dynamic Response Input and Output Data.. .... . B-I iv LIST OF FIGURES Figure Page S1 BJ Configuration . . . . . . . .... . . . . ....... . . . . . . . 4 2 Vee-Balloon Configuration .... ................ . 4 3 GAC No. 1649 Balloon Configuration ...... .......... 6 4 Balloon Tether Model in Longitudinal Plane ......... 7 5 Balloon Tether Model in Lateral Plane......... 8 6 Balloon Geometry and Applied Forces.......... 9 7 Longitudinal Dynamics of Tethered Balloon .- BJ Nominal 13 8 Longitudinal Dynamics of Tethered Balloon - BJ Nominal 13 9 Lateral Wind Gusts ......... .................. 17 I,1 Lateral Stability Characteristics of Nominal BJ Tethered Balloon at 10,000 Feet Altitude .... ....... 20 11 Lateral Case 28, Coordinates Response ..... ........ 21 12 Lateral Case 28, Power Spectral Density ..... ...... 22 13 Longitudinal Stability Characteristics of Nominal BJ Tethered Balloon at 10,000 Feet Altitude .... ....... 25 14 Lateral Case 4, Coordinates Response. .... . ....... 26 15 Lateral Case 4, Power Spectral D7nsify ....... ...... 27 Iv LIST OF TABLES Table Page I Summary of Balloon/Cable Systems ....... ........... 5 II Summary of Cable Solutions ......... .............. 6 III A. C-mparipon of A±ongitudinal Dynamic Cases .... 16 B. Compa-.ison of Lateral Dynamic Cases ........ ... 16 IV Sunwiry of Lateral Stability Case I ......... ... 19 V Peak Amplitudes for Lateral Dynamics Case 28. ..... 23 VI Snimmary of Longitudinal Stability Case 1 (Reference 2) ........ . ................... ... 24 VII Wind Gust Duration Effects on BJ Nominal Balloon in Longitudinal Plane ..... .................. .... 29 VIII Wind Gust Ramp Rise Time Effects on BJ Nominal Balloon in Latecial Plane ...... .................. ... 30 IX Effect of Increase irn Gust Velocity on BJ Nominal Balloon in Lateiil Plane .... ............... .... 31 X Effect of Trim Angle of Attack on Longitudinal Response of Three Balloon Types ............. ... 32 XI Effect of Lateral Wind Gusts on the Lateral Response of the Three Balloon Types ................. .... 32 XII Longitudinal Response of BJ Balloon with Tail Size Variation .......... ...................... ... 34 XIII Lateral Response of BJ and Vee Balloons with Tail Size Vdiiation ...... .................... .... 34 XIV Effect of Intermediate AltiLudes on Longitudinal Response of the BJ Balloon ................. .... 35 XV Effect of Intermediate Altitudes on Lateral Response of the BJ Balloon ...... .................. ... 35 XVI Effect of Reducing Winds on Longitudinal and Lateral Response ............. ....................... .7 XVII Effect of Nolaro and Amgal Tethers on Longitudinal Response ....... .................... ...... 38 vi Ii Tab le Page XVIII Effect of Nolaro and Amgal Tethers on Lateral Response 38 XIX Effect of Payload Location on Longitudinal Response . 39 XX 'Effect of Winch Altitude Location on Longitudinal ieusponse ........ ...................... ...... 39 XXI Effect of Various Design Operational Altitudes. . .. 40 XXII Structural Design Parameters .................. 41 XXIII Instrumentation Design Parameters ............ .... 41 XXIV Maximum Variations of Structural and Instrumentation Design Performance Parameters - Lolgitudinal Cases.. 42 XXV Maximum Variations of Structural and Instrumentation Design Performance Parameters - Lateral Cases . . . . 43 vi. SECTION I INTRODUCT ION The objective of this program is to investigate the dynamic behavior of tethered balloons and in so doing to establish design criteria for tethered balloons, tethers and payloads. The program is organized into three steps: (1) Definition of balloon systems for dynamic analysis (2) A study of stability characteristics of tethered ballon systems (3) A stuay of dynamic response of tethered balloon systems to wind gusts Reference I presents a definition of tethered balloon systems to be used for dynamic studies. These systems are summarized in this report, Reference 2 presents a development of the study of stability characteristics of the tethered balloon systems. This report presents further development of the equations of motion and solutions fo, the dynamic response of the tethered balloon system. Recommendations for the design of tethered balloon systems are presented in this report. The system is defined as a balloon tetherod at the end of a cable which is fixed to a stationary winch. The tether is represented by 3 straight links, each of the same length. The links are considered rigid and connected to each other by frictionless hinges., Goodyear Aerospace Corporation has employed this method of representation of the tethered system for other studies such as that described in Reference 3. For design purposes in the subject investigation the follouing design parameters were specified by AFCRL. Payload and Design Altitude Payload (Ib) Float Altitude (ft above MSL) P i h 1500 95,000 1000 10,000 600 20,000 Tethers Two specific tether constructionb ire to be considered, NOIARO by ýJolumbian Rope Company, and Amgal-MonitoL AA wire rope by United States Steel. A safety factor of 2.0 will be used with NOLARO; 1.5 for Amgal-Monitor. Tether design load will be Lased on a survival wind of 1.3 timet the operational wind at balloon altitude. Wind Profile The wind profile as specified is tabulated below: Operational Survival Altitude Wind Speed Wind Seed (ft above MSL) (knots) (knots) Sea Level 10 13. 5,000 31 40.3 10,000 40 52. 20,000 54 70.2 2 SECTION II BALLOON DESCRIPTION The ballGon systems evaluated in this stability investigation have been described in Reference 1. The three balloon types in these systems are the British BJ Balloon, the Vee-Balloon and the GAC No. 1649 Single Hull Balloon as depicted in Figures 1, 2, and 3. The nominal tethered systems character-istics defined in Reference I are suntarizeG in Tables I and II. Static and dynamic aerodynamic characteristics for the balloons have been determined from experimental data where available and by analycical techniques otherwise. The aerodynamic characteristics are presented and discussed in Reference 2 along with additional mass and suspension system geometries which were not included in Reference 1. Design parameters varied in the stability investigation include the trim angle of attack, vertical location of the suspension system confluence point below the balloons, the free-static lift, and the tail size. The trim angle of attack can be controlled by locating the confluence point of the suspension system. The location of this point is established by the two coordinates as shown in Figure 1. The fuselage staion from the nose and the waterline below the centerline of the balloon defines this point. The trim angle of attack change as a function of bridle apex point location has been calculated and is given in Reference 2. The free-static lift as used in this report is the excess buoyant lift provided by the balloon after the balloon physical weight and the weight of the tether in no wind is supported. Nominal tail sizes for each balloon are depicted in Figures 1, 2, and 3. For the British BJ Balloon, tail size is increased and decreased about the nominal by changing the linear dimension- of each of the three tails and main-taining similar proportions. The inters .tion of the trailing edge of the tails and the hull is maintained at the same point for all tail sizes. For the Vee-Balloon the horizontal tail geometry is maintained. The vertical tails are increased in size in two steps for tails below the hulls. Also investigated are the conditions where vertical tails are i cated symmetrically above and below the hulls. In addition to establishing aerodynamic characteristics for these tail configurations as noted in Reference 2, the increase in physical mass and additional mass was computed and used in the determination of stability characteristics. Ascent and descent studies for a specific balloon also incorporate the change in mass characteristics due to air density changes with altitude. Trademark, Gooyear Aerospace Corporation, Akron, Ohio ',4315 3 MAXIMUM DIAMETER I 0.287-Figure i, BJ Configuration /MAXIMUJ, DIAMET ER Figure 2.. Vee-Balloon Configuration 4 :ý -M ii 02 I -; ol N .30 0A , 4 V CC ~~~4 w 4 ~ -mA~~ 4, 4 uM ii V -6 .4 Cc w i N 0 0. ) 4j 0) N~~c 4 44N . .0r- (ni 4 0 ~~~ --4 . E-4 Table I1. Summary of Cable SoAutions Bulloor Type Altitude Hull Tether 8.5. O.D. htat z T Length Total Volume Type I I Tether (ft) (ft3) (3b) (ft) (lb/ft) (ft) (des) (Ib) (it) (ib) LkI 5,000 + 0 161.6 1,115 0 --46,000 NOLARO 3,813 0.02679 0.03211 0 h -5 S.L. 2.62 144.2 956 .692 183 t'- 1, 5W lb IO,OOG 0 165.7 6,115 0 -Vee-bal loon 5.000 80.000 NOLARtO 16,600 0.04875 0.11400 1,775 154.7 5,552 5,314 606 S.L. 4,594 146.6 4,991 11,060 1,260 h 1 I0,000 ft -" P -,O0 lb 10,000 0 167.5 10,350 0 -. 5,000 144,000 AAGAL 20,000 0.03642 0.30400 1,389 160.6 8,836 5,193 1.:80 S.L. 3.,52 153.9 7,300 10,605 3,230 10,000 0 163.4 2,287 0 - _,000 60,000 NOIA1o 7,540 0.03420 0.05600 2,456 143.6 2,036 5,600 3'4 I S.L.A _ 7.377 127.7 1,761 12,639 708 h " 10,OGC ft 10,000 0 165.3 2,975 0 - P -1.000 lb 5,000 75,000 ANGAL 6,750 0.02083 0.09970 1,8643 153.2 2,482 5,340 532 S.L. 5,072 140.3 1.989 11.305 1,128 10,000 0 168.5 3,812 0 - 51649 ,,,Thn ria 5,000 80,000 NOIA.O 12.700 0.04330 0.0900 1,702 153.0 3,374 5,298 476 S.L. 4.950 161.1 2,930 11,274 1.015 h = I0,000 ft -I I. P -1,000 lb 10,000 0 170.4 6.375 0 --5.000 133,000 AMGAL 14,800 0.03125 0.22000 1,193 161.8 5,281 5,145 1,132 S.L. 3,2?, 152.7 4,187 10,559 2,325 20,000 0 166.7 15,4% 0 8.3 15,000 1,646 155.9 13.897 5,272 1,710 h 20,000 ft 10,000 500,000 NOLARO 7,814 0.08281 0.32427 4,,495 143.7 12,171 11,039 3,580 Slb 000 9,106 130.2 10,702 17,856 5,790 S.L. 17,231 112.0 9,107 27,436 8,900 0.3L L Figure 3. GAC No. 1649 Balloon Configuration I6 SECTION III TECHNIQUES FOR STUDY OF DYNAMIC BEHAVIOR OF TETHERED BALLOONS 1. MATHEMATICAL MODELS A system of differential equations was developed (Reference 2 and Appendix A of this report) that describes the motion of the tethered balloon in three dimensions. The degrees of freedom associated with the motion are yaw, pitch and roll of the balloon abcut its dynamic mass center, and pitch "and yaw (lateral rotation) of the tether. There are a total of 3 + 2N degrees of freedom where N is the number of links used to simulate the tether. First consider the longitudinal degrees of freedom. The dependent variables shown in Figure 4 are 0 (pitch of the balloon) arnd •r (pitch of the "r"th link), where r is a particular link. All angles are shown positive. 0! -W2 Side View Figure 4 . Balloon Tether •dei in Longitudinal Plane LN 1 In Figure 4 VBR is the relative velocity of the balloon's center of gravity with respect to the air and is the resultint of the steady wind, the wind gust, the balloon translational motion and the velocity due to rotation cf the balloon about its center of mass. The angle of attack (a) is the angle that the relative wind forms with the longitudinal axis of the balloon. The lateral degrees of freedom are displayed in Figure 5 which gives the front and top views of the tethered balloon. The lateral degrees of freedom are: (yaw of balloon), 0 (roll of balloon), and Ur (yaw of Yrith link). All angles are shown positive. aN ,/ NTE PN-1 arA Za 2 z x Front View Top Viev Figure 5. Ball~on Tether Model in Lateral Plane Pertinent geometry of the tethered balloon and applied forces are identified in Figure 6. In order to separate the equations of motion into a longitudinal response and a lateral response, it was further assumed that the system was near equilibrium. This resulted in a set of equations describing the longitudinal motion which is coupled only in the pitching variables of the balloon and the pitching variables of the tether. However, the second set of equations for the lateral motion does not completely uncouple from the 8 L LS FBVA WB - Total weight of balloon including internal gases FEVA - Vertical aerodynamiic forces acting on balloon FBHA - Horizontal aerodynamic forces acting on balloon -MOB - Aerodynamic pitching reorient acting on balloon CA - Aerodynamic reference center (at center of hull volume) CL - Center of buoyancy of total displaced volume Cm - Apparent mass center and dynamic center in pitch (and yaw) CO - Center of gr vity of total wei ght of balloon Figure 6. Balloon Geometry V cnd Applied Forces longitudinal degrees of freedom because the equilibrium angles in the longi-tudinAl plane are not zero. Therefore, when solving the lateral degrees of freedom, it must be assumed that the longitudinal variables remain constant and equal to their equilibrium values. In both the longitudinal and lateral cases, the tether is simulated by three rigid links The number of uncoupled dynamic equations is four for the longitudinal response and five for the lateral response. f 2. GENERAL STABILITY THEORY The equilibrium configuration of a tethered ballon can be defined as "that position which demands that the summation of all applied moments equals zero. The equilibrium is said to be stable if, for any small disturbance, 9 I! the system ultimately returns to its equilibrium conditions. Two types of stability are of interest. In the first (statically stable), a small displace-ment of the system will create forces which tend to return the system to its equilibrium position. The second (dynamically stable) produces a motion which eventually restores equilibrium. If the motion is periodic, it is character-ized by a damped frequency and a damping ratio. Similar definitions apply for statically and dynamically unstable motions. A third possibility is for the system to be neutrally stable during which the motion neither diverges nor converges. It was necessary during this study to develop techniques to investigate and understand the stability of the system. To this end, characteristic equa-tions are derived. The general approach is as follows: (1) Derive the nonlinear equations of motion in three dimensions for each degree-of-freedom (2) Assume the motion is near equilibrium so that the equations can be linearized and separated into a longitudinal motion and a lateral motion (3) Laplace transform the linear equations from the time domain to the "S" domain assuming that the initial con-ditions are zero. This establishes a matrix equation of the following form: [A] {X (S)} = o~ where IX (S)l is the eigenvector and [A] is a square matrix whose elements are quadratics in S containing the physical properties of the system. (4) Expand the determinant of [A] such that the characteristics polynominal is obtained. Each root of the characteristic equatior represents a term in the general solution of the form, Aie Sit, where Si is the "i"th root and Ai is an amplitude, dependent on the initial conditions of the system. Both real and complex roots may appear where the complex roots occur in conjugate pairs. In general for "n" degrees of freedom, the characteristic equation will yield "2n" roots. Each pair of complex conjugate roots represents one oscillatory motion, while each real root represent. one aperiodic motion. First consider an oscillatory system. This motion is characterized by two roots of the form Si = X ± i Y, where X and Y are real numbers and i = VfT. Several important quantities can be found from the root. The natural frequency associated with this motion is Wn = JX2 + Y2 . The damping ratio is W = -. The damping frequency is A)d =(in n -4T= y It is also of interest to Rnow the time to half amplitude for a stable root or the time to double amplitude for an unstable root. This quaLl.jty can easily be found by considering one oscillatory motion. The general solution for free vibration is 10 Z Ce n (s dt+n) (2) where 0 is the phase angle dependent upon initial conditions C is a constant eependent upon initial conditions The second possibility is an aperiodic motion given by the expression Z CeX (3) where X is the real part of one root and the imaginary part (f) is zero If X is negative, Z approaches zero as time increases indefinitely and the motion is said to be overdamped. Like the oscillatory motion, roots which give overdamped motions will also occur in pairs. Howevtr, unlike the complex conjugate roots which result in one oscillatory motion, each real r-oot is a dinstinct motion. Therefore, it is possible for an "n" degree-of-freedom system to have "2n" distinct motions if the system is so heavily damped that all the roots to the characteristic equation are real. There is a third possible motion which is a borderline case. If two roots are real and equal, the system is said to be critically damped. The motion will be aperiodic 4nd both roots will give the same motion. The general solution to the motion of the system is a linear combina-tiou of all the motions defined by the roots to the characteristic equation. Associated with each root is a mode shape which gives the relative amplitudes of each degree of freedom when the system is responding to one particular root. It is of interest to establish these mode shapes so that each stability curve can be associated with a definite motion of the whole system. For example, one mode shape may show that the pitching motion of the balloon !S very large compared to the motion of the tether-3. STABILITY ANALYSIS Derivations of the equations of motion of the tethered balloon system and development of the characteristic equations for a tethered balloon system approximated with a three-link tether are given in Reference 2. The four linearized longitudinal equations are Laplace transformed, and an eighth order characteristic equation generated which specifies stability characteristics of the system. In like manner, the five linearized lat3ral equations can be reduced to a tenth order equation which gives stability information in the lateral degrees-of-freedom. The roots of these character-istic equations identify the natural frequencies, damped frequencies and damping ratios. 11 I Results of characteristic equations analysis can be displayed in the form of plots in a complex plane typically as shown in Figure 10 of this report. In these plots, the abscissa is the real part of the roots to the characteristic equation, the ordinate is the imaginary part. A negative real part means that mode of oscillation is ,-onverging or stable; a positive real part is a diverging mode. Only the first and second quadrants are displayed because the roots are complex conjugates which are symmetric to the real axes, except in the case of overdamped roots which lie on the real axes. In general, the following information is easily available for each root directly from the plots. The natural frequency is measured in rad/sec as the distance from the origin to the root; the damped frequency (rad/sec) is the value of the imaginary part of the root; the damping ratio is equal to the absolute value of the real part of the root divided by the natural frequency. DYNAMIC RESPONSE ANALYSIS The calculation of the balloon system response to specific disturbances is the subject of the dynamic response analysis. The most general motion the system can have is a linear superposition of the normal modes. Each aperiodic or nonoscillatory normal mode has one arbitrary constant (the initial value of an, ne of the variables) associated with it; and each periodic or oscillatory normal mode has two arbitrary constants (the amplitude and phase angle of any one of the variables) associated with it. The total number of arbitrary constants is then equal to the number of aperiodic modes plus twice the number of periodic modes; i.e. to the degree of the character-istic equation, or the order of the system. A specific disturbance will excite the normal modes in varying degrees and establish the values of the arbitrary constants. The dynamic response of tethered balloon systems to various wind dis-turbances is obtained by integrating numerically the longitudinal and lateral equations of motion to produce a time history of the dynamics. The start conditions, or equil brium conditions for the dynamic response computer pro-grams are obtained fi m the linearized stability computer programs (Reference 2). This approach to analysis has the advantage that wind gusts can be pro-duced and the actual motion of the sysLem can be observed. The major dis-advantage is that a greater amount of computer time is required when compared to evaluation of stability by investigating the roots of the characteristic equations. The equations of motion for the longitudinal dynamics of a tethered balloon system were initially derived in two forms (see Appendix A):, 1) inertia terms which contain products of angular velocities are neglectedp 2) inertia terms which contain products of angular velocities are included. The concept of neglecting products of angular velocities is associated with the assumption that angular velocities are small; and therefore, products of angular velocities are negligible. Numerical integrations were made with the computer to determille the effect of neglecting the inertia termis containing products of angular velociLies, Although the effect is obviously present (Figures 7 and 8), the overall difference. between the results of the two sets of equations is small. 12 LONGITUDINAL DYNAMICS OF TETN(REO BALLOON 6J NOMINAL g!. . . . . . .T ....... INCLUDES PRODUCTS OF ANGULAR VELOCITY TERMS --. DOES NOT INCLUDE PRODUCTS OF ANGULAR VELOCITY TERMS II C3 , 4 4 ' ; o o Ie 4e6 • o s 7 o TIME -S7CONDS Figure 7 LONGIIUDINAL DYNAMICS OF TETHERED BALLOON IJJ NOMINAL -...... INCLUDES PRODUCTS OF ANGULAR VELOCITY TERMS a DOES NOT INCLUDE PRODUCTS OF ANGULAR VELOCITY TERMS •J lt a- a' 0L • • . Cl --I ----o to so •'0 6 o0 ?a so to TIME -SECONDS Figure 8 13 Figure 7 shows the time history of the four generalized coordinates • , •2, 4ý, and 9. A horizortal gust of 20 feet per second with a duration of 20 secieds was applied. The dotted lines represent the angular aisplacement when the products of angular velocities are included in the inertia terms. Figure 8 presents the range and altitude of the balloon as a function of time and also the variation in tether tension at the winch and at the bridle con-fluence point. Since this gust velocity is considered to be high, it is concluded that the differenrve between the solid lines and the dotted lines is maximum. It was decided that dynamic simulation studies would be conductea with a model which neglects products of angular velocities for three reasons. First, the equations containing products of angular velocities ate shown to give only slightly different results. Second, it is desirable to keep the dynami: equa-tions compatible to the equations used in the stability study (the stability study used linearized equations). Third, i: is desirable to keep the longitu-dinal equations compatible to the l, teral equations (to derive the lateral equations of motion containing products of angular velocities in the inertia terms would be a very difficult task because of the number of terms involved.) 14 SECTION IV DYNAMIC RESPONSE OF TETHERED BALLOONS 1. DESIGN CONDITIONS INVESTIGATED The parametric study conditions found in Apendix B, Tables B-I and S~B-II and Reference 2 are the source of information for the results and recommendations of this report. Each parametric dynamic response study condition has a computer drawn plot of various balloon parameters versus time. These plots aie found in Appendix B. The parameters chosen for investigation during this study are placed in three major categories: (1) "Balloon on Station" performance parameters (2) "Structural Design" performance parameters (3) "On Board Instrumentation" Design parameters The "Balloon on Station" performance parameters are defined as balloon displacements from equilibrium and, specifically, are changes in balloon range, altitude, pitch angle, roll angle, yaw angle, and lateral displacement induced by a parameter variation, such as a change in trim angle. The "Structural Design" performance parameters of interest are the cable tension at winch and bridle, and the balloon translational and rotational accelerations. The "On Board Instrumentation" design parameters are defined as the balloon angular displacements and rates. The effects of gusts, trim angle, altitude, tail size, and other parameters on balloon performance parameters are reviewed in detail later in this section of the report. Table III lists the cases compared and the reason for comparison for the longitudinal and lateral cases. The operational winds used for the investigation, as a function of altitude, are listed in the Introduction to this report. Gusts used for dynamic response are added to the operational winds and are applied to the balloon. The gusts used in analysis are discussed next. Consider the nominal altitude condition of 10,000 feeE where the operational design wind was 67.5 feet per second. Longitudinal gusts were applied where the horizontal wind increased by 30 percent, for a discrete time, to a value of 1.3 x 67.5 = 87.6 fps. Gusts of 2, 10, and 20 seconds and infinite time durations were investigated with various rise times. A case was also investigated where the 30 percent gust increment was applied vertically upward. 1 15 Table III A. Comparison of Longitudinal Dynamic Cases Cases Balloon Compared Type Reason for Comparison 1,2,3,4,5 BJ Nominal Gusts and Wind Effects 3,21,18 All Comparison ot Balloon Types 3,6,7 BJ Nominal Trim Angle Effects 18,19,Zo VEE Balloon Trim Angle Effects 21,22,23 GAC Ba" ½on Trim Angle Effects 11,3,12 BJ Nominal Tail Size Effect 3,14,15 BJ Nominal Winching Effects 13,3 BJ Notinal Reduced Wind (40%) Effect 10,3 BJ Nominal Amgal vs NOLARO 16,3 BJ Nominal Effect of Payload on Underside of Balloon 17,3 BJ Nominal Winch at 5000 It, and 0.0 Ft. 8,3,9 BJ Operational Altitude Effects Trim angle changed by varying fore and aft location of bridle confluence point B. Comparison of Lateral Dynamic Cases Cases Balloon Compared Type Reason for Comparison 24,25,26 BJ Nominal Effects of Gi-st Ramp Time 27,26 BJ Nominal Effects of 50% Increase in Gust Velocity 40,37,25 BJ, VEE,GAC Comparison of Balloon Types 32,25,33 BJ Tail Size Effect 37,38,39 VEE Balloon Tail Size Effect 35,36 BJ Nominal Winch Effects 34,25 BJ Nominal Reduced Wind Effect (100%--140%) 31,25 BJ Nominal Effects of Amgal vs NOLARO 29,25,30 BJ Operational Altitude Effect' 16 The lateral gusts applied to the tethered balloon systems to investi-gate dynamic response may be generally interpreted in two ways as shown in Figure 9 which depicts the wind in a horizontal plane. Condition A - Build-up Condition B - Shift of wind from side in wind direction Vw steady longitudinal axis -~of ballJoonA - -Vg Vw steady VR Vw steady Vg= 12 and 18 fps a =10, 15°0 Vg = 1 tr 3 5 setonds a tr 1, 3, 5 seconds tr tr Figure 9 . lateral Wind Gusts In Conditior A the wind gust from the side builds up with time. This is approximately e4uivalent to the Condition B where the wind remains constant in magaiitude and changes heading direction with time. GENERAL DISCUSSION OF DYNAMIC RESPONSE OF TETHERED BALLOON SYSTEMS The non-linear differential equations defining the moticn of a tethered balloon system and the linearized equations used in studying the stability characteristics of the balloon system are given in Reference 2 together with the results of the stability analysis of various balloon systems. Appendix A L of this report expands the equations of motion of Reference 2 to include wind gusts. The response of the non-linear balloon system when subjected to wind gusts has been programmed for computer solution using numerical integration methods. Various balloon systems, subjected to different types of gusts, have been studied ising this computer program and th- results are found in ?' Appendix B. The response plots shown in Appendix B reveal that the balloon systems are highly damped. This agrees with the stability analyses of the balloon systems given in Reference 2. In fact, the motion is so highly damped that the lower frequency response modes are, in general, damped out within two cycles or less of motion. Thus, the high aerodynamic damping involved in the balloon systems, makes it very difficult to evaluate the response plots with regard to phase relationships between coordinates, damping characteristics, and the frequencies involved in the motion. It becomes evident after looking at the 17 response plots that only the amplitude and direction of the response can be obtained. A mathematical tool will be required to obtain information regarding phase relationships, natural frequencies, and danping characteristics of the modal motions involved. Such a tool would be ve-y valuable for evaluating flight data from tethered balloons. What then, is this mathematical tool? The resonse of a highly damped multidegree of freedom system to a known disturbance can be represented mathematically as the sum of the system's modal responses. Assuming the system follows the response of a viscously damped system, a modal response definition is given by-the following equation. Z = (c e -4i Wnit) sin ((jd. t + 0) The total response ZT of a particular coordinate is then given by =e i ni t) sin ( jd t+ (I) i=l d N represents the numbez of frequencies involved in the response Srepresents the damping ratio (i represents the natural frequency Wd represents the damped frequency 0i represents the phase angle C represents an amplitude constant Each modal res onse Z. has 4 unknowns, since Wd is related to Wn •2. by 3d = 1- -Thus any response curve which is defined by a sufficient number of points can be mathematically represented by Equation (i) and solved for the unknown constants using numerical curve fitting methods. No curve fitting program is presently available to the prcject for use with Equation (1); however,thece is a curve fitting program available which uses a Fourier Transiorm and performs a power spectral density (P.S.D.) analysis of the Fourier Transformation. This program is only valid for analyzing systems with no damping. The program still gives a general overall view of the frequencies involved in the motion and It is thought worthwhile to analyze the response of two dynamic cases and to compare the frequencies found with the frequencies obtained from the stability analysis of the same balloon configurations given in Reference 2. The dynamic cases chosen for frequency comparison are lateral dynamic response Case 28 and longitudinal dynamic response Case 4. The cases represent a nominal BJ balloon at 10,000 feet. Lateral response i& for a 10 second duration step side gust and longitudinal response is for a 20 second duration step horizontal gust. 18 a. COMPARISON OF THE NON-LINEAR LATERAL DYNAMICS CASE 28, WITH THE LINEARIZED STABILITY CASE 1 The small displacement motion of a forced non-linear system, in the time era after the forcing function is removed, should result in system motions similar to the sum of the mode shape times participation factor for each mode found in the linearized stability analysis of the system. To verify this, Lateral Dynamics Case 28 and Lateral Stability Case 1 (Reference 2) were chosen for comparison. A complete description of Lateral Dynamics Case 28 is found in Appendix B, and the Lateral Stability Case 1 description is found in Reference 2. For convenience the results of Lateral Stability Case I are summarized in Table IV and Figure 10 of this report. In order to analyze the coordinate motions of Case 28, a power spectral density analysis with Fourier Transform was performed on each of the lateral coordinate degrees of freedom; ar, , O1, o2 r G 3 . The input coordinate waveforms and PSD plots for each degree of freedom are given in Figures 11 and 12. The PSD plots reveal that there are two predominate modes involved in the system motion. These are underdamped oscillatory modes 1 and 2 with natural frequencies and damping ratios of in 1 = 0.0051 Hz and •1 = 0.511 for mode 1 and fn2 = 0.136 Hz and 42 = 0.207 for mode 2. Mode one (0.0051 Hz) is the predominate mode with mode 2 (0.136 Hz) of smaller ampli-tude superimposed on mode 1. The lateral wind gust disLurbance in Case 28 is removed after T = 11 seconds, therefore all comparisons should be made after the first peak is reached beyond 11 seconds. Tt should be noted that mode 2 is predominately a 0 or roll mode, and this is evident from the plots of the coordinates, since only the 0 coordinate plot has evidence of a 0.136 Hz mode. On the 0 plot mode 2 appears damped out of the 0 motion by the time T = 37 see. Table IV. Summary of Lateral Stability Case 1 Normalized Modes Mode Frequency Damping 3 Number Hz Ratio Ampl. 4 0 Ampl. 4 0 Ampl 40 Ampl. 40 Ampl. 40 1 0.0051 0.511 1.0 0.0 0.195 301 0.342 83.3 0.429 222 0.483 275 2 0.1363 0.207 1.0 0.0 9.12 252 0.016 17.7 0.051 215 0.102 58.9 3 0.000215 1.0 1.0 0.0 0.0082 0 8.86 180 2.51 180 0.487 180 4 0.0252 1.0 1.0 0.0 0.927 0 1.22 360 2.27 180 0.957 360 5 0.0578 1.0 1.0 0.0 1.58 0 2.02 360 3.59 180 1.54 360 0 0.1295 1.0 1.0 0.0 23.1 180 0.345 180 0.147 360 0.513 360 7 .1608 1.0 1.0 0.0 60.2 360 7.34 180 6.09 180 12.7 360 8 0.4320 1.0 1.0 0.0 0.988 360 0.00007 180 0.0006 360 0.0051 180 19 Lateral BJ-Eff,.ct of Trim Angle, j.12 Run Nos. 1,2,3 O T = 8.5" (NOMINAL) Ot T = 6.40 ,.08 O T =-1- o5" i. F) 0 -.04 0 ~T Sd 4B ® R -A2 1 -- 48 -Z 6- -2 Lateral BJ-Nominal, Run No. 1 '.7 M ode Shapes -I --6 -r -- • -2 -Z 2 -2 0 -/ A,4 - e -I A CNdr __, , Figure 10. Lateral Stability Characteristics of Nominal BJ Tethered Balloon at 10,000 ieet Altitude 20 (AS91 E LAT. psi CASE n LAI Pi01 too ?A so IN 00 o oo c1o l00 3o .oo00 0o s0 I. ,oo ]so 4o, L•s to LA T I.SC C iO oo SIGI 300 )S0 400 CASE ?9 LAI 3Io(?) CASE 01 LAT SILO3) I .r T Figure 11. Lateral Case 28, Coordinates Response 21 M 1S LAT. ?$I CASE 19 LT. fHI N Nz baS 0.0 0,7 0.9 FREQUENCY (HZ) FREQUENCY CIZ) LASE to LAI SICGI) foo r 'b, 0, 53 04 0S 04 07 0a FREQUENCY 0Ii?) CASE '0 LAY SIGC)) CAME to LAI SlCCG ) PSI) ?so O0 0f 0"3 0"4 OPs 0 1 05 O5 4 FREQUENCY WH) FREQUENCY (HZ) Figure 12. Lateral Case 28, Power Spectral Density 22 The aerodynamic damping characteristics of the system appear viscous in nature for the 0.0051 Hz mode. Table V lists, for Case 28, the time and amplitude of the various IL •coordinates for the first and second peaks after the gust is removed. Included in Table V is the period, amplitude ratio, and frequency as calcu-lated from the coordinate data of Case 28. The period and amplitude ratio for the coordinates 4f, 0, and a3 show very good agreement with a 0.0051 Hz viscously damped ( • = 0.51) system. The period between peaks of a 0.0051 Hz viscously damped systum with • 0.51 is given by " ;1 1 T 1 = 228 sec. r f 12• f 1( 0.0051 -O.5 The amplitude ratio between successive peaks is given by Xn+ i-2 I = .,2i (0.51) 0"025 Xn I I -10.5l1 002 e e Table V. Peak Amplitudes for Lateral Dynamics Case 28 First Peak Second Peak Period Frequency Ampl. Ratio Coordinate After Gtst After Gust T 1 Xn + 1 Time Ampl. Time Ampl. Sec. T Xn 1 29 -3.905 250 -0.099 221 0.0045 0.0253 61 -0.404 289 -0.00973 228 0.0044 0.0241 26 -0.475 197 0.0943 171 0.0059 0.199 a2 31 1.33 224 0.0630 193 0.0052 0.0474 0 76 -0.834 304 -0.0207 228 0.0044 0.0248 23 b. COMPARISON OF NON LTNEAR LONGITUDINAL DYNAMICS, CASE 4, WITH LINEARIZED LONGITUDINAL STABILITY, CASE 1 A description of Longitudinal Dynamics, Case 4, is found in Appendix B and Linearized Longitudinal Stability, Case 1, is defined in Reference 2. A summary of the Longitudinal Stability, Case 1, is given in Table VI and Figure 13 of this report. A Fourier and PSD analysis is performed on each of the Longitudinal Dynamics, Case 4, coordinate degrees of freedom 0, ti, t2, and 93" The input coordinate waveform and PSD plot for each degree of freedom is given in Figures 14 and 15 respectively. The need of a mathematical tool for analyzing a highly damped response is very evident in Longitudinal Dynamics, Case 4. The response of every coordinate is damped out, or nearly so, in less than one cycle. The Fourier and PSD non-damped harmonic analyis of this motion is not too meaningful. However, it does indicate that the motion is comprised of frequencies below 0.05 Hz with the most power below 0.01 Hz. Table VI. Summary of Longitudinal Stability Case 1 (Reference 2) Normalized Modes Mode Frequency Damping 0 41 2 43 Number Hz Ratio -N be____ Amp1., LZ Ampl. 0a Ampl. Lo Ampl. Lo 1 0.00554 -0.037 1.0 0.0 2.14 281. 1.27 238. 1.28 194.0 2 0.0728 0.653 1.0 0.0 0.131 197. 0.080 353. 0.058 39.7 3 0.0986 0.5! 1.0 0.0 0.0747 9.6 0.214 188. 0.145 7.4 4 0.0156 1.0 1.0 0.0 0.888 0.0 0.800 0.0 0.363 1 3. 5 0.1715 1.0 1.0 J 0.0 0.00033 0.0 0.0123 0.0 0.0066 180. C. YAW-ROLL COUPLING IN LATERAL DYNAMIC CASES Yaw-roll coupling for BJ and GAC single hull balloons subjected to lateral wind is evident when the V1 response plots are reviewed. The Vee balloon appears to have negligible coupling. As an example, observe the motion of coordinate V/ (Reference Appendix B) for Lateral Case 28. One would think that, when a balloon is in equilibrium with a steady head wind and then is subjected to a side gust from the left, the balloon would yaw into the resultant wind. This would be a negative yaw angle, however, Lateral Case 28 shows the balloons first yaw motion is positive or away from the resultant wind for a time of approximately 7 seconds and to iv o +4a before the balloon starts yawing toward the resultant wind. 24 -TW Longitudinal BJ-Effect of Trim Angle, [ 1 •Run Nos. 1,2,3 f.A 0 a T 8. 85 (NOMINAL) 0 U3 T 6.5 S• aT =10.5 ° gT (D(~ -I •• - / .o -. 9 - -60 Longitudinal BJ-Nominal, Run No. 1 Mode Shapes Figure 13, Longitudinal Stability Characteristics of Nominal BJ Tethered Balloon at 10,000 Feet Altitude 25 r. I I EASE 4 ONGl TMI TA E.ASE 4 LOW'.r• -11(1) •NEND T , CASE 4 .. NG IQ( EASE 4 ONG. (I¢3) so 100 ISO M0 ZOO 100 350 AD4000 5 OO tD 30 5 0 T T Figure 14. Lateral Casc 4, Coordinates Responsc 26 LAIG 4 LONG ICI1A EASE 4 LK I1(1) N N "D. 05. 0 03 0 0.0 01 0' OS0 0-1 0! 0 • OS 006 0 7 0a FRtOlUeNCY (HZ) FREQUENCY (MZ', L. E 4 LOON, mI Q) E.ASE 4 LONG W1(3) FISO S C3 NENO 0,) 00r 0 3 0) 05 00& 0( Ob '0 0) 0f 03 0' 0 S 0 6 0', OS FHREUQUENY ((HZ) Figure 15. Lateral Case 4, Power Spectral Density 27 To explain this motion refer to Equation 86 in Appendix A. Neg-lecting link acceleration terms results in the following equation. ~~~~~ 3 'M Rj Rkm " 2 31~ ýkm -'YB S" + Y7,B C 0 I+IMS [ -Rjm 3 3 + Rkm " 2~ [ML _Rkln 13]2 2s22 +[M Rm 1 3 ] + C2 @ [IBS2 + IZB C + s2 1 IYB -S2 CO IYZBI q Substituting a and b for the coefficients of 0 and , results in the following equation. F = a + b 4, The equation is solved for F -a0 where F o is the total externally applied balloon y oment. Case 28 parameters have the following values at T = 5.5 seconds (from digital printout of Case 28). Fp = -40039 ft-lb 0 = 3.060 °/sec' = 1.207°/sec2 IXB 116,000 Slug Ft 2 = 2.373 0/sec = 1.9890 /sec 2 I = 26,000 Slug Ft 0 = 0.5310 i = 0.4260 YB IZB 119,000 Slug Ft 2 IYZB 12,500 Slug Ft2 28 Inspection of the j coefficient, b, reveals it co be positive. Since F1 = -40039 ft-lb and will tend to turn the balloon into the resultant wind, the only term left to cause to be positive is the roll coupling term ao In concluflon the initial q motion away from the resultant wind is caused by roll (0) inertia coupling. EFFECTS OF DESIGN PARAMETERS ON TETHEPED BALLOON DESIGN a. Balloon on Station Performance Parameters (1) General. As listed in Part I of this section a number of tethered balloon design parameters and flight conditions were varied to esteblish their influence on a tethered balloon design. This subsection summarizes some of the major effects on tethered balloon dynamic behavior as established from the dynamic response calculations presented in Appendix B and an evaluation of these data. (2) Wind Gust Effects. "he response data in Table VII shows the effect of increasing the duration of a horizontal gust on the longiLudinal motions of the nominal BJ balloon design for 10,000 foot altitude. It is apparent from an examination of these data that the longer duration of the gust the greater the angle of attack of the balloon becomes. A new equilibrium con-dition is reached when the ;ust is left on indefinitely. As a consequence of the increased angle of attack and wind velocity the aerodynamic lift increases and a somewhat more favorable aerodynamic lift to drag ratio is obtained. This serves to increase the altitude of the balloon and reduce the downwind displacement of the balloon. A substantial increase in tether tension is also evident. A nominal BJ balloon operating at 10,000 feet is subjected to a 20.2 ft/sec, 20 second d&ration, horizontal gust in Longitudinal Case 4 and to a 20.2 ft/sec, 20 second duration, vertical gust in Longitudinal Case 4A. Table VII also summarizes the results of these two cases. Table VII. Wind Gust Duration Effects on BJ Nominal Balloon in Longitudinal Plane Altitude -10,000 Ft. Gust Velocity = 20.2 Ft/Sec Payload Weight -1,000 Lb. Gust Duration -t Trim Angle . 8.50 Balloon Displacements Cable Tensions At Yaximum Cable From Equilibrium Equilibrium Tensions ,=Cs G A A J A Case G Range Altitude Pitch Winch Bridle Winch Bridle No. Down/Up Sec. Ft., Ft, Deg., Lb. Lb. Lb, Lb. S 00 -65/800 690 5.7 1650 2210 3630 4200 2 2 -12/16 15 1.9 1650 2210 2250 2830 3 1 0 -52/73 82 2.9 1650 2210 2650 3210 4 20 -b8/146 121 4.7 1650 2210 3180 3740 4A 20 0/451 345 1._5 1650 2210 3582 3941 29 Specific comments for this comparison are: (1) The vertical gust displaced the balloon up range approxi-mately three times the horizontal gust up range displacement. (2) The vertical gust caused the balloon to rise three times higher than the horizontal gust. (3) The pitch angle change from a vertical gust is only 1/4 the pitch angle change due to a horizontal gust of equal magnitude. (4) Maximum cable tension increased by 400 pounds due to the vertical gust. It is noted that equilibrium cable tension is slightly differcnt from that listed in Table I. This is a result of different balloon trim angles and number of cable links used in the two analyses. The effects of gust ramp rise time on the response of a nominal BJ balloon is given in Table VIII. These cases represent BJ nominal balloons at 10,000 fe(It Lhat are subjected to a steady lateral gust with ramp rise times of 1, 3 and 5 seconds. Table VIII reveals that gust linear ramp rise times of 1, 3, and 5 seconds have little effect on balloon response, except for balloon roll where the 1 second rise time is 1.8 times larger than the 5 second rise time response. Table VIII. Wind Gust Ramp Rise Time Effects on BJ Nominal Balloon in Lateral Plane Balloon Displacements Cable Tension Maximum Cable Ramp Gust from Initial Equilibrium At Equilibrium Tensi n Case Rise Velocity Yaw Roll Let. At At At At No. Time -c +/- Displ; Winch Bridle Winch Bridle __ Deg. Deg. Ft, Lb. Lb. Lb. Lb. 24 1 sec t2 4 /-13.6 3.9/-0.4 460 1650 2210 1686 2258 25 3 -ec 12 4.3/-13.8 3.1/-0.35 460 1650 2210 1680 2250 26 5 se, 12 4 /-13.6 2.2/-0.4 460 1650 2210 1675 2244 The effects of a 50 percent increase in lateral gust velocity on the response of a nominal BJ balloon is given in Table IX. The balloon is opera-ting at 10,000 feet in a steady head wind when subjected to steady lateral gusts of 12 ft/sec and 18 ft/sec. Each gust has a ramp rise time of five seconds. Table IX reveals that a 50 percent increase in steady lateral gust velocity result. in approximately a 50 percent increase in balloon lateral displacement, roll, and yaw response. 30 Table IX. Effect of Increase in Gust Velocity on BJ Nominal Balloon in Lateral Plar, Balloon Displacements Cable Tension Maximum Cible from Initial Equilibrium At Equilibrium Tension Yaw Roll Lat. At At At At Case No. +- Displ. Winch Bridle Winch Bridle C o Deg, Deg. Lb. Lb. Lb. Lb. 26 4.0/-13.6 2.2/-0.4 460 1650 2210 1675 2244 27 6.0/-20.0 3.2/-0.6 710 1650 2210 1708 2285 (3) Comparison of Balloon Types. The effects of trim angle on the longitudinal response of BJ nominal, VEE, and GAC single hull balloons is given in Table X. The balloons, in equilibrium with a steady head wind at 10,000 feet, are subjected to a horizontal longitudinal gust of 20.2 feet/ second for ten seconds. Table X reveals that: (1) the down range displacements from equilibrium decreases with an increase in trim angle of attack (2) the GAC single hull balloon has the smallest displacement from equilibrium of the three balloons (3) the BJ nomi li balloon has the smallest cable tension of the three balloon tyr,es (4) cable tension increases with an increase in trim angle (5) zhanges in trim angle havea negligible effect on pitch response to gusts. The effects of lateral steady wind gusts on the lateral response of BJ nominal, VEE, and GAC single hull balloons are given in Table XI. The balloons are in equilibrium at 10,000 feet with a steady head wind and then subjected to a steady lateral horizontal gust of 12 ft/sec with a three second ramp rise time. Table XI reveals that: (1) the response of the three balloons is approximately the same in both yaw and roll response (2) the VEE balloon has 1/3 the lateral displacement of the other balloons (3) the VEE baltcon has 2.5 times more tether load than the BJ balloon as a result of the larger aerodynamic lift of this balloon. 31 Table X. Effect of Trim Angle of Attack on Longitudinal Response of Three Balloon Types Balloon Displacements Cable Tensions aximum Cable from Equilibrium At iquilibrium Tensions Type Case Trim Range At. Pitch At At At At Balloon No. Angle Down/Up Winch Bridle Winch Bridle Ft, Ft., Degree Lb, Lb. Lb. Lb. BJ 6 6.5 -65/65 65 2.8 1350 1880 2070 2610 Nom. 3 8.5 -52/73 82 2.9 1650 2210 2650 3210 7 10.5 -50/75 55 2.9 1950 2525 3220 3800 19 5.0 -35/135 80 4.2 3450 4500 7500 8600 VEE 18 7.0 -22/119 60 3.7 4620 5802 10100 11300 20 9.1 -19/105 40 3.3 6000 7200 12900 14000 CAC 22 6.0 -60/30 20 0.70 I 2200 I 3070 3150 4000 Single 21 17. -48/38 20 0.75 2696 3596 3940 4830 Hull1 23 10.0 -40/42 25 0.40 3230 4150 4850 5770 Table XI. Effect of Lateral Wind Gusts on the Lateral Response of the Three Balloon Types Balloon Displacements Cable Tensions Maximum Cable from Eruilibrium At Equilibrium Tensions Type Case Yaw R0oll Lat. At At At At Balloon No. +/- Displ. Winch Bridle Winch Bridle Deg. Deg., Ft. Lb. Lb, Lb. Lb. BJ 25 4.3/-13.8 3.1/-0.35 460 1650 2210 1680 2250 VEE 37 0.5/-15.5 3.1/-0.4 178 4620 5802 4683 5822 GAC 40 5.5/-14.5 2.7/-0.5 510 2696 3596 2734 3645 32 (4) Effect of Tail Size. The effect of tail size on the response of nominal BJ balloons subjected to longitudinal gusts is given in Table XII. The effects of tail size on the response of BJ nominal and VEE balloons subjected to lateral gusts are given in Table XIII. The longitudinal cases of Table XII are in equilibrium with a ste.4v headwind and then subjected to 20.2 ft/sec step gust of 10-second duration. The tail sizes investigated were 81%, 100% and 144% of nominal tail area. Table XII reveals that: (1) both up and down range displacements decrease with increasing tail size T(2) pitch angle response decreases with increasing tail size (3) the maximum tether load at bridle and winch decrease with increasing tail size. The lateral cases of Table XIII are in equilibrium with a steady head wind and then subjected to a lateral steady gust of 12 ft/sec with a ramp rise time of three seconds. Table XIII reveals that: (1) the BJ balloon has larger lateral displacement response and less roll response with increasing tail size (2) the VEE balloon has smaller response in yaw, roll, and lateral displacement with increasing tail size (3) tether loads for both types of balloons increase with an increase in tail size. (5) Effects on Balloon Behavior of Intermediate Altitudes. The effects of intermediate altitude operation on a nominal BJ balloon designed for 10,000 feet are given in Table XIV. The altitudes investigated are 1,000, 4,000 and 10,000 feet. The balloon is in equilibrium with a steady head wind and then is subjected to longitudinal gusts of 10-second duration with varying velocities of 20.2 ft/sec, 13.6 ft/sec, and 7.1 ft/sec for longitudinal dynamics cases 3, 14, and 15, respectively. The wind gust velocities are reduced proportional to the steady wind velocity reductions with altitude. Table XIV reveals the longitudinal responses all iecrease with a decrease in altitude andthe maximum tether load at brid~e is proportional to the gust velocity. The effects of intermediate altitude operation on the lateral response of a BJ balloon designed for 10,000 feet are given in Table XV. The balloon is in equilibrium with a steady head wind and then is subjected to lateral steady gusts of 12 ft/sec, 8 ft/sec, and I ft/sec corresponding to altitudes of 10,000, 4,000, and 1,000 feet. Table XV reveals that-(1) the lateral responses of the nominal BJ balloon subjected to lateral steady gusts increase with dpcreasing altitude (2) a review of the computer plots of Case 36 (1000 feet) in Appendix B indicates the response is unstable. A detailed discussion of Lateral Case 36 follows. 33 Table XII. Longitudinal Response of BJ Balloon With Till Size Variation Ball6nn Displacements Cable Tensions niximum Cable From Equilibrium At Equilibrium Tensions Trim Tail Case Range Alt, Pitch At At At At Angle Size No, Down/Up Winch Bridle Winch Bridle Deg. Ft. Ft. Deg. Lb. Lb. Lb. Lb. 11.0 81. 11 -61/88 70 4.6 1730 1880 3000 3550 8.5 100% 3 -52/73 82 2.9 1650 2210 2650 3210 7.4 144% 12 -50/55 40 1.3 1750 2320 2600 3170 Table XIII. Lateral Response of BJ and VEE Balloons With Tail Size Variation Balloon Displacements Cable Tensions Maximum Cable from Equilibrium at Equilibrium Tensions Type Tail Trim Case Yaw Roll Lat. At At At At Balloon Size Angle No., +/- +/- Displ, Winch Bridle Winch Bridle Deg. Deg, Ft., Lb. Lb. Lb. Lb, 81% 10.98" 32 3.4/-13.5 lI/-I 420 1700 2283 1818 2400 BI 100% 8.5, 25 4.3/-13.8 3.1/-0.35 460 1650 2210 1680 2250 144% 7.37 33 6.1/-13.3 2.3/-0.6 480 2383 2901 2435 2972 1007. 6.960 37 0.5/-15 3.1/-0.4 177 4650 4805 4683 4855 VEE 200% 7.23 38 0 /-Il 1.9/-0.2 78 4763 '904 4771 5909 300% 7.5 39 0 /-1l 0.7/-0.1 72 4874 6017 4883 6021 34 t Table XIV. Effect of Intermediate Altitudes on Longitudinal = Response of the BJ Balloon I Balloon Displacements Cable Tensions Maximum Cble from Equilibrium at Equilibrium Tension Case Altitude Gust Range Alt. Pitch At At At At No. Velocity Down/Up Winch Bridle Winch Bridle Feet Ft/Sec Ft. Ft. Deg. Lb., Lb. Lb., Lb. 3 10,000 20.2 -52/73 82 2.9 165(. 2210 2650 3210 14 4,000 13.6 -30/0 -10 0.6 1420 1650 2040 2260 15 1,000 7.1 -25/10 - 3 0.45 980 1020 1110 1160 Table XV. Effect of Intermediate Altitudes on Lateral Response of the BJ Balloon Balloon Displacements Cable Tensions Maximum Cable from Enuilibrium at Eq.'ilibrium Tensions Case Gust Alt. Yaw Roll Lat. A At At No. Velocity ý/- +/- Displ. W id Winch Bridle Ft/Sec Ft., Deg., Deg. Ft. Lb., Lb. Lb. Lb. 35 8.0 4,000 13. 1/-27.0 2.1/-1.26 490 1422 1645 1449 E 1667 35 Case 36 represents a nominal BJ Balloon designed for a 10,000-foot operational altitude and operating at 1,000 feet altitude. The balloon is in equilibrium with a steady headwind and then is subjected to a 1 foot/second st.eady lateral wind. A lateral stability analysis of this condition is found in Reference 2 Run No. 22. Reference 2 (Figure 42) reveals an unstable mode with a frequency of 0.011 Hz. The shape of this mode is also found in Reference 2 (Figure Ii) and has the following relationships between amplittdes and phase angles when the mode is normalized on the balloon yaw angle, Relative Phase Angle Coordinate Amplitude Degree Balloon Yaw 0 1.000 0.0 Balloon Roll 0 0.253 101.3 Link I Yaw O7 0.340 82.4 Link 2 Yaw a2 0.332 81.9 Link 3 Yaw a3 0.324 80.9 Data is taken from computer digital printout of Run 22 (this data not included in Reference 2). The stability analyses also reveals this mode has a rise time to double amplitude of 86.2 seconds. Turning to the comiter plots for lateral case 36 (see Appendix B), the following observations are made: (1) Tether links are in phase and of approximately equal amplitude. (2) The balloon roll angle is approximately 900 out of phase with respect to the balloon yaw angle. (3) The three links are approximately 83 degrees out of phase with respect to the balloon yaw angle. (4) The rise time to double amplitude for the balloon yaw angle is approximately 85 seconds. (5) The oscillation has a period of 93 seconds (0.0108 Hz). (6) Observations (I) through (5) indicate that lateral case 36 is predominantly the unstable 0.011 Hz mode found in the stability analysis run 22. (7) It should be noted also that the oscillations appear to be diverging, as would be expected, since the 0.011 Hz mode is predominate. 36 In conclusion lateral case 36 indicates a nominal BJ Balloon designed for a 10,000-foot operating altitude will have lateral instability problems. if it is operated at 1,000 feet. Lateral cases 35 and 37 fly the same balloon at 4,000 and 10,000 feet and reveal no diverging motion. GAr has investigated this type of instability problems in Reference 4. In this Reference the formulation of a method of analysis to predict the instability of short tethered balloons exposed to wind was derived and experimental tests were performed on towed models to verify the analysis. As a result of this work, it can be stated that lateral instability occurs when the inverted pendulum frequency of the balloon on the tether coincides with the balloon yaw frequency. Since the pendulum frequency is inversely proportional to the square root of the tether length, it is obvious that for a certain wind velocity a Lritical tether length exists. It should also be mentioned that Case 36 is not a winching condition, since the cable was a fixed length in the analysis. However, it does suggest that a possible stability problem will occur when the balloon is winched through the 1,000 foot altitude level. (6) Effects of Reduced Winds (40 Percent of Nominal Winds). The effects of reducing the steady equilibrium wind and gusts to 40 percent of nominal on response in the longitudinal and lateral planes are summarized in Table XVI. In comparing the effects of a 40 percent reduction in winds,it should be noted that the equilibrium trim angle decreased from 8.5 degrees at 100 per-cent of nominal wind to 2.96 degrees at 40 percent of nominal wind. The yaw response reveals no change in response, while the other responses are reduced as expected. Table XVI. Effect of Reducing Winds on Longitudinal and Lateral Response a) Longitudinal Response Balloon Displacements Cable Tensions Maximum Cable from Eouilibrium at Equilibrium Tensions Case Wind Trim { Range IAlt, Pitch At A At A No, And Angle Down/U Ft. Winch Brdle ich Bridle Gut . I Bride.. Deg. Ft. Ft eg. Lb,. Lb, Lb, I Lb. 3 Nominal j8.5 -52/73 82 12.9 1650 2210 2650 3210 13 40% Nom. 2.96" -38/0.0 8 0.63 J 315 875 355 910 b) Lateral Response F B4 0loon Displacements 1 Cable Tension Maximum Cable from Initial Equilibrium_ at Equilibrium Tension Case Wind Trim Yaw I ool L.at. At At At At No. And Angle +/- Displ. Winch Bridle Winch Bridle c Gue Deg.. Deg. Deg. Ft. Lb. Lb. Lb. I Lb. 25 Nomi tal 1 ~ 8_ 4 B/Ist 3.1/-().35 460 30 2201680 2250 34 402o. 9 49-13 98 O. -0. 3b5 816321.2 87. 37 (7) Effects of Nolaro versus Amgal Tethers. The effects of Nolaro and Amgal tethers on the longitudinal response of BJ type balloons are given in Table XVII. The BJ type balloons are at 10,000 feet in equilibrium with a steady head wind and then are subjected to a 20.2 ft/sec horizontal longitudinal gust of 10 seconds duration. The balloon using the Nolaro tether has a volume of 60,000 cu. ft., while the balloon using the Amgal tether has a volume of 75,000 cu. ft. The larger balloon is required because the Amgal tether is heavier than the Nolaro tether. The trim angle for both cases is 8.50. Table XVII reveals the Nolaro system has slightly more response and smaller tether loads than the Amgal system. Table XVIL Effect of Nolaro and Arigal Tethers on Longitudinal Response Balloon Displacements Cable Tensions Max C From Equilibrium At Equilibrium Tensions Range Cable Case Up/Down Alt Pitch Winch Bridle Winch Bridle Type No. Ft Ft. D-g. Lb Lb Lb Lb Nolaro 3 -52/'13 82 2.9 1650 2210 2650 3210 Amgal 10 -60/50 28 2. 1 1880 2870 3190 4180 The effects of Nolaro and Amgal tethers on the lateral response of 60 K. Ft 3 and 75K. F3 BJ type balloon sy,tems are given in Table XVIII. The balloon systems are in equilibrium with a steady head wind and then stbjected to a steady lateral horizontal gust of 12 ft/sec with a ramp rise time of three seconds. Table XVIII reveals that (1) Yaw and roll response of the Nolaro system are slightly larger than the response of the Amgal system (2) The lateral displacement of the Amgal system is 1.2 times the response of the Nolaro system (3) The Amgal system has larger tether loads as expected since the Amgal tether is heavier. Table XVIII. Effect of Nolaro and Amgal Tethers on Lateral Response Balloon Displacements -From Initial Cable Tension Max Cable Equilibrium At Equilibrium Tension Trim Yaw Roll Lat At At At At Case Cable Angle +/- +/- Displ. Winch Bridle Winch Bridle No. Type Deg. Deg. Deg. Ft. Lb Lb Lb Lb 25 Nolaro 8.5 4. 3/-14 3. 1/-. 35 460 1650 2210 1680 2250 31 Amgal 8.3 4 /-13 2. 1/-..4 560 1880 2872 1905 2.905 38 (8) Effects of Payload Location. The effects of payload location on the longitudinal response of a nominal BJ balloor, at 10,000 ft are given in Table XIX. Case three has the 1,000 pound payload located on the confluence point of the bridle while case sixteen has the payload mounted on the underside of the balloon. The balloons are in equilibrium with a steady head wind and then subjected to a head gust of 20.2 ft/sec with a duration of 10 seconds. Table XIX reveals the configuration with the payload at the bridle confluence point (case 3) has approximately 40 percent more longitudinal response than the payload mounted on the balloon bottom configuration. The tether lcads are approximately the same for both configurations. Table XIX. Effect of Payload Location on Longitudinal Response Balloon Displacements Cable Tension Max Cable From Equilibrium At Equilibrium Tension Range At At At At Case Payload Down/Up Alt Pitch IWinch Brile Winch Bridle No. Location Ft. Ft. Deg. Lb. Lb. Lb. T h 3 Confl. -52 /73 82 2.9 1650 2210 2650 3210 16 Underside -49 /56 48 2.1 1650 2210 2540 3100 (9) Effect of Winch Altitude Location. The effect of winch altitude location on the longitudinal re5ponse of a nominal BJ balloon remainirg at 10,000 feet MSL is g.ven in Table XX. The wirch is located at 0.0 ft and 5,000 ft for cases 3 and 17 respectively. The balloons are in equilibrium with a steady head wind and subjected to a 20.2 ft/sec gust cf 10 seconds duration. Table XX reveals that, (1) the range and altitude responses are reduced when the winch is at 5,000 fect, (2) the pitching res.ponse is slightly larger with the winch at 5,000 feet, (3) the tether load is maximum when the winch is :t 5,000 feet. Table XX. Effect of Winch Altitude Location on Longitudinal Response Balloon Displa "ernents Cable Tension Max Cable From Equilibri,.m At Equilibrium Tension Winch Range At At At I At Case Alt. Down/Up Alt. Pitch Winch Bridle Winch Bridle No. Ft. Ft. Ft. Deg. LIU. Lb. Lb, Lb. 3 0.0 -52 /73 82 Z.9 1650 2210 2650 3210 17 5000 -20 /47 24 3. 1 1940 2210 3670 3940 39 (10) Effects of Different Operational Altitudes. The effects of three different operational altitudes on the response of BJ type balloons are given in Table XXII. The design operational altitudes and corresponding balloon volumes are 5,000 ft. and 46,000 cu. ft., 10,000 ft. and 60,000 cu. ft., and 20,000 ft. and 500,000 cu. ft. The balloons are in equilibrium with steady head winds which vary according to wind velocity altitude profile. Longitudinal ten second duration gusts of 15.7 ft/ sec , 20.2 ft/sec, and 27.3 ft/sec are applied respectively to the 5,000, 10,000, and 20,000 ft. operational altitude configurations in the longitudinal cases. Lateral three second ramp rise time steady gusts of 9 ft/sec, 12 ft/sec, and 16 ft/sec are applied respectively to the 5,000, 10,000, and 20,000 ft. operational altitude configurations in the lateral cases. Table XXI reveals that (1) the range displacements from equilibrium increase with an increase in operational altitude (2) pitch displacements from equilibrium decreases with an increase in operational altitude (3) the lateral responses show little change due to operational altitude (4) the tether load increases WLLih altitude as expected. Table XXI. Effect of Various Design Operational Altitudes Balloon Displacement Cable Tensions Max Cable From Equilibrium At Equilibrium Tensions S1 r Balloon Equil. Gust Range Cas, Volume Alt. Vel. Down/Up Alt. Pitch Winch Bridle Winch Bridle No. 1Ft 3 Ft. Ft/Sec Ft. Ft. Deg. Lb. Lb. Lb. Lb. 8 46000 5000 15.7 -12 /38 17 3.8 800 960 1290 1840 3 60000 10000 20.2 -52 /73 82 2.9 1650 2210 2650 3210 500000 20000 27.3 -127 /70 80 1.45 8600 15030 10820 17300 Balloon Displacements From Initial Cable Tension Max Cable Equilibrium At Equilibrium Tension Trim Yaw Roll Lat At At At At SCase Angl e +/- +/- Di spl. Winch Bridle Winch Bridle No. T)e g. Deg. Deg. Ft. Lb. Lb. Lb. Lb. 29 5. 4D /14.5 3.8/-.5 560 803 845 829 866 25 8.50 4.V/-13.83.1/-.35 460 1650 2210 1680 2250 30 8.43 5. 7/-14 3.6/-1 670) 9612 15033 9735 152i7 40 B. Balloon Structural Design Performance Parameters The balloon structural design performance parameters of interest are the tensions at the winch and bridle and the translational and rotational accelerations of the balloon center of gravity. Tables XXIV and XXV list the maximum variation of these parameters from equilibrium for selected longitudinal and lateral cases. From a design limit load point of view, the values found in Table XXII represent the structural design parameters for the three balloon types subjected to longitudinal gusts of 20.2 ft/sec and 10 second duration and lateral steady gusts of 12 ft/sec. Table XXII. Structural Design Parameters Item Translational Rotational Cable Accelerations -Accelerations- Max. Tensions Balloo Ft/sec2 Deg/Secz Lbs. Type Hor. Vert. Lat. Pitch Yaw Roll Winch Bridle BJ Nom 3.8 2.6 8.3 3.4 1.9 2.8 3630 4200 VEE 3.5 2.9 3.0 5.6 1.4 1.2 12900 14000 GAC 3. 1 3.8 5.0 .4 .7 .5 4850 5770 C. Balloon Instrumentation Design Performance Parameters The balloon instrumentation design performance parameters of interest in designing an"on-board" instrumentation package are as follows: (1) displacements (horizontal, vertical, and lateral) (2) velocities (horizontal, vertical, and lateral) (3) angular displacements (pitch, yaw, and roll) (4) angular velocities (pitch, yaw, and roll) Tables XXIV and XXV list the maximum variation of these parameters from equilibrium for selected longitudinal and lateral cases. From an instrumentation design point of view, the values found in Table XXIII represent the instrumentation design parameters for the three balloon types sub 'ected to longitudinal gusts of 20.2 ft/sec and 10 second duration and lateral stpady gusts of 12 ft/sec. Table XXIII. Instrumentation Design Parameters Item Displacements Velocity Angular Displ. Angular Velocities Ft. Ft/Sec Degree Deg/Sec rBallon __ ___-_____ __ __ Type nor. Vert. Lat. Hor. Vert. Lat. Pitch Yaw Roll Pitch Yaw Roll BiNom 73 82 560 6.2 3,5 16.91 2.2 13.,8 3.9 1. 14 1, 74 2.2 VEE 135 80 I 177 9.3 4.1 !1.5 4,3 15.4 3. 1.,73 2. 4 1 , GAC 60 2-5 506 6.6 1. 1 8.5 ,8 14.3 2,7 10 1.5 41 C 0 o 000 00 0 0 o 1000 00 00 m vcO v.r 01 00 040 22 00 00 00 00 00 00 0 i- 0 N . aa O o q0- 0' in m A nnl r.;-m NNNý 0N1- 1 N0.0aIn0. Omi 100 N MNNNA ,ea 0~~ ~ 0a.a !.r a- r- O4N ., -m 0" 0-00. 010 A 1 -0 0 r- 0 -4 fN 0,fNU 000 , -on a. cc- V1I V~i v Mn mm..V Z- N~sU 0ONO' r- N a coo 5.. 1 N 0 i CD -0 ccsf do 10 0 v'IfINfl 0 tn c~--~ a, ___ 0_____0 0 m 0w; 9m O : 1[ W x 0 lAo ts- t-l U0 -w I.N fn 1 mW IAV Vs m V0N IQ . 4a, .0 oo -0 '. 00 ON , W 1. 'ANN 1t- I: o 4 : 1ý~5 .r-y NO1ýc% -m N 0Co' 5N-o V)45. AVs m (N C N J') N Vlux V 0 KJ U v, - -4 ->51. ow E <nN)~~~Vs ~ Z0 o~0. 00 0 0 u ~ A' 5 N) N) o oo o o 7 e d bb ds ? N oo -~ orv)rG 37T 3TDNY 3iDNV _j __ ~ d0 42 AN ' N' w0. Wno- M 4 4 q 04~ . 0 -q M1 D A0 . t4" m mN N .0 10 a.' 0.0 ' 0 0 ta .0M-a cy. fq.a On~o~ 0 m c '0g. 00- o 000I coo~a 0 0 0 w 0 N 4-a m v@ -10 r- In'a 0 0% V r1 4 0. -P r- t0 r-. m 1%40 flaa. a '0 ~ U2 cr m - N m -000 4.:.3 "'0 IaN . go~C 1'44 T9. '0 m. 0.-0 j.. 0 0 In F4 co4 t- .0 -a 0 -S I0M U aO,.4 w0 ' 0 0 a .~. w 4z ,C, N4 4IIn .0 r--a0 <0 0 0 0a ia 0 a o0 MW -0 4 > '0 -""I 5.'ll Llfll3 1S1 f~ 31,0 NS5 )1 3,150 a- a' 0 NNNv L N N M U4 [ 2. COMPARISON OF BALLOON TYPES The BJ balloon, with ram air filled ballonet and fins, provides the smallest size tethered balloon system to fly a lOOu-pound payload at 10,000 feet. However, it has the greatest downrange displacement from I the ground tether point. Equilibrium tether tensions are the lowest. Longitudinal excursions around equilibrium are modera.te, but lateral dis-placements due to side gusts are substantial as compared to the Vee-type balloon. Generally the Vee balloon investigated has a somewhat greater angle-of-attack excursion about static equilibrium for a given input, but with the high aerodynamic lift configuration, results in less overall downrange displacemnent from the tether point. Lateral excursion is a minimum for this balloon type, Overali, the Vee offers the least excursion with wind variation but at the expense of relativelly high tether tension. Elastic devices can be built into the suspension systeri to limit angle of attack and tether tension. However, analysis of elastic suspension is beyond the scope of the reported analysis. The Vee-type balloon then offers a configuration for applications where the tethered system should be nearer vertical with respect to f.•ie tether point, and where displacement from the tether point should be minimized. The GAC single-hull balloon provides a greater aerodynamic lift and aerodynamic lift-to-drag ratio and less downrange displacement from the tether point than the BJ balloon, ind not much greater than that 'or the Vee for the 10,000-foot altitude system (4950 feet as compared to 7377 feet for the BJ ano 4594 feet for the Vee). The equilibrium cable tension is moderate (3800 pounds). Longitudinal motions of this balloon are substentiially less for given input wind disturbances than for the other types., Lateral displacements are comparable to those for the BJ balloon for an equal disturbance. The first longitudinal and lateral modes of GAC single-hullballo'on type are notably more stable (see Figures 31 and 51 of Reference 2) than the other balloon types Structural and onboard payload design parameters for particular wind gust inputs are listed in Tables XXIV and XXV. INFLUENCE OF TRIM ANGLE ON LONGITUDINAL AND LATERAL DYNAMIC BEHAVIOR Considering the BJ balloon, longitudinal dynamic response da)-a indicates that trim angle-of-attack does not significantly change the response of the balloon in longitudinal translation and rotation., ln view of the lateral dynamic response, analysis indicates that the two lower modes of motion are excited (i.e., pendulum motion of the balloon and tether as a whole, and balloon roll). Trim angle chan,ýe does rnot substantially influence these modes for the BJ balloon. However, the vertical location of the confluence point affects the second mode (balloon roll).) The apex closer to the balloon results in greater damping of this mode while not greatly affecting the da[.ping of the first mode (Figure 36 of Reference 2). 45 Longitudinal dynamic response of the Vee balloon is somewhat more sensitive to design trim angle-of-attack. A tethered balloon system, employing the Vee balloon, should be designed to fly at moderate angle-cf-attack (say 7 degrees) inasmuch as longitudinal instability tendencies o-ist at low angle-of-attack. This tendency is indicatec in the linearized stabilit, analysis by the lack of damping in the roots of the characteristic equation for the lowest frequency mode of motion. From the stability analysis, first and second longitudinal and lateral modes of motion for the GAC single hull balloon are stable and well damped (Figures 31 and 51 of Reference 2). The first two longitudinal modes are, respectively, a pitching motion of tether and balloon, and a pitching of the balloon. The first two lateral modes are a coupled balloon yaw and lateral link rotation, and a coupled balloon yaw and roll and lateral link rotation. These modes of motion are relatively insensitive to trim angle of attack and vertical location of the bridle confluence point. Dynamic response data also verify that longitudinal linear and angular displacements from equilibrium are relatively insensitive to trim angle-of-attack changes. EFFECT OF TAIL SIZE ON LONGITUDINAL AND 'LATERAL DYNAMIC BEHAVIOR Tail size for the BJ balloon was varied by proportionally changing all dimensions to obtain tail area of 81%, 100% and 144% of the nominal design values. Stability analysis indicates that tail size increase to 144% is required to give longitudinal stab-lity for the first oscillatory mode,but that all tail sizes provide lateral stability in the first lateral mode. Also, the tether dynamic load factor reduces with larger tail sizes because of reduced angle-of-attack excursions. However, increased tail sizes increase the side displacements somewhat. It appears that 100% tails should be adequate since near neutrdl stability is present about the equilibrium point for the first longitudinal mode, and dynamic response data does not show displacement diver-gence for any of the tail sizes for the 180 seconds of response that were calculated. The Vee balloon investigation was confined to dete-mining the effect of increasing vertical tail area only. Consequently, longitudinal dynamic characteristics are not greatly affected. The vertical tails on the lower side of the two hulls are increased in area from the 100% nominal size to 300%. An examination of the roots of the characteristic equations for lateral stability (Figure 49 of Reference 2) indicates that the larger tails give somewhat greater damping and lower frequency for the first mode, and an inverse effect for the second mode. The lateral dytamic response data indicate that lateral displacements of yaw, roll, and side motion are all reduced with the larger tail sizes. Tail areas of 200% of nominal then appears to be most desirable. EFFECT OF ALTITUDES INTERMEDIATE TO TEE DESIGN ALTITUDE ON BEHAVIOR OF THE BJ TETHERED BALLOON SYSTEM As the BJ tethered balloon syF"em designed for 10,000 feet is brought to lower and lower altitudes, the wind speed it sees is reduced and the trim angle of attack changes inasmuch as the suspension bridle geometry remains unchanged. Generally, the longitudinal modes of motion become less damoed at the lower altitudes. Dynamic response data indicatcs that longitudinal response is diminished, as might be expected, with the lower steady winds and gusts used in the analysis. 46 r • " • ' ' : • • • • " •. .. s...•\| .• • • • .•.. _ .. .. • -• ••,• • ... The lateral stability analysis reveals an unstable lateral first mode consisting of coupled balloon yaw and lateral displacement of the 1000-foot altitude. Generally, damping of all lateral modes is reduced substantially at the lower altitudes. A divergent lateral motion is revealed in the dynamic response analysis at 1000 feet altitude for the 10,000-foot design. Comparing t dynamic response and stability data, it is concluded that the unstable first lateral mode consists of a coupled yaw lateral displacement. This mode cscillates at a frequency which is approximately equal to an inverted pendulum moac where the frequency is established by the pendulum length and the restoring forces consisting of the net buoyancy and the aerodynamic forces. The divergent motion was not revealed at the 4000-foot altitude. Design features which improve the damping of this mode, such as larger tails and/or automatically actuated stabilizing aerodynamic surfaces, are required. A more detailed investigation and analysis is required to make specific recommendations to improve the dynamic characteristics exposed by this exploratory analysis. EFFECTS OF REDUCED WIND ON THE BEHAVIOR OF THE BJ TETHERED BALLOON SYSTEM The longitudinal and lateral dynamic characteristics consisting of frequencies and damping qualities change considerably as the wind is reduced. The trim angle of attack steadily decreases as wind d•ecreases. Generally, damping of the tethered balloon system also reduces substantially as wind is reduced. As steady winds and wind gusts reduce yaw response is unchanged but other balloon motion responses reducL as might be expected. NOLARO VS AMGAL TETHERS The BJ balioon system employing a NOLARO type tether offers the advantage of a smaller balloon to meet the requirements of supporting a 1000-pcond payload at 10,000 feet. Dynamic response of the balloon longitudinal and lateral degrees of freedom are not greatly influenced by the different tethers with the exception of lateral displacements. Greater lateral displacements result with the system em~loying AMGAL. This may be attributed to the lesser lateral damping of the smaller diameter AMGAL tether. For 20,000 foot altitude designs, a single balloon tethered system can not be achieved with a reasonable size balloon using AMGAL tether material. Considering these factors, the NOLARO type tether generally will provide a more favorable tethered balloon system. EFFECT OF PAYLOAD LOCATION The configuration with payload located at the confluence point of the bridle has somewhat greater longitudinal response than that with the payload located at the underside of the balloon. However, longitudinal response changes are relatively minor and need not influence i decision on payload location. Lateral dynami- response characteristics were not determined for this design variation. EFFECT OF WINCH ALTITUDE LOCATION In this comparison the BJ balloon is still flown at 10,000 feet altitude, M.S.L., but the winch is at 5000 feet NM.S.L. Dynamic response is somewhat larger in the longitudinal plane with the winch at 5000 feet. Lateral dynamic respnnse characteristics were not determined for this design variation. Lateral stability analysis indicates that one of the overdamped modes becomes oscillatory when the winch is located at 5000 feet. 47 10. BJ TETHERED BALLOON SYSTEMS DESIGNED FOR VARIOUS ALTITUDES Tethered balloon system designs for altitudes of 5000, 10,000 and 20,000 feet have been defined. As commented elsewhere, it was not possible to obtain a reasonably sized single talloon system with AMGAL tether for the 20,000 foot condition. Dynamic response analysis indicates that longitudinal displacements are somewhat reduced for higher altitude designs and lateral displacements are somewhat increased. However, substantial changes have not been observed. Dynamic load factors for tether cable design are significantly less for the higher altitude design. Longitudinal stability analysis indicates that the second and third oscillatory modes, which are balloon pitching modes, reduce in frequency for the higher altitude designs. The lower frequencies might be somewhat attributed to the larger pitching inertia associated with the larger balloons. It is recommended that the mathematical techniques developed be investigated further by comparing analytical predictions of the tools with experimental data. After this correlation has been completed, more compre-hensive analysis of tethered balloon systems of specific interest with wind conditions and gusts to be expected in operations can be conducted. !I REFERENCES 1. Definition of Tethered Balloon Systems, Scientific Report No. I AFCRL-71-0213, 31 March 1971. Investigation of Stability Characteristics of Tethered Balloon Systems, Scientific Report No. 2, AFCRL-71-0406, 30 July 1971. Doyle, George R., Jr.: Mathematical Model for the Ascent and Descent of a High-Altitude Tethered Balloon; Journal of Aircraft, Vol. 6, No. 5, September-October 1969. GER-13668, Prediction of Flight Stability of Tethered Balloons, Goodyear Aerospace Corporation, Akron, Ohio, February 1968 49 APPENDIX A EQUATIONS OF MOTION FOR DYNAMIC SIMULATION A-I TABLE OF CONTENTS Section Page I INTRODUCTION ....... ................. . -3 II LONGITUDINAL EQUATIONS OF MOTION FOR SMALL ANGULAR VELOCITIES .......... A -20 Ill NON LINEAR LONGITUDINAL EQUATIONS OF MOTION .......... ................... A-30 IV LATERAL EQUATIONS OF MOTION FOR SMALL ANGULAR VELOCITIES .......... A-40 A-2 ISECTION I INTRODUCTION In Appendix A of the Second Scientific Report, the equations of motion of a tethered balloon were derived in three dimensions. I The initial derivation assumed that angular velocities were small so that products of angular velocities could be neglected. This assumption reduced the complexity of the equations (especially the lateral equations) by a considerable amount, but each equation remained norlinear and coupled in each degree of freedom. It is assumed that the reader has read Ref. 2 Appendix A, Section II and has a clear understanding of the derivation of the equation of motion, for this discussion will begin with the results of that section. Many of the terms used in this derivation are defined in the listing which follows. Reference 2 A-3 -2-FORTRAN STANDARD DESCRIPTION UNITS AA(4,4) A four by four array of variables used to define incremental velocities in the numerical integration rad/sec AAG(8) An array of eight variables signifying the angle of the gust to the horizon corre-sponding to TTG(8) deg AALP(16) An array of sixteen variables signifying the angle-of-attack of the balloon rad AALPD(16) An array of sixteen variables signifying the angle-of-attack of the balloon deg AALT(8) An array of eight variables signifying altitudes of steady state wind velocities ft AG ag Gust angle off the horizon interpolated from AAG(8) and TTG(8) deg AGR ag Gust angle off the horizon rad ALPB aB Angle-of-attack of balloon rad ALPBD &B d aB/dt rad/sec ALPBDD &B d a B/dt deg/sec ALPBDE 0B Angle-of-attack of balloon deg ALPSL A ratio of two angle-of-attack differences used in inter-polation of aerodynamic coefficients of the balloon at some a ALTSL A ratio of two altitude differ-ences used in interpolation of steady state wind velocities AXB iXB The inertial force acting on the balloon along the lateral axis of the balloon slug-ft/ 2 a sec AYB UYB The inertial force acting on the balloon along the longitudinal slur-ft/ axis of the balloon sec A-4 rJ 3 FORTRAN STANDARD DESCRIPTION UNITS AZB AZB The inertial force acting on the balloon along the slu -ft/ vertical axis of the balloon sec2 BBET(8) An array of eight variables signifying the sideslip angle of the balloon rad BBETD(8) An array of eight variables signifying the sidsslip angle of the balloon deg BET3 Sideslip angle of balloon (positive-wind to the right of balloon's centerline) rad BETBD d S/dt rad/sec BETBDD d ý/dt deg/sec SBETBDE Sideslip angle of balloon (positive-wind to the right of balloon's centerline) deg BETSL A ratio of two sideslip angle differences useu in inter-polation of aerodynamic coefficients of the balloon at some C(3) C Non-dimensional center-of-r pressure of "r" th link CCDB(16) An array of sixteen variatles signifying drag coefficients of the balloon corresponding to AALPD(16) CCLB(16) An array of sixteen variables signifying lift coeýfficients of the balloon in the longi-tudinal program corresponding to AALPD(16) CCLB(8) An array of eight variables signifying roll moment coef-ficients of the balloon in the lateral program corresponding to BBETD(8) CCLLB(16) An array of sixteen variables signifying lift coefficients of the balloon in the lateral program corresponding to AALPD(16) A-5 -4 FORTRAN STANDARD DESCRIPTION UNITS CCMB(16) An array of sixteen variables signifying pitch moment coefficients of the balloon corresponding to AALPD(16) CCNB(8) An array of eight variables signifying yaw moment coefficients of the balloon corresponding to BBETD(8) CDB CDB Drag coefficients of balloon CDC C DDrag coefficient of link DDC (infinite cylinder) CDTDB C D Balloon drag coefficient due to pitch velocity rad "CGAMB C YB Cos YB CLB C Lift coefficient of balloon BLB in longitudinal program CLB C XBalloon roll moment coefficient CLCB in lateral program CLLB C LLift coefficient of balloon in BLB lateral program CLPHDB C Balloon roll moment coefficient due to roll velocity rad CLPSDB C ZýB Balloon roll moment coefficient due to yaw velocity rad CLTDB CLbB Balloon lift coefficient due L to pitch velocity rad C mBalloon pitch moment CCmB coefficient CMTDB CmbB Balloon pitch moment coef-ficient due to pitch velocity rad CNB CnB Balloon yaw moment coefficient CNPHDB Cn Balloon yaw moment coefficient due to roll velocity rad A-6 S- ~5 -FORTRAN STANDARD DESCRIPTION UNITS CNPSDB C Balloon yaw moment coef-nB ficient due to yaw velocity rad COMI Input data defining com-puter run CPHI CO cos CPHI2 C22 Cos2 CPSI C 1 cos CSIG(3) Coi cos oi ,(i = 1, 2, 3) CSIG2(3) C2 G Cos2a. (i = 1, 2, 3) 1 1'i CTHE CO cos 0 CTHE2 C2 Cos2 2 CTPX(3) C(O+c.) cos(O + 4.) ,(i = 1, 2, 3) CTPX2(3) C2 (0+.) cos2 (0 + .i),(i = 1, 2, 3) CXI2(3) C i cos 4i ,(i = 1, 2,3) CXI(3) Cci cos 4i (i = 1, 2, 3) 2 2 CXMX(3,3) C(Ci- j) cos(4i -4.) (i = 1,2,3,; j= 1,2,3) CYB CYB Side force coefficient CYPIDB CY$B Side force coefficient of balloon due to roll velocity :ad-CYPSDB CY ýB Side force coefficient of balloon due to yaw velocity rad DB dB Aerodynamic ryprence length B of balloon (V -) ft DC Diameter of tether ft jCH(3) DC.i Horizontal aerodynamic force acting on the center-of-pressure of the "i"th link (i = 1,2,3) lbf A-7 J6 FORTRAN STANDARD DESCRIPTION UNITS DCS(3) DCSi Lateral aerodynamic force acting on the center-of-pressure or the "i"th link, (i = 1,2,3) lbf DCV(3) D CVi Vertical aerodynamic force acting on the center-of-pressure of the "i"th link, (i = 1,2,3) lbf DD(4,4) A four by four array of variables signifying the coefficients of the second derivatives in the equations of motion slug-ft DRAGB DRAGB Aerodynamic drag acting on balloon DT Integration time increment sec DTP Number of integrations be-tween data printouts OTPl Number of integrations be-tween data printouts when DT = W71 DTP2 Number of integrations be-tween data printouts when DT = DT2 DTI Integration timr. increment when T < TDTC sec DT2 Integration t.ime increment when T > TDTC sec DYPRB qB Effective dynamic pressure 2 acting on c.g. of balloon lbf/ft DYPRH(4) qHi Effective dynamic pressure acting on the "i"th link hinge)(i = 1,2,3,4). The first hinge is at the winch lbf/i't A-8 -7-FORTRAN STANDARD DESCRIPTION UNITS EE(4) An array of four variables signifying the generalized forces plus terms containing products of angular velocities if this opt'on is chosen. See FF ft-lbf EEPS(3,3) C.. A three by three array re-1J presenting the transformation matrix between the balloon body unit vectors and the inertia frame unit vectors FBHA BHA Horizontal aerodynamic forces acting on balloon lbf FBSA FBSA Lateral aerodynamic forces acting on balloon lb £ FBVA FBVA Vertical aerodynamic forces acting on balloon lbf FCH(3) FCHi Horizontal component of tension acting on the "i"th link, (i = 1,2,3) lbf FCN(3) FCNi Aerodynamic force acting normal to "i"th link, (i = 1,2,3) lbf FCS(3) FCSi Lateral component of tension acting on the "i"th link, (i = 1, 2,3) lb f FCV(3) FCvi Vertical component of tension acting on the "i"th link, (i = 1,2,3) lbf FF Input variable which gives the option of including inertia terms containing pro-ducts of angular velocities in the longitudinal program. If these terms are not wanted set FF = 0.. If the terms are to be included, set FF to any number except "0.". A-9 -8-FORTRAN STANDARD DESCRIPTION UNITS FPHI F• Generalized force acting on balloon's roll degree-of-freedom ft-lbf FPSI F Generalized force ac..ting on balloon's yaw degree-of-freedom ft-lbf FSIG(3) F.i Generalized force acting on the "i"th link's yaw degree-of-freedoml(i = 1,2,3) ft-lbf FTHE F Generalized force acting on balloon's pitch degree-of-freedom ft-lbf FWH Horizontal component of tether tension acting at winch lbf FWS Lateral component of tether tension acting rt winch lbf FWV Vertical component of tether tension acting at winch lbf Generalized force acting on the "i"th link's pitch degree-of-freedom,(i -1,2,3) ft-lbf G g Acceleration of gravity =2 32.17 ft/sec 2 ft/sec GAMB EB' FB In the longitudinal prcgram yB is the relative flight path angle of the balloon in the longitudinal plane; in the lateral program PF is the relative flight path angle of the balloon in the horizontal plane rad GAMBD B, rB d YB/dt or d PB/dt rad/sec GAMBDE YB, FB See GAMB deg GA±4BDD) Y P B, B dyB/dt or d B/dt deg/sec B, B IXB IXB Apparent pitch mment of 2 inertia of balloon slug-ft A-10 -9-FORTRAN STANDARD DESCRIPTION UNITS IYB IyB Apparent roll moment of 2 inertia of balloon slug-ft IYZB IyzB Apparent product of inertia of balloon with respect to roll and yaw body axis slag-ft IZB IZB !-parent yaw moment of 2 iaertia of balloon slug-ft L Length of one tether link ft LIFTB LIFTB Aerodynamic lift acting on balloon Lbf LS Ls Buoyant lift of balloon = weight of displaced air lbf M m Mass of one tether link slugs MAL Added mass of balloon along its longitudinal axis slugs MAS Added mass of balloon along its lateral axis slugs MAV Added mass of bal'oon along its vertical axis slugs ML ML Apparent mass of balloon along its longitudinal axis. Apparent mass equals total static mass associated with the balloon plus air mass being accelerated in a particular direction. slugs MPHIB M Aerodynamic rolling moment acting on balloon ft-lbf MPL mPL Mass of payload-located at bridle tether connection slugs MPSIB M Aerodynamic yawing moment acting on balloon ft-lbf MS VS Apparent mass of balloon along its lateral axis slugs MTHEB MeB Aerodynamic pitching moment acting on balloon ft-lbf A-I1 !--10-FORTRAN STANDARD DESCRIPTION UNITS MV MV Apparent mass of balloon along its vertical axis slugs PHI Roll angle of balloon (positive-right wing down) rad PHID dý/dt rad/sec PHIDD d 2/dt 2 rad/sec 2 PHIDDD d d 2 /dt 2 deg/sev 2 PHIDDE dO/dt deg/sec PTIIDEG 0 Roll angle of balloon (positive-right wing down) deg PHID2 2 $2 rad 2/sec2 PR Atmospheric pressure (not 2 used in program) lb/ft PSI ' Yaw angle of balloon (positive-nose to right) rad PSID 4 d'/dt rad/sec PSD 2 22 PSIDD d d/dt rad/sec2 PSIDDD d d 2 /dt 2 deg/sec 2 PSIDDE 4 di/dt deg/sec PSIDEG ' Yaw angle of balloon (positive-nose to right) dc-PSID2 42 42 rad 2/sec2 RHOB PB Air density at ZB slug/ft 3 RHOC(3) Pci Air density at Zcil(i=l,2; 3 ) slug/ft 3 RHOCH(3) Hi Air density at Z (i1,2,3,4) slug/ft Hi cHi, RJA RjA Distance along longitudinal axis of balloon from aero-dynamic reference center to bridle apex (positive toward nose) ft A-12 -11 -FORTRAN STANDARD DESCRIPTION UNITS I RJB R Distance along longitudinal 3B axis of balloon from center of buoyancy to bridle apex (positive toward nose) ft tRG R. Distance along longitudinal j3g axis of balloon from center of gravity to bridle apex (positive toward nose) ft RIM R. Distance along longitudinal axis of balloon from dynamic mass center to bridle apex (positive toward nose) ft RKA RkA Distance along vertical axis of balloon from aero-dynamic reference center to bridle apex (positive up) ft RKB RkB Distance along vertical axis of balloon from center of buoyancy to bridle apex (positive up) ft RKG Rkg Distance along vertical axis of balloon from center of gravity to bridle apex (positive up) ft RKM Rkm Distance along vertical axis of balloon from dynamic mass center to bridle apex (positive up) ft SB S Aerodynamic reference area of balloon (VT) ft SC(3) S Aerodynamic refe-rence area ci of one tether link = (2) (dc) 2 (i = 1,2,3) ft SGAMB SYB sin yB SIG(3) a. Yaw angle of "i"th link, (i = 1,2,3) (positive-counter-clockwise looking down range) rad SIGD(3) . d.i/dt (i = 1,2,3) rad/sec A-13 -12 -FORTRAN STANDARD DESCRIPTION UNITS SIGDD,3) or d2ai/dt2 ,(i = 1,2,3) rad/sec SIGDDD(3) zi d2ai/dt2 ,(i = 1,2,3) deg/sec2 SIGDDE(3) ai do.i/dt (i - 1,2,3) deg/sec SIGDEG(3) a. Yaw angle of "i"th link/ 1(i = 1,2,3) (positive-counterclockwise looking down range) deg 2 .22 SIGD2(3) a. (i = 1,2,3) rad /sec-SPHI so sin SPHI2 S 2 sin2 O SPSI Si sin SSIG(3) So. sin a. (i = 1,2,3) 1 1 STHE so sin 0 STHE2 S 0 sin 2e STPX(3) S(6 +Ci) sin (6+-i) ,(i = 1,2,3) STPX2(3) S2 (6+Ei) sin 0+ci) )(i = 1,2,3) SXI(3) SCi sin Ci , (i = 1,2,3) SX12(3) $2ci sin2 Ci (i = 1,2,3) SXMX(3,3) S(Ci-Cj) sin(i-•.) C (i=1,2,3; j=1,2,3) S2THE S28 sin 20 T t Flight time sec TDTC Flight time at which DT is changed from DTI to DT2 and DTP is changed from DTP1 to DTP2 sec TENS(3) T. Tension in tether at top of "i"th link,(i = 1,2,3) lbf TENSW T Tension in tether at winch lbf w f A-14 -13 -FORTRAN STANDARD DESCRIPTION UNITS TETH Total length of tether ft THE 0 Pitch angle of balloon (positive-nose up) rad THED a de/dt rad/sec 2C 2 THEDD 0 d2O/dt rad/sec THEDDD 8 d 20/dt deg/sec2 THEDDE 0 de/dt deg/sec THEDEG 0 Pitch angle of balloon (positive-nose up) deg THED2 62 2 rad 2/sec2 THEODE 0 Equilibrium pitch angle of balloon deg TSL A ratio of two time differences used in inter-polation of gust velocities and gust angles TTG(8) An array of eight variables signifying the time history of the gust sec TTT Flight time at which computer simulation ends sec VBR VBR Relative velocity of balloon's c.g. with respect to the air ft/sec VC(3) Vci Relative velocity of the c.p. of the "i"th link with re-spect to the air) (i-l,2,3) ft/sec VG V Gust velocity ft/sec g VGH VgH Component of gust in YtB direction ft/sec VGS V Component of gust in XB gs direction ft/sec A-15 -14 -FORTRAN STANDARD DESCRIPTION UNITS VGV V Component of gust in ZB gv direction ft/sec VHR(4) VHri Relative wind velocity of "i"th link hinge in the Y B direction,(i=l'2'3'4)" The first hinge is at the winch. ft/sec VS Speed of sound (not used in proaram) ft/sec VVG(8) An array of eight variables signifying gust velocity acting on balloon and corresponding to TTG(8) ft/sec VVW(8) An array of eight variables signifying steady state wind profile and correspond-ing to AALT(8) ft/sec VW V Steady state wind velocity at ZB ft/sec VWC(3) Vwci Steady state wind velocity at Z ci ,(i = 1,2,3) ft/sec VWH(4) VwHi Steady state wind velocity at ZcHi) (i = 1,2,3,4) ft/sec WB WB Total weight of balloon -includes balloon and bridle material, and enclosed gases lbs WPL Weight of payload located at bridle tether connection lb WTC Weight of tether lb XB X B Lateral displacement of balloon's c.g. in the inertia reference frame ft XBD XB dXB/dt ft/sec XJ3DD xB d 2 XB/dt 2 ft/sec2 XBDDR Lateral acceleration of balloon's c.g. relative 2 to air ft/sec A -16 -15-FORTRAN STANDARD DESCIPT ION UNITS XBDR Lateral velocity of balloon's c.g. relative to air ft/sec XCD(3) X Lateral velocity of c.p. of ci "i"the link (i = 1,2,3) ft/sec XCDD(3) X Lateral acceleration of c.g. 2 ci of "i"th link)(i = 1,2,3) ft/sec XCHD(4) XcHi Lateral velocity of "i"th link hinge,(i = 1,2,3,4). The first hinge is at the winch ft/sec XI(3) Fi Pitch angle of "i"th link, U = 1,2,3) (positive-rotated Sup from the horizon, clockwise) rac I XID(3) Fi dCi/dt ,(i = 1,2,3) rad/.-ec XIDD(3) d ýi/dt 2 (i = 1,2,3) rad/s~c2 ,, 2 XIDDDE(3) Ci d i./dt2•(i = 1,2,3) deg/sec XIDDEG(3) dFi/dt (i = 1,2,3) deg/sec XIDEG(3) Fi Pitch angle of "i"th link, I (i = 1,2,3) (positive-rotated up from the horizon, clockwise) deg SXID2(3) 2 F 2 (i = 1,2,3) rad 2/sec2 1 1 XIODEG(3) .io Equilibrium pitc.h angle of "i"th link• (i = 1,2,3) deg XPLDD XPL Lateral acceleration of 2 payload ft/sec YB YB Down range displacement of balloon's c.g. in the inertia reference frame ft YBD YB dYB/dt ft/sec YBDD YB d2YB/dt' ft/sec 2 A-17 -16 -FORTRAN STANDARD DESCRIPTION UNITS YBDDR Down range acceleration of balloon's c.g. relative to 2 ,23air ft/sec YBDR Down range velocity of balloon's c.g. relative to air ft/sec YCD(3) Yci Down range velocity of C.P. of "i"th linkg,(i = 1,2,3) ft/sec YCDD(3) Y ci Down range acceleration of c.g. of "i"th link. (i = 1,2,3) "ftisec YCHD(4) YcHi Down range velocity of "i"th link hinge,(i = 112,3,4) The first hinge is at the winch ft/sec YPLDD e Down range acceleration of payload ft/sec2 ZB ZB Altitude of balloons c.g. in inertia reference frame ft ZBD ZB dZB/dt ft/sec ZBDD ZB d 2 ZB/dt 2 ft/sec 2 ZBDDR Vertical acceleration of balloon's c.g. relative to2 air ft/sec2 ZBDR Vertical velocity of balloon's c.g. relative to air ft/sec ZC(3) Zci Altitude of C.P. of "i"th C link /(i = 1,2,3) ft ZCD(3) Zci Vertical velocity of C.P. of "i"th link,(i = 1,2,3) ft/sec ZCDD(3) Z Vertical accleration of c.g. j of "i"th link,(i = 1,2,3) ft/sec2 ZCH(4) Zc1i Altitude of "i"th link hinge,(i = 1,2,3,4) The first hinge is at the winch ft A-18 1-17--I -1 FORTRAN STANDARD DESCRIPTION UNITS ZCHD(4) ZHi Vertical velority of "i"th link hingel(i = 1,2,3,4) The first hinge is at the winch ft/sec ZPLDD z PL Vertical acceleration of 2 payload ft/sec A-19 SECTION II LONGITUDINAL EQUATIONS OF MOTION FOR SMALL ANGULAR VELOCITIES First consider the longitudinal equation of motion. The dependent variables (see Figure 1) are e (pitch of the balloon) and Ir (pitch of the "r" the link), where r is a particular link. The appropriate equations of motion in the Second Scientific Report, Appendix A, Section II are (80) and (97). Since these equations were derived in three dimensions, th.y contain the lateral degrees-of-freedom, •p(yaw of balloon), cp(roll of balloon), and ar (yaw of "r" th link). It will be assumed that the balloon remains in equilibrium in the lateral degrees-of-freedom. Therefore, a 0 -r -- 0. With this assumption Equation (80) and (97) are rewritten as follows: -2 ma fJd o'd) 1> X f-sJ~ .4 ' 0 "where $e = sin 6 and CA = cosO , etc. A-7O -V. NP FIGURE 1 -BALLOON TETHER MODEL IN LONGITUDINAL PLANE In order to solve Equations (1) and (2), it will be assumed that N = 3. Rewrite Equations (1) and (2) with this assumption and substitute in the trigometric identities for the sums of angles. [u,, ,pR +Aif, -r,) -J,,o •. [ ,s •,e U)l -, IpcL'3J I. 4,J& s 14+0)-4, &1 Rim c(f,.,,)] j F-2 A-21 For r 2 For r 2,3 M.4L I ~ S (f4+i) -A,~i ct J 0~6 IzL ,,,,,,,-)44, sfJ•÷• t ,,c s(fYo tg l ,- J#)] ,L~~ ~~223 sw ns (ft) 1 fff (6) The generalized forces (F,,, F, FI , Fr ) were derived in Section III, Appendix A of the Second Scientific Report. The following necessary equations are taken from that section. F+= L-j,,g -W ,; t F,,-c, [b,,,, oc' . .4[+&F,, 5•t,SMIfii4[,),AU- f,,(D,,, c' J I SReference 2 A - 3 ,7,, LJciwe 5F# DcV fi.J+[v -1 f fvv c1 -P& 2 +Li 4 -D J -3[IC' f4Dv fj (10) --LIF"8 (C RAe g(S Y ) (12) + (13) ._e R I-F7 /~OLD61c(8~~ (14) .F8, 7 S • C V[. . [ ,,o [. e,,+,I-) (16 C (18) --w , -z ,2,-P, Se r-1.tr,[ (17,) •, •] A-2 "JL 4v,-LIT (¢•e -Oag(S))(2 + (19) Vj" (20) 4 ,xh, [ -V .v ,)/( it + K, -v 1)] (22) oa= i-Y', (23) See Figure 1 In order to show the load distribution along the tether at any time, the tension in the tether is calculated at the end of each lirk. This is done by summing forces in the vertical and hcrizontal direction6 at each hinge. Before summing forces, consider the inertia forces acting on the balloon. FigLre 2 shows the balloon and with the apparent inertia forces acting along and normal to the centerline. :• ',/•(%o, i so) -e FIGURE 2 INERTIA FORCES ACTING ON BALLOON IN LONGITUDINAL PLANE A-25 2A,/0,(; C6-S .5oC)6 c +/n ,, c so) S - •2 [M•11",i1.s, ] t ,9 + -_],S 9 (24, •, • -,lit] so ,r., -,- .M F , 2 & /,'2, 61 (25) :ye The summation of forces acting on the balloon are: L OV A ,, -WO T (26) 2Fy-: TOR Fl?,,q (2/) where: TBV and TBH are the vertical and horizontal tension forces at the bridle-tether confluence point respectively. Note that the payload forces are not included in these oummations, but will be included shortly. Equating Equations (24) and (26), and Equations (25) and (27) gives the following expressions for TBV and TBH. Toy - Ls tF,3,. -Wa0 -Z I[ IV,( 2 67 YBs2~~ [. -410] 54Pcc, (28) Ti :F81 R 2L/4s.41SIge6 c 8Li +/0, 4L'v (29) Equations (28) and (29) give the tensions on the tether at a point just above the payload. Figure 3 shows the payload with all forces acting on it. A -2ý6 ,v FIGURE 3 APPLIED FORCES ACTING ON PAYLOAD IN LONGI'rUDINAL PLANE Summing forces in the vertical and horizontal direcJ..ons gives: j I , -P, T T,--I , Fr- ply,, (30) ~FY /':pIAP ~ FMToA (31) where F CV and FCH are the vertical and horizontal components of tensfon at the top of the tether below the payload. Substituting Equation (28) into (30) and (29) into (31) and solving for the vertical and horizontal components of tension at the top of the tether results in the following equations. A -27 o 5 [4~y t] I6 All' -A,0 67 "'1 (32) fcP Foq ?, Af Jc4 yaL +A6i41s CO 't9 l Y. (3 The totai tension at the top of the tether is: SA• F + Fc&v (34) Figure 4 shows the applied forces acting on the "N"th link FVF 2 v FIGURE 4 APPLIED FORCES ACTING ON "N"TH LINK IN LONGITUDINAL PLANE A -28 The aerodynamic forces DCHN and DCVN are the horizontal and vertical components respectively of the normal aerodynamic force (FCN4 ) acting at the c.p. of the "N"th link due to the relative motion of the "N"th link with respect to the air. This relative motion tdkes into account both the steady state wind and the horizontal and vertical inertial velocities of the link. Expressions for DCHN and DCVN are given in Equations (16) and (17). The normal force acting on the link is given by: Siunming vertical and horizontal forces on the "N" link: (37) Solving for the tension forces at the bottom of the "N"th link gives the following equations. F < q- ~(38) .. i Fc, •ce, C t ( -t -,Jý (40) F,,,,. -, f ,•, SF' , 'VAI (39) 7 2i . (42) The tension at the winch is found by setting v = 1. A-29 I SECTION III NON LINEAR LONGITUDINAL EQUATIONS OF MOTION Section II of this Pppendix contains the longitudinal equations of motion of a tethered balloon if angular velocities are small and products of angular velocities are negligible. This assumption was made in the derivation in Appendix A of the Second Scientific Report. It is now desirable to examine the equations of motion when large angular velccities are allowed. As in Section II, appropriate equations will be extracted from Appendix A of the Second Scientific Report. These equations are (41), (79) and (96). Equation (41) is the total kinetic energy of the system: Equation (79) and (96) are terms that are needed for Lagrange's equations. However, these equations are functions of both the longi-tudinal and lateral degrees-of-freedom. Th.erefore, before rewriting Equations (41), (79) and (96) of Appendix A set all lateral degrees-of-freedom equal to zero ( ¢ o 0). r The total kinetic energy of the system in the longitudinal plane is: 2 JLV (~)VhJ.b Is (43) where Reference 2 A-30 PL (45 -'K (46) Now write Equation (79) of Appendix A --NN Rewriting Equation (96) of Appendix A: -, -O 4:4 Langranue 's eqa~ations for the system are: d (49) (50) Reference 2 A- 1 First derive the terms in Equation (49). Using Equation (47) the following term is formed. AF~ +1,. •" (51) From Equation (43) -/I (52) From Equation (46) 4"" • ,'") . O(53) C~b4) "(54) -~ '~~'i (56) .- (57) A-i?. (58) & (60) Substitute Equations (53) to (60) into (52): 4.1 • : -.4 , " (61) t~j?~ O f.,t4) ~ f~~f)(62) Substituting Equations (51) and (62) into Eqation (49) yields the final nonlinear balloon pitching equation. NAe, F0 is the same for both the nonlinear equation that follows, and Equation (1) which assumed small angular velocities. The expression for F is given by Equation (7), N 4ý4 A-33 Now consider the tether equation of motion (50). From Equation (48) it follows: NS {(Ž'dS11Ž #f.c}-) -C~~}Jsf-. __ o, .=..': ,. Al 4(64 From Equation (43)" T A-34 r From Equat-ion (44): 4. PC S(66) A- (67) 1-; f.,, t, From Equation (45), Ž1/ (68) Q~s ,~ ;~(69) From Equation (4•,,: (70) -' iA f (72) A-35 Substituting Equations (67), (69), (71), (72), (54), and (55) into Equation (65) yields.. N 4g1 IV -MJ. 1 ,,+ S ff C. c(t+,-#,, (] 73) Substitute Equations (64) and (, .) into Equation (50) to give the final nonlinear tether pitching equation. F1 ýas already been foand in Equations (8), (9) and (10). /A-I .j 1 I) ~I f = V •-' -i'• •. sL I • ,w:Ali.j . 2 ~~ 4j ( 1 4/ 6e 4o11 A -36 IN 4' (' LW;)5(OA) X'f.bis( 4- F1 (74) Now assume the tether is composed of three links (N = 3). Rewriting the balloon pitching Eq.ation (63) gives: -, R.,,,. S. 164, 9:",.t) -,,,1X /y/., ". sRi, C) 16 + 3. ,,S;, It4f .I4f) For N = 3 and 1 1, Equation (74) becomes: rA-37 37 t", S ( f; -fl) }3 4=1 + , '=,/ €( =I For N = 3 and r = 2, Equation (74) becomes: .t1 2{[2j +4,,,,]c(g g)k,,,,s&,)ds(9) ,-,••+l••,¢J t121~ 69 0g/~ tig~~b)' !Tý +'/M,. +/0. .•).'A .•/+ f} f2 41al + IM,L] C f. -f,) t A. S (a + f2)s&.¢) ,,,(•),9 T C+ f,) 3 3 S f41s,,'c (624 f ,(. ) A-38 For N =3 and r =3, Equation (74) becomes: ,e. (78 T/+W~c.' ,1 •-, s (4+ /0 •,, . + " c1 A.4-3 +,,,-,if ns, ;,• . -(+ (78) f'.•.•s(e . A -39 I SECTION IV LATERAL EQUATIONS OF MOTION FOR SMALL ANGULAR VELOCITIES Sections II and III considered the equations of motion of a tethered balloon in the longitudinal plane. This section will consider the equations of motion of a tethered balloon when the longitudinal degrees-of-freedom are fixed at their equilibrium value and the system is allowed to move only in the lateral degrees-of-freedom. Figure 5 displays a front and top view of the tethered balloon. All angles are shown positive. The lat--ral degrees-of-freedom are: •p(yaw of balloon), S(roll of balloon), and ar (yaw of "r"th link). SNOSE OF -G-BALLOON J' / vi / TOP VIEW F•',NT VIE\V, FIGURE 5 -BALLOON TETHER MODEL IN LATERAL PLANE A-40 ! The appropriate equations of moticn are Equations (73), (86), and (108) of Appendix A of the Second Scientific Report. These three equations will be rewritten below remembering that O(pitch of balloon) and ý (pitch of "r"t r link) are constants. The first equation is the balloon ':• equation of motion. fly•, [ (,ese 5 Y, (O4A )]oS.7 5 -ll-sc:l ,2:1 ~ c~ (csso. (CO, C44'o SP) SA4 94 ("5 ai)-sY5( L ')I +1 ce~h s~)~ 5 t~)tS~lg -~ L' I~u The second equation is the balloon roll equation. + I•, (-•4 ,)c -o) (-s,, 4 3 -"+4I),,] 5. ll, ry,,C4 .cjt F, (80) Reference 2 A-41 The equation for the tether link yaw equation of motion Js given as follows: ih-i IN = 4 +I Al .... ,, (. ,o C -,)- 9 -. , A -84) In Equations (79), (80), and (8l;.there are several trig-nometric combinations of the balloon angles ýp, e, 6. TheF.e terms are the elements of a matrix (Equation (51)) given in Appendix A of the Second Scientific Report. The matrix (E. .) gives the relation between the balloon unit vectors and the inertia frame. S4) 5,PS4' +4 (,t 5 S& CS?-CO 4#Sr -5 (82) -465q~.S'&C The balloon yaw Equation (79) is rewritten as: 'V #j cc Sri t6" -,s,- s IJ)A e-21 +AIa .,f a- tR13 "in The balloon roll Equation (80) is rewritten as: Reference 2 A-43 #3e CAI + p S i5 r; ÷IL~ ~~ Yo 5 4. F.0 (04) The tether yaw Equiation (81) is rewritten as: OC,'~ mSri Jcat ~ f,, [,sr. b, . 450 .5rC(fA-f.1 AI ~~~~I LC cSxn 6, -T 4 SO 33 f- f4 kjA -50,'; S £3 6 d6dl-j~, Le"11cra 6 +S ,ct.,-s SfLc d1Af611cU Le, 5c c c, -5 s~i /4-,- 6,Z, 6#4' Cf j 5 l (85) IA-44 In order to solve Equationc (83), (84), and (85), assume N = 3. The balloon yaw equation is: 12- 631 6b +R4 6 Cirf,Lco61 d C4a L43-S 0; ]L,q ., 6CA,;]1 A -45 +I.+s-The balloon roll equation is' -54- , ui r. £,, saj6A s 'Asji, saicF +n9 IJ' £ 1 e7JF# (87)~ From Equation (85), three equations will be generated (one for each link) by assuming r = 1, r = 2, r = 3 consecutively. For r = 1 cSG cf, -Su 2C,1jeo-,, +soc ,c , ,6. ; - sr;, s I, A-46 tlcgi co-, +- Sri Sr2, A tALrit #SoJ Cf~-M£,,Sj5III Ci+S t SG14 d,6v+rceiws1f6,j +411,t jC4_i, ." t._ 4~ 6,IJE[Ci, 4,7, tS I CtI62 -S91 .fi d2 ) +4cj6,,+si Sar J~Si C.V f. J371If Cnlca; E3, 1597c so- 4~~ S 1' £di G For r =2, A -47 +it -4,21 3 9L; 46j, Val' L j' 2 22 -S j 4 +41. r1~ E3Tei C j, Si 161j L Soi 16 ~3 1, -5' . S111 (89) A -48 For r 3/ f/6c'' -fr f ef-o- f o Sal- C~4 -Sust,£ 40It 0', vi~ $~oref, &a-5 Sir S~~IU ý21Y pc - so.-S 4Us 3 ) ~Saii ri ~f- j j '-IC eCiA +Sif s-a- Sal 3 ,iI C6i Sfle5ui / fJ.mfi-zi, tS2jP62 SJIojfj; M Cu +Sc r C cj ) C6II-s S9 i 6/31 A -49 fA 5--~I[r 6,+SJ-C,16 9, 6 (90) The generalized forces (F1,, FV, FB 1 , F.G 2, F. 3 ) were derived in Section V,Appendix A of the Second Scientific Report. The follow0ing equations are taken from that section. 44jS [1-Rj, 4 4 Ih ~ tR4A 161,)1; '(4 4iJfi J611 (1 4:W4- x.-L,16,] +f-p,?.], + (92) ~~~t~~ Fiuo 44e6 +2~ MOOlO s; ~ The aerodynamic forces and moments are: r+- •'B 5 8 I:6VB, Cyb ),;r• + ' ""1v--, (96)•+ Reference 2 A-50 F" L V ~ ~ c- -, -vo; -,-ittC. wa4 'p~ (97) VOO(lOOe s i s so J-,gI,,/, )cjp8 (98) cs.. ý +T ce,.. (do• bcA in 5eoi tudia Coe Vt iC (99) I'rs actg n te b n a(100) Vag,- A is; ,, 1 W + • , (101) p (102) 9A-51 (103) 3 (104) As in the longitudinal dynamics, it is desirable to know the tension in the te~ther. This is lone by summing forces at each hinge puint. First consider the apparent inertia f.crces acting on the balloon along and normal to the centerline. X8 4,s el +,£,~~ (105) aye~ ?-M C2314£ i~ Ls (106) d142 A ivRi C3 'j L4 2+ (107) A-51 In Equations (105), (106), and (107), XB, YB and Z, are inertia accelerations of the balloon: 1AXB' 4yB, and '%ZB are inertia forces in the positive direction of the balloon's lateral, longitudinal, and vertical axes. Cij are elements of the matrix defined in Equation (82). Summing the apparent inertia forces in the lateral, longitudinal, and vertical directions gives: :F- 61d,, -1 L42, d.1, f 1 634 (108) 4 6;j 2, 4I 3 2 iba t (109) The forces acting on the balloon above the payload are: •Fx. Fu- Tat (111) Fy To -•- Fe m (13.2) IF, Ls-4v, FeVA --Ts, (113) Now equate Equations (108), (109), and (110) with Equations (111), (112), and (113) respectively, and the following expressions for TBS, TBH, and TBV are found. es 7 (6%,;c 4,5 - 6, a + 6 34 aiA) (114) To-~ k a.f 0 y I6~c, (116) If forces are summed at the payload location, the components of tension at the top of the tether results. tFS5v FOSA- 4tI h5VS 4 6)Od(117) Fco# FsA (6 1 4ays i6l 4 -g 6j~?sk6 o (118) The vertical and horizontal forces on the "N"th link are the same as they were in the longitudinal dynamics program. The force in the lateral direction is given in Equation (99). In general for the "r"th link. FeM (120) ~ -~, f~~n~.~'~l ~(122) if(~ 1 Y ~(123) The tension at the winch is found by setting r 1, A -53 r. APPENDIX B DYNAMIC RESPONSE INPUT AND OUTPUT DATA LONGITUDINAL AND LATERAL DYNAMIC SIMULATIONS CASES I THROUGH 40 B-i SECTION I INTRODUCTION Appendix B consists of the input data used in making tne computer run cases 1 through 40,and computer constructed plots of selected oct1-'at 7ariables for these cases. Table B-i lists input data for the longitudinal dynamic cases I through 23, and Table B-II lists the input data for the lateral dynamic cases 24 through 40. Tables B-Ill and B-IV list the balloon longitudinal and lateral static aerodynamic coefficients respectively. Computer constructed plots (3 plots per case are included on each page) for cases I through 40 follow Table B-IV. The parameters plotted with respect to time are listed below. Longitudinal Cases I through 23 Balloon pitch angle Link 3 pitch angle Plot Link 2 pitch angle 1 Link I pitch angle Tension at bridle Tension at winch Plot Balloon altitude 2 Balloon range Wind gust velocity Pitching velocity of balloon Plot Wind angle of attack 3 Relative velocity of balloon Lateral Cases 24 through 40 Balloon roll angle Balloon yaw angle Plot Balloon yaw rate 1 Balloon sideslip angle Balloon lateral displacement Link 3 yaw angle Plot Link 2 yaw angle 2 Link I yaw angle Wind gu.t velocity Relative velocity of Balloon Plot Tension at bridle 3 Tension at winch B-2 0 __ __0 00a_ _ 0 1. 0 0 0 .. O S~ 00 0ý ý 2ý co NC Ca Cc N 0Oi P 0 0 00 0 0 4Jg 0 00 00 -0 0 ;7 00 Ct i. -0 ! 0 0 Ca. 0 0 CiON 00 0 ci ciN4 i Ci 4 4 u 41. 0 -4. iA -,4N -4 ~~ 04 00 0 00 0 z 4 4 .41 0 0 Ný C 0 0.0 00 0 0 0ý 00 0 1, ___ E 80 0 80 O 0 0 440~~ uc00 0 0 -~ .. N 0 -. 04 -------~-3 ----1' ~..u (.' C5 C; X Nl hr_ _ _ _ _ N f- w _ 0 _ _ _ _ _ _ _m _ vN m 0 v A 'o__ _ _ot- _ _ _ _ _ _ _ _ _ w _ _N _ _ O vcc Hr OD Or 00 N 0 0 C- .0 co ' D v' C .: V; __ __ --0 , '0 -r, r v v 0 a 0 ' LA . 4 .0 m0 w0 . 0 .0 W; vi V; .16 .0 N 0 N 94 ----N -0 -r ---' ~ C .0m N' hr .0 a, Go.- N -C ) N 1 N v' ( 00 .0 NN ' N N N NrN hr . -----r -'hr 00 W ~0' .0 '0 ' N N . h N .0 BN-4 -. 0n 0- o0 100 r4, F4 0 B-5 d~ ~ 0 o0 0 0 v o 0 0 0 0 0 0 -0 0 1.) 0 0 0' 0 c p~~~ ~~~~ 'aI(1 -1N 0N0'0C 0 v' --0 -vL N IC. 0 0 0 0 0 t- 0 0 0 0 0 m 0 0 0 0 0 0 uO 0 0 0 0 0 to in N 0 0 N %0 L% V In 0'% Ln t- Ln -- -~ 0 0 0 0 CC 0 0S c) 0 0 0 ~I4 00 0 0 0 0 0 0 0 0 0 0 0 0 00 00 0 m 0w 0 ~~W 0 0 0 0 'a a 0 C 0 N v) L CI- F-Cý 0 0 0 0 00 0 0 0010 to 4 0. 0 0 0D 00 0 0v 0 0 0 0 00 0 N0 1 0 mC 1-c~0 C-~ 0' LA 14 rN ' 0 NA N' ' C C -C' C'- 0 14 -' '0 -CC0O 00 Cl -. 0 NC NJ) V VC V a N (n V 0 41 -n a' -4 F- ', r- N0 O O 0 r- v U 'a aC r- r o w 0 0 O 41_ _ _ 0- NA CC -00 Nf 0- 0 (n N N 'a 10 __ __ __ N -N 10 4 040 ' N (n ) V V .0 04 1410) 1 4 L _o >- 14 0D 00- C C 4 C ' 0.M N0 mo oA o 0 00 0___ 0 0 0 H4 0 0 4 -4) 4)3 0 B-6' Ln~~~~ N a6 tN 4- e r ýa 6r (n '0 N 0 0 t 1 10 t- U') r) e ' C' n Mn C V) N4 r r- ý C6 N3 N U N N N N NN '0 1" 0'4 fn CY 16 '06 16 1 0 0 C6 06 C6 r-CJI ' l0 N m6 10 N N r- 0 vJ o N N' I-' i fn rV) -Ln m cN '0 '0 i' cC' m t- Jn t- co cr t- C) l) ( ' in' 0 I- in 0 V 'o r- 0fn )0N 10 a LA f- C o a 0. -0 N0 r 0 1 3 -0 , r -'0 a, 0 N 0, 0 -~'D 0 N-I- 0n 0' r0- N aIn U) -00 00 LI) N0 v It) L0 Ln N- N- 00 0 0 '.0 10 L CID a r- r- '. ' n ND 0 'D m a, 0 f) 0 0 j U) 0' N a, i o~ w mn v 0 en NT ~ 0 10 NtN in 'D t OD 0' 0 : N (n' in '. -N CC:) a-ri A.! C. flOAmr4 'C". --- -I4J 0 _ 00 £4 a .o ------------- --------- N .0 - I"- U 14U . 00 0000.. 000 .0 .0 -.0 N E ---0. ___ £4 Sf1 0 .0 0. __ £4 --N N N 5,1 .0 140 IfS N -Q... .0 0 0 t- 0 OC 'IC N .0 -0 N 1 -0 C, N .0 555 45 U S m N 0 0 r- -Q. -.0 .0 .0 4 N N 4 0 0 N O 5 I , U .0 C4 .0 __ ---. 0" 0 -.. C N .0 I N .0 0 C U C '0 -.0 5" -N 4 0 55 0 '0 55 0 .0 .0 5" ... 4 IC" 0 0 0) N N N N . ZI. -U .5.4 o N 4 4 N 4 .0 N 4 __ 0 N 4 CCC 0 '0 --N 0 --0 '-C 0 H -H .0 r- o 4, -C' 0.1 r- .0 N 5'. 5' 0' U 0) -4 a'. 0 0 0 0 0 0 0 C" 0 0 0 0 4 _ .0 0 -' 0 4, a' 5- CC - 4 C-I .0 0" 4 0 4 -N CCC N C, 4 CCC o 0 0 0 0 0 0 0 0 0 4 0 -.0 .0 .0 CCC -0 N C" C'- C' .0 .0 4, 0 4, -N .0 '0 5- 0 0 0 5- .0 5- 5- CCC 0 5-CCC N CCC 0 4 .0 .0 U, --0'-------4 -z C" N .0 CCC 0' 0 0 0 0 4 0' 0 4 5- 5'- 0 1-14 0 '0 4 CCC '0 5- -0' O- "C Win. -n CC 5- -N 0 4 --I-. -N ---.4 N .0 .0 .0 N Co C, J14 4 .0 -, 0 Co CCC --' CO 5- -? 0 0 0 0 0 0 0 U -CC -5- C' CCC CC Co --C --is a a' , CC Co 0 5- 4, CC ---'4, Co 5C 9C Q 4, 4, a' 0 5 4' 0' 0 B-8 'T 0 ~'. 0 00 o0 >' 0> E- (LN~ ~ N O N -" 0 ) r 0 0 01 0 0 v 0n 0 0l c N~ N~ 1 N M' m N n m' m n en ( C4 ---t~ N-OD 0 0 01 0' -4 'V (nl 0 N 0 0 000 n t N 0 (n 00 00 ~ 4) M I N (1 v '0 00 a 0 %0~E0 n N~ OD In 0 0 N ' efn 0 0 0 N- 'tN ~ 0 '14 U) 0 0 (?n U) r N 1-4 0 0 0 0 0 0 0 H 0 J4 -N ti) 0 0 CO 00 v (ON m-N 0 U) N 0 1 -0 N 0 '.0 In C w 0f N- N U) 0 In (n N 0or N N N -. 0 0m c.N N 0 0 0 0 0 u tn 0) N- In) '0 0' 4 -. a, %0 0 It a' '.o -~ 44 -4 4 N C1 -4 In -L 0: a, 0'. Mn 00~ '.r- N N n r- --,D a'ý --M 0 In 0 -0 0'. N '-0 04) N N- 0 c(Y) '0 No Ln 0 ~ r N -0l N0 4 4 0' '-4,00 rq-H I 0L 4i (r1 0 0 ' 0 .1 -10 0' 0f ') '0 0'0 0 0 0 0 0 -4 0 00 N f ' -00 0 (Y t- -0 0 N- r- c0 H O N -4 0 --0 r- m- 00N-4 e pq0 '0 '0 0r 0 In 0 0' CO 0 0 0 0 0 0 4-4 r 0 C7 0 -0 0 -0 -Lj 4 0 a, (n 07 '.0 N. 0 '0 0 0 0 ' 0 0 0 nr -0 000- 0 Nr -0 NU') -v c-CD 0 CO 0-I -4 C- 0 In N i C U U U QU O QU U U U u Qu u tc pq pH Hz If) TABLE B-IV. LATERAL STATIC AERODYNAMIC COEFFICIENTS Sideslip Angle -20 0 20 C .7260 0.0 -. 726 y BJ C -. 0366 0.0 .0366 (Nominal) m C1 -. 1226 0.0 .1226 C .646 0.0 -. 646 BJ C .0279 0.0 -. 0279 (81 % Tail) M C1 .0907 0.0 090'e C .923 0.0 -. 923 Y BJ C -. 174 0.0 . 174 (144% Tail) m C -. 193 0.0 .193 C .299 0.0 -. 299 y VEE C -. 0014 0.0 .0014 (Nominal) m CI ,002 0.0 .002 C .502 0.0 -. 502 y "VEE C -. 154 0.0 .154 (200% Bottom m Tail) C .0471 0.0 .0741 C .706 0. 0 -. 706 y VEE C .3065 0.0 .3065 (300% Bottom m Tail) C1 .0768 0.0 .0768 C .6b3 0.0 -. 663 Y GAC C - 0768 0.0 .0768 (Nominal) m C1 -. 130i 0.0 .1305 B-I1 ;n -innw AL ý44f.'- 1+ I IT I r Milt t + + -4 + t f I -4- i 4-T Tj + n.r. Y TmFRFn Rklin'N rl L I I t I i I I I 7'r I r r I I -T + 1. T if 7 -F -ij _4 + -it -A --4 + - - -- -- - -T 4 p14 + 4 1 14 -W1 -7 e4 4 4 4-4 4 -4- --4- 4 -4- diff 7 1 1 1 , I I Tj TFT 1] t Ll nkfl TM r n; i= qLoOm-44 1411--rl X-44 1 ' t t -TT-ý-f ITIT -4 --4-m 7A + J-64 -t-4-I T 4-!T B- F 5 ljJ4.. . IL,NA Jl4S . 4-4- NOW 74, 2 W .Z 1' -'-41 L A -13 44 -,LNI Z 4L.34T 2 65 2, 1111~ ~ -------4-,--------.-----'- _____I 04 btINAL 10 Se L. LglT ~33 lo 4D12.o- lo Bi 4001MJAL .10 SEC. UUST5 0) B 4 oCIJi pA04W 2,S, MMB S T m Lt$W 10 -. . . 1$ 'b S U 'P 0 05 0 Sb B -1 5 0 S ~ S. h ~ i N. I . NA. C-" 44A" (VERTICAl. C.ST-j4 • •.j~ C , 4 ý (4AI ('VFRTICAL CUST) IT-B II 1''El LJA BA L i 4A V F I'-1- I' I 14+ H-t -47. -Z -i: -11- ==T- -7 -4 --4---~ 'WIND) Ifi1~ L ~ i 0 Ir -I'4 f. I i 7 1 -L Al L -I E r(ACMAL WID~ B-1 7 r ~ ~ ~ ~ 4 4~, r ~ -4-4 -- -4 --i -. m 4 t4, TI-i~ t 1 ý4 . . .. I IT 4 -r K7 1f tIi 44E IYII. 4p2 F-l2 I-4 L. I + -T, --7!--4-~Aff-T 1-4 --1 7_ I T, -7 77I + I- T- , ~ t -- -44 -~ B-18 iLtl4IJU01tAL OVAgic: at LtIHLRLD BALLOON I2NGITUD~N.'h QYMAMI..S OF !ETHLRED BALLqiO 3 iiiS.ID StC. 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1710	https://www.directlinesoftware.com/survey.htm	Home Land Record Reference Surveying Units and Terms 2 Oct 2019 Here is our list of units of measure, surveying terms, surveyors' slang and abbreviations, water descriptions, and trees. If you don't see your favorite obscure units or terms, please let us know. We're happy to add to our list. Units of Measure Acre - The (English) acre is a unit of area equal to 43,560 square feet, or 10 square chains, or 160 square poles. It derives from a plowing area that is 4 poles wide and a furlong (40 poles) long. A square mile is 640 acres. The Scottish acre is 1.27 English acres. The Irish acre is 1.6 English acres. Arpent - Unit of length and area used in France, Louisiana, and Canada. As a unit of length, approximately 191.8 feet (180 old French 'pied', or foot). The (square) arpent is a unit of area, approximately .845 acres, or 36,802 square feet. Chain - Unit of length usually understood to be Gunter's chain, but possibly variant by locale. A 100 foot chain is also sometimes used by American surveyors. See also Rathbone's chain. The name comes from the heavy metal chain of 100 links that was used by surveyors to measure property bounds. Colpa - Old Irish measure of land equal to that which can support a horse or cow for a year. Approximately an Irish acre of good land. Compass - One toise. Cuerda - Traditional unit of area in Puerto Rico. Equal to about .971 acres. Known as the "Spanish acre". Engineer's Chain - A 100 foot chain containing 100 links of one foot apiece. Furlong - Unit of length equal to 40 poles (220 yards). Its name derives from "furrow long", the length of a furrow that oxen can plow before they are rested and turned. See Gunter's chain. Ground - A unit of area equal to 2400 sq. ft., or 220 sq. meters, used in India. Gunter's Chain - Unit of length equal to 66 feet, or 4 poles. Developed by English polymath Edmund Gunter early in the 1600's, the standard measuring chain revolutionized surveying. Gunter's chain was 22 yards long, one tenth of a furlong, a common unit of length in the old days. An area one chain wide by ten chains long was exactly an acre. In 1595 Queen Elizabeth I had the mile redefined from the old Roman value of 5000 feet to 5280 feet in order for it to be an even number of furlongs. A mile is 80 chains. Hectare - Metric unit of area equal to 10,000 square meters, or 2.471 acres, or 107,639 square feet. Hide - A very old English unit of area, a hide was of variable size depending on locale and the quality of the land. It was the amount of land to support a family, and ranged from 60 to 180 acres. After the Norman conquest in 1066 it became standardized at around 120 acres. Hundred - An adminstrative area larger than a village and smaller than a county. In England it was 100 hides in size, and the term was used for early settlements in Virginia, Maryland, and Delaware. Labor - The labor is a unit of area used in Mexico and Texas. In Texas it equals 177.14 acres (or 1 million square varas). League (legua) - Unit of area used in the southwest U.S., equal to 25 labors, or 4428 acres (Texas), or 4439 acres (California). Also, a unit of length-- approximately three miles. Link - Unit of length equal to 1/100 chain (7.92 inches). Morgen - Unit of area equal to about .6309 acres. It was used in Germany, Holland and South Africa, and was derived from the German word Morgen ("morning"). It represented the amount of land that could be plowed in a morning. Out - An 'out' was ten chains. When counting out long lines, the chain carriers would put a stake at the end of a chain, move the chain and put a stake at the end, and so on until they ran "out" of ten stakes. Perch - See pole . Point - A point of the compass. There are four cardinal points (North, South, East, West), and 28 others yielding 32 points of 11.25 degrees each. A survey line's direction could be described as a compass point, as in "NNE" (north northeast). To improve precision, the points would be further subdivided into halves or quarters as necessary, for example, "NE by North, one quarter point North". In some areas, "and by" meant one half point, as in "NE and by North". Pole - Unit of length and area. Also known as a perch or rod. As a unit of length, equal to 16.5 feet. A mile is 320 poles. As a unit of area, equal to a square with sides one pole long. An acre is 160 square poles. It was common to see an area referred to as "87 acres, 112 poles", meaning 87 and 112/160 acres. Pueblo - A Spanish grant of less than 1000 acres. Rancho - A Spanish grant of more than 1000 acres. Rathbone's Chain - A measuring chain two poles, or 33 feet, in length. Rod - See pole Rood - Unit of area usually equal to 1/4 acre. Toise - Traditional French unit of length equal to 6 old French 'pieds' or feet, or 6.4 English feet. Vara - Unit of length (the "Spanish yard") used in the U.S. southwest. The vara is used throughout the Spanish speaking world and has values around 33 inches, depending on locale. The legal value in Texas was set to 33 1/3 inches early in the 1900's. Virgate - An old English unit of area, equal to one quarter of a hide. The amount of land needed to support a person. Standard Surveying Terms A Frame - A measuring device built in the shape of an A. The distance between the legs is 6.6 feet (one tenth of a chain. To measure the acreage of a small square parcel, multiply the width and height in "A's" and move the decimal point three places to the left. For example, a square that is 6 A's wide and 4 A's tall is .024 acres. Aliquot - The description of fractional section ownership used in the U.S. public land states. A parcel is generally identified by its section, township, and range. The aliquot specifies its precise location within the section, for example, the northwest quarter of the southeast quarter. Auditor's map - was made by the County Surveyor at the request of the auditor for tax purposes. Many were made in the 1800's. Very little field work was done. The map was created bu the use of various documents, piecing together other surveys, a few rough measurements in the field, etc. Generally, not accurate. Azimuth - The number of degrees from north (or other reference direction) that a line runs, measured clockwise. Back sight - After measuring from point A to B, reading the heading from B back to A. Various factors can cause the headings to not be exactly the reverse of one another. Baseline - In the U.S. Public land surveying system, a surveyed east-west (i.e. latitudinal) reference line, often hundreds of miles in length, from which tiers of townships are are surveyed to the north and south. There are approximately two dozen baselines in the lower 48 states. See also meridian. Bearing - See azimuth. Bearings taken with a compass will be referenced to magnetic north unless otherwise noted. Benchmark - A survey mark made on a monument having a known location and elevation, serving as a reference point for surveying. Call - Any feature, landmark, or measurement called out in a survey. For example, "two white oaks next to the creek" is a call. So is "North 3 degrees East 120 poles". Chain carrier - An assistant to the surveyor, the chain carriers moved the surveying chain from one location to another under the direction of the surveyor. This was a position of some responsibility, and the chain carriers took an oath as "sworn chain carriers" that they would do their job properly. Chord - The straight line connecting the end points of an arc. Condition - See Conditional line. Conditional line - An agreed line between neighbors that has not been surveyed, or which has been surveyed but not yet granted. Corner - The beginning or end point of any survey line. The term corner does not imply the property was in any way square. Declination - The difference between magnetic north and geographic (true) north. Surveyors used a compass to determine the direction of survey lines. Compasses point to magnetic north, rather than true north. This declination error is measured in degrees, and can range from a few degrees to ten degrees or more. Surveyors may have been instructed to correct their surveys by a particular declination value. The value of declination at any point on the earth is constantly changing because the location of magnetic north is drifting. More information about historical values of declination is available. First station - See Point of Beginning Flag - A bright plastic ribbon tied to a lath stake. Used to mark points along a survey line. Gore - A thin triangular piece of land, the boundaries of which are defined by surveys of adjacent properties. Loosely, an overlap or gap between properties. See also strip. Landmark - A survey mark made on a 'permanent' feature of the land such as a tree, pile of stones, etc. Line Tree - Any tree that is on a property line, specifically one that is also a corner to another property. Merestone - A stone that marks a boundary. See monument. Meridian - In the U.S. public land surveying system, a surveyed north-south (i.e. longitudinal) reference line, often hundrends of miles in length, from which ranges are surveyed to the east and west. There are approximately two dozen meridians in the lower 48 states. See also baseline. Mete - In the context of surveying, a measure, i.e. the direction and distance of a property line. Metes and Bounds - An ancient surveying system that describes the perimeter of a parcel of land in terms of its bearings and distances and its relationship to natural features and adjacent parcels. Monument - A permanently placed survey marker such as a stone shaft sunk into the ground. Open line - A survey line, usually the final one, that is not measured and marked (blazed) by the surveyor but is instead calculated. Point of Beginning - The starting point of the survey Point of intersection - The point where two non-parallel lines intersect. More specifically, the point where two tangents to a curved line intersect. Plat - A drawing of a parcel of land. More specifically, the drawing created by the surveyor that shows the field work, with bearings, distances, etc. Plot plan - A diagram showing the proposed or existing use of a specific parcel of land. Plunge - 1) Inversion of a transit in order to make measurements that cancel errors in the transit, or to extend a line over an obstacle. 2) The angle a falling line makes with the horizontal. Protraction - in the rectangular survey system, the representation of a boundary or corner not run, marked, or fixed by the field survey as evidenced by the field notes. For example, a surveyed section might be protracted into lots by someone in the office. Quarter corner - in the public land surveying system, a point halfway between the corners of a section. A section can be divided into four equal quarters by connecting its quarter corner points. A section's quarter corners are identified by the section line they are located on (north, south, east, west). Range - In the U.S. public land surveying system, a north-south column of townships, identified as being east or west of a reference longitudinal meridian, for example, Range 3 West. See township. Riser - a tree branch or other similar object stuck in the ground and flagged to mark a survey point. Searles Spiral - A surveying technique used by railroad surveyors in the the late 1800s and early 1900s whereby they approximate a spiral by use of multiple curved segments. Section - In the U.S. public land surveying system, an area one mile square. See aliquot. Standard Corner - a corner that is on a standard parallel or base line Strip - A rectangular piece of land adjoining a parcel, created when a resurvey turns up a tiny bit larger than the original survey. The difference is accounted for by temperature or other effects on measuring chains. See also gore. Tangent line - A line that touches a circle at exactly one point and which makes a right angle with the circle's radius. For example, a circle that fills a square has four tangent points and the square's sides are tangent lines. An arc (curve) in a survey is part of a larger circle. One can construct tangent lines at the end points of the arc. Tie line - A survey line that connects a point to other surveyed lines. Tier - In the U.S. public land surveying system, an east-west row of townships identified as being north or south of a latitudinal baseline. Total station - A survey instrument that combines a theodolite and distance meter. Township - In the U.S. public land surveying system, an area six miles square, containing 36 sections. The townships are organized in tiers and ranges, identified with respect to a baseline and meridian. For example, Township 13 North Range 6 West describes a township's location. Traverse - 1) any line surveyed across a parcel, 2) a series of such lines connecting a number of points, often used as a base for triangulation. Trend - the bearing of a line along a falling course. Trocha - Spanish for 'path'. In the southeast U.S. it is used for a cut or cleared survey line. Witness Tree - Generally used in the U.S. public land states, this refers to the trees close to a section corner. The surveyor blazed them and noted their position relative to the corner in his notebook. Witness trees are used as evidence for the corner location. Zenith angle - An angle measured from a vertical reference. Zero degrees is a vertical line pointing up, 90 degrees is horizontal, and 180 degrees is straight down. Surveyors' Slang Surveying, like any profession, has its special terms and slang. Some are just humorous, some help distinguish similar sounds (e.g. eleven and seven), and some are just plain strange! Balls - Slang for numeric .00, as in 4-balls (4.00) Beep - Verb. To use a magnetic detector to look for iron pipe, etc. Blood - To slowly raise the levels rod in order that the instrument man can read the foot markings. Boot - To raise the levels rod some number of inches so as to be visible to the instrument man, e.g. "Boot 6!" means "raise it 6 inches." Blue topping - In road or grading work the surveyor sets stakes and paints their tops blue to represent the required elevation. Graders then work to just cover the blue tops of the stakes. Box - Data collector. Bug - To use a magnetic locator to search for an iron pipe. Bullseye - Zero degrees of inclination. Burn - See shoot Burn one - Measure from the one foot mark on the tape rather than from the end of the tape in order to increase the accuracy of the measurement. Bust - Closure error, i.e. the amount by which the survey fails to perfectly close. Cap - A metal or plastic cover on the end of a rebar or pipe, typically stamped or printed with the surveyor's license number or other identifier. Cut line - To clear vegetation for a line of sight between two survey control points. Double nickel - Slang for .55, as in 6-double nickel (6.55) Dummy or dummy-end - The base or zero end of a tape or chain, as in "hold dummy at the face of the curb." Dump - Download data from the data collector. EDM - Electromagnetic Distance Measurement device, the instrument used by modern surveyors that replaces the use of measurement chains. It determines distance by measuring the time it takes for laser light to reflect off a prism on top of a rod at the target location. Ginney - A wooden dowel 6-9 inches in length with a sharpened end. Set in the ground to mark survey points. Glass - The EDM prism. Gun - Originally, a transit, but potentially any measurement instrument in use, e.g. theodolite, EDM, or Total Station. Hours - Degrees Hub and Tack - A 2" by 2" stake that is set in the ground and that contains a nail ("tack") that precisely marks the point being set. Jigger - Transit (Australia and New Zealand) Legs - Tripod Pogo - Prism pole Pole - Approximate unit of measure (about 0.1 foot) used for stake out, e.g. "Move a pole to the left and drive that hub in" Punk - See railroad. Railroad - Slang for eleven, as in 42-railroad (42.11) Rodman - The person holding the rod with the EDM prism. This person is the modern version of a chain carrier or chain man. Shoot - Measure distance with an EDM Spike - Usually a 60 penny nail used to mark survey points in hard ground. Stob - In the southeast U.S., a wooden stake or post, but in modern surveying, a piece of rebar used to mark a property boundary. Tie - To locate something with the transit or other measuring device. Top - Slang for eleven. See railroad. Trip - Slang for triple digits, as in trip5 means 555, and 43trip7 means 43.777 Turn - The rodman is told to stay in place while the gun or level is moved to a new location. Wave - To slowly move the levels rod back and forth in order to confirm that a measurement was made when the rod was truly vertical. Zero - Zero degrees, minutes, and seconds. A perfect zero. Surveyors' Abbreviations You might find the following abbreviations on a plat drawing. When a surveyor resurveys a tract they will "find" evidence of earlier surveys and will "set" new markers of their own. Thus F and S are common in abbreviations. The term "held" may be applied to markers found or set in the presence of other nearby pipes, rebar, etc. It means "this is the true corner". BRL - Building restriction line. BS - Back sight BSL - Building setback line. CIP - Capped iron pin CL - Center line Con Mon F - Concrete monument found CRB - Capped rebar EBL - East boundary line. Eastbound Lane. EIP - Existing iron pipe FD, FND - Found GPPM - General property parcel map IPF - Iron pipe/pin found IPPM - Individual property parcel map IPS - Iron pipe/pin set IRF - Iron rod found IRS - Iron rod set L.O.D. - Limit of Disturbance. The area to be cleared, graded, etc. LS - Licensed/Land Surveyor # MAG - New concrete nails are magnetic nails and are stamped with MAG on the head and are easier to find with metal detectors. MBS - Minimum building setback NBL - North boundary line. Northbound Lane. N/F - Now or formerly NIP - New iron pin NMS - No monument set NPP - Nail in power pole NTCFP - Nail on top of corner fence post NTFP - Nail on top of fence post NTS - Not to scale OHE - Overhead Electric PC - Point of curvature. The point at which a straight line begins to curve, i.e. the point of tangency to the curve. See PT. PCC - Point of compound curvature. The point where curves of different radii meet. PDE - Public drainage easement PI - Point of intersection PK - Point Known, PK nail PK nail - A concrete nail made by Parker Kaelon, stamped PK, that marks a survey point. See also hub and tack. PL - Property line POB - Point of beginning. The starting point of a survey. POL - Point on line. Used in situations where the end (final) point cannot be seen from the transit. Also used when the final point falls in water. PRC - Point of reverse curve. The point in an S-type compound curve where two curves of different polarity meet. PSDE - Private storm drain easement. PT - Point of tangency. The point at which a curve ends and straight survey line begins. See PC. R/C - Rod and cap, or rebar and cap RP - Radius point R/W - Right of way SBL - South boundary line. Southbound lane. SC - Standard corner SCM - Set concrete monument SMN - Set mag nail SR - Steel rebar SRS - Steel rod set (rebar or other steel) STE - Sight triangle easement UE - Utility easement WBL - West boundary line. Westbound lane. WC - Witness corner Water Terms Arroyo - A small steep-walled (usually) dry watercourse with a flat floor. A gulch or gully. Chiefly in the U.S. southwest. Bank - Edge of a stream. Bed and banks - For property lines that cross a body of water, this term is used to explicitly refer to the bottom of the water. Bottom - Land along a river. Branch - Small stream. Brook - Small stream. Creek - Small stream. Drain - Small dry stream or gully. Draughts of - (pronounced drafts). See waters of. Drean - See drain. Ford - Shallow part of a stream or river where one could cross. Fork - Meeting point of two streams. "In the fork of" means between two branches. Gut - A narrow passage between hills. A stream in such a passage. A drain. Head - The source of a stream. Headwaters - The smallest streams that combine to make a larger stream. Kill - (Dutch) Creek. Lower - Toward the mouth of a stream. Further down along its course. Opposite of upper. Meander - "with the meanders of the stream" means the survey line follows the twists and turns of the stream. Mouth - The place where a stream enters another, larger stream. Narrows - Narrow part of a stream. River - Large stream. Run - Small stream. Shoal - Shallows. Spring - A pool or other source of water that feeds a stream. Swale - A low, generally marshy tract of land, either natural or manmade, e.g. for managing water runoff. Swamp - In the southeastern U.S., a stream, particularly one that has has swampy parts. A marsh. Thalweg - 1. An imaginary line connecting the lowest points of a valley. 2. The line connecting the lowest points of a stream's channel. 3. The surface midline of a channel. Thread of a creek. A figurative expression used to signify the center line of the main channel of a stream when the flow rate is low. Upper - Toward the head of a stream. Further up along its course. Opposite of lower. Vly - (Dutch) Swampy lowland. Waters ("watters") of - In the drainage of. On the branches of. Trees Alder - Ash - has tough, straight-grained wood Aspen - a type of poplar Basswood - see linden Beech - smooth gray bark and small edible nuts Birch, (burch) - Black gum - see tupelo Blackjack - a type of small oak Black oak - Black walnut - Box elder - Box oak - Buckeye - Buffaloberry - Cedar - Cherry - Chestnut - American chestnut has been virtually destroyed by blight. Chestnut oak - has leaves resembling a chestnut Chittamwood - see Wooly Bumelia Cottonwood - Dogwood - Elder - Elm - Fir - Gambrel oak - see Oakbrush Gum - subtypes: black, sweet Hackberry - has cherry-like fruit Hawthorn - Hazel - Hemlock - Hickory, hiccory, hickry - has edible nuts and hard wood Hornbeam - has hard, heavy wood Ironwood - see hornbeam Juniper - Larch - Laurel - Lightwood - highly resinous pine, suitable for stakes Live oak - Lowerwood - transcription error for sourwood Magnolia Maple, (maypole) Mountain birch - Oak, (oake) - subtypes: black, box, chestnut, live, pin, post, red, scrub, shrub, Spanish, swamp white, white Oakbrush - scrub oak prevalent in southern Colorado west of the divide Pawpaw - Persimmon - has plum-like fruit Pine - Pin oak - Pohiccory - see hickory Ponderosa pine - Poplar, popular - Post oak - wood used for posts Red cedar - Red oak - Sapling, (saplin) - young tree Sassafras - bark used in medicines and beverages Scrub oak - usually found in dry, rocky soil Serviceberry - (sarvisberry) Sour gum - see tupelo Sourwood - sorrel tree Spanish oak - Spruce - Sugar tree - sugar maple Sumac - (shumac) Swamp white oak - heavy, hard wood used in shipbuilding, furniture, etc. Sweet gum - hard reddish brown wood used for furniture Sycamore - Tamarack - an American larch having reddish brown bark Tamarisk - small shrub found in the southwest Tupelo - Walnut - black White oak - Wooly Bumelia - leaves resemble a live oak with a fine fur-like fuzz on the underside. Yew - Sources You can find definitions for most of these units, terms, and words in any good unabridged dictionary. There are also books dealing with units of measure and surveying. Wm. Johnston, "For Good Measure". Untitled. Book listing State and Federal Laws relating to measures. Funk & Wagnall's Unabridged Dictionary, 1963 Webster's Unabridged Dictionary, 1959 Oxford English Dictionary Robert's Dictionnaire de la Langue Francaise, 1979 Discussions with Mr. Galtjo Geertsema, Land Surveyor; Ms. Patricia Law Hatcher, lecturer on land records. "Land and Property Research in the U.S", Wade Hone, 1997 Contributions from surveyors. \| \| \| --- \| \| Copyright 2010 Direct Line Software \| 71 Neshobe Rd. \| \| \| Newton, MA 02468 \| \| \| ph: 617 527-9566 \| \| \| deeds@directlinesoftware.com \|
1711	https://www.youtube.com/watch?v=5y2a0Zhgq0U	Simple Explanation of the Minimax Algorithm with Tic-Tac-Toe Science Buddies 401000 subscribers 1226 likes Description 91400 views Posted: 8 Jan 2024 This video explains the fundamentals behind the Minimax algorithm and how it can be utilized in two-player turn-taking games such as Tic-Tac-Toe. Science Buddies also hosts a library of instructions for over 1,500 other hands-on science projects, lesson plans, and fun activities for K-12 parents, students, and teachers! Visit us at to learn more. 0:00 Introduction 0:07 Basics of Tic-Tac-Toe 0:24 Minimax Algorithm 0:58 Key Components of Minimax 1:08 Evaluation Function 1:47 Maximizing and Minimizing Player 2:02 Steps of Minimax 2:15 Base Case 2:29 Recursive Exploration 2:55 Backtracking 3:42 Conclusion STEM #sciencebuddies Connect with Science Buddies: X: FACEBOOK: INSTAGRAM: PINTEREST: 16 comments Transcript: Introduction hey everybody it's Tracy from Science Buddies today we're going to explore how the Minimax algorithm works with a classic game tic tac toe before we start Basics of Tic-Tac-Toe on the algorithm let's revisit the basics of tic tac toe Tic Tac Toe consists of a 3X3 grid and the goal of the game is to get three in a row now imagine if we could teach a computer to not just play the game but to play it with an unbeatable strategy the Mini Max Minimax Algorithm algorithm is a decision-making algorithm and what makes it so powerful is its ability to think ahead Minimax is a recursive algorithm meaning it doesn't just look at the current state of the game it anticipates future moves and imagines the entire sequence of moves that follow this recursive nature allows Minx to consider all possible moves and outcomes creating a branching tree of possibilities called the game tree mini Max's ultimate goal is to find the optimal move for a player maximizing their advantage while minimizing the opponent's potential gains let's go over the key components Key Components of Minimax of the Minimax algorithm which are the evaluation function and the maximizing and minimizing player at the heart of the algorithm lies the evaluation Evaluation Function function which is responsible for assigning a score to each possible board State this score is the algorithm's way of quantifying how favorable or unfavorable a given state is when the evaluation function looks at the board it interprets the current state of the game a positive score implies an advantage for player o the algorithm sees potential Victory conversely a negative score signals an advantage for player x a path that might lead to victory for the opponent a neutral score of zero is given when neither player wins and the game ends in a draw the evaluation function acts as a guide for the algorithm to choose the most favorable outcome for the player it's Maximizing and Minimizing Player representing let's talk about the maximizing and minimizing player in the context of tic tac toe for our purposes we assigned o as the maximizing player and X as the minimizing player but you can switch the rows as long as long as you stay consistent throughout the entire game now that we understand the Steps of Minimax foundation of Minx let's delve into a step-by-step process which includes the base case recursive exploration and backtracking every time we run the Minimax algorithm it will first check Base Case the base case the base case is indicated by what we call a terminal State meaning that the game ended in a win a loss or a draw at this stage the evaluation function assigns a score based on the outcome with the base case set in place Recursive Exploration the algorithm then starts exploring the game tree recursively every Branch represents a possible move and every node signifies a potential board State at each turn minax alternates between players as it navigates through the tree explores all conceivable moves each time considering the potential outcomes this process is similar to a player thinking several moves considering the consequences of each decision as Minimax Backtracking reaches the end of its exploration it engages in a strategic backtracking maneuver the algorithm starts to backtrack through the game tree carrying with it the scores assigned to each node as it ascends minimac strategically chooses the path that leads to the optimal outcome for the player it represents in this case which path do you think o would take you might think it's either one of these two paths since they end in a win for o but that's not exactly the case you have to remember that Mini Max assumes that the other player is playing to win meaning that even though this scenario is possible X is unlikely to let o win that easily when X could easily win itself in the previous step so we end up in this Branch o can at least guarantee itself a draw and the win if X is not playing optimally and with that we reach the end of this Conclusion explanation of the minia max algorithm we learned that the Minx algorithm is a decision-making tool for two-player turn-taking games it systematically explores a game Tree by anticipating future moves and employs an evaluation function to assign scores to board States finally it strategically chooses optimal paths through recursive exploration and backtracking for a selection of minax projects please refer to the links in the description below and for a thousand other projects for all areas of Science and Engineering visit our website www.sciencebuddies.org
1712	https://www.youtube.com/watch?v=ic5b6pZ9jM0	GCSE Maths Foundation OCR June 2022 paper 3 Question 25 percentage decrease Get Great @ Maths 370 subscribers 4 likes Description 152 views Posted: 20 Jan 2024 This video will show you how to complete a GCSE exam question on how to find percentage decrease. maths #math #mathematics #ukmaths #gcse #gcsemaths #ocr #foundationmaths #exampaper #examquestion #examrevision #revision #mathsrevision #mathsexam #mathspaper #june2022 #june22 #paper2 #j560 #tips #formulas #percentage #subscribe #follow #getgreatatmaths DON'T MISS ANOTHER VIDEO...SMASH THIS SUBSCRIBE BUTTON 👉🏻 The video is an OCR J560 GCSE UK Maths Foundation Paper 3 June 2022 Calculator question 25. The exam question, mark scheme, model solution and working are shown to show how to gain full marks on a GCSE maths exam paper. Once again thank you for all you support, its incredible what we're building together and amazing to have you as part of the team. If you haven't already hit that notification bell and be the first to see every video. Let us know in the comments if you've subbed and clicked that bell and we'll comment back 😉 🙌🏻 Also FOLLOW us on: Transcript: going to need to be able to answer this question is the formula for the percentage decrease. And to find our percentage decrease, we need the original amount times by your multiplier. And your multiplier is calculated by doing your percentage which is out of 100. Take away the percentage that is given. Now the reason it's take away the percentage is because it is a reduction. If it was an increase, you would add it. But because it is a decrease, you are subtracting. and then you divide it by 100 to make it a decimal. So this is called the multiplier here, this section here. And it is calculated by doing 100 take away the percentage divided by 100. So if you multiply your original amount, in this case that is the £18,000 by your multiplier, then that will give you the new price. Press pause, have a go at working out the answer, and then press play when you're ready to hear the answer. Well done guys. Hopefully you've managed to work out the value of R. So what we've got to do is we've got to do this formula here for percentage decrease and we're going to be doing it twice because if we do it once it will tell us how much the price of the car is after it's been reduced by 30%. And then we do it again and then that will tell me the price after the special offer. Well, we know the price after the special offer is 9,450 but it will help us work out this value of R. So we're first of all going to put the values into this formula. So the car price in the sale. Okay. So my sale price is going to be the normal price the original amount which is this £18,000 times by my multiplier which is going to be 100 take away the percentage which was 30% divided by 100. Now you can literally type this in the calculator because it is a calculator paper and it will give you one mark. If you find it useful, you could put brackets around this fraction here to make it easier to type in, but it's not necessary. Once you've typed that in on your calculator, remember to use your fraction button here, the two button, the two squares above each other. Then just press equals and it will give you 12,600. And that is your second mark. Now, from here, we've got to go again. We've got a special offer. Now, now the special offer, the car is reduced by R%. So this is the original amount now. So 12,600. I'm now going to multiply it by r% or I'm going to work out my multiply when the percentage is r%. And that is going to equal the 9,450. So my special offer is the answer we just got for 12,600 times by my new multiplier which will be 100 take away r because the new amount has been reduced by r divided by 100. And we know our special offer is going to be 9,450. So that's going to equal the 12,600 100 - r all over 100 for your third mark. From here, we need to work out what this 100us r is. So we're going to get rid of this times by 12,600 and we're going to get rid of the divide by 100. Now you could do this in two steps or you could do it in one step. It is personal preference, but as you've got a calculator, it's fairly easy to do it in one step. So, first of all, to get rid of this 12,600, we're going to do the opposite, which is divide by 12,600. So, on your calculator, you need to type in 9,450 divided by 12,600. Then, we need to get rid of this divide by 100. So, the opposite to divide by 100 is multiply by 100. and the answer you've now got on your calculator. If you times that by 100 and press equals, you should find you get 75. So this is your fourth mark. So if you get 75 is equal to 100 minus r, we have got four marks. From here we now need to find the value of r. So 100 take away something is going to be 75. Or we can think about solving it. So to get rid of minus r from both sides, I'm going to add r to both sides. To get rid of plus 75, I'm going to subtract 75 from both sides. So when if I do 100 take away the 75, that will give me the value of r for your fifth and final mark. Remember, write your answer on the dotted line and give yourself a tick where the answers are awarded. So everywhere I've written one mark, you need to write down where the method marks are awarded for the question. So you know how to get the method marks in the exam to make sure you get the full five marks for this question. The mark scheme shows there are five marks for this question. B2 are two method marks for getting the 12,600. M1 is a method mark for getting to this stage here. OE means or equivalent. You can see the method that I used uh was doing 100 take away 30 / 100. Now 100 take away 30 is 70. So I did an equivalent method because I never wrote down 70 over 100. Um so if you got that answer there, you would get an equivalent mark for getting your method mark for this stage here. M0 is no marks. So you've got no marks if you just wrote down 70% of 18,000. Down here we've got two method marks for using a different method to what I did. So this method here is using the percentage um change formula. So the original amount take away the final answer divided by the original make it a percentage times by 100. So the percentage change formula has been used here. The method I used is down here on the right hand side. B2 M1 is for leading to getting an answer of 75 which is what we did. Uh so from rearrange this formula here still gives you the same method marks. It's just an alternative method. It is personal preference as to which way you prefer to do the answer. Just beware if you didn't get any if you just wrote down the final answer. Um so you got zero because you didn't write down any method marks and you got the answer of 25%. SC2 means you will get two method marks awarded um for getting an answer of 25 with no or insufficient working. So if you're working wasn't good enough or you didn't have any working, you can only get a maximum of two marks for the question even though you got the answer correct. Down here um you will get one mark for getting an answer of 0.25 0.75 or 75% with no or insufficient working. So these are where method marks were awarded if you didn't show your working. But on a five mark question, it is really worth showing you're working out, especially when it says make sure you show you're working, otherwise you won't get the method marks in the exam. If you found this video useful, please share, like, and subscribe so we can all get
1713	https://math.stackexchange.com/questions/595483/finding-indicated-trigonometric-value-in-specified-quadrant	trigonometry - Finding indicated trigonometric value in specified quadrant - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding indicated trigonometric value in specified quadrant Ask Question Asked 11 years, 9 months ago Modified11 years, 9 months ago Viewed 15k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. If I have csc θ θ = - 10 3 10 3 and have to find tan θ θ in quadrant III, would I use 1 + cot 2 θ cot 2⁡θ = csc 2 θ csc 2⁡θ then find reciprocal which would be tan θ θ? If so, I get 3 91−−√91 3 91 91 as tan, but that doesn't seem right as if I wanted to get cot θ θ from tan now, it would be different. I hope that made sense. Any help would be appreciated! trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Dec 6, 2013 at 13:29 amWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges asked Dec 6, 2013 at 13:05 o.oo.o 335 3 3 gold badges 5 5 silver badges 15 15 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. csc θ=1 sin θ=−10 3⟹sin θ=−3 10=opposite hypotenuse csc⁡θ=1 sin⁡θ=−10 3⟹sin⁡θ=−3 10=opposite hypotenuse tan θ=sin θ cos θ=opposite adjacent tan⁡θ=sin⁡θ cos⁡θ=opposite adjacent All you need is to find cos θ cos⁡θ in the third quadrant to compute tangent. Use the right angle that θ θ forms with the x-axis and the Pythagorean Theorem: 3 2+adjacent 2=10 2⟹adjacent=91−−√3 2+adjacent 2=10 2⟹adjacent=91 or else use the identity: sin 2 θ+cos 2 θ=1 sin 2⁡θ+cos 2⁡θ=1 knowing that cos θ<0 cos⁡θ<0 in the third quadrant. However, you are also correct having used your method: tan θ=3 91−−√=3 91−−√91 tan⁡θ=3 91=3 91 91 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Dec 6, 2013 at 13:29 answered Dec 6, 2013 at 13:16 amWhyamWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges 5 Oh wow, I was actually correct. I tried using the theorem you gave me, but I kept getting a positive answer. I get sqrt(91)/10. When I move 9/100 over to other side I subtracted it by 100/100 then took square root. Am I doing basic math wrong? D:o.o –o.o 2013-12-06 13:30:17 +00:00 Commented Dec 6, 2013 at 13:30 cos θ cos⁡θ, in Quadrant III, is negative, (remember, the identity above is in cos 2 θ cos 2⁡θ, so cos θ=±cos 2 θ−−−−−√cos⁡θ=±cos 2⁡θ. So cos θ=−91−−√10 cos⁡θ=−91 10, giving us tan θ=sin θ cos θ=−3/10−91−−√/10=3 91−−√⋯tan⁡θ=sin⁡θ cos⁡θ=−3/10−91/10=3 91⋯ amWhy –amWhy 2013-12-06 13:34:51 +00:00 Commented Dec 6, 2013 at 13:34 Oh I see. Thank you so much for your help! Makes a lot more sense now.o.o –o.o 2013-12-06 13:36:23 +00:00 Commented Dec 6, 2013 at 13:36 And no, you're not doing basic math wrong, it's just a matter of knowing which quadrant you're working in, and the corresponding signs of the trig functions. You're welcome!amWhy –amWhy 2013-12-06 13:36:38 +00:00 Commented Dec 6, 2013 at 13:36 @amWhy: Needs a TU w/G! +1 Amzoti –Amzoti 2013-12-07 00:14:52 +00:00 Commented Dec 7, 2013 at 0:14 Add a comment\| You must log in to answer this question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 8Finding csc θ csc⁡θ given cot θ cot⁡θ 3How do you find the sine and cosine from the tangent? 12Proving that csc θ cot θ−cot θ csc θ=tan θ sin θ csc⁡θ cot⁡θ−cot⁡θ csc⁡θ=tan⁡θ sin⁡θ 1Trigonometric ratios involving negative angles 1Finding the exact values of trig functions in a quadrant 0Evaluating a Trigonometric Expression involving Periodicity 0If tan t=1 4 tan⁡t=1 4 and the terminal point for t is in Quadrant III, find sec t+cot t sec⁡t+cot⁡t. 0Rearranging a trigonometric expression 1Finding which quadrant θ θ lies in given the value of only tan−1 θ tan−1⁡θ 2Why do we use this complicated algorithm for finding values of trigonometric functions? 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1714	https://www.quora.com/In-how-many-ways-can-4-people-be-chosen-from-a-set-of-10-and-be-seated-in-a-row-of-4-chairs	In how many ways can 4 people be chosen from a set of 10 and be seated in a row of 4 chairs? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Arrangement Product Counted Combinations of Events Probability and Mathemati... Counting Numbers Orderly Arrangement Permutations Arrangements 5 In how many ways can 4 people be chosen from a set of 10 and be seated in a row of 4 chairs? All related (43) Sort Recommended Assistant Bot · Sep 6 Count ordered selections: choose 4 people out of 10 and arrange them in 4 chairs. Method 1 (combination × permutation): Choose 4: C(10,4) Arrange them in 4 chairs: 4! So total = C(10,4) × 4! = (10 choose 4)·24. Compute: C(10,4) = 210, so 210 × 24 = 5040. Method 2 (direct permutation): Number of ordered 4-tuples from 10 = P(10,4) = 10·9·8·7 = 5040. Answer: 5040. Upvote · Related questions More answers below In how many ways can 10 people be seated on a bench if only 4 seat are available? In how many ways can 10 students be arranged in a row? In how many ways can 4 students be seated in a row of 20 chairs? In how many ways can 5 people be seated in a row of 9 chairs? In how many ways can 8 students be arranged in a row for a class picture if Paul and Peter want to stand next to each other and John and Joey also want to be together? Matt Probert Software developer and IT guy. Building and operating tech. · Author has 60 answers and 1.8M answer views ·Updated 1y Originally Answered: In how many ways can 4 people be seated in a row of 4 seats? · Let’s think about it step by step as if you were seating the people. Chair 1, how many “options” (people) do you have to put in chair #1? 4 Chair 2, same question. We already seated 1 person in chair #1, so there are only 3 people left. So we have 3 possibilities (people) for chair #2. How many variations so far for the first two chairs? 4 3 = 12. Why multiply? Well for any SINGLE choice for chair one, there are 3 choices for chair two. If we label our people as A, B, C, D and our chairs as 1, 2, 3, 4… Chair 1 Chair 2 A B, C or D B A, C or D C A, B or D D A, B Continue Reading Let’s think about it step by step as if you were seating the people. Chair 1, how many “options” (people) do you have to put in chair #1? 4 Chair 2, same question. We already seated 1 person in chair #1, so there are only 3 people left. So we have 3 possibilities (people) for chair #2. How many variations so far for the first two chairs? 4 3 = 12. Why multiply? Well for any SINGLE choice for chair one, there are 3 choices for chair two. If we label our people as A, B, C, D and our chairs as 1, 2, 3, 4… Chair 1 Chair 2 A B, C or D B A, C or D C A, B or D D A, B or C You can write this out more verbosely of all possibilities and you will see there are 12 (43) possible combinations. If we wanted to know how many ways we could seat 2 of 4 people in 2 chairs, we would be done. If we continue… Chair 3 We already picked 1 of 4 for chair #1 1 of 3 for chair #2 …so we are left with 2 choices (people) we can pick for chair #3. Chair 1 Chair 2 Chair 3 A B C or D A C B or D A D B or C B A C or D B C A or D B D A or C C... D... Total possibilities for first three chairs: 432 = 24. If we only wanted to know how many ways we could seat 3 of 4 people in 3 chairs, we would be done. Chair 4 Well if we go all the way to the fourth chair, there is only 1 person left (since we have four chairs and four people). So there isn’t any choice left. We made all our choices on the first chairs: Chair1: choose 1 of 4 people Chair2: choose 1 of 3 people Chair3: choose 1 of 2 people Chair4: choose 1 of 1 people (no choice) So the final number of possibilities is 4321 = 24. Permutations In math, changes of orderings are called permutations. They are the same things (e.g. people) just in a different order. You could picture the chairs as being in a row if that helps. The question could also be about order of 4 people in a line and the answer and math would be the same. The answer to the general question: How many ways can we permute n things? …where n can be any integer number and things can be people or books or whatever you like) is n! (“n factorial”). example: 4! = 4321 You saw in the intermediate steps that if we only wanted to seat 2 of the 4 and wanted to know how many possibilities, we came up with 43. This could be written more generally as: 4 3 2 1 4! 4 3 = —------------ = -------- 2 1 (4 - 2)! If n are the total number of items/people we can choose from and c are the number we want to pick, then you get to the general form: n! / (n - c)!. Note that 0! = 1, so if we have 4 (n) and want 4 (c), the bottom component becomes 1 and we are back to n!. Combinations What if you don’t care about order? e.g. you have 4 people and you want to choose 2 for your basketball game. It doesn’t matter which order you pick them, just which two you get. Because we aren’t talking about seating chart or order in line or priority to get a new kidney. i.e. AB is the same as BA (remember we gave our four people letters to identify them earlier). We can re-use everything we did with permutations…but now we just need to “remove” the ordering aspect. We know that if we have 4 people and choose 2 of them for seats we had 43 = 12 possibilities where the order matters. In order to not care about order, we have to ask: How many ways can those 2 chosen people be seated in those 2 seats? That looks the same as the “how many ways can we seat 4 people in 4 seats?” question we already answered. It is. We can seat 2 people in 2 seats 2! = 21 = 2 ways. So if we take the answer to: How many ways can we select and seat 2 of 4 people where order matters? (4!/(4 – 2)! = 12) and divide by How many ways can we seat 2 people in 2 seats (2!/(2–2)! = 2! = 2) We get: How many ways can we choose 2 out of 4 people to receive chairs (where we don’t care about the order)? Or how many possibilities do I have if I want to pick 2 people out of 4 to be on my basketball team? 4!/(4 – 2)!/2! = 6 To generalize this, let’s go back to using n for number of people/things we are picking/choosing from and c for the number of people/things we want to pick. When order matters, we had: n! / (n - c)! = number of permutations of c items selected from a group of size n. If we wanted to take out the order factor, we needed to divide that by the number of permutations of cout of c items: c! / (c - c)! = c! / 0! = c! / 1 = c!. …e.g. how many ways can those 2 people be seated in those 2 seats? …so the final result (how many ways to choose cthings from a group of n where we don’t care about order)… n! ------------- = n choose c c! (n - c)! These are called combinations and are very useful when you want to know how many possibilities you have not caring about order (for instance playing cards in your hand, you don’t care which order you draw them in, just what you get). Notice the only difference to permutations (where we care about order) is the presence of the c! divisor that removes the variations in ordering. So if you remove that you are back to ordered permutations instead of unordered combinations. How many total unique (don’t care about order) 5-card hands dealt from a 52 card deck? n=52``c=5 52! 52! -------------- = -------- = 2,598,960 5! (52 - 5)! 5! 47! How many unique hands where you draw 4 aces? Well if you take the 4 aces and you choose all 4 there is only 1 way to do that n=4, c=4 4! / (4! 0!) = 1 …but there are 48 cards left in the deck and you need to pick one of those 48 for your fifth card. Note 1! = 1 just like 0! = 1. 48! 484746...21 ------------- = ---------------- = 48 1! (48 - 1)! 4746...21 So there are 1 48 = 48 5 card hands where you have all four aces. What is the probability of getting dealt 4 aces from a 52 card deck in 5-card poker? 48 2,598,960 --------- = 0.0000184.. or 1 chance in --------- = 54,145 2,598,960 48 …pretty good right? Powerball Select five numbers between 1 and 69 for the white balls, then select one number between 1 and 26 for the red Powerball. So the total possible combinations (order does not matter) of numbers in the Powerball game are the possible combinations of the 5 white balls MULTIPLIED by the possible combinations of the red ball. (69 choose 5) (26 choose 1) 69! 26! -------------- -------------- 5! (69 - 5)! 1! (26 - 1)! = 11238513 26 = 292,201,338 So for a single ticket (single number combination) your odds of winning (and possibly sharing) the jackpot are 1 in 292201338 or 0.00000034%. If they just had 6 white balls (instead of 5 white, 1 red), the number of combinations are… (69 choose 6) 69! -------------- = 119,877,472 6! (69 - 6)! So over 2x better odds for the players with 6 white balls. By separating the red ball out into a “special” 6th slot, they are able to significantly decrease the odds through the multiplication of 26 red-ball combinations without adding many more balls (ie 7 or 8 white balls). Fewer balls probably has a psychological effect on how players perceive their chances of winning…as well as reducing the time taken to fill out tickets…so they can play more. Upvote · 9 4 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 36 Charles Marshall Diploma from Brainerd Senior High School (Graduated 2000) ·7y There is 10 people. Each time you pick a person, it reduces how many for the next pick by one. You need to do this four times. 10987 = 5040. So there is 5040 ways to pick 4 people from a starting pool of 10. You have 4 chairs with 4 people that can sit in them. Each time you pick a person to sit down, it reduces how many for the next pick by one. You need to do this four times. 4321 = 24. So there is 24 ways to place 4 people into 4 chairs. Combining the two: 504024= 120960. So there is 120,960 ways to pick 4 people from a pool of 10 and then place them into 4 chairs. Upvote · 9 7 9 2 Arif Aulakh B.S. in Applied Mathematics, University of Toronto ·7y In order to solve this problem, we only need to find out how many ways that 4 people can be chosen from the set of 10, since every possible group will be seated in the row of 4 chairs. The question would be different, if perhaps, we were asking how many ways can 4 people be chosen from a set of 10 and be seated in a row of 2. The number of ways that 4 people can selected from a set of 10 is (n k)(n k), where n=10 n=10 and k=4 k=4 where the combinatorial (n k)=n!k!(n−k)!(n k)=n!k!(n−k)! Therefore, the number of ways 4 people can be seated is: (10 4)=10 4!(6!)(10 4)=10 4!(6!) =210=210 Therefore, there a Continue Reading In order to solve this problem, we only need to find out how many ways that 4 people can be chosen from the set of 10, since every possible group will be seated in the row of 4 chairs. The question would be different, if perhaps, we were asking how many ways can 4 people be chosen from a set of 10 and be seated in a row of 2. The number of ways that 4 people can selected from a set of 10 is (n k)(n k), where n=10 n=10 and k=4 k=4 where the combinatorial (n k)=n!k!(n−k)!(n k)=n!k!(n−k)! Therefore, the number of ways 4 people can be seated is: (10 4)=10 4!(6!)(10 4)=10 4!(6!) =210=210 Therefore, there are 210 ways in which 4 people can be chosen from a set of 10 and be seated in a row of 4 seats. Upvote · Related questions More answers below There are 8 boys and 7 girls in a team. In how many ways can they be arranged on a work table if boys and girls are in separate groups? In how many ways can you seat 5 people out of 10 in a row of 5 chairs? In how many ways can 6 students be seated in a row of 6 seats if 2 of the students insist on sitting beside each other? How many ways can 3 ladies and 3 gents can be seated around a round table so that any two and only two ladies sit together? In how many ways can five boys and four girls be seated in a row of nine chairs when the girls must sit on adjacent chairs? Greg Weidman B.E. in Electrical Engineering&Mathematics, Villanova University (Graduated 1992) · Author has 379 answers and 10.7M answer views ·4y Originally Answered: Eight students are lined up to be seated. How many ways can 4 of them be seated in a row of 4 chairs? · You have now asked 9 questions that are clearly taken word for word from your homework. The people on quora like to be helpful, but that's kind of crossing a line. That's not the way this site works. Here's my answer, which is not the answer to your homework: If we are selecting four students out of eight, and seating them on the floor with four chairs in a row, then we can start with the number of ways you can arrange students and chairs in a row: 8!. However, that assumes we can distinguish one chair from another. If we cannot, then we divide this by 4!, which is the number of ways the four ch Continue Reading You have now asked 9 questions that are clearly taken word for word from your homework. The people on quora like to be helpful, but that's kind of crossing a line. That's not the way this site works. Here's my answer, which is not the answer to your homework: If we are selecting four students out of eight, and seating them on the floor with four chairs in a row, then we can start with the number of ways you can arrange students and chairs in a row: 8!. However, that assumes we can distinguish one chair from another. If we cannot, then we divide this by 4!, which is the number of ways the four chairs can be arranged. If the students we selected are numbered 1 2 3 4, and every chair is the same, then this includes arrangements like 3CC4C1C2. We need to multiply this by the number of ways we can select 4 students out of 8, which is denoted 8S4, and can be calculated as 8! Divided by 4!×4!. Overall, we get (8!)^2/(4!)^3, which is 117,600. If you want the students to sit in the chairs, the first part is easier and is left as an exercise. Upvote · 9 1 Sponsored by VAIZ.com Tool Like Asana — Try For Free Our Better Alternative! VAIZ — Better Asana Alternative That Boosts Productivity with Built-in Doc Editor & Task Management. Sign Up 99 14 Doug Skilton Enjoy recreational math · Author has 2.1K answers and 3.9M answer views ·Updated 7y There are 10!/6! = 5,040 distinct ways of arrangement (permutations) of groups of 4, selected from 10 people, and seated in a row of 4 chairs. Most of these include all the same people but ordered differently. If it is the number of distinct combinations being sought then the figure 5,040 must be divided by 4! = 5,040/4! = 210 distinct combinations, none of which include all the same people. Upvote · Aryan Vatsal NSO and IMO - Bronze Medallist · Author has 133 answers and 162.9K answer views ·3y Originally Answered: In how many ways can 4 people be seated in a row of 4 seats? · There are 4 people who have to be seated in 4 places. So, this is a question of Permutation. So, number of ways in which the people can be seated are - 4321 = 4! = 24 Thus, there are 24 possible ways. Hope this answers your question. Good day to you. Upvote · 9 2 Sponsored by SpeechText.AI AI-powered legal transcription service. Convert speech to text from any audio/video format, including TRM files. Fully compliant with GDPR. Start Now 9 3 Armando Flores Former Consultant/Senior Engineer, IBM/Lexmark Retired · Author has 2.1K answers and 1.6M answer views ·4y Originally Answered: Eight students are lined up to be seated. How many ways can 4 of them be seated in a row of 4 chairs? · Eight students are lined up to be seated. How many ways can 4 of them be seated in a row of 4 chairs? Continue Reading Eight students are lined up to be seated. How many ways can 4 of them be seated in a row of 4 chairs? Upvote · 9 2 Ernest Leung B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views ·1y Originally Answered: In how many ways can 4 people be seated in a row of 4 seats? · In how many ways can 4 people be seated in a row of 4 seats? Method 1: No. of ways for 4 persons to taken 4 seats in a row = ₄P₄ = 24 Method 2: • The first one takes any one of the 4 seats. • The second one takes any one of the 3 seats left, • The third one takes any one of the 2 seats left. • The last one takes the only 1 seat left. No. of ways for 4 persons to taken 4 seats in a row = 4 × 3 × 2 × 1 = 24 Upvote · Sponsored by Wordeep Proofreading Grammar correction solution using artificial intelligence. Have a look at our real-time AI based proofreading and grammar correction service. You will be amazed! Learn More 99 95 Lorn Olsen MA Actuarial Math from the University of Michigan · Author has 287 answers and 168.5K answer views ·7y Because the problem doesn’t care what order the people are seated in the chairs, this is a Combination problem and not a Permutation problem. The combination of 10 things taken 4 at a time is calculated as: nCr = n! / ( r! (n! - r! ) ) 10C4 = 10! / ( 4! 6! ) = 210 Upvote · 9 1 Stalin Armijos Statistics 101 · Author has 210 answers and 658.5K answer views ·7y There are 10C4 ways to choose 4 people from a group of 10. Then, there are 4! ways for them to be seated in a row of 4 chairs. Therefore, 10C4 4! = 5040 Upvote · 9 1 Shrey Bhatt Former Graduate Management Trainee at Rebel Foods (Formerly Faasos) (2019–2021) · Author has 112 answers and 101.6K answer views ·4y Originally Answered: Eight students are lined up to be seated. How many ways can 4 of them be seated in a row of 4 chairs? · So 8 students can be arranged by taking 4 at a time can be done in 8P4 ways 8P4 = 1680 ways Upvote · Alberto Cid M.S.E. in Telecommunications Engineering&Data Transmission, Technical University of Madrid (Graduated 2008) · Author has 2K answers and 3.8M answer views ·7y Related In how many ways can you seat 5 people out of 10 in a row of 5 chairs? Some people gave the answer, and the operations (calculations) made to reach that number… I’ll try to explain why: It’s a row of 5 chairs, and, so, the order is important: it’s not the same ABCDE and EDCBA. In the former, A is on the first chair, and in the latter A is on the last chair of the row. Imagine the first one to seat is the first chair in the row. How many different people could you choose? There are 10 people, you could choose any of those 10, so there are 10 cases. Once you chose one person (out of 10) for the first chair… how many can you choose for the second? Only 9, because yo Continue Reading Some people gave the answer, and the operations (calculations) made to reach that number… I’ll try to explain why: It’s a row of 5 chairs, and, so, the order is important: it’s not the same ABCDE and EDCBA. In the former, A is on the first chair, and in the latter A is on the last chair of the row. Imagine the first one to seat is the first chair in the row. How many different people could you choose? There are 10 people, you could choose any of those 10, so there are 10 cases. Once you chose one person (out of 10) for the first chair… how many can you choose for the second? Only 9, because you already chose one for the first chair… So, for rows of two chairs you can make 109 ways = 90… Let’s see: imagine digits from 0 to 9 represent the 10 different people. Yo name one person as 0, the next as 1… and so on till the last named as 9, that’s 10 people. If you make numbers of two digits, they would be 100, from 00 to 99. The first digit indicates the person in the first chair and the second digit is the person in the second chair… but this way we allowed numbers like 00 or 55 that have the same person on two chairs, which is impossible! So, if you exclude those you’ll have it: 100–10 = 90. This was the case with 2 chairs, and I hope it’s clear. With 5 chairs you continue with the same logic: For the first chair you have 10 possible people. For the 2nd chair you have 9 possible people left. For the 3rd chair you have 8 possible people left. For the 4th chair you have 7 possible people left. For the 5th chair you have 6 possible people left. And you multiply: 109876 = 10!/5! Another way to think the problem. Imagine the order didn’t matter… How many groups (sets) of 5 people can you make getting from a total of 10? This is called combinations of 10 objects (elements) taken in groups of 5… it’s written as C(10,5) or 10C5 … C(10,5) = 10! / [ 5! (10–5)! ] Now, in each of those groups of 5 … we want to make an order for the chairs… How many ways we have to make an order with a group of 5? This is called a permutation… P(5)=5! It’s the same idea we said before: you take 1 of five for the first chair, you have 5 to get, you get 1 of the remaing 4 and that’s 4 possible… Total: 54321 And the answer to que question is: C(10,5) P(5) = 10! / [ 5! (10–5)! ] 5! = 10! / 5! = 109876 It’s no suprise that the result is the same. Upvote · 9 2 Related questions In how many ways can 10 people be seated on a bench if only 4 seat are available? In how many ways can 10 students be arranged in a row? In how many ways can 4 students be seated in a row of 20 chairs? In how many ways can 5 people be seated in a row of 9 chairs? In how many ways can 8 students be arranged in a row for a class picture if Paul and Peter want to stand next to each other and John and Joey also want to be together? There are 8 boys and 7 girls in a team. In how many ways can they be arranged on a work table if boys and girls are in separate groups? In how many ways can you seat 5 people out of 10 in a row of 5 chairs? In how many ways can 6 students be seated in a row of 6 seats if 2 of the students insist on sitting beside each other? How many ways can 3 ladies and 3 gents can be seated around a round table so that any two and only two ladies sit together? In how many ways can five boys and four girls be seated in a row of nine chairs when the girls must sit on adjacent chairs? How many ways can four people be seated in chairs in one row of 13 chairs, from left to right? How many different combinations would there be for four friends who play a game sitting in four chairs. If the four of us change chairs in every possible seating arrangement, how many arrangements (or combinations) would that be? In a group of 5 people, 2 do not go along well with each other. In how many ways can the group be seated in a round table such that those two persons are not seated next to each other? In how many ways can 5 people sit in 5 chairs such that "Person A" doesn't seat next to "Person B"? There are 6 boys and 5 girls. In how many ways can they be arranged in a row such that no two girls are together and no two boys are together? Related questions In how many ways can 10 people be seated on a bench if only 4 seat are available? In how many ways can 10 students be arranged in a row? In how many ways can 4 students be seated in a row of 20 chairs? In how many ways can 5 people be seated in a row of 9 chairs? In how many ways can 8 students be arranged in a row for a class picture if Paul and Peter want to stand next to each other and John and Joey also want to be together? There are 8 boys and 7 girls in a team. In how many ways can they be arranged on a work table if boys and girls are in separate groups? 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1715	https://royalsocietypublishing.org/doi/10.1098/rsos.211631	Open AccessResearch articles Stochastic rounding: implementation, error analysis and applications Matteo Croci Matteo Croci Oden Institute, University of Texas at Austin, Austin, TX, 78712, USA Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Massimiliano Fasi Massimiliano Fasi Department of Computer Science, Durham University, Durham, DH1 3LE, UK Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Nicholas J. Higham Nicholas J. Higham Department of Mathematics, The University of Manchester, Manchester, M13 9PL, UK Contribution: Conceptualization, Data curation, Formal analysis, Funding acquisition, Investigation, Methodology, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Theo Mary Theo Mary Sorbonne Université, CNRS, LIP6, Paris, 75005, France Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author and Mantas Mikaitis Mantas Mikaitis Department of Mathematics, The University of Manchester, Manchester, M13 9PL, UK mantas.mikaitis@manchester.ac.uk Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Project administration, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author Matteo Croci Matteo Croci Oden Institute, University of Texas at Austin, Austin, TX, 78712, USA Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Massimiliano Fasi Massimiliano Fasi Department of Computer Science, Durham University, Durham, DH1 3LE, UK Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Nicholas J. Higham Nicholas J. Higham Department of Mathematics, The University of Manchester, Manchester, M13 9PL, UK Contribution: Conceptualization, Data curation, Formal analysis, Funding acquisition, Investigation, Methodology, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author , Theo Mary Theo Mary Sorbonne Université, CNRS, LIP6, Paris, 75005, France Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Software, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author and Mantas Mikaitis Mantas Mikaitis Department of Mathematics, The University of Manchester, Manchester, M13 9PL, UK mantas.mikaitis@manchester.ac.uk Contribution: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Project administration, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing Google Scholar Find this author on PubMed Search for more papers by this author Published:09 March 2022 Review history Review history is available via Web of Science at Abstract Stochastic rounding (SR) randomly maps a real number x to one of the two nearest values in a finite precision number system. The probability of choosing either of these two numbers is 1 minus their relative distance to x. This rounding mode was first proposed for use in computer arithmetic in the 1950s and it is currently experiencing a resurgence of interest. If used to compute the inner product of two vectors of length n in floating-point arithmetic, it yields an error bound with constant √𝑛⁢𝑢 with high probability, where u is the unit round-off. This is not necessarily the case for round to nearest (RN), for which the worst-case error bound has constant nu. A particular attraction of SR is that, unlike RN, it is immune to the phenomenon of stagnation, whereby a sequence of tiny updates to a relatively large quantity is lost. We survey SR by discussing its mathematical properties and probabilistic error analysis, its implementation, and its use in applications, with a focus on machine learning and the numerical solution of differential equations. 1. Introduction Rounding is the act of mapping a given number to one having a certain number of digits in a given base. For illustration, consider the task of rounding, in base 10, a 2-significant-digit number to 1 significant digit. If we round to the closest 1-digit number, for example, then 1.4 rounds to 1 and 1.7 rounds to 2. We denote the rounding operator by fl, thus we write fl(1.4) = 1 and fl(1.7) = 2. This rounding rule, called round to nearest (RN), is deterministic: the value of fl(x) depends only on x, and repeating the rounding yields the same result. Suppose we want to compute 1 + 0.1 in 1-digit base-10 arithmetic. With RN we obtain fl(1 + 0.1) = 1. Another option is to round to either of the two nearest 1-digit numbers with a probability that depends on the distances to those numbers. If in our example we define fl(1 + 0.1) as 1 with probability 0.9 and as 2 with probability 0.1, then the expected result is 0.9 × 1 + 0.1 × 2 = 1.1, which is the exact answer. This probabilistic rounding is called stochastic rounding (SR), and this simple example demonstrates one way in which it can be useful in practice: by occasionally rounding up the value of a sum, SR avoids the phenomenon of stagnation, whereby a long sum of small quantities—0.1 in this example—is lost to rounding. SR was first proposed over 60 years ago, but until recently had proved useful only in rather specialized contexts. In the last 5 years or so, however, this rounding mode has enjoyed a resurgence of interest, mainly because of the increasing availability of low-precision floating-point arithmetic in hardware and the recognition that, compared with RN, SR can produce errors that grow more slowly with the problem size. By its very nature, SR ensures that rounding errors are random and so encourages cancellation of errors, and while this effect benefits all precisions, it is particularly important at low precisions, where numbers have perhaps only 3 or 4 significant decimal digits and error growth can quickly destroy all accuracy. The worst-case error bounds for SR are a factor 2 larger than those for RN, however, and so SR does not benefit all computations. The aim of this work is to survey SR, describing \| \| \| \| \| \| \| --- --- --- \| \| — \| its basic properties (§2), \| \| \| \| \| \| — \| its history (§3), \| \| \| \| \| \| — \| floating-point arithmetics to which it might be applied (§4), \| \| \| \| \| \| — \| how SR compares with RN as regards its basic properties, including ways in which SR is less satisfactory than RN (§5), \| \| \| \| \| \| — \| its probabilistic rounding error analysis (§6), \| \| \| \| \| \| — \| how it can be implemented (§7), and \| \| \| \| \| \| — \| how and why it is being used in applications (§8). \| \| \| \| \| 2. What is stochastic rounding? Let F denote a finite subset of ℝ. We denote by fl any rounding operator that maps numbers in ℝ to either of the two nearest numbers in F. For 𝑥∈ℝ, define the two rounding candidates ⌊𝑥⌋=max⁡{𝑦∈𝐹 : 𝑦≤𝑥}and⌈𝑥⌉=min⁡{𝑦∈𝐹 : 𝑦≥𝑥}, so that ⌊𝑥⌋≤𝑥≤⌈𝑥⌉, with equality throughout if x ∈ F. Note that when 𝑥∉𝐹, the two numbers ⌊𝑥⌋ and ⌈𝑥⌉ are adjacent in F. For 𝑥∈ℝ∖𝐹, SR is defined by fl⁡(𝑥)={⌈𝑥⌉,with probability 𝑞⁡(𝑥),⌊𝑥⌋,with probability 1−𝑞⁡(𝑥),2.1 where q(x) ∈ [0, 1]. The simplest choice is q(x) = 0.5, in which case we round up or down with equal probability, independently of x. As is customary in the literature [1,2], we call this less commonly used form of SR mode 2 SR. Another choice is to set in (2.1) 𝑞⁡(𝑥)=𝑥−⌊𝑥⌋⌈𝑥⌉−⌊𝑥⌋,2.2 which means that we round x to the next larger or smaller number y ∈ F with probability 1 minus the distance between x and y divided by ⌈𝑥⌉−⌊𝑥⌋. See figure 1 for an illustration. The choice (2.2) yields mode 1 SR, which is the most interesting SR mode from a numerical point of view. Unless otherwise stated, here SR means mode 1 SR. For the rest of this paper, we take F to be a floating-point number system, unless otherwise stated, as this is the case of greatest interest, but much of what we say is applicable to fixed-point arithmetic. 3. Early history of stochastic rounding To the best of our knowledge, the earliest proposal of SR was in a one-paragraph abstract of a communication presented by Forsythe in 1949 at the 52nd meeting of the American Mathematical Society . The abstract claims that SR can be used to reduce the accumulation of round-off errors observed by Huskey in solving a simple system of ordinary differential equations (ODEs). The numerical integration that Forsythe and Huskey consider entails a sum of real values which is further reduced to a sum of integers, most likely intended as fixed-point representations of reals. The suggestion is to perform this rounding by random round-off, a suggestive name for mode 1 SR. The abstract concludes by stating that numerical tests on some unspecified IBM equipment confirm that SR can eliminate the ‘peculiarities’ noticed by Huskey on the ENIAC.1 The first hardware implementation of SR we are aware of was described by Barnes et al. in 1951. The authors describe a digital computer with 8-digit decimal arithmetic and explain that using SR rather than RN in multipliers and dividers simplified the implementation. As their implementation rounds up or down with equal probability, this constitutes an early example of mode 2 SR. A note by Forsythe, originally written in 1950 and reprinted in 1959 (see , footnote 1), provides more details about the proposal to round stochastically when solving ODEs. The document suggests to implement mode 1 SR for decimal arithmetic as follows: On a decimal machine, instead of adding a 5 in the most significant position of the digits to be dropped (ordinary rounding off), one adds a random decimal digit to each of the digital positions to be dropped. As with ordinary rounding off, the addition carry-over determines whether the rounding off is ‘up’ or ‘down’. It is not clear whether this excerpt refers to a hardware implementation or to a modification that could be done in software on the computers of the time. This technique has been used in recent hardware implementations for rounding binary numbers [7,8]. In a 1966 paper, Hull & Swenson test various probabilistic rounding error models by comparing the results of stochastically rounded operations with the expected error predicted by the models. According to the description provided at the beginning of the section ‘Simulation of the Models’ [9, p. 109], however, the implementation of SR that Hull & Swenson consider differs from the one we examine. In order to round stochastically the result of an arithmetic operation, they first perform the operation in double precision arithmetic, then add a pseudo-random number between −1/2 and 1/2 of the unit in the last place of the most significant half of the double precision result. Subsequent calculations use the modified double precision value, which presumably includes the original quantity in its least significant half and the added random quantity. Despite the different spirit, we mention this contribution here as it is one of the earliest manuscripts we are aware of that considers non-deterministic rounding modes. 4. Floating-point arithmetics Before describing the finer details of SR, we recall some necessary background on floating-point arithmetic. We discuss the formats in the IEEE 754 standard for floating-point arithmetic and two other formats of practical interest, bfloat16 and TensorFloat-32. 4.1. IEEE 754 standard floating-point arithmetics The IEEE standard 754 for floating-point arithmetic was first released in 1985 and then revised in 2008 and 2019 . The standard dictates the encoding rules for binary and decimal floating-point data types, the precision and exponent range of some standard formats, and the accuracy requirements of basic arithmetic operations. It also prescribes how to handle exceptional cases and specifies a set of recommended mathematical functions that software and hardware floating-point libraries should provide in order to ensure a consistent numerical behaviour. Table 1 reports the parameters for four binary floating-point data types defined in the latest revision of the standard. Most hardware implements the data types binary32 and binary64, commonly known as single and double precision, respectively. Of the remaining formats, binary16 is defined only as a storage format, but it has been implemented as an arithmetic in hardware by several manufacturers. One other IEEE format is binary128, which is not listed in the table. While binary128 is mainly supported in software, it is also available in hardware on the IBM Power9 and z13 processors. Table 1. Parameters of various binary floating-point formats: number of digits of precision including the implicit bit (p), smallest positive representable normal (fmin) and subnormal (smin) numbers, and largest positive number (fmax). The ‘binaryxy’ formats are from the IEEE 754 standard. View inlineView popup Table 1. Parameters of various binary floating-point formats: number of digits of precision including the implicit bit (p), smallest positive representable normal (fmin) and subnormal (smin) numbers, and largest positive number (fmax). The ‘binaryxy’ formats are from the IEEE 754 standard. Collapse \| \| bfloat16 \| binary16 \| binary32 \| binary64 \| \| p \| 8 \| 11 \| 24 \| 53 \| \| fmin \| 2−126 \| 2−14 \| 2−126 \| 2−1022 \| \| smin \| 2−133 \| 2−24 \| 2−149 \| 2−1074 \| \| fmax \| 2127(2 − 2−7) \| 215(2 − 2−10) \| 2127(2 − 2−23) \| 21023(2 − 2−52) \| We now briefly recall some key aspects of IEEE floating-point number systems and the definitions and main properties of normalization and subnormal numbers. We focus on binary formats, since most commercially available hardware implements only binary arithmetic. A binary floating-point number x has the form (−1)𝑠×𝑚×2𝑒−𝑝+1, where s is the sign bit, p is the precision, m ∈ [0, 2p − 1] is the integer significand, and e ∈ [emin, emax], with emin = 1 − emax, is the integer exponent. In order for x ≠ 0 to have a unique representation, the number system is normalized so that the most significant bit of m—the implicit bit in IEEE 754 parlance—is always set to 1 if \|𝑥\|≥2𝑒min. Therefore, all floating-point numbers with m ≥ 2p−1 are normalized. Numbers that have absolute value below that of the smallest normalized number 2𝑒min are said to be subnormal: they have exponent e = emin, integer significand m < 2p−1, and therefore precision lower than that of normalized values (between 1 and p − 1 bits). Subnormal numbers provide the means to represent values in the subnormal range (−2𝑒min,2𝑒min), and are necessary in order to ensure that a floating-point number system satisfies Sterbenz’s lemma (described in §5) and has desirable properties such as gradual underflow. Because of the variable precision, however, subnormal numbers require special treatment in both software and hardware implementations of floating-point arithmetics. This is likely to cause performance and chip area overhead, and as a result it is not uncommon for hardware manufacturers not to support subnormal numbers. Two important numbers related to the precision p are the machine epsilon 𝜀𝑀=21−𝑝,4.1 which is the spacing of the floating-point numbers just to the right of 1, and the unit round-off 𝑢=2−𝑝=12⁢𝜀𝑀,4.2 which is an upper bound on the relative error incurred when a real value is rounded to a precision-p floating-point representation using RN. For further details, we refer the reader to [15, ch. 1] and [16, ch. 2]. The latest revision of the IEEE 754 standard defines six rounding modes, which are listed in table 2. Four rounding modes are required for a floating-point arithmetic to be compliant: round to nearest with ties to even (RN), round towards positive (or towards +∞, or up, RU), round towards negative (or towards −∞, or down, RD), and round towards zero (RZ). In the rest of the paper we refer to these modes collectively as ‘standard rounding’. Table 2. Rounding modes defined in the 2019 revision of the IEEE 754 standard . View inlineView popup Table 2. Rounding modes defined in the 2019 revision of the IEEE 754 standard . Collapse \| rounding mode \| description \| \| to nearest with ties to even (RN) \| round to a nearest floating-point value and if the two nearest floating-point values are equally close, round to the one with an even least significant digit. This is a default rounding mode. \| \| to nearest with ties to away \| round to a nearest floating-point value and if the two nearest floating-point values are equally close, round to the number with larger magnitude. Only required for decimal floating-point data types. \| \| to nearest with ties to zero \| round to a nearest floating-point value and if the two nearest floating-point values are equally close, round to the number with smaller magnitude. Only required for augmented operations [12, sec. 9.5]. \| \| towards positive (RU) \| round to a nearest floating-point value that is no less than the argument. \| \| towards negative (RD) \| round to a nearest floating-point value that is no larger than the argument. \| \| towards zero (RZ) \| round to a nearest floating-point value that is no larger in magnitude than the argument. \| The IEEE 754-2019 standard recommends extended or extendable precisions [12, §3.7] to enhance the basic formats listed in table 1. As an example, Intel provides an 80-bit extended precision format that has a 15-bit exponent and a 64-bit significand—the bit to the left of the radix point being stored explicitly in this case, as opposed to the IEEE 754 formats, which rely on the implicit bit convention and use the value of the exponent field to determine the leading bit of the significand. Arithmetic operations can be performed in higher precision and the results need not be rounded to binary64 until the final result of a computation leaves the higher-precision registers. Note that the use of 80-bit arithmetic is susceptible to double rounding, whereby a value may be rounded incorrectly to the final floating-point format when it is rounded to an intermediate format (extended precision, in this case) first [17,18]. Boldo & Melquiond have shown, using the Coq proof assistant, that double-rounding issues can be avoided if the extended precision format uses a deterministic rounding mode called round to odd [19,20]. 4.2. Non-IEEE arithmetics Among the non-IEEE floating-point formats implemented in recent hardware, we are particularly interested in those based on binary32: bfloat16 and TensorFloat-32, which lower the precision p from 23 to 8 and 11 bits, respectively. The main idea behind these formats is to reduce the memory and hardware arithmetic costs without narrowing the dynamic range; this contrasts with the aim behind the binary16 format, which allocates to the exponent field fewer bits than binary32 and therefore has a more limited dynamic range. Bfloat16, which was originally proposed by Google and formalized by Intel , is available on the Armv8 architecture , on the NVIDIA Ampere chips and on some Intel microarchitectures . TensorFloat-32 is a format used internally in the tensor cores (matrix multiply-accumulate units) of the NVIDIA Ampere microarchitecture . This 19-bit format is meant to be a low-precision replacement for binary32, but is not used for data storage and is not available in any other arithmetic unit on these GPUs. 5. Stochastic rounding versus round to nearest In RN, which is the default rounding mode in most floating-point arithmetics, fl(x) is the number in F nearest to x, with some tie-breaking strategy for handling the case where x is equidistant from the next and previous floating-point numbers. While SR and RN share some properties, they also differ in some important respects. We first describe three properties that SR and RN have in common. In fact, the first two properties hold for any rounding mode. \| \| \| \| \| \| \| --- --- --- \| \| — \| If x ∈ F then fl(x) = x, that is, rounding a floating-point number leaves it unchanged. \| \| \| \| \| \| — \| If x and y are floating-point numbers with y/2 ≤ x ≤ 2y then fl(x − y) = x − y (assuming subnormal numbers are supported). This result, known as Sterbenz’s lemma [15, thm. 2.5], [25, thm. 1.8.2], relies on the fact that x − y is exactly representable. \| \| \| \| \| \| — \| In base-2 arithmetic, any floating-point numbers x and y such that x ≤ y satisfy the inequalities 𝑥≤fl⁡(𝑥+𝑦)2≤𝑦, which do not hold for all bases [1, §3.1]. \| \| \| \| \| SR and RN differ in some key properties, however. For example, if 𝑥∉𝐹 then in general fl(\|𝑥\|)≠\|fl⁡(𝑥)\| and fl( − x) ≠ −fl(x) for SR. Moreover, SR is not monotonic: x < y does not imply that fl(x) ≤ fl(y), as can be seen by considering any pair of reals x and y lying between two adjacent floating-point numbers. Several results that describe how simple identities for real numbers are (partially) preserved for floating-point numbers under RN are well known. Generally similar, but weaker, results hold for SR. For a binary format with RN, for example, we have that fl⁡(√𝑥2)=\|𝑥\| if x is a floating-point number [15, Prob. 2.20], barring underflow and overflow. For SR, however, if x is a floating-point number in the interval (1, 2) then fl⁡(√𝑥2)∈{\|𝑥\|−𝜀𝑀,\|𝑥\|,\|𝑥\|+𝜀𝑀}, where ɛM is the machine epsilon defined in (4.1). A consequence of this fact is that the inequality fl⁢(𝑥/√𝑥2+𝑦2)≤1, which is always satisfied by RN [15, Prob. 2.21], is not necessarily true when SR is used. Kahan proved that if m and n are integers such that \|𝑚\|<2𝑝−1 (where p is the precision) and n = 2i + 2j for some i and j then fl(n × fl(m/n)) = m with RN [26, thm. 7]. Thus for example, fl(5 × fl(m/5)) = fl(17 × fl(m/7)) = m. Under SR, however, we can say only that fl(n × fl(m/n)) is either m, the next smaller floating-point number, or the next larger floating-point number. Full details of the above results, as well as other properties that differ between SR and RN, are given by Connolly et al. [1, §3]. Before replacing RN with SR it is vital to consider whether a certain computation relies on properties of RN that go beyond the standard model of floating-point arithmetic (6.2) below, and if so, whether these properties remain true for SR. The solution of the quadratic equation ax2 + bx + c = 0 is a striking example of the subtle issues that may occur when switching from RN to SR: if evaluated using SR, the discriminant b2 − 4ac can be negative even when b2 > 4ac. This is a consequence of the non-monotonicity of SR, and it could lead one to incorrectly conclude that a quadratic equation has no real solutions when in fact it has two (almost) identical real roots. The results in this section suggest that SR is less attractive than RN. It is the rounding error results described in the next section that account for the interest in SR. 6. Rounding error analysis with SR If 𝑥∈ℝ lies within the range of the floating-point number system F, it can be shown that for RN one has fl⁡(𝑥)=𝑥⁢(1+𝛿),\|𝛿\|<𝑢,6.1 where the unit round-off u is defined in (4.2). Rounding error analysis is usually based on the standard model of floating-point arithmetic [15, eqn (2.4)], which assumes that the elementary arithmetic operations are rounded to nearest (as is the case for IEEE standard arithmetic with the default rounding mode), so that if no overflow or underflow occurs they satisfy fl⁡(𝑥 op 𝑦)=(𝑥 op 𝑦)⁢(1+𝛿),\|𝛿\|≤𝑢, op∈{+,−,×,/}.6.2 Analogous models can be devised for unitary operations, such as the square root, or ternary operations, such as the fused multiply-add. The model (6.2) is customarily used in rounding error analysis, and is based on the premise that the result of a floating-point elementary arithmetic operation should be as accurate as the correctly rounded infinitely precise result. Error analysis based on (6.2) is referred to as worst-case analysis because it can only use \|δ\| ≤ u and cannot exploit possible cancellation of rounding errors across multiple operations or instances where δ = 0. For RN, the bound in (6.1) can be tightened to \|δ\| ≤ u/(1 + u) < u [16, thm. 2.3], [27, eqn (18)]. The model (6.2) can be improved accordingly, but this has no impact on the analysis discussed here. When multiple floating-point operations are performed in a sequence, rounding errors accumulate. For example, if s = x1y1 + x2y2 + x3y3 is computed in floating-point arithmetic, the computed ̂𝑠 satisfies ̂𝑠= ((𝑥1⁢𝑦1⁢(1+𝛿1)+𝑥2⁢𝑦2⁢(1+𝛿2))⁢(1+𝛿3)+𝑥3⁢𝑦3⁢(1+𝛿4))⁢(1+𝛿5)= 𝑥1⁢𝑦1⁢(1+𝛿1)⁢(1+𝛿3)⁢(1+𝛿5)+𝑥2⁢𝑦2⁢(1+𝛿2)⁢(1+𝛿3)⁢(1+𝛿5)+𝑥3⁢𝑦3⁢(1+𝛿4)⁢(1+𝛿5), for some δ1, …, δ5 of magnitude at most u. It is clear from this example that rounding error analysis for vector and matrix operations involves dealing with multiple terms of the form ∏𝑛𝑖=1(1+𝛿𝑖). The following lemma [15, Lem. 3.1] bounds the distance between 1 and the product of n terms of the form (1 + δi)±1 by means of the quantity 𝛾𝑛=𝑛⁢𝑢1−𝑛⁢𝑢, a ubiquitous constant in rounding error analysis. Lemma 6.1. If \|δi\| ≤ u and ρi = ±1 for i = 1, …, n, and nu < 1, then 𝑛∏𝑖=1(1+𝛿𝑖)𝜌𝑖=1+𝜃𝑛,\|𝜃𝑛\|≤𝛾𝑛.6.3 Under SR, we define the elementary floating-point operations +,−,×,/ to be the stochastically rounded exact ones. Therefore, for SR, (6.2) holds with u replaced by 2u: fl⁡(𝑥 op 𝑦)=(𝑥 op 𝑦)⁢(1+𝛿),\|𝛿\|≤2⁢𝑢, op∈{+,−,×,/}.6.4 Standard rounding error analysis based on the model (6.2) clearly remains valid for (6.4), with u replaced by 2u. This means that one will necessarily obtain larger worst-case error bounds for SR than for RN. However, SR injects randomness into the rounding errors and so we can intuitively expect it to cause the final error in a computation to be smaller than in the worst case, and hence possibly smaller than for RN in some circumstances. In the next subsection, we explain why this intuition is correct. 6.1. Probabilistic error analysis Modelling rounding errors as random variables to obtain probabilistic error bounds is an old idea going back to von Neumann & Goldstine , Henrici [29–31] and Hull & Swenson , among others. This line of thought has led to the following rule of thumb: a realistic bound on the rounding error of a linear algebra algorithm can be obtained by replacing, in a worst-case error bound, all constants that depend on the dimensions of matrices and vectors by their square roots. The key idea is to exploit the statistical effects of these random variables in the propagation of rounding errors; see, for example, [32, p. 318]. Higham & Mary provided the first rigorous proof of the validity of this criterion: they showed that for random independent zero-mean rounding errors δi, the constant γn in (6.3), can be replaced by ̃𝛾𝑛⁡(𝜆)=exp⁡(𝜆⁢√𝑛⁢𝑢+𝑛⁢𝑢21−𝑢)−1=𝜆⁢√𝑛⁢𝑢+𝑂⁡(𝑢2), with a modest constant λ > 0 with high probability. Subsequently, Higham & Mary and Ipsen & Zhou obtained a probabilistic error bound for inner products that only requires mean independence of rounding errors, an assumption weaker than independence. Connolly et al. derived the following probabilistic version of Lemma 6.1 under these assumptions [1, thm. 4.6]. Here, 𝔼⁡[𝑋] denotes the expectation of the random variable X. Theorem 6.2. Let δ1, δ2, …, δn be random variables of mean zero such that 𝔼⁡(𝛿𝑖∣𝛿𝑖−1,…,𝛿1)=𝔼⁡(𝛿𝑖)=0, for i = 2, …, n. If \|δi\| ≤ u and ρi = ±1, for i = 1, …, n, then for any constant λ > 0, 𝑛∏𝑖=1(1+𝛿𝑖)𝜌𝑖=1+𝜃𝑛,\|𝜃𝑛\|≤̃𝛾𝑛⁡(𝜆)6.5 holds with probability at least 1 − 2exp ( −λ2/2). For RN, the assumptions in theorem 6.2 on the zero mean and mean independence of rounding errors are often satisfied in practice, but it is easy to construct examples where either of the assumptions fails and the backward error almost attains the worst-case bound (6.3) (and thus exceeds the probabilistic bound (6.5) by a factor √𝑛). Importantly, however, SR always satisfies the conditions of theorem 6.2, as shown by the following result [1, lem. 5.2]. Theorem 6.3. Let the computation of interest generate rounding errors δ1, δ2, …, in that order. If SR is used then the δi are random variables of mean zero such that 𝔼⁡(𝛿𝑖∣𝛿𝑖−1,…,𝛿1)=𝔼⁡(𝛿𝑖)⁢(=0). It follows that the probabilistic bound (6.5) holds unconditionally for SR provided that is u replaced by 2u in view of (6.4). Hence for SR, the rule of thumb that one can replace nu in a worst-case error bound by √𝑛⁢𝑢 is a rule. The probabilistic bound is most favourable when the worst-case bound is approximately attained. One situation in which this happens is when many tiny increments are applied to a relatively large quantity. If ϕ ∈ F is updated by increments h1, h2, …, which have magnitude smaller than half of the spacing of the floating-point numbers around ϕ, then using RN gives ϕ = fl(ϕ + h1) = fl(fl(ϕ + h1) + h2) = ···, and the information in the updates is lost. This phenomenon, known as stagnation, commonly occurs in practical applications. It arises, for example, in neural networks, when parameter updates become very small, or in numerical methods for ODEs and partial differential equations (PDEs), when a very small time step is chosen. SR avoids stagnation, as some of the updates produce rounding that changes the partial sum. This can be seen from the following result [1, thm. 6.2]. Theorem 6.4 (inner products). Let y = aTb, where 𝑎,𝑏∈ℝ𝑛, be evaluated in floating-point arithmetic. Under SR, the computed ̂𝑦 satisfies 𝔼⁡(̂𝑦)=𝑦 regardless of the order in which the sums of products are evaluated. Taking bi ≡ 1 in the theorem, we see that the expected value of a sum is the true value under SR. As a simple example, suppose we run the code x = 1; for i = 1 : 10 do x = x + ɛM/4 end in floating-point arithmetic. Since the spacing of the floating-point numbers between 1 and 2 is ɛM = 2u, with RN every addition rounds down and the computed result is ̂𝑥=1. With SR, however, each addition has a probability 1/4 of rounding up, giving an increment of ɛM. Hence the expected result is 1 + 10 · (1/4) · ɛM, which is the exact result, albeit not a floating-point number in the working precision. We emphasize that the benefits of SR are not restricted to curing stagnation. This rounding mode ensures zero-mean rounding errors, so can produce smaller errors than RN in situations where RN systematically produces rounding errors of one sign (see [33, §4.2.2] for an example). 7. Implementation Here, we discuss how to implement SR. 7.1. SR expressed in terms of other rounding modes We can express the SR operator in terms of other rounding operators by writing, for 𝑥∉𝐹, fl⁡(𝑥)={RA⁡(𝑥),with probability 𝑞⁡(𝑥),RZ⁡(𝑥),with probability 1−𝑞⁡(𝑥),7.1 where RA denotes the operator that rounds away from zero and q(x) ∈ [0, 1]. For mode 1 SR, we can rewrite (2.2) as 𝑞⁡(𝑥)=𝑥−RZ⁡(𝑥)RA⁡(𝑥)−RZ⁡(𝑥).7.2 In order to implement (2.1) or (7.1) in practice, we need to define a discrete version of the SR operator. Given a positive integer k, which controls the number of bits used to approximate the continuous definition (2.1), let P be a random precision-k floating-point number drawn from the uniform distribution over the interval [0, 1).2 We have that, for 𝑥∉𝐹, fl⁡(𝑥)=⎧{⎨{⎩RA⁡(𝑥),if 𝑃<𝑥−RZ⁡(𝑥)RA⁡(𝑥)−RZ⁡(𝑥),RZ⁡(𝑥),if 𝑃≥𝑥−RZ⁡(𝑥)RA⁡(𝑥)−RZ⁡(𝑥),7.3 where SR, RA, RZ round to some precision p. It is worth noting that the choice of the optimal k for implementing SR is one of the main open questions surrounding mode 1 SR. A lower value of k makes a hardware implementation cheaper but is expected to reduce the accuracy benefit that SR may potentially bring: setting k = 1, for example, gives mode 2 SR. While (7.3) is an accessible definition of SR, most of the implementations discussed in this section do not take the comparison-based approach that (7.3) would suggest. On the contrary, they add bits from a random stream to the part of the number that will be truncated. The equivalence between this is the idea, which we will discuss in much greater detail in §§7.3 and 7.4, and (7.3) is shown in [37, §4]. Table 3 compares the features of SR in a number of implementations available. Table 3. Summary of SR implementations. Here, p is the target precision of rounding; k is the precision of the random number used for SR; the ‘type’ column has ‘I’ for integer or fixed-point arithmetic and ‘F’ for floating-point arithmetic; the ‘op.’ column indicates the class of operations supported or ‘any’ if rounding of any operation is supported; the ‘H/S’ column has H if SR is in hardware and S if software; and in the ‘applications’ column ML and QC stand for machine learning and quantum computing, respectively. View inlineView popup Table 3. Summary of SR implementations. Here, p is the target precision of rounding; k is the precision of the random number used for SR; the ‘type’ column has ‘I’ for integer or fixed-point arithmetic and ‘F’ for floating-point arithmetic; the ‘op.’ column indicates the class of operations supported or ‘any’ if rounding of any operation is supported; the ‘H/S’ column has H if SR is in hardware and S if software; and in the ‘applications’ column ML and QC stand for machine learning and quantum computing, respectively. Collapse \| reference \| type \| p \| k \| op. \| H/S \| applications \| \| Barnes et al. (1951) \| I \| 8 \| 1 \| ×, / \| H \| general \| \| Gupta et al. (2015) \| I \| 18 \| 30 \| dot prod. \| H \| ML \| \| Davies et al. (2018) \| — \| 7 \| — \| ×, + \| H \| ML \| \| Higham & Pranesh (2019) \| F \| ≤64 \| 64 \| any \| S \| general \| \| Hopkins et al. (2020) \| I \| 32 \| 2–32 \| × \| S \| ODE solve \| \| Mikaitis (2020) ([8,40] \| I, F \| ≤32 \| 32 \| any \| H \| general \| \| Meurant (2020) \| I, F \| ≤64 \| 64 \| any \| S \| general \| \| Fasi & Mikaitis (2020) \| F \| ≤64 \| 64 \| any \| S \| general \| \| Croci & Giles (2020) \| F \| ≤32 \| 32–64 \| +, × , / \| S \| general/PDEs \| \| Fasi & Mikaitis (2021) \| F \| p \| p \| +, ×,/,√ \| S \| general \| \| Paxton et al. (2021) \| F \| ≤32 \| 32–64 \| any \| S \| general \| \| Klöwer (2021) \| F \| ≤32 \| 32–64 \| any \| S \| general \| \| Krishnakumar & Zeng (2021) \| I \| n \| m = n \| × \| — \| QC \| 7.2. Proposed IEEE 754 style properties of SR The definition in the previous section does not cover edge cases such as overflow, underflow and rounding of infinities and NaNs (not-a-number). In the following, we propose our definition of SR for these edge cases by giving some properties of SR analogous to those of the rounding modes defined in the IEEE 754 standard . \| \| \| \| \| \| \| --- --- --- \| \| — \| If the exact number is in the range of the target format, SR should be performed as though the number was originally held in p + k bits and then rounded to p bits according to (2.1). Here, the extra k bits refer to the precision of SR, as well as the number of random bits required. \| \| \| \| \| \| — \| Overflows: if the exact number lies between the maximum representable number ±fmax and the neighbouring value that is not representable in the target format and will be treated as ±∞, SR is performed as though the value is representable, to preserve the statistical information about the round-off bits. \| \| \| \| \| \| — \| When the exact number is smaller than the smallest value representable in the target format, SR should round stochastically to one of the two neighbouring floating-point values in the target format, either zero or the smallest representable value, maintaining the sign. \| \| \| \| \| \| — \| When subnormals are disabled or not supported in the target format and the exact value is in the range of underflow, SR should round either to zero or to the smallest normalized value, again without changing the sign. \| \| \| \| \| \| — \| ±∞ and ±0 should not be rounded (changed) by the SR operation. NaNs with payloads that cannot be represented in the target format should not be stochastically rounded: a NaN with an implementation-defined payload may be returned, as per IEEE 754 [12, §6.2.3] (relevant in a mixed-precision setting, for example converting binary64 to binary32). \| \| \| \| \| \| — \| As in the standard rounding operations [12, §4.3], inexact, underflow and overflow exceptions should be signalled by the SR operation. \| \| \| \| \| 7.3. Modifying basic floating-point algorithms to include SR Now we discuss how to modify classical algorithms for addition and multiplication of floating-point numbers [16, ch. 7], [27, §4.2.1], [47, ch. 8] in order to obtain algorithms that support SR and can readily be implemented in software or hardware. Algorithms for other operations to include SR such as fused multiply-add (FMA) or division can be similarly derived by modifying the original algorithms for the IEEE 754 arithmetic operations [16,47]. To the best of our knowledge, these algorithms are new in that no general methods to round to precision p taking normalization into account had so far been proposed in the literature. Addition. The sum r = \circ(x + y), where \circ ∈ {RN, RZ, RD, RU} and x and y are binary floating-point numbers, can be computed as we now explain. Let 𝑥=(−1)𝑠𝑥×𝑚𝑥×2𝑒𝑥−𝑝+1 and 𝑦=(−1)𝑠𝑦×𝑚𝑦×2𝑒𝑦−𝑝+1 be two normalized precision-p floating-point numbers. We assume that sx = sy = 0, which implies that x and y are positive, in order to avoid considering sign interactions which may transform the addition into a subtraction. This restriction does not affect our main observations pertaining to the implementation of SR. The role that the sign of the operands plays in this algorithm is discussed, for instance, in [16, §7.3]. We make additional observations about subtraction when necessary. The sum 𝑟=∘(𝑥+𝑦)=(−1)𝑠𝑟×𝑚𝑟×2𝑒𝑟−𝑝+1 is computed as follows: \| \| \| \| \| \| \| --- --- --- \| \| (i) \| If ey > ex, swap x and y to ensure that ex ≥ ey. \| \| \| \| \| \| (ii) \| Alignment of the significands: compute 𝑚𝑦×2−(𝑒𝑥−𝑒𝑦) by shifting my to the right by ex − ey places. Set er = ex. It is not necessary to keep all the bits that are shifted out: maintaining only two bits plus a third sticky bit suffices—see below. \| \| \| \| \| \| (iii) \| Sum of the significands: compute 𝑚𝑡=𝑚𝑥+𝑚𝑦×2−(𝑒𝑥−𝑒𝑦). At this step, mt is an exact sum of the significands. \| \| \| \| \| \| (iv) \| Normalization of the result: since 0 ≤ mt < 2p+1, we may need to normalize the result by shifting mt to the right by one place (if mt ≥ 2p) and increasing er by 1 (note that a shift left may be needed with subtraction—see below). \| \| \| \| \| \| (v) \| Rounding: the significand of the rounded sum, mr, is computed by rounding the normalized exact sum mt to p significant bits according to ∘, and renormalizing if required. At this point, r is the correctly rounded sum of x and y. \| \| \| \| \| In order to perform the rounding at step (v) of the algorithm, it may seem necessary to preserve all the bits that, being after the bit in the pth position, are dropped off during steps (ii) and (iv). It can be shown, however, that for \circ ∈ {RN, RZ, RD, RU} it suffices to keep only the first two discarded bits after the one in position p plus an extra specially computed bit. These are the guard bit G, in position p + 1, the round bit R, in position p + 2, and the sticky bit T, in position p + 3, which is a logical OR of all the bits after the (p + 2)nd. Together, these three bits are called in short the GRT bits [47, §8.4.3]. In the algorithm above, they are formed in step (ii) and updated in step (iv) if normalization is required. We now explain how the algorithm should be modified in order to include SR as an option in step (v). Figure 2 demonstrates stochatic rounding of mt pictorially. We use the same notation as in §7.1, and use k to denote the number of bits used for rounding, or equivalently the number of bits in the random number used to perform the rounding. In step (ii), instead of computing the GRT bits from the shifted-out bits of my, we keep the k trailing bits beyond that in position p. Depending on the implementation, it might be necessary to manipulate appropriately those extra bits when subtraction is considered. An alignment of more than p + k − 1 bits is unnecessary, as in that case SR with k bits would not have any effect and the largest summand x would be returned unchanged. If a shift is required in step (iv), the whole p + k-bit significand has to be shifted in order to keep the trailing k bits correct after the normalization. In order to perform step (v), it is necessary to generate k bits from a stream of uniformly distributed random bits. This operation is expensive, but can be performed asynchronously at any point before reaching the last step, as it does not require any information about x or y. Finally, the k bits from the random stream are added to the k bits immediately following the first p bits of the normalized mt; if this operation leads to a carry-out, we increment the top p bits of mt by 1 and truncate the bits after the first p bits to form the rounded significand mr. Implementing SR by adding random bits to the fraction is almost universally used in the software and hardware implementations discussed in §7.4 and §7.5 below. We need to consider normalization and whether the k bits required for rounding could be altered by shifting. Shifting is necessary in three cases. First, when the addition of the significands causes a carry-out, mt is normalized by means of a one-place shift to the right, which does not violate the bottom k bits. Secondly, when the difference of exponents ex − ey is larger than 1, the smaller operand is aligned so that there are multiple zeros at the front and consequently only a left shift by one position may be required on effective subtraction. For this reason, one extra bit is needed to make sure that a 1 that drops off the p + k bits is shifted in correctly by the left shift. Thirdly, when the exponent difference is 1 cancellation may occur on effective subtraction and multiple left shifts may be required to normalize the result. Since the alignment was performed by shifting right by only one place; however, there is no risk of any bits being shifted beyond the (p + k)th position, and therefore no incorrect bits will be shifted in during the normalization. Therefore, only one extra bit is necessary, and the width of mt should be p + k + 1 bits. If the sum of two floating-point numbers is subnormal, then it is exact and no rounding is required [16, thm. 4.2]. If one of the inputs is subnormal, then the significand alignment step requires minor modifications as per [16, §7.3.3], while the rest of the algorithm for the floating-point addition with SR remains the same. Addition of floating-point numbers can result in the following exceptions: overflow, underflow, inexact and NaN [47, p. 425]. With SR these may be handled as discussed in §7.2. Multiplication. Given two normalized positive floating-point numbers x and y as in the previous section, the product r = \circ (x × y) can be computed as follows: \| \| \| \| \| \| \| --- --- --- \| \| (i) \| Product of the significands: compute the 2p-bit integer mt = mx × my. The fact that 2p−1 ≤ mx, my < 2p implies that 22p−2 ≤ mt < 22p. \| \| \| \| \| \| (ii) \| Sum of the exponents: compute er = ex + ey. At this point, we have the exact product 𝑚𝑡×2𝑒𝑟−2⁢𝑝+2. \| \| \| \| \| \| (iii) \| Normalization of the result: if mt ≥ 22p−1 we need to normalize the result by shifting the significand by one place to the right and increasing er by 1. \| \| \| \| \| \| (iv) \| Rounding: the normalized exact product mt is rounded to p significant bits according to ∘, giving mr. At this point, mr is the rounded product of x and y. \| \| \| \| \| As with addition, the G and T bits are required to perform the rounding in step (iv) (the R bit is not required since left shifts cannot occur here). After the normalization in step (iii), the unrounded result will be in the top p bits, whereas the bottom p bits will be used for rounding. In order to implement SR with k random bits (in this case k ≤ p, as mt has at most 2p bits and there is no need to consider larger values of k) we need to add a k-bit random number to the top k of the bottom p bits of the internal significand mt: a carry will increment the pth bit of the top segment of mt, causing the number to round up. The significand of the stochastically rounded result will consist of the top p bits of mt. For subnormal values, a few modifications have to be incorporated in the multiplication algorithm. First of all, if both inputs are subnormal, then the product will be in the underflow range and SR can be handled as in §7.2. If the inputs are normalized but yield a subnormal result, it is necessary to shift mt right by more than one place in step (iii), depending on how much the product’s exponent differs from that of subnormals in the target format. If only one of the inputs is subnormal, then the product can be either normal or subnormal. Two approaches are taken in this case: either the subnormal input is normalized before the multiplication, or the product’s significand is normalized by a left shift. We do not go into details, as this does not change the SR algorithm (SR is performed in step (iv) after the product has been normalized or denormalized as required), and refer the reader to [16, §7.4.2]. Multiplication of floating-point numbers can result in the following exceptions: overflow, underflow, inexact and NaN [47, p. 438]. With SR these may be handled as discussed in §7.2. 7.4. Simulation of SR in software SR can be simulated in software in a straightforward fashion by relying on high-precision floating-point arithmetic. The computation is performed in higher precision and the high-precision result is rounded using (7.1), where q(x) in (7.2) is based on the (higher-precision) approximation to x rather than on its exact value. This approach is easy to implement as long as higher-than-working-precision arithmetic is available, be it in hardware, for instance when emulating binary32 rounding using binary64 arithmetic, or in software, through arbitrary precision libraries such as the GNU Multiprecision Library (GNU MPFR) . In practice, once the high-precision solution has been computed, the rounding step can be performed in several ways. The Matlab function chop3 and the FLOATP_Toolbox4 for Matlab leverage the Matlab random number generator to draw a random number r from the uniform distribution over the open interval (0, 1) and choose the rounding direction depending on whether r is larger or smaller than q(x). These software packages directly implement SR through floating-point comparison operations as in (7.3). Most software favours the use of integer random numbers, integer arithmetic and bit manipulation. The implementation of SR in the QPyTorch5 package , for example, rounds stochastically a binary32 number y to a floating-point format with precision p < 23 as follows. First, it generates a 32-bit integer q by zeroing out the leading (p − 1) + 9 bits of a 32-bit random integer. These zeros are introduced in positions that correspond to the sign of y (1 bit), its exponent (the next 8 bits), and the p most significant bits of its significand (the following p − 1 bits, in view of the implicit bit conversion). Next, the algorithm computes 𝑛=̃𝑦+𝑞, where ̃𝑦 is the binary representation of y seen as an integer, uses a bitmask to zero out the 24 − p trailing digits of n, as 24 is the number of precision bits in a binary32 number, and finally returns the value thus obtained as a 32-bit floating-point number. An example illustrating this rounding technique is provided in figure 3. The implementations in the CPFloat6 C library use an analogous technique when rounding binary32 as well as binary64 floating-point numbers to lower precision. The same approach is followed by Verificarlo7 , an instrumentation tool which uses the GNU Compiler Collection (GCC) quad format as extended precision for binary64, and binary64 as extended precision for binary32. The approach of simulating SR through extended precision is also used in the mcaquad back-end of the Valgrind tool Verrou8 [51,52]. Verrou offers a second back-end, called verrou, which performs SR without using higher precision. The algorithms in the verrou back-end use double-double arithmetic, and emulate high-precision computations implicitly by representing numbers with at least 106 bits of precision as the unevaluated sum of two binary64 values. These techniques are based on reduction operations [12, §9.4], also known as error-free transformations, and are used to approximate the distance between the exact result of a computation and the two rounding candidates in binary64 arithmetic. These are then used to approximate the rounding probabilities. Fasi & Mikaitis propose a similar but more general approach for implementing in software stochastically rounded elementary arithmetic operations: addition/subtraction, multiplication, division and extraction of the square root. For each operation, they propose two algorithms, one that uses only RN, and one that combines it with RZ, RU and RD. These algorithms compute first the result of the elementary arithmetic operation and then the error induced by the rounding. The rounding direction is chosen by adding a pre-generated random number to the computed error, and then adding this quantity to the result of the operation—Fasi & Mikaitis show that this procedure is equivalent to the direct implementation of SR via (7.3), which relies on a comparison-based approach. In numerical experiments, both types of algorithms are faster than a C implementation that relies on the GNU MPFR library, and the RN-only versions are faster on x86 architectures, where switching the rounding mode incurs a high performance penalty [16, §12.3.2]. Klöwer’s Julia software package StochasticRounding.jl9 defines three new Julia floating-point types that automatically include SR. These correspond to bfloat16, binary16 and binary32 (table 1), and use the Xoroshiro128Plus fast pseudo-random number generator (PRNG).10 Composability and type flexibility in Julia enable SR computations in single and half precision in a large number of numerical software and mathematical libraries. Automatic application of SR operations is extremely advantageous from a user standpoint, as it allows significant code simplification. To implement (7.3), the implementation is also based on the approach of adding random bits to the part of the floating-point number that will be truncated. In terms of fixed-point arithmetic with SR, Hopkins et al. and Mikaitis have recently implemented a set of rounding and multiplication operations11 and used them on low-power ARM integer processors. Multiplication routines for various fixed-point formats in the ISO 18037 embedded C standard were developed by exploiting the fact that ARM processors return the full-precision result of integer multiplication using two registers: multiplying two 32-bit fixed-point values, for example, returns the exact 64-bit result with all the information of integer and fraction bits of products preserved. The bottom bits of the fraction can then be used to round the results to one of the standard fixed-point formats stochastically. In this implementation, the comparisons between the random numbers and the round-off bits are implemented directly as in (7.3), except using integer arithmetic. 7.5. Overview of available devices and patents Now we review hardware designs discussed in the literature, some of which are already available in commercial hardware. The Graphcore Intelligence Processing Unit (IPU) is a highly parallel machine learning accelerator that supports SR for binary32 and binary16 arithmetic [54, §2.1], [55, ch. 10], (table 1). The patent filed by Graphcore reveals some technical details that are not specified in the documentation but may reflect the hardware implementation of the IPU. The document explains how binary32 values are stochastically rounded to binary16 precision in hardware by using a PRNG, also implemented in hardware—this kind of conversion might be performed in the IPU, although this is not reported. The algorithm begins by generating a 24-bit random number, that is, one random bit for each bit in the significand of a binary32 value. It then uses 13 or more of those bits to round a number to binary16. The number of random bits that are actually used depends on whether rounding will result in a normal or subnormal number. The actual rounding is performed by adding these random bits to the part of the significand that will be discarded. This operation may generate a carry bit to be added to the least significant bit of the significand of the input truncated to 11 significant bits. A non-zero carry bit will cause the absolute value of the number to round up. This type of implementation is used in most of the references mentioned below. Two patents filed by IBM disclose methods for implementing floating-point adders and multipliers that use SR. The authors demonstrate the techniques on an 8-bit data type, but mention binary32 and binary64 as examples of other formats to which the approach can be applied. The procedures require a fixed number of random bits be loaded into a register, but the patents do not explicitly mention how these bits should be generated. SR is performed by using the bits that drop off when fitting the result of the adder or multiplier into the 8-bit format, and as in the case of the Graphcore IPU, the rounding step is implemented by adding the random bit stream to the round-off bits, in both the adder and the multiplier . It is not mentioned, however, whether the sum is normalized before being rounded, and if so whether the bits that are to be shifted out during the normalization are also taken into account when rounding—in the algorithms outlined in §7.3 we propose that the rounding step should be performed after the normalization for a correct implementation. A patent from AMD describes methods and circuits to use SR in conjunction with integer adders or accumulators . The document shows the design of (i) a mixed-precision adder that computes the sum of a 32-bit and a 16-bit number by using a 16-bit random number supplied through a third input lane, and (ii) a 32-bit accumulator which takes as inputs both the next 16-bit value to accumulate and a 16-bit random number, and returns a 16-bit stochastically rounded sum—once again, SR is implemented by adding random bits to the round-off digits, as was the case for the Graphcore IPU. The pseudo-random numbers in the proposed SR unit are generated with a linear-feedback shift register (LFSR), but no specific algorithm is mentioned. A patent from NVIDIA demonstrates a method to round stochastically floating-point values to lower precision, using a fixed, programmable, or computable rounding bit position . The authors explain how to round binary64 values to binary32, and binary32 values to binary16 and bfloat16. A distinctive feature of this design is that it performs SR by using the bottom bits of the significand of the number to be rounded without relying on a random number generator [61, fig. 2B]. For example, the 23-bit fraction of a binary32 number can be rounded to the 10-bit fraction of a binary16 value by setting k = 8, taking the bottom 8 bits of the fraction in place of a random bit stream, aligning and adding them to the significand as in figure 2, and finally setting the 13 least significant bits to zero. The authors note that this method for performing SR has an advantage over using real random numbers, since it is deterministic and cheaper to implement. They do not mention, however, whether replacing the random number with part of the input causes SR to lose any of its desirable properties. Gupta et al. discuss the hardware prototype of a fixed-point matrix multiplier based on a two-dimensional (2D) systolic array architecture and demonstrate experimental results from a field-programmable gate array (FPGA) implementation. Each node of the systolic array is a multiply-and-accumulate digital signal processing (MACC DSP) unit that multiplies two integers and accumulates the result into an internal register. Each element of the matrix product is produced by a single MACC DSP unit. The hardware is generalized, but the authors report results for an implementation in which each MACC DSP unit accepts inputs of at most 18 bits and accumulates the partial results in an internal 48-bit register. When the matrix product is computed, each 48-bit element is passed through an SR unit (there is one for each column of the 2D array of MACC DSP units) to produce the 18-bit rounded and saturated results. The pseudo-random numbers needed to implement SR are generated using an LFSR. The 30 random bits are added to the 30 least significant bits of the 48-bit internal register. This may cause a carry to propagate to the 18 most significant bits of the result. Finally, the trailing 30 bits are set to zero, thereby producing the rounded number. The Intel Loihi and the SpiNNaker2 [8,40,62] digital neuromorphic processors include SR. The Intel Loihi processor has multiply-accumulate hardware that computes a 7-bit approximation to x[t] = α · x[t − 1] + δ · s[t] (with s[t] ∈ {0, 1}, α a decay factor, and δ an impulse amount added at each step). It is not specified where SR is applied in this computation, what precision and type of random numbers are used, and how SR is implemented. The SpiNNaker2 SR accelerator rounds and saturates 64-, 32- or 16-bit to 32- or 16-bit fixed-point numbers with SR. As a special case, it also includes rounding from IEEE 754 binary32 to bfloat16. The random bits needed for rounding are produced using the 32-bit hardware pseudo-random number generator available on SpiNNaker2 . The number of bits to be used for rounding is programmable for fixed-point formats (it can be anything between 1 and 32 bottom bits of the input) and is fixed to 16 for binary32-to-bfloat16 rounding. As in the implementations mentioned above, SR is performed by adding the random bits to the round-off bits (figure 2). 8. Applications In applications, SR can replace existing rounding modes (usually RN) either globally or in certain parts of an algorithm, and either true random numbers or pseudo-random numbers can be used. The latter are often preferred as they ensure reproducibility of the result. In this section, we review applications where SR has been applied, and in some cases provide Matlab experiments for demonstration, which are made available on GitHub.12 8.1. Numerical linear algebra For most numerical linear algebra algorithms, rounding error analysis builds on Lemma 6.1, or some variation of it, thus these algorithms can benefit from the smaller bound guaranteed for SR by the probabilistic error analysis of Theorem 6.2. For inner products, in particular, we have the following result [1, thm. 4.8]. Theorem 8.1 (inner products). Let y = aTb, where 𝑎,𝑏∈ℝ𝑛, be evaluated in floating-point arithmetic with SR. Then for any constant λ > 0 the computed ̂𝑦 satisfies ̂𝑦=(𝑎+Δ⁢𝑎)𝑇⁢𝑏=𝑎𝑇⁢(𝑏+Δ⁢𝑏),\|Δ⁢𝑎\|≤̃𝛾𝑛⁡(𝜆)⁢\|𝑎\|, \|Δ⁢𝑏\|≤̃𝛾𝑛⁡(𝜆)⁢\|𝑏\|8.1 with probability at least 1 − 2nexp ( − λ2/2) regardless of the order of evaluation. The worst-case error bound corresponding to (8.1) has the same form but with the constant ̃𝛾𝑛⁡(𝜆) replaced by γn, which is roughly a factor √𝑛 larger [15, §3.1]. As a special case, we can take bi ≡ 1 and deduce that ∣𝑛∑𝑖=1𝑎𝑖−fl⁡(𝑛∑𝑖=1𝑎𝑖)∣≤̃𝛾𝑛⁢𝑛∑𝑖=1\|𝑎𝑖\|. Figure 4 plots the relative errors for the sum ∑𝑛𝑖=1fl⁢(1/𝑖) computed in binary16 (table 1) with RN and SR for a range of n. Note that the summands are already converted to binary16 (with RN), so the only errors are in the summation. This example models a very slowly growing sum of decaying summands. We see that SR has much smaller errors than RN for larger n and that the errors for SR are mostly well within the probabilistic bound with λ = 1. Matrix products are considered in the following result [1, thm. 4.9], in which we denote by aj the jth column of a matrix A. Theorem 8.2 (matrix–matrix products). Let C = AB with 𝐴∈ℝ𝑚×𝑛 and 𝐵∈ℝ𝑛×𝑝 be evaluated in floating-point arithmetic with SR. For any λ > 0, the jth column of the computed ̂𝐶 satisfies ̂𝑐𝑗=(𝐴+Δ⁢𝐴𝑗)⁢𝑏𝑗,\|Δ⁢𝐴𝑗\|≤̃𝛾𝑛⁡(𝜆)⁢\|𝐴\|, 𝑗=1,…,𝑛,8.2 with probability at least 1 − 2mnexp ( − λ2/2), and hence \|𝐶−̂𝐶\|≤̃𝛾𝑛⁡(𝜆)⁢\|𝐴\|⁢\|𝐵\|,8.3 with probability at least 1 − 2mnpexp ( −λ2/2). The worst-case error bounds corresponding to (8.2) and (8.3) have the same form but with ̃𝛾⁡(𝜆) replaced by γn [15, §3.5]. This result is illustrated in figure 5, which plots the backward error for computing a matrix–vector product y = Ax where 𝐴∈ℝ100×𝑛 has entries drawn from the uniform distribution over [0, 10−3] and 𝑥∈ℝ𝑛 has entries sampled from the uniform distribution over and [0, 1]. We see that RN attains its worst-case rate of error growth and hits a relative error of 1, whereas SR has slower error growth and maintains some accuracy for all n. In fact, stagnation (see §6.1) occurs in this example when 𝑛≳103 for binary16 and when 𝑛≳102 for bfloat16, as shown by the increased rate of error growth from these points onwards for RN. As this rounding error analysis and the examples illustrate, SR is especially useful for large-scale and/or low-precision computations. 8.2. Machine learning The use of SR in neural networks is not a new idea. Höhfeld & Fahlman [63,64] proposed it in 1992, calling it probabilistic rounding. Today, SR is being used in machine learning in conjunction with half precision arithmetic, not least because of its ability to avoid the problem of stagnation that affects RN. Gupta et al. show that SR can be used for training deep neural networks in 16-bit fixed-point arithmetic with little or no degradation in the classification accuracy. Su et al. successfully train neural networks in 8-bit fixed-point arithmetic using SR and offer some suggestions as to why SR is beneficial in this context. Muller & Indiveri use SR (called randomized rounding or online stochastic) to map continuous neural network weights to a discrete low-precision fixed-point representation. It is shown that with SR the networks can perform well with lower precision weights than required with the standard rounding. Essam et al. combine SR with dynamic precision fixed-point arithmetic formats (variable integer scaling factors) in training neural networks. Na et al. implement SR with dynamic fixed-point arithmetic in hardware for neural network applications. Used to train neural networks with 24-bit fixed-point numbers, SR provides performance similar to that of binary32, but with lower hardware area and energy costs. The authors also show that without SR even 64-bit arithmetic is not enough to train the kinds of neural networks they used. Joardar et al. implement a 32- to 16-bit fixed-point arithmetic SR unit in an in-memory computing device for neural network training, based on resistive random access memory (ReRAM), which uses an LFSR 16-bit pseudo-random number generator. They report that SR added negligible ReRAM cell area overhead with each SR circuit adding less than 1%. SR was implemented by adding a 16-bit pseudo-random number to the bottom 16 bits of a 32-bit addition result and then truncating the result to 16 bits—the same technique as used in most of the hardware designs mentioned in §7. SR has also been applied in machine learning with floating-point arithmetic. Wang et al. train neural networks in 8-bit floating-point arithmetic with SR, obtaining factor 2–4 speed-ups over 32-bit training. Zamirai et al. find that either of SR and compensated summation [15, §4.3] enables training in bfloat16 to match 32-bit training. Xia et al. use mode 2 SR and show faster convergence in training using 16-bit fixed-point numbers. They modify mode 2 SR, in that instead of rounding only the values that are not representable, all values are rounded, including those that are exactly representable in the current floating-point format. Mellempudi et al. show that training neural networks using SR with 8-bit floating-point numbers yields performance comparable to that of binary32. Ortiz et al. compare 12-bit fixed- and floating-point formats with and without SR with binary32 arithmetic in the training of neural networks. They find that SR can be very useful in improving the accuracy: in their experiments with a 12-bit fixed-point format, using RN produced a training accuracy of just 10%, while switching to SR produced a training accuracy on par with that of binary32 arithmetic [74, table 2]. Zhang et al. use SR with the block floating-point (BFP) [76, p. 26] number representation technique in a training system for deep neural networks. Their system exploits BFP of variable precision over the whole training process, based on the number of training iterations and neural network layers used. In this work, SR is applied in conversion of floating-point values to a lower precision BFP representation [75, §III]. There is some indication [75, fig. 4] that SR is implemented, similarly to most implementations reviewed here, by adding random bit streams to parts of numbers later truncated (as shown in figure 2). A hardware implementation of the proposed training method is also presented, with a matrix multiplier in BFP with SR [75, §V]. To perform SR in hardware, the authors use 8-bit random binary streams from an LFSR which they sum with the significands of BFP numbers [75, §V-C], followed by truncation of the bottom bits. The authors report a 2-6× speed-up over prior approaches when using variable precision BFP for this application. At the same time, they report having similar accuracy to previous approaches, which is attributed to the combination of variable precision BFP and SR. We note that the survey by Wang et al. of custom hardware for deep learning includes a review of work that uses SR. Another survey by Lee et al. reviews arithmetic- and implementation-level techniques in deep learning and includes various works that use SR. There are many more examples of SR being of use in low-precision machine learning applications. For more details, see the references cited in the papers discussed above. 8.3. Numerical verification software Numerical verification software uses SR to explore the propagation of rounding errors in applications: a particular computation is repeated multiple times and the distribution of errors from these runs is used to draw conclusions about the sensitivity of a code to rounding errors. Mode 2 SR, known as stochastic arithmetic , is used for example in the CADNA library . An approach that includes as options both mode 1 and mode 2 SR is Monte Carlo arithmetic [81,82], a method used by tools such as Verificarlo and Verrou [51,52]. Monte Carlo arithmetic is more general than SR, not least because as well as randomly rounding the result of a floating-point operation it can also randomly perturb its inputs and output. 8.4. Ordinary differential equations The analysis of rounding errors in ODEs typically follows the classical convergence theory of time-stepping methods [29,83], in which the global error introduced by the ODE integration procedure is expressed in terms of the local errors introduced at each time step, and the global error is bounded in terms of the step size h, which we will assume is fixed. These local errors comprise both (local) truncation errors and (local) rounding errors. It is clear that unless the contribution to the global error from rounding errors decays to zero at the same rate as the contribution from truncation error, the overall convergence of the method can be impacted. Unfortunately, the analysis by Henrici [29,83] shows that the global error of an order-p ODE solver under RN is O(uh−1 + hp). The O(uh−1) term is often overlooked in the literature, and indeed in binary64 arithmetic the unit round-off u is usually small enough to make it negligible for the step sizes h of interest, but the O(uh−1) term cannot be neglected when computations are performed in reduced precision arithmetic. While not explicitly mentioning SR, early work by Henrici in the 1960s [29,83] and by Arató in the 1980s considers rounding errors arising in ODE solvers as independent (rather than mean-independent) random variables of zero mean. Henrici indicates that, whenever rounding errors have this random structure, the term O(uΔt−1) can be replaced with a term characterized by a milder growth in Δt. The analysis by Arató in rewrites the problem of estimating the global rounding error as the solution of a stochastic differential equation. It is curious that stochastic differential equations have not yet appeared in the actual analysis of SR errors for ODEs and PDEs. It has been shown experimentally that SR can alleviate the accumulation of rounding errors in ODE solvers. Hopkins et al. and Mikaitis use, on an ODE that models neurons in two configurations, four different solvers including RK2 Midpoint and RK3 Heun. They compare the results obtained in fixed-point arithmetic with those obtained using the same solvers run in binary32 and binary64 arithmetics. For the fixed-point solvers, they consider three rounding variants in the multiplication operation: bit truncation, RN and mode 1 SR. In the experiments, 32-bit fixed-point arithmetic with SR in multipliers shows accuracy similar to that of binary64 arithmetic in all cases, while fixed-point arithmetic with RN and bit truncation, as well as binary32 arithmetic, accumulate significant errors in the progression of the ODE system, ending up with a very different neuron behaviour. Floating-point arithmetics (binary16, bfloat16, binary32) with SR in adds and multiplies have been considered by Fasi & Mikaitis . ODEs exhibiting exponential decay were solved with Euler, midpoint and Heun solvers. For very small time steps, where rounding errors dominate the overall error of the solution, using SR produced a final solution error lower than that of RN. Figure 6 shows this for the solution errors with the forward Euler method solved in various arithmetics. The ODE system {𝑢′⁡(𝑡)=𝑣⁡(𝑡),𝑢⁡(0)=1,𝑣′⁡(𝑡)=−𝑢⁡(𝑡),𝑣⁡(0)=0,8.4 whose solution is the unit circle in the (u, v) plane, was solved using binary16 and bfloat16 arithmetics for increasingly smaller integration steps [37, §8.3.2]. The results in figure 7 for the forward Euler method with h = 2π/n demonstrate that with RN the computed solutions are not meaningful for very small integration steps, while with SR the computed solution reproduces the unit circle quite well. 8.5. Partial differential equations Little is known about the interplay between SR and the typical algorithmic components of PDE solvers, namely sparse iterative solvers, preconditioning, optimization and time-stepping methods. Croci & Giles analyse the effects of RN and SR in the solution of the heat equation with Runge–Kutta methods and finite differences, and explain how the numerical scheme should be implemented in order to reduce rounding errors. The analysis for RN yields the same O(uΔt−1) rate in all dimensions as in Henrici’s work on ODEs [29,83], and in related work on the heat equation by Jézéquel . On the other hand, using SR yields considerably smaller error bounds. In fact, Croci & Giles prove that the leading-order component of the rounding errors introduced by SR are zero-mean random variables that are independent in space and mean-independent in time. Thanks to this lack of correlation, much milder blow-up rates are obtained for the global rounding errors. These rates are essentially O(uΔt−1/4) in one dimension (1D), O(u\|log(Δt)\|1/2) in two dimensions (2D), and O(u) in three dimensions (3D). Interestingly, rounding errors become asymptotically smaller as the dimension increases: the larger the dimension, the more error cancellation occurs because of the spatial independence of the SR errors. The lack of error correlation and the zero-mean property are not solely responsible for the success of SR for this problem. Croci & Giles also show that the RN solution is prone to stagnation, and in fact the phenomenon may occur from the very first step if Δt is small enough to cause the RN solution to never move away from the initial condition. On the other hand, SR is resilient to stagnation, which did not affect the SR solution in numerical experiments. In unpublished work, these results have been extended to linear parabolic PDEs and the finite-element method and, together with numerical experimentation in binary16 and bfloat16 precision, show that while RN can fail to compute meaningful solutions, SR computations always exhibit near-machine-precision accuracy for sufficiently small time steps and mesh sizes. We expect similar results as in the parabolic case to hold for hyperbolic PDEs, with the exception perhaps of the stagnation behaviour. Here, we consider a diffusion equation with Dirichlet boundary conditions, {𝑢𝑡⁡(𝑡,𝑥)=∇⋅(𝐷⁡(𝑥)∇𝑢⁡(𝑡,𝑥))+𝑓⁡(𝑥),𝑥∈𝒟=(0,1)𝑑,𝑡∈[0,1],𝑢⁡(0,𝑥)=𝑢0⁡(𝑥),𝑢⁡(𝑡,𝑠)=1,𝑠∈∂𝒟,𝑡∈[0,1],8.5 where 𝐷⁡(𝑥)=⎧{⎨{⎩12⁢(sin⁡(𝜋⁢𝑥)2+1),in 1D,13⁢[sin⁡(𝜋⁢𝑥1)2+1sin⁡(𝜋⁢𝑥1)⁢cos⁡(𝜋⁢𝑥2)sin⁡(𝜋⁢𝑥1)⁢cos⁡(𝜋⁢𝑥2)cos⁡(𝜋⁢𝑥2)2+1],in 2D, and f(x) is chosen so that the exact solution to (8.5) at steady state is 𝑢⁡(∞,𝑥)={16⁢(𝑥⁢(1−𝑥))2+1,in 1D,(16⁢𝑥1⁢𝑥2⁢(1−𝑥1)⁢(1−𝑥2))2+1,in 2D. By using bfloat16 arithmetic with RN and SR, in figure 8 we show the effect of stagnation on the numerical steady-state solution of this problem in 1D as we vary the initial condition. We solve (8.5) using the finite-element method with piecewise linear basis functions and the forward and backward Euler schemes. We note that the RN solution always stagnates close to the initial condition, while SR successfully captures the correct steady-state solution. In figure 9, we plot the relative (i.e. normalized by the unit round-off) global rounding errors for both RN and SR against the theoretical bounds of Croci & Giles . While the RN error indeed grows linearly with Δt−1 until stagnation, the SR error increases very mildly in 1D and almost unnoticeably in 2D. For a similar 3D problem (not shown), the errors are just bounded by a multiple of the unit round-off. It therefore seems that SR is able to control the growth of rounding errors without requiring more accurate summation techniques such as that in . Paxton et al. investigate experimentally the effects of RN and SR in chaotic ODE and PDE systems related to climate modelling: the Lorenz system, heat diffusion, a nonlinear shallow-water model for turbulent flow over a ridge, and a coarse resolution global atmospheric model with simplified parametrizations. They simulate these models in various precisions using from 62- to 10-bit floating-point formats and compare their results via the Wasserstein distance, a metric used to measure the discrepancy between probability distributions. They find that SR can effectively mitigate the growth of rounding errors in both simple heat diffusion and turbulent models. Furthermore, they report the occurrence of stagnation when RN is used to solve the heat equation, confirming the results in . Overall, the findings by Paxton et al. show that reduced precision with SR is a valid alternative to standard binary64 precision computations. The authors also suggest that SR might become relevant in next-generation climate models. 8.6. Quantum mechanics In quantum mechanics, an integer variant of SR has been used by several authors in order to estimate the dominant eigenvalues of Hamiltonian matrices using Monte Carlo versions of the power iteration. The goal is to compute the ground state eigenvector φ0 of a Hamiltonian matrix H as a linear combination of a set of basis states \|0〉 , …, \|n〉. The coefficients of the linear combination, or basis-state amplitudes, are the inner products ci = 〈i\|φ0〉. At each step of the power method, the coefficient ci is approximated by an integer 𝑛(𝑘)𝑖, and for all i the approximations at step k + 1 are computed from those at step k. Once the iteration has converged numerically, the basis-state amplitude for the state \|i〉 is estimated as the average value of 𝑛(𝑘)𝑖 over k. Nightingale & Blöte are the first to suggest the use of SR for the solution of this problem. They use a random-walk model, and in their work the integers 𝑛(𝑘)𝑖 count the number of random walkers that are in state \|i〉 at iteration k. The integer SR function used in this work is fl⁡(𝑥)={[𝑥]+1,with probability 𝑥−[𝑥],[𝑥],with probability 1−(𝑥−[𝑥]),8.6 where [x] denotes the integer part of 𝑥∈ℝ, defined by [𝑥]={⌊𝑥⌋,𝑥≥0,⌈𝑥⌉,𝑥<0. Allton et al. improve on this idea by suggesting a new scheme, called stochastic truncation, further developed by Hamer et al. and Hamer & Court . All these variants of the stochastic truncation method use essentially the same rounding function (8.6). Price et al. propose a variation of (8.6) which essentially applies integer SR only to a specific interval fl⁡(𝑥)=⎧{⎨{⎩𝑥,𝑥≥1, 1,1>𝑥>𝑃,0,𝑃≥𝑥>0, where P is a random number from the interval [0, 1]. This rounding operator keeps the ‘exact’ value of x for large x but allows some of the values below 1 to be stochastically replaced by 0. 8.7. Quantum computing Krishnakumar & Zeng show how to implement mode 1 and 2 SR for quantum computing applications and demonstrate that mode 1 provides accuracy or circuit complexity improvements. Mode 1 SR in this work is called quantum rounding. It is shown that quantum rounding can be implemented by using the fact that quantum computing has a probabilistic component—measuring a state of a quantum register can return different results with certain probabilities. The authors show that a quantum rounding circuit can be made to round with proportional probabilities according to the mode 1 stochastic rounding in (2.1). Once such a circuit is used multiple times to measure the value of a quantum register (as is commonly done in quantum computing in order to improve confidence in the results), the average value will be more accurate because of the properties of mode 1 SR. The authors show that implementing fixed-point multiplication in a fault-tolerant quantum setting requires two to three times less resources for the same accuracy targets, compared with when RN is used. 8.8. Other applications Various other applications use SR in one way or another. We give overviews of a few such applications. In digital signal processing, SR goes under the name random rounding and has been considered for fixed-point arithmetic. Callahan demonstrates a simple 16-bit filter that is more accurate with SR than with the standard rounding. Two hardware implementations of SR are also demonstrated, and one of them interestingly does not require random number generation but uses a value that is perturbed on each rounding operation. Bargh et al. and Tran et al. address the problem of preserving privacy when publicly releasing datasets. Their goal is to find the best ways to minimize the disclosure of personal information and share only data that does not infringe peoples’ privacy. One of the aspects considered is how to transform sensitive information in specific cells of tabulated data. In [93, §4.2.2], rounding is discussed as an alternative to suppression, which is the simple removal of values that are at risk of disclosing private information, a process which may potentially delete useful data. In this research SR, under the name of random rounding [95, §5.4.3], is used to round numerical data to one of the two nearest integer multiples of a given base. In base 10, for example, the number 26 would be rounded to 20 or to 30 with probabilities 40% and 60%, respectively. SR is useful here as it does not always increase large values and decrease small ones as RN would . Being unbiased, moreover, SR can hide the information about the original data , and may even provide protection against differencing, where sensitive information can be extracted from the differences in multiple tables . Rounding to integer in a stochastic fashion is also considered by Gösgens et al. in the study of models for the spread of infections, by Matter & Potgieter , to solve a problem of resource allocation, and by Hörl & Balac , for exploring travel demand in cities using transport simulations. Wu explores SR and a modification of it called dither rounding in the context of stochastic computing. Dither rounding is more complex than SR, as it requires keeping track of the number of rounding operations performed. However, Wu shows that dither rounding can achieve similar accuracy, but with lower variance, in matrix multiplication and in machine learning algorithms for digit classification. There is some connection between SR and the technique of dither that is a common component in audio analogue-to-digital and digital-to-analogue conversion [100–102]. In analogue-to-digital conversion, dither relates to the randomization the analogue signal undergoes before being converted to a low-precision quantized digital representation . The same term is used by the authors to refer to the randomization at the other end, when converting a digital signal back to analogue. For example, the following excerpt from discusses a method of dither of the digital signal that is similar to an implementation of SR mode 1 where a set of random bits are added to the part of fraction to be truncated. If a digital manipulation (such as a gain reduction) is performed, there may be a tendency to take the intermediate higher precision numbers generated by the multiplication and simply truncate or round them to the bit width of the system. This will in many cases leave the signal improperly dithered. [···] The fractional truncated bits have some influence on the dither, in keeping with their relative position. If cost or processing time were no object, then any digital manipulation should be carried out with full accuracy, and the dither carry bit (0 or 1) can be determined by an appropriate digital random number added to the bits to be truncated. In practice such schemes would probably work well by considering only the first 3 or 4 bits to be truncated. See also [101, fig. 9] for a diagram that sketches an implementation of SR mode 1 in an integer multiplier. 9. Conclusion Hardware units supporting SR are not yet widely available but have started to appear: as we discussed in §7.5, Graphcore and Intel are producing processors with SR built in. Patents from AMD, NVIDIA, IBM and other computing companies describing implementations of SR in fixed- or floating-point arithmetic units show that this rounding mode could become more widely available in the future. When hardware is not available, simulation in software of arithmetics with SR can be used to explore its behaviour. Multiple simulators have been developed, as discussed in §7.4. These are available in various forms for Matlab, C, Julia and Python. Rounding error analysis with SR, discussed in §6, shows that compared with the standard rounding modes, SR guarantees probabilistic error bounds significantly smaller than the worst-case bounds and it also avoids the problem of stagnation (§6.1), where small values are lost to rounding when they are added to an increasingly large accumulator. This explains the success of SR in the applications described in §8. We covered work using SR in various forms, in numerical linear algebra, machine learning, ODE and PDE solvers, quantum computing and other areas. The wide array of applications in which SR has been tried and led to improved accuracy demonstrates that it is a useful technique to consider when arithmetics with standard rounding modes are insufficiently accurate. SR provides a useful alternative to extended-precision registers, arbitrary-precision libraries, multi-word representations and arithmetics, compensated algorithms and other means for improving accuracy. Data accessibility Data and relevant code for this research work are stored in GitHub: and have been archived within the Zenodo repository: Authors' contributions All authors collected and reviewed the literature, wrote drafts of the survey, and approved the final version for publication. M.C.: conceptualization, data curation, formal analysis, investigation, methodology, software, validation, visualization, writing—original draft, writing—review and editing; M.F.: conceptualization, data curation, formal analysis, investigation, methodology, software, validation, visualization, writing—original draft, writing—review and editing; N.J.H.: conceptualization, data curation, formal analysis, funding acquisition, investigation, methodology, resources, software, supervision, validation, visualization, writing—original draft, writing—review and editing; T.M.: conceptualization, data curation, formal analysis, investigation, methodology, software, validation, visualization, writing—original draft, writing—review and editing; M.M.: conceptualization, data curation, formal analysis, investigation, methodology, project administration, software, supervision, validation, visualization, writing—original draft, writing—review and editing. All authors gave final approval for publication and agreed to be held accountable for the work performed therein. Competing interests We declare we have no competing interests. Funding M.C., N.J.H. and M.M. were supported by Engineering and Physical Sciences Research Council grant no. EP/P020720/1. M.F. was supported by Wenner-Gren Foundations grant no. UPD2019-0067. N.J.H. was also supported by the Royal Society. T.M. was supported by the InterFLOP project (ANR-20-CE46-0009) of the French National Agency for Research. Acknowledgements We thank Claude-Pierre Jeannerod, the academic editor, and the two anonymous referees for their feedback, which helped us improve the presentation of the work. Footnotes 1 Electronic numerical integrator and computer—the first programmable, general-purpose digital computer, made in 1945. 2 We remark that floating-point numbers are not uniformly distributed in [0, 1], but here we need the precision-k random floating-point number to be sampled from that interval uniformly in the sense of real numbers. This is not a consideration required in the hardware algorithms in §7.3 since there SR is performed at the bit level, using uniformly distributed integers. Usage of other random number distributions in SR has been explored by Xia et al. . 3 See 4 See 5 See 6 See 7 See 8 See 9 See 10 See 11 See 12 See © 2022 The Authors. Published by the Royal Society under the terms of the Creative Commons Attribution License which permits unrestricted use, provided the original author and source are credited. 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Li C, Du P, Li K, Liu Y, Jiang H and Quan Z (2022) Accurate Goertzel Algorithm: Error Analysis, Validations and Applications, Mathematics, 10.3390/math10111788, 10:11, (1788) This Issue March 2022 Volume 9Issue 3 Article Information PubMed:35291325 Published by:Royal Society Online ISSN:2054-5703 History: Manuscript received14/10/2021 Manuscript accepted04/02/2022 Published online09/03/2022 License: © 2022 The Authors. Published by the Royal Society under the terms of the Creative Commons Attribution License which permits unrestricted use, provided the original author and source are credited. 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1716	https://web.ma.utexas.edu/users/m408s/m408d/CurrentWeb/LM11-5-2.php	\| \| \| \| \| --- --- \| \| Home Integration by PartsIntegration by Parts Examples Integration by Parts with a definite integral Going in Circles Tricks of the Trade Integrals of Trig FunctionsAntiderivatives of Basic Trigonometric Functions Product of Sines and Cosines (mixed even and odd powers or only odd powers) Product of Sines and Cosines (only even powers) Product of Secants and Tangents Other Cases Trig SubstitutionsHow Trig Substitution Works Summary of trig substitution options Examples Completing the Square Partial FractionsIntroduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of IntegrationSubstitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper IntegralsType 1 - Improper Integrals with Infinite Intervals of Integration Type 2 - Improper Integrals with Discontinuous Integrands Comparison Tests for Convergence Modeling with Differential EquationsIntroduction Separable Equations A Second Order Problem Euler's Method and Direction FieldsEuler's Method (follow your nose) Direction Fields Euler's method revisited Separable EquationsThe Simplest Differential Equations Separable differential equations Mixing and Dilution Models of GrowthExponential Growth and Decay The Zombie Apocalypse (Logistic Growth) Linear EquationsLinear ODEs: Working an Example The Solution in General Saving for Retirement Parametrized CurvesThree kinds of functions, three kinds of curves The Cycloid Visualizing Parametrized Curves Tracing Circles and Ellipses Lissajous Figures Calculus with Parametrized CurvesVideo: Slope and Area Video: Arclength and Surface Area Summary and Simplifications Higher Derivatives Polar CoordinatesDefinitions of Polar Coordinates Graphing polar functions Video: Computing Slopes of Tangent Lines Areas and Lengths of Polar CurvesArea Inside a Polar Curve Area Between Polar Curves Arc Length of Polar Curves Conic sectionsSlicing a Cone Ellipses Hyperbolas Parabolas and Directrices Shifting the Center by Completing the Square Conic Sections in Polar CoordinatesFoci and Directrices Visualizing Eccentricity Astronomy and Equations in Polar Coordinates Infinite SequencesApproximate Versus Exact Answers Examples of Infinite Sequences Limit Laws for Sequences Theorems for and Examples of Computing Limits of Sequences Monotonic Covergence Infinite SeriesIntroduction Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral TestPreview of Coming Attractions The Integral Test Estimates for the Value of the Series Comparison TestsThe Basic Comparison Test The Limit Comparison Test Convergence of Series with Negative TermsIntroduction, Alternating Series,and the AS Test Absolute Convergence Rearrangements The Ratio and Root TestsThe Ratio Test The Root Test Examples Strategies for testing SeriesStrategy to Test Series and a Review of Tests Examples, Part 1 Examples, Part 2 Power SeriesRadius and Interval of Convergence Finding the Interval of Convergence Power Series Centered at x=a Representing Functions as Power SeriesFunctions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin SeriesThe Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor PolynomialsTaylor Polynomials When Functions Are Equal to Their Taylor Series When a Function Does Not Equal Its Taylor Series Other Uses of Taylor Polynomials Functions of 2 and 3 variablesFunctions of several variables Limits and continuity Partial DerivativesOne variable at a time (yet again) Definitions and Examples An Example from DNA Geometry of partial derivatives Higher Derivatives Differentials and Taylor Expansions Differentiability and the Chain RuleDifferentiability The First Case of the Chain Rule Chain Rule, General Case Video: Worked problems Multiple IntegralsGeneral Setup and Review of 1D Integrals What is a Double Integral? Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates \| \| IntroductionWith the exceptions of geometric series, where r may be negative, or the rare series with telescoping partial sums, the convergence tests we have worked with so far only work with positive-termed series. When the terms in a series can be positive or negative, things get more complicated; the sequence {sn} of partial sums may not be monotonic, so it can be bounded yet divergent. This module will introduce the Alternating Series Test, which works on series in which the terms have alternating signs.Alternating Series and the Alternating Series Test An alternating series is a series ∑n=1∞an where an has alternating signs. Notice that if an has alternating signs, we will be able to let bn=\|an\|, and write an=(−1)nbn or an=(−1)n−1bn. For instance, ∑n=1∞(−1)n−1n=1−12+13−14+…=∑n=1∞(−1)n−11n has terms an=(−1)n−1n and bn=1n. \| \| \| Alternating Series Test (AST): If the alternating series ∑n=1∞(−1)n−1bn=b1−b2+b3−b4+⋯ satisfies 1. bn>0, 2. bn+1≤bn for all n, and 3. limn→∞bn=0, then the series converges. \| In other words, if the absolute values of the terms of an alternating series are non-increasing and converge to zero, the series converges. This is easy to test; we like alternating series. To see how easy the AST is to implement, DO: Use the AST to see if ∑n=1∞(−1)n−11n converges. This series is called the alternating harmonic series. This is a convergence-only test. In order to show a series diverges, you must use another test. The best idea is to first test an alternating series for divergence using the Divergence Test. If the terms do not converge to zero, you are finished. If the terms do go to zero, you are very likely to be able to show convergence with the AST. Warning: The converse of the AST is not true; we have series that are alternating and convergent and do not satisfy the AST. For example, if we take the terms of ∑1n2=1+12+14+19+116+125+⋯, and exchange the first two terms, then the second two, etc., and then put in alternating signs, we get 12−1+19−14+125−116+⋯, which does not satisfy the conditions of the AST since bn+1≤bn does not hold for all n. However, this series is convergent (we will be able to prove its convergence later using the ideas of Absolute Convergence). The following video will explain how the AST works, give more details on the alternating harmonic series, and look at the values of some interesting alternating series. << Prev Next >> \|
1717	https://math.stackexchange.com/questions/1068254/let-z-1-z-2-and-z-3-be-complex-vertices-of-an-equilateral-triangle-show	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$. Ask Question Asked Modified 2 years, 1 month ago Viewed 7k times $\begingroup$ Prompt: Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$. Question: I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$. What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane? Is there any intuition to be gained here? complex-analysis intuition Share edited Dec 14, 2014 at 21:46 Semiclassical 18.7k44 gold badges3939 silver badges9999 bronze badges asked Dec 14, 2014 at 21:25 mathjacksmathjacks 3,67422 gold badges3333 silver badges5050 bronze badges $\endgroup$ 4 $\begingroup$ It does show that geometric properties of the plane can be expressed concisely in terms of complex numbers. $\endgroup$ Tim Raczkowski – Tim Raczkowski 2014-12-14 21:36:18 +00:00 Commented Dec 14, 2014 at 21:36 1 $\begingroup$ This expression has appeared here before; I recall seeing some elegant interpretation of the formula on some other similar question, but I can't find it. $\endgroup$ Milo Brandt – Milo Brandt 2014-12-14 21:52:17 +00:00 Commented Dec 14, 2014 at 21:52 8 $\begingroup$ Does this answer your question? Equilateral Triangle from three complex points $\endgroup$ UmbQbify – UmbQbify 2020-07-25 20:49:52 +00:00 Commented Jul 25, 2020 at 20:49 1 $\begingroup$ To those voting to close as a duplicate, please consider that this question asks about the "intuition to be gained here" and the linked duplicate target is just about the computation. $\endgroup$ KReiser – KReiser 2020-07-26 01:39:37 +00:00 Commented Jul 26, 2020 at 1:39 Add a comment \| 8 Answers 8 Reset to default 14 $\begingroup$ You can make a nice, somewhat abstract argument that connects this; in particular, equilateral triangles have various nice properties - like: For fixed $z_1,z_2$, there are exactly two $z_3$ such that $(z_1,z_2,z_3)$ is an equilateral triangle. If $z_1$ and $z_2$ are real, then those $z_3$ are reflections around the real axis. Any permutation of an equilateral triangle is an equilateral triangle - so $(z_1,z_2,z_3)$ being equilateral implies $(z_1,z_3,z_2)$ and $(z_2,z_3,z_1)$ are also equilateral. The map $z\mapsto az+b$ preserves equilateral triangles. If we make a leap to trying to think about an analogous polynomial $P(z_1,z_2,z_3)$ such that, at any root, $(z_1,z_2,z_3)$ is an equilateral triangle, then the above conditions translate to: $P(z_1,z_2,z_3)$ has degree two in each of its variables. It has only real coefficients. $P(z_1,z_2,z_3)$ is a symmetric polynomial. $P(z_1,z_2,z_3)$ is homogenous. Together, these yield that if such a $P$ existed, $P$ would have to be a sum of the only two symmetric and homogenous polynomials of degree two, meaning, for some $\alpha$ and $\beta$: $$P(z_1,z_2,z_3)=\alpha(z_1^2+z_2^2+z_3^2)+\beta(z_1z_2+z_2z_3+z_3z_1)$$ but, since we desired $P(z,z,z)=0$ to be a solution, we can calculate that clearly $\alpha=-\beta$ and, dividing out by $\alpha$ would give that the only polynomial that could feasibly represent equilateral triangles would be: $$P(z_1,z_2,z_3)=z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1.$$ Then, if you showed that if $P(z_1,z_2,z_3)=0$ implied $P(z_1+c,z_2+c,z_3+c)=0$, which could be done algebraically, or perhaps through some clever manipulation, you would have that the roots of this polynomial satisfied all the conditions we wanted of equilateral triangles - and those conditions suffice to uniquely define the set of equilateral triangles, which proves that $P$'s roots are all equilateral triangles. The point of this answer is, of course, to draw a connection between the symmetries defining the original problem and how the carry over to symmetries in the algebra and, indeed, suffice to show the statement. You could do this the other way round to learn geometric symmetries from the algebraic form of $P$ as well. Share edited Dec 14, 2014 at 22:33 answered Dec 14, 2014 at 22:28 Milo BrandtMilo Brandt 62k55 gold badges113113 silver badges197197 bronze badges $\endgroup$ 2 2 $\begingroup$ How does first condition correspond to "P(z1,z2,z3) has degree two in each of its variables. It has only real coefficients. " $\endgroup$ Clemens Bartholdy – Clemens Bartholdy 2021-01-26 17:17:11 +00:00 Commented Jan 26, 2021 at 17:17 1 $\begingroup$ Indeed, it doesn't have to be necessarily of degree two: if you take the 10-th power of the polynomial he defined, the roots stays the same. Nevertheless, the polynomial can't be linear because of the observation he made, so the minimal degree polynomial has to be at least of degree two. Since you can find such polynomial, 2 is enough. $\endgroup$ Andrea Marino – Andrea Marino 2021-02-02 14:41:15 +00:00 Commented Feb 2, 2021 at 14:41 Add a comment \| 9 $\begingroup$ Here's an example of what I mean in my comment. The fact that all sides are equal, and that the angles between them are equal can be expressed quite elegantly as $$\frac{z_3-z_2}{z_2-z_1}=\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$ The desired equation now follows from some elementary algebra. I believe this result can be generalized to any regular n-gon. Share answered Dec 14, 2014 at 22:13 Tim RaczkowskiTim Raczkowski 20.6k22 gold badges2626 silver badges6767 bronze badges $\endgroup$ Add a comment \| 4 $\begingroup$ If $z_1=w_1+x$, $z_2=w_2+x$ and $z_3=w_3+x$, the identity becomes $$ w_1^2+2w_1x+x^2+ w_2^2+2w_2x+x^2+ w_3^2+2w_3x+x^2=\ w_1w_2+w_1x+w_2x+x^2+ w_2w_3+w_2x+w_3x+x^2+ w_3w_1+w_3x+w_1x+x^2 $$ that becomes $$ w_1^2+w_2^2+w_3^2=w_1w_2+w_2w_3+w_3w_1 $$ so, taking $x=-z_3$, we can assume $z_3=0$ and we need to prove that if $z_1$, $z_2$ and $0$ are the vertices of an equilateral triangle, then $$ z_1^2+z_2^2=z_1z_2. $$ If $z_1=w_1u$ and $z_2=w_2u$ (with $u\ne0$), the identity above becomes $w_1^2+w_2^2=w_1w_2$, so we can assume $z_1=e^{i\pi/6}$, which means that $z_2=e^{i(\pi/6+\pi/3)}$ or $z_2=e^{i(\pi/6-\pi/3)}$. If $z_2=e^{i(\pi/6+\pi/3)}=e^{i\pi/2}=i$, then a rotation brings us in the situation with $z_1=e^{-i\pi/6}$ and $z_2=e^{i\pi/6}$. So we are reduced to proving that $$ e^{i\pi/3}+e^{-i\pi/3}=1 $$ which is obvious. Share answered Dec 14, 2014 at 22:49 egregegreg 246k2121 gold badges155155 silver badges353353 bronze badges $\endgroup$ Add a comment \| 4 +100 $\begingroup$ @Milo Brandt is completely correct. Now, there aren't any things left to know about the equilateral triangles, but there are a lot of amazing things with the complex numbers. I am talking about the relationships of complex numbers and the coordinate geometry... With the help of complex numbers, it becomes very easy for us to calculate or prove things like this. Imagine proving the same thing with the coordinates/coordinate geometry. It makes our work easier. Now, since you want to know about the equilateral triangles, these properties,or I should say, these results may help you: let the circum centre of the triangle be $z_0$ , then, $3z^2_0 = z^2_1 + z^2_2 + z^2_3$ The angle between the points must be $60°$ or $Ï/3^c$(most obvious). Thus, $arg((z_3 - z_1)/(z_2 - z_1)) = Ï/3^c$. And similar for all other angles. Best ever property for equilateral triangle: The triangle with vertices as $z_1,z_2$ and $z_3$ can only be equilateral if and only if: $1/(z_1 - z_2) + 1/(z_2 - z_3) + 1/( z_3 - z_1) =0$. I am going to prove the last point for your convenience... Let, $a = z_1 - z_2$ $ b = z_2 - z_3$ $ c = z_3 - z_1$ Thus, $a + b + c = 0$ Also, $1/a + 1/b + 1/c = 0$. [ GIVEN IN QUESTION] Thus, $1/a + (b+c)/bc = 0$ $1/a - a/bc = 0$ [ as $a + b + c =0$] Thus, $a^2 = bc$ $\|a^3\| = \|abc\|$ Similarly, $\|b^3\| = \|abc\|$ and $\|c^3\| = \|abc\|$ Now, add these equations $\|a^3\| + \|b^3\| + \|c^3\| = 3\|abc\|$ Which is only possible when $\|a\| + \|b\| + \|c\| = 0$ or $ \|a\| = \|b\| = \|c\| $ But since $\|a\|,\|b\|,\|c\|>0$ , Thus $\|a\| = \|b\| = \|c\|$. Thus the triangle is an equilateral triangle! Hence, Proved! Share answered Jan 31, 2021 at 19:03 Nandeesh BhatraiNandeesh Bhatrai 63633 silver badges1414 bronze badges $\endgroup$ 1 $\begingroup$ +1, good answer $\endgroup$ Amrut Ayan – Amrut Ayan 2024-06-25 13:30:14 +00:00 Commented Jun 25, 2024 at 13:30 Add a comment \| 3 $\begingroup$ This equation can be restated as $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$. With $z:=z_1-z_2$, some $\omega=(\exp\pm2\pi i/3)$ satisfies $z_2-z_3=\omega z,\,z_3-z_1=\omega^2z$. Since $\omega^3=1\ne\omega$, $\omega^2+\omega+1=\frac{\omega^3-1}{\omega-1}=0$. The LHS is then $z^2(1+\omega^2+\omega^4)=z^2(1+\omega^2+\omega)=0$. Share answered Jan 31, 2021 at 21:21 J.G.J.G. 118k88 gold badges7979 silver badges146146 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ Does it tell us something unique about equilateral triangles? In the complex plane the sides of any triangle are represented by numbers $a,b,c$ where $$a+b+c=0.$$ The equation you are interested in is equivalent to the equation $$a^2+b^2+c^2=0$$ and therefore tells us that equilateral triangles have the very special property that all quadratic symmetric polynomials of the sides are constant. This gives us an elegant tool for investigating these triangles with minimal algebraic manipulation. I will give just two examples. The sides of an equilateral triangle satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Simply divide the symmetric quadratic equation $ab+ba+cb=0$ by $abc$. The equation $a^2+b^2+c^2=0$ only represents equilateral triangles $abc-c^3=c(ab+bc+ca)-c^2(a+b+c)=0$. Then $a^3,b^3,c^3$ are all equal (to $abc$) and therefore $a,b,c$ all have the same length. Share edited Feb 2, 2021 at 18:49 answered Feb 2, 2021 at 18:43 user502266user502266 $\endgroup$ 0 Add a comment \| 1 $\begingroup$ The significance of the expression $$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1\tag1$$ is that it is the most concise and easiest to apply in analyzing an equilateral triangle in the complex plane. There may be some suspicion on the statement. It is tempting, for example, to point out that the expression $$\|z_1-z_2\| = \|z_2-z_3\| = \|z_3-z_1\|\tag2 $$ is simpler. However, it is only so in appearance, not in complex algebraic operation. It is actually the following complex conjugate expression in disguise $$(z_1-z_2)(\bar z_1-\bar z_2 )= (z_2-z_3)(\bar z_2-\bar z_3 ) =( z_3-z_1)(\bar z_3-\bar z_1 )\tag3$$ which is more involved than (1). Besides, the expression (2) is in fact a system of two equations, which is again more involved than the single equation (1). In practical application, it is more straightforward to check the condition (1) in determining whether $z_1$, $z_2$ and $z_3$ form an equilateral triangle than, say, checking the condition (2). Share edited Feb 4, 2021 at 0:15 answered Jan 31, 2021 at 21:06 QuantoQuanto 123k99 gold badges196196 silver badges297297 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ An analogous condition is $(z_1 + z_2 + z_3)^2 = 3(z_1 z_2 + z_1 z_3+z_2 z_3)$, or $$\left(\frac{z_1+z_2+z_3}{3}\right)^2 = \frac{z_1 z_2 + z_1 z_3 + z_2 z_3}{3}$$ that is, the equation with roots $z_k$, $k=1,3$, is of the form $$z^3 - 3 a z^2 + 3 a^2 z + b = 0$$ or $$(z-a)^3 = c$$ This also indicates how to generalize the statement for $n$ complex numbers ( impose $n-2$ equalities). Share answered Aug 14, 2023 at 0:49 orangeskidorangeskid 57.3k33 gold badges5151 silver badges121121 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 14 Equilateral Triangle from three complex points How does this equality on vertices in the complex plane imply they are vertices of an equilateral triangle? 3 Book recommendation for geometric intuition 2 Conditions of equilateral triangle in complex plane/equivalence of angles between segments 2 Prove that iff the vertices represent an equilateral triangle, then $a^2+\beta^2+\gamma^2-\alpha\beta - \gamma\alpha - \beta\gamma=0$ Proving quadratic formula by considering a function of coefficents 2 Geometric significance of the terms of the relation $a^2+b^2+c^2=ab+bc+ca$ that holds when complex $a$, $b$, $c$ form an equilateral triangle? Related 14 Equilateral Triangle from three complex points 1 Conditions To Make Complex Numbers $z_1, z_2, z_3, z_4$ Vertices of a Square 1 Geometrical Applications of Complex Numbers 0 Finding vertices when circumcentre is given for equilateral triangle by complex number method 1 Why is this result true for complex numbers? 0 maximum product of distances of a point and vertices of equilateral triangle 2 On showing that three complex points form an equilateral triangle Hot Network Questions What "real mistakes" exist in the Messier catalog? Lingering odor presumably from bad chicken Best solution to prevent loop between tables for granular relations What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters? How to understand the reasoning behind modern Fatalism? How can I get Remote Desktop (RD) to scale properly AND set maximum windowed size? 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1718	https://www.tiger-algebra.com/en/solution/combination-without-repetition/c%2810%2C4%29/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Combinations without repetition Other Ways to Solve Step-by-step explanation 1. Find the number of terms in the set n represents the total number of items in the set: c(n,k) c(10,4) n=10 2. Find the number of items selected from the set k represents the number of items selected from the set: c(n,k) c(10,4) k=4 3. Calculate the combinations using the formula Plug n (n=10) and k (k=4) into the combination formula: C(n,k)=n!k!(n-k)! C(10,4)=10!4!(10-4)! C(10,4)=10!4!·6! C(10,4)=10·9·8·7·6·5·4!4!·6! C(10,4)=10·9·8·7·6·56! C(10,4)=10·9·8·7·6·56·5·4·3·2·1 C(10,4)=210 There are 210 ways that 4 items chosen from a set of 10 can be combined. How did we do? Why learn this Combinations and permutations If you have 2 types of crust, 4 types of toppings, and 3 types of cheese, how many different pizza combinations can you make? If there are 8 swimmers in a race, how many different sets of 1st, 2nd, and 3rd place winners could there be? What are your chances of winning the lottery? All of these questions can be answered using two of the most fundamental concepts in probability: combinations and permutations. Though these concepts are very similar, probability theory holds that they have some important differences. Both combinations and permutations are used to calculate the number of possible combinations of things. The most important difference between the two, however, is that combinations deal with arrangements in which the order of the items being arranged does not matter—such as combinations of pizza toppings—while permutations deal with arrangements in which the order the items being arranged does matter—such as setting the combination to a combination lock, which should really be called a permutation lock because the order of the input matters. What these two concepts have in common, is that they both help us understand the relationships between sets and the items or subsets that make up those sets. As the examples above illustrate, this can be used to better understand many different types of situations. Combinations and permutations Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
1719	https://www.sciencedirect.com/science/article/pii/S0169534724001393	Life histories are not just fast or slow - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Abstract Keywords Life history diversity The fast–slow continuum Multiple axes to grind: beyond the fast–slow continuum Life history theory and the fast–slow continuum Challenges for measurement and analysis of life history traits Towards a hypothesis-driven, empirically sufficient era of life history theory Concluding remarks Acknowledgments Declaration of interests References Glossary Show full outline Cited by (16) Figures (2) Tables (1) Table 1 Trends in Ecology & Evolution ----------------------------- Volume 39, Issue 9, September 2024, Pages 830-840 Opinion Life histories are not just fast or slow Author links open overlay panel Iain Stott 1 2 10@, Roberto Salguero-Gómez 3 10@, Owen R.Jones 2, Thomas H.G.Ezard 4, Marlène Gamelon 5, Shelly Lachish 3, Jean-Dominique Lebreton 6, Emily G.Simmonds 7 8, Jean-Michel Gaillard 5 11, Dave J.Hodgson 9 11 Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights Life history studies support the fast–slow continuum as the dominant but not unique axis structuring life history variation. Other important axes of life history variation associated with development and reproductive tactics exist, and exploration of further axes, clusters, and boundaries of life history variation is needed. Existing life history analyses are venturing far from theory and could benefit from a stronger focus on hypothesis testing rather than exploration. We make recommendations to identify the structuring axes of life history variation through data choices and analytical methods of dimensionality reduction with recourse to a comprehensive model of life history. Abstract Life history strategies, which combine schedules of survival, development, and reproduction, shape how natural selection acts on species’ heritable traits and organismal fitness. Comparative analyses have historically ranked life histories along a fast–slow continuum, describing a negative association between time allocation to reproduction and development versus survival. However, higher-quality, more representative data and analyses have revealed that life history variation cannot be fully accounted for by this single continuum. Moreover, studies often do not test predictions from existing theories and instead operate as exploratory exercises. To move forward, we offer three recommendations for future investigations: standardizing life history traits, overcoming taxonomic siloes, and using theory to move from describing to understanding life history variation across the Tree of Life. Previous article in issue Next article in issue Keywords life history theory life history strategy life history trait reproductive tactic phylogenetic comparative analysis Life history diversity Any individual organism is born, survives, develops, possibly reproduces, and ultimately dies. The organism’s resulting life history is the pathway it takes on this journey and comprises key life history traits (see Glossary; e.g., age at maturation or death, rate of growth or reproduction). There are countless pathways through life, so the Tree of Life boasts an astounding diversity of life history strategies (Box 1), defined by allocations of limited resources to the competing needs of survival, development, or reproduction. Selection on these trade-offs should maximize fitness within the constraints of body size, lifestyle, resource availability, and environmental conditions. Central aims of evolutionary ecology include quantifying life history strategies, identifying ecological and evolutionary patterns in strategies across species, and understanding the evolution of life histories in a rapidly changing world . Box 1 Life history strategies Are carrots (Daucus carota) similar to blue-ringed octopuses (Hapalochlaena lunulata) (Figure I)? They both start life small and grow fast, but few survive to adulthood, produce many offspring in one breeding event at around 2 years old, and are semelparous. By contrast, humans (Homo sapiens sapiens) and blue whales (Balaenoptera musculus) grow slowly and mature late but are likely to survive to adulthood, usually produce one offspring every 2–3 years, and have remarkably similar lifespans: The oldest known living human, Jeanne Calment, reached 122 years of age, and blue whales may live up to 110 years . The sacred fig tree (Ficus religiosa) and giant barrel sponge (Xestospongia muta) are like carrots and blue-ringed octopuses in that they start life small, are unlikely to survive to adulthood, and, once mature, produce many offspring. X. muta and F. religiosa are also like humans and blue whales, however, because they are iteroparous over long lifetimes: One sacred fig ‘Jaya Sri Maha Bohi’ in Sri Lanka is the oldest known human-planted tree, which at around 2300 years old is similar in age to the oldest known Caribbean giant barrel sponge . Although distantly related organisms may share similar life history strategies, closely related organisms can show remarkably different life histories. The Dasyuridae family, to which the Tasmanian devil (Sarcophilus harrisii) belongs, is one of the most diverse marsupial families , with dasyurid species showing different life history strategies, despite being closely related. Maturity usually occurs at ~11 months, but, although in some species all males die by 12 months old following a single synchronous mating, others reproduce many times annually and live for several years . Plant life history strategies are often even more labile than animal strategies: pine trees (Pinus spp.) vary vastly in lifespan, despite having similar morphology and physiology: The Virginia pine (P. virginiana) rarely lives over 150 years , whereas the bristlecone pine (P. longaeva) holds the record of nonclonal longevity at 4850 years old . The Tree of Life abounds with examples of distantly related organisms sharing convergent life history strategies, despite divergent body size and lifestyle, and vice versa. Evolutionary ecologists seek to understand, across broad taxonomic groups, how and why diverse life history strategies, such as those exemplified here, are distributed across phylogenies, space, and time. 1. Download: Download high-res image (553KB) 2. Download: Download full-size image Figure I. Carrots (Daucus carota) and blue-ringed octopuses (Hapalochlaena lunulata) are distantly related organisms with very different forms and functions. Yet, these organisms have very similar life history strategies: As illustrated on the inset graph, relative rates of survival (red) start low but increase with size into adulthood, and, at around 2 years of age, both organisms die following a single reproductive event. Increasing numbers of studies use comparative methods to address these central aims because life history databases are rapidly growing in number, volume, taxonomic cover, and accessibility [e.g., 2., 3., 4.]. This rapid increase has taken place thanks to contributions from citizen science and remote sensing, novel monitoring technologies, and open science. Although welcome, these new developments are not without challenges. The first two central aims of quantifying life histories and identifying patterns among species are often addressed, but choices of data and analytical approaches vary, often between taxa, thus precluding reliable comparisons between studies. Moreover, the third central aim of understanding life history evolution is often not addressed, because approaches are mostly correlative rather than rooted in testing hypotheses from the existing theory. We propose a roadmap for future enquiries into life history diversity across species. We include several theoretical, empirical, and analytical recommendations to move toward research underpinned by testable hypotheses rooted in current evolutionary theories rather than exploratory approaches to data analysis. In this context, we overview evidence that the fast–slow continuum remains the main structural axis describing life history diversity, providing a useful context for identifying and explaining novel patterns describing substantial life history variation. However, we also argue that life histories are more than just fast or slow. The fast–slow continuum Life history strategies are defined by covariation, and therefore trade-offs, among life history traits. For instance, an organism that allocates a lot to growth or reproduction early in life will likely pay a toll on future survival . Across species, life history strategies are typically structured along an axis defined by the trade-off between allocation of time to reproduction and development versus survival. This trade-off underpins the ubiquitous fast–slow continuum, which is usually the dominant axis structuring life history variation across species, including in mammals [6., 7., 8.], birds , insects [9,10], fish , and plants . There is evidence that the fast–slow continuum is strongly associated with ecological and environmental patterns and processes, with a frequently observed association being that slow species have higher risk of extinction, on average, than fast species. For instance, slow fishes have higher sensitivity of population growth to sea temperatures , slow mammals and birds have higher extinction rates , and slow species display more negative responses to human disturbances than their fast counterparts [15,16]. On the other hand, fast species are more sensitive to certain projected environmental outcomes of climate change . These findings are key to protecting future ecosystem function and helping develop cost-effective conservation plans . Even in community ecology, the fast–slow continuum explains interspecific patterns in trophic ecology better than established trade-off theories , and integration of the fast–slow continuum into models of disease ecology has recently been advocated to provide more mechanistic explanations . Multiple axes to grind: beyond the fast–slow continuum The fast–slow continuum alone is not an adequate universal pattern of species’ life history strategies: Many extant life history strategies fall outside its predictions. Marine turtles and most trees illustrate a common outlier: species characterized by long lives, late maturations, high survival, and high reproductive outputs (where fast–slow would predict low) . More generally, organisms with vital rates closely linked to size and development but weakly to age can be exceptions to the fast–slow ‘rule’ . A growing body of evidence shows that life history strategies are also defined using an axis of variation involving developmental or reproductive patterns. Such axes include altricial versus precocial offspring , high to low recruitment rates , or semelparity to iteroparity [6,10,12]. In some cases, axes other than reproduction or development explain substantial life history variation, such as age-specific distributions of mortality and reproduction . However, the exact meaning of additional axes is hindered by the vast heterogeneity of life history traits analyzed in this context . A key research priority of ecology and evolution is to understand life history variation where survival and reproduction are closely linked to factors other than age. These factors include body size, especially in species with retrogression ; life stage, especially in species with diapause ; and metabolic rate . Another key research priority is to identify potential clusters of life history variation and the limits of the variation among viable life histories . A strong theoretical steer is important in the process of identifying and explaining life history patterns; indeed, even theoretical understanding of the origin and maintenance of the fast–slow continuum has room for improvement. Life history theory and the fast–slow continuum To look for meaningful life history patterns beyond the fast–slow continuum, we need a clearer theoretical understanding of how the fast–slow continuum fits into wider life history theory. An organism’s life history strategy determines its fitness, because life history reflects the rates of survival and reproduction that determine numerical representation in future generations . The associations among traits that define a life history strategy not only result from selection acting on heritable phenotypic traits but also filter how selection contributes to the phenotype (Box 2). This idea is underpinned by the canonical Euler-Lotka identity, which comprises age-based schedules of survival and reproduction, from birth to oldest age , and characterizes how survival and reproduction over the life course determine a population’s growth rate and long-term viability. However, the identity’s simplicity masks the diversity of combinations of vital rate schedules that can achieve the same population growth, does not include temporal or spatial variation in vital rates, and does not account for life histories with size- or stage-dependent vital rates . Box 2 A model life history: from resource acquisition to allocation and fitness 1. Download: Download high-res image (405KB) 2. Download: Download full-size image Figure I. A schematic view of a framework to examine and quantify life history strategies, showing the relationships between different trait types. Colors illustrate what currencies traits are measured in: energy (orange), space (red), and time (blue). An organism’s life history strategy is shaped by interacting traits, which fulfill different needs of the life history and which are measured in different ways (Figure I). At the core of the life history strategy is resource allocation to competing developmental traits: supporting survival (somatic maintenance) versus reproduction (the germline) versus growth in either . The key unit of measurement, or currency, of developmental traits is amounts and rates of energy allocation. Processes of resource allocation can be difficult to measure directly, but the results of resource allocation are often evident in the phenotype. Phenotypic traits encompass any observable organismal traits, often part of the morphology, behavior, and physiology of the organism. Phenotypic traits may be measurable using many currencies; we identify key metrics of use of space, such as home range size (behavioral) or specific root length (morphological); expenditure of energy, such as metabolic rate (physiological) and foraging efficiency (behavioral); and time provides a means to track these as rates, such as body mass growth. Other types of phenotypic traits and currencies may exist. Phenotypic traits are responsible for acquiring resources and so will also counterinfluence developmental traits . The consequences of resource acquisition and allocations is the ability to survive and reproduce across the lifespan. Survival and fertility distributions determine life history traits, which are almost exclusively measured in time. Life history traits can be durations (e.g., average age at maturity); rates (e.g., number of offspring per year); or temporal probabilities linked to survival, development, and reproduction (e.g., probability of surviving to maturity). Life history traits are often converted to expectations at population or species level, especially where this makes it possible to measure average phenotypes (e.g., mean life expectancy) for events that occur only once in an organism’s life cycle (e.g., death). Life history traits ultimately govern an organism’s fitness and its population size, structure, and dynamics by removing or adding individuals to future generations. Demographic and evolutionary processes act together to mold genotype frequencies in populations and species in future generations, given constraints imposed by evolutionary history. Genotype naturally influences all organismal traits (assuming some heritability). Traits can all covary with and feed back on one another directly or indirectly and are subject to constraints imposed by evolutionary history, physical laws, Bauplan, and the environment . Natural selection acts to favor those life history strategies that maximize the organism’s fitness in its environment, given those constraints . Theory based on the principle of allocation of limited resources predicts life history variation both among species (via allometric constraints and macroevolutionary processes) and within species (via natural selection in fluctuating or heterogeneous environments). Trade-offs between lifespan and growth/reproduction are more likely among species because resource allocation is more variable than resource acquisition at this level, whereas within species, the reverse is true . Species are therefore arranged along a fast–slow continuum as a result of interspecific trade-offs, whereas positive associations between lifespan and reproductive output often define the primary axis of variation among individuals within species, at least for birds and mammals . In many clades, resource allocation theory may oversimplify trade-offs that act on complex suites of (co)varying traits underpinning vital rates, which could further depend on population density, nonresource environmental factors, and phylogenetic constraints . The macroevolution of life histories of species in fluctuating environments could follow similar principles to evolutionary models of heritable trait variation among individuals within populations. Models of density-dependent selection in fluctuating environments conclude that life history strategies are selected to maximize a function of both rate of increase and lifetime reproductive success . This prediction resonates with a fast–slow axis, depending on fluctuating environmental pressures, instead of settling to either extreme. The stochasticity of rates of survival will further influence selection on life cycle ‘speed’: Uncertainty in adult survival is likely to favor early reproduction, whereas uncertainty in juvenile survival will favor iteroparity, higher-quality offspring, and longer life or alternatively favor bet-hedging and seed-bank strategies . Several unique life history strategies could achieve the same lifetime reproductive success. This perhaps explains frequent observation of the second axis of variation, focused on distribution of reproduction over age. Selection to optimize survival, development, and reproduction involves a sequential series of traits over ages or states, which has to date been overlooked (Box 2); therefore, if selection acts on a combination of rates of survival and reproduction, tensions exist among multiple strategies for producing offspring. Yet, our current theories of life history evolution, including the Euler-Lotka identity, do not tell us how organisms should achieve maximum fitness. Different life stages may be under different selection pressures on their vital rates. Not all stages of any given life history should be equally fast or slow, especially given organisms usually grow in size toward reproductive maturity, and in some cases past, such as in indeterminate growers. Future analyses should test whether some stages are consistently faster or slower than others or whether the time spent in a given stage depends on species features and/or environmental context. Vital rates could be selected to promote population resilience alongside long-term rate of increase. In stochastic environments, fitness will depend on selection for life histories that confer resilience, incorporating resistance to and/or recovery from numerical change following demographic disturbances . Slow life history traits may result from selection for resistance through allocation of resources to defend against typical disturbances. Fast life histories may result from selection to quickly recover from disturbances. If a negative association between resistance and recovery exists, certain disturbance regimes might help to explain the fast–slow continuum, but other patterns of covariation, dependent on disturbance regimes, could select for different interspecific patterns in life history. Challenges for measurement and analysis of life history traits The choice of trait data used to evaluate interspecific life history patterns and theories is nontrivial, with important consequences for the interpretation of analyses. Life history trait data vary greatly in abundance, accessibility, and quality (Table 1) . Quantifying life history traits requires significant effort to collect high-quality longitudinal data in order to observe events, especially when rare, and infer population-level rates (Table 1). The gold standard is detailed schedules of age- or stage-specific vital rates reliably estimated from long-term individually based monitoring of populations in the wild , so available trait data are taxonomically biased . Common sets of life history traits can be derived from these schedules across species (Table 1). However, because the traits derived from these schedules are mathematical functions of the same underlying schedules of survival, development, and reproduction, correlations among traits are a combination of the real and the artefactual . Table 1. Comparison of different types of demographic data in the context of life history strategy comparative analyses. \| Data type \| Strengths \| Weaknesses \| Source(s) \| Examples/Refs \| --- --- \| Simple traits, based on expert natural historian knowledge and opinion \| • Easy to measure • Available for most widely known species \| • Difficult to verify • Often missing measurement types (e.g., mean, median or mode) • Usually lack measures of variation • Terminology and meaning is often taxon-specific \| Field guides, books \| Typically lifespan, age at maturity, gestation interval, clutch size, frequency of reproduction, mass at birth and at maturity \| \| Simple traits, derived from published measurements \| • Verifiable using primary literature • Often available in open databases • Often include population-level replicates or estimates of variation \| • Available for fewer species, especially in more charismatic taxa • Verifiable sources may be difficult to find \| Primary literature, open databases \| Databases containing many thousands of species for mammals, fish, reptiles, birds, amphibians, flowering plants (e.g., [4,52]) \| \| Life cycle models with age- or stage-based schedules of survival and reproduction (e.g., life tables, projection matrices, integral projection models) \| • Quantifies whole lifespan • Popular for both plants and animals • Verifiable using primary literature • Often available in open databases • Often include population-level replicates or estimates of variation • Large toolbox of methods to derive diverse life history trait measurements • Can generate derived life history traits (see below) • Can also be used to generate measures of population performance and resilience (e.g., [35,55]) \| • Not available for most species, mostly concerns tetrapods with many broad taxa neglected • Data labor-intensive to collect • Often synthesized from multiple sources (sometimes even interspecific) • Vital rates measured with variable precision and often contain errors in inference or parameterization • Vary in length/dimension • Require expertise to handle data and calculate derived measures (usually programming) \| Primary literature, open databases \| Databases available for plants and animals [2,3,53], and detailed data for humans \| \| Life history traits derived from life cycle models using algebraic and computational methods \| • Benefit from all advantages of life cycle models as above • Overcome the issue that models vary in length/dimension • Standardized sets of measurements amenable to comparative analysis • Measures include entire life cycle • Measures include many which are not observable (e.g., life expectancy, generation time) \| • Suffer from disadvantages of life cycle models as above: taxonomic breadth, data requirements and sources, measurement error • Often assume conditions not met in real systems (e.g., density-independent population growth, stable age/stage structure) • Possible to conflate life history and demographic traits derived from models (e.g., asymptotic or transient population growth) \| Derivation from life cycle models \| \| \| Selection pressures on traits, describing the ‘importance’ of vital rates to fitness using the derivative of the latter with respect to the former (e.g., elasticity or sensitivity) \| • Infer the filter that converts vital rates into fitness and imposes selective outcomes • Provide well-established framework for life history variation in plants \| • Selection pressures are not life history traits per se • Often assume the same conditions as life cycle models as above • Elasticities usually have some constraints across vital rates; hence, life history trade-offs are inevitable \| Derivation from life cycle models \| \| To help standardize the trait sets used to describe life histories, we identify three key ways (currencies; Box 2) to quantify life history data, although other currencies will exist. Many studies use dimension-reduction approaches to capture information that defines life history strategies, which condense trait sets into fewer, usually orthogonal, composite variables (Box 3). Past comparative analyses have been performed on heterogeneous life history trait currencies. Mixing currencies confounds the interpretation of the structuring axes of life history variation because orthogonality can emerge from having different units. Single-currency suites of life history variables may aid in interpretation of dimension-reduction analysis. An awareness of how each trait is measured and consequences for analytical outcomes is important. Even within currencies, the units of measurement vary; for example, demographic measures mix rates, durations, ages, and frequencies. Currencies can also be confused by measuring traits in one currency as proxies for another, such as using morphological or behavioral sexual displays as proxies for allocation of energy to reproduction . Using trait sets with common currencies, alongside appropriate analytical approaches, will yield optimal results. Box 3 Statistical methods for comparative analyses of life histories A set of measurable, correlated traits can be described using a smaller number of emergent, orthogonal variables representing dominant axes of life history variation, such as the fast–slow continuum. Principal component analysis (PCA) is one common statistical method to achieve this goal (e.g., [3,7,9]). PCA is fairly naive in that it does not model measurement error or nonlinearities among traits, but some guidance exists regarding how many dimensions explain meaningful amounts of variation . It is less clear how to compare dominant axes across independent analyses with data sets comprising different traits, taxa, and sampling methodologies. Relationships between two different multivariate data sets (e.g., climatic and life history data) can be explored using canonical correlation analysis (CCA), which yields two sets of emergent uncorrelated variables by calculating axes with highest correlation between variable sets. This approach recognizes that dominant axes of variation (e.g., fast–slow) may not yield strong relationships with other ecological or evolutionary processes. Unlike PCA and CCA, factor analysis (FA) treats measured variables as functions of latent variables, with associated measurement or residual error. Because FA does not require latent variables to be uncorrelated, it offers solutions to some problems of PCA and CCA , although it comes with issues including occasional emergence of spurious latent variables. We suggest that FA, and broader structural equation models, of which FA is a special case , are better suited to the testing of hypothetical rather than data-driven axes of variation. Rather than measure axes of variation, cluster analyses (CAs) measure multidimensional boundaries of life history variation. CA may help understand, after standardizing for rankings on the fast–slow continuum (using, e.g., generation time), which life history strategies do not exist and why, as much as the clustered patterning of those that do. Hierarchical CA has been used to identify substructure in life history variation and could prove useful to apply more widely in comparative life history theory. Depending on the researcher’s perspective, a given environmental, phylogenetic, or morphological variable may drive life history variation or be a nuisance covariate to deal with statistically . Comparative analyses must account for nonindependence due to shared evolutionary history . Phylogenetic methods exist for PCA and FA but not yet for CA. Correlates of traits (e.g., organism size) can be included in analyses, although this risks deriving axes defined by non–life history traits, or regressed against life history to yield residuals for analysis, but this can introduce statistical biases . Towards a hypothesis-driven, empirically sufficient era of life history theory To overcome the contemporary challenges concerning life history data and analyses, we make the recommendations outlined in the following sections. Agreeing on a universal set of life history traits, derived from schedules of survival, development, and reproduction Such data can implicitly be measured in a single currency of rates per unit of time . This time-unit harmonization will strengthen the links between life history, demography, and fitness , given their explicit treatment of time in the Euler-Lotka equation. Such rates still need credible transformation during statistical analysis (Box 3). We further encourage the development of theories of life history evolution using state variables linked to other currencies, with energy being a primary candidate. Empirical analyses should first be restricted to traits measured in the same currency, then multivariate analyses (e.g., canonical analyses) could be used to look for covariation among dimension-specific axes of life history variation. Filling gaps in life history trait data, including for microbes, fungi, nonvascular plants and invertebrates, and especially marine species There are challenges inherent to this recommendation; for many organisms, we lack good working definitions of life cycles, let alone what constitutes an individual, reproduction, or even death. Variation in lifestyle (e.g., sedentary versus mobile), Bauplan (e.g., modular versus unitary), growth pattern (e.g., determinate versus indeterminate), and reproductive modes (e.g., sexual versus asexual) further complicate the comparative landscape. Attempts to fill data gaps should prioritize measures to facilitate broad comparisons of life history across all taxa. Concurrently, there is a need for data on vital rate variation within species across all taxa . Time series of vital rate data exist for relatively few species and are necessary for comprehensive assessments of, for example, density-dependent mechanisms driving vital rate variation . Embracing comparative analyses covering broad taxa and levels of biological organization Besides some exceptions [42,43], multivariate studies seeking to understand axes of life history variation across kingdoms of life emerged only recently [17,44,45]. Previous studies encompassing a broad taxonomic range were limited to bivariate analyses . Naturally, taxon-specific inquiry will still play an important role, especially in data-rich taxa. However, limiting analyses to separate groups of organisms implies a perceived wisdom that gross differences in morphologies, physiologies, and lifestyles of different groups inevitably create different selection pressures on their life history strategies. This presumption should be backed up with empirical analyses of whether and how the dominant axes of life history variation change among taxa. Existing evidence is they do not . Achieving consensus regarding evolutionary ancestry across all species to implement robust phylogenetic analyses Recent phylogenetic advances have yielded trees encompassing ever-larger taxonomic groups [e.g., 46,47]. The details of ancestral relationships in many parts of these trees, particularly in deeper evolutionary time, remain debated, however, with multiple gaps to fill. We require better consensus on the best way to analytically incorporate phylogenetic covariance. The classic assumption of Brownian trait evolution is often not supported, and more realistic models such as Ornstein-Uhlenbeck do not solve the need to incorporate evolution toward trait optima and account for shifts in trait optima across taxa . There is a strong potential to find patterning in life history strategies common across all species, but we anticipate being surprised by deep-rooted differences in how natural selection has shaped strategies in particular clades. Adopting analytical approaches that infer explicit links between life history traits and emergent axes of life history variation We see a role for the wider use of factor analysis (FA) to help discover life history axes as latent factors of observed vital rates. Further development of FA algorithms will be key to help test hypotheses derived from life history theory, particularly phylogenetically controlled confirmatory FA methods. We also urge the development and application of canonical correlation analysis to reveal associations between the multivariate life history traits of species and their multivariate suites of demographic, phenotypic, and ecological features. Advancing theoretical frameworks for the macroevolution of life histories Great theoretical progress has been made in our understanding of selection on rates of survival and reproduction in stable, stochastic, and fluctuating environments . To understand variation in lifetime schedules of vital rates, there is now a need to focus more on both age and stage structuring in these models. Necessary advances include incorporation of resource allocation to growth, investigation into whether multiple life histories can maximize fitness under some circumstances, exploration of life history structure within discrete life stages, and the consideration of new theoretical frameworks such as selection for demographic resilience. Concluding remarks The fast–slow continuum is the major structuring axis of life history variation but not the only one. More hypothesis-driven research should guide the study of multivariate life history traits and the ecological and evolutionary drivers and limits. Likely candidates include different reproductive or developmental tactics (see Outstanding questions). We argue that more time spent on theory and hypothesis development will lead to more targeted efforts to better understand how life history strategies are shaped by the environment, ancestry, and Bauplan. Outstanding questions Is there a universal characterization of life histories across the tree of life? Does the fast–slow continuum explain the majority of life history variation across all taxa? It cannot, because it captures patterns across life history trajectories and does not account for age- or stage-specific variation in traits. It provides the broad picture in tetrapods (especially mammals and birds), which are relatively well studied, and only in recent years has the fast–slow continuum been shown to be important in sessile and clonal organisms (vascular plants, corals). Whether the fast–slow continuum is the major structuring axis of life history variation at the trajectory level has not been studied in most invertebrates other than insects, where the fast–slow continuum has been identified, but seems to be less structuring than in tetrapods. Are other axes of life history variation consistent among taxa? Traits relating to reproductive or developmental tactics have repeatedly emerged as the second most important axis of life history variation. However, heterogeneity in the set of traits analyzed so far prevents us from concluding taxonomic universality. Are there clearly defined clusters and boundaries of life history variation across taxa that occupy distinct spaces in multidimensional life history trait space? Biomechanical constraints limit the range of covariation among life history traits. Intuitively, certain life history strategies should not exist if they defy physical or evolutionary laws. How sensitive are examinations of life history trait variation to particular choices of data and analytical approaches? If the impact of these choices on research outcomes is significant, cognizance of such consequences is of paramount importance. Acknowledgments T. Coulson and three anonymous reviewers provided constructive feedback on earlier versions of this work. I.S. was supported by Marie Skłodowska-Curie Award (MSCA) WHYAGE (746235). R.S.G. was supported by an NERC Independent Research Fellowship (NE/M018458/1) and NERC Pushing the Frontiers (NE/X013766/1). J.M.G. was supported by the ‘DivInT’ ANR program (ANR-22-CE02-0020). 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Many life-history traits are related to body size with larger species typically exhibiting a slower pace of life and lower fecundity. However, soil-living organisms may exhibit size-independent life-history strategies due to habitat space constraints, but this has never been tested. Here, we synthesize life-history traits in springtails (Insecta: Collembola) and mites (Acari: Oribatida, Astigmata, Mesostigmata), the most abundant microarthropods worldwide, living mainly in litter and the pore space of soil. We related life-history traits to body size and individual metabolic rate, and showed that life-history traits of soil microarthropods display a trade-off between lifespan and reproductive rate, spanning a continuum from fast to slow life-history strategies. Oribatida exhibit remarkably slow life-histories and long lifespans with lower reproductive rates than Collembola, Astigmata and Mesostigmata. Despite fresh body mass of soil microarthropods varying by three orders of magnitude, fast and slow life-history strategies occurred in all size classes suggesting largely size-independent life-history strategies. Overall, these findings indicate a soil animal economics spectrum that bears key implications for understanding local biodiversity and the coexistence of soil animal species, such as how Collembola and Oribatida coexist worldwide. ### Age-dependent decline in sperm quality and function in a naturally short-lived vertebrate 2025, BMC Ecology and Evolution ### Nematode mind: Exploring the role of the RNA interference pathway in learning, memory and beyond 2025, Philosophical Transactions of the Royal Society B Biological Sciences ### Patterns and Mechanisms of Niche Partitioning Between Related Parasitoids (Hymenoptera) Sharing the Same Host Species 2025, Insects ### Toward an Understanding of Cultural Orientations Through the Lens of Human Behavioral Ecology 2025, Evolutionary Behavioral Sciences ### More social species live longer, have longer generation times and longer reproductive windows 2024, Philosophical Transactions of the Royal Society B Biological Sciences View all citing articles on Scopus Glossary Altricialspecies where newborns are helpless and require parental care. Bauplanthe generalized structural body plan that characterizes a group of organisms and especially a major taxon. Includes the set of anatomical and physiological structures that allows living functions to acquire energy from the environment, to assimilate that energy, and to transform assimilated energy into tissues. Diapausespontaneous interruption of development in some animals, marked by a strong reduction in metabolic activity. Fast–slow continuumaxis of variation in species’ life history strategies structured along two extremes: ‘fast’ species develop fast, mature early, reproduce a lot, and die young, whereas ‘slow’ species develop slowly, mature late, reproduce little, and live a long life. We use the popularized ‘fast–slow’ form; however, Stearns originally conceived it as ‘slow–fast.’ Fitnesshow good a particular genotype is at leaving offspring in the next generation relative to other genotypes in the same population. Generation timemean age of reproduction in a population. Other definitions exist, such as the average time between two consecutive generations. Genotypethe genetic makeup of an organism. Indeterminate groweran organism whose Bauplan is not genetically determined, so it continues to grow after it has achieved sexual maturity (e.g., all plants, some mushrooms, fish, amphibians, reptiles, and many mollusks). Iteroparityreproductive schedule whereby multiple reproductive events occurs during the life cycle of an organism. Life expectancythe average time period that an organism of a given age may expect to live in a population. Life expectancy at birth corresponds to the mean age at death in a population. Life history strategyLife history strategies are defined as covariations among life history traits (e.g., long life, late maturity, and long development usually display a strong positive covariation and correspond to a strategy displayed by slow-living organisms). Life history traitkey moments along the life cycle of an organism describing survival, development, and/or reproduction functions. Examples include times to events such as age at maturity or life expectancy and rates such as growth or reproduction. Lifetime reproductive successcumulative number of offspring recruited over the lifetime of an individual. It is a dimensionless number but provides an approximation of the population growth rate (and thereby gets a time dimension) when divided (on a log scale) by generation time. Precocialspecies where newborns have an advanced state (e.g., with open eyes at birth in mammals) and are able to feed themselves and move independently almost immediately. Retrogressionthe ability to regress to a smaller, younger, or less developed stage with time. Semelparityreproductive schedule whereby a single reproductive event occurs during the life cycle of an organism. In fatal semelparity, death occurs right after reproduction. Vital or demographic ratekey demographic process that shapes the dynamics of a population. At minimum, these rates include survival, development (in stage-based models), and reproduction but can also include dispersal, dormancy, etc. 10 Shared first authors 11 Shared senior authors @ X: @iainmstott (I. Stott); @Rob_SalGo (R. Salguero-Gómez). © 2024 The Authors. Published by Elsevier Ltd. Recommended articles Co-cultures: exploring interspecies culture among humans and other animals Trends in Ecology & Evolution, Volume 39, Issue 9, 2024, pp. 821-829 Cédric Sueur, Michael A.Huffman ### Problem-solving ability: a link between cognition and conservation? Trends in Ecology & Evolution, Volume 39, Issue 7, 2024, pp. 609-611 Amanda R.Ridley, Elizabeth M.Speechley ### On the hunt for facilitation in symbiont communities Trends in Ecology & Evolution, Volume 39, Issue 9, 2024, pp. 793-796 Fletcher W.Halliday, …, Isabelle Stiver ### On the multiscale dynamics of punctuated evolution Trends in Ecology & Evolution, Volume 39, Issue 8, 2024, pp. 734-744 Salva Duran-Nebreda, …, Sergi Valverde View PDF ### Blindingly transparent – anonymity in an era of openness: a reply to Cardini Trends in Ecology & Evolution, Volume 39, Issue 8, 2024, p. 702 Shinichi Nakagawa, Malgorzata Lagisz ### Science writing: avoid the peril of ‘revealing too much’ Trends in Ecology & Evolution, Volume 39, Issue 8, 2024, p. 701 Andrea Cardini Show 3 more articles Article Metrics Citations Citation Indexes 16 Captures Mendeley Readers 93 View details About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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1720	https://es.scribd.com/doc/282961146/Teoria-de-La-Elasticidad-Timoshenko-6-Metodos-Elasto-Energi	Teoria de La Elasticidad - Timoshenko - 6 Métodos Elasto-Energi \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 500 views 36 pages Teoria de La Elasticidad - Timoshenko - 6 Métodos Elasto-Energi TEORÍA DE LA ELASTICIDAD - TIMOSHENKO - CAP 6 Full description Uploaded by Rommel Castillo Go to previous items Go to next items Download Save Save Teoria de La Elasticidad - Timoshenko - 6 Métodos ... 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1721	https://brainly.com/question/13935148	[FREE] Rewrite the following equation in slope-intercept form. Identify the slope and the y-intercept. x - 3y = - brainly.com Advertisement Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +79,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +14,2k Ace exams faster, with practice that adapts to you Practice Worksheets +8,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Rewrite the following equation in slope-intercept form. Identify the slope and the y-intercept. x−3 y=0 Slope-intercept form: y=m x+b Slope (m): slope= Y-intercept (b): b= 2 See answers Explain with Learning Companion NEW Asked by aareyiahking • 11/20/2019 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 2202 people 2K 5.0 2 Upload your school material for a more relevant answer y=3 1x slope: 3 1 b: 0 Answered by zipzer0 •15 answers•2.2K people helped Thanks 2 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 2202 people 2K 5.0 2 College Physics 1e - OpenStax Physics for AP® Courses 2e - Kenneth Podolak, Henry Smith Physics 2e - Paul Peter Urone, Roger Hinrichs Upload your school material for a more relevant answer The equation x−3 y=0 can be rewritten in slope-intercept form as y=3 1x. The slope is 3 1 and the y-intercept is 0. Explanation To rewrite the equation x−3 y=0 in slope-intercept form, we need to isolate y. Step 1: Start with the original equation: x−3 y=0 Step 2: Move x to the other side: −3 y=−x Step 3: Divide both sides by −3: y=3 1x Now we have the equation in slope-intercept form, which is y=m x+b, where m is the slope and b is the y-intercept. From our equation y=3 1x: The slope (m) is 3 1. The y-intercept (b) is 0, since there is no constant term added to the equation. Therefore: Slope (m): 3 1 Y-intercept (b): 0 Examples & Evidence An example of the slope and y-intercept is that for every increase of 1 in x, y increases by 3 1. This means the line representing this equation rises one unit vertically for every three units it runs horizontally. The process of isolating y demonstrates how to solve for the slope and y-intercept, verifying that the rewritten form is correct. Thanks 2 5.0 (1 vote) Advertisement Community Answer This answer helped 20844202 people 20M 2.0 0 The slope of the equation x - 3y = 0 is 1/3 and the y-intercept is 0. Slope-Intercept Form of a Linear Equation The slope-intercept form of a linear equation is given by the equation y = mx + b, where m represents the slope and b represents the y-intercept. Slope of the Equation To find the slope of the equation x - 3y = 0, we need to rearrange the equation in slope-intercept form. Start by isolating y on one side of the equation: Subtract x from both sides: -3y = -x Divide both sides by -3: y = (1/3)x From this equation, we can see that the slope of the equation is 1/3. Y-Intercept The y-intercept represents the point where the line intersects the y-axis. In the slope-intercept form, the y-intercept is represented by the value of b. There is no constant term in the equation x - 3y = 0, which means that the y-intercept is 0. Learn more about Slope-Intercept Form of Linear Equations here: brainly.com/question/30058826 SPJ2 Answered by robinkumar9613 •22.4K answers•20.8M people helped Thanks 0 2.0 (1 vote) Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1 Is this equation correct? 6 3⋅7 3=4 2 3 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
1722	https://medium.com/@ashishku0359/combinatorics-a-comprehensive-exploration-0dc7d95f8571	Sitemap Open in app Sign in Sign in Combinatorics: A Comprehensive Exploration Ashish Kumar 6 min readJul 24, 2024 Combinatorics is a fascinating and essential branch of mathematics that deals with counting, arrangement, and combination of objects. Its principles and techniques are vital in various fields such as computer science, optimization, and probability theory. This article offers an in-depth exploration of combinatorial concepts, including fundamental counting principles, permutations, combinatorial designs, graph theory, the pigeonhole principle, and the inclusion-exclusion principle. We will also discuss their applications in algorithm design and optimization. 1. Basics of Counting Principles and Permutations 1.1 Fundamental Counting Principle The fundamental counting principle, also known as the multiplication principle, is a foundational concept in combinatorics. It states that if there are nnn ways to perform one task and mmm ways to perform another task independently, then there are n×mn \times mn×m ways to perform both tasks in sequence. 1.2 Permutations Permutations refer to the different ways to arrange a set of objects where the order matters. The number of permutations of n objects taken r at a time is given by: 1.3 Combinations Combinations refer to the selection of items where the order does not matter. The number of combinations of n objects taken r at a time is calculated using: 2. Understanding Combinatorial Designs and Graphs 2.1 Combinatorial Designs Combinatorial designs are arrangements of elements within sets according to specific criteria. These designs have applications in various fields, including experimental design and coding theory. a. Balanced Incomplete Block Designs (BIBDs) A Balanced Incomplete Block Design is a design where each pair of distinct elements appears together in exactly one block. BIBDs are characterized by parameters (v,k,λ) where: v is the number of elements. k is the number of elements in each block. λ is the number of blocks in which each pair of elements appears together. Example: Consider a design with parameters (7,3,1). This means there are 7 elements, each block contains 3 elements, and each pair of elements appears together in exactly one block. b. Projective Planes A projective plane is a type of combinatorial design where every pair of lines intersects in exactly one point, and every point lies on exactly n+1 lines. Projective planes are used in various applications, including coding theory and cryptography. Example: The finite projective plane of order 2 consists of 7 points and 7 lines, with each line containing exactly 3 points and each point lying on exactly 3 lines. 2.2 Graph Theory Graph theory studies graphs, which are mathematical structures used to model pairwise relationships between objects. A graph consists of vertices (nodes) and edges (connections between nodes). a. Types of Graphs Undirected Graphs: Edges have no direction. For example, a social network graph where friendships are mutual. Directed Graphs (Digraphs): Edges have a direction, representing one-way relationships. For example, a road map with one-way streets. Weighted Graphs: Edges have weights representing costs or distances. For example, a network of cities connected by roads with varying lengths. b. Key Concepts in Graph Theory Paths and Cycles: A path is a sequence of edges connecting vertices, while a cycle is a path that starts and ends at the same vertex. Connectivity: A graph is connected if there is a path between any pair of vertices. Otherwise, it is disconnected. Graph Coloring: The process of assigning colors to vertices so that no two adjacent vertices share the same color. This concept is used in scheduling and map coloring problems. Example: Get Ashish Kumar’s stories in your inbox Join Medium for free to get updates from this writer. Consider a graph representing a network of cities where each city is a vertex and each road is an edge. Analyzing this graph helps in optimizing routes and understanding network connectivity. 3. Exploring the Pigeonhole Principle and Inclusion-Exclusion Principle 3.1 Pigeonhole Principle The pigeonhole principle is a simple yet powerful tool in combinatorics. It states that if more items are distributed among fewer containers than there are items, then at least one container must contain more than one item. Formal Statement: If n items are put into mmm containers, and n>m, then at least one container must hold more than one item. Example: If you have 10 pairs of socks and only 9 drawers, then at least one drawer must contain more than one pair of socks. This principle is used in various proofs and problem-solving scenarios to demonstrate the existence of certain conditions. 3.2 Inclusion-Exclusion Principle The inclusion-exclusion principle is used to calculate the number of elements in the union of multiple sets by considering overlaps. For two sets A and B, the principle is: 4. Applications in Algorithm Design and Optimization 4.1 Algorithm Design Combinatorics plays a crucial role in designing and analyzing algorithms. Key areas include: a. Complexity Analysis Understanding the time and space complexity of algorithms often involves combinatorial reasoning. For example, analyzing sorting algorithms such as quicksort or mergesort involves counting the number of possible permutations of elements and understanding the impact on computational efficiency. b. Graph Algorithms Graph algorithms are fundamental in computer science and are based on combinatorial principles. For instance: Dijkstra’s Algorithm: Finds the shortest path in a weighted graph. Kruskal’s Algorithm: Computes the minimum spanning tree of a graph. Ford-Fulkerson Algorithm: Solves the maximum flow problem in a network. c. Randomized Algorithms Randomized algorithms use probabilistic methods to solve problems more efficiently. Combinatorial techniques such as the probabilistic method and the pigeonhole principle are used to analyze their performance and ensure that they produce good results with high probability. 4.2 Optimization Combinatorial optimization involves finding the best solution from a finite set of possible solutions. Key techniques include: a. Integer Programming Integer programming is an optimization technique where some or all variables are required to be integers. It is used to solve problems such as scheduling, resource allocation, and network design. Combinatorial techniques help in formulating and solving integer programming problems. b. Dynamic Programming Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems. It involves combinatorial techniques to enumerate possible solutions and compute optimal values efficiently. For example, the knapsack problem and the traveling salesman problem can be tackled using dynamic programming. c. Approximation Algorithms For NP-hard problems, finding exact solutions may be infeasible. Approximation algorithms provide near-optimal solutions using combinatorial insights and heuristics. Examples include: Greedy Algorithms: Used for problems like interval scheduling and activity selection. Local Search Algorithms: Used for problems like the traveling salesman problem and the knapsack problem. Conclusion Combinatorics is a rich and diverse field that provides essential tools and techniques for counting, arranging, and optimizing various mathematical and real-world problems. By understanding and applying combinatorial principles such as counting principles, permutations, combinatorial designs, graph theory, the pigeonhole principle, and the inclusion-exclusion principle, one can analyze complex problems, design efficient algorithms, and optimize solutions effectively. Whether in mathematics, computer science, or optimization, the concepts of combinatorics play a crucial role in advancing both theoretical and practical applications. Math Mathematics Education ## Written by Ashish Kumar 0 followers ·2 following Mentor, Teacher, education and more No responses yet Write a response What are your thoughts? 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1723	https://www.geeksforgeeks.org/physics/problems-on-self-and-mutual-inductance/	Problems on Self and Mutual Inductance Last Updated : 29 Nov, 2021 Suggest changes 3 Likes Electromagnetic induction, often known as induction, is a process in which a conductor is placed in a certain position and the magnetic field varies or remains stationary as the conductor moves. A voltage or EMF (Electromotive Force) is created across the electrical conductor as a result of this. Michael Faraday discovered this electromagnetic induction rule. He built up a leading wire similar to the diagram below, which he linked to a device that measured the voltage across the circuit. The voltage in the circuit is measured when a bar magnet passes through the device. The significance of this is that it is a method of creating electrical energy in a circuit by employing magnetic fields rather than batteries. The principle of electromagnetic induction is used by equipment such as generators, transformers, and motors. Inductor and Inductance A resistor is an electrical component that opposes and regulates the flow of electricity. Resistance is the ability of a conductor to resist current flow. While a capacitor is a device that briefly stores charge and energy, capacitance refers to a capacitor's ability to store energy. An inductor is nothing more than a coil. To make a coil, a conducting wire is tightly coiled. It will produce its constant magnetic field when a direct current is supplied through it. Instead, if we use an alternating current or a continually changing source, the current flowing through the coil will change. It will induce its emf in the opposite direction to the supply due to shifting magnetic flux. This is referred to as inductance. So, inductance refers to a conductor's ability to resist changing current, but an inductor is an electrical device that does so. Self-Inductance The battery will produce a constant current in the coil if the rheostat resistance is kept constant. A continuous magnetic field will be induced inside the coil due to the constant current provided to it. As the resistance of the rheostat is changed, the current flowing through the coil will also change. Since the current is changing, they will be a changing magnetic flux inside the coil. Due to the effect of changing magnetic flux, an emf will be induced inside this coil, trying to oppose the magnetic flux. Hence, due to the induced emf, the direction of current induced will be opposite to that of current supplied. The current flowing in the coil determines the flux induced in the coil: ϕ ∝ I As a result, the flux-to-current ratio must be constant, as this will define the coil's capacity to create magnetic flux in relation to the current provided. Self-inductance is the name for this constant (L). An inductor is the result of this process. If there are N turns on the coil, then Differentiating this equation in terms of time on both sides d/dt (Nϕ)=d/dt (LI) N dϕ/dt=L dI/dt But, according to Faraday’s law of EMI, emf induced in a coil is, ε=–N dϕ/dt The negative sign indicates that the induced emf is in the opposite direction of the present rate of change. ε=–L dI/dt L=–ε /dI/dt As a result, self-inductance may alternatively be defined as the emf induced per unit rate of current change within the coil minus the emf induced per unit rate of current change. The SI unit of inductance is: Wb/A=Vs/A=Henry (H) Magnetic Energy Stored in an Inductor An inductor stores energy in the form of a magnetic field because it may induce its emf. Let us calculate the magnetic energy stored in an inductor using the following equation: The rate of work done for a current I in the circuit may be represented as: dW/dt=ε I Substituting the following equation for the coil's induced emf: dW/dt=–L dI/dt I dW=–LI dI Integrating both sides of this equation: ∫dW=∫–LI dI W=–1/2 LI2 Work done is the inverse of energy stored. So, if the inductor's initial magnetic energy is zero, the energy stored in the inductor is: U = 1/2 LI2 Mutual Inductance Self-inductance is similar to mutual inductance. Instead, it consists of only two coils. One of them is given current, while the other is given emf. Coil-1 is linked to a supply and a rheostat in the diagram, whereas coil-2 is connected to a galvanometer. Coil-1 will begin to flow with a fluctuating current. It will be induced with a shifting magnetic flux as a result of this. Because the coils are so near together, the changing magnetic flux will be connected to the other. Because the magnetic flux through coil-2 is constantly changing with respect to time, an emf opposing this magnetic flux will be induced in coil-2, causing a current to flow in the opposite direction of the provided current. As a consequence, we may claim that the current delivered in coil-1 causes the flux via coil-2. ϕ2 ∝ I1 As a result, the flux through coil-2 and the current delivered in coil-1 will have a constant ratio. Mutual inductance in coil-2 owing to coil-1 is the name for this constant. M21=ϕ2 / I1 We may define mutual inductance between two coils as the ratio of flux associated with one coil per unit current delivered in the other coil using the following equation. ϕ2=M21 I1 Taking the preceding equation and differentiating it with regard to time d/dt(ϕ2)=d/dt(M21 I1) Using Faraday’s law of EMI ε2=–M21 dI1/dt M21=–ε2/dI1/dt As a result, mutual inductance may alternatively be defined as the emf induced per unit rate of change in current in the inducing coil minus the emf induced per unit rate of change in current in the inducing coil. Applications of Inductors Choking circuits mostly employ inductors (choke coil). In a transformer, the mutual inductance concept is applied. As the formula suggests, they are employed to store energy. To create alternating current, it's used in an LC oscillator circuit. Used in power converters as well (ac-ac or dc-dc). Sample Questions Question 1: When a current of 2mA is supplied to a coil with 100 turns, a magnetic flux of magnitude 0.2Wb is linked with it. Find the self-inductance of this coil. Answer: Current supplied to the coil I=2mA Number of turns in the coil (N)=100 Magnetic flux linked with the coil (ϕ)=0.2Wb Self-inductance of a coil is given by the equation L=Nϕ/I Substituting the values L=100×0.2/2 ∴ L=10H This coil has a self-inductance of 10H. Question 2: Determine the energy stored in an inductor of inductance 100mH when a current of 0.2A is passed through it. Answer: The inductance of the inductor (L)=100mH Current passed through it (I)=0.2A Energy stored in an inductor is given by the equation U=1/2LI2 Substituting the values U=1/2×100×10–3×0.2×0.2 ∴U=2mJ The energy stored in this coil is 2mJ. Question 3: What is the principle of mutual induction? Answer: Mutual induction is based on the electromagnetic induction concept. When one coil's magnetic field connects to the other, the second coil generates its own emf. Mutual induction is the term for this procedure. Question 4: Two long solenoids having length L and cross-sectional area A are placed in such a way that their axis coincide. Find the mutual inductance of this system. Answer: A long solenoid's magnetic field is provided by: B=μ0NI/L As a result, if electricity is delivered through one of the solenoids, the magnetic flux within it will be connected to the other. As a result, this system's mutual induction will be: M=B1A/L ∴ M=μ0N1N2A/L Question 5: What are the applications of inductors? Answer: Inductors are commonly used in choking circuits (choke coil). The mutual inductance notion is used in transformers. They are used to store energy, as the formula says. It's utilised in an LC oscillator circuit to generate alternating current. It's also used in power converters (ac-ac or dc-dc). Question 6: What is the use of mutual induction? Answer: Mutual induction is used in transformers, generators, motors, etc. Question 7: What is the use of Inductance? Answer: Inductance is required to induce back emf and keep current flowing even after the switch is turned off. Question 8: A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10−3 Wb. find self-inductance. Answer: Φ=4×10−3 Wb/turn N= Number of turns =500 I= Current flowing =2A N×Φ=L×I ∴500×4×10−3 =L×2 ∴L= 500×4×10−3/2 ∴L=1H ∴ Self-inductance of solenoid is 1.0H Question 9: A solenoid (air core) has 400 turns, is 20 cm long and has a cross-section of 4cm2. Then find the coefficient of self-induction. Answer: n=400turns ,l=20cm ,A=4cm2 L= 1⋅25×10−6Hm−1×400×400×4×100 / 100×100×20 =4×10−4H Question 10: When a current of 4A between two coils changes to 12A in 0.5s in primary and induces an emf of 50mV in the secondary. Calculate the mutual inductance between the two coils. Answer: M= e2/ di2/dt =(50 x 10-3) / (8/0.5) =3.125 x 10-3 H =3.125 mH A amanarora3dec Improve Article Tags : School Learning Physics Class 12 Physics-Class-12 Explore Physics Tutorial 15+ min read Mechanics Rest and Motion 10 min readForce 11 min readWhat is Pressure? 7 min readFriction 8 min readInertia Meaning 8 min readNewton's Laws of Motion \| Formula, Examples and Questions 9 min readUniversal Law of Gravitation 15 min readWhat is Gravity? 12 min readLaw of Conservation of Energy 11 min readFree Body Diagram 10 min readInclined Plane 10 min readWork Done 12 min readConservative Forces - Definition, Formula, Examples 7 min readEnergy 10 min readFrame of Reference 6 min read Kinematics Kinematics \| Definition, Formula, Derivation, Problems 10 min readWhat is Motion? 9 min readDistance and Displacement 5 min readSpeed and Velocity 13 min readAcceleration 9 min readWhat is Momentum Equation? 6 min readEquations of Motion: Derivations and Examples 11 min readUniform Circular Motion 9 min readProjectile Motion 15+ min readRelative Motion 10 min read Rotational Mechanics Concepts of Rotational Motion 10 min readAngular Motion 7 min readAngular Frequency 10 min readRotational Kinetic Energy 7 min readTorque 10 min readAngular Momentum 10 min readCentre of Mass 15 min readCentre of Gravity 8 min readRadius of Gyration 11 min readMoment of Inertia 15+ min read Fluid Mechanics Mechanical Properties of Fluids 11 min readWhat is Viscosity? 10 min readBuoyant Force 13 min readArchimedes Principle 12 min readPascal's Law 10 min readReynolds Number 6 min readStreamline Flow 7 min readLaminar and Turbulent Flow 9 min readBernoulli's Principle 14 min readPoiseuilles Law Formula 4 min readStoke's Law 11 min read Solid Mechanics What is Stress? 9 min readStress and Strain 12 min readStress-Strain Curve 11 min readElasticity and Plasticity 9 min readModulus of Elasticity 12 min readModulus of Rigidity 11 min readYoung's Modulus 8 min readBulk Modulus Formula 7 min readShear Modulus and Bulk Modulus 7 min readPoisson's Ratio 9 min readStress, Strain and Elastic Potential Energy 9 min read Thermodynamics Basics Concepts of Thermodynamics 12 min readZeroth Law of Thermodynamics 7 min readFirst Law of Thermodynamics 8 min readSecond Law of Thermodynamics 10 min readThermodynamic Cycles 15 min readThermodynamic State Variables and Equation of State 5 min readEnthalpy: Definition, Formula and Reactions 12 min readState Functions 7 min readCarnot Engine 5 min readHeat Engine - Definition, Working, PV Diagram, Efficiency, Types 14 min read Wave and Oscillation Introduction to Waves - Definition, Types, Properties 11 min readWave Motion 12 min readOscillation 8 min readOscillatory Motion Formula 3 min readAmplitude Formula 6 min readWhat is Frequency? 9 min readAmplitude, Time Period and Frequency of a Vibration 5 min readEnergy of a Wave Formula 7 min readSimple Harmonic Motion 15+ min readDisplacement in Simple Harmonic Motion 10 min read Sound Production and Propagation of Sound 6 min readWhat are the Characteristics of Sound Waves? 7 min readSpeed of Sound 12 min readReflection of Sound 9 min readRefraction of Sound 5 min readHow do we hear? 7 min readAudible and Inaudible Sounds 10 min readExplain the Working and Application of SONAR 8 min readNoise Pollution 8 min readDoppler Effect - Definition, Formula, Examples 7 min readDoppler Shift Formula 3 min read Electrostatics Electrostatics 13 min readElectric Charge 8 min readCoulomb's Law 9 min readElectric Dipole 11 min readDipole Moment 6 min readElectrostatic Potential 12 min readElectric Potential Energy 15+ min readPotential due to an Electric Dipole 7 min readEquipotential Surfaces 9 min readCapacitor and Capacitance 11 min read
1724	https://math.libretexts.org/Courses/Cosumnes_River_College/Math_373%3A_Trigonometry_for_Calculus/08%3A_Non-Right_Triangle_Trigonometry/8.01%3A_The_Law_of_Sines	8.1: The Law of Sines - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode Chapter 8: Non-Right Triangle Trigonometry Math 373: Trigonometry for Calculus { "8.1.01:_Resources_and_Key_Concepts" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.1.02:_Homework" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "8.01:The_Law_of_Sines" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.02:_The_Law_of_Cosines" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.03:_Areas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.04:_Vectors-A_Geometric_Approach" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.05:_Vectors-_An_Algebraic_Approach" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Triangles_and_Circles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Coordinate_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Radian_Measure_and_the_Circular_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Graphs_of_the_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Analytic_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Trigonometric_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Non-Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Polar_Coordinates_and_Complex_Numbers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_The_Language_of_Mathematics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 20 Jul 2025 23:05:13 GMT 8.1: The Law of Sines 145911 145911 Roy Simpson { } Anonymous Anonymous 2 false false [ "article:topic", "Law of Sines", "stage:final", "license:ccbysa", "showtoc:no", "parallax", "author@Roy Simpson", "authorname:roysimpson", "licenseversion:12", "Ambiguous case" ] [ "article:topic", "Law of Sines", "stage:final", "license:ccbysa", "showtoc:no", "parallax", "author@Roy Simpson", "authorname:roysimpson", "licenseversion:12", "Ambiguous case" ] Search site Search Search Go back to previous article Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. 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Chapter 8: Non-Right Triangle Trigonometry 6. 8.1: The Law of Sines Expand/collapse global location 8.1: The Law of Sines Last updated Save as PDF Page ID 145911 Roy Simpson Cosumnes River College ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{#1}}} ) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\AA}{\unicode[.8,0]{x212B}}) ( \newcommand{\vectorA}{\vec{#1}} % arrow) ( \newcommand{\vectorAt}{\vec{\text{#1}}} % arrow) ( \newcommand{\vectorB}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}) ( \newcommand{\vectorC}{\textbf{#1}}) ( \newcommand{\vectorD}{\overrightarrow{#1}}) ( \newcommand{\vectorDt}{\overrightarrow{\text{#1}}}) ( \newcommand{\vectE}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} ) ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{#1}}} ) (\newcommand{\avec}{\mathbf a}) (\newcommand{\bvec}{\mathbf b}) (\newcommand{\cvec}{\mathbf c}) (\newcommand{\dvec}{\mathbf d}) (\newcommand{\dtil}{\widetilde{\mathbf d}}) (\newcommand{\evec}{\mathbf e}) (\newcommand{\fvec}{\mathbf f}) (\newcommand{\nvec}{\mathbf n}) (\newcommand{\pvec}{\mathbf p}) (\newcommand{\qvec}{\mathbf q}) (\newcommand{\svec}{\mathbf s}) (\newcommand{\tvec}{\mathbf t}) (\newcommand{\uvec}{\mathbf u}) (\newcommand{\vvec}{\mathbf v}) (\newcommand{\wvec}{\mathbf w}) (\newcommand{\xvec}{\mathbf x}) (\newcommand{\yvec}{\mathbf y}) (\newcommand{\zvec}{\mathbf z}) (\newcommand{\rvec}{\mathbf r}) (\newcommand{\mvec}{\mathbf m}) (\newcommand{\zerovec}{\mathbf 0}) (\newcommand{\onevec}{\mathbf 1}) (\newcommand{\real}{\mathbb R}) (\newcommand{\twovec}{\left[\begin{array}{r}#1 \ #2 \end{array}\right]}) (\newcommand{\ctwovec}{\left[\begin{array}{c}#1 \ #2 \end{array}\right]}) (\newcommand{\threevec}{\left[\begin{array}{r}#1 \ #2 \ #3 \end{array}\right]}) (\newcommand{\cthreevec}{\left[\begin{array}{c}#1 \ #2 \ #3 \end{array}\right]}) (\newcommand{\fourvec}{\left[\begin{array}{r}#1 \ #2 \ #3 \ #4 \end{array}\right]}) (\newcommand{\cfourvec}{\left[\begin{array}{c}#1 \ #2 \ #3 \ #4 \end{array}\right]}) (\newcommand{\fivevec}{\left[\begin{array}{r}#1 \ #2 \ #3 \ #4 \ #5 \ \end{array}\right]}) (\newcommand{\cfivevec}{\left[\begin{array}{c}#1 \ #2 \ #3 \ #4 \ #5 \ \end{array}\right]}) (\newcommand{\mattwo}{\left[\begin{array}{rr}#1 \amp #2 \ #3 \amp #4 \ \end{array}\right]}) (\newcommand{\laspan}{\text{Span}{#1}}) (\newcommand{\bcal}{\cal B}) (\newcommand{\ccal}{\cal C}) (\newcommand{\scal}{\cal S}) (\newcommand{\wcal}{\cal W}) (\newcommand{\ecal}{\cal E}) (\newcommand{\coords}{\left{#1\right}_{#2}}) (\newcommand{\gray}{\color{gray}{#1}}) (\newcommand{\lgray}{\color{lightgray}{#1}}) (\newcommand{\rank}{\operatorname{rank}}) (\newcommand{\row}{\text{Row}}) (\newcommand{\col}{\text{Col}}) (\renewcommand{\row}{\text{Row}}) (\newcommand{\nul}{\text{Nul}}) (\newcommand{\var}{\text{Var}}) (\newcommand{\corr}{\text{corr}}) (\newcommand{\len}{\left\|#1\right\|}) (\newcommand{\bbar}{\overline{\bvec}}) (\newcommand{\bhat}{\widehat{\bvec}}) (\newcommand{\bperp}{\bvec^\perp}) (\newcommand{\xhat}{\widehat{\xvec}}) (\newcommand{\vhat}{\widehat{\vvec}}) (\newcommand{\uhat}{\widehat{\uvec}}) (\newcommand{\what}{\widehat{\wvec}}) (\newcommand{\Sighat}{\widehat{\Sigma}}) (\newcommand{\lt}{<}) (\newcommand{\gt}{>}) (\newcommand{\amp}{&}) (\definecolor{fillinmathshade}{gray}{0.9}) Note to the Instructor- All of the Law of Sines Learning Objectives The Law of Sines Theorem: Law of Sines Example ( \PageIndex{ 1 } ) Advice: Save Calculator Work for the Last Step Checkpoint ( \PageIndex{ 1 } ) Solving Triangles with the Law of Sines Example ( \PageIndex{ 2 } ) Checkpoint ( \PageIndex{ 2 } ) Finding an Angle Example ( \PageIndex{ 3 } ) Caution: Do Not Use Rounded Values to Calculate Checkpoint ( \PageIndex{ 3 } ) Note:Unpacking Law of Sines Subtleties Example ( \PageIndex{ 4 } ) Checkpoint ( \PageIndex{ 4 } ) Example (\PageIndex{5}) Checkpoint (\PageIndex{5}) Visualizing the Law of Sines Cases The Law of Sines - Ambiguous Case The "No Solution" Case The "One Solution" Cases The "Two Solutions" Case Applications Requiring the Law of Sines Example (\PageIndex{6}) Checkpoint (\PageIndex{6}) Example ( \PageIndex{ 7 } ) Checkpoint ( \PageIndex{ 7 } ) Example ( \PageIndex{ 8 } ) Checkpoint ( \PageIndex{ 8 } ) Footnotes Note to the Instructor- All of the Law of Sines This is a very thorough section of all cases for the Law of Sines and their applications. Please pay close attention to my tabular method. Learning Objectives Use the Law of Sines to find a missing side in an oblique triangle. Solve an application using the Law of Sines. Use the Law of Sines to find all possible solutions for the ambiguous case. Solve an application involving the ambiguous case. Hawk A.I. Section-Specific Tutor Please note that, to access the Hawk A.I. Tutor, you will need a (free) OpenAI account. We have learned to use trigonometric ratios to solve right triangles. However, the ratios are only valid for the sides of right triangles. Can we find unknown sides or angles in an oblique triangle? Figure ( \PageIndex{ 1 } ) In this section and the next, we find relationships between the sides and angles of oblique triangles. These relationships are called the Law of Sines and the Law of Cosines. We use what we already know about right triangles to derive these new rules. The Law of Sines Theorem: Law of Sines If the angles of a triangle are (A), (B), and (C), and the opposite sides are (a), (b), and (c), respectively, then [\dfrac{\sin\left( A \right)}{a}=\dfrac{\sin\left( B \right)}{b}=\dfrac{\sin\left( C \right)}{c},\nonumber ]or equivalently,[\dfrac{a}{\sin\left( A \right)}=\dfrac{b}{\sin\left( B \right)}=\dfrac{c}{\sin\left( C \right)}.\nonumber ] Proof Consider the oblique triangle in the figure below. By drawing in the altitude (h) of the triangle, we create two right triangles, (\triangle B C D) and (\triangle A B D). We can write expressions in terms of (h) for (\sin \left(A\right)) and for (\sin \left(C\right)). Looking at (\triangle B C D), we see that[\dfrac{h}{a}=\sin\left( C\right).\nonumber ]Looking at (\triangle A B D), we see that[\dfrac{h}{c}=\sin \left(A\right).\nonumber ]Now we solve each of these equations for (h):[ \begin{array}{crcl} \quad & \dfrac{h}{a} = \sin\left( C \right) & \implies & h = a \, \sin\left( C \right) \[6pt] \text{and} &\dfrac{h}{c}=\sin\left( A \right) & \implies & h = c \, \sin\left( A \right) \[6pt] \end{array} \nonumber ]Therefore,[ \begin{array}{rrclcl} & a\, \sin\left( C \right) & = & c\, \sin\left( A \right) &&\[6pt] \implies & \dfrac{\sin\left( C \right)}{c} & = & \dfrac{\sin\left( A \right)}{a} & \quad & \left( \text{divide both sides by }ac \right) \[6pt] \end{array} \nonumber ]We have derived a relationship between the angles (A) and (C) and their opposite sides, (a) and (c). If we know any three of these quantities, we can find the fourth. In a similar way, by drawing in the altitude from the vertex (C), we can show that[\dfrac{\sin\left( A \right)}{a}=\dfrac{\sin\left( B \right)}{b}. \nonumber ]Putting both results together, we have proved the Law of Sines. The Law of Sines is true for any triangle, whether acute, right, or obtuse. In the following example, we use the Law of Sines to find a distance. Example ( \PageIndex{ 1 } ) Two observers on the shore sight a ship at an unknown distance from the shoreline. The observers are 400 yards apart at points (A) and (B), and they each measure the angle from the shoreline to the ship, as shown below. How far is the ship from the observer at (A)? Figure ( \PageIndex{ 2 } ) Solution First, note that (\triangle A B C) is not a right triangle, so we cannot use the trigonometric ratios directly to find the sides of this triangle. Let ( d ) be the unknown distance opposite (\angle B=79.4^{\circ}). To use the Law of Sines, we must know another angle and the side opposite that angle. Luckily, we know that side (c = 400), and we can compute the angle at the ship, (\angle C).[\angle C=180^{\circ}-\left(79.4^{\circ}+83.2^{\circ}\right)=17.4^{\circ}.\nonumber ]Now we apply the Law of Sines, using angles (B) and (C).[\begin{array}{rrclcl} & \dfrac{b}{\sin\left( B \right)} & = & \dfrac{c}{\sin\left( C \right)} & \quad & \left(\text{Law of Sines}\right) \[6pt] \implies & \dfrac{d}{\sin \left(79.4^{\circ}\right)} & = & \dfrac{400}{\sin \left(17.4^{\circ}\right)} & \quad & \left(\text{substitute}\right) \[6pt] \implies & d& = & \dfrac{400 \sin\left( 79.4^{ \circ } \right)}{\sin\left( 17.4^{ \circ } \right)} & \quad & \left(\text{multiply both sides by }\sin\left( 79.4^{ \circ } \right)\right) \[6pt] \implies & d & \approx & 1315 && \[6pt] \end{array}\nonumber ]The ship is about 1315 yards from the observer at (A). In Example ( \PageIndex{ 1 } ), we saved all calculator work for the last step. That is, we did not approximate the values of ( \sin\left( 79.4^{ \circ } \right) ) and ( \sin\left( 17.4^{ \circ } \right) ), and then use those approximations to get a further approximation - doing so would propagate errors. Advice: Save Calculator Work for the Last Step As much as possible, only reach for technology in any mathematical work at the very last moment. Another piece of advice is to collect information into tables when working through an application. This habit is especially helpful with the current material (and the material in the next section). In Example ( \PageIndex{ 1 } ), we could have collected the information into a table where the columns are the angles and their opposite side lengths. When missing information, we place a temporary question mark. Therefore, we would have started with the table[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline\angle A = 83.2^{ \circ } & a = ? \[6pt] \hline \angle B = 79.4^{ \circ }& b = ? \[6pt] \hline \angle C = ? & c = 400 \[6pt] \hline\end{array} \nonumber ]Looking at that first column, we realize these are the interior angles of a triangle and, as such, should sum to ( 180^{ \circ} ). Hence,[ \angle C = 180^{ \circ }- 83.2^{ \circ } - 79.4^{ \circ } = 17.4^{ \circ }. \nonumber ]Thus, our table becomes[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline \angle A = 83.2^{ \circ } & a = ? \[6pt] \hline \angle B = 79.4^{ \circ }& b = ? \[6pt] \hline\angle C = 17.4^{ \circ } & c = 400 \checkmark \[6pt] \hline \end{array} \nonumber ]That checkmark (( \checkmark )) is not an accident. Recall,the Law of Sines tells us[ \dfrac{a}{\sin\left( A \right)}=\dfrac{b}{\sin\left( B \right)}=\dfrac{c}{\sin\left( C \right)}.\nonumber ]If we have the sine of an angle and the length of its opposite side, then we know the value of these ratios. That is, once we have a complete row in our table, we can try to use the Law of Sines.1Since our goal is to find the side length opposite ( \angle B ) (which, for the sake of building the table, I call ( b )), we work with the equation[\dfrac{b}{\sin\left( B \right)}=\dfrac{c}{\sin\left( C \right)}. \nonumber ]We substitute the values from the table to get[\dfrac{b}{\sin \left(79.4^{\circ}\right)} = \dfrac{400}{\sin \left(17.4^{\circ}\right)} \nonumber ]and finish as we did in Example ( \PageIndex{ 1 } ). Checkpoint ( \PageIndex{ 1 } ) Delbert and Francine are 40 feet apart on one side of a river. They make angle measurements to a pine tree on the opposite shore, as shown below. What is the distance from Francine to the pine tree? Figure ( \PageIndex{ 3 } ) Answer About 114.8 feet Solving Triangles with the Law of Sines To apply the Law of Sines to find a side, we must know one angle of the triangle and its opposite side (either (a) and (A), or (b) and (B), or (c) and (C) ), and one other angle. Then, we can find the side opposite that angle. Example ( \PageIndex{ 2 } ) In the triangle shown below, (A=37^{\circ}), (B=54^{\circ}), and (a=11). Find (b). Solve the triangle. Figure ( \PageIndex{ 4 } ) Solution We begin by building our table of given information (using question marks where we aren't explicitly given information).[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 37^{ \circ } & a = 11 \checkmark \[6pt] \hline B = 54^{ \circ } & b = ? \[6pt] \hline C = ? & c = ? \[6pt] \hline \end{array} \nonumber ]We already have a complete row! This means we can immediately try using the Law of Sines; however, I cannot resist finding the value of ( C ) when it is so easy to compute.[ C = 180^{ \circ} - 54^{ \circ } - 37^{ \circ } = 89^{ \circ }. \nonumber ]Hence, our table becomes[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 37^{ \circ } & a = 11 \checkmark \[6pt] \hline B = 54^{ \circ } & b = ? \[6pt] \hline C = 89^{ \circ }& c = ? \[6pt] \hline \end{array} \nonumber ] We use the Law of Sines with (a) and angle (A) to find (b).[\begin{array}{rrclcl} & \dfrac{a}{\sin\left( A \right)} & = & \dfrac{b}{\sin\left( B \right)} & \quad & \left( \text{Law of Sines} \right) \[6pt] \implies &\dfrac{11}{\sin \left(37^{\circ}\right)} & = & \dfrac{b}{\sin \left(54^{\circ}\right)} & \quad & \left( \text{substitute} \right)\[6pt] \implies &\dfrac{11 \sin\left( 54^{ \circ } \right)}{\sin \left(37^{\circ}\right)} & = & b & \quad & \left( \text{multiply both sides by }\sin\left( 54^{ \circ } \right) \right) \[6pt] \end{array} \nonumber ]Evaluating this expression with a calculator, we find that (b \approx 14.79). We fill this in our table for display purposes only(we will not use the rounded value for computing).[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 37^{ \circ } & a = 11 \checkmark \[6pt] \hline B = 54^{ \circ } & b \approx 14.79 \[6pt] \hline C = 89^{ \circ }& c = ? \[6pt] \hline \end{array} \nonumber ] 2. At this point, we are very close to having the triangle solved. We only need the length of side ( c ). Using the given side, (a),is safer than using the value we calculated for (b).[ \begin{array}{rrclcl} & \dfrac{a}{\sin\left( A \right)} & = & \dfrac{c}{\sin\left( C \right)} & \quad & \left( \text{Law of Sines}\right) \[6pt] \implies &\dfrac{11}{\sin \left(37^{\circ}\right)} & = & \dfrac{c}{\sin \left(89^{\circ}\right)} & \quad & \left( \text{substitute} \right)\[6pt] \implies &\dfrac{11 \sin\left( 89^{ \circ } \right)}{\sin\left(37^{\circ}\right)} & = & c & \quad & \left( \text{multiply both sides by }\sin\left( 89^{ \circ } \right) \right) \[6pt] \end{array} \nonumber ]So (c \approx 18.28). Now our table is complete, and we have solved the triangle.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 37^{ \circ } & a = 11 \checkmark \[6pt] \hline B = 54^{ \circ } & b \approx 14.79 \[6pt] \hline C = 89^{ \circ }& c \approx 18.28\[6pt] \hline \end{array} \nonumber ] Checkpoint ( \PageIndex{ 2 } ) In the triangle below, (A=65^{\circ}), (C=42^{\circ}), and (c=16). Solve the triangle. Figure ( \PageIndex{ 5 } ) Answer (B=73^{\circ}, b \approx 22.87, a \approx 21.68) Finding an Angle We can also use the Law of Sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them. Example ( \PageIndex{ 3 } ) In the triangle below, (B=55^{\circ}), (a=5), and (b=11). Solve the triangle. Figure ( \PageIndex{ 6} ) Solution Again, let's start by filling out a table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = ?& a = 5\[6pt] \hline B = 55^{ \circ } & b = 11\checkmark \[6pt] \hline C = ?& c = ? \[6pt] \hline \end{array} \nonumber ]This is the first time where we cannot simply find the remaining angles using the fact that the sum of angles in a triangle is ( 180^{ \circ} ). Luckily, we still have a complete row so we can use the Law of Sines, and we will do so to find (\sin\left( A \right)).[ \begin{array}{rrclcl} &\dfrac{\sin\left( A \right)}{a} & = & \dfrac{\sin\left( B \right)}{b} & \quad & \left( \text{Law of Sines} \right) \[6pt] \implies &\dfrac{\sin\left( A \right)}{5} & = & \dfrac{\sin \left(55^{\circ}\right)}{11} & \quad & \left( \text{substitute} \right) \[6pt] \implies &\sin\left( A \right) & = & \dfrac{5 \sin \left(55^{\circ}\right)}{11} & \quad & \left( \text{multiply both sides by }5 \right) \[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{5 \sin \left(55^{\circ}\right)}{11} \right) & \quad & \left( \text{finding the reference angle - } \right. \[6pt] &&&&& \left. \text{see }\textbf{Unpacking Law of Sines Subtleties}\right) \[6pt] \end{array} \nonumber ]We have a reference angle, and we know that each angle in a triangle must be positive but less than ( 180^{ \circ } ). Therefore, ( A ) is either in ( \mathrm{QI} ) or ( \mathrm{QII} ).2Here's where a little extra logic comes into play. Since the length of side ( b ) is greater than the length of side ( a ), it must be the case that ( B \gt A ). Since ( B = 55^{ \circ } ), ( 0 \lt A \lt 55^{ \circ } ). So[A = \sin^{-1}\left(\dfrac{5 \sin \left(55^{\circ}\right)}{11}\right) \approx 21.9^{\circ}.\nonumber ]Let's update our table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A \approx 21.9^{ \circ } & a = 5\[6pt] \hline B = 55^{ \circ } & b = 11\checkmark \[6pt] \hline C = ?& c = ? \[6pt] \hline \end{array} \nonumber ]Now that we know two angles, we can find the third.[C = 180^{\circ} - B - A\approx 103.1^{\circ}. \nonumber ]Again,we update the table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A \approx 21.9^{ \circ } & a = 5\[6pt] \hline B = 55^{ \circ } & b = 11\checkmark \[6pt] \hline C \approx 103.1^{ \circ }& c = ? \[6pt] \hline \end{array} \nonumber ]Finally, we use the Law of Sines to find side (c).[ \begin{array}{rrclcl} &\dfrac{b}{\sin\left( B \right)} & = & \dfrac{c}{\sin\left( C \right)} & \quad & \left(\text {Law of Sines}\right) \[6pt] \implies & \dfrac{b \, \sin\left( C \right)}{\sin\left( B \right)} & = & c & \quad & \left( \text{multiplying both sides by }\sin\left( C \right) \right) \[6pt] \end{array} \nonumber ]Substituting in the values of ( b ), ( B), and ( C ) (stored in the calculator)and evaluating this expression with a calculator gives (c \approx 13.1). Thus, our final table is the following:[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A \approx 21.9^{ \circ } & a = 5\[6pt] \hline B = 55^{ \circ } & b = 11\checkmark \[6pt] \hline C \approx 103.1^{ \circ }& c \approx 13.1\[6pt] \hline \end{array} \nonumber ] Caution: Do Not Use Rounded Values to Calculate Example ( \PageIndex{ 3 } ) has a lot going on; however, the simplest thing to cover at this point is proper etiquette when it comes to calculations. While I am displaying the approximations in the tables within Example ( \PageIndex{ 3 } ), any computations relying on these are done using the exact or stored values in the calculator.For example, when I do any computation relying upon ( A ), I either use ( \sin^{-1}\left(\frac{5 \sin \left(55^{\circ}\right)}{11}\right))or the unrounded approximation stored in my calculator. This is done to avoid the propagation of rounding errors. All modern calculators can store results, and I suggest you learn how to use that feature. For example, on the calculator I use (the TI-30XS), there is a button labeled ( \fbox{$\mathrm{sto}\rightarrow$} ) and another button labeled ( \fbox{$\mathrm{X}^{\mathrm{yzt}}{\mathrm{abc}}$} ). To store the result of( \sin^{-1}\left(\frac{5 \sin \left(55^{\circ}\right)}{11}\right)), I use the following button sequence:[ \fbox{$\mathrm{sto}\rightarrow$} \quad \fbox{$\mathrm{X}^{\mathrm{yzt}}{\mathrm{abc}}$} \, \left( \text{tapping this a few times until }a\text{ is displayed} \right)\quad \fbox{$\mathrm{enter}$} \nonumber ]When I need to access that stored value, I tap(\fbox{$\mathrm{X}^{\mathrm{yzt}}_{\mathrm{abc}}$} )until I get the variable I stored it as, and then hit ( \fbox{$\mathrm{enter}$} ). Checkpoint ( \PageIndex{ 3 } ) Sketch a triangle with (C=93^{\circ}),(a=7), and (c=11). Use the Law of Sines to find another angle of the triangle. Solve the triangle, and label your sketch with the results. Answers 1. (A \approx 39.5^{\circ}) 2. (b \approx 8.13, B \approx 47.5^{\circ}) In Examples ( \PageIndex{ 1 } ) and ( \PageIndex{ 2 } ), we used the Law of Sines to find missing side lengths, but we didn't need the Law of Sines to find angles. In Example ( \PageIndex{ 3 } ) (and Checkpoint ( \PageIndex{ 3 } )), on the other hand,we had no choice but to use the Law of Sines to find a missing angle - this is where we need to be careful when using the Law of Sines. Note:Unpacking Law of Sines Subtleties It's imperative to recognize that all angles within a triangle must be positive but less than ( 180^{ \circ } ). Thus, the sine of any angle within a triangle will always be positive. Since the side lengths are also positive, the arcsine we use to find the missing angle will always return an acute angle (i.e., an angle between ( 0^{ \circ } ) and ( 90^{ \circ } )). "Why do we need to be aware of this?" When using the Law of Sines to find a missing angle in a triangle, there could be two such triangles satisfying the given conditions - one containing the acute returned by the arcsine,and the other containing its supplement. "How do we know when we have two triangles for a given set of information?" When using the Law of Sines to find a missing angle, we consider the angle returned by the arcsine to be a reference angle. To help with this discussion, let's call this reference angle ( \widehat{\theta} ). ( \widehat{\theta}) gives two possibilities for the true value of the angle we seek. The first is ( \theta = \widehat{\theta}). The second is its supplement, ( \theta^{\prime} = 180^{ \circ } - \widehat{\theta}) (the "prime" notation is traditionally used for the supplementary angle). While ( \theta ) is the solution that gives one triangle, ( \theta^{\prime} ) is a candidate for a second value (and, therefore, a second triangle). This is because the sine is positive and, therefore, its argument could also be between ( 90^{ \circ } ) and( 180^{ \circ } ). With our candidate angles, ( \theta ) and ( \theta^{\prime} ), in hand, we then check whether the other given angle works with each of these computed values. Sometimes, the provided information works with both candidate values; however,at the very least, the acute angle always works. To illustrate this concept, let's look back at Example ( \PageIndex{ 3 } ). We found (\widehat{A}=\sin^{-1}\left(\frac{5 \sin \left(55^{\circ}\right)}{11} \right) \approx 21.9^{ \circ }). Thus, we immediately knew ( A \approx 21.9^{ \circ } ) worked and would yield a triangle; however, what about the supplement of this angle? In ( \mathrm{QII}), our alternate option for ( A ) is ( A^{\prime} = 180^{ \circ } - \widehat{A} \approx 158.1^{ \circ }); however, if we tried to fit this obtuse angle into a triangle, we would compute the final missing angle to be[ C = 180^{ \circ } - A-B \approx 180^{ \circ } - 158.1^{ \circ } -55^{ \circ }= -33.1^{ \circ }.\nonumber ]This is impossible because angles within a triangle are never negative. Thus, we only have one triangle (given by the acute angle ( A \approx 21.9^{ \circ })). Again, sometimes both angles produce (different) triangles, and sometimes only the acute angle will work. You should always check whether both angles provide solutions. Because there can be more than one solution for a triangle when using the Law of Sines to find a missing angle, we call this situation the ambiguous case for the Law of Sines. Example ( \PageIndex{ 4 } ) Solve the triangle in which (B=14.4^{\circ}), (a=8), and (b=3). Sketch the resulting triangle. Solution We start, as usual, by building a table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = ?& a = 8\[6pt] \hline B = 14.4^{ \circ } & b = 3\checkmark \[6pt] \hline C = ?& c = ? \[6pt] \hline \end{array} \nonumber ]Other than the numbers, this looks like it's going to mimic what we did in Example ( \PageIndex{ 3 } ).[\begin{array}{rrclcl} &\dfrac{\sin\left( A \right)}{a} & = & \dfrac{\sin\left( B \right)}{b} & \quad & \left(\text{Law of Sines}\right) \[6pt] \implies &\sin\left( A \right)& = & \dfrac{a \, \sin\left(B\right)}{b} & \quad & \left(\text{multiplying both sides by }a\right) \[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a \sin\left(B\right)}{b} \right)& \quad & \left( \text{finding the reference angle} \right) \[6pt] \end{array} \nonumber ]Grabbing a calculator, substituting in the known values of (B), (a), and (b), we get ( \widehat{A} \approx 41.5^{ \circ }) (again, the exact value is stored in the calculator to be used for other calculations). As stated previously, the acute angle will always work because if ( A \approx 41.5^{ \circ } ), then ( C= 180^{\circ} - A - B \approx 124.1^{\circ} ); however, before throwing all of this into a table, let's check to see if the supplementary (obtuse) angle works as well. Let[ A^{\prime} = 180^{ \circ } - A \approx 138.5^{ \circ }.\nonumber ]Then[C^{\prime} = 180^{\circ} - A^{\prime} - B \approx 27.1^{\circ}.\nonumber ]Since this is not a negative angle, this creates another workable triangle! Thus, if ( A \approx 41.5^{ \circ } ), there is room in the triangle for the final, third angle ( C \approx 124.1^{ \circ } ), and if( A^{\prime} \approx 138.5^{ \circ } ), there is room in another triangle for the final, third angle ( C^{\prime} \approx 27.1^{ \circ } ). In these ambiguous cases, we build two tables of information, one for the acute angle and one for the obtuse angle.[ \begin{array}{ccc} \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A \approx 41.5^{ \circ }& a = 8\[6pt] \hline B = 14.4^{ \circ } & b = 3\checkmark \[6pt] \hline C\approx 124.1^{ \circ }& c = ? \[6pt] \hline \end{array} & \quad &\begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A^{\prime} \approx 138.5^{ \circ}& a = 8\[6pt] \hline B = 14.4^{ \circ } & b = 3\checkmark \[6pt] \hline C^{\prime} \approx 27.1^{ \circ }& c^{\prime} = ? \[6pt] \hline \end{array} \[6pt] \end{array} \nonumber ]In either case, we need to compute the final missing sides, ( c) and ( c^{\prime} ), using the Law of Sines. Remember, when using the Law of Sines, we want to use as much of the given information as possible because it's cleaner than our approximated information. Moreover, we use the non-rounded values and save rounding for the end.[ \begin{array}{rrclcl} & \dfrac{c}{\sin\left( C \right)} & = & \dfrac{b}{\sin\left( B \right)} & \quad & \left( \text{Law of Sines} \right) \[6pt] \implies & c & = & \dfrac{b \, \sin\left(C\right)}{\sin\left(B\right)} & \quad & \left( \text{multiply both sides by }\sin\left( C\right) \right) \[6pt] \end{array}\nonumber ]This formula is the same for ( c^{\prime} ) (just replace ( C ) with ( C^{\prime} )). Hence, we grab a calculator, substitute the necessary values, and get our approximations (listed in the tables below).[ \begin{array}{ccc} \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A \approx 41.5^{ \circ }& a = 8\[6pt] \hline B = 14.4^{ \circ } & b = 3\checkmark \[6pt] \hline C\approx 124.1^{ \circ }& c \approx 9.99\[6pt] \hline \end{array} & \quad &\begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A^{\prime} \approx 138.5^{ \circ}& a = 8\[6pt] \hline B = 14.4^{ \circ } & b = 3\checkmark \[6pt] \hline C^{\prime} \approx 27.1^{ \circ }& c^{\prime} \approx 5.5 \[6pt] \hline \end{array} \[6pt] \end{array} \nonumber ]The following figure shows both triangles. Figure ( \PageIndex{ 7 } ) Checkpoint ( \PageIndex{ 4 } ) Suppose that (C=29.7^{\circ}),(b=8), and (c=5). Find two possible values for angle (B). Solve the triangle for both values of (B), and sketch both solutions. Answer 1. (B=52.4^{\circ}) or (B=127.6^{\circ}) 2. (A=97.9^{\circ}, a=10) Figure ( \PageIndex{ 8 } ) or (A=22.7^{\circ}, a=3.9) Figure ( \PageIndex{ 9 } ) Example (\PageIndex{5}) Solve the triangle in which(C=73^{\circ}), (a=15), and (c=2). Sketch your result. Solution Building a table, we get the following:[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = ?& a = 15\[6pt] \hline B = ?& b = ? \[6pt] \hline C = 73^{ \circ }& c = 2\checkmark \[6pt] \hline \end{array} \nonumber ]Seeing that we have a complete row, we use that to help us find the missing angle from the first row (angle ( A )).[\begin{array}{rrclcl} &\dfrac{\sin\left( A \right)}{a} & = & \dfrac{\sin\left( C \right)}{c} & \quad & \left(\text{Law of Sines}\right) \[6pt] \implies &\sin\left( A \right)& = & \dfrac{a \, \sin\left(C\right)}{c} & \quad & \left(\text{multiplying both sides by }a\right) \[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a \, \sin\left(C\right)}{c} \right)& \quad & \left( \text{finding the reference angle} \right) \[6pt] \end{array} \nonumber ]Grabbing a calculator, substituting in the known values of (C), (a), and (c), we find the calculator returns an error! What happened? Consider the following value:[\dfrac{a \, \sin\left(C\right)}{c}= \dfrac{15 \sin\left( 73^{ \circ } \right)}{2} \approx 7.2\nonumber ]This is what ( \sin\left( A \right) ) is supposed to equal; however, we know that the sine function is always between ( -1) and ( 1 ). What does this mean? This means there is no such triangle where(C=73^{\circ}), (a=15), and (c=2). We can see this from a sketch. Figure ( \PageIndex{ 10 } ) The angle ( C ) is large, but the length of its opposite side is relatively small (when compared to side length ( a )). If we imagine side ( c ) being able to swing freely, we can see it will never intersect side ( b ). Therefore, forming a triangle with the given information is impossible. Example ( \PageIndex{ 5 } ) illustrates the final possibility when using the Law of Sines to find a missing angle. In summary, when using the Law of Sines to find a missing angle within a triangle, one of three possibilities occurs: There is a single triangle satisfying the given information. In this case, the angle will directly result from the arcsine when using the Law of Sines. Two triangles satisfy the given information. In this case, one triangle will be the direct result of the arcsine, and the second triangle will contain the supplement of that angle. There is no such triangle satisfying the given information. In this case, your calculator will return a domain error when computing the arcsine. Checkpoint (\PageIndex{5}) You are told that, in a triangle, ( B = 30^{ \circ} ), ( b=7 ), and ( c = 15 ). Without a calculator, explain how you know no such triangle satisfies the given information. Answer While using the Law of Sines to find a missing angle in a triangle, you must compute[ sin^{-1}\left( \dfrac{15\sin\left( 30^{ \circ } \right)}{7}\right) = \sin^{-1}\left( \dfrac{15}{14} \right). \nonumber ]This means that ( \sin\left( C \right) = \frac{15}{14} ), which is not possible. Visualizing the Law of Sines Cases Up to this point, we have computationally justified that the Law of Sines results in one, two, or no triangles for a given set of information; however, the visual of why this is true is often more impactful. If we know two sides(a) and (b)of a triangle and the acute angle (\alpha) opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of (a) in relation to (b) and (\alpha), as shown below. The Law of Sines - Ambiguous Case In each of the following cases, ( h = b \sin\left( \alpha \right) ) is the altitude of the triangle. Therefore, ( b \gt h). The "No Solution" Case Condition: (a \lt h) Reason: (a)is shorter than the altitude of the (non-existent) triangle. Therefore, it is impossible for side ( a ) to intersect the base of the "triangle." The "One Solution" Cases Condition #1: (a =h) Reason: (a) is exactly the right length to make a right triangle. Condition #2: ( a \gt b\gt h ) Reason: If( a \gt b ), then one triangle is created as side ( a ) swings downward to intersect the base; however, as that side continues to swing, it is too long to strike the base again. Hence, there can be only one triangle. The "Two Solutions" Case Condition: (b \gt a \gt h) Reason: Since ( a \gt h ), side ( a ) will strike the base a it swings downward, but before becoming vertical. Moreover, because ( a \lt b ), side ( a ) will strike the base once more as it continues to swing clockwise. Applications Requiring the Law of Sines The Law of Sines is used quite a bit in applications within Trigonometry. While I could not cover all possible applications, I have chosen two that commonly occur. Example (\PageIndex{6}) A vertical communications tower is located at the top of a steep hill, as shown below. The angle of inclination of the hill is (56^{ \circ }). A guy wire is to be attached to the top of the tower and the ground 175 yards downhill from the base of the tower. The angle formed by the guy wire is (26^{ \circ }). Find the length of the cable required for the guy wire. Round your answer to the nearest yard. Figure ( \PageIndex{ 11 } ) Solution While it's nice that we have been given a sketch of the situation, I find it's best to redraw the sketch without the "fluff," labeling vertices in uppercase and side lengths in lowercase (as we have been doing this entire time). Figure ( \PageIndex{ 12 } ) We now build our table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 26^{ \circ } & a = ?\[6pt] \hline B = ?& b = ? \[6pt] \hline C = ?& c = 175 \[6pt] \hline \end{array} \nonumber ]We do not have a complete row in this table, so we might think the Law of Sines is not going to be useable; however, looking at the provided sketch, we see one additional piece of information that we have not used. The green line (side ( c )) makes an angle of ( 56^{ \circ } ) with the horizontal. If we draw a horizontal line at vertex ( B ), we can label the angle between the horizontal and the extended green line as ( 56^{ \circ} ) (see Figure ( \PageIndex{ 13 } ) below). Since the communications tower (side ( a)) is vertical, the angle between the extended green line and the tower is the complement of ( 56^{ \circ } ) (again, see Figure( \PageIndex{ 13 } ) below). This is ( 34^{ \circ } ). Figure ( \PageIndex{ 13 } ) Finally, we can see that the angle within the triangle at vertex ( B ) is the supplement of this ( 34^{ \circ } ) angle. Hence, ( B = 180^{ \circ } - 34^{ \circ } = 146^{ \circ }). Let's go ahead and place this information in our table.[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 26^{ \circ } & a = ?\[6pt] \hline B = 146^{ \circ }& b = ? \[6pt] \hline C = ?& c = 175 \[6pt] \hline \end{array} \nonumber ]Again, we do not have a complete row; however, now that we have two angles, we can easily find the third.[ C = 180^{ \circ } - A - B = 8^{ \circ } \nonumber ]Filling in the table with this information, we get the following:[ \begin{array}{\|c\|c\|} \hline \textbf{Angle} & \textbf{Opposite Side} \[6pt] \hline A = 26^{ \circ } & a = ?\[6pt] \hline B = 146^{ \circ }& b = ? \[6pt] \hline C = 8^{ \circ }& c = 175 \, \checkmark \[6pt] \hline \end{array} \nonumber ]Finally, we have a complete row in the table, which tells us that the Law of Sines is now available. We are being asked to find side length ( b ). Therefore,[\begin{array}{rrclcl} &\dfrac{b}{\sin\left( B \right)}& = & \dfrac{c}{\sin\left( C \right)} & \quad & \left(\text{Law of Sines}\right) \[6pt] \implies &b& = & \dfrac{c \, \sin\left(B\right)}{\sin\left( C \right)} & \quad & \left(\text{multiplying both sides by }\sin\left( B \right)\right) \[6pt] \end{array} \nonumber ]Substituting in the values of ( B ), ( C ), and ( c ), we get[ b = \dfrac{175 \sin\left( 146^{ \circ } \right)}{\sin\left( 8^{ \circ }\right)} \approx 703. \nonumber ]Hence, the length of the guy wire is approximately 703 yards. Checkpoint (\PageIndex{6}) Lap and Piyali are on the north side of a river that runs east to west. Lap is standing 225 meters due west of Piyali. They spot a treasure chest on the south shore of the river. For Lap, the angle between the chest and Piyali is ( 81^{ \circ } ), and for Piyali, the angle between the chest and Lap is ( 58^{ \circ } ) (see Figure ( \PageIndex{ 14 } ) below). How far is Lap from the chest? Round your answer to the nearest meter. Figure ( \PageIndex{ 14 } ) Answer 291 meters. The following example (and the subsequent Checkpoint)illustrates that the Law of Sines can be used when you might otherwise use Right Triangle Trigonometry. Example ( \PageIndex{ 7 } ) Ron wants to measure the height of a castle controlled by hostile forces. When he is as close as he can get to the castle, the angle of elevation to the top of the wall is (18.5^{\circ}). He then retreats 20 yards and measures the angle of elevation again; this time, it is (15.9^{\circ}). How tall is the castle? Figure ( \PageIndex{ 15 } ) Solution Notice that (h) is one side of the right triangle (\triangle ADC). If we find its hypotenuse, labeled (r) in Figure ( \PageIndex{ 15 } ), we can use the sine ratio to find (h). To find (r), we consider a second triangle, (\triangle ABC), as shown in Figure ( \PageIndex{ 16 } ). Figure ( \PageIndex{ 16 } ) In this triangle, we know side (BC = 20) and would like to find side (AC = r). We can use the Law of Sines to find (r), but first,we must calculate the other angles of the triangle. Now, the angle opposite (r), (\angle ABC), is the complement of (18.5^{\circ}), so[ \angle ABC = 180^{\circ} - 18.5^{\circ} = 161.5^{\circ} \nonumber]The angle opposite the 20-yard side, (\angle BAC), is[\angle B A C=180^{\circ}-\left(161.5^{\circ}+15.9^{\circ}\right)=2.6^{\circ} \nonumber]Now we can apply the Law of Sines to find (r). We have[\begin{array}{rrclcl} & \dfrac{r}{\sin \left(161.5^{\circ}\right)} & = & \dfrac{20}{\sin \left(2.6^{\circ}\right)} & \quad & \left( \text{Law of Sines} \right) \[6pt] \implies & r & = & \dfrac{20 \sin \left(161.5^{\circ}\right)}{\sin \left(2.6^{\circ}\right)} & \quad & \left( \text{multiplying both sides by }\sin\left( 161.5^{ \circ} \right)\right) \[6pt] \end{array} \nonumber ]Finally, using the right triangle (\triangle A D C), we can write[ \begin{array}{rrclcl} & \dfrac{h}{r} & = & \sin \left(15.9^{\circ}\right) & \quad & \left( \text{right triangle definition of the sine}\right) \[6pt] \implies & h & = & r \, \sin \left(15.9^{\circ}\right) & \quad & \left( \text{multiplying both sides by }r \right) \[6pt] && = &\dfrac{20 \sin \left(161.5^{\circ}\right)}{\sin \left(2.6^{\circ}\right)} \cdot \sin\left( 15.9^{ \circ } \right) && \[6pt] && \approx & 38.33 && \[6pt] \end{array} \nonumber ]The castle is about (38.33) yards tall. Checkpoint ( \PageIndex{ 7 } ) Solve the problem in the previous example again, but instead of finding (r), find the length (A B), and then use (\triangle A B D) to find (h). Answer (A B=120.79), the castle is about (38.33) yards tall. If you look at a nearby object and alternately close your left and right eyes, the object seems to jump in position. This apparent change occurs because your eyes view the object from two positions spaced several centimeters apart. If the object at point (O) is straight ahead of one eye, it appears to be at some angle (p) away from the line of sight of the other eye. The angle (p) is called the parallax of the object. Use Figure ( \PageIndex{ 17 } ) to see that (p) is also the angle between the directions to your two eyes when viewed from point (O). (What fact from Geometry justifies this statement?) Figure ( \PageIndex{ 17 } ) Astronomers use parallax to determine the distance from Earth to stars and other celestial objects. Two observers on Earth at a known distance apart both measure the direction to the star. The difference in angle between those two directions is the parallax. Example ( \PageIndex{ 8 } ) Astronomers 1000 kilometers apart observe an asteroid with a parallax of (0.001^{\circ}). How far is the asteroid from Earth? Solution We let (x) represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with a base of approximately (1000 ) km and equal sides of length (x), as shown below. Figure ( \PageIndex{ 18 } ) The base angles of the triangle are both (\frac{180^{\circ}-0.001^{\circ}}{2}=89.999^{\circ}). Thus,[\begin{array}{rrclcl} & \dfrac{x}{\sin \left(89.999^{\circ}\right)} & = & \dfrac{1000}{\sin \left(0.001^{\circ}\right)} & \quad & \left( \text{Law of Sines}\right) \[6pt] \implies & x & = & \dfrac{1000 \sin\left(89.999^{\circ}\right)}{\sin \left(0.001^{\circ}\right)} & \quad & \left( \text{multiplying both sides by }\sin\left( 89.999^{ \circ } \right) \right) \[6pt] \implies & x & \approx & 57,000,000 && \[6pt] \end{array} \nonumber ]The asteroid is about 57 million kilometers from Earth (roughly one-third of the distance to the Sun). Checkpoint ( \PageIndex{ 8 } ) Two observers 800 kilometers apart observe an object with a parallax of (0.0005^{\circ}). How far is the object from Earth? Answer About (91,673,247) km Footnotes 1I say "try" because sometimes solving a triangle using the Law of Sines will not work. We will see those cases momentarily. 2In all honesty, since ( A ) is an angle within a triangle and not an angle in standard position within a prescribed coordinate system, it is incorrect to say that ( A \in \mathrm{QI} ) or ( A \in \mathrm{QII}); however, I am okay with the "looseness" of this language. This page titled 8.1: The Law of Sines is shared under a CC BY-SA 12 license and was authored, remixed, and/or curated by Roy Simpson. Back to top Chapter 8: Non-Right Triangle Trigonometry 8.1.1: Resources and Key Concepts Was this article helpful? Yes No Recommended articles Article typeSection or PageAuthorRoy SimpsonLicenseCC BY-SALicense Version1.2Show Page TOCnoStageFinal Tags Ambiguous case author@Roy Simpson Law of Sines parallax © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
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1726	https://www.sciencedirect.com/science/article/pii/S0925772112000260	Designing and proving correct a convex hull algorithm with hypermaps in Coq - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Abstract Keywords 1. Introduction 2. Related work 3. Hypermaps, combinatorial oriented maps and their specification in Coq 4. Geometric setting 5. Convex hull and incremental algorithm 6. Designing the incremental algorithm in Coq 7. Extracting our Coq program into OCaml 8. Topologic properties 9. Geometric properties 10. Conclusion References Show full outline Cited by (16) Figures (52) Show 46 more figures Computational Geometry Volume 45, Issue 8, October 2012, Pages 436-457 Designing and proving correct a convex hull algorithm with hypermaps in Coq☆ Author links open overlay panel Christophe Brun, Jean-François Dufourd, Nicolas Magaud Show more Outline Add to Mendeley Share Cite rights and content Under an Elsevier user license Open archive Abstract This article presents the formal design of a functional algorithm which computes the convex hull of a finite set of points incrementally. This algorithm, specified in Coq, is then automatically extracted into an OCaml program which can be plugged into an interface for data input (point selection) and graphical visualization of the output. A formal proof of total correctness, relying on structural induction, is also carried out. This requires to study many topologic and geometric properties. We use a combinatorial structure, namely hypermaps, to model planar subdivisions of the plane. Formal specifications and proofs are carried out in the Calculus of Inductive Constructions and its implementation: the Coq system. Previous article in issue Next article in issue Keywords Convex hull Hypermaps Formal specifications Computer-aided proofs Coq system 1. Introduction Our general aim is to lead a formal survey in geometric modeling and computational geometry in order to improve the programming techniques and ensure the algorithms correctness. In this paper, we present a formal case study in computational geometry on a classical problem which involves elementary geometric objects: computing the incremental convex hull of a finite collection of planar points. The originality of the means we use to achieve our purpose relies on one hand on the fact that the specifications and the formal proofs of programs are expressed in the formalism of the Calculus of Inductive Constructions implemented in the Coq system, and on the other hand on the fact that we work in a topology-based geometric modeling framework where the planar subdivisions are described by combinatorial oriented maps . But, in order to be more general for the subsequent applications, we first use combinatorial hypermaps and then specialize them into combinatorial oriented maps. A (two-dimensional) hypermap is a simple algebraic structure consisting of a finite set whose elements are called darts and of two permutations on this set. It allows to model surface subdivisions (into vertices, edges and faces) and to distinguish between the topologic and geometric aspects of the studied objects. For years, we have formally described hypermaps in order to handle subdivisions and their transformations as well as to prove topologic properties of surfaces , . Our hypermap specification is done by structural induction, which makes the constructive definition of operations and the proofs of surface properties easier. However, for the moment, we almost exclusively worked on the combinatorial topology of surfaces, but we want to deal with geometric embeddings as well. That is the reason why we begin by studying a classical plane problem which is not only rich enough to highlight many interesting problems, but also simple enough to reveal them easily and completely. The geometric aspects we consider are particularly simple but fundamental in computational geometry. The embedding is straightforward and maps subdivision vertices into points, edges into line segments and faces are represented as polygonal frontiers. However, the question of the plane orientation is crucial. In our framework, it is captured using Knuthʼs axiom system for orientation . This system defines orientation according to the order in which a triple of points is enumerated in the plane (either clockwise or counterclockwise). One of its advantages is that it allows to isolate the required numerical tests and in a first step to elude the difficult numerical accuracy problems. In fact, we do not address these issues in this first attempt which instead focuses on the correctness of data structures and related operations. Real numbers are idealized using the axiom system provided in the Coq library. In this setting, our work consists in designing a functional convex hull algorithm, automatically extracting a program in OCaml augmented with input handling and a graphical display of the result, and formally proving its total correctness. This proof consists in checking the termination of the algorithm as well as highlighting several useful topologic and geometric properties. All our specification and proof development was interactively assisted by the Coq proof assistant , , . In Section 2, we list and briefly survey some related works about formalized proofs in combinatorial topology and computational geometry. In Section 3, we recall some basic mathematical definitions and properties on the combinatorial oriented hypermaps and then specify them in Coq. In Section 4, we propose a numerical model of the plane based on real numbers and slightly enlarge Knuthʼs axiom system for orientation. In Section 5, we describe our functional algorithm to build a convex hull incrementally. In Section 6, we formalize this algorithm in the Coq proof assistant. In Section 7, we explain how to extract from this description an operational program in OCaml. In Section 8, we present and prove the topologic properties required to establish the correctness of this algorithm. In Section 9, we do the same with the geometric properties. Finally, some conclusions and future works are given in Section 10. In the following, the Coq notions required to understand the developments are progressively introduced, but the details of the proofs are out of the scope of this paper. 2. Related work 2.1. Convex hull computation and subdivision modeling The computation of the convex hull of a finite set of points is one of the first and most important concepts studied in computational geometry. It has several different definitions in the literature. There are also several construction methods, such as the incremental algorithm, Jarvisʼ march or Grahamʼs scan , , , , , . In a two-dimensional setting, the convex hull is a polygon, but its construction often requires to handle broken lines or even several polygons (e.g. in the divide and conquer approach). More generally, geometric algorithms deal with irregular subdivisions of surfaces into vertices, edges and faces. Even if they may be fairly simple in most computational geometry algorithms, these subdivisions are worth being handled consistently. That is why, in our work, we have a strong focus on these subdivisions and their properties. Nowadays, a good way of studying geometric objects is to distinguish between their topological structure and their embedding. Topology, as a combinatorial tool, may be defined via a concrete datatype such as half edges, winged edges or quad-edges. A widespread approach in computer science is to encourage abstract representations of data. Combinatorial oriented maps of dimension two allow to describe, in an algebraic way, general subdivisions of closed orientable surfaces, which is exactly what we require for our study of convex hulls. However, we prefer to work first with combinatorial hypermaps which are more general, homogeneous in the two dimensions and easier to specialize depending on our needs. Thus, we shall be able to reuse our work dealing with hypermaps to study more complex subdivisions. Then, we constrain the hypermaps in order to capture combinatorial oriented maps exactly. Combinatorial oriented maps and their extensions have been studied extensively and led to several implementations in geometric modeling, , . One of the implementations was carried out in the library CGAL . In addition to topology, we need to embed hypermaps (and maps) into the oriented Euclidean plane to be able to formalize what a convex hull is. As usual, our embedding consists in mapping vertices into points of the plane, all other objects being obtained by linearization. We thus rely on the axiom system for geometric computation and orientation proposed by D. Knuth in his book “Axioms and Hulls” . This axiom system, based on orientation properties of triples of points in the plane, allows to isolate numerical accuracy issues in computations and let us focus on the logic tests required in the algorithms. This approach is particularly well-suited to carry out formal proofs of correctness of the considered algorithms. 2.2. Assisted proofs in computational geometry Formal proofs in the field of computational geometry, especially focusing on convex hull algorithms have been carried out. Pichardie and Bertot use the Coq proof system to develop a formal proof of correctness of the incremental algorithm as well as Jarvisʼ march . They also consider Knuthʼs axiom system but they simply represent convex hulls as lists of points which lead to several technicalities and is likely to prevent any further extension, especially in a three-dimensional setting. Meikle and Fleuriot use the Isabelle proof system to formally prove the correctness of a program computing convex hulls using Grahamʼs scan. Their approach relies on Hoare logic, which prevents them from having a simple functional description of the program. None of the above-mentioned works relies on any topological structure. However, hypermaps have been used highly successfully to model planar subdivisions in the formalization and proof of the four-color theorem in Coq by Gonthier et al. , . Their specification allows to prove some significant results including a proof of the Jordan curve theorem which forms the cornerstone of the proof of the four-color theorem. However, their specification approach as well as the proof techniques (using reflection in Coq) are fairly different from the methodology we follow in this paper. An in-depth comparison can be found in . At Strasbourg University, the library specifying hypermaps, onto which our present work on convex hulls is built, was successfully used to prove some basic results such as the genus theorem, Euler formula for polyhedra and a discrete version of the Jordan curve theorem . This library was also used to carry out a formal proof of correctness of a functional algorithm to perform image segmentation by merging adjacent faces and to develop a time-optimal C-program . 3. Hypermaps, combinatorial oriented maps and their specification in Coq In this section, we follow the presentation carried out in but restrict ourselves to the notions relevant to this work. We introduce the notions of combinatorial hypermaps and maps to represent our input data, intermediate computations, and the resulting convex hull. We start with mathematical definitions and then explain how to formalize such definitions in the framework of Coq. 3.1. Mathematical aspects 3.1.1. Definitions Hypermaps are one of the most general structures to describe finite surface subdivisions topologically. Definition 1 Hypermap and combinatorial oriented map (1)A (two-dimensional) hypermap is an algebraic structure M=(D,α 0,α 1), where D is a finite set, the elements of which are called darts, and where α 0, α 1 are permutations on D. (2)When α 0 is an involution without fixpoint on D (i.e. ∀x∈D, α 0(α 0(x))=x and α 0(x)≠x), then M is called a combinatorial oriented map. (3)For each dimension k∈{0,1}: if y=α k(x), y is the k-successor of x, x is the k-predecessor of y, and x and y are said to be k-linked together. Thus the combinatorial oriented maps are a subclass of the class of hypermaps. The notion of hypermaps is well suited to carry out formal proofs . However combinatorial oriented maps are much easier to use in geometric modeling , , , , . Example 1 In Fig. 1, as functions α 0 and α 1 on D={1,…,11} are permutations (i.e. one-to-one correspondences), M=(D,α 0,α 1) is a hypermap. It represents a subdivision of the plane with a triangle and a rectangle adjacent to one another; it also features a hanging line segment as well as an isolated segment. 1. Download: Download full-size image Fig. 1. An example of hypermap. In our drawings of hypermaps on surfaces, we represent each dart as a simple curved segment (a line segment in the plane) oriented from a bullet to a small stroke: 0-linked (resp. 1-linked) darts share the same small stroke (resp. bullet). By convention, we always adopt that k-successors turn counterclockwise on the plane around small strokes and bullets. Note that our hypermap definition allows the void map (i.e. D=∅) and fixpoints with respect to k. 3.1.2. Cells of hypermaps The topological cells of a hypermap (i.e. its vertices, edges, faces and connected components) can be combinatorially defined, mainly through the classical notion of orbit. Definition 2 Orbits and hypermap cells (1)Let f be a permutation in a finite set D. The orbit of x∈D for f is the dart sequence 〈f〉(x)=(x,f(x),f 2(x),…,f p−1(x)), where p, called the period of the orbit, is the smallest integer such that f p(x)=x. (2)In an hypermap M=(D,α 0,α 1), 〈α 0〉(x) is the 0-orbit or edge of dart x, 〈α 1〉(x) its 1-orbit or vertex, 〈ϕ〉(x) its face, for ϕ=α 1−1∘α 0−1. (3)The connected component of x in M, denoted by 〈α 0,α 1〉(x), is the set of darts which are accessible from x by any composition sequence of α 0 and α 1. Faces are defined, through ϕ=α 1−1∘α 0−1, for a dart traversal also in counterclockwise order, when the hypermap is drawn on a surface. Then, every face which encloses a bounded (resp. unbounded) region on its left is called internal (resp. external). Example 2 In Fig. 1, the hypermap M contains 5 edges (strokes), 6 vertices (bullets), 4 faces and 2 connected components. For instance, 〈α 0〉(7)=(7,6,8) is the edge of dart 7, 〈α 1〉(7)=(7,3,4) its vertex. Regarding ϕ which is equal to α 1−1∘α 0−1, we have ϕ(2)=7, ϕ(7)=9 and ϕ(9)=2. Then the (internal) face of 2 is 〈ϕ〉(2)=(2,7,9). In addition the (external) face of 3 is 〈ϕ〉(3)=(3,10,11,8,5). Since α 0 and α 1 are permutations, it is clear that, for Π=〈α 0〉,〈α 1〉,〈α 1−1∘α 0−1〉 or 〈α 0,α 1〉, y∈〈Π(x)〉 is equivalent to x∈〈Π(y)〉. In a combinatorial oriented map, each edge is composed of exactly 2 darts. This is standard practice in geometric modeling to represent orientable surface subdivisions , , , , . We shall adopt this approach later in this work. 3.1.3. Planarity and Euler formula Let d,e,v,f and c be the numbers of darts, edges, vertices, faces and connected components of a hypermap M=(D,α 0,α 1). Definition 3 Euler characteristic, genus, planarity (1)The Euler characteristic of M is χ=v+e+f−d. (2)The genus of M is g=c−χ/2. (3)When g=0, M is said to be planar. Example 3 In Fig. 1, the Euler characteristic of M is χ=6+5+4−11=4 and its genus g=2−χ/2=0. Consequently, the hypermap M is planar. A planar hypermap satisfies the property χ=2×c, which is a generalization of the well-known Euler formula. 3.1.4. Embedding We only consider embedding issues for combinatorial oriented maps. For this class of hypermaps, the embedding into the plane is a mapping of vertices into distinct points, edges into straight lines connecting two points (being two embedded vertices), and faces as possibly open regions of the plane. For more details on embeddings and on the planarity, the reader is referred to , . 3.2. Specifications in Coq Coq , is the implementation of the Calculus of Inductive Constructions, which is a type theory as well as a powerful higher-order intuitionistic logical framework designed to formalize and prove mathematical properties in an interactive way. All the definitions of the previous section are formalized in this framework. 3.2.1. Preliminary specifications We first define a type for the dimensional indexes 0 and 1 of an hypermap. It consists in an inductive type dim: Inductive dim : Set := zero : dim \| one : dim. All objects being typed in Coq, dim has the type Set of all concrete types. Its constructors are the constants zero and one. For each inductive type, the generic equality predicate = is built-in but its decidability is not, because the logic of Coq is intuitionistic. For dim, the latter can be established as the following lemma (note that in Coq ~ stands for logic negation, + or \/ for disjunction and /\ for conjunction): Lemma eq_dim_dec : forall (i:dim)(j:dim), {i=j} + {~i=j}. Once it is made, its proof is an object of the sum type {i=j}+{~i=j}, i.e. a function, named eq_dim_dec, which tests whenever its two arguments are equal or not. This lemma is interactively proved with some tactics, the reasoning being a simple structural induction on both i and j, which boils down to a simple case analysis here. Indeed, from each inductive type definition, Coq generates an induction principle, usable either to prove propositions or to build total functions on the type. Next, we identify the type dart and its equality decidability eq_dart_dec with the built-in type of natural numbers nat and eq_nat_dec. Finally, to manage exceptions, a nil dart is a renaming of 0: 1. Download: Download full-size image We choose a constructive point of view for hypermaps, which is close to the usual incremental building of surface subdivisions in geometric modeling rather than considering an observational point of view with an already built set of darts equipped with all its permutations, as it is done in . 3.2.2. Free maps The hypermaps are now approached by a general notion of free map, thanks to a free algebra of terms of inductive type fmap with 3 constructors, V, I and L, respectively for the empty (or void) map, the insertion of a dart, and the linking of two darts: 1. Download: Download full-size image Example 4 The hypermap M in Fig. 1 can be modeled by the free map represented in Fig. 2 where the 0- and 1-links by L are represented by arcs of circles, and where the orbits remain open. For instance, a submap of the hypermap M of Fig. 2, consisting of darts 3,2,9 and 10 is represented by the following term in Coq : (L (L (L (I (I (I (I V 3 p3) 2 p2) 10 p10) 9 p9) zero 3 2) one 10 2) zero 10 9). 1. Download: Download full-size image Fig. 2. Hypermap example with its incompletely linked orbits. When darts are inserted into a free map, they come together with an embedding point which is a couple of real numbers. As the reader can see from Fig. 2, some geometrical consistency properties must be enforced. For instance, the points p2 and p10 respectively associated with 2 and 10 must be equal. Coq also generates an induction principle on free maps. In the following, the use of the constructors will be constrained by preconditions to avoid meaningless free maps. The corresponding subtype of the hypermaps will be characterized by an invariant, called inv_hmap, systematically used in conjunction with fmap (see Section 3.2.3 for details). Next, observers of free maps can be defined. The predicate exd expresses that a dart exists in a hypermap. Its definition is recursive, which is indicated by the keyword Fixpoint. It proceeds by pattern matching on m written match m with.... The attribute {struct m} allows Coq to verify that the recursive calls are performed on smaller fmap terms, thus ensuring termination. The result is either False or True, the two basic constants of Prop, the built-in type of propositions. Note that terms are in prefix notation and that _ is a place holder. Proving the decidability exd_dec of exd is straightforward by induction on m. 1. Download: Download full-size image Then, a restriction of function α k, denoted A, is defined. It is designed so that its orbits cannot be closed. However, because Coq only allows total functions to be defined, A is extended with the nil dart when it will otherwise close the orbit (the inverse A_1 being similar): 1. Download: Download full-size image Predicates succ and pred express that a dart has a k-successor and a k-predecessor (non-nil), with the decidability properties succ_dec and pred_dec: Definition succ (m:fmap)(k:dim)(d:dart) : Prop := A m k d <> nil. Example 5 In Fig. 2, A m zero 6 = 8, A m zero 4 = nil, succ m zero 6, ~succ m zero 4, A_1 m one 9 = 8, pred m one 9. 3.2.3. Hypermaps As said previously, preconditions written as predicates are introduced for operators I and L. The precondition prec_I for I states that the nil dart cannot be inserted into a free map and that a dart x can only be inserted if it does not already belong to the free map. The precondition prec_L for L verifies that the darts x and y we want to link to one another are actually already inserted in the free map, that x has no successor at the involved dimension and that y has no predecessor at this dimension either. Finally, it also prevents a link from x to y from being added if it would close the orbit of x. 1. Download: Download full-size image If I and L are only used when the appropriate precondition holds, the built free map necessarily has open orbits. Such a condition was required to make merging orbits by concatenation easier. It also reduces the number of links required in the computation of the convex hull. Overall the built free map satisfies the invariant: 1. Download: Download full-size image Such a hypermap was already drawn in Fig. 2. In fact, thanks to other operations namely cA and cA_1, it can always be considered as a true hypermap exactly equipped with operations α k. Indeed, the operations cA and cA_1 close A and A_1; thus we can do as if the k-orbits were closed. In addition, for any k (A m k) and (cA m k) extend the function α k to darts which do not belong to the map m and return the dart nil. Example 6 In Fig. 2, cA m one 4 = 7, cA_1 m one 7 = 4, cA m one 11 = nil, cA_1 m one 11 = nil. In addition, when the input dart does not belong to the map, we have cA m zero 12 = nil and cA_1 m zero 12 = nil. Fundamental properties we prove are: for any m and k, (A m k) and (A_1 m k) are injections inverse of each other, and (cA m k) and (cA_1 m k) are permutations inverse of each other, and are closures. The reader interested in the technical details is referred to our formal proof development . Finally, traversals of faces are based on a function F and its closure cF (see for details), which correspond to ϕ as defined in Definition 2. Properties similar to the ones of A, cA are proved for F, cF and their inverses F_1, cF_1. Example 7 In Fig. 2, F m 4 = nil, cF m 4 = 6. Further topologic properties may be considered while proving the correctness of our convex hull algorithm. In addition, invariants dealing with geometry must be defined. So we now present the geometric setting in which our computations take place. 4. Geometric setting Convex hull computations do not only rely on topology but also on geometric properties of the involved points. In this article, we choose to work with Cartesian geometry in two dimensions and we consider each point p to be a couple of reals which are its coordinates in the plane (i.e. p=(x p,y p) with x p,y p∈R). To compute convex hulls incrementally, we need a predicate to determine the orientation of three points in the plane. As in and , we follow Knuthʼs approach to handle orientation in the plane. We first specify the orientation predicate with its properties and then implement it when the plane is represented by R 2. 4.1. Specification The predicate ccw(p,q,r) expresses whether the points p,q,r are enumerated clockwise or not. Fig. 3 exemplifies this orientation predicate ccw for such a triple of points. The example on the left (a) denotes a case where the triple (p,q,r) is oriented counterclockwise. The one in the middle (b) denotes a case where the three points are collinear. Finally, the one on the right (c) denotes a case where the triple (p,q,r) is oriented clockwise. 1. Download: Download full-size image Fig. 3. The orientation predicate. In fact, Knuth chooses not to handle degenerate cases. He assumes that three points are always in general position, i.e. no two of them coincide and they do not all lie on the same line. In our work, we assume the same, therefore the case (b) of Fig. 3 cannot happen. The orientation predicate is specified as follows: Property 1 Geometric orientation predicate P.1(cyclicity): ∀p,q,r,c c w(p,q,r)⇒c c w(q,r,p). P.2(symmetry): ∀p,q,r,c c w(p,q,r)⇒¬c c w(p,r,q). P.3(non-degeneracy): ∀p,q,r,¬c o l l i n e a r(p,q,r)⇒c c w(p,q,r)∨c c w(p,r,q). P.4(interiority): ∀p,q,r,t,c c w(t,q,r)∧c c w(p,t,r)∧c c w(p,q,t)⇒c c w(p,q,r). P.5(transitivity): ∀p,q,r,s,t,c c w(t,s,p)∧c c w(t,s,q)∧c c w(t,s,r)∧c c w(t,p,q)∧c c w(t,q,r)⇒c c w(t,p,r). P.5 bis(dual transitivity): ∀p,q,r,s,t,c c w(s,t,p)∧c c w(s,t,q)∧c c w(s,t,r)∧c c w(t,p,q)∧c c w(t,q,r)⇒c c w(t,p,r). Note that even if the collinearity case does not happen, a complete axiomatization requires to have an additional predicate collinear which expresses that three points lie on the same line. Properties 1, 2, and 3 are immediate to understand. Properties 4, 5, and 5 bis are illustrated in Fig. 4. Dotted lines correspond to premises and solid lines to conclusions of these properties. 1. Download: Download full-size image Fig. 4. Properties 4, 5 and 5 bis of Knuthʼs orientation predicate ccw. All these properties are required not only to design an algorithm which works fine and without bugs for any configuration of points in general position, but also to carry out its proof of correctness. We shall use this specification of the orientation predicate as an interface in our implementation of the convex hull algorithm. However, to make sure it is consistent, we do also prove all the above mentioned properties hold in our setting. 4.2. Implementing Knuthʼs orientation predicate for R 2 Our implementation uses the following concrete definition of ccw. Definition 4 Orientation of a triple of points Let (p,q,r) be a triple of points in the plane whose coordinates in R are (x p,y p), (x q,y q) et (x r,y r). The orientation predicate is defined according to the sign of the determinant det(p,q,r).det(p,q,r)=\|x p y p 1 x q y q 1 x r y r 1\|. This means, if det(p,q,r)>0, c c w(p,q,r) holds (and p,q,r are enumerated counterclockwise), whereas, if det(p,q,r)⩽0, c c w(p,q,r) does not hold (and p,q,r are enumerated clockwise or are collinear). Real numbers are described in Coq using an axiom system . Basic operations (+, -, ×, /) are specified and their more advanced properties are derived from this abstract specification. Thus, the function det can be easily implemented as follows: 1. Download: Download full-size image From this definition, we derived the orientation predicate ccw: Definition ccw (p q r : point) : Prop := (det p q r > 0). From this definition and properties of real numbers, we formally prove in Coq that all the properties of the specification hold. In addition, the orientation property is decidable, meaning it can be used in conditional expressions of algorithms. The theorem ccw_dec expresses this decidability property and is formally proved in Coq. Lemma ccw_dec : forall (p q r : point), {ccw p q r}+{~ccw p q r}. We now have a framework to handle the orientation predicate in a formal way. No issue related to numerical computations shall be considered in the rest of this article. We shall only consider we have a decidable predicate ccw available, which satisfies the above-mentioned specification and can be used to determine the orientation of a triple of points in the plane. 5. Convex hull and incremental algorithm In this section, we introduce the convex hull concept and we describe the incremental convex hull algorithm whose formal correctness shall be proved. 5.1. Convex hull definition The computation of planar convex hulls in one of the first problems that was studied in computational geometry. Many definitions leading to different algorithms were proposed in the literature , , , . In this work, we choose a definition well-suited for our topological hypermap model, for using Knuthʼs orientation predicate ccw and for the incremental algorithm we will study. Let P be a set of points in the plane. Like most of the authors, we assume that points are in general position, i.e. no two points coincide and no three ones are collinear. Definition 5 Convex hull The convex hull of S is the convex polygon P whose vertices t i, numbered in a counterclockwise order traversal for i=1,…,n with n+1=1, are points of S such that, for each edge [t i t i+1] of P and for each point p of S different from t i and from t i+1, ccw(t i,t i+1,p) holds. In other words, every point p of S different from t i and t i+1 lies on the left of the oriented line generated t i t i+1→. Fig. 5 shows a characterization of a convex hull using predicate ccw. On the left (a), we have a finite set P of points. In the middle (b), we have a convex polygon T with its grayed interior (which shall be formally defined later in the article). On the right (c), arrows denote oriented lines t i t i+1→ derived from edges [t i t i+1] of T. All featured points p,p′,… do lie on the left of these oriented lines. 1. Download: Download full-size image Fig. 5. Characterizing a convex hull. 5.2. Incremental algorithm The incremental algorithm computes the convex hull of P by building it step by step. At each step, a new point of P is considered and a new convex hull is computed. It takes as input the current hull (the one built with all the already-processed points). Then, either the new point lies inside the already-built polygon and the algorithm moves on to the next step, or it lies outside of the polygon and the algorithm will have to remove some edges and add two new ones to build a new convex polygon. This corresponds to the usual naive algorithm which is found in most books of computational geometry (e.g. in ). Since we assume that points are in general position, the new point can never be on the already-built polygon, i.e. be equal to a previously-added point or lie on an existing edge. The incremental algorithm can be decomposed into three functions named CH, CHI and CHID in the code. •The first function CH initiates the incremental computation of the convex hull. For a single point, the convex hull is the point itself. For more than one point, the algorithm starts with an initial set containing only two points and computes a first convex hull which is simply an edge linking the two points. Then it calls the function CHI with this first convex hull and the remaining points to be treated. •The second function, CHI, takes every element s of the initial set P and calls the insertion function CHID to build a new convex hull. It proceeds by case analysis. Then, for each new point s in P, it extends the already-built convex polygon using the insertion operation CHID. •The last one, CHID, computes the convex hull of a convex polygon T and an extra point s, i.e. it inserts s into the already-built convex hull polygon T. It uses tests based on Knuthʼs orientation predicate ccw. According to Definition 5, we know that the interior of polygon T is defined by the points x of the plane such that c c w(t i,t i+1,x) for any edge [t i t i+1] of T. In addition, the line generated by t i t i+1→ divides the plane remainder into two open half planes characterized by the value of c c w(t i,t i+1,x) for every point x. Therefore, one can easily locate the point s with respect to each edge [t i t i+1] of the polygon T. We simply have to evaluate c c w(t i,t i+1,s). Repeating this test for all i=1,…,n, this tells us whether s lies inside or outside T. If s lies inside T, the convex hull of (T∪s) is the same as the one of T. Otherwise, s necessarily lies outside T, the algorithm removes edges of T which are visible from s and creates two new edges [t l s] and [s t r] to connect s to the leftmost vertex t l and to the rightmost vertex t r. All these notions are defined precisely in the following definitions and illustrated in Fig. 6. 1. Download: Download full-size image Fig. 6. Computing a new convex hull from a convex polygon T and a new point s. Definition 6 Visible edges, leftmost vertex, rightmost vertex Let T be a planar convex polygon with at least two vertices and s be a point of the plane. (1)The edge [t i t i+1] of T is visible from s whenever ¬c c w(t i,t i+1,s) holds. (2)The vertex t l of T is the leftmost vertex with respect to s if c c w(t l−1,t l,s) and ¬c c w(t l,t l+1,s) hold. (3)The vertex t r of T is the rightmost vertex with respect to s if ¬c c w(t r−1,t r,s) and c c w(t r,t r+1,s) hold. Note that we shall have to prove the equivalence of the existence of t l and t r later in this article. Indeed, when s is inside the polygon, t l and t r do not exist. Otherwise, when s is outside, both of them exist. No other cases shall be consider as s cannot be collinear with two of the vertices of the convex polygon. In addition, we shall prove the uniqueness of these two vertices t l and t r when they exist. This algorithm shall be formalized according to our data structures, namely hypermaps. 6. Designing the incremental algorithm in Coq 6.1. Data representation The initial set of points of the plane from which the convex hull is computed is represented as an object of type fmap which is constrained to be a combinatorial oriented map where each point is represented by an isolated linkless dart whose embedding is the point coordinates (see Fig. 7(a)). 1. Download: Download full-size image Fig. 7. Representation of the input and output of the algorithm as maps of type fmap. The final convex hull is a polygon represented as an object of type fmap which is constrained in order to be a combinatorial oriented map. Each polygon vertex is represented by a topologic vertex (two distinct darts with the same embedding linked at dimension one) and each edge is represented by a topologic edge (two distinct darts with different embedding linked at dimension zero). This is illustrated in Fig. 7(b). We shall see that all intermediate computations are also represented by combinatorial oriented maps possibly with isolated darts. Therefore, in the remainder of the paper, we shall only consider a subtype of objects of type fmap which are actually combinatorial oriented maps. As the incremental computation of the convex hull relies on orientation tests in the plane, one must direct the polygon counterclockwise. This is achieved by always linking darts in the same direction, links being represented by small arrows in the drawings. Furthermore, darts representing points which are inside the convex hull are kept in the final map where they are isolated nonlinked darts (see Fig. 7(b)). These darts can be erased if required. 6.2. Precondition In the formalization, it is necessary to make the precondition the input map m must satisfy before the application of CH more precise. This precondition has four predicates and it is defined as follows: 1. Download: Download full-size image The hypermap m of course has to verify the hypermap invariant inv_hmap which is explained in Section 3.2.3. It must have no link at all between darts (no L constructor), which is the property the predicate linkless expresses. The predicate well_emb expresses that the geometric embedding must be sound, i.e. all input darts must have different embeddings. The well_emb predicate captures this property although it also ensures some additional technical properties when links occur (see Section 9.2). In this first experiment, we assume no three darts having different embeddings can be embedded into three collinear points. For a map, the corresponding predicate noncollinear is specified as follows: 1. Download: Download full-size image In the above definition, fpoint m d is the point onto which the dart d is embedded in the map m. 6.3. Classifying the darts Following our implementation decisions to design the incremental algorithm, it appears one can classify darts into three different kinds. To make the characterization more visual, we choose to do it using three different colors (blue, red and black) depending on the links the darts are involved in. Black darts are isolated darts with no links at all. Blue darts are those with exactly one predecessor at dimension one and exactly one successor at dimension zero. Red darts are those with exactly one predecessor at dimension zero and exactly one successor at dimension one. Note that the meaning of these colors is completely different from those used in . We remind the reader that in our definition of hypermaps in Coq, a given dart cannot have more than one successor and one predecessor at each dimension. Our classification of darts is presented in Fig. 8 (which is also readable in black and white). 1. Download: Download full-size image Fig. 8. The three kinds of darts and their role in our description of the convex hull. In Coq, predicates black_dart, blue_dart and red_dart respectively express that a dart x is black, blue or red in a given map m: 1. Download: Download full-size image The three dart kinds appear in our description of the convex hull: black darts are drawn as full lines, blue darts as dashed lines and red darts as dotted lines (see Fig. 8(b)). Their decidability is proved by the functions black_dart_dec, blue_dart_dec, red_dart_dec which can be used for branching in the code depending on the dart color. 6.4. Postconditions We summarize the main properties we expect from our convex hull computation. Overall several topologic properties are required as well as the fundamental property that the built polygon is actually convex. As far as geometric properties are concerned, we expect that the free map returned by the function CH actually verifies the convex property (see Definition 5). It relies on Knuthʼs orientation predicate ccw and can be transcripted into Coq using darts as follows: 1. Download: Download full-size image 6.5. Visible, leftmost and rightmost darts We emphasized in Section 5.2 the role of the visibility of an edge from a point as well as the role of the leftmost and rightmost visible vertices. As we work with darts, the visibility of an edge is expressed on any one of its darts, and the leftmost and rightmost vertices are replaced by two darts, the actual leftmost one and the actual rightmost one (see Fig. 10 for a graphical description). First, we define the predicates visible and invisible using the classification of darts and Knuthʼs orientation predicate ccw. The predicate invisible is exactly the negation of visible: 1. Download: Download full-size image As usual, decidability properties visible_dec and invisible_dec are proved. Then, following Definition 6, we specify two predicates left_dart and right_dart which state that a dart is the leftmost or the rightmost dart of m with respect to a point p: 1. Download: Download full-size image These predicates decidability is proved by the two lemmas left_ dart_dec and right_dart_dec. Note that, by convention, the leftmost dart always is a blue one and the rightmost dart is a red one. In addition, we shall have to prove the equivalence of the existence and the uniqueness of these two vertices. We will go back to these crucial questions in Section 9.1. 6.6. Programming the incremental algorithm with Coq In this section, we write in Coq our incremental algorithm by structural recursion on free maps. • We first define the main function CH which computes the whole convex hull of a finite set of points in the plane represented by a map m. If the initial map m is empty, it returns the empty map V. If m has only one dart, it returns a map with only one isolated dart. If it has at least two darts, it proceeds as follows: the function CH builds a first convex polygon for two of the involved darts using CH2 (Fig. 9) and then calls the recursive functionCHI. Since CH input is reduced to a dart set, no other case must be considered. 1. Download: Download full-size image 1. Download: Download full-size image Fig. 9. A convex hull of two darts built by the CH2 function. 1. Download: Download full-size image Fig. 10. CHID behavior. Note that max_dart m returns the largest dart (in fact, darts are integers) of the map m. As we will see, this function helps to simulate the generation of new darts, i.e. darts which do not appear in m. Note that CH2 uses two new darts (see above). Therefore the call to CHI, which also needs a new dart, is done with the parameter (max_dart m)+3. • Given two distinct darts x1 and x2, the function CH2 builds the combinatorial oriented map shown in Fig. 9. To do that, it introduces two new darts, namely, max+1 and max+2, and links them conveniently with x1 and x2. Instead of having a simple edge as presented in Section 5.2, we actually have a flattened polygon (in Fig. 9, edges are curved for visibility reasons) consisting in four darts and their links. This allows us to handle the case of two points in the same way as the general one. 1. Download: Download full-size image • Finally, function CHI takes the darts of m one-by-one and builds for each one a new convex hull using CHID and the parameter max. Then the recursive call of CHI is with parameter max+1. 1. Download: Download full-size image Now, we describe our function CHID. 6.7. A step of convex hull building As already hinted in the previous section, function CHID computes the convex hull of a convex polygon represented by a map m (of type fmap) and a new point represented by a dart x. It works by structural recursion on m by studying each dart and each link separately. Darts are processed in random order (one dictated by the structure of the fmap term) while reconstructing the polygon (instead of traversing them in the sequential order dictated by the counterclockwise traversal of the polygon). Because m is modified at each recursive call, CHID keeps a reference map mr, which is the same as m when CHID is first called. This reference map is useful to perform tests and will never be modified during the whole execution of the function. At each step of the computation, m is a submap of the reference map mr according to the following definition: 1. Download: Download full-size image Initially, m is equal to mr and at each recursive call, we formally prove in Coq the property submap m mr still holds. As previously said, there are two cases in CHID. If the new point lies inside the convex polygon, the function CHID simply inserts the dart x into the map m without any links. If it lies outside the convex polygon, the function CHID removes the edges of the polygon which are visible from the new point and creates two new ones connecting it with the leftmost vertex and the rightmost vertex of the polygon. In fact, CHID works constructively and not destructively: it always rebuilds from scratch the hypermap result by adding darts and their new links. If a link does not have to be reintegrated, it is quite simply forgotten. In this context, a recursive call to (CHID m mr x p max) unfolds as follows (see Fig. 11). Let us explain this in detail: •If m is the empty map (line 04), CHID simply returns the dart x with no links. • If m matches (I m0 x0 p0) (line 05), CHID checks the dart kind of x0 in mr. –If x0 is a blue dart in mr (line 06), the program tests whether x0 belongs to an edge of mr which is invisible from the new dart x embedded into point p (line 07). This test is achieved using the predicate invisible_dec. If the edge of x0 is invisible from p (line 08), the dart is simply kept in the map. Otherwise, the program tests (line 09) whether x0 is the new leftmost dart of mr with respect to x. If x0 is the leftmost dart (line 10), it remains in the map. In addition, a new dart max (embedded into p) is inserted and linked to x at dimension one. Finally x0 and max are linked at dimension zero. Otherwise, x0 is simply kept in the map (line 12). –If x0 is a red dart in mr, a similar reasoning step is performed (lines 13–18). –If x0 is a black dart in mr, it is kept in the map (line 19). •If m matches (L m0 zero x0 y0) (line 20), CHID tests whether the edge formed by x0 and y0 is invisible from x embedded into p (line 21). If it is invisible from p, the link at dimension zero between x0 and y0 is kept (line 22). Otherwise, it is not added again in the result map (line 23). •Similar steps apply if m matches (L m0 one x0 y0) (lines 24–29). 1. Download: Download full-size image Fig. 11. CHID function in Coq. 7. Extracting our Coq program into OCaml The Coq proof assistant features an extraction mechanism which automatically generates certified programs in OCaml or Haskell from proofs and specifications developed in Coq. It uses the Curry-Howard isomorphism between functional programming and natural deduction. This paradigm states that: “proof = program” and “proposition = type”. We use this feature to extract an OCaml program which computes convex hulls from our specification of Fig. 11. Coq datatypes such as fmap are automatically extracted into standard OCaml datatypes. However, some basic definitions or axioms can be manually translated into OCaml terms when the extraction mechanism does not know how to translate them. We reproduce all the manual translation commands required below. 1. Download: Download full-size image We choose to map Coq real numbers into OCaml floating-point numbers to be able to quickly extract our Coq implementation into a prototype program in OCaml. However, we must underline that such a translation (which actually is an approximation) is unsound. Indeed, among other issues, adding floating-point numbers in OCaml is not an associative operation. In addition, such a translation may lead to errors in the evaluation of the geometric predicates as underlined in . A more sensible extraction we would use in the event we want to insert this proved-correct program into a modeler would be to consider rational numbers in Coq rather than real numbers. This shall be sufficient for our purposes and the extraction of rational numbers from their implementation in Coq into the one in OCaml will be straightforward and more importantly, it would be safe. The extracted program only contains the code of functions CH, CHI, CH2, and CHID which computes the convex hull. One then has to create a graphical interface in order to be able to select points of the plane, transform this input into a map, let the extracted function CH compute the convex hull and transform the resulting map into a polygonal line (together with some remaining isolated points inside) which can be displayed on the screen. For convenience, translation functions from lists of points to maps (list_to_fmap) and from Peanoʼs integers to binary integers and vice-versa (i2n and n2i) are also provided: 1. Download: Download full-size image Fig. 12 presents a snapshot of the graphical interface. All links are symbolized with small circles, with a small dot inside for vertices and nothing for edges. 1. Download: Download full-size image Fig. 12. Graphical interface. From our algorithm written in Coq, we manage to automatically derive a program which can actually run on a computer. The next step is to make sure the algorithm is correct, i.e. it really computes convex hulls. This consists in proving several topologic and geometric properties. Several consistency properties (e.g. free maps at stake are meaningful ones all the way) are required. We shall also prove that the output verifies the definition of convex hull given in Section 5.1. 8. Topologic properties Initially, we decided to prove topologic properties first, and then focus on geometric ones. However, for some of the topologic properties, geometric properties inevitably interfere. We cannot reason on topologic issues without taking into account some basic geometric facts. This section focuses on proofs of topologic properties, even though they do sometimes rely on geometric properties. Relevant topologic properties are the hypermap property invariant preservation, the property that the hypermap describing the convex hull (at some stage of the computation) is always a polygon, the preservation of the initial darts in the computed convex hull and the planarity property for the convex hull computed by the algorithm. 8.1. Dart kinds and their evolution throughout the algorithm As presented in Section 6.3, it is possible to classify darts handled by our algorithm into three kinds (blue, red, or black). As it proceeds by structural induction on a map, the insertion function CHID considers darts one after another in random order. In addition, links are not studied at the same time as the darts they do link. Consequently, during recursive calls, not only some darts can shift from one kind to another but also darts may not belong to any kind anymore (e.g. when a dart loses only one of its links). They do end up in intermediate states during the execution of the insertion function CHID. For example, blue darts can either lose their incoming link at dimension zero, or their outgoing link at dimension one. If both links are removed, then they actually become black darts. To make proofs easier, especially in the proof of planarity (see Section 8.5), we define four intermediate kinds darts can belong to: namely half_blue_succ for a half blue dart with only a successor (at dimension one), half_blue_pred for a half blue dart with only a predecessor (at dimension zero): 1. Download: Download full-size image We can immediately transpose these definitions for red darts. These new kinds, representing intermediate states and possible changes of kinds for blue and half blue darts, are presented in Fig. 13. It works exactly the same for red darts as well. All these changes happening to the input combinatorial oriented map explain why we must have a reference map: the initial one, right before the first call to the insertion function (see Section 6.7 for details) to test the existence of darts and links between darts. 1. Download: Download full-size image Fig. 13. Possible changes of kinds for blue darts while CHID is executed. To consider all possible cases which can happen during recursive calls to this function, we establish 78 lemmas which express these changes in dart kinds. Here are two examples of such lemmas: • The first lemma blue_dart_CHID_1 expresses that, if a dart d is blue in the reference map mr and belongs to the current map m, then it remains in the result map (CHID m mr x tx px max): 1. Download: Download full-size image • The second lemma blue_dart_CHID_11 expresses that, if a dart d, different from the new dart x, is blue in the reference map mr, belongs to the current map m, and is visible from the point to be inserted px but that its predecessor at dimension one is not visible, then its successor at dimension zero in the map (CHID m mr x tx px max) is a new dart max: 1. Download: Download full-size image 8.2. Hypermap invariant preservation The first important theorem we want to prove is that the invariant inv_hmap holds all the way from the initial map m to the final one (CH m). This predicate inv_hmap states that darts must belong to the map before being linked together and one dart cannot be inserted twice in the same map (see Section 3.2.3). 1. Download: Download full-size image The proof proceeds by induction of the free map m and relies both on the lemmas of the previous section about the way darts may change kinds during the execution of the algorithm, and on the technical proofs of uniqueness of the leftmost and rightmost darts (see Section 9.1). This illustrates that topologic properties do depend on geometric properties in geometric algorithms such as computing convex hulls. Indeed, the key property to establish the above-mentioned theorem is that whether a dart d is kept in the current combinatorial oriented map (topologic property) simply relies on whether it is visible or not with respect to the point being inserted into the convex hull (geometric property). 8.3. The convex hull is a topologic polygonal set The predicate inv_poly on free maps expresses what a topologic polygonal set is. Informally, it consists of a set of polygons and isolated darts. In our algorithm, we expect to actually obtain a polygon together with some isolated darts. In a map verifying inv_poly, all darts are either black, blue, or red darts. No dart is partially linked. Therefore, for each dart d in the map, it either is black and isolated with no links or it is blue or r ed, meaning it belongs to the connected component which forms what we expect to be the convex hull. 1. Download: Download full-size image We prove that the free map returned by the function CH verifies the invariant inv_poly and therefore that it is a topologic polygonal set. 1. Download: Download full-size image The proof proceeds the same way as in the proof of the property inv_hmap but also uses the equivalence of the existence of the leftmost and rightmost darts (see Section 9.1). 8.4. Initial darts preservation Another fundamental property to prove is that darts which belong to the initial map do belong to the final map (denoting the convex hull) with their embeddings. 1. Download: Download full-size image In addition, only new (red) darts which are inserted during the convex hull computation can be removed from the map representing the final convex hull. According to CHID behavior, we note that darts extracted from the initial map and inserted into the convex hull are either black if they do lie inside the already-built convex hull, or blue if they do lie outside. Conversely, all black and blue darts of the resulting map are in the initial map. In addition, all red darts are new darts created using function max_dart (see Section 6.6). Then, to prove the initial darts are still present in the resulting map, it simply remains to be proved that black and blue darts are kept in the resulting map each time a new dart is inserted. 8.5. Planarity So far, we proved that our algorithm eventually produces a polygon and some isolated darts. We must still formally verify that this resulting polygon is actually planar. The planarity property planar is defined as follows , : Definition planar (m:fmap) := genus m = 0. We prove that, if m verifies the preconditions presented in Section 6.2, then the result of the incremental computation of the convex hull (CH m) is planar. 1. Download: Download full-size image This proof uses the planarity criteria established in , as well as the classification of the darts. One of the planarity characterizing lemmas is presented below: 1. Download: Download full-size image The predicate eqc (resp. expf), proposed in , respectively express that two darts belong to the same connected component (resp. to the same face). This lemma characterizes what is required for a free map m in which we link x to y at dimension zero to be planar. Such a free map will be planar if and only if the map m is planar and either x and y do not belong to the same connected components, or there exists a path in a face from the image of x by the closure function cA_1 at dimension one to y. This characterization obviously requires some preconditions, namely that m verifies the inv_hmap property and that x and y verify the precondition for 0-linking two darts together (prec_L). 8.6. Connected components and face numbering The last two topological properties we need to establish are that the number of connected components is equal to 1 and that the number of faces in the map returned by CH is equal to 2 (plus the number of isolated darts). Note this only holds when the initial map contains at least two darts. These two properties are stated using functions nc and nf computing the number of connected components and the number of faces of a map (see for details). These two properties are shown to be equivalent to one another. However completing the proofs of these properties would be very tedious in Coq and is not done in this work. 1. Download: Download full-size image This first property states that, as soon as the number of darts in a free map m is greater or equal to 2, the number of connected components in the computed convex hull (CH m) is equal to 1 plus the number of isolated darts (black darts which lie inside the convex hull). The proof would proceed by induction on the structure of the free maps and would lead to numerous and intricate cases to handle. 1. Download: Download full-size image The second property states that, as soon as the number of darts in a free map m is greater or equal to 3, the number of faces in the computed convex hull (CH m) is equal to 2 (inside and outside) plus the number of isolated darts (black darts which lie inside the convex hull). As said before, these two properties are equivalent thanks to the Euler formula. However neither of these two properties, nor the equivalence were formally proved in Coq. Indeed, such proofs are doable but very tedious. There are no theoretical issues involved but we would have to handle an exponential increase in the amount of cases to prove. The main difficulty is to do numbering of darts, vertices, edges, faces and connected components during the computations of the insertion function CHID. This would require to use the inductive definitions of the predicates expf and eqc at several levels in patterns of CHID like (L (L (I (I (CHID m0 mr x t p max) x0 p0). We now focus on the geometric properties required to prove that our algorithm actually computes a convex hull. 9. Geometric properties Key geometric properties are that darts are embedded in a consistent way in the plane and that the computed free map is actually a convex hull. 9.1. Uniqueness and equivalence of existence of the leftmost and rightmost vertices The first properties we establish are technical ones dealing with the uniqueness and the equivalence of the existence of two darts: the leftmost one and the rightmost one. • To prove the uniqueness, one assumes there are two leftmost darts and then proves that they are equal. We use the convexity properties as well as visibility ones for both darts. This leads to six triples of darts whose orientation contradict either the property 5 or 5 bis of Knuth. The theorem for the leftmost dart is the following one: 1. Download: Download full-size image A similar theorem is proved for the rightmost dart. • The proof of the equivalence of the existence of the leftmost dart and the rightmost dart are expressed by the theorem exd_left_right_dart (and its reciprocal) which states that if a leftmost dart exists in m, then there also exists a rightmost dart in m. Note that both darts may not exist (when the considered dart lies inside the already constructed convex hull). Therefore, the strongest property we can show is that whenever one of the these two darts exist, then the other exists as well. 1. Download: Download full-size image The proof proceeds by iteration on the darts belonging to the face. It relies on the property that the face is bounded, i.e. its number of darts is finite and can be computed which was proved using noetherian induction in . Concretely, it means the number of iterations of cF on a dart d required to come back to dart d again is known in advance. One of the most significant lemmas required to prove the above-mentioned theorem is the following one: 1. Download: Download full-size image Note that the k-iterate of a function f from a dart d, f k(d), is written Iter f k d in our framework. This lemma states that if there exists a blue dart visible from the point p we want to insert, then as long as we move around the face and do not find the rightmost dart, all traversed darts are visible from p. This property is illustrated in Fig. 14. 1. Download: Download full-size image Fig. 14. Face iteration from dart d: all traversed darts are visible from p. 9.2. Embedding We prove that darts are well embedded with respect to their links. To do that, we first define the property well_emb which was already used in the precondition of the function CH in Section 6.2. For each dart in the hypermap, its embedding must be different from those of its successor and predecessor at dimension zero but the same that those of its successor and predecessor at dimension one. In addition, all other darts must have a different embedding: 1. Download: Download full-size image This definition is illustrated in Fig. 15. On the left-hand side, we have a blue dart x with its 0-successor x0 and its 1-predecessor x_1, and on the right-hand side, a red dart x with its 1-successor x1 and its 0-predecessor x_0. 1. Download: Download full-size image Fig. 15. Correct embedding of the darts with respect to their links. Then, we establish the theorem: 1. Download: Download full-size image To prove it, we proceed as usual using our classification of darts (see Section 8.1) and prove the darts do keep their embeddings during recursive calls to the insertion function CHID. 9.3. Convexity Theorem convex_CH states the fundamental geometric property of the convex hull computation. It expresses that, provided the initial free map m verifies the preconditions, the final free map (CH m) is actually convex. More precisely, the final free map represents a polygon being the convex hull and some isolated darts inside this convex hull as well: 1. Download: Download full-size image The convex property was defined in Section 6.4. The proof of this theorem proceeds using the invariant propertywell_emb as well as properties of Knuthʼs orientation predicate. 10. Conclusion This work is a first experiment to see how our ideas on designing and certifying geometric algorithms work. The specification of a convex hull computation algorithm constitutes some sort of benchmark to check whether our library on hypermaps and geometric predicates is adequate with respect to our specification and proof goals. What we achieve is designing a functional algorithm and formally proving its total correctness with the Coq proof system. The termination of the algorithm is immediate because of its inductive construction. Properties justifying the partial correctness are all proved with the exception of the uniqueness of the polygonal convex hull, which remains complicated and shall be one of our next challenges. Fig. 16 provides some key figures about the size of the development and makes a distinction between the size of the specifications and the size of the proofs (the ratio is almost 1 to 10). The basic library corresponds to the already existing specifications and proofs presented in . The amount of specifications and proofs developed for this formal proof of correctness of the incremental convex hull algorithm is summarized in the second column. The Coq files of the development are available online . We manage to extract and make usable an OCaml program which, given a set of points in the plane, computes the convex hull (using the extracted code) and displays its result on the screen. This confirms the functional algorithm we designed is close to an actual implementation. 1. Download: Download full-size image Fig. 16. Key figures about the size of our Coq development. A worthwhile extension would be to derive an efficient program in a procedural language, where hypermaps would be represented with a concrete datatype such as linked lists, as in . In the short term, this could be performed by hand, as we did for our image segmentation algorithm in . In the longer term, one expects to automate this process and formally prove the correctness of the actual program we use, as this was carried out in for a square root computation algorithm for arbitrarily large numbers. In both cases, such refinements of programs should be studied by decomposing them into a sequence of elementary transformations. A procedural implementation mimicking the strategy of our functional algorithm on a linked list would be really close to the classical incremental algorithm and as efficient as it, namely O(n 2) in the worst case, n being the number of points in the input. One may object that it is still far from the optimal complexity which is O(n log(n)), but any implementation of the incremental algorithm has this drawback. The next step in our work is to study a variant of our incremental algorithm in Coq. In this variant, at each step when a new point is inserted, searching for the leftmost and rightmost darts would be performed by a traversal of a single face of the current polygon instead of studying darts in random order. The hypermap update can be performed by generating two new darts, unlinking of a few darts and relinking some others when the point we intend to add is outside the convex hull. Consequently, we would be very close to the common implementation of a convex hull incremental algorithm. In addition, all convex hull algorithms such as Graham scan or Jarvis march could be revisited using our library on hypermaps and their operations. Switching to a three-dimensional setting with a polyhedral convex hull would then be another challenge. This means handling general polyhedral surfaces and at this stage, the use of hypermaps of dimension 2 is even more meaningful. Finally, other computational geometry algorithms must be investigated. Currently, one of the authors studies triangulations and an algorithm to build Delaunay diagrams. Last but not least, even if our handling of geometric predicates using Knuthʼs orientation predicate is very convenient, we sooner or later must deal with numerical accuracy. This becomes a key issue in computational geometry as advocated in , which, for a large part, deals with an orientation predicate and an algorithm to compute a convex hull incrementally. More specifically, some efficient techniques to statically verify the validity of some numeric predicates in algorithms are developed using interval arithmetics and the Gappa tool . This tool was successfully used in computational geometry for validating the orientation predicate. Our approach was relevant at the beginning of our investigations, but results on exact and/or lazy arithmetics and their formalizations (e.g. ) should be integrated in our formal development. Special issue articles Recommended articles References Y. Bertot, P. Castéran Interactive Theorem Proving and Program Development CoqʼArt: The Calculus of Inductive Constructions Texts in Theoretical Computer Science, An EATCS Series, Springer-Verlag, Berlin/Heidelberg (2004) 469 pp Google Scholar Y. Bertot, N. Magaud, P. Zimmermann A proof of GMP square root Journal of Automated Reasoning, 29 (3–4) (2002), pp. 225-252 Special Issue on Automating and Mechanising Mathematics: In Honour of N.G. de Bruijn. An earlier version is available as an INRIA research report RR4475 View in ScopusGoogle Scholar Y. Bertrand, J.-F. 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Dufourd A hypermap framework for computer-aided proofs in surface subdivisions – Genus theorem and Euler formula SACʼ07: Proceedings of the 22nd ACM Symposium on Applied Computing, ACM Press, New York (2007), pp. 757-761 CrossrefView in ScopusGoogle Scholar J.-F. Dufourd Design and formal proof of a new optimal image segmentation program with hypermaps Pattern Recognition, 40 (11) (2007), pp. 2974-2993 View PDFView articleView in ScopusGoogle Scholar J.-F. Dufourd Discrete Jordan curve theorem: A proof formalized in Coq with hypermaps STACSʼ08: Proceedings of the 25th International Symposium on Theoretical Aspects of Computer Science (2008), pp. 253-264 View in ScopusGoogle Scholar J.-F. Dufourd Polyhedra genus theorem and Euler formula: A hypermap-formalized intuitionistic proof Theoretical Computer Science, 403 (2–3) (2008), pp. 133-159 View PDFView articleView in ScopusGoogle Scholar J.-F. Dufourd An intuitionistic proof of a discrete form of the Jordan curve theorem formalized in Coq with combinatorial hypermaps Journal of Automated Reasoning, 43 (1) (2009), pp. 19-51 CrossrefView in ScopusGoogle Scholar H. Edelsbrunner Algorithms in Combinatorial Geometry Springer-Verlag, New York (1987) Google Scholar J. Edmonds A combinatorial representation for polyhedral surfaces Notices of the American Mathematical Society, 7 (1960) Google Scholar E. Flato, D. Halperin, I. Hanniel, O. Nechushtan, E. Ezra The design and implementation of planar maps in CGAL WAEʼ99: Proceedings of the 3rd International Workshop on Algorithm Engineering, Lecture Notes in Computer Science, vol. 1668, Springer-Verlag, London (1999), pp. 154-168 CrossrefView in ScopusGoogle Scholar G. Gonthier, A computer-checked proof of the four colour theorem, Technical report, Microsoft Research, Cambridge, 2005. Google Scholar G. Gonthier Formal proof – the four-colour theorem Notices of the American Mathematical Society, 55 (11) (2008), pp. 1382-1393 Google Scholar G. Gonthier, A. Mahboubi, A small scale reflection extension for the Coq system, Tech. Rep. RR-6455, INRIA, 2008, hal.inria.fr/inria-00258384/. Google Scholar L.J. Guibas, J. Stolfi Primitives for the manipulation of general subdivisions and the computation of Voronoi diagrams STOCʼ83: Proceedings of the Fifteenth Annual ACM Symposium on Theory of Computing, ACM, New York (1983), pp. 221-234 View in ScopusGoogle Scholar G. Huet, G. Kahn, C. Paulin-Mohring, The Coq proof assistant – a tutorial, version 8.1, INRIA, France, 2007, Google Scholar A. Jacques Constellations et graphes topologiques Combinatorial Theory and Applications, 2 (1970), pp. 657-673 Google Scholar N. Julien, Certified exact real arithmetic using co-induction in arbitrary integer base, in: FLOPS, 2008, pp. 48–63. Google Scholar L. Kettner, K. Mehlhorn, S. Pion, S. Schirra, C. Yap Classroom examples of robustness problems in geometric computations Computational Geometry: Theory and Applications (CGTA), 40 (1) (2008), pp. 61-78 View PDFView articleView in ScopusGoogle Scholar D. Knuth Axioms and Hulls Lecture Notes in Computer Science, vol. 606, Springer-Verlag, Berlin/Heidelberg (1992) 109 pp Google Scholar P. Lienhardt Topological models for boundary representation: a comparison with n-dimensional generalized maps Computer-Aided Design, 23 (1) (1991), pp. 59-82 View PDFView articleView in ScopusGoogle Scholar M. Mantyla, R. Sulonen GWB: A solid modeler with Euler operators IEEE Computer Graphics and Applications, 2 (7) (1982), pp. 17-31 View in ScopusGoogle Scholar L. Meikle, J. Fleuriot Mechanical theorem proving in computational geometry H. Hong, D. Wang (Eds.), Automated Deduction in Geometry, Lecture Notes in Computer Science, vol. 3763, Springer, Berlin/Heidelberg (2006), pp. 1-18 5th International Workshop, ADG 2004, Gainesville, FL, USA, September 16–18, 2004 CrossrefView in ScopusGoogle Scholar G. Melquiond, Gappa : Génération automatique de preuves de propriétés arithmétiques, version 0.12.3, INRIA, France, 2010, Google Scholar D. Pichardie, Y. Bertot Formalizing convex hull algorithms R.J. Boulton, P.B. Jackson (Eds.), Theorem Proving in Higher Order Logics, Lecture Notes in Computer Science, vol. 2152, Springer, Berlin/Heidelberg (2001), pp. 346-361 14th International Conference, TPHOLs 2001, Edinburgh, Scotland, UK, September 3–6, 2001 CrossrefView in ScopusGoogle Scholar F.P. Preparata, M.I. Shamos Computational Geometry: An Introduction (5th printing), Monographs in Computer Science, Springer-Verlag, New York (1993) 398 pp Google Scholar R. 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First, sets in d-dimensional Euclidean space R d are mapped into an infinite-dimensional Banach space C(S) (whose elements are functions) via support functions, and then set-based classification in R d is converted into function-based classification in C(S). Second, we define the hyperplane via the Riesz representation theorem in Banach space, and discuss the Hausdorff distance of hyperpalnes and maximum margin principle (MMP) in C(S). Based on MMP, we construct an optimal problem and discuss some of its properties. Thereafter, we establish an SFM to solve set-based classification. Experiments about water quality evaluation and set-valued data classifications show the superiority of SFM. ### Automatic inspection system of surface defects on optical IR-CUT filter based on machine vision 2014, Optics and Lasers in Engineering Citation Excerpt : In our system, edge crack is described by computing convex hull and convexity defects of a contour, which is applied in the field of gesture recognition and tracking. The computation of the convex hull of a finite set of points is one of the first and most important concepts studied in computational geometry . The convex hull has several different definitions and several construction methods . Show abstract The paper presents an automatic surface defects inspection system for optical Infrared Cut-off (IR-CUT) filter, which is applied in all kinds of color cameras and video devices. The system involves illumination and imaging module, moving module, flipping module and machine vision algorithm. To highlight all the defected regions, the improved dark-field illumination technique is utilized in the imaging module. In order to accurately localize the region of optical IR-CUT filter in the captured image, stationary wavelet transform (SWT) is introduced to template matching algorithm. The introduction of SWT provides a more accurate estimate of the variances in the image and further facilitates the identification of the defected regions. The defects extraction method in this paper avoids the use of complicated learning process from a set of samples. Convexity theory is implemented on the algorithm of defects classification of edge crack. Experimental results on a variety of optical IR-CUT filter samples, including non-defective samples, samples with defects of stain, scratch and edge crack, have shown the efficiency (1.05 s per sample) and accuracy (96.44%) of the proposed system. Moreover, defects extraction performances of different filters are compared in this paper. The research and application of the system will greatly liberate the human workforce and inspire ideas to detect the defects of some other small optical elements. ### Verified Textbook Algorithms: A Biased Survey 2020, Lecture Notes in Computer Science Including Subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics ### Tests and proofs for custom data generators 2018, Formal Aspects of Computing ### A Verified ODE Solver and the Lorenz Attractor 2018, Journal of Automated Reasoning ### A verified algorithm for geometric zonotope/hyperplane intersection 2015, Cpp 2015 Proceedings of the 2015 ACM Conference on Certified Programs and Proofs Co Located with Popl 2015 View all citing articles on Scopus ☆ This work is partly supported by the research project GALAPAGOS, French ANR. Copyright © 2012 Elsevier B.V. All rights reserved. Part of special issue Geometric Constraints and Reasoning Edited by Xiao-Shan Gao, Robert Joan-Arinyo, Dominique Michelucci Download full issue Other articles from this issue Body-and-cad geometric constraint systems October 2012 Kirk Haller, …, Neil White View PDF ### A case study in formalizing projective geometry in Coq: Desargues theorem October 2012 Nicolas Magaud, …, Pascal Schreck View PDF ### A generalized Malfatti problem October 2012 Ching-Shoei Chiang, …, Paul Rosen View PDF View more articles Recommended articles No articles found. Article Metrics Citations Citation Indexes 16 Captures Mendeley Readers 9 View details About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. 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1727	https://math.stackexchange.com/questions/2232150/calculate-the-sum-of-an-infinite-series/2232235	calculate the sum of an infinite series - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more calculate the sum of an infinite series [closed] Ask Question Asked 8 years, 5 months ago Modified8 years, 5 months ago Viewed 151 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Closed. This question is off-topic. It is not currently accepting answers. This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Closed 8 years ago. Improve this question Here is the series: ∑k=0∞(−1)k 2 k+1.∑k=0∞(−1)k 2 k+1. I don't know how to start at all. Thank you for your help. sequences-and-series Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 13, 2017 at 11:05 Petch Puttichai 637 1 1 gold badge 6 6 silver badges 17 17 bronze badges asked Apr 13, 2017 at 10:35 KateKate 23 1 1 gold badge 1 1 silver badge 4 4 bronze badges 0 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Let S=∑∞k=0 x 2 k+1 2 k+1 S=∑k=0∞x 2 k+1 2 k+1 d S d x=∑k=0∞(x 2)k=1 1−x 2 d S d x=∑k=0∞(x 2)k=1 1−x 2 for \|x 2\|<1\|x 2\|<1 Integrate both sides to get S=ln 1+x 1−x+K S=ln⁡1+x 1−x+K x=0⟹0=ln 1+K⟺K=0 x=0⟹0=ln⁡1+K⟺K=0 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 13, 2017 at 10:44 lab bhattacharjeelab bhattacharjee 279k 20 20 gold badges 213 213 silver badges 337 337 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. (−1)k 2 k+1=−i⋅i 2 k+1 2 k+1(−1)k 2 k+1=−i⋅i 2 k+1 2 k+1 Now for −1≤x<1,−1≤x<1, ln(1+x)=x−x 2 2+x 3 3−x 4 4+⋯ln⁡(1+x)=x−x 2 2+x 3 3−x 4 4+⋯ ln(1+x)−ln(1−x)=?ln⁡(1+x)−ln⁡(1−x)=? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 13, 2017 at 10:38 lab bhattacharjeelab bhattacharjee 279k 20 20 gold badges 213 213 silver badges 337 337 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I'm a student, so I may not have the mathematical rigour necessary, but I'll try to help. It turns out that the Taylor Series for arctan(x)arctan⁡(x) is this: arctan(x)=∑k=0∞(−1)k(x 2 k+1)2 k+1=x 1−x 3 3+x 5 5−x 7 7+...arctan⁡(x)=∑k=0∞(−1)k(x 2 k+1)2 k+1=x 1−x 3 3+x 5 5−x 7 7+... So if x=1, you get: arctan(1)=∑k=0∞(−1)k 2 k+1=1 1−1 3+1 5−1 7+...arctan⁡(1)=∑k=0∞(−1)k 2 k+1=1 1−1 3+1 5−1 7+... Which is exactly your series. Since arctan(1)=π 4 arctan⁡(1)=π 4, your series converges to π 4 π 4. So ∑k=0∞(−1)k 2 k+1=π 4∑k=0∞(−1)k 2 k+1=π 4 Some other things: How do I know your series converges? Well, because of Leibniz's rule for alternating series, which you'll find here. How do I know that's the Taylor Series for arctan(x)arctan⁡(x)? If you take the derivative of arctan(x)arctan⁡(x), which is 1 1+x 2 1 1+x 2, and now you take its Taylor series and integrate it, you get the Taylor Series for arctan(x)arctan⁡(x). Here's a Quora Post that explains how to obtain it: (The links are spaced out because Quora won't let me otherwise) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 13, 2017 at 16:18 CommunityBot 1 answered Apr 13, 2017 at 11:59 elcocodrilotitoelcocodrilotito 77 5 5 bronze badges 1 Welcome to Math.SE! For the symbols, you can take a look at this meta question. I suggest you edit your answer to write the formula directly. You can also "hide the links" behind a description in the writing tools, to make it more convenient.Arnaud D. –Arnaud D. 2017-04-13 12:29:37 +00:00 Commented Apr 13, 2017 at 12:29 Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions sequences-and-series See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked -3Calculating the following series: ∑∞k=0(−1)k 2 k+1∑k=0∞(−1)k 2 k+1 43Why is arctan(x)=x−x 3/3+x 5/5−x 7/7+…arctan⁡(x)=x−x 3/3+x 5/5−x 7/7+…? 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1728	https://www.tutorialspoint.com/signals_and_systems/pdf/region_of_convergence.pdf	Copyright © tutorialspoint.com REGION OF CONVERGENCE REGION OF CONVERGENCE The range variation of σ for which the Laplace transform converges is called region of convergence. Properties of ROC of Laplace Transform ROC contains strip lines parallel to jω axis in s-plane. If x is absolutely integral and it is of finite duration, then ROC is entire s-plane. If x is a right sided sequence then ROC : Re{s} > σo. If x is a left sided sequence then ROC : Re{s} < σo. If x is a two sided sequence then ROC is the combination of two regions. ROC can be explained by making use of examples given below: Example 1: Find the Laplace transform and ROC of Example 2: Find the Laplace transform and ROC of R RO OC C t t t t x(t) = e u(t) −at L. T[x(t)] = L. T[e u(t)] = −at 1 S+a Re > −a ROC : Res >> −a x(t) = u(−t) eat L. T[x(t)] = L. T[ u(t)] = eat 1 S−a Res < a ROC : Res < a Example 3: Find the Laplace transform and ROC of For For Referring to the above diagram, combination region lies from –a to a. Hence, Causality and Stability For a system to be causal, all poles of its transfer function must be right half of s-plane. A system is said to be stable when all poles of its transfer function lay on the left half of s-plane. x(t) = u(t) + u(−t) e−at eat L. T[x(t)] = L. T[ u(t) + u(−t)] = + e−at eat 1 S+a 1 S−a Re{s} > −a 1 S+a Re{s} < a 1 S−a ROC : −a < Res < a A system is said to be unstable when at least one pole of its transfer function is shifted to the right half of s-plane. A system is said to be marginally stable when at least one pole of its transfer function lies on the jω axis of s-plane. ROC of Basic Functions f F ROC ROC: Re{s} > 0 ROC:Re{s} > 0 ROC:Re{s} > 0 ROC:Re{s} > a t s u(t) 1 s tu(t) 1 s2 u(t) tn n! sn+1 u(t) eat 1 s −a ROC:Re{s} > -a ROC:Re{s} < a ROC:Re{s} < -a ROC:Re{s} > a ROC:Re{s} > a ROC:Re{s} > -a ROC:Re{s} > -a ROC:Re{s} < a ROC:Re{s} < a ROC:Re{s} < -a ROC:Re{s} < -a u(t) e−at 1 s + a u(t) eat − 1 s −a e−at u(−t) − 1 s + a t u(t) eat 1 (s −a)2 tneat u(t) n! (s −a)n+1 te−at u(t) 1 (s + a)2 tn e−at u(t) n! (s + a)n+1 teat u(−t) − 1 (s −a)2 tn eat u(−t) − n! (s −a)n+1 te−at u(−t) − 1 (s + a)2 tn e−at u(−t) − n! (s + a)n+1 cos e−at bt s + a (s + a)2 + b2 sin e−at bt b (s + a)2 + b2
1729	https://byjus.com/maths/types-of-angles/	Published Time: 2019-09-24T15:30:22+05:30 In geometry, there are various types of angles, based on measurement. The names of basic angles are Acute angle, Obtuse angle, Right angle, Straight angle, reflex angle and full rotation. An angle is geometrical shape formed by joining two rays at their end-points. An angle is usually measured in degrees. There are various types of angles in geometry. Angles form the core part of the geometry in mathematics. They are the fundamentals that eventually lead to the formation of the more complex geometrical figures and shapes. Types Of Angles in Geometry What are Angles When two rays combine with a common endpoint and the angle is formed. The two components of an angle are “sides” and “vertex”. The side can be categorized into terminal sides and initial sides (or vertical sides) as shown in the image below. Representation of an Angle These two rays can combine in multiple fashions to form the different types of angles in mathematics. Let us begin by studying these different types of angles in geometry. Parts of Angle Vertex – Point where the arms meet. Arms – Two straight line segments form a vertex. Angle – If a ray is rotated about its endpoint, the measure of its rotation is called angle between its initial and final position. For More Information On Parts Of An Angle, Watch The Below Video: 14,180 Classification of Angles Angles can be classified into two main types: Based on Magnitude Based on Rotation Six Types of Angles In Maths, there are mainly 5 types of angles based on their direction. These five angle types are the most common ones used in geometry. These are: Acute Angles Obtuse Angles Right Angles Straight Angles Reflex Angles Full Rotation Angle Types Based on Magnitude The images above illustrate certain types of angles. Acute Angle Acute Angle An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. The figure above illustrates an acute angle. Obtuse Angle Obtuse Angle An obtuse angle is the opposite of an acute angle. It is the angle which lies between 90 degrees and 180 degrees or in other words; an obtuse angle is greater than 90 degrees and less than 180 degrees. The figure above illustrates an obtuse angle. Right Angle Right Angle A right angle is always equal to 90 degrees. Any angle less than 90 degrees is an acute angle whereas any angle greater than 90 degrees is an obtuse angle. The figure above illustrates a right angle or a 90-degree angle. Straight Angle Straight Angle A straight angle is 180 degrees when measured. The figure above illustrates a straight angle or a 180-degree angle. You can see that it is just a straight line because the angle between its arms is 180 degrees. Reflex Angle Since this measurement is less than 90 degrees, the arms form an acute angle. But what about the angle on the other side? What is the larger angle that is complementary to the acute angle called? It is called a reflex angle. The image below illustrates a reflex angle. Reflex Angle Any angle that has a measure which is greater than 180 degrees but less than 360 degrees (which coincides with 0 degrees) is a reflex angle. Full Rotation An angle equal to 360 degrees is called full rotation or full angle. It is formed when one of the arms takes a complete rotation to form an angle. Angle Types Based on Rotation Based on the direction of measurement or the direction of rotation, angles can be of two types: Positive Angles Negative Angles Positive Angles Positive angles are those angles which are measured in a counterclockwise direction from the base. In most cases, positive angles are used to represent angles in geometry. From the origin, if an angle is drawn in the (+x, +y) plane, it forms a positive angle. Negative Angles Negative angles are those angles which are measured in a clockwise direction from the base. From the origin, if an angle is drawn towards the (x, -y) plane, it forms a negative angle. Pair of Angles When two angles are paired, then there exist different angles, such as Complementary angles Supplementary angles Linear Pair Adjacent angles Vertically Opposite angles Complementary and Supplementary Angles Apart from the aforementioned types, there are two more angle types which are complementary angles and supplementary angles. If the sum of two angles is equal to 180°, then they are supplementary angles, and if the sum is equal to 90°, then they are called complementary angles. Linear Pair When the non-common arms of adjacent angles are just opposite to each other, or they extend in the opposite direction, then they are called linear pairs. By linear it is clear that they form a straight line. Adjacent Angles When two angles are connected with one common arm and have one common vertex and also the non-common arms are either side of the common arm, then they are called adjacent angles. Vertical Angles When two lines intersect each other at a single point (called vertex), then the angle formed on either side of the common vertex is called vertical angles or vertically opposite angles. Angles Formed By Transversal A line that cuts or intersects two or more lines at different points, is called a transversal. It is therefore, there are angles formed at the point of intersection. They are: Interior angles Exterior angles Pairs of Alternate interior angles Pairs of Alternate exterior angles Pairs of Corresponding angles Pairs of interior angles on the same side of the transversal Video Lesson on Types of Angles 9,04,174 Frequently Asked Questions – FAQs Q1 What are the six different angles in geometry based on measurement? The six different angles in geometry based on magnitude are: Acute angle, Obtuse angle, Right angle, Straight angle, Reflex angle and full angle Q2 What is an acute angle in geometry? Give examples. The angle that measures less than 90 degrees is called acute angle. Examples: 30°, 45°, 60°, 85° are acute angles. Q3 What is obtuse angle? Give an example. The angle that measures greater than 90 degrees is obtuse. For example, 145° is an obtuse angle. Q4 What is reflex angle?Give an example. Reflex angle is an angle that measures more than 180° and is less than 360°. For example, 270° is a reflex angle. Q5 What do you mean by zero angle? When both the arms of an angle overlap each other and measure the angle of 0°, then it is called the zero angle. Q6 Is obtuse angle and reflex angle same? Obtuse angle is different from reflex angle because obtuse lies between 90 degrees and 180 degrees but reflex is alway more than 180 degrees. Test your Knowledge on Types Of Angles Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Start Quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted View Quiz Answers and Analysis X Login To View Results Mobile Number Send OTP Did not receive OTP?Request OTP on Voice Call Login To View Results Name Email ID Grade City View Result Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Send OTP Did not receive OTP?Request OTP on Voice Call Website Post My Comment PreetiJuly 8, 2020 at 1:04 pm Nice learning with byj’us. 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1730	http://www.physicsbootcamp.org/doppler-effect.html	Physics Bootcamp Samuel J. Ling (\require{cancel}\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} ) Section 15.6 Doppler Effect So far in our discussion, there was no motion of detector or source of wave with respect to the medium. Hence, in all the foregoing studies, we have assumed that the frequency of wave detected is same as that of the source. However, if either the source or the detector moves with respect to the medium, there will be a difference in the frequency detected at the detector and that produced at the source. The effect is called the Doppler effect, named after the Austrian physicist Christian Johann Doppler (1803-1853), who in 1842 proposed the effect in his book titled, “Uber das farbige Lict der Dopperlterne,” which means “the colored light of double stars”. Doppler effect also provides explanation for the high pitch of an incoming train whistle and low pitch of a train moving away. To be sure, the incoming train whistle also appears louder since the whistle is blown in the direction of the platform, but here we are not concerned with the loudness, but the pitch or the frequency. This is also the reason why the spectrum of light is shifted to the lower frequency, the so-called red-shift, if a star or a galaxy is moving away. To understand the Doppler effect we will first consider two instances, one in which the source is stationary with respect to the medium while the detector moves at a constant velocity, and the other in which the detector is stationary with respect to the medium and the source moves with a constant velocity. Then we will combine the results to figure out what would happen when both the source and the detector move with respect to the medium. To keep it simple the relative motion will be kept along the line joining the source and the detector, which will be taken to coincide with the (x)-axis with positive (x)-axis pointed from detector to source. Subsection 15.6.1 Doppler Effect When Detector is Moving and Source is Stationary Let (v_D) be the speed of the detector while the source is fixed with respect to the medium, and (v) be the speed of the wave in the still medium as shown in Figure 15.11. Note that if the detector is moving away from the source with a speed greater than the speed (v) of the wave in the medium, then the wave will never catch up. Therefore, we will impose the restriction here that if the detector is moving away from the source its speed be less than the wave velocity. \begin{equation} v_D \lt v\ \ \textrm{if detector moving away.} \end{equation} We do not need to put any restriction if the detector is moving towards the source. Let (f_0) be the frequency of the wave emitted by the source. That is, there will be a new wave front every (1/f_0) second. Wavefronts from the source will travel towards the detector at speed (v) with the distance between the wavefronts equal to (\lambda =v/f_0\text{,}) which is the wavelength (\lambda\text{.}) For nonrelativistic speeds, e.g. (v) and (v_D\lt \lt c\text{,}) the speed of light in vacuum, the speed of the detector with respect to the moving wave front will be (v+v_D) if moving toward the source, and (v-v_D) if moving away from the source. Hence, the distance (\lambda) between the wavefronts will be covered by the detector in different time then (1/f_0\text{.}) Let (T_D) be the time the detector encounters the successive wavefronts. Then (T_D) will be \begin{equation} T_D = \frac{\lambda}{v\pm v_D}, \end{equation} \begin{align} + \amp\ \ \text{when detector moves towards the source,} \ - \amp\ \ \text{when detector moves away from the source.} \end{align} The frequency of the wave detected by the detector will be the inverse of this time period. Therefore, the frequency detected by the moving detector will be \begin{equation} f = \frac{1}{T_D} = \left(\frac{v\pm v_D}{v}\right) f_0, \end{equation} where \begin{align} + \amp \ \ \text{when detector moves towards the source,} \ - \amp \ \ \text{when detector moves away from the source.} \end{align} Thus, when the detector moves towards the stationary source, the frequency at the detector is higher than that produced by the source, and when the detector is moving away from the stationary source, the frequency detected is lower than the one produced. This explains why the pitch appears higher if you run towards a stationary horn and lower if you run away from the stationary horn. Subsection 15.6.2 Doppler Effect When Detector is Stationary and Source is Moving Consider a source of wave that moves with a speed (v_S) with respect to the medium in which the wave travels at the speed (v) while the detector is fixed in its place with respect to the medium as shown in Figure 15.12. Even though the source is moving, it still emits wave at the same frequency (f_0\text{.}) Waves move out in expanding spheres centered at the the instantaneous location of the source which is changing in time due to the movement of the source. Hence, to an observer stationary with the medium, the wave fronts will be spaced differently on the two sides of the source. In front of the source, they will appear more closely spaced and in the back more widely separated, because as the source moves, it reduces the distance between a previously emitted wavefront and itself before laying the next wavefront’s source point. Therefore, wavelength will be different in front of the source than behind it. \begin{equation} \lambda = \begin{cases} (v-v_S)/f_0\amp \ \text{ when distance decreasing, }\ (v+v_S)/f_0\amp \ \text{ when distance increasing } \end{cases} \end{equation} Since the speed of the wave is (v) in the medium, the frequency observed at the detector will be (v/\lambda) . \begin{equation} f = \left(\frac{v}{v\pm v_S}\right) f_0, \end{equation} where \begin{align} + \amp\ \ \text{ when distance increasing,}\ -\amp\ \ \text{ when distance decreasing. } \end{align} Subsection 15.6.3 Doppler Effect when Both Source and Detector Moving It is quite common that both source and detector move with respect to the medium. In that case we can imagine a stationary frame of reference in the medium, and then perform a two-step process. First we find the frequency in the imagined stationary frame using the source-moving transformation, and then find the frequency detected by the detector-moving transformation. Let (f_0) be the frequency of the source, and (v_S\text{,}) (v_D) and (v) be speeds of the source, the detector and wave with respect to the stationary medium. Let (f_M) be the frequency in the imagined frame stationary with respect to the medium, and (f) be the frequency observed in the detector. The following two-step process will give the frequency observed by the detector in terms of the frequency of produced at the source. \begin{equation} f_M = \left(\frac{v}{v\pm v_S}\right) f_0, \end{equation} where sign is chosen as follows. \begin{align} + \amp \ \ \text{ behind the source.} \ - \amp \ \ \text{ in front of the source.} \end{align} \begin{equation} f = \left(\frac{v\pm v_D}{v}\right) f_M, \end{equation} where signs are chosen as follows. \begin{align} + \amp \ \ \text{ when detector moves towards the source.} \ - \amp \ \ \text{ when detector moves away from the source.} \end{align} We can combine these equations to obtain one equation giving us the frequency detected to the frequency produce, as long as we remember the meaning of different signs. \begin{equation} f = \left( \frac{v\pm v_D}{v\pm v_S}\right)f_0 \end{equation} Note: Because of the nonrelativistic calculation, we cannot use the formulas derived here for light; you will need to use special theory of relativity for correct relation called the relativistic Doppler effect. We quote the relation without derivation. \begin{equation} f = \sqrt{\frac{c- v_S}{c+v_S}} f_0 \end{equation} where the sign of (v_S) is as follows. \begin{align} v_S \gt 0 \amp \ \ \text{ if source moving towards receiver.} \ v_S \lt 0 \amp \ \ \text{ if source moving away from receiver.} \end{align} Exercises 15.6.4 Exercises 1. Doppler Effect for a Moving Runner. A car horn at rest sounds a frequency of (400\text{ Hz}\text{.}) (a) You are running towards the horn at a speed of (10\text{ m/s}\text{,}) what frequency will you hear? (b) What frequency will you hear if you were running away from the horn? Assume the speed of sound in air to be approximately (343\text{ m/s}\text{.}) Hint. Use the formulas for moving detector Answer. (a) (412\text{ Hz} \text{,}) (b) (388\text{ Hz}\text{.}) Solution. (a) Since source is fixed and you are moving towards the source, you will hear a higher frequency sound. \begin{equation} f = \left( \frac{v + v_D}{v}\right)\, f_0 = \frac{343+10}{343}\times 400 = 412\text{ Hz}. \end{equation} (b) Since source is fixed and you are moving away the source, you will hear a lower frequency sound. \begin{equation} f = \left( \frac{v - v_D}{v}\right)\, f_0 = \frac{343-10}{343}\times 400 = 388\text{ Hz}. \end{equation} 2. Doppler Effect for Sound Emitted by a Moving Car. A car horn at rest sounds a frequency of (400\text{ Hz}\text{.}) (a) If the car is moving towards you at speed 10 m/s, what frequency will you hear? (b) If the car is moving away from you at speed 10 m/s, what frequency will you hear? Assume the speed of sound in air to be approximately 343 m/s. Hint. Using moving source formula. Answer. (a) (412\text{ Hz} \text{,}) (b) (385\text{ Hz}\text{.}) Solution. (a) Since detector is fixed and souce is moving towards the detector, you will hear a higher frequency sound. \begin{equation} f = \left( \frac{v}{v - v_S}\right)\, f_0 = \frac{343}{343-10}\times 400 = 412\text{ Hz}. \end{equation} (b) Since detector is fixed and source is moving away the detector, you will hear a lower frequency sound. \begin{equation} f = \left( \frac{v}{v + v_S}\right)\, f_0 = \frac{343}{343+10}\times 400 = 385\text{ Hz}. \end{equation} 3. Detemining Speed of the Source From Doppler Effect Data. While sitting on your porch you hear a fire engine siren at a frequency (1620\text{ Hz}) when approaching towards you, and a frequency (1590\text{ Hz}) when receding from you. From this data, determine the speed of the fire engine assuming it was constant and the frequency of the siren when fire engine heard by the driver of the fire engine. Use 343 m/s for the speed of sound. Hint. Use moving source formulas with unknown (f_0) and (v_S\text{.}) Answer. (3.2\text{ m/s}\text{,}) (1605\text{ Hz}\text{.}) Solution. (a) and (b). Let (f_0) be the frequency of the fire engine heard by the driver. Let (u) be the speed of the fire engine on the road. Therefore, frequencies you will hear when engine is approaching and when it is receding will have to obey the following equations. \begin{align} \amp 1620 = \frac{343}{343-u}\, f_0, \ \amp 1590 = \frac{343}{343+u}\, f_0. \end{align} We solve these equations to find (u) and (f_0\text{.}) Solve for (u\text{,}) we divide them to get \begin{equation} \frac{1620}{1590} = \frac{343+u}{343-u}. \end{equation} \begin{equation} (1620 + 1590) u = (1620 - 1590)\times 343\ \ \rightarrow\ \ u = 3.206\text{ m/s}. \end{equation} Put this in one of the Doppler equtions to get (f_0\text{.}) \begin{equation} f_0 = \frac{343-3.226}{343} \times 1600 = 1605\text{ Hz}. \end{equation} 4. Doppler Effect When Both Source and Detector are Moving. A humming bird is flying towards you with speed (10\text{ m/s}) while you are running towards it at (5\text{ m/s}\text{,}) both speeds with respect to the ground. You hear sound of frequency (4500\text{ Hz}\text{.}) What is the frequency of the sound the Humming bird is making? Use (343\text{ m/s}) for speed of sound. Hint. Use the formula that has both moving source and moving detector. Answer. (4306\text{ Hz}\text{.}) Solution. Because you are moving towards the bird, your movement will increase frequency heard. And because, the humming bird is moving towards you, that will also increase frequency you will hear. Therefore, the formula you will use will be \begin{equation} f = \left(\frac{v+v_D}{v-v_S}\right) f_0 \end{equation} We want the value of (f_0\text{.}) Putting in the numbers we get \begin{equation} 4500 = \left(\frac{343+5}{343-10}\right) f_0. \end{equation} This gives \begin{equation} f_0 = 4306\text{ Hz}. \end{equation} 5. Adjustments for Resonance in Air Column from Moving Tuning Fork (JEE, 2020). A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of (2\text{ ms}^{−1}) in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is (320\text{ ms}^{−1}\text{,}) find the smallest value of the percentage change required in the length of the pipe. (Indian JEE, 2020) Hint. Use Doppler effect of moving source. Answer. Solution. Let us denote (f_k) for the frequency of the (k^\text{th}) resonance when air column has length (L\text{.}) Adjustable air columns are usually made by filling tubes with water. This gives column open at one end, where tuning fork is placed, while other end is the surface of water, which acts as a closed end. The condition of resonance with nonmoving tuning fork will be \begin{equation} f_k = k\,f_1,\ \ f_1 = \frac{v}{4L}. \end{equation} Now, when the tuning fork is moving with speed (V_t) towards the air column, the resonances will be shifted to \begin{equation} f_k^\prime = \frac{v}{v-V_t}f_k = k\,f_1^\prime, \end{equation} \begin{equation} f_1^\prime = \frac{v}{v-V_t} f_1 = \frac{v}{v-V_t}\; \frac{v}{4L} \end{equation} Clearly (f_k^\prime \ne f_k) if (L) is not changed. The new length needed (L^\prime) will give the prior resonance if \begin{equation} \frac{v}{v-V_t}\; \frac{v}{4L^\prime} = \frac{v}{4L}. \end{equation} Therefore, we need \begin{equation} \frac{L^\prime}{L} = \frac{v}{v-V_t} = \frac{1}{ 1 - V_t/v}. \end{equation} Percentage change will be \begin{equation} \frac{L^\prime - L}{L} = \frac{1}{ 1 - V_t/v} - 1 = \frac{1}{ v/V_t - 1}. \end{equation} Now, we plug in the numerical values to get \begin{equation} \frac{1}{ v/V_t - 1} = \frac{1}{ 320/2 - 1} = 0.0063. \end{equation} That is (0.63\%\text{.}) PrevTopNext Found typo?
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permutations of length N that have exactly K inversions. An inversion is defined as a pair of indices (i, j) such that i < j and p(i) > p(j). This problem is significant in combinatorial mathematics and has applications in sorting algorithms and sequence analysis. Potential pitfalls include misunderstanding the definition of inversions and not accounting for all possible permutations. Approach To solve this problem, we can use dynamic programming. The idea is to build a table where dp[n][k] represents the number of permutations of length n with exactly k inversions. We start with a naive approach of generating all permutations and counting inversions, but this is not feasible for larger values of N due to its factorial time complexity. Instead, we use a dynamic programming approach to optimize the solution: Initialize a 2D array dp where dp[i][j] represents the number of permutations of length i with exactly j inversions. Base case: dp = 1 (an empty permutation has zero inversions). Transition: For each length i and each number of inversions j, calculate dp[i][j] by considering the position of the last element in the permutation. Algorithm Here is a step-by-step breakdown of the dynamic programming approach: Initialize a 2D array dp with dimensions (N+1) x (K+1) and set all values to 0. Set dp = 1. For each i from 1 to N: For each j from 0 to K: For each x from 0 to i-1: Update dp[i][j] by adding dp[i-1][j-x] if j-x >= 0. Code Implementation ``` public class PermutationsWithKInversions { public static int countPermutations(int N, int K) { // Initialize the dp array int[][] dp = new int[N + 1][K + 1]; // Base case: one way to arrange 0 elements with 0 inversions dp = 1; // Fill the dp array for (int i = 1; i <= N; i++) { for (int j = 0; j <= K; j++) { for (int x = 0; x < i; x++) { if (j - x >= 0) { dp[i][j] += dp[i - 1][j - x]; } } } } // The answer is the number of permutations of length N with exactly K inversions return dp[N][K]; } public static void main(String[] args) { int N = 4; int K = 2; System.out.println(countPermutations(N, K)); // Output: 5 } } ``` Complexity Analysis The time complexity of this approach is O(NK) because we have three nested loops iterating over N, K, and i. The space complexity is also O(NK) due to the dp array. Edge Cases Consider edge cases such as: N = 0 or K = 0: The result should be 1 if both are 0, otherwise 0. K > N(N-1)/2: This is impossible, so the result should be 0. Testing To test the solution comprehensively, consider a variety of test cases: Simple cases: N = 1, K = 0; N = 2, K = 1. Edge cases: N = 0, K = 0; N = 5, K = 10. Complex cases: Larger values of N and K within constraints. Thinking and Problem-Solving Tips When approaching such problems, it is crucial to: Understand the problem definition and constraints thoroughly. Start with a brute-force solution to understand the problem better. Look for patterns and optimize using dynamic programming or other techniques. Practice similar problems to improve problem-solving skills. Conclusion In this blog post, we discussed how to solve the problem of counting permutations with exactly K inversions using dynamic programming. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in computer science. Additional Resources For further reading and practice, consider the following resources: Inversion (discrete mathematics) - Wikipedia Counting Inversions - GeeksforGeeks LeetCode - Coding Challenge Platform Previous Lesson Prev Lesson Next Lesson Next Lesson × Go From “I Suck at Coding” to Landing Your Dream Job Our interactive tutorials and AI-assisted learning will help you master problem-solving skills and teach you the algorithms to know for coding interviews. Start Coding for FREE
1732	https://www.studocu.com/en-us/messages/question/10635802/find-the-solution-of-the-system-of-equations4-x-minus-2-y-equals-minus-64x2y6x-plus	[Solved] Find the solution of the system of equations 4 x minus 2 y equals - Calculus III (MATH 240) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent Find the solution of the system of equations. 4, x, minus, 2, y, equals, minus, 6 4x−2y= −6 x, plus, 4, y, equals, minus, 24 x+4y= −24 Calculus III (MATH 240) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist University of Pennsylvania Calculus III Question Find the solution of the system of equations 4 x minus 2 y equals University of Pennsylvania Calculus III Question LS ### Luke 9 months ago Find the solution of the system of equations. 4, x, minus, 2, y, equals, minus, 6 4x−2y= −6 x, plus, 4, y, equals, minus, 24 x+4y= −24 Like 0 Related documents MATH 240 - Final Exam Solutions - Fall 2020 Calculus III Practice materials 86%(7) Seminar assignments - linear algebra problems Calculus III Assignments 100%(2) Lecture 11.1.2: Understanding First-Order Systems in Differential Equations Calculus III Lecture notes None Week 11 Rec - week 11 recitation Calculus III Lecture notes None MATH 240 - Notes 11.1.1: Systems of Differential Equations Intro Calculus III Lecture notes None Lecture 11.2.1: First-order Systems with Constant Coefficients Calculus III Lecture notes 100%(3) Week 2 Practice Problems Calculus III Practice materials None Answer Created with AI 9 months ago Sure, let's solve this system of linear equations step by step. Step 1: Write down the system of equations The system of equations is: js 4x - 2y = -6 x + 4y = -24 Step 2: Simplify the eq... Like 0 This is a preview Go Premium and unlock full access to AI trained answers and study materials Optimised for study questions Access all study materials Get Unlimited Downloads Upload Share your documents to unlock Free Trial Get 7 days of free Premium Already Premium?Log in AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Calculus III MATH 240 University of Pennsylvania 82 Documents Go to course 5 Lecture 11.2.2L: Solutions to Linear Systems with Complex Eigenvalues Calculus III 100%(3) 2 Lecture 11.2.1: First-order Systems with Constant Coefficients Calculus III 100%(3) 2 6.1 Linear Transformations Calculus III 100%(3) 89 Math 240 Notes Calculus III 100%(3) Discover more from: Calculus III MATH 240University of Pennsylvania82 Documents Go to course 5 Lecture 11.2.2L: Solutions to Linear Systems with Complex Eigenvalues Calculus III 100%(3) 2 Lecture 11.2.1: First-order Systems with Constant Coefficients Calculus III 100%(3) 2 6.1 Linear Transformations Calculus III 100%(3) 89 Math 240 Notes Calculus III 100%(3) 5 Week 13 Rec - week 13 recitation Calculus III 100%(3) 20 MATH 240 - Final Exam Solutions - Fall 2020 Calculus III 86%(7) Related Answered Questions There are no questions yet. Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. Copyright © 2025 StudeerSnel B.V., Keizersgracht 424-sous, 1016 GC Amsterdam, KVK: 56829787, BTW: NL852321363B01 Cookies give you a personalised experience We’re not talking about the crunchy, tasty kind. These cookies help us keep our website safe, give you a better experience and show more relevant ads. We won’t turn them on unless you accept. Want to know more or adjust your preferences? Reject all Accept all cookies Manage cookies
1733	https://it.gvda-instrument.com/info/three-steps-to-use-the-sugar-meter-84081634.html	Tre passaggi per utilizzare il misuratore di zucchero - Conoscenza admin@gvda-instrument.com +86-18822802390 Lingua Italiano English Cymraeg Norsk हिंदी magyar România limbi Čeština Indonesia Български Latviešu русский Türkçe Casa Prodotti Strumenti elettronici Strumenti di misura Strumenti di misura di livello Metal Detector Industriale Strumento analizzatori Misuratore di umidità della temperatura Strumenti di misura fisica Altri prodotti più venduti Chi siamo OEM Contattaci Risorsa Conoscenza Multimter digitale Ph meter Misuratore di velocità Binocolo per la visione notturna Misuratore di altezza del cavo ad ultrasuoni Termometro a infrarossi Rilevatore di metalli Esposimetro Lux Livello digitale Rilevatore di gas Alimentazione DC Misuratore Brix Microscopio digitale Tester di prese Penna tester di tensione Stazione di saldatura Regrattometro digitale Telemetro laser Tester di radiazione di campo elettromagnetico Rivelatore da parete Misuratore di umidità del legno Fonometro Pinza amperometrica digitale Anemometro Misuratore di spessore del rivestimento Manometro differenziale Misuratore di ossigeno disciolto Diventare distributore VR Casa>Conoscenza> Contenuto Conoscenza Categoria di prodotto Strumenti elettronici Multimetro digitale Multimetro digitale ricaricabile Smart True RMS 10000 conteggi Misuratore di morsetto digitale Penna tester di tensione Socket Tester Fonometro Saldatore Rilevatore di radiazioni Tester per cavi Spelafili Nuovo Alimentatore CC Nuovo Strumenti di misura Distanziometro laser Livello laser Telemetri da golf Binocolo per la visione notturna microscopio digitale Staffa di livello laser Regrattometro digitale Misuratore di vibrazioni Strumenti di misura di livello Livello digitale Goniometro Metal Detector Industriale Strumento analizzatori Misuratore di umidità del grano Misuratore di umidità del legno Tester di qualità dell'acqua Rilevatore di gas Misuratore di umidità della temperatura Termometro digitale per carne Termometro a infrarossi Igrometro digitale Strumenti di misura fisica Spessimetro ad ultrasuoni Strumento di misurazione dell'altezza Misuratore di velocità Luxmetro digitale Altri prodotti più venduti Multimetro digitale (1) Pinza amperometrica digitale (2) Misuratore di umidità del legno (3) Termometro a infrarossi-1 Contattaci Contatto: MS Judy Yan WhatsApp/WeChat/Mob.: 86-18822802390 Email: marketing@gvdasz.com admin@gvda-instrument.com Tel Telefono: 86-755-27597356 Aggiungi: Stanza 610-612, Huachuangda Affari Edificio, Distretto 46, Cuizhu Strada, Xin'an Street, Bao'an, Shenzhen Tre passaggi per utilizzare il misuratore di zucchero May 29, 2023 Visualizza: 104 Tre passaggi per utilizzare il misuratore di zucchero Il principio dell'idrometro è lo stesso di quello del misuratore di zucchero. Misura il peso specifico di una soluzione specifica dall'altezza che galleggia nella soluzione. Ciò che viene misurato è la densità relativa, che è la densità della soluzione rispetto all'acqua. Il peso specifico dell'acqua è 1.000. La lettura è citata dallo standard Alla temperatura (20 gradi), la densità di un liquido dipende dalla temperatura, quindi la lettura dell'idrometro/metro Brix deve essere corretta al peso specifico alla temperatura standard. È solo che le unità del misuratore di zucchero e dell'idrometro sono diverse, ma le unità possono essere convertite. Formula di conversione sperimentale accurata: grado P=(-668.962) più (1262.45peso specifico)-776.43(seconda potenza del peso specifico) più (182.94terza potenza del peso specifico gravità) Formula di stima: Contenuto di zucchero ≈【(peso specifico—1)1000】/4 Ad esempio, il peso specifico del mosto misurato a 20 gradi è 1,048 e il contenuto esatto di zucchero è 11,9 gradi P. Secondo la formula di stima, il contenuto di zucchero è di circa 12 gradi P. Come usare il misuratore di zucchero (1) Pulizia del misuratore di zucchero Il misuratore di zucchero deve essere pulito con il mosto da testare e non può essere lavato con acqua o altri liquidi, in modo da evitare cambiamenti nella concentrazione del mosto e ottenere risultati di misurazione accurati tanto quanto possibile. Allo stesso modo, per miscelare uniformemente la temperatura del mosto nel cilindro graduato, anche l'asta di agitazione a spirale utilizzata deve essere lavata con il mosto da misurare. (2) Raffreddamento del mosto Prendi una piccola quantità di mosto e mettila in un cilindro di metallo e usa un'asta di agitazione a spirale per raffreddare rapidamente il mosto a circa 20 gradi (perché anche se c'è compensazione della temperatura, anche la compensazione della temperatura è limitata dall'intervallo di temperatura). Naturalmente, durante il raffreddamento, è necessario assicurarsi che il mosto non possa essere diluito ed evitare l'aumento della concentrazione causato dall'evaporazione dell'acqua nel mosto. (3) Lettura del misuratore del contenuto di zucchero Tenere con attenzione l'estremità superiore del misuratore del contenuto di zucchero, posizionarlo lentamente sulla scala del valore stimato, attendere un momento e dopo che il misuratore del contenuto di zucchero è stabile, leggere la posizione del misuratore del contenuto di zucchero tubo dalla scala convessa dove il mosto entra in contatto con il valore di visualizzazione del misuratore del contenuto di zucchero. Quindi controllare il valore di correzione corrispondente alla scala di temperatura nella metà inferiore del misuratore di zucchero. Se la temperatura del mosto misurato è superiore a 20 gradi, aggiungere il valore di correzione al valore visualizzato del misuratore di zucchero; Sottrai il valore di correzione. Se la lettura di un misuratore di precisione è 11,6 gradi P e il valore di correzione sotto i 20 gradi è 0,2 gradi P, allora la concentrazione del mosto è 11.6-0.2=11.4 gradi P, quello è, a 20 gradi, 100 kg di mosto contengono 11,4 kg di percolato. Le caratteristiche e il principio di funzionamento del misuratore di zucchero Un paio di Riepilogo della conoscenza del misuratore di zucchero IL prossimo Articolo Potrebbe piacerti anche Rilevatore sotterraneo di tesori d'... Rilevatore di gas combustibile Alla... Misuratore di salinità digitale per... Telemetro laser da golf da caccia d... Spessimetro ad ultrasuoni in vetro ... Rilevatore portatile di fughe di ga... 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1734	https://www.ck12.org/geometry/Triangle-Inequality-Theorem/rwa/Triangle-Inequality-Theorem/	Triangle Inequality Theorem Sum of the lengths of any two sides of a triangle is greater than the third side. Notes/Highlights \| Color \| Highlighted Text \| Notes \| \| --- --- \| \| Show More \| \| \| \| Image Attributions ShowHide Details Description Difficulty Level Date Created: Last Modified: Language Concept Nodes: ShowHide Resources
1735	https://www.quora.com/Can-you-explain-the-concept-of-locus-in-relation-to-a-variable-point-moving-on-a-circle	Something went wrong. Wait a moment and try again. Variable Change Locus Map Mathematical Concepts Coordinate Plane Variable Geometry Variable Inputs Locus (mathematics) 5 Can you explain the concept of locus in relation to a variable point moving on a circle? Michael Paglia Former Journeyman Wireman IBEW · Author has 33.3K answers and 5.2M answer views · 1y I think standard type loci are in ellipses A circle is a special ellipse Its the radius in a circle In an ellipse the 2 distances from the 2 loci are constant As one gets smaller other is bigger I keep thinking Kepler He has the speed and distance thing with planets I think standard type loci are in ellipses A circle is a special ellipse Its the radius in a circle In an ellipse the 2 distances from the 2 loci are constant As one gets smaller other is bigger I keep thinking Kepler He has the speed and distance thing with planets Related questions What is the locus point of a circle? How can I find the locus of a point of intersection? What do you mean by the statement “the locus of the center of a circle which touches a given line at a given point”? What is the locus of a point which moves so that its absolute is always 9? What is the locus of a circle? Sarthak Prasad Studied Electrical Engineering at National Institute of Technology Silchar (Graduated 2022) · 8y Related I don't understand what a locus is in math. Could you explain and give an example on how to use it? There are various geometrical figures like circles, lines, ellipses etc. which we see in our daily life around us but the all follow some specific definitions when projected on a Cartesian Plane. For instance, a line maybe just a LINE, a circle just a CIRCLE and an ellipse a circle which is elongated for you but geometry follows some definitions for them. In geometry a line: • is straight (no curves), • has no thickness, and. • extends in both directions without end (infinitely). If it does have ends it is called a "Line Segment". A Circle is the set of all points in a plane that are at a given There are various geometrical figures like circles, lines, ellipses etc. which we see in our daily life around us but the all follow some specific definitions when projected on a Cartesian Plane. For instance, a line maybe just a LINE, a circle just a CIRCLE and an ellipse a circle which is elongated for you but geometry follows some definitions for them. In geometry a line: • is straight (no curves), • has no thickness, and. • extends in both directions without end (infinitely). If it does have ends it is called a "Line Segment". A Circle is the set of all points in a plane that are at a given distance from a given point, the centre; equivalently it is the curve traced out by a point that moves so that its distance from a given point is constant. An ellipse is a curve in a plane surrounding two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. When you talk of a locus you actually talk of the set of points which lie on the specific figure. In other words, when you plot all the points of the locus you will get the specific figure. Eg. If we take a locus of a straight line in the form y=mx+c like y=5x+1. Nw take any random value of x ,say 2. If 2 is the abscissa of any point on the line y=5x+1 then its ordinate will be calculated using the given equation of the Straight line. y=5x+1 => y=5(2) +1=11 Our results mean that the point (2,11) lies on the straight line whose equation is y=5x+1. If you take a few more values of x and calculate y or vice versa and plot these points on a cartesian plane and join them you will get a line whose equation is the one given to us, that is, y=5x+1. Now lets talk about a circle. It is the set of all points in a plane that are at a given distance from a given point, the centre. If we talk of the locus of points which are at a distance of 5 units from the origin, we will have a circle of radius 5 units. Equation of such circle will be (x^2)+(y^2)=5^2=25. If you take the value of abscissa as 0 then the value of ordinate will be calculated using the given equation. (x^2)+(y^2)=25 => (0^2)+(y^2)=25 =>(y^2)=25 => \|y\|=5. It means that the points (0,5) and (0,-5) lie on the given circle Since they follow the condition of being 5 units away from the origin. If you keep on taking some random values of x and calculate y or vice versa you will get some points which when plotted graphically will form a circle of radius 5 units but since a circle is a closed figure unlike a line which extends infinitely, not all values will get you points which could be plotted. For instance, if you take the value of abscissa as 10 you get the value of ordinate as (-75)^(1/2) which isn’t a real number and hence can’t be plotted graphically. According to me, you should try solving some questions on locus by plotting everything graphically on a graph paper. It will help you understand the concept easily. Rick Marcus Former R&D Engineer at Stanford University (1983–1989) · Author has 823 answers and 114.9K answer views · 1y A circle is the locus of points all of which are equal distance from a point which is called the ‘center’. All points on the circle are of equal distance from the center. Robbie Goodwin Author has 11.8K answers and 13.1M answer views · 8y Related I don't understand what a locus is in math. Could you explain and give an example on how to use it? A location is the place where a thing is at a given time; a single, specific point, such as might be made by sticking one end of a pair of compasses into a piece of paper. Does that make sense? If you really can’t use a dictionary or Google, a locus is the set of all possible points that might meet certain conditions. A circle is the locus of all points on a plane that are a fixed distance from a fixed point; all the places the other end of the compasses could occupy as it moved round the hole in the paper and the distance between the compasses’s ends didn’t change. A straight line is the locus o A location is the place where a thing is at a given time; a single, specific point, such as might be made by sticking one end of a pair of compasses into a piece of paper. Does that make sense? If you really can’t use a dictionary or Google, a locus is the set of all possible points that might meet certain conditions. A circle is the locus of all points on a plane that are a fixed distance from a fixed point; all the places the other end of the compasses could occupy as it moved round the hole in the paper and the distance between the compasses’s ends didn’t change. A straight line is the locus of all points in a plane equidistant from two fixed points. Related questions What is the definition of a locus of a variable point? What is the general guideline for finding the locus of points that is an intersection of two variable lines? A tangent is drawn from the point (-a, 0) to a variable circle, centre (a, 0) . What is the locus of the point of contact? What is the relationship between the locus of a fixed point and the movement of another fixed point on a circle? Can you explain the concept of a locus for a point that moves based on its distance from two other points? Gopal Menon B Tech in Chemical Engineering, Indian Institute of Technology, Bombay (IITB) (Graduated 1975) · Author has 10.2K answers and 15.1M answer views · 6y Related What does locus in a circle mean? What does locus in a circle mean. Locus is a set of points that meet some specified geometrical conditions. In case of a circle the geometrical condition is that all these points should be equidistant from a fixed point, known as the centre of the circle. Jan Bolluyt Studied Mathematics & Physics (till 12th) (Graduated 1970) · Author has 985 answers and 73.8K answer views · Mar 28 Related What's the locus of a moving point which is equidistant from the points (2a, 2b)? First you find the midpoint on the line segment that goes through 2A and 2B which to we geometry teachers looks like one point. Once you find the midpoint, draw a perpendicular line through the midpoint and you are done. First you find the midpoint on the line segment that goes through 2A and 2B which to we geometry teachers looks like one point. Once you find the midpoint, draw a perpendicular line through the midpoint and you are done. Bheema Mudda Former Rtd Director at Government of India (2002–2007) · Author has 4.8K answers and 2.8M answer views · 3y Related I don't understand what a locus is in math. Could you explain and give an example on how to use it? locus is the path of point object that follows assigned condition eg locus of a point always equidistant from a given point here locus is circle of a point which traces circle & maintaining equidistant from the centre of circle which is called radius eg 2 parallel line : locus of point equidistant from a line . Here locus [path] of point is a straight line always keeping the same distance from another line Sam Verma b.sc agriculture from I. K. Gujral Punjab Technical University (Graduated 2019) · Author has 517 answers and 297K answer views · 4y Related What is the locus of a circle? We can characterize locus in math as the arrangement of the multitude of focuses that fulfill given conditions. Notice that as per the definition, a locus is certifiably not a solitary point yet a bunch of focuses. To determine a locus, you will need to do the following two steps Use the condition given to draw purposes of the locus until you see an example. Connect the points and describe the locus fully. A great example of locus and we are all very familiar with it is the one resulting in a circle such as the circle shown in the figure above. A circle is the locus of points at a given distance fr We can characterize locus in math as the arrangement of the multitude of focuses that fulfill given conditions. Notice that as per the definition, a locus is certifiably not a solitary point yet a bunch of focuses. To determine a locus, you will need to do the following two steps Use the condition given to draw purposes of the locus until you see an example. Connect the points and describe the locus fully. A great example of locus and we are all very familiar with it is the one resulting in a circle such as the circle shown in the figure above. A circle is the locus of points at a given distance from a given point and whose center is the given point and whose radius is the given distance. In the model above, we show the locus with red ran lines. Notice the utilization of the green portion to show that there is a solitary point E on the locus that is equidistant from the meeting lines. By a similar token, for the dark fragment, there is a solitary point F on the locus that is equidistant from the meeting lines. The locus of all points equidistant from two given points A and B is the perpendicular of the line segment joining A and B. We show the locus of all focuses equidistant to focuses An and B with the red ran line. We utilize two green portions to show where 1 single point or point C is equidistant to point An and B. The locus of focuses equidistant from two given equal lines is the line corresponding to the two given lines and situated between these given lines. The locus of focuses equidistant from two concentric circles is the circle concentric with the given circles and found halfway between them. In our example immediately above, the locus is shown with the red dashed circle. The locus of points equidistant from the sides of a given angle is the bisector of the angle. Harish Chandra Rajpoot authored 'Advanced Geometry' on research articles in Mathematics & Radiometry · Author has 1.6K answers and 7.7M answer views · 7y Related If a variable circle touches two given circles externally, then what is the locus of the centre of the variable circle? Let [math]a[/math] & [math]b[/math] be the constant radii of given circles with centers at A & B respectively & [math]r[/math] be the radius of variable circle with center C (as a parametric point) then we have [math]AB=a+b[/math] , [math]AC=a+r[/math] , [math]BC=b+r[/math] [math]AC-BC=a+r-(b+r)=a-b=\text{constant}\quad (\forall \ \ a\ne b)[/math] The above result shows that the center C of variable circle moves in such a way that the difference of its distances from two fixed points A & B (i.e. centers of given circles of unequal radii) is constant hence by definition, the locus of center C of variable circle is a hyperbola. But if the radii of given (fixed) circles with centers A & B Let [math]a[/math] & [math]b[/math] be the constant radii of given circles with centers at A & B respectively & [math]r[/math] be the radius of variable circle with center C (as a parametric point) then we have [math]AB=a+b[/math] , [math]AC=a+r[/math] , [math]BC=b+r[/math] [math]AC-BC=a+r-(b+r)=a-b=\text{constant}\quad (\forall \ \ a\ne b)[/math] The above result shows that the center C of variable circle moves in such a way that the difference of its distances from two fixed points A & B (i.e. centers of given circles of unequal radii) is constant hence by definition, the locus of center C of variable circle is a hyperbola. But if the radii of given (fixed) circles with centers A & B are equal i.e. [math]a=b[/math] then [math]AC-BC=a-a=0[/math] The above result shows that the center C of variable circle moves in such a way that the difference of its distances from two fixed points A & B (i.e. centers of given circles of equal radii ) is zero i.e. the center C of variable circle is always equidistant from two fixed points A & B i.e. center C moves on the perpendicular bisector of line joining the centers A & B hence, the locus of center C of variable circle is a straight line which is perpendicular bisector of line joining the centers of given circles. Robert Paxson BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views · 3y Related What is the locus of a point whose sum of distances from two fixed perpendicular lines is constant? Let the two perpendicular lines be the coordinate axes, [math]x=0[/math] and [math]y=0[/math], and any point [math]P[/math] on the locus, math[/math], then we have: [math]d_1+d_2=c[/math] [math]\|x\|+\|y\|=c[/math] where [math]c>0[/math] is a constant. This is the equation of a square that looks like this for [math]c=1[/math]: In general, let [math]y=mx+b[/math] and [math]y=-\frac{1}{m}x+f[/math], and any point [math]P[/math] on the locus, math[/math], then we have: [math]d_1+d_2=c[/math] and, noting that the distance from a point math[/math] to a line [math]px+qr+s=0[/math] is: [math]d=\frac{\|px_0+qy_0+s\|}{\sqrt{p^2+q^2}}[/math] we have, for [math]y=mx+b[/math], or [math]mx-y+b=0[/math]: [math]d_1=\frac{\|mx-y+b\|}{\sqrt{m^2+(-1)^2}}[/math] [math]d_1=\frac{\|mx-y+b\|}{\sqrt{m^2+1}}[/math] and for [math]y=-\frac{1}{m}x+f[/math], or [math]x+my-mf=0[/math]: [math]d_2=\frac{\|x+[/math] Let the two perpendicular lines be the coordinate axes, [math]x=0[/math] and [math]y=0[/math], and any point [math]P[/math] on the locus, math[/math], then we have: [math]d_1+d_2=c[/math] [math]\|x\|+\|y\|=c[/math] where [math]c>0[/math] is a constant. This is the equation of a square that looks like this for [math]c=1[/math]: In general, let [math]y=mx+b[/math] and [math]y=-\frac{1}{m}x+f[/math], and any point [math]P[/math] on the locus, math[/math], then we have: [math]d_1+d_2=c[/math] and, noting that the distance from a point math[/math] to a line [math]px+qr+s=0[/math] is: [math]d=\frac{\|px_0+qy_0+s\|}{\sqrt{p^2+q^2}}[/math] we have, for [math]y=mx+b[/math], or [math]mx-y+b=0[/math]: [math]d_1=\frac{\|mx-y+b\|}{\sqrt{m^2+(-1)^2}}[/math] [math]d_1=\frac{\|mx-y+b\|}{\sqrt{m^2+1}}[/math] and for [math]y=-\frac{1}{m}x+f[/math], or [math]x+my-mf=0[/math]: [math]d_2=\frac{\|x+my-mf\|}{\sqrt{1^2+m^2}}[/math] [math]d_2=\frac{\|x+my-mf\|}{\sqrt{m^2+1}}[/math] Therefore, the equation for the locus is: [math]\frac{\|mx-y+b\|}{\sqrt{m^2+1}}+\frac{\|x+my-mf\|}{\sqrt{m^2+1}}=c[/math] [math]\|mx-y+b\|+\|x+my-mf\|=c\sqrt{m^2+1}[/math] where [math]c>0[/math]. If [math]c=1[/math], [math]m=2[/math], [math]b=3[/math] and [math]f=4[/math], then we get a square that looks like this: If [math]c=\sqrt{2}[/math], [math]m=1[/math], [math]b=0[/math] and [math]f=0[/math], then we get a square that looks like this: where the perpendicular construction lines are shown as dashed blue and dashed red. The center of the square is at the point: math[/math] The distance from the center of the square to a corner is [math]c[/math]. The length of a side of the square is [math]\sqrt{2}c[/math]. Dave Benson trying to make maths easy. · Author has 6K answers and 2M answer views · 2y Related How can you show that the locus of point p(x,y) which moves on x²+y²-6x+4y-12=0 is a circle? General equation for any conic section is Ax²+Bxy+Cy²+Dx+Ey+F = 0, & A, B, C, D, E, F are constants. If A and C are non zero and equal, and both have the same sign, then it will be a circle. So x²+y²-6x+4y-12 = 0 is a circle and any point p(x,y) moving on it will be a circle. Compare equation of circle x²+y²+2gx+2fy+c = 0 & centre (g,f) = C(3,-2) radius r = √(g²+f²-c) = √{3²+(-2)²-(-12)} = √(9+4+12) =√25 = 5 General equation for any conic section is Ax²+Bxy+Cy²+Dx+Ey+F = 0, & A, B, C, D, E, F are constants. If A and C are non zero and equal, and both have the same sign, then it will be a circle. So x²+y²-6x+4y-12 = 0 is a circle and any point p(x,y) moving on it will be a circle. Compare equation of circle x²+y²+2gx+2fy+c = 0 & centre (g,f) = C(3,-2) radius r = √(g²+f²-c) = √{3²+(-2)²-(-12)} = √(9+4+12) =√25 = 5 Michael Paglia Former Journeyman Wireman IBEW · Author has 33.3K answers and 5.2M answer views · 1y Related What is the locus of points that lie on three mutually perpendicular diameters of a circle? How do you have more than 2 perpendicular diameters Do you understand locus points? A circle is not the way to understand locus Its only the radius to the Circumference That is the center Ellipses have 2 loci They are near the ends The distance from the 2 to the “ Circumference “ of the Ellipse is constant As a point moves around it One distance gets smaller Closer to that point And bigger from the other Always the same total Kepler used this thinking with speed and distance for the planets going away or coming close to the sun His involved the speed if planets from the suns gravity with distance too I want How do you have more than 2 perpendicular diameters Do you understand locus points? A circle is not the way to understand locus Its only the radius to the Circumference That is the center Ellipses have 2 loci They are near the ends . The distance from the 2 to the “ Circumference “ of the Ellipse is constant As a point moves around it One distance gets smaller Closer to that point And bigger from the other Always the same total Kepler used this thinking with speed and distance for the planets going away or coming close to the sun His involved the speed if planets from the suns gravity with distance too I want to say total area of the “sector" formed from one point to another in same amount of time Something like that Subhasish Debroy Former SDE at Bharat Sanchar Nigam Limited (BSNL) · Author has 6.6K answers and 5.8M answer views · 7y Related What does locus in a circle mean? I don't know if I get your question! If it is related to defining circle then it is — if a point is moving around a fixed point in such way that it is keeping a fixed/constant distance from the fixed point through the course of it's entire journey then locus of the equation of the moving point is called a circle. Related questions What is the locus point of a circle? How can I find the locus of a point of intersection? What do you mean by the statement “the locus of the center of a circle which touches a given line at a given point”? What is the locus of a point which moves so that its absolute is always 9? What is the locus of a circle? What is the definition of a locus of a variable point? What is the general guideline for finding the locus of points that is an intersection of two variable lines? A tangent is drawn from the point (-a, 0) to a variable circle, centre (a, 0) . What is the locus of the point of contact? What is the relationship between the locus of a fixed point and the movement of another fixed point on a circle? Can you explain the concept of a locus for a point that moves based on its distance from two other points? What is a locus, and how do you find the locus of a given point? Can you explain the concepts of locus and variable? What is the locus of a point whose distance from a fixed point is in constant ratio to the tangent drawn from it to a given circle? What is the process for finding the locus of a variable point in geometry? What is the locus of a point which moves so that its abscissa is always 9? Related questions What is the locus point of a circle? How can I find the locus of a point of intersection? What do you mean by the statement “the locus of the center of a circle which touches a given line at a given point”? What is the locus of a point which moves so that its absolute is always 9? What is the locus of a circle? What is the definition of a locus of a variable point? What is the general guideline for finding the locus of points that is an intersection of two variable lines? A tangent is drawn from the point (-a, 0) to a variable circle, centre (a, 0) . What is the locus of the point of contact? What is the relationship between the locus of a fixed point and the movement of another fixed point on a circle? Can you explain the concept of a locus for a point that moves based on its distance from two other points? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1736	https://www.physicsforums.com/threads/calculating-frequency-of-sin-x.987263/	Log in More options Style variation System Light Dark Contact us Close Menu You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Forums Physics Classical Physics Electromagnetism Calculating frequency of sin(x) Thread starter madeinusa Start date Tags : Frequency AI Thread Summary The discussion centers on the confusion surrounding the calculation of the frequency of the sine function, specifically sin(x). Participants clarify that while the period T is indeed 2π, the argument of the sine function must be dimensionless, meaning it should be expressed as sin(ωt) rather than sin(t) if t represents time. The correct frequency calculation involves using angular frequency ω, leading to f = ω/2π, which is valid if ω is appropriately defined. The conversation highlights the distinction between mathematical and physical interpretations of periodic functions, emphasizing the need for dimensionless arguments in trigonometric functions. Ultimately, the consensus is that using sin(t) as a time-dependent function is mathematically incorrect. 1 madeinusa : 4 : 0 Greetings,I'm confused about calculating frequency of sin(x) function.Applying basic equations I get this:T= 2pif = 1/T = 1/(2pi) = 1/6.28 = 0.15923 HzI know it's wrong, but what am I missing?Also, how can the period be in radians, I always thought that T should be in seconds. Physics news on Phys.org New adaptive optics system promises sharper gravitational-wave observations Physics-informed AI learns local rules behind flocking and collective motion behaviors New perspectives on light-matter interaction: How virtual charges influence material responses 2 DrClaude Mentor : 8,477 : 5,693 \sin(x)## is not a function of time, at least, not explicitly. So let's make it explicit:$$\sin(\omega t)$$or$$\sin(2 \pi \nu t)$$where ##\omega## is the angular frequency, expressed in radians per second (often simply written as 1/s, without the radian part) and ##\nu = \omega / 2 \pi## is the frequency, expressed in Hz, you are after. 3 madeinusa : 4 : 0 so for sin(x), ω = 1then my calculation f = ω/2π = 1/6.28 = 0.15923 Hz should be correct? 4 weirdoguy : 1,323 : 1,307 madeinusa said: what am I missing? The whole meaning and purpose of calculating a frequency. Why are you using ##\sin(x)##? What is ##x##? Besides, period in physics is something a little different than in pure maths. 5 madeinusa : 4 : 0 according to this: Example: sin(x) This is the basic unchanged sine formula. A = 1, B = 1, C = 0 and D = 0 So amplitude is 1, period is 2π, there is no phase shift or vertical shift: so if the period T is 2pi then frequency f should be 1/T = 1/(2pi) = 1/6.28 = 0.15923 Hz 6 weirdoguy : 1,323 : 1,307 This discuss maths, not physics as far as I can see. Period in maths is something different than in physics. 7 madeinusa : 4 : 0 OK, how about this: sin(t) , t is timeω should be 1 so the frequeny f = ω/2π = 1/6.28 = 0.15923 Hz Is this correct? 8 weirdoguy : 1,323 : 1,307 Yes (if ##\omega## is expressed in ##\frac{1}{s}##). Last edited: 9 ZapperZ Staff Emeritus Science Advisor Homework Helper Insights Author : 32,814 : 4,725 madeinusa said: OK, how about this: sin(t) , t is time Thus is mathematically wrong. The argument for any trig function must be dimensionless. You cannot use sin(t) if t is time. It must be of the form sin(ωt). Zz. 10 sophiecentaur Science Advisor Homework Helper : 30,081 : 7,377 madeinusa said: OK, how about this: sin(t) , t is timeω should be 1 so the frequeny f = ω/2π = 1/6.28 = 0.15923 HzIs this correct? I think you would find this a lot easier if you try to ignore your 'private' way of looking at the topic and just followed the mainstream. You could "how about" a lot if ideas but you will make faster progress if you try to go along with the usual treatment - it was developed for a good reason and yields some useful understanding. PS this is not just a "Do it this way and stop thinking" message. People can often feel victimised when they try things out on PF but people have some pretty fair ideas on this site. 11 FactChecker Science Advisor Homework Helper : 9,271 : 4,577 ZapperZ said: Thus is mathematically wrong. The argument for any trig function must be dimensionless. You cannot use sin(t) if t is time. It must be of the form sin(ωt). It is common to express any function of time as f(t). The sin function is as good as any function. IMO, there is no reason to insist that it be treated differently. In that context, the period is 0.159. 12 sophiecentaur Science Advisor Homework Helper : 30,081 : 7,377 FactChecker said: The sin function is as good as any function. True but the units are not intuitive and how do you deal conveniently with a mixture of several frequencies at once??. It's hardly surprising that 'everyone' uses ωt, is it? Using Sin(t) is standing up in a hammock - I think we can all agree - if pushed. FactChecker said: It is common to express any function of time as f(t) The function Exp(-t) is another one which, refusing to include a scale constant inside the brackets really does make life difficult and for no good reason. Where would we be without our e and our ω (unless we were pure mathematicians, of course). 13 George Jones Staff Emeritus Science Advisor Gold Member : 7,643 : 1,601 FactChecker said: It is common to express any function of time as f(t). The sin function is as good as any function. IMO, there is no reason to insist that it be treated differently. In that context, the period is 0.159. It is certainly true the sine function is periodic in its argument, whatever the interpretation of its argument. It is very unusual in physics, however, for the argument of sine to have physical units. For example, if ##t## has the units of seconds, what are the units of $$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - ... ?$$ If ##t## has units of seconds, then, on the right, the physical units of the first term are ##s##, the units of the second term are ##s^3##, etc. How do we sum these terms to get a physically sensible answer? If ##t## has no physical units, then the sum makes physical sense. 14 gmax137 Science Advisor Education Advisor : 3,132 : 3,604 sophiecentaur said: standing up in a hammock Thanks, sophie, I never heard that before 15 sophiecentaur Science Advisor Homework Helper : 30,081 : 7,377 The full joke made me laugh a lot - back in about 1963! it keeps well. Similar threads Are frequency and energy related? Replies : 4 Views : 2K Alternator torque when connected to a phase-shifted AC source Replies : 3 Views : 2K Fundamental frequency of a rectangular-cavity whistle Replies : 1 Views : 2K How should I show that solutions can be expressed as a Fourier series? Replies : 3 Views : 1K Given the frequency and wavespeed, find k Replies : 7 Views : 2K The reciprocal relationship between frequency and period Replies : 13 Views : 8K Graph Sin(2x+6): Is Parent Function the Best Option? Replies : 6 Views : 1K Finding frequency of an AC current sin wave? 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1737	https://www.youtube.com/watch?v=JF40Cq-iiP0	How to do a PROOF in SET THEORY - Discrete Mathematics TrevTutor 303000 subscribers 4772 likes Description 293388 views Posted: 9 Sep 2021 We learn how to do formal proofs in set theory using intersections, unions, complements, and differences. 0:00 - [Intro] 0:49 - [Language of Set Theory] 3:31 - [Proof #1] 6:15 - [Proof #2] 11:12 - [Proof #3] 14:25 - [Proof #4] SetTheory #Proofs #DiscreteMath Support me on Patreon: Visit my website: Subscribe on YouTube: --Playlists-- Discrete Mathematics 1: Discrete Mathematics 2: --Recommended Textbooks-- Discrete and Combinatorial Mathematics (Grimaldi): Discrete Mathematics (Johnsonbaugh): Discrete Mathematics and Its Applications (Rosen): Book of Proof (Hammack): Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding. Special Thanks to the following incredible supporters for supporting the channel and making these videos possible! - Nikita Tsyganov 146 comments Transcript: Intro one of the first proofs that you'll probably have to do in your discrete math course is proving that two sets are equivalent for example at the top here we might have to prove that de morgan's law holds so we usually shorten this as dem and you might be thinking well how can i do that because normally we just use a set law or we use a venn diagram and then we're done but instead what you're supposed to do is prove it step by step using what's called a direct proof so you would show that if you have x in this set then x would be in the other set on the right and vice versa and usually that's a pretty tough place to start so we're going to go step by step on what each step means in terms of regular languages we'll do some examples and by the end of this video this proof will be very easy for you okay so it's good to remind ourselves of Language of Set Theory the actual language of set theory and what things mean so if we have a set a we can say that okay x is an element of that set and with a picture usually what that means i'll use a different color for this is that we have some element x right in there so this is a really easy thing to work with what's more interesting is when we get into things like the complement of a you might see this as a bar or maybe ac or a with a little prime symbol above it and what this means in here is that you have a inside the circle and you have the complement of a outside so x is in this outside set an a complement now what this means in terms of set proofs is two things it could mean that x is an element of a complement and alternatively they could also mean that x is not an element of a so we can use these two facts when looking at say a complement as a set okay if we take a look at a union b well what does this mean this means that x we take a look at a and b we have our element x somewhere it can be in any of these spaces so in natural language we would say that x is an element of a or x is an element of b and then in a set theory proof normally you have two different paths to go down you have a path to consider what if x is an a what if x is in b and normally you split at that point we have a intersection b so in terms of the language here let's say we have our two sets a and b we know specifically that x is going to be right in the center portion so i should really just draw it right in here and the language we use for this is we say that x is an element of a and x is an element of b that's how we can think of things in natural language uh the last one which is a minus b or the difference of a and b uh sometimes you'll read it like this sometimes you'll see it as a right slash we draw our two sets a and b here it means that x is in a but x is not in b so x specifically is in this part somewhere so specifically we can rewrite this as x as an element of a and x is not an element of b so now that we know how to take these signs and convert them into natural human language we can start working with proofs piece by piece so let's start with something relatively Proof #1 simple let's prove that the intersection of a and b is a subset of a union b okay so how do we do a direct proof with this well we need to have a starting point and if we're ever doing a subset proof we always want to make an assumption about the left hand side we want to say something like okay suppose x is an element of a intersection b so let's say that x is on the left now we're going to prove using natural language and breaking down the sets that x is going to be on this side as well x is going to be an element of a union b so let's think back to the language uh if x is an a intersection b there's a way we can break this down we can say then x is in a and x is in b okay so we need to get to the fact that x is in a union b and we know that this means or so we should take one of these two it doesn't really matter which one we take but we should take it to the next step so then we can say um if x isn't a then we know i'll just write this out in all language then we know that x isn't a or x is in b so because x is an a or x is in b we know that x is an element of a union b because this is just taking the natural language and then converting it back into set theory and this would be the end of the proof at this point so a little bit daunting at first but there's not too many things we could do alternatively if we chose x is equal to b or x is in b sorry we could do the same step as before if x is in b then we know that x is in b or x is an a so whenever we have something just on its own we can always do or with any other set because or just require that it's in one of those two okay and just to confirm that this works we can always draw pictures so we know that x is in the center of these two sets let me just draw these a little bit bigger here's a here's b we know that x is in there because x is in a and b and here's what x uh here's what a or b is a union b it's in the whole thing so if x is in the middle if x is at a and b then obviously x is in a or b okay that's the first one Proof #2 let's do one that's a little bit more complicated so again this is a nice subset proof so what we're going to do is we're going to make an assumption we're going to assume that x is in that left side set so let's assume that x is an a minus b intersection c so there's a lot more going on here but it's just going to be the same thing we're just breaking apart the language and the sets so if x is in a minus b and c remember we can say that x is an a and x is not in the minus set so and x is not in b intersection c okay so we want to get that x is in this set over here a minus b union a minus c and this looks nothing like what we have so far so we're going to have to be a little bit clever with this right side so let's think about this um we're saying that x is not in b intersection c so what this means is that x could be over here and b or x could be over here in c or if we think about the entire universe x might be anywhere outside of it so what we do know is that from this x is going to be in the complement of b so that's everything outside of b or x is going to be in the complement of c in other words it's going to be in something outside of c so uh to draw this in a picture just so we can really understand what this means i'll draw this again with b and c and the outline if x is in b complement this means it's somewhere here so i'm just trying to fill in all of this area here so we can see it's not in that center point it's not in this b and c area and if we think about x being in c complement that means it's uh somewhere in one of these areas so you can see the only thing that isn't covered is b and c which is exactly what we want because we don't want x in there okay so i'm going to keep this over from the last time x is an a and x is in b complement or x is in c complement okay fantastic uh now what can we do well again we're trying to get a minus b or a minus c so if we know that x is an a and x is in one of those sets we can distribute this with natural language we can say that x is an a and x is in b complement or x is an a and x is in c complement so one of those will be true okay well what's really nice here is that we're just a couple steps away from taking a look at that top set and getting a match so we could say x is an a and remember another way to write this one if x isn't b complement this means that x is not in b so that's one condition or x is an a and we can do the same thing with x's and c complement that's the same thing as saying that x is not in c okay if we bracket this off and we take a look at the thing we're trying to prove up top maybe we can see a similarity now so now that we have x is an a and x is not in b we can transfer this over and say x is in a minus b because this is just the definition or x is in a minus c because again we have that same definition there and now we can take this natural language or and we can combine these two we can say x is an a minus b or is the same thing as union and x would be in a minus c there as well so now we've proven that if x is in this set a minus b intersection c then it follows that x is in this final set a minus b union a minus c and this is exactly what we wanted to prove up top so this one was more complicated but we're just taking the language that we used in that first list in the first slide and breaking it down to do this so just step by step piece by piece it worked out so let's go back to this original question let's prove de morgan's law and Proof #3 what we notice is that we have an equality sign so we need to do two things we need to prove that the complement of a union b is a subset of a complement intersection b complement and we also need to prove the reverse that a intersection b complement is a subset of a union b complement so let's call this one let's call this two and let's start with one so what we'll do is we're going to assume that x is on that left side so x is an element of a union b complement so if x is an a union b complement this then means that x is not in a union b and if x is not an a or x is not in b in natural language we can write this as x is not an a and x is not in b so these two are the same thing x is not an a or b is the same thing as saying x is not an a and x is not in b okay well this is nice so if x is not an a that's the same thing as saying x is an a complement and if x is not in b that's the same thing as saying x is in b complement oh if i take a look at the top of the page i'm just one step away from putting it back into set so then we can say that x is in a complement intersection b complement all we've done is we've taken this and and converted it into intersection okay so that's one direction what's nice about this is that we can really just go in the opposite direction if we wanted to so for simplicity we could say if we start at 2 we work all the way back up to 1 backwards but let's just do it piece by piece anyway so i think i can fit this on the right side of the screen next to the original so let's just keep everything on the same page let's just use a different color so we don't get these confused so i'm going to assume for number two that x is in a complement intersection b complement okay so that means that x is an a complement and x is in b complement okay so this means in this case uh that x is not in a and x is not in b so if x isn't in either a or b then we can write that x is not in a union b which means that x is in a union b complement so we've now proven both sides so we should complete this and say therefore a union b complement is equal to a complement intersection b complement so we had to do both ways because we're looking for equality and all these give us our subsets when we go from one to the other okay i want to do one more question with Proof #4 you and try to beat you or try to beat me if i can and try to beat me if you can so if a is a subset of b then a minus c is a subset of b minus c so this one again is not too challenging but if this is the first time you're doing it this is a good exercise so we're given this information this this is true we can use this as a fact so what we want to do is we want to prove that a minus c is a subset of b minus c so i'm going to start with an assumption and assume that x is in that set so i will assume that x is an element of a minus c okay so natural language just means that x is in a and x is not in c okay that's good um but now what do we do somehow we have to get b and not c so really what we want in our step is we want to get to this step x is not in c and x is in b well we can use this fact here since we know that a is a subset of b this tells us that x is an a means x is in b so this is our bridging step so because x is an a means that x is x is in b then we can go to this step right here x is in b and x is not in c therefore x is an element of b minus c so we have our assumption here x is an a minus c and we've gotten to the fact that x is in b minus c therefore this whole thing has been proven so hopefully you can do some set theory proofs now i'll have another video with cartesian products and power sets but this is a great place to get you started and if you have any problems again just ask below and i'm sure someone or me or another commenter is happy to answer you
1738	https://www.ixl.com/math/grade-6/find-the-whole-given-a-part-and-a-percent	IXL \| Find the whole given a part and a percent \| 6th grade math SKIP TO CONTENT [x] - [x] IXL Learning Sign in- [x] Remember Sign in nowJoin now IXL Learning Learning Math Skills Lessons Videos Games Fluency Zone New! Language arts Skills Videos Games Science Social studies Spanish Recommendations Recommendations wall Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Student awards Assessment Analytics Takeoff Inspiration Learning All Learning Math Language arts Science Social studies Spanish Recommendations Skill plans Learning Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Assessment Analytics Takeoff Inspiration Membership Sign in Math Math Language arts Language arts Science Science Social studies Social studies Spanish Spanish Recommendations Recommendations Skill plans Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Awards Big Ideas Math (2019) - 6th grade Find the whole given a part and a percent 5S4 Share skill Copy the link to this skill share to facebook share to twitter Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. Let's go! 3 Find the whole given a part and a percent 5S4 Share video Copy the link to this video share to facebook share to twitter You are watching a video preview. Become a member to get full access! You've reached the end of this video preview, but the learning doesn't have to stop! Join IXL today! Become a memberSign in Incomplete answer You did not finish the question. Do you want to go back to the question? Go back Submit Learn with an example Fill in the missing number. 50% of = 32 Submit Back to practice ref_doc_title. Back to practice Learn with an example Learn with an example question Fill in the missing number. 50% of = 30 key idea A percent is 1 part out of 100. 1% is equal to 1 100 or 0.01. Proportions can be used to solve percent problems. part whole=percent 100 solution The part is 30. The percent is 50%. Let n represent the whole. Write a proportion to solve for n. 30 n=50 100 50 n=30 · 100 50 n=3,000 50 n ÷ 50=3,000 ÷ 50 n=60 50% of 60 = 30 You could also find the answer by writing and solving an equation: part=percentage of whole 30=0.5 · n 30=0.5 n 30 ÷ 0.5=0.5 n ÷ 0.5 60=n 50% of 60 = 30 Looking for more? Try this: Lesson: Percents Learn about percents! What are percents? Relating percents, decimals, and fractions Finding percents of numbers Finding percents using proportions Using percents in the real world Lesson: Percents Learn with an example Excellent! You got that right! Continue Learn with an example Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Questions answered Questions 0 Time elapsed Time 00 00 02 hr min sec SmartScore out of 100 IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more 0 You met your goal! Teacher tools Group Jam Live Classroom Leaderboards Work it out Not feeling ready yet? These can help:Percents of numbersPercents of numbers - Sixth grade CSBDivide decimalsDivide decimals - Sixth grade Y7CDivide fractionsDivide fractions - Sixth grade LNFLesson: Percents Learn about percents! What are percents? Relating percents, decimals, and fractions Finding percents of numbers Finding percents using proportions Using percents in the real world Lesson: Percents Company \| Membership \| Blog \| Help center \| User guides \| Tell us what you think \| Testimonials \| Careers \| Contact us \| Terms of service \| Privacy policy © 2025 IXL Learning. All rights reserved. Follow us First time here? 1 in 4 students uses IXL for academic help and enrichment. Pre-K through 12th grade Sign up nowKeep exploring
1739	https://math.stackexchange.com/questions/57429/functions-similar-to-log-but-with-results-between-0-and-1	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Functions similar to Log but with results between 0 and 1 Ask Question Asked Modified 3 years, 7 months ago Viewed 47k times 27 $\begingroup$ I need a function similar to Log but it should produce numbers between 0 and 1 Something like: f(0)=0 f(1)=0.1 f(2)=0.15 f(3)=0.17 f(100)=0.8 f(1000)=0.95 f(1000000000)=0.99999999 I need this in my program that I am programming and I can use only standard functions like log, exp, etc... Any help would be appreciated. functions Share edited Aug 14, 2011 at 17:36 J. M. ain't a mathematician 76.7k88 gold badges222222 silver badges347347 bronze badges asked Aug 14, 2011 at 17:02 410503410503 38111 gold badge33 silver badges44 bronze badges $\endgroup$ 7 $\begingroup$ I've got a question: f(1000000001)=1 is ok or do you need something like $f(n)_{n \rightarrow \infty}<1$ ? $\endgroup$ ulead86 – ulead86 2011-08-14 17:09:20 +00:00 Commented Aug 14, 2011 at 17:09 5 $\begingroup$ $f(x) = 1-\mathrm{e}^{-x}$ ? $\endgroup$ Sasha – Sasha 2011-08-14 17:09:43 +00:00 Commented Aug 14, 2011 at 17:09 $\begingroup$ Or take the partial sums $a_1=\frac {1}{2^n}$ or some other geometric series, maybe with a weighing factor. $\endgroup$ gary – gary 2011-08-14 17:13:19 +00:00 Commented Aug 14, 2011 at 17:13 $\begingroup$ "I need this in my program that I am programming" - what exactly are you trying to do that makes you need such a function? $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2011-08-14 17:13:53 +00:00 Commented Aug 14, 2011 at 17:13 1 $\begingroup$ There's a lot of sigmoidal functions to choose from: arctangent, hyperbolic tangent, the logistic function... $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2011-08-14 17:32:16 +00:00 Commented Aug 14, 2011 at 17:32 \| Show 2 more comments 3 Answers 3 Reset to default 25 $\begingroup$ Sasha's suggestion of $f(x) = 1-\exp(-x)$ is good, but doesn't fit your example values too well, even if you scale $x$ appropriately. However, some other similar function, such as $f(x) = 1-1/(1+x)$, might work even better. Share edited Aug 14, 2011 at 17:50 Jonas Meyer 55.9k99 gold badges216216 silver badges310310 bronze badges answered Aug 14, 2011 at 17:25 Ilmari KaronenIlmari Karonen 26.9k44 gold badges7171 silver badges110110 bronze badges $\endgroup$ 3 19 $\begingroup$ In case you're using this, you can generalize the example to $f(x) = x/(a+x)$; this way, you can adjust $a$ to have a slower or faster growing curve. $\endgroup$ Gerben – Gerben 2011-08-14 19:03:08 +00:00 Commented Aug 14, 2011 at 19:03 1 $\begingroup$ How would one scale either of those functions to get the value 1 at some predefined x, say x=100? $\endgroup$ ed22 – ed22 2020-08-27 09:35:19 +00:00 Commented Aug 27, 2020 at 9:35 1 $\begingroup$ @ed22: Both of these functions will asymptotically approach 1 as $x$ grows but will never reach it, like the OP asked for. Of course, you could just define $g(x) = f(x) \mathbin/ f(100)$ so that $g(100) = 1$, with $g(x)$ then approaching an arbitrary asymptote of $1/f(100)$ as $x$ grows further beyond $100$. But depending on how you want $g(x)$ to behave for $x$ between $0$ and $100$, that may not actually do what you want. If you can describe the behavior you want, you could maybe ask a new question about it. $\endgroup$ Ilmari Karonen – Ilmari Karonen 2020-08-27 10:12:50 +00:00 Commented Aug 27, 2020 at 10:12 Add a comment \| 9 $\begingroup$ I like the hyperbolic tan (and it likes me) $$f(x) = \tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{e^{2x}-1}{e^{2x}+1} = 1 - \frac{2}{e^{2x}+1}$$ $f$ is strictly increasing and satisfies $f'(0) = 1$, $f(0) = 0$, $f(\infty) = 1$. If you want to map $(-\infty, \infty)$ into $[0, 1)$ (instead of $[0, \infty)$), use $$f(x) = \frac{\tanh(x)+1}{2} = 1 - \frac{1}{e^{2x}+1}.$$ I have seen this called the "logistic" curve, or "s-shaped" curve. Share answered Aug 15, 2011 at 3:47 marty cohenmarty cohen 111k1010 gold badges8888 silver badges186186 bronze badges $\endgroup$ Add a comment \| 8 $\begingroup$ If you want it to involve log, try $f(x) = 1 - \log(a)/\log(a+bx)$ for suitable positive numbers $a$ and $b$. Share edited Aug 14, 2011 at 19:03 Did 285k2727 gold badges334334 silver badges613613 bronze badges answered Aug 14, 2011 at 18:59 Robert IsraelRobert Israel 472k2828 gold badges376376 silver badges714714 bronze badges $\endgroup$ 1 1 $\begingroup$ The beauty of this answer is that you can interpret $\log$ here as either the common or natural one... :D $\endgroup$ J. M. ain't a mathematician – J. M. ain't a mathematician 2011-08-14 19:02:24 +00:00 Commented Aug 14, 2011 at 19:02 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functions See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked Term for functions with results between 0 and 1? 0 Function similar to log(a/b) but with results between [0, 1] or [-1, 1] 0 A linear function based on remaining time returning values between 0 and 1 Related Function that magnifies small changes and compresses large changes Program to find closest function to fit arbitrary data a function which is 1. easy to compute and 2. equal to 1 inside a interval, and quickly transitions to zero towards the boundaries 1 Looking for two asymptotic functions 3 a Function to Push Numbers Away From a Central Number 0 How to map logarithmic scale onto linear space? 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1740	https://www.oed.com/dictionary/hierarchy_n	Skip to main content Advanced search First published 1898; not fully revised hierarchynoun Factsheet What does the noun hierarchy mean? There are six meanings listed in OED's entry for the noun hierarchy, one of which is labelled obsolete. See ‘Meaning & use’ for definitions, usage, and quotation evidence. hierarchy has developed meanings and uses in subjects including Christianity (Middle English) ecclesiastical (mid 1500s) logic (mid 1600s) sciences (mid 1600s) Entry status OED is undergoing a continuous programme of revision to modernize and improve definitions. This entry has not yet been fully revised. See meaning & use How common is the noun hierarchy? About 20occurrences per million words in modern written English \| \| \| --- \| \| 1750 \| 2.5 \| \| 1760 \| 2.8 \| \| 1770 \| 2.9 \| \| 1780 \| 3.4 \| \| 1790 \| 3.3 \| \| 1800 \| 3.8 \| \| 1810 \| 4.5 \| \| 1820 \| 4.5 \| \| 1830 \| 4.8 \| \| 1840 \| 5.7 \| \| 1850 \| 5.2 \| \| 1860 \| 4.6 \| \| 1870 \| 4.8 \| \| 1880 \| 4.1 \| \| 1890 \| 4.0 \| \| 1900 \| 3.7 \| \| 1910 \| 3.3 \| \| 1920 \| 3.7 \| \| 1930 \| 5.1 \| \| 1940 \| 7.0 \| \| 1950 \| 9.7 \| \| 1960 \| 14 \| \| 1970 \| 18 \| \| 1980 \| 20 \| \| 1990 \| 24 \| \| 2000 \| 23 \| \| 2010 \| 23 \| See frequency How is the noun hierarchy pronounced? British English /ˈhʌɪ(ə)rɑːki/ HIGH-uh-rar-kee U.S. English /ˈhaɪ(ə)ˌrɑrki/ HIGH-uh-rar-kee See pronunciation Where does the noun hierarchy come from? Earliest known use Middle English The earliest known use of the noun hierarchy is in the Middle English period (1150—1500). OED's earliest evidence for hierarchy is from around 1380, in English Wycliffite Sermons. hierarchy is a borrowing from French. Etymons: Old French ierarchie. See etymology Nearby entries hieraco-, comb. form hiera picra, n.1379– hierarch, adj. & n.1486– hierarchal, adj.1641– hierarchic, adj.1681– hierarchical, adj.1471– hierarchically, adv.1624– hierarchism, n.1846– hierarchist, n.1640– hierarchize, v.1884– hierarchy, n.c1380– hieratic, adj.1656– hieratica, n.1832– hieratical, adj.1656– hieratico-, comb. form hieraticopolitical, adj.1685– hieratite, n.1882– hiero-, comb. form hierocracy, n.1794– hierocratic, adj.1851– hierocratical, adj.1801– Browse more nearby entries Etymology Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Meaning & use Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Pronunciation Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Forms Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Frequency Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Compounds & derived words Thank you for visiting Oxford English Dictionary To continue reading, please sign in below or purchase a subscription. After purchasing, please sign in below to access the content. View our subscription options Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Sign in Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Entry history for hierarchy, n. hierarchy, n. was first published in 1898; not fully revised. hierarchy, n. was last modified in September 2025. Revision of the OED is a long-term project, and oed.com is a living text, updated every three months. Entries which have not been fully revised may include partial interim updates, including: corrections and revisions to definitions, especially to improve clarity, accuracy, or intelligibility; new or updated quotation evidence, and reverified or redated bibliographical information; new or updated pronunciations (transcriptions and audio files); new or revised etymological information and improved coverage of variant spellings; new senses or phrases added in print and online updates since OED2 (1989). Revisions and additions of this kind were last incorporated into hierarchy, n. in September 2025. Earlier versions of this entry were published in: OED First Edition (1898) Find out more OED Second Edition (1989) Find out more View hierarchy in OED Second Edition Cite Permanent link: Chicago 18 Oxford English Dictionary, “,” , . Copied to clipboard MLA 9 “” Oxford English Dictionary, Oxford UP, , . Copied to clipboard APA 7 Oxford University Press. (n.d.). In Oxford English dictionary. Retrieved , from Copied to clipboard hieraco-, comb. form hiera picra, n.1379– hierarch, adj. & n.1486– hierarchal, adj.1641– hierarchic, adj.1681– hierarchical, adj.1471– hierarchically, adv.1624– hierarchism, n.1846– hierarchist, n.1640– hierarchize, v.1884– hierarchy, n.c1380– hieratic, adj.1656– hieratica, n.1832– hieratical, adj.1656– hieratico-, comb. form hieraticopolitical, adj.1685– hieratite, n.1882– hiero-, comb. form hierocracy, n.1794– hierocratic, adj.1851– hierocratical, adj.1801– Browse more nearby entries
1741	https://www.ebsco.com/research-starters/religion-and-philosophy/reliability	Research Starters Home EBSCO Knowledge Advantage TM Reliability Reliability is a crucial concept in research and data collection, referring to the consistency with which an instrument measures a characteristic or attribute. It is essential for ensuring that the data collected is meaningful and can be trusted. Reliability is quantified by examining the observed variability in scores, which can arise from true differences among respondents as well as errors in measurement. Various methods, including parallel forms, test-retest, and split-half techniques, are used to estimate reliability. Factors influencing reliability can be personal, such as the participant's understanding of questions or their emotional state during assessment. For instance, temporary conditions like fatigue can affect responses, leading to variability that does not reflect true opinions or characteristics. Additionally, external factors, such as the environment in which a survey is conducted, can impact participant responses. To achieve valid results, a data collection instrument must be both reliable and valid; reliability alone does not guarantee that an instrument measures what it is intended to measure. Ultimately, ensuring high reliability in research instruments is vital for drawing accurate conclusions and making informed decisions based on the data gathered. Published in: 2024 By: Wienclaw, Ruth A. Go to EBSCOhost and sign in to access more content about this topic. Reliability To yield useable data, surveys, assessment tools, and other data collection instruments need to be both reliable and valid. Reliability is a measure of the degree to which such instruments consistently measure a characteristic or attribute. Statistically, reliability is a measure of the observed variability in obtained scores on an instrument. Variability can come both from true variance (such as differences in opinions, knowledge, or other characteristics of the individual) or from error variance. The total variability of a data collection or assessment instrument is the sum of the true variability and the variability due to error. Reliability can be estimated through the use of parallel forms of the instrument, repeated administration of the same form of the instrument, subdivision of the instrument into two parallel groups of items, and analysis of the covariance among the individual items. In the case of data collection or assessment instruments, reliability—the degree to which a data collection or assessment instrument consistently measures a characteristic or attribute—is essential for the instrument to be valid. In other words, one must be confident that the instrument measures what it purports to measure. No matter how well-written a data collection or assessment instrument appears to be on its face, it cannot be valid unless it is reliable. If a measure is not reliable, it does not consistently measure the same thing. In other words, it is not measuring the construct that it was designed to measure, so the instrument is neither reliable nor valid. Therefore, both validity and reliability are essential when conducting survey research, so that the data collected in the study will actually give researchers the information they are trying to gather. Without both reliability and validity, the data collected are meaningless, and no conclusions can be drawn. True Data Variance vs. Data Error Even in the physical sciences, two sets of measures performed on the same individuals never exactly duplicate each other. To the extent that this is true, the measurement instrument is unreliable, whether it is a physical scale used to measure the weight of a chemical compound or a paper-and-pencil survey used to measure a person's attitude toward something. For example, on a scale of 1 to 10, what one person describes as a 10 another person may call a 9.5. This does not necessarily mean that their opinions are different, just that the two people are expressing them differently. Some of the total observed variance (the square of the standard deviation) in scores is due to true variance, or real differences in the way that people are responding to the question. The other part of the total variance is due to error. Factors Affecting Reliability Personal Factors There are many reasons why a data collection instrument may not be reliable and thus may contribute to the error variance. In general, what social scientists try to measure are lasting and general characteristics of individuals related to the underlying construct that the assessment instrument is trying to measure. However, other types of characteristics that are not part of the underlying construct, such as the individual's test-taking techniques and general ability to comprehend instructions, may also be measured. In addition to the permanent characteristics of individuals, there are also temporary characteristics that can affect their responses to questions on data collection instruments. These might include such factors as general health, fatigue, or emotional strain, all of which can affect the way that an individual responds to a question -- a phenomenon familiar to anyone who has had to take a test in school when he or she was ill. Similarly, external conditions such as heat, light, ventilation, or even momentary distraction can impact one's responses in a way that does not reflect the underlying theoretical construct. Further, the subject's motivation can also impact the reliability of a data collection instrument. For example, for the most part teachers assume that their students are motivated to do well on any data collection instruments (e.g., a mid-term exam) given them. However, the same assumption cannot be made when asking a random sample of individuals to answer questions on the data collection or assessment instrument. For instance, it is often difficult to get shoppers to cooperate in opinion surveys because they are intent on accomplishing their errands so that they can go home. The motivation that may be offered to entice participation in the survey, such as a crisp new dollar bill or a carton of instant macaroni and cheese, is nothing compared to the motivation of students to do well in a course. Difficulty in Understanding the Data Collection Instrument Another source of variability in the way people respond to a data collection instrument may be individual differences in the way that people interpret the questions on the instrument. Care must always be taken in the development of a data collection instrument to write to a level that can be understood by all the people who will answer the questions. The questions need to be written unambiguously and with proper spelling, grammar, and punctuation to help reduce the possibility of low reliability because people do not understand what the questions are asking. For example, a child could easily take a question about a person who "lives near" him or her to mean the family members in the immediate household rather than a neighbor. Similarly, a question about how much one likes "sweet tea" means a different thing in the southern United States, where it refers to iced tea sweetened with simple syrup, than it does in Great Britain. The use of clear, concise language and operational definitions can help increase the reliability of the instrument. Reliability problems may also stem from individual differences in the way that people interpret responses to a data collection instrument. Even in cases where the end points of the scale are operationally defined with clear examples, people who moderately dislike something could possibly vary their answers between 20 and 40 on a scale of 100, yet all mean the same thing. Similarly, some people never give a perfect score to anything on a rating scale because they believe that there is always room for improvement. Inaccurate Measurements Another potential cause of lack of reliability is when the data collection instrument is not valid and is actually measuring more than one thing. For example, a researcher might set up an experiment to determine whether men or women are more likely to stop and assist a stranger on the street who needs help. This could be done by having a confederate drop a sheaf of loose papers and counting how many times a man stops to help and how many times a woman stops to help. If more men stop than women, the researcher will probably conclude that men are more likely to help a stranger than are women. However, these results might not be replicable, as a great number of factors can affect a person’s willingness to stop and help. Similarly, great variability and concomitant low reliability can be found when data are collected through the use of a structured interview. Even when the questions are always asked with the same wording, differences between interviewers and how they are perceived by the people answering the questions can result in a situation where the same data collection instrument has widely disparate responses because of the interviewing styles of different interviewers. Testing the Efficacy of Data Collection Instruments In general terms, reliability is defined as the degree to which a data collection or assessment instrument consistently measures a characteristic or attribute. There are several ways that the reliability of such an instrument can be estimated. The first of these methods involves the administration of two parallel forms of the instrument under specified conditions. The statistical correlation of the results of the two administrations is calculated to determine the degree of variance between the forms. However, it is important to note that it is typically difficult to develop two equivalent forms of the same assessment instrument that both have equal discriminability. As a result of this difficulty, a second method of determining reliability, called test-retest reliability, is frequently used. In this approach, the same form of the data collection instrument is administered twice to the same sample of individuals, and the correlation between the two scores is calculated to determine the reliability. A third approach to estimating reliability is to subdivide a single instrument into two presumably parallel groups of items. All of the items on the instrument are given to one sample of individuals at one time, and then the items are split out and treated as if they were two separate instruments. Each group is scored separately, and the resulting scores are correlated. This approach is called split-half reliability. Finally, an analysis of covariance can be calculated among the individual items on the assessment instrument to determine the true score and error variances. Applications Case Study: Presenting Child Need In an example of the application of reliability and validity assessment to a real-world problem, Forrester, Fairtlough, and Bennet (2007) examined the inter-rater reliability of methods to describe the needs of children to children's services in England and Wales. Although it is important to look at the unique characteristics of each case when determining what type of help a child needs, it is also important to be able to speak about "need" using a common language that will reliably discriminate between various classifications of need so that children can be given the help necessary for their well-being. Several typologies of need identification have been developed, but in order to maximize their usefulness, they must yield reliable results. Forrester, Fairtlough, and Bennet (2007) examined 200 consecutive file studies of closed referrals, first analyzing the files to classify the presenting needs or potential needs of the child in each case and then grouping them into clusters of variables for issues that occurred more than once. Fifty randomly selected cases were then tested to determine the relationship between the variables. The patterns were statistically analyzed using cross-tabulation and Spearman's rank correlation coefficient. Based on this analysis, the variables were reduced to a final list of ten, plus an "other" category. The Results The authors reached four main conclusions about the reliability of descriptions of children's need used by social services. First, it was found that descriptions of need that relied on a "main" need were not as reliable as other approaches, and that patterns of incidence could not be described adequately using the construct of "main" need. Although such an approach may simplify data presentation, it does not adequately describe the complexity of a child's situation, nor does it give any indication of the seriousness of the need. Second, although it was found that other approaches to describing need were more reliable than the "main" need approach, they, too, were not without their problems. In these typologies, classifications such as "dysfunctional family" or "unstable or otherwise detrimental family" were vague and had low levels of reliability. The authors urged that such terms be better defined in order to increase reliability. Third, the meaning of the legal definition of need had low levels of agreement between the raters. In part, this appeared to be a result of the fact that the legal definition emphasized seriousness of the need rather than presence of the need, as is the case in the other definitions and typologies. The authors concluded that typologies needed to be developed for the full range of referrals to children's services. In addition, they cautioned that the concept of "main" need that was used in both research and government policy was unreliable and not a good indicator of a child's situation or problem. The use of this concept could lead to misclassification and inappropriate intervention for at-risk children. Third, some specific categories that were currently used, such as "dysfunctional family," required better definition to increase reliability of assessments. Finally, future typologies of need should be tested for inter-rater reliability before being implemented. It is only through a reliable instrument that at-risk children can be consistently identified and their needs appropriately assessed. The Children's Physical Environment Rating Scale In another example of a reliability study, Moore and Sugiyama (2007) examined the reliability and validity of a new scale to be used for assessing the physical environment of early childhood educational facilities. The literature links the physical environment of such facilities to cognitive and social development during early childhood. The Children's Physical Environment Rating Scale (CPERS) comprises 124 items clustered into fourteen scales that focus on planning, overall architectural quality, indoor activity spaces, and outdoor play areas. The reliability of the CPERS was tested for inter-rater reliability, test-retest reliability, and internal consistency. Inter-rater reliability was tested in forty-six childhood development centers in Sydney, Australia. Each center was assessed by two of seven raters through several cycles of field testing. The resulting data were statistically analyzed to determine the degree of agreement between raters for each item and Cronbach's generalizability coefficient G for each subscale. These analyses showed a high degree of agreement and generalizability between the raters on the items on the CPERS. Based on this result, the authors concluded that the CPERS is a reliable instrument that can consistently be used to rate the physical environment of an early childhood facility, both for research purposes and in general. In addition, the authors examined the degree to which scores on the CPERS were stable over time. Each of eleven early childhood development centers was assessed once and then reassessed again three to five weeks later. The results were analyzed using Cronbach's G. The results showed a high degree of test-retest reliability, indicating that scores on the CPERS are stable over time and are consistent measures. Finally, internal consistency of the scales in the CPERS was assessed simultaneously but independently by two raters similar to center directors who might use the scale on a routine basis to assess eleven centers. The results of the ratings were analyzed using Cronbach's alpha to show the internal consistency of each subscale. In general, the results showed that the CPERS has very high internal consistency and is highly reliable for use in assessing early childhood centers. Conclusion In order for a data collection or assessment instrument to be valid and test what it purports to measure, it must be designed to be reliable and consistently measure a characteristic or attribute. If an instrument is not both reliable and valid, the resulting data are not of use to the researcher. There are many potential sources of variability in the results of data collection and assessment instruments. These include lasting and general characteristics of the individual, lasting but specific characteristics of the individual, temporary but general characteristics of the individual, temporary and specific characteristics of the individual, and systematic or chance factors affecting the administration of the instrument. In addition, the data from every assessment instrument will also contain some degree of variability that is attributable to error. The total variability of a data collection or assessment instrument is the sum of the true variability and the variability due to error. Reliability can be estimated through the use of parallel forms of the instrument, repeated administration of the same form of the instrument, subdivision of the instrument into two presumably parallel groups of items, and analysis of the covariance among the individual items. Terms & Concepts Confederate: A person who assists a researcher by pretending to be part of the experimental situation while actually only playing a rehearsed part meant to stimulate a response from the research subject. Correlation: The degree to which two events or variables are consistently related. Correlation may be positive (as the value of one variable increases, the value of the other variable increases), negative (as the value of one variable increases, the value of the other variable decreases), or zero (the values of the two variables are unrelated). Correlation does not imply causation. Data: In statistics, quantifiable observations or measurements that are used as the basis of scientific research. Operational Definition: A definition that is stated in terms that can be observed and measured. Reliability: The degree to which a data collection or assessment instrument consistently measures a characteristic or attribute. An assessment instrument cannot be valid unless it is reliable. Sample: A subset of a population. A random sample is a sample that is chosen at random from the larger population with the assumption that it will reflect the characteristics of the larger population. Standard Deviation: A measure of variability that describes how far the typical score in a distribution is from the mean of the distribution. Survey: (a) A data collection instrument used to acquire information on the opinions, attitudes, or reactions of people; (b) a research study in which members of a selected sample are asked questions concerning their opinions, attitudes, or reactions, and the responses are analyzed and used to extrapolate from the sample to the underlying population. Survey Research: A type of research in which data about the opinions, attitudes, or reactions of the members of a sample are gathered using a survey instrument. The phases of survey research are goal setting, planning, implementation, evaluation, and feedback. Unlike experimental research, survey research does not allow for the manipulation of an independent variable. Validity: The degree to which a survey or other data collection instrument measures what it purports to measure. A data collection instrument cannot be valid unless it is reliable. Content validity is a measure of how well assessment instrument items reflect the concepts that the instrument developer is trying to assess. Construct validity is a measure of how well an assessment instrument measures what it is intended to measure as defined by another assessment instrument. Face validity is when an assessment instrument appears to measure what it is trying to measure. Cross validity is the validation of an assessment instrument with a new sample to determine if the instrument is valid across situations. Predictive validity refers to how well an assessment instrument predicts future events. Essay by Ruth A. Wienclaw, PhD Dr. Ruth A. Wienclaw holds a PhD in industrial/organizational psychology with a specialization in organization development from the University of Memphis. She is the owner of a small business that works with organizations in both the public and private sectors, consulting on matters of strategic planning, training, and human-systems integration. Bibliography Bowen, N. K. (2008). Cognitive testing and the validity of child-report data from the elementary school success profile. Social Work Research, 32(1), 18-28. Retrieved July 11, 2024, from EBSCO Online Database Academic Search Complete. Compton, D., Love, T. P., & Sell, J. (2012). Developing and assessing intercoder reliability in studies of group interaction. Sociological Methodology, 42(1), 348-364. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Forrester, D., Fairtlough, A., & Bennet, Y. (2007). Describing the needs of children presenting to children's services: Issues of reliability and validity. Journal of Children's Services, 2(2), 48-59. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Hendrick, T. M., Fischer, A. H., Tobi, H., & Frewer, L. J. (2013). Self-reported attitude scales: Current practice in adequate assessment of reliability, validity, and dimensionality. Journal of Applied Social Psychology, 43(7), 1538-1552. Retrieved November 8, 2013, from EBSCO Online Database SocINDEX with Full Text. 0290&site=ehost-live Moore, G. T. & Sugiyama, T. (2007). The Children's Physical Environment Rating scale (CPERS): Reliability and validity for assessing the physical environment of early childhood educational facilities. Children, Youth and Environments, 17(4), 24-53. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Nunnally, J. C. (1978). Psychometric theory (2nd ed.). New York: McGraw-Hill Book Company. Peterson, R. A., & Yeolib, K. (2013). On the relationship between coefficient alpha and composite reliability. Journal of Applied Psychology, 98(1), 194-198. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Teixeira de Melo, A., Alarcão, M., & Pimentel, I. (2012). Validity and reliability of three rating scales to assess practitioners' skills to conduct collaborative, strength-based, systemic work in family-based services. American Journal of Family Therapy, 40(5), 420-433. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Suggested Reading Allen, M. J. & Yen, W. M. (1979). Introduction to measurement theory. Monterey, CA: Brooks/Cole Publishing Company. Bulloch, S. (2013). Seeking construct validity in interpersonal trust research: A proposal on linking theory and survey measures. Social Indicators Research, 113(3), 1289-1310. Retrieved July 11, 2024, from EBSCO Online Database SocINDEX with Full Text. Conley, T. B. (2006). Court ordered multiple offender drunk drivers: Validity and reliability of rapid assessment. Journal of Social Work Practice in the Addictions, 6(3), 37-51. Retrieved July 11, 2024, from EBSCO Online Database Academic Search Complete. Dunn, T. W., Smith, T. B., & Montoya, J. A. (2006). Multicultural competency instrumentation: A review and analysis of reliability generalization. Journal of Counseling & Development, 84(4), 471-482. Retrieved July 11, 2024, from EBSCO Online Database Academic Search Complete. Gillaspy, J. A. Jr. & Campbell, T. C. (2007). Reliability and validity of scores from the Inventory of Drug Use Consequences. Journal of Addictions & Offender Counseling, 27(1), 17-27. Retrieved July 11, 2024, from EBSCO Online Database Academic Search Complete. Lemke, E. & Wiersma, W. (1976). Principles of psychological measurement. Chicago: Rand McNally College Publishing Company. Lewis, C. A. & Cruise, S. M. (2006). Temporal stability of the Francis Scale of Attitude toward Christianity among 9- to 11-year-old English children: Test-retest data over six weeks. Social Behavior and Personality, 34(9), 1081-1086. Retrieved July 11, 2024, from EBSCO Online Database Academic Search Complete. Related Topics Validity
1742	https://fiveable.me/ap-pre-calc/unit-1/polynomial-functions-rates-change/study-guide/tQN39nNwYGsKoKj1	Polynomial Functions and Rates of Change - AP Precalc Study Guide \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintables upgrade 📈AP Pre-Calculus Unit 1 Review 1.4 Polynomial Functions and Rates of Change All Study Guides AP Pre-Calculus Unit 1 – Polynomial and Rational Functions Topic: 1.4 📈AP Pre-Calculus Unit 1 Review 1.4 Polynomial Functions and Rates of Change Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA 📈AP Pre-Calculus Unit & Topic Study Guides AP Precalculus Exam Unit 1 – Polynomial and Rational Functions Unit 1 Overview: Polynomial and Rational Functions 1.1 Change in Tandem 1.2 Rates of Change 1.3 Rates of Change in Linear and Quadratic Functions 1.4 Polynomial Functions and Rates of Change 1.5 Polynomial Functions and Complex Zeros 1.6 Polynomial Functions and End Behavior 1.7 Rational Functions and End Behavior 1.8 Rational Functions and Zeros 1.9 Rational Functions and Vertical Asymptotes 1.10 Rational Functions and Holes 1.11 Equivalent Representations of Polynomial and Rational Expressions 1.12 Transformations of Functions 1.13 Function Model Selection and Assumption Articulation 1.14 Function Model Construction and Application Unit 2 – Exponential and Logarithmic Functions Unit 3 – Trigonometric and Polar Functions Unit 4 – Functions Involving Parameters, Vectors, and Matrices Exam Skills practice questions print guide report error 🤓A Refresher on Polynomial Functions Wanna see what a polynomial function looks like? You’ve probably seen a lot of them throughout your math classes. more resources to help you study practice questionscheatsheetscore calculator A picture displaying the various types of polynomial graphs. Image Courtesy of Wikiversity Looks familiar, right? What do polynomial functions mean, though? 🤔 A polynomial function is a type of mathematical function that is represented by a sum of terms, where each term consists of a constant coefficient multiplied by a variable raised to a non-negative integer power. 💡 Side note: The degree of the polynomial is determined by the highest power of the variable in the polynomial. A nonconstant polynomial function of x is one that is not a constant, and is represented by the general form $p(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0$, where $n$ is a positive integer, $a_i$ is a real number for each $i$ from 0 to $n$, and $a_n$ is nonzero. The leading term is $a_nx^n$ and the leading coefficient is $a_n$. A constant function is also considered a polynomial function of degree zero. 📊Minima and Maxima When analyzing the behavior of a polynomial function, it’s important to understand how it changes as the input variable increases or decreases. One key concept is the idea of a "switching point," where the function changes from increasing to decreasing or vice versa. At these points, the function will have a local maximum or minimum output value, which is the highest or lowest point on the function in the immediate vicinity of the switching point. In some cases, the polynomial function may have a restricted domain, which means that it only applies within a certain range of input values. At the endpoint of this range, the function may also have a local maximum or minimum. These points are important because they represent the highest or lowest possible output values for the function within its domain. When considering all of the local maxima or minima of a polynomial function, it is possible to identify the greatest or least of these values. These points are known as the global or absolutemaximum and minimum, respectively, and they represent the highest and lowest points on the entire function, regardless of its domain or any other constraints. 🌐 A graph showing the local & global maximums and minimums. Image Courtesy of Wikimedia Commons Polynomial functions are mathematical functions that can be expressed as a sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient. These functions are widely used in mathematics and science to model a variety of real-world phenomena. 🧐Where Are The Zeros? One important concept in the study of polynomial functions is the location of their zeros, which are the input values that make the function output zero. If a polynomial function has two distinct real zeros, then there must be at least one point between them where the function has a local maximum or local minimum. 0️⃣ This is because the function must change sign as it passes through each zero, and in order for this to happen, it must have a point of inflection where its direction changes from increasing to decreasing, or vice versa. This point of inflection is necessarily a local maximum or local minimum. Here’s another way to define inflection points: when a polynomial function changes from being concave up to concave down, it means that the function is increasing at an increasing rate, and then begins to increase at a decreasing rate. This corresponds to a point of inflection where the rate of change of the function is at its maximum. Similarly, when a polynomial function changes from being concave down to concave up, it means that the function is decreasing at a decreasing rate, and then begins to decrease at an increasing rate. This corresponds to a point of inflection where the rate of change of the function is at its minimum. 🔢Even Degree Polynomials Moreover, the behavior of polynomial functions of even degree is also of particular interest. Polynomial functions of even degree have an even number of turning points, which are the points where the function changes direction from increasing to decreasing or vice versa. Four different polynomial graphs with even number degrees. Image Courtesy of Jill Williams Some trends to note: If the leading coefficient of an even degree polynomial is positive, the function will have a global minimum. 🔽 If the leading coefficient is negative, the function will have a global maximum. 🔼 This is because the function will approach negative infinity as the input variable goes to either positive or negative infinity, and since it is continuous, it must reach a maximum or minimum point somewhere in between. Frequently Asked Questions How do I find the degree and leading coefficient of a polynomial function? Put the polynomial in standard form (combine like terms so terms are in descending powers of x). Then: - Degree: the highest exponent n with a nonzero coefficient. For example p(x) = -4x^5 + 3x^2 + 1 has degree 5. A nonzero constant has degree 0. - Leading term: the term with that highest power, a_n x^n. - Leading coefficient: the coefficient a_n of the leading term. In p(x) above the leading coefficient is -4. If the polynomial is given in factored form, add multiplicities of factors to get the degree (e.g., (x-1)^2(x+2) has degree 3) and multiply the constants from each factor to find the leading coefficient (include any overall scalar). Knowing degree and leading coefficient helps you predict end behavior and turning points (see CED 1.4.A and Topic 1.4 study guide for examples)—study guide: ( For more practice, try problems at ( What's the difference between a local maximum and a global maximum in polynomial functions? A local (or relative) maximum is a point where the polynomial’s output is higher than all nearby outputs—it’s where the graph switches from increasing to decreasing. A global (or absolute) maximum is the single highest output value over the entire domain (or over the given, possibly restricted, domain); it’s the greatest of all local maxima. For polynomials remember: local extrema occur where the function changes increasing/decreasing (or at an included endpoint) (CED 1.4.A.2). Also, between any two distinct real zeros there’s at least one local max or min (CED 1.4.A.3). And note end behavior: even-degree polynomials can have a global max or min depending on the leading coefficient, while odd-degree polynomials have no global max and no global min on all real numbers (CED 1.4.A.4 & 1.4.A.1). For more practice and examples see the Topic 1.4 study guide ( and unit review ( I'm confused about what a point of inflection is - can someone explain it simply? A point of inflection is where the graph of a polynomial changes its concavity—that is, it switches from “cup up” (concave up) to “cup down” (concave down) or vice versa. In rate-of-change language (AP CED 1.4.A.5): it’s where the rate of change of the function (the first derivative) goes from increasing to decreasing or from decreasing to increasing. Quick checks you can use: - Look at the graph: an inflection point is where the curve’s bend flips. - Algebraically: f''(x) = 0 or undefined at the point is a candidate, but you must confirm f'' actually changes sign there. Example: f(x) = x^3 has f''(x) = 6x, so x = 0 is an inflection point because f'' changes from negative to positive. On the AP exam you’ll be asked to describe concavity and whether the rate of change is increasing/decreasing (so tie inflection points to changes in f' behavior). For more practice and examples, check the Topic 1.4 study guide ( and lots of practice problems ( When does a polynomial function have a global maximum vs a global minimum? Short answer: it depends on degree and the leading coefficient. - If the polynomial is constant, every value is both a global max and min (they’re equal). - If the degree is odd (n odd), the polynomial has opposite end behavior on the two sides (one → −∞, the other → +∞), so it has no global maximum or minimum on all real numbers. - If the degree is even (n even), the ends point the same way, so there is a global extremum: - leading coefficient a_n > 0 → both ends up → +∞ so the graph has a global minimum (lowest y-value), but no global maximum. - leading coefficient a_n < 0 → both ends down → −∞ so the graph has a global maximum (highest y-value), but no global minimum. Also note: on a restricted domain (interval), a polynomial can have global max/min at endpoints or local turning points (CED 1.4.A.2). For more examples and AP-style practice, see the Topic 1.4 study guide ( and practice problems ( How do I identify where a polynomial function is increasing or decreasing? Look at the rate of change (slope) of the polynomial to tell where it’s increasing or decreasing. Quick steps you can use on any polynomial p(x): - Find the critical inputs where the instantaneous rate of change is zero: compute p′(x) (or use difference/graph if you haven’t learned derivatives) and solve p′(x)=0. For polynomials p′ exists everywhere, so only roots of p′ matter. - Make a sign chart for p′(x): pick a test x in each interval between consecutive roots of p′. If p′(x)>0 on an interval, p is increasing there; if p′(x)<0, p is decreasing there. Those sign changes mark local maxima/minima (turning points). - Use multiplicity: if a zero of p(x) has even multiplicity the graph touches and may not change sign; if odd it crosses—this affects turning behavior between zeros (CED 1.4.A.3). - For exam graphs, you can also use slopes of secant lines (average rate of change) or inspect concavity/inflection to support answers (CED 1.4.A.2, 1.4.A.5). For AP prep, practice making sign charts and connecting p′ sign to local/global extrema. Review Topic 1.4 study guide ( and try practice problems ( What's the formula for finding local maxima and minima of polynomial functions? Short answer: find where the rate of change (the derivative) is zero or changes sign. For polynomials (which are smooth everywhere) local maxima/minima occur only at critical points where f′(x) = 0. Then use a sign test or the second-derivative test to decide max vs. min. Step-by-step: 1. Differentiate: f(x) = a_n x^n + … + a_0 → f′(x). 2. Solve f′(x) = 0 to get critical points. 3. Test each critical point: - First-derivative (sign) test: if f′ goes + → − at x0, f has a local max; if − → +, a local min. - Or second-derivative test: compute f″(x0). If f″(x0) > 0 → local min (concave up). If f″(x0) < 0 → local max (concave down). If f″(x0) = 0, the test is inconclusive—use the sign test. Note: by CED 1.4.A.2 and 1.4.A.5, maxima/minima are exactly where the polynomial switches increasing/decreasing or where concavity changes. For practice and AP-style problems, see the Topic 1.4 study guide ( and more practice problems ( I don't understand why there has to be a max or min between every two zeros - why is this true? Good question—here’s a clean way to see it. Let f be a nonconstant polynomial and let a < b be two distinct real zeros, so f(a)=f(b)=0. Because polynomials are continuous on [a,b], either f(x)=0 for every x in (a,b) (then there are infinitely many zeros), or not. If not, then \|f\| attains a positive maximum value on [a,b] by the Extreme Value Theorem, and that maximum must occur at some interior point c in (a,b) (it can’t be at the endpoints because f(a)=f(b)=0). At that interior c, f(c) is either the largest positive value (a local maximum) or the most negative value (a local minimum). By differentiability of polynomials, f′(c)=0 as Rolle’s theorem also predicts. So between any two distinct real zeros you get at least one interior point that’s a local max or min (CED 1.4.A.3, uses continuity/differentiability). If you want a quick review of this topic, check the Topic 1.4 study guide ( and try a few practice problems ( How do I know if a polynomial function changes from concave up to concave down? Look for where the concavity (the “bend”) changes—that’s an inflection point. For a polynomial p(x): 1. Compute p′(x) and then p″(x). 2. Solve p″(x) = 0 (polynomials have no points where p″ is undefined). 3. Check whether p″ actually changes sign at each solution (use a sign chart or test values left and right). If p″ changes sign, the graph changes concavity there (concave up → concave down or vice versa) and you have an inflection point. Quick example: p(x) = x^3. Then p″(x) = 6x. Solve 6x = 0 → x = 0. Since p″<0 for x<0 and p″>0 for x>0, the graph changes from concave down to concave up at x = 0. This is exactly what the CED calls an inflection point (1.4.A.5). For more practice and examples, see the Topic 1.4 study guide ( and the AP Precalculus practice set ( What does it mean when they say the rate of change is increasing or decreasing? “Rate of change” means how fast the output (y) changes with the input (x)—think slope of the graph. When they say the rate of change is increasing or decreasing, they’re talking about how that slope itself behaves. - Rate of change increasing: the slopes are getting larger as x increases (derivative f′(x) is increasing). Graphically the curve is concave up—it’s bending upward. - Rate of change decreasing: the slopes are getting smaller (f′(x) is decreasing). Graph is concave down—it’s bending downward. Key AP links to CED ideas: where the rate of change changes from increasing to decreasing (or vice versa) you have a point of inflection (1.4.A.5). Where the rate of change crosses zero (slope changes sign) you often get a local max or min (1.4.A.2). Questions on the exam sometimes ask you to describe “increasing at a decreasing rate” (slope positive but decreasing)—practice that wording on FRQ-style items (see Topic 1.4 study guide: For more review and practice, check the Unit 1 overview ( and tons of practice problems ( When do I use the first derivative vs the second derivative for polynomial functions? Use the first derivative when you care about rates of change and where the function increases or decreases. For a polynomial p(x), p′(x) gives the instantaneous rate of change; solve p′(x)=0 (critical points) to find candidates for local maxima or minima (CED 1.4.A.2). Use sign changes of p′ to decide if p switches from increasing to decreasing (so a local max) or vice versa (local min). Use the second derivative when you care about concavity and how the rate of change itself is changing. p′′(x) tells you concavity: p′′(x)>0 means concave up (rate of change increasing), p′′(x)<0 means concave down (rate of change decreasing). Points where p′′ changes sign are inflection points (CED 1.4.A.5). Also use p′′ at a critical point: if p′(c)=0 and p′′(c)>0 → local min; p′′(c)<0 → local max (when p′′ exists). For AP practice and worked examples on these ideas, see the Topic 1.4 study guide ( and try problems at ( How do I find all the critical points of a polynomial step by step? 1) Take the derivative p'(x). For a polynomial this always exists (no undefined points). 2) Factor p'(x) completely (use algebra or your calculator). 3) Solve p'(x) = 0 to get candidate critical points (these are the only ones for polynomials). 4) Determine multiplicity of each root from the factorization: - If the root has odd multiplicity → p' changes sign → a local max or min (turning point). - If even multiplicity → p' does not change sign → no local extremum (graph “touches” the axis). 5) (Optional quick test) Use the second derivative p''(x): - If p''(c) > 0 at a critical point c → local minimum. - If p''(c) < 0 → local maximum. - If p''(c) = 0, the test is inconclusive; use sign-chart for p' or higher derivatives. 6) If the domain is restricted, also check endpoints for local/global extrema. Show your algebra/derivative work (AP free-response expects steps). For extra practice and AP-aligned examples, see the Topic 1.4 study guide ( and try problems at ( I missed class - what's the difference between absolute and relative extrema? Relative (local) extrema are highs or lows compared to nearby points: a relative maximum at x0 means p(x0) ≥ p(x) for all x in some small interval around x0 (the graph switches from increasing to decreasing there), and a relative minimum means p(x0) ≤ p(x) nearby (switch from decreasing to increasing). Absolute (global) extrema are the largest or smallest outputs over the entire domain: the absolute maximum is the greatest of all relative maxima (or endpoint values), and the absolute minimum is the least of all relative minima (CED 1.4.A.2). For polynomials, check turning points (where increasing/decreasing changes) and endpoints if the domain is restricted; between any two real zeros there’s at least one local extremum (CED 1.4.A.3). On the AP exam you may be asked to identify or compare local vs. global extrema from graphs, tables, or formulas—practice doing both. For a quick refresher, see the Topic 1.4 study guide ( and try problems on the AP Precalculus practice page ( Why do even degree polynomials always have either a global max or global min? Because the highest-degree term dominates for large \|x\|, an even-degree polynomial has the same end behavior on both sides: if the leading coefficient a_n > 0 then a_n x^n → +∞ as x → ±∞, and if a_n < 0 then a_n x^n → −∞ as x → ±∞. That means the graph either rises to +∞ on both ends (so the function must have a lowest point somewhere—a global minimum) or falls to −∞ on both ends (so it must have a highest point somewhere—a global maximum). Polynomials are continuous, so between the two ends the function attains an extreme value (usually at a critical point or at a turning point). This is exactly the CED fact: “Polynomial functions of an even degree will have either a global maximum or a global minimum” (1.4.A.4). For extra practice and examples, check the Topic 1.4 study guide ( and the Unit 1 overview ( How do I solve word problems involving polynomial rates of change? Start by translating the story into a polynomial model p(x) (or build one from given data). Always state units. Then use these steps: 1. Identify what rate you need. Average rate of change on [a,b] = (p(b)−p(a))/(b−a) with units—AP FRQs expect this and your work shown. 2. For instantaneous behavior, use the first derivative idea: p'(x) gives the instantaneous rate. If you can use differentiation, compute p'(x) and evaluate; if not, approximate with slopes of secant lines over tiny intervals or use your graphing calculator (Part A of the exam allows it). 3. Determine increasing/decreasing: sign of p'(x) (positive ⇒ increasing). Find local extrema where p' changes sign (or where slopes switch). 4. For whether the rate of change is increasing or decreasing, use p''(x) or look at concavity: concave up ⇒ rate of change increasing; concave down ⇒ rate of change decreasing. Points where concavity changes are inflection points. 5. Always interpret answers in context with units and give decimal answers to three places when using a calculator (AP guidance). Need worked examples and practice? Check the Topic 1.4 study guide ( and try problems from the AP Precalculus practice set ( Can someone explain how to find inflection points using calculus? Inflection points are where a polynomial’s graph changes concavity (concave up ↔ concave down). Using calculus: 1. Compute f''(x) (the second derivative). For a polynomial this is another polynomial. 2. Solve f''(x) = 0 (or where f'' is undefined—not an issue for polynomials). Those x-values are candidates. 3. Test each candidate by checking the sign of f'' just left and right of the point. If f'' changes sign, x is an inflection point; if it doesn’t (e.g., f'' has an even-multiplicity root), it’s not. Example: p(x)=x^3 → p''(x)=6x, p''(0)=0 and changes sign, so x=0 is an inflection point at (0,0). Link to the CED concept: inflection points are where the rate of change of the function’s rate of change changes (CED 1.4.A.5). 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1743	https://www.somersetkey.com/ourpages/auto/2017/12/13/63309824/geo%20ch%205%20sec%204%20teach.pdf	© Houghton Mifflin Harcourt Publishing Company Name Class Date Explore Constructing Triangles Given Three Side Lengths Two triangles are congruent if and only if a rigid motion transformation maps one triangle onto the other triangle. Many theorems can also be used to identify congruent triangles. Follow these steps to construct a triangle with sides of length 5 in., 4 in., and 3 in. Use a ruler, compass, and either tracing paper or a transparency. A Use a ruler to draw a line segment of length 5 inches. Label the endpoints A and B. B Open a compass to 4 inches. Place the point of the compass on A, and draw an arc as shown. C Now open the compass to 3 inches. Place the point of the compass on B, and draw a second arc. D Next, find the intersection of the two arcs. Label the intersection C. Draw _ AC and _ BC . Label the side lengths on the figure. E Repeat steps A through D to draw △DEF on a separate piece of tracing paper. The triangle should have sides with the same lengths as △CAB. Start with a segment that is 4 in. long. Label the endpoints D and E as shown. Resource Locker A B 4 in. 5 in. A B 4 in. 3 in. 5 in. A B C 4 in. 3 in. 5 in. D E F 4 in. 3 in. 5 in. Module 5 255 Lesson 4 5.4 SSS Triangle Congruence Essential Question: What does the SSS Triangle Congruence Theorem tell you about triangles? y ; GE_MNLESE385795_U2M05L4.indd 255 02/04/14 12:54 AM Common Core Math Standards The student is expected to: G-CO.B.8 Explain how the criteria for triangle congruence (... SSS) follow from the definition of congruence in terms of rigid motions. Also G-CO.B.7, G-CO.C.10, G-SRT.B.5 Mathematical Practices MP.7 Using Structure Language Objective Have small groups of students complete a triangle congruence chart. COMMON CORE COMMON CORE HARDCOVER PAGES 221230 Turn to these pages to find this lesson in the hardcover student edition. SSS Triangle Congruence ENGAGE Essential Question: What does the SSS Triangle Congruence Theorem tell you about triangles? If three sides of one triangle are congruent to three sides of another triangle, you can conclude that the triangles are congruent. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Point out that the “structural beams” mentioned in the Preview refer to the steel sides of the triangles. Then preview the Lesson Performance Task. 255 HARDCOVER Turn to these pages to find this lesson in the hardcover student edition. © Houghton Mifflin Harcourt Publishing Company Name Class Date Explore Constructing Triangles Given Three Side Lengths Two triangles are congruent if and only if a rigid motion transformation maps one triangle onto the other triangle. Many theorems can also be used to identify congruent triangles. Follow these steps to construct a triangle with sides of length 5 in., 4 in., and 3 in. Use a ruler, compass, and either tracing paper or a transparency.  Use a ruler to draw a line segment of length 5 inches. Label the endpoints A and B.  Open a compass to 4 inches. Place the point of the compass on A, and draw an arc as shown.  Now open the compass to 3 inches. Place the point of the compass on B, and draw a second arc.  Next, find the intersection of the two arcs. Label the intersection C. Draw _ AC and _ BC . Label the side lengths on the figure.  Repeat steps A through D to draw △DEF on a separate piece of tracing paper. The triangle should have sides with the same lengths as △CAB. Start with a segment that is 4 in. long. Label the endpoints D and E as shown. Resource Locker G-CO.B.8 Explain how the criteria for triangle congruence (... SSS) follow from the definition of congruence in terms of rigid motions. Also G-CO.B.7, G-CO.C.10, G-SRT.B.5 COMMON CORE A B 4 in. 5 in. A B 4 in. 3 in. 5 in. A B C 4 in. 3 in. 5 in. D E F 4 in. 3 in. 5 in. Module 5 255 Lesson 4 5 . 4 SSS Triangle Congruence Essential Question: What does the SSS Triangle Congruence Theorem tell you about triangles? DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A GE_MNLESE385795_U2M05L4.indd 255 02/04/14 12:55 AM 255 Lesson 5 . 4 L E S S O N 5 . 4 InDesign Notes 1. This is a list opy Notes This is a list Bold, Italic, Strickthrough. InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company  Compare △CAB and △DEF. Are they congruent? How do you know? Reflect 1. Discussion When you construct △CAB, how do you know that the intersection of the two arcs is a distance of 4 inches from A and 3 inches from B? 2. Compare your triangles to those made by other students. Are they all congruent? Explain. Explain 1 Justifying SSS Triangle Congruence You can use rigid motions and the converse of the Perpendicular Bisector Theorem to justify this theorem. SSS Triangle Congruence Theorem If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. Example 1 In the triangles shown, let ¯ AB ≅ ¯ DE , ¯ AC ≅ ¯ DF , and ¯ BC ≅ ¯ EF . Use rigid motions to show that △ABC ≅ △DEF. A B D E F C Yes. Triangle CAB can be mapped to △DEF by a translation that maps point B to point F, and then a rotation about point F. The arcs are sections of circles sets of points, respectively, 3 inches and 4 inches from A. The other circle is the set of points that is 3 inches away from B. The intersection has the properties of both circles. Yes, because there is a sequence of rigid motions that maps any one of the triangles onto any of the others. Module 5 256 Lesson 4 y ; y ; GE_MNLESE385795_U2M05L4.indd 256 23/03/14 2:43 AM Math Background The Side-Side-Side Triangle Congruence Theorem is often presented as the first of the Triangle Congruence Theorems because it is easy to demonstrate concretely. In this course, it is the third theorem presented because the justification requires students to apply the Perpendicular Bisector Theorem. Reinforce this justification throughout the lesson. EXPLORE Constructing Triangles Given Three Side Lengths INTEGRATE TECHNOLOGY Students may use geometry software to explore the concept of constructing a triangle with given side lengths. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Give each student a piece of dry spaghetti. They should measure and break the spaghetti so that they have three pieces that are the same length as the sides of the triangle in the exercise. Instruct them to make a triangle with these pieces on top of the triangle they drew. Have them compare the triangles and explain why they are congruent. Then ask them to consider whether it is possible to create a triangle from the spaghetti pieces that is not congruent to the triangle they drew in the exercise. EXPLAIN 1 Justifying SSS Triangle Congruence INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Some justifications in this exercise may be difficult for some students to understand. Have students model each step with the triangles they have drawn on tracing paper. Encourage students to restate the steps in their own words. Students should fully understand each step before moving on to the next one. QUESTIONING STRATEGIES Draw two segments that share an endpoint. How many different segments could be drawn to turn this figure into a triangle? There is only one segment. PROFESSIONAL DEVELOPMENT SSS Triangle Congruence 256 © Houghton Mifflin Harcourt Publishing Company A Transform △ABC by a translation along ‾ ⇀ AD followed by a rotation about point D, so that _ AB and _ DE coincide. The segments coincide because they are the same length. Does a reflection across _ AB map point C to Therefore, point D lies on the perpendicular point F? To show this, notice that DC = DF, bisector of _ CF by the converse of the which means that point D is equidistant from perpendicular bisector theorem. Because EC = EF, point C and point F. point E also lies on the perpendicular bisector of _ CF . Since point D and point E both lie on the perpendicular bisector of _ CF and there is a unique line through any two points, ‹ − › DE is the perpendicular bisector of _ CF . By the definition of reflection, the image of point C must be point F. Therefore, △ABC is mapped onto △DEF by a translation, followed by a rotation, followed by a reflection, and the two triangles are congruent. B Show that △ABC ≅ △PQR. Triangle ABC is transformed by a sequence of rigid motions to form the figure shown below. Identify the sequence of rigid motions. (You will complete the proof on the following page.) 1. 2. 3. A B D E F C A B D E F C A B D E F C A B P Q R C A B P Q R C Translation along ‾ ⇀ AP Reflection across _ PQ . Rotation about P so that _ PQ and _ AB coincide. Module 5 257 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 257 5/22/14 5:38 PM COLLABORATIVE LEARNING Peer-to-Peer Activity Give each student five index cards. Students should prepare a card that shows two labeled triangles for each of the following situations: • The triangles can be proved congruent by SSS. • The triangles can be proved congruent by SAS. • The triangles can be proved congruent by ASA. • The triangles can be proved noncongruent. • There is not enough information to determine congruence. Collect and shuffle them. Divide students into pairs and give each pair ten cards. Have them sort their cards into the above categories. 257 Lesson 5 . 4 © Houghton Mifflin Harcourt Publishing Company Complete the explanation by filling in the blanks with the name of a point, line segment, or geometric theorem. Because _ QR ≅ , point Q is equidistant from and . Therefore, by the converse of the Theorem, point Q lies on the of _ RC . Similarly, _ PR ≅ . So point lies on the perpendicular bisector of . Because two points determine a line, the line ‹ − › PQ is the . By the definition of reflection, the image of point C must be point . Therefore, △ABC ≅ △PQR because △ABC is mapped to by a translation, a rotation, and a . Reflect 3. Can you conclude that two triangles are congruent if two pairs of corresponding sides are congruent? Explain your reasoning and include an example. Your Turn 4. Use rigid motions and the converse of the perpendicular bisector theorem to explain why △ABC ≅ △ADC. A B D C _ QC _ PC _ RC P R △PQR R (or C ) C (or R) Perpendicular Bisector perpendicular bisector reflection perpendicular bisector of _ RC No; you need a third piece of information to ensure a rigid motion maps one triangle to the other, such as congruent included angles or another pair of congruent sides. _ AB ≅ _ AD and _ CB ≅ _ CD , so A is equidistant from B and D, and C is equidistant from B and D. By the converse of the Perpendicular Bisector Theorem, ‹ − › AC is the perpendicular bisector of _ BD . By the definition of a reflection, point D is the image of point B reflected across _ AC . The reflection also maps _ AC onto _ AC , so △ABC ≅ △ADC. Module 5 258 Lesson 4 y ; y ; GE_MNLESE385795_U2M05L4 258 6/3/14 8:16 AM DIFFERENTIATE INSTRUCTION Manipulatives Have students create a triangle and a quadrilateral using strips of construction paper or tagboard and brass fasteners. Then have students attempt to change the angles in each without bending the strips of paper. Students should notice that the triangle always remains the same but that they can create many different quadrilaterals. Discuss how this activity illustrates the Side-Side-Side Triangle Congruence Theorem. SSS Triangle Congruence 258 InDesign Notes 1. This is a list InDesign Notes 1. This is a list InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company Explain 2 Proving Triangles Are Congruent Using SSS Triangle Congruence You can apply the SSS Triangle Congruence Theorem to confirm that triangles are congruent. Remember, if any one pair of corresponding parts of two triangles is not congruent, then the triangles are not congruent. Example 2 Prove that the triangles are congruent or explain why they are not congruent.  AB = DE = 1.7 m, so _ AB ≅ _ DE . BC = EF = 2.4 m, so _ BC ≅ _ EF . AC = DF = 2.3 m, so _ AC ≅ _ DF . The three sides of △ABC are congruent to the three sides of △DEF. △ABC ≅ △DEF by the SSS Triangle Congruence Theorem.  DE = = 20 cm, so . DH = = 12 cm, so . EH = = 24 cm, so . The three sides of △DEH are congruent to the three sides of , so the two triangles are congruent by . Your Turn Prove that the triangles are congruent or explain why they are not congruent. 5. 6. A B C D E F 2.3 m 2.3 m 1.7 m 1.7 m 2.4 m 2.4 m D E F H G 20 cm 20 cm 24 cm 24 cm 12 cm 12 cm M N P Q R S 54 in. 54 in. 28 in. 38 in. 32 in. 38 in. K G L J the SSS Triangle Congruence Theorem FG _ DE ≅ _ FG _ DH ≅ _ FH _ EH ≅ GH FH GH △FGH The corresponding sides _ MN and _ QR are It is given that _ GK ≅ _ GL and _ JK ≅ _ JL , not congruent. and _ GJ ≅ _ GJ by the Reflexive Property. not congruent. Therefore, the triangles are Module 5 259 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 259 21/03/14 11:50 AM LANGUAGE SUPPORT Connect Context The phrase Given Three Side Lengths in the title of the Explore section may confuse some English Learners. Explain that it indicates that you will be constructing triangles when you are given the lengths of the sides. Remind students that in math, they are often given partial information to help solve a problem, and that the information is called a given. EXPLAIN 2 Proving Triangles Are Congruent Using SSS Triangle Congruence INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Although the example emphasizes the SSS Triangle Congruence Theorem, it is important for students to keep in mind that the triangles are congruent because one can be mapped onto the other by one or more rigid motions. Maintain this connection by asking which segments of one triangle map onto certain segments of another triangle. QUESTIONING STRATEGIES When do you use the SSS Triangle Congruence Theorem instead of the ASA or SAS Triangle Congruence Theorems to determine whether two triangles are congruent? When you know three pairs of corresponding congruent sides and no pairs of corresponding congruent angles, you cannot use a theorem that involves an angle. LANGUAGE SUPPORT Have students work in small groups. Have them complete a chart like the following, highlighting the sides and angles that are congruent in each pair of triangles. Triangle Congruence Theorems Theorem Definition Picture 259 Lesson 5 . 4 © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©MaximImages/Alamy Explain 3 Applying Triangle Congruence You can use the SSS Triangle Congruence Theorem and other triangle congruence theorems to solve many real-world problems that involve congruent triangles. Example 3 Find the value of x for which you can show the triangles are congruent.  Lexi bought matching triangular pendants for herself and her mom in the shapes shown. For what value of x can you use a triangle congruence theorem to show that the pendants are congruent? Which triangle congruence theorem can you use? Explain. _ AB ≅ _ JK and _ AC ≅ _ JL , because they have the same measure. So, if _ BC ≅ _ KL , then △ABC ≅ △JKL by the SSS Triangle Congruence Theorem. Write an equation setting the lengths equal and solve for x. 4x - 6 = 3x - 4; x = 2  Adeline made a design using triangular tiles as shown. For what value of x can you use a triangle congruence theorem to show that the tiles are congruent? Which triangle congruence theorem can you use? Explain. Notice that _ PQ ≅ _ MN and ≅ _ MO , because they have the same measure. If _ NO ≅ _ QR , then △MNO ≅ by the Triangle Congruence Theorem. Write an equation setting the lengths equal and solve for x. v Your Turn 7. Craig made a mobile using geometric shapes including triangles shaped as shown. For what value of x and y can you use a triangle congruence theorem to show that the triangles are congruent? Which triangle congruence theorem can you use? Explain. A K L J B C 3 cm 3 cm 3 cm 3 cm (4x - 6) cm (3x - 4) cm M N O P Q R 4 in. 4 in. 4 in. 4 in. 4 in. (3x - 11) in. F G H T U V 60º 30º (8x - 12) cm (7y + 4)º 12 cm _ PR △PQR SSS m∠H = 30° by the Triangle Sum Theorem, so ∠H ≅ ∠V. ∠G ≅ ∠U, they are right angles. If ¯ GH ≅ ¯ UV , then △FGH ≅ △TUV by the ASA. x = 3; y = 8 3x - 11 = 4, 3x = 15, x = 5 Module 5 260 Lesson 4 y ; y ; GE_MNLESE385795_U2M05L4 260 10/14/14 7:14 PM EXPLAIN 3 Applying Triangle Congruence AVOID COMMON ERRORS Because students are concentrating on sides when using the SSS Triangle Congruence Theorem, they may not write corresponding angles in the correct order in the congruence statement. Discuss methods they can use to make sure they get the order right. QUESTIONING STRATEGIES How do you know whether angles in one triangle are congruent to the angles in the other triangle? You know the triangles are congruent by SSS, so you know that corresponding angles are also congruent because of CPCTC. CONNECT VOCABULARY Have students collaboratively write a definition for a triangle, providing supports. For example: “ A triangle is a _, It has _ sides. It has 3 __. A right triangle has a ___.” SSS Triangle Congruence 260 © Houghton Mifflin Harcourt Publishing Company Elaborate 8. An isosceles triangle has two sides of equal length. If we ask everyone in class to construct an isosceles triangle that has one side of length 8 cm and another side of length 12 cm, how many sets of congruent triangles might the class make? 9. Essential Question Check-In How do you explain the SSS Triangle Congruence Theorem? Use a compass and a straightedge to complete the drawing of △DEF so that it is congruent to △ABC. 1. On a separate piece of paper, use a compass and a ruler to construct two congruent triangles with the given side lengths. Label the lengths of the sides. 2. 3 in., 3.5 in., 4 in. 3. 3 cm, 11 cm, 12 cm Evaluate: Homework and Practice • Online Homework • Hints and Help • Extra Practice A B D F C E 11 cm 12 cm 12 cm 3 cm 3 cm 11 cm 4 in. 4 in. 3 in. 3 in. 3.5 in. 3.5 in. The class could make at most 2 sets of congruent triangles. One set would have sides of length 8 cm, 8 cm, and 12 cm. The second set would have sides of length 12 cm, 12 cm, and 8 cm. Possible answer: Two triangles are congruent if a series of rigid motion transformations maps one triangle onto the other. For two triangles that have pairs of congruent sides, begin by translating and then rotating one triangle so that it shares one side with the other triangle. Then apply the converse of the Perpendicular Bisector Theorem to show that a reflection completes the mapping. Module 5 261 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 261 6/9/15 2:29 AM ELABORATE QUESTIONING STRATEGIES Two triangles appear to be congruent. You know that three pairs of corresponding parts are congruent, but you have no information about the other corresponding parts. How can you determine whether the triangles really are congruent? Sample answer: If the congruent parts are two pairs of corresponding angles and the included side, or two pairs of corresponding sides and the included angle, you can use the ASA or SAS Triangle Congruence Theorem. If the congruent parts are three pairs of corresponding sides, you can use the SSS Triangle Congruence Theorem. Otherwise, check whether there is a sequence of rigid motions that map one triangle onto the other. SUMMARIZE THE LESSON When do you use the SSS Triangle Congruence Theorem? What information do you need in order to use this theorem? You use the SSS Triangle Congruence Theorem to prove that two triangles are congruent by using three pairs of congruent sides. You need to know that all three pairs of corresponding sides are congruent to use it. 261 Lesson 5 . 4 InDesign Notes 1. This is a list opy Notes This is a list Bold, Italic, Strickthrough. InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company Identify a sequence of rigid motions that maps one side of △ABC onto one side of △DEF. 4. 5. 6. 7. In each figure, identify the perpendicular bisector and the line segment it bisects, and explain how to use the information to show that the two triangles are congruent. 8. 9. A B D E F C A B F E D C A B F E D C A B F E D C A B D C D E F G Possible answer: A translation along ‾ ⇀ CF , and then a clockwise rotation about F so that ― CB coincides with ― FE . Possible answer: A translation along ‾ ⇀ AD , and then a clockwise rotation about D so that ― AB coincides with ― DE . Possible answer: A translation along ‾ ⇀ AD , and then a counterclockwise rotation about D so that ― AB coincides with ― DE . Possible answer: A translation along ‾ ⇀ CF , and then a counterclockwise rotation about F so that ― CA coincides with ― FD . ‹ − › BC is the perpendicular bisector of ― AD . This shows that A maps to D by a reflection across ‹ − › BC . ‹ − › DF is the perpendicular bisector of ― EG . This shows that E maps to G by a reflection across ‹ − › DF . Module 5 262 Lesson 4 y ; y ; GE_MNLESE385795_U2M05L4.indd 262 21/03/14 11:50 AM Exercise Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–3 2 Skills/Concepts MP.4 Modeling 4–9 2 Skills/Concepts MP.1 Problem Solving 10–13 2 Skills/Concepts MP.2 Reasoning 14–20 3 Strategic Thinking MP.1 Problem Solving 21–27 2 Skills/Concepts MP.3 Logic EVALUATE ASSIGNMENT GUIDE Concepts and Skills Practice Explore Constructing Triangles Given Side Lengths Exercises 1–3 Example 1 Justifying SSS Triangle Congruence Exercises 4–9 Example 2 Proving Triangles Are Congruent Using SSS Triangle Congruence Exercises 10–13 Example 3 Applying Triangle Congruence Exercises 14–20 AVOID COMMON ERRORS Some students may have trouble applying SSS to adjacent triangles. Adjacent triangles share a side, so you can apply the Reflexive Property to get a pair of congruent sides. SSS Triangle Congruence 262 InDesign Notes 1. This is a list InDesign Notes 1. This is a list InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company Prove that the triangles are congruent or explain why this is not possible. 10. 11. 12. 13. 14. Carol bought two chairs with triangular backs. For what value of x can you use a triangle congruence theorem to show that the triangles are congruent? Which triangle congruence theorem can you use? Explain. R Q S T 6 m 6 m 9 m 9 m Y N X M Z P 51º 50º 50º 50º G (9x - 21) in. H 25 in. 25 in. I T U 25 in. 25 in. 15 in. V J K W X Y L 12 cm 10 cm 8 cm 8 cm J K W X Y L 9 cm 10 cm 7 cm 8 cm Congruent, by SSS congruence; ― RS ≅ ― TQ because they have the same measure, ― RT ≅ ― RT by the reflexive property, and _ ST ≅ _ QR because they have the same measure. Not congruent, because only one triangle has an interior angle of 51° so there is no rigid motion that will map one to the other. Possibly congruent, depending on the lengths of the unlabeled sides. Not congruent. The diagrams show a total of 4 different side lengths, but two congruent triangles have only 3 different side lengths. X = 4; _ GH ≅ _ TU and _ GI ≅ _ TV , because they have the same measure. So, if _ HI ≅ _ UV , then ▵GHI ≅ △TUV by the SSS Triangle Congruence Theorem. Module 5 263 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 263 21/03/14 11:50 AM COMMUNICATE MATH Introduce the word criterion and its plural, criteria. Ask students if they can define criterion and give an example of its use. For instance, you might discuss a college’s criteria for accepting students (a completed application, a minimum grade-point average, a minimum SAT score, and so on). Tell students that they have developed three criteria for determining whether two triangles are congruent, the ASA, SAS, and SSS Triangle Congruence Theorems. Discuss how they can determine whether two given triangles meet any of these criteria. 263 Lesson 5 . 4 InDesign Notes 1. This is a list opy Notes This is a list Bold, Italic, Strickthrough. InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company 15. For what values of x and y can you use a triangle congruence theorem to show that the triangles are congruent? Which triangle congruence theorem can you use? Explain. Find all possible solutions for x such that △ABC is congruent to △DEF. One or more of the problems may have no solution. 16. △ABC: sides of length 6, 8, and x. △DEF: sides of length 6, 9, and x - 1. 17. △ABC: sides of length 3, x + 1, and 14. △DEF: sides of length 13, x - 9, and 2x - 6 18. △ABC: sides of length 17, 17, and 2x + 1. △DEF: sides of length 17, 17, and 3x - 9 19. △ABC: sides of length 19, 25, and 5x - 2. △DEF: sides of length 25, 28, and 4 - y 20. △ABC: sides of length 8, x - y, and x + y △DEF: sides of length 8, 15, and 17 21. △ABC: sides of length 9, x, and 2x - y △DEF: sides of length 8, 9, and 2y - x 22. These statements are part of an explanation for the SSS Triangle Congruence Theorem. Write the numbers 1 to 6 to place these strategies in a logical order. The statements refer to triangles ABC and DEF shown here. Rotate the image of △ABC about E, so that the image of _ BC coincides with _ EF . Apply the definition of reflection to show D is the reflection of A across ‾ ⇀ EF . Conclude that △ABC ≅ △DEF because a sequence of rigid motions maps one triangle onto the other. Translate △ABC along ‾ ⇀ BE . Define ‾ ⇀ EF as the perpendicular bisector of the line connecting D and the image of A. Identify E, and then F, as equidistant from D and the image of A. 9 cm (14x - 47) cm 60º (15y)º 30º C E G F 30º A B A E D F B C x = 4, y = 4; ∠A ≅ ∠E, ∠C ≅ ∠G, and ― AC ≅ _ EG , ASA x = 9 no solution x = 10 x = 6, y = -15 x = 16, y = 1; x = 16, y = -1 Possible solution: x = 8, y = 8 2 5 6 1 4 3 Module 5 264 Lesson 4 y ; y ; GE_MNLESE385795_U2M05L4.indd 264 21/03/14 11:50 AM INTEGRATE TECHNOLOGY Students may find it useful to use geometry software to model exercises they are struggling with. SSS Triangle Congruence 264 InDesign Notes 1. This is a list InDesign Notes 1. This is a list InCopy Notes 1. This is a list © Houghton Mifflin Harcourt Publishing Company 23. Determine whether the given information is sufficient to guarantee that two triangles are congruent. Select the correct answer for each lettered part. A. The triangles have three pairs of congruent corresponding angles. sufficient not sufficient B. The triangles have three pairs of congruent corresponding sides. sufficient not sufficient C. The triangles have two pairs of congruent corresponding sides and one pair of congruent corresponding angles. sufficient not sufficient D. The triangles have two pairs of congruent corresponding angles and one pair of congruent corresponding sides. sufficient not sufficient E. Two angles and the included side of one triangle are congruent to two angles and the included side of the other triangle. sufficient not sufficient F. Two sides and the included angle of one triangle are congruent to two sides and the included angle of the other triangle. sufficient not sufficient 24. Make a Conjecture Does a version of SSS congruence apply to quadrilaterals? Provide an example to support your answer. 25. Are two triangles congruent if all pairs of corresponding angles are congruent? Support your answer with an example. A. the triangles have the same shape but not necessarily the same size. B. SSS Triangle Congruence Theorem C. These conditions do not produce a unique triangle. D. The Triangle Sum Theorem will show that the third pair of angles is congruent, so the triangles are congruent by the ASA Triangle Congruence Theorem. E. ASA Triangle Congruence Theorem F. SAS Triangle Congruence Theorem No; Possible example: a variety of non-congruent rhombuses, including No; At least one pair of sides must be congruent. For example, two a square, may have 4 sides of the same length. triangles that have three 60 degree angles may not have at least one pair of congruent corresponding sides, such as a triangle with side lengths of 4 inches and another triangle with side lengths of 8 inches. Module 5 265 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 265 21/03/14 11:50 AM JOURNAL Have students compare and contrast the SSS and SAS theorems and support their answers with a sketch that illustrates each theorem. 265 Lesson 5 . 4 © Houghton Mifflin Harcourt Publishing Company H.O.T. Focus on Higher Order Thinking 26. Explain the Error Ava wants to know the distance JK across a pond. She locates points as shown. She says that the distance across the pond must be 160 ft by the SSS Triangle Congruence Theorem. Explain her error. 27. Analyze Relationships Write a proof. Given: ∠BFC ≅ ∠ECF, ∠BCF ≅ ∠EFC _ AB ≅ _ DE , _ AF ≅ _ DC Prove: △ABF ≅ △DEC Statements Reasons Lesson Performance Task Mike and Michelle each hope to get a contract with the city to build benches for commuters to sit on while waiting for buses. The benches must be stable so that they don’t collapse, and they must be attractive. Their designs are shown. Judge the two benches on stability and attractiveness. Explain your reasoning. L 160 ft 210 ft 210 ft 190 ft 190 ft M N J K A B F E D C Mike Michelle The distance is 160 ft, but the justification should be the SAS Triangle Congruence Theorem. 1. ∠BFC ≅ ∠ECF, ∠BCF ≅ ∠EFC 2. ― FC ≅ ― FC 3. △BFC ≅ △ECF 6. △ABF ≅ △DEC 4. ― FB ≅ ― CE 5. ― AB ≅ ― DE , ― AF ≅ ― DC 1. Given 2. Reflexive Property of Congruence 3. ASA Triangle Congruence Theorem 4. CPCTC 5. Given 6. SSS Triangle Congruence Theorem Answers will vary. While Mike's bench will probably be chosen as the more attractive, students should note that Michelle's is more stable. There are many quadrilaterals with the same side lengths as Mike's bench. But, by the SSS Congruence Theorem, the triangles in Michelle's design cannot change shape. Module 5 266 Lesson 4 y ; GE_MNLESE385795_U2M05L4.indd 266 6/9/15 2:29 AM EXTENSION ACTIVITY Have students construct models of Mike and Michelle’s benches from sturdy pieces of cardboard or photo-print paper cut into long, narrow strips and attached at the corners with brads. Students can explore further by making other polygons, checking their stability, and explaining why, for example, there is no SSSSS Pentagon Congruence Theorem. A 5-sided polygon is not stable. This means that there are many different shapes that can be constructed from the same 5 sides of a pentagon. INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Ask students to consider the changes in Mike’s bench if it were to transform from the original shape to complete collapse. Which properties of the bench would change and how would they change? Which properties would not change? Sample answer: Change: During collapse, the area inside the parallelogram would decrease continually; the measures of the angles would change, with the measures of one pair of opposite angles increasing continually and the measures of the other pair decreasing continually. Not change: The lengths of the sides and the perimeter of the parallelogram would not change. AVOID COMMON ERRORS Students may argue that Mike’s bench may be solidly constructed with screws, nails, and glue, so that it would not collapse. Stress that the bench is a real-world object introduced here to model an abstract geometrical principle, and that it is not a perfect model. The important conclusion to draw is the one relating to quadrilaterals, not to benches. Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. SSS Triangle Congruence 266
1744	https://gromacs.bioexcel.eu/t/isothermal-compressibility-of-liquid-water-at-pressure-1-atm/1621	Isothermal Compressibility of Liquid Water at pressure = 1 atm GROMACS version: 2020.2 GROMACS modification: No Here post your question Hi, I have a quick question about determining the Isothermal Compressibility of Liquid Water at pressure = 1 atm on Gromacs simulation. I found a paper below about the Isothermal Compressibility of Liquid Water at 1 atm. On page 121, there is a table (table II) showing the Isothermal Compressibility of Water at different temperatures. For example, for the temperature = 5 ℃, the suggested compressibility value is 4.9e-5 bar (~atm). Would it be possible to just use this value for the compresibility value in .mdp file when I do the pressure coupling if I want to do a simulation (water box with a protein) at 1 atm and 5 ℃ for example? Thank you for your help. Hi, you can change the value of isothermal compressibility, but it is good to check what happen to the water model when you do that. For example you can check by running a water box at the new condition and check that the property you get are correct (e.i density). Note the standard biomolecular force field are effective poentials that are parameterized for room temperature and 1 atm, thus one can not assume that they work under other condition. Best regards Alessandra Hi Alessandra, That sounds very reasonable. I’ll first try checking that before I run my simulation. Thank you so much for your help! Best Regard, Karin Related topics \| Topic \| \| Replies \| Views \| Activity \| --- --- \| Isothermal compressibility User discussions \| 1 \| 539 \| March 11, 2022 \| \| How to estimate anisotropic isothermal compressibility in GROMACS User discussions \| 0 \| 983 \| September 8, 2020 \| \| On compressibility under pressure coupling section of mdp User discussions mdp-parameters \| 1 \| 795 \| December 13, 2022 \| \| How to determine compressibility? User discussions \| 4 \| 2710 \| August 16, 2022 \| \| Is it possible to compute isothermal area compressibility modulus in GROMACS? User discussions \| 2 \| 1133 \| June 16, 2021 \| Powered by Discourse, best viewed with JavaScript enabled
1745	https://ro-journal.biomedcentral.com/articles/10.1186/s13014-019-1214-3	Advertisement Grading of MRI–detected skull-base invasion in nasopharyngeal carcinoma with skull-base invasion after intensity-modulated radiotherapy Radiation Oncology volume 14, Article number: 10 (2019) Cite this article 2817 Accesses 17 Citations Metrics details Abstract Background The aim of this study is to evaluate the prognostic value of grading MRI–detected skull-base invasion in nasopharyngeal carcinoma (NPC) with skull-base invasion after intensity-modulated radiotherapy (IMRT). Methods This study is a retrospective chart review of 469 non-metastatic NPC patients with skull-base invasion. Patients were classified as extensive skull-base invasion (ESBI) group and limited skull-base invasion (LSBI) group. Results Multivariate analysis showed that the skull-base invasion (LSBI vs. ESBI) was an independent prognostic predictor of progression free survival (PFS). The estimated 5-year local failure free survival (LFFS), distant metastasis free survival (DMFS), PFS, and overall survival (OS) rates for patients in the T3-LSBI and T3-ESBI group were 92.9% versus 93.5, 89.8% versus 86.1, 81.6% versus 76.4, and 93.5% versus 86.3%, respectively (P > 0.05). Conclusion Grading of MRI-detected skull-base invasion is an independent prognostic factor of NPC with skull-base invasion. It is scientific and reasonable for skull-base invasion as a single entity to be classified as T3 classification. Introduction Nasopharyngeal carcinoma (NPC) is endemic in China and over 33,000 new patients were diagnosed in 2012 . According to the 8th edition of the American Joint Committee on Cancer (AJCC) staging system for NPC, T classification is based on the anatomical extent of the primary tumor and which has been proposed in the era of intensity-modulated radiotherapy (IMRT) [2, 3]. In the 8th edition of the AJCC staging system for NPC, skull-base invasion is classified as T3 disease . With respect to the prognostic value of magnetic resonance imaging (MRI)-detected skull-base invasion for NPC, there are limited reports, especially, for patients treated by IMRT [4,5,6]. The aim of this study is to grade MRI–detected skull-base invasion in NPC with skull-base invasion and evaluate the prognostic value of the grading in the era of IMRT. Materials and methods Patients and patient workup This study was approved by the Institutional Review Board to identify the patients diagnosed with NPC in our center. Because this study was a retrospective study, consent was not obtained and patient records were anonymized and de-identified prior to analysis. The medical records of consecutive 695 patients with previously untreated, biopsy-proven, non-metastatic NPC that was treated with IMRT between January 2007 and February 2012 in our center were retrospectively evaluated. Of these, 469 patients with skull-base invasions were included in this study. All patients were restaged according to the 8th edition of the AJCC staging system. The pretreatment workup included a complete history and physical examination, hematology, and biochemistry profiles, fiber-optic nasopharyngoscopy, MRI of the head and neck, bone scintigraphy, computed tomography (CT) scan of the chest and abdominal region, and dental check. MR imaging All patients underwent MRI on a 1.5- or 3.0-T system (Magnetom Symphony/ Verio, Siemens Healthcare, Erlangen, Germany) with a head-and-neck combined coil. The scan range covered from the suprasellar cistern to the inferior margin of the sternoclavicular joint. All patients underwent T1 weighted and fat-suppressed T2 weighted sequences. After bolus injection of 0.2 ml/kg gadopentetate dimeglumine, contrast-enhanced T1-weighted images were obtained. Two radiologists independently evaluated all scans, and any disagreements were resolved by consensus. Skull-base invasion was diagnosed using the following criteria: (1) a defect in the low signal intensity of the bone cortex on T1-weighted image and (2) high signal intensity marrow replacement by low signal intensity tissue on T1-weighted image (an obvious enhancement in the enhanced scan) . Patients were classified as limited skull-base invasion (LSBI) group if they had invasion of one or more of these sites including the pterygoid process, base of sphenoid bone, petrous apex, clivus, and foramen lacerum. Patients were classified as extensive skull-base invasion (ESBI) group if they had invasion of one or more of these sites including the medial pterygoid plate, foramen ovale, pterygopalatin fossa, foramen rotundum, foramen magnum, hypoglossal canal, lateral pterygoid plate and jugular foramen [4, 6]. Treatment All patients received definitive IMRT. A detailed description of IMRT has been previously reported . Briefly, using the simultaneous integrated boost technique, the dose prescribed was 69–70.4 Gy, 63–67.2 Gy, 60–60.8 Gy and 54–54.4 Gy in 30–32 fractions delivered over 6 weeks at the periphery of the planning target volume (PTV) of primary tumor, PTV of metastatic lymph nodes, PTV of high-risk clinical target volume, and PTV of low-risk clinical target volume, respectively. Most patients (n = 459, 97.9%) received platinum-based neoadjuvant, concurrent, or adjuvant chemotherapy. Follow-up and statistical analysis Follow-up was calculated from the first day of treatment to the date of the event or the last follow-up visit. All patients were followed up after the completion of radiotherapy: 1 month after the completion of IMRT, every 3 months in the first 2 years, every 6 months from Year 3 to Year 5, and annually thereafter. The Statistical Package for Social Sciences, version 17.0 (SPSS Inc., Chicago, IL, USA), software was used for statistical analysis. The χ2, and Fisher exact t tests were used to compare the differences between the extensive skull-base invasion (ESBI) group and limited skull-base invasion (LSBI) group. The local failure free survival (LFFS), distant metastasis free survival (DMFS), progression free survival (PFS), and overall survival (OS) were estimated by use of the Kaplan–Meier method. LFFS, DMFS, PFS and OS were measured from Day 1 of treatment to the date of the event. Multivariate analysis was performed by using the Cox proportional hazards model. All statistical tests were two sided, and P < 0.05 was considered to be statistically significant. Results Grading of MRI-detected skull-base invasion Incidence of skull-base invasion of each site in the 469 patients is shown in Table 1. Of the 469 patients, 185 patients were classified into the LSBI group, and 284 patients were classified into the ESBI group. The patient characteristics of the LSBI group and ESBI group are shown in Table 2. Treatment outcomes The median follow-up period was 61 months (range, 2–116 months). By the last follow-up, 22.8% (107/469) of patients developed treatment failure and more patients developed treatment failure in the ESBI group (26.1% vs. 17.8%, p = 0.038). The details of treatment failure are listed in Table 3. The estimated 5-year LFFS, DMFS, PFS, and OS rates for the whole group were 91.9, 86.1, 76.6 and 87.5%, respectively. The estimated 5-year LFFS, DMFS, PFS, and OS rates for patients in the LSBI and ESBI group were 92.6% versus 90.8% (P = 0.296), 90.0% versus 84.2% (P = 0.116), 81.3% versus 73.6% (P = 0.032), and 93.1% versus 84.0% (P = 0.024), respectively (Fig. 1). Kaplan–Meier survival curves for patients in the ESBI and LSBI groups. (ESBI, extensive skull-base invasion; LSBI, limited skull-base invasion; LFFS, local failure free survival; DMFS, distant metastasis free survival; PFS, progression free survival; OS, overall survival) Univariate and multivariate analyses The value of various potential prognostic factors including age, sex, skull-base invasion, T classification, N classification and concurrent chemotherapy on predicting LFFS, DMFS, PFS, and OS were evaluated. Univariate analysis by log-rank test showed that skull-base invasion (LSBI vs. ESBI) was associated with PFS (P = 0.042), and OS (P = 0.024) (Table 4). Multivariate analysis by Cox proportional-hazards model showed that the skull-base invasion (LSBI vs. ESBI) was an independent prognostic predictor of PFS (HR 1.523, 95%CI 1.006–2.306, P = 0.047). (Table 5). T-classification category of the grading in patients with T3 classification According to the 8th AJCC staging system, 257 patients were classified as T3 classification. Of these, 83 (32.3%) patients developed extensive skull-base invasion (T3-ESBI) and 174 (67.7%) didn’t (T3-LSBI). The estimated 5-year LFFS, DMFS, PFS, and OS rates for patients in the T3-LSBI and T3-ESBI group were 92.9% versus 93.5% (P = 0.997), 89.8% versus 86.1% (P = 0.562), 81.6% versus 76.4% (P = 0.280), and 93.5% versus 86.3% (P = 0.299), respectively. The estimated 5-year LFFS, DMFS, PFS, and OS rates for the patients with T4 classification were 89.5, 83.2, 72.6 and 83.2%, respectively (Fig. 2) No significant difference was observed in terms of LFFS, DMFS, PFS, and OS between patients with T3-ESBI and those with T4 classification (P > 0.05). When extensive skull-base invasion was classified as T3 classification, the segregation of survival curves between the T3 and T4 classifications was clearly displayed. Kaplan–Meier survival curves for patients in the T3-ESBI, T3-LSBI and T4 groups. (ESBI, extensive skull-base invasion; LSBI, limited skull-base invasion; LFFS, local failure free survival; DMFS, distant metastasis free survival; PFS, progression free survival; OS, overall survival) Discussion In this study, we observed a high incidence of skull-base invasion in NPC and that grading of skull-base invasion is an independent prognostic factor of PFS in NPC after IMRT. MRI is recommended as the preferred modalities for NPC staging and has proven to be more sensitive in detecting early infiltration of tumor cells into the bone marrow [8,9,10]. Based on MRI, skull base erosion may be observed in 50–70% of NPC [4,5,6, 9, 10]. In this study, 469/695 (67.5%) patients with skull-base invasions were reported. In the era of IMRT, MRI-detected skull-base invasion was not observed to be an independent prognostic factor for NPC. However, the classification of skull-base invasion (LSBI vs. ESBI) was an independent prognostic factor in T3 (according to the 7th edition of the AJCC staging system) NPC patients in terms of the 5-year OS (P = 0.028), DMFS (P = 0.032), and PFS rates (P = 0.002) . The result of this study indicates that LSBI was associated with a better prognosis in terms of PFS compared to ESBI. Foramen ovale, foramen rotundum, hypoglossal canal and jugular foramen all belong to the ESBI group and are neural foramina. These areas were frequently related with MRI-detected cranial nerve involvement, which was associated with distant metastasis and poor survival . As distant metastasis is the most commonly failure pattern for NPC treated by IMRT, especially, for patients with ESBI [6, 12]. Although most ESBI patients (277/284; 97.5%) in our study were treated by chemoradiotherapy, they still had an unsatisfactory survival rate. Further studies including more intensive systemic approach or newer agents are needed to improve treatment outcome for these patients. In the 7th edition of the AJCC staging system for NPC, patients with skull-base invasion were classified as T3, and this classification remains in the 8th edition of the AJCC staging system. No significant difference was observed in terms of LFFS, DMFS, PFS, and OS between patients with T3-ESBI and those with T3-LSBI (p > 0.05), which was probably associated with the aid of IMRT, MRI, and the use of chemotherapy [12,13,14]. In addition, when ESBI was classified as T3 classification, the segregation of survival curves between the T3 and T4 classifications was clearly displayed. In a sense, this study demonstrated that it was more suitable for skull-baseinvasion as a single entity to be classified as T3 classification. There are several limitations in the current study, including the inclusion of patients treated at a single center and the retrospective nature of the study design. The effect of skull-base invasion on the prognosis and staging of patients with NPC should be further confirmed by other cohorts from different centers. Conclusion Grading of MRI-detected skull base erosion is an independent prognostic factor of NPC treated by IMRT. Our results confirm that it is scientific and reasonable for skull-base invasion as a single entity to be classified as T3 classification in the AJCC staging system for NPC. Abbreviations American Joint Committee on Cancer Confidence interval Computed tomography Distant metastasis free survival Extensive skull-base invasion Hazard ratio Intensity-modulated radiotherapy Local failure free survival Limited skull-base invasion Nasopharyngeal carcinoma Overall survival Progression free survival Planning target volume References IARC. 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Article Google Scholar Cao C, et al. Magnetic resonance imaging-detected intracranial extension in the T4 classification nasopharyngeal carcinoma with intensity-modulated radiotherapy. Cancer Res Treat. 2017;49(2):518–25. Article Google Scholar Lee AW, et al. Management of nasopharyngeal carcinoma: current practice and future perspective. J Clin Oncol. 2015;33(29):3356–64. Article Google Scholar Download references Funding The present study was supported by Zhejiang Province Medical and Health Science and Technology Project [No. 2017182785, 2018244679]. Availability of data and materials Our data cannot be made publicly available for ethical reasons. Data are from the present study whose authors may be contacted at chenxiaozhong2016@163.com or Department of Radiation Oncology, Zhejiang Cancer Hospital, Hangzhou, China. Author information Authors and Affiliations Department of Radiation Oncology, Zhejiang Cancer Hospital, Hangzhou, 310022, China Yanru Feng, Caineng Cao, Qiaoying Hu & Xiaozhong Chen Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Contributions Conceived and designed the experiments: YF CC QH XC. Performed the experiments: YF CC QH XC. Analyzed the data: YF CC QH XC. Contributed reagents/materials/ analysis tools: YF CC QH XC. Wrote the paper: YF CC XC. Gave many suggestions in the formation of the manuscript: CC QH XC. All authors read and approved the final manuscript. Corresponding author Correspondence to Xiaozhong Chen. Ethics declarations Ethics approval and consent to participate This study obtained approval from the Independent Ethics Committee of the Zhejiang Cancer Hospital to identify patients diagnosed with nasopharyngeal carcinoma in our center. 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Reprints and permissions About this article Cite this article Feng, Y., Cao, C., Hu, Q. et al. Grading of MRI–detected skull-base invasion in nasopharyngeal carcinoma with skull-base invasion after intensity-modulated radiotherapy. Radiat Oncol 14, 10 (2019). Download citation Received: 22 November 2018 Accepted: 08 January 2019 Published: 17 January 2019 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. 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1746	https://arxiv.org/pdf/2404.13997	Engineering Edge Orientation Algorithms Henrik Reinstädtler # Heidelberg University, Germany Christian Schulz # Heidelberg University, Germany Bora Uçar # Ñ CNRS and LIP, ENS de Lyon, France UMR5668 (CNRS, ENS de Lyon, Inria, UCBL1) France Abstract Given an undirected graph G, the edge orientation problem asks for assigning a direction to each edge to convert G into a directed graph. The aim is to minimize the maximum out degree of a vertex in the resulting directed graph. This problem, which is solvable in polynomial time, arises in many applications. An ongoing challenge in edge orientation algorithms is their scalability, particularly in handling large-scale networks with millions or billions of edges efficiently. We propose a novel algorithmic framework based on finding and manipulating simple paths to face this challenge. Our framework is based on an existing algorithm and allows many algorithmic choices. By carefully exploring these choices and engineering the underlying algorithms, we obtain an implementation which is more efficient and scalable than the current state-of-the-art. Our experiments demonstrate significant performance improvements compared to state-of-the-art solvers. On average our algorithm is 6.59 times faster when compared to the state-of-the-art. 2012 ACM Subject Classification Mathematics of computing → Graph algorithms; Mathematics of computing → Paths and connectivity problems Keywords and phrases edge orientation, pseudoarboricity, algorithm engineering Acknowledgements We acknowledge support by DFG grant SCHU 2567/3-1. The authors acknowledge support by the state of Baden-Württemberg through bwHPC. 1 Introduction Graphs and networks play a vital role in our connected society for modelling and understanding complex problems. A graph consists of a set of vertices connected by edges, which may be directed or undirected, depending on the specific modeling requirements. For some applications, such as stabilizing telecommunication networks [ 29 ], it is necessary to orient each edge of an undirected graph, thereby converting it into a directed graph. In telecommunication networks a lower number of outgoing edges equates to a higher fault tolerance, as not too many connections would be affected by a fault in one of the connection hubs modelled as vertex. One frequently used quality metric for an orientation is the maximum out-degree of a vertex. Given an undirected graph G, the edge orientation problem asks for an orientation of G in which the maximum out-degree of a vertex is minimized. The edge orientation problem has a wide range of applications [ 19 ]. Besides stabilizing telecom-munication networks, other applications include storing optimal graphs [ 1] or analysis of structural rigidity [ 30 ]. In map labeling [ 23 ] dense areas of maps can only have one label. In order to formalize this concept, one related task is to identify the densest sub-graph, which is the set of vertices, that has the highest edge to vertex ratio. This task is called the max-density task and has further applications in bioinformatics, web analysis and scheduling [ 19 ]. For more applications and their details we refer the reader to Georgakopoulos and Politopoulos . There are different approaches to solve the edge orientation problem exactly. One of the simplest algorithms is by Venkateswaran [ 29 ] which solves the problem in O(m2) time. The core idea of this algorithm is to repeatedly find paths with a breadth-first-search from a vertex having high out-degree to a vertex having low out-degree. Once such a path is found, the orientation of each of the edges on arXiv:2404.13997v1 [cs.DS] 22 Apr 2024 2 Engineering Edge Orientation Algorithms the path is reversed. Thus by this operation the degree of the high out-degree vertex is reduced and conversely, the degree of the low-degree vertex is increased. Other approaches solve the problem by using a flow based formulation [ 2, 24 ]. Kowalik [ 24 ] gives an approximation scheme for this problem. These approaches analyse the theoretical aspects, without in depth practical study. The current best complexity bounds and the only implementation in Java of an exact version of Kowalik’s flow based solution are presented by Blumenstock [ 6 ]. However, an ongoing challenge in edge orientation algorithms is their scalability, particularly in handling large-scale networks with millions or billions of edges efficiently. As real-world networks continue to expand in size and complexity, there is a need for algorithms capable of scaling effectively to such massive datasets. Contribution. We introduce a novel framework of algorithms inspired by Venkateswaran work [ 29 ]of improving paths to tackle the edge orientation problem. Our experimental results show the advantages of this approach over conventional flow-based formulations. Additionally, we offer an alternative proof of correctness for this framework. In addition to a verbatim implementation of the original algorithm, we provide an accelerated version in which we engineer all components of the algorithm. This includes efficient pruning of the search space, batch search of improving paths and better initialization routines that are used as starting point of the algorithm. Experiments show that our algorithm can scale to huge instances and that the performance improvements of our algorithm over the state-of-the-art are very large. For example our fastest algorithm is on average 6.59 times faster than our (faster) C++ reimplementation of the state-of-the-art solver. This paper is organized as follows: We first introduce basic concepts and provide an extensive review of related work in Section 2 and 3. Section 4 provides an alternative proof of correctness using a flow-based formulation on a bipartite representation of the graph for the algorithm by Venkateswaran. We then proceed to discuss several speed-up techniques, including eager depth-first search, and explore various variations of breadth-first and depth-first search in Section 4. These new approaches are extensively benchmarked in Section 5, followed by our conclusions in Section 6. 2 Preliminaries Orientation Problem. An undirected graph G = ( V, E) consists of a vertex set V containing n vertices and an edge set E ⊆ V 2 of m edges. An orientation O assigns a direction to each edge in G and results in a directed graph. We denote the direction of an edge by O(e) = u → v, if e is oriented from u to v. For an edge e = {u, v} oriented as u → v, we say that e is an outgoing edge of u and incoming edge of v. The out-degree d(O, v) of a vertex v in an orientation O is defined as the number of outgoing edges of v, that is, d(O, v) = \|{ u : v → u ∈ O}\| . If the orientation is clear from the context, we write d(v). Similarly, the in degree of a vertex is the number of its incoming edges. We call the vertices with the largest out-degree in an orientation peak vertices . A path in an orientation is a finite sequence of vertices u1, . . . , un, where there is an edge oriented ui → ui+1. We call a path simple if no vertex is contained twice in the path. In an orientation, changing the orientation of an edge from v → u to u → v is called flipping. A path can be flipped by flipping all edges contained in it once. For a given graph G, the edge orientation problem asks for an orientation O such that d⋆ = max v∈V d(O, v) is minimized. (Pseudo-)arboricity. A forest is a collection of edges, where each vertex is only connected by one path. The edges of an undirected graph can be partitioned into disjoint forests and the minimum number of forests is known as the arboricity . A more relaxed version of this problem is to decompose the graph into pseudoforests, where in every connected component there can be at most one cycle. The minimum number of pseudoforests partitioning the edges is called pseudoarboricity .The pseudoarboricity is known to be equivalent to the maximum out-degree of an optimal edge H. Reinstädtler, C. Schulz and B. Uçar 3 orientation [ 29 ]. The average density of a graph is defined as the ratio of edges to vertices. A sub-graph contains only edges between a subset of vertices. The maximum average density of any sub-graph is closely related to the pseudoarboricity. As shown by Picard and Queyranne [ 26 ] and Venkateswaran [ 29 ] the pseudoarboricity is equal the ceiling of the maximum average density of any sub-graph. Picard and Queyranne [ 26 ] show the arboricity is equal to either the pseudoarboricity or the pseudoarboricity plus one. Integral Flows. Given a directed graph G = ( V, E), a source vertex s ∈ V , a target vertex t ∈ V ,and an edge capacity function c : E → N, a flow is a function f : E → N that satisfies two conditions: (i) the flow does not exceed the capacity in any edge; (ii) and the inflow in every vertex equals the outflow, except for s and t, which have respectively positive out and in flows. Given a flow, one can define a residual capacity for each edge, which is equal to edge’s original capacity minus the current flow. A residual network for a given flow consists of all edges with a positive residual capacity. If the flow on an edge is equal its capacity, the edge is called saturated. When changing the capacity of an edge, an edge can become under or over-saturated and the flow needs to be updated. Bipartite b-Matching. In a bipartite graph the vertex set can be partitioned in disjoint sets S, T such that all edges contain one vertex from S and T each. A matching M is a set of edges no two of which share a vertex. A maximum matching is a matching with the largest number of edges. For a given positive integer b(v) for each vertex v, the b-matching problem asks for a set F of edges with the largest cardinality such that the vertex v is included in at most b(v) in F. A b-matching in a bipartite graph (S ∪ T, E) can be found by modeling it as a flow problem as follows. First, add a source vertex s and a sink t to the graph. Then add edges between s every vertex in v in S with capacity b(v) as well as edges between each vertex w in T with capacity b(w) from w to t. The original edges are assigned unit capacity. The saturated edges of an integral max-flow from s to t are the edges in an optimal b-matching. 3 Related Work Edge Orientation Task and Pseudoarboricity. There has been a wide variety of approaches to solve the edge orientation task and the identical pseudoarboricity task. One algorithm is by Algorithm 1 Algorithm by Venkateswaran 1: procedure VENKATESWARAN (G = ( V, E)) 2: O ← an arbitrary orientation of G 3: k ← max v∈V d(O, v) 4: S ← { v ∈ V \| d(O, v) = k} 5: T ← { v ∈ V \| d(O, v) ≤ k − 2} 6: while BFS finds path P=s, . . . , t from S to T in O do 7: Flip P in O 8: Remove s from S 9: Remove t from T if d(O,t) = k − 1 10: if S empty then 11: k ← k − 1 12: S ← { v ∈ V \| d(O, v) = k} 13: T ← { v ∈ V \| d(O, v) ≤ k − 2} return k4 Engineering Edge Orientation Algorithms Venkateswaran [ 29 ] and underlies the algorithms proposed in this work. Venkateswaran gives an algorithm for computing an extremal orientation to minimize the maximum in degree. The algorithm can be easily translated to the out-degree setting, as shown in Appendix Algorithm 1. After arbitrary initializing an orientation, the algorithm starts to search for improvements by finding paths between the set of vertices S with max out-degree k and the set of vertices T with out-degree strictly lower than k − 1. If S is empty, then k is reduced by one, and the sets S and T are reinitialized. If no path is found, the current k is returned as optimal. The correctness of this algorithm follows from the density of the sub-graph induced by the vertices visited by the failing BFS starting from S. These have a degree of at least k − 1 and there is at least one with degree k, totaling an average density greater than k − 1, leading to a pseudoarboricity of k by the sub-graph density argument. It is proven that the running time of this algorithm is O(m2), since each path can be found in O(m) and the number of improvements can be bounded by m as well. The argument is that the total number of paths for one vertex is bounded by its out-degree, which again is bounded by the number of edges in total. A 2-approximation of the pseudoarboricity or the maximum average density can be found in linear time [ 10 , 19 ] by repeatedly deleting min degree vertices. Georgakopoulos and Politopoulos [ 19 ]provide an algorithm for finding the densest subset in a more general setting of set systems, improving on ideas by Goldberg [ 20 ] by pruning the graph during their binary search scheme. The algorithm can be used to compute the max out-degree, but does not find a suitable orientation. Asahiro et al. [ 2] give a solution with running time O(m3/2 log d⋆) based on flows for the problem and offer more results for approximating the related weighted orientation problem. Kowalik [ 24 ]give an approximation scheme using flows that can be used to calculate exact solutions as well. They construct a virtual graph with a source and a sink vertex. Given an orientation O and a test value d′ they add an arc between the source vertex and vertices with higher than d′ out-degree with capacity d(O, v) − d′. Each vertex with lower than d′ out-degree is connected to the sink with capacity d′ − d(O, v). For each oriented edge u → v, there is an arc v to u with capacity 1 . The flow is then computed and all edges with a corresponding exhausted arc are flipped. If there is an arc from the source without flow, the test failed and the maximum out-degree must be higher. With a binary search the optimal value d⋆ and a suitable orientation can be found in log d⋆ steps. Blumenstock [ 6] presents bounds for the aforementioned flow based solution using Dinic’s al-gorithm and almost unit capacity networks. The best general worst case bound for the problem is O(m3/2√log log d⋆). Moreover, Blumenstock [ 6] also presents the first practical evaluation of algorithms for the pseudoarboricity problem by implementing the flow based algorithms by Kowa-lik [ 24 ] and Georgakopoulos and Politopoulos [ 19 ]. Experiments shows that Georgakopoulos and Politopoulos’s method using unit capacity networks is faster than other flow based solutions, including Kowalik’s algorithm. However, subsequent analysis revealed a subtle implementation error in the implementation of the algorithm which caused it to output incorrect results in rare cases. Once corrected, the performance advantage of this method is no longer apparent. We use both of these algorithms as the state-of-the-art in our experiments. Bipartite Matching and Network Flows. Finding a b-matching can be done by transforming the problem into an uncapacitated matching problem as described by Gabow [ 18 ]: vertices (and their respective edges) are replicated according to their capacity. Bipartite matching is a well-studied and understood part of computer science. The best known worst-case algorithm for solving bipartite matching is Hopcroft-Karp [ 22 ] with a complexity of O(m · √n). It can be seen as a specialization of Dinic’s algorithm for general flows. For graphs with capacity 1 per vertex, like in the bipartite matching case, the complexity of Dinic’s algorithm is O(n1/2m) according to Tarjan and Even [ 17 ]matching the bound from Hopcroft-Karp. For general flows Dinic’s algorithm has a complexity of O(n2m) and better alternatives like Goldberg and Tarjans algorithm are available in practice. H. Reinstädtler, C. Schulz and B. Uçar 5 4 Edge Orientation Framework and Engineering Techniques 312abctcba321scapacity 1 capacity dLR Figure 1 Example bipartite repr. transformation. We give an interpretation of the working of Algorithm 1 using maximum flows and b-matchings in bipartite graphs. This interpret-ation allows us to identify key aspects for en-gineering to improve the practical running time of the algorithm.For a given graph G = ( V, E),we construct a bipartite graph BG = ( L ∪ R, EB) with \|E\| + \|V \| vertices, and 2\|E\| edges. Roughly speaking, L will represent the edges of the ori-ginal graph and R will represent the vertices of the original graph. For each edge e = {u, v} ∈ E, we have a vertex ℓ(e) ∈ L. For each vertex v ∈ V , we have a vertex r(v) ∈ R. For each e = {u, v} ∈ E,we have two edges in EB: one between ℓ(e) and r(u) and another between ℓ(e) and r(v). In this bipartite graph, an edge orientation can be described by an assignment of each L vertex to one of its two neighbors in R. Here, each L-vertex is assigned to a unique R-vertex, and each R-vertex can have multiple R-vertices assigned to it. The edge orientation problem can thus be solved by finding an assignment of L vertices to R vertices which minimizes the maximum number of L vertices assigned to an R-vertex. This can be solved by a network flow formulation. Our network flow formulation starts with BG and adds two special vertices s and t to BG. The vertex s is connected to all L vertices, and all R-vertices are connected to the vertex t. Then, a capacity of 1 is attributed to all edges between s and L-vertices, and also to all edges of BG; that is from a vertex ℓ(e) to the vertices r(u) and r(v) for e = {u, v} ∈ E. If we attribute a capacity of d to edges (r,t) and have a feasible flow with a total flow value of \|E\| (out of s), then all edges can be oriented while having an out-degree of at most d (see below). An example for a simple graph G of three vertices and three edges, the flow graph obtained by adding s and t to BG can be found in Figure 1. Due to the capacity of 1 for each (s, ℓ ) edge, every edge in the original graph will be oriented in only one direction. For an edge e = {u, v} in the original graph, the edge is oriented from u → v,if in BG there is a flow from ℓ(e) to r(u) and v → u, if there is a flow from ℓ(e) to r(v). Each edge is assigned a direction, since we search for a flow in which all \|E\| edges in the flow graph of the form (s, ℓ ) are saturated. Clearly, the smallest value of d allowing this will be the optimal d⋆. When a smaller value is tested and the out flow in s is not equal \|E\|, some edges are not assigned a direction. Since this is a classical capacitated network flow problem with integer capacities, an integer solution can be found with augmenting path-based algorithms [ 12 , Theorem 26.10]. A binary search for the optimal d will have in the worst case log dG steps, where dG is the maximum degree of a vertex in G. Instead of running a flow algorithm from scratch, the results of previous searches can be reused. When increasing the value of d, the previous flow can be used as a starting point. If we decrease the test value, the flow in oversaturated edges starting in R must be balanced first by reducing the flow along a path starting in s to t. Afterwards any augmenting path needs to be found. Correspondence to Venkateswaran’s Algorithm. Given an arbitrary orientation for the out-degree, we can construct the flow in the bipartite representation as follows: For each edge e = {u, v},oriented u → v we set the flow from ℓ(e) to r(u) to 1 and ℓ(e) to r(v) to 0. In the residual network there is a residual capacity of 1 between r(v) and ℓ(e). The obtained flow network is saturated in all edges (s, l), l ∈ L, since each edge is assigned a direction and therefore has a flow from ℓ(e) to either u or v. The initial test value for the max out-degree is set to the maximum out-degree of the orientation and repeatedly decreased, until some edges from s to L become unsaturated. In each decreasing step there is an oversaturated edge (r(v),t), that needs to be decreased by one, at most for every v and along to a path to s the flow needs to be reduced by one. This path has 6 Engineering Edge Orientation Algorithms length 2, since one arbitrary edge (ℓ(e), r(v)) with flow greater zero will be chosen and its flow as well as the flow on the single incoming edge (s, ℓ (e)) reduced. Afterwards an augmenting path needs to be found. This path must include ℓ(e) as second vertex since all other edges (s, L) are saturated. Instead of first reducing and then finding an augmenting path, we can combine both steps. In other words, the flow algorithm needs to find distinct paths in the residual network between all vertices in R with oversaturated edges to t and vertices in R, that have edges towards edges with enough residual capacity. This can be done using breadth-first-search algorithms. The vertices in R with oversaturated edges are exactly the vertices in S, since we decreased the test value by one. Similarly, the last vertices before t in augmenting paths need to be vertices r(v) ∈ R with residual capacity in (r(v),t) and enough unsaturated incoming edges. These vertices are exactly those vertices T in Venkateswaran algorithm. Due to the bipartite structure paths must include the vertices in L. The edge vertices L in a residual network of BG can be only traversed in one direction by the BFS, exactly like the BFS in Venkateswaran. The only difference, compared to Venkateswaran, is that in our network, the path is twice as long, since each edge in G consists out of two subedges in BG. When in our model a too low test value is set, augmenting paths cease to exist. 4.1 The Proposed Framework Algorithm 2 Orientation by Exhaustive Search. 1: procedure EXHAUSTIVE SEARCH (G = ( V, E), O) 2: repeat 3: for v ∈ V with d(v, O) = max w∈V d(O, w) do 4: if P = FIND PATH (G, O, v) exists then 5: Flip P until no path has been flipped in last iteration 6: return O We now present our techniques and framework to improve the performance of Venkateswaran’s algorithm in practice. We introduce data structures and algorithmic choices inspired by the maximum cardin-ality matching problems [ 15 , 16 , 22 , 27 ]. Engineering techniques include a simple orientation improving algorithm with com-plexity O(m) that yields a much better start-ing point for the path finding algorithm in practice. Moreover, we enhance the efficiency of path searches by either reusing information or batching the searches and exploring paths in reverse order. Our general algorithm framework is shown in Algorithm 2. It is an adapted version of the algorithm by Venkateswaran. Instead of constructing the sets S and T explicitly we iterate over all vertices that currently have the largest out-degree and try to find an improving path, which is a path in the orientation from one of these vertex to a vertex with lower out-degree. If such a path is found, it is flipped, resulting in a lower out-degree for the start vertex. The algorithm continues this process until no such path exists. The correctness of this general algorithm follows from Venkateswaran’s algorithm. It has an equivalent termination criterion as the original algorithm, since when it finishes when no further improving path starting at peak vertices can be found. Instead of searching from all peak vertices at the same time, we search for every peak vertex separately. Like the algorithm by Venkateswaran this algorithm has a time complexity of O(m2) since we need to find at most m improvements by the argument of Venkateswaran [ 29 ]. Furthermore, each improvement can be found with one DFS or BFS in O(m), yielding its total complexity. 4.2 Engineering Techniques We now discuss how we can improve the techniques to compute orientations. First, we discuss the linear 2-approximation data reduction and come up with a fast initialization algorithm. Secondly, we review different ways of finding improving paths and general ideas to optimize this search. H. Reinstädtler, C. Schulz and B. Uçar 7 4.2.1 Two Approximation Data Reduction The linear 2-approximation proposed by Charikar [ 10 ] can be used as data reduction. It computes a 2 –approximation dapprox using a bucket priority queue and vertices with degree dapprox 2 or less can be removed safely due to the densest sub-graph argument. In other words, since the dapprox is a 2–approximation, the optimum outdegree must be at least dapprox 2 . Vertices with a degree that is smaller or equal to this value can therefore be removed iteratively. The edges of these vertices are oriented outwards. The approximation needs to process every edge of the graph at least once and maintaining the needed priority queue is not negligible. Therefore, running it always as a preprocessing step can be costly, if the resulting guess is small and does not remove much of the graph. Instead, we propose to run the 2-approximation conditionally on the density of a graph ρ = mn . If the density is smaller than some parameter c, we directly skip to find improving paths, instead of wasting time to compute a low approximation. In experiments we tune this parameter. 4.2.2 Fast Initialization Algorithm 3 FastImprove Algorithm 1: procedure FAST IMPROVE (G = ( V, E), O) 2: for v ∈ V do 3: for e = v → u ∈ O do 4: if d(O, u) < d(O, v) − 1 then 5: Flip( e) 6: for e = u → v ∈ O do 7: if d(O, u) − 1 > d(O, v) then 8: Flip( e) 9: return O Venkateswaran’s algorithm starts with an arbitrary orientation and thus has worst-case complexity O(m2). However, it is evident that starting with a good orientation can lead to a faster algorithm compared to an arbitrary initialization. The F AS - T IMPROVE -Algorithm in Algorithm 3 makes one pass over an arbitrary orientation and flips an edge v → u, if such a flipping improves the largest out-degree of v and u. Since only direct improvements are found, the algorithm does not find the optimal solution. However, it can improve a given orienta-tion significantly. In our implementation, vertices are represented by their IDs which are numbers in the range 0, . . . , n − 1 . Our algorithm initializes an orientation with u → v for u > v and then applies the FastImprove algorithm to quickly improve it. The complexity of one pass is O(m). In the experimental evaluation in Section 5, we show that this simple algorithm can tremendously speed up computation time of the overall algorithm. 4.2.3 Path Finding Algorithms There are two basic ways to find paths in a directed graph: either breadth or depth first search. Breadth first search works by using a queue and adding neighboring vertices following first in first out order. Depth first search is usually implemented using a stack and a visited array. The visited array stores whether a vertex has been visited before, while the stack keeps track of the current path. The next not visited neighboring vertex is always added to the stack and marked as visited, the neighbors are explored recursively using the stack, i. e., depth first. Once all neighbors are explored the previous vertex of the stack is (re-)visited and its remaining neighbors are visited. Batched BFS. The batched BFS approach finds improving paths similar to the original idea by Venkateswaran of an adapted BFS. It does not fit the previous described framework in Algorithm 2. Instead of starting a BFS from every vertex, we start the BFS by putting all peak vertices in the queue and flip a path once we have found an improving path by breadth first search. We continue the search until all vertices have been reached or, in case of success, we found an improvement for all peak vertices. The search is continued until for none of the vertices we find an improving path. This yields a correct result, since we inserted all peak vertices at the start and thus will find a shortest improving 8 Engineering Edge Orientation Algorithms path, if there exists one. However, we will not always find a path for all peak vertices at once. DFS. We now look at ways to improve depth-first-search. Our ideas include checking neighboring vertices eagerly, ensuring independent paths by visiting peak vertices only as starting vertices of paths, reusing the visited array and eagerly ordering path searches. Early Check. The first improvement is to check the out-degree of all neighbors before continuing exploring them in the recursion. This increases the overall complexity by a constant factor, but in practice speed up the exploration massively by preventing unneeded recursion. DFS with Independent Paths. We propose for the DFS to not continue traversal via other peak vertices while searching improving paths. There is no added benefit in adding a peak vertex to a path during search as each peak vertex needs its own improving path. More precisely, they have to be distinct, since flipping one of the paths would change the orientation of the joint part of the two paths, effectively stopping the second path from being valid. Hence, we need to find independent paths from both of the vertices regardless. DFS with Shared Visited Array. A DFS can reuse information stored in the visited array for multiple consecutive searches for improving paths. Here, we reset the visited array used in a classical DFS only after one pass over all peak vertices and not after every path search. Thus, we enhance efficiency by retaining the visited array’s state across multiple consecutive searches to identify im-proving paths, rather than resetting it after each search. This method strategically excludes previously explored and non-improving sub-graphs from subsequent searches, significantly reducing computa-tional redundancy. By maintaining the visited array across searches, we ensure that once a path from a peak vertex is improved, subsequent searches do not redundantly explore the same paths or sub-graphs already deemed non-improving. dd-1 d-2 Figure 2 Left: Layered example orient-ation in a graph with nine edges and vertices. Right: Visualization of the onion-like struc-ture of the orientation problem. Figure 2 (Left) gives a simple example orientation, where not resetting the visited array is helpful: If the first path from the bottom left to the top left vertex is found and flipped, we do not want the search starting in the middle vertex to traverse via left tree that was already explored during the first DFS. Moreover, this example orientation shows, why this DFS approach is better than the BFS described in the earlier section. The BFS would find at most two improving paths, because it eagerly finds pre-decessors. It would assign the middle vertices (yellow) only two distinct predecessors at most due to the graph structure, leaving one peak vertex (purple) without suc-cessors and leading to two paths for the bottom left vertex. However, we are only interested in one path per peak vertex. Thus, this leads to more path searches than necessary. Eager Path Search. The FastImprove algorithm only uses direct improvements and reduces their number, leaving longer improving paths to be discovered. Therefore, our problem has an onion-like structure, multiple layers of vertices with the same out-degree are surrounding some core peak vertices after the initialization. By our classical approach we would find improvements from core to the lower degree outer layers and slowly growing the number of peak vertices. This leads to prolonged paths, as there will be multiple paths needed to be found while decreasing the out-degree of a peak vertex by one at a time. An example for this is shown in Figure 2 (Right), the vertices with out-degree d will need to be improved multiple times via the vertices with out-degree d − 1 . Instead of slowly searching consecutively, we propose to first find paths for the outer layers of the problem. We define i as the number of layers we reduce eagerly in the reverse order of their out-degree,i. e., d − i, d − i + 1, and so on. Moreover, if only fewer vertices than some parameter c have an out-degree of d − i + k in the orientation, we propose to repeatedly search up to k improvements with a DFS. Intuitively, this H. Reinstädtler, C. Schulz and B. Uçar 9 reduces the number of times the algorithm needs to collect these vertices and find paths. As both ideas are combined with the idea of the reused visited array, this method does not return an exact solution always: Suppose d is the current max out-degree, then it can occur, that there is no improvement for d − 1, but there would exist some path from d out-degree vertices to d − 2 out-degree vertices through d − 1 vertices. Since the d − 1 vertices are checked first, there is no improving path found and the visited information is kept leading to all d − 1 vertices marked visited. The searches started in d can not traverse over these vertices and the search is unsuccessful. Therefore, it requires to run either a DFS or a BFS afterwards. We propose to decide based on the maximum size of the layers and the resulting max out-degree whether to run a batched BFS or DFS. Moreover, the choice of i and c is crucial, in Section 5 we test static values for c and i. Additionally, we test for i a dynamic value of pmax d − ρ with ρ = mn as average density of the graph and max d being the out-degree of the starting orientation. The maximum out-degree of the starting orientation max d and ρ are the natural lower and upper bounds for this problem. We chose the square root as a natural damping function in order to not explore too many layers eagerly. 5 Experimental Evaluation Methodology. We implemented our algorithms and data structures in C++ 20 . We compiled our program and all competitors using g++-12.1 with full optimization turned on ( -O3 flag). In our experiments, we use two types of machines provided by a cluster for our experiments. Both machines are equipped with 20 -core Intel Xeon Gold 6230 processors running at 2.10 GHz and having a cache of 27.5 MB. Machine type A has two sockets and 96 GB of RAM, machine type B has four sockets with 3 TB of RAM. For the five graphs with more than 1 billion edges we use the machine type B. We run each algorithm on each instance five times and use the arithmetic mean of the running time of these independent runs in the experiments. Up to four experiments were run in parallel on machine type A and 6 on machine type B. The order of the experiments was random. The experiments that did not finish within 5 hours were only repeated once. The running times reported include the setup of data structure needed by the algorithms, but does not include the initial loading of the graphs from the hard drive. For comparisons, we use performance profiles as proposed by Dolan and Moré [ 14 ]. We plot which fraction of instances is solved by an algorithm within τtopt , for τ ≥ 1 and topt being the best running time reported by any algorithm on a given instance. A higher fraction at a lower τ means that more instances are solved within this τ, implicating a better performance. Instances. We use graphs from the SuiteSparse Sparse Matrix Collection [ 13 ] with more than 1 mil-lion vertices for benchmarking. The set contains 67 graphs from a wide range of applications. Statist-ics on the instances, including the number of edges, the number of vertices, the minimum and the maximum degree of a vertex, and the number of connected components, are shown in Table 4 in the Appendix. The largest graph in our set has more than five billion edges. Based on the average density of the graphs, we chose nine representative instances ( bold in Table 4) for evaluating different parameters for the proposed algorithms and running preliminary experiments faster. These represent-ative instances are called test set in the following and are not included in the geometric means and performance profiles in the final experiment. State-of-the-art. Blumenstock [ 6] provided us with a Java code implementing Kowalik’s exact algorithm. We ported that implementation with Dinic’s algorithm to C++ and validated that our implementation is on average 1.42 times faster than the Java implementation (compiled with Open-JDK 17 and run sequentially) on the instances used by Blumenstock [ 6]. A more detailed comparison 10 Engineering Edge Orientation Algorithms of the running time of these implementations is reported in Appendix Table 2. We have also imple-mented Georgakopoulos and Politopoulos’s approach to be able to do a more conclusive comparison than what was available in [ 6 ]. We can run this algorithm with Dinic’s algorithm or Push-Relabel algorithm for better performance. We use a Push-Relabel implementation in our final experiments for comparison since it is 1.6 times faster than Dinic’s algorithm on our implementation. The algorithm by Georgakopoulos and Politopoulos computes only the pseudoarboricity, i.e., the objective value of an orientation, and not an orientation itself. As described by Blumenstock [ 6] our implementation computes the pseudoarboricity and uses Kowalik’s reorientation scheme once to obtain an orientation. In the following, we refer to our implementation of Georgakopoulos and Politopoulos [ 19 ] by G&P and to our implementation of the exact Kowalik [ 24 ] by Kowalik . Preliminary experiments on the test set showed that using the 2-approximation initialization before Kowalik and G&P results in, respectively, 1.90 and 3.02 times faster running time than not using this initialization. Therefore, we run the final experiment with 2-approximation for both of the state-of-the-art methods. We note that our experiments with G&P and Kowalik are more comprehensive and conclusive than what were available in earlier work . Implemented Algorithms. For our main approaches we implemented a vanilla DFS ( DfsNaive )and the algorithms devised in Section 4.2. These include Fast Improvement ( FastImprove ), DFS with improvements ( Dfs ), BFS ( Bfs ) and the eager path search ( EPS ). Our implementation of 2-approximation ( TwoA ) and the 2-approximation conditioned on the average density ( TwoADens ) can be run before all algorithms considered here. 5.1 Parameter Study for the Proposed Algorithms In the subsequent sections, we adjust our parameters and investigate the different algorithmic com-ponents described in this paper using the nine graphs selected specifically for parameter tuning. Naive Methods. We conducted a comparative analysis of several DFS implementations: one without any of the techniques outlined in Section 4.2.3 ( DfsNaiveWithoutShared ), one that incorporates all the proposed techniques ( Dfs ), another that reuses the visited array ( DfsNaiveShared ), and the batched BFS ( Bfs ). All these implementations utilized the FastImprove approach, and we present only the average execution times. Our findings indicate that the Dfs method is the most efficient, being significantly faster than the others. Specifically, the DfsNaiveShared is 2.02 times slower, and the Bfs is notably slower by a factor of 10.38. The slowest was the DfsNaiveWithoutShared , which was 206.443 times slower than the optimized Dfs . Based on these results, we will exclusively use the optimized Dfs , i.e. DFS with improvements, for the subsequent parameter tuning experiments. Fast Improve. To demonstrate the effectiveness of the FastImprove technique, we conducted experiments on the small dataset, running Dfs both with and without the technique. The variant lacking FastImprove failed to complete within 5 hours on the instance from the MAWI set, which notably has an extremely high maximum degree of approximately 63M. In stark contrast, the FastImprove -enabled variant completed the same instance in just 18.84 seconds. Across the remaining eight instances, the non-FastImprove variant performed consistently slower, averaging 27% longer processing times. Given that omitting FastImprove led to a timeout on one instance and consistently slower performance on others, we use the FastImprove algorithm for all instances. Eager Path Search. We now optimize the parameters for our eager path search. In the eager path search we search for improving paths not only for the peak vertices, but for layers with lower H. Reinstädtler, C. Schulz and B. Uçar 11 Figure 3 (a) The running time of EPS+Dfs and EPS+Bfs normal-ized by EPS+Dfs plotted against the resulting min max out-degree. Low out-degree solutions are solved faster by Dfs as finishing method. 10 110 210 3 d 10 0 10 1 10 2 trel EPS+Bfs EPS+Dfs (b) The running time of EPS with or without 2-approximation plotted against the average density ρ=mn of the graph (normalized by EPS ). 10 010 1 ρ 10 0 trel EPS TwoA+EPS out-degree. The parameters are the number of layers eagerly searched and the maximum size of a layer that will be searched for even more eagerly. Furthermore, we probe which method ( Bfs or Dfs )should be used to finalize the orientation. We tested the number of layers eagerly searched for values i ∈ { 2, 5, 10 , pmax d − ρ} with ρ = mn and max d being the current maximum out-degree of the starting orientation. We used the optimized Dfs to compute the final orientation. For the sampled set the dynamic approach i = pmax d − ρ yields the best results on average. The static approach with i = 5 is only 2 % slower, while the approaches with i = 2 and 10 are 7.3 resp. 8.3 % slower. We select the dynamic approach for our final experiment and now progress into the second parameter the eager size. If a layer size is smaller than the eager size the algorithm tries to find multiple improvements consecutively, instead of requiring the vertices to be collected again. In this experiment we combined the dynamic approach and DFS with the eager size parameter c ∈ { 10 , 100 , 1000 , 10000 }. The best geometric mean we can report for c = 100 , for c = 10 , 1000 , 10000 we report a 1.5%, 0.6 % resp. 0.16 % higher geometric mean for the running time. Therefore, we select c = 100 for our final experiment. Finally, we investigate when to run either the Bfs or Dfs after the eager path search. Figure 3a shows the running time divided by the optimal value plotted against the min max out-degree. We observe that for lower out-degree ( < 10 ) it is more beneficial to run the optimized Dfs . On one instance with low out-degree using Bfs is 60 times slower. However, on instances with higher out-degree the Bfs is consistently faster by a small relative margin. We run Dfs , if the out-degree is less than 10, and otherwise Bfs . Two Approximation. We now look into the efficiency of the 2 -approximation as outlined in Sec-tion 4.2.1. Specifically, we investigate the effectiveness of not always executing the two-approximation prior to the eager path search. The outcomes of this study are illustrated in Figure 3b, where we plot the average density of the graphs against their average running time. The experiment indicates that for graphs with low average density, pre-running the two-approximation does not yield beneficial results. We run the 2-approximation exclusively for graphs with an average density exceeding 10 to achieve reasonable reductions on such graphs. Final Algorithm. The final configuration of our algorithm RapidPathOrientation (RPO) is as follows: After using the 2 -approximation conditionally on the average density ( mn > 10 ) and executing the FastImprove algorithm, it runs the EPS with dynamic layer count of p(max d − ρ). If less than 100 vertices are in a layer they are relaxed more eagerly. Finally, in order to produce a correct result a Bfs is run for out-degree higher than 10 and the specialized Dfs for lower degree. 12 Engineering Edge Orientation Algorithms Figure 4Running time performance profile on 58 instances for state-of-the-art algorithms TwoA+Kowalik ,TwoA+G&P and our RapidPathOrientation (RPO ) approach. On the right hand side we present a detailed zoom to values up to 6.5. 10 010 110 210 310 4 τ 0.0 0.2 0.4 0.6 0.8 1.0 Fraction of Instances 123456 τ 0.0 0.2 0.4 0.6 0.8 1.0 Fraction of Instances RPO TwoA+G&P (PR) TwoA+Kowalik 5.2 Comparison with State-of-the-Art We now compare our approach to the state-of-the-art on the whole dataset. In Figure 4 the per-formance profile for our algorithm ( RapidPathOrientation ) is shown in comparison to the algorithm by Kowalik [ 24 ] ( TwoA+Kowalik ) and Georgakopoulos & Politopoulos [ 19 ] ( TwoA+G&P ) with combined with the 2–approximation on 58 problem instances. On average our algorithm is 6.59 times faster than TwoA+Kowalik and 36.27 times faster than TwoA+G&P ; moreover on 88 % of instances it is the fastest approach. As seen in the right plot of Figure 4, RapidPathOrientation solves all instances within a factor of τ < 1.56 of the fastest algorithm (on an instance) and the competitor approaches solve around 50 % of instances within a factor of τ = 6. Since the profile of RapidPathOrientation is higher than both TwoA+Kowalik and TwoA+G&P for all values of τ, we conclude that the proposed approach is faster than the reimplementation of the state-of-the-art methods. Moreover, as shown in the preliminary experiment, the speed up over the Java implementations is even more pronounced. We report detailed per instance results of the running times in Appendix Table 5. Table 1 shows the running times averaged per sub dataset. We now give a more detailed discussion of the performance of the different algorithm for the subgroups of our dataset. In general, it can be observed that competing algorithms can solve instances from the MAWI subset faster. Out of all the instances for which our approach is not the fastest, our algorithm experiences the highest relative slowdown on the mawi_’130 instance. Here, RapidPathOrientation needs 56% more time than both competing algorithms. This is due to the fact that the 2-approximation is very effective on the MAWI dataset, however the 2=approximation is not run by our algorithm on these instances as they have a low average density ( mn < 2 ). When running our approach on these instances with the 2-approximation algorithm as a preprocessing algorithm, we achieve similar running times. The performance differences are very big especially on the low density and low out-degree instances of the flowipm22 set. For example, on the spielman_k600 instance TwoA+G&P is 243.10 times and TwoA+Kowalik is 65.46 times slower than our new approach. In general, this is due to the fact that the FlowIPM22 instances have a high maximum degree compared to their optimum out-degree of 3. The competitors can not limit their search space properly and have to perform many reorientations using their flow algorithm. Similarly, we can observe advantages of our approach on random generated graphs in the dimacs10 set, like the delaunay_n24 instance. RapidPathOrientation is around 80 times faster than TwoA+Kowalik and 221 times faster than TwoA+G&P . Moreover, the TwoA+G&P algorithm has issues with some instances from numerical backgrounds ( huge ) in the dimacs10 instances, resulting in a running time much higher than the other approaches (up to 16393.74 times slower). The TwoA+G&P algorithm has to solve more flows to converge according to its binary search scheme, while its pruning mechanism is not that efficient on these instances. The pruning re-H. Reinstädtler, C. Schulz and B. Uçar 13 moves vertices with unsaturated edges connected to the target vertex, when a new lower bound for the pseudoarboricity is accepted, which does not happen often on those instances. In gen-eral, our new algorithm RapidPathOrientation solves instances from numerical backgrounds and road networks, represented in the dimacs10 set, significantly faster than the previous approaches. Table 1 Geometric mean running time in seconds grouped by sub data set and for all instances. TwoA+G&P TwoA+Kowalik RPO flowipm22 104.62 33.57 0.57 gap 12 267.57 4808.19 2628.25 genbank 70.14 70.63 36.69 law [7, 8] 5.02 2.97 1.82 mawi 8.40 8.39 10.07 snap 8.86 5.15 1.85 sybrandt 5340.81 3064.19 2226.32 dimacs10 337.96 21.49 1.86 all 144.85 26.31 3.99 On the first instance from the gap dataset, gap-urand , TwoA+G&P requires 81481.60 seconds to finish, 15.71-fold of what our new approach and 6.31-fold of what TwoA+Kowalik require to finish. On the gap-kron instance, both competitors are 1.38 ( TwoA+G&P ) resp. 1.34 ( TwoA+Kowalik ) times slower. On all five instances from genbank set, our new approach is about 1.9 times faster than each of the com-petitors. There are no significant deviations in this dataset. On the single instance from the law dataset we report that RapidPathOrienta-tion is 1.63 times faster than TwoA+Kowalik and 2.76 times faster than TwoA+G&P . The snap dataset contains social graphs and road networks. On the three road networks our ap-proach requires less than 1/10th of the run-ning time in comparison to the fastest compet-itor ( TwoA+Kowalik ). For two of the social graphs ( com-youtube ,as-skitter ) all approaches require nearly exactly the same time. The com-friendster instance is solved by TwoA+G&P about 1.13 times faster than our approach. On the com-livejournal the TwoA+Kowalik is faster by a factor of 1.08. The sybrandt set contains the two biggest instances in our dataset. For the moliere’16 TwoA+Kowalik is the fastest algorithm, our approach is 14% slower. RapidPathOrientation solves the agatha’15 instance 2.16 times faster than the fastest competitor TwoA+Kowalik . 6 Conclusion We have proposed a new framework for algorithms solving the edge orientation problem based on the ideas of Venkateswaran [ 29 ] and gave a new flow-based proof. We have investigated a vast variety of engineering techniques for the problem. Our techniques include a fast improvement heuristic, specialized depth-first-search, scheduling path searches more eagerly as well as running a data reduction based on a 2-approximation algorithm. Experiments have shown that with our final algorithm outperforms the fastest exact state-of-art algorithm by a factor of 6.59 on average. Especially on low density and low out-degree instances, like road networks and instances from numerical backgrounds, our algorithm outperforms its competitors. Only on low density, high out-degree instances the competitors have an advantage and compute an orientation slightly faster. 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A Additional Tables Table 2: Running times [s] comparing C++ with Java implementation of exact Kowalik [ 24 ] by Blumenstock [ 6] on the data set presented in . com-Amazon 5 2.47 1.81 com-Youtube 46 14.96 8.15 com-DBLP 57 1.70 1.15 com-LiveJournal 194 86.85 70.19 com-Orkut 228 382.50 303.49 d⋆ Java C++ Table 3: Average statistics for the sub data sets. # = number of instances in data set + number of instances in test set. flowipm22 5 2.9 × 10 7 3.0 × 10 7 1.0 8.0 × 10 2 2.00 gap 2 1.3 × 10 8 2.1 × 10 9 3.6 × 10 7 7.9 × 10 5 3.00 genbank 5 1.3 × 10 8 1.4 × 10 8 1.5 × 10 4 3.2 × 10 1 1.00 law [7, 8] 1 1.1 × 10 6 5.6 × 10 7 4.5 × 10 4 1.1 × 10 4 0.00 mawi 4 1.0 × 10 8 1.1 × 10 8 1.8 × 10 6 9.5 × 10 7 0.75 snap 8 1.0 × 10 7 2.5 × 10 8 3.3 × 10 3 1.5 × 10 4 0.62 sybrandt 2 1.1 × 10 8 4.6 × 10 9 1.3 × 10 4 7.4 × 10 6 0.50 dimacs10 31 9.4 × 10 6 2.3 × 10 7 8.2 × 10 3 4.3 × 10 3 1.61 test set 9 1.3 × 10 7 3.1 × 10 7 2.0 × 10 5 7.0 × 10 6 1.78 # n m components max d min d Table 4: Instance statistics for the 67 graphs used in experiments. σ Relative size of 2-approximation, ∆ maximum degree, δ minimum degree. Instances in the test set are bold . 333sp dimacs10 3.7 × 10 6 1.00 1.1 × 10 7 1.00 2.99 1.0 2.8 × 10 1 2n σn m σm m/n comp. ∆ δ Continued on next page REFERENCES 17 Table 4: Instance statistics for the 67 graphs used in experiments. σ Relative size of 2-approximation, ∆ maximum degree, δ minimum degree. Instances in the test set are bold . (Continued) adaptive dimacs10 6.8 × 10 6 1 1.4 × 10 7 1 2.00 1.0 4.0 2 agatha’15 sybrandt 1.8 × 10 8 0.03 5.8 × 10 9 0.10 31.50 1.3 × 10 1 1.3 × 10 7 1 as365 dimacs10 3.8 × 10 6 1.00 1.1 × 10 7 1.00 2.99 1.0 1.4 × 10 1 2 asia_osm dimacs10 1.2 × 10 7 0.75 1.3 × 10 7 0.76 1.06 1.0 9.0 1 channel-’b050 dimacs10 4.8 × 10 6 1.00 4.3 × 10 7 1.00 8.89 1.0 1.8 × 10 1 6 as-skitter snap 1.7 × 10 6 0.00 1.1 × 10 7 0.01 6.54 7.6 × 10 2 3.5 × 10 4 1 belgium_osm dimacs10 1.4 × 10 6 0.86 1.5 × 10 6 0.87 1.08 1.0 1.0 × 10 1 1 com-friendster snap 6.6 × 10 7 0.00 1.8 × 10 9 0.01 27.53 1.0 5.2 × 10 3 1 com-livejournal snap 4.0 × 10 6 0.00 3.5 × 10 7 0.01 8.67 1.0 1.5 × 10 4 1 com-orkut snap 3.1 × 10 6 0.01 1.2 × 10 8 0.05 38.14 1.0 3.3 × 10 4 1 com-youtube snap 1.1 × 10 6 0.00 3.0 × 10 6 0.03 2.63 1.0 2.9 × 10 4 1 delaunay_n20 dimacs10 1.0 × 10 6 1 3.1 × 10 6 1 3.00 1.0 2.3 × 10 1 3 delaunay_n21 dimacs10 2.1 × 10 6 1 6.3 × 10 6 1 3.00 1.0 2.3 × 10 1 3 delaunay_n22 dimacs10 4.2 × 10 6 1 1.3 × 10 7 1 3.00 1.0 2.3 × 10 1 3 delaunay_n23 dimacs10 8.4 × 10 6 1 2.5 × 10 7 1 3.00 1.0 2.8 × 10 1 3 delaunay_n24 dimacs10 1.7 × 10 7 1 5.0 × 10 7 1 3.00 1.0 2.6 × 10 1 3 europe_osm dimacs10 5.1 × 10 7 0.80 5.4 × 10 7 0.81 1.06 1.0 1.3 × 10 1 1 gap-kron gap 1.3 × 10 8 0.00 2.1 × 10 9 0.10 15.73 7.1 × 10 7 1.6 × 10 6 0 gap-urand gap 1.3 × 10 8 1.00 2.1 × 10 9 1.00 16.00 1.0 6.8 × 10 1 6 germany_osm dimacs10 1.2 × 10 7 0.81 1.2 × 10 7 0.82 1.07 1.0 1.3 × 10 1 1 g’b’_osm dimacs10 7.7 × 10 6 0.70 8.2 × 10 6 0.72 1.05 1.0 8.0 1 hollywood-2009 law 1.1 × 10 6 0.00 5.6 × 10 7 0.06 49.46 4.5 × 10 4 1.1 × 10 4 0 hugebubbles’00 dimacs10 1.8 × 10 7 1 2.7 × 10 7 1 1.50 1.0 3.0 2 hugebubbles’10 dimacs10 1.9 × 10 7 1 2.9 × 10 7 1 1.50 1.0 3.0 2 hugebubbles’20 dimacs10 2.1 × 10 7 1 3.2 × 10 7 1 1.50 1.0 3.0 2 hugetrace’00 dimacs10 4.6 × 10 6 1 6.9 × 10 6 1 1.50 1.0 3.0 2 hugetrace’10 dimacs10 1.2 × 10 7 1 1.8 × 10 7 1 1.50 1.0 3.0 2 hugetrace’20 dimacs10 1.6 × 10 7 1 2.4 × 10 7 1 1.50 1.0 3.0 2 hugetric’00 dimacs10 5.8 × 10 6 1 8.7 × 10 6 1 1.50 1.0 3.0 2 hugetric’10 dimacs10 6.6 × 10 6 1 9.9 × 10 6 1 1.50 1.0 3.0 2 hugetric’20 dimacs10 7.1 × 10 6 1 1.1 × 10 7 1 1.50 1.0 3.0 2 italy_osm dimacs10 6.7 × 10 6 0.80 7.0 × 10 6 0.81 1.05 1.0 9.0 1n σn m σm m/n comp. ∆ δ Continued on next page 18 REFERENCES Table 4: Instance statistics for the 67 graphs used in experiments. σ Relative size of 2-approximation, ∆ maximum degree, δ minimum degree. Instances in the test set are bold . (Continued) kmer_a2a genbank 1.7 × 10 8 0.00 1.8 × 10 8 0.00 1.06 5.4 × 10 3 4.0 × 10 1 1 kmer_p1a genbank 1.4 × 10 8 0.00 1.5 × 10 8 0.00 1.07 7.9 × 10 3 4.0 × 10 1 1 kmer_u1a genbank 6.8 × 10 7 0.00 6.9 × 10 7 0.00 1.02 4.4 × 10 4 3.5 × 10 1 1 kmer_v1r genbank 2.1 × 10 8 0.00 2.3 × 10 8 0.00 1.09 9.0 8.0 1 kmer_v2a genbank 5.5 × 10 7 0.00 5.9 × 10 7 0.00 1.06 1.5 × 10 4 3.9 × 10 1 1 kron_g500-’20 dimacs10 1.0 × 10 6 0.01 4.5 × 10 7 0.13 42.55 2.5 × 10 5 1.3 × 10 5 0 kron_g500-’21 dimacs10 2.1 × 10 6 0.00 9.1 × 10 7 0.10 43.41 5.5 × 10 5 2.1 × 10 5 0 mawi’000 mawi 3.6 × 10 7 0.00 3.7 × 10 7 0.00 1.03 7.5 × 10 5 3.2 × 10 7 1 mawi’030 mawi 6.9 × 10 7 0.00 7.2 × 10 7 0.00 1.04 1.3 × 10 6 6.3 × 10 7 1 mawi’130 mawi 1.3 × 10 8 0.00 1.4 × 10 8 0.00 1.05 2.2 × 10 6 1.2 × 10 8 1 mawi’330 mawi 2.3 × 10 8 0.00 2.4 × 10 8 0.00 1.06 4.0 × 10 6 2.1 × 10 8 0 mawi’345 mawi 1.9 × 10 7 0.00 1.9 × 10 7 0.00 1.02 4.4 × 10 5 1.6 × 10 7 1 moliere’16 sybrandt 3.0 × 10 7 0.00 3.3 × 10 9 0.00 110.41 2.5 × 10 4 2.1 × 10 6 0 m6 dimacs10 3.5 × 10 6 1 1.1 × 10 7 1 3.00 1.0 1.0 × 10 1 3 naca0015 dimacs10 1.0 × 10 6 1 3.1 × 10 6 1 3.00 1.0 1.0 × 10 1 3 n’l’_osm dimacs10 2.2 × 10 6 0.83 2.4 × 10 6 0.84 1.10 1.0 7.0 1 nlr dimacs10 4.2 × 10 6 1 1.2 × 10 7 1 3.00 1.0 2.0 × 10 1 3 packing-’b050 dimacs10 2.1 × 10 6 0.94 1.7 × 10 7 0.95 8.15 1.4 × 10 1 1.8 × 10 1 0 rgg_n_2_20_s0 dimacs10 1.0 × 10 6 0.02 6.9 × 10 6 0.02 6.57 3.0 3.6 × 10 1 0 rgg_n_2_21_s0 dimacs10 2.1 × 10 6 0.27 1.4 × 10 7 0.30 6.91 8.0 3.7 × 10 1 0 rgg_n_2_22_s0 dimacs10 4.2 × 10 6 0.14 3.0 × 10 7 0.15 7.24 5.0 3.6 × 10 1 0 rgg_n_2_23_s0 dimacs10 8.4 × 10 6 0.07 6.4 × 10 7 0.07 7.57 5.0 4.0 × 10 1 0 rgg_n_2_24_s0 dimacs10 1.7 × 10 7 0.03 1.3 × 10 8 0.04 7.90 2.0 4.0 × 10 1 0 road_central dimacs10 1.4 × 10 7 0.71 1.7 × 10 7 0.76 1.20 1.0 8.0 1 roadnet-ca snap 2.0 × 10 6 0.81 2.8 × 10 6 0.87 1.40 8.7 × 10 3 1.2 × 10 1 0 roadnet-pa snap 1.1 × 10 6 0.80 1.5 × 10 6 0.86 1.41 3.0 × 10 3 9.0 0 roadnet-tx snap 1.4 × 10 6 0.78 1.9 × 10 6 0.85 1.38 1.4 × 10 4 1.2 × 10 1 0 road_usa dimacs10 2.4 × 10 7 0.71 2.9 × 10 7 0.76 1.20 1.0 9.0 1 spielman_k200 flowipm22 2.7 × 10 6 1 2.7 × 10 6 1 1.01 1.0 4.0 × 10 2 2 spielman_k300 flowipm22 9.0 × 10 6 1 9.1 × 10 6 1 1.01 1.0 6.0 × 10 2 2 spielman_k400 flowipm22 2.1 × 10 7 1 2.1 × 10 7 1 1.00 1.0 8.0 × 10 2 2 spielman_k500 flowipm22 4.2 × 10 7 1 4.2 × 10 7 1 1.00 1.0 1.0 × 10 3 2n σn m σm m/n comp. ∆ δ Continued on next page REFERENCES 19 Table 4: Instance statistics for the 67 graphs used in experiments. σ Relative size of 2-approximation, ∆ maximum degree, δ minimum degree. Instances in the test set are bold . (Continued) spielman_k600 flowipm22 7.2 × 10 7 1 7.2 × 10 7 1 1.00 1.0 1.2 × 10 3 2 venturilevel3 dimacs10 4.0 × 10 6 0.83 8.1 × 10 6 0.83 2.00 1.0 6.0 2n σn m σm m/n comp. ∆ δ Table 5: Average running time per instance. Sorted by d⋆. 5 repetitions. adaptive 2 5508.26 8.80 1.40 asia_osm 2 56.45 14.28 1.16 belgium_osm 2 7.17 1.72 0.17 europe_osm 2 288.53 88.28 6.65 germany_osm 2 61.71 13.60 1.62 g’b’_osm 2 33.52 7.72 0.96 hugebubbles’00 2 70 001.31 31.00 4.27 hugebubbles’10 2 31 454.70 44.19 8.66 hugebubbles’20 2 27 227.08 46.97 8.49 hugetrace’00 2 2901.07 7.22 0.98 hugetrace’10 2 8391.65 22.87 3.50 hugetrace’20 2 7916.97 31.78 5.27 hugetric’00 2 3568.83 9.04 1.22 hugetric’10 2 4805.79 13.60 2.46 hugetric’20 2 7161.24 14.68 2.67 italy_osm 2 31.45 8.07 0.76 n’l’_osm 2 10.50 2.15 0.22 roadnet-ca 2 11.33 3.44 0.25 roadnet-pa 2 5.71 1.72 0.14 road_central 2 92.38 26.98 4.60 road_usa 2 145.56 34.94 3.56 spielman_k200 2 14.16 5.42 0.08 spielman_k300 2 51.19 16.52 0.38 spielman_k400 2 123.57 39.95 0.64 spielman_k500 2 250.26 79.07 1.24 spielman_k600 2 559.14 150.55 2.30 d⋆ TwoA+G&P TwoA+Kowalik RapidPathOrientation Continued on next page 20 REFERENCES Table 5: Average running time per instance. Sorted by d⋆. 5 repetitions. (Continued) 333sp 3 2408.29 95.47 3.19 as365 3 2820.13 148.65 3.90 delaunay_n20 3 72.27 19.73 0.47 delaunay_n21 3 168.78 48.29 1.01 delaunay_n22 3 413.96 123.88 2.20 delaunay_n23 3 950.79 322.33 4.82 delaunay_n24 3 2275.57 816.72 10.28 m6 3 1415.00 132.64 4.08 naca0015 3 613.64 24.18 0.86 kmer_v1r 3 132.97 133.62 58.33 nlr 3 1500.28 176.09 4.70 roadnet-tx 3 7.13 2.26 0.15 venturilevel3 3 25.09 3.44 0.59 kmer_u1a 5 40.80 40.71 21.67 kmer_v2a 8 32.63 33.10 18.30 channel-’b050 9 1031.67 10.47 0.66 packing-’b050 9 76.12 4.54 0.41 kmer_a2a 10 110.12 109.49 59.56 kmer_p1a 10 87.13 89.17 48.26 rgg_n_2_20_s0 12 0.88 0.77 0.25 rgg_n_2_21_s0 12 7.86 3.46 0.59 rgg_n_2_22_s0 13 9.89 6.09 1.11 rgg_n_2_23_s0 14 13.85 9.49 2.22 rgg_n_2_24_s0 14 22.65 18.12 5.24 gap-urand 17 81 481.60 12 910.44 5183.51 mawi’345 40 2.26 2.29 2.08 com-youtube 46 0.71 0.58 0.58 mawi’000 58 4.41 4.38 4.48 mawi’030 73 8.16 8.20 9.67 mawi’130 78 14.67 14.67 22.85 as-skitter 90 1.08 1.03 1.04 mawi’330 93 33.95 33.65 48.18 agatha’15 97 22 506.38 9267.07 4291.18 d⋆ TwoA+G&P TwoA+Kowalik RapidPathOrientation Continued on next page REFERENCES 21 Table 5: Average running time per instance. Sorted by d⋆. 5 repetitions. (Continued) com-livejournal 194 4.26 4.14 4.51 com-orkut 228 27.61 14.00 9.18 moliere’16 232 1267.39 1013.19 1155.05 com-friendster 274 917.42 1056.45 1039.20 kron_g500-’20 908 13.62 5.73 2.65 hollywood-2009 1 104 5.02 2.97 1.82 kron_g500-’21 1 178 21.94 10.85 5.78 gap-kron 2 369 1846.96 1790.70 1332.63 d⋆ TwoA+G&P TwoA+Kowalik RapidPathOrientation
1747	https://math.stackexchange.com/questions/5085211/maximal-size-of-a-set-with-restricted-subset-sum-divisibility-by-m-and-n	Skip to main content Maximal size of a set with restricted subset‐sum divisibility by m and n Ask Question Asked Modified 30 days ago Viewed 47 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Given integers m,n≥2, determine the largest positive integer k such that there exist k integers not divisible by m with the following property: for any nonempty subset of these numbers, if their sum is divisible by n, then their sum must also be divisible by mn. My attempt I first looked at the case gcd(m,n)=1. In that situation any sum divisible by n can never pick up an extra factor of m since none of the summands is a multiple of m. Thus the condition is vacuous and one can take arbitrarily many integers. This suggests that a finite bound on k only arises when d=gcd(m,n)>1. Assume d>1 and write n=dn′ and m=dm′. I tried to prove k≤d−1 by working modulo d. Given any d integers a1,…,ad none divisible by m, their residues modulo d lie in the set {1,2,…,d−1}. By the pigeonhole principle one can select a nonempty subset I⊂{1,…,d} with ∑i∈Iai≡0(modd). Since d∣n, this subset sum is divisible by n, but it is not divisible by m=dm′ because each ai is not 0(modm). This contradicts the required property, so no set of size d can work, giving k≤d−1. For the construction of d−1 integers, I chose one representative from each nonzero residue class modulo d. Concretely, pick aj=jn′ for j=1,2,…,d−1. Each aj is not divisible by m=dm′, and any subset sum divisible by n=dn′ must involve all d residue classes mod d to hit 0(modd), which is impossible with fewer than d elements. Hence no nonempty subset sum can be a multiple of n without already being a multiple of mn. This gives an example of size d−1. I feel this captures the right bound, but I’m not completely confident in the details: Why does any sum divisible by n require all d distinct residue classes mod d? I also tried applying zero‐sum theorems in cyclic groups (like Erdős–Ginzburg–Ziv), but those guarantee a zero‐sum of fixed length n rather than forcing divisibility by both n and m. Any pointers on formalizing the extremal construction or alternate combinatorial tools would be appreciated combinatorics number-theory modular-arithmetic divisibility Share CC BY-SA 4.0 Follow this question to receive notifications asked Jul 24 at 7:12 LegyenLegyen 8144 bronze badges 2 I can't follow your argument in the case gcd(m,n)=1. If (m,n)=(2,3), say...can you exhibit a large "good" k? – lulu Commented Jul 24 at 9:45 In my (2,3) case, look at your set (mod6). Every element of your set is odd, hence one of {1,3,5}(mod6). Now 3 is impossible, as any such element would be a counterexample. But if there were three elements that were 1(mod6), then their sum would be odd but divisible by 3. Same if there were 3 that were 5(mod6). Or have I misunderstood the question? – lulu Commented Jul 24 at 9:52 Add a comment \| 0 Reset to default You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics number-theory modular-arithmetic divisibility See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 3 Any subset of size 6 from the set S={1,2,4,…9} must contain two elements whose sum is 10. 3 Subsets and Divisibility 0 Proving that among any 2n−1 integers, there's always a subset of n which sum to a multiple of n Hot Network Questions How can I separate the 2 wheels in this model into individual objects? 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1748	https://www.weizmann.ac.il/math/klartag/sites/math.klartag/files/uploads/santalo.pdf	The Santal´ o point of a function, and a functional form of Santal´ o inequality S. Artstein, B. Klartag, V. Milman∗ Abstract Let L(f) denotes the Legendre transform of a function f : Rn →R. We generalize a theorem of K. Ball about even functions, and prove that for any measurable function f ≥0, there exists a translation ˜ f(x) = f(x −a) such that Z Rn e−˜ f Z Rn e−L( ˜ f) ≤(2π)n. (1) If we select a as to minimize the left hand side of (1), then equality in (1) is satisﬁed if and only if e−f is the distribution of a Gaussian random variable. This inequality immediately implies Santal´ o inequal-ity for convex bodies, as well as a new concentration inequality for the Gaussian measure. 1 Introduction In this paper we present some developments in the study of the ge-ometry of log-concave functions. This approach is further continued in [KM]. Log-concave functions have been investigated intensively by many authors, and we view this paper as a step in the “geometriza-tion” of log-concave functions. We ﬁnd that the intuition coming from the study of convex bodies enables one to formulate functional inequal-ities which turn out to be of independent interest. Also, the functional inequalities can sometimes be applied to functions related to a convex body, and give back strong inequalities for convex bodies. This scheme was pursued for example in [K]. ∗The ﬁrst and third authors were partially supported by a BSF grant. The second named author was supported by NSF grant DMS-0111298 and the Bell Companies Fel-lowship. 1 A measure µ on Rn is called log-concave if for any measurable A, B ⊂Rn and any parameter 0 < λ < 1, µ (λA + (1 −λ)B) ≥µ(A)λµ(B)1−λ, (2) where A + B = {a + b : a ∈A, b ∈B} is the Minkowski sum of A and B and λA = {λa : a ∈A} is the λ-homothety of A. The ﬁrst example of a log-concave measure is the standard Lebesgue measure V oln on Rn. The log-concavity of the Lebesgue measure fol-lows from the Brunn-Minkowski inequality (see (5) below). Similarly, a uniform measure on a convex body is log-concave. More examples for log-concave measures stem from Brunn’s concavity principle [Br]. This principle states that any lower dimensional marginal of a uniform mea-sure on a convex body is a log-concave measure. Moreover, marginals of uniform measures on convex bodies are essentially the only source for log-concave measures, as these marginals form a dense subset in the class of all log-concave measures. A function f : Rn →[0, ∞) is called log-concave if log f is concave on the support of f. The two notions are closely related, as was shown in [Bo]: a measure µ on Rn whose support is not contained in any aﬃne hyperplane is log-concave if and only if it is absolutely continuous with respect to the Lebesgue measure, and its density is a log-concave function. Therefore the standard Gaussian measure on Rn with density ( √ 2π)−n/2 exp{−\|x\|2/2}, is a log-concave measure, where \| · \| is the standard euclidean norm in Rn. One of the fundamental tools in convex geometry is that of duality. For a convex body K ⊂Rn (a compact convex set with the origin in its interior), its polar is deﬁned by K◦= {x ∈Rn : sup y∈K ⟨x, y⟩≤1}, where ⟨x, y⟩is the standard scalar product in Rn. Generalizing the theory of convex bodies to log-concave functions, one of the ﬁrst tasks is to understand what is the correct deﬁnition of the dual of a function. A hint may arise from the fact that log-concave functions are essentially marginals of convex bodies, and we may ﬁnd a way to induce the notion of duality from convex bodies to their marginals, the log-concave measures. This idea leads, as explained in Section 3, to the following deﬁnition. For a log-concave function f : Rn →[0, ∞) we deﬁne its dual (or polar) as f ◦(x) = inf y∈Rn h e−⟨x,y⟩/f(y) i . (3) To better understand this deﬁnition recall the classical Legendre trans-form (see e.g. [Ar]): For a function ϕ : Rn →R, its Legendre transform 2 is deﬁned by Lϕ(x) = sup y∈Rn [⟨x, y⟩−ϕ(y)] . The above deﬁnition of polarity is simply Legendre transform in the logarithm, as −log f ◦= L(−log f). Deﬁnition (3) makes sense for any non-negative function f, not necessarily log-concave. As the Leg-endre transform of any function is a convex function, f ◦is always log-concave. Also, if ϕ is convex and lower semi-continuous (see [R]) then LLϕ = ϕ. Translating to our language, if f is a log-concave upper semi continuous function, then f ◦◦= f. The semi continuity assump-tions are rather technical. Any log-concave function may be modiﬁed on a set of Lebesgue measure zero, to become upper semi continuous and log-concave. As an example, consider a norm ∥· ∥K on Rn with unit ball K. It is easy to check that for the function ϕ(x) = 1 2∥x∥2 K (which is convex) one has (Lϕ)(x) = 1 2∥x∥2 K◦; that is, Legendre transform has a clear relation to usual duality of convex bodies. Denote by ∥· ∥∗the dual norm. Then the functions exp{−1 2∥x∥2} and exp{−1 2 (∥x∥∗)2} are dual functions, in the sense of 3. In fact, for every pair 1 < p, q < ∞with 1 p + 1 q = 1 the relation exp −1 p∥x∥p ◦ = exp −1 q (∥x∥∗)q (4) holds true. Also, (1K)◦= exp{−∥x∥K◦}, where 1K denotes the indicator function of the convex body K, that is, a function which is 1 on K and 0 on its complement. Finally, it is not diﬃcult to see (e.g. Lemma 2.2) that the only self dual function is the Gaussian g(x) = exp{−1 2\|x\|2}, and therefore it will play the role which, in the theory of convex bodies, is played by the unique self dual body - the euclidean ball, which we denote by Dn. Another fundamental tool in convex geometry is Minkowski addi-tion. To deﬁne the right analogue for addition and multiplication by scalar of functions, we consider another known operation, which is re-lated to the Legendre transform, namely the Asplund product. Given two functions f, g : Rn →[0, ∞), their Asplund product is deﬁned by (f ⋆g)(x) = sup x1+x2=x f(x1)g(x2). It is easy to check that 1K ⋆1T = 1K+T and (f ⋆g)◦= f ◦g◦, so the dual of an Asplund product is the usual product of the dual functions. We argue that Asplund product of log-concave functions is the right analogue for Minkowski addition of convex bodies. 3 To deﬁne the λ-homothety of a function f(x), which we denote (λ · f)(x) we use (λ · f)(x) = f λ x λ . Note that for a log-concave f one has indeed f ⋆f = 2 · f, and that (λ · f)◦= (f ◦)λ. To check whether the deﬁnitions of Minkowski addition, of homo-thety and of duality for log-concave functions are meaningful and make sense, it is natural to ask whether the basic inequalities for convex bod-ies such as the Brunn-Minkowski inequality and the Santal´ o inequality remain true. The role of “volume” will be played of course by the integral. We discuss ﬁrst the Brunn-Minkowski inequality. In its dimension free form, Brunn-Minkowski inequality states that for any two bodies A, B ⊂Rn and for any 0 < λ < 1 one has V ol(λA + (1 −λ)B) ≥V ol(A)λV ol(B)1−λ. (5) We have ready made a functional analogue of the Brunn-Minkowski inequality, namely the Pr´ ekopa-Leindler inequality (see e.g. [P]). In the above notation it states precisely that Theorem 1.1 (Pr´ ekopa-Leindler) Given f, g : Rn →[0, ∞) and 0 < λ < 1, Z (λ · f) ⋆((1 −λ) · g) ≥ Z f λ Z g 1−λ . (notice that the multiplication is in homothety sense and not standard multiplication!) The standard Brunn-Minkowski inequality follows directly from Pr´ ekopa-Leindler by considering indicator functions of sets. Let us turn to Santal´ o inequality. Santal´ o inequality for convex bodies says that for a centrally-symmetric set K ⊂Rn (i.e. K = −K) with a ﬁnite positive volume, one has V ol(K)V ol(K◦) ≤(V ol(Dn))2. (6) Equality holds if and only if K is an ellipsoid. A non-centrally sym-metric version also holds true, although there one needs to choose the right center with respect to which the polarity is deﬁned, as there is no natural “zero” and in principal the polar body may be unbounded. More precisely, Santal´ o proved that (see [MeP2] for a simpler proof) Theorem 1.2 For any convex body K there exists a point x0 such that, denoting ˜ K = K −x0, one has Vol( ˜ K)Vol( ˜ K◦) ≤(Vol(Dn))2. (7) 4 It is possible to set x0 as the center of mass of K. The minimum over x0 of the left hand side is equal (Vol(Dn))2 if and only if K is an ellipsoid. The point x0 for which the minimal product in (7) is attained, is called the Santal´ o point of the body K. This point is zero if and only if the barycenter of the dual body lies at the origin. The corresponding inequality for log-concave functions can be stated as follows Theorem 1.3 Let f : Rn →[0, ∞) be any function such that 0 < R f < ∞. Then, for some vector x0, deﬁning ˜ f(x) = f(x −x0), one has Z Rn ˜ f Z Rn ˜ f ◦≤(2π)n. (8) If f is log-concave, we may choose x0 = R xf(x) R f(x) , the center of mass of f. The minimum over x0 of the left hand side product equals (2π)n if and only if f is a gaussian. Note that (8) is precisely inequality (1) stated in the abstract. The Santal´ o point of a function f is the point x0 that minimizes R Rn (f(x −x0))◦dx. As in the case of convex bodies, this point is zero if and only if the barycenter of f ◦lies at the origin. The main part of this paper is devoted to the proof of Theorem 1.3. It has recently come to our attention that a statement equivalent to inequality (8), for the case of an even function f, appeared in the Ph.D. thesis of K. Ball [Ba1]. We reproduce his proof in Section 2 below for the convenience of the reader. As one would expect, Santal´ o inequality for convex bodies follows easily from Theorem 1.3. Indeed, a standard computation gives that (also in the case of non-symmetric K, which includes the origin, in which case ∥x∥K is deﬁned by inf{r : x ∈rK}) Z Rn e− ∥x∥2 K 2 dx = cnVol(K), (9) where cn = (2π)n/2/Vol(Dn) (see e.g. [P] p. 11). Using together (9), (8), and (4) we get Santal´ o inequality (7). The rest of the paper is organized as follows. In Section 2 we present the argument which appeared in [Ba1]. In Section 3 we elaborate on the deﬁnition of the polar function, and in Section 4 we prove inequality (8). The equality case in Theorem 1.3 is settled in Section 5, and in Section 6 we present applications to Gaussian concentration. We thank G. Schechtmann for pointing to us the connection of our inequality with the paper of B. Maurey [Mau]. 5 2 The even case Theorem 2.1 (K. Ball, [Ba1]) Let ϕ : Rn →[0, ∞) be an even convex function. Assume that 0 < R e−ϕ < ∞. Then Z e−ϕ Z e−Lϕ ≤(2π)n. (10) Proof: Note that for any x, y ∈Rn, ϕ(x) + Lϕ(y) ≥⟨x, y⟩. (11) Abbreviate [ϕ < t] = {x ∈Rn : ϕ(x) < t}. If Lϕ(x) < s, then for any y ∈[ϕ < t] by (11) we have ⟨x, y⟩< s + t. Hence for any s, t ∈R, [Lϕ < s] ⊂(s + t)[ϕ < t]◦. Since ϕ is even, the set [ϕ < t] is centrally-symmetric, and by Santal´ o inequality (6), V ol([Lϕ < s]) ≤(s + t)nV ol([ϕ < t]◦) ≤(s + t)n V ol(Dn)2 V ol([ϕ < t]). Denote f(t) = V ol([ϕ < t]) and g(s) = V ol([Lϕ < s]). Then for any s, t ∈R, e−sg(s)e−tf(t) ≤e−(s+t)(s + t)nV ol(Dn)2. In our notation, if F(t) = e−tf (t), G(s) = e−sg(s) and H(u) = V ol(Dn)e−u(2u) n 2 , then ( 1 2 · F) ⋆( 1 2 · G) ≤H. The one dimensional Pr´ ekopa-Leindler inequality (Theorem 1.1 here) implies that Z R e−tV ol([ϕ < t]) Z R e−sV ol([Lϕ < s]) = Z F Z G ≤ Z H 2 . A straightforward calculation (e.g. [P], page 11) yields that R H 2 = (2π)n. Since R Rn e−ψ = R R e−tV ol([ψ < t])dt for any function ψ, the proof is complete. □ We are not aware of a straightforward generalization of Ball’s ar-gument to the non-even case. The diﬃculty lies in the fact that if f is not even, the one has to apply diﬀerent translations to [ϕ < t] for diﬀerent values of t. It will be useful for us later on to note that in the one-dimensional case, Ball’s proof demonstrates that gaussians are the only even functions satisfying equality in (10). For a convex even ϕ : R →R, we say that it is a maximizer if R e−ϕ R e−Lϕ = 2π. Lemma 2.2 Assume that ϕ is a maximizer. Then there exist a > 0 and b ∈R such that ϕ(t) = b + at2. 6 Proof: Note ﬁrst that if ϕ is a maximizer then for any b ∈R and a > 0 also the function t 7→ϕ(ta)+b is a maximizer. Indeed, L (ϕ(ta) + b) = Lϕ t a −b and hence Z e−(ϕ(ta)+b) Z e−L(ϕ(ta)+b) = Z e−ϕ Z e−Lϕ = 2π. Choosing the correct a and b, we may assume that R e−ϕ = R e−Lϕ and that ϕ(0) = Lϕ(0) = 0. In order for equality in (10) to hold, the functions F, G and H from the above proof must satisfy the equality conditions in Pr´ ekopa´ a-Leindler inequality. These appear, e.g., in [Bar] and state that there exists x ∈R such that F R F = G(·−x) R G . According to our assumptions R F = R G, hence F is a translation of G. Also since ϕ and Lϕ are convex even functions, their minimum is ϕ(0) = Lϕ(0) = 0. Therefore F(t), G(t) = 0 for t < 0 and F(t), G(t) > 0 for t > 0. Since F is a translation of G, necessarily F(t) = G(t) and V ol([ϕ < t]) = V ol([Lϕ < t]) for any t ∈R. Since the functions are one-dimensional and even, we conclude that ϕ = Lϕ. By the deﬁnition of the Legendre transform, for any s, t ∈R, ϕ(s) + ϕ(t) ≥st and in particular, ϕ(t) ≥t2 2 and hence Lϕ(t) ≤t2 2 . Since ϕ = Lϕ, we conclude that ϕ(t) = t2 2 . □ Remark: We want to remark that another proof of K. Ball, for Santal´ o inequality for convex bodies (see [Ba1],[MeP1]) can be directly gener-alized to our setting. The proof uses Steiner symmetrizations, and we can deﬁne the Steiner symmetrization of a function: For a function f(x) deﬁned on Rn and a hyperplane H = e⊥in Rn we denote, for y ∈H and t ∈R, f(y,+)(t) = f(y + te), f(y,−)(t) = f(y −te) and deﬁne its Steiner symmetrization by (SHf)(y + te) = 1 2 · f(y,+) ⋆ 1 2 · f(y,−) (t). It can be shown that for an even function f the product s(f) = R f R f ◦ grows when a Steiner symmetrization is performed, and this eventually reduces the question to a one dimensional one. 3 Duality in the space of log-concave func-tions A function f : Rn →[0, ∞) is called s-concave if f 1 s is concave on the support of f. (Note the perhaps non-standard deﬁnition, with 7 1 s replaced by s). Any s-concave function, for s > 0, is also log-concave. It is also possible to approximate any log-concave function f : Rn →[0, ∞) with an s-concave function fs as follows: fs(x) = 1 + log f(x) s s + (12) where x+ = max{x, 0}. The log-concavity of f implies the s-concavity of fs. Note also that fs ≤f for any s > 0, and since a log-concave func-tion is continuous on its support one has fs s→∞ − →f locally uniformly on Rn. Let s > 0 be an integer. By the Brunn concavity principle a func-tion on Rn is s-concave if and only if it is a marginal of a uniform measure on a convex body in Rn+s. One is thus tempted to deﬁne the polar of an s-concave function f, as the marginal of K◦where K ⊂Rn+s is some convex body whose marginal is f. This approach is problematic, as the body K whose marginal is f is not unique. How-ever, having this idea in mind, we deﬁne Lsf(x) = inf {y:f(y)>0} 1 −⟨x,y⟩ s s + f(y) . Clearly Lsf ≤f ◦. Deﬁne Ks(f) = {(x, y) ∈Rn × Rs : √sx ∈suppf, \|y\| ≤f 1/s(√sx))}. Up to some rescaling, the function f is the marginal on Rn of the uniform measure on the body Ks(f). In particular we have, denoting the volume of the unit euclidean ball in Rs by κs = V ol(Ds) = πs/2 Γ( s 2 +1), that V ol(Ks(f)) = Z Rn κsf √sx dx = κs sn/2 Z Rn f. The following Lemma clariﬁes the relations between our various deﬁ-nitions. Lemma 3.1 For any f : Rn →[0, ∞), (Ks(f))◦= Ks(Ls(f)). Proof: Let (x, y) ∈Rn × Rs. Then (x, y) ∈(Ks(f))◦if and only if for any (u, v) ∈Rn × Rs such that √su ∈suppf and \|v\| ≤f 1/s(√su) one has ⟨x, u⟩+ ⟨y, v⟩≤1. 8 This is equivalent to the fact that for any u′ ∈suppf one has ⟨x, u′⟩/√s+ \|y\|f 1/s(u′) ≤1. We conclude that (x, y) ∈(Ks(f))◦if and only if \|y\|s ≤ inf u′∈suppf 1 −⟨x,u′⟩ √s s f(u′) . Whenever the inﬁmum is non-negative, it equals Lsf(√sx), and it is non-negative if and only if √sx ∈suppLsf. This completes the proof. □ If f is upper semi continuous, s-concave, and f(0) > 0, then Ks(f) is closed, convex and contains the origin. Hence Ks(f)◦◦= Ks(f). By Lemma 3.1 we conclude that LsLsf = f as Ks(f) determines f and Ks(f) = (Ks(f)◦)◦= Ks(Ls(f))◦= Ks(Ls(Ls(f))). However, for log-concave functions which are not s-concave, the trans-form Ls is less natural. For a ﬁxed s > 0, the class of s-concave functions is rather restricted among the log-concave functions. This class becomes larger and larger as s approaches inﬁnity, and the in-creasing union of all these classes is a dense subset in the space of log-concave functions. In view of this, it is natural that the notion of duality for log-concave functions we deﬁned earlier, is the limit of Ls when s →∞. The proofs of the following lemmas are given in the Appendix. Lemma 3.2 Let f, f1, f2, ... : Rn →[0, ∞) be log-concave functions such that fn →f on a dense subset A ⊂Rn. Then, 1. R fn → R f. 2. If R f < ∞then R xfn → R xf. 3. f ◦ n →f ◦locally uniformly on the interior of the support of f ◦. Lemma 3.3 Let f : Rn →[0, ∞) be a log-concave function. Assume that xs ∈Rn and xs s→∞ − →0. Then for e fs(x) = fs(x −xs), lim inf s→∞ Z Rn Ls e fs ≥ Z Rn f ◦. where fs is the s-concave function associated to f via (12). Remark: A variant of our duality notions appeared many years ago in [Mi1] and [Mi2]. Let f : Rn−1 →R be a positive convex function. Let (y, t) be coordinates in Rn = Rn−1 × R. Then for any µ > 0, there 9 exists a unique norm ∥· ∥on Rn such that y, 1 √µ = 1+µf(y) √µ for y ∈Rn−1. Then, x, 1 √µ ∗ = 1+µTµf(x) √µ where (Tµf) (x) = sup y∈Rn−1 ⟨x, y⟩−f(y) 1 + µf(y) . Note that limµ→0 Tµ = L. 4 Proof of functional Santal´ o We are now ready for the proof of Theorem 1.3. The ﬁrst Corollary is a direct consequence of the Santal´ o inequality for convex bodies and the considerations of the previous section. Corollary 4.1 Let f be an s-concave function on Rn, with 0 < R f < ∞, and whose center of mass is at the origin (i.e. R xf(x) = 0). Then Z Rn f Z Rn Ls(f) ≤snκ2 n+s κ2 s with equality if and only if f is a marginal of the uniform distribution of an (n + s) dimensional ellipsoid. Proof: Noticing that the center of mass of the convex body Ks(f) remains at the origin, by Theorem 1.2 Z f Z Ls(f) = sn κ2 s Vol(Ks(f))Vol((Ks(f))◦) ≤snκ2 n+s κ2 s . Since 0 is the Santal´ o point of Ks(f)◦, equality holds if and only if Ks(f)◦is an ellipsoid, i.e., if Ks(f) is an ellipsoid and f is a marginal of the uniform distribution of an ellipsoid. □ Using the convergence of Lsf to f ◦we can infer the functional Santal´ o inequality. Proof of Theorem 1.3: Since for any function f ◦◦≥f, we may assume to begin with that f : Rn →[0, ∞) is log-concave. Translating f if necessary, we also assume that the barycenter of f is at the origin. Denote by xs the barycenter of fs, and denote e fs = fs(x −xs). By Lemma 3.2, xs →0 as s →∞. By Lemma 3.2, Lemma 3.3 and Corollary 4.1, Z f Z f ◦≤lim inf s→∞ Z e fs Z Ls e fs ≤lim s→∞ snκ2 n+s κ2 s = (2π)n. □ 10 Let f : Rn →[0, ∞) be a log-concave function with 0 < R f < ∞. Denote F(z) = R Rn (f(x −z))◦dx, and assume that R f ◦= minz∈Rn F(z). Since (f(x −z))◦= e−⟨x,z⟩f ◦(x), then F(z) = R e−⟨x,z⟩f ◦(x)dx. We conclude that 0 = ∇F(z)\|z=0 = − Z Rn xf ◦(x)dx. The diﬀerentiation under the integral sign is allowed as the derivative is locally bounded and integrals converge (see the proof of Lemma 3.3 below). Hence, if the Santal´ o point of a function lies at the origin, so is the center of gravity of the dual function. 5 The equality case Let f : Rn →[0, ∞) be a log-concave function, such that R f ◦< ∞ and R xf ◦(x)dx = 0. By Theorem 1.3, R f R f ◦≤(2π)n. We say that f is a maximizer if Z Rn f Z Rn f ◦= (2π)n. In this section we prove the following: Proposition 5.1 The function f is a maximizer if and only if it is a gaussian function, i.e. f(x) = ce−⟨Ax,x⟩for a positive-deﬁnite A and some number c > 0. Let fn : Rn →[0, ∞) and gm : Rm →[0, ∞) be maximizers. Then the function h : Rn × Rm →[0, ∞) deﬁned by h(x, y) = fn(x)gm(y) is also a maximizer. Indeed, h is log-concave, h◦(x, y) = f ◦ n(x)g◦ m(y) and hence the barycenter of h◦is at the origin. Also, we have Z Rn+m h Z Rn+m h◦= Z Rn fn Z Rn f ◦ n Z Rm gm Z Rm g◦ m = (2π)n+m. Next we show that the set of maximizers is also closed under the oper-ation of taking a marginal. We start with a variant of a result due to Meyer and Pajor [MeP2]. Recall that the Steiner symmetrization of a body K with respect to a hyperplane H = v⊥is the body T such that for any x ∈H the segment T ∩[x + Rv] is centered at x and has the same length as K ∩[x + Rv] Lemma 5.2 Let v⊥= H ⊂Rn be a hyperplane (v ∈Sn−1). Let K ⊂Rn be a convex body. Then the Steiner symmetrization of K with respect to H = v⊥, which we denote by T ⊂Rn, satisﬁes min z V ol((T −z)◦) ≥min z V ol((K −z)◦). 11 Proof: We use the terminology and results of [MeP2]. For a ∈Rn denote, as in [MeP2], the set (K −a)◦+ a by Ka. Denote also Ht = {x : ⟨x, v⟩= t}, and choose any z ∈H such that z + Rv intersects the interior of K. We claim that there exists a scalar v(z) ∈R such that Hv(z) is a medial hyperplane for Kz+v(z)v, in the sense that it partitions the body into two parts of equal volume. Indeed, put f(t) = V ol(Kz+tv∩H+ t ) V ol(Kz+tv) , where H+ t = {x : ⟨x, v⟩≥t}. Since K is convex, K ∩(z + Rv) is a segment of the form [z + t1v, z + t2v] for t1 < t2. Then f(t) →0 as t →t1 and f(t) →1 as t →t2. By continuity, there exist v(z) ∈R such that f(v(z)) = 1/2. Let T be the Steiner symmetrization of K with respect to H. By Lemma 7 from [MeP2], for any z ∈H, V ol(T z) ≥V ol(Kz+v(z)v) (13) (notice that symmetrizing with respect to parallel hyperplanes results in translates of the same body). Let z0 ∈Rn be such that V ol(T z0) = minx V ol(T x). Since T is invariant under reﬂections with respect to v necessarily z0 ∈H (see Lemma 2 in [MeP2]). We see that min x V ol (T x) = V ol(T z0) ≥V ol(Kz0+v(z0)v) ≥min x V ol(Kx). □ Remark: The lemma implies that the expression V ol(K)minxV ol(Kx) increases when a Steiner symmetrization is applied. Also, the Santal´ o point of a Steiner symmetrization of K with respect to H, is the orthog-onal projection onto H of the Santal´ o point of K. By taking the limit of an appropriate sequence of Steiner symmetrizations, all with respect to vectors in E⊥, we conclude that the expression V ol(K)minxV ol(Kx) increases also when a Schwartz symmetrization is applied. Recall that the Schwartz symmetral of K with respect to E⊥is the body T such that for any x ∈E, the body T ∩(x + E⊥) is a euclidean ball centered at x and V ol(T ∩(x + E⊥)) = V ol(K ∩(x + E⊥)). Lemma 5.3 Let f : Rn →[0, ∞) be a maximizer. Let m < n and let E ⊂Rn be an m-dimensional subspace. Deﬁne g : E →[0, ∞) by g(x) = Z x+E⊥f. Then g is a maximizer. Proof: Let xs be the Santal´ o point of fs (with respect to Ls-duality) and denote ˜ fs(x) = f(x −xs). By Corollary 4.1, for any s > 0, Z ˜ fs Z Ls ˜ fs ≤(2π)n. 12 We claim that it is impossible that \|xs\| →∞. Indeed, otherwise since fs ≤f we have ˜ fs →0 locally uniformly and by Lemma 3.3 we have lim inf R Ls ˜ fs = ∞. Hence we may choose a converging subsequence, xsk →x0. Without loss of generality, we assume that xs →x0. Denote ˜ f(x) = f(x −x0). Since f is a maximizer, lim inf s→∞ Z ˜ fs Z Ls ˜ fs ≥ Z ˜ f Z ˜ f ◦≥ Z f Z f ◦= (2π)n. Hence lim inf s→∞ sn κ2 s V ol(Ks(fs))V ol((Ks(fs) −xs)◦) ≥(2π)n. This expression only becomes larger when xs is replaced by any other point. Let s0 be such that for s > s0, V ol(Ks(fs)) min z V ol((Ks(fs) −z)◦) > [(2π)n −ε] κ2 s sn . Let gs : E →[0, ∞) be the marginal of the uniform measure on Ks(fs). Then gs is s + n −m-concave. The body Ks+n−m(gs) is the Schwartz symmetral with respect to E⊥of the body Ks(fs). By Lemma 5.2 and the remark following it V ol(Ks+n−m(gs))V ol(Ks+n−m(gs)◦) ≥V ol(Ks(fs)) min z V ol((Ks(fs)−z)◦) which in turn exceeds [(2π)n −ε] κ2 s sn , and hence Z E gs Z E Ls+n−m(gs) > [(2π)n −ε] κ2 s sn (s + n −m)m κ2 s+n−m s→∞ − →(2π)m− ε (2π)n−m . Since ε > 0 was an arbitrary number, we conclude that lim inf s→∞ Z E gs Z E (gs)◦≥lim inf s→∞ Z E gs Z E Ls+n−m(gs) ≥(2π)m, as Ls+n−mgs ≤(gs)◦. Since gs →g pointwise, and all functions are log-concave, by Lemma 3.2 we conclude that R E g R E g◦≥(2π)m. It remains to prove that the Santal´ o point of g lies at the origin. Since fs →f, by Lemma 3.2 we have R xf ◦ s →0 and the Santal´ o point of fs tends to zero. Since the Santal´ o point of gs is an orthogonal projection of the Santal´ o point of fs, the lemma follows. □ Lemma 5.4 Let f : R →[0, ∞) be a log-concave function whose barycenter is at zero, and assume that R f R f ◦= 2π. Then f is a gaussian. 13 Proof: Deﬁne g(x) = f(−x). By the above f(x)g(y) is a 2-dimensional maximizer, and so is its marginal on the line l = {(x, x) ∈R2; x ∈R}, which is in turn equivalent to the convolution ˜ f(x) = [f ∗g] (x) = Z R f(y)f(y −x)dy. The function ˜ f is an even one dimensional maximizer, and by Lemma 2.2, ˜ f must be a gaussian. According to Cram´ er Theorem (e.g. Page 1 of [LO]), if the convolution of two densities is gaussian, then both of them are gaussian. We conclude that f is gaussian. □ Proof of Proposition 5.1: By Lemma 5.3, all one-dimensional marginals of f are maximizers, and hence by Lemma 5.4 all of the one dimensional marginals of f are gaussians. A classical fact is that if all marginals of a function are gaussians, then the function itself is a gaussian. This completes the proof. □ 6 Applications In the remarkable paper [Mau], following a not less remarkable result by M. Talagrand [T], B. Maurey deﬁned the property (τ) (for Tala-grand) of a couple (µ, w), where µ is a probability measure on Rn and w a positive function on Rn. A couple (µ, w) is said to satisfy (τ) if for every bounded function ϕ on Rn Z e−ϕdµ Z einf{ϕ(x−y)+w(y):y∈Rn} ≤1 (where we agree that 0·∞≤1). The expression in the exponent of the second integral is called the inf convolution of ϕ and w, and denoted by ϕ□w. Note that −log (f ⋆g) = (−log f) □(−log g). Maurey shows, in particular, that the couple (γn, \|x\|2/4) satisﬁes property (τ), where γn denotes the n-dimensional gaussian density. As a corollary he gets the gaussian concentration inequality: for every 1-Lip. function ϕ on Rn, for two independent standard gaussian vectors X and Y , and any λ > 0 Ee λ √ 2 (ϕ(X)−ϕ(Y )) ≤eλ2/2. This inequality is optimal in the sense that if ϕ is a linear functional, it is in fact an equality. We will show below that for an even function ϕ a stronger concen-tration inequality holds true, namely the factor √ 2 on the left hand side can be omitted. Also, as a consequence of our Theorem 1.3 we 14 will get a similar concentration inequality for general functions, with-out √ 2, however with an additional linear functional inserted, which corresponds to the right choice of a Santal´ o point. We begin by deﬁning the corresponding property which we call (even τ). A couple (µ, w) is said to satisfy (even τ) if for every bounded even function ϕ on Rn Z e−ϕdµ Z eϕ□wdµ ≤1. Theorem 6.1 The pair (γn, \|x\|2/2) satisﬁes (even τ). Proof We need to show Z e−(\|x\|2/2+ϕ(x))dx Z e−(\|x\|2/2−ϕ□w) ≤(2π)n. Let f = e−(\|x\|2/2+ϕ(x)), and check that L(−log f) = \|x\|2/2 −inf y ( \|x −y\| 2 2 + ϕ(y) ) = \|x\|2/2 −(ϕ□w)(x). Thus the inequality we need is precisely R f R f ◦≤(2π)n for even functions, which follows from K. Ball’s Theorem 2.1. □ From Theorem 6.1 we can deduce a strong gaussian concentration inequality for even functions. Namely we prove Corollary 6.2 For every ϕ on Rn which is 1-Lip. and even, for X, Y independent normalized gaussian vectors in Rn, and any λ > 0, Eeλ(ϕ(X)−ϕ(Y )) ≤eλ2/2. Proof The proof is identical to the one in [Mau]. We repeat it for the convenience of the reader. Denote ψ = (λϕ)□w, where ϕ is 1-Lip. and even, and w(x) = \|x\|2/2. Then for some y ψ(x) = λϕ(y) + \|x −y\|2/2 ≥ λϕ(x) −λ\|x −y\| + \|x −y\|2/2 ≥ λϕ(x) −λ2/2. Hence Eeλ(ϕ(X)−ϕ(Y )) ≤ eλ2/2Eeψ(X)Ee−λϕ(Y ) = eλ2/2Ee(λϕ□w)(X)Ee−λϕ(Y ) ≤eλ2/2. This completes the proof. □ 15 We can similarly deduce theorems regarding general functions ϕ, but with an extra correcting linear factor, corresponding to the Santal´ o point. Namely we can prove Corollary 6.3 For every ϕ on Rn which is 1-Lip, for X, Y indepen-dent normalized gaussian vectors in Rn, and any λ > 0, there exists a linear functional φ0 such that Eeλ(ϕ(X)−ϕ(Y ))−φ0(X) ≤eλ2/2. This follows from the following Theorem 6.4 For every ϕ on Rn there exists an x0 ∈Rn such that Z e−ϕdγn Z e(ϕ□w)(x)−⟨x0,x⟩dγn(x) ≤1. The proofs are similar to the ones described above, and are omitted. Note that property (even τ) (as opposed to the more standard property (τ), say) does not seem to easily tensorize. This is the reason we need the full strength of the high dimensional functional version of Santal´ o inequality. 7 Appendix: Proofs of convergence theo-rems for log-concave functions Proof of Lemma 3.2: For the ﬁrst part of the lemma, ﬁrst recall that the convergence is locally uniform on the interior of the support of f (see e.g. Theorem 10.8 in [R]). Hence if R f = ∞, the lemma follows by restricting the integrals to large enough compact sets. Assume that R f < ∞. By log-concavity, sup\|x\|>R f(x) →0 as R →∞. Pick a point x0 with f(x0) > 0. We may assume without loss of generality that x0 = 0 and f(x0) > 1. Let R be such that \|x\| = R ⇒f(x) < 1 e. Then for n large enough, say n > n0, fn(0) > 1, \|x\| = R ⇒fn(x) < 1 e. (14) Notice that on {x : \|x\| > R} we have fn ≤e−\|x\| R . Indeed, 1 e > fn R x \|x\| ≥fn(0)1−R \|x\| fn(x) R \|x\| ≥fn(x) R \|x\| . Of course R \|x\|>R e−\|x\| R is ﬁnite, and by the dominated convergence the-orem, Z \|x\|>R fn → Z \|x\|>R f. 16 In the compact domain {x; \|x\| ≤R} the convergence is uniform, and the ﬁrst part of the lemma follows. Similarly, since also R \|x\|>R \|x\|(1/4)\|x\|/R < ∞, the dominated con-vergence theorem also implies the second part of the lemma. For the third assertion, denote gn = −log fn and g = −log f, convex functions. Then gn →g pointwise and locally uniformly on the interior of the support of g. It is enough to show that Lgn →Lg on a dense set. Let x be a smooth point of Lg. Then there exists a unique supporting hyperplane to Lg at x. Hence g(y) −⟨x, y⟩has a unique minimum in y, say y0. Fix ε > 0. By the uniform convergence, for large enough n, \|y −y0\| = ε = ⇒ gn(y) −⟨x, y⟩> gn(y0) −⟨x, y0⟩. Hence, the convex function gn(y) −⟨x, y⟩has a local minimum in the ball of radius ε around y0, and for a large enough n, Lgn(x) = sup \|y−y0\|<ε ⟨x, y⟩−gn(y) n→∞ − →Lg(x) as the convergence is uniform on the ball. Since for log-concave func-tions the set of smooth points is dense (see Theorem 25.5 in [R]), the third part of the lemma follows. □ Proof of Lemma 3.3: Fix y ∈Rn. Then for any y′ ∈Rn, log f(y′) ≤−⟨y, y′⟩−log f ◦(y) and hence, if −⟨y, xs⟩+ log f ◦(y) < s, e fs(y′) = fs(y′ −xs) ≤ 1 −⟨y, y′ −xs⟩+ log f ◦(y) s s + = 1 + ⟨y, xs⟩−log f ◦(y) s s " 1 −1 s y 1 + ⟨y,xs⟩−log f ◦(y) s , y′ +#s + . We conclude that Ls e fs y 1+ ⟨y,xs⟩−log f◦(y) s ≥ 1 + ⟨y,xs⟩−log f ◦(y) s −s ≥ e−⟨y,xs⟩f ◦(y). Denote gs = Ls e fs and ys = y 1+ ⟨y,xs⟩−log f◦(y) s . Then for any y in the support of f ◦, (since xs →0) ys s→∞ − →y, lim inf s→∞gs(ys) ≥f ◦(y). (15) Since gs are log-concave, by Theorem 10.9 in [R], there exists a subse-quence sn and a log-concave function g, such that gsn →g converges 17 locally uniformly on the interior of the support of g. By (15), and since the convergence is locally uniform, g ≥f ◦. Hence, by Lemma 3.2, lim inf n→∞ Z gsn = Z g ≥ Z f ◦. We conclude that any increasing sequence sn contains a subsequence snk such that lim inf k→∞ Z Lsnk e fsnk = lim inf k→∞ Z gsnk ≥ Z f ◦. This concludes the proof. □ References [Ar] V. I. Arnold, Mathematical methods of classical mechanics. Translated from the Russian by K. Vogtmann and A. Weinstein. Second edition. Graduate Texts in Mathematics, 60. Springer-Verlag, New York, 1989. [Ba1] K. Ball, PhD dissertation, Cambridge. [Ba2] K. Ball, Logarithmically concave functions and sections of con-vex sets in Rn. Studia Math. 88 (1988), no. 1, 69–84. [Bar] F. Barthe, On a reverse form of the Brascamp-Lieb inequality. Invent. Math. 134 (1998), no. 2, 335–361. [Bo] C. Borell, Convex set functions in d-space. Period. Math. Hungar. 6 (1975), no. 2, 111–136. [Br] H. Brunn, Referat ¨ Uber eine Arbeit: Exacte Grundlagen f¨ ur einer Theorie der Ovale, (S.B. Bayer, ed.) Akad. Wiss. (1894) 93–111. [K] B. Klartag, An isomorphic version of the slicing problem, to ap-pear to J. Funct. Anal. [KM] B. Klartag, V.D. Milman, Geometry of log-concave functions and measures, preprint. [LO] J. V. Linnik, I.V. Ostrovs’kiˇ ı, Decomposition of random variables and vectors. Translated from the Russian. Translations of Math-ematical Monographs, Vol. 48. American Mathematical Society, Providence, R. I. (1977). [Mau] B. Maurey, Some deviation inequalities. Geom. Funct. Anal. 1 (1991), no. 2, 188–197. [MeP1] M. Meyer, A. Pajor, On Santal´ os inequality. Geometric aspects of functional analysis (1987–88), 261–263, Lecture Notes in Math., 1376, Springer, Berlin, 1989. 18 [MeP2] M. Meyer, A. Pajor, On the Blaschke-Santal´ o inequality. Arch. Math., Vol 55, 82–93 (1990). [Mi1] V.D. Milman, A certain transformation of convex functions and a duality of the β- and δ-characteristics of a B-space, M.R. v.41, 797; Dokl Akad, Nauk SSSR, 33-35 (1969); Soviet Math. Dok. 10, 789-792 (1969). [Mi2] V.D. Milman, Duality of some geometric characteristics of Ba-nach spaces, Teor. Funksii, Funksional Analiz i Prilozhen, 18, 120-137 (1973). [P] G. Pisier, The volume of convex bodies and Banach space geome-try. Cambridge Tracts in Mathematics, 94. Cambridge University Press, Cambridge, 1989. [R] R. Rockafellar, Tyrrell Convex analysis. Princeton Mathematical Series, No. 28 Princeton University Press, Princeton, N.J. 1970. [Sa] L. A. Santal´ o, An aﬃne invariant for convex bodies of n-dimensional space. (Spanish) Portugaliae Math. 8, (1949). 155– 161. [T] M. Talagrand, A new isoperimetric inequality and the concen-tration of measure phenomenon, Geom. Aspects of Funct. Anal. (1989–90), Lecture Notes in Math., 1469, Springer, Berlin, (1991) 94–124. Shiri Artstein, Department of Mathematics, Princeton University, Prince-ton NJ 08544-1000 USA, and School of Mathematics, Institute for Ad-vanced Study, 1 Einstein Drive, Princeton NJ, 08450 USA. Email address: artstein@math.princeton.edu and artst@math.ias.edu. Boaz Klartag, School of Mathematics, Institute for Advanced Study, 1 Einstein Drive, Princeton NJ, 08450 USA. Email address: klartag@math.ias.edu. Vitali Milman, School of Mathematical Science, Tel Aviv University, Ramat Aviv, Tel Aviv, 69978, Israel. Email address: milman@post.tau.ac.il 19
1749	https://mathworld.wolfram.com/PrimeCircle.html	TOPICS Prime Circle A prime circle of order is a free circular permutation of the numbers from 1 to with adjacent pairs summing to a prime. The number of prime circles for , 2, ..., are 1, 1, 1, 2, 48, 512, ... (OEIS A051252). The prime circles for the first few even orders are given in the table below. \| \| \| prime circles \| \| 2 \| \| 4 \| \| 6 \| \| 8 \| , \| See also Circular Permutation Explore with Wolfram\|Alpha More things to try: zig number (3+(1-sqrt(2))/(1+sqrt(2)))^(1/3) divisors 3600 References Filz, A. "Problem 1046." J. Recr. Math. 14, 64, 1982.Filz, A. "Problem 1046." J. Recr. Math. 15, 71, 1983.Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 105-106, 1994.Sloane, N. J. A. Sequence A051252 in "The On-Line Encyclopedia of Integer Sequences." Referenced on Wolfram\|Alpha Prime Circle Cite this as: Weisstein, Eric W. "Prime Circle." From MathWorld--A Wolfram Resource. Subject classifications
1750	https://www.themathdoctors.org/where-do-logarithms-come-from/	Skip to content Where Do Logarithms Come From? April 26, 2024 April 25, 2024 / Algebra, Calculus / History, Logarithms / By Dave Peterson Having answered many questions recently about logarithms, I realized we haven’t yet covered the basics of that topic. Here we’ll introduce the concept by way of its history, and subsequently we’ll explore how they work. What does that button do? We’ll start with this 1998 question: 'Log' Button In math we have been using the 'log' button on our calculators to solve problems that involve compound interest, etc., and our teacher says we'll learn what the 'log' button does next year. But we want to know now! The teacher says it has something to do with the opposite of exponents, but I still have no clue. I sometimes struggle with the order in which to explain things, because when you need to use a new concept, sometimes the students aren’t ready for the concept on which it is based, so they can’t fully understand it (and perhaps don’t really need to). So we may have to tell a student, just use it for now. But I’m with the student: Some sort of explanation would be nice. Doctor Sam answered Brad in that spirit: Well, logarithms can be a big topic and if you are interested you might try reading about them in another math book. But for a quick introduction, here goes: First, the word logarithm is really a synonym for the word exponent or power. I tell students, “the logarithm is the exponent”; what I mean by that is that when we see that, say, (10^4=10,000), we can say that 4 (the exponent) is the logarithm of 10,000. And that’s what the teacher means by “the opposite of exponents”: When we find a logarithm, we are finding what exponent is needed to get that number. We’re working backward. Second, there are lots of different kinds of logarithms depending upon the base you are using. The LOG key on your calculator is probably "base 10 logarithm". Try this: make a table of values of different powers of 10, like this: power x: -2 -1 0 1 2 3 ... 10^x: 0.01 0.1 1 10 100 1000 ... Ordinarily we read these as "10 to the power 2 equals 100" or "10 to the power 3 equals 1000." This is fine when you need to calculate a power of ten, but what if you know the answer and need to find the exponent? To solve 10^x = 10000, for example, you have to think "ten to what power is ten thousand?" That's not too hard. But what if the problem is 10^x = 25? So far we just have a name (“logarithm”, or “log” for short) for what we want to do. The table gives us a method, but only in the easiest cases. LOGARITHMS were invented (back in the sixteenth century I think) to answer these kinds of questions. LOG(25) means "the power of 10 that produces 25" and LOG(1000) = 3 because "the power of 10 that produces 1000 is 3." You can make a little table of logarithms just by switching the two rows of my table above: 10^x: 0.01 0.1 1 10 100 1000 ... power x: -2 -1 0 1 2 3 ... only now the second row is called the logarithm. So here it is one more time: n: 0.01 0.1 1 10 100 1000 ... LOG(n): -2 -1 0 1 2 3 ... The trouble is, when we want to fill in numbers in between, none of the arithmetic you’ve learned can do it; there is no simple formula for the log. (That’s why we need a name, and a calculator button for it!) What we’re doing here is finding the inverse of the exponential function: $$\text{If }y = 10^x\text{, then }x=\log(y)$$ Back in the sixteen hundreds (in fact even back in the dark ages when I went to school in the 1960's) we didn't have pocket calculators with LOG keys. Companies published books of tables, page after page of the powers of 10 that would give you almost any number you wanted. It's a lot easier to press a key! Here’s an example of such a table, from the book I used in college (CRC Standard Mathematical Tables): The table gives the “mantissa”, or decimal part, of logarithms, which is the log of a number from 1 to 10. For example, to find the logarithm of 1.1223 I find 112 in the left column and 2 on the top, leading me to 04 999; then I see that the next number to the right, 04 038, is an increase of 39, and the little table at the right tells me that 3/10 of 39 is 11.7. So I add (04999+11.7=05010.7), and move the decimal point to the start, (0.050107). A calculator shows it to be (0.0501089\dots). The table claims to be accurate to five places, so we’d round our answer to (0.05011), which agrees. Yes, calculators are easier … By the way, when we look up numbers in the table in reverse, finding the number whose log is a given value, they called this the “anti-log“, because they thought of the log as the main thing, what the table was made for — but in a modern view, it’s the exponential that’s primary and the log is its inverse. So what was called the anti-log is, in fact, just the exponential function. I do want to mention that the little table I used gives powers of 10, and so these are called base ten logarithms (or COMMON LOGARITHMS). But almost any number can be used as the base. If you are interested in solving problems with powers of 2 (175 = 2^x) it would be helpful to have a base two logarithm key on your calculator. You probably don't have one, but computer scientists who use powers of two a lot do. We’ll see more about these other bases soon. Common logarithms as a calculation tool Here’s a look at the history of logarithms, from 1996: Logarithms: History and Use I have been asked to explain logarithms from a non-numerical sense to non-math-oriented people. It doesn't seem to be enough for me to show the equation and how it works, they want to know why. Any thoughts? Also, do you have short anecdotal history for the development of the concept of logarithm? Finally, why is it called a "logarithm"? logos = reason, arithmos = number. Doctor Anthony answered the history aspect of Linda’s question, starting with two different purposes for logs: It is a very great economy of effort if we can reduce multiplication to the addition of two numbers. The possibility of adding numbers that can be looked up in tables compiled "forever," as Napier remarked, instead of carrying out a lengthy process of multiplication, was suggested in two ways that were quite independent. The first arose in connection with the preparation of trig. tables for use in navigation. The second was closely connected with the laborious calculation involved in reckoning compound interest on investments. Tables of trigonometric functions already existed, and it was found that they could be used as a calculating tool: In 1593 two Danish mathematicians suggested the use of trig. tables for shortening calculations. They used the formula: sin(A)cos(B) = (1/2)sin(A+B) + (1/2)sin(A-B) Thus to multiply 0.173650.99027, you look up in tables and find 0.17365 = sin(10), 0.99027 = cos(8) and the above formula gives sin(10)cos(8) = (1/2)(sin(18) + sin(2)) From tables sin(18) = 0.30902 sin(2) = 0.03490 sin(18) + sin(2) = 0.34392 and (1/2)(sin(18)+sin(2)) = 0.17196 Giving 0.173650.99027 = 0.17196 This formula is called the product-to-sum identity. Four table lookups and an addition took less work (and was less error-prone) than doing the multiplication by hand. This device probably suggested to Napier, who is usually called the inventor of logarithms, a simple method for multiplying by a process of addition. Napier had been working on his invention of logarithms for twenty years before he published his results, and this would place the origin of his ideas at about 1594. He had been thinking of the sequences which had been published now and then of successive powers of a given number. In such sequences it was obvious that sums and differences of indices of the powers corresponded to products and quotients of the powers themselves; but a sequence of integral powers of a base, such as 2, could not be used for computations because the large gaps between successive terms made interpolation too inaccurate. So to keep the terms of a geometric progression of INTEGRAL powers of a given number close together it was necessary to take as the given number something quite close to 1. Napier therefore chose to use 1 - 10^(-7) or 0.9999999 as his given number. To achieve a balance and to avoid decimals, Napier multiplied each power by 10^7. That is, if N = 10^7[1 - 1/10^7]^L, then L is Napier's logarithm of the number N. Thus his logarithm of 10^7 is 0. At first he called his power indices "artificial numbers", but later he made up the compound of the two Greek words Logos (ratio) and arithmos (number). For more details on the word, see here; and here (including how “reason” and “ratio” are related). A given difference in logarithms corresponds to a given ratio of numbers. Napier did not think of a base for his system, but nevertheless his tables were compiled through repeated multiplications, equivalent to powers of 0.9999999 Obviously the number decreases as the index or logarithm increases. This is to be expected because he was essentially using a base which is less than 1. A more striking difference between his logarithms and ours lies in the fact that his logarithm of a product or quotient was not equal to the sum or difference of the logarithms. If L1 = log(N1) and L2 = log(N2), then N1 = 10^7(1-1/10^7)^L1 and N2 = 10^7(1-1/10^7)^L2, so that N1N2/10^7 = 10^7(1-1/10^7)^(L1+L2), so that the sum of Napier's logarithms will be the logarithm not of N1N2 but of N1N2/10^7. Similar modifications hold, of course, for logarithms of quotients, powers and roots. These differences are not too significant, for they merely involve shifting a decimal point. So the first “logarithms” weren’t quite what we think of now. Napier's work was published in 1614 and was taken up enthusiastically by Henry Briggs, a professor of Geometry at Oxford. He visited Napier and discussed improvements and modifications to Napier's method of logarithms. Briggs proposed that powers of 10 should be used with log(1) = 0 and log(10) = 1. Napier was nearing the end of his life, and the task of making up the first table of common logarithms fell to Briggs. Instead of taking powers of a number close to 1, as had Napier, Briggs began with log(10) = 1 and then found other logarithms by taking successive roots. By finding sqrt(10) = 3.162277 for example, Briggs had log(3.162277) = 0.500000, and from 10^(3/4) = sqrt(31.62277) = 5.623413 he had log(5.623413) = 0.7500000. Continuing in this manner, he computed other common logarithms. Briggs published his tables of logarithms of numbers from 1 to 1000, each carried out to 14 places of decimals, in 1617. Briggs also introduced the words "mantissa" for the positive fractional part and "characteristic" for the integral part (positive or negative). These last two terms were important in using tables, but not so much today. The first tables of logarithms contained inaccuracies which were noticed and corrected from time to time. The labor expended in constructing them was enormous, and it stimulated the search for better methods of calculating them. This gave a new impetus to the study of infinite series, for example sqrt(2) = (1 - (1/2))^(-1/2) which gives rise to an infinite, convergent series when expanded according to the binomial theorem. This work culminated in the extremely important exponential series, where e = Limit {1 + 1/n}^n as n -> infinity. It is easy to show that e^x = Limit {1 + 1/n}^(nx) generates the series shown below: e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ... to infinity, and e = 1 + 1 + 1/2! + 1/3! + 1/4! + .... = 2.718281828... e is now used as the base of logarithms in almost all advanced work. Modern calculators use something related to this when you push the button. And this number e leads us to the natural logarithm: Natural logarithms and compound interest Doctor Anthony mentioned a second purpose for logarithms, related to compound interest. That comes up in this 1998 question: History and Applications of the Natural Logarithm We have been using e in continuous growth problems, where we are examining the growth potential of natural disease populations. I'm so surprised at how often this number comes up in other applications, though. Where did e come from? Who first derived it? Why is it so common in the field of biology? The question is about the number (e\approx2.71828\dots), which is the base of the natural logarithm, called (\log_e(x)) or (\ln(x)) (and often just written as (\log(x)) at higher levels, where it is the default). Doctor Rob answered: John Napier, the inventor of logarithms, is credited with discovering this constant. Leonhard Euler is credited with popularizing the use of the letter e for this number. There are two reasons for its frequent appearance in mathematical contexts. The first is that it is the limit, as n grows without bound, of (1 + 1/n)^n. This particular definition lends itself to problems involving continuous compounding of interest. Think of adding 1/n of the current total to that current total n times. This formula would represent compound interest at a rate of 100%, compounded n times a year. As we compound more often, so that our interest earns interest, we earn more; so that with continuous compounding, rather than doubling in a year, the investment would be multiplied by 2.718. In fact, with daily compounding, the amount after a year would be $$\left(1+\frac{1}{365}\right)^{365}=2.714567\dots$$ Hourly compounding would yield $$\left(1+\frac{1}{365\cdot24}\right)^{365\cdot24}=2.718126\dots$$ The result is that when interest on principal P is compounded continuously at an annual rate of r, the amount after t years is $$A=Pe^{rt}$$ The natural logarithm, the log with base e, is the inverse function, so to find the time at which a certain amount is obtained, we take the logarithm of the ratio (\frac{A}{P}): $$t=\frac{1}{r}\log_e\left(\frac{A}{P}\right)$$ The second is that e is the only constant such that the slope of the graph of y = e^x at 0 is 1; or, in other words, the function e^x is the solution of the differential equation dy/dx = y, with initial conditions y = 1 when x = 0. Similarly, the solution of dy/dx = ay, with initial conditions y = 1 when x = 0, is e^(ax). Here are the graphs of exponential functions with several different bases (2, e, 10), showing that the slope of (y=e^x) at (x=0) is 1: This is the reason that e, and natural logs, are ubiquitous in calculus and related higher mathematics. The logarithm is the inverse function, obtained by swapping the roles of x and y: This is the biological connection. You are talking about the instantaneous rate of growth of something being proportional to the amount of the something. That something can be a population of bacteria in a culture, or a population of cockroaches in a garbage dump. Here, "k" represents the fraction of them reproducing at any given time x. If "k" is negative, it could represent the fraction of organisms dying at any given time. Unrestricted growth follows an exponential function, derived from that differential equation, namely $$N=N_0e^{kx}$$ This is the model for radioactive decay of radium, uranium, and so on. This is also the approximate model for inflation -- exponential growth. Radioactive decay is exponential growth is the same, with a negative rate: $$N=N_0e^{-kx}$$ The way to calculate e is by using as many terms as necessary for the accuracy you need in the following infinite series: e = 1 + 1 + 1/2 + 1/(23) + 1/(234) + 1/(2345) + 1/(23456) + ... The terms shown above are already enough to obtain a value for e that is roughly 2.718. It’s worth pointing out that we can actually use any base we want to describe these growth and decay problems; we’ll see that in another post. But the natural exponential, and the natural logarithm, are what arise directly from the differential equation, and directly display the growth rate. Relationship between common and natural logs We’ll close with a 2005 question to tie things together: Common Logarithms and Natural Logarithms I am currently studying logarithms and I saw that logarithms can take the form of ln or log. What is the difference between the two? I think it's very confusing because I looked it up in my math book and they state log e x = ln x. Then they state log 10 x = log x. I am confused about how e and 10 work with ln and log. I think that if I see a problem such as y = ln(x-1) and the book asks to find the inverse, I have to change it to log, but I don't know how to do that. I simply don't understand ln and log's relationship. Doctor Tom answered, starting with the common logarithm: Hi Katie, You're confused because it really is confusing! When logarithms are first introduced, it is much easier for students to think about logarithms in base 10. In base 10, log(.01) = -2 log(.1) = -1 log(1) = 0 log(10) = 1 log(100) = 2 log(1000) = 3 an so on--it just sort of counts the zeros, or, more accurately, the log is the power of 10 that creates the number you are taking the log of. Since 100 is 10^2, the log of 100 is 2. Using common logarithms, we only need a table for values between 1 and 10, because $$\log(a\cdot10^n)=\log(a)+\log(10^n)=\log(a)+n,$$ so we can separately find the “characteristic” (the whole number part) and the “mantissa” (the fractional part) of the logarithm. For example, $$\log(112.23)=\log(1.1223\cdot10^2)=\log(1.1223)+\log(10^2)=\log(1.1223)+2\approx0.05011+2=2.05011$$ Engineers also tend to use log base 10 for most calculations for the same reason. You can just look at the size and know the magnitude of the number. if the log (base 10) is 5.46, without even thinking, you know the number is "between" 10^5 and 10^6, so it's between 100,000 and 1,000,000. So in introductory texts and in engineering books, when you see "log", it usually means "log base 10". If they DO want to talk about other bases, they either put a little subscript after the "log", like "log_e". (I can't draw a subscript, so I'm using the "_" to mean that the next character should be smaller and written as a subscript). So (\log) is just a nickname for the common log, (\log_{10}); and (\log_{e}) is the natural log, nicknamed (\ln). Since log base e is often important for engineers, particularly electrical engineers, they often use "ln" instead of "log_e" since it's quicker to write, and it's a mnemonic for "logarithm, natural", or "natural logarithm". Now for mathematicians, the "natural log" really IS much more natural, so since that's the ONLY type of logarithm they use, they often just write "log" instead of "ln". I know this seems really confusing, but if you know what you're doing, you can almost always tell that it's a natural log just by looking at the equation. Of course if there is a chance of confusion, everybody always writes them like "log_10" or "log_e" to make it obvious what's going on. So “log” means the default in whatever field you are in. (This can cause confusion if you look around online, because you may stumble into a world other than that of your class, in which, say, “log” is used where your teacher would use “ln”. Be careful! Since you are just starting to learn about logarithms, you will always see "log" as meaning "log_10". Computer scientists often have a very good use for "log base 2", or "log_2", and it comes up so often there that they often write "lg" instead of "log_2". I've never seen them shorten "lg" or "log_2" to just "log" like the mathematicians, however, so you're never in any danger there. In fact, it is not uncommon in computer science today for “log” to mean base 2. Wikipedia says, Many disciplines write log x as an abbreviation for logbx when the intended base can be inferred based on the context or discipline (or when the base is indeterminate or immaterial). In computer science, log usually refers to log2, and in mathematics log usually refers to loge . In other contexts, log often means log10. When you solve a problem, check how the symbol is used in your source. We’ll see more about logarithms next time. 3 thoughts on “Where Do Logarithms Come From?” Pingback: Why Do Logarithms Work That Way? – The Math Doctors Pingback: Why Can’t a Logarithm Have a Negative Base? – The Math Doctors Pingback: Derangements: How Often is Everything Wrong? – The Math Doctors Leave a Comment Cancel Reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
1751	https://www.nature.com/scitable/definition/meiosis-88/	This page has been archived and is no longer updated Register \| Sign In meiosis Meiosis is a type of cell division that reduces the number of chromosomes in the parent cell by half and produces four gamete cells. This process is required to produce egg and sperm cells for sexual reproduction. During reproduction, when the sperm and egg unite to form a single cell, the number of chromosomes is restored in the offspring. Meiosis begins with a parent cell that is diploid, meaning it has two copies of each chromosome. The parent cell undergoes one round of DNA replication followed by two separate cycles of nuclear division. The process results in four daughter cells that are haploid, which means they contain half the number of chromosomes of the diploid parent cell. Meiosis has both similarities to and differences from mitosis, which is a cell division process in which a parent cell produces two identical daughter cells. Meiosis begins following one round of DNA replication in cells in the male or female sex organs. The process is split into meiosis I and meiosis II, and both meiotic divisions have multiple phases. Meiosis I is a type of cell division unique to germ cells, while meiosis II is similar to mitosis. Meiosis I, the first meiotic division, begins with prophase I. During prophase I, the complex of DNA and protein known as chromatin condenses to form chromosomes. The pairs of replicated chromosomes are known as sister chromatids, and they remain joined at a central point called the centromere. A large structure called the meiotic spindle also forms from long proteins called microtubules on each side, or pole, of the cell. Between prophase I and metaphase I, the pairs of homologous chromosome form tetrads. Within the tetrad, any pair of chromatid arms can overlap and fuse in a process called crossing-over or recombination. Recombination is a process that breaks, recombines and rejoins sections of DNA to produce new combinations of genes. In metaphase I, the homologous pairs of chromosomes align on either side of the equatorial plate. Then, in anaphase I, the spindle fibers contract and pull the homologous pairs, each with two chromatids, away from each other and toward each pole of the cell. During telophase I, the chromosomes are enclosed in nuclei. The cell now undergoes a process called cytokinesis that divides the cytoplasm of the original cell into two daughter cells. Each daughter cell is haploid and has only one set of chromosomes, or half the total number of chromosomes of the original cell. Meiosis II is a mitotic division of each of the haploid cells produced in meiosis I. During prophase II, the chromosomes condense, and a new set of spindle fibers forms. The chromosomes begin moving toward the equator of the cell. During metaphase II, the centromeres of the paired chromatids align along the equatorial plate in both cells. Then in anaphase II, the chromosomes separate at the centromeres. The spindle fibers pull the separated chromosomes toward each pole of the cell. Finally, during telophase II, the chromosomes are enclosed in nuclear membranes. Cytokinesis follows, dividing the cytoplasm of the two cells. At the conclusion of meiosis, there are four haploid daughter cells that go on to develop into either sperm or egg cells. Further Exploration Concept Links for further exploration cell division \| replication \| metaphase \| anaphase \| telophase \| linkage \| chromosome \| cytokinesis \| haploid \| prometaphase \| principle of segregation \| principle of independent assortment \| spindle fibers \| gamete \| DNA \| chromatin \| nucleus \| cytoplasm \| eukaryote \| prophase \| recombination \| principle of segregation \| Principles of Inheritance Related Concepts (23) Genetics Gene Inheritance and Transmission Gene Expression and Regulation Nucleic Acid Structure and Function Chromosomes and Cytogenetics Evolutionary Genetics Population and Quantitative Genetics Genomics Genes and Disease Genetics and Society Cell Biology Cell Origins and Metabolism Proteins and Gene Expression Subcellular Compartments Cell Communication Cell Cycle and Cell Division Scientific Communication Career Planning Loading ... LearnCast You have authorized LearnCasting of your reading list in Scitable. Do you want to LearnCast this session? Yes No This article has been posted to your Facebook page via Scitable LearnCast. Change LearnCast Settings © 2014 Nature Education About\| Contact\| Press Room\| Sponsors\| Terms of Use\| Privacy Notice\| Glossary\| Catalog\| Home\| Library\| Blogs Scitable Chat
1752	https://www.savemyexams.com/dp/physics/ib/23/sl/revision-notes/the-particulate-nature-of-matter/thermal-energy-transfers/temperature-and-kinetic-energy/	IBPhysicsDPSLRevision NotesThe Particulate Nature of MatterThermal Energy TransfersTemperature & Kinetic Energy Temperature & Kinetic Energy (DP IB Physics): Revision Note Author Katie M Last updated DP Physics: SLDP Physics: HL Temperature & Kinetic Energy Particles in gases usually have a range of speeds The average kinetic energy of the particles Ekcan be calculated using the equation Where: Ek= average kinetic energy of the particles in joules (J) kB = 1.38 × 10–23 J K–1 (Boltzmann's constant) T = absolute temperature in kelvin (K) This tells us that the absolute temperature of a body is directly proportional to the average kinetic energy of the molecules within the body Relationship between absolute temperature and average random kinetic energy of molecules Worked Example The surface temperature of the Sun is 5800 K and contains mainly hydrogen atoms. Calculate the average speed of the hydrogen atoms, in km s−1, near the surface of the Sun. Answer: Step 1: List the known quantities Temperature, T = 5800 K Mass of a hydrogen atom = mass of a proton, mp = 1.673 × 10−27 kg Boltzmann constant, kB = 1.38 × 10−23 J K−1 Step 2: Equate the equations relating kinetic energy with temperature and speed Average kinetic energy of a molecule: Kinetic energy: Step 3: Rearrange for average speed and calculate Average speed: v = 11 980 m s−1 = 12 km s−1 Unlock more, it's free! Join the 100,000+ Students that ❤️ Save My Exams the (exam) results speak for themselves: Test yourselfFlashcards Did this page help you? Previous:Temperature ScalesNext:Internal Energy Author:Katie M Expertise: Physics Content Creator Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.
1753	https://math.stackexchange.com/questions/3149575/max-and-min-inequality	real analysis - Max and Min Inequality - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Max and Min Inequality Ask Question Asked 6 years, 6 months ago Modified6 years, 6 months ago Viewed 264 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I have to show that for a,b,c,d∈R a,b,c,d∈R \|a∨b−c∨d\|≤\|a−c\|∨\|b−d\|\|a∨b−c∨d\|≤\|a−c\|∨\|b−d\| I know this can be showed using cases, but I need help with a proof that doesn't involve cases. I found out this inequality can be written as: \|a∨b−c∨d\|≤(a∨c−a∧c)∨(b∨d−b∧d)\|a∨b−c∨d\|≤(a∨c−a∧c)∨(b∨d−b∧d) Since a∨c=(a+c)+\|a−c\|2 a∨c=(a+c)+\|a−c\|2 and a∧c=(a+c)−\|a−c\|2 a∧c=(a+c)−\|a−c\|2 I guess the proof uses this form of the inequality, but I don't know how to write such proof. real-analysis inequality absolute-value Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 15, 2019 at 17:39 MMMMMM asked Mar 15, 2019 at 17:26 MMMMMM 53 5 5 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. This is basically just repeated applications of the triangle inequality; namely, we have \|max(a,b)−max(c,d)\|=∣∣∣(a−c)+(b−d)+\|a−b\|−\|c−d\|2∣∣∣≤(\|a−c\|+\|b−d\|)+\|\|a−b\|−\|c−d\|\|2.\|max(a,b)−max(c,d)\|=\|(a−c)+(b−d)+\|a−b\|−\|c−d\|2\|≤(\|a−c\|+\|b−d\|)+\|\|a−b\|−\|c−d\|\|2. But, \|\|a−b\|−\|c−d\|\|≤\|\|a−c\|+\|b−c\|−(\|c−b\|+\|b−d\|)\|=\|\|a−c\|−\|b−d\|\|.\|\|a−b\|−\|c−d\|\|≤\|\|a−c\|+\|b−c\|−(\|c−b\|+\|b−d\|)\|=\|\|a−c\|−\|b−d\|\|. That should give the desired result. Edit: For the last step, I used \|a−b\|=\|a−c+c−b\|≤\|a−c\|+\|b−c\|.\|a−b\|=\|a−c+c−b\|≤\|a−c\|+\|b−c\|. You use that same trick on both terms in the absolute value. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 15, 2019 at 18:26 answered Mar 15, 2019 at 17:58 Gary MoonGary Moon 2,565 1 1 gold badge 10 10 silver badges 13 13 bronze badges 2 What form of the triangle inequality you used at that last step?MMM –MMM 2019-03-15 18:22:36 +00:00 Commented Mar 15, 2019 at 18:22 I added an edit to (hopefully) address your question.Gary Moon –Gary Moon 2019-03-15 18:26:35 +00:00 Commented Mar 15, 2019 at 18:26 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis inequality absolute-value See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 3Prove that for real numbers x x, if x 2−5 x+4≥0 x 2−5 x+4≥0, then either x≤1 x≤1 or x≥4 x≥4. 1Direct proof of inequality between arithmetic and harmonic mean 3inequality with min on both sides 0Solving a quadratic inequality geometrically 7Maximum difference inequality 1Triangle Inequality Issue 4How to show that min{α,max{β,γ}}=max{min{α,β},min{α,γ}}.min{α,max{β,γ}}=max{min{α,β},min{α,γ}}. 0Calculation question about max(x,min(y,z)) = min(max(x,y),max(x,z)) Hot Network Questions ConTeXt: Unnecessary space in \setupheadertext What is a "non-reversible filter"? 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1754	https://www.datacamp.com/tutorial/minkowski-distance	Skip to main content Minkowski Distance: A Comprehensive Guide Minkowski distance is a way of measuring the straight or curved path between two points, depending on a chosen parameter that affects the shape. Keep reading to learn about the fundamentals, applications, and comparisons of Minkowski distance in various fields. Oct 9, 2024 · 11 min read Distance metrics form the backbone of numerous algorithms in data science and machine learning, enabling the measurement of similarity or dissimilarity between data points. In this guide, we'll explore the foundations of Minkowski distance, its mathematical properties, and its implementations. We'll examine how it relates to other common distance measures and demonstrate its use through coding examples in Python and R. Whether you're developing clustering algorithms, working on anomaly detection, or fine-tuning classification models, understanding Minkowski distance can enhance your approach to data analysis and model development. Let’s take a look. What is Minkowski Distance? Minkowski distance is a versatile metric used in normed vector spaces, named after the German mathematician Hermann Minkowski. It's a generalization of several well-known distance measures, making it a fundamental concept in various fields such as math, computer science, and data analysis. At its core, Minkowski distance provides a way to measure the distance between two points in a multi-dimensional space. What makes it particularly useful is its ability to encompass other distance metrics as special cases, primarily through a parameter p. This parameter allows the Minkowski distance to adapt to different problem spaces and data characteristics. The general formula for Minkowski distance is: Where: x and y are two points in an n-dimensional space p is a parameter that determines the type of distance (p ≥ 1) \|xi - yi\| represents the absolute difference between the coordinates of x and y in each dimension Minkowski distance is useful for two main reasons. For one, it gives you flexibility to toggle between Manhattan or Euclidean distance as needed. Secondly, it recognizes that not all datasets (think high-dimensional spaces) are well-suited to either purely Manhattan or purely Euclidean distance. In practice, the parameter p is typically chosen by incorporating a train/test validation workflow. By testing different values of p during cross-validation, you can determine which value provides the best model performance for your specific dataset. How Minkowski Distance Works Let’s take a look at how Minkowski distance relates to other distance formulas, and then walk through an example. Generalization of other distance metrics The first thing to consider is how the Minkowski distance formula contains within it the formulas for Manhattan, Euclidean, and Chebyshev distances. Manhattan Distance (p = 1): When p is set to 1, the Minkowski distance becomes Manhattan distance. Also known as city block distance or L1 norm, Manhattan Distance measures the sum of absolute differences. Euclidean Distance (p = 2): When p is set to 2, Minkowski distance becomes Euclidean distance. Euclidean distance is the most common distance metric, representing the straight-line distance between two points. Chebyshev Distance (p → ∞): Chebyshev distance, also known as chessboard distance, measures the maximum difference along any dimension. Working through an example To truly grasp the functionality and power of Minkowski distance, let's work through an example. This exploration will help us understand how the parameter p affects the calculation and interpretation of distances in multi-dimensional spaces. Let's consider two points in a 2D space: Point A: (2, 3) Point B: (5, 7) We'll calculate the Minkowski distance between these points for different p values. The parameter p in the Minkowski distance formula controls the metric's sensitivity to differences in individual components: When p=1: All differences contribute linearly. When p=2: Larger differences have a more significant impact due to squaring. When p>2: Even more emphasis is placed on larger differences. When p→∞: Only the maximum difference among all dimensions matters. As p increases, the Minkowski distance generally decreases, approaching the Chebyshev distance. This is because higher p values give more weight to the largest difference and less to smaller differences. To visualize how different values of p affect the distance calculation between our points A(2, 3) and B(5, 7), let's examine the following graph: Observing the graph, we can see how the distance measure changes as p increases: The Manhattan distance (p=1), represented by the green path, yields the longest path, as it strictly follows the grid. The Euclidean distance (p=2), shown by the orange straight line, provides a direct, straight-line path. The Chebyshev distance (p=∞), depicted by the red dashed lines, focuses solely on the largest coordinate difference, creating a path that moves maximally in one dimension before addressing the other. The Minkowski distance with p=3 in purple shows a slight curve, hinting at the transition between Euclidean and Chebyshev distances. This visualization helps us understand why different p values might be chosen for various applications. For instance, Manhattan distance might be more appropriate in city navigation problems, while Euclidean distance is often used in physical space calculations. Higher p values, like in the Minkowski p=3 case, can be useful in scenarios where larger differences should be emphasized, and Chebyshev distance might be preferred when the maximum difference in any dimension is the most critical factor. Applications of Minkowski Distance The Minkowski distance, with its adjustable parameter p, is a flexible tool used across various fields. By changing p, we can tailor how we measure the distance between points, making it suitable for different tasks. Below are four applications where Minkowski distance plays an important role. Machine learning and data science In machine learning and data science, Minkowski distance is fundamental for algorithms that rely on measuring the similarity or dissimilarity between data points. One prominent example is the k-Nearest Neighbors (k-NN) algorithm, which classifies data points based on the categories of their nearest neighbors. By using Minkowski distance, we can adjust the parameter p to change how we calculate the "closeness" between points. Pattern recognition Pattern recognition involves identifying patterns and regularities in data, such as handwriting recognition or facial feature detection. In this context, Minkowski distance measures the difference between feature vectors representing patterns. For example, in image recognition, each image can be represented by a vector of pixel values. Calculating the Minkowski distance between these vectors allows us to quantify how similar or different the images are. By adjusting p, we can control the sensitivity of the distance measure to differences in specific features. A lower p might consider overall differences across all pixels, while a higher p could emphasize significant differences in certain regions of the image. Anomaly detection Anomaly detection aims to identify data points that deviate significantly from the majority, which is crucial in areas like fraud detection, network security, and fault detection in systems. Minkowski distance is used to measure how far a data point is from others in the dataset. Points with large distances are potential anomalies. By choosing an appropriate p, analysts can improve the sensitivity of anomaly detection systems to the kinds of deviations that are most relevant to their specific context. Computational geometry and spatial analysis In computational geometry and spatial analysis, Minkowski distance is used to compute distances between points in space, which is the basis for many geometric algorithms. For example, collision detection in these domains relies on Minkowski distance to determine when objects are close enough to interact. By adjusting p, developers can create diverse collision boundaries, ranging from angular (lower p) to rounded (higher p). Beyond collision detection, Minkowski distance can be useful in spatial clustering and shape analysis. Varying the value of p allows researchers to emphasize different aspects of spatial relationships, from city block distances to overall shape similarities. Mathematical Properties of Minkowski Distance The Minkowski distance is not only a versatile tool in practical applications but also an important concept in mathematical theory, particularly in the study of metric spaces and norms. Metric space properties The Minkowski distance satisfies the four essential properties required for a function to be considered a metric in a metric space: Non-negativity: The Minkowski distance between any two points is always non-negative, d(x,y)≥0. This is evident as it is the p-th root of a sum of non-negative terms (absolute values raised to the power p). Identity of Indiscernibles: The Minkowski distance between two points is zero if and only if the two points are identical. Mathematically, d(x,y) = 0 if and only if x=y. This follows because the absolute difference between identical components is zero. Symmetry: The Minkowski distance is symmetric, meaning d(x,y)=d(y,x). This property holds because the order of subtraction in the absolute value terms does not affect the outcome. Triangle Inequality: The Minkowski distance satisfies the triangle inequality, which states that for any three points x, y, and z, the distance from x to z is at most the sum of the distance from x to y and from y to z; formally, d(x,z)≤d(x,y)+d(y,z). This property is less intuitive to prove directly from the formula and generally requires more advanced mathematics but essentially ensures that taking a direct path between two points is the shortest route. Norm generalization The Minkowski distance acts as a general framework that unifies various ways to measure distances in mathematical spaces through the concept of norms. In simple terms, a norm is a function that assigns a non-negative length or size to a vector in a vector space, essentially measuring how "long" the vector is. By adjusting the parameter p in the Minkowski distance formula, we can smoothly transition between different norms, each providing a unique method for calculating vector length. For example, when p=1, the Minkowski distance becomes the Manhattan norm, measuring distance as the sum of absolute differences along each dimension—imagine navigating a grid of city streets. With p=2, it turns into the Euclidean norm, calculating the straight-line ("as-the-crow-flies") distance between points. As p approaches infinity, it converges to the Chebyshev norm, where the distance is determined by the largest single difference among dimensions. This flexibility allows the Minkowski distance to adapt to various mathematical and practical contexts, making it a versatile tool for measuring distances in different scenarios. Calculating Minkowski Distance in Python and R Let's explore implementations of Minkowski distance calculations using both Python and R. We'll examine readily available packages and libraries that can achieve this. Python example To calculate Minkowski distance in Python, we can use the SciPy library, which provides efficient implementations of various distance metrics. Here's an example that calculates Minkowski distance for different p values: ``` import numpy as np from scipy.spatial import distance Example points point_a = [2, 3] point_b = [5, 7] Different p values p_values = [1, 2, 3, 10, np.inf] print("Minkowski distances using SciPy:") for p in p_values: if np.isinf(p): # For p = infinity, use Chebyshev distance dist = distance.chebyshev(point_a, point_b) print(f"p = ∞, Distance = {dist:.2f}") else: dist = distance.minkowski(point_a, point_b, p) print(f"p = {p}, Distance = {dist:.2f}") Powered By ``` By running this code, readers can observe how the distance changes with different p values, reinforcing the concepts discussed earlier in the article. ``` Minkowski distances using SciPy: p = 1, Distance = 7.00 p = 2, Distance = 5.00 p = 3, Distance = 4.50 p = 10, Distance = 4.02 p = ∞, Distance = 4.00 Powered By ``` This code demonstrates: How to use SciPy's distance functions for Minkowski and Chebyshev distances. Calculation of distances for various p values, including infinity. The relationship between Minkowski distance and other metrics (Manhattan, Euclidean, Chebyshev). R example For R, we will utilize the dist() function from the stats library: ``` Define the Minkowski distance function using stats::dist minkowski_distance <- function(x, y, p) { points <- rbind(x, y) if (is.infinite(p)) { # For p = Inf, use method = "maximum" for Chebyshev distance distance <- stats::dist(points, method = "maximum") } else { distance <- stats::dist(points, method = "minkowski", p = p) } return(as.numeric(distance)) } Example usage point_a <- c(2, 3) point_b <- c(5, 7) Different p values p_values <- c(1, 2, 3, 10, Inf) cat("Minkowski distances between points A and B using stats::dist:\n") for (p in p_values) { distance <- minkowski_distance(point_a, point_b, p) if (is.infinite(p)) { cat(sprintf("p = ∞, Distance = %.2f\n", distance)) } else { cat(sprintf("p = %g, Distance = %.2f\n", p, distance)) } } Powered By ``` This code demonstrates: How to create a function minkowski_distance using the dist() function from stats. Handling of different p values, including infinity for Chebyshev distance. Calculation of Minkowski distance for various p values. Formatting output to display distances rounded to 2 decimal places. The output of this code will be: ``` Minkowski distances between points A and B using stats::dist: p = 1, Distance = 7.00 p = 2, Distance = 5.00 p = 3, Distance = 4.50 p = 10, Distance = 4.02 p = ∞, Distance = 4.00 Powered By ``` This R implementation provides a counterpart to the Python example, allowing readers to see how Minkowski distance can be calculated in different programming environments. Conclusion Minkowski distance provides a flexible and adaptable approach to measuring distances in multi-dimensional spaces. Its ability to generalize other common distance metrics through the parameter p makes it a valuable tool across various fields in data science and machine learning. By adjusting p, practitioners can tailor their distance calculations to the specific characteristics of their data and the requirements of their projects, potentially enhancing results in tasks ranging from clustering to anomaly detection. As you apply Minkowski distance in your own work, we encourage you to experiment with different p values and observe their impact on your results. For those looking to deepen their understanding and skills, we recommend exploring the Designing Machine Learning Workflows in Python course and considering our Data Scientist Certification career program. These resources can help you build on your knowledge of distance metrics and apply them effectively in various scenarios. Become an ML Scientist Upskill in Python to become a machine learning scientist. Start Learning for Free Author Vinod Chugani As an adept professional in Data Science, Machine Learning, and Generative AI, Vinod dedicates himself to sharing knowledge and empowering aspiring data scientists to succeed in this dynamic field. Minkowski Distance FAQs What is Minkowski distance? Minkowski distance is a generalized metric used to measure the distance between two points in multi-dimensional space. It's defined by a parameter 'p' which allows it to encompass other distance metrics as special cases, making it highly versatile for various applications in data science and machine learning. How does Minkowski distance relate to Euclidean and Manhattan distances? Minkowski distance is a generalization of both Euclidean and Manhattan distances. When the parameter p=2, it becomes Euclidean distance, and when p=1, it becomes Manhattan distance. This flexibility allows Minkowski distance to adapt to different problem spaces and data characteristics. What happens when p approaches infinity in Minkowski distance? As p approaches infinity, Minkowski distance converges to Chebyshev distance. In this case, the distance is determined by the maximum difference along any single dimension, which is useful in certain optimization problems and worst-case scenario analyses. Can Minkowski distance be used with categorical data? Minkowski distance is primarily designed for numerical data. For categorical data, other distance measures like Hamming distance or Gower distance are more appropriate. However, if categorical data is properly encoded into numerical form, Minkowski distance can be applied. Is Minkowski distance affected by the scale of features? Yes, Minkowski distance is sensitive to the scale of features. If features have different scales, it's important to normalize or standardize the data before applying Minkowski distance to ensure all features contribute appropriately to the distance calculation. Are there any limitations to using Minkowski distance? While versatile, Minkowski distance assumes that all dimensions contribute equally to the distance, which may not always be appropriate. It can be computationally intensive for large datasets, especially with fractional p values. Additionally, it may not capture complex, non-linear relationships in the data that some other distance measures might reveal. Topics Data Science Python Learn with DataCamp Track Machine Learning Scientist in Python 0 min Discover machine learning with Python and work towards becoming a machine learning scientist. Explore supervised, unsupervised, and deep learning. See Details Start Course Course Ensemble Methods in Python 4 hr 11.2K Learn how to build advanced and effective machine learning models in Python using ensemble techniques such as bagging, boosting, and stacking. See Details Start Course Course Data Structures and Algorithms in Python 4 hr 28.7K Explore data structures such as linked lists, stacks, queues, hash tables, and graphs; and search and sort algorithms! See Details Start Course See More Related Tutorial Understanding Euclidean Distance: From Theory to Practice Explore how Euclidean distance bridges ancient geometry and modern algorithms, with coding examples in Python and R, and learn about its applications in data science, machine learning, and spatial analysis. Vinod Chugani Tutorial Understanding Chebyshev Distance: A Comprehensive Guide Learn how Chebyshev distance offers a unique approach to spatial problems. Uncover its applications in robotics, GIS, and game development with coding examples in Python and R. Vinod Chugani Tutorial What is Cosine Distance? Explore cosine distance and cosine similarity. Discover calculations, applications, and comparisons with other metrics. Learn to implement in R and Python using numpy. Vinod Chugani Tutorial What is Manhattan Distance? Learn how to calculate and apply Manhattan Distance with coding examples in Python and R, and explore its use in machine learning and pathfinding. Vinod Chugani Tutorial Mean Shift Clustering: A Comprehensive Guide Discover the mean shift clustering algorithm, its advantages, real-world applications, and step-by-step Python implementation. Compare it with K-means to understand key differences. Vidhi Chugh Tutorial Understanding Skewness And Kurtosis And How to Plot Them A comprehensive visual guide into skewness/kurtosis and how they effect distributions and ultimately, your data science project. Bex Tuychiev See More See More
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find, then convert the year and month to decimal Ask Question Asked 7 years, 7 months ago Modified4 years, 10 months ago Viewed 22k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. 5 months in role 1 year 11 months in role 5 years 1 month in role 2 years 8 months in role 1 year in role 3 years 9 months in role Hello, I am trying to convert the list above to a decimal date. For example. I want [3 years 9 months in role] to be shown in another cell as 3.75 but I am really struggling to figure out a formula. So far I have =LEFT(A2,(FIND(" ",A2,1)-1)) but this obviously just grabs the first value before the space. This formula needs to take into account the variations above (sometimes it's only month shown, etc). Thanks in advance for your help. excel Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications asked Feb 8, 2018 at 19:26 B FredB Fred 5 1 1 gold badge 1 1 silver badge 3 3 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Use this array formula: =TRIM(MID(" 0 Year " & A1,MAX(IFERROR(FIND("}}}",SUBSTITUTE(" 0 year " & A1,"year","}}}",{1,2})),0))-3,2))+TRIM(MID(" " & A1 & " 0 month",MIN(IFERROR(FIND("}}}",SUBSTITUTE(" " & A1 & " 0 month","month","}}}",{1,2})),999))-3,2))/12 Being an array formula it needs to be confirmed with ctrl-Shift-Enter instead of Enter when exiting edit mode. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 8, 2018 at 19:40 Scott CranerScott Craner 153k 10 10 gold badges 52 52 silver badges 88 88 bronze badges 1 Comment Add a comment B Fred B FredOver a year ago That is exactly it! I've spent a better part of the morning trying to do this. Really appreciate your help! This makes perfect sense. 2018-02-08T19:57:21.9Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. Power query solution. Assumes a single column Table in Excel named ("Table1") with a Column Header of "Text". In case it helps! ```xml let Source = Excel.CurrentWorkbook(){[Name="Table1"]}[Content], Month = Table.AddColumn(Source, "Month", each if Text.Contains([Text], "month") = true then try Text.Middle([Text], Text.PositionOf([Text], "month") -3, 2) otherwise Text.Middle([Text], Text.PositionOf([Text], "month") -2, 2) else 0), Year = Table.AddColumn(Month, "Year", each if Text.Contains([Text], "year") = true then try Text.Middle([Text], Text.PositionOf([Text], "year") -3, 2) otherwise Text.Middle([Text], Text.PositionOf([Text], "year") -2, 2) else 0), ToNumber = Table.TransformColumnTypes(Year,{{"Month", Int64.Type}, {"Year", Int64.Type}}), Calc = Table.AddColumn(ToNumber, "Calc", each [Year]+([Month]/12)) in Calc ``` Run code snippet Edit code snippet Hide Results Copy Expand Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 8, 2018 at 20:22 JJBJJB 53 4 4 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. First convert total years and months together into number of months. Eg: 4 years 5 months = 53 months. Then divide the total months by 12 to get the decimal calculation. Eg: 53 / 12 = 4.41 years. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 16, 2020 at 7:13 FairozFairoz 9 1 1 bronze badge Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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1758	https://www.vedantu.com/question-answer/dimethyl-ether-forms-adduct-to-the-bf3-but-silyl-class-12-chemistry-cbse-5ff0330079d9ab0919cdb1a9	Talk to our experts 1800-120-456-456 Dimethyl ether forms adduct to the [B{F_3}], but silyl ether does not form adduct. Repeaters Course for NEET 2022 - 23 © 2025.Vedantu.com. All rights reserved
1759	https://direns.minesparis.psl.eu/Sites/Thopt/en/co/combustion.html	content menu page footer Online course and simulator for engineering thermodynamics General Navigation: Thermodynamics fundamentals Methodological guides Technologies Global issues Education Software Glossary Stack: Online course and simulator for engineering thermodynamics Thermodynamics fundamentals Thermodynamics of combustion Menu: Basic concepts and qualitative approach of thermodynamic laws Thermodynamics of combustion Energy exchanges in a process Laws of thermodynamics Substance properties Thermodynamics software packages Content: Thermodynamics of combustion Most ash-free fuels are hydrocarbons. By neglecting traces of nitrogen and sulphur, their molecule consists mainly of carbon C and hydrogen H atoms as well as some O oxygen. By reducing the formulation to a carbon atom, the general formula of a fuel is therefore CHyOx. As the condition x <= y/2 is always checked, the CHyOx formula becomes: CHa + x H2O a thus represents the so-called hydrogen available for combustion related to the complete oxidation of a carbon unit. The water in the fuel does not participate in the combustion reaction. It ends up in the fumes. The complete combustion of a CHa formula fuel with pure oxygen would be governed by this equation: it takes 1 oxygen molecule to form CO2, and a/4 oxygen molecules to form a/2 H2O. Combustion of CHa with oxygen However, in energy systems, combustions are almost always carried out with air as an oxidizer. Since dry air is approximately 79% nitrogen and 21% oxygen, the nitrogen/oxygen ratio is 79/21 = 3.762. That's why the air can be represented by the expression O2 + 3.762 N2 A complete combustion with air of a CHa formula fuel is therefore governed by this equation: Combustion of CHa with atmospheric air Stoichiometric combustion is a combustion carried out with the exact amount of oxidizer to completely oxidize the fuel. It is the one that leads to the highest end-of-combustion temperature. We will consider it as a reference combustion This equation means that the stoichiometric combustion of a CHa fuel mole requires (1 + a/4) oxygen moles and produces 1 mole of carbon dioxide and a/2 moles of water. If the oxidizer is air, 3.76 (1 + a/4) nitrogen moles are also involved, but because they do not react with the fuel, they end up in the burnt gases. Nitrogen is said to remain inert. When combustion is non- stoichiometric, it can be characterized in several ways: Either by excess air e, which as the name suggests, represents the amount of air in excess Or by the air factor lambda, which is equal to 1 + the excess air Or by the richness R, ratio of the number of moles (or mass) of fuel contained in a specified amount of mixture, to the number of moles (or mass) of fuel in the stoichiometric mixture. R = 1 corresponds to the stoichiometric mixture, R < 1 to excess air, and R > 1 to excess fuel These three factors are linked by simple relationships : R = 1/(1 + e) and lambda = 1 + e = 1/R We will use preferentially lambda in the following, because it is the term multiplier of air in the combustion equation. If lambda it is greater than 1, i.e. excess air, the complete combustion with air of a CHa formula fuel is governed by this equation : Combustion of CHa with air in excess Thermoptim uses a generalization of this equation for more complex fuels than CHa. Comparison of last two equations helps to understand what happens when air is available in excesss. The fuel reacts with oxygen as in the stoichiometric reaction, and all excess air is found without reacting in the burnt gases. lambda is the term that multiplies the number of air moles in the combustion equation. For lambda = 1, the reaction is stoichiometric. This small calculator allows you to study the influence of a and lambda on this equation. The table below shows you the number of moles of the products and their total, how the molar fractions are calculated, and their values for the chosen setting.You can test the two oxidizers used previously, oxygen or atmospheric air. Combustion calculator Higher and lower heating values During combustion, the maximum amount of energy release is obtained when the water in the fumes is cooled enough to be liquefied, which requires a very low temperature. The value of the full reaction heat then takes the name of higher heating value, or HHV. In the most general case where all the water produced remains in the vapor state, it is given the name of lower heating value or LHV Vaporization enthalpy are far from negligible (for water, it is worth about 45 MJ/kmole at 0°C). There are therefore significant differences between the HHV and the LHV fuel values, and it is important to specify which one is being used. The values of the LHV of fuels typically used in energy installations are quite close to each other. It can be shown that the LHV of hydrocarbons decreases from 44 MJ/kg to 40 MJ/kg as distillation products are becoming heavier. In conclusion, therefore, to determine the energies involved in a complete combustion reaction, the heating value of the fuel is used, specifying whether it is the higher heating value (HHV) or the lower heating value (LHV). Combustions in Thermoptim In Thermoptim, a combustion process is represented by a combustion chamber component with two inlet processes, on the one hand the oxidizer, here air at the compressor outlet, connected on the blue port, and on the other hand the fuel connected on the red port. Combustion process in the diagram editor The burnt gases come out through the green port, here connected to the turbine. The composition of the fuel is defined in the downstream point of the "fuel" process, whose subsance must be a pure or compound gas. The natural gas composition in Montoir de Bretagne is given here : Natural gas composition The screen of a combustion has many parameters. Combustion process The three main methods of calculation are: The "Calculate lambda" option determines the average air factor (lambda >= 1), based on the value of the end-of-combustion temperature set Tc. The "Calculate T" option determines Tc from the set lambda value. In both cases, the flow of the "fuel" process is adjusted so that the ratio of the volume flows of oxidizer and fuel is equal to the air factor. he choice "Set fuel flow" determines lambda and Tc based on the characteristics of the fuel and the oxidizer. The mass flow of the process being evaluated (combustion) is equal to the sum of fuel and fuel flows, which means that the combustion process behaves, hydraulically, like a flow mixer. For other settings, when combustion takes place in an open system, the pressure set by the upstream point is generally chosen, which means that the combustion chamber is isobaric. For simple use of Thermoptim, these settings are enough. Incomplete combustion, dissociation, quenching temperature We assumed in the foregoing that the combustion was complete, whereas sometimes this is not the case, especially at high temperatures, and unburned gases appear in the fumes. This is called dissociation. If, starting from such a situation, the temperature of the reactive environment is gradually lowered, it is found that from a certain threshold its composition stabilizes and no longer varies. It is said that the reaction is quenched and the quenching temperature is called the value of this threshold. Thermoptim can take this phenomenon into account when the dissociation mode is checked, and if it is shown on the one hand the rate of CO2 dissociation in CO, and on the other hand the value of the quenching temperature. Combustion process with dissociation In this example, the CO2 dissociation rate was set at 5%, and the quenching temperature value was 900 °C. The impact of this change in setting on the composition of the fumes is illustrated by Figures 5 and 6: Carbon monoxide CO and hydrogen H2 appear in the event of dissociation. Flue gas composition without dissociation Flue gas composition with dissociation A combustion calculator with more setting possibilities than the small tool we presented at the beginning of this page is available here. copyright R. Gicquel v2025.1 \|
1760	https://pubmed.ncbi.nlm.nih.gov/6317159/	Gestational trophoblastic tumors metastatic to the lung. Radiologic--clinical correlations - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Radiologic--clinical correlations A S Hendin PMID: 6317159 DOI: 10.1002/1097-0142(19840101)53:1<58::aid-cncr2820530111>3.0.co;2-q Item in Clipboard Gestational trophoblastic tumors metastatic to the lung. Radiologic--clinical correlations A S Hendin. Cancer.1984. Show details Display options Display options Format Cancer Actions Search in PubMed Search in NLM Catalog Add to Search . 1984 Jan 1;53(1):58-61. doi: 10.1002/1097-0142(19840101)53:1<58::aid-cncr2820530111>3.0.co;2-q. Author A S Hendin PMID: 6317159 DOI: 10.1002/1097-0142(19840101)53:1<58::aid-cncr2820530111>3.0.co;2-q Item in Clipboard Cite Display options Display options Format Abstract The relationship between human chorionic gonadotropin (HCG) titers in urine and the number of discrete metastatic pulmonary nodules was studied in patients with persistent trophoblastic disease, after evacuation of hydatidiform mole, and before starting chemotherapy. A significant difference in HCG titers was found between patients with 0 to 2 nodules and patients with 5 or more nodules. A weak linear relationship was found. The distribution of 57 discrete pulmonary nodules in 13 patients was plotted by lung zones (upper, middle, and lower thirds). Twenty-eight percent of the nodules were in the upper third of the lungs. Two patients had solitary apical nodules. This differs from the characteristic predominantly basilar distribution of blood-borne metastases of other neoplasms. Pulmonary spread may have occurred during curettage of moles, when the patients were recumbent and pulmonary blood flow was redistributed to the upper portions of the lungs. PubMed Disclaimer Similar articles [Study on the indication of surgical resection of pulmonary metastasis of malignant trophoblastic tumor].Zhang Y, Xiang Y, Ren T, Wan XR, Yang XY.Zhang Y, et al.Zhonghua Fu Chan Ke Za Zhi. 2005 Feb;40(2):83-6.Zhonghua Fu Chan Ke Za Zhi. 2005.PMID: 15840284 Chinese. [Importance of arteriography in complicated trophoblastic disease].Tariel D, Body G, Kerhuel J, Lemarie E, Rouleau P, Lansac J.Tariel D, et al.J Gynecol Obstet Biol Reprod (Paris). 1984;13(6):643-50.J Gynecol Obstet Biol Reprod (Paris). 1984.PMID: 6097610 French. Utility of diagnostic imaging in the staging of gestational trophoblastic disease.Williams AG Jr, Mettler FA Jr, Wicks JD.Williams AG Jr, et al.Diagn Gynecol Obstet. 1982 Summer;4(2):159-63.Diagn Gynecol Obstet. 1982.PMID: 6284468 Evaluation of the criteria used to make the diagnosis of nonmetastatic gestational trophoblastic neoplasia.Kohorn EI.Kohorn EI.Gynecol Oncol. 1993 Feb;48(2):139-47. doi: 10.1006/gyno.1993.1025.Gynecol Oncol. 1993.PMID: 8381374 Review. From the archives of the AFIP. Gestational trophoblastic disease: radiologic-pathologic correlation.Wagner BJ, Woodward PJ, Dickey GE.Wagner BJ, et al.Radiographics. 1996 Jan;16(1):131-48. doi: 10.1148/radiographics.16.1.131.Radiographics. 1996.PMID: 10946695 Review. See all similar articles MeSH terms Adult Actions Search in PubMed Search in MeSH Add to Search Chorionic Gonadotropin / urine Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Lung Neoplasms / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Lung Neoplasms / secondary Actions Search in PubMed Search in MeSH Add to Search Lung Neoplasms / urine Actions Search in PubMed Search in MeSH Add to Search Neoplasm Metastasis Actions Search in PubMed Search in MeSH Add to Search Neoplasm Seeding Actions Search in PubMed Search in MeSH Add to Search Pregnancy Actions Search in PubMed Search in MeSH Add to Search Radiography Actions Search in PubMed Search in MeSH Add to Search Trophoblastic Neoplasms / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Uterine Neoplasms / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Substances Chorionic Gonadotropin Actions Search in PubMed Search in MeSH Add to Search LinkOut - more resources Medical MedlinePlus Health Information [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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1761	https://en.wikipedia.org/?title=Multi-linear_regression&redirect=no	Multi-linear regression - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Multi-linear regression [x] Add languages Add links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Add interlanguage links Print/export Download as PDF Printable version In other projects From Wikipedia, the free encyclopedia Redirect to: Linear regression Retrieved from " This page was last edited on 11 June 2010, at 06:17(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search Multi-linear regression Add languagesAdd topic
1762	https://www.scribd.com/document/559684783/ch-16	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Upload 0 ratings0% found this document useful (0 votes) 496 views28 pages CH 16 Uploaded by Narendran Kumaravel You are on page 1/ 28 Waveguides (by Milton L. Kult) 16.1. Introduction The electromagnetic waves of Chapter 14 can be guided in a given direction of propagation using several different methods.For instance, the two-conductor transmission line, supporting what are essentially plane waves at megahertz frequencies, wasconsidered in Chapter 15 . The present chapter is restricted to single-conductor (hollow-pipe) waveguides , of rectangular orcircular cross section, which operate in the gigahertz (microwave) range. These devices too support "plane waves"—in thesense that the wavefronts are planes perpendicular to the direction of propagation. However, the boundary conditions at theinner surface of the pipe force the fields to vary over a wavefront. 16.2. Transverse and Axial Fields The waveguide is positioned with the longitudinal direction along the z axis. In general, the guide walls have σ = ∞ (perfectconductor) and the dielectric-filled hollow has σ = 0 (perfect dielectric), μ = μ μ , and ϵ = ϵ ϵ . It is further supposed that ρ = 0(no free charge) in the dielectric. The dimensionsfor the cross section are inside dimensions. In Fig. 16-1( a ) the a × b rectangular waveguide is shown in a Cartesian coordinate system, Fig. 16-1( b ) shows the circular or cylindrical waveguide ofradius a in a cylindrical coordinate system. Figure 16-1 As in Chapter 14 the time dependence e will be assumed for the electromagnetic field in the dielectric core; this time factorwill be suppressed everywhere in the analysis (as in phasor notation). Thus we have the following expressions for the fieldvector F (which stands for either E or H ), assuming wave propagation in the + z direction. c 0 r 0 r jωt Rectangular coordinates. F = F( x , y ) e − jkz whereF( x , y ) = F x ( x , y )a x + F y ( x , y )a y + F z ( x , y )a z ≡ F T ( x , y )+ F z ( x , y )a z Cylindrical coordinates. F = F( r , ϕ ) e − jkz whereF( r , ϕ ) = F r ( r , ϕ )a r + F ϕ ( r , ϕ )a ϕ + F z ( r , ϕ )a z ≡ F T ( r , ϕ )+ F z ( r , ϕ )a z © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information. Download to read ad-free Because the dielectric is lossless ( σ = 0), the wave propagates without attenuation; hence, the wave number k = 2 π / λ (in rad/m)is constrained to be real and positive. Note: In the other chapters of this book, unbounded dielectric media are considered, for which the wave number, notated β ,depends on frequency and dielectric properties only. However, as will soon appear, the wave number in a bounded dielectricdepends additionally on the geometry of the boundary. This important distinction is emphasized by the employment of a newsymbol, k , in the present chapter.The reason for decomposing the field vector into a transverse vector component F and an axial vector component F a is two-fold. On the one hand, the boundary conditions apply to E and H alone (see Problems 16.1 and 16.2 ). On the other hand, aswill now be shown, the complete E and H fields in the waveguide are known once either Cartesian component D or H is known. 16.2.1. Transverse Components from Axial Components. Assume a rectangular coordinate system. Maxwell's equation ( 2 ) of Section 14.2 yields the three scalar equations(1a)(1b)(1c)Maxwell's equation ( 1 ) of Section 14.2 , with σ = 0, gives three additional scalar equations:(2a)(2b) T z zT T z z − jωμH x = jkE y + ∂ E z ∂ y − jωμH y = − jkE x − ∂ E z ∂ x − jωμH z = −∂ E y ∂ x ∂ E x ∂ y jωϵE x = jkH y + ∂ H z ∂ y jωϵE y = − jkH x − ∂ H z ∂ x jωϵE z = −∂ H y ∂ x ∂ H x ∂ y © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information. Download to read ad-free (2c)Now eliminate H between ( 1a ) and ( 2b ), and H between ( 1b ) and ( 2a ), to obtain(3a)(3b)in which k ≡ ω μ ϵ − k . The parameter k (also in rad/m) functions as a critical wave number ; see Problem 16.3 . Finally, slide(3b) and (3a) back into (2a) and (2b), to find(3c)(3d)By exciting the waveguide in suitable fashion it is possible to force either E or H (but not both) to vanish identically. Thenonvanishing axial component will then determine all other components via Equations ( 3 ).See Problems 16.4 and 16.5 for the analogous results in cylindrical coordinates. 16.3. TE and TM Modes; Wave Impedances The two types of waves found in Section 16.2 are referred to as transverse electric (TE) or transverse magnetic (TM) waves,according as E ≡ 0 or H ≡ When carrying such waves, the guide is said to operate in a TE or TM mode .For any transverse electromagnetic wave, the wave impedance (in ohms) is defined as(4)(compare Chapter 14 ). For a waveguide in a TE mode, ( 1a ) and ( 1b ) imply x y E y = − + jkk 2 c ∂ E z ∂ y jωμk 2 c ∂ H z ∂ xE x = − − jkk 2 c ∂ E z ∂ x jωμk 2 c ∂ H z ∂ y 2 c 22 c H y = − − jkk 2 c ∂ H z ∂ y jωϵk 2 c ∂ E z ∂ xH x = − + jkk 2 c ∂ H z ∂ x jωϵk 2 c ∂ E z ∂ y z z z z η ≡ \|E T \|\|H T \| © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information. Download to read ad-free or(5)Because ( 4 ) only involves lengths of two-dimensional vectors, η must be independent of the coordinate system. Problem 16.6 confirms the value of η by recalculating it in cylindrical coordinates. In Problem 16.7 it is shown (using rectangularcoordinates) that(6) 16.4. Determination of the Axial Fields All that remains for a complete description of the TE and TM modes is the determination of the respective axial fields: F = H for TE; F = E for TM. The good word is that F e , being a Cartesian component of F (in either rectangular or cylindricalcoordinates), must satisfy the scalar wave equation found in Section 14.2 ,(7)together with appropriate boundary conditions which are inferred from the boundary conditions on the components of F .[ Warning : Transverse components such as H e are not Cartesian components and do not obey a scalar wave equation.] 16.4.1. Explicit Solutions for TE Modes of a Rectangular Guide. The wave equation ( 7 ) becomeswhere, as previously defined, k = ω μ ϵ − k . Solving by separation of variables ( Section 9.7 ),(8) \|E T \| 2 = \| E x \| 2 + ∣∣ E y ∣∣ 2 = ( ) 2 ( ∣∣ H y ∣∣ 2 \| H x \| 2 ) = ( ) 2 \|H T \| 2 ωμk TE ωμk TE η TE = ωμk TE TE η = k TM z z z z z − 2 ( F z e − jkz ) = − 2 ( F z e − jkz ) T ϕ − jkz + k 2 c TE H z = 0∂ 2 H z ∂ x 2 ∂ 2 H z ∂ y 2 2 c TE22TE H z ( x , y ) = ( x k x x + x k x x A y cos k y y + B y sin k y y ) © McGraw-Hill Education. All rights reserved. Any use is subject to the Terms of Use, Privacy Notice and copyright information. 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1763	https://fiveable.me/key-terms/intermediate-algebra/percent-concentration	new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Intermediate Algebra Percent Concentration 📘intermediate algebra review key term - Percent Concentration Citation: MLA Definition Percent concentration is a measure of the amount of a substance present in a solution or mixture, expressed as a percentage of the total volume or mass of the solution. It is a crucial concept in understanding and solving mixture and uniform motion applications. 5 Must Know Facts For Your Next Test Percent concentration is often used to describe the composition of a solution, where the solute is measured as a percentage of the total solution. In mixture problems, percent concentration is used to determine the relative amounts of different components in a mixture and how they affect the overall properties of the mixture. Uniform motion applications may involve percent concentration, such as calculating the concentration of a substance in a moving solution or the rate of change in concentration over time. Percent concentration can be calculated using the formula: Percent concentration = (Mass or Volume of Solute / Total Mass or Volume of Solution) × 100. Understanding percent concentration is essential for solving problems related to the composition and behavior of solutions, mixtures, and moving systems. Review Questions Explain how percent concentration is used in mixture problems to determine the relative amounts of different components. In mixture problems, percent concentration is used to describe the composition of the mixture and the relative amounts of the different components. By knowing the percent concentration of each component, you can determine the proportions of the substances in the mixture and how they contribute to the overall properties of the mixture. For example, if a mixture contains 60% water and 40% alcohol, the percent concentration of each component can be used to calculate the volume or mass of each substance in the mixture, which is crucial for solving mixture-related problems. Describe how percent concentration is relevant in uniform motion applications, such as calculating the concentration of a substance in a moving solution or the rate of change in concentration over time. In uniform motion applications, percent concentration is important for understanding the behavior and composition of moving solutions or mixtures. For instance, if a substance is being transported in a moving solution, the percent concentration of the substance can be used to determine its concentration at different points along the motion path. Additionally, the rate of change in the percent concentration over time can be calculated, which is useful for understanding the dynamics of the system and how the composition is evolving as the solution or mixture moves. Knowing the percent concentration in these types of problems is crucial for accurately modeling and predicting the behavior of the system. Analyze how the formula for calculating percent concentration, $\text{Percent concentration} = \left(\frac{\text{Mass or Volume of Solute}}{\text{Total Mass or Volume of Solution}}\right) \times 100$, can be rearranged and applied to solve different types of mixture and uniform motion problems. The formula for percent concentration can be rearranged and applied in various ways to solve mixture and uniform motion problems. For example, if you know the total mass or volume of a solution and the percent concentration of a solute, you can use the formula to calculate the mass or volume of the solute. Conversely, if you know the mass or volume of the solute and the total mass or volume of the solution, you can use the formula to calculate the percent concentration. This flexibility allows the percent concentration formula to be used to determine the composition of mixtures, the concentration of substances in moving solutions, and the changes in concentration over time, which are all crucial for solving a wide range of mixture and uniform motion applications. Related terms Solute: The substance that is dissolved in a solution, typically in smaller amounts compared to the solvent. Solvent: The substance in which the solute is dissolved, typically in larger amounts compared to the solute. Concentration: The measure of the amount of a substance present in a solution or mixture, typically expressed as a ratio or percentage. 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1764	https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/23%3A_Electromagnetic_Induction_AC_Circuits_and_Electrical_Technologies/23.08%3A_Electric_Generators	Skip to main content 23.8: Electric Generators Last updated : Feb 20, 2022 Save as PDF 23.7: Eddy Currents and Magnetic Damping 23.9: Back Emf Page ID : 2711 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives By the end of this section, you will be able to: Calculate the emf induced in a generator. Calculate the peak emf which can be induced in a particular generator system. Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in "Induced Emf and Magnetic Flux." We will now explore generators in more detail. Consider the following example. Example 23.8.123.8.1: Calculating the Emf Induced in a Generator Coil The generator coil shown in Figure 23.8.123.8.1 is rotated through one-fourth of a revolution (from θ=0∘θ=0∘ to θ=90∘θ=90∘ ) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced? Strategy: We use Faraday’s law of induction to find the average emf induced over a time ΔtΔt: emf=−NΔΦΔt. emf=−NΔΦΔt.(23.8.1) We know that N=200N=200 and Δt=15.0msΔt=15.0ms, and so we must determine the change in flux ΔΦΔΦ to find emf. Solution: Since the area of the loop and the magnetic field strength are constant, we see that ΔΦ=Δ(BAcosθ)=ABΔ(cosθ). ΔΦ=Δ(BAcosθ)=ABΔ(cosθ).(23.8.2) Now, Δ(cosθ)=−1.0Δ(cosθ)=−1.0, since it was given that θθ goes from 0∘0∘ to 90∘90∘. Thus ΔΦ=−ABΔΦ=−AB, and emf=NABΔt. emf=NABΔt.(23.8.3) The area of the loop is A=πr2=(3.14...)(0.0500m)2=7.85×10−3m2A=πr2=(3.14...)(0.0500m)2=7.85×10−3m2. Entering this value gives emf=200(7.85×10−3m2)(1.25T)1.50×10−3s=131V. emf=200(7.85×10−3m2)(1.25T)1.50×10−3s=131V.(23.8.4) Discussion: This is a practical average value, similar to the 120 V used in household power. The emf calculated in the example is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width \w and height l in a uniform magnetic field, as illustrated in Figure 23.8.2. Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be emf=Blv, where the velocity v is perpendicular to the magnetic field B. Here the velocity is at an angle θ with B, so that its component perpendicular to B is vsinθ (Figure 23.8.2). Thus in this case the emf induced on each side is emf=Blvsinθ, and they are in the same direction. The total emf around the loop is then emf=2Blvsinθ. This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity ω. The angle θ is related to angular velocity by θ=ωt, so that emf=2Blvsinωt. Now, linear velovity v is related to angular velocity ω by v=rω. Here r=ω/2, so that v=(w/2)ω, and emf=2Blw2ωsinωt=(w)Bωsinωt. Noting that the area of the loop is A=w, and allowing for N loops, we find that emf=NABωsinωt is the emf induced in a generator coil of N turns and area A rotating at a constant angular velocity ω in a uniform magnetic field B. This can also be expressed as emf=emf0sinωt, where emf0=NABω is the maximum (peak) emf. Note that the frequency of the oscillation is f=ω/2π, and the period is T=1/f=2π/ω. Figure 23.8.3 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal. The fact that the peak emf, emf0=NABω, makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater ω), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties. One of the authors as a juvenile found it amusing to ride his bicycle fast enough to burn out his lights, until he had to ride home lightless one dark night. Figure shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements of multiple coils and split rings can produce smoother DC, although electronic rather than mechanical means are usually used to make ripple-free DC. Figure 23.8.4: Split rings, called commutators, produce a pulsed DC emf output in this configuration. Example 23.8.2: Calculating the Maximum Emf of a Generator Calculate the maximum emf, emf0, of the generator that was the subject of Example. Strategy Once ω, the angular velocity, is determined, emf0=NABω can be used to find emf0. All other quantities are known. Solution Angular velocity is defined to be the change in angle per unit time: ω=ΔθΔt. One-fourth of a revolution is π/2 radians, and the time is 0.0150 s; thus, ω=π/2rad0.0150s=104.7rad/s. 104.7 rad/s is exactly 1000 rpm. We substitute this value for ω and the information from the previous example into emf0=NABω, yielding emf0=NABω=200(7.85×10−3m2)(1.25T)(104.7rad/s)=206V. Discussion The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be. In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. 23.8.5 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Back Emf, we shall further explore the action of a motor as a generator. Summary An electric generator rotates a coil in a magnetic field, inducing an emfgiven as a function of time by emf=NABωsinωt, where A is the area of an N-turn coil rotated at a constant angular velocity ω in a uniform magnetic field B. The peak emf (emf_0) of a generator is emf0=NABω. electric generator : a device for converting mechanical work into electric energy; it induces an emf by rotating a coil in a magnetic field emf induced in a generator coil : emf=NABωsinωt, where A is the area of an N-turn coil rotated at a constant angular velocity ω in a uniform magnetic field B, over a period of time t peak emf : (emf_0=NABω)0=NABω 23.7: Eddy Currents and Magnetic Damping 23.9: Back Emf
1765	https://clinicaltrials.gov/study/NCT01515943	Study Details \| NCT01515943 \| Convergence Insufficiency Treatment Study (CITS) \| ClinicalTrials.gov Hide glossary Glossary Study record managers: refer to the Data Element Definitions if submitting registration or results information. Search for terms Accepts healthy volunteers A type of eligibility criteria that indicates whether people who do not have the condition/disease being studied can participate in that clinical study. Active comparator arm An arm type in which a group of participants receives an intervention/treatment considered to be effective (or active) by health care providers. Adverse event An unfavorable change in the health of a participant, including abnormal laboratory findings, that happens during a clinical study or within a certain amount of time after the study has ended. This change may or may not be caused by the intervention/treatment being studied. Age or age group A type of eligibility criteria that indicates the age a person must be to participate in a clinical study. This may be indicated by a specific age or the following age groups: The age groups are: Child (birth-17) Adult (18-64) Older Adult (65+) All-cause mortality A measure of all deaths, due to any cause, that occur during a clinical study. Allocation A method used to assign participants to an arm of a clinical study. The types of allocation are randomized allocation and nonrandomized. Arm A group or subgroup of participants in a clinical trial that receives a specific intervention/treatment, or no intervention, according to the trial's protocol. Arm type A general description of the clinical trial arm. It identifies the role of the intervention that participants receive. Types of arms include experimental arm, active comparator arm, placebo comparator arm, sham comparator arm, and no intervention arm. Baseline characteristics Data collected at the beginning of a clinical study for all participants and for each arm or comparison group. These data include demographics, such as age, sex/gender, race and ethnicity, and study-specific measures (for example, systolic blood pressure, prior antidepressant treatment). Canceled submission Indicates that the study sponsor or investigator recalled a submission of study results before quality control (QC) review took place. If the submission was canceled on or after May 8, 2018, the date is shown. After submission of study results, a study record cannot be modified until QC review is completed, unless the submission is canceled. Certain agreements Information required by the Food and Drug Administration Amendments Act of 2007. In general, this is a description of any agreement between the sponsor of a clinical study and the principal investigator (PI) that does not allow the PI to discuss the results of the study or publish the study results in a scientific or academic journal after the study is completed. Certification A sponsor or investigator may submit a certification to delay submission of results information if they are applying for FDA approval of a new drug or device, or new use of an already approved drug or device. A sponsor or investigator who submits a certification can delay results submission up to 2 years after the certification/extension first submitted date, unless certain events occur sooner. See Delay Results Type in the Results Data Element definitions for more information about this certification. Certification/extension first posted The date on which information about a certification to delay submission of results or an extension request was first available on ClinicalTrials.gov. ClinicalTrials.gov does not indicate whether the submission was a certification or extension request. There is typically a delay between the date the study sponsor or investigator submitted the certification or extension request and the first posted date. Certification/extension first submitted The date on which the study sponsor or investigator first submitted a certification or an extension request to delay submission of results. A sponsor or investigator who submits a certification can delay results submission up to 2 years after this date, unless certain events occur sooner. There is typically a delay between the date the certification or extension request was submitted and the date the information is first available on ClinicalTrials.gov (certification/extension first posted). Certification/extension first submitted that met QC criteria The date on which the study sponsor or investigator first submitted a certification or an extension request that is consistent with National Library of Medicine (NLM) quality control (QC) review criteria. The sponsor or investigator may need to revise and submit a certification or extension request one or more times before NLM's QC review criteria are met. It is the responsibility of the sponsor or investigator to ensure that the study record is consistent with the NLM QC review criteria. Meeting QC criteria for an extension request does not mean that the National Institutes of Health (NIH) has determined that the request demonstrates good cause. The process for review and granting of extension requests by the NIH is being developed. Clinical study A research study involving human volunteers (also called participants) that is intended to add to medical knowledge. There are two types of clinical studies: interventional studies (also called clinical trials) and observational studies. Clinical trial Another name for an interventional study. ClinicalTrials.gov identifier (NCT number)The unique identification code given to each clinical study upon registration at ClinicalTrials.gov. The format is "NCT" followed by an 8-digit number (for example, NCT00000419). Collaborator An organization other than the sponsor that provides support for a clinical study. This support may include activities related to funding, design, implementation, data analysis, or reporting. Condition/disease The disease, disorder, syndrome, illness, or injury that is being studied. On ClinicalTrials.gov, conditions may also include other health-related issues, such as lifespan, quality of life, and health risks. Contact The name and contact information for the person who can answer enrollment questions for a clinical study. Each location where the study is being conducted may also have a specific contact, who may be better able to answer those questions. Cross-over assignment A type of intervention model describing a clinical trial in which groups of participants receive two or more interventions in a specific order. For example, two-by-two cross-over assignment involves two groups of participants. One group receives drug A during the initial phase of the trial, followed by drug B during a later phase. The other group receives drug B during the initial phase, followed by drug A. So during the trial, participants "cross over" to the other drug. All participants receive drug A and drug B at some point during the trial but in a different order, depending on the group to which they are assigned. Data Monitoring Committee (DMC)A group of independent scientists who monitor the safety and scientific integrity of a clinical trial. The DMC can recommend to the sponsor that the trial be stopped if it is not effective, is harming participants, or is unlikely to serve its scientific purpose. Members are chosen based on the scientific skills and knowledge needed to monitor the particular trial. Also called a data safety and monitoring board, or DSMB. Early Phase 1 (formerly listed as Phase 0)A phase of research used to describe exploratory trials conducted before traditional phase 1 trials to investigate how or whether a drug affects the body. They involve very limited human exposure to the drug and have no therapeutic or diagnostic goals (for example, screening studies, microdose studies). Eligibility criteria The key requirements that people who want to participate in a clinical study must meet or the characteristics they must have. Eligibility criteria consist of both inclusion criteria (which are required for a person to participate in the study) and exclusion criteria (which prevent a person from participating). Types of eligibility criteria include whether a study accepts healthy volunteers, has age or age group requirements, or is limited by sex. Enrollment The number of participants in a clinical study. The "estimated" enrollment is the target number of participants that the researchers need for the study. Exclusion criteria A type of eligibility criteria. These are reasons that a person is not allowed to participate in a clinical study. Expanded access A way for patients with serious diseases or conditions who cannot participate in a clinical trial to gain access to a medical product that has not been approved by the U.S. Food and Drug Administration (FDA). Also called compassionate use. There are different expanded access types. For more information, see FDA Expanded Access: Information for Patients. Expanded access status Available:Expanded access is currently available for this investigational treatment, and patients who are not participants in the clinical study may be able to gain access to the drug, biologic, or medical device being studied. No longer available:Expanded access was available for this intervention previously but is not currently available and will not be available in the future. Temporarily not available:Expanded access is not currently available for this intervention but is expected to be available in the future. Approved for marketing: The intervention has been approved by the U.S. Food and Drug Administration for use by the public. Expanded access type Describes the category of expanded access under U.S. Food and Drug Administration (FDA) regulations. There are three types of expanded access: Individual Patients: Allows a single patient, with a serious disease or condition who cannot participate in a clinical trial, access to a drug or biological product that has not been approved by the FDA. This category also includes access in an emergency situation. Intermediate-size Population: Allows more than one patient (but generally fewer patients than through a Treatment IND/Protocol) access to a drug or biological product that has not been approved by the FDA. This type of expanded access is used when multiple patients with the same disease or condition seek access to a specific drug or biological product that has not been approved by the FDA. Treatment IND/Protocol: Allows a large, widespread population access to a drug or biological product that has not been approved by the FDA. This type of expanded access can only be provided if the product is already being developed for marketing for the same use as the expanded access use. Experimental arm An arm type in which a group of participants receives the intervention/treatment that is the focus of the clinical trial. Extension request In certain circumstances, a sponsor or investigator may request an extension to delay the standard results submission deadline (generally one year after the primary completion date). The request for an extension must demonstrate good cause (for example, the need to preserve the scientific integrity of an ongoing masked trial). All requests must be reviewed and granted by the National Institutes of Health. This process for review and granting of extension requests is being developed. See Delay Results Type in the Results Data Element definitions for more information. Facility name The name of the hospital or institution where a clinical study takes place. Note that not all study records include this information. To use this filter, you can enter some or all of a facility name, or type a few letters and select from the list that appears. Factorial assignment A type of intervention model describing a clinical trial in which groups of participants receive one of several combinations of interventions. For example, two-by-two factorial assignment involves four groups of participants. Each group receives one of the following pairs of interventions: (1) drug A and drug B, (2) drug A and a placebo, (3) a placebo and drug B, or (4) a placebo and a placebo. So during the trial, all possible combinations of the two drugs (A and B) and the placebos are given to different groups of participants. FDAAA 801 Violations A FDAAA 801 Violation is shown on a study record when the U.S. Food and Drug Administration (FDA) has issued a Notice of Noncompliance to the responsible party of an applicable clinical trial. A Notice of Noncompliance indicates that the FDA has determined the responsible party was not in compliance with the registration or results reporting requirements for the clinical trial under the Food and Drug Administration Amendments Act of 2007, Section 801 (FDAAA 801). The National Library of Medicine (NLM) is required by FDAAA 801 to add information to a study record about any FDAAA 801 Violation. This information is provided by the FDA. There are three categories of information that may be included: Violation: Shown when the FDA issues a Notice of Noncompliance and posts the Notice of Noncompliance on its designated webpage. There are three types of violations: Failure to submit required clinical trial information Submission of false or misleading clinical trial information Failure to submit primary and secondary outcomes Correction: Shown when the FDA confirms that the responsible party has updated the study record to correct the violation and posts the correction notice on its designated webpage. Because of the time for FDA review and processing, there may be a delay between the date when the study record was updated and the addition of correction information to the FDAAA 801 Violation information. Penalty: Shown when the FDA imposes a penalty for the violation and posts the penalty notice on its designated webpage. First posted The date on which the study record was first available on ClinicalTrials.gov after National Library of Medicine (NLM) quality control (QC) review has concluded. There is typically a delay of a few days between the date the study sponsor or investigator submitted the study record and the first posted date. First submitted The date on which the study sponsor or investigator first submitted a study record to ClinicalTrials.gov. There is typically a delay of a few days between the first submitted date and the record's availability on ClinicalTrials.gov (the first posted date). First submitted that met QC criteria The date on which the study sponsor or investigator first submits a study record that is consistent with National Library of Medicine (NLM) quality control (QC) review criteria. The sponsor or investigator may need to revise and submit a study record one or more times before NLM's QC review criteria are met. It is the responsibility of the sponsor or investigator to ensure that the study record is consistent with the NLM QC review criteria. Food and Drug Administration Amendments Act of 2007, Section 801 (FDAAA 801)U.S. Public Law 110-85, which was enacted on September 27, 2007. Section 801 of FDAAA amends Section 402 of the U.S. Public Health Service Act to expand ClinicalTrials.gov and create a clinical study results database. For more information on FDAAA 801, see the Clinical Trial Reporting Requirements page on this site. Funder type Describes the organization that provides funding or support for a clinical study. This support may include activities related to funding, design, implementation, data analysis, or reporting. Organizations listed as sponsors and collaborators for a study are considered the funders of the study. ClinicalTrials.gov refers to four types of funders: U.S. National Institutes of Health Other U.S. Federal agencies (for example, Food and Drug Administration, Centers for Disease Control and Prevention, or U.S. Department of Veterans Affairs) Industry (for example: pharmaceutical and device companies) All others (including individuals, universities, and community-based organizations) Gender-based eligibility A type of eligibility criteria that indicates whether eligibility to participate in a clinical study is based on a person's self-representation of gender identity. Gender identity refers to a person's own sense of gender, which may or may not be the same as their biological sex. Group/cohort A group or subgroup of participants in an observational study that is assessed for biomedical or health outcomes. Human subjects protection review board A group of people who review, approve, and monitor the clinical study's protocol. Their role is to protect the rights and welfare of people participating in a study (referred to as human research subjects), such as reviewing the informed consent form. The group typically includes people with varying backgrounds, including a community member, to make sure that research activities conducted by an organization are completely and adequately reviewed. Also called an institutional review board, or IRB, or an ethics committee. For more information, see Who can join clinical research? on this site. Inclusion criteria A type of eligibility criteria. These are the reasons that a person is allowed to participate in a clinical study. Informed consent A process used by researchers to communicate to potential and enrolled participants the risks and potential benefits of participating in a clinical study. For more information, see Who can join clinical research? on this site. Informed consent form (ICF)The document used in the informed consent or process. Intervention model The general design of the strategy for assigning interventions to participants in a clinical study. Types of intervention models include: single group assignment, parallel assignment, cross-over assignment, and factorial assignment. Intervention/treatment A process or action that is the focus of a clinical study. Interventions include drugs, medical devices, procedures, vaccines, and other products that are either investigational or already available. Interventions can also include noninvasive approaches, such as education or modifying diet and exercise. Interventional study (clinical trial)A type of clinical study in which participants are assigned to groups that receive one or more intervention/treatment (or no intervention) so that researchers can evaluate the effects of the interventions on biomedical or health-related outcomes. The assignments are determined by the study's protocol. Participants may receive diagnostic, therapeutic, or other types of interventions. Investigator A researcher involved in a clinical study. Related terms include site principal investigator, site sub-investigator, study chair, study director, and study principal investigator. Last update posted The most recent date on which changes to a study record were made available on ClinicalTrials.gov. There may be a delay between when the changes were submitted to ClinicalTrials.gov by the study's sponsor or investigator (the last update submitted date) and the last update posted date. Last update submitted The most recent date on which the study sponsor or investigator submitted changes to a study record to ClinicalTrials.gov. There is typically a delay of a few days between the last update submitted date and when the date changes are posted on ClinicalTrials.gov (the last update posted date). Last update submitted that met QC criteria The most recent date on which the study sponsor or investigator submitted changes to a study record that are consistent with National Library of Medicine (NLM) quality control (QC) review criteria. It is the responsibility of the sponsor or investigator to ensure that the study record is consistent with the NLM QC review criteria. Last verified The most recent date on which the study sponsor or investigator confirmed the information about a clinical study on ClinicalTrials.gov as accurate and current. If a study with a recruitment status of recruiting; not yet recruiting; or active, not recruiting has not been confirmed within the past 2 years, the study's recruitment status is shown as unknown. Location A place where a research site for a clinical study can be found. Location information can be searched using a facility name, a city, state, zip code, or country. A location where a study is being conducted may also include contact information. Location countries Countries in which research facilities for a study are located. A country is listed only once, even if there is more than one facility in the country. The list includes all countries as of the last update submitted date; any country for which all facilities were removed from the study record are listed under removed location countries. Masking A clinical trial design strategy in which one or more parties involved in the trial, such as the investigator or participants, do not know which participants have been assigned which interventions. Types of masking include: open label, single blind masking, and double-blind masking. NCT number A unique identification code given to each clinical study record registered on ClinicalTrials.gov. The format is "NCT" followed by an 8-digit number (for example, NCT00000419). Also called the ClinicalTrials.gov identifier. No intervention arm An arm type in which a group of participants does not receive any intervention/treatment during the clinical trial. Observational study A type of clinical study in which participants are identified as belonging to study groups and are assessed for biomedical or health outcomes. Participants may receive diagnostic, therapeutic, or other types of interventions, but the investigator does not assign participants to a specific interventions/treatment. A patient registry is a type of observational study. Observational study model The general design of the strategy for identifying and following up with participants during an observational study. Types of observational study models include cohort, case-control, case-only, case-cross-over, ecologic or community studies, family-based, and other. Other adverse event An adverse event that is not a serious adverse event, meaning that it does not result in death, is not life-threatening, does not require inpatient hospitalization or extend a current hospital stay, does not result in an ongoing or significant incapacity or interfere substantially with normal life functions, and does not cause a congenital anomaly or birth defect; it also does not put the participant in danger and does not require medical or surgical intervention to prevent one of the results listed above. Other study IDs Identifiers or ID numbers other than the NCT number that are assigned to a clinical study by the study's sponsor, funders, or others. These numbers may include unique identifiers from other trial registries and National Institutes of Health grant numbers. Other terms In the search feature, the Other terms field is used to narrow a search. For example, you may enter the name of a drug or the NCT number of a clinical study to limit the search to study records that contain these words. Outcome measure For clinical trials, a planned measurement described in the protocol that is used to determine the effect of an intervention/treatment on participants. For observational studies, a measurement or observation that is used to describe patterns of diseases or traits, or associations with exposures, risk factors, or treatment. Types of outcome measures include primary outcome measure and secondary outcome measure. Parallel assignment A type of intervention model describing a clinical trial in which two or more groups of participants receive different interventions. For example, a two-arm parallel assignment involves two groups of participants. One group receives drug A, and the other group receives drug B. So during the trial, participants in one group receive drug A "in parallel" to participants in the other group, who receive drug B. Participant flow A summary of the progress of participants through each stage of a clinical study, by study arm or group/cohort. This includes the number of participants who started, completed, and dropped out of the study. Patient registry A type of observational study that collects information about patients' medical conditions and/or treatments to better understand how a condition or treatment affects patients in the real world. Phase The stage of a clinical trial studying a drug or biological product, based on definitions developed by the U.S. Food and Drug Administration (FDA). The phase is based on the study's objective, the number of participants, and other characteristics. There are five phases: Early Phase 1 (formerly listed as Phase 0), Phase 1, Phase 2, Phase 3, and Phase 4. Not Applicable is used to describe trials without FDA-defined phases, including trials of devices or behavioral interventions. Phase 1 A phase of research to describe clinical trials that focus on the safety of a drug. They are usually conducted with healthy volunteers, and the goal is to determine the drug's most frequent and serious adverse events and, often, how the drug is broken down and excreted by the body. These trials usually involve a small number of participants. Phase 2 A phase of research to describe clinical trials that gather preliminary data on whether a drug works in people who have a certain condition/disease (that is, the drug's effectiveness). For example, participants receiving the drug may be compared to similar participants receiving a different treatment, usually an inactive substance (called a placebo) or a different drug. Safety continues to be evaluated, and short-term adverse events are studied. Phase 3 A phase of research to describe clinical trials that gather more information about a drug's safety and effectiveness by studying different populations and different dosages and by using the drug in combination with other drugs. These studies typically involve more participants. Phase 4 A phase of research to describe clinical trials occurring after FDA has approved a drug for marketing. They include postmarket requirement and commitment studies that are required of or agreed to by the study sponsor. These trials gather additional information about a drug's safety, efficacy, or optimal use. Phase Not Applicable Describes trials without FDA-defined phases, including trials of devices or behavioral interventions. Placebo An inactive substance or treatment that looks the same as, and is given in the same way as, an active drug or intervention/treatment being studied. Placebo comparator arm An arm type in which a group of participants receives a placebo during a clinical trial. Primary completion date The date on which the last participant in a clinical study was examined or received an intervention to collect final data for the primary outcome measure. Whether the clinical study ended according to the protocol or was terminated does not affect this date. For clinical studies with more than one primary outcome measure with different completion dates, this term refers to the date on which data collection is completed for all the primary outcome measures. The "estimated" primary completion date is the date that the researchers think will be the primary completion date for the study. Primary outcome measure In a clinical study's protocol, the planned outcome measure that is the most important for evaluating the effect of an intervention/treatment. Most clinical studies have one primary outcome measure, but some have more than one. Primary purpose The main reason for the clinical trial. The types of primary purpose are: treatment, prevention, diagnostic, supportive care, screening, health services research, basic science, and other. Principal investigator (PI)The person who is responsible for the scientific and technical direction of the entire clinical study. Protocol The written description of a clinical study. It includes the study's objectives, design, and methods. It may also include relevant scientific background and statistical information. Quality control (QC) review National Library of Medicine (NLM) staff perform a limited review of submitted study records for apparent errors, deficiencies, or inconsistencies. NLM staff identify potential major and advisory issues and provide comments directly to the study sponsor or investigator. Major issues identified in QC review must be addressed or corrected (see First submitted that met QC criteria and Results first submitted that met QC criteria). Advisory issues are suggestions to help improve the clarity of the record. NLM staff do not verify the scientific validity or relevance of the submitted information. The study sponsor or investigator is responsible for ensuring that the studies follow all applicable laws and regulations. Randomized allocation A type of allocation strategy in which participants are assigned to the arms of a clinical trial by chance. Recruitment status Not yet recruiting: The study has not started recruiting participants. Recruiting: The study is currently recruiting participants. Enrolling by invitation: The study is selecting its participants from a population, or group of people, decided on by the researchers in advance. These studies are not open to everyone who meets the eligibility criteria but only to people in that particular population, who are specifically invited to participate. Active, not recruiting: The study is ongoing, and participants are receiving an intervention or being examined, but potential participants are not currently being recruited or enrolled. Suspended: The study has stopped early but may start again. Terminated: The study has stopped early and will not start again. Participants are no longer being examined or treated. Completed: The study has ended normally, and participants are no longer being examined or treated (that is, the last participant's last visit has occurred). Withdrawn: The study stopped early, before enrolling its first participant. Unknown: A study on ClinicalTrials.gov whose last known status was recruiting; not yet recruiting; or active, not recruiting but that has passed its completion date, and the status has not been last verified within the past 2 years. Registration The process of submitting and updating summary information about a clinical study and its protocol, from its beginning to end, to a structured, public Web-based study registry that is accessible to the public, such as ClinicalTrials.gov. Removed location countries Countries that appeared under location countries but were removed from the study record by the sponsor or investigator. Reporting group A grouping of participants in a clinical study that is used for summarizing the data collected during the study. This grouping may be the same as or different from a study arm or group. Responsible party The person responsible for submitting information about a clinical study to ClinicalTrials.gov and updating that information. Usually the study sponsor or investigator. Results database A structured online system, such as the ClinicalTrials.gov results database, that provides the public with access to registration and summary results information for completed or terminated clinical studies. A study with results available on ClinicalTrials.gov is described as having the results "posted." Note: The ClinicalTrials.gov results database became available in September 2008. Older studies are unlikely to have results available in the database. Results delayed Indicates that the sponsor or investigator submitted a certification or extension request. Results first posted The date on which summary results information was first available on ClinicalTrials.gov after National Library of Medicine (NLM) quality control (QC) review has concluded. There is typically a delay between the date the study sponsor or investigator first submits summary results information (the results first submitted date) and the results first posted date. Some results information may be available at an earlier date if Results First Posted with QC Comments. Results first posted with QC comments The date on which summary results information was first available on ClinicalTrials.gov with quality control review comments from the National Library of Medicine (NLM) identifying major issues that must be addressed by the sponsor or investigator. As of January 1, 2020, initial results submissions for applicable clinical trials (ACTs) that do not meet quality control review criteria will be publicly posted on ClinicalTrials.gov with brief standardized major comments. Accordingly, the Results First Posted with QC Comments date may be earlier than the Results First Posted date for an ACT with summary results information that is not consistent with NLM quality control review criteria. Results first submitted The date on which the study sponsor or investigator first submits a study record with summary results information. There is typically a delay between the results first submitted date and when summary results information becomes available on ClinicalTrials.gov (the results first posted date). Results first submitted that met QC criteria The date on which the study sponsor or investigator first submits a study record with summary results information that is consistent with National Library of Medicine (NLM) quality control (QC) review criteria. The sponsor or investigator may need to revise and submit results information one or more times before NLM's QC review criteria are met. It is the responsibility of the sponsor or investigator to ensure that the study record is consistent with the NLM QC review criteria. Results returned after quality control review The date on which the National Library of Medicine provided quality control (QC) review comments to the study sponsor or investigator. The sponsor or investigator must address major issues identified in the review comments. If there is a date listed for results returned after quality control review, but there is not a subsequent date listed for results submitted to ClinicalTrials.gov, this means that the submission is pending changes by the sponsor or investigator. Results submitted to ClinicalTrials.gov Indicates that the study sponsor or investigator has submitted summary results information for a clinical study to ClinicalTrials.gov but the quality control (QC) review process has not concluded. The results submitted date indicates when the study sponsor or investigator first submitted summary results information or submitted changes to summary results information. Submissions with changes are typically in response to QC review comments from the National Library of Medicine (NLM). If there is a date listed for results submitted to ClinicalTrials.gov, but there is not a subsequent date listed for results returned after quality control review, this means that the submission is pending review by NLM. Secondary outcome measure In a clinical study's protocol, a planned outcome measure that is not as important as the primary outcome measure for evaluating the effect of an intervention but is still of interest. Most clinical studies have more than one secondary outcome measure. Serious adverse event An adverse event that results in death, is life-threatening, requires inpatient hospitalization or extends a current hospital stay, results in an ongoing or significant incapacity or interferes substantially with normal life functions, or causes a congenital anomaly or birth defect. Medical events that do not result in death, are not life-threatening, or do not require hospitalization may be considered serious adverse events if they put the participant in danger or require medical or surgical intervention to prevent one of the results listed above. Sex A type of eligibility criteria that indicates the sex of people who may participate in a clinical study (all, female, male). Sex is a person's classification as female or male based on biological distinctions. Sex is distinct from gender-based eligibility. Sham comparator arm An arm type in which a group of participants receives a procedure or device that appears to be the same as the actual procedure or device being studied but does not contain active processes or components. Single group assignment A type of intervention model describing a clinical trial in which all participants receive the same intervention/treatment. Sort studies by The Sort studies by option is used to change the order of studies listed on the Search Results page. You can sort by Relevance or Newest First: Relevance: Studies that best match your search terms appear higher in the search results list. This is the default display for all searches. Newest First: Studies with the most recent First posted dates appear higher in the search results list. Sponsor The organization or person who initiates the study and who has authority and control over the study. Statistical analysis plan (SAP)The written description of the statistical considerations and methods for analyzing the data collected in the clinical study. Status Indicates the current recruitment status or the expanded access status. Study completion date The date on which the last participant in a clinical study was examined or received an intervention/treatment to collect final data for the primary outcome measures, secondary outcome measures, and adverse events (that is, the last participant's last visit). The "estimated" study completion date is the date that the researchers think will be the study completion date. Study design The investigative methods and strategies used in the clinical study. Study documents Refers to the type of documents that the study sponsor or principal investigator may add to their study record. These include a study protocol, statistical analysis plan, and informed consent form. Study IDs Identifiers that are assigned to a clinical study by the study's sponsor, funders, or others. They include unique identifiers from other trial study registries and National Institutes of Health grant numbers. Note: ClinicalTrials.gov assigns a unique identification code to each clinical study registered on ClinicalTrials.gov. Also called the NCT number, the format is "NCT" followed by an 8-digit number (for example, NCT00000419). Study record An entry on ClinicalTrials.gov that contains a summary of a clinical study's protocol information, including the recruitment status; eligibility criteria; contact information; and, in some cases, summary results. Each study record is assigned a ClinicalTrials.gov identifier, or NCT number. Study registry A structured online system, such as ClinicalTrials.gov, that provides the public with access to summary information about ongoing and completed clinical studies. Study results A study record that includes the summary results posted in the ClinicalTrials.gov results database. Summary results information includes participant flow, baseline characteristics, outcome measures, and adverse events (including serious adverse events). Study start date The actual date on which the first participant was enrolled in a clinical study. The "estimated" study start date is the date that the researchers think will be the study start date. Study type Describes the nature of a clinical study. Study types include interventional studies (also called clinical trials), observational studies (including patient registries), and expanded access. Submitted date The date on which the study sponsor or investigator submitted a study record that is consistent with National Library of Medicine (NLM) quality control (QC) review criteria. Title The official title of a protocol used to identify a clinical study or a short title written in language intended for the lay public. Title acronym The acronym or initials used to identify a clinical study (not all studies have one). For example, the title acronym for the Women's Health Initiative is "WHI." Type of intervention A process or action that is the focus of a clinical study. Interventions include drugs, medical devices, procedures, vaccines, and other products that are either investigational or already available. Interventions can also include noninvasive approaches, such as education or modifying diet and exercise. U.S. Agency for Healthcare Research and Quality (AHRQ)An agency within the U.S. Department of Health and Human Services. AHRQ's mission is to produce evidence to make health care safer, higher quality, more accessible, equitable, and affordable, and to work within the U.S. Department of Health and Human Services and with other partners to make sure that the evidence is understood and used. U.S. Food and Drug Administration (FDA)An agency within the U.S. Department of Health and Human Services. The FDA is responsible for protecting the public health by making sure that human and veterinary drugs, vaccines and other biological products, medical devices, the Nation's food supply, cosmetics, dietary supplements, and products that give off radiation are safe, effective, and secure. Unknown A type of recruitment status. It identifies a study on ClinicalTrials.gov whose last known status was recruiting; not yet recruiting; or active, not recruiting but that has passed its completion date, and the status has not been verified within the past 2 years. Studies with an unknown status are considered closed studies. 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Read our full for details. ClinicalTrials.gov is a website and online database of clinical research studies and information about their results. The National Library of Medicine (NLM) maintains the website. The study sponsor or investigator submits information about their study to ClinicalTrials.gov and is responsible for the safety, science, and accuracy of any study they list. Before joining a study, talk to your health care professional about possible risks and benefits. To learn more about taking part in studies, read . ClinicalTrials.gov ID Sponsor Information provided by Last Update Posted What would you like to download? File Format Study Details Researcher View Results Posted Record History On this page Study Overview Contacts and Locations Participation Criteria Study Plan Collaborators and Investigators Publications Study Record Dates More Information Study Overview Contacts and Locations This section provides contact details for people who can answer questions about joining this study, and information on where this study is taking place. To learn more, please see the . Participation Criteria Researchers look for people who fit a certain description, called eligibility criteria. Some examples of these criteria are a person's general health condition or prior treatments. For general information about clinical research, read Learn About Studies( Eligibility Criteria Description Inclusion Criteria: Age: 9 to <18 years Patient has access to a computer with a CD/DVD drive and internet access for the next 12 weeks Best-corrected visual acuity of ≥20/25 in each eye at distance and near Exophoria at near at least 4 pd greater than at distance Reduced positive fusional convergence at near (<20 pd or fails Sheard's criterion that the PFV measures less than twice the magnitude of the near phoria). PFV is recorded as the prism magnitude where vision is first blurred (or break if no blur is reported). Near point of convergence of ≥6 cm break Randot Preschool stereoacuity of at least 400 seconds of arc CI Symptom Survey score ≥16 No use of a plus add for near or base-in prism for at least 2 weeks preceding enrollment Patient must be wearing appropriate refractive correction (spectacle or contact lenses) for at least 2 weeks prior to enrollment if refractive error is present (based on a cycloplegic refraction within the last 6 months) that meets the following: Myopia more than -0.75D spherical equivalent (SE) in either eye Hyperopia more than +2.00D SE in either eye SE anisometropia >1.00D Astigmatism > 1.00D or > 1.50D of meridional difference in either eye Refractive correction for patients meeting the above refractive error criteria must meet the following guidelines: SE anisometropia must be within 0.25D of the full anisometropic correction. Astigmatism cylinder must be within 0.50D of full correction and axis must be within 5 degrees. For hyperopia, the spherical component can be reduced by up to 1.50D at investigator discretion provided the reduction is symmetrical and results in residual (i.e., uncorrected) SE hyperopia that does not exceed +2.00D. For myopia, the SE must be within 0.25D of the full myopic correction. Parent and patient understand the protocol and are willing to accept randomization. Parent has home phone (or access to phone) and is willing to be contacted by Jaeb Center staff. Relocation outside of area of an active PEDIG site within the next 15 months is not anticipated. Exclusion Criteria: ≥2 logMAR line difference in best-corrected visual acuity between the two eyes Constant or intermittent exotropia at distance; constant exotropia at near Any esotropia at distance or near Distance exophoria > 10 pd History of strabismus surgery Anisometropia ≥2.00D in any meridian between the eyes Prior intraocular or refractive surgery Primary vertical heterophoria greater than 1 pd Diseases known to affect accommodation, vergence, and ocular motility such as multiple sclerosis, Graves orbitopathy, myasthenia gravis, diabetes mellitus, or Parkinson disease Current use of any ocular or systemic medication known to affect accommodation or vergence such as anti-anxiety agents (e.g., Librium or Valium), anti-arrhythmic agents (e.g., Cifenline, Cibenzoline), anti- cholinergics (e.g., Motion sickness patch (scopolamine), bladder spasmolytic drugs (e.g., Propiverine), hydroxychloroquine, chloroquine, phenothiazines (e.g., Compazine, Mellaril, Thorazine), tricyclic antidepressants (e.g., Elavil, Nortriptyline, Tofranil) Near point of accommodation >20 cm in the right eye Manifest or latent nystagmus evident clinically History of chronic headaches unrelated to reading activity Active symptomatic allergic conjunctivitis Developmental disability, mental retardation, attention deficit hyperactivity disorder (ADHD), or learning disability diagnosis that in the investigator's discretion would interfere with treatment or evaluation Household member or sibling already enrolled in the CITS OR previously enrolled in the CITT Household member is an eye professional, ophthalmic technician, ophthalmology or optometry resident, orthoptist, or optometry student, or employed in an eye care setting Acquired brain injury Previous office- or home-based vision therapy, orthoptics, home-based near- target push-ups, pencil push-ups, or home-based computer therapy Ages Eligible for Study 9 Years to 17 Years (Child) Sexes Eligible for Study All Accepts Healthy Volunteers No Study Plan This section provides details of the study plan, including how the study is designed and what the study is measuring. Expand all / Collapse all How is the study designed? Design Details Primary Purpose :Treatment Allocation :Randomized Interventional Model :Parallel Assignment Masking :Triple (Participant Care Provider Outcomes Assessor) Arms and Interventions \| Participant Group/Arm \| Intervention/Treatment \| --- \| \| Participant Group/Arm \| Active Comparator: Computer-based therapy (CBT) The CBT group will be assigned active home-based computer vergence/accommodative therapy (15 minutes/day) plus placebo yoked prism flipper therapy (5 minutes/day) for a total of 20/minutes per day, 5 days per week for the 12-week treatment phase. \| Intervention/Treatment \| Other: Active home-based computer vergence/accommodative therapy At enrollment, subjects will be prescribed 15 minutes/day of active home-based computer therapy for 5 days/week during the 12 week treatment phase. Active home-based computer therapy will be provided the Home Therapy System (HTS) computer software and will include both fusional vergence and accommodative therapy. Subjects will perform the computer therapy while wearing red/blue glasses and accommodative therapy will be performed using the HTS accommodative flippers. Please refer to the procedures manual for further details. Other Names: Home Therapy System (HTS, Home Vision Therapy, Inc.) Procedure: Placebo yoked prism flippers Subjects will be prescribed 5 minutes/day of placebo yoked prism flipper therapy for 5 days/week during the 12 week treatment phase. This procedure utilizes a 4 pd base right/4 pd base left prism flipper and "Accommodative Hopping Cards". Subjects view the text on the card through the prisms at 40 cm, perform the appropriate task and then flip the prism flipper to the other side after each word before proceeding to the next line. Please refer to the procedures manual for further details. The task remains constant, but the nature of the procedure changes with time: Weeks 1-4: View the target through prism flippers Weeks 5-8: Wear red-blue filter glasses while viewing the text through prism flippers Weeks 9-12: Wear polaroid glasses while viewing the text through prism flippers \| \| Participant Group/Arm \| Active Comparator: Near target push-up (NTP) The NTP group will be assigned placebo home-based computer vergence/accommodative therapy (5 minutes/day) plus near target push-ups (15 minutes/day) for a total of 20/minutes per day, 5 days per week for the 12-week treatment phase. \| Intervention/Treatment \| Procedure: Near target push-ups At enrollment, subjects will be prescribed 15 minutes/day (3 sessions of 5 minutes each) of near target push-ups (NTP) for 5 days/week during the 12-week treatment phase. An alphabet pencil will be used as the target and an index card placed in the background will provide physiological diplopia control. With the pencil positioned at arm's length directly between the subject's eyes, the subject will slowly bring the pencil toward his/her nose while focusing on the small letter on the pencil. When the subject is no longer able to maintain a single image of the pencil, he/she will slowly move the target away from the nose until the pencil becomes single again. This procedure will be repeated several times. Please refer to the procedures manual for further details. Other Names: Pencil push-ups Other: Placebo home-based computer vergence/accommodative therapy At enrollment, subjects will be prescribed either 5 minutes/day (NTP group) or 15 minutes/day (Placebo group) of placebo home-based computer therapy for 5 days/week during the 12 week treatment phase. Placebo computer-based therapy will be provided by the Home Therapy System (HTS) computer software. The vergence procedures are similar to the active version, however, the tasks will be modified to ensure no demand on the vergence system and no accommodative therapy is included in the placebo version. Please refer to the procedures manual for further details. Other Names: Home Therapy System (HTS, Home Vision Therapy, Inc.) \| \| Participant Group/Arm \| Placebo Comparator: Placebo The placebo group will be assigned placebo home-based computer vergence/accommodative therapy (15 minutes/day) plus placebo yoked prism flipper therapy (5 minutes/day) for a total of 20/minutes per day, 5 days per week for the 12-week treatment phase. \| Intervention/Treatment \| Other: Placebo home-based computer vergence/accommodative therapy At enrollment, subjects will be prescribed either 5 minutes/day (NTP group) or 15 minutes/day (Placebo group) of placebo home-based computer therapy for 5 days/week during the 12 week treatment phase. Placebo computer-based therapy will be provided by the Home Therapy System (HTS) computer software. The vergence procedures are similar to the active version, however, the tasks will be modified to ensure no demand on the vergence system and no accommodative therapy is included in the placebo version. Please refer to the procedures manual for further details. Other Names: Home Therapy System (HTS, Home Vision Therapy, Inc.) Procedure: Placebo yoked prism flippers Subjects will be prescribed 5 minutes/day of placebo yoked prism flipper therapy for 5 days/week during the 12 week treatment phase. This procedure utilizes a 4 pd base right/4 pd base left prism flipper and "Accommodative Hopping Cards". Subjects view the text on the card through the prisms at 40 cm, perform the appropriate task and then flip the prism flipper to the other side after each word before proceeding to the next line. Please refer to the procedures manual for further details. The task remains constant, but the nature of the procedure changes with time: Weeks 1-4: View the target through prism flippers Weeks 5-8: Wear red-blue filter glasses while viewing the text through prism flippers Weeks 9-12: Wear polaroid glasses while viewing the text through prism flippers \| What is the study measuring? Primary Outcome Measures \| Outcome Measure \| Measure Description \| Time Frame \| --- \| Treatment Group Comparison of the Percentage of Participants Classified as an Overall Success at 12 Weeks - HB-C Versus HB-PU \| Pairwise treatment group comparison (HB-C versus HB-PU) of the percentages of participants meeting success criteria using binomial regression adjusting for baseline covariates of CISS score (<28 points vs ≥ 28 points), mean NPC break (<10 cm vs ≥ 10 cm), and mean PFV blur (≥ 15 PD vs <15 PD) using linear contrasts with Bonferroni adjustment for multiple comparisons (Type I error rate = 2.5%). Overall success was defined as meeting all of the following criteria at 12 weeks: 1. Convergence Insufficiency Symptom Survey (CISS): 12-week score <16 points and at least 9-point improvement from baseline at 12 weeks 2. Near point of convergence (NPC) break: 12-week/baseline mean NPC break <0.763 and a mean 12-week NPC break <6 cm 3. Positive fusional vergence (PFV) blur: 12-week/baseline mean PFV blur >1.419 and a mean 12-week PFV break >15 pd \| 12 weeks after randomization (baseline) \| \| Treatment Group Comparison of the Percentage of Participants Classified as an Overall Success at 12 Weeks - HB-C Versus HB-P \| Pairwise treatment group comparison (HB-C versus Placebo) of the percentages of participants meeting success criteria using binomial regression adjusting for baseline covariates of CISS score (<28 points vs ≥ 28 points), mean NPC break (<10 cm vs ≥ 10 cm), and mean PFV blur (≥ 15 PD vs <15 PD) using linear contrasts with Bonferroni adjustment for multiple comparisons. Overall success was defined as meeting all of the following criteria at 12 weeks: 1. Convergence Insufficiency Symptom Survey (CISS): 12-week score <16 points and at least 9-point improvement from baseline at 12 weeks 2. Near point of convergence (NPC) break: 12-week/baseline mean NPC break <0.763 and a mean 12-week NPC break <6 cm 3. Positive fusional vergence (PFV) blur: 12-week/baseline mean PFV blur >1.419 and a mean 12-week PFV break >15 pd \| 12-weeks after randomization (baseline) \| Secondary Outcome Measures \| Outcome Measure \| Measure Description \| Time Frame \| --- \| Number of Participants Classified as Having Met Success Criteria Based on CI Signs/Symptoms at 12 Weeks by Treatment Group \| The number of subjects classified as a success based on signs/symptoms at the 12-week visit. Success is based on the Convergency Insufficiency Symptom Survey (CISS) defined as improvement of 9 or more points from baseline and a 12-week score of <16 points. \| 12 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria Based on the Mean NPC Break at 12 Weeks by Treatment Group \| The number of subjects who are classified as a success based on the mean NPC break at the 12-week visit. Success is based on the mean NPC (near point of convergence) break is defined as having a mean NPC break of <6 cm at 12 weeks and a 12-week to baseline ratio of <0.763 for mean NPC break. \| 12 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria Based on the Mean PFV Blur at 12 Weeks by Treatment Group \| The number of subjects who are classified as a success based on the mean PFV blur at the 12-week visit. Success is based on the mean PFV (positive fusional vergence) blur, defined as having a mean PFV blur of >15 pd at 12 weeks and a 12-week to baseline ratio of >1.419 for mean PFV blur. \| 12 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria for Both Clinical Measures at 12 Weeks by Treatment Group \| The number of subjects classified as a success based on clinical measures of CI (mean NPC break & mean PFV blur) at the 12-week visit. Clinical success is defined according to whether both criteria (below) are met as follows: 1. Near point of convergence (NPC) break: 12-week/baseline mean NPC break <0.763 and a 12-week mean NPC break <6 cm 2. Positive fusional vergence (PFV) blur: 12-week/baseline mean PFV blur >1.419 and a 12-week mean PFV blur >15 pd \| 12-weeks after randomization (baseline) \| \| Number of Participants Classified as Having Improved in All 3 Outcomes Measures From Baseline to 12 Weeks by Treatment Group \| Improvement in all 3 outcome measures at 12 weeks will be defined as follows: 1. Convergency Insufficiency Symptom Survey (CISS): Improvement of 9 or more points since baseline 2. Near point of convergence (NPC) break: 12-week/baseline mean NPC break <0.763 3. Positive fusional vergence (PFV) blur: 12-week/ baseline mean PFV blur >1.419 (Note: All 3 criteria must be met in order to be classified as an "improver" at the 12-week primary outcome visit). \| 12 weeks after randomization (baseline) \| \| Number of Participants Classified as an Overall Success Based on the 3 Outcomes Measures From Baseline to 6 Weeks by Treatment Group \| To be considered an overall success, each of the following criteria must be met for the 3 outcome measures at 6 weeks: 1. Convergence Insufficiency Symptom Survey (CISS): 6-week score <16 points and at least 9-point improvement from baseline at 6 weeks 2. Near point of convergence (NPC) break: 6-week/baseline mean NPC break <0.763 and a mean 6-week NPC break <6 cm 3. Positive fusional vergence (PFV) blur: 6-week/baseline mean PFV blur >1.419 and a mean 6-week PFV break >15 pd (Note: All 3 criteria must be met in order to be classified as an overall success at the 6-week visit). \| 6 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria Based on CI Signs/Symptoms at 6 Weeks by Treatment Group \| The number of subjects classified as a success based on signs/symptoms at the 6-week visit. Success is based on the Convergency Insufficiency Symptom Survey (CISS) defined as improvement of 9 or more points from baseline and a 6-week score of <16 points. \| 6 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria Based on the Mean NPC Break at 6 Weeks by Treatment Group \| The number of subjects who are classified as a success based on the mean NPC break at the 6-week visit. Success is based on the mean NPC (near point of convergence) break is defined as having a mean NPC break of <6 cm at 6 weeks and a 6-week to baseline ratio of <0.763 for mean NPC break. \| 6 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria Based on the Mean PFV Blur at 6 Weeks by Treatment Group \| The number of subjects who are classified as a success based on the mean PFV blur at the 6-week visit. Success is based on the mean PFV (positive fusional vergence) blur, defined as having a mean PFV blur of >15 pd at 6 weeks and a 6-week to baseline ratio of >1.419 for mean PFV blur. \| 6 weeks after randomization (baseline) \| \| Number of Participants Classified as Having Met Success Criteria for Both Clinical Measures at 6 Weeks by Treatment Group \| The number of subjects classified as a success based on clinical measures of CI (mean NPC break & mean PFV blur) at the 6-week visit. Clinical success is defined according to whether both criteria (below) are met as follows: 1. Near point of convergence (NPC) break: 6-week/baseline mean NPC break <0.763 and a 6-week mean NPC break <6 cm 2. Positive fusional vergence (PFV) blur: 6-week/baseline mean PFV blur >1.419 and a 6-week mean PFV blur >15 pd \| 6-weeks after randomization (baseline) \| \| Number of Participants Classified as Having Improved in All 3 Outcomes Measures From Baseline to 6 Weeks by Treatment Group \| Improvement in all 3 outcome measures at 6 weeks will be defined as follows: 1. Convergency Insufficiency Symptom Survey (CISS): Improvement of 9 or more points since baseline 2. Near point of convergence (NPC) break: 6-week/baseline mean NPC break <0.763 3. Positive fusional vergence (PFV) blur: 6-week/baseline mean PFV blur >1.419 (Note: All 3 criteria must be met in order to be classified as an "improver" at the 6-week visit). \| 6 weeks after randomization (baseline) \| Collaborators and Investigators This is where you will find people and organizations involved with this study. Publications Study Record Dates These dates track the progress of study record and summary results submissions to ClinicalTrials.gov. Study records and reported results are reviewed by the National Library of Medicine (NLM) to make sure they meet specific quality control standards before being posted on the public website. More Information Record History About About ClinicalTrials.gov Release Notes Site Map Help Give us feedback Glossary Customer Support Legal Disclaimer Terms and Conditions About About ClinicalTrials.gov Release Notes Site Map Help Give us feedback Glossary Customer Support Legal Disclaimer Terms and Conditions National Library of Medicine 8600 Rockville Pike, Bethesda, MD 20894 ClinicalTrials.gov An official website of the U.S. Department of Health and Human Services( National Institutes of Health( National Library of Medicine( and National Center for Biotechnology Information( About HHS About NIH About NLM About NCBI Accessibility support FOIA requests No FEAR Act data Office of the Inspector General Privacy policy HHS Vulnerability Disclosure Looking for U.S. government information and services? Visit USA.gov( Revision: v3.1.0 HHS Vulnerability Disclosure
1766	https://en.wikipedia.org/wiki/Epicyclic_gearing	Jump to content Search Contents 1 Overview 2 History 3 Requirements for non-interference 4 Gear speed ratios of conventional epicyclic gearing 5 Accelerations of standard epicyclic gearing 6 Torque ratios of standard epicyclic gearing 7 Fixed carrier train ratio 7.1 Spur gear differential 8 Gear ratio of reversed epicyclic gearing 9 Compound planetary gears 10 Power splitting 11 Advantages 12 3D printing 13 Gallery 14 See also 15 References 16 External links Epicyclic gearing العربية Azərbaycanca Български Català Čeština Dansk Deutsch Eesti Español فارسی Français Հայերեն Hrvatski עברית ქართული Lombard Magyar Bahasa Melayu Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Русский Suomi Svenska Türkçe Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Two gears mounted so the center of one gear revolves around the center of the other For the ring gear used in starter motors, see Starter ring gear. An epicyclic gear train (also known as a planetary gearset) is a gear reduction assembly consisting of two gears mounted so that the center of one gear (the "planet") revolves around the center of the other (the "sun"). A carrier connects the centers of the two gears and rotates, to carry the planet gear(s) around the sun gear. The planet and sun gears mesh so that their pitch circles roll without slip. If the sun gear is held fixed, then a point on the pitch circle of the planet gear traces an epicycloid curve. An epicyclic gear train can be assembled so the planet gear rolls on the inside of the pitch circle of an outer gear ring, or ring gear, sometimes called an annulus gear. Such an assembly of a planet engaging both a sun gear and a ring gear is called a planetary gear train. By choosing to hold one component or another—the planetary carrier, the ring gear, or the sun gear—stationary, three different gear ratios can be realized. Overview [edit] Epicyclic gearing or planetary gearing is a gear system consisting of one or more outer, or planet, gears or pinions, revolving about a central sun gear or sun wheel. Typically, the planet gears are mounted on a movable arm or carrier, which itself may rotate relative to the sun gear. Epicyclic gearing systems also incorporate the use of an outer ring gear, which meshes with the planet gears. Planetary gears (or epicyclic gears) are typically classified as simple or compound planetary gears. Simple planetary gears have one sun, one ring, one carrier, and one planet set. Compound planetary gears involve one or more of the following three types of structures: meshed-planet (there are at least two more planets in mesh with each other in each planet train), stepped-planet (there exists a shaft connection between two planets in each planet train), and multi-stage structures (the system contains two or more planet sets). Compared to simple planetary gears, compound planetary gears have the advantages of larger reduction ratio, higher torque-to-weight ratio, and more flexible configurations. The axes of all gears are usually parallel, but for special cases like pencil sharpeners and differentials, they can be placed at an angle, introducing elements of bevel gear (see below). Further, the sun, planet carrier and ring axes are usually coaxial. Epicyclic gearing is also available which consists of a sun, a carrier, and two planets which mesh with each other. One planet meshes with the sun gear, while the second planet meshes with the ring gear. For this case, when the carrier is fixed, the ring gear rotates in the same direction as the sun gear, thus providing a reversal in direction compared to standard epicyclic gearing. History [edit] Around 500 BC, the Greeks invented the idea of epicycles, of circles travelling on the circular orbits. With this theory Claudius Ptolemy in the Almagest in 148 AD was able to approximate planetary paths observed crossing the sky. The Antikythera Mechanism, circa 80 BC, had gearing which was able to closely match the Moon's elliptical path through the heavens, and even to correct for the nine-year precession of that path. (The Greeks interpreted the motion they saw, not as elliptical, but rather as epicyclic motion.) In the 2nd century AD treatise The Mathematical Syntaxis (a.k.a. Almagest), Claudius Ptolemy used rotating deferent and epicycles that form epicyclic gear trains to predict the motions of the planets. Accurate predictions of the movement of the Sun, Moon, and the five planets, Mercury, Venus, Mars, Jupiter, and Saturn, across the sky assumed that each followed a trajectory traced by a point on the planet gear of an epicyclic gear train. This curve is called an epitrochoid.[citation needed] Epicyclic gearing was used in the Antikythera Mechanism, circa 80 BC, to adjust the displayed position of the Moon for the ellipticity of its orbit, and even for its orbital apsidal precession. Two facing gears were rotated around slightly different centers; one drove the other, not with meshed teeth but with a pin inserted into a slot on the second. As the slot drove the second gear, the radius of driving would change, thus invoking a speeding up and slowing down of the driven gear in each revolution.[citation needed] Richard of Wallingford, an English abbot of St. Albans monastery, later described epicyclic gearing for an astronomical clock in the 14th century. In 1588, Italian military engineer Agostino Ramelli invented the bookwheel, a vertically revolving bookstand containing epicyclic gearing with two levels of planetary gears to maintain proper orientation of the books. French mathematician and engineer Desargues designed and constructed the first mill with epicycloidal teeth c. 1650. Requirements for non-interference [edit] In order that the planet gear teeth mesh properly with both the sun and ring gears, assuming equally spaced planet gears, the following equation must be satisfied: where are the number of teeth of the sun gear and the ring gear, respectively and is the number of planet gears in the assembly and is a whole number If one is to create an asymmetric carrier frame with non-equiangular planet gears, say to create some kind of mechanical vibration in the system, one must make the teething such that the above equation complies with the "imaginary gears". For example, in the case where a carrier frame is intended to contain planet gears spaced 0°, 50°, 120°, and 230°, one is to calculate as if there are actually 36 planetary gears (10° equiangular), rather than the four real ones. Gear speed ratios of conventional epicyclic gearing [edit] The gear ratio of an epicyclic gearing system is somewhat non-intuitive, particularly because there are several ways in which an input rotation can be converted into an output rotation. The four basic components of the epicyclic gear are: Sun gear: The central gear Carrier frame: Holds one or more planetary gear(s) symmetrically and separated, all meshed with the sun gear Planet gear(s): Usually two to four peripheral gears, all of the same size, that mesh between the sun gear and the ring gear Ring gear, Moon gear, Annulus gear, or Annular gear: An outer ring with inward-facing teeth that mesh with the planetary gear(s) The overall gear ratio of a simple planetary gearset can be calculated using the following two equations, representing the sun-planet and planet-ring interactions respectively: where : are the angular velocities of the ring gear, sun gear, planetary gears, and carrier frame respectively, and are the number of teeth of the ring gear, the sun gear, and each planet gear respectively. from which we can derive the following: and only if In many epicyclic gearing systems, one of these three basic components is held stationary (hence set for whichever gear is stationary); one of the two remaining components is an input, providing power to the system, while the last component is an output, receiving power from the system. The ratio of input rotation to output rotation is dependent upon the number of teeth in each of the gears, and upon which component is held stationary. Alternatively, in the special case where the number of teeth on each gear meets the relationship the equation can be re-written as the following: where : is the sun-to-planet gear ratio. These relationships can be used to analyze any epicyclic system, including those, such as hybrid vehicle transmissions, where two of the components are used as inputs with the third providing output relative to the two inputs. In one arrangement, the planetary carrier (green in the diagram above) is held stationary, and the sun gear (yellow) is used as input. In that case, the planetary gears simply rotate about their own axes (i.e., spin) at a rate determined by the number of teeth in each gear. If the sun gear has teeth, and each planet gear has teeth, then the ratio is equal to For instance, if the sun gear has 24 teeth, and each planet has 16 teeth, then the ratio is ⁠−+24/ 16 ⁠, or ⁠−+3/ 2 ⁠; this means that one clockwise turn of the sun gear produces 1.5 counterclockwise turns of each of the planet gear(s) about its axis. Rotation of the planet gears can in turn drive the ring gear (not depicted in diagram), at a speed corresponding to the gear ratios: If the ring gear has teeth, then the ring will rotate by turns for each turn of the planetary gears. For instance, if the ring gear has 64 teeth, and the planets 16 teeth, one clockwise turn of a planet gear results in ⁠16/ 64 ⁠, or ⁠1/ 4 ⁠ clockwise turns of the ring gear. Extending this case from the one above: One turn of the sun gear results in turns of the planets One turn of a planet gear results in turns of the ring gear So, with the planetary carrier locked, one turn of the sun gear results in turns of the ring gear. The ring gear may also be held fixed, with input provided to the planetary gear carrier; output rotation is then produced from the sun gear. This configuration will produce an increase in gear ratio, equal to If the ring gear is held stationary and the sun gear is used as the input, the planet carrier will be the output. The gear ratio in this case will be which may also be written as This is the lowest gear ratio attainable with an epicyclic gear train. This type of gearing is sometimes used in tractors and construction equipment to provide high torque to the drive wheels. In bicycle hub gears, the sun is usually stationary, being keyed to the axle or even machined directly onto it. The planetary gear carrier is used as input. In this case the gear ratio is simply given by The number of teeth in the planet gear is irrelevant. Accelerations of standard epicyclic gearing [edit] From the above formulae, we can also derive the accelerations of the sun, ring and carrier, which are: Torque ratios of standard epicyclic gearing [edit] In epicyclic gears, two speeds must be known in order to determine the third speed. However, in a steady state condition, only one torque must be known in order to determine the other two torques. The equations which determine torque are: where: — Torque of ring (annulus), — Torque of sun, — Torque of carrier. For all three, these are the torques applied to the mechanism (input torques). Output torques have the reverse sign of input torques. These torque ratios can be derived using the law of conservation of energy. Applied to a single stage this equation is expressed as: In the cases where gears are accelerating, or to account for friction, these equations must be modified. Fixed carrier train ratio [edit] A convenient approach to determine the various speed ratios available in a planetary gear train begins by considering the speed ratio of the gear train when the carrier is held fixed. This is known as the fixed carrier train ratio. In the case of a simple planetary gear train formed by a carrier supporting a planet gear engaged with a sun and ring gear, the fixed carrier train ratio is computed as the speed ratio of the gear train formed by the sun, planet and ring gears on the fixed carrier. This is given by In this calculation the planet gear is an idler gear. The fundamental formula of the planetary gear train with a rotating carrier is obtained by recognizing that this formula remains true if the angular velocities of the sun, planet and ring gears are computed relative to the carrier angular velocity. This becomes, This formula provides a simple way to determine the speed ratios for the simple planetary gear train under different conditions: The carrier is held fixed, ωc=0, The ring gear is held fixed, ωr=0, The sun gear is held fixed, ωs=0, Each of the speed ratios available to a simple planetary gear train can be obtained by using band brakes to hold and release the carrier, sun or ring gears as needed. This provides the basic structure for an automatic transmission. Spur gear differential [edit] A spur gear differential is constructed from two identical coaxial epicyclic gear trains assembled with a single carrier such that their planet gears are engaged. This forms a planetary gear train with a fixed carrier train ratio R = −1. In this case, the fundamental formula for the planetary gear train yields, or Thus, the angular velocity of the carrier of a spur gear differential is the average of the angular velocities of the sun and ring gears. In discussing the spur gear differential, the use of the term ring gear is a convenient way to distinguish the sun gears of the two epicyclic gear trains. Ring gears are normally fixed in most applications as this arrangement will have a good reduction capacity. The second sun gear serves the same purpose as the ring gear of a simple planetary gear train but clearly does not have the internal gear mate that is typical of a ring gear. Gear ratio of reversed epicyclic gearing [edit] Some epicyclic gear trains employ two planetary gears which mesh with each other. One of these planets meshes with the sun gear, the other planet meshes with the ring gear. This results in different ratios being generated by the planetary and also causes the sun gear to rotate in the same direction as the ring gear when the planet carrier is the stationary. The fundamental equation becomes: where which results in: : when the carrier is locked, : when the sun is locked, : when the ring gear is locked. Compound planetary gears [edit] "Compound planetary gear" is a general concept and it refers to any planetary gears involving one or more of the following three types of structures: meshed-planet (there are at least two or more planets in mesh with each other in each planet train), stepped-planet (there exists a shaft connection between two planets in each planet train), and multi-stage structures (the system contains two or more planet sets). Some designs use "stepped-planet" which have two differently-sized gears on either end of a common shaft. The small end engages the sun, while the large end engages the ring gear. This may be necessary to achieve smaller step changes in gear ratio when the overall package size is limited. Compound planets have "timing marks" (or "relative gear mesh phase" in technical term). The assembly conditions of compound planetary gears are more restrictive than simple planetary gears, and they must be assembled in the correct initial orientation relative to each other, or their teeth will not simultaneously engage the sun and ring gear at opposite ends of the planet, leading to very rough running and short life. In 2015, a traction based variant of the "stepped-planet" design was developed at the Delft University of Technology, which relies on compression of the stepped planet elements to achieve torque transmission. The use of traction elements eliminates the need to have "timing marks" as well as the restrictive assembly conditions as typically found. Compound planetary gears can easily achieve larger transmission ratio with equal or smaller volume. For example, compound planets with teeth in a 2:1 ratio with a 50T ring gear would give the same effect as a 100T ring gear, but with half the actual diameter. More planet and sun gear units can be placed in series in the same housing (where the output shaft of the first stage becomes the input shaft of the next stage) providing a larger (or smaller) gear ratio. This is the way most automatic transmissions work. In some cases multiple stages may even share the same ring gear which can be extended down the length of the transmission, or even be a structural part of the casing of smaller gearboxes. During World War II, a special variation of epicyclic gearing was developed for portable radar gear, where a very high reduction ratio in a small package was needed. This had two outer ring gears, each half the thickness of the other gears. One of these two ring gears was held fixed and had one tooth fewer than did the other. Therefore, several turns of the "sun" gear made the "planet" gears complete a single revolution, which in turn made the rotating ring gear rotate by a single tooth like a cycloidal drive.[citation needed] Power splitting [edit] More than one member of a system can serve as an output. As an example, the input is connected to the ring gear, the sun gear is connected to the output and the planet carrier is connected to the output through a torque converter. Idler gears are used between sun gear and the planets to cause the sun gear to rotate in the same direction as the ring gear when the planet carrier is stationary. At low input speed, because of the load on the output, the sun will be stationary and the planet carrier will rotate in the direction of the ring gear. Given a high enough load, the turbine of the torque converter will remain stationary, the energy will be dissipated and the torque converter pump will slip. If the input speed is increased to overcome the load the converter turbine will turn the output shaft. Because the torque converter itself is a load on the planet carrier, a force will be exerted on the sun gear. Both the planet carrier and the sun gear extract energy from the system and apply it to the output shaft. Advantages [edit] Planetary gear trains provide high power density in comparison to standard parallel axis gear trains. They provide a reduction in volume, multiple kinematic combinations, purely torsional reactions, and coaxial shafting. Disadvantages include high bearing loads, constant lubrication requirements, inaccessibility, and design complexity. The efficiency loss in a planetary gear train is typically about 3% per stage. This type of efficiency ensures that a high proportion (about 97%) of the energy being input is transmitted through the gearbox, rather than being wasted on mechanical losses inside the gearbox. The load in a planetary gear train is shared among multiple planets; therefore, torque capability is greatly increased. The more planets in the system, the greater the load ability and the higher the torque density. The planetary gear train also provides stability due to an even distribution of mass and increased rotational stiffness. Torque applied radially onto the gears of a planetary gear train is transferred radially by the gear, without lateral pressure on the gear teeth. In a typical application, the drive power connects to the sun gear. The sun gear then drives the planetary gears assembled with the external gear ring to operate. The whole set of planetary gear system revolves on its own axis and along the external gear ring where the output shaft connected to the planetary carrier achieves the goal of speed reduction. A higher reduction ratio can be achieved by doubling the multiple staged gears and planetary gears which can operate within the same ring gear. The method of motion of a planetary gear structure is different from traditional parallel gears. Traditional gears rely on a small number of contact points between two gears to transfer the driving force. In this case, all the loading is concentrated on a few contacting surfaces, making the gears wear quickly and sometimes crack. But the planetary speed reducer has multiple gear contacting surfaces with a larger area that can distribute the loading evenly around the central axis. Multiple gear surfaces share the load, including any instantaneous impact loading, evenly, which makes them more resistant to damage from higher torque. The housing and bearing parts are also less likely to be damaged from high loading as only the planet carrier bearings experience significant lateral force from the transmission of torque, radial forces oppose each other and are balanced, and axial forces only arise when using helical gears. 3D printing [edit] \| \| \| This section does not cite any sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (February 2021) (Learn how and when to remove this message) \| Planetary gears have become popular in the maker community, due to their inherent high torque capabilities and compactness/efficiency. Especially within 3D printing, they can be used to rapidly prototype a gear box, to then be manufactured with machining technologies later. A geared-down motor must turn farther and faster in order to produce the same output movement in the 3D printer which is advantageous if it is not outweighed by the slower movement speed. If the stepper motor has to turn farther then it also has to take more steps to move the printer a given distance; therefore, the geared-down stepper motor has a smaller minimum step-size than the same stepper motor without a gearbox. While down-gearing improves precision in unidirectional motion, it adds backlash to the system and so reduces its absolute positioning accuracy. Since herringbone gears are easy to 3D print, it has become very popular to 3D print a moving herringbone planetary gear system for teaching children how gears work. An advantage of herringbone gears is that they don't fall out of the ring and don't need a mounting plate, allowing the moving parts to be clearly seen. Gallery [edit] Split ring, compound planet, epicyclic gears of a car rear-view mirror positioner. This has a ratio from input sun gear to output black ring gear of −5/352. Reduction gears on Pratt & Whitney Canada PT6 gas turbine engine. One of three sets of three gears inside the planet carrier of a Ford FMX Ravigneaux transmission See also [edit] Hypocycloidal gearing Antikythera mechanism – ancient mechanical astronomical computer Continuously variable transmission (CVT) Cycloidal drive Epicycloid Ford Model T – had a 2 speed planetary transmission. Gearbox Harmonic drive Hub gear, for bicycles, etc. NuVinci continuously variable transmission Ravigneaux planetary gearset Lepelletier gear mechanism Rohloff Speedhub – 14-ratio bicycle hub gearbox Simpson planetary gearset Sturmey Archer – First major manufacturer of bicycle hubs using planetary gears Uni Wheel - a wheel that incorporates a planetary gear system References [edit] ^ a b c J. J. Uicker, G. R. Pennock and J. E. Shigley, 2003, Theory of Machines and Mechanisms, Oxford University Press, New York. ^ a b B. Paul, 1979, Kinematics and Dynamics of Planar Machinery, Prentice Hall. ^ Machinery, Volume 19. University of California. 1913. p. 979. ^ Hillier, V.A.W. (2001). "Planetary gearing and unidirectional clutches". Fundamentals of Motor Vehicle Technology (4th ed.). Cheltenham, UK: Nelson Thornes. p. 244. ISBN 0-74-870531-7. ^ Harrison, H.; Nettleton, T. (1994). Principles of Engineering Mechanics (2nd ed.). Oxford, UK: Butterworth-Heinemann. p. 58. ISBN 0-34-056831-3. ^ "What are planetary gears and how do they work". Power Transmission Components. Retrieved 2024-09-25. ^ Wright, M.T. (2007). "The Antikythera Mechanism reconsidered" (PDF). Interdisciplinary Science Reviews. 32 (1): 27–43. Bibcode:2007ISRv...32...27W. doi:10.1179/030801807X163670. S2CID 54663891. Retrieved 20 May 2014. ^ a b Coy, J.J.; Townsend, D.P.; Zaretsky, E.V. (1985). Gearing (PDF) (Report). NASA Reference Publication. Vol. 1152. AVSCOM Technical Report 84-C-15. ^ Randl, Chad (15 May 2008). Revolving Architecture: A history of buildings that rotate, swivel, and pivot. New York, NY: Princeton Architectural Press. p. 19. ISBN 978-156898681-4. OCLC 1036836698. ISBN 1568986815 ^ Musson, A.E.; Robinson, Eric H. (1969). Science and Technology in the Industrial Revolution. Toronto, ON: University of Toronto Press. p. 69. ISBN 9780802016379. OCLC 1036858215. ^ "How to derive and calculate epicyclic gear ratio equations in planetary gear systems". buseco.net. ^ Miller, John M. (May 2006). "Hybrid electric vehicle propulsion system architectures of the e-CVT type". IEEE Transactions on Power Electronics. 21 (3): 756–767. Bibcode:2006ITPE...21..756M. doi:10.1109/TPEL.2006.872372. S2CID 4986932. ^ P. A. Simionescu (1998-09-01). "A Unified Approach to the Assembly Condition of Epicyclic Gears". Journal of Mechanical Design. 120 (3): 448–453. doi:10.1115/1.2829172. ^ "Archimedes Drive". ^ "Unconventional Gear Profiles in Planetary Gearboxes". ^ "52514 Caterpillar Tractor D8 Bulldozer Powershift Transmission Promotional Film". 10 August 2022. ^ Lynwander, P., 1983, Gear Drive Systems: Design and Application. Marcel Dekker, New York ^ Smith, J. D., 1983, Gears and Their Vibration: A Basic Approach to Understanding Gear Noise. Marcel Dekker, New York and MacMillan, London ^ MECHTEX. "Planetary Gearbox \| Introduction and Construction". MECHTEX. Retrieved 2025-01-03. ^ Amer, Mohammed; Lin, Chung-Cheng; Ismail, Hasan; Wu, Shin-Hung (2024-02-01). "Planetary Gearbox Design and Development using Additive Manufacturing". An-Najah University Journal for Research: 52 – via ResearchGate. ^ "Which factors affect the accuracy of the planetary gearbox? \| GearKo Reducers" (in Chinese (China)). Retrieved 2025-01-03. External links [edit] Wikimedia Commons has media related to Epicyclic gears. Kinematic Models for Design Digital Library (KMODDL), movies and photos of hundreds of working mechanical-systems models at Cornell. "Epicyclic gearing animation in SVG" "Animation of Epicyclic gearing" The "Power Split Device" The "Interactive Planetary Gearset tutorial" Prius Gearbox Planetary Gearbox Short Cuts for Analyzing Planetary Gearing Archived 2021-02-25 at the Wayback Machine \| v t e Gears \| \| Systems \| Spur gear systems Worm drive Rack and pinion Epicyclic (planetary) gearing Sun and planet gear Harmonic drive Cycloidal drive Non-circular gear \| \| Shapes \| Spur Bevel Crown Spiral bevel Helical Herringbone \| \| Geartooth profiles \| Involute Cycloid \| \| Mechanics \| Transmission Differential Coupling Train Bicycle gearing Continuously variable transmission Offset \| \| Examples \| \| \| \| --- \| \| Bicycles \| Cogset Derailleur Hub gear Shaft-driven bicycle Sprocket \| \| Horology \| Wheel train \| \| \| See also \| Ball screw Leadscrew Jackscrew Belt drive Chain drive Gear manufacturing Freewheel Spur gear corrected tooth \| \| v t e \| \| Part of the Automobile series \| \| Automotive engine \| Diesel engine Electric Fuel cell Hybrid (Plug-in hybrid) Internal combustion engine Petrol engine Steam engine \| \| Transmission \| Automatic transmission Chain drive Direct-drive Clutch Constant-velocity joint Continuously variable transmission Coupling Differential Direct-shift gearbox Drive shaft Dual-clutch transmission Drive wheel Automated manual transmission Electrorheological clutch Epicyclic gearing Fluid coupling Friction drive Gearshift Giubo Hotchkiss drive Limited-slip differential Locking differential Manual transmission Manumatic Parking pawl Park-by-wire Preselector gearbox Semi-automatic transmission Shift-by-wire Torque converter Transaxle Transfer box Transmission control unit Universal joint \| \| Wheels and tires \| Wheel hub assembly Wheel + Rim + Alloy wheel + Hubcap Tire + Off-road + Racing slick + Radial + Rain + Run-flat + Snow + Spare + Tubeless \| \| Hybrid \| Electric motor Hybrid vehicle drivetrain Electric generator Alternator \| \| Portal Category \| \| Authority control databases \| \| International \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Epicyclical gearing Gears Hidden categories: CS1 Chinese (China)-language sources (zh-cn) Articles with short description Short description is different from Wikidata All articles with unsourced statements Articles with unsourced statements from September 2019 Articles with unsourced statements from November 2011 Articles needing additional references from February 2021 All articles needing additional references Commons category link is on Wikidata Webarchive template wayback links Epicyclic gearing Add topic
1767	https://www.cracksat.net/sat/sentence-completion/	SAT Sentence Completion: Practice tests and explanations_CrackSAT.net Toggle navigation Toggle navigation Home SAT Tests SAT Downloads SAT Books PSAT College Admission AP Prep ACT Prep Go SAT Sentence Completion: Practice tests and explanations SAT Sentence Completion Practice Tests SAT sentence completion test 1 SAT sentence completion test 2 SAT sentence completion test 3 SAT sentence completion test 4 SAT sentence completion test 5 SAT sentence completion test 6 SAT sentence completion test 7 SAT sentence completion test 8 SAT sentence completion test 9 SAT sentence completion test 10 SAT sentence completion test 11 SAT sentence completion test 12 SAT sentence completion test 13 SAT sentence completion test 14 SAT sentence completion test 15 SAT sentence completion test 16 SAT sentence completion test 17 SAT sentence completion test 18 SAT sentence completion test 19 SAT sentence completion test 20 SAT sentence completion test 21 SAT sentence completion test 22 SAT sentence completion test 23 SAT sentence completion test 24 SAT sentence completion test 25 SAT sentence completion test 26 SAT sentence completion test 27 SAT sentence completion test 28 SAT sentence completion test 29 SAT sentence completion test 30 SAT sentence completion test 31 SAT sentence completion test 32 SAT sentence completion test 33 SAT sentence completion test 34 SAT sentence completion test 35 SAT sentence completion test 36 SAT sentence completion test 37 SAT sentence completion test 38 SAT sentence completion test 39 SAT sentence completion test 40 SAT sentence completion test 41 SAT sentence completion test 42 SAT sentence completion test 43 SAT sentence completion test 44 SAT sentence completion test 45 SAT sentence completion test 46 SAT sentence completion test 47 SAT sentence completion test 48 SAT sentence completion test 49 SAT sentence completion test 50 SAT sentence completion test 51 SAT sentence completion test 52 SAT sentence completion test 53 SAT sentence completion test 54 SAT sentence completion test 55 SAT sentence completion test 56 SAT sentence completion test 57 SAT sentence completion test 58 SAT sentence completion test 59 SAT sentence completion test 60 SAT sentence completion test 61 SAT sentence completion test 62 SAT sentence completion test 63 SAT sentence completion test 64 SAT sentence completion test 65 SAT sentence completion test 66 SAT sentence completion test 67 SAT sentence completion test 68 SAT sentence completion test 69 SAT sentence completion test 70 SAT sentence completion test 71 More Information Privacy Contact us Copyright Link to us Sitemap SAT Passage-based Reading: Practice tests and explanations SAT Sentence Completion: Practice tests and explanations SAT Improving Sentences: Practice tests and explanations Toggle navigation Digital SAT Test Reading and Writing Math SAT Test sat prep sat skills SAT Grammar Old SAT Practice Tests SAT Math sat math multiple choice sat math grid-ins SAT Reading SAT Writing and Language SAT Subject Tests sat math 1&2 subject test sat physics subject test sat chemistry subject test sat biology subject test sat us history subject test sat world history subject test sat literature subject test sat French subject test sat Spanish subject test SAT Downloads SAT Books PSAT College Admission College Essay Samples College Essay Tips SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. All content of site and practice tests copyright © 2022 Max. sitemap Home Privacy Copyright Contact Link to us 1.7k Shares Share Share Share Share Tweet Share
1768	https://www.imada.sdu.dk/~rolf/Edu/DM815/E10/2dcollisions.pdf	2-Dimensional Elastic Collisions Without Trigonometry This document is intended to introduce you to solving 2-dimensional elastic collision problems for circles without complicated trigonometry. It is much easier to use vectors to solve 2-dimensional collision problems than using trigonometry. If you don't know the basics of using vectors, try reading about it at For solving collision problems, you will need to know how to add and subtract vectors, find unit vectors, multiply a vector by a scalar, and compute a dot product. If you aren't familiar with vectors, that may sound complicated, but keep in mind that all you really need is addition, subtraction, multiplication, division, and taking square roots; no complicated trigonometry. Physics of elastic collisions in one dimension An elastic collision is a collision in which kinetic energy is conserved. That means there is no energy lost as heat or sound during the collision. In the real world, there are no perfectly elastic collisions on an everyday scale of size. But you can get the sense of an elastic collision by imagining a perfect pool ball which doesn't waste any energy when it collides. In an elastic collision, both kinetic energy and momentum are conserved (the total before and after the collision remains the same). Momentum is the product of mass and velocity: p=m⋅v The kinetic energy of an object is one-half times its mass times the square of its velocity: KE=1 2 mv 2 Notation: Throughout this document, m is mass and v is velocity. Subscripts 1 and 2 refer to one of the two colliding objects. An apostrophe after a variable means that the value is taken after the collision (called prime; i.e., v' is “v prime”). Now it is easy to write the conservation of momentum and kinetic energy as two equations: Conservation of momentum: m1v1m2v2=m1v1'm2v2' Conservation of kinetic energy: 1 2 m1v1 21 2 m2v2 2=1 2 m1v1' 21 2 m2v2' 2 Combining these two equations and doing a lot of algebra gives the final (after collision) velocities of objects 1 and 2: v1'= v1m1−m22m2v2 m1m2 v2'=v2m2−m12m1v1 m1m2 This result allows us to find the velocity of two objects after undergoing a one-dimensional elastic collision. We will use this result later in the 2-dimensional case. Elastic collisions in two dimensions We will follow a 7-step process to find the new velocities of two objects after a collision. The basic goal of the process is to project the velocity vectors of the two objects onto the vectors which are normal (perpendicular) and tangent to the surface of the collision. This gives us a normal component and a tangential component for each velocity. The tangential components of the velocities are not changed by the collision because there is no force along the line tangent to the collision surface. The normal components of the velocities undergo a one-dimensional collision, which can be computed using the one-dimensional collision formulas presented above. Next the unit normal vector is multiplied by the scalar (plain number, not a vector) normal velocity after the collision to get a vector which has a direction normal to the collision surface and a magnitude which is the normal component of the velocity after the collision. The same is done with the unit tangent vector and the tangential velocity component. Finally the new velocity vectors are found by adding the normal velocity and tangential velocity vectors for each object. 1. Find unit normal and unit tangent vectors. The unit normal vector is a vector which has a magnitude of 1 and a direction that is normal (perpendicular) to the surfaces of the objects at the point of collision. The unit tangent vector is a vector with a magnitude of 1 which is tangent to the circles' surfaces at the point of collision. First find a normal vector. This is done by taking a vector whose components are the difference between the coordinates of the centers of the circles. Let x1, x2, y1, and y2 be the x and y coordinates of the centers of the circles. Then the normal vector n is:  n=〈x2−x1 , y2−y1〉 Next, find the unit vector of n, which we will call un. This is done by dividing by the magnitude of n:  un=  n ∣ n∣=  n nx 2n y 2 Next we need the unit tangent vector. This is easy to find from the unit normal vector. Just make the x component of the unit tangent vector equal to the negative of the y component of the unit normal vector, and make the y component of the unit tangent vector equal to the x component of the unit normal vector:  ut=〈−uny ,unx〉 2. Create the initial (before the collision) velocity vectors, v1 and v2. These are just the x and y components of the velocities put into vectors:  v1=〈v1 x ,v1 y〉(and similarly for v2). Note that this step really isn't necessary if the velocities are already represented as vectors. This step assumes that the velocities are initially represented as x and y components. 3. Keep in mind that after the collision, the tangential component of the velocities is unchanged and the normal component of the velocities can be found using the one-dimensional collision formulas presented above. So we need to resolve the velocity vectors, v1 and v2, into normal and tangential components. To do this, project the velocity vectors onto the unit normal and unit tangent vectors by taking the dot product of the velocity vectors with the unit normal and unit tangent vectors. Let v1n be the scalar (plain number, not a vector) velocity of object 1 in the normal direction. Let v1t be the scalar velocity of object 1 in the tangential direction. Similarly, let v2n and v2t be for object 2. These values are found by projecting the velocity unit normal unit tangent vectors onto the unit normal and unit tangent vectors, which is done by taking the dot product: v1n=  un⋅ v1 v1t=  ut⋅ v1 v2n=  un⋅ v2 v2t=  ut⋅ v2 4. Find the new tangential velocities (after the collision). This is the simplest step of all. The tangential components of the velocity do not change after the collision because there is no force between the circles in the tangential direction during the collision. So, the new tangential velocities are simply equal to the old ones: v1t'=v1t v2t'=v2t Remember that the apostrophe after the variable name means “prime,” which in turn means “after the collision.” 5. Find the new normal velocities. This is where we use the one-dimensional collision formulas. The velocities of the two circles along the normal direction are perpendicular to the surfaces of the circles at the point of collision, so this really is a one-dimensional collision. v1n'=v1nm1−m22m2v2n m1m2 v2n'=v2nm2−m12m1v1n m1m2 6. Convert the scalar normal and tangential velocities into vectors. This is easy—just multiply the unit normal vector by the scalar normal velocity and you get a vector which has a direction that is normal to the surfaces at the point of collision and which has a magnitude equal to the normal component of the velocity. It is similar for the tangential component.  v1n'=v1n'⋅ un  v1t'=v1t'⋅ ut  v2n'=v2n'⋅ un  v2t'=v2t'⋅ ut 7. Find the final velocity vectors by adding the normal and tangential components for each object:  v1'=  v1n' v1t'  v2'=  v2n' v2t' Now we have the final (after collision) velocity of each object as vectors. To see these formulas in action, check out: Bouncescope ( This program allows you to simulate lots of balls bouncing around elastically. The mass, size, velocity, and color of each ball can be customized, or you can add random balls. To check for updated versions of this document, visit: ©2006 Chad Berchek Vobarian Software
1769	http://www.askthenerd.com/NOW/CH17/17_1-8.pdf	C H A P T E R17 MASTERING ORGANIC CHEMISTRY Predicting the Products of Aromatic Substitution Reactions Understanding the Mechanisms of Aromatic Sub-stitution Reactions Predicting the Effect of a Substituent on the Rate and Regiochemistry of an Electrophilic Aromatic Substitution Reaction Synthesizing Aromatic Compounds L L L L Aromatic Substitution Reactions Look for this logo in the chapter and go to OrganicChemistryNow at for tutorials, simulations, problems, and molecular models. M ost of the reactions discussed in this chap-ter involve the attack of an electrophile on an aromatic compound. Although the initial step of the mechanism resembles that of the electrophilic ad-dition reactions of carbon–carbon double bonds dis-cussed in Chapter 11, the ﬁnal product here results from substitution of the electrophile for a hydrogen on the aro-matic ring rather than addition. Therefore, these reactions are called electrophilic aromatic substitutions. First, the general mechanism for these reactions is pre-sented. This is followed by a speciﬁc example, the substi-tution of a nitro group onto a benzene ring. Then the effect of a group that is already present on the ring on the rate of the reaction and its regiochemistry is discussed in detail. Next, reactions that add halogens, sulfonic acid groups, alkyl groups, and acyl groups to the aromatic ring are pre-sented. In each case the required reagents, the mechanism for generating the electrophile, the usefulness, and the lim-itations of the reactions are discussed. These reactions are very important and constitute the majority of the chapter. Next, three different mechanisms for nucleophilic substitutions on aromatic rings are presented. These are followed by several other reactions that are useful in syn-thesis because they interconvert groups attached to aro-matic rings. Finally, the use of combinations of all of these reactions to synthesize a variety of substituted aro-matic compounds is discussed. 17.1 Mechanism for Electrophilic Aromatic Substitution A general mechanism for the electrophilic aromatic sub-stitution reaction is outlined in Figure 17.1. The process begins by reaction of the electrophile with a pair of pi electrons of the aromatic ring, which acts as the nucle-ophile, in a fashion very similar to the addition reactions described in Chapter 11, which begin by reaction of an electrophile with the pi electrons of an alkene. This results in the formation of a carbocation called an arenium ion. Removal of a proton from the arenium ion by some weak base that is present restores the aromatic ring and results in the substitution of the electrophile for a hydrogen on the aromatic ring. It is instructive to examine the energetics of this reaction and compare them to those of the addition reactions of Chapter 11 (see Figure 17.2). Because of its aromatic reso-nance energy, benzene is considerably more stable than the alkene. Because of reso-nance, the arenium ion that is produced in the electrophilic aromatic substitution reaction is more stable than the carbocation produced in the addition reaction. However, the are-nium ion is no longer aromatic, so its stabilization relative to the carbocation is less than the stabilization of benzene relative to the alkene. Because the transition states for both of these reactions resemble the carbocation intermediates (recall the Hammond postu-late), the transition state leading to the arenium ion must have lost most of its aromatic stabilization also. This causes the activation energy for the electrophilic aromatic substi-tution reaction, Gs ‡, to be larger than the activation energy for the addition reaction, Ga ‡. In other words, the loss of aromatic resonance energy that occurs on going to the transition state for substitution results in a higher activation barrier. The substitution re-action usually requires much stronger electrophiles than the addition reaction. 672 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS E H The electrophile reacts with a pair of pi electrons of the aromatic ring. This step resembles the first step of the reaction in which electrophiles react with alkenes, described in Chapter 11. This is the slow step of the reaction. Rather than add as a nucleophile to the positive carbon, a base removes a proton from the arenium ion to regenerate the aromatic ring with its large resonance stabilization. This carbocation is called an arenium ion. Although it has considerable resonance stabilization, it has lost the aromatic resonance energy of benzene because the cycle of orbitals does not extend entirely around the ring. Note that the resonance structures have the positive charge located on the carbons ortho and para to the carbon that is bonded to the electrophile. slow Arenium ion B H E H H H H H H H H H H + E H H H H H H E H + H H H H + E H H H H H + 2 1 2 1 Figure 17.1 MECHANISM OF A GENERAL ELECTROPHILIC AROMATIC SUBSTITUTION REACTION. The arenium ion, like any carbocation, has two reaction pathways available. It could react with a nucleophile to give an addition product, or it could lose a proton to some base in the system to give a substitution product. If addition were to occur, the product would be a cyclohexadiene, which is no longer aromatic because it has only two dou-ble bonds in the ring. It has lost at least 35 kcal/mol (146 kJ/mol) of stabilization, the difference in resonance energy between an aromatic ring and a cyclohexadiene. Obvi-ously, the formation of the aromatic ring in the substitution product is greatly favored. All of the electrophilic aromatic substitution reactions follow this same general mechanism. The only difference is the structure of the electrophile and how it is gener-ated. Let’s look at a speciﬁc example, the nitration of benzene. This reaction is accom-plished by reacting benzene with nitric acid in the presence of sulfuric acid: H2SO4 Benzene Nitrobenzene HNO3 H2O NO2 17.1 I MECHANISM FOR ELECTROPHILIC AROMATIC SUBSTITUTION 673 Energy H±C±C±H E W W H H W W Nu CœC E+ H H H H ± ± ± ± Reaction coordinate Electrophilic addition reaction Electrophilic aromatic substitution reaction Ga ‡ Gs ‡ H±C±C±H E W W H W H E+ E Arenium ion E H + + a b Figure 17.2 REACTION ENERGY DIAGRAMS FOR AN ELECTROPHILIC ADDITION TO AN ALKENE AND AN ELECTROPHILIC AROMATIC SUBSTITUTION REACTION. b a The mechanism for this reaction is presented in Figure 17.3. The electrophile, NO2 (ni-tronium ion) is generated from the nitric acid by protonation of an OH group. Water then acts as a leaving group to generate the electrophile. The rest of the mechanism is identical to that outlined in Figure 17.1. 17.2 Effect of Substituents When a substituted benzene is nitrated, the substituent on the ring has an effect on the rate of the reaction. In addition, the NO2 electrophile can attach ortho, meta, or para to the substituent. For example, when toluene is nitrated, it is found to react 17 times faster than benzene. Substitution occurs primarily ortho and para to the methyl group. The methyl group accelerates the reaction compared to benzene and directs the in-coming electrophile to the ortho and para positions. Both the rate enhancement and the NO2 NO2 37% 3% 60% p-Nitrotoluene m-Nitrotoluene o-Nitrotoluene Toluene HNO3 H2SO4 CH3 CH3 NO2 CH3 CH3 674 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Figure 17.3 MECHANISM OF THE NITRATION OF BENZENE. The OH of the nitric acid is protonated in the first step. The NO2+ electrophile reacts with the benzene in the same manner as outlined in Figure 17.1. Some base in the reaction mixture, such as water, removes a proton from the arenium ion to regenerate the aromatic ring. Water then acts as a leaving group in the second step to generate the nitronium ion electrophile, NO2+. H±O±S±OH O X X O . . . . H±O±N + + + – – – + + + + O O œ ± . . . . . . . . . . . . . . H±O±N W H O O œ ± . . . . . . . . . . . . H3O NO2 . . O œ H O W N . . . . . . . . . . . . H2O O N O . . . . . . . . ± ± ± ± 2 4 1 2 3 3 4 1 O H H regiochemistry of the reaction can be understood by examination of the three possible arenium ions produced by attack of the electrophile at the positions ortho, meta, and para to the methyl group. Consider the case of attack at an ortho position ﬁrst: In one of the resonance structures, the positive charge is located on the carbon bonded to the methyl substituent. As we are well aware, the methyl group will stabilize the car-bocation, so this arenium ion is somewhat lower in energy (more stable) than the are-nium ion produced in the nitration of benzene itself. Consider next the arenium ion produced by attack of the electrophile at a position meta to the methyl group: In this case, none of the resonance structures has the positive charge located on the car-bon bonded to the methyl group. This ion has no extra stabilization when compared to the arenium ion formed from benzene. Finally, consider the case of attack of the electrophile at the para position: This arenium ion is similar to that produced by attack at the ortho position in that the positive charge is located on the carbon bonded to the methyl group in one of the reso-nance structures. Therefore, it is more stable than the arenium ion formed from benzene. Arenium ion from para attack Most stable resonance structure H NO2 + + CH3 H NO2 + CH3 H NO2 CH3 Arenium ion from meta attack H NO2 CH3 + H NO2 + CH3 H NO2 CH3 + Arenium ion from ortho attack Most stable resonance structure H NO2 CH3 H NO2 + H NO2 CH3 + + CH3 17.2 I EFFECT OF SUBSTITUENTS 675 Although all arenium ions are unstable, reactive intermediates, those resulting from ortho and para attack of the electrophile on toluene, but not that from meta attack, are more stable than the arenium ion produced from benzene itself. Therefore, for toluene the transition states leading to the ortho and para ions are at lower energy than is the transition state leading to the meta ion or the transition state that is formed in the nitra-tion of benzene. Thus, the methyl group accelerates the attack of the electrophile at the ortho and para positions. The methyl is an activating group (it makes the aromatic ring react faster), and it is an ortho/para directing group. The effects of other groups can be understood by similar reasoning. When the elec-trophile bonds ortho or para to the substituent, the positive charge is located on the car-bon that is bonded to the substituent in one of the resonance structures. If the substituent is one that can stabilize the carbocation, then it accelerates the reaction and directs the incoming electrophile to the ortho and para positions. If, on the other hand, the sub-stituent is one that destabilizes the carbocation, then it slows the reaction and directs the electrophile to the meta position so that the positive charge is never on the carbon di-rectly attached to the substituent. The meta arenium ion is less destabilized than the or-tho or para ions in this case. Some other examples will help clarify this reasoning. The nitration of methoxybenzene (anisole) proceeds 10,000 times faster than does nitration of benzene and produces predominantly the ortho and para isomers of ni-troanisole. From these results it can be seen that the methoxy group is also an ortho/para director and is a much stronger activating group than is the methyl group. Examination of the resonance structure for the para arenium ion that has the positive charge located on the carbon bonded to the methoxy group explains these results. Two electronic effects are operating in this case: an inductive effect and a resonance ef-fect. Because of the high electronegativity of the oxygen, the methoxy group withdraws electrons by its inductive effect (see Section 4.5). If this were the only effect operating, Especially stable resonance structure OCH3 . . . . + H NO2 OCH3 . . + H NO2 67% 2% 31% Methoxybenzene (anisole) NO2 HNO3 H2SO4 OCH3 OCH3 OCH3 NO2 NO2 OCH3 676 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS then it would be a deactivating group. However, in this case there is also a resonance ef-fect. As shown in the resonance structure on the right, the methoxy group is a resonance electron-donating group. This resonance structure is especially stable because the octet rule is satisﬁed for all of the atoms. A similar, especially stable resonance structure can be written for the arenium ion that is produced by reaction of the electrophile at the or-tho position. However, when the electrophile reacts at the meta position, the positive charge is never located on the carbon bonded to the methoxy group, so this especially stable resonance structure cannot be formed. Overall, the resonance effect dominates the inductive effect in this case so the methoxy group is a strongly activating group and an ortho/para director. With a few exceptions that are discussed shortly, any group that has an unshared pair of electrons on the atom bonded to the ring, represented by the general group Z in the following equation, has a similar resonance effect and acts as an activating group and an ortho/para director: PROBLEM 17.1 Show all of the resonance structures for the arenium ion that is produced by attack of the NO2 electrophile at the ortho position of anisole. Which of these structures is es-pecially stable? PROBLEM 17.2 Show all of the resonance structures for the arenium ion that is produced by attack of the NO2 electrophile at the meta position of anisole. Is there an especially stable res-onance structure in this case? PROBLEM 17.3 Explain why these compounds react faster than benzene in electrophilic aromatic sub-stitution reactions and give predominantly ortho and para products: Now let’s consider an example where the substituent slows the reaction. The nitra-tion of nitrobenzene occurs approximately 107 times more slowly than the nitration of benzene and gives predominantly the meta-isomer of dinitrobenzene. a) NH2 . . b) c) CH2CH3 Z . . Z . . E+ + Especially stable resonance structure E Z+ H E H 17.2 I EFFECT OF SUBSTITUENTS 677 The nitro group is deactivating and directs the incoming electrophile to the meta po-sition. Again, examination of the arenium ions provides an explanation for these re-sults. First, consider the arenium ion produced by attack of the electrophile at an ortho position: The nitro group is an electron-withdrawing group both by its inductive effect and by its resonance effect. The ﬁrst resonance structure is especially destabilized because the positive charge is located directly adjacent to the electron-withdrawing nitro group. Thus, the presence of the nitro group on the ring dramatically slows attack of an elec-trophile at the ortho position. Now, consider attack of the electrophile at a meta position: This time there is no resonance structure that has the positive charge on the carbon bonded to the nitro group. The arenium ion is still destabilized by the electron-withdrawing effect of the nitro group, but this ion is not destabilized as much as the ion produced by attack of the electrophile at the ortho position because the positive charge is never as close to the nitro group. The arenium ion produced by attack of the electrophile at the para position resem-bles that produced by attack at the ortho position in that it also has a resonance struc-ture that has the positive charge on the carbon that is bonded to the nitro group. The destabilization of this ion is comparable to that of the ortho ion. Arenium ion from meta attack + +N œ ± . . O . . . .O . . . . – H NO2 + +N œ ± . . O . . . .O . . . . – H NO2 + +N œ ± . . O . . . .O . . . . – H NO2 Arenium ion from ortho attack H NO2 + + H NO2 + N œ ± . . O . . . .O . . . . – +N œ ± . . O . . . .O . . . . – +N œ ± . . O . . . .O . . . . – H NO2 + NO2 NO2 6% HNO3 H2SO4 NO2 NO2 NO2 NO2 NO2 92% 2% 678 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Overall, then, the nitro group slows the reaction, but it slows attack at the meta po-sition less than it slows attack at the ortho and para positions. It is a deactivating group and a meta-directing group. In general, any group that withdraws electrons from the ring by an inductive and/or resonance effect behaves similarly. PROBLEM 17.4 Explain why these compounds react more slowly than benzene in electrophilic aromatic substitution reactions and give predominantly meta products: So far, groups have been either activating and ortho/para directors or deactivating and meta directors. The halogens are exceptions to this generalization. They are slightly deactivating compared to benzene but still direct to the ortho and para positions. For ex-ample, chlorobenzene is nitrated 17 times slower than benzene and produces predomi-nantly ortho- and para-chloronitrobenzene. Why do the halogens have this unusual behavior? Like the methoxy group, the in-ductive and resonance effects of the halogens are in competition. In the case of the halo-gens, however, the inductive electron-withdrawing effect is slightly stronger than the resonance electron-donating effect. The high electronegativity of ﬂuorine is responsible for its inductive effect being stronger than its resonance effect. The other halogens are weaker resonance electron-donating groups because their p orbitals do not overlap well with the 2p AO of the ring carbon, owing to the longer length of the carbon–halogen bond and the size of the 3p, 4p, or 5p AO. As a result, the halogens are weakly deacti-vating groups. But because resonance electron donation is most effective at the ortho and para positions, these positions are deactivated less than the meta position. There-fore, the halogens are slightly deactivating ortho/para directors. The effect of almost any substituent can be understood on the basis of similar rea-soning. Table 17.1 lists the effect on both the reaction rate and the regiochemistry of the substituents most commonly found on benzene rings. Rather than just memorizing this table, try to see the reasons why each group exhibits the behavior that it does. The strongly activating, ortho/para directors all have an unshared pair of electrons on the atom attached to the ring that is readily donated by resonance. Alkyl and aryl groups NO2 NO2 64% 1% 35% Chlorobenzene HNO3 H2SO4 Cl Cl NO2 Cl Cl b) +N(CH3)3 Cl– a) CF3 17.2 I EFFECT OF SUBSTITUENTS 679 stabilize a positive charge on an adjacent carbon, so they are activating ortho/para di-rectors, although they are less activating than the preceding groups. The deactivating meta-directing groups all have a positive or partial positive charge on the atom attached to the ring. The halogens are unusual because they are weakly deactivating, ortho/para directors. PROBLEM 17.5 Predict the effect of these substituents on the rate and regiochemistry of electrophilic aromatic substitution reactions: c) ±C±Cl O X b) ±S±CH3 O X X O a) ±S±CH3 . . . . 680 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Table 17.1 Effect of Substituents on the Rate and Regiochemistry of Electrophilic Aromatic Substitution Reactions Substituent Rate Effect Regiochemistry Comments ±NR3 ±NH3 + + ±NO2 ±CN ±SO3H ±CX3 ±COR O X O X ±CNH2 ±CH O X O X ±CR O X ±COH Halogens ±X . . . . . . ±R ±Ar Alkyl groups Aryl groups ±NHCCH3 O X . . ±OCCH3 . . . . ±OR . . . . ±SR . . . . O X ±NR2 . . ±NHR . . ±NH2 . . ±OH . . . . Strongly activating Moderately activating Weakly activating Weakly deactivating Moderately deactivating Strongly deactivating ortho and para ortho and para ortho and para ortho and para meta meta Resonance donating effect is stronger than inductive withdrawing effect Resonance donating effect is stronger than inductive withdrawing effect, but not as much so as above Weak inductive or resonance donors Resonance donating effect controls regio-chemistry but is weaker than inductive withdrawing effect that controls rate Inductive and resonance withdrawers Inductive and/or resonance withdrawers Finally, let’s compare the amount of ortho product to the amount of para product pro-duced in a reaction of an aromatic compound that has an ortho/para director on the ring. There are two ortho positions and only one para position, so if statistics were the only important factor, the ratio of ortho to para products should be 2 to 1. However, attack at the ortho position can be disfavored by the steric effect of the group. Obviously, this de-pends partly on the size of the group and the size of the electrophile. In addition, some reactions are more sensitive to steric effects than others. In general, then, the ratio of or-tho to para product ranges from 2 to 1 in favor of the ortho product to predominantly para for reactions involving bulky substituents or reactions that are very sensitive to steric hin-drance. An example of this steric effect can be seen by comparing the following reaction to the nitration of toluene presented earlier (page 674). The major product from toluene is the ortho- isomer (60% ortho and 37% para). In contrast, the bulky group of t-butyl-benzene causes the major product from its nitration to be the para-isomer. PRACTICE PROBLEM 17.1 Show the products of the reaction of ethylbenzene with nitric acid and sulfuric acid. Solution The ethyl group, like other alkyl groups, is weakly activating and directs to the ortho and para positions. The small amount of meta product that is formed is usually not shown. NO2 NO2 HNO3 H2SO4 CH2CH3 CH2CH3 CH2CH3 CH2CH3 HNO3 H2SO4 CH3 W W H3CCCH3 H3CCCH3 CH3 W W ± NO2 CH3 W W H3CCCH3 W NO2 82% 18% t-Butylbenzene HNO3 H2SO4 17.2 I EFFECT OF SUBSTITUENTS 681 PROBLEM 17.6 Show the products of these reactions: 17.3 Effect of Multiple Substituents The situation is more complicated if there is more than one substituent on the ben-zene ring. However, it is usually possible to predict the major products that are formed in an electrophilic aromatic substitution reaction. When the substituents di-rect to the same position, the prediction is straightforward. For example, consider the case of 2-nitrotoluene. The methyl group directs to the positions ortho and para to itself—that is, to positions 4 and 6. The nitro group directs to positions meta to itself—that is, also to positions 4 and 6. When the reaction is run, the products are found to be almost entirely 2,4-dinitrotoluene and 2,6-dinitrotoluene, as expected: If the groups direct to different positions on the ring, usually the stronger activating group controls the regiochemistry. Groups that are closer to the top of Table 17.1 con-trol the regiochemistry when competing with groups lower in the table. In the case of 3-nitrotoluene the methyl group directs to positions 2, 4, and 6 while the nitro group di-rects to position 5. Because the methyl group is a stronger activating group than the ni-tro group, it controls the regiochemistry: Note that none of the product where the new nitro group has been added to position 2, between the two groups, is formed. In general, the position between two groups that are meta to each other is not very reactive because of steric hindrance by the groups on ei-ther side of this position. O2N O2N 1% 60% 38% 3-Nitrotoluene 1 3 2 4 5 6 NO2 CH3 NO2 NO2 NO2 NO2 CH3 CH3 CH3 HNO3 H2SO4 O2N O2N 1% 70% 29% 2-Nitrotoluene 1 2 3 4 5 6 NO2 CH3 NO2 CH3 NO2 CH3 NO2 CH3 NO2 HNO3 H2SO4 b) Br HNO3 H2SO4 a) NO2 HNO3 H2SO4 682 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS PROBLEM 17.7 Explain which positions would be preferentially nitrated in the reaction of these com-pounds with nitric acid and sulfuric acid: PROBLEM 17.8 Show the major products of these reactions: 17.4 Nitration The reagents and the mechanism for the nitration of an aromatic ring have already been discussed. The reaction is very general and works with almost any substituent on the ring, even strongly deactivating substituents. (85%) Methyl benzoate CO2CH3 CO2CH3 NO2 HNO3 H2SO4 15°C b) OCH3 Br HNO3 H2SO4 a) CCH3 CH2CH2CH3 HNO3 H2SO4 O X f) OH Cl e) NHCCH3 O X NO2 d) CH2CH3 CH2CH3 c) CH3 OCH3 b) NHCCH3 CCH3 O X X O a) CH2CH3 NO2 17.4 I NITRATION 683 The concentration of the NO2 electrophile is controlled by the strength of the acid that is used in conjunction with nitric acid. Milder conditions, such as nitric acid without sulfuric acid, nitric acid and acetic acid, or nitric acid in acetic anhydride, are employed when the ring is strongly activated, as illustrated in the following example: The nitro group that is added to the ring in these reactions is a deactivating group. This means that the product is less reactive than the reactant, so it is easy to add only one nitro group to the ring. However, it is possible to add a second nitro group, if so desired, by using more vigorous conditions. Thus, the reaction of ben-zoic acid using the same conditions as shown earlier for methyl benzoate (HNO3, H2SO4, 15°C) results in the formation of the mononitration product, m-nitrobenzoic acid. Under more drastic conditions (higher temperature and higher sulfuric acid concentration), two nitro groups can be added, as illustrated in the following equation: The following example illustrates a problem that sometimes occurs with amino sub-stituents: The dimethylamino group is a strong activator and an ortho/para director, yet the ma-jor product from the reaction is the meta-isomer. This unexpected result is due to the 18% 63% N,N-Dimethylaniline NO2 NO2 HNO3 H2SO4 N CH3 H3C ± ± N CH3 H3C ± ± N CH3 H3C ± ± (58%) Benzoic acid CO2H CO2H NO2 O2N HNO3 H2SO4 145°C (76%) 1,3,5-Trimethylbenzene CH3 CH3 H3C CH3 NO2 CH3 H3C HNO3 CH3CO2H 684 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS basicity of the amino group. In the strongly acidic reaction mixture, the nitrogen is protonated: The NH(CH3)2 group deactivates the ring and directs to the meta position. The major product, the meta-isomer, results from the reaction of the protonated amine. The minor product, the para-isomer, results from the reaction of a very small amount of unproto-nated amine. Although its concentration in the strongly acidic solution is extremely small, the unprotonated amine is many orders of magnitude more reactive than its con-jugate acid, so some of the para product is formed. It is fairly common for the electron pair on an amino group, which is a good Lewis base, to react with a Lewis acid under the strongly electrophilic conditions of these sub-stitution reactions. This changes the substituent from a strong activating group to a strong deactivating group. As a result, the reaction often has the undesired regiochem-istry, and in some cases the desired reaction may not occur at all. Because the exact re-sult is difﬁcult to predict or control, the amino substituent is usually modiﬁed to decrease its reactivity. The strategy is similar to that employed in the Gabriel synthesis (see Section 10.6). A “protecting group” that makes the electrons on the nitrogen less basic is bonded to the amino group. After the desired substitution reaction has been ac-complished, the protecting group is removed and the amino group is regenerated. The most common method to decrease the reactivity of an unshared pair of electrons on an atom is to attach a carbonyl group to that atom. Therefore, the amine is ﬁrst reacted with acetyl chloride to form an amide. (This reaction and its mechanism are described in detail in Section 19.6. To help you remember the reaction for now, note that the nitrogen nucle-ophile attacks the carbonyl carbon electrophile, displacing the chloride leaving group.) Because of delocalization of the nitrogen’s electron pair onto the carbonyl oxygen, the electrons of the acetylamino group are less available for delocalization into the ring by resonance. (This is why the acetylamino group is a weaker activator than the amino group.) In addition, the electron pair on the nitrogen of an amide is much less basic, and reactions with Lewis acids in the substitution reactions are not usually a problem. How-ever, the acetylamino group is still an activator and an ortho/para director, so substitu-. . CH3CCl O X NH CCH3 + – O W ± ± . . . . . . NH±CCH3 O ± ± . . . . . . Acetylamino group Aniline NH2 H3C±N±CH3 + H W H±O±S±OH O X X O . . . . HSO4– CH3 H3C ± ± N . . 17.4 I NITRATION 685 tion reactions work well. After the substitution has been completed, the acetyl group can be removed by hydrolysis of the amide bond (This reaction is very similar to the imide hydrolysis employed in the Gabriel synthesis [Section 10.6] and the ester hydrolysis used in the acetate method for the preparation of alcohols [Section 10.2].) An example of the use of this strategy is illustrated in the following synthesis. (Note that the acetyl-amino group controls the regiochemistry of the reaction, so it is a stronger activator than the methoxy group in this reaction.) PROBLEM 17.9 Show the products of these reactions: 17.5 Halogenation Chlorine and bromine can be substituted onto an aromatic ring by treatment with Cl2 or Br2. With all but highly activated aromatic rings (amines, phenols, polyalkylated rings), a Lewis acid catalyst is also required to make the halogen electrophile strong enough to accomplish the reaction. The most common catalysts are the aluminum and iron halides, AlCl3, AlBr3, FeCl3, and FeBr3. An example is provided by the following equation: Cl2 HCl (58%) Cl AlCl3 Chlorobenzene HNO3 KOH H2O c) O X CH3CCl NH2 b) O X CCH3 HNO3 H2SO4 a) OCH3 HNO3 CH3CO2H OCH3 NHCCH3 O X O X (77%) CH3CCl p-Methoxyaniline NH2 OCH3 NHCCH3 O X OCH3 NO2 NH2 OCH3 NO2 HNO3 KOH H2O 686 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS As shown in the following equation, the Lewis acid, AlCl3 in this case, bonds to one of the atoms of Cl2 to produce the electrophilic species: A pair of pi electrons of the aromatic ring then bonds to the electrophilic chlorine as AlCl4 leaves. (AlCl4 is a weaker base and a better leaving group than Cl.) The re-mainder of the mechanism is the same as that illustrated in Figure 17.1. This substitution reaction provides a general method for adding chlorine or bromine to an aromatic ring. Because both are deactivating substituents, the product is less re-active than the starting aromatic compound, so it is possible to add a single halogen. The reaction works with deactivated substrates, as illustrated in the following example: As mentioned previously, halogenation of highly reactive substrates can be accom-plished without the use of a Lewis acid catalyst. Thus, the bromination of mesitylene (1,3,5-trimethylbenzene) is readily accomplished by reaction with bromine in carbon tetrachloride: With very reactive compounds, such as anilines and phenols, it is often difﬁcult to stop the reaction after only one halogen has added to the ring. In such cases the prod-uct that is isolated usually has reacted at all of the activated positions: CO2H Cl Cl Cl2 H2O (78%) NH2 CO2H NH2 2-Aminobenzoic acid CCl4 (82%) Br Br2 HBr CH3 CH3 H3C CH3 CH3 H3C Mesitylene (1,3,5-trimethylbenzene) Br2 FeBr3 (75%) NO2 NO2 Nitrobenzene Br Cl3Al –AlCl4 H + Cl . . . . . . Cl±Cl . . . . . . . . . . . . Cl3Al±Cl±Cl. . . . . . . . . . + – 17.5 I HALOGENATION 687 Click Mechanisms in Motion to view Electrophilic Aromatic Bromination. If this is a problem, the solution again is to decrease the reactivity of the ring by mod-iﬁcation of the activating group. The carbonyl protecting group is removed after the halogenation is accomplished. PROBLEM 17.10 Show all of the steps in the mechanism for this reaction: PROBLEM 17.11 Show the products of these reactions: CH3 f) Br2 Br AlBr3 e) NO2 CH3 Br2 FeBr3 d) F Cl2 AlCl3 c) NO2 Cl2 AlCl3 b) OH 2 Br2 a) CH3 Br2 FeBr3 Br2 HBr NO2 NO2 Br FeBr3 CH3 NHCCH3 O X O X (67%) CH3CCl p-Methylaniline NH2 CH3 CH3 CH3 NHCCH3 O X Br Br NH2 Br2 KOH H2O 688 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS 17.6 Sulfonation A sulfonic acid group can be substituted onto an aromatic ring by reaction with con-centrated sulfuric acid as shown in the following example: Although the electrophile varies depending on the exact reaction conditions, it is often sulfur trioxide, SO3, that is formed from sulfuric acid by the loss of water. The mecha-nism for the addition of this electrophile proceeds according to the following equation: Again, this is a general reaction that works for deactivated as well as activated rings. The SO3H group that is added is a deactivator, so the reaction can be halted after the substitution of a single group. In contrast to the other reactions that have been presented so far, this substitution is readily reversible. Reaction of a sulfonic acid in a mixture of water and sulfuric acid re-sults in removal of the sulfonic acid group. In this case a proton is the electrophile. An example is provided by the following equation: PROBLEM 17.12 Show all of the steps in the mechanism for this reaction: SO3H H2O H2SO4 H H2SO4 SO3H H2SO4 (56%) H2O H2SO4 2-Nitroaniline NH2 NO2 NH2 NO2 S O O O œ œ . . . . . . . . . . . . ± ± ± H + OœS O X O . . . . . . . . . . . . . . O±S±OH O X X O . . . . . . OœS±O O X . . . . . . . . . . . . . . OœS±O±H O X . . . . . . . . . . . . H2SO4 Benzenesulfonic acid – – – H2SO4 H2O (71%) Cl Cl SO3H Chlorobenzene p-Chlorobenzenesulfonic acid 17.6 I SULFONATION 689 PROBLEM 17.13 Show the products of these reactions: 17.7 Friedel-Crafts Alkylation Developed by C. Friedel and J. M. Crafts, the reaction of an alkyl halide with an aro-matic compound in the presence of a Lewis acid catalyst, usually AlCl3, results in the substitution of the alkyl group onto the aromatic ring: In most cases the electrophile is the carbocation that is generated when the halide acts as a leaving group. The role of the aluminum chloride is to complex with the halogen to make it a better leaving group. From the point of view of the alkyl halide, the mech-anism is an SN1 reaction with the pi electrons of the aromatic ring acting as the nucle-ophile (see Figure 17.4). Although the most common method for generating the electrophile for the alkyla-tion reaction employs an alkyl halide and aluminum trichloride, it can be generated in other ways also. For example, the reaction in the following equation uses the reaction of an alcohol and an acid to produce the carbocation: TsOH (87%) H2O CH3 CH2±OH CH2 CH3 H3C CH3 (71%) CH3CHCH2CH3 Cl W HCl CH3CHCH2CH3 AlCl3 d) CH3 OH H2SO4 c) H2SO4 b) CH3 H2SO4 a) SO3H H2SO4 690 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Alternatively, the carbocation can be generated by protonation of an alkene. This reac-tion resembles the additions to alkenes discussed in Chapter 11. An example is provided by the following equation: Several limitations occur with the Friedel-Crafts alkylation reaction. First, the alkyl group that is added to the ring is an activating group. This causes the alkylated product to be more reactive (by a factor of about 2) than the starting aromatic compound. There-fore, a signiﬁcant amount of product where two or more alkyl groups have been added is commonly formed. The best solution to this problem is to use a large excess of the aromatic compound that is to be alkylated. This can easily be accomplished for com-pounds that are readily available, such as benzene or toluene, by using them as the sol-vent for the reaction. Note that the Friedel-Crafts alkylation is the only one of these electrophilic aromatic substitution reactions in which the product is more reactive than the starting material. All of the other reactions put deactivating groups on the ring, so they do not suffer from the problem of multiple substitution. H2SO4 (68%) H H H H H 17.7 I FRIEDEL-CRAFTS ALKYLATION 691 The aluminum trichloride bonds with an electron pair on the chlorine of the alkyl halide to form a Lewis acid–base adduct. This changes the leaving group to AlCl4–, which is a weaker base and a better leaving group than chloride anion. The carbocation acts as an electrophile and reacts with a pair of pi electrons of the aromatic ring. Or this can be viewed as an SN1 reaction, with the weakly nucleophilic aromatic ring attacking the carbocation. The remainder of the mechanism is identical to the general mechanism outlined in Figure 17.1. A base in the reaction mixture, such as AlCl4–, removes a proton to produce the final product, HCl, and AlCl3, which can begin the process anew. The AlCl4– leaves, producing a carbocation intermediate. AlCl3 + + + PhCH2±Cl±AlCl3 . . . . HCl CH2Ph AlCl3 Cl3Al±Cl. . . . . . AlCl4 PhCH2 PhCH2±Cl . . . . . . – – – CH2Ph H 1 1 2 2 3 3 4 4 Active Figure 17.4 MECHANISM OF THE FRIEDEL-CRAFTS ALKYLATION REACTION. T est yourself on the concepts in this ﬁgure at OrganicChemistryNow. A second limitation is that aromatic compounds substituted with moderately or strongly deactivating groups cannot be alkylated. The deactivated ring is just too poor a nucleophile to react with the unstable carbocation electrophile before other reactions oc-cur that destroy it. The ﬁnal limitation is one that plagues all carbocation reactions: rearrangements. Be-cause the aromatic compound is a weak nucleophile, the carbocation has a lifetime that is longer than is the case in most of the other reactions involving this intermediate, allowing ample time for rearrangements to occur. An example is provided by the following equation: Despite these limitations, alkylation of readily available aromatic compounds, such as benzene and toluene, using carbocations that are not prone to rearrange, is a useful re-action. Intramolecular applications of this reaction have proven to be especially valuable. PROBLEM 17.14 Show all of the steps in the mechanism for the formation of both products in this reaction: PROBLEM 17.15 Show the products of these reactions: b) NHCCH3 O X CH3CCH3 CH2 X H2SO4 a) OCH3 CH3CHCH3 Cl W AlCl3 CH2CH2CH3 CH3CHCH3 CH3CH2CH2Cl HCl AlCl3 (85%) OH SnCl4 O O (55%) CH3 Ph CH3 Ph H2SO4 HO CH3CH2CH2CH2 Cl W AlCl3 0°C Butylbenzene sec-Butylbenzene 34% 66% CH2CH2CH2CH3 CH3CHCH2CH3 692 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS PROBLEM 17.16 Show syntheses of these compounds from benzene: PRACTICE PROBLEM 17.2 Explain which of these routes would provide a better method for the preparation of p-nitrotoluene: Solution Route A works ﬁne. Toluene is readily nitrated, and the methyl group is an ortho/para director. The only problem is that both the desired compound and its ortho-isomer are produced and must be separated. (This is a common problem, and we usually assume that the separation can be accomplished, although it is not always easy in the labora-tory.) Route B is unsatisfactory because the Friedel-Crafts alkylation reaction does not work with deactivated compounds such as nitrobenzene. Furthermore, even if the al-kylation could be made to go, the nitro group is a meta director, so the desired product would not be formed. Route A HNO3 H2SO4 Route B CH3Cl AlCl3 CH3 NO2 p-Nitrotoluene CH3 NO2 b) a) f) Cl AlCl3 NO2 e) Cl AlCl3 d) OH H2SO4 c) Cl AlCl3 17.7 I FRIEDEL-CRAFTS ALKYLATION 693 694 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Focus On Synthetic Detergents, BHT, and BHA A soap is the sodium salt of carboxylic acid attached to a long, nonpolar hydrocarbon chain. When a soap is placed in hard water, the sodium cations exchange with cations such as Ca2 and Mg2. The resulting calcium and magnesium salts are insoluble in water and precipitate to form “soap scum.” 2 CH3(CH2)16CO2 Na Ca2 [CH3(CH2)16CO2 ]2 Ca2 2 Na Precipitates Synthetic detergents were invented to alleviate this problem. Rather than use the anion derived from a carboxylic acid with a large nonpolar group, detergents employ the anion derived from a sulfonic acid attached to a large nonpolar group. The calcium and magnesium salts of these sulfonic acids are soluble in water, so detergents do not precipitate in hard water and can still accomplish their cleaning function. Two of the reactions that are used in the industrial preparation of detergents are elec-trophilic aromatic substitution reactions. First, a large hydrocarbon group is attached to a ben-zene ring by a Friedel-Crafts alkylation reaction employing tetrapropene as the source of the carbocation electrophile. The resulting alkylbenzene is then sulfonated by reaction with sul-furic acid. Deprotonation of the sulfonic acid with sodium hydroxide produces the detergent. A detergent Tetrapropene HCl H2SO4 40°C + NaOH SO3 –Na+ SO3H The exact structure of the alkyl group on the benzene ring is not important as long as it is large enough to confer the necessary hydrophobic character. Tetra-propene was used in the early versions of detergents because it was readily and cheaply available from the treatment of propene with acid. In this reaction, four propenes combine to form tetrapropene through carbocation intermediates. (In addition to the compound shown in the equation, an isomer with the double bond between carbon 2 and carbon 3 is also formed. If you are interested in the mechanism for this reaction, it is a variation of the cationic polymerization mechanism described later in Section 24.3.) 17.7 I FRIEDEL-CRAFTS ALKYLATION 695 However, the detergent prepared from tetrapropene caused a problem in sewage treatment plants. The microorganisms that degrade such compounds start from the end of the hydrocarbon chain and seem to have trouble proceeding through tertiary car-bons. The presence of several tertiary carbons in the tetrapropene chain slows its biodegradation to the point at which a significant amount passes through a treatment plant unchanged. This causes the resulting effluent and the waterways into which it is discharged to become foamy, an environmentally unacceptable result. To solve this problem, most modern detergents are prepared from straight-chain alkenes. The resulting linear alkylbenzenesulfonate detergents are more easily de-graded, and our rivers are no longer foamy. An example of a typical alkylation is shown in the following equation: PROBLEM 17.17 What isomeric alkyl benzene should also be formed in this reaction? Butylated hydroxytoluene (BHT) and butylated hydroxyanisole (BHA) are antioxidants that are added to foods and many other organic materials to inhibit decomposition caused by reactions with oxygen. Perhaps you have seen these compounds listed among the ingredients on your cereal box at breakfast. (The mecha-nism of operation for these antioxidants is described in Section 21.8.) Both of these compounds are prepared by Friedel-Crafts alkylation reactions. BHT is synthesized by the reaction of p-methylphenol with 2-methylpropene in the presence of an acid catalyst. 2 CH3CœCH2 CH3 W AlCl3 HCl OH OH CH3 CH3C± ±CCH3 W W H3C H3C W W CH3 CH3 CH3 Butylated hydroxytoluene (BHT) p-Methylphenol (p-hydroxytoluene) 2-Methylpropene (isobutylene) CH3(CH2)6CHœCH(CH2)5CH3 HCl AlCl3 CH3(CH2)6CH(CH2)6CH3 4 H3PO4 205°C 1000 psi Continued 17.8 Friedel-Crafts Acylation The reaction of an aromatic compound with an acyl chloride in the presence of a Lewis acid (usually AlCl3) results in the substitution of an acyl group onto the aromatic ring. An example of this reaction, known as the Friedel-Crafts acylation, is provided by the following equation: The electrophile, an acyl cation, is generated in a manner similar to that outlined in Fig-ure 17.4 for the generation of the carbocation electrophile from an alkyl halide. First the Lewis acid, aluminum trichloride, complexes with the chlorine of the acyl chloride. Then AlCl4 leaves, generating an acyl cation. The acyl cation is actually more stable than most other carbocations that we have encountered because it has a resonance struc-ture that has the octet rule satisﬁed for all of the atoms: AlCl3 CH3±C±Cl O X + + + _ _ . . . . . . CH3±C±Cl±AlCl3 O X . . . . AlCl4 CH3±CœO. . . . CH3±C O. . ± ± ± Acyl cation CH3C±Cl O X O X Benzene Acetyl chloride Acetophenone CCH3 (61%) HCl AlCl3 696 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS Addition of a proton to 2-methylpropene produces the t-butyl carbocation, which then alkylates the ring. Conditions are adjusted so that two t-butyl groups are added. BHA is prepared in a similar manner by the reaction of p-methoxyphenol with 2-methylpropene and an acid catalyst. In this case conditions are adjusted so that only one t-butyl group is added. Because the hydroxy group and the methoxy group are both activating groups, a mixture of products is formed in this case. CH3CœCH2 CH3 W CCH3 W W CH3 CH3 CCH3 W W CH3 CH3 Butylated hydroxyanisole (BHA) p-Methoxyphenol (p-hydroxyanisole) OCH3 OH OCH3 OH OCH3 OH AlCl3 HCl 2-Methylpropene Click Coached Tutorial Problems to quiz yourself on Mechanisms of Electrophilic Aromatic Substitution. The Friedel-Crafts acylation reaction does not have most of the limitations of the alkylation reaction. Because of the stability of the acyl cation, rearrangements do not occur in this reaction. In addition, the acyl group that is added to the ring is a deactiva-tor, so the product is less reactive than the starting aromatic compound. Therefore, there is no problem with multiple acyl groups being added to the ring. However, like alkyla-tions, the acylation reaction does not work with moderately or strongly deactivated substrates—that is, with rings that are substituted only with meta directing groups. As a consequence of fewer limitations, the Friedel-Crafts acylation reaction is more useful than the alkylation reaction. Anhydrides can be used in place of acyl chlorides as the source of the electrophilic acyl cation: As seen in this example, the acylation reaction is more sensitive to steric effects than the other reactions that have been discussed so far and tends to give predomi-nantly the para product. Some additional examples are provided by the following equations: Finally, the intramolecular version of the Friedel-Crafts acylation reaction has proved to be very valuable in the construction of polycyclic compounds, as illustrated in the following equation: (83%) ClCH2CCl O X (55%) CH3CCl O X O X CCH3 NHCCH3 O X X O CCH2Cl NHCCH3 O X CH3 CH3CHCH3 p-Isopropyltoluene AlCl3 AlCl3 CH3 CH3CHCH3 (79%) CH3COCCH3 O X O X CH3COH O X X O CCH3 Br Br AlCl3 17.7 I FRIEDEL-CRAFTS ALKYLATION 697 For intramolecular reactions, treatment of a carboxylic acid with sulfuric acid or polyphosphoric acid is sometimes used to generate the acyl cation electrophile. This method is usually too mild for intermolecular acylations but works well for intramole-cular examples, as shown in the following equation: PROBLEM 17.18 Explain why the Friedel-Crafts acylation of p-isopropyltoluene shown on the previous page results in the substitution of the acyl group at the position ortho to the methyl group. PROBLEM 17.19 Show the products of these reactions: e) PhCH2CCl O X O X OCCH3 AlCl3 d) HO2C H3PO4 c) O O O AlCl3 b) CH3CCl O X O X CCH3 AlCl3 a) CH2CH3 PhCCl O X AlCl3 (90%) O X COH O O O CH3 AlCl3 H2SO4 CH3 O CH3 O O (91%) O X C AlCl3 Cl O W 698 CHAPTER 17 I AROMATIC SUBSTITUTION REACTIONS PROBLEM 17.20 Suggest syntheses of these compounds using Friedel-Crafts acylation reactions: 17.9 Electrophilic Substitutions of Polycyclic Aromatic Compounds Polycyclic aromatic compounds also undergo electrophilic aromatic substitution reac-tions. Because the aromatic resonance energy that is lost in forming the arenium ion is lower, these compounds tend to be more reactive than benzene. For example, the bromi-nation of naphthalene, like that of other reactive aromatic compounds, does not require a Lewis acid catalyst: Naphthalene also undergoes the other substitution reactions described for benzene. For example, it is acylated under standard Friedel-Crafts conditions: Note that both the bromination and the acylation of naphthalene result in the sub-stitution of the electrophile at the 1 position. None of the isomeric product with the electrophile bonded to the 2 position is isolated in either case. The higher reactivity of the 1 position can be understood by examination of the resonance structures for the are-nium ion. When the electrophile adds to the 1 position, the arenium ion has a total of seven resonance structures, whereas only six exist for the arenium ion resulting from addition of the electrophile to the 2 position. (92%) HCl CCH3 O X CH3CCl O X AlCl3 (75%) Naphthalene Br2 HBr Br 1 8 2 7 3 6 4 5 c) O b) OCH3 CCH3 X O a) CH3 CCH2CH2CH3 X O 17.9 I ELECTROPHILIC SUBSTITUTIONS OF POLYCYCLIC AROMATIC COMPOUNDS 699 Click Coached Tutorial Problems for additional practice showing the products of Electrophilic Aromatic Substitution Reactions.
1770	https://baike.baidu.com/item/%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F/5792503	网页新闻贴吧知道网盘图片视频地图文库资讯采购百科帮助首页秒懂百科特色百科知识专题加入百科百科团队权威合作最小正周期播报讨论上传视频收藏查看我的收藏 0有用+1 0 本词条由“科普中国”科学百科词条编写与应用工作项目审核。如果一个函数f（x）的所有周期中存在一个最小的正数，那么这个最小的正数就叫做f（x）的最小正周期（minimal positive period）.例如，正弦函数的最小正周期是2π . 根据上述定义，我们有：对于正弦函数y=sinx, 自变量x只要并且至少增加到x+2π时，函数值才能重复取得正弦函数和余弦函数的最小正周期是2π。 y=Asin(ωx+φ), T=2π/ω（其中ω必须>0）中文名 : 最小正周期外文名 : minimal positive period 领域 : 数学算法实例 : 函数f(x)±g(x)最小正周期的求法基本概念 : 函数f（x）所有周期中最小的正数应用 : 图像分析、信号处理目录 1算法实例 ▪定义法 ▪公式法 ▪最小公倍数法 ▪图象法 ▪恒等变换法 2补充问题算法实例播报编辑函数f(x)±g(x)最小正周期的求法定义法概念：根据周期函数和最小正周期的定义，确定所给函数的最小正周期。例1、求函数y=\|sinx\|+\|cosx\|的最小正周期. 解:∵ =\|sinx\|+\|cosx\| =\|－sinx\|+\|cosx\| =\|cos(x+π/2)\|+\|sin(x+π/2)\| =\|sin(x+π/2)\|+\|cos(x+π/2)\| =f(x+π/2) 对定义域内的每一个x，当x增加到x+π/2时，函数值重复出现，因此函数的最小正周期是π/2.（如果f（x+T）=f（x），那么T叫做f（x）的周期）。例2 、求函数的最小正周期。解：把看成是一个新的变量z,那么2sinz的最小正周期是2π。由于。所以当自变量x增加到x+4π且必须增加到x+4π时，函数值重复出现。 ∴函数的最小正周期是4π。公式法这类题目是通过三角函数的恒等变形，转化为一个角的一种函数的形式，用公式去求，其中正余弦函数求最小正周期的公式为T=2π/\|ω\| ，正余切函数T=π/\|ω\|。函数f(x)=Asin(ωx+φ)和f(x)=Acos(ωx+φ)(A≠0,ω>0）的最小正周期都是；函数f(x)=Atan(ωx+φ)和f(x)=Acot(ωx+φ)(A≠0,ω>0）的最小正周期都是，运用这一结论，可以直接求得形如y=Af(ωx+φ)(A≠0,ω>0）一类三角函数的最小正周期（这里“f”表示正弦、余弦、正切或余切函数）。例3、求函数y=cotx－tanx的最小正周期. 解：y=1/tanx－tanx=(1－tan^2· x)/tanx=2(1－tan^2·x)/(2tanx)=2cot2x ∴T=π/2 函数为两个三角函数相加，若角频率之比为有理数，则函数有最小正周期。最小公倍数法设f(x)与g(x)是定义在公共集合上的两个三角周期函数，T1、T2分别是它们的周期，且T1≠T2，则f(x)±g(x)的最小正周期T1、T2的最小公倍数，分数的最小公倍数=T1,T2分子的最小公倍数/T1、T2分母的最大公约数。求几个正弦、余弦和正切函数的最小正周期，可以先求出各个三角函数的最小正周期，然后再求期最小公倍数T,即为和函数的最小正周期。例4、求函数y=sin3x+cos5x的最小正周期. 解：设sin3x、cos5x的最小正周期分别为T1、T2，则T1=2π/3,T2=2π/5 ，所以y=sin3x+cos5x的最小正周期T=2π/1=2π. 例5、求y=sin3x+tan2x/5 的最小正周期. 解：∵sin3x与tan2x/5 的最小正周期是2π/3与5π/2,其最小公倍数是10π/1=10π. ∴y=sin3x+tan2x/5的最小正周期是10π. 说明：几个分数的最小公倍数，我们约定为各分数的分子的最小公倍数为分子，各分母的最大公约数为分母的分数。图象法概念：作出函数的图象，从图象上直观地得出所求的最小正周期。例6、求y=\|sinx\|的最小正周期. 解：由y=\|sinx\|的图象可知y=\|sinx\|的周期T=π. 图示例7、求下函数的最小正周期。（1）（2）解：（1）先作出函数的图象（见图1）观察图象，易得所求的周期为T=π/3。图示（2）先作出的图象（见图2）观察图象，易得所求的周期为T=π。恒等变换法概念：通过对所给函数式进行恒等变换，使其转化为简单的情形，再运用定义法、公式法或图象法等求出其最小正周期。 (1) f(x)=sin(x+π/3)cos(x-π/3) (2) f(x)=sin6x+cos6x (3) f(x)= 解 (1) ∴最小正周期为T= π (2) f(x)=sin6x+cos6x =(sin2x+cos2x)(sin4x-sin2xcos2x+cos4x) =(sin4x-sin2xcos2x+cos4x) =(sin2x+cos2x)2-3sin2xcos2x =1-3/4sin2x =5/8+3/8cos4x ∴最小正周期为T=π/2 (3) 它与－cos2x的周期相同，故得 f(x)的最小正周期为T=π 补充问题播报编辑函数f(x)=sin2x-4sin³xcosx(x∈R)的最小正周期为（ B ） A.π/4 B.π/2 C.π D.2π 成长任务编辑入门编辑规则本人编辑内容质疑在线客服官方贴吧意见反馈举报不良信息未通过词条申诉投诉侵权信息封禁查询与解封 ©2025 Baidu 使用百度前必读 \| 百科协议 \| 隐私政策 \| 百度百科合作平台 \| 京ICP证030173号
1771	https://stackoverflow.com/questions/35517051/split-a-list-of-numbers-into-n-chunks-such-that-the-chunks-have-close-to-equal	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Split a list of numbers into n chunks such that the chunks have (close to) equal sums and keep the original order Ask Question Asked Modified 5 years, 11 months ago Viewed 15k times This question shows research effort; it is useful and clear 26 Save this question. Show activity on this post. This is not the standard partitioning problem, as I need to maintain the order of elements in the list. So for example if I have a list ``` [1, 6, 2, 3, 4, 1, 7, 6, 4] ``` and I want two chunks, then the split should give ``` ``` for a sum of 17 on each side. For three chunks the result would be ``` ``` for sums of 12, 12, and 10. Edit for additional explanation I currently divide the sum with the number of chunks and use that as a target, then iterate till I get close to that target. The problem is that certain data sets can mess the algorithm up, for example trying to divide the following into 3:- ``` [95, 15, 75, 25, 85, 5] ``` Sum is 300, target is 100. The first chunk would sum to 95, second would be sum to 90, third would sum to 110, and 5 would be 'leftover'. Appending it where it's supposed to be would give 95, 90, 115, where a more 'reasonable' solution would be 110, 100, 90. end edit Background: I have a list containing text (song lyrics) of varying heights, and I want to divide the text into an arbitrary number of columns. Currently I calculate a target height based on the total height of all lines, but obviously this is a consistent underestimate, which in some cases results in a suboptimal solution (the last column is significantly taller). python algorithm partitioning Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Feb 20, 2016 at 14:55 Ng Oon-Ee asked Feb 19, 2016 at 23:35 Ng Oon-EeNg Oon-Ee 1,27311 gold badge1515 silver badges2828 bronze badges 4 4 Heights? What are heights? – erip Commented Feb 19, 2016 at 23:36 Also, do you want this for two sublists or arbitrary sublists? – erip Commented Feb 19, 2016 at 23:40 1 do you think the problem could be reworded as Split a list in n sublists such that the sum of values differ by a minimum? do you need the sublists or the indexes? – Pynchia Commented Feb 19, 2016 at 23:41 I think this is a very interesting problem and I might have a greedy approach that runs in O(n) for any given number of chunks. I'll report back tomorrow. – timgeb Commented Feb 20, 2016 at 0:37 Add a comment \| 8 Answers 8 Reset to default This answer is useful 15 Save this answer. Show activity on this post. This approach defines partition boundaries that divide the array in roughly equal numbers of elements, and then repeatedly searches for better partitionings until it can't find any more. It differs from most of the other posted solutions in that it looks to find an optimal solution by trying multiple different partitionings. The other solutions attempt to create a good partition in a single pass through the array, but I can't think of a single pass algorithm that's guaranteed optimal. The code here is an efficient implementation of this algorithm, but it can be hard to understand so a more readable version is included as an addendum at the end. ``` def partition_list(a, k): if k <= 1: return [a] if k >= len(a): return [[x] for x in a] partition_between = [(i+1)len(a)/k for i in range(k-1)] average_height = float(sum(a))/k best_score = None best_partitions = None count = 0 while True: starts = +partition_between ends = partition_between+[len(a)] partitions = [a[starts[i]:ends[i]] for i in range(k)] heights = map(sum, partitions) abs_height_diffs = map(lambda x: abs(average_height - x), heights) worst_partition_index = abs_height_diffs.index(max(abs_height_diffs)) worst_height_diff = average_height - heights[worst_partition_index] if best_score is None or abs(worst_height_diff) < best_score: best_score = abs(worst_height_diff) best_partitions = partitions no_improvements_count = 0 else: no_improvements_count += 1 if worst_height_diff == 0 or no_improvements_count > 5 or count > 100: return best_partitions count += 1 move = -1 if worst_height_diff < 0 else 1 bound_to_move = 0 if worst_partition_index == 0\ else k-2 if worst_partition_index == k-1\ else worst_partition_index-1 if (worst_height_diff < 0) ^ (heights[worst_partition_index-1] > heights[worst_partition_index+1])\ else worst_partition_index direction = -1 if bound_to_move < worst_partition_index else 1 partition_between[bound_to_move] += move direction def print_best_partition(a, k): print 'Partitioning {0} into {1} partitions'.format(a, k) p = partition_list(a, k) print 'The best partitioning is {0}\n With heights {1}\n'.format(p, map(sum, p)) a = [1, 6, 2, 3, 4, 1, 7, 6, 4] print_best_partition(a, 1) print_best_partition(a, 2) print_best_partition(a, 3) print_best_partition(a, 4) b = [1, 10, 10, 1] print_best_partition(b, 2) import random c = [random.randint(0,20) for x in range(100)] print_best_partition(c, 10) d = [95, 15, 75, 25, 85, 5] print_best_partition(d, 3) ``` There may be some modifications to make depending on what you are doing with this. For example, to determine whether the best partitioning has been found, this algorithm stops when there is no height difference among partitions, it doesn't find anything better than the best thing it's seen for more than 5 iterations in a row, or after 100 total iterations as a catch-all stopping point. You may need to adjust those constants or use a different scheme. If your heights form a complex landscape of values, knowing when to stop can get into classic problems of trying to escape local maxima and things like that. Output ``` Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 1 partitions The best partitioning is With heights Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 2 partitions The best partitioning is With heights [17, 17] Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 3 partitions The best partitioning is With heights [12, 12, 10] Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 4 partitions The best partitioning is With heights [7, 9, 8, 10] Partitioning [1, 10, 10, 1] into 2 partitions The best partitioning is With heights [11, 11] Partitioning [7, 17, 17, 1, 8, 8, 12, 0, 10, 20, 17, 13, 12, 4, 1, 1, 7, 11, 7, 13, 9, 12, 3, 18, 9, 6, 7, 19, 20, 17, 7, 4, 3, 16, 20, 6, 7, 12, 16, 3, 6, 12, 9, 4, 3, 2, 18, 1, 16, 14, 17, 7, 0, 14, 13, 3, 5, 3, 1, 5, 5, 13, 16, 0, 16, 7, 3, 8, 1, 20, 16, 11, 15, 3, 10, 10, 2, 0, 12, 12, 0, 18, 20, 3, 10, 9, 13, 12, 15, 6, 14, 16, 6, 12, 9, 9, 16, 14, 19, 1] into 10 partitions The best partitioning is With heights [100, 95, 94, 92, 90, 87, 100, 93, 102, 102] Partitioning [95, 15, 75, 25, 85, 5] into 3 partitions The best partitioning is With heights [110, 100, 90] ``` Edit Added the new test case, [95, 15, 75, 25, 85, 5], which this method handles correctly. Addendum This version of the algorithm is easier to read and understand, but is a bit longer due to taking less advantage of built-in Python features. It seems to execute in a comparable or even slightly faster amount of time, however. ``` partition list a into k partitions def partition_list(a, k): #check degenerate conditions if k <= 1: return [a] if k >= len(a): return [[x] for x in a] #create a list of indexes to partition between, using the index on the #left of the partition to indicate where to partition #to start, roughly partition the array into equal groups of len(a)/k (note #that the last group may be a different size) partition_between = [] for i in range(k-1): partition_between.append((i+1)len(a)/k) #the ideal size for all partitions is the total height of the list divided #by the number of paritions average_height = float(sum(a))/k best_score = None best_partitions = None count = 0 no_improvements_count = 0 #loop over possible partitionings while True: #partition the list partitions = [] index = 0 for div in partition_between: #create partitions based on partition_between partitions.append(a[index:div]) index = div #append the last partition, which runs from the last partition divider #to the end of the list partitions.append(a[index:]) #evaluate the partitioning worst_height_diff = 0 worst_partition_index = -1 for p in partitions: #compare the partition height to the ideal partition height height_diff = average_height - sum(p) #if it's the worst partition we've seen, update the variables that #track that if abs(height_diff) > abs(worst_height_diff): worst_height_diff = height_diff worst_partition_index = partitions.index(p) #if the worst partition from this run is still better than anything #we saw in previous iterations, update our best-ever variables if best_score is None or abs(worst_height_diff) < best_score: best_score = abs(worst_height_diff) best_partitions = partitions no_improvements_count = 0 else: no_improvements_count += 1 #decide if we're done: if all our partition heights are ideal, or if #we haven't seen improvement in >5 iterations, or we've tried 100 #different partitionings #the criteria to exit are important for getting a good result with #complex data, and changing them is a good way to experiment with getting #improved results if worst_height_diff == 0 or no_improvements_count > 5 or count > 100: return best_partitions count += 1 #adjust the partitioning of the worst partition to move it closer to the #ideal size. the overall goal is to take the worst partition and adjust #its size to try and make its height closer to the ideal. generally, if #the worst partition is too big, we want to shrink the worst partition #by moving one of its ends into the smaller of the two neighboring #partitions. if the worst partition is too small, we want to grow the #partition by expanding the partition towards the larger of the two #neighboring partitions if worst_partition_index == 0: #the worst partition is the first one if worst_height_diff < 0: partition_between -= 1 #partition too big, so make it smaller else: partition_between += 1 #partition too small, so make it bigger elif worst_partition_index == len(partitions)-1: #the worst partition is the last one if worst_height_diff < 0: partition_between[-1] += 1 #partition too small, so make it bigger else: partition_between[-1] -= 1 #partition too big, so make it smaller else: #the worst partition is in the middle somewhere left_bound = worst_partition_index - 1 #the divider before the partition right_bound = worst_partition_index #the divider after the partition if worst_height_diff < 0: #partition too big, so make it smaller if sum(partitions[worst_partition_index-1]) > sum(partitions[worst_partition_index+1]): #the partition on the left is bigger than the one on the right, so make the one on the right bigger partition_between[right_bound] -= 1 else: #the partition on the left is smaller than the one on the right, so make the one on the left bigger partition_between[left_bound] += 1 else: #partition too small, make it bigger if sum(partitions[worst_partition_index-1]) > sum(partitions[worst_partition_index+1]): #the partition on the left is bigger than the one on the right, so make the one on the left smaller partition_between[left_bound] -= 1 else: #the partition on the left is smaller than the one on the right, so make the one on the right smaller partition_between[right_bound] += 1 def print_best_partition(a, k): #simple function to partition a list and print info print ' Partitioning {0} into {1} partitions'.format(a, k) p = partition_list(a, k) print ' The best partitioning is {0}\n With heights {1}\n'.format(p, map(sum, p)) tests a = [1, 6, 2, 3, 4, 1, 7, 6, 4] print_best_partition(a, 1) print_best_partition(a, 2) print_best_partition(a, 3) print_best_partition(a, 4) print_best_partition(a, 5) b = [1, 10, 10, 1] print_best_partition(b, 2) import random c = [random.randint(0,20) for x in range(100)] print_best_partition(c, 10) d = [95, 15, 75, 25, 85, 5] print_best_partition(d, 3) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 22, 2016 at 23:21 answered Feb 20, 2016 at 1:53 Shawn SullivanShawn Sullivan 16666 bronze badges 8 Thank you @Shawn Sullivan, your comment on single pass possibly being impossible echoes my thoughts on looking at everyone's solutions. I've tried related single pass methods and it always seems to come up short. I'll have to digest your solution a bit first... – Ng Oon-Ee Commented Feb 20, 2016 at 15:01 Cool, let me know if you have any questions on how it works. I also made a shorter version of the algorithm by turning some of the for loops into other expressions and worked through the truth tables for the conditional at the end to make the partition adjustment able to be expressed with one line. I posted that too in case you're interested, although it's a little harder to read the code. – Shawn Sullivan Commented Feb 21, 2016 at 8:05 @NgOon-Ee I've made a few more improvements to Edit 2 which improve the code, but it's still a bit tougher to follow than the original IMO. However, as long as it's clear how this approach works, I consider the Edit 2 code my current answer. I've left the original version mostly as-is in case it's easier to understand, but if this answer is considered best, I'd make the second implementation the primary answer. – Shawn Sullivan Commented Feb 21, 2016 at 11:29 While I think I'll probably use the other answer by timgeb, this is clearly the 'correct' answer due to the unpredictable nature of the problem. I also think the second implementation should be made the primary answer, with the first implementation as an addendum for easier understanding (even that took me quite a while to look through really). – Ng Oon-Ee Commented Feb 21, 2016 at 15:31 1 @ShawnSullivan: Thanks a lot! Adaptation for Python 3: gist.github.com/laowantong/ee675108eee64640e5f94f00d8edbcb4 – Aristide Commented Feb 6, 2018 at 10:58 \| Show 3 more comments This answer is useful 5 Save this answer. Show activity on this post. Here's the best O(n) greedy algorithm I got for now. The idea is to greedily append items from the list to a chunk until the sum for the current chunk exceeds the average expected sum for a chunk at that point. The average expected sum is updated constantly. This solution is not perfect, but as I said, it is O(n) and worked not bad with my tests. I am eager to hear feedback and suggestions for improvement. I left my debug print statements in the code to provide some documentation. Feel free to comment them in to see what's going on in each step. CODE ``` def split_list(lst, chunks): #print(lst) #print() chunks_yielded = 0 total_sum = sum(lst) avg_sum = total_sum/float(chunks) chunk = [] chunksum = 0 sum_of_seen = 0 for i, item in enumerate(lst): #print('start of loop! chunk: {}, index: {}, item: {}, chunksum: {}'.format(chunk, i, item, chunksum)) if chunks - chunks_yielded == 1: #print('must yield the rest of the list! chunks_yielded: {}'.format(chunks_yielded)) yield chunk + lst[i:] raise StopIteration to_yield = chunks - chunks_yielded chunks_left = len(lst) - i if to_yield > chunks_left: #print('must yield remaining list in single item chunks! to_yield: {}, chunks_left: {}'.format(to_yield, chunks_left)) if chunk: yield chunk yield from ([x] for x in lst[i:]) raise StopIteration sum_of_seen += item if chunksum < avg_sum: #print('appending {} to chunk {}'.format(item, chunk)) chunk.append(item) chunksum += item else: #print('yielding chunk {}'.format(chunk)) yield chunk # update average expected sum, because the last yielded chunk was probably not perfect: avg_sum = (total_sum - sum_of_seen)/(to_yield - 1) chunks_yielded += 1 chunksum = item chunk = [item] ``` TEST CODE ``` import random lst = [1, 6, 2, 3, 4, 1, 7, 6, 4] lst = [random.choice(range(1,101)) for _ in range(100)] chunks = 3 print('list: {}, avg sum: {}, chunks: {}\n'.format(lst, sum(lst)/float(chunks), chunks)) for chunk in split_list(lst, chunks): print('chunk: {}, sum: {}'.format(chunk, sum(chunk))) ``` TESTS with your list: ``` list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 17.0, chunks: 2 chunk: [1, 6, 2, 3, 4, 1], sum: 17 chunk: [7, 6, 4], sum: 17 list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 11.33, chunks: 3 chunk: [1, 6, 2, 3], sum: 12 chunk: [4, 1, 7], sum: 12 chunk: [6, 4], sum: 10 list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 8.5, chunks: 4 chunk: [1, 6, 2], sum: 9 chunk: [3, 4, 1], sum: 8 chunk: , sum: 7 chunk: [6, 4], sum: 10 list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 6.8, chunks: 5 chunk: [1, 6], sum: 7 chunk: [2, 3, 4], sum: 9 chunk: [1, 7], sum: 8 chunk: , sum: 6 chunk: , sum: 4 ``` TESTS with random lists of length 100 and elements from 1 to 100 (printing of the random list omitted): ``` avg sum: 2776.0, chunks: 2 chunk: [25, 8, 71, 39, 5, 69, 29, 64, 31, 2, 90, 73, 72, 58, 52, 19, 64, 34, 16, 8, 16, 89, 70, 67, 63, 36, 9, 87, 38, 33, 22, 73, 66, 93, 46, 48, 65, 55, 81, 92, 69, 94, 43, 68, 98, 70, 28, 99, 92, 69, 24, 74], sum: 2806 chunk: [55, 55, 64, 93, 97, 53, 85, 100, 66, 61, 5, 98, 43, 74, 99, 56, 96, 74, 63, 6, 89, 82, 8, 25, 36, 68, 89, 84, 10, 46, 95, 41, 54, 39, 21, 24, 8, 82, 72, 51, 31, 48, 33, 77, 17, 69, 50, 54], sum: 2746 avg sum: 1047.6, chunks: 5 chunk: [19, 76, 96, 78, 12, 33, 94, 10, 38, 87, 44, 76, 28, 18, 26, 29, 44, 98, 44, 32, 80], sum: 1062 chunk: [48, 70, 42, 85, 87, 55, 44, 11, 50, 48, 47, 50, 1, 17, 93, 78, 25, 10, 89, 57, 85], sum: 1092 chunk: [30, 83, 99, 62, 48, 66, 65, 98, 94, 54, 14, 97, 58, 53, 3, 98], sum: 1022 chunk: [80, 34, 63, 20, 27, 36, 98, 97, 7, 6, 9, 65, 91, 93, 2, 27, 83, 35, 65, 17, 26, 41], sum: 1022 chunk: [80, 80, 42, 32, 44, 42, 94, 31, 50, 23, 34, 84, 47, 10, 54, 59, 72, 80, 6, 76], sum: 1040 avg sum: 474.6, chunks: 10 chunk: [4, 41, 47, 41, 32, 51, 81, 5, 3, 37, 40, 26, 10, 70], sum: 488 chunk: [54, 8, 91, 42, 35, 80, 13, 84, 14, 23, 59], sum: 503 chunk: [39, 4, 38, 40, 88, 69, 10, 19, 28, 97, 81], sum: 513 chunk: [19, 55, 21, 63, 99, 93, 39, 47, 29], sum: 465 chunk: [65, 88, 12, 94, 7, 47, 14, 55, 28, 9, 98], sum: 517 chunk: [19, 1, 98, 84, 92, 99, 11, 53], sum: 457 chunk: [85, 79, 69, 78, 44, 6, 19, 53], sum: 433 chunk: [59, 20, 64, 55, 2, 65, 44, 90, 37, 26], sum: 462 chunk: [78, 66, 32, 76, 59, 47, 82], sum: 440 chunk: [34, 56, 66, 27, 1, 100, 16, 5, 97, 33, 33], sum: 468 avg sum: 182.48, chunks: 25 chunk: [55, 6, 16, 42, 85], sum: 204 chunk: [30, 68, 3, 94], sum: 195 chunk: [68, 96, 23], sum: 187 chunk: [69, 19, 12, 97], sum: 197 chunk: [59, 88, 49], sum: 196 chunk: [1, 16, 13, 12, 61, 77], sum: 180 chunk: [49, 75, 44, 43], sum: 211 chunk: [34, 86, 9, 55], sum: 184 chunk: [25, 82, 12, 93], sum: 212 chunk: [32, 74, 53, 31], sum: 190 chunk: [13, 15, 26, 31, 35, 3, 14, 71], sum: 208 chunk: [81, 92], sum: 173 chunk: [94, 21, 34, 71], sum: 220 chunk: [1, 55, 70, 3, 92], sum: 221 chunk: [38, 59, 56, 57], sum: 210 chunk: [7, 20, 10, 81, 100], sum: 218 chunk: [5, 71, 19, 8, 82], sum: 185 chunk: [95, 14, 72], sum: 181 chunk: [2, 8, 4, 47, 75, 17], sum: 153 chunk: [56, 69, 42], sum: 167 chunk: [75, 45], sum: 120 chunk: [68, 60], sum: 128 chunk: [29, 25, 62, 3, 50], sum: 169 chunk: [54, 63], sum: 117 chunk: [57, 37, 42], sum: 136 ``` As you can see, as expected it gets worse the more chunks you want to generate. I hope I was able to help a bit. edit: The yield from syntax requires Python 3.3 or newer, if you are using an older version just turn the statement into a normal for loop. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Feb 20, 2016 at 2:51 timgebtimgeb 79k2020 gold badges129129 silver badges150150 bronze badges 4 Thanks for this, but the edge cases I was talking about (consistent underestimation) still faces an issue with this method. Added an example data set which causes the problem, with this method it actually yields [95, 15], , and [25, 85, 5], which is not a bad guess but still not as good as [95, 15], [75, 25], and [85, 5] – Ng Oon-Ee Commented Feb 20, 2016 at 14:59 @NgOon-Ee yeah, my solution is more tailored towards giving good guesses, not perfect ones. I'm not sure how much better it can get while staying greedy and within O(n). I'll have to think about this some more. One idea I'm having is to use my solution to get the chunks and then make another pass over the chunks to optimize them, switching out first/last elements. Maybe you can try to attack this if you need it fast. In theory, you should get very good guesses with a few extra passes over the chunks. – timgeb Commented Feb 20, 2016 at 19:49 @NgOon-Ee please try this in the else clause: chunksum = chunksum - avg_sum + item instead of chunksum = item. Comment out/delete the line where avg_sum is updated. This seems to give better results for some cases, for example [95, 15], [75, 25] and [85, 5] for a three-split of [95, 15, 75, 25, 85, 5]. – timgeb Commented Feb 21, 2016 at 10:57 Thanks, this solution is probably the most user-friendly. Unfortunately the more I look at it the more I realize it just postpones the inevitable mistake, mostly because the problem itself is ill-defined for such single pass methods as Shawn Sullivan stated. I'll upvote this, but based on that technicality I think his answer is more correct. – Ng Oon-Ee Commented Feb 21, 2016 at 15:29 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. Simple and concise way using numpy. Assuming ``` import numpy.random as nr import numpy as np a = (nr.random(10000000)1000).astype(int) ``` Then, assuming you need to divide the list into p parts with approximately equal sums ``` def equisum_partition(arr,p): ac = arr.cumsum() #sum of the entire array partsum = ac[-1]//p #generates the cumulative sums of each part cumpartsums = np.array(range(1,p))partsum #finds the indices where the cumulative sums are sandwiched inds = np.searchsorted(ac,cumpartsums) #split into approximately equal-sum arrays parts = np.split(arr,inds) return parts ``` Importantly, this is vectorised: ``` In : %timeit parts = equisum_partition(a,20) 53.5 ms ± 962 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) ``` You could checking the quality of the splitting, ``` partsums = np.array([part.sum() for part in parts]).std() ``` The splits are not great, but I suspect they are optimal given that the ordering is not changed. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Jan 3, 2019 at 14:33 answered Jan 3, 2019 at 14:28 Milind RMilind R 70111 gold badge88 silver badges2222 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. This is a minorly edited version of @Milind R's numpy-approach (BTW a big thanks, sir). Namely, I realized that in a real-life-scenario, the partitions suggested by the script may end up being sub-optimal, if the elements are not "uniformly" spread in the array in terms of their values. To counter this I "uniformified" the array by rearranging the elements of the as 'smallest', 'largest', 'second smallest', 'second largest', etc. The down part is that this makes the script considerably (~5x) slower. ``` import numpy.random as nr import numpy as np a = (nr.random(10000000)1000).astype(int) ``` The edited partitioning algorithm: ``` def equisum_partition(arr,p, uniformify=True): #uniformify: rearrange to ['smallest', 'largest', 'second smallest', 'second largest', etc..] if uniformify: l = len(arr) odd = l%2!=0 arr = np.sort(arr) #add a dummy element if odd length if odd: arr = np.append(np.min(arr)-1, arr) l = l+1 idx = np.arange(l) idx = np.multiply(idx, np.subtract(1, np.multiply( np.mod(idx, 2), 2)) ) arr = arr[idx] #remove the dummy element if odd: arr = arr[1:] #cumulative summation ac = arr.cumsum() #sum of the entire array partsum = ac[-1]//p #generates the cumulative sums of each part cumpartsums = np.array(range(1,p))partsum #finds the indices where the cumulative sums are sandwiched inds = np.searchsorted(ac,cumpartsums) #split into approximately equal-sum arrays parts = np.split(arr,inds) return parts ``` In the original answer's example this doesn't play too much of a role since due the randomness of the example array. With uniformify: ``` %%time parts = equisum_partition(a,20) partsums = np.array([part.sum() for part in parts])# partsums.std() Wall time: 624 ms 266.6111212984185 ``` Without uniformify: ``` %%time parts = equisum_partition(a,20, uniformify=False) partsums = np.array([part.sum() for part in parts])# partsums.std() Wall time: 105 ms 331.19071544957296 ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Sep 30, 2019 at 12:46 answered Sep 30, 2019 at 9:51 mjkvaakmjkvaak 54966 silver badges77 bronze badges 0 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I think a good approach would be to sort the input list. Then add the smallest and largest to one list. The second smallest and second largest to the next list and so on, until all elements are added to the list. ``` def divide_list(A): A.sort() l = 0 r = len(A) - 1 l1,l2= [],[] i = 0 while l < r: ends = [A[l], A[r]] if i %2 ==0: l1.extend(ends) else: l2.extend(ends) i +=1 l +=1 r -=1 if r == l: smaller = l1 if sum(l1) < sum(l2) else l2 smaller.append(A[r]) return l1, l2 myList = [1, 6, 2, 3, 4, 1, 7, 6, 4] print divide_list(myList) myList = [1,10,10,1] print divide_list(myList) ``` Output ``` ([1, 7, 2, 6], [1, 6, 3, 4, 4]) ([1, 10], [1, 10]) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Feb 20, 2016 at 0:23 Garrett RGarrett R 2,6621313 silver badges1515 bronze badges 3 1 given the numbers represent words/song lyrics I think the original order of the elements matter – Pynchia Commented Feb 20, 2016 at 0:40 what about division into three? – danidee Commented Feb 20, 2016 at 0:48 4 Op says order matters – erip Commented Feb 20, 2016 at 1:02 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. This is coming kind of late but i came up with a function that does what you need it takes a second parameter that tells it how it should split the list ``` import math my_list = [1, 6, 2, 3, 4, 1, 7, 6, 4] def partition(my_list, split): solution = [] total = sum(my_list) div = total / split div = math.ceil(div) criteria = [div] (total // div) criteria.append(total - sum(criteria)) if sum(criteria) != total else criteria temp = [] pivot = 0 for crit in criteria: for count in range(len(my_list) + 1): if sum(my_list[pivot:count]) == crit: solution.append(my_list[pivot:count]) pivot = count break return solution print(partition(my_list, 2)) # Outputs print(partition(my_list, 3)) # Outputs ``` it would fail for 4 divisions, because you obviously stated in your question that you want to maintain the order ``` 4 divisions = [9, 9, 9, 7] ``` and your sequence can't match that Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Feb 20, 2016 at 1:39 danideedanidee 9,63422 gold badges3939 silver badges5858 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Here is some code that returns 2-ples of slice indexes for each sublist. ``` weights = [1, 6, 2, 3, 4, 1, 7, 6, 4] def balance_partitions(weights:list, n:int=2) -> tuple: if n < 1: raise ValueError("Parameter 'n' must be 2+") target = sum(weights) // n results = [] cost = 0 start = 0 for i, w in enumerate(weights): delta = target - cost cost += w if cost >= target: if i == 0 or cost - target <= delta: results.append( (start, i+1) ) start = i+1 elif cost - target > delta: # Better if we didn't include this one. results.append( (start, i) ) start = i cost -= target if len(results) == n-1: results.append( (start, len(weights)) ) break return tuple(results) def print_parts(w, n): result = balance_partitions(w, n) print("Suggested partition indices: ", result) for t in result: start,end = t sublist = w[start:end] print(" - ", sublist, "(sum: {})".format(sum(sublist))) print(weights, '=', sum(weights)) for i in range(2, len(weights)+1): print_parts(weights, i) ``` Output is: ``` [1, 6, 2, 3, 4, 1, 7, 6, 4] = 34 Suggested partition indices: ((0, 6), (6, 9)) - [1, 6, 2, 3, 4, 1] (sum: 17) - [7, 6, 4] (sum: 17) Suggested partition indices: ((0, 4), (4, 7), (7, 9)) - [1, 6, 2, 3] (sum: 12) - [4, 1, 7] (sum: 12) - [6, 4] (sum: 10) Suggested partition indices: ((0, 3), (3, 5), (5, 7), (7, 9)) - [1, 6, 2] (sum: 9) - [3, 4] (sum: 7) - [1, 7] (sum: 8) - [6, 4] (sum: 10) Suggested partition indices: ((0, 2), (2, 4), (4, 6), (6, 7), (7, 9)) - [1, 6] (sum: 7) - [2, 3] (sum: 5) - [4, 1] (sum: 5) - (sum: 7) - [6, 4] (sum: 10) Suggested partition indices: ((0, 2), (2, 3), (3, 5), (5, 6), (6, 7), (7, 9)) - [1, 6] (sum: 7) - (sum: 2) - [3, 4] (sum: 7) - (sum: 1) - (sum: 7) - [6, 4] (sum: 10) Suggested partition indices: ((0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 9)) - [1, 6] (sum: 7) - (sum: 2) - (sum: 3) - (sum: 4) - (sum: 1) - (sum: 7) - [6, 4] (sum: 10) Suggested partition indices: ((0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)) - [1, 6] (sum: 7) - (sum: 2) - (sum: 3) - (sum: 4) - (sum: 1) - (sum: 7) - (sum: 6) - (sum: 4) Suggested partition indices: ((0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9)) - (sum: 1) - (sum: 6) - (sum: 2) - (sum: 3) - (sum: 4) - (sum: 1) - (sum: 7) - (sum: 6) - (sum: 4) ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Feb 20, 2016 at 2:19 aghastaghast 15.4k44 gold badges3030 silver badges5858 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Here's how I might attack this problem for the case of two desired sublists. It's probably not as efficient as it could be, but it's a first cut. ``` def divide(l): total = sum(l) half = total / 2 l1 = [] l2 = [] for e in l: if half - e >= 0 or half > abs(half - e): l1.append(e) half -= e else: l2.append(e) return (l1, l2) ``` You can see it in action here: ``` (l1, l2) = divide([1, 6, 2, 3, 4, 1, 7, 6, 4]) print(l1) [1, 6, 2, 3, 4, 1] print(l2) [7, 6, 4] (l1, l2) = divide([1,1,10,10]) print(l1) [1, 1, 10] print(l2) ``` I'll leave other cases to you as an exercise. :) Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 20, 2016 at 0:52 answered Feb 19, 2016 at 23:50 eriperip 17.1k1111 gold badges7272 silver badges131131 bronze badges 4 Please explain the downvote. Can't learn anything if there's no feedback. – erip Commented Feb 19, 2016 at 23:56 1 I didn't downvote you, but I"m trying to understand how this works. It looks like you greedily add to l1 until you get to more than half of the total. Then you add to l2. What if you had a list like [1,1,10,10]. Wouldn't this produce [1,1] [10,10] ? – Garrett R Commented Feb 19, 2016 at 23:59 Whoops, indeed! Need to check if next element will cause less of a difference of half. Will update soon – erip Commented Feb 20, 2016 at 0:28 Thanks, I'm using something very similar to this right now (almost identical except for naming of variables and I handle more than two sublists), but the problem is when the data tends to provide smaller than expected lists (I've added an example for that) then it tends to overshoot. – Ng Oon-Ee Commented Feb 20, 2016 at 14:44 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python algorithm partitioning See similar questions with these tags. 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1772	https://archive.nptel.ac.in/content/storage2/courses/104103023/module1/lec3/5.html	=============== Module 1 : Oxidation Reactions Lecture 3 : Chromium Oxidants 1 2 3 4 5 6 7 CrO 3 in acetic anhydride or acetic acid has been found to transform alkenes to epoxides (Scheme 9). Scheme 9 Scheme 10 1.3.8 Oxidation of Aromatic Side Chains The selective oxidation of alkyl chains attached to an aromatic ring can be carried out with CrO 3 in acetic anhydride. Thus, p-nitrotoluene is converted into p-nitrobenzylidine diacetate, which on hydrolysis in the presence of acid gives p-nitrobenzaldehyde (Scheme 11). Scheme 11 Chromyl chloride can oxidize o-, m-, or p- xylenes to give tolualdehyde in 70-80% yield (Scheme 12). Scheme 12 Several studies focus on the oxidation of aromatic alkyl side chains to give carboxylic acids (Scheme 13). Scheme 13 1 2 3 4 5 6 7
1773	https://www.ck12.org/flexi/cbse-math/length-conversion/convert-1-foot-into-meters./	Flexi answers - Convert 1 foot into meters. \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Length Conversion Question Convert 1 foot into meters. Flexi Says: There are 0.3048 meters in 1 foot. A foot (ft) is part of the US customary system of measurement, whereas a meter (m) is the base unit of length defined by the International System of Units (SI). It is defined as the length of the path traveled by light in a vacuum during a time interval of 1/299,792,458 of a second. Convert feet (ft) to meters (m) and vice versa! Click here to Explore an Easy-To-Use Interactive Tool. To explore more, click here! Analogy / Example Try Asking: What is the height of 1.8 meters in feet?What is the height of 155 cm in feet?How many feet is equal to 1 meter? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
1774	https://chemistry.stackexchange.com/questions/142097/a-problem-regarding-solubility-and-concentration	equilibrium - A problem regarding solubility and concentration - Chemistry Stack Exchange Join Chemistry By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Chemistry helpchat Chemistry Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more A problem regarding solubility and concentration Ask Question Asked 4 years, 11 months ago Modified4 years, 11 months ago Viewed 126 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. A solution is 0.10 M 0.10 M B a(N O X 3)X 2 B a(N O X 3)X 2 and 0.10 M 0.10 M S r(N O X 3)X 2 S r(N O X 3)X 2. If solid N a X 2 C r O X 4 N a X 2 C r O X 4 is added to the solution, what is the [B a X+][B a X+] when S r C r O X 4 S r C r O X 4 begins to precipitate? K s p=1.2×10−10 for B a C r O X 4 K s p=1.2×10−10 for B a C r O X 4 K s p=3.5×10−5 for S r C r O X 4 K s p=3.5×10−5 for S r C r O X 4 The solution given in my textbook is: K s p 1=[S r X 2+][C r O X 4 X 2−]K s p 1=[S r X 2+][C r O X 4 X 2−] [C r O X 4 X 2−]=3.5×10−5 0.1=3.5×10−4 M(1)(1)[C r O X 4 X 2−]=3.5×10−5 0.1=3.5×10−4 M K s p 2=[B a X 2+][C r O X 4 X 2−]K s p 2=[B a X 2+][C r O X 4 X 2−] [C r O X 4 X 2−]X t o t a l≈[C r O X 4 X 2−]from S r C r O X 4[C r O X 4 X 2−]X t o t a l≈[C r O X 4 X 2−]from S r C r O X 4 [B a X 2+]=1.2×10−10 3.5×10−4=3.4×10−7 M[B a X 2+]=1.2×10−10 3.5×10−4=3.4×10−7 M First of all, in equation (1)(1) why did they put [S r X 2+][S r X 2+] same as the initial concentration (before the addition of N a X 2 C r O X 4 N a X 2 C r O X 4)? When N a X 2 C r O X 4 N a X 2 C r O X 4 is added, [S r X 2+][S r X 2+] will react with C r O X 4 X 2−C r O X 4 X 2− (obtained from N a X 2 C r O X 4 N a X 2 C r O X 4) to form precipitate. The concentration of [S r X 2+][S r X 2+] will undoubtedly change until the equilibrium is established. This given solution is totally confusing, I would appreciate if you could tell me the proper way of solving this problem. I tried this on my own, but after calculating the concentration of C r O X 4 X 2−C r O X 4 X 2−, I couldn't make sense out of this question. equilibrium solubility precipitation salt Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 28, 2020 at 14:54 Mithoron 4,706 14 14 gold badges 41 41 silver badges 63 63 bronze badges asked Oct 28, 2020 at 12:53 Eyy bossEyy boss 570 7 7 silver badges 15 15 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. It is supposed addition of solid N a X 2 C r O X 4 N a X 2 C r O X 4 would cause negligible volume change, so we can afford to consider the same [S r X 2+][S r X 2+] as the initial one. The task is rather theoretical, supposing the time point when [S r X 2+][C r O X 4 X 2−][S r X 2+][C r O X 4 X 2−] just reached K s p,S r C r O X 4 K s p,S r C r O X 4, while K s p,S r C r O X 4 K s p,S r C r O X 4 and [S r X 2+][S r X 2+] are given. The proper computation procedure is already presented in you solved task. What else is unclear ? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 28, 2020 at 14:19 PoutnikPoutnik 48k 4 4 gold badges 59 59 silver badges 118 118 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I am going to assume that you already understand the theory behind this equilibrium problem. I wanted to first say that you are absolutely correct in that the [Sr2+] will "undoubtedly change until the equilibrium is established". The short answer to your question is that the solution in the textbook used an assumption that we often use in equilibrium calculations for simplification. Different textbooks have different rules, but the general idea is that, for an equilibrium constant K, such that K = [x][I + x], we can approximate K ~ [x][I] given that K << [I]. I have often seen that the condition for K << [I] holds for when [I] is at least 10^2 times greater than K. So, basically, your textbook just simplified the work. Instead of writing [Sr2+] = 0.10 + x, they wrote the initial conc (0.10) because Ksp = 3.510^-5 << 0.10 to avoid solving an annoying quadratic. Hope that helps. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 28, 2020 at 14:15 ryh8ryh8 1 1 1 bronze badge Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Before we begin, there are two implicit assumptions: That there is not a significant change in the volume of the solution by either adding N a X 2 C r O X 4 N a X 2 C r O X 4 nor the precipitation of the B a C r O X 4 B a C r O X 4 That concentrations can be used instead of activities. First of all, in equation (1)(1) why did they put [S r X 2+][S r X 2+] same as the initial concentration (before the addition of N a X 2 C r O X 4 N a X 2 C r O X 4)? The problem statement asks for the [B a X 2+][B a X 2+] when S r C r O X 4 S r C r O X 4 "begins to precipitate." Given that "some" S r C r O X 4 S r C r O X 4 must precipitate to reach the point where S r C r O X 4 S r C r O X 4 has already begun to precipitate, let's define some variables: [S r X 2+]X i n i t[S r X 2+]X i n i t -- The initial concentration of S r X 2+S r X 2+, 0.10 molar. [S r X 2+]X p p t[S r X 2+]X p p t -- the concentration of S r X 2+S r X 2+ removed by ppt. [S r X 2+]X f i n a l=[S r X 2+]X i n i t−[S r X 2+]X p p t[S r X 2+]X f i n a l=[S r X 2+]X i n i t−[S r X 2+]X p p t -- the concentration of S r X 2+S r X 2+ after some S r C r O X 4 S r C r O X 4 has precipitated. Now here comes the rub. Chemistry isn't like math. The value of π π has been calculated to millions of digits. [S r X 2+]X i n i t[S r X 2+]X i n i t was given as 0.10 molar. So there are only two significant figures in this value. If 0.01% of the S r X 2+S r X 2+ is used to form the initial precipitate then [S r X 2+]X f i n a l≈([S r X 2+]X i n i t=0.10 M)[S r X 2+]X f i n a l≈([S r X 2+]X i n i t=0.10 M) Note that the problem could have been written differently. The problem could have asked for [B a X 2+][B a X 2+] when S r C r O X 4 S r C r O X 4 reaches its K s p K s p (or some variation thereof). This would be right at the equilibrium point, before any S r C r O X 4 S r C r O X 4 has formed. When N a X 2 C r O X 4 N a X 2 C r O X 4 is added, [S r X 2+][S r X 2+] will react with C r O X 4 X 2−C r O X 4 X 2− (obtained from N a X 2 C r O X 4 N a X 2 C r O X 4) to form precipitate. The concentration of [S r X 2+][S r X 2+] will undoubtedly change until the equilibrium is established. Obviously when S r C r O X 4 S r C r O X 4 nucleates there is some finite size of the S r C r O X 4 S r C r O X 4 particle which will be stable in an aqueous solution. So if we use a nanoliter of the cation solution then the given answer will be wrong. Thus there is a implied assumption that a macro volume of the solution is being used. "Macro" in this case meaning that creating a detectable amount of the S r C r O X 4 S r C r O X 4 does not change the [S r X 2+][S r X 2+] from its initial nominal value of 0.10 molar. I would appreciate if you could tell me the proper way of solving this problem. The book gives the proper answer albeit without a full explanation of the additional assumptions. The notion is that B a C r O X 4 B a C r O X 4 is more insoluble than S r C r O X 4 S r C r O X 4, so most of the B a X 2+B a X 2+ will be removed before any S r X 2+S r X 2+ starts to precipitate. I'll add that this assumption isn't really true. As the B a C r O X 4 B a C r O X 4 precipitates, it will incorporate some of the S r X 2+S r X 2+ into the B a C r O X 4 B a C r O X 4 precipitate. Again, not an appreciable amount compare to what is left in solution since initially [S r X 2+]=[B a X 2+][S r X 2+]=[B a X 2+]. However if initially [S r X 2+]=0.01⋅[B a X 2+][S r X 2+]=0.01⋅[B a X 2+] then there would be a significant loss of S r X 2+S r X 2+ in the solution. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 28, 2020 at 19:31 answered Oct 28, 2020 at 18:59 MaxWMaxW 22.5k 2 2 gold badges 38 38 silver badges 81 81 bronze badges Add a comment\| Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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1775	https://pubmed.ncbi.nlm.nih.gov/10690563/	Protein metabolism in the extremely low-birth weight infant - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Protein metabolism in the extremely low-birth weight infant S C Kalhan1,S Iben Affiliations Expand Affiliation 1 Robert Schwartz, MD, Center for Metabolism & Nutrition, Case Western Reserve University School of Medicine, Cleveland, Ohio, USA. sck@po.cwru.edu PMID: 10690563 DOI: 10.1016/s0095-5108(05)70005-1 Item in Clipboard Review Protein metabolism in the extremely low-birth weight infant S C Kalhan et al. Clin Perinatol.2000 Mar. Show details Display options Display options Format Clin Perinatol Actions Search in PubMed Search in NLM Catalog Add to Search . 2000 Mar;27(1):23-56. doi: 10.1016/s0095-5108(05)70005-1. Authors S C Kalhan1,S Iben Affiliation 1 Robert Schwartz, MD, Center for Metabolism & Nutrition, Case Western Reserve University School of Medicine, Cleveland, Ohio, USA. sck@po.cwru.edu PMID: 10690563 DOI: 10.1016/s0095-5108(05)70005-1 Item in Clipboard Full text links Cite Display options Display options Format Abstract Although extensive data are available on the impact of nutrient and protein administration on growth, plasma amino acids, and nitrogen balance in the newborn and growing infants, relatively few studies have carefully examined the dynamic aspects of protein metabolism in vivo and particularly in the micropremie or ELBW infant. These studies show that the very preterm infants, either because of immaturity or because of the intercurrent illness, have high rates of protein turnover and protein breakdown. This high rate of proteolysis is not as responsive to nutrient administration. Intervention strategies aimed at promoting nitrogen accretion, such as insulin, human growth hormone, or glutamine, have not thus far resulted in enhanced protein accretion and growth. This may be, in part, due to limitations in delivery of adequate calorie and nitrogen. PubMed Disclaimer Similar articles Vitamin metabolism and requirements in the micropremie.Greer FR.Greer FR.Clin Perinatol. 2000 Mar;27(1):95-118, vi. doi: 10.1016/s0095-5108(05)70008-7.Clin Perinatol. 2000.PMID: 10690566 Review. Effect of enteral glutamine or glycine on whole-body nitrogen kinetics in very-low-birth-weight infants.Parimi PS, Devapatla S, Gruca LL, Amini SB, Hanson RW, Kalhan SC.Parimi PS, et al.Am J Clin Nutr. 2004 Mar;79(3):402-9. doi: 10.1093/ajcn/79.3.402.Am J Clin Nutr. 2004.PMID: 14985214 Clinical Trial. Bone mineral metabolism in the micropremie.Rigo J, De Curtis M, Pieltain C, Picaud JC, Salle BL, Senterre J.Rigo J, et al.Clin Perinatol. 2000 Mar;27(1):147-70. doi: 10.1016/s0095-5108(05)70011-7.Clin Perinatol. 2000.PMID: 10690569 Review. Protein/amino acid metabolism and nutrition in very low birth weight infants.Kalhan S, Bier D, Yaffe S, Catz C, Grave G.Kalhan S, et al.J Perinatol. 2001 Jul-Aug;21(5):320-3. doi: 10.1038/sj.jp.7210550.J Perinatol. 2001.PMID: 11536026 Review. [Strategies for nutrition of the preterm infant with low and very low birth weight].Sluncheva B.Sluncheva B.Akush Ginekol (Sofiia). 2010;49(2):33-9.Akush Ginekol (Sofiia). 2010.PMID: 20734675 Bulgarian. See all similar articles Cited by Influence of the calcium concentration in the presence of organic phosphorus on the physicochemical compatibility and stability of all-in-one admixtures for neonatal use.Ribeiro Dde O, Lobo BW, Volpato NM, da Veiga VF, Cabral LM, de Sousa VP.Ribeiro Dde O, et al.Nutr J. 2009 Oct 26;8:51. doi: 10.1186/1475-2891-8-51.Nutr J. 2009.PMID: 19857269 Free PMC article. Higher versus lower protein intake in formula-fed low birth weight infants.Fenton TR, Premji SS, Al-Wassia H, Sauve RS.Fenton TR, et al.Cochrane Database Syst Rev. 2014 Apr 21;2014(4):CD003959. doi: 10.1002/14651858.CD003959.pub3.Cochrane Database Syst Rev. 2014.Update in: Cochrane Database Syst Rev. 2020 Jun 23;6:CD003959. doi: 10.1002/14651858.CD003959.pub4.PMID: 24752987 Free PMC article.Updated. Higher versus lower protein intake in formula-fed low birth weight infants.Fenton TR, Al-Wassia H, Premji SS, Sauve RS.Fenton TR, et al.Cochrane Database Syst Rev. 2020 Jun 23;6(6):CD003959. doi: 10.1002/14651858.CD003959.pub4.Cochrane Database Syst Rev. 2020.PMID: 32573771 Free PMC article. High Amino Acid Intake in Early Life Is Associated With Systolic but Not Diastolic Arterial Hypertension at 5 Years of Age in Children Born Very Preterm.Rozé JC, Bacchetta J, Lapillonne A, Boudred F, Picaud JC, Marchand-Martin L, Bruel-Tessoulin A, Harambat J, Biran V, Nuyt AM, Darmaun D, Ancel PY.Rozé JC, et al.J Am Heart Assoc. 2024 Jan 2;13(1):e032804. doi: 10.1161/JAHA.123.032804. Epub 2023 Dec 29.J Am Heart Assoc. 2024.PMID: 38156453 Free PMC article. Effect of intravenous amino acids on glutamine and protein kinetics in low-birth-weight preterm infants during the immediate neonatal period.Kadrofske MM, Parimi PS, Gruca LL, Kalhan SC.Kadrofske MM, et al.Am J Physiol Endocrinol Metab. 2006 Apr;290(4):E622-30. doi: 10.1152/ajpendo.00274.2005. Epub 2005 Nov 1.Am J Physiol Endocrinol Metab. 2006.PMID: 16263773 Free PMC article.Clinical Trial. Publication types Research Support, U.S. Gov't, P.H.S. 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1776	https://onlinelibrary.wiley.com/doi/pdf/10.1002/9781118635360.app1	Error - Cookies Turned Off Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Skip to Main Content Search term Advanced SearchCitation Search Login / Register Individual login Institutional login REGISTER Journal Articles Actions Cookies disabled Cookies are disabled for this browser. Wiley Online Library requires cookies for authentication and use of other site features; therefore, cookies must be enabled to browse the site. Detailed information on how Wiley uses cookies can be found in our Privacy Policy. Additional links About Wiley Online Library Privacy Policy Terms of Use About Cookies Manage Cookies Accessibility Wiley Research DE&I Statement and Publishing Policies Developing World Access Help & Support Contact Us Training and Support DMCA & Reporting Piracy Sitemap Opportunities Subscription Agents Advertisers & Corporate Partners Connect with Wiley The Wiley Network Wiley Press Room Copyright © 1999-2025 John Wiley & Sons, Inc or related companies. All rights reserved, including rights for text and data mining and training of artificial intelligence technologies or similar technologies. Log in to Wiley Online Library Email or Customer ID Password Forgot password? NEW USER >INSTITUTIONAL LOGIN > Change Password Old Password New Password Too Short Weak Medium Strong Very Strong Too Long Your password must have 10 characters or more: a lower case character, an upper case character, a special character or a digit Too Short Password Changed Successfully Your password has been changed Create a new account Email Returning user Forgot your password? Enter your email address below. Email Please check your email for instructions on resetting your password. If you do not receive an email within 10 minutes, your email address may not be registered, and you may need to create a new Wiley Online Library account. Request Username Can't sign in? Forgot your username? Enter your email address below and we will send you your username Email Close If the address matches an existing account you will receive an email with instructions to retrieve your username
1777	https://www.anthem.com/medpolicies/abc/active/gl_pw_c185825.html	CG-MED-53 Cervical Cancer Screening Using Cytology and Human Papillomavirus Testing Clinical UM Guideline Subject: Cervical Cancer Screening Using Cytology and Human Papillomavirus Testing Guideline #: CG-MED-53Publish Date: 01/30/2025 Status: ReviewedLast Review Date: 11/14/2024 Description This document addresses cervical cancer screening and testing for human papillomavirus (HPV) to assess cervical cancer risk. Cervical cancer screening is comprised of cervical cytology with Papanicolaou testing (also known as a ‘Pap test’) and testing for HPV DNA. Pap tests are used to identify pre-cancerous or cancerous tissues present on the cervix. Screening for HPV aids in identifying individuals at higher risk for developing cervical cancer. Note: This document addresses the use of cervical cancer screening in the general population. It does not address the use of cervical cancer screening technologies or procedures for the work-up or surveillance of either individuals with known precancerous lesions or a known history of cervical cancer. Note: For additional information on cervical cancer screening, please see: ADMIN.00002 Preventive Health Guidelines MED.00087 Optical Detection for Screening and Identification of Cervical Cancer Clinical Indications MedicallyNecessary: Cervical cancer screening with cytology is considered medically necessary for individuals who are 21 years of age or older. Screening for the presence of HPV is considered medically necessary for individuals who are 30 years of age or older. Not Medically Necessary: Cervical cancer screening with cytology is considered not medically necessary when the criteria above have not been met. Screening for the presence of HPV is considered not medically necessarywhen the criteria above have not been met. The term “individual” in this document refers to any person with an intact cervix, regardless of gender identity. Coding The following codes for treatments and procedures applicable to this guideline are included below for informational purposes. Inclusion or exclusion of a procedure, diagnosis or device code(s) does not constitute or imply member coverage or provider reimbursement policy. Please refer to the member's contract benefits in effect at the time of service to determine coverage or non-coverage of these services as it applies to an individual member. When services may be Medically Necessary when criteria are met: CPT 87623 Infectious agent detection by nucleic acid (DNA or RNA); Human Papillomavirus (HPV), low-risk types (eg, 6, 11, 42, 43, 44) 87624 Infectious agent detection by nucleic acid (DNA or RNA); Human Papillomavirus (HPV), high-risk types (eg, 16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, 68), pooled result 87625 Infectious agent detection by nucleic acid (DNA or RNA); Human Papillomavirus (HPV), types 16 and 18 only, includes type 45, if performed 87626 Infectious agent detection by nucleic acid (DNA or RNA); Human Papillomavirus (HPV), separately reported high-risk types (eg, 16, 18, 31, 45, 51, 52) and high-risk pooled result(s) 88141 Cytopathology, cervical or vaginal (any reporting system), requiring interpretation by physician 88142 Cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation; manual screening under physician supervision 88143 Cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation; with manual screening and rescreening under physician supervision 88147 Cytopathology smears, cervical or vaginal; screening by automated system under physician supervision 88148 Cytopathology smears, cervical or vaginal; screening by automated system with manual rescreening under physician supervision 88150 Cytopathology, slides, cervical or vaginal; manual screening under physician supervision 88152 Cytopathology, slides, cervical or vaginal; with manual screening and computer-assisted rescreening under physician supervision 88153 Cytopathology, slides, cervical or vaginal; with manual screening and rescreening under physician supervision 88164 Cytopathology, slides, cervical or vaginal (the Bethesda System); manual screening under physician supervision 88165 Cytopathology, slides, cervical or vaginal (the Bethesda System); with manual screening and rescreening under physician supervision 88166 Cytopathology, slides, cervical or vaginal (the Bethesda System); with manual screening and computer-assisted rescreening under physician supervision 88167 Cytopathology, slides, cervical or vaginal (the Bethesda System); with manual screening and computer-assisted rescreening using cell selection and review under physician supervision 88174 Cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation; screening by automated system, under physician supervision 88175 Cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation; with screening by automated system and manual rescreening or review, under physician supervision 0502U Human papillomavirus (HPV), E6/E7 markers for high-risk types (16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, 66, and 68), cervical cells, branched-chain capture hybridization, reported as negative or positive for high risk for HPV QuantiVirus™ HPV E6/E7 mRNA Test for Cervical Cancer, DiaCarta, Inc, DiaCarta, Inc HCPCS G0123 Screening cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation, screening by cytotechnologist under physician supervision G0124 Screening cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation, requiring interpretation by physician G0141 Screening cytopathology smears, cervical or vaginal, performed by automated system, with manual rescreening, requiring interpretation by physician G0143 Screening cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation, with manual screening and rescreening by cytotechnologist under physician supervision G0144 Screening cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation, with screening by automated system, under physician supervision G0145 Screening cytopathology, cervical or vaginal (any reporting system), collected in preservative fluid, automated thin layer preparation, with screening by automated system and manual rescreening under physician supervision G0147 Screening cytopathology smears, cervical or vaginal, performed by automated system under physician supervision G0148 Screening cytopathology smears, cervical or vaginal, performed by automated system with manual rescreening P3000 Screening papanicolaou smear, cervical or vaginal, up to 3 smears, by technician under physician supervision P3001 Screening papanicolaou smear, cervical or vaginal, up to 3 smears, requiring interpretation by physician Q0091 Screening papanicolaou smear; obtaining, preparing and conveyance of cervical or vaginal smear to laboratory ICD-10 Diagnosis All diagnoses When services are Not Medically Necessary: For the procedure codes listed above when criteria are not met. Discussion/General Information According to the American Cancer Society (ACS) (2024a), about 14,000 new cases of invasive cervical cancer will be diagnosed in 2024 with approximately 4300 deaths occurring from the disease. Cervical cancer screening is a highly effective method of identifying squamous cell cervical cancer. When identified early, cervical cancer can be treated and results in high survival rates. The 5-year relative survival rate for localized cervical cancer is 90% (ACS, 2024b). Cervical cancer screening is comprised of cervical cytology with Papanicolaou testing (also known as a ‘Pap smear’ or “Pap test”) and testing for human papillomavirus (HPV) DNA. Pap tests are used to identify pre-cancerous or cancerous cells present on the cervix. When such cells are found, excision treatments can be used to completely remove the cancerous tissue. The detection of HPV DNA is used as an indication of the cancerous potential of a lesion and the potential risk of developing cervical cancer in the future. According to the ACS, nearly all cases of cervical cancer test positive for HPV DNA. However, not all HPV types result in the development of cervical cancer. Two types of HPV, type 16 and type 18 have been found to be associated with 65% to 75% of all cervical cancers. Another 10 HPV types are associated with the remaining cases. The United States Preventive Services Task Force (USPSTF, 2018) recommends regular cervical cancer screening for eligible individuals aged 21 to 65 years old (Grade A recommendation). Cervical cancer screening every 3 years with cervical cytology alone is recommended in individuals aged 21 to 29 years. For individuals aged 30 to 65 years, any of the following screening options are recommended: Every 3 years with cervical cytology; or Every 5 years with primary high-risk human papillomavirus (hrHPV) testing; or Every 5 years with both hrHPV testing and with cytology (co-testing). The evidence summary upon which the guideline was based (Melnikow, 2018) stated that there was consistent evidence from clinical trials that primary hrHPV screening increased detection of cervical abnormalities (i.e., high-grade dysplasia or more severe) in the initial round of screening by approximately 2 to 3 times compared with cytology alone. The American College of Obstetricians and Gynecologists (ACOG) (2021; reaffirmed 2024) and the American Society for Colposcopy and Cervical Pathology (ASCCP) (2021) have endorsed the 2018 USPSTF guidelines. The 2021 ACOG Practice Advisory replaced the organization’s 2016 Practice Bulletin. The USPSTF (2018) guideline recommends (Grade A recommendation) cervical cancer screening for average risk individuals beginning at age 21. This age criterion is based on data that cervical cancer before age 21 is rare, disease progression is slow, and there is high likelihood that abnormal cervical cytology in individuals less than 21 would regress. As such, the task force consensus that screening individuals less than 21 years of age could result in more harm than benefit is based on the potential of overtreatment contributing to future adverse pregnancy outcomes. In addition, the USPSTF (2018) guideline recommends beginning HPV testing at age 30. The USPSTF recommendations refer both to primary HPV testing and HPV co-testing. Several HPV test kits have been approved by the Food and Drug Administration (FDA) as primary HPV tests in individuals age 25 and older. These include the Cobas® HPV test (Roche Diagnostics), which was approved in 2014 as a primary screening test (previously approved as co-test along with cervical cytology) the Onclarity™ HPV assay (Beckton, Dickenson and Company), which was FDA-approved in 2018. A number of other test kits have been approved for co-testing with cervical cytology (Fontham, 2020). There is a Grade D recommendation (meaning, the USPSTF recommends against routinely providing to asymptomatic individuals and found at least fair evidence that testing is ineffective or that harms outweigh benefits) against cervical screening in the following populations: Individuals younger than 21 years of age; Women who have had a hysterectomy; Women older than 65 years of age. In July 2020, the ACS published an updated cervical cancer screening guideline recommending cervical cancer screening beginning at age 25 (Fontham, 2020). For individuals aged 25 to 65, the ACS recommends a primary HPV test specifically approved as a primary screening test by the FDA every 5 years. If primary HPV testing is not available, they recommend either screening with an HPV test and Pap test every 5 years or cervical cytology every 3 years. As was the case in the development of other guidelines, the committee considered the balance of likely benefits and harms according to the age at screening initiation. The ACS recommendation used the same decision model as the USPSTF 2018 recommendation, which suggests that decreasing the age of HPV testing from 30 to 25 will result in additional colposcopies, but more life-years saved. The ACS recommendations reflect the impact of HPV vaccination, first introduced in 2007 and the entry of vaccinated cohorts, now in their 20s, into the screening-eligible age range. Cytology-based screening is much less efficient in vaccinated populations, as abnormal cytology disproportionately identifies minor abnormalities resulting from HPV types that are associated with lower cancer risk. The 2021 ACOG Practice Advisory included the following statement regarding the 2020 ACS guidelines: Despite the demonstrated efficacy and efficiency of primary hrHPV testing, uptake of this screening method has been slow because of the limited availability of FDA-approved tests and the significant laboratory infrastructure changes required to switch to this screening platform. Limited access to primary hrHPV testing is of particular concern in rural and under-resourced communities and among communities of color, which have disproportionately high rates of cervical cancer incidence, morbidity, and mortality. Although cytology-based screening options are still included in the ACS guidelines in acknowledgement of these barriers to widespread access and implementation, ACS strongly advocates phasing out cytology-based screening options in the near future. Until primary hrHPV testing is widely available and accessible, cytology-based screening methods should remain options in cervical cancer screening guidelines. Although HPV self-sampling has the potential to greatly improve access to cervical cancer screening, and there is an increasing body of evidence to support its efficacy and utility, it is still investigational in the United States. Special populations HIV-infection In 2024, the Centers for Disease Control and Prevention (CDC), the National Institutes of Health (NIH), and the HIV Medicine Association of the Infectious Diseases Society of America, in their Guidelines for the Prevention and Treatment of Opportunistic Infections in Adults and Adolescents with HIV, published an updated recommendation on cervical cancer screening for individuals who are HIV-positive. The document states, “new recommendation for cervical cancer screening to start at age 21 based on the HIV/AIDS Cancer Match Study with no reported cases of cervical cancer below the age of 25.” The guideline cited the HIV/AIDS Cancer Match Study which analyzed data from a population of 164,084 women with HIV and found no cases of invasive cervical cancer in women under 25 years-old over 69,900 person-years of follow-up (standardized incidence ratio [SIR], 0; 95% confidence interval [CI], 0 to 7.1). SIR is a ratio of the observed incidence of a disease to the expected incidence of a disease in the general population. The authors stated, “The rationale for beginning screening at age 21 is to provide a 3- to 5-year window prior to age 25, when the risk of ICC in WWH [women with HIV] exceeds that of the general population.” Match study data were published by Stier and colleagues in 2021. Immunosuppression without HIV-infection In 2019, Moscicki and colleagues published an expert panel guideline on cervical cancer screening recommendations for immunocompromised women without HIV infection. The document noted that the relevant literature was reviewed and, given the small amount of data available, recommendations were “largely based on expert opinion”. For individuals who have had solid organ transplants or allogeneic hematopoietic stem cell transplants, cervical cytology is recommended for individuals under age 30. For individuals 30 years of age or older, co-testing with HPV is preferred, but cytology alone is acceptable. If cytology alone is used, annual testing is recommended unless the results of 3 consecutive tests are negative, in which case cytology can be performed every 3 years. If co-testing with HPV is performed, a baseline co-test with cytology and HPV is recommended and, if the cytology test is normal and HPV is negative, co-testing can be performed every 3 years. If transplants occur before the individual is 21 years old, the recommendation is to begin screening within 1 year of sexual debut. The Moscicki publication cited an analysis of the U.S. Scientific Registry of Transplant Recipients (USSRTR) study (Madeleine, 2013). This analysis reported on the incidence of HPV-related cancers in a cohort of 187,679 solid organ transplant recipients. Risk data were presented in standardized incidence ratios (SIRs), which is a ratio of the observed incidence of a disease to the expected incidence of a disease in the general population. The reported SIR was 3.3 (95% CI, 2.6 to 4.2) for in situ cervical cancer, and 1.0 (95% CI, 0.8 to 1.3) for invasive cervical cancer. This indicated a greater than 3-fold increased risk of in situ cervical cancers and no increased risk for invasive cancers. Additionally, compared to subjects 50 years of age and older, subjects 18 to 34 years of age had a significantly greater risk of in situ cervical cancer (incidence rate ratio [IRR]=4.7). The median age at diagnosis was 38 years for in situ cervical cancer and 44.5 years for invasive cervical cancer and the median time from transplant to cervical diagnosis was 2.6 years for in situ cervical cancer and 3.8 years for invasive cervical cancer. Diethylstilbestrol (DES) The NIH (2021) recommends an annual medical examination that includes a pelvic examination and a Pap test that analyzes cells gathered from the cervix and the vagina for women who were exposed to DES in utero. DES use was discontinued in the United States in the 1970s and thus there are no longer any women who were exposed to DES in utero who are under 21 years old. Definitions High-risk human papillomavirus (HrHPV): Types of HPV that have been linked to an increased risk of cervical cancer. There are more than 100 types of HPV and at least 14 of these, including HPV 14 and 18, are known to cause cancer. Screening: The testing of persons, in either the general population or those at high risk, for specific diseases or conditions in the absence of signs or symptoms of disease. Surveillance: The ongoing systematic active observation or testing of a medical condition with the purpose of detecting changes that warrant new or additional interventions to prevent and control its worsening or spreading. References Peer Reviewed Publications: Madeleine MM, Finch JL, Lynch CF, et al. HPV-related cancers after solid organ transplantation in the United States. Am J Transplant. 2013; 13(12):3202-3209. Moscicki AB, Flowers L, Huchko MJ et al. Guidelines for cervical cancer screening in immunosuppressed women without HIV Infection. J Low Genit Tract Dis. 2019; 23(2):87-101. Ogilvie GS, van Niekerk D, Krajden M, et al. Effect of screening with primary cervical HPV testing vs cytology testing on high-grade cervical intraepithelial neoplasia at 48 months: the HPV FOCAL randomized clinical trial. JAMA. 2018; 320(1):43-52. Stier EA, Engels E, Horner MJ et al. Cervical cancer incidence stratified by age in women with HIV compared with the general population in the United States, 2002-2016. AIDS.2021; 35(11):1851-1856. Vahteristo M, Leinonen MK, Sarkeala T, et al. Similar effectiveness with primary HPV and cytology screening - long-term follow-up of randomized cervical cancer screening trial. Gynecol Oncol. 2024; 180:146-151. Winer RL, Lin J, Anderson ML, et al. Strategies to increase cervical cancer screening with mailed human papillomavirus self-sampling kits: A randomized clinical trial. JAMA. 2023; 330(20):1971-1981. Government Agency, Medical Society, and Other Authoritative Publications: American Cancer Society (ACS). 2024a. Key statistics for cervical cancer. Available at: Accessed on September 29, 2024. American Cancer Society (ACS). 2024b. Survival rates for cervical cancer. Available at: Accessed on September 29, 2024. American College of Obstetricians and Gynecologists. Reaffirmed April 2024, Practice Advisory: Updated Cervical Cancer Screening Guidelines. Available at: Accessed on September 29, 2024. Centers for Disease Control and Prevention (CDC), the National Institutes of Health (NIH), and the HIV Medicine Association of the Infectious Diseases Society of America. 2024. Guidelines for the Prevention and Treatment of Opportunistic Infections in Adults and Adolescents with HIV. Available at: Accessed on September 29, 2024. Fontham ETH, Wolf AMD, Church TR et al. Cervical cancer screening for individuals at average risk: 2020 guideline update from the American Cancer Society. CA Cancer J Clin. 70(5):321-346. Marcus JZ, Cason P, Downs LS Jr et al. The ASCCP Cervical Cancer Screening Task Force endorsement and opinion on the American Cancer Society updated cervical cancer screening guidelines. J Low Genit Tract Dis. 2021;25(3):187-191. Melnikow J, Henderson JT, Burda BU, et al. Screening for cervical cancer with high-risk human papillomavirus testing: updated evidence report and systematic review for the US Preventive Services Task Force. JAMA. 2018; 320(7):687-705. National Institutes of Health (NIH). 2021. Diethylstilbestrol (DES) and Cancer. Available at: Accessed on September 29, 2024. US Preventive Services Task Force, Curry SJ, Krist AH, Owens DK, Barry MJ, et al. Screening for cervical cancer: US Preventive Services Task Force recommendation statement. JAMA. 2018; 320(7):674-686. Index Cervical cancer The use of specific product names is illustrative only. It is not intended to be a recommendation of one product over another, and is not intended to represent a complete listing of all products available. History StatusDateAction Reviewed 11/14/2024 Medical Policy & Technology Assessment Committee review. Revised Discussion/General Information and References sections. Updated Coding section with 01/01/2025 CPT changes; revised descriptor for 87624, added 87626 replacing 0500T deleted as of 01/01/2025. 10/01/2024 Updated Coding section with 10/01/2024 CPT changes, added 0502U. Revised 11/09/2023 MPTAC review. Reformatted Clinical Indications. Updated Description, Discussion/General Information and References sections. Reviewed 11/10/2022 MPTAC review. Updated Discussion/General Information and References sections. Revised 11/11/2021 MPTAC review. Removed bullet points in MN statements and removed criteria on chronically immunosuppressed individuals. Updated Discussion/General Information and References sections. Revised 11/05/2020 MPTAC review. Updated Discussion/General Information and References sections. Corrected minor typographical error in first NMN statement. Reformatted Coding section Reviewed 11/07/2019 MPTAC review. Reviewed 01/24/2019 MPTAC review. Updated Discussion and References sections. Revised 01/25/2018 MPTAC review. Revised 01/17/2018 Hematology/Oncology Subcommittee review. Added “Using Cytology and” to title for clarification. Clarified MN and NMN statements regarding scope of document addressing only screening for HPV, not other types of HPV testing. 01/01/2018 The document header wording updated from “Current Effective Date” to “Publish Date.” Updated Coding section with 01/01/2018 CPT changes; added 0500T, removed 88154 deleted 12/31/2017. Revised 08/03/2017 MPTAC review. Revised 07/10/2017 Hematology/Oncology Subcommittee review. Revised title. Added reference to ADMIN.00006 Preventive Health Guidelines to description section. Added MN and NMN statements regarding HPV testing. Changed the term ‘women’ to ‘individual’ in clinical indications section. Updated Coding, Rationale and References section. Reviewed 11/03/2016 MPTAC review. Reviewed 11/02/2016 Hematology/Oncology Subcommittee review. Updated References section. New 11/05/2015 MPTAC review. 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Alternatively, commercial or FEP plans or lines of business which determine there is not a need to adopt the guideline to review services generally across all providers delivering services to Plan’s or line of business’s members may instead use the clinical guideline for provider education and/or to review the medical necessity of services for any provider who has been notified that his/her/its claims will be reviewed for medical necessity due to billing practices or claims that are not consistent with other providers, in terms of frequency or in some other manner. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, or otherwise, without permission from the health plan. © CPT Only - American Medical Association
1778	https://www.lehman.edu/faculty/anchordoqui/517_4.pdf	Special Relativity Luis Anchordoqui Saturday, September 25, 2010 IV: Relativistic Kinematics Luis Anchordoqui Saturday, September 25, 2010 Transformation of velocities Let a particle move (in our reference frame S) u = dx dt Its velocity is a distance dx in time dt In a reference frame S-bar which moves with , it has only moved a distance ☛ this distance has been moved in a time : it’ s velocity in the moving reference frame is therefore This is it: the velocity u-bar is not simply smaller by v but must be corrected by the denominator which admittedly ~1 when both the particle or the moving frame is much slower than the light velocity d¯ x = γ(dx −vdt) d¯ t = γ dt −v c2 dx v ¯ t ¯ ux = d¯ x d¯ t = γ(dx −vdt) γ(dt −dxv/c2) = dx −vdt dt −dxv/c2 = dx/dt −v 1 −(dx/dt)v/c2 = ux −v 1 −uxv/c2 Saturday, September 25, 2010 For the components of velocity transverse to the motion of S’ In invariant terms (i.e. independent of the coordinate system), component of velocity parallel to component of velocity perpendicular to then v v u⊥ u ¯ u⊥= u⊥ γ(1 −uv/c2) ¯ uy = d¯ y d¯ t = dy γ(dt −dxv/c2) = uy γ(1 −uxv/c2) ¯ uz = d¯ z d¯ z = dz γ(dt −dxv/c2) = uz γ(1 −uxv/c2) Take ¯ u= u−v 1 −uv/c2 Saturday, September 25, 2010 A bullet is shot from a spaceship that speeds with 1/2c Example: We deﬁne the space-ship’ s reference frame to be the resting frame then is the velocity in our reference frame. We move away from the space-ship with v=-1/2c and the bullet is shot with u=+3/ 4c ¯ u so, the bullet has only 90% of the light velocity in our reference frame (with Galileo’ s velocity addition rule, it would have (3/ 4 + 1/2)c = 5/ 4 c) Now, the space-ship emits a light pulse also in our reference frame the light pulse has velocity c !!! relative to our reference frame. The velocity of that bullet is 3/ 4c in the space-ship’ s reference frame in the direction of motion What is the velocity of that bullet in our reference frame? Saturday, September 25, 2010 Drag effect If the speed of light in the liquid at rest is u’, and the liquid is set to move with velocity v, then the speed of light relative to the outside was found to be of the form Neglecting terms the velocity addition formula yields Flowing air drags sound along with it ➤ To what extent a ﬂowing transparent liquid will drag light along with it? On the basis of an ether theory, it would be conceivable that there is no drag at all, since light is a disturbance of the ether and not of the liquid yet experiments indicated that there was a drag: the liquid seemed to force the ether along with it but only partially u = u+ kv k = 1 −1/n2 n = c/u O(v2/c2) u = u+ v 1 + uv/c2 ≈(u+ v) 1 −uv c2 ≈u+ v 1 −u2 c2 = u+ kv Saturday, September 25, 2010 The Doppler effect is very important when describing the effects of relativistic motion in astrophysics Doppler effect Consider a source which emits a period of radiation over a time it takes to move from P₁ to P₂ The effect is the combination of both relativistic time dilation and time retardation ∆t 2.4 Doppler effect !"# %&''(#) ##+, -. /#)0 -1'&),23, 4"#3 5#.+)-6-37 ,"# #8 " #6$ : % #6$ e +&. : e !&$&6.#)/#) &; &< ' # Saturday, September 25, 2010 If is emitted circular frequency of the radiation ωem ∆t= 2π ωem in the rest frame, then and the time between the two events in the observer’ s frame is: ∆t = Γ∆t= Γ 2π ωem However, this is not the observed time between the events because there is a time difference involved in radiation emitted from P₁ and P₂ Let ☛ D = distance to observer from P₂ and ☛ t₁ = time of emission of radiation from P₁ ☛ t₂ = time of emission of radiation from P₂ (γ ≡Γ) Saturday, September 25, 2010 Then, the times of reception and are trec 1 trec 2 Hence the period of the pulse received in the observer’ s frame is trec 2 = t2 + D c Therefore !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!9: "##$ !"#$#%&$#' %( $#) % $#) + %( & ' , , , , -¤ ¦ £ ¥ % & (6% e )&. -' , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , -¤ ¦ £ ¥ + / %( % + ! " ( ' , , , ,6% e )&. + / 6% ( ' , , , , e )&. + ¤ ¦ £ ¥ / "#$% !" % ( + / (+ + (+ %% ( (" % ( ;/ t)<, 7 7 7 7 7 7 7 7 7 7 7 K ;/ t#2 7 7 7 7 7 7 7 7 7 7 = ' ( 7 7 7 7 e '), > ¤ ¦ £ ¥ ? t)<, t#2 K = ' ( 7 7 7 7 e '), > ¤ ¦ £ ¥ 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 t#2 K = e '), > ! " 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 ? ? K trec 1 = t1 + D + V ∆t cos θ c Saturday, September 25, 2010 ❖ the factor is a pure relativistic effect ❖ the factor ❖ In terms of linear frequency is the result of time retardation !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!9 !"#$ %&'()$ $ +,$ &$ -.#$ #/&(+0+,(+'$ #%%#'(1$ ("#$ %&'() $+,$("#$#,./($)%$(+2#$#(&3&(+)45$64$(#2,$)%$/+4 #&$%#8.#4'9: "#& K = ( e '), ¤ ¦ K = e '), > ! " i)<, i#2 K = e '), > ! " 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 ? Γ (1 −βcosθ) The factor is known as the Doppler factor and ﬁgures prominently in the theory of relativistically beamed emission !"#$%&'() " +,$-.)/.$&,$("#$0)112#$%&'()$&.3$%+45#,$1)6+.#.(27$+.$ ("#)7$)%$#2&(+8+,(+'&227$9#&6#3$#6+,,+).: 2.5 Apparent transverse velocity Derivation b @ K @ e '), A ! " = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = B Saturday, September 25, 2010 Apparent transverse velocity A relativistic effect which is extremely important in high energy astrophysics and which is analyzed in a very similar way to the Doppler effect, relates to the apparent transverse velocity of a relativistically moving object !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!9: " #6$ ! % #6$ e "#$ ! e %#&#'$()() &+ &, ' " #6$ e $-. ! Consider an object which moves from P₁ to P₂ in a time in the observer’ s frame. In this case need not be the time between the beginning and end of a periodic. Indeed, in practice, is usually of order a year ∆t ∆t ∆t Saturday, September 25, 2010 As before, the time difference between the time of receptions of photons emitted at P₁ and P₂ are given by The apparent distance moved by the object is Hence, the apparent velocity of the object is: l⊥= V ∆t sin θ ∆trec = ∆t 1 −V c cos θ In terms of βapp = Vapp c βapp = β sin θ 1 −β cos θ β = V c ☛ ☛ Then non-relativistic limit is just as we would expect However, note that this result is not a consequence of the Lorentz transformation, but a consequence of light travel time effects as a result of the ﬁnite speed of light Vapp = V ∆t sin θ ∆t 1 −V c cos θ = V sin θ 1 −V c cos θ Vapp c = V c sin θ 1 −V c cos θ ☛ Vapp = V sin θ Saturday, September 25, 2010 Consequences For angles close to the line of sight the effect of this equation can be dramatic. First determine the angle for which the apparent velocity is a maximum dβapp dθ = (1 −β cos θ)β cos θ −β sin θβ sin θ (1 −β cos θ)2 = β cos θ −β2 (1 −β cos θ)2 This derivative is zero when cos θ = β At the maximum If then and the apparent velocity of an object can be larger then the speed of light βapp = β sin θ 1 −β cos θ = β 1 −β2 1 −β2 = β 1 −β2 = Γβ Γ 1 β ≈1 Saturday, September 25, 2010 We actually see such effects in AGN. Features in jets apparently move at faster than light speed (after conversion of the angular motion to a linear speed using the redshift of the source) This was originally used to argue against the cosmological interpretation of quasar redshifts. However, as you can see such large apparent velocities are an easily derived feature of large apparent velocities !"# !"$ !"$% !"$# &'()+(,+ +,(-+./-0(1+02304/)53+./'15+ /66/66 e Plots of for various indicated values of as a function of βapp β θ Saturday, September 25, 2010 The following images are from observations of 3C 273 over a period of 5 years from 1977 to 1982 They show proper motions in the knots C₃ and C₄ of 0.79 ± 0.03 mas/yr and 0.99 ± 0.24 mas/yr respectively These translate to proper motions of 5.5 ± 0.2h⁻¹c and 6.9 ± 1.7h⁻¹c respectively Unwin et al., ApJ 289 (1985) 109 !"#$% &'()' +% ,-./% 012/ "#$/%345 Saturday, September 25, 2010 Apparent length of a moving rod The Lorentz-Fitzgerald contraction gives us the relationship between the proper lengths of moving rods An additional factor enters when we take into account time retardation !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!97 2.6 Apparent length of a moving rod !"#$%&'#()+,-).#'/01$2&()'/2)-&($.-3#4$54$)"#$'#0/)-&(4"-6 7#)8##($)"#$6'&6#'$0#(.)"4$&9$:&3-(.$'&14;$<($/11-)-&(/0 9/2)&'$#()#'4$8"#($8#$)/=#$-()&$/22&5()$)-:#$'#)/'1/)-&(; 6" # 6" e 2&4 > e !&$&74#'3#' $? $@ % & 6" e 4-( > ' ( Saturday, September 25, 2010 Consider a rod of length in the observer’ s frame Now the apparent length of the rod is affected by the fact that photons which arrive at the observer at the same time are emitted at different times P₁ corresponds to when the trailing end of the rod passes at time t₁ and P₂ corresponds to when the leading end of the rod passes at time t₂ Equating the arrival times for photons emitted from P₁ and P₂ at times t₁ and t₂ respectively When the trailing end of the rod reaches P₂ the leading end has to go a further distance which it does in t₂ - t₁ secs ∆x −L L = Γ−1L0 t1 + D + ∆x cos θ c = t2 + D c ⇒t2 −t1 = ∆x cos θ c Saturday, September 25, 2010 Hence and the apparent projected length is This is another example of the appearance of the ubiquitous Doppler factor And these can also be recovered by considering the differential form of the reverse Lorentz transformations ∆x −L = V ∆x cos θ c ⇒∆x = L 1 −V c cos θ Lapp = ∆x sin θ = L sin θ 1 −β cos θ = L0 Γ(1 −β cos θ) = δL0 Saturday, September 25, 2010 Aberration !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!9: 5 "88 " e # 5v "88v "v ev #v $ % $v %v Because of the law of transformation of velocities, a velocity vector make different angles with the direction of motion From the above laws for transformation of velocities The difference from the non-relativistic case is the factor of Γ tanθ = v⊥ v = v ⊥ Γ(v + V = vsin θ Γ(vcos θ + V) ) Saturday, September 25, 2010 The most important case of this is when v = v’ = c and and the angles made by the light rays in the two frames satisfy We put v = c cos θ v= c cos θ v⊥= c sin θ v ⊥= c sin θ β = V c c cos θ = c cos θ+ V 1 + V c cos θ⇒cos θ = cos θ+ β 1 + β cos θ c sin θ = c sin θ Γ 1 + V c cos θ ⇒sin θ = sin θ Γ(1 + β cos θ) Saturday, September 25, 2010 Half-angle formula There is a useful expression for aberration involving half-angles Using the identity The aberration formulae can be written as tan θ 2 = sin θ 1 + cos θ tan θ 2 = 1 −β 1 + β 1 2 tan θ 2 Saturday, September 25, 2010 Isotropic radiation source Consider a source of radiation which emits isotropically in its rest frame and which is moving with velocity V with respect to an observer (in frame S) ➤ The source is at rest in S’ Consider rays emitted at right angles to the direction of motion The angle of these rays in S is This has These rays enclose half the light emitted by the source, so that in the reference frame of the observer the light is emitted in a forward cone (when is large) of half-angle θ ≈Γ−1 Γ θ = ±π 2 sin θ = ± 1 Γ !"#$%&'()#%+,-)./$,"0,1%2(34-"5",-"0%&66(0-,!!!! 78!9: !"#$%&'())$"+(,')"-)(&%.%"#)/0%,0)'1%.$)%$".("2%,334)%#)%.$ ('$.)-(1')#&)/0%,0)%$)1"5%#6)/%.0)5'3",%.4) )/%.0)('$2',. .")#)"7$'(5'()8%#)-(1') 9:);0')$"+(,')%$).)('$.)%#) )/0%,0 %$)1"5%#6)/%.0)5'3",%.4) )/%.0)('$2',.).") : " # #v " # # #v $ % $v %v " < K = = = $%# < > Saturday, September 25, 2010
1779	https://www.sciencedirect.com/topics/mathematics/arithmetic-progression	Skip to Main content My account Sign in Arithmetic Progression In subject area:Mathematics Arithmetic progression is defined as a sequence of numbers in the form A = {an + b, n = 1, 2, …}, where 'a' is the common difference and 'b' is a constant. AI generated definition based on: Encyclopedia of Physical Science and Technology (Third Edition), 2003 How useful is this definition? Add to Mendeley Also in subject area: Computer Science Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Mathematical Induction 2011, Mathematical Analysis and Proof (Second Edition)David S.G. Stirling 3.2Arithmetic Progressions An arithmetic progression is a sequence which starts with a number a and then increases with a fixed step d : a, a + d, a + 2d, … . Let sn denote the sum of the first n terms, so that We wish to find a simple expression for sn. We can do this by writing down sn in its natural order and then with the terms in the reverse order. Then, adding corresponding terms, We have proved the following little result. (The polite name for a little result, usually a step towards something more interesting, is a Lemma.) Lemma 3.1 The sum of the first n terms of the arithmetic progression a, a + d, a + 2d, … is . ℕ The result just proved and others like it are often much clearer if we use the Σ notation for summation. The general term of the arithmetic progression above is a + (k – 1)d, this being the kth term, so that a + (a + d) + … + (a + (n – 1)d) is the sum of the terms when k = 1, k = 2, …, and k = n all added together, and we write it as More generally if f is some formula involving k, ∑nk=1 f (k) denotes f (1) + f (2) + .... + f (k), the sum of the values obtained when k is successively substituted by 1, 2, …, n, or whatever values are indicated by the Σ sign. Thus ∑5k = 3 f(k) = f(3) + f(4) + f(5). Notice that the k here is a “dummy” variable and that ∑nk = 1 f(k) does not depend on k; we could replace k throughout by any other symbol which suits us except one which already has another meaning in this expression. ∑nn = 1 n will not do, since the symbol n has two meanings. The other common simple type of sequence is the geometric progression, which is a sequence of the form a, ar, ar2, ar3, … r being called the common ratio. Again there is a formula for the sum of the first n terms. Let sn = a + ar + … + arn−1, the sum of the first n terms. Then Therefore, provided r ≠ 1 (to avoid dividing by zero), (Notice the form of this: the first term of the progression multiplies the whole expression and the power of r in the numerator is the number of terms. If r > 1 it is usually neater to write so that the denominator is positive. In the case where r = 1 the formula makes no sense, but in this case all the terms summed are equal so the answer is easy.) We have proved: Lemma 3.2 If r ≠ 1 and n is a natural number This result has an extremely useful consequence (which we call a corollary). Suppose we wish to factorise xn – yn. Then (assuming x ≠ 0) . Here we need to notice that we can use the formula in Lemma 3.2 with r = y/x and a = 1. Then So and multiplying through by xn gives () We proved this for all real numbers satisfying the two conditions x ≠ 0 and x ≠ y. In the two excluded cases, x = 0 and x = y, the result is trivial to check, so we notice independently that the result is true for them and the restriction can be removed. The main virtue of () is that xn – yn has a factor of x – y, although the expression for the second factor is also useful. These two examples, the arithmetic and geometric progressions, are useful but they give a slightly misleading impression in that it is easy to prove the formulae directly once you spot the trick. A more widely applicable way of proving that a formula is true for all natural numbers is to use mathematical induction. Notice that we wish to prove that a result is true for all natural numbers, so we could not do this by testing each value in turn. View chapterExplore book Read full chapter URL: Book2011, Mathematical Analysis and Proof (Second Edition)David S.G. Stirling Chapter Sequences and Series 2015, Introduction to Actuarial and Financial Mathematical MethodsS.J. Garrett 5.3.1Arithmetic and geometric progressions In what follows, we will need to be able to classify a sequence (either finite or infinite) as either an arithmetic progression or a geometric progression. However, as we shall see, a sequence need not necessarily be either. An arithmetic progression is a sequence in which consecutive terms differ by a common difference. For example, the sequence {4,6,8,10,…} is an infinite arithmetic progression with common difference 2. Mathematically, we can express an arithmetical progression as (5.1) That is, the nth term is given by xn = a + (n − 1)d for , where d is the common difference and a is the first term in the sequence. In contrast, a geometric progression is a sequence in which consecutive terms are related by a common factor. For example, the sequence {1,2,4,8,16,…} is a geometric progression with common factor 2. Mathematically, we can express a geometric progression as (5.2) That is, the nth term is given by xn = arn−1 for , where r is the common factor and a is the first term in the sequence. Not all sequences can be classified as being either arithmetic or geometric progression. For example, the series defined by nth term xn = 1 + 4(n − 1) − 0.1n−1 can be considered as having two components, one arithmetic and one geometric. Furthermore, it may be difficult or indeed impossible to express a sequence defined with a recursive formula in terms of arithmetic and geometric components. The techniques that follow are therefore limited in their application, but the skills development from a solid practical understanding of them are very useful. Example 5.9 Classify the following as either arithmetic or geometric progressions (or some combination of both). State the starting value and common difference/factor in each case. a. : {10,14,18,22,…,50} b. : {−1,1,−1,1,…} c. : nth term given by xn = 0.2 × 4n for n = 1,…,10 d. : nth terms given by xn = 2n − 0.1n for n = 1,2,…. e. : {2,5,9,15,…,16,413} Solution a. : This is a finite arithmetic progression with common difference 4 and starting value 10. The nth term is given as xn = a + (n − 1)d = 10 + 4(n − 1) for n = 1,…,11. b. : This is an infinite geometric progression with common factor − 1 and starting value − 1. The nth term is given by xn = arn−1 = (−1) × (−1)n−1. c. : This is a finite geometric progression with common factor 4. However, it is not in the standard form of Eq. (5.2) and should be rewritten as xn = 0.8 × 4n−1 for n = 1,…,10. The starting value is then clearly 0.8. d. : This can be considered as a combination of the two types of progression. The first component is 2n = arn−1 = 2 × 2n−1 and so is an infinite geometric progression with starting value 2 and common factor 2. The second component is − 0.1n = −0.1 + (−0.1)(n − 1), an infinite arithmetical progression with starting value − 0.1 and common difference − 0.1. e. : Although it might not be immediately obvious, this is a finite series with nth term given by xn = 1 + 2(n − 1) + 2n−1 for n = 1,…,15. The series is therefore formed from a finite arithmetic progression with starting value 1 and common difference 2, and a geometric progression with starting value 1 and common factor 2. View chapterExplore book Read full chapter URL: Book2015, Introduction to Actuarial and Financial Mathematical MethodsS.J. Garrett Chapter Series and Integrals 2013, Guide to Essential Math (Second Edition)S.M. Blinder 7.1Some Elementary Series An arithmetic progression is a sequence such as 1, 4, 7, 10, 13, 16. As shown in Section 1.2, the sum of an arithmetic progression with terms is given by (7.1) where is the first term, is the constant difference between terms, and is the last term. A geometric progression is a sequence which increases or decreases by a common factor , for example, 1, 3, 9, 27, 81, … or 1, 1/2, 1/4, 1/8, 1/16, … The sum of a geometric progression is given by, (7.2) When , the sum can be carried to infinity to give (7.3) We had already found from the binomial theorem that (7.4) By applying the binomial theorem successively to , then to each you can show that (7.5) Therefore serves as a generating function for the Fibonacci numbers: (7.6) An infinite geometric series inspired by one of Zeno’s paradoxes is (7.7) Zeno’s paradox of motion claims that if you shoot an arrow, it can never reach its target. First it has to travel half way, then half way again—meaning 1/4 of the distance—then continue with an infinite number of steps, each taking it closer. Since infinity is so large, you’ll never get there! What we now understand that Zeno possibly didn’t (some scholars believe that his argument was meant to be satirical) was that an infinite number of decreasing terms can add up to a finite quantity. The integers 1 to to add up to (7.8) while the sum of the squares is given by (7.9) and the sum of the cubes by (7.10) View chapterExplore book Read full chapter URL: Book2013, Guide to Essential Math (Second Edition)S.M. Blinder Chapter Appendices 2013, Mathematics for Physical Chemistry (Fourth Edition)Robert G. Mortimer Appendix BSome Mathematical Formulas and Identities 1. : The arithmetic progression of the first order to n terms, 2. : The geometric progression to n terms, 3. : The definition of the arithmetic mean of , 4. : The definition of the geometric mean of , 5. : The definition of the harmonic mean of : If is the harmonic mean, then 6. : If for all values of x, then 7. : . 8. : . 9. : . 10. : . 11. : . 12. : . 13. : . 14. : . 15. : . 16. : . 17. : . 18. : . 19. : . 20. : . 21. : . 22. : . 23. : . 24. : . 25. : . 26. : . 27. : . 28. : . 29. : . 30. : . 31. : . 32. : . 33. : . 34. : . 35. : . 36. : . 37. : . 38. : . 39. : . 40. : . 41. : . 42. : . 43. : . 44. : Relations obeyed by any triangle with angle A opposite side a, angle B opposite side b, and angle C opposite side c: (a) : . (b) : . (c) : . View chapterExplore book Read full chapter URL: Book2013, Mathematics for Physical Chemistry (Fourth Edition)Robert G. Mortimer Chapter FORMS OF DIMENSIONLESS RELATIONS 2007, Applied Dimensional Analysis and Modeling (Second Edition) 13/9 Area of a triangle whose side-lengths form an arithmetic progression. Given a triangle whose side-lengths a, b, and c form an arithmetic progression, i.e., where d ≫ 0 is the common difference (see Fig. 13-22). (a) : What are the relevant variables in an expression for the area of such a triangle? (b) : Construct a complete set of dimensionless variables based on the set obtained in (a). (c) : Prove that the relationship among the dimensionless variables established in (b) cannot be a monomial one. View chapterExplore book Read full chapter URL: Book2007, Applied Dimensional Analysis and Modeling (Second Edition) Chapter Sequences 2017, Engineering Mathematics with Examples and ApplicationsXin-She Yang Abstract This chapter introduces some important concepts about sequences, including simple arithmetic progression, geometric sequence, sum of sequences and series. View chapterExplore book Read full chapter URL: Book2017, Engineering Mathematics with Examples and ApplicationsXin-She Yang Chapter Sequences 2017, Engineering Mathematics with Examples and ApplicationsXin-She Yang • : Sequences will be discussed in detail, including both arithmetic progression and geometric progression. • : The difference equation for Fibonacci sequences and its solution techniques will be explained in detail. The link with the golden ratio will also be highlighted. • : Show the ways of calculating the sums of both arithmetic and geometric series, especially the sum of infinite geometrical series. View chapterExplore book Read full chapter URL: Book2017, Engineering Mathematics with Examples and ApplicationsXin-She Yang Chapter Handbook of Dynamical Systems 2006, Handbook of Dynamical SystemsVitaly Bergelson, ... Máté Wierdl Theorem 1.17 [130,131]. For any r ∈ ℕ and any finite partition , one of the Ci contains arbitrarily long arithmetic progressions. We remark that one cannot, in general, expect to get in Theorem 1.17 an infinite arithmetic progression in one of the Ci. Indeed, let us represent ℕ as the union of disjoint intervals of increasing length and alternately color them red and blue. This obviously gives a two-coloring ℕ = R ∪ B without an infinite monochromatic progression. Another remark is that Theorem 1.17 implies the following, ostensibly stronger, finitistic version. View chapterExplore book Read full chapter URL: Handbook2006, Handbook of Dynamical SystemsVitaly Bergelson, ... Máté Wierdl Chapter Sequences and Series 2015, Introduction to Actuarial and Financial Mathematical MethodsS.J. Garrett 5.3.2Evaluating finite sums We begin by working toward a closed-form expression for the sum of an arithmetical progression of k terms. This expression will depend on the progression’s starting value and common difference. We denote this sum as SA,k, with the “A” indicating that it is the sum of an arithmetic progression and the k indicating the number of terms in the series. Using Eq. (5.1), we are seeking an expression for Using the basic properties of summations, we can manipulate this as follows The problem therefore becomes one of finding an expression for . An evaluation of this sum is easily obtained by adding the sequence to its reverse, term by term. That is, We therefore have (5.3) This is a useful result in itself and can be used to complete the required the derivation of the closed-form expression, (5.4) This expression can be used to evaluate any summation that can be manipulated into the standard form of a finite arithmetic progression, as given in Eq. (5.1). Example 5.10 Use an algebraic method to evaluate the following summations. a. b. : 4 + 3 + ⋯ − 15 c. d. Solution a. : This is already in the standard form of the sum of an arithmetic progression with a = 50, d = 4, and k = 10. Equation (5.4) leads to SA,10 = 5 × (100 + 9 × 4) = 680. b. : The nth term in the sequence begin summed is given by xn = 4 − (n − 1) for n = 1,…,20. It is therefore an arithmetic progression with a = 4, d = −1, and k = 20, and so SA,20 = 10 × (8 − 19 × 1) = −110. c. : We can write this as which is a sequence formed from a standard arithmetic progression with a = 10, d = 10, and k = 200. The sum is then obtained as SA,200 = 100 × (20 + 199 × 10) = 201,000. d. : This is the sum of an arithmetic progression with common difference 1, but some manipulation is required to express it in standard form. Using the substitution i = n − 15, we proceed as and so have a standard form of the sum of an arithmetic progression with a = 18, d = 1, and k = 10. The sum is therefore SA,10 = 5 × (36 + 9) = 225. Example 5.11 Determine an expression for the following series for general k ∈{2,3,4,…}. Evaluate Solution Using the properties of logarithms and some further manipulation, we can rewrite the series as Note that we have also used Eq. (5.3). We can then determine thatNote that the second summation requires further manipulation. In particular, we need to express the summation as a series. This can be done by setting n = i + 9 and soNote that we have again used Eq. (5.3). We now consider the sum of a finite geometric progression. This is denoted SP,k, such that where a is the starting value, r the common factor, and . A simple way to obtain the closed-form expression is to subtract the sequence formed by rSG,k from SG,k. In particular, we have which leads to (5.5) This expression enables one to calculate the sum of a finite number of terms in a geometric expression when r≠1. This restriction on r reflects that Eq. (5.5) is a rational function with r = 1 a root of the denominator. Example 5.12 Use an algebraic method to evaluate the following summations. a. b. : 1 + 10 + 100 + ⋯ + 100,000 c. d. Solution a. : This is already in the standard form of the sum of a geometric progression with a = 10, r = 2, and k = 5. Equation (5.5) then leads to . b. : The nth term in the sequence can be represented by xn = 1 × 10n−1 for n = 1,…,6. It is then a geometric progression with a = 1, r = 10, and k = 6, and . c. : We can write this as which is the sum of a standard geometric progression with a = 0.5, r = 0.5, and k = 500. The sum is then obtained as . d. : This is the sum of a geometric progression with common ratio 0.4, but some manipulation is required to express it in standard form. Using the substitution i = n + 11 we proceed as and so have a standard form with a = 0.4−9, r = 0.4, and k = 36. The value is . Example 5.13 Evaluate the following summations. a. b. Solution a. : We note that which is seen to be the sum of a standard arithmetic progression (with a = 2, d = 2, and k = 15) and a standard geometric progression (with a = 2, r = 2, and k = 15). These two components can be evaluated separately as b. : We can manipulate this into the sum of an arithmetic and a geometric progression, using the standard expressions for the sums. View chapterExplore book Read full chapter URL: Book2015, Introduction to Actuarial and Financial Mathematical MethodsS.J. Garrett Chapter Recursion 2023, Discrete MathematicsAli Grami 14.1Sequences A sequence with its discrete structure presents an ordered list of terms, where repetitions are allowed. Sequences are an important data structure in computer science and engineering. A function whose domain is a subset of the set of integers, generally consecutive nonnegative integers, is called a sequence. A sequence is a function whose domain is either all the integers between two given integers or all the integers greater than or equal to a given integer. The function , denoted by and called a term of the sequence, is the image of the integer . The integer of the term is called an index. The term is the general term of a sequence, and it is often used to define a sequence. Note that the notation describes the sequence. For instance, the list of terms of the sequence , where , namely , , , …, starts with , corresponding to . Hence the sequence is as follows: . The terms of a sequence are usually ordered in increasing order of subscripts. The first term of the sequence is called the initial term. A sequence is finite if its domain is finite; otherwise, it is infinite. Note that a finite sequence of terms is also called a string or an n-tuple, its last term is known as the final term, and the length of a string is the number of terms in the string. For instance, the set of two-digit integers that are positive represents a finite sequence of 90 terms, where 10 is its initial term and 99 is its final term; the set of prime numbers represents an infinite sequence, where the initial term is 2 and there is no final term, as there are infinitely many prime numbers. The most widely known sequences are geometric and arithmetic progressions. A geometric progression is a sequence with the general term , where the initial term and the common ratio are real numbers, and is an integer. The set of terms of a geometric progression is thus as follows: . An arithmetic progression is a sequence with the general term , where the initial term and the common difference are real numbers, and is an integer. The set of terms of an arithmetic progression is thus as follows: . It is often required to add or multiply a number of terms in a sequence whose terms are as follows: , where and are both integers and . The sum of the terms and the product of the terms can be written in a compact form as follows, respectively: Note that the symbol , the capital Greek letter sigma, is the summation notation, the dummy variable is the summation index, the symbol , which is the capital Greek letter pi, is the product notation, the dummy variable is the product index, and the integers and are the lower limit and the upper limit of each index, respectively. Note that a series is an extended sum of terms, such as the sum of the first thousand positive integers. The terms of a sequence can be specified either by providing a formula for each term of the sequence as a function of its position or by expressing each term as a combination of the previous terms. The focus of this chapter is to find an explicit formula or a general formula, called a closed-form formula, for the terms of the sequence specified through a recurrence relation. View chapterExplore book Read full chapter URL: Book2023, Discrete MathematicsAli Grami Related terms: Rational Number Trivial Solution Initial Value Sampling Interval Decimal Place Negative Sign Recurrence Relation Brahmagupta Furstenberg Summation View all Topics
1780	https://rucsm.org/physics/labdescriptions/1016.pdf	Atwood’s Machine EX-5501 ScienceWorkshop Page 1 of 21 Written by Chuck Hunt 2012 Newton’s Second Law – Atwood’s Machine Equipment: Included: 1 Photogate/Pulley System ME-6838A 1 Mass and Hanger Set ME-8979 1 Universal Table Clamp ME-9376B 1 60-cm Long Threaded Rod ME-8977 1 Multi Clamp ME-9507 1 Braided Physics String SE-8050 Required (Not included): 1 850 Universal Interface UI-5000 1 PASCO Capstone UI-5400 1 Balance or Scale SE-8707 1 Calipers (suggested: Digital Calipers) SE-8710 Introduction: The purpose of this activity is to study the relationship between net force, mass, and acceleration as stated by Newton’s 2nd Law, using an Atwood’s Machine apparatus, built with a PASCO Super Pulley. The Super Pulley has very low friction and small mass. The Photogate Head, when attached to a Super Pulley, is used to measure the velocity of both masses as one moves up and the other moves down. The slope of the graph of velocity vs. time is the acceleration of the system. Careful measurement when there is no net force allows the student to compensate for friction. The effect of the motion of the pulley on the measured experimental acceleration will also be examined. This will provide a way to roughly compensate for the (small) effect of the pulley. Atwood’s Machine EX-9973 ScienceWorkshop Page 2 of 21 Simplified Theory: The acceleration of a system is directly proportional to the net applied force and inversely proportional to the system’s mass, as stated by Newton’s 2nd Law of Motion: a = Fnet/Msystem . Atwood's Machine consists of two unequal masses connected by a single string that passes over an ideally massless and frictionless pulley as in Figure 1. When released, the heavier object accelerates downward while the lighter object accelerates upward. The free-body diagrams below show the forces acting on each of the masses. T is the tension in the string, assumed to be the same for both masses. This is a good assumption as long as it is a single strand of string and a relatively light pulley system. With m1 > m2, m1 is descending mass and m2 is the ascending mass. The magnitude of the acceleration, a, is the same for each mass, but the masses accelerate in opposite directions. We adopt the convention that down is positive. Then the equation of motion of the descending mass, m1, is m1g – T = m1a . The equation of motion for the ascending mass, m2, is m2g –T = m2(-a) . Eliminating T between the two equations yields: (m1 – m2)g= (m1 + m2)a Which may be solved for the acceleration a = (m1 – m2)g /(m1 + m2) = Fnet/Msystem Eq. (1) Figure 1: Atwood’s Machine Figure 2: Free Body Diagram for Atwood’s Machine Atwood’s Machine EX-5501 ScienceWorkshop Page 3 of 21 Written by Chuck Hunt 2012 Setup: 1. Attach the PASCO Super Pulley to the Photogate using the 15 cm treaded black rod as shown in Figure 3. Attach one end of the photogate wire to the telephone jack on the photogate and the other end to Digital Input 1 on the 850 Universal Interface. 2. Attach the Multi Clamp to a table and arrange the rods as shown in Figure 4. 3. Cut 1.5 m piece of the braided string and tie loops on each end to hold a mass hanger. Determine the mass of the string in grams and enter it in column 1 (String Mass) of the Acceleration Data table under the Analysis tab. Enter the value in each of the first four rows. 4. Add a single 50 g mass to one mass hanger (5 g) for a total mass of 55 g. We will call this mass hanger m1. To a second mass hanger (call it m2) add one 20 g, one 10 g, two 5 g, two 2 g, and one 1g mass for a total of 50 g (including the 5 g hanger). 5. Adjust the height of the pulley so that when mass hanger m1 is touching the floor, mass hanger m2 is a few centimeters below the pulley. The Photogate should be horizontal so the string does not pull sideways on the pulley. Atwood’s Machine EX-9973 ScienceWorkshop Page 4 of 21 Friction & String Mass Compensation: 1. Add an additional 5 g to m2. The two masses should now balance. If there were no friction and the string were massless, then if we gave m1 a push downward, it would continue at constant speed. 2. Move m1 to its highest point. Click RECORD. Give m1 a gentle push downward and release it. 3. When m1 stops moving or strikes the floor, click STOP. The masses may touch as they pass each other. If so, click the Delete Last Run button at the bottom of the screen and repeat the run. 4. Click the Data Summary button at the left of the screen. Double click on the current run (probably Run #1) and re-label it “Equal mass run”. Click Data Summary again to close the panel. 5. Your Speed vs Time graph should look like Figure 5. Answer Question 1 on the Conclusions page. 6. Add 0.5 g to m1 and repeat. Label this run “0.5 g run”. Then add another 0.5 g and repeat again until you get a graph like Figure 6. Since you only have 0.5 g increments, your graph may not be quite as symmetric as Figure 6. (Of course, you could add small paper clips to get as close as possible.) Note that the slowing/speeding up is now in the hundredths of a meter per second rather than tenths of a meter per second. Enter the amount of mass (in grams) you added to compensate for friction in column 2 (Friction Mass) of the Acceleration Data table under the Analysis tab in all four rows. Answer question 2 on the Conclusions page. 7. Leave the extra mass on m1 to compensate for friction. We will include the extra mass (+ any paper clips) and the mass of the string in the total mass of the system, but will not include them in the mass difference, m1 – m2 . 8. Remove the extra 5 g from m2! Atwood’s Machine EX-5501 ScienceWorkshop Page 5 of 21 Written by Chuck Hunt 2012 Procedure: (sensors set at 10 Hz) 1. Transfer 2 g to m1 from m2. Note that this keeps the total mass constant. 2. Move m1 to its highest point. Click RECORD. Give m1 a gentle push downward and release it. 3. When m1 strikes the floor, click STOP. It is better if you can catch m2 just before m1 hits the floor. Otherwise, the m1 hanger may come loose and you’ll end up chasing masses around the floor. The masses may touch as they pass each other. If so, click the Delete Last Run button at the bottom of the screen and repeat the run. 4. Click the Data Summary button at the left of the screen. Double click on the current run and re-label it “55v50 g run”. 5. Transfer 2 g from m2 to m1. Note that this keeps the total mass constant. Repeat steps 1-3. Label this run “57v48 g run”. 6. Transfer another 2 g from m2 to m1 and repeat. Label this run “59v46 g run”. 7. Transfer another 2 g from m2 to m1 and repeat. Label this run “61v44 g run”. Atwood’s Machine EX-9973 ScienceWorkshop Page 6 of 21 Data: 1. Click on the black triangle by the Run Select icon on the graph toolbar and select “55v50 g run”. 2. Click the Scale to Fit icon on the left of the graph toolbar. 3. Click the Selection icon and adjust the handles on the selection box to highlight the linear portion of the data. Use as much of the data as possible but be sure not to include any points where the masses were not moving freely. 4. Click the black triangle by the Curve Fit icon and select “Linear”. The slope on the linear portion of a speed versus time plot is the acceleration. Record the value in column 8, row 1 of the Acceleration Data table under the Analysis tab. 5. Repeat for the “57v48 g run” data and record it in row two. Then do the other two runs. Atwood’s Machine EX-5501 ScienceWorkshop Page 7 of 21 Written by Chuck Hunt 2012 Analysis: 1. Click open the Calculator at the left of the screen. Examine line 3 and verify that the calculation of “Theory a” in column 7 agrees with Equation 1 from Theory. 2. Compare the “Theory a” column to the “Exp. a” column. The “Theory a” numbers should be consistently high. The percent by which “Theory a” is high is calculated in column 9 (“% high”). 3. What could be the source of the systematic error? If you answered “the pulley”, then you are correct. The pulley also has mass, and it accelerates. We will deal with rotating systems later in the course. For now we just assert that the pulley may be treated as an additional constant mass added to the total mass of the system. The effective total mass of the pulley wheel (which does not equal it’s mass) may be estimated from our four runs. 4. Calculate the average percentage by which “Theory a” is high (the average ot the “% high” column). 5. Multiply the percentage (divided by 100) by the “Total mass” in column 5. Enter the value in all four rows of column 10 (“Effect. Rot. m). 6. The corrected total mass (“Corr Tot m), corrected acceleration (“Corr a”), and the percent by which the experimental a (“Exp. a) disagrees with the corrected acceleration (“% diff”) now show in last three columns. Atwood’s Machine EX-9973 ScienceWorkshop Page 8 of 21 Conclusions: 1. For the “Equal mass run”, step 5 from the Friction tab: a. Why are the masses slowing down? b. Why is the rate at which the masses slow down decreasing with time (the slope is less negative)? Hint: does the string weight tend to speed the masses up or slow them down, or both? 2. For the run where the speed stayed as constant as possible, step 6 under the Friction tab: a. Why do the masses first slow down and then speed up? Hint: remember the string. b. You may not see the periodic oscillation that shows in Figure 6. If you do see it, try to explain it. 3. Does your data support Newton’s Second Law of Motion? Support your answer! 4. The PASCO specs on the pulley wheel list its actual mass as 5.5 g. If all the mass were at the same distance as the string is from the axis of the wheel, the effective rotational mass would be 5.5 g. Some of the mass is actually further from the axis which increases the effective mass, but quite a bit of the mass is closer to the axis. The true effective mass is probably 4.5+/-1.0 g. How well does this agree with your results? Atwood’s Machine EX-5501 ScienceWorkshop Page 9 of 21 Written by Chuck Hunt 2012 8. PROCEDURE A: Total Mass Constant NOTE: The procedure is easier to perform if one person handles the apparatus and a second person handles the computer. 1. Place the illustrated combination of mass-pieces on the hangers. Record the masses 1 m and 2 m in the Data Table, including the 5-g mass of the hanger itself. 20 g 20 g 10 g 5 g 5 g 20 g 10 g 10 g 5 g 5 g 5 g m1 descending mass m2 ascending mass 2. Move the 1 m mass hanger upward until the 2 m mass hanger almost touches the floor. Hold the string to prevent the system from moving just yet. 3. Click ‘Start’ to begin recording data and then release the string. 4. Stop the motion by gently grabbing the string, before the hangers reach the pulley at the top or the floor at the bottom. This will prevent the small pieces from jumping out of the hangers. 5. Click ‘Stop’ to end the data recording. Do not worry if the data on the graph has sections of “extra data” from the moments when the masses were still not moving, or if they collided with each other, or from after you stopped them. 6. Repeat two more times with the same masses 1 m and 2 m . 7. In the computer: Select Run #1 from the Data Menu in the Graph display. (If multiple data runs are showing, first select No Data from the Data Menu and then select Run #1.) Click the “Scale-to-fit” button to rescale the Graph axes to fit the data. 8. Click the ‘Fit’ menu button and select ‘Linear’. The computer will make a linear fit of all the data and tell you the slope “m” of the best-fitting line. 9. If there are parts of the plot that are not needed, place the cursor on the lower-left edge of the “good data.” Click, hold, and drag the cursor up until only the part of the plot needed is highlighted. (See the diagram below.) Atwood’s Machine EX-9973 ScienceWorkshop Page 10 of 21 By highlighting only a portion of the graph, the linear fit calculation uses only those points. 10. In a graph of Velocity vs. Time, the slope is the acceleration. Record the value of the slope as the “Experimental Acceleration” in the Data Table. 11. Repeat for each of the data runs and then calculate the average experimental acceleration for this set of masses 1 m and 2 m . 12. For the next set of data, move a mass-piece from the 2 m hanger to the 1 m hanger. This will make the descending mass larger than before, the ascending mass smaller than before, while still keeping the total mass of the system unchanged. (This process changes the net force by changing the difference in weight of the hanging masses, but without changing the total mass in the system.) Record the new masses 1 m and 2 m in the Data Table. Repeat the data collection process. 13. Repeat the above step to create two more mass combinations. For each run, the net force changes but the total mass of the system remains constant. Measure the experimental acceleration three times for each mass combination to get a good average. Atwood’s Machine EX-5501 ScienceWorkshop Page 11 of 21 Written by Chuck Hunt 2012 PROCEDURE B: Net Force Constant 1. Place the illustrated combination of mass-pieces on the hangers. Record the masses 1 m and 2 m in the Data Table, including the 5-g mass of the hanger itself. 20 g 20 g 10 g 5 g 20 g 5 g m1 descending mass m2 ascending mass 20 g 2. Add 5 grams to each mass hanger. Record the new 1 m and 2 m in the Data Table. Notice that the addition of the extra mass only changed the total mass of the system, but without altering the difference between the ascending and descending masses. Record data and measure the experimental acceleration three times to calculate an average 3. For the next trial, place more mass on each hanger, the same amount on each. This will increase the total mass of the system, but without changing the difference between 1 m and 2 m . 4. Measure the acceleration of each mass combination three times to calculate the average. 5. If you want to perform Procedure C, the examination of the effect of the pulley, dismantle the equipment and use a scale to measure the mass of the string. Record it in the space provided in the Data Table for Procedure C. PROCEDURE A & B: Calculations 1. For each of the data runs, calculate the total mass 2 1 m m  and record it in the Data Table. 2. For each of the data runs, using the measured mass values, calculate and record the net force:  g m m Fnet 2 1   3. Using the total mass and the net force, calculate the theoretical acceleration using Newton’s 2nd Law of Motion: total M F a net  4. For each data run, calculate and record the percent difference between the experimental acceleration and the theoretical acceleration: Atwood’s Machine EX-9973 ScienceWorkshop Page 12 of 21 100 2 difference % exp exp             theo theo a a a a (The percent difference compares the difference between the values to the average of the values.) Atwood’s Machine EX-5501 ScienceWorkshop Page 13 of 21 Written by Chuck Hunt 2012 PROCEDURE C: The Rotational Inertia and the Friction of the Pulley System THEORY Theoretical Assumptions One of the main assumptions of the Atwood’s Machine theoretical analysis is that the pulley-system has no significant mass and that its motion has no effect on the measured acceleration of the system. However, it is obvious to the observer that the pulley moves together with the hanging masses and is thus also part of “the system.” (So is the string, by the way.) In order to account for this “extra mass,” let’s introduce a term to the total mass of the system: m m m M     2 1 actual total . Here m will be the excess mass unaccounted for in the previous theoretical analysis. We expect m to be (hopefully) very small. The other main assumption is that the system is frictionless. This is also an idealization. Even for the specially designed type of bearing of the pulley there may be a small amount of friction that may drag the pulley. The strings may also contribute to the drag if they slip and rub against the pulley. The presence of any frictional force will reduce the net force. m1 m2 m1 > m2 a a = a/r linear motion linear motion rotational motion r pulley-system Let’s introduce a general frictional term to the calculation of the net force: f g m g m f F F        ) ( 2 1 assumed net actual net Here assumed net F is the net force we used in the ideal case, the difference in weight between the masses 1 m and 2 m . Notice that the presence of any excess mass and the presence of any frictional force must result in a measured value of acceleration that is lower than the ideal theoretical value predicted in procedures A & B. The Effect of the Rotation of the Pulley The pulley rotates with constant angular acceleration . As long as there is no slippage, the linear acceleration of any point on the rim of the pulley must be the same as the linear acceleration of the masses and of the string. That is, r a   . Atwood’s Machine EX-9973 ScienceWorkshop Page 14 of 21 The body diagram below shows the tensional forces acting on the pulley, of radius r and moment of inertia I . No friction is considered in this part of the analysis. T1 > T2 r T2 T1 = a/r I Let 1 T be the tension from the side of the string connected to 1 m , and let 2 T be the tension from the side of the string connected to 2 m . These tensional forces provide the torques that make the pulley rotate. Because 1 T tries to make the pulley rotate clockwise, while 2 T tries to make the pulley rotate counterclockwise, the torques they apply are in opposite directions, but the torque from 1 T must be greater, since we know the system in the end rotates and accelerates clockwise (in the direction of the larger, descending mass). The equation of rotational motion for pulley is:  I r T r T   2 1 . Applying the no-slippage condition, this equation becomes: r a I r T r T   2 1 Now, from previous analysis (see the ‘Theory’ section for Procedure A) we know that a m g m T 1 1 1   , and that g m a m T 2 2 2   . Substituting into the previous equation, then simplifying and re-arranging, we see that the linear acceleration of the system must be: 2 2 2 1 assumed net 2 1 2 1 r I r I m m F m m g m g m a        . If the pulley-system has any significant rotational inertia, then its mass contribution ( m) to the system has the form 2 r I m   . Notice that if I is very small this expression reduces to the theoretical formula used in the analysis of Procedures A and B. If there is also friction opposing the motion of the pulleys, the same analysis can be used to show that the effect is a reduction to the net force. 2 2 1 (assumed) net r I m m f F a      . The Analysis The experimentally measured acceleration must include all effects of excess mass and friction. Let’s say that the measured acceleration for each run is actually given by m M f F a      assumed total aasumed net Atwood’s Machine EX-5501 ScienceWorkshop Page 15 of 21 Written by Chuck Hunt 2012 Rearranging this equation, f a m a M F      assumedl tota assumed net , where m and f  are the unknowns we are attempting to find. PROCEDURE C: Calculations 1. Transfer the following data from Procedures A and B into the Data Table for Procedure C: the measured experimental acceleration of each run (this will be a ), the net force of each run (this will now be assumed net F ), and the total mass of each run (this will now be assumed total M ). 2. Calculate the quantity a M F assumedl total assumed net  for each data run and enter the results in the table. 3. Make a plot of the quantity a M F assumedl total assumed net  in the y-axis versus the experimental acceleration, a , in the x-axis. Notice that the equation should be linear: b x m y f a m a M F            assumed total assumed net The excess mass m will be the slope of the line. The frictional force f  will be the y-intercept. 4. Draw the best-fitting line for your plot and calculate the slope and the y-intercept. Record the values as m and f , respectively. Don’t forget the units. 5. Use calipers to measure the diameter of the Super Pulley. Determine the radius of the pulley and enter it in the Data Table. 6. Use the calculated excess-mass m to calculate the rotational inertia of the pulley: m r I   2 Atwood’s Machine EX-9973 ScienceWorkshop Page 16 of 21 Lab Report: Newton’s Second Law – Atwood’s Machine Name: _________ DATA TABLE for PROCEDURE A: Total Mass Constant Run Descending 1 m Ascending 2 m Experimental exp a (Slope of the v vs. t plot.) Trial 1 Trial 2 Trial 3 Average 1 2 3 4 [g] [g] [m/s2] Run Descending 1 m Ascending 2 m Total Mass total M   2 1 m m  Net Force net F  g m m 2 1  Experimental exp a (Average) Theoretical theo a total net M F Percent difference 1 2 3 4 [g] [g] [g] [N] [m/s2] [m/s2] Atwood’s Machine EX-5501 ScienceWorkshop Page 17 of 21 Written by Chuck Hunt 2012 DATA TABLE for PROCEDURE B: Net Force Constant Run Descending 1 m Ascending 2 m Experimental exp a (Slope of the v vs. t plot.) Trial 1 Trial 2 Trial 3 Average 5 6 7 8 [g] [g] [m/s2] Run Descending 1 m Ascending 2 m Total Mass total M   2 1 m m  Net Force net F  g m m 2 1  Experimental exp a (Average) Theoretical theo a total net M F Percent difference 5 6 7 8 [g] [g] [g] [N] [m/s2] [m/s2] Atwood’s Machine EX-9973 ScienceWorkshop Page 18 of 21 DATA for PROCEDURE C: The Effect of the Pulley Mass of string used in Procedures A & B: __ g Diameter of the Super Pulley: _ cm Radius of the Super Pulley,  r __ cm Run From Previous Activities Calculate a M F assumedl total assumed net  Experimental Acceleration a Net Force assumed net F  g m m 2 1  Total Mass assumed total M   2 1 m m  1 2 3 4 5 6 7 8 [m/s2] [N] [kg] [N] From the plot: Slope, m= _ kg y-intercept, f = __ N Calculation: Rotational Inertia of the pulley, I = __ Atwood’s Machine EX-5501 ScienceWorkshop Page 19 of 21 Written by Chuck Hunt 2012 QUESTIONS PROCEDURE A: Total Mass Constant 1. Look at the data: as the net force increased, what happened to the acceleration? Did it increase, decrease or stay constant? 2. Did a change in the net force produce a change in acceleration by the same factor? Do your results agree with Newton’s 2nd Law? 3. Use this grid to make a plot of Net Force vs. Experimental Acceleration and draw the best fitting line. Experimental Acceleration Atwood’s Machine EX-9973 ScienceWorkshop Page 20 of 21 4. Calculate the slope of the best-fitting line. What does the slope of the best-fit line represent? (Hint: what are the units of the slope?) QUESTIONS PROCEDURE B: Net Force Constant 1. Look at the data: as the total mass increased, what happened to the acceleration? Did it increase, decrease or stay constant? 2. Did a change in the total mass produce a change in acceleration by the same factor? Do your results agree with Newton’s 2nd Law? QUESTIONS PROCEDURE C: The Rotational Inertia of the Pulleys 1. The motion and mass of the string that moves the system was never considered in any part of the theoretical analysis. Looking at your results, is it reasonable to ignore the mass of the string as part of the “total mass of the system”? Discuss. Atwood’s Machine EX-5501 ScienceWorkshop Page 21 of 21 Written by Chuck Hunt 2012 2. Looking back at the ideal case: was it safe to assume that the system is essentially frictionless? 3. What has a larger impact on the percent differences found in procedures A & B, the small excess mass or the small amount of friction? Discuss based on your results. 4. Assuming frictional forces only act to oppose the linear motion of the masses, and that each mass receives the same amount of friction f , prove that f f 2   .
1781	https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/resources/lec1/	2 R. STANLEY, HYPERPLANE ARRANGEMENTS LECTURE 1 Basic definitions, the intersection poset and the characteristic polynomial 1.1. Basic definitions The following notation is used throughout for certain sets of numbers: N nonnegative integers P positive integers Z integers Q rational numbers R real numbers R+ positive real numbers C complex numbers [m] the set {1, 2, . . . , m } when m ≤ N We also write [tk]ψ(t) for the coefficient of tk in the polynomial or power series ψ(t). For instance, t24 = 6. A finite hyperplane arrangement A is a finite set of affine hyperplanes in some vector space V ∪= Kn, where K is a field. We will not consider infinite hyperplane arrangements or arrangements of general subspaces or other objects (though they have many interesting properties), so we will simply use the term arrangement for a finite hyperplane arrangement. Most often we will take K = R, but as we will see even if we’re only interested in this case it is useful to consider other fields as well. To make sure that the definition of a hyperplane arrangement is clear, we define a linear hyperplane to be an (n − 1)-dimensional subspace H of V , i.e., H = {v ≤ V : κ v = 0},· where κ is a fixed nonzero vector in V and κ v is the usual dot product: · (κ1, . . . , κ n) · (v1, . . . , v n) = � κivi. An affine hyperplane is a translate J of a linear hyperplane, i.e., J = {v ≤ V : κ v = a},· where κ is a fixed nonzero vector in V and a ≤ K. If the equations of the hyperplanes of A are given by L1(x) = a1, . . . , L m(x) = am, where x = (x1, . . . , x n) and each Li(x) is a homogeneous linear form, then we call the polynomial QA(x) = (L1(x) − a1) · · · (Lm(x) − am) the defining polynomial of A. It is often convenient to specify an arrangement by its defining polynomial. For instance, the arrangement A consisting of the n coordinate hyperplanes has QA(x) = x1x2 · · · xn. Let A be an arrangement in the vector space V . The dimension dim( A) of A is defined to be dim( V ) (= n), while the rank rank( A) of A is the dimension of the space spanned by the normals to the hyperplanes in A. We say that A is essential if rank( A) = dim( A). Suppose that rank( A) = r, and take V = Kn . Let LECTURE 1. BASIC DEFINITIONS 3 Y be a complementary space in Kn to the subspace X spanned by the normals to hyperplanes in A. Define W = {v ≤ V : v · y = 0 ≡y ≤ Y }. If char( K) = 0 then we can simply take W = X. By elementary linear algebra we have (1) codim W (H ⊕ W ) = 1 for all H ≤ A. In other words, H ⊕ W is a hyperplane of W , so the set AW := {H ⊕W : H ≤ A} is an essential arrangement in W . Moreover, the arrangements A and AW are “essentially the same,” meaning in particular that they have the same intersection poset (as defined in Definition 1.1). Let us call AW the essentialization of A, denoted ess( A). When K = R and we take W = X, then the arrangement A is obtained from AW by “stretching” the hyperplane H ⊕ W ≤ AW orthogonally to W . Thus if W � denotes the orthogonal complement to W in V , then H� ≤ AW if and only if H� � W � ≤ A. Note that in characteristic p this type of reasoning fails since the orthogonal complement of a subspace W can intersect W in a subspace of dimension greater than 0. Example 1.1. Let A consist of the lines x = a1, . . . , x = ak in K2 (with coordinates x and y). Then we can take W to be the x-axis, and ess( A) consists of the points x = a1, . . . , x = ak in K. Now let K = R. A region of an arrangement A is a connected component of the complement X of the hyperplanes: X = Rn − � H. H⊆A Let R(A) denote the set of regions of A, and let r(A) = #R(A), the number of regions. For instance, the arrangement A shown below has r(A) = 14. It is a simple exercise to show that every region R ≤ R(A) is open and convex (continuing to assume K = R), and hence homeomorphic to the interior of an n-dimensional ball Bn (Exercise 1). Note that if W is the subspace of V spanned by the normals to the hyperplanes in A, then R ≤ R(A) if and only if R ⊕ W ≤ R(AW ). We say that a region R ≤ R(A) is relatively bounded if R ⊕ W is bounded. If A is essential, then relatively bounded is the same as bounded. We write b(A) for 4 R. STANLEY, HYPERPLANE ARRANGEMENTS the number of relatively bounded regions of A. For instance, in Example 1.1 take K = R and a1 < a2 < < ak . Then the relatively bounded regions are the · · · regions ai < x < ai+1 , 1 → i → k − 1. In ess( A) they become the (bounded) open intervals (ai, a i+1 ). There are also two regions of A that are not relatively bounded, viz., x < a1 and x > ak. A (closed) half-space is a set {x ≤ Rn : x κ ⊂ c} for some κ ≤ Rn , c ≤ R. If · H is a hyperplane in Rn, then the complement Rn − H has two (open) components ¯whose closures are half-spaces. It follows that the closure R of a region R of A is a finite intersection of half-spaces, i.e., a (convex) polyhedron (of dimension n). A ¯bounded polyhedron is called a (convex) polytope . Thus if R (or R) is bounded, ¯then R is a polytope (of dimension n). An arrangement A is in general position if {H1, . . . , H p} ∗ A, p → n ⊆ dim( H1 ⊕ · · · ⊕ Hp) = n − p {H1, . . . , H p} ∗ A, p > n = .⊆ H1 ⊕ · · · ⊕ Hp � For instance, if n = 2 then a set of lines is in general position if no two are parallel and no three meet at a point. Let us consider some interesting examples of arrangements that will anticipate some later material. Example 1.2. Let Am consist of m lines in general position in R2 . We can compute r(Am) using the sweep hyperplane method. Add a L line to Ak (with AK ∅ { L} in general position). When we travel along L from one end (at infinity) to the other, every time we intersect a line in Ak we create a new region, and we create one new region at the end. Before we add any lines we have one region (all of R2). Hence r(Am) = #intersections + #lines + 1 �m� = + m + 1. 2 Example 1.3. The braid arrangement Bn in Kn consists of the hyperplanes Bn : xi − xj = 0, 1 → i < j → n. Thus Bn has ⎜n� hyperplanes. To count the number of regions when K = R, note 2 that specifying which side of the hyperplane xi − xj = 0 a point (a1, . . . , a n) lies on is equivalent to specifying whether ai < aj or ai > aj . Hence the number of regions is the number of ways that we can specify whether ai < aj or ai > aj for 1 → i < j → n. Such a specification is given by imposing a linear order on the ai’s. In other words, for each permutation w ≤ Sn (the symmetric group of all permutations of 1, 2, . . . , n ), there corresponds a region Rw of Bn given by Rw = {(a1, . . . , a n) ≤ Rn : aw(1) > aw(2) > · · · > aw(n)}. Hence r(Bn ) = n!. Rarely is it so easy to compute the number of regions! Note that the braid arrangement Bn is not essential; indeed, rank( Bn) = n − 1. When char( K) = 2 the space W ∗ Kn of equation (1) can be taken to be ⇔ W = {(a1, . . . , a n) ≤ Kn : a1 + + an = 0}.· · · The braid arrangement has a number of “deformations” of considerable interest. We will just define some of them now and discuss them further later. All these arrangements lie in Kn, and in all of them we take 1 → i < j → n. The reader who 5LECTURE 1. BASIC DEFINITIONS likes a challenge can try to compute their number of regions when K = R. (Some are much easier than others.) • generic braid arrangement : xi − xj = aij , where the aij ’s are “generic” (e.g., linearly independent over the prime field, so K has to be “sufficiently large”). The precise definition of “generic” will be given later. (The prime field of K is its smallest subfield, isomorphic to either Q or Z/p Z for some prime p.) • semigeneric braid arrangement : xi −xj = ai, where the ai’s are “generic.” • Shi arrangement : xi − xj = 0, 1 (so n(n − 1) hyperplanes in all). • Linial arrangement : xi − xj = 1. • Catalan arrangement : xi − xj = −1, 0, 1. • semiorder arrangement : xi − xj = −1, 1. • threshold arrangement : xi + xj = 0 (not really a deformation of the braid arrangement, but closely related). An arrangement A is central if H⊆A H = �. Equivalently, A is a translate ⇔ of a linear arrangement (an arrangement of linear hyperplanes, i.e., hyperplanes passing through the origin). Many other writers call an arrangement central, rather than linear, if 0 ≤ H⊆A H. If A is central with X = H⊆A H, then rank( A) = codim( X). If A is central, then note also that b(A) = 0 [why?]. There are two useful arrangements closely related to a given arrangement A. If A is a linear arrangement in Kn, then projectivize A by choosing some H ≤ A to be the hyperplane at infinity in projective space P n−1 . Thus if we regard K P n−1 = {(x1, . . . , x n) : xi ≤ K, not all xi = 0}/ ∪,K where u ∪ v if u = κv for some 0 = κ ≤ K, then ⇔ = P n−2 H = ({(x1, . . . , x n−1, 0) : xi ≤ K, not all xi = 0}/ ∪) ∪ K . The remaining hyperplanes in A then correspond to “finite” (i.e., not at infinity) projective hyperplanes in P n−1 . This gives an arrangement proj( A) of hyperplanes K in A r A 1 proj( ). Hence (proj( )) = P n−1 . When K = R, the two regions R and −R of A become identified in K r(A). When n = 3, we can draw P 2 as a disk with 2 R antipodal boundary points identified. The circumference of the disk represents the hyperplane at infinity. This provides a good way to visualize three-dimensional real linear arrangements. For instance, if A consists of the three coordinate hyperplanes x1 = 0, x2 = 0, and x3 = 0, then a projective drawing is given by 2 13 The line labelled i is the projectivization of the hyperplane xi = 0. The hyperplane at infinity is x3 = 0. There are four regions, so r(A) = 8. To draw the incidences among all eight regions of A, simply “reflect” the interior of the disk to the exterior: 6 R. STANLEY, HYPERPLANE ARRANGEMENTS 23 14 13 24 12 34 Figure 1. Aprojectivization of the braid arrangement B4 2 1312 Regarding this diagram as a planar graph, the dual graph is the 3-cube (i.e., the vertices and edges of a three-dimensional cube) [why?]. For a more complicated example of projectivization, Figure 1 shows proj( B4) (where we regard B4 as a three-dimensional arrangement contained in the hyper plane x1 + x2 + x3 + x4 = 0 of R4), with the hyperplane xi = xj labelled ij , and with x1 = x4 as the hyperplane at infinity. 7LECTURE 1. BASIC DEFINITIONS We now define an operation which is “inverse” to projectivization. Let A be an (affine) arrangement in Kn, given by the equations L1(x) = a1, . . . , Lm(x) = am. Introduce a new coordinate y, and define a central arrangement cA (the cone over A) in Kn × K = Kn+1 by the equations L1(x) = a1y, . . . , Lm(x) = amy, y = 0. For instance, let A be the arrangement in R1 given by x = −1, x = 2, and x = 3. The following figure should explain why cA is called a cone. −1 32 It is easy to see that when K = R, we have r(cA) = 2r(A). In general, cA has the “same combinatorics as A, times 2.” See Exercise 1. 1.2. The intersection poset Recall that a poset (short for partially ordered set) is a set P and a relation → satisfying the following axioms (for all x, y, z ≤ P ): (P1) (reflexivity) x x→ (P2) (antisymmetry) If x → y and y → x, then x = y. (P3) (transitivity) If x → y and y → z, then x z.→ Obvious notation such as x < y for x → y and x = y, and y ⊂ x for x → y will be ⇔ used throughout. If x → y in P , then the (closed) interval [x, y ] is defined by [x, y ] = {z ≤ P : x → z → y}. Note that the empty set � is not a closed interval. For basic information on posets not covered here, see [18 ]. Definition 1.1. Let A be an arrangement in V , and let L(A) be the set of all nonempty intersections of hyperplanes in A, including V itself as the intersection over the empty set. Define x → y in L(A) if x ∀ y (as subsets of V ). In other words, L(A) is partially ordered by reverse inclusion. We call L(A) the intersection poset of A. Note. The primary reason for ordering intersections by reverse inclusion rather than ordinary inclusion is Proposition 3.8. We don’t want to alter the well-established definition of a geometric lattice or to refer constantly to “dual geometric lattices.” The element V ≤ L(A) satisfies x ⊂ V for all x ≤ L(A). In general, if P is a poset then we denote by ˆ 0 for all 0 an element (necessarily unique) such that x ⊂ ˆ8 R. STANLEY, HYPERPLANE ARRANGEMENTS Figure 2. Examples of intersection posets x ≤ P . We say that y covers x in a poset P , denoted x � y, if x < y and no z ≤ P satisfies x < z < y. Every finite poset is determined by its cover relations. The (Hasse) diagram of a finite poset is obtained by drawing the elements of P as dots, with x drawn lower than y if x < y , and with an edge between x and y if x � y. Figure 2 illustrates four arrangements A in R2, with (the diagram of) L(A) drawn below A. A chain of length k in a poset P is a set x0 < x1 < < xk of elements of · · · P . The chain is saturated if x0 � x1 � � xk . We say that P is graded of rank · · · n if every maximal chain of P has length n. In this case P has a rank function rk : P ∃ N defined by: • rk( x) = 0 if x is a minimal element of P . • rk( y) = rk( x) + 1 if x � y in P . If x < y in a graded poset P then we write rk( x, y ) = rk( y) − rk( x), the length of the interval [x, y ]. Note that we use the notation rank( A) for the rank of an arrangement A but rk for the rank function of a graded poset. = Kn Proposition 1.1. Let A be an arrangement in a vector space V ∪ . Then the intersection poset L(A) is graded of rank equal to rank (A). The rank function of L(A) is given by rk( x) = codim( x) = n − dim( x), where dim (x) is the dimension of x as an affine subspace of V . Proof. Since L(A) has a unique minimal element ˆ = V , it suffices to show that 0 (a) if x�y in L(A) then dim( x)−dim( y) = 1, and (b) all maximal elements of L(A) have dimension n − rank( A). By linear algebra, if H is a hyperplane and x an affine subspace, then H ⊕ x = x or dim( x) − dim( H ⊕ x) = 1, so (a) follows. Now suppose that x has the largest codimension of any element of L(A), say codim( x) = d. Thus x is an intersection of d linearly independent hyperplanes (i.e., their normals are linearly independent) H1, . . . , H d in A. Let y ≤ L(A) with e = codim( y) < d . Thus y is an intersection of e hyperplanes, so some Hi (1 → i d) is linearly independent → from them. Then y ⊕ Hi = � and codim( y ⊕ Hi) > codim( y). Hence y is not a⇔ maximal element of L(A), proving (b). �9LECTURE 1. BASIC DEFINITIONS 1−1 −1 −1 112−2 −1 1 Figure 3. An intersection poset and M¨ obius function values 1.3. The characteristic polynomial A poset P is locally finite if every interval [x, y ] is finite. Let Int( P ) denote the set of all closed intervals of P . For a function f : Int( P ) ∃ Z, write f (x, y ) for f ([ x, y ]). We now come to a fundamental invariant of locally finite posets. Definition 1.2. Let P be a locally finite poset. Define a function μ = μP : Int( P ) ∃ Z, called the M¨ obius function of P , by the conditions: μ(x, x ) = 1, for all x ≤ P (2) μ(x, y ) = � μ(x, z ), for all x < y in P. − x⊇z<y This second condition can also be written � μ(x, z ) = 0, for all x < y in P. x⊇z⊇y 0, then we write μ(x) = μ(ˆIf P has a ˆ 0, x ). Figure 3 shows the intersection poset L of the arrangement A in K3 (for any field K) defined by QA(x) = xyz (x + y), together with the value μ(x) for all x ≤ L. A important application of the M¨ obius inversion for obius function is the M¨ mula . The best way to understand this result (though it does have a simple direct proof) requires the machinery of incidence algebras. Let I(P ) = I(P, K ) denote the vector space of all functions f : Int( P ) ∃ K. Write f (x, y ) for f ([ x, y ]). For f, g ≤ I(P ), define the product f g ≤ I(P ) by f g (x, y ) = � f (x, z )g(z, y ). x⊇z⊇y It is easy to see that this product makes I(P ) an associative Q-algebra, with mul tiplicative identity ζ given by � 1, x = y ζ(x, y ) = 0, x < y. Define the zeta function α ≤ I(P ) of P by α(x, y ) = 1 for all x → y in P . Note that the M¨ obius function μ is an element of I(P ). The definition of μ (Definition 1.2) is � 10 R. STANLEY, HYPERPLANE ARRANGEMENTS equivalent to the relation μα = ζ in I(P ). In any finite-dimensional algebra over a field, one-sided inverses are two-sided inverses, so μ = α−1 in I(P ). Theorem 1.1. Let P be a finite poset with M¨ obius function μ, and let f, g : P ∃ K. Then the following two conditions are equivalent: f (x) = � y∗x g(y), for all x ≤ P g(x) = � μ(x, y )f (y), for all x ≤ P. y∗x Proof. The set KP of all functions P ∃ K forms a vector space on which I(P ) acts (on the left) as an algebra of linear transformations by (�f )( x) = � �(x, y )f (y), y∗x where f ≤ KP and � ≤ I(P ). The M¨ obius inversion formula is then nothing but the statement αf = g √ f = μg. We now come to the main concept of this section. Definition 1.3. The characteristic polynomial ψA(t) of the arrangement A is de fined by (3) ψA(t) = � μ(x)tdim( x). x⊆L(A) For instance, if A is the arrangement of Figure 3, then ψA(t) = t3 − 4t2 + 5t − 2 = (t − 1) 2(t − 2) . Note that we have immediately from the definition of ψA(t), where A is in Kn , that ψA(t) = tn − (# A)tn−1 + .· · · Example 1.4. Consider the coordinate hyperplane arrangement A with defining polynomial QA(x) = xn. Every subset of the hyperplanes in A has ax1x2 · · · different nonempty intersection, so L(A) is isomorphic to the boolean algebra Bn of all subsets of [n] = {1, 2, . . . , n }, ordered by inclusion. Proposition 1.2. Let A be given by the above example. Then ψA(t) = (t − 1) n . Proof. The computation of the M¨ obius function of a boolean algebra is a standard result in enumerative combinatorics with many proofs. We will give here a naive proof from first principles. Let y ≤ L(A), r(y) = k. We claim that (4) μ(y) = (−1) k . The assertion is clearly true for rk( y) = 0, when y = ˆ 0. We need to 0. Now let y > ˆ show that (5) �(−1) rk( x) = 0. x⊇y11 LECTURE 1. BASIC DEFINITIONS ion The number of x such that x → y and rk( x) = i is ⎜k�, so (5) is equivalent to the i well-known identity (easily proved by substituting q = −1 in the binomial expans of (q + 1) k ) � i k =0 (−1) i⎜k� = 0 for k > 0. �i
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(Editor) Section 9.5 Calculus and Polar Functions The previous section defined polar coordinates, leading to polar functions. We investigated plotting these functions and solving a fundamental question about their graphs, namely, where do two polar graphs intersect? We now turn our attention to answering other questions, whose solutions require the use of calculus. A basis for much of what is done in this section is the ability to turn a polar function (r=f(\theta)) into a set of parametric equations. Using the identities (x=r\cos(\theta)) and (y=r\sin(\theta)\text{,}) we can create the parametric equations (x=f(\theta)\cos(\theta)\text{,}) (y=f(\theta)\sin(\theta)) and apply the concepts of Section 9.3. Subsection 9.5.1 Polar Functions and (dy/dx) We are interested in the lines tangent to a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is (\frac{dy}{dx}\text{.}) Given (r=f(\theta)\text{,}) we are generally not concerned with (r\,'=\fp(\theta)\text{;}) that describes how fast (r) changes with respect to (\theta\text{.}) Instead, we will use (x=f(\theta)\cos(\theta)\text{,}) (y=f(\theta)\sin(\theta)) to compute (\frac{dy}{dx}\text{.}) Using Key Idea 9.3.2 we have \begin{equation} \frac{dy}{dx} = \frac{dy}{d\theta}\Big/\frac{dx}{d\theta}\text{.} \end{equation} Each of the two derivatives on the right hand side of the equality requires the use of the Product Rule. We state the important result as a Key Idea. Key Idea 9.5.2. Finding (\frac{dy}{dx}) with Polar Functions. Let (r=f(\theta)) be a polar function. With (x=f(\theta)\cos(\theta)) and (y=f(\theta)\sin(\theta)\text{,}) \begin{equation} \frac{dy}{dx} = \frac{\fp(\theta)\sin(\theta) +f(\theta)\cos(\theta) }{\fp(\theta)\cos(\theta) -f(\theta)\sin(\theta) }\text{.} \end{equation} Example 9.5.3. Finding (\frac{dy}{dx}) with polar functions. Consider the limaçon (r=1+2\sin(\theta)) on ([0,2\pi]\text{.}) Find the equations of the tangent and normal lines to the graph at (\theta=\pi/4\text{.}) Find where the graph has vertical and horizontal tangent lines. Solution We start by computing (\frac{dy}{dx}\text{.}) With (\fp(\theta) = 2\cos(\theta)\text{,}) we have \begin{align} \frac{dy}{dx} \amp = \frac{2\cos(\theta) \sin(\theta) + \cos(\theta) (1+2\sin(\theta) )}{2\cos^2(\theta) -\sin(\theta) (1+2\sin(\theta) )}\ \amp = \frac{\cos(\theta) (4\sin(\theta) +1)}{2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) }\text{.} \end{align} When (\theta=\pi/4\text{,}) (\frac{dy}{dx}=-2\sqrt{2}-1) (this requires a bit of simplification). In rectangular coordinates, the point on the graph at (\theta=\pi/4) is ((1+\sqrt{2}/2,1+\sqrt{2}/2)\text{.}) Thus the rectangular equation of the line tangent to the limaçon at (\theta=\pi/4) is \begin{equation} y=(-2\sqrt{2}-1)\big(x-(1+\sqrt{2}/2)\big)+1+\sqrt{2}/2 \approx -3.83 x+8.24\text{.} \end{equation} The limaçon and the tangent line are graphed in Figure 9.5.4. The normal line has the opposite-reciprocal slope as the tangent line, so its equation is \begin{equation} y \approx \frac{1}{3.83}x+1.26\text{.} \end{equation} 2. To find the horizontal lines of tangency, we find where (\frac{dy}{dx}=0\text{;}) thus we find where the numerator of our equation for (\frac{dy}{dx}) is 0. \begin{equation} \cos(\theta) (4\sin(\theta) +1)=0 \Rightarrow \cos(\theta) =0 \text{ or } 4\sin(\theta) +1=0\text{.} \end{equation} On ([0,2\pi]\text{,}) (\cos(\theta) =0) when (\theta=\pi/2,\,3\pi/2\text{.}) Setting (4\sin(\theta) +1=0) gives (\theta=\sin^{-1}(-1/4)\approx -0.2527 = -14.48^\circ\text{.}) We want the results in ([0,2\pi]\text{;}) we also recognize there are two solutions, one in the third quadrant and one in the fourth. Using reference angles, we have our two solutions as (\theta =3.39) and (6.03) radians. The four points we obtained where the limaçon has a horizontal tangent line are given in Figure 9.5.4 with black-filled dots. To find the vertical lines of tangency, we set the denominator of (\frac{dy}{dx}=0\text{.}) \begin{align} 2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 .\ \end{align} Convert the (\cos^2(\theta)) term to (1-\sin^2(\theta)\text{:}) \begin{align} 2(1-\sin^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0\ 4\sin^2(\theta) + \sin(\theta) -2 \amp = 0.\ \end{align} Recognize this as a quadratic in the variable (\sin(\theta)\text{.}) Using the quadratic formula, we have \begin{align} \sin(\theta) \amp = \frac{-1\pm\sqrt{33}}{8}\text{.} \end{align} We solve (\sin(\theta) = \frac{-1+\sqrt{33}}8) and (\sin(\theta) = \frac{-1-\sqrt{33}}8\text{:}) \begin{align} \sin(\theta) \amp =\frac{-1+\sqrt{33}}8 \amp \sin(\theta) \amp = \frac{-1-\sqrt{33}}{8}\ \theta \amp = \sin^{-1}\left(\frac{-1+\sqrt{33}}8\right) \amp \theta \amp = \sin^{-1}\left(\frac{-1-\sqrt{33}}8\right)\ \theta \amp = 0.6349 \amp \theta \amp = -1.0030 \end{align} In each of the solutions above, we only get one of the possible two solutions as (\sin^{-1}(x)) only returns solutions in ([-\pi/2,\pi/2]\text{,}) the (4)th and (1)st quadrants. Again using reference angles, we have: \begin{equation} \sin\theta = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6349,\,2.5067 \text{ radians } \end{equation} and \begin{equation} \sin(\theta) = \frac{-1-\sqrt{33}}8 \Rightarrow \theta = 4.1446,\,5.2802 \text{ radians. } \end{equation} These points are also shown in Figure 9.5.4 with white-filled dots. When the graph of the polar function (r=f(\theta)) intersects the pole, it means that (f(\alpha)=0) for some angle (\alpha\text{.}) Thus the formula for (\frac{dy}{dx}) in such instances is very simple, reducing simply to \begin{equation} \frac{dy}{dx} = \tan\alpha\text{.} \end{equation} This equation makes an interesting point. It tells us the slope of the tangent line at the pole is (\tan\alpha\text{;}) some of our previous work (see, for instance, Example 9.4.9) shows us that the line through the pole with slope (\tan\alpha) has polar equation (\theta=\alpha\text{.}) Thus when a polar graph touches the pole at (\theta=\alpha\text{,}) the equation of the tangent line at the pole is (\theta=\alpha\text{.}) Example 9.5.5. Finding tangent lines at the pole. Let (r=1+2\sin(\theta)\text{,}) a limaçon. Find the equations of the lines tangent to the graph at the pole. Solution We need to know when (r=0\text{.}) \begin{align} 1+2\sin(\theta) \amp = 0\ \sin(\theta) \amp = -1/2\ \theta \amp = \frac{7\pi}{6},\,\frac{11\pi}6\text{.} \end{align} Thus the equations of the tangent lines, in polar, are (\theta = 7\pi/6) and (\theta = 11\pi/6\text{.}) In rectangular form, the tangent lines are (y=\tan(7\pi/6)x) and (y=\tan(11\pi/6)x\text{.}) The full limaçon can be seen in Figure 9.5.4; we zoom in on the tangent lines in Figure 9.5.6. Subsection 9.5.2 Area When using rectangular coordinates, the equations (x=h) and (y=k) defined vertical and horizontal lines, respectively, and combinations of these lines create rectangles (hence the name “rectangular coordinates”). It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral. When using polar coordinates, the equations (\theta=\alpha) and (r=c) form lines through the origin and circles centered at the origin, respectively, and combinations of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles. Consider Figure 9.5.7.(a) where a region defined by (r=f(\theta)) on ([\alpha,\beta]) is given. (Note how the “sides” of the region are the lines (\theta=\alpha) and (\theta=\beta\text{,}) whereas in rectangular coordinates the “sides” of regions were often the vertical lines (x=a) and (x=b\text{.})) Partition the interval ([\alpha,\beta]) into (n) equally spaced subintervals as (\alpha = \theta_1 \lt \theta_2 \lt \cdots \lt \theta_{n+1}=\beta\text{.}) The length of each subinterval is (\Delta\theta = (\beta-\alpha)/n\text{,}) representing a small change in angle. The area of the region defined by the (i)th subinterval ([\theta_i,\theta_{i+1}]) can be approximated with a sector of a circle with radius (f(c_i)\text{,}) for some (c_i) in ([\theta_i,\theta_{i+1}]\text{.}) The area of this sector is (\frac12f(c_i)^2\Delta\theta\text{.}) This is shown in Figure 9.5.7.(b), where ([\alpha,\beta]) has been divided into 4 subintervals. We approximate the area of the whole region by summing the areas of all sectors: \begin{equation} \text{ Area } \approx \sum_{i=1}^n \frac12f(c_i)^2\Delta\theta\text{.} \end{equation} This is a Riemann sum. By taking the limit of the sum as (n\to\infty\text{,}) we find the exact area of the region in the form of a definite integral. Figure 9.5.8. Video presentation of Theorem 9.5.9 Theorem 9.5.9. Area of a Polar Region. Let (f) be continuous and non-negative on ([\alpha,\beta]\text{,}) where (0\leq \beta-\alpha\leq 2\pi\text{.}) The area (A) of the region bounded by the curve (r=f(\theta)) and the lines (\theta=\alpha) and (\theta=\beta) is \begin{equation} A \,=\, \frac12\int_\alpha^\beta f(\theta)^2 \, d\theta\, =\, \frac12\int_\alpha^\beta r^{\,2} \, d\theta \end{equation} The theorem states that (0\leq \beta-\alpha\leq 2\pi\text{.}) This ensures that region does not overlap itself, which would give a result that does not correspond directly to the area. Example 9.5.10. Area of a polar region. Find the area of the circle defined by (r=\cos(\theta)\text{.}) (Recall this circle has radius (1/2\text{.})) Solution This is a direct application of Theorem 9.5.9. The circle is traced out on ([0,\pi]\text{,}) leading to the integral \begin{align} \text{ Area } \amp = \frac12\int_0^\pi \cos^2(\theta)\, d\theta\ \amp = \frac12\int_0^\pi \frac{1+\cos(2\theta)}{2}\, d\theta\ \amp = \frac14\big(\theta +\frac12\sin(2\theta)\big)\Bigg\|_0^\pi\ \amp = \frac14\pi\text{.} \end{align} Of course, we already knew the area of a circle with radius (1/2\text{.}) We did this example to demonstrate that the area formula is correct. Example 9.5.11. Area of a polar region. Find the area of the cardioid (r=1+\cos(\theta)) bound between (\theta=\pi/6) and (\theta=\pi/3\text{,}) as shown in Figure 9.5.12. Figure 9.5.12. Finding the area of the shaded region of a cardioid in Example 9.5.11 Solution This is again a direct application of Theorem 9.5.9. \begin{align} \text{ Area } \amp = \frac12\int_{\pi/6}^{\pi/3} (1+\cos(\theta) )^2\, d\theta\ \amp = \frac12\int_{\pi/6}^{\pi/3} (1+2\cos(\theta) +\cos^2(\theta) )\, d\theta\ \amp = \frac12\left(\theta+2\sin(\theta) +\frac12\theta+\frac14\sin(2\theta)\right)\Bigg\|_{\pi/6}^{\pi/3}\ \amp = \frac18\big(\pi+4\sqrt{3}-4\big) \approx 0.7587\text{.} \end{align} Area Between Curves. Our study of area in the context of rectangular functions led naturally to finding area bounded between curves. We consider the same in the context of polar functions. Consider the shaded region shown in Figure 9.5.13. We can find the area of this region by computing the area bounded by (r_2=f_2(\theta)) and subtracting the area bounded by (r_1=f_1(\theta)) on ([\alpha,\beta]\text{.}) Thus \begin{equation} \text{ Area } \,= \,\frac12\int_\alpha^\beta r_2^{\,2}\, d\theta - \frac12\int_\alpha^\beta r_1^{\,2}\, d\theta = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta\text{.} \end{equation} Figure 9.5.13. Illustrating area bound between two polar curves Key Idea 9.5.14. Area Between Polar Curves. The area (A) of the region bounded by (r_1=f_1(\theta)) and (r_2=f_2(\theta)\text{,}) (\theta=\alpha) and (\theta=\beta\text{,}) where (f_1(\theta)\leq f_2(\theta)) on ([\alpha,\beta]\text{,}) is \begin{equation} A = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\, d\theta\text{.} \end{equation} Example 9.5.15. Area between polar curves. Find the area bounded between the curves (r=1+\cos(\theta)) and (r=3\cos(\theta)\text{,}) as shown in Figure 9.5.16. Figure 9.5.16. Finding the area between polar curves in Example 9.5.15 Solution We need to find the points of intersection between these two functions. Setting them equal to each other, we find: \begin{align} 1+\cos(\theta) \amp = 3\cos(\theta)\ \cos(\theta) \amp =1/2\ \theta \amp = \pm \pi/3 \end{align} Thus we integrate (\frac12\big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)) on ([-\pi/3,\pi/3]\text{.}) \begin{align} \text{ Area } \amp = \frac12\int_{-\pi/3}^{\pi/3} \big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)\, d\theta\ \amp = \frac12\int_{-\pi/3}^{\pi/3} \big( 8\cos^2(\theta) -2\cos(\theta) -1\big)\, d\theta\ \amp = \frac12\big(2\sin(2\theta) - 2\sin(\theta) +3\theta\big)\Bigg\|_{-\pi/3}^{\pi/3}\ \amp = \pi\text{.} \end{align} Amazingly enough, the area between these curves has a “nice” value. Example 9.5.17. Area defined by polar curves. Find the area bounded between the polar curves (r=1) and (r=2\cos(2\theta)\text{,}) as shown in Figure 9.5.18. Figure 9.5.18. The region bounded by the functions in Example 9.5.17 Solution We need to find the point of intersection between the two curves. Setting the two functions equal to each other, we have \begin{equation} 2\cos(2\theta) = 1 \Rightarrow \cos(2\theta) = \frac12 \Rightarrow 2\theta = \pi/3 \Rightarrow \theta=\pi/6\text{.} \end{equation} Figure 9.5.19. Breaking the region bounded by the functions in Example 9.5.17 into its component parts In Figure 9.5.19, we zoom in on the region and note that it is not really bounded between two polar curves, but rather by two polar curves, along with (\theta=0\text{.}) The dashed line breaks the region into its component parts. Below the dashed line, the region is defined by (r=1\text{,}) (\theta=0) and (\theta = \pi/6\text{.}) (Note: the dashed line lies on the line (\theta=\pi/6\text{.})) Above the dashed line the region is bounded by (r=2\cos(2\theta)) and (\theta =\pi/6\text{.}) Since we have two separate regions, we find the area using two separate integrals. Call the area below the dashed line (A_1) and the area above the dashed line (A_2\text{.}) They are determined by the following integrals: \begin{equation} A_1 = \frac12\int_0^{\pi/6} (1)^2\, d\theta\qquad A_2 = \frac12\int_{\pi/6}^{\pi/4} \big(2\cos(2\theta)\big)^2\, d\theta\text{.} \end{equation} (The upper bound of the integral computing (A_2) is (\pi/4) as (r=2\cos(2\theta)) is at the pole when (\theta=\pi/4\text{.})) We omit the integration details and let the reader verify that (A_1 = \pi/12) and (A_2 = \pi/12-\sqrt{3}/8\text{;}) the total area is (A = \pi/6-\sqrt{3}/8\text{.}) Subsection 9.5.3 Arc Length As we have already considered the arc length of curves defined by rectangular and parametric equations, we now consider it in the context of polar equations. Recall that the arc length (L) of the graph defined by the parametric equations (x=f(t)\text{,}) (y=g(t)) on ([a,b]) is \begin{equation} L = \int_a^b \sqrt{\fp(t)^2 + g'(t)^2}\, dt = \int_a^b \sqrt{x'(t)^2+y'(t)^2}\, dt\text{.}\label{eq_polar_arclength}\tag{9.5.1} \end{equation} Now consider the polar function (r=f(\theta)\text{.}) We again use the identities (x=f(\theta)\cos(\theta)) and (y=f(\theta)\sin(\theta)) to create parametric equations based on the polar function. We compute (x'(\theta)) and (y'(\theta)) as done before when computing (\frac{dy}{dx}\text{,}) then apply Equation (9.5.1). The expression (x'(\theta)^2+y'(\theta)^2) can be simplified a great deal; we leave this as an exercise and state that \begin{equation} x'(\theta)^2+y'(\theta)^2 = \fp(\theta)^2+f(\theta)^2\text{.} \end{equation} This leads us to the arc length formula. Theorem 9.5.20. Arc Length of Polar Curves. Let (r=f(\theta)) be a polar function with (\fp) continuous on ([\alpha,\beta]\text{,}) on which the graph traces itself only once. The arc length (L) of the graph on ([\alpha,\beta]) is \begin{equation} L = \int_\alpha^\beta \sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta = \int_\alpha^\beta\sqrt{(r\,')^2+ r^2}\, d\theta\text{.} \end{equation} Example 9.5.21. Arc length of a limaçon. Find the arc length of the limaçon (r=1+2\sin(t)\text{.}) Solution With (r=1+2\sin(t)\text{,}) we have (r\,' = 2\cos(t)\text{.}) The limaçon is traced out once on ([0,2\pi]\text{,}) giving us our bounds of integration. Applying Theorem 9.5.20, we have \begin{align} L \amp = \int_0^{2\pi} \sqrt{(2\cos\theta)^2+(1+2\sin\theta)^2}\, d\theta\ \amp = \int_0^{2\pi} \sqrt{4\cos^2\theta+4\sin^2\theta +4\sin\theta+1}\, d\theta\ \amp = \int_0^{2\pi} \sqrt{4\sin\theta+5}\, d\theta\ \amp \approx 13.3649\text{.} \end{align} Figure 9.5.22. The limaçon in Example 9.5.21 whose arc length is measured The final integral cannot be solved in terms of elementary functions, so we resorted to a numerical approximation. (Simpson's Rule, with (n=4\text{,}) approximates the value with (13.0608\text{.}) Using (n=22) gives the value above, which is accurate to 4 places after the decimal.) Subsection 9.5.4 Surface Area The formula for arc length leads us to a formula for surface area. The following Theorem is based on Theorem 9.3.21. Theorem 9.5.23. Surface Area of a Solid of Revolution. Consider the graph of the polar equation (r=f(\theta)\text{,}) where (\fp) is continuous on ([\alpha,\beta]\text{,}) on which the graph does not cross itself. The surface area of the solid formed by revolving the graph about the initial ray ((\theta=0)) is: \begin{equation} \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\sin(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta\text{.} \end{equation} 2. The surface area of the solid formed by revolving the graph about the line (\theta=\pi/2) is: \begin{equation} \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\cos(\theta)\sqrt{\fp(\theta)^2+f(\theta)^2}\, d\theta\text{.} \end{equation} Example 9.5.24. Surface area determined by a polar curve. Find the surface area formed by revolving one petal of the rose curve (r=\cos(2\theta)) about its central axis, as shown in Figure 9.5.25. (a) (b) Figure 9.5.25. Finding the surface area of a rose-curve petal that is revolved around its central axis Solution We choose, as implied by the figure, to revolve the portion of the curve that lies on ([0,\pi/4]) about the initial ray. Using Theorem 9.5.23 and the fact that (\fp(\theta) = -2\sin(2\theta)\text{,}) we have \begin{align} \text{ Surface Area } \amp = 2\pi\int_0^{\pi/4} \cos(2\theta)\sin(\theta)\sqrt{\big(-2\sin(2\theta)\big)^2+\big(\cos(2\theta)\big)^2}\, d\theta\ \amp \approx 1.36707\text{.} \end{align} The integral is another that cannot be evaluated in terms of elementary functions. Simpson's Rule, with (n=4\text{,}) approximates the value at (1.36751\text{.}) This chapter has been about curves in the plane. While there is great mathematics to be discovered in the two dimensions of a plane, we live in a three dimensional world and hence we should also look to do mathematics in 3D — that is, in space. The next chapter begins our exploration into space by introducing the topic of vectors, which are incredibly useful and powerful mathematical objects. Exercises 9.5.5 Exercises Terms and Concepts 1. Given polar equation (r=f(\theta)\text{,}) how can one create parametric equations of the same curve? 2. With rectangular coordinates, it is natural to approximate area with ; with polar coordinates, it is natural to approximate area with . In the following exercises, find (\lz{y}{x}) (in terms of (\theta)). Then find the equations of the tangent and normal lines to the curve at the indicated (\theta)-value. 3. For the polar equation (r=1\text{,}) find (\lz{y}{x}) in terms of (\theta\text{.}) Type ‘theta’ for (\theta\text{.}) (\lz{y}{x}=) At (\theta = \pi/4\text{,}) the tangent line has equation and the normal line has equation . 4. For the polar equation (r=\cos(\theta)\text{,}) find (\lz{y}{x}) in terms of (\theta\text{.}) Type ‘theta’ for (\theta\text{.}) (\lz{y}{x}=) At (\theta = \pi/4\text{,}) the tangent line has equation and the normal line has equation . 5. (r=1+\sin(\theta)\text{;})(\theta = \pi/6) 6. (\ds r=1-3\cos(\theta)\text{;})(\theta = 3\pi/4) 7. For the polar equation (r=\theta\text{,}) find (\lz{y}{x}) in terms of (\theta\text{.}) Type ‘theta’ for (\theta\text{.}) (\lz{y}{x}=) At (\theta = \pi/2\text{,}) the tangent line has equation and the normal line has equation . 8. For the polar equation (r=\cos(3\theta)\text{,}) find (\lz{y}{x}) in terms of (\theta\text{.}) Type ‘theta’ for (\theta\text{.}) (\lz{y}{x}=) At (\theta = \pi/6\text{,}) the tangent line has equation and the normal line has equation . 9. For the polar equation (r=\sin(4\theta)\text{,}) find (\lz{y}{x}) in terms of (\theta\text{.}) Type ‘theta’ for (\theta\text{.}) (\lz{y}{x}=) At (\theta = \pi/3\text{,}) the tangent line has equation and the normal line has equation . 10. (\ds r=\frac1{\sin(\theta) -\cos(\theta) }\text{;})(\theta = \pi) In the following exercises, find the values of (\theta) in the given interval where the graph of the polar function has horizontal and vertical tangent lines. 11. (\ds r=3\text{;})([0,2\pi]) 12. (\ds r=2\sin(\theta)\text{;})([0,\pi]) 13. (\ds r=\cos(2\theta)\text{;})([0,2\pi]) 14. Find the values of (\theta) in ([0,2\pi)) where the graph of (r=1+\cos(\theta)) has horizontal and vertical tangent lines. Horizontal tangent lines for (\theta\in\bigg{)(\bigg}) Vertical tangent lines for (\theta\in\bigg{)(\bigg}) In the following exercises, find the equation of the lines tangent to the graph at the pole. 15. (\ds r=\sin(\theta)\text{;})([0,\pi]) 16. (\ds r=\sin(3\theta)\text{;})([0,\pi]) In the following exercises, find the area of the described region. 17. Enclosed by the circle: (\ds r=4\sin(\theta)) 18. Enclosed by the circle (\ds r=5) 19. Find the area enclosed by one petal of (r=\sin(3\theta)\text{.}) 20. Enclosed by one petal of the rose curve (r=\cos (n\,\theta)\text{,}) where (n) is a positive integer. 21. Find the area enclosed by the cardioid (r=1-\sin(\theta)\text{.}) 22. Enclosed by the inner loop of the limacon (\ds r=1+2\cos(\theta)) 23. Find the area enclosed by the outer loop of the limacon (r=1+2\cos(\theta)) (including area enclosed by the inner loop). 24. Find the area enclosed between the inner and outer loop of the limacon (r=1+2\cos(\theta)\text{.}) 25. Find the area enclosed by (r=2\cos(\theta)\text{,}) (r=2\sin(\theta)\text{,}) and the (x)-axis, as shown: The area is . 26. Enclosed by (r=\cos(3 \theta)) and (r=\sin(3\theta)\text{,}) as shown: 27. Find the area enclosed by (r=\cos(\theta)) and (r=\sin(2\theta)\text{,}) as shown: The area is . In the following exercises, answer the questions involving arc length. 28. Use the arc length formula to compute the arc length of the circle (r=2\text{.}) 29. Use the arc length formula to compute the arc length of the circle (r=4\sin(\theta)\text{.}) 30. Use the arc length formula to compute the arc length of (r=\cos \theta+\sin \theta\text{.}) 31. Use the arc length formula to compute the arc length of the cardioid (r=1+\cos\theta\text{.}) (Hint: apply the formula, simplify, then use a Power-Reducing Formula to convert (1+\cos \theta) into a square.) 32. Approximate the arc length of one petal of the rose curve (r=\sin(3\theta)) with Simpson’s Rule and (n=4\text{.}) 33. Let (x(\theta) = f(\theta)\cos(\theta)) and (y(\theta)=f(\theta)\sin(\theta)\text{.}) Show, as suggested by the text, that \begin{equation} x\,'(\theta)^2+y\,'(\theta)^2 = \fp(\theta)^2+f(\theta)^2\text{.} \end{equation} In the following exercises, answer the questions involving surface area. 34. Use Theorem 9.5.23 to find the surface area of the sphere formed by revolving the circle (r=2) about the initial ray. 35. Use Theorem 9.5.23 to find the surface area of the sphere formed by revolving the circle (r=2\cos(\theta)) about the initial ray. 36. Find the surface area of the solid formed by revolving the cardioid (r=1+\cos(\theta)) about the initial ray. 37. Find the surface area of the solid formed by revolving the circle (r=2\cos(\theta)) about the line (\theta=\pi/2\text{.}) 38. Find the surface area of the solid formed by revolving the line (r=3\sec(\theta)\text{,}) (-\pi/4\leq\theta\leq\pi/4\text{,}) about the line (\theta=\pi/2\text{.}) 39. Find the surface area of the solid formed by revolving the line (r=3\sec\theta\text{,}) (0\leq\theta\leq\pi/4\text{,}) about the initial ray.
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Intro to Physics Units 1h 29m Introduction to Units 26m + Unit Conversions 18m + Solving Density Problems 13m + Dimensional Analysis 10m + Counting Significant Figures 5m + Operations with Significant Figures 14m 2. 1D Motion / Kinematics 3h 56m Vectors, Scalars, & Displacement 13m + Average Velocity 32m + Intro to Acceleration 7m + Position-Time Graphs & Velocity 26m + Conceptual Problems with Position-Time Graphs 22m + Velocity-Time Graphs & Acceleration 5m + Calculating Displacement from Velocity-Time Graphs 15m + Conceptual Problems with Velocity-Time Graphs 10m + Calculating Change in Velocity from Acceleration-Time Graphs 10m + Graphing Position, Velocity, and Acceleration Graphs 11m + Kinematics Equations 37m + Vertical Motion and Free Fall 19m + Catch/Overtake Problems 23m 3. 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Capacitors & Dielectrics 2h 2m Capacitors & Capacitance 8m + Parallel Plate Capacitors 19m + Energy Stored by Capacitor 15m + Capacitance Using Calculus 7m + Combining Capacitors in Series & Parallel 15m + Solving Capacitor Circuits 29m + Intro To Dielectrics 18m + How Dielectrics Work 2m + Dielectric Breakdown 4m 27. Resistors & DC Circuits 3h 8m Intro to Current 6m + Resistors and Ohm's Law 14m + Power in Circuits 11m + Microscopic View of Current 8m + Combining Resistors in Series & Parallel 37m + Kirchhoff's Junction Rule 4m + Solving Resistor Circuits 31m + Kirchhoff's Loop Rule 1h 14m 28. Magnetic Fields and Forces 2h 23m Magnets and Magnetic Fields 21m + Summary of Magnetism Problems 9m + Force on Moving Charges & Right Hand Rule 26m + Circular Motion of Charges in Magnetic Fields 11m + Mass Spectrometer 33m + Magnetic Force on Current-Carrying Wire 22m + Force and Torque on Current Loops 17m 29. Sources of Magnetic Field 2h 30m Magnetic Field Produced by Moving Charges 10m + Magnetic Field Produced by Straight Currents 27m + Magnetic Force Between Parallel Currents 12m + Magnetic Force Between Two Moving Charges 9m + Magnetic Field Produced by Loops andSolenoids 42m + Toroidal Solenoids aka Toroids 12m + Biot-Savart Law (Calculus) 18m + Ampere's Law (Calculus) 17m 30. Induction and Inductance 3h 38m Intro to Induction 5m + Magnetic Flux 12m + Faraday's Law 28m + Lenz's Law 22m + Motional EMF 18m + Transformers 8m + Mutual Inductance 24m + Self Inductance 18m + Inductors 7m + LR Circuits 22m + LC Circuits 35m + LRC Circuits 14m 31. Alternating Current 2h 37m Alternating Voltages and Currents 18m + RMS Current and Voltage 9m + Phasors 20m + Resistors in AC Circuits 9m + Phasors for Resistors 7m + Capacitors in AC Circuits 16m + Phasors for Capacitors 8m + Inductors in AC Circuits 13m + Phasors for Inductors 7m + Impedance in AC Circuits 18m + Series LRC Circuits 11m + Resonance in Series LRC Circuits 10m + Power in AC Circuits 5m 32. Electromagnetic Waves 2h 14m Intro to Electromagnetic (EM) Waves 23m + The Electromagnetic Spectrum 7m + Intensity of EM Waves 22m + Wavefunctions of EM Waves 19m + Radiation Pressure 24m + Polarization & Polarization Filters 30m + The Doppler Effect of Light 6m 33. Geometric Optics 2h 57m Ray Nature Of Light 10m + Reflection of Light 9m + Index of Refraction 16m + Refraction of Light & Snell's Law 22m + Total Internal Reflection 5m + Ray Diagrams For Mirrors 35m + Mirror Equation 19m + Refraction At Spherical Surfaces 9m + Ray Diagrams For Lenses 22m + Thin Lens And Lens Maker Equations 24m 34. Wave Optics 1h 15m Diffraction 8m + Diffraction with Huygen's Principle 14m + Young's Double Slit Experiment 24m + Single Slit Diffraction 27m 35. Special Relativity 2h 10m Inertial Reference Frames 14m + Special Vs. Galilean Relativity 17m + Consequences of Relativity 52m + Lorentz Transformations 45m Fluid Mechanics Density Fluid Mechanics Density: Videos & Practice Problems Video Lessons Practice Worksheet Topic summary Fluids, including liquids and gases, can change shape to fit their containers, characterized by density (ρ), defined as mass (m) divided by volume (v). Common densities include freshwater (1000 kg/m³) and air (1.2 kg/m³). Specific gravity (SG) relates a material's density to freshwater's density, expressed as SG = ρ_material / 1000. Understanding these concepts aids in solving problems involving buoyancy and weight, where weight (W) is calculated using W = mg, with g as the acceleration due to gravity. 1 concept Intro to Density Video duration: 8m Play a video: Intro to Density Video Summary Fluids, encompassing both liquids and gases, exhibit unique properties that distinguish them from solids. Unlike solids, which maintain a fixed shape, fluids can adapt their shape to fit the container they occupy. A fundamental concept in understanding fluids is density, which is defined as the mass of a substance divided by its volume. The Greek letter ρ (rho) represents density, and the formula can be expressed as: Here, m denotes mass, measured in kilograms (kg), and V represents volume, typically expressed in cubic meters (m³). Volume is calculated by multiplying three dimensions: length, width, and height, which can also be referred to as depth. Thus, the formula for volume can be summarized as: In practical applications, if the density of a material and its dimensions are known, one can easily calculate the mass using the rearranged formula: It is important to note that objects made from the same material will have the same density, regardless of their size. For instance, a small block of wood and a large block of wood will both have the same density, even though their masses and volumes differ. This consistency in density is crucial when analyzing mixtures of liquids, as liquids with higher densities will settle at the bottom of a container, while those with lower densities will rise to the top. To illustrate these concepts, consider a scenario where we need to calculate the total weight of air in a large warehouse modeled as a rectangular prism with dimensions of 100 meters in length, 100 meters in width, and 10 meters in height. The density of air is approximately 1.225 kg/m³, and the acceleration due to gravity is taken as 10 m/s². First, we calculate the volume of the warehouse: ³ Next, we find the mass of the air using the density: ³ ³ Finally, to determine the weight of the air, we apply the weight formula: ² This calculation reveals that the weight of the air in the warehouse is 1,225,000 newtons, highlighting the significant mass that can be contained within a large volume. Understanding these principles of fluid mechanics is essential for various applications in science and engineering. Study Smarter with Worksheets. Follow along with each video using our printable worksheets 2 concept Density Values & Conversions Video duration: 8m Play a video: Density Values & Conversions Video Summary Understanding density is crucial for solving various problems related to materials such as water, saltwater, blood, air, oil, and wood. Density is defined as mass per unit volume, and it is commonly expressed in several units. For instance, freshwater has a density of 1000 kilograms per cubic meter (kg/m³), which means that a volume of 1 cubic meter of freshwater has a mass of 1000 kilograms. This can also be expressed as 1 kilogram per liter (kg/L) or 1 gram per cubic centimeter (g/cm³). The conversion between these units is essential for calculations, especially when dealing with different substances. For example, saltwater has a density of approximately 1030 kg/m³, which can be converted to 1.03 kg/L or 1.03 g/cm³. Similarly, whole blood has a density of about 1060 kg/m³. The conversions rely on two key factors: 1 cm³ = 1 mL and 1 m³ = 1000 L. These relationships allow for seamless transitions between different units of measurement. When considering air at sea level, its density is significantly lower at about 1.2 kg/m³, which is approximately 800 times less dense than water. This lower density explains why we can move freely through air. Other materials like oil and wood typically have densities around 800 kg/m³, making them less dense than freshwater. To illustrate the application of density in problem-solving, consider a scenario where you have 500 mL of a liquid with a density of 2.2 g/cm³. To find the mass, you can use the formula for density, which states that density (ρ) equals mass (m) divided by volume (V): ρ = m/V. Rearranging this gives m = ρ × V. Here, the volume is 500 mL, which is equivalent to 500 cm³ (since 1 mL = 1 cm³). Substituting the values, you get: m = 2.2 g/cm³ × 500 cm³ = 1100 g To convert grams to kilograms (the SI unit), divide by 1000, resulting in 1.1 kg. To find the weight (w), use the equation w = mg, where g is the acceleration due to gravity, approximately 10 m/s² for simplification. Thus, the weight is: w = 1.1 kg × 10 m/s² = 11 N This example demonstrates how to apply density concepts and unit conversions to solve real-world problems effectively. 3 concept Specific Gravity Video duration: 4m Play a video: Specific Gravity Video Summary Specific gravity (SG) is a term that relates to the density of a material, defined as the ratio of the density of that material to the density of fresh water. The formula for specific gravity can be expressed as: SG = Since the density of fresh water is a constant value of 1,000 kg/m³, the specific gravity can also be simplified to: SG = This means that specific gravity is a dimensionless number, as the units cancel out when dividing two densities. For example, if the density of a material is 2,000 kg/m³, its specific gravity would be: SG = This indicates that the material is twice as dense as water. Conversely, if a material has a specific gravity of 0.7, its density can be calculated as: Density = SG × 1000 = Specific gravity does not directly relate to gravitational force; rather, it serves as a comparative measure of density relative to water. Understanding specific gravity is essential in various applications, such as determining whether an object will float or sink in water. To illustrate the application of specific gravity in calculations, consider a wooden cube with a specific gravity of 0.8 and a weight of 16,000 Newtons. To find the volume of the cube, we first need to determine its mass using the relationship between weight (W), mass (m), and gravitational acceleration (g): W = m × g Rearranging this gives: m = Assuming g = 10 m/s² for simplicity, we find: m = Next, we calculate the density of the wooden cube using its specific gravity: Density = SG × 1000 = Now, we can find the volume (V) using the density formula: Density = ⟹ V = Substituting the values gives: V = This calculation demonstrates how specific gravity can be used to derive other important properties of materials, reinforcing its significance in scientific and engineering contexts. 4 Problem A wooden door is 1 m wide, 2.5 m tall, 6 cm thick, and weighs 400 N. What is the density of the wood in g/cm3? (use g = 10 m/s2) A 0.0267 g/cm3 B 0.267 g/cm3 C 267 g/cm3 2670 g/cm3 Problem Suppose an 80 kg (176 lb) person has 5.5 L of blood (1,060 kg/m3 ) in their body. How much of this person's total mass consists of blood? What percentage of the person's total mass is blood? A 0.0519 kg B 0.0583 kg C 5.19 kg D 5.83 kg Problem You want to verify if a 70-g crown is in fact made of pure gold (19.32 g/cm3 ), so you lower it by a string into a deep bucket of water that is filled to the top. When the crown is completely submerged, you measure that 3.62 mL of water has overflown. Is the crown made of pure gold? A B Do you want more practice? More sets Density Fluid Mechanics 6 problems 19. Fluid Mechanics - Part 1 of 2 3 topics 8 problems Patrick 19. Fluid Mechanics - Part 2 of 2 4 topics 7 problems Chapter Go over this topic definitions with flashcards More sets Density definitions Fluid Mechanics 15 Terms Density quiz Fluid Mechanics 10 Terms Take your learning anywhere! Prep for your exams on the go with video lessons and practice problems in our mobile app. Here’s what students ask on this topic: The formula for density (ρ) is defined as the mass (m) of an object divided by its volume (V). Mathematically, it is expressed as: In calculations, if you know the mass and volume of an object, you can determine its density by dividing the mass by the volume. For example, if an object has a mass of 10 kg and a volume of 2 m³, its density would be: ³ To convert between different units of density, you need to use appropriate conversion factors. Common conversions include: 1 kg/m³ = 0.001 g/cm³ 1 g/cm³ = 1000 kg/m³ 1 kg/m³ = 0.001 kg/L For example, to convert 1000 kg/m³ to g/cm³, you multiply by 0.001: ³ ³ ³ Specific gravity (SG) is a dimensionless quantity that compares the density of a material to the density of freshwater. It is calculated using the formula: For example, if a material has a density of 2000 kg/m³, its specific gravity would be: This means the material is twice as dense as freshwater. Density plays a crucial role in determining the buoyancy of an object. An object will float in a fluid if its density is less than the density of the fluid. Conversely, it will sink if its density is greater. This is because buoyant force, which acts upward, is equal to the weight of the fluid displaced by the object. Mathematically, buoyant force (F_b) is given by: ρ Where ρ_fluid is the density of the fluid, V_displaced is the volume of fluid displaced, and g is the acceleration due to gravity. Common densities of materials include: Freshwater: 1000 kg/m³ Saltwater: 1030 kg/m³ Air: 1.2 kg/m³ Oil: ~800 kg/m³ Wood: ~600-800 kg/m³ These densities are typically expressed in kilograms per cubic meter (kg/m³), but can also be converted to grams per cubic centimeter (g/cm³) or kilograms per liter (kg/L) using appropriate conversion factors. Your Physics tutor Patrick Ford Physics and Maths lead instructor
1784	https://www.albert.io/blog/context-clues-sat-reading-and-writing-review/	Context Clues: SAT® Reading and Writing Review \| Albert Blog & Resources Skip to content Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources Log In Sign Up ➜ SAT® Reading and Writing Context Clues: SAT® Reading and Writing Review The Albert Team Last Updated On: April 16, 2025 What We Review Toggle - [x] Understanding Context Clues for the SAT® Reading and Writing Test What Are Context Clues? Types of Context Clues Connotation vs. Denotation Strategies for Using Context Clues Effectively Practice Exercise Quick Reference Vocabulary Chart Conclusion Sharpen Your Skills for SAT® Reading and Writing Need help preparing for your SAT® Reading and Writing exam? Understanding Context Clues for the SAT® Reading and Writing Test Context clues are often overlooked, but they play an essential role in reading comprehension and vocabulary skills. On the SAT® Reading and Writing Test, words in context can be tricky. Yet, with the right strategies, these questions can become much easier. Therefore, learning how to spot and interpret context clues can quickly boost test performance. This article explores various types of context clues, highlights the difference between connotation vs. denotation, and offers useful practice exercises. Start practicing SAT® Reading and Writing on Albert now! What Are Context Clues? Context clues are hints or pieces of information found near a word or phrase. These clues help the reader figure out the meaning even if the word is unfamiliar. When a passage includes an unfamiliar term, the surrounding sentences or phrases can guide you to the correct definition. As a result, context clues help readers understand complex material more quickly. When facing a tough vocabulary question on the SAT®, rely on the clues in the passage. Rather than skipping the word, look for details that point to its meaning. Clues might include examples, definitions, or even opposites of the unknown word. Types of Context Clues There are several types of context clues. Recognizing them can make reading passages less intimidating. Below are five main types: definition, synonym, antonym, examples, and inference clues. 1. Definition Clues A definition clue is one of the easiest to spot. Writers sometimes provide a direct explanation of a word. Often, this can appear in parentheses or immediately after the word. Example Sentence: “The arboretum, a place where various trees are cultivated, is a great place to visit.” Step-by-Step Solution: Identify the unfamiliar word: “arboretum.” Notice the phrase that follows: “a place where various trees are cultivated.” Recognize that this phrase provides a direct definition. Conclude that “arboretum” is a location dedicated to growing different types of trees. 2. Synonym Clues Synonyms are words with similar meanings. In many sentences, a writer includes a synonym near the unfamiliar term to help clarify it. Therefore, watch for a word that resembles the meaning of the unknown word. Example Sentence: “The immigrant felt a deep nostalgia for his homeland, filled with yearning for his past.” Step-by-Step Solution: Spot the word “nostalgia.” Observe the phrase “filled with yearning for his past.” See that “yearning” closely matches “nostalgia.” Realize that “nostalgia” refers to a strong longing for something from the past. 3. Antonym Clues Antonym clues involve a contrast. Since antonyms have opposite meanings, clues in the sentence might signal a deliberate difference. Words like “unlike,” “however,” or “in contrast” often introduce this type of clue. Example Sentence: “Unlike her gregarious brother, who loved social gatherings, she was quite reserved.” Step-by-Step Solution: Notice the word “gregarious.” Identify the contrast word: “Unlike.” See that “gregarious” is the opposite of “reserved.” Infer that “gregarious” means outgoing, sociable, or friendly. 4. Examples Clues Example clues describe the unknown word by listing specific instances. Writers often introduce these clues with phrases such as “such as,” “for example,” or “including.” Example Sentence: “The naturalist enjoyed observing wildlife, such as otters, red-tailed hawks, and elk.” Step-by-Step Solution: Look at the word “wildlife.” Notice the examples that follow: “otters, red-tailed hawks, and elk.” Realize they are animals living in nature. Understand that “wildlife” refers to animals that exist in their natural habitat. 5. Inference Clues Inference clues are slightly less direct. The meaning of the word can be understood by looking at the overall context of the sentence or paragraph. Therefore, pay close attention to the main idea. Example Sentence: “After the tumultuous events of the day, he felt a sense of calm wash over him.” Step-by-Step Solution: Identify the key word: “tumultuous.” Observe the contrast: “tumultuous events” versus “a sense of calm.” Infer that “tumultuous” suggests chaos or disorder. Conclude that the day was filled with intense or disruptive happenings. Ready to boost your SAT® Reading and Writing scores? Explore our plans and pricing here! Connotation vs. Denotation Understanding “connotation vs. denotation” helps you choose the best word in a given context. Denotation is the straightforward meaning of a word, as found in dictionaries. Connotation, however, is the emotional or cultural feeling a word carries. Paying attention to both helps when selecting precise words on the SAT®. Example of Denotation vs. Connotation: Word: “Home” Denotation: The place where one lives. Connotation: Safety, comfort, and warmth. Step-by-Step Solution: Recognize “home” refers to a physical location. Note the emotional aspect of safety and comfort. Understand how the connotation can shift the tone of a sentence. Conclude that word choice can affect the emotional response of readers. When a test question asks for the most precise word or phrase, always consider the connotation and the denotation. If the passage aims for a nostalgic feel, a word with a positive connotation might be best. But if the passage intends to sound clinical, the purely literal term might be more suitable. Strategies for Using Context Clues Effectively Read Surrounding Sentences: Writers often place context clues in the same sentence or nearby paragraphs. Therefore, do not rush; read carefully for references that point to the meaning of an unknown term. Look for Signal Words: Terms like “unlike,” “for instance,” or “in contrast” can guide you toward definitions, synonyms, or antonyms. If a sentence includes “or” before a new phrase, that phrase might be a restatement of the unknown word. Practice Regularly: Reading various texts, such as articles and essays, develops familiarity with common vocabulary. This practice also enhances reasoning skills when encountering complex passages on the SAT®. Practice Exercise Read the paragraph below. Each underlined word or phrase can be understood using context clues. Fill in the blank with the best meaning. Then, check the step-by-step solution. Paragraph: “It was a(n) unpredictable day, marked by sudden changes in weather. First, the sky was overcast (), but within minutes it became bright and clear (). Despite the erratic (______) weather, people continued their outdoor activities.” “unpredictable” = (a) uncertain or changeable, (b) standard or typical “overcast” = (a) covered with clouds, (b) delightfully sunny “bright and clear” = (a) rainy and gloomy, (b) full of sunlight and no clouds “erratic” = (a) unpredictable or inconsistent, (b) calm and steady Step-by-Step Solution “unpredictable” The paragraph states there are “sudden changes in weather.” Therefore, “unpredictable” means (a) uncertain or changeable. “overcast” Contrasted with “bright and clear,” it likely involves clouds. Hence, “overcast” is (a) covered with clouds. “bright and clear” This is the opposite of overcast. The best choice is (b) full of sunlight and no clouds. “erratic” The paragraph mentions quick shifts in weather. So, “erratic” means (a) unpredictable or inconsistent. Quick Reference Vocabulary Chart Copy and paste the table below into a Google Doc for easy reference: WordDefinition Context Clue Hints or information surrounding a word to derive meaning. Denotation The literal meaning of a word. Connotation The emotional or implied meaning of a word. Inference A conclusion drawn from the given context. Synonym A word with a similar meaning. Antonym A word with the opposite meaning. Example A specific instance that illustrates a concept. Conclusion Context clues are powerful tools for understanding unfamiliar words in a passage. By recognizing how surrounding phrases and sentences work together, it becomes easier to decode new vocabulary. This skill is especially important for tackling words in context on the SAT® Reading and Writing Test. Furthermore, remembering the distinction between connotation vs. denotation ensures the use of precise language that fits each situation. With regular practice and an awareness of these strategies, students can greatly improve their comprehension, vocabulary, analysis, and synthesis skills. Whether facing simple definition clues or more subtle inference clues, careful reading and consistent exposure to various texts will strengthen confidence. Ultimately, context clues open new doors to understanding complex materials, leading to success on the SAT® and beyond. Sharpen Your Skills for SAT® Reading and Writing Are you preparing for the SAT® Reading and Writing test? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world SAT® Reading and Writing problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! Text Structure and Purpose: SAT® Reading and Writing Review Compare and Contrast: SAT® Reading and Writing Review Central Idea: SAT® Reading and Writing Review Need help preparing for your SAT® Reading and Writingexam? 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1785	https://chemistrytalk.org/weak-acids-and-bases/	Skip to content Weak Acids and Weak Bases Core Concepts In this tutorial, you will learn about weak acids and weak bases, including why they are considered weak and examples of each. We’ll also show you a list of weak acids, and the weakest bases. Topics Covered in Other Articles Strong Acids and Bases Lewis Acid and Bases Acid Base Theories Acid Base Chemistry Properties of Acids and Bases What makes an Acid or Base ‘Weak’? Weak Acids When thinking about what makes about an acid weak, think about its dissociation property. When they are dissolved in a solution, weak acids do not completely dissociate into their different constituent ions. Furthermore, when a weak acid is dissolved in water, an equilibrium is developed between the concentrations of the weak acid itself and its constituent ions. A weak acid is not completely ionized in solution. For example, hydrofluoric acid, HF, is a weak acid. When dissolved in water, HF ion exist in equilibrium with H+, which reacts with water to form hydronium, and F– ions. Since the acid does not completely dissociate into its ionic components, it is a weak acid. HA + H2O ⇌ H3O+ + A– There are only several strong acids. Once you learn and memorize the strong acids, you may assume all other acids are weak acids. Furthermore, you can distinguish a strong acid from a weak one by looking at the acid dissociation constant, Ka, or pKa values. Strong acids will tend to have a high Ka and low pKa value while weak acids have small Ka and high pKa values. Weak Bases The concept is very similar when thinking about a weak base; however, in this case, a weak base is a base that cannot fully ionize or fully accept hydrogen ions in a solution. The base needs to dissociate in solution in order for it to react with the acid to form an acid base pair; if the base does not break into its constituent ions, it is not able to react the way it should. When dissolved in water, the solution contains a small amount of hydroxide ions and a large amount of the undissociated base. A weak base does not completely ionize. For example, with ammonium hydroxide, NH4OH, the counterpart’s of the base are NH4+ and OH–; the NH4+ supposedly accepts the acidic hydrogen in solution; however, in a weak base these ion do not fully dissociate so there is a mixture of NH4OH and the acid base pair. B + H2O ⇌ BH+ + OH– You can distinguish a strong base from a weak one by its base dissociation constant, Kb. This variable mathematically represents the base strength; the weaker the base, the smaller the Kb value. Example – Is Acetic Acid Strong or Weak? Acetic acid has a Ka value of 1.8 x 10-5. Since the Ka is quite small, acetic acid is a weak acid, and it does not completely ionize in water. Examples of Weak Acids and Bases List of Weak Acids Weak acids remain mainly as a complete molecule in solution. Only a small fraction of the acid molecule “breaks up”, or dissociates, into ions. However, these “weak acids” can still be highly corrosive and dangerous. This is a short list of weak acid examples – but there are many more! Formic Acid (HCOOH) Acetic Acid (CH3COOH) Benzoic Acid (C6H5COOH) Hydrofluoric Acid (HF) Phosphoric Acid (H3PO4) Sulfurous Acid (H2SO3) Carbonic Acid (H2CO3) Nitrous Acid (HNO2) Hydrocyanic Acid (HCN) Hydrosulfuric Acid (H2S) Citric Acid (C6H8O7) List of Weak Bases Here is a list of some weak base examples. Many weak bases are simply slightly soluble hydroxides, like magnesium hydroxide, while others are organic compounds. Ammonium Hydroxide (NH4OH) Aniline (C6H5NH2) Ammonia (NH3) Methylamine (CH3NH2) Ethylamine (CH3CH2NH2) Aluminum hydroxide (Al(OH)3) Magnesium Hydroxide (Mg(OH)2) Pyridine (C5H5N) Sodium Bicarbonate (NaHCO3) What is the weakest base? Water can be considered both a weak acid and a weak base, because it ionizes slightly to give both hydrogen ions and hydroxide ions. It is definitely one of the weakest bases! Salts of strong acids and strong bases create solutions are considered neutral. Examples are sodium perchlorate, sodium chloride, and sodium nitrate. We would classify these compounds as “neutral” rather than as weak bases. Weak Acids and Bases in Real World Applications Weak acids and bases are commonly used in organic chemistry and biochemistry to create a buffer solution, in titrations, or to catalyze a specific reaction. Acetic Acid is an active component seen in vinegar. Benzoic Acid occurs naturally in many plants and is usually used for food preservation. Ammonia is seen in household cleaning supplies, bug repellents, fertilizer and more. Weak acids and bases form buffers in the blood to keep it at a regulated pH Strong vs. Concentrated & Weak vs. Dilute You should be careful when thinking about a strong acid or base versus a concentrated one, and a weak acid or base versus a dilute one. A concentrated acid means a solution without much water and a diluted acid means a solution with a high amount of solvent. Just because a base has a high concentration does not mean it is strong; and just because an base is very dilute, does not not mean it is a weak and non dangerous base. Further Reading What is pKa? What is pH? Buffer Solution Equilibrium Constant Acid Base Neutralization Reaction Polyprotic acids
1786	http://opac.mk.ua/cgi-bin/irbis64r_11/cgiirbis_64.exe?LNG=&C21COM=S&I21DBN=ELKAT&P21DBN=ELKAT&S21FMT=infow_wh&S21ALL=%28%3C.%3ES%3D%D0%BC%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0%3C.%3E%29&Z21ID=&S21SRW=GOD&S21SRD=UP&S21STN=1&S21REF=5&S21CNR=1010&FT_REQUEST=&FT_PREFIX=	Електронний каталог Миколаївської обласної універсальної наукової бібліотеки Главная Упрощенный режим Описание Шлюз Z39.50Авторизация Фамилия Пароль ЕЛЕКТРОННИЙ КАТАЛОГ МИКОЛАЇВСЬКОЇ ОБЛАСНОЇ УНІВЕРСАЛЬНОЇ НАУКОВОЇ БІБЛІОТЕКИ Базы данных#### Електронний каталог книг - результаты поиска### Вид поиска Електронний каталог книг БД бібліографічних описів статей з періодичних виданьБД розпоряджень голови Миколаївської державної адміністраціїКаталог документів іноземними мовамиБД Рішення і розпорядження Миколаївської обласної радиБД "Краєзнавство"БД статей з періодики до 1917 рокуБД "Грамплатівки"БД біблографічних описів нот з періодичних виданьКорпоративний каталог районних ЦБ області\| Область поиска \| \| \| \| \| \| Найдено в других БД:БД бібліографічних описів статей з періодичних видань (7)Каталог документів іноземними мовами (5)БД "Краєзнавство" (13)БД статей з періодики до 1917 року (2) Формат представления найденных документов: полныйинформационныйкраткий Отсортировать найденные документы по: авторузаглавиюгоду изданиятипу документа Поисковый запрос:(<.>S=математика<.>) Общее количество найденных документов : 738 Показаны документы с 1 по 738 1. - [x]Вид документа : Многотомное издание Шифр издания : 51/В 65 Автор(ы) : Войтяховский, Ефим Дмитриевич Заглавие : Теоретической и практической курс чистой математики.../ Ефим Дмитриевич Войтяховский. Т.3: Тригонометрия Выходные данные : Москва: В вольн.тип.у Хр.Клаудия, 1787 Колич.характеристики :,286 c., л.черт. Цена : Б.ц. Предметные рубрики:Математика Геомеирія Тригонометрія Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 2. - [x]Вид документа : Многотомное издание Шифр издания : 51/Б 40 Автор(ы) : Безу,Этьен. Заглавие : Курс математики господина Безу.../ Безу,Этьен ; Пер.: Загорского. Ч.2,: содержащая геометрию и плоскую тригонометрию Выходные данные : Москва: Унив.тип., У Ридигера и Клаудия, 1798 Колич.характеристики :302 с. Примечания : К кн.припл.1 кн.: Безу Э.Сферическая тригонометрия,(служащая продолжением второй части Математического Курса г.Безу) переведена для благородного юношества,воспитывающегосяв Университетском пансионе. - М.,1799. - 58 с., л ил. Цена : Б.ц. Предметные рубрики:Математика Фізико-математичні науки Фізико-математичні дисципліни Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 3. - [x]Вид документа : Однотомное издание Шифр издания : 514/З-47 Автор(ы) : Зейлигер Д. Н. Заглавие : Из области геометрии и механики Выходные данные : Б.м., 18-- Колич.характеристики :С. 155-190 с., л. черт. Приплетено: 1. Зейлигер Д. Н. Механика подобно-изменяемой системы. Вып. 1 : Теория векторов/ Д. Н. Зейлигер. - С. 1-122, л. черт. с. - // Записки математического отд. - Т. 11 2. Зейлигер Д. Н. Механика подобно-изменяемой системы. Вып. 2: Теория винтов/ Д. Н. Зейлигер. - С. 149-221, л. черт. с. - // Записки математического отд. - Т. 11 3. Зейлигер Д. Н. Механика подобно-изменяемой системы. Вып. 3: Статика подобно-изменяемой системы/ Д. Н. Зейлигер. - С. 11-108, л. черт. с. - // Записки математического отд. - Т. 13 Примечания : // Записки математического отд. - Т. 11, 13 Цена : Б.ц. УДК : 514+531.01](081) Предметные рубрики: Фізико-математичні науки Математика Геометрія Механіка Класична механіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 4. - [x]Вид документа : Многотомное издание Шифр издания : 51/К 36 Автор(ы) : Кестнер, Абрагам Готгельф Заглавие : Начальные основания прикладной математики,сочиненные Авраамом Готгельфом Кестнером/ Абрагам Готгельф Кестнер. Ч.2: содержащая астрономию,географию,хронологию,гномонику,артиллерию,фортификацию и гражданскую архитектуру Выходные данные : Санкт-Петербург: Изд. комис. об учреждении нар. училищ при Имп.Акад.Наук, 1803 Колич.характеристики :V,648 c., л.ил. Цена : Б.ц. Предметные рубрики: Прикладна математика Математика Фізико-математичні науки Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 5. - [x]Вид документа : Однотомное издание Шифр издания : 517/Г 95 Автор(ы) : Гурьев, Семён Емельянович Заглавие : Основания трансцендентной геометрии кривых поверхностей Выходные данные : Б.м., 1806 Колич.характеристики :106 с., л.черт. Цена : Б.ц. Предметные рубрики: Геометрія Математика Геометричні форми Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 6. - [x]Вид документа : Многотомное издание Шифр издания : 51/В 65 Автор(ы) : Войтяховский, Ефим Дмитриевич Заглавие : Курс чистой математики в пользу и употребление юношества и упражняющихся в математике/ Ефим Дмитриевич Войтяховский. - 6-е изд. Т.1,: исправленный,новыми правилами дополненный и многими примерами умноженный Выходные данные : Москва: В Тип.Решетникова, 1817 Колич.характеристики :XVI,293, с. Цена : Б.ц. Предметные рубрики:Математика Фізико-математичні науки Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 7. - [x]Вид документа : Однотомное издание Шифр издания : 51/Ф 83 Автор(ы) : Франкер Заглавие : Курс чистой математики,содержащей арифметику,начальную алгебру,основания геометрии,прямолинейную тригонометрию и аналитическую геометрию : пер. с фр. Выходные данные : Москва: В Тип.А.Семена, 1819 Колич.характеристики :IV,VIII,551 c.,XII л.ил. Цена : Б.ц. Предметные рубрики:Математика Арифметика Алгебра Геометрія Тригонометрія Аналітична геометрія Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 8. - [x]Вид документа : Однотомное издание Шифр издания : 517.5/Е 16 Автор(ы) : Евклид Заглавие : Эвклидовых начал восемь книг, а именно : первые шесть, одиннадцатая и двенадцатая, содержащая в себе Основания геометрии Выходные данные : Санкт-Петербург: Тип. департамента нар. просвещения, 1819 Колич.характеристики :XVI, 480 с., л. черт. Цена : Б.ц. Предметные рубрики: Геометрія Математика Елементарна геометрія Экземпляры :ОРК(1) Свободны: ОРК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 9. - [x]Вид документа : Многотомное издание Шифр издания : 511/Б 91 Автор(ы) : Бурдон Заглавие : Арифметика/ пер.с фр.(с 9-го изд.) В.Ч. Ч.1 Выходные данные : СПб.: Тип.Отд.Корп.Внутр.Стражи, 1834 Колич.характеристики :,VII,II,271, с Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 10. - [x]Вид документа : Многотомное издание Шифр издания : 511кр/Л 27 Автор(ы) : Латышев Л. Заглавие : Основания арифметики для употребления в Черноморской штурманской роте: [в 2 ч.]/ сост. Л. Латышев. - Изд. 2-е, несколько испр. против первого. Ч. 1 - 2 Выходные данные : Николаев: Тип. Черном. гидрогр. депо, 1842 - Колич.характеристики :256, 278 с Серия: Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 11. - [x]Вид документа : Однотомное издание Шифр издания : 52/В 75 Автор(ы) : Воробьев, Петр Заглавие : Записки из математической географии, сферической и частью практической астрономии, составленные и изданные в 1844 году учителем училища корпуса топографов подполковником Воробьевым Выходные данные : Б.м., 1844 Колич.характеристики :,IX,334 с.,3 л.черт.: табл. Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 12. - [x]Вид документа : Однотомное издание Шифр издания : 512/Л 27 Автор(ы) : Латышев Л. Заглавие : Основания алгебры : для учащихся Черномор. штурман. роты Выходные данные : Б.м., 1850 Колич.характеристики :378 с Цена : Б.ц. Предметные рубрики:математика елементарна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 13. - [x]Вид документа : Однотомное издание Шифр издания : 517/Д 29 Автор(ы) : Деларю, Даниил Заглавие : Курс дифференциального исчисления и теории алгебраических функций/ Даниил Деларю. Т. 1 Выходные данные : Санкт-Петербург: Тип. Имп. АН, 1869 Колич.характеристики :XVI, 2012 с Приплетено: 1. Дмитриев А. Начальные основания сферической геометрии и сферической тригонометриии: с 2-я табл. чертежей и с 3-я политипажами/ А. Дмитриев. - Санкт-Петербург: В тип. А. Якобсона, 1874. - IV, 72 с. 2. Ceppe A. Тригонометрия. Вып. 2 : Сферическая тригонометрия/ пер. Ев. Гутора. - Воронеж: Изд. Е. С. Гутора и С. А. Степанцова. Тип. Г. М. Веселовского, 1870. - 66 с., 1 л. черт. с. Примечания : Посвящ. памяти Михаила Васильевича Остроградского Цена : Б.ц. УДК : 517.2+514.116.3](075) Предметные рубрики:Математика Диференціальне числення Функції (мат.) Геометрія Тригонометрія Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 14. - [x]Вид документа : Многотомное издание Шифр издания : 530/У 55 Автор(ы) : Умов, Николай Алексеевич Заглавие : Курс математической физики/ сост. Н. А. Умов. Вып. 1: Введение Выходные данные : Одесса: Тип. Ульриха и Шульце, 1878 Колич.характеристики :139 с.: рис. Серия: Примечания : Из XXVI т. Зап. Имп. Новорос. Ун-та Цена : Б.ц. Предметные рубрики: математична фізика математика фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 15. - [x]Вид документа : Однотомное издание Шифр издания : 517/С 77 Автор(ы) : Старков К. Заглавие : К вопросу об интегрировании совокупных диффернциальных уравнений : Излож. мемуара Коши, по поводу сделаного к нему доп. профессором Имп. Новорос. ун-та Е. Ф. Сабининым Выходные данные : Одесса, 1878 Колич.характеристики :78 с Приплетено: 1. Старков А. Общий интеграл уравнения с частными производными n-го порядка вида/ А. Старков. - Одесса, 1878. - С. 1-8 с. 2. Starkoff A. Theorie des equations generales/ A. Starkoff. - Одесса, 1885. - С. 143- 200 с. 3. Старков А. Интегрирование рациональной дроби с мнимыми корнями в знаменателе/ А. Старков. - С. 87-92 с. 4. Сонин Н. Я. Об одной задаче вариационного исчесления: (ст. 2-я)/ Н. Я. Сонин. - С. 87-107 с. 5. Лиин В. Н. Новое построение Мориса Д'окань для определения отношения скоростей в направляющих механизмах Поселье и Гарта: чит. в заседании 2 нояб. 1884 г./ В. Н. Лиин. - С. 109-111, 1 л. черт. с. 6. Старков А. Об одном линейном дифференциальном уравнении 3-го порядка/ А. Старков. - С. 13-22 с. 7. Старкова А. Об одной задаче вариационного исчесление/ А. Старкова. - С. 23-46 с. 8. Старкова А. О некоторых особенностях в постановке задачи Ньютона о поверхности наименьшего сопротивления: (из сообщения, сделаного в заседании Мат. отд. Новорос. о-ва Естествоиспытателей 2-го нояб. 1884 г.)/ А. Старкова. - С. 47-56, 1 л. черт. с. 9. Старков А. К теории линейных дифференциальных уравней/ А. Старков. - С. 191-200 10. Старков А. К вопросу о поверхности наименьшего сопротивления при движении в несжигаемой жидкости/ А. Старков. - С. 49-54 с. 11. Отыскание уравнение кривой, образующей при вращении поверхностей наименьшего сопротивления. - С. 55-136 с. Цена : Б.ц. УДК : 517.9(081) Предметные рубрики: Рівняння Математичні терміни Математика Алгебраїчні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 16. - [x]Вид документа : Однотомное издание Шифр издания : 52/Ш 34 Автор(ы) : Шведов, Федор Никифорович Заглавие : Теория [исчисления] форм комет Выходные данные : Б.м., 1879 Колич.характеристики :52, 147, 109 с.: л. табл. Примечания : Часть текста на фр. Цена : Б.ц. Предметные рубрики: Астрономія Математика Небесна механіка Содержание : Независимое от исчисления Ламберта определение albedo белого картона/ А. Кононович. Способы вычисления орбит двойных звезд/ А. Кононович. Способы, в которых отыскание элементов истинной орбиты основано на предварительном определении видимой орбиты ; Мысли о прошедшем и будущем нашей планеты, прочтенные 6 декабря 1875 года в пользу славян, пострадавших при восстании в Герцоговине/ Головинский. Теория солнечных неравенств в движении Луны/ А. Краснов. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 17. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 91 Автор(ы) : Буняковский В. Я. Заглавие : О наибольших величинах в вопросах относящихся к нравственному ожиданию : чит. в заседании Физ.-мат. отд. 20 наяб. 1879 Выходные данные : Санкт-Петербург: Тип. Имп. АН, 1879 Колич.характеристики :22 с Приплетено: 1. Буняковский В. Я. О некоторых частных случаях интегрируемости в конечном виде дифференциала и других выражений подобного вида/ В. Буняковский. - Санкт-Петербург: Тип. Имп. Ан, 1863. - 32 с. - Прил. к 3-му т. Записок Имп. АН, №2 2. Буняковский В. Я. Геометрические соображения о наивыгоднейшем размещении громоотводов/ В. Буняковский. - Санкт-Петербург: В тип. Имп. АН, 1863. - 31, 2 л. черт. с. - Прил. к 4-му т. Записок Имп. АН, №3 3. Буняковский В. Я. О самосчетах и о новом их применении: чит. в заседании Физ.-мат. отд. 20 апр. 1876/ В. Я. Буняковский. - Санкт-Петербург: Тип. Имп. АН, 1876. - 28, 1 л. черт. с. - Прил. к 27-му т. Записок Имп. АН, №4 4. Буняковский В. Я. О сумовании численных таблиц по приближению/ В. Я. Буняковский. - Санкт-Петербург: В Тип. Имп. АН , 1867. - 41, 1 л. черт. с. - Прил. к 12-му т. Записок Имп. АН, №4 5. Буняковский В. Я. Заметка об одной формуле относящейся к теории чисел: чит. в заседании Физ.-Мат. отьд. Имп. АН 2 дек. 1886 г./ В. Я. Буняковский. - Санкт-Петербург: Тип. Имп. АН, 1887. - 16 с. - Прил. к 55-му т. Записок Имп. АН, №5 6. Буняковский В. Я. О соединениях особенного рода, встречающихся в вопросе о дефектах/ В. Я. Буняковский. - Санкт-Петербург: Тип. Имп. АН, 1871. - 71 с. - Прил. к 20-му т. Записок Имп. АН, №2 7. Буняковский В. Я. Об одном видоизменении способа, известного под названием Эратосфенова решета: чит. в заседании Физ.-мат. отд. 13 апр. 1882/ В. Я. Буняковский. - Санкт-Петербург: Тип. Имп. АН, 1882. - 32 с. - Прил. к 41-му т. Записок Имп. АН, №3 Примечания : Прил. к 36-му т. Записок Имп. АН, №1 Цена : Б.ц. УДК : 51(081) Предметные рубрики:Математика Фізико-математичні науки Математичне мислення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 18. - [x]Вид документа : Однотомное издание Шифр издания : 517/В 23 Автор(ы) : Ващенко-Захарченко М. Е. Заглавие : Начала Евклида с пояснительным введением и толкованиями Выходные данные : Б.м., 1880 Колич.характеристики :XV, 747 с.: ил. Цена : Б.ц. Предметные рубрики: геометрія історія геометрії математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 19. - [x]Вид документа : Однотомное издание Шифр издания : 517/С 23 Заглавие : Сборник [статей по математике] Выходные данные : Б.м., 1883 Приплетено: 1. Поссе К. А. О трансцендентности чисел/ К. А. Поссе. - С. 50-60 с. 2. Макаров Н. Построение перспектив точек, линий и плоских фигур, лежащих на наклонных плоскостях, без выстраевания вспомогательных ортогональных проэкций/ Н. Макаров. - С. 92-110 с. 3. Евиевич И. Несколько слов о замене выражения. Линейные выражения вида/ И. Евиевич. - С. 228-237 с. 4. Ленц Р. Ртутный инверсор/ Р. Ленц. - С. 240-246 с. 5. Паукер Г. Начало возможных перемещений/ Г. Паукер. - С. 304-345 с. 6. Ильина А. Заметка об интегрировании дифференциальных уравнений вида/ А. Ильина. - С. 234-248 с. 7. Красновский М. Интегрирование уравнения движения упругой пластинки при некоторых частных предположениях относительно ее контура/ М. Красновский. - С. 192-231 с. 8. Занчевский И. М. О трехшестной сочлененной системе/ И. М. Занчевский. - С. 32-42 с. 9. Сонина Н. Я. Об одной задаче вариационного исчисления/ Н. Я. Сонина. - С. 2-11 с. 10. Сонина Н. Я. Обобщение одной формулы Абеля/ Н. Я. Сонина. - С. 144-150 с. 11. Умова Н. Геометрическое значение интегралов Френеля/ Н. Умова. - С. 58-86 с. 12. Новикова П. М. Признак устойчивости движения и его связь с одним из признаков maximuma или minimuma простого определенного интеграла/ П. М. Новикова. - 152-172 с. Цена : Б.ц. УДК : 517.2(082) Предметные рубрики:Математика Диференціальні обчислення Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 20. - [x]Вид документа : Однотомное издание Шифр издания : 513/В 23 Автор(ы) : Ващенко-Захарченко М. Е. Заглавие : Элементарная геометрия в объёме гимназического курса Выходные данные : Б.м., 1883 Колич.характеристики :XXIX, 355 с.: ил Цена : Б.ц. Предметные рубрики: геометрія математика геометричні форми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 21. - [x]Вид документа : Однотомное издание Шифр издания : 530/У 55 Автор(ы) : Умов, Николай Алексеевич Заглавие : Из лекций математической физики Выходные данные : Б.м., 1883 Колич.характеристики :71 с Примечания : Содерж. : 1. Теория бесконечно малых колебаний консервативной системы около положения устойчивого равновесия. 2. Колебания системы с одной степенью свободы. Созвучие и абсорбция Цена : Б.ц. Предметные рубрики: математична фізика математика фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 22. - [x]Вид документа : Многотомное издание Шифр издания : 51/В 23 Автор(ы) : Ващенко-Захарченко М. Е. Заглавие : История математики. Исторический очерк развития геометрии/ М. Е. Ващенко-Захарченко. Т. 1 Выходные данные : К.: В тип. Имп. Ун-та св. Владимира, 1883 Колич.характеристики :XII, 684 с Серия: Цена : Б.ц. Предметные рубрики: геометрія історія геометрії математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 23. - [x]Вид документа : Многотомное издание Шифр издания : 5(06)/З-32 Заглавие : Записки Уральского общества любителей естествознания. - Парал. назв. : нем. Т. 9, вып. 1 Выходные данные : , 1885 Колич.характеристики :17 с.: л. табл. Цена : Б.ц. ББК : 5(06) Предметные рубрики: природничі науки математика форми життя Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 24. - [x]Вид документа : Однотомное издание Шифр издания : 511/Ш 82 Автор(ы) : Шохор-Троцкий С.И. Заглавие : Методика арифметики с приложением сборника упражнений по арифметике для учащих Выходные данные : Б.м., 1886 Колич.характеристики :215,146, IV с., Цена : Б.ц. Предметные рубрики: Методика викладання Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 25. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 73 Автор(ы) : Богаевский, Валериан Заглавие : Основания псевдосферической геометрии : сочинение Выходные данные : Б.м., 1888 Колич.характеристики :40 с Приплетено: 1. Богаевский Упругость кристаллических тел: сочинение/ Валериан Богаевский. - Киев: Тип. В. И. Завадского, 1889. - 24 с. 2. Паолис Основания алгеброических форм/ Риккардо де Паолис. - Киев: Тип. В. И. Завадского, 1889. - 54 с. 3. Основные законы механической теории теплоты/ Валериан Богаевский. - Киев: Тип. В. И. Завадского, 1889. - 24 с. 4. Теория геодезической кривизны/ Валериан Богаевский. - Киев: Тип. В. И. Завадского, 1890. - 24 с. 5. Свойства несовершенных газовых/ Валериан Богаевский. - Киев: Тип. В. И. Завадского, 1890 6. Основания псевдосферической геометрии: сочинение/ Валериан Богаевский. - Киев: Тип. Имп. ун-та Св. Владимира, 1888. - 40 с. Цена : Б.ц. УДК : 51(081) Предметные рубрики: Фізико-математичні науки Математика Геометрія Математична фізика Алгебра Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 26. - [x]Вид документа : Однотомное издание Шифр издания : 5(09)/В 78 Заглавие : VIII Сьезд русских естествоиспытателей и врачей в С.-Петербурге от 28 декабря 1889 г. до 7 января 1890 г. : с рис. и 2-мя гипсометр. картами Европ. России Выходные данные : Б.м., 1890 Колич.характеристики :LXXVIII, 130, 77, 64, 188, 46, 94, 38, 32, 59, 108, 36, 28, IV с.: ил. Цена : Б.ц. ББК : 5(09) Предметные рубрики: природничі науки природознавство фізико-математичні науки математика астрономія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 27. - [x]Вид документа : Многотомное издание Шифр издания : 513/З-28 Автор(ы) : Занчевский И. М. Заглавие : Геометрические места в теории осей вращения/ И. М. Занчевский . Вып. 1 Выходные данные : Одесса: Тип. А. Шульце, 1891 Колич.характеристики :XII, 82 с., л. cхем. Серия: Примечания : Отт. из Зап. Мат. отд. Новорос. О-ва естествоиспытателей. - 1892. - Т. 14 Цена : Б.ц. Предметные рубрики: геометрія математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 28. - [x]Вид документа : Однотомное издание Шифр издания : 511/Ш 82 Автор(ы) : Шохор-Троцкий С.И. Заглавие : Методика арифметики с приложением методического сборника задач для учащих в начальных школах . -3-е изд. ,знач. испр.и заново обраб. Выходные данные : Б.м., 1892 Колич.характеристики :IV, 204 с Цена : Б.ц. Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 29. - [x]Вид документа : Однотомное издание Шифр издания : 510/Ш 82 Автор(ы) : Шохор-Троцкий С.И. Заглавие : Цель и средства преподавания низшей математики с точки зрения требований общего образования Выходные данные : Б.м., 1892 Колич.характеристики :116 с Цена : Б.ц. Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 30. - [x]Вид документа : Однотомное издание Шифр издания : 51/Л 53 Автор(ы) : Леффлер, Анна-Карлотта Заглавие : Софья Ковалевская : Воспоминания А. К. Леффлер Выходные данные : Б.м., 1893 Колич.характеристики :ІІ,315 с Цена : Б.ц. Предметные рубрики:Математика Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 31. - [x]Вид документа : Однотомное издание Шифр издания : 51/Л 64 Автор(ы) : Литвинова Е. Ф. Заглавие : Н. И. Лобачевский : Его жизнь и научная деятельность : биогр. очерк Выходные данные : Б.м., 1895 Колич.характеристики :80 с.: портр. Серия: ЖЗЛ : Жизнь замечательных людей . Биографическая библиотека Ф. Павленкова Цена : Б.ц. Предметные рубрики:Математика-- історія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 32. - [x]Вид документа : Однотомное издание Шифр издания : 016/У 42 Заглавие : Указатель книг по учебной математической литературе, с 1884 г. по 1895 г. Выходные данные : Б.м., 1896 Колич.характеристики :44 стб. Цена : Б.ц. Предметные рубрики: галузева бібліографія математика бібліографічні видання бібліографічні посібники Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 33. - [x]Вид документа : Однотомное издание Шифр издания : 511/Щ 64 Автор(ы) : Щербина, Константин Моисеевич Заглавие : О преподавании систематического курса обыкновенных дробей Выходные данные : Б.м., 19-- Колич.характеристики :22 с Цена : Б.ц. УДК : 511.13(07) Предметные рубрики:Математика Арифметика Дроби Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 34. - [x]Вид документа : Однотомное издание Шифр издания : 517/П 91 Автор(ы) : Пфейффер Г. В. Заглавие : Интегрированые системы линейных дифференциальных уравнений с постоянными коэффициентами Выходные данные : Б.м., 19-- Колич.характеристики :49 с Приплетено: Абрамович О гипергеометрических функциях с одной особенной по виду точкой/ К. Ф. Абрамович. - С. 1-32, 33-118 с. Цена : Б.ц. УДК : 517.9+514.7](082) Предметные рубрики:Математика Рівняння Лінійні рівняння Диференціальні рівняння Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 35. - [x]Вид документа : Многотомное издание Шифр издания : 5/Ф 50 Заглавие : Физико-математический ежегодник, посвященный вопросам математики, физики, химии и астрономии в элементарном изложении/ изд. кружка авт. "Сб. в помощь самоообразованию". № 1, 1900 Выходные данные : Москва: Т-во тип. А. И. Мамонтова, 1900 - Колич.характеристики :591 с.: 128 рис в тексте и 8 табл. Серия: Цена : Б.ц. Предметные рубрики: Фізико-математичні науки Природничі науки Математика Фізика Астрономія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 36. - [x]Вид документа : Однотомное издание Шифр издания : 51/Ц 61 Автор(ы) : Циммерман В. А. Заглавие : Десятичные приближения чисел и способы приближенного вычисления суммы, разности, произведения и частного Выходные данные : Б.м., 1901 Колич.характеристики :38 с Цена : Б.ц. Предметные рубрики:математика математичні задачі обчислювальна математика елементарна математика наближені обчислювання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 37. - [x]Вид документа : Однотомное издание Шифр издания : 512/Н 65 Автор(ы) : Никульцев П. Заглавие : Нахождение частного при делении на 9, 99, 999, 11, 101, 1001 Выходные данные : Б.м., 1903 Колич.характеристики :6 с Примечания : Прил. к Циркулярам по Моск. уч. окр. за 1903 г. Цена : 20 к. УДК : 512.12 Предметные рубрики:Математика Математичні дії Елементарна алгебра Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 38. - [x]Вид документа : Однотомное издание Шифр издания : 51/А 47 Автор(ы) : Алексеев В.Г. Заглавие : Математика как основание критики научно-философского мировозрения Выходные данные : Юрьев: В тип.К.Маттисена, 1903 Колич.характеристики :50 с Примечания : Из Сб.учеб.-лит.о-ва при Имп.Юрьев.ун-те,Т.7 Цена : Б.ц. Предметные рубрики:математика Экземпляры :1 яр.(1) Свободны: 1 яр.(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 39. - [x]Вид документа : Многотомное издание Шифр издания : 511/Ш 82 Автор(ы) : Шохор-Троцкий С. И. Заглавие : Методика арифметики/ С. И. Шохор-Троцкий. - 7-е изд.,испр.и знач.доп. Ч.1: Для учителей одноклассных начальных школ Выходные данные : Санкт-Петербург: Тип. С.-Петерб. т-ва печ. и изд. дела "Труд", 1903 Колич.характеристики :XV, 816, IV с Цена : Б.ц. Предметные рубрики:Математика Арифметика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 40. - [x]Вид документа : Однотомное издание Шифр издания : 513/К 73 Автор(ы) : Котурницкий П. В. Заглавие : Квинкункс и его применения в естествознании и ткацком деле Выходные данные : Б.м., 1903 Колич.характеристики :240 с., л. черт. Цена : Б.ц. Предметные рубрики: Геометрія Математика Геометричні побудови Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 41. - [x]Вид документа : Многотомное издание Шифр издания : 517/Х-19 Автор(ы) : Хандриков, Митрофан Заглавие : Элементы математического анализа: в 3 т./ Митрофан Хандриков. Т. 1-2 Выходные данные : К.: [Тип. Имп. Ун-та св. Владимира АО печ. и изд. дела Н. Т. Корчак-Новицкого], 1905-1907 Колич.характеристики :1387 с Серия: Цена : Б.ц. Предметные рубрики: Математичний аналіз аналіз математика варіаційне числення векторний аналіз функціональний аналіз інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 42. - [x]Вид документа : Однотомное издание Шифр издания : 51/Д 64 Автор(ы) : Долгушин П. А. Заглавие : Вычисления по приближению: для учащихся в ст. кл. сред. шк./ П. А. Долгушин. - Отт. из журн. : Унив. изв. за 1908 г. Вып. 1 Выходные данные : К.: Тип. Имп. ун-та Св. Владимира акционер. о-ва печ. и изд. дела Н. Т. Корчак-Новицкого, 1908 - Колич.характеристики :, ІІ, 73 с. Цена : Б.ц. Предметные рубрики: обчислення наближені обчислення математика навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 43. - [x]Вид документа : Многотомное издание Шифр издания : 517/Х-19 Автор(ы) : Хандриков, Митрофан Заглавие : Элементы математического анализа: в 3 т./ Митрофан Хандриков. Т. 3: Анализ бесконечно малых Выходные данные : К.: [Тип. Имп. Ун-та св. Владимира АО печ. и изд. дела Н. Т. Корчак-Новицкого], 1908 Колич.характеристики :VI, C. 1391-1946, II, IX Цена : Б.ц. Предметные рубрики: Математичний аналіз аналіз математика варіаційне числення векторний аналіз функціональний аналіз інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 44. - [x]Вид документа : Однотомное издание Шифр издания : 51/Щ 64 Автор(ы) : Щербина К. М. Заглавие : Математика в русской средней школе : обзор трудов и мнений по вопр. об улучш. прогр. математики в сред. шк. за послед. 9 лет (1899 - 1907) Выходные данные : Б.м., 1908 Колич.характеристики :152 с Цена : Б.ц. Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 45. - [x]Вид документа : Однотомное издание Шифр издания : 513/К 12 Автор(ы) : Каган В. Ф. Заглавие : Задача обоснования геометрии в современной постановке : речь, произнес. при защите дис. на степ. магистра чист. математики Выходные данные : Б.м., 1908 Колич.характеристики :35, с Цена : Б.ц. Предметные рубрики:математика геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 46. - [x]Вид документа : Многотомное издание Шифр издания : 512/Ф 59 Автор(ы) : Финисов П. К. Заглавие : Решения алгебраических задач : по сб., сост. Н.А. Шапошниковым и Н. К. Вальцовым/ П. К. Финисов. - 3-е изд. Ч.1 : (для кл. 3 и 4) Выходные данные : К. ; Пб. ; Х.: Юж.-Рус. книгоиздательство Ф. А. Иогансона. Тип. И. И. Чоколова, 1909 Колич.характеристики :324, III с Серия: Цена : Б.ц. Предметные рубрики: алгебра математика математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 47. - [x]Вид документа : Однотомное издание Шифр издания : 517/Г 75 Автор(ы) : Граве, Димитрий Александрович Заглавие : Элементарный курс теории чисел : лекции, чит. в ун-те св. Владимира Выходные данные : Б.м., 1909 Колич.характеристики :, 240, , IV с.: табл. Цена : Б.ц. Предметные рубрики:математика чисел теорія аналітичні функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 48. - [x]Вид документа : Многотомное издание Шифр издания : 514/Б 61 Автор(ы) : Билибин Н. Заглавие : Курс тригонометрии/ Н. Билибин. Ч.1: Прямолинейная тригонометрия : (Решение треугольников) Выходные данные : СПб.: Тип.М.Стасюлевича, 1909 Колич.характеристики :ІХ,411 с.: ил. Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 49. - [x]Вид документа : Многотомное издание Шифр издания : 51/С 63 Заглавие : Сообщения Харьковского математического общества: вторая серия. Т. 11, № 1-6 Выходные данные : Х.: Тип. и Литогр. М. Зильберберг и сыновья, 1910 Колич.характеристики :312 с Серия: Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 50. - [x]Вид документа : Многотомное издание Шифр издания : 016 : 51/Р 89 Заглавие : Русская математическая библиография: в 2-х вып./ под ред. Д. М. Синцова. Вып. 1: Список сочинений по чистой и прикладной математике, напечатанных в России в 1908 году Выходные данные : Одесса: Книгоиздательство "Матезис". Тип. Акц. Юж.-Рус. О-ва Печ. Дома, 1910 Колич.характеристики :76, ІІІ с Серия: Цена : (В пер.) Предметные рубрики:Математика Бібліографічні посібники Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 51. - [x]Вид документа : Однотомное издание Шифр издания : 513/Р79 Автор(ы) : Роу Саидра Заглавие : Геометрические упражнения с куском бумаги : пер. с англ. Выходные данные : Б.м., 1910 Колич.характеристики :IX, 173 с.: ил Цена : Б.ц. Предметные рубрики: геометрія математика геометричні форми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 52. - [x]Вид документа : Однотомное издание Шифр издания : 517/А 72 Автор(ы) : Антонов Л. М. Заглавие : Таблицы квадратов, кубов, степеней и упрощенного умножения с объяснением : Пособие для учащих, учащихся, коммерсантов, инженеров, техников и пр. Выходные данные : Б.м., 1910 Колич.характеристики :308 с Цена : Б.ц. Предметные рубрики:математика чисел теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 53. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 79 Автор(ы) : Больцано, Бернард Заглавие : Парадокcы бесконечного, изданные по посмертной рукописи автора Др. Фр. Пржигонским Выходные данные : Б.м., 1911 Колич.характеристики :VI, 120 с Серия: Библиотека классиков точного знания; 4 Цена : Б.ц. Предметные рубрики:математика фізико-математичні науки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 54. - [x]Вид документа : Однотомное издание Шифр издания : 517.2/К 56 Автор(ы) : Ковалевский, Гергард Заглавие : Основы дифференциального и интегрального исчислений Выходные данные : Б.м., 1911 Колич.характеристики :503 с Коллективы : пер. с нем. под ред. С. О. Шатуновского Цена : Б.ц. Предметные рубрики:Математика числення диференціальне числення інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 55. - [x]Вид документа : Многотомное издание Шифр издания : 510/Б 82 Автор(ы) : Борель, Эмиль Заглавие : Элементарная математика/ Эмиль Борель ; пер. с нем. П. Штеккеля ; под ред. Б. Ф. Кагана. Ч.1: Арифметика и алгебра Выходные данные : Одесса: Матезис ; Тип."Техник", 1911 Колич.характеристики :LXIV, 433 с., л. табл. Цена : Б.ц. Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 56. - [x]Вид документа : Однотомное издание Шифр издания : 51/Ш 95 Автор(ы) : Шуберт, Герман Заглавие : Математические развлечения и игры Выходные данные : Б.м., 1911 Колич.характеристики :ХІІ, 357 с Коллективы : пер. с 3-го нем. изд. И. Л. Левинтова под ред. и с добавл. "Вестн. Опыт. Физики и Элемент. Мат." Цена : Б.ц. Предметные рубрики:Математика ігри дозвілля Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 57. - [x]Вид документа : Однотомное издание Шифр издания : 51/Ф 81 Автор(ы) : Фосс А. Заглавие : О сущности математики Выходные данные : Б.м., 1911 Колич.характеристики :116 с Цена : 30 грн. УДК : 51 Предметные рубрики:Математика Природничі науки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 58. - [x]Вид документа : Однотомное издание Шифр издания : 51/С 23 Заглавие : Сборник статей по математике Выходные данные : Санкт-Петербург, 1912 Колич.характеристики :1 с Приплетено: 1. Ващенко-Захарченко М. Ю. Опыт изложения дифференциального и интегрального исчислений без помощи бесконечно-малых и пределов/ М. Ю. Ващенко-Захарченко. - VII, 77 с. 2. Черный С. Д. О числе возможных решений задачи о вычислении параболлических орбит по способу Ольберса/ С. Д. Черный. - 44 с. 3. Суслов Г. К. Закон центра инерции и закон моментов/ Г. К. Суслов. - 16 с. 4. Воронца П. В. Значение понятия о геометрической производной в кинематике точки/ П. В. Воронца. - 16 с. Цена : Б.ц. УДК : 51(082) Предметные рубрики:Математика Фізико-математичні науки Математичні терміни Алгебра Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 59. - [x]Вид документа : Многотомное издание Шифр издания : 016 : 51/Р 89 Заглавие : Русская математическая библиография: в 2-х вып./ под ред. Д. М. Синцова. Вып. 2: Список сочинений по чистой и прикладной математике, напечатанных в России в 1909 году Выходные данные : Одесса: Книгоиздательство "Матезис". Тип. Акц. Юж.-Рус. О-ва Печ. Дома, 1912 Колич.характеристики :XV, 92 с Цена : (В пер.) Предметные рубрики:Математика Бібліографічні посібники Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 60. - [x]Вид документа : Однотомное издание Шифр издания : 511/Ф 53 Автор(ы) : Филиппов А. О. Заглавие : Четыре арифметические действия. Числа натуральные Выходные данные : Б.м., 1912 Колич.характеристики :87 с Цена : Б.ц. Предметные рубрики: арифметика математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 61. - [x]Вид документа : Однотомное издание Шифр издания : 513/Ф 95 Автор(ы) : Фурре Е. Заглавие : Очерк истории элементарной геометрии Выходные данные : Б.м., 1912 Колич.характеристики :48 с.: ил Коллективы : пер. с фр. А. И. Бакова Серия: Библиотека элементарной математики ; 2 Цена : Б.ц. Предметные рубрики:математика геометрія історія геометрії елементарна геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 62. - [x]Вид документа : Однотомное издание Шифр издания : 51/С 23 Заглавие : Сборник статей по математике Выходные данные : Санкт-Петербург, 1912 Приплетено: 1. Слешинский И. В. О судимости непрерывных дробей/ И. В. Слешинский. - С. 201-127 с. 2. Слешинский И. В. Доказательство существования некоторых пределов/ И. В. Слешинский. - С. 129-140 с. 3. Слешинский И. В. К теории способа наименьших квадратов/ И. В. Слешинский. - С. 201-264 с. 4. Циммерман О разложении в непрерывную дробь функции, определяемой дифференциальным уравнением вида.../ Владимир Циммерман. - С. 1-139, ІІ с. 5. Синцов Д. Заметки об уравнениях, аналогичных уравнению Риккати/ Д. Синцов. - С. 1-17 с. 6. Сикстель В. Основные теоремы сферической геометрии/ В. Сикстель. - С.18-41, 1 л. черт. с. 7. Граве П. О геометрическом представлении эллиптических интегралов и функций/ П. Граве. - С. 159-197 с. 8. Синцов Д. Новый вид управления линейчатых поверхностей/ Д. Синцов Цена : Б.ц. УДК : 51(082) Предметные рубрики:Математика Фізико-математичні науки Геометрія Алгебра Рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 63. - [x]Вид документа : Однотомное издание Шифр издания : 51/Ю 50 Автор(ы) : Юнг, Дж. Заглавие : Как преподавать математику? : преподавание математики в сред. и нач. шк. : в 2 ч. Выходные данные : Б.м., 1912 Колич.характеристики :XVI,425,IX с Цена : Б.ц. Предметные рубрики:Математика Математика-- теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 64. - [x]Вид документа : Многотомное издание Шифр издания : 513(03)/Э 68 Заглавие : Энциклопедия элементарной математики: рук. для препод. и изуч. элементар. математику : в 3 т./ сост. Г. Вебер, И. Вельштейн, В. Якобсталь ; пер. с нем. под ред., предисл. и с примеч. В. Кагана . - Изд. 2-е. Т. 1, кн. 1: Основания геометрии Выходные данные : Одесса: Изд. "Матезис". Тип. под фирм. "Вест. Винодел.", 1913 Колич.характеристики :XII, 360 с.: черт. Серия: Цена : Б.ц. Предметные рубрики: Геометрія Математика Елементарна геометрія Елементарна математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 65. - [x]Вид документа : Однотомное издание Шифр издания : 51(03)/Г 75 Автор(ы) : Граве, Димитрий Александрович Заглавие : Энциклопедия математики : очерк ее соврем. положении : 152 черт. в тексте Выходные данные : Б.м., 1913 Колич.характеристики :Х, 601 с Цена : Б.ц. Предметные рубрики:математика чисел теорія ймовірностей теорія математична фізика довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 66. - [x]Вид документа : Многотомное издание Шифр издания : 517/Ч 35 Автор(ы) : Чезаро, Эрнесто Заглавие : Элементарный учебник алгебраического анализа и исчисления бесконечно малых: с многочисл. прим. для упражнения / Эрнесто Чезаро ; предисл. авт. ; пер. с нем. с примеч. и доп. К. А. Поссе. Ч. 1 : с 28 черт. Выходные данные : Одесса: "Матезис". Тип.Акц. Юж.-Рус. О-ва Печ. Дела, 1913 Колич.характеристики :XV, 632 с Серия: Цена : Б.ц. Предметные рубрики: Алгебра Математика Числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 67. - [x]Вид документа : Однотомное издание Шифр издания : 513/К 12 Автор(ы) : Каган В. Ф. Заглавие : О преобразовании многогранников : докл. , прочит. в общ. собр. Первого Всерос. Съезда преподавателей математики Выходные данные : Б.м., 1913 Колич.характеристики :27 с.: черт. Цена : Б.ц. Предметные рубрики: геометрія геометричні форми математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 68. - [x]Вид документа : Однотомное издание Шифр издания : 513/К 85 Автор(ы) : Крыжановский Д. А. Заглавие : О максимальных и минимальных свойствах плоских фигур Выходные данные : Б.м., 1913 Колич.характеристики :100 с.: черт. Серия: Элементарные этюды по физике и математике; 1 Цена : Б.ц. Предметные рубрики: геометрія математика геометричні форми елементарна геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 69. - [x]Вид документа : Однотомное издание Шифр издания : 511/Ш 75 Автор(ы) : Шмулевич П. К. Заглавие : Курс теоретической арифметики : в объеме программ вступ. экзаменов в спец. высш. учеб. заведениях . -Изд. 8-е Выходные данные : Б.м., 1914 Колич.характеристики :101, с.: ил. Цена : Б.ц. Предметные рубрики: Арифметика Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 70. - [x]Вид документа : Многотомное издание Шифр издания : 513(03)/Э 68 Заглавие : Энциклопедия элементарной математики: рук. для преподающих и изуч. элементар. математику : в 3 т./ сост. Г. Вебер, И. Вельштейн, В. Якобсталь ; пер. с нем. под ред., предисл. и с примеч. В. Кагана. - 2-е изд. Т. 2: Энциклопедия элементарной геометрии Выходные данные : , 1914 Колич.характеристики :VIII, 321 с.: черт. Цена : Б.ц. Предметные рубрики: тригонометрія математика елементарна геометрія довідкові видання стереометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 71. - [x]Вид документа : Многотомное издание Шифр издания : 517/Ч 35 Автор(ы) : Чезаро, Эрнесто Заглавие : Элементарный учебник алгебраического анализа и исчисления бесконечно малых: с многочисл. прим. для упраж. : предисл. авт./ Эрнесто Чезаро ; пер. с нем. с примеч. и доп. К. А. Поссе. Ч. 2 : с 71 черт. Выходные данные : Одесса: Изд. "Матезис". Тип. Акц. Юж.-Рус. О-ва Печ. Дела, 1914 Колич.характеристики :VIII, 480 с Цена : Б.ц. Предметные рубрики: Алгебра Математика Числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 72. - [x]Вид документа : Однотомное издание Шифр издания : 517.2/К1 Автор(ы) : Кашин Н. В. Заглавие : Основания математического анализа : учеб. кн. для ст. кл. сред. шк. Выходные данные : Б.м., 1916 Колич.характеристики :XIII, 621 с.: ил. Цена : Б.ц. Предметные рубрики: Математичний аналіз Аналіз Математика Функціональний аналіз Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 73. - [x]Вид документа : Однотомное издание Шифр издания : 511/Ш 82 Автор(ы) : Шохор-Троцкий С.И. Заглавие : Методика арифметики : для учителей сред. учеб. заведений . -4-е изд., пересм. Выходные данные : Б.м., 1916 Колич.характеристики :XIV, 524 с Цена : Б.ц. Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 74. - [x]Вид документа : Однотомное издание Шифр издания : 51/81 Автор(ы) : Фосс А. Заглавие : Сущность математики Выходные данные : Б.м., 1923 Колич.характеристики :117 с Цена : Б.ц. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 75. - [x]Вид документа : Однотомное издание Шифр издания : 51/П 58 Автор(ы) : Попов Г.Н. Заглавие : Очерки по истории математики : для учащихся,студ.рабфаков и любителей математики Выходные данные : Б.м., 1923 Колич.характеристики :166 с Цена : Б.ц. Предметные рубрики:математика історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 76. - [x]Вид документа : Однотомное издание Шифр издания : 5/В 75 Автор(ы) : Воронов Н. Н. Заглавие : Образцы конкурсных задач по алгебре, геометрии, тригонометрии и физике, предлагавшихся на экзаменах в Киевских и Московских ВУЗ'ах и ВТУЗ'ах в годы: 1927, 1928 и 1929. С подобными решениями и главнейшими формулами . -Изд. 3-е, доп. и испр. Выходные данные : Б.м., 1930 Колич.характеристики :210 с.: ил Цена : (В пер.) : 5 р. Предметные рубрики: фізико-математичні дисципліни фізико-математичні науки математика фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 77. - [x]Вид документа : Однотомное издание Шифр издания : 511/Т 60 Автор(ы) : Торндайк Э. Л. Заглавие : Психология арифметики Выходные данные : М. ; Л. : Учпедгиз, 1932 Колич.характеристики :302 с Цена : (В пер.) : 2 р. 55 к. Предметные рубрики:математика методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 78. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 87 Автор(ы) : Брилинг С. Р. Заглавие : Математический справочник : для студентов, инженеров и техников . -Изд. 2-е доп Выходные данные : Б.м., 1933 Колич.характеристики :170 с.: ил. ISBN, Цена (в пер.): 4 р. Предметные рубрики:математика довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 79. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 87 Автор(ы) : Бритман М. С. Заглавие : Исследование методов интегрирования дифференциальных уравнений второго порядка Выходные данные : Б.м., 1933 Колич.характеристики :62 с Цена : 2 руб. 50 к. Предметные рубрики:Математика Рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 80. - [x]Вид документа : Однотомное издание Шифр издания : 51/Б 87 Автор(ы) : Бритман М. С. Заглавие : Разложение функций в ряд Фурье Выходные данные : Б.м., 1933 Колич.характеристики :16 с Цена : 65 к. Предметные рубрики:Математика Тригонометричні функції Ряди (мат.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 81. - [x]Вид документа : Однотомное издание Шифр издания : 517.2/Л 77 Автор(ы) : Лопиталь де Г. Ф. Заглавие : Анализ бесконечно малых Выходные данные : Б.м., 1935 Колич.характеристики :431 с Цена : (В пер.): 8 руб. Предметные рубрики:математика диференціальне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 82. - [x]Вид документа : Однотомное издание Шифр издания : 51/Э 32 Заглавие : Леонард Эйлер,1707-1783 : сб.ст. и материалов к 150-летию со дня смерти Выходные данные : Б.м., 1935 Колич.характеристики :238 с Серия: Труды института истории наууки и техники/ Серия 2 ; вып.1 Примечания : Парал. тит. л.: фр. Цена : (В пер.): 15 р. Предметные рубрики:математика - історія математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 83. - [x]Вид документа : Однотомное издание Шифр издания : 518/Д 33 Автор(ы) : Денисюк И. Н. Заглавие : Что такое номограмма и как ею пользоваться Выходные данные : Б.м., 1935 Колич.характеристики :55, с.: ил. Цена : 1 р. 20 к. Предметные рубрики: номограмма математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 84. - [x]Вид документа : Однотомное издание Шифр издания : 513/А 46 Автор(ы) : Александров И. Заглавие : Геометрические задачи на построение и методы их решения : учеб.пособие . -2-е изд. под ред.и с доп.С.Ю.Калецкого Выходные данные : Б.м., 1936 Колич.характеристики :200 с.: черт. Цена : (В пер.): 7 р 20 к. Предметные рубрики:математика геометрія тригонометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 85. - [x]Вид документа : Однотомное издание Шифр издания : 518/Т 12 Заглавие : Таблицы Барлоу квадратов,кубов,корней квадратных,корней кубичных и обратных величин целых чисел от 1 до 10000 : с добавл.7 номогр.для облегчения интерполяции и для подсчета погреш. . -Изд.2-е Выходные данные : Б.м., 1936 Колич.характеристики :233 с.: табл. Примечания : Парал. тит. л.: англ. Цена : (В пер.): 6 р.50 к. Предметные рубрики:математика обчислювальна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 86. - [x]Вид документа : Однотомное издание Шифр издания : 51/К 62 Автор(ы) : Кольман Э. Заглавие : Предмет и метод современной математики Выходные данные : Б.м., 1936 Колич.характеристики :315 с Цена : (В пер.): 8 р.25 к. Предметные рубрики:математика фізико-математичні науки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 87. - [x]Вид документа : Однотомное издание Шифр издания : 518/С 74 Заглавие : Справочник по номографии Выходные данные : Б.м., 1937 Колич.характеристики :276 с Цена : (В пер.): 8 р.25 к. Предметные рубрики: номографія математика довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 88. - [x]Вид документа : Однотомное издание Шифр издания : 517/Ф 59 Автор(ы) : Фиников С. П. Заглавие : Проективно-дифференциальная геометрия Выходные данные : Б.м., 1937 Колич.характеристики :263 с Цена : (В пер.) : 4 р 75 к. Предметные рубрики: геометрія математика проективна геометрія диференціальна геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 89. - [x]Вид документа : Многотомное издание Шифр издания : 517/С 23 Заглавие : Сборник задач по высшей математике: учеб.пособие/ под ред.Н.М.Гюнтера и Р.О.Кузьмина. Ч.1 . -Изд.9-е,испр. Выходные данные : Л. ; М.: ОНТИ НКТП СССР, 1937 Колич.характеристики :272 с Цена : (В пер.): 3 р.55 к. Предметные рубрики: вища математика математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 90. - [x]Вид документа : Многотомное издание Шифр издания : 6П5/С 74 Заглавие : Справочное руководство по машиностроению/ под общ.ред.В.М.Майзель. Т.1: Математика Выходные данные : Х.: ОНТИ : ДНТВУ : НКТП, 1937 Колич.характеристики :764 с.: табл. Цена : (В пер.): 12 р.50 к. Предметные рубрики: машинознавство математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 91. - [x]Вид документа : Однотомное издание Шифр издания : 513/Д 28 Автор(ы) : Декарт, Ренэ Заглавие : Геометрия с приложением избранных работ П. Ферма и переписки Декарта Выходные данные : Москва; Ленинград: Научтехиздат НКТП СССР, 1938 Колич.характеристики :296 с.: ил Серия: Классики естествознания. Физика, механика, математика, астрономия Цена : 8 р. Предметные рубрики: Геометрія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 92. - [x]Вид документа : Однотомное издание Шифр издания : 51/А 14 Автор(ы) : Абельсон И.Б. Заглавие : Две прогрессии Выходные данные : Б.м., 1938 Колич.характеристики :163 с.: черт. Серия: Научно-популярная серия "Академия Наук - стахановцам"/ под общ.ред.В.Л.Комарова Цена : (В пер.): 4 р. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 93. - [x]Вид документа : Многотомное издание Шифр издания : 51/Г 91 Автор(ы) : Грэнвиль В. Заглавие : Курс дифференциального и интегрального исчислений: учеб. для вузов/ Н. Лузин. Ч.2: Интегральное исчесление . -Изд. 6-е Выходные данные : М.; Л.: ГОНТИ, 1938 Колич.характеристики :239 с.: ил Цена : (В пер.) : 4 р 50 к. Предметные рубрики: Інтегральне числення математика диференціальне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 94. - [x]Вид документа : Многотомное издание Шифр издания : 517/С 23 Заглавие : Сборник задач по высшей математике: учеб. пособие для втузов и вузов/ под ред. Н. М. Гюнтера и Р. О. Кузьмина. - Изд. 10-е, испр. Ч. 2 Выходные данные : Л. ; М.: ГОНТИ НКТП СССР, 1938 Колич.характеристики :296 с Цена : (В пер.) : 5 р. 65 к. Предметные рубрики: вища математика математика математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 95. - [x]Вид документа : Многотомное издание Шифр издания : 517/С 23 Заглавие : Сборник задач по высшей математике: учеб.пособие/ под ред.Н.М.Гюнтера и Р.О.Кузьмина. Ч.3 . -Изд.2-е, испр. Выходные данные : Л. ; М.: ОНТИ НКТП СССР, 1938 Колич.характеристики :339 с Цена : (В пер.): 4 р.75 к. Предметные рубрики: вища математика математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 96. - [x]Вид документа : Однотомное издание Шифр издания : 51/Г 97 Автор(ы) : Гутерман И. В. Заглавие : Таблицы умножения трехзначных чисел на двузначные Выходные данные : Б.м., 1939 Цена : (В пер.) : 2 р. 80 к. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 97. - [x]Вид документа : Многотомное издание Шифр издания : 518/М 34 Заглавие : Математические таблицы/ табл.перераб.И.А.Гутерман и К.Г.Ивановой ; под ред.А.М.Журавского. Т.1 Выходные данные : М.: В/О Союзгоручет, 1939 Колич.характеристики :339 с Цена : (В пер.): 5 р.25 к. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 98. - [x]Вид документа : Однотомное издание Шифр издания : 51/С 23 Заглавие : Сборник,посвященный памяти академика Дмитрия Александровича Граве Выходные данные : Б.м., 1940 Колич.характеристики :326 с., л.портр. Цена : (В пер.): 10 р. Предметные рубрики:математика математики Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 99. - [x]Вид документа : Многотомное издание Шифр издания : 518/М 34 Заглавие : Математические таблицы/ табл.сост.И.А.Гутерман и К.Г.Ивановой ; под ред.А.М.Журавского. Т.2 Выходные данные : М.: БИЗ ЦУНХУ Госплана СССР, 1940 Колич.характеристики :232 с Цена : (В пер.): 3 р.70 к. Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 100. - [x]Вид документа : Однотомное издание Шифр издания : 518/Х-86 Автор(ы) : Хохлов, Александр Иванович Заглавие : Карманные математические таблицы : пятизначные Выходные данные : Б.м., 1940 Колич.характеристики :207 табл. Цена : (В пер.) : 4 р. Предметные рубрики: Обчислювання Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 101. - [x]Вид документа : Однотомное издание Шифр издания : 51/Л 68 Заглавие : Николай Иванович Лобачевский (1793-1856) : сб. ст. Выходные данные : Б.м., 1943 Колич.характеристики :83 с.: ил Цена : 4 р. Предметные рубрики:математика історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 102. - [x]Вид документа : Однотомное издание Шифр издания : 518/А 92 Заглавие : Атлас номограмм : номограммы сконструированы и выполнены в НИИ математики МГУ под рук. Е. Б. Карпина Выходные данные : Б.м., 1943 Колич.характеристики :37 с Цена : 17 р. Предметные рубрики: номографія математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 103. - [x]Вид документа : Многотомное издание Шифр издания : 517/Ч-34 Автор(ы) : Чебышев, Пафнутий Львович Заглавие : Полное собрание сочинений/ Пафнутий Львович Чебышев ; АН СССР. Т. 1: Теория чисел Выходные данные : Москва ; Ленинград: Изд-во АН СССР, 1944 Колич.характеристики :342 с ISBN, Цена (В пер.): 23 р. Предметные рубрики: Чисел теорій Математика Груп теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 104. - [x]Вид документа : Однотомное издание Шифр издания : 51/К 56 Автор(ы) : Ковалевская, Софья Васильевна Заглавие : Воспоминания детства и автобиографические очерки Выходные данные : Б.м., 1945 Колич.характеристики :227 с.: ил Цена : (В пер.) : 20 р. Предметные рубрики:математика історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 105. - [x]Вид документа : Многотомное издание Шифр издания : 5/М 82 Заглавие : Московский университет памяти Исаака Ньютона,1643 - 1943. Московский университет памяти Исаака Ньютона,1643-1943 Выходные данные : Б.м., 1946 Колич.характеристики :106 с., л.портр. Цена : 9 р. Предметные рубрики: природничі науки фізико-математичні науки фізика хімія математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 106. - [x]Вид документа : Однотомное издание Шифр издания : 517/В 49 Автор(ы) : Виноградов С. П. Заглавие : Краткий курс высшей математики : учебник . -9-е изд.,перераб.Б.В.Кутузовым Выходные данные : Б.м., 1946 Колич.характеристики :295 с Цена : (В пер.): 10 р.80 к. Предметные рубрики: Вища математика Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 107. - [x]Вид документа : Однотомное издание Шифр издания : 513/К 72 Автор(ы) : Костин В. И. Заглавие : Н. И. Лобачевский и его геометрия Выходные данные : Б.м., 1947 Колич.характеристики :73, с.: ил. Цена : 1 р. 60 к. Предметные рубрики:Математика Геометрія обчислювання Історія геометрії Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 108. - [x]Вид документа : Однотомное издание Шифр издания : 517/А 46 Автор(ы) : Александров, Павел Сергеевич Заглавие : Комбинаторная топология Выходные данные : Б.м., 1947 Колич.характеристики :660 с Цена : (В пер.) : 31 р.70 к. Предметные рубрики: Топологія Математика Вузли (мат.) Графи (мат.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 109. - [x]Вид документа : Однотомное издание Шифр издания : 517.1/Т 46 Автор(ы) : Тичмарш Е. К. Заглавие : Дзета-функция Римана Выходные данные : Б.м., 1947 Колич.характеристики :155 с Примечания : Парал. тит. л.: англ. Цена : (В пер.) : 11 р. Предметные рубрики: чисел теорія математика аналітичні функції дзета функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 110. - [x]Вид документа : Однотомное издание Шифр издания : 517.11/Д 29 Автор(ы) : Делоне, Борис Николаевич Заглавие : Петербургская школа теории чисел Выходные данные : Б.м., 1947 Колич.характеристики :419, с.: ил. Коллективы : АН СССР Серия: Научно-популярная серия Цена : (В пер.) : 20 р. Предметные рубрики:Математика Чисел теорія Історія математики Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 111. - [x]Вид документа : Многотомное издание Шифр издания : 512/В 17 Автор(ы) : Ван Дер Варден Б. Л. Заглавие : Современная алгебра/ Б. Л. Ван Дер Варден. Ч.2 Выходные данные : М. ; Л.: ОГИЗ. Гостехиздат, 1947 Колич.характеристики :260 с Цена : (В пер.): 13 р. Предметные рубрики: лінійна алгебра математика груп теорія (мат.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 112. - [x]Вид документа : Однотомное издание Шифр издания : 513/Э 30 Автор(ы) : Эйзенхарт Л. П. Заглавие : Риманова геометрия Выходные данные : Б.м., 1948 Колич.характеристики :315 с Примечания : Парал. тит. л.: англ. Цена : (В пер.) : 17 р.30 к. Предметные рубрики: Геометрія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 113. - [x]Вид документа : Однотомное издание Шифр издания : 51/К 12 Автор(ы) : Каган, Вениамин Фёдорович Заглавие : Великий русский ученый Н. И. Лобачевский и его место в мировой науке Выходные данные : Б.м., 1948 Колич.характеристики :83 с., л.портр. Цена : 1 р.50 к. Предметные рубрики:математика історія математики математики геометрія історія геометрії Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 114. - [x]Вид документа : Однотомное издание Шифр издания : 51/М 34 Заглавие : Материалы для биографии Н.И.Лобачевского Выходные данные : Б.м., 1948 Колич.характеристики :826 с., л.портр.: ил. Цена : (В пер.): 70 р. Предметные рубрики:математика історія математики математика геометрія історія геометрії Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 115. - [x]Вид документа : Однотомное издание Шифр издания : 51/К 12 Автор(ы) : Каган, Вениамин Фёдорович Заглавие : Лобачевский . -Изд. 2-е., доп. Выходные данные : Б.м., 1948 Колич.характеристики :503 с., л.ил.: ил. Серия: Научно-популярная серия : биогр. Цена : (В пер.) : 25 р. Предметные рубрики: геометрія історія геометрії математика історія математики Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 116. - [x]Вид документа : Многотомное издание Шифр издания : 512/А 45 Заглавие : Алгебраический реферативный сборник: за 1941-1946 гг. ; [в 3-х вып.]/ ред. А. Г. Курош. Вып. 1: Алгебра полиномов. Линейная алгебра. Теория полей. Теория колец и алгебр Выходные данные : М.: Иностр. лит., 1948 - Колич.характеристики :248 с Цена : (В пер.): 18 р.25 к. Предметные рубрики: алгебра математика лінейна алгебра вища алгебра груп теорія (мат.) топологічна алгебра кілець і полів теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 117. - [x]Вид документа : Однотомное издание Шифр издания : 51/К 56 Автор(ы) : Ковалевская, Софья Васильевна Заглавие : Научные работы Выходные данные : Б.м., 1948 Колич.характеристики :305 с Цена : (В пер.): 22 р.50 к. Предметные рубрики:математика історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 118. - [x]Вид документа : Многотомное издание Шифр издания : 512/А 45 Заглавие : Алгебраический реферативный сборник: за 1941-1946 гг. ; [в 3-х вып.]/ ред.А.Г.Курош. Вып. 3: Обобщение групп,колец и структур. Топологическая алгебра. Группы и алгебры Ли.Алгебра анализа книги Выходные данные : М.: Иностр.лит., 1948 - Колич.характеристики :167 с Цена : (В пер.): 18 р.25 к. Предметные рубрики: алгебра математика лінейна алгебра вища алгебра груп теорія (мат.) топологічна алгебра кілець і полів теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 119. - [x]Вид документа : Многотомное издание Шифр издания : 512/А 45 Заглавие : Алгебраический реферативный сборник: за 1941-1946 гг. ; [в 3-х вып.]/ ред.А.Г.Курош. Вып. 2: Теория групп. Теория структур Выходные данные : М.: Иностр.лит., 1948 - Колич.характеристики :198 с Цена : (В пер.): 15 р.25 к. Предметные рубрики: алгебра математика лінейна алгебра вища алгебра груп теорія (мат.) топологічна алгебра кілець і полів теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 120. - [x]Вид документа : Однотомное издание Шифр издания : 53/Г 85 Автор(ы) : Гринберг Г.А. Заглавие : Избранные вопросы математической теории электрических и магнитных явлений Выходные данные : М. ; Л.: Изд-во Акад. Наук СССР, 1948 Колич.характеристики :727 с Цена : (В пер.) : 48 р. 50 к. Предметные рубрики: математична фізика математика фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 121. - [x]Вид документа : Многотомное издание Шифр издания : 517/Г 99 Автор(ы) : Гюнтер Н. М. Заглавие : Сборник задач по высшей математике: учеб.пособие для вузов/ Н. М. Гюнтер, Р.О.Кузьмин. Т.2 . -Изд-12-е,испр. Выходные данные : Л. ; М.: Гостехиздат, 1949 Колич.характеристики :223 с Цена : (В пер.): 7 р. Предметные рубрики: вища математика математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 122. - [x]Вид документа : Многотомное издание Шифр издания : 517/С 50 Автор(ы) : Смирнов, Владимир Иванович Заглавие : Курс высшей математики: учебник/ Владимир Иванович Смирнов. Т.3,ч.1 . -Изд.4-е Выходные данные : Л. ; М.: Гостехиздат, 1949 Колич.характеристики :335 с Цена : (В пер.): 11 р.20 к. Предметные рубрики: вища математика математика навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 123. - [x]Вид документа : Однотомное издание Шифр издания : 517/В 75 Автор(ы) : Вороной Г. Ф. Заглавие : Собрание сочинений: в 3 т./ Г. Ф. Вороной. Т. 2 Выходные данные : К.: Изд-во АН УССР, 1952 Колич.характеристики :390, с.: портр. Цена : (В пер. ) 28 грн 80 к. Предметные рубрики: чисел теорія математика аналітичні функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 124. - [x]Вид документа : Однотомное издание Шифр издания : 51/З-11 Заглавие : З досвіду роботи вчителів математики і креслення Выходные данные : Б.м., 1959 Колич.характеристики :28 с Цена : 10 к. Предметные рубрики:Математика Креслення Вчителі математики Географич. рубрики: Миколаїв, місто (Мк) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 125. - [x]Вид документа : Однотомное издание Шифр издания : 4И(анг)/А 64 Заглавие : Англо-русский словарь математических терминов Выходные данные : Б.м., 1962 Колич.характеристики :371 с. Коллективы : АН СССР ; Мат. ин-т им. В. А. Стеклова Цена : (В пер.) : 2 р 12 к. Предметные рубрики: Англійська мова Словники Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 126. - [x]Вид документа : Многотомное издание Шифр издания : 530/П 78 Заглавие : Проблемы математической физики. Вып.1: Спектральная теория и волновые процессы Выходные данные : Л.: Изд-во Ленингр.ун-та, 1966 Колич.характеристики :132 с Цена : 60 к. Предметные рубрики: математична фізика математика фізика спектроскопія хвилі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 127. - [x]Вид документа : Многотомное издание Шифр издания : 530/П 78 Заглавие : Проблемы математической физики. Вып.8: Дифференциальные уравнения. Спектральная теория. Теория распространения волн Выходные данные : Л.: Изд-во Ленингр.ун-та, 1976 Колич.характеристики :176 с Цена : 1 р.5 к. Предметные рубрики: математична фізика математика фізика спектроскопія хвилі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 128. - [x]Вид документа : Многотомное издание Шифр издания : Б 27665 22.16/Д 13 Автор(ы) : Давидов, Микола Олексійович, Давидов, Микола Олексійович Заглавие : Курс математичного аналізу: підрруч./ Микола Олексійович Давидов. Т. 2: Функції багатьох змінних і диференціальні рівняння . -2-е изд., перероб. і доп. Выходные данные : : Вища школа Б.м., 1978 - 1991 Колич.характеристики :366 с Серия: ISBN, Цена 5-11-002284-4: (в опр.) : 3 р 20 к., р. ББК : 22.161я73 Предметные рубрики:Математика Аналіз Диференціальні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 129. - [x]Вид документа : Однотомное издание Шифр издания : 6П5.1/О-28 Заглавие : Общетехнический справочник . -2-е изд. перераб. и доп. Выходные данные : Москва: Машиностроение, 1982 Колич.характеристики :415 с.: ил. Серия: Серия справочников для рабочих Цена : (В пер.) : 1 р. 70 к. Предметные рубрики: Техніка Математика Технічна механіка Електротехніка Допуски (техн.) Посадки (техн.) Вимірювання технічні Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 130. - [x]Вид документа : Многотомное издание Шифр издания : Пр 52-61 22.19/В 94 Заглавие : Вычислительная и прикладная математика: республиканский междуведомственный научный сборник. Вып. 61 Выходные данные : Киев: Вища шк., 1987 Колич.характеристики :128 с Коллективы : Киев. орденов Ленина и Окт. революции Гос. ун-т им. Т. Г. Шевченко Серия: Цена : 1 р. 50 к. ББК : 22.19 Предметные рубрики: Природничі науки Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 131. - [x]Вид документа : Многотомное издание Шифр издания : Пр 52-63 Заглавие : Вычислительная и прикладная математика: республиканский междуведомственный научный сборник. Вып. 63 Выходные данные : Киев: Вища шк., 1987 Колич.характеристики :123 с Коллективы : Киев. орденов Ленина и Окт. революции Гос. ун-т им. Т. Г. Шевченко Цена : 1 р. 40 к. ББК : 22.19 Предметные рубрики: Природничі науки Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 132. - [x]Вид документа : Многотомное издание Шифр издания : Пр 52-62 Заглавие : Вычислительная и прикладная математика: республиканский междуведомственный научный сборник. Вып. 62 Выходные данные : Киев: Вища шк., 1987 Колич.характеристики :133 с Коллективы : Киев. орденов Ленина и Окт. революции Гос. ун-т им. Т. Г. Шевченко Цена : 1 р. 50 к. ББК : 22.19 Предметные рубрики: Природничі науки Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 133. - [x]Вид документа : Однотомное издание Шифр издания : Г 19 22.1/С 88 Автор(ы) : Стюарт, Иен Заглавие : Тайны катастрофы Выходные данные : Б.м., 1987 Колич.характеристики :76 с Цена : 1 р 50 к. ББК : 22.17 Предметные рубрики:Математика Техніка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 134. - [x]Вид документа : Однотомное издание Шифр издания : А 5775 22.1/О 66 Автор(ы) : Орлов, Всеволод Борисович, Скороход , Нина Сергеевна, Сосинский, Алексей Брониславович Заглавие : Русско-англо-немецко-французский математический словарь : основ. термины : ок. 3 000 терминов Выходные данные : Б.м., 1987 Колич.характеристики :304 с Цена : 6 000 крб. ББК : 22.1я2 Предметные рубрики: Словники Математика Англійська мова Французька мова Російська мова Термінологія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 135. - [x]Вид документа : Однотомное издание Шифр издания : Б 865 22.1/А 61 Автор(ы) : Амелькин В. В. Заглавие : Дифференциальные уравнения в приложениях Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1987 Колич.характеристики :160 с. Цена : 20 к. ББК : 22.151.61 Предметные рубрики: Диференційні рівняння Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 136. - [x]Вид документа : Однотомное издание Шифр издания : Б 219 22.15/В 19 Автор(ы) : Васильев, Анатолий Михайлович Заглавие : Теория дифференциально-геометрических структур : учеб. пособие для вузов Выходные данные : Б.м., 1987 Колич.характеристики :192 с Примечания : Библиогр.: с. 188-190 Цена : 40 к. ББК : 22.151.6я73 Предметные рубрики: Природничі науки Навчальні видання Математика Геометричні структури Диференційні числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 137. - [x]Вид документа : Однотомное издание Шифр издания : Б 88 22.1/Б 74 Автор(ы) : Богомолов, Николай Васильевич, Сергиенко, Людмила Юльевна Заглавие : Сборник дидактических заданий по математике : учеб. пособие Выходные данные : Москва: Выс. шк., 1987 Колич.характеристики :192 с.: ил. Цена : Б.ц. ББК : 22.1я723 Предметные рубрики:Математика Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 138. - [x]Вид документа : Однотомное издание Шифр издания : Б 779 22.16/Л 64 Автор(ы) : Литвинов, Вильям Григорьевич Заглавие : Оптимизация в эллиптических граничных задачах с приложением к механике Выходные данные : Б.м., 1987 Колич.характеристики :366 с Примечания : Библиогр.: с. 353-366 Цена : (в пер.) : 3 р 80 к. ББК : 22.161.62 Предметные рубрики:Математика Оптимізація математична Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 139. - [x]Вид документа : Однотомное издание Шифр издания : Б 1229 22.17/В 19 Автор(ы) : Васильев, Всеволод Викторович, Ралдугин, Евгений Александрович Заглавие : Электронные модели задач на графах Выходные данные : Б.м., 1987 Колич.характеристики :152 с Коллективы : АН УССР; Институт проблем моделирования в энергетике Примечания : Библиогр.: с. 150-152 Цена : 1 р. 30 к. ББК : 22.174.2 Предметные рубрики: Природничі науки Математика Моделі задач-- електронні Теорія графов Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 140. - [x]Вид документа : Однотомное издание Шифр издания : Б 4712 22.1/М 34 Заглавие : Математика сегодня : науч.-метод. сб. Выходные данные : Б.м., 1987 Колич.характеристики :231 с Цена : 1 р. 80 к. ББК : 22.1 Предметные рубрики: Природничі науки Математика Математичні проблеми Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 141. - [x]Вид документа : Однотомное издание Шифр издания : Б 2743 22.1/З-13 Автор(ы) : Заварыкин В. М., Житомирский В. Г., Лапчик М. П. Заглавие : Техника вычислений и алгоритмизаций : учеб. пособие Выходные данные : Б.м., 1987 Колич.характеристики :159 с Примечания : Библиогр.: с. 157 Цена : 25 к. ББК : 22.12 Предметные рубрики: Природничі науки Навчальні видання Математика Мікропроцесорна техніка Використання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 142. - [x]Вид документа : Однотомное издание Шифр издания : Г 19. 22.17/С 88 Автор(ы) : Стюарт, Иен Заглавие : Тайны катастрофы Выходные данные : Б.м., 1987 Колич.характеристики :76 ил Цена : (В пер.) : 1 р. 50 к. ББК : 22.17 Предметные рубрики: Природничі науки Математика Теорія катастроф Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 143. - [x]Вид документа : Однотомное издание Шифр издания : Б 8406 22.1/Т 35 Автор(ы) : Теребилов, Олег Федорович Заглавие : Логика математического мышления : монография Выходные данные : Б.м., 1987 Колич.характеристики :191 с. Примечания : Бібліогр. в підрядк. прим. Цена : 65 к. ББК : 22.12 Предметные рубрики:Математика Логіка мислення Математичне мислення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 144. - [x]Вид документа : Однотомное издание Шифр издания : Б 2895 22.18/Т 33 Заглавие : Теория управляющих систем : сб. науч. трудов Выходные данные : Б.м., 1987 Колич.характеристики :284 с. Примечания : Бібліогр. в кінці ст. Цена : 2 р. 50 к. ББК : 22.181 Предметные рубрики:Математика Управляючі счистеми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 145. - [x]Вид документа : Однотомное издание Шифр издания : Б 278 22.1/Е 72 Автор(ы) : Ермаков С. М., Жиглявский А. А. Заглавие : Математическая теория оптимального эксперимента : учеб. пособие для вузов Выходные данные : Б.м., 1987 Колич.характеристики :319 с Примечания : Библиогр.: с. 315-316 Цена : (в пер.) : 1 р. ББК : 22.18 Предметные рубрики: Природничі науки Навчальні видання Математика Математична теорія Експеримент Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 146. - [x]Вид документа : Однотомное издание Шифр издания : Б 1735 22.16/С 17 Автор(ы) : Самойленко, Анатолий Михайлович, Перестюк, Николай Алексеевич Заглавие : Дифференциальные уравнения с импульсным воздействием Выходные данные : Б.м., 1987 Колич.характеристики :287 с Примечания : Библиогр.: с. 280-285 Цена : (В пер.) : 2 р. 80 к. ББК : 22.16 Предметные рубрики: Природничі науки Навчальні видання Математика Диференціальні рівняння Теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 147. - [x]Вид документа : Однотомное издание Шифр издания : Б 272 22.1/Ж 91 Автор(ы) : Журбенко И. Г. Заглавие : Анализ стационарных и однородных случайных систем : учеб. пособие Выходные данные : Б.м., 1987 Колич.характеристики :239 с Примечания : Библиогр.: с. 233-237 Цена : (в пер.) : 85 к. ББК : 22.72я8я73 Предметные рубрики: Природничі науки Математика Навчальні видання Стаціонарні системи Однорідні системи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 148. - [x]Вид документа : Однотомное издание Шифр издания : Б 3124 22.19/О-75 Автор(ы) : Сулима И. М. Заглавие : Основные численные методы и их реализация на микрокалькуляторах : учеб. пособие для вузов Выходные данные : Б.м., 1987 Колич.характеристики :312 с Примечания : Библиогр.: с. 304 Цена : (В пер.) : 75 к. ББК : 22.192я73 Предметные рубрики: Природничі науки Навчальні видання Математика Числові методи Калькулятори Задачі Вирішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 149. - [x]Вид документа : Однотомное издание Шифр издания : Б 1416 22.1/Д 69 Автор(ы) : Дороговцев А. Я. Заглавие : Математический анализ : сборник задач Выходные данные : Б.м., 1987 Колич.характеристики :407 с Цена : (в пер.) : 1 р. 10 к. ББК : 22.161.73 Предметные рубрики: Природничі науки Математика Математичний аналіз Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 150. - [x]Вид документа : Однотомное издание Шифр издания : Б 5124 74.26/М 54 Заглавие : Методика преподавания математики в средней школе: частная методика Выходные данные : Б.м., 1987 Колич.характеристики :416 с Серия: Учебное пособие для педагогических институтов Примечания : Библиогр.: с.410-415 Цена : 1 р 30 к. ББК : 74.262я73 Предметные рубрики: Методика Педагогіка Математика Школознавство Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 151. - [x]Вид документа : Однотомное издание Шифр издания : Б 5342 22.19/Б 30 Автор(ы) : Бахвалов, Николай Сергеевич, Жидков, Николай Петрович, Кобельков, Георгий Михайлович Заглавие : Численные методы : учеб. пособие для вузов Выходные данные : Б.м., 1987 Колич.характеристики :600 с Примечания : Библиогр.: с. 593-595 Цена : (в пер.) : 1 р. 60 к. ББК : 22.192я73 Предметные рубрики: Природничі науки Навчальні видання Математика Чисельні методи Задачі-- розв'язування Обчислювальна техніка - комп'ютери Алгоритми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 152. - [x]Вид документа : Однотомное издание Шифр издания : Б 2349 22.19/М 34 Заглавие : Математические модели и вычислительные методы : сб. трудов Выходные данные : Б.м., 1987 Колич.характеристики :270 с Примечания : Библиогр.: с 269-270 Цена : (В пер.) : 2 р. 90 к. ББК : 22.19 + 22.181 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні моделі Математичне моделювання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 153. - [x]Вид документа : Однотомное издание Шифр издания : Б 3415 22.1/Ш 37 Автор(ы) : Шикин, Евгений Викторович Заглавие : Линейные пространства и отображения : учеб. пособие для вузов Выходные данные : Б.м., 1987 Колич.характеристики :310 с.: ил. Примечания : Библиогр.: с. 307 Цена : (в пер.) : 1 р. ББК : 22.143я73 Предметные рубрики:Математика Лінейний простір Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 154. - [x]Вид документа : Однотомное издание Шифр издания : Б 3767 22.1/М 34 Автор(ы) : Любарский Г. Я. Заглавие : Математическое моделирование и эксперимент Выходные данные : Б.м., 1987 Колич.характеристики :159 с Коллективы : АН УССР Примечания : Библиогр.: с. 155-156 Цена : (В пер.) : 1 р. 70 к. ББК : 22.1 Предметные рубрики: Природничі науки Математика Математичне моделювання Експерименти Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 155. - [x]Вид документа : Однотомное издание Шифр издания : Б 3776 22.13/Ш 56 Автор(ы) : Шидловский, Андрей Борисович Заглавие : Трансцендентные числа Выходные данные : Б.м., 1987 Колич.характеристики :448 с Примечания : Библиогр.: с. 436-447 Цена : (В пер.) : 4 р. 60 к. ББК : 22.137 Предметные рубрики: Природничі науки Математика Трансцендентні числа Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 156. - [x]Вид документа : Однотомное издание Шифр издания : Б 32469 22.1/Б 73 Автор(ы) : Богатырёв, Геннадий Иванович, Боковнев, Олег Александрович Заглавие : Математика для подготовительных курсов техникумов на базе 8 классов средней школы : учеб. пособие для сред. спец. учеб. заведений . -2-е изд., перераб. Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1988 Колич.характеристики :408 с. ISBN, Цена 5-02-013744-8: 9 р. 90 к. ББК : 22.1я723 Предметные рубрики:Математика Алгебра Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 157. - [x]Вид документа : Однотомное издание Шифр издания : Б 17515 22.2/М 34 Заглавие : Математические вопросы теории распространения волн. 18 : сборник работ Выходные данные : Б.м., 1988 Колич.характеристики :184 с Коллективы : ленинградское отд. Матем. ин-та им. В. А. стеклова АН СССР Серия: Записки научных семинаров Ломи; т. 173 Цена : 1 р 90 к. ББК : 22.236.35 Предметные рубрики:Математика Хвильові рухи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 158. - [x]Вид документа : Однотомное издание Шифр издания : Б 7472. 22.17/К 89 Автор(ы) : Кузнецов, Олег Петрович, Адельсон-Бельский, Георгий Максимович Заглавие : Дискретная математика для инженера . -2-е изд., пераб. и доп. Выходные данные : Б.м., 1988 Колич.характеристики :480 с Примечания : Библиогр.: с. 473-475 ISBN, Цена 5-283-01568-7: (в пер.) : 1 р 80 к. ББК : 22.174 Предметные рубрики:Математика Дискретна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 159. - [x]Вид документа : Однотомное издание Шифр издания : Б 8538 22.181/Ж 78 Автор(ы) : Жожикашвили В. А., Вишневский В. М. Заглавие : Сети массового обслуживания : теория и применение к сетям ЭВМ Выходные данные : Б.м., 1988 Колич.характеристики :191 с Примечания : Библиогр.: с. 185-190 ISBN, Цена 5-256-0015-Х: 65 к. ББК : 22.181 Предметные рубрики: Природничі науки Математика Мережі комп'ютерні Обслуговування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 160. - [x]Вид документа : Однотомное издание Шифр издания : Г 160 22/К 65 Автор(ы) : Конфорович А. Г., Андриевская А. М. Заглавие : История развития математики : альбом Выходные данные : Киев: Выща шк., 1988 Колич.характеристики :95 с.: ил Примечания : Библиогр.: с. 95 Цена : 3 р 80 к. ББК : 22.1г Предметные рубрики:Математика Фізико-математичні науки Фізико-математичні дисципліни Історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 161. - [x]Вид документа : Однотомное издание Шифр издания : Б 12373 22.1/П 62 Заглавие : Пособие по математике для поступающих в вузы : учеб. пособие . -3-е изд. перераб. Выходные данные : Б.м., 1988 Колич.характеристики :719 с.: ил. ISBN, Цена 5-02-013745-6: (в пер.) : 1 р 70 к. ББК : 22.1 Предметные рубрики:Математика Рівняння Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 162. - [x]Вид документа : Однотомное издание Шифр издания : А 1830 22.1/М 59 Автор(ы) : Микиша, Анатолий Михайлович, Орлов, Всеволод Борисович Заглавие : Толковый математический словарь : около 2500 терминов Выходные данные : Б.м., 1988 Колич.характеристики :244 с ISBN, Цена 5-200-00-246-х: 90 к. ББК : 22.1я21 Предметные рубрики: Природничі науки Довідкові видання Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 163. - [x]Вид документа : Однотомное издание Шифр издания : Б 7607 22.19/Н 64 Автор(ы) : Никольский, Сергей Михайлович Заглавие : Квадратные формулы . -2-е изд., доп. Выходные данные : Б.м., 1988 Колич.характеристики :255 с Примечания : Библиогр.: с. 248-249 ISBN, Цена 5-02-013786-3: (в пер. ) : 2 р 10 к. ББК : 22.192.1 Предметные рубрики:Математика Квадрат Формули Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 164. - [x]Вид документа : Однотомное издание Шифр издания : Б 12370 74.26/Э-75 Автор(ы) : Эрдниев, Пюрвя Мучкаевич, Эрдниев, Батыр Пюрвяевич Заглавие : Теория и методика обучения математике в начальной школе Выходные данные : Б.м., 1988 Колич.характеристики :208 с ISBN, Цена 5-7155-0121-0: 1 р 20 к. ББК : 74.262 Предметные рубрики: Педагогіка Математика Методика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 165. - [x]Вид документа : Однотомное издание Шифр издания : Б 4712. 22.1/М 34 Заглавие : Математика сегодня : науч.-метод. сб. Выходные данные : Б.м., 1988 Колич.характеристики :166 с Цена : 80 к. ББК : 22.1 Предметные рубрики: Природничі науки Математика Математичні думки Математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 166. - [x]Вид документа : Однотомное издание Шифр издания : А 2127 22.1/В 62 Автор(ы) : Воднев , Владимир Трофимович, Наумович, Адольф Федорович, Наумович, Нил Федорович Заглавие : Математический словарь высшей школы : общая часть . -Изд. 2-е Выходные данные : Б.м., 1988 Колич.характеристики :528 с Примечания : Библиогр.: с. 527 Цена : (в пер.) : 3 р. ББК : 22.1 Предметные рубрики: Природничі науки Довідкові видання Вища школа Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 167. - [x]Вид документа : Однотомное издание Шифр издания : Б 5913 22.16/К 89 Автор(ы) : Кузнецов, Альберт Николаевич, Батрак, Юрий Артемович, Иванова, Александра Георгиевна Заглавие : Интеграл по мере в техническом вузе : учеб. пособие Выходные данные : Б.м., 1988 Колич.характеристики :69 с.: ил. Примечания : Библиогр.: с. 67 Цена : 30 к. ББК : 22.161.12я73 Предметные рубрики:Математика Інтеграл Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 168. - [x]Вид документа : Многотомное издание Шифр издания : М 194-1 22.16/К 88 Автор(ы) : Кудрявцев, Лев Дмитриевич Заглавие : Курс математического анализа: учеб. для вузов : в 3 т./ Лев Дмитриевич Кудрявцев. - Изд. 2-е, перераб. и доп. Т. 1 Выходные данные : Москва: Высш. шк., 1988 Колич.характеристики :712 с Серия: ISBN, Цена 5-06-001290-5: 1 р. 60 к. ББК : 22.161я73 Предметные рубрики:Математика Математичний аналіз Диференціальне числення Гомологічна алгебра Інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 169. - [x]Вид документа : Многотомное издание Шифр издания : М 194-2 22.16/К 88 Автор(ы) : Кудрявцев, Лев Дмитриевич Заглавие : Курс математического анализа: учеб. для вузов : в 3 т./ Лев Дмитриевич Кудрявцев. - Изд. 2-е, перераб. и доп. Т. 2 Выходные данные : Москва: Высш. шк., 1988 Колич.характеристики :575, с.: ил. ISBN, Цена 5-06-001452-5: 1 р. 40 к. ББК : 22.161я73 Предметные рубрики:Математика Математичний аналіз Диференціальне числення Гомологічна алгебра Інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 170. - [x]Вид документа : Однотомное издание Шифр издания : Б 9972 22.181/М 31 Автор(ы) : Маслов, Виктор Павлович, Мясников, Вениамин Петрович, Данилов, Владимир Григорьевич Заглавие : Математическое моделирование аварийного блока Чернобыльской АЭС Выходные данные : Б.м., 1988 Колич.характеристики :144 с Примечания : Библиогр.: с. 144 Цена : 1 р 50 к ББК : 22.181+31.47 Предметные рубрики:Математика Моделювання Чорнобильська АЕС Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 171. - [x]Вид документа : Однотомное издание Шифр издания : Б 7967 22.18/Б 94 Автор(ы) : Бухтияров, Алексей Михайлович, Фролов, Геннадий Дмитриевич, Олюнин, Виктор Юрьевич Заглавие : Сборник задач по программированию на языке ПЛ/1 : учеб. пособие для вузов . -Изд. 3-е, испр. Выходные данные : Б.м., 1988 Колич.характеристики :320 с ISBN, Цена 5-02-013795-2: 70 к. ББК : 22.18 Предметные рубрики: Природничі науки Математика Програмування Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 172. - [x]Вид документа : Однотомное издание Шифр издания : Б 12563 22.18/Ш 42 Автор(ы) : Шелест, Александр Евгеньевич Заглавие : Микрокалькуляторы в физике : справ. пособие Выходные данные : Б.м., 1988 Колич.характеристики :272 с Примечания : Библиогр.: с. 267-268 ISBN, Цена 5-02-013793-6: 90 к. ББК : 22.183.4 Предметные рубрики: Природничі науки Довідкові видання Математика Фізика Мікрокалькулятори Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 173. - [x]Вид документа : Однотомное издание Шифр издания : Б 16184 20.1/А 47 Автор(ы) : Алексеев, Валерий Михайлович Заглавие : Элементарная математика. Решение задач : учеб. пособие . -2-е изд., перераб. и доп. Выходные данные : Киев: Выща шк., 1989 Колич.характеристики :383 с.: ил. Примечания : Библиогр. : с. 380 ISBN, Цена 5-11-001328-4: 85 к. ББК : 22.1 Предметные рубрики: Елементарна математика Математика Алгебра Тригонометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 174. - [x]Вид документа : Однотомное издание Шифр издания : Б 16583 22.1/В 19 Автор(ы) : Васильева, Аделаида Борисовна, Тихонов, Николай Андреевич Заглавие : Интегральные уравнения : учеб. пособие для вузов Выходные данные : Б.м., 1989 Колич.характеристики :158 с Примечания : Библиогр.: с 155-156 ISBN, Цена 5-211-00344-6: 35 к. ББК : 22.161.67я73 Предметные рубрики: Природничі науки Навчальні видання Математика Фізика Інтеграли Рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 175. - [x]Вид документа : Однотомное издание Шифр издания : Б 21233 22.17/Л 55 Автор(ы) : Лидбеттер, Малькольм, Линдгрен, Георг, Хольгер, Ротсен Заглавие : Экстремумы случайных последовательностей и процессов Выходные данные : Б.м., 1989 Колич.характеристики :392 с Примечания : Библиогр.: с. 361-383 ISBN, Цена 0-387-90731-9: (в пер.) : 3 р. 20 к. ББК : 22.17 Предметные рубрики: Природничі науки Математика Теорія Випадкові процеси Екстремальні значення-- теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 176. - [x]Вид документа : Однотомное издание Шифр издания : Б 13683 22.10/Л 68 Автор(ы) : Лобанова, Людмила Володимирівна, Фінкельштейн, Ленід Петрович Заглавие : Вибрані задачі елементарної математики Выходные данные : Б.м., 1989 Колич.характеристики :96 с Серия: Бібліотечка фізико-математичної школи. Математика Примечания : Бібліогр.: с. 94-95 ISBN, Цена 5-11-001325: 20 к. ББК : 22.10 Предметные рубрики:Математика Елементарна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 177. - [x]Вид документа : Однотомное издание Шифр издания : Б 14876 22.1/Б 17 Автор(ы) : Базылев, Вячеслав Тимофеевич Заглавие : Геометрия дифференцируемых многообразий : учеб. пособие для вузов Выходные данные : Б.м., 1989 Колич.характеристики :222 с Примечания : Библиогр.: с. 218-219 ISBN, Цена 5-06-000462-7: 53 к. ББК : 22.151.62я73 Предметные рубрики: Природничі науки Навчальні видання Математика Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 178. - [x]Вид документа : Однотомное издание Шифр издания : Б 17345 Автор(ы) : Бабин , Анатолий Владимирович, Вишик, Марк Иосифович Заглавие : Аттракторы эволюционных уравнений Выходные данные : Б.м., 1989 Колич.характеристики :294 с Коллективы : АН СССР ; Ин-т проблем механики Примечания : Библиогр.: с. 287-292 ISBN, Цена 5-02-006632-Х: (в пер.) : 4 р. ББК : 22.161.61 Предметные рубрики: Природничі науки Математика Математичні методи Диференційні рівняння Аттактори Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 179. - [x]Вид документа : Однотомное издание Шифр издания : Б 17609 22.1/В 15 Автор(ы) : Валуцэ, Иван Иванович, Дилигул, Григорий Дмитриевич Заглавие : Математика для техникумов на базе средней школы : учеб пособия для ср. учеб. завед. . -2-е изд., перераб. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :575 с ISBN, Цена 5-02-013930-0: (в пер.) : 1 р. 40 к. ББК : 22.1я723 Предметные рубрики: Природничі науки Навчальні видання Математика Вектори Комплексні числа Похідна Інтеграл Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 180. - [x]Вид документа : Однотомное издание Шифр издания : Б 17279 22.15/Д 50 Автор(ы) : Кованцов Н. И. Заглавие : Дифференциальная геометрия, топология, тензорный анализ : сборник задач : учеб. пособие . -2-е изд., перераб. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :398 с ISBN, Цена 5-11-001416-7: (в пер.) : 95 к. ББК : 22.15 Предметные рубрики: Природничі науки Навчальні видання Математика Топологія Тензорний аналіз Риманова геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 181. - [x]Вид документа : Однотомное издание Шифр издания : Б 17683 22.17/Ж 24 Автор(ы) : Жалдак М. И., Квитко А. Н. Заглавие : Теория вероятностей с элементами информатики : практикум : учеб. пособие Выходные данные : Б.м., 1989 Колич.характеристики :264 с Примечания : Библиогр.: с. 260 ISBN, Цена 5-11-001356-Х: (в пер.) : 70 к. ББК : 22.17 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Інформатика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 182. - [x]Вид документа : Однотомное издание Шифр издания : Б 17772 22.1/М 34 Автор(ы) : Шеврин Л. Н., Гейн А. Г., Коряков И. О., Волков М. В. Заглавие : Математика : учебник-собеседник для 5-6 классов средней школы Выходные данные : Б.м., 1989 Колич.характеристики :495 с.: ил. Серия: Библиотека учителя математики ISBN, Цена 5-09-001537-6: (в пер.) : 90 к. ББК : 22.1я72 Предметные рубрики:Математика Алгебра Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 183. - [x]Вид документа : Однотомное издание Шифр издания : б 18158 22.19/М 30 Автор(ы) : Марчук, Гурий Иванович Заглавие : Методы вычислительной математики : учеб. пособие . -3-е изд., перераб. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :618 с Примечания : Библиогр.: с. 573-608 ISBN, Цена 5-020014222-0: (в пер.) : 1 р 60 к. ББК : 22.19я73 Предметные рубрики:Математика Диференціальні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 184. - [x]Вид документа : Однотомное издание Шифр издания : Б 13580 22.1/М 21 Автор(ы) : Малюта, Александр Николаевич Заглавие : Гиперкомплексные динамические системы : монография Выходные данные : Львов: Выща шк. ; Изд-во при Львов. ун-те, 1989 Колич.характеристики :120 с. Примечания : Библиогр. : с. 113-116 ISBN, Цена 5-11-000571-0: 1 р. 70 к. ББК : 22.18 Предметные рубрики: Кібернетика Математика Програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 185. - [x]Вид документа : Однотомное издание Шифр издания : Б 22344 22.172/П 75 Автор(ы) : Айвазян С. А., Бухштабер В. М., Енюков И. С., Мешалкин Л. Д. Заглавие : Прикладная статистика : Классификация и снтжение размерности : справ. изд. Выходные данные : Б.м., 1989 Колич.характеристики :607 с.: ил. Примечания : Библиогр.: с. 569-586 ISBN, Цена 5-279-00054-Х: (в пер. ) : 2 р. ББК : 22.172я2 Предметные рубрики:Математика Статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 186. - [x]Вид документа : Однотомное издание Шифр издания : Б 18369 22.1/Ц 97 Автор(ы) : Цыпкин, Александр Геннадиевич, Пинский, Александр Иосифович Заглавие : Справочник по методам решения задач по математике для средней школы . -2-е изд., перераб. и доп. Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1989 Колич.характеристики :576 с. ISBN, Цена 5-02-013792-8: 2 р. ББК : 22.1 Предметные рубрики:Математика Задачі Вища математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 187. - [x]Вид документа : Однотомное издание Шифр издания : Б 15844 22.1/С 23 Автор(ы) : Дыбов П. Т. Заглавие : Сборник задач по математике для поступающих в вузы . -Изд. 2-е, испр. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :271 с ISBN, Цена 5-06-000549-6: 70 к. ББК : 22.1 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 188. - [x]Вид документа : Однотомное издание Шифр издания : Б 16126 22.1/С 32 Автор(ы) : Сергеев, Игорь Николаевич, Олехник , Слав Николаевич, Гашков, Сергей Борисович Заглавие : Примени математику Выходные данные : Б.м., 1989 Колич.характеристики :240 с. ISBN, Цена 5-02-013946-7: 55 к. ББК : 22.1 Предметные рубрики:Математика Прикладна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 189. - [x]Вид документа : Однотомное издание Шифр издания : Б 15995 22.1/Ш 64 Автор(ы) : Ширяев, Альберт Николаевич Заглавие : Вероятность : учеб. пособие . -2-е изд., перераб. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :640 с Примечания : Библиогр.: с. 630-633 ISBN, Цена 5-02-013955: (в пер.) : 1 р 70 к. ББК : 22.171я73 Предметные рубрики:Математика Ймовірностей теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 190. - [x]Вид документа : Многотомное издание Шифр издания : М 194-3 22.16/К 88 Автор(ы) : Кудрявцев, Лев Дмитриевич Заглавие : Курс математического анализа: учеб. для вузов : в 3 т./ Лев Дмитриевич Кудрявцев. - Изд. 2-е, перераб. и доп. Т. 3 Выходные данные : Москва: Высш. шк., 1989 Колич.характеристики :351, с.: ил. ISBN, Цена 5-06-001516-5: 95 к. ББК : 22.161я73 Предметные рубрики:Математика Математичний аналіз Диференціальне числення Гомологічна алгебра Інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 191. - [x]Вид документа : Многотомное издание Шифр издания : Мб 319-1 22.172/С 74 Заглавие : Справочник по прикладной статистике: в 2 т./ под ред. Э. Ллойда, У. Ледермана. Т. 1 Выходные данные : Москва: Финансы и статистика, 1989 Колич.характеристики :508, с Серия: Примечания : парал. тит. л. : англ. - Библиогр. в конце гл. ISBN, Цена 5-279-00245-3: 2 р 50 к. ББК : 22.172я2 Предметные рубрики:Математика Математична статистика Статистика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 192. - [x]Вид документа : Однотомное издание Шифр издания : Б 16984 22.11/Ш 63 Автор(ы) : Шипачев, Виктор Семенович Заглавие : Основы высшей математики : учеб. пособие для вузов Выходные данные : Б.м., 1989 Колич.характеристики :479 с.: ил. ISBN, Цена 5-06-000048-6: (в пер.) : 1 р 10 к. ББК : 22.11 Предметные рубрики:Математика Вища математика Інтеграл Диференційні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 193. - [x]Вид документа : Многотомное издание Шифр издания : Мб 275-1 22.1/К 48 Автор(ы) : Клейн, Феликс Заглавие : Лекции о развитиии математики в ХIХ столетии: в 2 т. / Феликс Клейн ; подгот. к печати Р. Курантом, О. Нейгебауером ; пер. с нем. Н. М. Нагорного ; под. ред. М. М. Постникова. Т. І Выходные данные : Москва: Наука , 1989 - Колич.характеристики :456 с Серия: ISBN, Цена 5-02-013920-3: 2 р 50 к. ББК : 22.1 г Предметные рубрики:Математика Історія Фізико-математичні науки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 194. - [x]Вид документа : Однотомное издание Шифр издания : Б 16964 22.1/Ф 88 Автор(ы) : Фридман, Лев Моисеевич, Турецкий, Евсей Наумович Заглавие : Как научиться решать задачи : кн. для учащихся ст. классов срд. шк. . -2-е изд. дораб. Выходные данные : Б.м., 1989 Колич.характеристики :192 с.: ил. ISBN, Цена 5-09-000596-6: (в пер. ) : 50 к. ББК : 22.1 Предметные рубрики:Математика Методика розрахунків Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 195. - [x]Вид документа : Однотомное издание Шифр издания : Б 14767 22.16/Н 64 Автор(ы) : Никольский, Сергей Михайлович Заглавие : Элементы математического анализа : учеб. пособие . -2-е изд., перераб. и доп. Выходные данные : Б.м., 1989 Колич.характеристики :222 с.: ил. ISBN, Цена 5-02-013957-2: 35 к. ББК : 22.161я73 Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 196. - [x]Вид документа : Однотомное издание Шифр издания : Б 12972 22.1/Н 64 Автор(ы) : Никольская, Инна Львовна, Семенов, Ефим Евстафьевич Заглавие : Учимся рассуждать и доказывать : кн. для учащихся 6-10 кл. сред. шк. Выходные данные : Б.м., 1989 Колич.характеристики :192 с.: ил. ISBN, Цена 5-09-000591-5: (в пер.) : 1 р. ББК : 22.1 Предметные рубрики:Математика Шкільні знання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 197. - [x]Вид документа : Однотомное издание Шифр издания : Б 22174 22.161/Х 12 Автор(ы) : Хавин, Виктор Петрович Заглавие : Основы математического анализа. Дифференциальное и интегральное исчисление функций одной вещественной переменной : учеб. пособие Выходные данные : Б.м., 1989 Колич.характеристики :446 с Примечания : Библиогр.: с. 440-441 ISBN, Цена 5-288-00201-0: (в пер.) : 1 р 30 к. ББК : 22.161 Предметные рубрики:Математика Диференціальні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 198. - [x]Вид документа : Однотомное издание Шифр издания : Б 18499 Заглавие : Пакеты прикладных программ : математическое моделирование Выходные данные : Б.м., 1989 Колич.характеристики :128 с Коллективы : АН СССР Серия: Алгоритмы и алгоритмические языки Примечания : Библиогр.: с. 126 ISBN, Цена 5-02-007178-1: 55 к. ББК : 22.18 Предметные рубрики: Природничі науки Математика Математичне моделювання Прикладні програми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 199. - [x]Вид документа : Однотомное издание Шифр издания : Б 12938 22.18/Б 23 Автор(ы) : Банди, Брайан Заглавие : Основы линейного программирования Выходные данные : Б.м., 1989 Колич.характеристики :176 с Примечания : Библиогр.: с. 168-169 ISBN, Цена 5-256-00186-8: 70 к. ББК : 22.18 Предметные рубрики: Природничі науки Математика Лінійне програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 200. - [x]Вид документа : Однотомное издание Шифр издания : Б 14899 22.1/П 70 Автор(ы) : Прасолов, Виктор Васильевич, Шарыгин, Игорь Федорович Заглавие : Задачи по стереометрии Выходные данные : Б.м., 1989 Колич.характеристики :286 с.: ил. Серия: Библиотека математического кружка; вып. 19 Примечания : Библиогр.: с. 286 ISBN, Цена 5-02-013921: 75 к. ББК : 22.1 Предметные рубрики:Математика Стереометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 201. - [x]Вид документа : Однотомное издание Шифр издания : Б 15624 22.18/С 21 Автор(ы) : Сафонов, Владимир Олегович Заглавие : Языки и методы программирования в системе "Эльбрус" Выходные данные : Б.м., 1989 Колич.характеристики :389 с Примечания : Библиогр.: с. 386-389 ISBN, Цена 5-02-013983-1: (в пер.) : 2 р. 10 к. ББК : 22.183.392 Предметные рубрики: Природничі науки Математика Система "Ельбрус" Мови програмування Ель-76 Паскаль Технології програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 202. - [x]Вид документа : Однотомное издание Шифр издания : Б 26843 22.17/Л 43 Автор(ы) : Емеличев В. А. Заглавие : Лекции по теории графов Выходные данные : Б.м., 1990 Колич.характеристики :384 с Примечания : Библиогр.: с. 375-376 ISBN, Цена 5-02-013992-0: (в пер.) : 1 р. ББК : 22.174я73 Предметные рубрики: Природничі науки Математика Теорія графов Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 203. - [x]Вид документа : Однотомное издание Шифр издания : Б 24975 22.1/В 55 Автор(ы) : Вишенський, Володимир Андрійович, Перестюк, Миколай Олексійович, Самойленко, Анатолій Михайлович Заглавие : Збірник задач з математики : навчальний посібник для підготов. від. вузів Выходные данные : Б.м., 1990 Колич.характеристики :328 с ISBN, Цена 5-11-001702-6: 1 крб 10 к. ББК : 22.1 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 204. - [x]Вид документа : Однотомное издание Шифр издания : Б 25530 22.19/В 75 Автор(ы) : Воробьева, Галина Николаевна Заглавие : Практикум по вычислительной математике : учеб. пособие . -Изд. 2-е, перераб. и доп. Выходные данные : Б.м., 1990 Колич.характеристики :208 с Примечания : Библиогр.: с. 205 ISBN, Цена 5-06-001544-0: 35 к. ББК : 22.19 Предметные рубрики: Природничі науки Навчальні видання Математика Обчислення Практикум Лабораторні работи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 205. - [x]Вид документа : Однотомное издание Шифр издания : Б 30738 22.1/Г 70 Автор(ы) : Городецкий, Василий Васильевич, Нагнибида, Николай Иванович, Настасиев, Павел Павлович Заглавие : Методы решения задач по функциональному анализу : учеб. пособие Выходные данные : Киев: Выща шк., 1990 Колич.характеристики :479 с. Примечания : Библиогр. : с. 473-174 ISBN, Цена 5-11-002126-0: 2 р. 10 к. ББК : 22.16 Предметные рубрики:Математика Прикладна математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 206. - [x]Вид документа : Однотомное издание Шифр издания : Б 23310 22.171/К 56 Автор(ы) : Коваленко, Игорь Николаевич, Гнеденко, Борис Владимирович Заглавие : Теория вероятностей : учебник Выходные данные : Б.м., 1990 Колич.характеристики :328 с Примечания : Библиогр.: с. 321-322 ISBN, Цена 5-11-001842-1: (в пер. ) : 95 к. ББК : 22.171я73 Предметные рубрики:Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 207. - [x]Вид документа : Однотомное издание Шифр издания : Б 21755 22.18/Д 13 Автор(ы) : Давыдов, Эрик Георгиевич Заглавие : Исследование операций : учеб. пособие Выходные данные : Б.м., 1990 Колич.характеристики :383 с ISBN, Цена 5-06-001004-Х: (в пер.) : 1 р 10 к. ББК : 22.18 Предметные рубрики:Математика Дослідження Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 208. - [x]Вид документа : Однотомное издание Шифр издания : Б 20947 22.17/К 59 Автор(ы) : Козлов, Михаил Васильевич Заглавие : Элементы теории вероятностей в примерах и задачах Выходные данные : Б.м., 1990 Колич.характеристики :344 с Примечания : Библиогр.: с. 342 ISBN, Цена 5-211-00312-8: (в пер.) : 1 р 10 к. ББК : 22.17 Предметные рубрики:Математика Теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 209. - [x]Вид документа : Однотомное издание Шифр издания : Б 22712 22.14/Л 36 Автор(ы) : Левин, Михаил Григорьевич Заглавие : Программное обеспечение для решения задач линейной алгебры на СМ ЭВМ Выходные данные : Б.м., 1990 Колич.характеристики :199 с Примечания : Библиогр.: с. 195-196 ISBN, Цена 5-376-00648: 65 к. ББК : 22.143с51я2 Предметные рубрики: Природничі науки Довідкові видання Математика Лнійна алгебра Програмне забезпечення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 210. - [x]Вид документа : Однотомное издание Шифр издания : Б 21054 22.14/Л 86 Автор(ы) : Лурье, Михаил Владимирович, Александров, Борис Иванович Заглавие : Задачи на составление уравнений : учеб. руководство . -3-е изд., перераб. Выходные данные : Б.м., 1990 Колич.характеристики :95 с ISBN, Цена 5-02-014250-6: 20 к. ББК : 22.14 Предметные рубрики:Математика Рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 211. - [x]Вид документа : Однотомное издание Шифр издания : Б 22934 22.1/Л 64 Автор(ы) : Литлвуд Дж. Заглавие : Математическая смесь . -5-е изд., испр. Выходные данные : Б.м., 1990 Колич.характеристики :140 с ISBN, Цена 5-02-014332-4: 0 р 40 к. ББК : 22.1 Предметные рубрики:Математика Історія математики Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 212. - [x]Вид документа : Однотомное издание Шифр издания : Б 23522 22.1/Ш 96 Автор(ы) : Шунда, Никифор Миколайович, Томусяк А. А., Войцеховський А. П. Заглавие : Вступний курс математики : навч. посіб для пед. ін-тів Выходные данные : Б.м., 1990 Колич.характеристики :152 с.: іл. ISBN, Цена 5-11-002332-8: 25 к. ББК : 22.1я73 Предметные рубрики:Математика Рівняння Елементарні функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 213. - [x]Вид документа : Многотомное издание Шифр издания : МБ 428-1 22.1/Т 78 Заглавие : Труды Ленинградского математического общества/ Ленинград. гос. ун-т ; редкол. О. А. Ладыженская [и др.]. Т. 1 Выходные данные : Ленинград: Изд-во Ленинград. ун-та, 1990 Колич.характеристики :247 с Цена : 3 р 20 к. ББК : 22.1я54 Предметные рубрики:Математика Математична фізика Диференційні рівняння Топологія Алгебра Теорія управління Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 214. - [x]Вид документа : Многотомное издание Шифр издания : Мб 411-2 22.152/Б 53 Автор(ы) : Бессе, Артур Ланселот Заглавие : Многообразия Эйнштейна: в 2 т./ Артур Ланселот Бессе ; пер. с англ. Д. В. Алексеевского. Т. 2 Выходные данные : Москва: Мир, 1990 Колич.характеристики :703 с Примечания : Библиогр. : с. 653-681 ISBN, Цена 5-03-002066-7: (в пер.) : 3 р 10 к. ББК : 22.152.3 Предметные рубрики: Топологія Математика Алгебраїчна геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 215. - [x]Вид документа : Многотомное издание Шифр издания : Мб 411-1 22.152/Б 53 Автор(ы) : Бессе, Артур Ланселот Заглавие : Многообразия Эйнштейна: в 2 т./ Артур Ланселот Бессе ; пер. с англ. Д. В. Алексеевского. Т. 1 Выходные данные : Москва: Мир, 1990 Колич.характеристики :318 с Серия: ISBN, Цена 5-03-002065-9: 2 р 50 к. ББК : 22.152.3 Предметные рубрики: Топологія Математика Алгебраїчна геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 216. - [x]Вид документа : Однотомное издание Шифр издания : Б 24936 22.14/К 77 Автор(ы) : Крамор, Виталий Семенович Заглавие : Повторяем и систематизируем школьный курс алгебры и начал анализа Выходные данные : Б.м., 1990 Колич.характеристики :415 с ISBN, Цена 5-09-001295-4: (в пер.) : 95 к. ББК : 22.14 Предметные рубрики: Природничі науки Навчальні видання для самоосвіти-- середня школа Математика Алгебра Початок аналізу Шкільний курс Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 217. - [x]Вид документа : Однотомное издание Шифр издания : Б 19314 22.161/З-15 Автор(ы) : Вавилов В. В. Заглавие : Задачи по математике : начала анализа : справочное пособие Выходные данные : Б.м., 1990 Колич.характеристики :607 с ISBN, Цена 5-02-014201-8: (в пер.) : 2 р. 40 к. ББК : 22.161 Предметные рубрики: Природничі науки Довідкові видання Математика Початок аналізу Задачі-- завдання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 218. - [x]Вид документа : Однотомное издание Шифр издания : Б 21564 22.1/М 48 Автор(ы) : Мельников, Иван Иванович, Сергеев, Игорь Николаевич Заглавие : Как решать задачи по математике на вступительных экзаменах Выходные данные : Б.м., 1990 Колич.характеристики :303 с ISBN, Цена 5-211-00319-5: (в пер.) : 95 к. ББК : 22.1 Предметные рубрики: Природничі науки Математика Вступні іспити Задачі Рішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 219. - [x]Вид документа : Многотомное издание Шифр издания : 22.161/Н 64 Автор(ы) : Николский, Сергей Михайлович Заглавие : Курс математического анализа : В 2 т.: учеб. для вузов/ Сергей Михайлович Николский. ?XXX? Выходные данные : Москва: Наука ; л. ред. физ.-мат. лит., 1990 - 1991 ISBN, Цена 5-02-014424-Х: Б.ц. ББК : 22.161 Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 220. - [x]Вид документа : Шифр издания : Пб-311-1 22.1/Д 79 Автор(ы) : Дубровский, Владимир Натанович, Калинин, Анатолий Тимофеевич Заглавие : Математические заголовки . вып. 1: До и после кубика Выходные данные : Б.м., 1990 Колич.характеристики :144 с.: ил. ISBN, Цена 5-07-000097-7: . 50 к. ББК : 22.1 Предметные рубрики:Математика Головоломки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 221. - [x]Вид документа : Однотомное издание Шифр издания : Б 28903 22.1/Б 87 Автор(ы) : Брадіс, Володимир Модестович Заглавие : Чотиризначні математичні таблиці : для середньої школи Выходные данные : Київ: Радянська школа, 1990 Колич.характеристики :94 с. ISBN, Цена 5-330-01059-4: 45 к. ББК : 22.1 Предметные рубрики:Математика Математичні таблиці Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 222. - [x]Вид документа : Однотомное издание Шифр издания : Б 21957 22.1/С 23 Автор(ы) : Вуколов Э. А. Заглавие : Сборник задач по математике для втузов : методы оптимизации уравнения в частных производных, интегральные уравнения : учеб. пособие для вузов . -Изд. 2-е, перераб. Выходные данные : Б.м., 1990 Колич.характеристики :304 с Примечания : Библиогр.: с. 299 ISBN, Цена 5-02-014338-3: (в пер.) : 90 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Рівняння Похідна Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 223. - [x]Вид документа : Однотомное издание Шифр издания : Б 24178 22.161.5/С 71 Автор(ы) : Лигун А. А. Заглавие : Специальные вопросы теории приближений и оптимального управления распределенными системами : учеб. пособие Выходные данные : Б.м., 1990 Колич.характеристики :208 с Серия: Новое в науке и технике - студентам и учащимся; Вып. 21 Примечания : Библиогр.: с. 207-208 ISBN, Цена 5-11-002588-6: 35 к. ББК : 22.161.547я73 Предметные рубрики: Природничі науки Навчальні видання Математика Функції Тригонометрія Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 224. - [x]Вид документа : Однотомное издание Шифр издания : Б 26255 22.162/Б 48 Автор(ы) : Березанский, Юрий Макарович, Ус, Георгий Федорович, Шефтель, Зиновий Григорьевич Заглавие : Функциональный анализ : курс лекций : учеб. пособие для вузов Выходные данные : Б.м., 1990 Колич.характеристики :600 с Примечания : Библиогр.: с. 589-593 ISBN, Цена 5-11-001329-2: (в пер.) : 1 р. 80 к. ББК : 22.162 Предметные рубрики: Природничі науки Навчальні видання Математика Функціональний аналіз Інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 225. - [x]Вид документа : Многотомное издание Шифр издания : Мб 423-1 2.161/Н 64 Автор(ы) : Никольский, Сергей Михайлович Заглавие : Курс математического анализа: учеб. для вузов/ Сергей Михайлович Никольский. - Изд. 4-е, перераб. и доп. Т. 1 Выходные данные : Москва: Наука, 1990 Колич.характеристики :528 с Серия: ISBN, Цена 5-02014424-Х: (в опр.) :1 р 40 к. ББК : 22.161я73 Предметные рубрики: Математичний аналіз Аналіз Математика Навчальні видання для вищої школи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 226. - [x]Вид документа : Многотомное издание Шифр издания : Мб 319-2 22.172/С 74 Заглавие : Справочник по прикладной статистике: в 2 т./ под ред. Э. Ллойда, У. Ледермана. Т. 2 Выходные данные : Москва: Финансы и статистика, 1990 Колич.характеристики :525, с Примечания : Парал. тит. л. : англ. - Библиогр. в конце гл. ISBN, Цена 5-279-00246-1: 2 р 70 к. ББК : 22.172я2 Предметные рубрики:Математика Математична статистика Статистика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 227. - [x]Вид документа : Однотомное издание Шифр издания : Б 22869 22.161/М 29 Автор(ы) : Мартыненко, Владимир Семенович Заглавие : Операционное исчесление : учеб. пособие . -4-е изд., перераб. и доп. Выходные данные : Б.м., 1990 Колич.характеристики :359 с Примечания : Библиогр.: с. 357 ISBN, Цена 5-11-001846-4: (в пер.) : 1 р. ББК : 22.161 Предметные рубрики:Математика Обчислення (матем.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 228. - [x]Вид документа : Однотомное издание Шифр издания : Б 26202 22.1/К 66 Автор(ы) : Кордемский, Борис Анастасьевич Заглавие : Математическая смекалка . -8-е изд. Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1991 Колич.характеристики :576 с.: ил. ISBN, Цена 5-02-014955-1: 5 р. ББК : 22.1 Предметные рубрики:Математика Математичні ігри Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 229. - [x]Вид документа : Однотомное издание Шифр издания : Б 28152 22.1/К 89 Автор(ы) : Кузнецов, Альберт Николаевич Заглавие : Решение нестандартных задач в курсе математического анализа : метод. указания Выходные данные : Б.м., 1991 Колич.характеристики :54 с.: ил. Примечания : Библиогр.: с. 52-53 Цена : Б.ц. ББК : 22.161кр. Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 230. - [x]Вид документа : Однотомное издание Шифр издания : Б 26642 26.2/Г 68 Автор(ы) : Гордин, Владимир Александрович Заглавие : Математика, компьютер, прогноз, погоды Выходные данные : Б.м., 1991 Колич.характеристики :223 с.: ил. Примечания : Библиогр.: с. 214 ISBN, Цена 5-286-00417-2: 75 к. ББК : 26.2 Предметные рубрики:Математика Погода Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 231. - [x]Вид документа : Однотомное издание Шифр издания : Б 30715 22.1/Ш 26 Автор(ы) : Шарыгин, Игорь Федорович Заглавие : Математический винегрет Выходные данные : Б.м., 1991 Колич.характеристики :64 с ISBN, Цена 5-85934-002-8: 2 р. 60 к. ББК : 22.10 Предметные рубрики: Природничі науки Математика Математичні ігри Математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 232. - [x]Вид документа : Однотомное издание Шифр издания : Б 29662 22.2/А 86 Автор(ы) : Арутюнян , Нагуш Хачатурович, Манжуров , Александр Владимирович, Наумов, Вячеслав Энгельсович Заглавие : Контактные задачи механики растущих тел : монография Выходные данные : Москва: Наука, 1991 Колич.характеристики :175 с.: ил. Примечания : Библиогр. : с. 168-171 ISBN, Цена 5-02-006752-0: 2 р. 40 к. ББК : 22.251 Предметные рубрики:Математика Механіка Деформування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 233. - [x]Вид документа : Однотомное издание Шифр издания : Б 31288 22.16/В 49 Автор(ы) : Виноградова, Ирина Андреевна, Олехник , Слав Николаевич, Садовничий, Виктор Антонович Заглавие : Математический анализ в задачах и упражнениях : учеб. пособие для вузов Выходные данные : Б.м., 1991 Колич.характеристики :352 с ISBN, Цена 5-211-01559-2: (в пер.) : 4 р. 05 к. ББК : 22.16 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Задачі Вправи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 234. - [x]Вид документа : Однотомное издание Шифр издания : Б 27695 22.17/Л 64 Автор(ы) : Литтл Р. Дж. А, Рубин Д.б. Заглавие : Статистический анализ данных с пропусками Выходные данные : Б.м., 1991 Колич.характеристики :334 с Примечания : Библиогр.: с. 330-332 ISBN, Цена 5-279-00443-Х: (в пер.) : 3 р 10 к. ББК : 22.172.6 Предметные рубрики:Математика Статистичні розрахунки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 235. - [x]Вид документа : Однотомное издание Шифр издания : Б 27460 22.1/Б 59 Автор(ы) : Бибиков, Юрий Николаевич Заглавие : Курс обыкновенных дифференциальных уравнений : учеб. пособие Выходные данные : Москва: Высш. шк. , 1991 Колич.характеристики :303 с.: ил. ISBN, Цена 5-06-00106-6: 1 р. 20 к. ББК : 22.161 Предметные рубрики: Диференціальні рівняння Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 236. - [x]Вид документа : Однотомное издание Шифр издания : Б 26428 22.1/Г 56 Автор(ы) : Гнеденко, Борис Владимирович Заглавие : Введение в специальность математика Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1991 Колич.характеристики :240 с. Примечания : Библиогр. : с. 236 ISBN, Цена 5-02-014273-5: 2 р. 50 к. ББК : 22.1 Предметные рубрики:Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 237. - [x]Вид документа : Однотомное издание Шифр издания : Б 30976 22.1/Л 64 Автор(ы) : Литвиненко, Виктор Николаевич, Мордкович, Александр Григорьевич Заглавие : Практикум по элементарной математике : Алгебра. Тригонометрия : учеб. пособие для вузов Выходные данные : Б.м., 1991 Колич.характеристики :349 с ISBN, Цена 5-09-003393-5: (в пер.) : 2 р 56 к. ББК : 22.10я73 Предметные рубрики:Математика Алгоритми Тригонометрія Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 238. - [x]Вид документа : Однотомное издание Шифр издания : Б 28766 22.16/Н 52 Автор(ы) : Лебеда, Валентина Федоровна, Неделько, Евгений Юрьевич, Руденко, Надежда Андреевна, Щеглов, Александр Александрович Заглавие : Неопределенный и определенный интеграл : метод. указания Выходные данные : Б.м., 1991 Колич.характеристики :53 с.: ил. Цена : Б.ц. ББК : 22.161:12 Предметные рубрики:Математика Інтеграл Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 239. - [x]Вид документа : Однотомное издание Шифр издания : Б 32234 22.1/С 40 Автор(ы) : Симонов, Александр Яковлевич, Бакаев, Дмитрий Сергеевич, Эпельман, Александр Гиршевич Заглавие : Система тренировочных задач и упражнений по математике Выходные данные : Москва: Просвещение , 1991 Колич.характеристики :208 с.: ил. ISBN, Цена 5-09-002848-6: 5 р. 8 к. ББК : 22.1 Предметные рубрики: Задачі-- завдання Математика Навчальний посібник Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 240. - [x]Вид документа : Однотомное издание Шифр издания : Б 29920 22.1/С 17 Автор(ы) : Самусенко, Анатолий Васильевич, Казаченок, Виктор Владимирович Заглавие : Математика. Типичные ошибки абитуриентов : справ. пособие Выходные данные : Б.м., 1991 Колич.характеристики :188 с ISBN, Цена 5-339-00585-2: Б.ц. ББК : 22.1я2 Предметные рубрики: Природничі науки Довідкові видання Математика Іспити Конкурсні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 241. - [x]Вид документа : Однотомное издание Шифр издания : Б 26488 22.19/З-13 Автор(ы) : Заварыкин В. М., Житомирский В. Г., Лапчик М. П. Заглавие : Численные методы : учеб. пособие для вузов Выходные данные : Б.м., 1991 Колич.характеристики :176 с Примечания : Библиогр.: с. 173 ISBN, Цена 5-09-000599-0: (в пер.) : 80 к. ББК : 22.19 Предметные рубрики: Природничі науки Навчальні видання Математика Обчислювальна техніка-- використання Алгебра Лінійне програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 242. - [x]Вид документа : Многотомное издание Шифр издания : Мб-423-2 22.161/Н 64 Автор(ы) : Николский, Сергей Михайлович Заглавие : Курс математического анализа : В 2 т.: учеб. для вузов/ Сергей Михайлович Николский. . Т. 2 . -4-е изд., перераб. и доп. Выходные данные : Москва: Наука ; л. ред. физ.-мат. лит., 1991 Колич.характеристики :543 с Серия: ISBN, Цена 5-02-014424-Х: 2 р 88 к., р. ББК : 22.161 Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 243. - [x]Вид документа : Однотомное издание Шифр издания : Б 31254 22.1/Л 63 Автор(ы) : Лисичкин, Виктор Тимофеевич, Соловейчик , Иосефа Львовна Заглавие : Математика : учебник Выходные данные : Б.м., 1991 Колич.характеристики :480 с ISBN, Цена 5-06-00192-7: (в пер.) : 2 р. 90 к. ББК : 22.1 Предметные рубрики: Природничі науки Навчальні видання Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 244. - [x]Вид документа : Многотомное издание Шифр издания : Мб 423-2 22.161/Н 64 Автор(ы) : Никольский, Сергей Михайлович Заглавие : Курс математического анализа: учеб. для вузов/ Сергей Михайлович Никольский. - Изд. 4-е, перераб. и доп. Т. 2 Выходные данные : Москва: Наука, 1991 Колич.характеристики :543 с ISBN, Цена 5-02-014425-8: (в опр.) : 2 р 80 к. ББК : 22.161я73 Предметные рубрики: Математичний аналіз Аналіз Математика Навчальні видання для вищої школи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 245. - [x]Вид документа : Однотомное издание Шифр издания : Б 32438 22.15/Л 79 Автор(ы) : Лосев, Николай Васильевич Заглавие : 200 олимпийских задач по начертательной геометрии Выходные данные : Б.м., 1992 Колич.характеристики :145 с.: ил. ISBN, Цена 5-06-001920-9: 13 р 78 к ББК : 22.15 Предметные рубрики:Математика Геометрія нарисна Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 246. - [x]Вид документа : Однотомное издание Шифр издания : 81.432.1/К 73 Автор(ы) : Котов, Виктор Николаевич, Котова, Галина Николаевна, Король, Анатолий Николаевич Заглавие : Русско-украинско-английский математический словарь словосочетаний Выходные данные : Б.м., 1992 Колич.характеристики :152 с. ISBN, Цена 5-7707-2990-2: 150 р. ББК : 81.432.1 Предметные рубрики: Англійська мова Словники Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 247. - [x]Вид документа : Однотомное издание Шифр издания : Б 34137 22.18/Ш 92 Автор(ы) : Штоиер, Ральф Заглавие : Многокритериальная оптимизация. Теория. вычисления и приложения Выходные данные : Б.м., 1992 Колич.характеристики :504 с Примечания : Библиогр.: с. 489-498 ISBN, Цена 5-256-01016-6: (в пер.) : 160 крб. ББК : 22.18 Предметные рубрики:Математика Оптимізація Обчислення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 248. - [x]Вид документа : Однотомное издание Шифр издания : Пб 487-83 22.1/Х-77 Автор(ы) : Хонсбергер, Росс Заглавие : Математические изюминки Выходные данные : Москва: Наука : Гл. ред. физ.-мат. лит., 1992 Колич.характеристики :176 с. Серия: Библиотечка "Квант"; Вып. 83 ISBN, Цена 5-02-014406-1: 250 крб. ББК : 22.10 Предметные рубрики:Математика Задачі-- Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 249. - [x]Вид документа : Однотомное издание Шифр издания : Б 23979 22.1/С 77 Автор(ы) : Стариков, Виталий Тихонович Заглавие : Сборник задач с производственным содержанием по математике для профтехучилищ сельскохозяйственного профиля : учеб. пособие . -Изд. 2-е, перераб. и доп. Выходные данные : Б.м., 1992 Колич.характеристики :124 с Примечания : Библиогр.: с. 123 ISBN, Цена 5-339-00753-7: 12 р. 48 к. ББК : 22.1я722 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 250. - [x]Вид документа : Однотомное издание Шифр издания : Б 33219 22.16/К 89 Автор(ы) : Кузнецов, Альберт Николаевич, Лебедева, Валентина Федоровна Заглавие : Методические указания к решению задач повышенной трудности по курсу математического анализа втуза Выходные данные : Б.м., 1992 Колич.характеристики :175 с.: ил. Примечания : Библиогр.: с. 175 Цена : Б.ц. ББК : 22.161кр80 Предметные рубрики:Математика Методичні вказівки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 251. - [x]Вид документа : Однотомное издание Шифр издания : Б 23208 22.16/С 17 Автор(ы) : Самойленко, Анатолий Михайлович, Ткач , Борис Петрович Заглавие : Численно-аналитические методы в теории периодических уравнений с частными производными Выходные данные : Б.м., 1992 Колич.характеристики :208 с Коллективы : АН Украины; Ин-т математики Примечания : Библиогр.: с. 201-208 ISBN, Цена 5-12-002223-5: (в пер.) : 4 р. 06 к. ББК : 22.161.68 Предметные рубрики: Природничі науки Математика Диференціальні рівняння Лінійні рівняння Нелінійні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 252. - [x]Вид документа : Однотомное издание Шифр издания : Б 32025 22.1/І-26 Автор(ы) : Ігначков, Вадим Сергійович, Ігначкова, Алла Вікторівна Заглавие : Математика для вступників у вузи : навч. посіб. Выходные данные : Б.м., 1992 Колич.характеристики :173 с. Примечания : Бібліогр. : с. 171 ISBN, Цена 5-11-001083-3: ( в опр. ) : 120 крб ББК : 22.1 Предметные рубрики:Математика Навчальний посібник для вступників Математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 253. - [x]Вид документа : Однотомное издание Шифр издания : Б 34010 22.1/Г 69 Автор(ы) : Горнштейн П. И., Поляк Н. Н., Тульчинский В. К. Заглавие : Решение конкурсных задач по математике из сборника под редакцией М. И. Сканави Выходные данные : Москва: Текст ; ОКО, 1992 Колич.характеристики :246 с. ISBN, Цена 5-87660-003-2: 80 крб. ББК : 22.1 Предметные рубрики:Математика Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 254. - [x]Вид документа : Однотомное издание Шифр издания : Б 33688-1 22.1/М 34 Заглавие : Математика : тесты : задания для проверки знаний, умений и навіков віпускников общеобраз. школ, лицеев и гимназий. Ч. 1 Выходные данные : Б.м., 1993 Колич.характеристики :93 с Цена : 240 крб. ББК : 22.1в64 Предметные рубрики:Математика Тестування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 255. - [x]Вид документа : Однотомное издание Шифр издания : Б 34835 22.1/М 34 Заглавие : Материалы вступительных экзаменов : задачи по математике и физике Выходные данные : Б.м., 1993 Колич.характеристики :320 с ISBN, Цена 5-85843-002-3: 2500 крб. ББК : 22.1я723 + 22.3я729 Предметные рубрики: Природничі науки Навчальні видання Екзаменаційні питання Математика Фізика Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 256. - [x]Вид документа : Однотомное издание Шифр издания : Б 35024 22.1/У 45 Автор(ы) : Вишенський В. А., Ганюшків О. Г., Карташов М. В., Михайловський В. І., Призва Г. Й., Ядренко М. Й. Заглавие : Українські олімпіди : довідник Выходные данные : Київ: Вищ. шк., 1993 Колич.характеристики :415 с.: іл. ISBN, Цена 5-11-001859-6: 1000 крб. ББК : 22.1 Предметные рубрики:Математика Олімпіади-- математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 257. - [x]Вид документа : Однотомное издание Шифр издания : Б 35323 22.17/К 68 Автор(ы) : Королюк, Володимир Семенович Заглавие : Стохастичні моделі систем : навч. посіб. Выходные данные : Б.м., 1993 Колич.характеристики :137 с Примечания : Бібліогр.: с. 135 ISBN, Цена 5-325-00382-8: 2330 крб. ББК : 22.171.5я73 Предметные рубрики: Природничі науки Навчальні видання Математика Прикладна математика Стохастичні еволюції Марковські процеси Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 258. - [x]Вид документа : Многотомное издание Шифр издания : Б 34592 22.1/М 13 Автор(ы) : Мазур, Константин Иванович Заглавие : Решебник всех конкурсных задач по математике сборника/ Константин Иванович Мазур ; под. ред. М. И. Сканави. . Вып. 2 : Прогрессии. алгебраические уравнения. Логарифмы. показательные и логарифмические уравнения Выходные данные : : "Украинская энциклопедия" им. М. П. Бажана; Малое предприятие "Тандем"; Научная ред. Главной ред-гии Книги Памяти Украины Б.м., 1993 Колич.характеристики :409 с Серия: ISBN, Цена 5-88500-063-8: 35000 , р. ББК : 22.1я2 Предметные рубрики:Математика Рівняння Алгебрічні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 259. - [x]Вид документа : Многотомное издание Шифр издания : Б 1416-1 22.1/Д 69 Автор(ы) : Дороговцев А. Я. Заглавие : Математичний аналіз: у двох частинах : навч. посібник/ А. Я. Дороговцев. Ч. 1 Выходные данные : Київ: Либідь, 1993 Колич.характеристики :320 с Серия: ISBN, Цена 5-325-00380-1: (в пер.) : 30000 крб. ББК : 22.1 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Інтеграли Функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 260. - [x]Вид документа : Однотомное издание Шифр издания : Б 39604 22.1/В 11 Заглавие : Віхи розвитку математичного природознавства : збірник наукових праць Выходные данные : Б.м., 1994 Колич.характеристики :132 с Коллективы : НАН України ; Ін-т математики Цена : 1 грн 50 к. ББК : 22.1г(4 УКр) Предметные рубрики: Природничі науки Наукові праці Математика Природознавство Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 261. - [x]Вид документа : Многотомное издание Шифр издания : Б 35625 22.16/Ш 66 Автор(ы) : Шкіль, Микола Іванович Заглавие : Математичний аналіз : підручник ; у 2 ч./ под ред. Г. П.Трофимчука. . Ч. 1 . -2-е изд., перероб. і доп. Выходные данные : : Вища школа Б.м., 1994 Колич.характеристики :423 с Серия: Примечания : Бібліогр.: с. 418 ISBN, Цена 5-11-004414-7: (в пер.) : 9000 крб., р. ББК : 22.16 Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 262. - [x]Вид документа : Однотомное издание Шифр издания : Б 44250 22.1/Ц 32 Автор(ы) : Цейтлін, Георгій Овсійович Заглавие : Алгебра логіки та конструювання програм. Елементи дискретної математики Выходные данные : Київ: Наук. думка, 1994 Колич.характеристики :83 с. Примечания : Бібліогр. : с. 77-78 ISBN, Цена 5-12-003444-6: 10 к. ББК : 22.1 Предметные рубрики:Математика Алгебра-- логіка Дискретна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 263. - [x]Вид документа : Однотомное издание Шифр издания : В 2180 22.18/Е 60 Автор(ы) : Емельченков Е. П., Кристалинский Р. Е., Щедров Г. П. Заглавие : BASIC разговор с компьютером : иллюстрированнный курс программирования Выходные данные : Б.м., 1994 Колич.характеристики :78 с.: ил. ISBN, Цена 5-85440-144-4: (в пер.) : 161000 крб. ББК : 22.183.492 Предметные рубрики: Природничі науки Математика Курс програмування Мова програмування-- Бейсик Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 264. - [x]Вид документа : Многотомное издание Шифр издания : 22.1я2/М 13 Автор(ы) : Мазур, Константин Иванович Заглавие : Решебник всех конкурсных задач по математике сборника/ Константин Иванович Мазур ; под. ред. М. И. Сканави. . Вып. 3 : Тригонометрические уравнения. Неравенства Выходные данные : : "Украинская энциклопедия" им. М. П. Бажана; Малое предприятие "Тандем"; Научная ред. Главной ред-гии Книги Памяти Украины Б.м., 1994 ISBN, Цена 5-88500-063-8: 206000, р. ББК : 22.1я2 Предметные рубрики:Математика Рівняння Тригонометричні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 265. - [x]Вид документа : Однотомное издание Шифр издания : Б 43211 22.1/Ж 76 Автор(ы) : Жлуктенко В. І., Бєгун А. В., Клименко Р. М. Заглавие : Конкурсні задачі з математики для вступників до економічних вузів та факультетів Выходные данные : Б.м., 1994 Колич.характеристики :240 с ISBN, Цена 5-7707-6406-6: 1 грн 30 к. ББК : 22.1я729 Предметные рубрики: Природничі науки Навчальні видання Математика Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 266. - [x]Вид документа : Однотомное издание Шифр издания : 81.432.1/К 56 Автор(ы) : Коваленко, Евгений Григорьевич, Коваленко, Евгений Григорьевич Заглавие : Англо-русский математический словарь/ Евгений Григорьевич Коваленко. Т. 2: Limiting-Z Выходные данные : Москва: Эрика, 1994 Колич.характеристики :919 с. ISBN, Цена 5-86455-004-3: (В пер.) : 750 000 крб. ББК : 81.432.1-4 Предметные рубрики: Англійська мова Словники Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 267. - [x]Вид документа : Однотомное издание Шифр издания : 81.432.1-4/К 56 Автор(ы) : Коваленко, Евгений Григорьевич Заглавие : Англо-русский математический словарь/ Евгений Григорьевич Коваленко. Т. 1: A-Limiter Выходные данные : Москва: Эрика, 1994 Колич.характеристики :448 c. ISBN, Цена 5-86455-003-5: (В пер.) : 750 000 крб. ББК : 81.432.1-4 Предметные рубрики: Англійська мова Словники Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 268. - [x]Вид документа : Однотомное издание Шифр издания : Б 41675 22.16/Ш 96 Автор(ы) : Шунда, Никифор Миколайович, Томусяк, Андрій Андрійович Заглавие : Практикум з математичного аналізу : Інтегральне числення. Ряди : навч. посіб. для пед. навч. закладів Выходные данные : Б.м., 1995 Колич.характеристики :541 с Примечания : Бібліогр.: с. 539 ISBN, Цена 5-11-004276-4: (опр. ) : 3 грн 75 к. ББК : 22.161.12я73 Предметные рубрики:Математика Інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 269. - [x]Вид документа : Однотомное издание Шифр издания : Б 38013 22.1/И-29 Автор(ы) : Идельсон, Александр Владимирович, AБлюмкина, Инесса Абрамовна Заглавие : Математика : Учеб. пособ. для вуз. Выходные данные : Б.м., 1995 Колич.характеристики :55 с. ISBN, Цена 5-7310-0513-3: 95000 крб ББК : 22.1 Предметные рубрики:Математика Навчальні видання Вища математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 270. - [x]Вид документа : Однотомное издание Шифр издания : В 2312 22.171/Л 64 Автор(ы) : Лифшиц М. А. Заглавие : Гауссовские случайные функции Выходные данные : Б.м., 1995 Колич.характеристики :246 с Примечания : Библиогр.: с. 232-246 ISBN, Цена 5-7707-7539-4: 250000 грн ББК : 22.171.54 Предметные рубрики:Математика Функції (мат.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 271. - [x]Вид документа : Однотомное издание Шифр издания : Б 38328 22.1/Ч-48 Автор(ы) : Черкасов О. Ю., Якушев А. Г. Заглавие : Математика. Скорая помощь абитуриентам . -Изд. 2-е, испр. и доп. Выходные данные : Б.м., 1995 Колич.характеристики :348 с. ISBN, Цена 5-7611-0008-8: 310000 крб. ББК : 22.1я729 Предметные рубрики:Математика Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 272. - [x]Вид документа : Однотомное издание Шифр издания : Б 42032-1 22.1/П 54 Автор(ы) : Поляк Н. Н., Мерзляк, Аркадий Григорьевич Заглавие : Решение конкурсных задач по математике из сборника под ред. М. И. Сканави : Гл. 10-11 (гр. Б) Выходные данные : Б.м., 1995 Колич.характеристики :168 с.: ил. ISBN, Цена 5-85757-036-0: 4 грн 38 к. ББК : 22.11я729 Предметные рубрики:Математика Збірник задач Вища математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 273. - [x]Вид документа : Однотомное издание Шифр издания : Б 38149 22.1/Г 44 Автор(ы) : Гетманцев, Володимир Данилович, Саушкін, Олександр Федорович Заглавие : Математика : алгебра і початки аналізу : посібник Выходные данные : Київ: Либідь, 1995 Колич.характеристики :256 с. Примечания : Бібліогр. : с. 255 ISBN, Цена 5-325-00684-3: 320 000 крб. ББК : 22.14я7 Предметные рубрики: Алгебра Математичний аналіз Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 274. - [x]Вид документа : Однотомное издание Шифр издания : Б 39219 22.1/Я 95 Автор(ы) : Ячменев Л. Т. Заглавие : Математика в примерах и задачах : учеб. пособие для поступ. в вузы Выходные данные : Москва: ТОО Инжиниринго-Колсантинговая Компания "ДеКА", 1996 Колич.характеристики :287 с. ISBN, Цена 5-86006-051-3: 9 грн. 13 к. ББК : 22.1я729 Предметные рубрики:Математика Математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 275. - [x]Вид документа : Однотомное издание Шифр издания : Б 39376 22.1/Л 11 Автор(ы) : Ліо, Кі Заглавие : Ломиголовки : ігри без партнера Выходные данные : Б.м., 1996 Колич.характеристики :150 с Серия: Бібліотечка журналу "У світі математики" Цена : 2 грн ББК : 22.1я9 Предметные рубрики: Природничі науки Математика Математичні ігри Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 276. - [x]Вид документа : Однотомное издание Шифр издания : Б 35722. 22.1/З-41 Автор(ы) : Єгерев, Віктор Костянтинович, Зайцев, Володими Валентинович, Кордемський, Борис Анатолійович, Маслова, Тамара Миколаївна Заглавие : Збірник задач з математики для вступників до вузів . -3-тє вид., стер. Выходные данные : Б.м., 1996 Колич.характеристики :445 с.: іл. ISBN, Цена 5-11-004701-4: 3 грн 75 к. ББК : 22.1 Предметные рубрики:Математика Вища математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 277. - [x]Вид документа : Однотомное издание Шифр издания : Б 42032-2 22.1/П 54 Автор(ы) : Поляк Н. Н., Мерзляк, Аркадий Григорьевич Заглавие : Решение конкурсных задач по математике из сборника под ред. М. И. Сканави : Гл. 12 (гр. Б) Выходные данные : Б.м., 1996 Колич.характеристики :224 с.: ил. ISBN, Цена 5-89149-063-0: 4 грн 38 к. ББК : 22.11я729 Предметные рубрики:Математика Збірник задач Вища математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 278. - [x]Вид документа : Однотомное издание Шифр издания : Б 38257 22. 2/К 21 Автор(ы) : Карагодова, Олена Олександрівна, Черняк, Олександр Іванович Заглавие : Задачі вступних іспитів з математики з розв`язками : посібник Выходные данные : Б.м., 1996 Колич.характеристики :185 с ISBN, Цена 5-7770-0647-8: 380 000 крб. ББК : 22.1 я729 Предметные рубрики: Природничі науки Навчальні видання Математика Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 279. - [x]Вид документа : Однотомное издание Шифр издания : А 6247 22.1/Б 24 Автор(ы) : Барвінський, Анатолій Федорович, Гладунський, Василь Назарович, Гладунська, Ганна Андріївна Заглавие : Малий математичний довідник : навчальний посібник . -Вид. 2-е, перероб. Выходные данные : Б.м., 1997 Колич.характеристики :204 с Примечания : Бібліогр.: с. 201 ISBN, Цена 5-7773-0971-2: 3 грн 31 к. ББК : 22.1я2 Предметные рубрики: Природничі науки Навчальні видання Довідкові видання Математика Арифметика Алгебра Геометрія Задачі-- розв'язування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 280. - [x]Вид документа : Однотомное издание Шифр издания : Б 25573 22.1/З-41 Заглавие : Збірник математичних задач розвивального характеру для молодших та середніх класів : навч. посіб. Выходные данные : Б.м., 1998 Колич.характеристики :36 с Цена : 1 грн. 14 к. ББК : 22.10я721 Предметные рубрики: Навчальні видання Навчальні посібники Математика Задачі (навч.) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 281. - [x]Вид документа : Однотомное издание Шифр издания : Б 43614 22.18 кр/О-75 Автор(ы) : Місюра Н. В. Заглавие : Основи методів обчислень на ЕОМ : навч. посібник Выходные данные : Б.м., 1998 Колич.характеристики :55 с Примечания : Бібліогр.: с. 53 Цена : 3 грн. ББК : 22.183.49я73 кр Предметные рубрики: Природничі науки Навчальні видання Математика Комп'ютерна математика Основи чисельних методів Мова програмування -- PASKAL Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 282. - [x]Вид документа : Многотомное издание Шифр издания : Пв 112-1 74.58кр/С 88 Заглавие : Студентські наукові студії: Зб. наук. робіт/ Нац. ун-т "Києво-Могилянська Акад.", Миколаїв. філія ; редкол. : П. М. Тригуб та ін. Вип. 1 Выходные данные : Миколаїв: Вид-во МФ НаУКМА, 1998 Колич.характеристики :52 с.: іл.,портр. Примечания : Бібліогр.в кінці робіт ISBN, Цена 966-7458-01-6: 4 грн 17 к. ББК : 74.584(4УКР-4МИК)кр Предметные рубрики: Доповіді Конференції Краєзнавство Історичні науки Філософські науки Економічні науки Математика Політична економія Содержание : Діяльність Миколаївської міської Думи (1822-1835 рр.)/ Р. Валесюк. Миколаївська міська Дума у 1835-1841 рр./ І. Веліховська, Д. Яблоновський. Миколаївська міська Дума у 1848-1861 рр./ І. Веліховська. Миколаївська міська лікарня у 90-х рр. ХІХ ст./ Ю. Щербакова. Діяльність Миколаївської міської Думи у 1900-1905 рр./ І. Власенко. Миколаївський Голова О. М. Соковнін/ Г. Прилепська. Міський Голова І. Я. Баптизманський/ Г. Прилепська. Міський Голова М. П. Леонтович/ О. Романишин. Діяльність міського Голови В. П. Костенка/ А. Третьякова. Українці в Антарктиді/ С. Біліченко. Україна в сучасному геополітичному просторі/ М. Руденко. Взаємовідносини економіки і політики в суспільстві/ Т. Балушкіна. Етичні аспекти проблеми смерті (міні-соціологічне дослідження)/ О. Іванов, В. Холод. Соціальна обумовленість виникнення "Fantasy"/ О. Діордієв. Проблеми філософського перекладу/ М. Руденко. Як готувати промову/ Г. Бородаєнко. Жіночі образи у давньослов'янській міфології як носії опоетизованої інофрмації інформації про мистецтво й науку - риторику/ В. Гординець. Порівняльна характеристика слов'янської та грецької міфології/ Л. Зубова. Приватна власність: "крадіжка" чи необхідність?/ Є. Варакіна. Соціальний вимір інфляції/ Д. Буц. Економічні процеси в сучасній Україні: первісне нагромадження капіталу?/ О. Балабушко. Приватна власність на землю: за і проти/ Т. Боцула. Роль держави в сучасній економіці/ А. Тарнопольський. Імовірні організаційні форми задоволення потреб у медичних послугах/ А. Тарасенко. Мотивації праці/ Т. Загородній. Коефіцієнт ефективності виконання "Шести правил логістики"/ В. Дяків. Шість правил логістики: кількісний підхід/ О. Політуха. Кількісна оцінка рівня логістичного сервісу/ О. Політуха. Міркування про математику в порівнянні з іншими науками/ О. Політуха. Горілка як рушійна сила економічного прогресу/ Г. Рубський. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 283. - [x]Вид документа : Однотомное издание Шифр издания : В 3111 22.1 Заглавие : 100 великих имен в математике, физике и географии Выходные данные : Б.м., 1998 Колич.характеристики :496 с ISBN, Цена 5-900411-42-7: (в пер.) : 7 грн 65 к. ББК : 22.1г (о)д + 22.3г(о)д + 26.8г(о)д Предметные рубрики: Природничі науки Відомі вчені Математика Фізика Географія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 284. - [x]Вид документа : Однотомное издание Шифр издания : Б 41597 22.1/Л 99 Автор(ы) : Ляшко, Сергей Иванович, Клюшин, Дмитрий Анатольевич, Тригуб, Александр Семенович Заглавие : Моделирование и оптимизация подземного массопереноса Выходные данные : Б.м., 1998 Колич.характеристики :239 с.: табл. Примечания : Библиогр.: с. 222-237 ISBN, Цена 966-00-0477-Х: 6 грн 00 к. ББК : 22.161+22.311 Предметные рубрики:Математика Численні методи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 285. - [x]Вид документа : Однотомное издание Шифр издания : Б 41596 22.1/Л 99 Автор(ы) : Ляшко, Сергей Иванович Заглавие : Обобщенное управление линейными системами Выходные данные : Б.м., 1998 Колич.характеристики :465 с.: табл. Примечания : Библиогр.: с.454-465 ISBN, Цена 966-00-0478-8: 8 грн. ББК : 22.161.626 Предметные рубрики:Математика Лінейні системи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 286. - [x]Вид документа : Однотомное издание Шифр издания : Б 41627 22.15/К 49 Автор(ы) : Климко М. В., Лапій О. П., Павлівський Б. М. Заглавие : Довідник з геометрії. Означення, властивості, співвідношення Выходные данные : Б.м., 1998 Колич.характеристики :119 с.: табл. Цена : 2 грн 13 к. ББК : 22.151я721 Предметные рубрики:Математика Геометрія Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 287. - [x]Вид документа : Продолжающееся издание Шифр издания : Пв 112-6 74.58(4УКР)кр/С 88 Заглавие : Студентські наукові студії: Молодіж. наук. журн./ Миколаїв. держ. гуманіт. ун-т ім. П. Могили ; редкол.: Л. П. Клименко (голова) [та ін.]. - Частина тексту та резюме : англ. - Бібліогр. в кінці ст. Вип. 1(6)'2003 Выходные данные : Миколаїв: Вид-во МФ НаУКМА, 1998 - - Колич.характеристики :164 с.: портр. ISSN: 1609-8099 Цена : 15 грн 20к. ББК : 74.584(4УКР-4МИК)кр + 95.4кр Предметные рубрики: Технічні науки Екологія Психологія Математика Економічні науки Кібернетика Політологія Соціологія Содержание : Особливості застосування нечітких регуляторів до побудови комп'ютеризованих систем керування складними технічними об'єктами/ А. Кляцко. Засоби технічного захисту інформації в системі АКОР/ М. Скачков. Розробка САПР технологічних процесів інструментального цеху/ Є. Бабич. Розробка системи автоматизації управління матеріальними ресурсами на підприємстві/ Е. Коновалов. Розробка апаратно-програмних засобів оповіщення про надзвичайну ситуацію на хімічнонебезпечних об'єктах у м. Миколаєві/ В. Кучмій. "GraphMaker": прийоми програмування на С++/ І. Скалецький. Захист даних в Internet методами криптографії/ В. Берест. Побудова кардіодіагностичної системи/ Ю. Фоменко. Експертний підхід до стратегії фінансування процесів проведення виборчих кампаній на основі ідентифікації прихованих впливів/ І. Терентієва. Підходи щодо вирішення задачі автоматизованого класифікування друкованих видань/ С. Григор'єва. Автоматизація процесу складання розкладу занять/ М. Дворецький. Особливості впровадження програмного забезпечення в бухгалтерії МФ НаУКМА/ А. Третьяков. Розробка програмного засобу для реалізації АРМ "Комплектатор" для бібліотеки НаУКМА/ В. Уханова. Система аналізу монетарної політики України/ А. Перкаль. Комплексна оцінка рамсарського угіддя "Тилігульський лиман"/ О. Болтенко. Застосування геоінформаційних технологій у здійсненні гідрологічного моніторингу регіонального ландшафтного парку "Гранітно-степове Побужжя"/ В. Верцинський. Перспективи розширення екомережі в басейні річки Інгулець/ В. Кисіль. Санітарно-хімічна та санітарно-мікробіологічна оцінка якості води в р. Інгулець та її вплив на здоров'я населення/ Т. Недашковська. Вовк звичайний у дикій природі та в умовах зоопарку/ С. Яновська. Умови утримання і розмноження білорукого гібона в Миколаївському зоопарку/ М. Маслова. Спостереження за папугами виду корелл за допомогою прихованої камери (Миколаївський зоопарк)/ С. Бурик, М. Єршова, С. Яновська. Метод біологічної очистки води за допомогою біоінженерних споруд "Біоплато"/ Т. Золотова. Особливості розвитку філософії у взаємозв'язку з математикою у світі та в Україні/ М. Фоміна. Розвиток релігієзнавства як науки/ А. Боровий. Розвиток науки психології/ І. Качинська. Деякі аспекти психології людини/ Н. Мироненко. Внесок українських вчених у розвиток логіки/ Я. Філіппова. Загальні концепції виникнення та формування математичної логіки/ І. Мельник. Розвиток історії України як науки/ П. Богачов. Внесок українських вчених у розвиток історії економічних вчень/ І. Кремповська. Економіка як сучасна наука/ І. Басараб. Соціологія - наука про суспільство/ Т. Чебан. Дослідження науки мовознавства/ В. Подгорна. Розвиток науки політології/ І. Шандуренко. Внесок українських вчених у розвиток політології/ І. Абрамова. Розвиток науки "Історія держави і права України"/ О. Гула. Місце теорії держави та права / А. Коса. Внесок українських вчених у розвиток адвокатури/ С. Дубовська. Розвиток кібернетики як науки/ В. Матіїв. Розвиток та становлення кібернетики в Україні/ Г. Жовнеренко. Розвиток науки географії та основні проблеми навколишнього середовища/ В. Ситник. Становлення та функціонування хімії як фундаментальної науки знань/ В. Нестеренко. Біологія як наука. Внесок українських вчених в її розвиток/ О. Юр'єва. Генетика: історія, сучасність, перспективи/ О. Анедченко. Організація охорони здоров'я в Україні/ І. Лаптєв. Народна медицина як пласт науки, культури і мудрості/ В. Бєлан. Экземпляры : всего : ДПК(1), ЗРФ(1) Свободны: ДПК(1), ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 288. - [x]Вид документа : Однотомное издание Шифр издания : Б 43787 22.1/С 81 Автор(ы) : Стойлова, Любовь Петровна Заглавие : Математика : учебник для вузов Выходные данные : Б.м., 1999 Колич.характеристики :424 с ISBN, Цена 5-7695-0456-0: (в пер.) : 15 грн 40 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Логіка Алгебра Натуральні числа Геометричні фігури та величини Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 289. - [x]Вид документа : Однотомное издание Шифр издания : Б 45613 22.1/У 93 Автор(ы) : Ушаков, Рудольф Петрович Заглавие : Повторювальний курс математики : [посіб. для серед. закл. освіти] Выходные данные : Київ: Техніка, 1999 Колич.характеристики :502, с.: іл., табл. ISBN, Цена 966-575-195-6: 10 грн. ББК : 22.1я721 Предметные рубрики:Математика Экземпляры : всего : х.(1), ЗРФ(2) Свободны: х.(1), ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 290. - [x]Вид документа : Многотомное издание Шифр издания : Б 42546-1. 22.11/Д 18 Автор(ы) : Данко, Павел Ефимович, Попов, Александр Георгтевич, Кожевникова, Татьяна Яковлевна Заглавие : Высшая математика в упражнениях и задачах: в 2-х ч./ Павел Ефимович Данко. - 5-е изд., испр. . Выходные данные : : Высшая школа Б.м., 1999 Колич.характеристики :304 с Серия: ISBN, Цена 5-06-003070-9: 3 грн 16 к., р. ББК : 22.11я43 Предметные рубрики:Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 291. - [x]Вид документа : Однотомное издание Шифр издания : Б 43443 22.1/А 87 Автор(ы) : Архипов, Геннадий Иванович, Садовничий, Виктор Антонович, Чубариков, Владимир Николаевич Заглавие : Лекции по математическому анализу : учебник для вузов Выходные данные : Б.м., 1999 Колич.характеристики :695 с Примечания : Библиогр.: с.684-685 ISBN, Цена 5-06-003596-4: (в пер.) : 8 грн. 70 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Функції Інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 292. - [x]Вид документа : Многотомное издание Шифр издания : 22.11/Д 18 Автор(ы) : Данко, Павел Ефимович, Попов, Александр Георгтевич, Кожевникова, Татьяна Яковлевна Заглавие : Высшая математика в упражнениях и задачах: в 2-х ч./ Павел Ефимович Данко. - 5-е изд., испр. . Выходные данные : : Высшая школа Б.м., 1999 Колич.характеристики :416 с Серия: Примечания : Библиогр.: с. 416 ISBN, Цена 5-06-003070-9: 3 грн 16 к., р. ББК : 22.11я43 Предметные рубрики:Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 293. - [x]Вид документа : Однотомное издание Шифр издания : Б 43843 22.1/К 60 Автор(ы) : Колемаев, Владимир Алексеевич, Калинина, Вера Николаевна Заглавие : Теория вероятностей и математическая статистика : учебник Выходные данные : Б.м., 1999 Колич.характеристики :301 с Серия: Высшее образование Примечания : Библиогр.: с. 299-301 ISBN, Цена 5-86225-924-4: (в пер. ) : 14 грн 70 к. ББК : 22.17я73 Предметные рубрики:Математика Статистика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 294. - [x]Вид документа : Однотомное издание Шифр издания : Б 43470 22.1/Ш 63 Автор(ы) : Шипачев, Виктор Семенович Заглавие : Математический анализ : учеб. пособие Выходные данные : Б.м., 1999 Колич.характеристики :176 с.: ил. ISBN, Цена 5-06-003510-7: 3 грн 60 к. ББК : 22.161я73 Предметные рубрики:Математика Диференційні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 295. - [x]Вид документа : Однотомное издание Шифр издания : Б 44406 28кр/К 17 Автор(ы) : Калінін, Михайло Іванович, Єлісєєв В. В. Заглавие : Біометрія : підруч. для студ. вузів біол. і екол. напрямків Выходные данные : Миколаїв: Вид-во МФ НаУКМА, 2000 Колич.характеристики :202 с Коллективы : Нац. ун-т "Києво-Могилян. акад." Миколаїв. філ. Примечания : Бібліогр.: с. 201-202 ISBN, Цена 966-7458-22-9: 11, 11 грн 24к., 4 грн., грн 24 к. ББК : 28кр Предметные рубрики: Біометрія Біологічні дослідження Математика Экземпляры : всего : ЗРФ(1), х.(2) Свободны: ЗРФ(1), х.(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 296. - [x]Вид документа : Однотомное издание Шифр издания : Б 50244 22.1/В 54 Автор(ы) : Вітлінський, Вальдемар Володимирович , Наконечний, Степан Ількович , Терещенко, Тетяна Опанасівна Заглавие : Математичне програмування : посібник для самост. вивч. дисципліни Выходные данные : Київ: КНЕУ, 2001 Колич.характеристики :248 с Коллективы : Київ. нац. екон. ун-т Примечания : Бібліогр.: с. 245-246 ISBN, Цена 966-574-263-9: 20 грн 41к. ББК : 22.183.4я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне програмування Практичні задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 297. - [x]Вид документа : Многотомное издание Шифр издания : Мб 997-1 22.161/Ф 65 Автор(ы) : Фихтенгольц, Григорий Михайлович Заглавие : Курс дифференциального и интегрального исчисления: учеб. для вузов физ. и мех.-мат. специальностей: В 3 т./ Григорий Михайлович Фихтенгольц. - 8-е изд. Т.1 Выходные данные : Москва ; Санкт-Петербург: Невский диалект, 2001 Колич.характеристики :679 с.: ил. ISBN, Цена 5-9221-0156-0: (в пер.): 68 грн 90к. ББК : 22.161.11я73+22.161.12я73 Предметные рубрики:Математика Раціональні числа Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 298. - [x]Вид документа : Однотомное издание Шифр издания : Б 47962 22.17/Л 47 Автор(ы) : Леонтьев, Владимир Константинович Заглавие : Избранные задачи комбинаторного анализа : монография Выходные данные : Б.м., 2001 Колич.характеристики :179, с. Примечания : Библиогр.: с. 181 ISBN, Цена 5-7038-1862-1: ББК : 22.174.1я9 Предметные рубрики: Природничі науки Математика Комбінаторна математика Дискретна математика Інформатика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 299. - [x]Вид документа : Однотомное издание Шифр издания : Б 45697 22.1/П 31 Автор(ы) : Пехлецкий, Игорь Дмитриевич Заглавие : Математика : учеб. для образоват. учреждений сред. проф. образования Выходные данные : Москва: Мастерство, 2001 Колич.характеристики :298, с. ил. Серия: Среднее профессиональное образование ISBN, Цена 5-294-00055-5: 17 грн. 50 к. ББК : 22.1я722 Предметные рубрики:Математика Лінійна алгебра Диференціальні обчислення Теорія ймовірностей Математична статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 300. - [x]Вид документа : Однотомное издание Шифр издания : В 3827 22.1/Д 93 Автор(ы) : Дьяконов В. Заглавие : Mathematica 4 : учеб. курс Выходные данные : Санкт-Петербург: Питер, 2001 Колич.характеристики :654 с.: ил. Примечания : Библиогр.: с.628-632. - На обл.подзаг.: Система компьютер. математики с широкими возможностями. ISBN, Цена 5-272-00275-Х: 23 грн 80 к. ББК : 22.183.49я7+32.973.2-018я7 Предметные рубрики:Математика Комп'юерна математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 301. - [x]Вид документа : Многотомное издание Шифр издания : Б 60388-2 22.17/Ж 76 Автор(ы) : Жлуктенко, Володимир Іванович та ін Заглавие : Теорія ймовірностей і математична статистика: навч. посіб. : У 2 ч./ В. І. Жлуктенко, С. І. Наконечний, С. С. Савіна. Ч.2: Математична статистика Выходные данные : Київ: КНЕУ, 2001 Колич.характеристики :333, с.: іл. Коллективы : Київ. нац. екон. ун-т ISBN, Цена 966-574-265-5: (в опр.): 30 грн 40 к. ББК : 22.172я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 302. - [x]Вид документа : Многотомное издание Шифр издания : Мб 997-2 22.161/Ф 65 Автор(ы) : Фихтенгольц, Григорий Михайлович Заглавие : Курс дифференциального и интегрального исчисления: учеб. для вузов физ. и мех.-мат. специальностей/ В 3 т. - 8-е изд. Т.2 Выходные данные : Москва ; Санкт-Петербург: Невский диалект, 2001 Колич.характеристики :863 с.: ил. ISBN, Цена 5-9221-0157-9: (в пер.): 88 грн 70к. ББК : 22.161.11я73+22.161.12я73 Предметные рубрики:Математика Інтеграл Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 303. - [x]Вид документа : Многотомное издание Шифр издания : Мб 997-3 22.161/Ф 65 Автор(ы) : Фихтенгольц, Григорий Михайлович Заглавие : Курс дифференциального и интегрального исчисления: учеб. для вузов физ. и мех.-мат. специальностей/ В 3 т. - 8-е изд. Т.3 Выходные данные : Москва ; Санкт-Петербург: Невский диалект, 2002 Колич.характеристики :727 с.: ил. ISBN, Цена 5-9221-0158-7: (в пер.): 74 грн 45к. ББК : 22.161.11я73+22.161.12я73 Предметные рубрики:Математика Інтеграли Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 304. - [x]Вид документа : Однотомное издание Шифр издания : Б 50957 22.17/Ч-49 Автор(ы) : Черняк, Олександр Іванович, Обушна, Олена Миколаївна, Ставицький , Андрій Володимирович Заглавие : Теорія ймовірностей та математична статистика : Збірник задач . -2-ге вид.,випр. Выходные данные : Б.м., 2002 Колич.характеристики :199 с Серия: Вища освіта XXI століття Примечания : Бібліогр.: с.198-199 ISBN, Цена 966-620-117-8: 18 грн. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 305. - [x]Вид документа : Однотомное издание Шифр издания : Б 50395 22.17/Б 25 Автор(ы) : Барковський, Віктор Володимирович , Барковська, Ніна Володимирівна, Лопатін, Олексій Костянтинович Заглавие : Теорія ймовірностей та математична статистика : навч. посіб. для вузів . -3-тє вид., переробл. і допов. Выходные данные : Київ: ЦУЛ, 2002 Колич.характеристики :447 с.: іл. Серия: Математичні науки Примечания : Бібліогр.: с. 446-447 ISBN, Цена 966-7938-49-2: (в опр.): 31 грн 75к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Комп'ютерний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 306. - [x]Вид документа : Однотомное издание Шифр издания : Б 4216. 22.1/С 23 Автор(ы) : Романко В. К. Заглавие : Сборник задач по дифференциальным уравнениям и вариационному исчислению Выходные данные : Б.м., 2002 Колич.характеристики :256 с Серия: Технический университет Примечания : Библиогр.: с. 256 ISBN, Цена 5-932208-120-1: (в пер.) : 49 грн 50 к. ББК : 22.161 Предметные рубрики: Природничі науки Навчальні видання Математика Диференціальні рівняння Рівняння Варіаційне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 307. - [x]Вид документа : Однотомное издание Шифр издания : Б 59457 22.18/П 16 Автор(ы) : Пантелеев, Андрей Владимирович, Летова, Татьяна Александровна Заглавие : Методы оптимизации в примерах и задачах : учеб. пособие для вузов Выходные данные : Б.м., 2002 Колич.характеристики :544 с.: ил. Серия: Прикладная математика для ВТУЗов Примечания : Библиогр.: с. 543-544 ISBN, Цена 5-06-004137-9: (в пер.) : 119 грн 62 к. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні методи Оптимізація-- математична -- методи Алгоритм рішення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 308. - [x]Вид документа : Однотомное издание Шифр издания : Б 50722 22.16/В 67 Автор(ы) : Волков, Игорь Куприянович, Канатников, Анатолий Николаевич Заглавие : Интегральные преобразования и операционное исчисление : учеб. для техн. вузов . -2-е изд. Выходные данные : Б.м., 2002 Колич.характеристики :227 с Серия:Математика в техническом университете; вып. 11 Примечания : Библиогр.: с. 220-222 ISBN, Цена 5-7038-1273-9: (в пер.): 25 грн 56к. ББК : 22.161.2я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійна алгебра Інтеграл -- перетворення Приклади Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 309. - [x]Вид документа : Однотомное издание Шифр издания : Б 50232 22.161/М 34 Автор(ы) : Рудавський Ю. К. Заглавие : Математичний аналіз : навч. посібник Выходные данные : Б.м., 2002 Колич.характеристики :307 с.: іл. Коллективы : Нац. ун-т "Львів. політехніка" Примечания : Бібліогр.: с. 307 ISBN, Цена 966-553-259-6: 19 грн 20 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Математичні моделі Макроекономіка Мікроекономіка Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 310. - [x]Вид документа : Однотомное издание Шифр издания : Б 48815 22.1/К 30 Автор(ы) : Кац, Марк Заглавие : Вероятность и смежные вопросы в физике . -2-е изд., стер. Выходные данные : Москва: Едиториал УРСС, 2002 Колич.характеристики :272, с. Примечания : Назв. парал. : англ. ISBN, Цена 5-354-00185-4: 42 грн. 84 к. ББК : 22.136 + 22.311 Предметные рубрики: Теорія ймовірностей Статистична механіка Математика Фізика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 311. - [x]Вид документа : Однотомное издание Шифр издания : Б 48415 22.1/И 20 Автор(ы) : Иванова, Елена Евгеньевна Заглавие : Дифференциальное исчисление функций одного переменного : Учеб. для техн. вуз. . -2-е изд. Выходные данные : Москва: Изд-во МГТУ им. Н. Э. Баумана, 2002 Колич.характеристики :407 с.: ил. Серия:Математика в техническом университете; вып. 2 Примечания : Библиогр .: с. 395-397 ISBN, Цена 5-7038-1271-2: ( в пер. ) : 32 грн 87 к. ББК : 22.161.11я73 Предметные рубрики: Навчальні видання Математика Диференціали Нелінійні рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 312. - [x]Вид документа : Однотомное издание Шифр издания : Б 48348 22.17/Г 96 Автор(ы) : Гусак, Алексей Адамович, Бричикова, Елена Алексеевна Заглавие : Теория вероятностей : cправ. пособие к решению задач . -3-е изд.,стер. Выходные данные : Минск: Тетра Системс, 2002 Колич.характеристики :287, с. ISBN, Цена 985-470-046-1: (в пер.) : 19 грн 22 к. ББК : 22.171я2 Предметные рубрики:Математика Ймовірностей теорія Довідкові видання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 313. - [x]Вид документа : Однотомное издание Шифр издания : Б 48100 22.1/А 17 Автор(ы) : Абчук, Владимир Авраамович Заглавие : Математика для менеджеров и экономистов : учебник Выходные данные : Санкт-Петербург: Изд-во Михайлова В. А., 2002 Колич.характеристики :524 с. Серия: Высшее профессиональное образование Примечания : Библиогр. : с. 523-524 ISBN, Цена 5-8016-0177-5: 55 грн. 44 к. ББК : 22.1я73 Предметные рубрики:Математика Содержание : Место и роль математики в менеджменте и экономике ; Элементарная математика и логический анализ ; Элементы линейной алгебры и аналитической геометрии ; Теория вероятностей и маматематическая статистика ; Теория массового обслуживания (теория очередей) ; Статистические методы исследования операций ; Оптимизационные методы исследования операций (математическое программирование) ; Прикладные задачи оптимального распределения ресурсов ; Игровые методы исследования операций (теории игр и статистических решений) Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 314. - [x]Вид документа : Однотомное издание Шифр издания : Б 48265 22.161/В 67 Автор(ы) : Волевич, Леонид Романович, Гиндикин, Семен Григорьевич Заглавие : Метод многогранника Ньютона в теории дифференциальных уравнений в частных производных Выходные данные : Москва: Эдиториал УРСС, 2002 Колич.характеристики :309, с. Примечания : Библиогр.: с. 305-309 ISBN, Цена 5-8360-0329-7: (в пер.) : 107 грн 57 к. ББК : 22.161.62 Предметные рубрики: Природничі науки Математика Диференційні рівняння Багатокутник Ньютона Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 315. - [x]Вид документа : Однотомное издание Шифр издания : В 4242 22.1/Ш 13 Автор(ы) : Шабунин, Михаил Иванович, Сидоров, Юрий Викторович Заглавие : Теория функций комплексного переменного : учеб. для вузов Выходные данные : Б.м., 2002 Колич.характеристики :246, с.: ил. Серия: Технический университет Примечания : Библиогр.: с. 247 ISBN, Цена 5-94774-005-2: (в пер.) : 32 грн 40 к. ББК : 22.161.55я73 Предметные рубрики: Природничі науки Навчальні видання Математика Функції Теорія функцій комплексної змінної Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 316. - [x]Вид документа : Однотомное издание Шифр издания : Б 48372 22.1/Ш 25 Автор(ы) : Шарапов, Олександр Дмитрович, Семьонов, Дмитро Євгенович, Дербенцев, Василь Джорджевич Заглавие : Дискретний аналіз : навч.-метод. посіб. . -Навч.вид. Выходные данные : Б.м., 2002 Колич.характеристики :125, с. Примечания : Бібліогр.: с. 124 ISBN, Цена 966-574-380-5: 16 грн 65 к. ББК : 22.174я73 Предметные рубрики: Природничі науки Навчальні видання Математика Дискретний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 317. - [x]Вид документа : Однотомное издание Шифр издания : В 4239 22.1/Ш 26 Автор(ы) : Шарма Дж. Н., Сингх К. Заглавие : Уравнения в частных производных для инженеров : [учебник] Выходные данные : Москва: Техносфера, 2002 Колич.характеристики :318 с. Серия: Мир математики Примечания : Библиогр.: с. 316 ISBN, Цена 5-94836-004-0: (в пер.) : 79 грн 26к. ББК : 22.161.62я73+22.161.67я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інтеграли Рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 318. - [x]Вид документа : Однотомное издание Шифр издания : Б 48916 22.1/Ш 59 Автор(ы) : Шилов, Георгий Евгеньевич Заглавие : Математический анализ. Функции одного переменного . -2-е изд., стер. Выходные данные : Москва: Лань, 2002 Колич.характеристики :878 с Серия: Учебники для вузов. Специальная литература ISBN, Цена 5-9511-00208: (в пер.) : 61 грн 92 к. ББК : 22.161я73 Предметные рубрики:Математика Теорія множини Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 319. - [x]Вид документа : Многотомное издание Шифр издания : Б 48393-1 22.1/М 34 Автор(ы) : Заглавие : Математичний аналіз у задачах і прикладах: Навч. посіб. для пед. вузів : В 2 ч./ Л. І. Дюженкова [та ін.]. Ч. 1 Выходные данные : Київ: Вища шк., 2002 Колич.характеристики :461, с. ISBN, Цена 966-642-034-1: (в опр.) : 29 грн 95 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Задачі Приклади Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 320. - [x]Вид документа : Однотомное издание Шифр издания : Б 47343 22.1/Т 33 Автор(ы) : Самсонов Б. Б., Плохов Е. М., Филоненков А. И., Кречет т. В. Заглавие : Теория информации и кодирование : учеб. пособие Выходные данные : Ростов-на-Дону: Феникс, 2002 Колич.характеристики :287 с. Серия: Высшее образование (Учебники и учебные пособия ISBN, Цена 5-222-02240-4: 11 грн 14к. ББК : 22.182я73+32.811я73 Предметные рубрики:Математика Кодування Теорія інформації Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 321. - [x]Вид документа : Многотомное издание Шифр издания : Мб 970-2 22.16/Ф 65 Автор(ы) : Фихтенгольц, Григорий МИхайлович Заглавие : Основы математического анализа: [В 2 т.] : учеб. для вузов/ Григорий МИхайлович Фихтенгольц. - 7-е изд. Т.2 Выходные данные : Москва: ФИЗМАТЛИТ, 2002 - Колич.характеристики :439 с. ISBN, Цена 5-9221-0197-8: (в пер.): 45 грн 90к. ББК : 22.161я73 Предметные рубрики:Математика Чисельні методи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 322. - [x]Вид документа : Однотомное издание Шифр издания : Б 47698 22.1/Ф 91 Автор(ы) : Фролькис, Виктор Абрамович Заглавие : Введение в теорию и методы оптимизации для экономистов : [учебник] . -2-е изд. Выходные данные : СПб. [и др.]: Питер, 2002 Колич.характеристики :314 с. Серия: Учебное пособие Примечания : Библиогр.: с. 314. - На обл.: Моделирование рынков. Методы анализа запасов. Решение транспорт. задач. ISBN, Цена 5-318-00780-5: 20 грн 2 к. ББК : 22.183я73 Предметные рубрики:Математика Метод оптимізації Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 323. - [x]Вид документа : Многотомное издание Шифр издания : Мб 970-1 22.16/Ф 65 Автор(ы) : Фихтенгольц, Григорий Михайлович Заглавие : Основы математического анализа: [В 2 т.] : учеб. для вузов/ Григорий Михайлович Фихтенгольц. - 7-е изд. Т.1 Выходные данные : Москва: ФИЗМАТЛИТ, 2002 - Колич.характеристики :415 с. ISBN, Цена 5-9221-0196-Х: (в пер.): 45 грн 90к. ББК : 22.161я73 Предметные рубрики:Математика Диферанціальне числення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 324. - [x]Вид документа : Однотомное издание Шифр издания : Б 47331 22.11/Д 95 Автор(ы) : Дюженкова, Любов Іванівна , Дюженкова, Ольга Юріївна, Михалін, Геннадій Олександрович Заглавие : Вища математика : приклади і задачі : посібник Выходные данные : Київ: Вид.центр "Академія", 2002 Колич.характеристики :622, с. Серия: Альма-Матер Примечания : Бібліогр.: с.623 ISBN, Цена 966-580-138-4: (в опр.): 26 грн 88к. ББК : 22.11я73 Предметные рубрики:Математика Диференціальні числення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 325. - [x]Вид документа : Однотомное издание Шифр издания : Б 47553 22.1/П 69 Заглавие : Практикум по высшей математике для экономистов : учеб. пособие для вузов по экон. специальностям Выходные данные : Москва: ЮНИТИ, 2002 Колич.характеристики :422, с. ISBN, Цена 5-238-00459-1: (в пер.) : 54 грн 72 к. ББК : 22.11я73 Предметные рубрики:Математика Вища математика Навчальні видання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 326. - [x]Вид документа : Однотомное издание Шифр издания : Пб 672-6(02) 63.3(4УКР)кр/П 32 Автор(ы) : Південнослов'янський ін-т Київського Славістичного ун-ту Заглавие : Школа-семінар дослідника-початківця : IV іст.-культурол. читання, присвяч. міжнар. Дню слов'ян. писемності, освіти, культури : Огляд результатів наук.-дослід. роботи студ. за 2000-2001 навч. рік Выходные данные : Миколаїв: [ПСІ КСУ], 2002 Колич.характеристики :92 с. Примечания : Бібліогр. в кінці доп. Цена : 4 грн 60к. ББК : 63.3(4УКР)-7кр+74584(4УКР)кр Предметные рубрики: Науково-дослідна робота Філософія Психологія Педагогiка Історія Математика Економіка Етнографія Содержание : Південнослов'янський пансіон в Миколаєві/ О. Румянцева. Поляки на Миколаївщині/ Г. Зламан. Психодиагностика ведущих тенденций масс/ В. Грушевский. Динамика психогенного состояния женщины во время беременности/ Л. Полящук. Образ женщины-злодейки в современной рок-поэзии/ А. Гнатюк. Телесные аспекты самоидентичности личности/ Л. Каратеева. Некоторые особенности психологии страха/ З. Ергашова. Генотип и среда в формировании индивидуально-психологических особенностей человека/ А. Гончарук. Герменевтика: сущность, методы и перспективы/ Т. Халахур. Мыслительный процесс: переход от внутренней речи к внешней/ М. Мурдускина. Психология стрессов/ Е. Олиевская. Память и её свойства/ К. Гладун. Профессиональное самоопределение в юношеском возрасте в современном обществе/ Е. Пустовойт. Індивідуалізація навчання як новий етап розвитку освітніх технологій/ Н. Овчаренко. Роль гри у навчанні іноземній мові/ О. Мулявка. Определённый интеграл в задачах экономики/ М. Рейфман, Т. Лошицкая. Транспортная задача и её реализация на ПК/ И. Гарбуз, Е. Кондратюк. Основы лизинга и его особенности в Украине/ С. Донченко. Сущность франчайзинга и опыт его применения в Украине/ В. Крохина. Экономическая сущность страхования и особенности его развития в Украине/ К. Терентьев. Оценка финансового состояния предприятия/ А. Бузулан. Экономический анализ финансово-хозяйственной деятельности предприятия (на примере ОАО НП "Эра", г. Николаев)/ Т. Чайка. Анализ финансовой устойчивости предприятий/ В. Мироненко. Залежність національних (на прикладі України) та світових відсоткових ставок/ М. Безкоровайний. Экземпляры :ДПК(1) Свободны: ДПК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 327. - [x]Вид документа : Однотомное издание Шифр издания : Б 47834 22.1/В 75 Автор(ы) : Воронов, Михаил Владимирович, Мещерякова, Галина Пантелеевна Заглавие : Математика для студентов гуманитарных факультетов : учебник для вузов Выходные данные : Ростов-на-Дону: Феникс, 2002 Колич.характеристики :374, с. Коллективы : Санкт-Петербургский гос. ун-т технологии и дизайна Серия: Учебники, учебные пособия Примечания : Библиогр.: с. 361-365 ISBN, Цена 5-222-02212-9: (в пер.): 20 грн 52 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія множин Дискретна математика Математичний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 328. - [x]Вид документа : Однотомное издание Шифр издания : Б 60871 22.1/П 27 Автор(ы) : Перельман, Яков Исидорович Заглавие : Живая математика : мат. рассказы и головоломки Выходные данные : Б.м., 2003 Колич.характеристики :268, с.: ил. Серия: Занимательная наука ISBN, Цена 5-17-021303-4: 16 грн. ББК : 22.1я9 Предметные рубрики:Математика Математичні ігри Головоломки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 329. - [x]Вид документа : Однотомное издание Шифр издания : Б 54775 22.1/К 12 Автор(ы) : Каган, Михаил Лазаревич, Самохин, Михаил Васильевич Заглавие : Математика в инженерном вузе: алгебра и геометрия : учебник для вузов Выходные данные : Москва: Стройиздат, 2003 Колич.характеристики :207 с.: ил. Примечания : Библиогр.: с. 206 ISBN, Цена 5-274-01847-5: 42 грн 71 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Векторна алгебра Аналітична геометрія Лінійна алгебра Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 330. - [x]Вид документа : Однотомное издание Шифр издания : Б 60146 22.1/К 65 Автор(ы) : Конфорович, Андрій Григорович Заглавие : Кентаври Уранії : Худож. оповіді про математику Выходные данные : Б.м., 2003 Колич.характеристики :143 с.: іл. ISBN, Цена 966-8066-11-1: 5 грн. ББК : 22.1г Предметные рубрики:Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 331. - [x]Вид документа : Однотомное издание Шифр издания : В 5436 22.1/К 72 Автор(ы) : Костевич, Леонид Степанович Заглавие : Математическое программирование. Информационные технологии оптимальных решений : учеб. пособие для вузов Выходные данные : Минск: Новое знание, 2003 Колич.характеристики :424 с Примечания : Библиогр.: с. 419 ISBN, Цена 985-6516-83-8: (в пер.) : 57 грн 95 к. ББК : 22.183.4я73+32.973.26-018я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне програмування Методи оптимальних рішень Таблиці Excel Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 332. - [x]Вид документа : Однотомное издание Шифр издания : Б 53245 22.1/К 67 Автор(ы) : Корн, Гранино А., Корн, Тереза М. Заглавие : Справочник по математике для научных работников и инженеров : определения, теоремы, формулы . -6-е изд., стер. Выходные данные : Санкт-Петербург и др.: Лань, 2003 Колич.характеристики :831 с.: ил. Примечания : Библиогр. : с. 796-800 ISBN, Цена 5-8114-0485-9: 88 грн. 81 к. ББК : 22.1я2 Предметные рубрики: Довідкові видання-- математика Математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 333. - [x]Вид документа : Однотомное издание Шифр издания : В 5434 22.18/М 34 Автор(ы) : Харин Ю. С. Заглавие : Математические и компьютерные основы криптологии : учеб. пособие Выходные данные : Б.м., 2003 Колич.характеристики :381 с Примечания : Библиогр.: с. 371-378 ISBN, Цена 985-475-016-7: (в пер.): 53 грн 81 к. ББК : 22.181я73+32.811.4я73 Предметные рубрики: Природничі науки Навчальні видання Математика Криптологія Захист інформації Математичні основи Комп'ютерні основи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 334. - [x]Вид документа : Однотомное издание Шифр издания : Б 51291 22.18/Д 79 Автор(ы) : Дубровін, Валерій Іванович, Субботін, Сергій Олександрович Заглавие : Методи оптимізації та їх застосування в задачах навчання нейронних мереж : навч. посіб. для вищ. навч. закл. Выходные данные : Запоріжжя: Б.в., 2003 Колич.характеристики :135 с.: іл. Примечания : Бібліогр.: с.123-132 ISBN, Цена 966-7809-24-2: 5 грн. ББК : 22.18я73 Предметные рубрики:Математика Нейронні мережі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 335. - [x]Вид документа : Однотомное издание Шифр издания : Б 51153 22.19/Б 89 Автор(ы) : Брычков, Юрий Александрович , Маричев, Олег Игоревич, Прудников, Анатолий Платонович Заглавие : Таблицы неопределённых интегралов : учеб. изд. для вузов . -2-е изд., испр. Выходные данные : Москва: Физматлит, 2003 Колич.характеристики :200 с. ISBN, Цена 5-9221-0331-8: 30 грн. 89 к. ББК : 22.194.6я7 Предметные рубрики:Математика Інтеграли Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 336. - [x]Вид документа : Однотомное издание Шифр издания : Б 49372 22.1/Л 42 Автор(ы) : Лейфура, Валентин Миколайович , Голодницький, Григорій Ісакович , Файст, Йосип Іванович Заглавие : Математика : підручник Выходные данные : Б.м., 2003 Колич.характеристики :638, с ISBN, Цена 966-575-195-6: ББК : 22.11я73 Предметные рубрики: Природничі науки Навчальні видання Математика Економічна освіта Проценти Координати та вектори Рівняння Лінійне програмування Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 337. - [x]Вид документа : Однотомное издание Шифр издания : Б 42958 22.16/А 94 Автор(ы) : Афанасьев, Валерий Николаевич , Колмановский , Владимир Борисович, Носов, Валерий Романович Заглавие : Математическая теория конструирования систем управления : учебник для вузов . -Изд. 3-е,испр. и доп. Выходные данные : Москва: Высш. шк., 2003 Колич.характеристики :613, с Примечания : Библиогр.: с. 599-606 ISBN, Цена 5-06-004162-Х: (в пер.): 54 грн. 11к. ББК : 22.161.8я73 + 22.193.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична теорія Системи управління Содержание : Устойчивость управляемых систем ; Управление детерминированными системами ; Оптимальное управление динамическими системами при случайных возмущениях ; Расчет систем управления Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 338. - [x]Вид документа : Однотомное издание Шифр издания : Б 51382-1 22.18кр/К 14 Автор(ы) : Казарезов, Анатолій ЯковичВерланов, Юрій Юрійович Заглавие : Дослідження операцій: навч. посіб./ Анатолій Якович Казарезов, Юрій Юрійович Верланов. Ч.1: Математичне програмування Выходные данные : Миколаїв: Вид-во МДГУ ім. П. Могили, 2003 Колич.характеристики :82 с.: іл. Примечания : Бібліогр.: с. 81-82 ISBN, Цена 966-7458-66-0: 5 грн 60к. ББК : 22.183я73кр Предметные рубрики: Природничі науки Навчальні видання Математика Математичне программування Лінійне програмування Задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 339. - [x]Вид документа : Однотомное издание Шифр издания : Б 4223 22.17/Ч-68 Автор(ы) : Чистяков, Владимир Павлович Заглавие : Курс теории вероятностей . -Изд. 6-е, испр. Выходные данные : Б.м., 2003 Колич.характеристики :269 с Примечания : Библиогр.: с. 266 ISBN, Цена 5-9511-0008-9: (в пер.) : 27 грн. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 340. - [x]Вид документа : Однотомное издание Шифр издания : Б 50967 22.17кр/С 60 Автор(ы) : Соловйов, Станіслав Миколайович, Чупріков, Олександр Леонідович Заглавие : Розрахунки очікуваної точності при механічній обробці деталей на металорізальних верстатах : навч. посібник Выходные данные : Б.м., 2003 Колич.характеристики :134, с.: іл. Коллективы : Укр. держ. мор. техн. ун-т ім. адм. Макарова Примечания : Бібліогр.: с. 75-76 ISBN, Цена 5-87848-103-0: 5 грн. ББК : 22.17я73кр+34.63-125я73кр Предметные рубрики: Природничі науки Навчальні видання Математика Технологія-- машинобудування Математичні розрахунки Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 341. - [x]Вид документа : Однотомное издание Шифр издания : Б 51042 22.14/Ш 33 Автор(ы) : Швачич, Геннадій Григорович Заглавие : Сучасні інформаційні технології в математиці для економістів : підручник Выходные данные : Б.м., 2003 Колич.характеристики :236 с Коллективы : Нац. металург. акад. України, Дніпропетр. ун-т економіки та права Примечания : Бібліогр.: с. 233 ISBN, Цена 966-8253-83-3: 24 грн 50 к. ББК : 22.14с5+32.965я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інформаційні технології Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 342. - [x]Вид документа : Однотомное издание Шифр издания : Б 49467 22.1/Н 45 Автор(ы) : Нейгебауер, Отто Эдуард Заглавие : Точные науки в древности . -2-е изд., стер. Выходные данные : Москва: Едиториал УРСС, 2003 Колич.характеристики :223, с.: ил. Серия: Академия фундаментальных исследований. История математики Примечания : Библиогр. в конце гл. ISBN, Цена 5-354-00263-Х: 44 грн 24к. ББК : 22.1г+22.6г Предметные рубрики:Математика Елліністична наука Математика-астрономія Экземпляры : всего : ЗРФ(1), х.(1) Свободны: ЗРФ(1), х.(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 343. - [x]Вид документа : Однотомное издание Шифр издания : Б 49534 22.1/О-11 Заглавие : О квадратуре круга . -2-е изд., стер. Выходные данные : Москва: Едиториал УРСС, 2003 Колич.характеристики :155 с. ISBN, Цена 5-354-00218-4: 30 грн 31к. ББК : 22.1г Предметные рубрики:Математика Історія математики Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 344. - [x]Вид документа : Однотомное издание Шифр издания : Б 48810 22.1/И 85 Автор(ы) : Исаков, Валерьян Николаевич Заглавие : Элементы численных методов : учеб. пособие Выходные данные : Москва: Академия, 2003 Колич.характеристики :188, с.: ил. Серия: Высшее образование Примечания : Библиогр.: с. 185-186 ISBN, Цена 5-7695-0795-0: (в пер.): 25 грн 9 к. ББК : 22.193я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія погрішностей Рівняння Диференційні рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 345. - [x]Вид документа : Многотомное издание Шифр издания : Мб 996-3 22.11/Ч-49 Автор(ы) : Черненко, Владимир Дмитриевич Заглавие : Высшая математика в примерах и задачах: учеб. пособие для вузов: в 3 т./ Владимир Дмитриевич Черненко. - (Учебное пособие для вузов). Т.3 Выходные данные : Санкт-Петербург: Политехника, 2003 - Колич.характеристики :476 с.: ил. Примечания : Библиогр. : с. 465 ISBN, Цена 5-7325-0769-8: (в пер.) : 113 грн 20к. ББК : 22.11я73 Предметные рубрики:Математика Вища математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 346. - [x]Вид документа : Однотомное издание Шифр издания : В 4333 22.1/Г 32 Автор(ы) : Гельман, Виктор Яковлевич Заглавие : Решение математических задач средствами Excel : практикум Выходные данные : Санкт-Петербург и др.: Питер, 2003 Колич.характеристики :236 с.: ил. Серия: Учебник для вузов Примечания : Библиогр.: с.236 ISBN, Цена 5-94723-315-0: 23 грн. 33 к. ББК : 22.1с51я73 + 32.973.26-018.2я73 Предметные рубрики:Математика Збірник задач Аналітична геометрія Лінійна алгебра Математичний аналіз Теорія ймовірностей Математична статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 347. - [x]Вид документа : Однотомное издание Шифр издания : Б 51847 22.151/П 75 Автор(ы) : Привалов, Иван Иванович Заглавие : Аналитическая геометрия : учебник . -32-е изд. Выходные данные : Санкт-Петербург [и др.]: Лань, 2003 Колич.характеристики :299 с.: ил. Серия: Учебники для вузов. Специальная литература ISBN, Цена 5-8114-0518-9: (в пер.) : 36 грн 72 к. ББК : 22.151.54я73 Предметные рубрики:Математика Геометрія Навчальні видання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 348. - [x]Вид документа : Однотомное издание Шифр издания : Б 51874 22.161/Б 50 Автор(ы) : Берман, Георгий Николаевич Заглавие : Сборник задач по курсу математического анализа : учеб. пособие Выходные данные : Санкт-Петербург: Профессия, 2003 Колич.характеристики :432 с.: ил. ISBN, Цена 5-93913-009-7: (в пер.): 33 грн 98 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Збірник задач Вправи Функції-- графіки Похідна Інтеграли Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 349. - [x]Вид документа : Многотомное издание Шифр издания : Мб 996-2 22.11/Ч-49 Автор(ы) : Черненко, Владимир Дмитриевич Заглавие : Высшая математика в примерах и задачах: учеб. пособие для вузов: в 3 т./ Владимир Дмитриевич Черненко. - (Учебное пособие для вузов). Т.2 Выходные данные : Санкт-Петербург: Политехника, 2003 - Колич.характеристики :476, с.: ил. Примечания : Библиогр. : с. 477 ISBN, Цена 5-7325-0768-Х: (в пер.) : 113 грн 20к. ББК : 22.11я73 Предметные рубрики:Математика Вища математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 350. - [x]Вид документа : Однотомное издание Шифр издания : Б 40943 22.1/К 79 Автор(ы) : Кремер, Наум Шевелевич , Константинова, Ольга Григорьевна , Фридман, Мира Нисоновна Заглавие : Математика для поступающих в экономические вузы : учеб. пособие для подгот. отд-ний вузов экон. профиля . -4-е изд., перераб. и доп. Выходные данные : Москва: ЮНИТИ-ДАНА, 2003 Колич.характеристики :588, с Коллективы : под ред. Н. Ш. Кремера Примечания : Библиогр.: с. 597-601 ISBN, Цена 5-238-00486-9: ББК : 22.1я729 Предметные рубрики:Математика Алгебра Рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 351. - [x]Вид документа : Однотомное издание Шифр издания : Б 51407 22.161/С 17 Автор(ы) : Самойленко, Анатолій Михайлович, Кривошея, Сергій Арсенович, Перестюк, Микола Олексійович Заглавие : Диференціальні рівняння в задачах : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2003 Колич.характеристики :501, с Примечания : Бібліогр.: с. 499-500 ISBN, Цена 966-06-0292-8: (в опр.) : 25 грн 60 к. ББК : 22.161.6я73 Предметные рубрики: Природничі науки Навчальні видання Математика Диференціальні рівняння Задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 352. - [x]Вид документа : Многотомное издание Шифр издания : Мб 996-1 22.11/Ч-49 Автор(ы) : Черненко, Владимир Дмитриевич Заглавие : Высшая математика в примерах и задачах: учеб. пособие для вузов: в 3 т./ Владимир Дмитриевич Черненко. - (Учебное пособие для вузов). Т.1 Выходные данные : Санкт-Петербург: Политехника, 2003 - Колич.характеристики :703, с.: ил. Примечания : Библиогр. : с. 704 ISBN, Цена 5-7325-0767-1: (в пер.) : 113 грн 37к. ББК : 22.11я73 Предметные рубрики:Математика Вища математика Задачі-- математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 353. - [x]Вид документа : Однотомное издание Шифр издания : Б 58665 22.171/К 70 Автор(ы) : Коршунов, Дмитрий Алексеевич , Фосс, Сергей Георгиевич, Эйсымонт, Инна Михайловна Заглавие : Сборник задач и упражнений по теории вероятностей : учеб. пособие Выходные данные : Б.м., 2004 Колич.характеристики :191 с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 174-175 ISBN, Цена 5-8114-0587-1: (в пер.) : 42 грн 60 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Збірник задач Вправи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 354. - [x]Вид документа : Многотомное издание Шифр издания : Б 21402-1 22.1/С 23 Автор(ы) : Заглавие : Сборник задач по математике для втузов: В 4 ч./ под ред.: А. В. Ефимова, А. С. Поспелова. Ч. 1 Выходные данные : Москва: Физ.-мат. л-ра, 2004 Колич.характеристики :288 с ISBN, Цена 5-94052-034-0: (в пер.): 65 грн 41 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Векторна алгебра Аналітична геометрія Лінійні рівняння Лінійна алгебра Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 355. - [x]Вид документа : Однотомное издание Шифр издания : Б 57941 22.1/М 18 Автор(ы) : Малаховский, Владислав Степанович Заглавие : Числа знакомые и незнакомые : учеб. пособие Выходные данные : Калининград: Янтар. сказ, 2004 Колич.характеристики :184 с Примечания : Библиогр. : с. 178-180 ISBN, Цена 5-7406-0817-1: 30 грн. 17 к. ББК : 22.13я7 Предметные рубрики: Чисел теорія Математика Вища математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 356. - [x]Вид документа : Однотомное издание Шифр издания : Б 59290 22.1/Ж 24 Автор(ы) : Жалдак, Мирослав Іванович Заглавие : Математика (тригонометрія,геометрія,елементи стохастики) з комп'ютерною підтримкою : Навч.посіб. Выходные данные : К.: МАУП, 2004 Колич.характеристики :455 с.: іл. Примечания : Бібліогр.: с.455 ISBN, Цена 966-608-492-9: 29 грн 95к. УДК : 514.11:519.24](075) ББК : 22.151.0я729+22.17я729 Предметные рубрики:математика Экземпляры :т/о(1) Свободны: т/о(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 357. - [x]Вид документа : Однотомное издание Шифр издания : Б 58999 22.1/С 85 Автор(ы) : Стрижаков, Андрей Владимирович, Мартиросов, Александр Леонидович, Кубарев, Александр Евгеньевич Заглавие : Начертательная геометрия : учеб. пособие для вузов Выходные данные : Б.м., 2004 Колич.характеристики :316, с.: ил. Серия: Высшее образование Примечания : Библиогр.: с. 313 ISBN, Цена 5-222-05098-Х: (в пер.) : 24 грн 76 к. ББК : 22.151.34я73 Предметные рубрики: Природничі науки Навчальні видання Математика Нарисна геометрія Креслення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 358. - [x]Вид документа : Однотомное издание Шифр издания : Б 59536 22.1/С 17 Автор(ы) : Самойленко, Анатолій Михайлович, Петришин, Роман Іванович Заглавие : Математичні аспекти теорії нелінійних коливань : монографія Выходные данные : Б.м., 2004 Колич.характеристики :473, с Коллективы : НАН України ; Ін-т математики Примечания : Бібліогр.: с. 455-471 ISBN, Цена 966-00-0109-6: (в опр.) : 46 грн 41 к. ББК : 22.161.68я9+22.236.35я9 Предметные рубрики: Природничі науки Математика Нелінійні коливні системи Математичне обгрунтування Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 359. - [x]Вид документа : Однотомное издание Шифр издания : Б 58961 22.1/К 59 Автор(ы) : Козлов, Владимир Николаевич Заглавие : Математика и информатика : учеб. пособие для вузов Выходные данные : СПб [и др.]: ПИТЕР, 2004 Колич.характеристики :265 с.: ил. Серия: Учебное пособие Примечания : Библиогр. в конце гл. ISBN, Цена 5-469-00382-5: (в пер.) : 31 грн 14 к. ББК : 22.1я73+22.18я73 Предметные рубрики:Математика Інформатика Навчальні видання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 360. - [x]Вид документа : Многотомное издание Шифр издания : Б 21402-2 22.1/С 23 Автор(ы) : Заглавие : Сборник задач по математике для втузов: В 4 ч./ под ред.: А. В. Ефимова, А. С. Поспелова. Ч. 2 Выходные данные : Москва: Физ.-мат. л-ра, 2004 Колич.характеристики :431 с ISBN, Цена 5-94052-035-9: (в пер.): 75 грн 53 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Збірник задач Диференційне числення Інтеграли Рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 361. - [x]Вид документа : Однотомное издание Шифр издания : Б 59017 22.1/И 26 Автор(ы) : Игошин, Владимир Иванович Заглавие : Математическая логика и теория алгоритмов : учеб. пособие для вузов Выходные данные : Москва: Академия, 2004 Колич.характеристики :446, с.: ил. Серия: Высшее профессиональное образование. Педагогические специальности Примечания : Библиогр.: с.435-442 ISBN, Цена 5-7695-1363-2: (в пер.): 78 грн 10 к. ББК : 22.122я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична логіка Теорія алгоритмів Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 362. - [x]Вид документа : Однотомное издание Шифр издания : Б 57370 22.1/В 15 Автор(ы) : Валєєв, Кім Галямович, Джалладова, Ірада Агаверді Заглавие : Математичний практикум : Навч.посіб. Выходные данные : К.: КНЕУ, 2004 Колич.характеристики :682 с Примечания : Бібліогр.: с.681 ISBN, Цена 966-574-665-0: (в опр.): 31 грн 40к. УДК : 51(076.5) ББК : 22.1я73 Предметные рубрики:математика Экземпляры :т/о(1) Свободны: т/о(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 363. - [x]Вид документа : Однотомное издание Шифр издания : В 5334 22.1/П 60 Автор(ы) : Поршнев, Сергей Валентинович Заглавие : Вычислительная математика : курс лекций : учеб. пособие для вузов Выходные данные : Санкт-Петербург: БХВ-Петербург, 2004 Колич.характеристики :304 с.: ил. Примечания : Библиогр.: 303-304 ISBN, Цена 5--94157-400-2: (в пер.) : 41 грн 20 к. ББК : 22.19я73+32.973.26я73 Предметные рубрики:Математика Диференціальне числення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 364. - [x]Вид документа : Однотомное издание Шифр издания : Б 41918 22.1/Т 19 Автор(ы) : Тарасик, Владимир Петрович Заглавие : Математическое моделирование технических систем : учеб. для вузов Выходные данные : Минск: Дизайн ПРО, 2004 Колич.характеристики :639 с. Примечания : Библиогр. : с. 620-622 ISBN, Цена 985-452-080-3: (в пер.) : 204 грн 16к. ББК : 22.1я73+34.42я73 Предметные рубрики:Математика Математичне моделювання Математика-- Технічні системи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 365. - [x]Вид документа : Однотомное издание Шифр издания : Б 54188 22.19/С 17 Автор(ы) : Самарский, Александр Андреевич, Вабищевич, Петр Николаевич Заглавие : Численные методы решения обратных задач математической физики Выходные данные : Б.м., 2004 Колич.характеристики :478 с Примечания : Библиогр.: с. 475-476 ISBN, Цена 5-354-00156-0: (в пер.) : 105 грн 96 к. ББК : 22.193+22.311 Предметные рубрики: Природничі науки Навчальні видання Математика Математична фізика Рішення задач Диференційні рівняння Методи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 366. - [x]Вид документа : Однотомное издание Шифр издания : Б 54621 22.172/М 28 Автор(ы) : Мармоза, Анатолій Тимофійович Заглавие : Практикум з математичної статистики : навч. посіб. для вищ. навч.закл. Выходные данные : Київ: Кондор, 2004 Колич.характеристики :257 с. Примечания : Бібліогр. : с. 253 ISBN, Цена 966-7982-30-0: 26 грн. 78 к. ББК : 22.172я73 Предметные рубрики: Математична статистика Математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 367. - [x]Вид документа : Однотомное издание Шифр издания : Б 55096 22.1/С 78 Заглавие : Статистический анализ многомерных объектов произвольной природы : введение в статистику качеств Выходные данные : Б.м., 2004 Колич.характеристики :381 с Коллективы : Нац. ин-т бизнеса, Центр тестирования проф. образования при Моск. гос. ун-те печати Примечания : Библиогр.: с. 8-9 ISBN, Цена 5-98405-006-4: (в пер.) : 86 грн 49 к ББК : 22.172 Предметные рубрики: Природничі науки Математика Статистичний аналіз Масові явища Математичні методи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 368. - [x]Вид документа : Однотомное издание Шифр издания : Б 55119 22.1/Ч-59 Автор(ы) : Чижов, Евгений Борисович Заглавие : Введение в философию математических пространств Выходные данные : Б.м., 2004 Колич.характеристики :294 с Серия: Relata Refero Примечания : Библиогр. в конце гл. ISBN, Цена 5-354-00661-9: (в пер.) : 64 грн 48 к. ББК : 22.1в1+87.251 Предметные рубрики: Природничі науки Математика Математичний простір Філософія Філософські школи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 369. - [x]Вид документа : Однотомное издание Шифр издания : Б 54731 22.16/И 32 Автор(ы) : Макаров М. Б. Заглавие : Избранные задачи по вещественному анализу : учеб. пособие для вузов . -Изд. 2-е, перераб. и доп. Выходные данные : Б.м., 2004 Колич.характеристики :623 с.: ил. Примечания : Библиогр.: с. 614-619 ISBN, Цена 5-7940-0104-6: (в пер.) : 68 грн 4 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Аналіз-- класичний і сучасний Задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 370. - [x]Вид документа : Однотомное издание Шифр издания : Б 54776 22.1/К 93 Автор(ы) : Курант, Рихард, Роббинс, Герберт Заглавие : Что такое математика?. (Элементарный очерк идей и методов) : монография . -3-е изд., испр. и доп. Выходные данные : Москва: МЦНМО, 2004 Колич.характеристики :564 с.: ил. Примечания : Библиогр. : с. 551-556 ISBN, Цена 5-900916-45-6: ББК : 22.1я9 Предметные рубрики:Математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 371. - [x]Вид документа : Однотомное издание Шифр издания : Б 53123 22.1/П 59 Автор(ы) : Порошкин, Александр Григорьевич Заглавие : Лекции по функциональному анализу : учеб. пособие для студентов по мат. специальностям и направлениям подгот. Выходные данные : Москва: Вузов. кн., 2004 Колич.характеристики :431 с Примечания : Библиогр.: с. 410-413 ISBN, Цена 5-9502-0050-0: 93 грн 45 к. ББК : 22.162я73 Предметные рубрики:Математика Функціональний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 372. - [x]Вид документа : Однотомное издание Шифр издания : А 6790 22.1/Ш 20 Автор(ы) : Шанкин, Генрих Петрович Заглавие : Ценность информации : вопросы теории и приложений Выходные данные : Москва: Филоматис, 2004 Колич.характеристики :126 с Примечания : Библиогр.: с. 123-126 ISBN, Цена 5-98111-022-8: (в пер.) : 25 грн 20 к. ББК : 22.182 Предметные рубрики: Природничі науки Математика Криптографія Безпека інформації Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 373. - [x]Вид документа : Однотомное издание Шифр издания : Б 51850 22.1/Л 24 Автор(ы) : Лаппо, Лев Дмитриевич , Морозов, Александр Валерьевич , Попов, Максим Александрович Заглавие : Математика. ЕГЭ : Контрольные измерительные материалы. Методические указания по подготовке. Тестовые задания : учеб.-метод. пособие Выходные данные : Б.м., 2004 Колич.характеристики :222 с.: ил. ISBN, Цена 5-94692-827-9: ББК : 22.1я721+74.262я721 Предметные рубрики: Природничі науки Навчальні видання Методичні матеріали Математика Державний екзамен-- математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 374. - [x]Вид документа : Однотомное издание Шифр издания : Б 52234 22.13/Х-47 Автор(ы) : Хинчин, Александр Яковлевич Заглавие : Цепные дроби . -4-е изд., стер. Выходные данные : Москва: Едиториал УРСС, 2004 Колич.характеристики :111, с ISBN, Цена 5-354-00551-5: 21 грн 93к. ББК : 22.134 Предметные рубрики:Математика Дроби Метрична теорія дробів Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 375. - [x]Вид документа : Однотомное издание Шифр издания : В 5135 22.1/Я 49 Автор(ы) : Якушева, Галина Михайловна, Васильева, Алла Васильевна, Лобанова Д. Б. Заглавие : Большая математическая энциклопедия : для школьников и студентов Выходные данные : Москва: Филол.о-во "Слово": ОЛМА-ПРЕСС, 2004 Колич.характеристики :639 с.: ил. ISBN, Цена 5-8123-0092-5: (в пер.) : 91 грн 20к. ББК : 22.1я2 Предметные рубрики:Математика Экземпляры :т/о(1) Свободны: т/о(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 376. - [x]Вид документа : Однотомное издание Шифр издания : Б 36327 22.1/В 24 Автор(ы) : Ашихмин В. Н. Заглавие : Введение в математическое моделирование : учеб. пособие для вузов Выходные данные : Москва: Логос, 2004 Колич.характеристики :439 с.: ил. Серия: Новая университетская библиотека Примечания : Библиогр.: с. 431-435 ISBN, Цена 5-94010-272-7: (в пер.): 83 грн 16к. ББК : 22.181я73 Предметные рубрики: Природничі науки Навчальні видання Математика Моделювання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 377. - [x]Вид документа : Однотомное издание Шифр издания : Б 53650 22.151/С 60 Автор(ы) : Соломонов, Константин Николаевич, Бусыгина, Елена Борисовна , Чиченева, Ольга Николаевна Заглавие : Начертательная геометрия : учеб. для вузов Выходные данные : Б.м., 2004 Колич.характеристики :152, с.: ил. Серия: Высшее образование Примечания : Библиогр.: с. 154 ISBN, Цена 5-16-001870-0: 19 грн 76 к. ББК : 22.151.34я73 Предметные рубрики: Природничі науки Навчальні видання Математика Нарисна геометрія Метод проекцій Пряма Креслення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 378. - [x]Вид документа : Однотомное издание Шифр издания : Б 34607 22.1/С 21 Автор(ы) : Сафрай, Владимир Михайлович Заглавие : Справочник по высшей математике для студентов ВУЗов с примерами решения задач Выходные данные : Б.м., 2004 Колич.характеристики :356 с.: ил. Примечания : Библиогр.: с. 344 ISBN, Цена 5-902403-14-6: (в пер.): 39 грн 60 к. ББК : 22.11я2 Предметные рубрики: Природничі науки Довідкові видання Математика Вища математика Задачі Рішення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 379. - [x]Вид документа : Однотомное издание Шифр издания : Б 55592 22.2/Я 46 Автор(ы) : Якоби, Карл Густав Якоб Заглавие : Лекции по динамике . -Изд. 2-е, стер. Выходные данные : Б.м., 2004 Колич.характеристики :270, с ISBN, Цена 5-354-00689-9: 53 грн 70 к. ББК : 22.236 Предметные рубрики: Фізика Математика Динаміка Диференційні рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 380. - [x]Вид документа : Однотомное издание Шифр издания : Б 55582 22.1/С 50 Автор(ы) : Смирнов, Олег Александрович Заглавие : Математика : Учеб.пособие для поступающих в вузы Выходные данные : М.: Филол.о-во "Слово": ООО "ЭКСМО", 2004 Колич.характеристики :574 с Примечания : Авт.указан на оброте тит.л. ISBN, Цена 5-8123-0150-6: 24 грн 40к. ББК : 22.1я723 Предметные рубрики:математика Экземпляры :х.(1) Свободны: х.(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 381. - [x]Вид документа : Однотомное издание Шифр издания : Б 56156 22.1кр/К 88 Автор(ы) : Кузнецов, Альберт Миколайович Заглавие : Основи математики : Конспект лекцій для лінгвістів Выходные данные : Миколаїв: Вид-во НУК, 2004 Колич.характеристики :31, с.: іл. Коллективы : Нац.ун-т кораблебудування ім.адмірала Макарова,Каф.вищ.математики Примечания : Бібліогр.: с.32 : ББК : 22.11я73кр Предметные рубрики:математика Экземпляры :т/о(1) Свободны: т/о(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 382. - [x]Вид документа : Многотомное издание Шифр издания : Б 56032-1 22.1кр/Є 60 Автор(ы) : Ємельянова,Тетяна Василівна Заглавие : Математичні методи дослідження операцій: Коспект лекцій/ А. В. Варшамов, М. Б. Бондаренко. Ч.1 Выходные данные : Миколаїв: Вид-во НУК, 2004 Колич.характеристики :59 с Примечания : Бібліогр.: с. 58 Цена : 4 грн 10к. ББК : 22.183.5я73кр Предметные рубрики: Природничі науки Навчальні видання Математика Математичні методи-- застосування Дослідження Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 383. - [x]Вид документа : Однотомное издание Шифр издания : Б 55265 22.18/М 42 Автор(ы) : Медведєв, Микола Георгійович, Колодінська, Олена Вікторівна Заглавие : Дослідження операцій : навч. посібник Выходные данные : Б.м., 2004 Колич.характеристики :157 с Коллективы : Європ. ун-т Примечания : Бібліогр.: с. 154 ISBN, Цена 966-301-063-0: 10 грн 56 к. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Дослідження операцій Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 384. - [x]Вид документа : Однотомное издание Шифр издания : Б 55610 22.17/К 79 Автор(ы) : Кремер, Наум Шевелевич Заглавие : Теория вероятностей и математическая статистика : учеб. для вузов по экон. специальностям . -2-е изд., перераб. и доп. Выходные данные : Москва: ЮНИТИ-ДАНА, 2004 Колич.характеристики :573 с Примечания : Библиогр.: с. 533-534. - Парал.тит. л.: англ. ISBN, Цена 5-238-00573-3: ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Математика Статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 385. - [x]Вид документа : Многотомное издание Шифр издания : Б 60388-1 22.17/Ж 76 Автор(ы) : Жлуктенко, Володимир Іванович,Наконечний,степан Ількович Заглавие : Теорія ймовірностей і метаматична статистика: Навч.-метод.посіб.: У 2 ч./ Степан Ількович Наконечний. Ч.1: Теорія ймовірностей Выходные данные : Київ: КНЕУ, 2005 Колич.характеристики :303 с.: іл. Коллективы : Київ. нац. екон. ун-т Примечания : Бібліогр.: с. 294 ISBN, Цена 966-574-153-5: (в опр.): 28 грн 50 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Математика Навчальні видання Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 386. - [x]Вид документа : Однотомное издание Шифр издания : Б 61025 22.1/Л 79 Автор(ы) : Лось, Валерій Миколайович, Тихієнко, Віталій Петрович Заглавие : Математика: навчаємо міркувати: Розв'язування нестандартних задач : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2005 Колич.характеристики :309 с.: іл. Примечания : Бібліогр.: с.308-309 ISBN, Цена 966-7982-81-5: УДК : 51(076.1)(075.8) ББК : 22.1я73 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 387. - [x]Вид документа : Однотомное издание Шифр издания : Б 58216 22.19/Б 24 Автор(ы) : Барахнин, Владимир Борисович, Шапеев, Василий Павлович Заглавие : Введение в численный анализ : учеб. пособие Выходные данные : Санкт-Петербург; Москва; Краснодар: Лань, 2005 Колич.характеристики :106, с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с.105 ISBN, Цена 5-8114-0604-5: 26 грн 87 к. ББК : 22.193я73 Предметные рубрики: Природничі науки Навчальні видання Математика Чисельний аналіз Функції Інтегрування Рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 388. - [x]Вид документа : Однотомное издание Шифр издания : Б 1069 22.19/С 17 Автор(ы) : Самарский, Александр Андреевич Заглавие : Введение в численные методы : учеб. пособие для вузов . -Изд. 3-е, стер. Выходные данные : Б.м., 2005 Колич.характеристики :288 с.: ил. Серия: Классический университетский учебник Примечания : Библиогр.: с. 281 ISBN, Цена 5-8114-0602-9: (в пер.) : 53 грн 78 к. ББК : 22.192я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні моделі Числені методи Теорія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 389. - [x]Вид документа : Однотомное издание Шифр издания : В 5801 22.1/Ч-49 Автор(ы) : Черноруцкий, Игорь Георгиевич Заглавие : Методы принятия решений : учеб. пособие для вузов Выходные данные : Санкт-Петербург: БХВ-Петербург, 2005 Колич.характеристики :408 с.: ил. Примечания : Библиогр.: с. 395-398 ISBN, Цена 5-94157-481-9: (в пер.) : 60 грн 10 к. ББК : 22.183.1с51я73 Предметные рубрики: Природничі науки Навчальні видання Математика Прийняття рішень -- теорія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 390. - [x]Вид документа : Однотомное издание Шифр издания : Б 57775 22.17/А 66 Автор(ы) : Андрухаев, Хазерталь Махмудович Заглавие : Сборник задач по теории вероятностей : учеб. пособие для вузов . -Изд. 2-е, испр. и доп. Выходные данные : Москва: Высш. шк., 2005 Колич.характеристики :173, с.: ил. Примечания : Библиогр.: с.172 ISBN, Цена 5-06-004747-4: 48 грн. 95 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Содержание : События и их вероятности ; Схема Бернулли ; Случайные величины ; Элементы математической статистики Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 391. - [x]Вид документа : Однотомное издание Шифр издания : Б 57341 22.1/К 82 Автор(ы) : Кричевец, Анатолий Николаевич , Шикин, Евгений Викторович , Дьячков, Аркадий Георгиевич Заглавие : Математика для психологов : учебник . -2-е изд., испр. Выходные данные : Москва: Флинта: Моск. психол.-соц. ин-т, 2005 Колич.характеристики :371 с ISBN, Цена 5-89349-400-8: ББК : 22.1я73 Предметные рубрики:Математика Лінейні рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 392. - [x]Вид документа : Однотомное издание Шифр издания : Б 61827 22.1/Б 24 Автор(ы) : Барашков, Александр Сергеевич Заглавие : Математика : учеб. пособие для вузов Выходные данные : Б.м., 2005 Колич.характеристики :478, с Серия: Высшее образование ISBN, Цена 5-8123-0297-9: (в пер.): 36 грн. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Аналітична геометрія Лінійна алгебра Математичний аналіз Інтегральне числення Функції кількох перемінних Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 393. - [x]Вид документа : Однотомное издание Шифр издания : В4217 22.15/П 12 Автор(ы) : Павлова, Алина Абрамовна Заглавие : Начертательная геометрия : учеб. для вузов . -2-е изд., перераб. и доп. Выходные данные : Б.м., 2005 Колич.характеристики :300, с.: ил. Серия: Учебник для вузов Примечания : Библиогр.: с. 301 ISBN, Цена 5-691-01389-0: (в пер.) : 37 грн 84 к. ББК : 22.151.34я73 Предметные рубрики: Природничі науки Навчальні видання Математика Нарисна геометрія Креслення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 394. - [x]Вид документа : Однотомное издание Шифр издания : Б 39092 22.17/Ш 34 Автор(ы) : Шведов, Алексей Сергеевич Заглавие : Теория вероятностей и математическая статистика : учеб. пособие для вузов . -2-е изд., перераб. и доп. Выходные данные : Б.м., 2005 Колич.характеристики :252, с Серия: Учебники Высшей школы экономики Примечания : Библиогр.: с. 247-249 ISBN, Цена 5-7598-0214-3: (в пер.) : 54 грн 38 к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 395. - [x]Вид документа : Однотомное издание Шифр издания : Б 58256 22.16/С 72 Автор(ы) : Спивак, Майкл Заглавие : Математический анализ на многообразиях : учеб. пособие . -Изд. 2-е Выходные данные : Б.м., 2005 Колич.характеристики :158, с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 155 ISBN, Цена 5-8114-0646-0: 34 грн 34 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Диференціали Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 396. - [x]Вид документа : Однотомное издание Шифр издания : Б 58918 22.1/С 87 Автор(ы) : Струченков, Валерий Иванович Заглавие : Методы оптимизации : Основы теории, задачи, обучающие компьютерные программы : учеб. пособие Выходные данные : Б.м., 2005 Колич.характеристики :254, с.: ил. Примечания : Библиогр.: с. 254-255 ISBN, Цена 5-472-00465-9: 32 грн 13 к. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Комп'ютерні програми-- навчання Оптимізація-- математична -- методи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 397. - [x]Вид документа : Однотомное издание Шифр издания : Б 59049 22.1/М 34 Заглавие : Математика и грамматика русского языка для начальных классов (в таблицах и схемах) Выходные данные : М.: ООО "Лист Нью", 2005 Колич.характеристики :127 с.: ил. ISBN, Цена 5-7871-0145-6: 8 грн 39к. ББК : 22.1я721+81.411.2-922 Предметные рубрики:математика Экземпляры :х.(1) Свободны: х.(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 398. - [x]Вид документа : Однотомное издание Шифр издания : Б 59984 22.1/Ж 86 Автор(ы) : Жуков, Александр Владимирович , Самовол, Петр Исакович, Аппельбаум , Марк Виленович Заглавие : Элегантная математика : Задачи и решения: учеб. пособие Выходные данные : Москва: КомКн., 2005 Колич.характеристики :205 с.: ил. ISBN, Цена 5-474-00132-3: 39 грн 85 к. ББК : 22.1 Предметные рубрики: Природничі науки Математика Задачі-- розв'язування Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 399. - [x]Вид документа : Однотомное издание Шифр издания : Б 58211 22.1/Ж 79 Автор(ы) : Жолков, Сергей Юрьевич Заглавие : Математика и информатика для гуманитариев : учеб .для вузов . -Изд. 2-е, испр. и доп. Выходные данные : Москва: Альфа-М, 2005 Колич.характеристики :527 с Примечания : Библиогр.: с. 519 ISBN, Цена 5-98281-049-5: (в пер.): 64 грн 80 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інформатика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 400. - [x]Вид документа : Однотомное издание Шифр издания : Б 60407 22.1/К 82 Автор(ы) : Кривилев, Александр Владимирович Заглавие : Основы компьютерной математики с использованием системы MATLAB : учеб. пособие Выходные данные : Б.м., 2005 Колич.характеристики :483, с., 8 с. цв. ил.: ил. Примечания : К кн. прил. КД-1236 ISBN, Цена 5-94558-013-9: (в пер.) : 59 грн 49 к. ББК : 22.1с51я73+32.973.26-018.2я73 Предметные рубрики: MATLAB Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 401. - [x]Вид документа : Однотомное издание Шифр издания : Б 56366 22.1/П 82 Автор(ы) : Просветов, Георгий Иванович Заглавие : Математика для юристов : Задачи и решения: учеб.-метод. пособие Выходные данные : Москва: Изд-во РДЛ, 2005 Колич.характеристики :207 с Примечания : Библиогр.: с. 201 ISBN, Цена 5-93840-070-8: 18 грн 14 к. ББК : 22.1я73 Предметные рубрики:Математика Рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 402. - [x]Вид документа : Однотомное издание Шифр издания : Б 58224 22.1/К 78 Автор(ы) : Крахин, Александр Васильевич Заглавие : Математика для юристов : учеб. пособие Выходные данные : Москва: Флинта: МПСИ, 2005 Колич.характеристики :197 с Коллективы : РАО, Моск. психол.-соц. ин-т Серия: Библиотека студента Примечания : Библиогр.в конце гл. ISBN, Цена 5-89349-799-6: ББК : 22.1я73 Предметные рубрики:Математика Право Статистика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 403. - [x]Вид документа : Многотомное издание Шифр издания : В4237-2 22.151/П 12 Автор(ы) : Павлова, Алина Абрамовна Заглавие : Начертательная геометрия: Практикум для вузов : В 2 ч./ Ирина Владимировна Глазкова. - (Практикум для вузов). Ч. 2 Выходные данные : Москва: ВЛАДОС, 2005 Колич.характеристики :95 с.: ил. ISBN, Цена 5-691-00876-5: 20 грн 10 к. ББК : 22.151.34я73 Предметные рубрики: Природничі науки Навчальні видання Практикум Математика Нарисна геометрія Креслення Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 404. - [x]Вид документа : Однотомное издание Шифр издания : Б 63214 22.1/С 23 Заглавие : Сборник задач по аналитической геометрии и линейной алгебре : учеб. пособие для вузов . -Изд. 2-е, перераб. и доп. Выходные данные : Б.м., 2005 Колич.характеристики :372 с Серия: Классический университетский учебник Примечания : Библиогр.: с. 371-372 ISBN, Цена 5-94010-375-8: (в пер.) : 67 грн 32 к. ББК : 22.151.54я73+22.143я73 Предметные рубрики: Природничі науки Навчальні видання Математика Аналітична геометрія Лінійна алгебра Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 405. - [x]Вид документа : Однотомное издание Шифр издания : Б 63243 22.1/Л 73 Автор(ы) : Локшин, Александр Александрович , Сагомонян, Елена Артуровна, Саакян, Ася Сергеевна Заглавие : Графики функций . -2-е изд., испр. и доп. Выходные данные : Б.м., 2005 Колич.характеристики :72, с.: ил. Примечания : Библиогр.: с.73 ISBN, Цена 5-9502-0091-8: ББК : 22.161.5 Предметные рубрики: Функції (мат.) Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 406. - [x]Вид документа : Однотомное издание Шифр издания : Б 65848 22.17/К 78 Автор(ы) : Кравчук, Анатолій Федорович Заглавие : Дискретний аналіз : навч. посіб. для вищ. навч. закл. . -2-ге вид., допов. Выходные данные : Б.м., 2005 Колич.характеристики :331 с.: іл. Коллективы : Кіровоград. нац. техн. ун-т Примечания : Бібліогр.: с. 328-331 ISBN, Цена 966-8515-88-9: 30 грн 30 к. ББК : 22.174я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне моделювання Математична логіка Дискретний аналіз Економіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 407. - [x]Вид документа : Однотомное издание Шифр издания : Б 62254 22.1/А 50 Автор(ы) : Алиев, Рзахан Гюльмагомедович Заглавие : Уравнения в частных производных : учеб. пособие для вузов . -2-е изд., доп. Выходные данные : Б.м., 2005 Колич.характеристики :319 с.: ил. Серия: Учебник для вузов Примечания : Библиогр. : с. 309-310 ISBN, Цена 5-472-01053-5: 41 грн. 40 к. ББК : 22.16я73 Предметные рубрики: Математичний аналіз Лінійні рівняння Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 408. - [x]Вид документа : Многотомное издание Шифр издания : В4237-1 22.151/П 12 Автор(ы) : Павлова, Алина Абрамовна Заглавие : Начертательная геометрия: Практикум для вузов: В 2 ч./ Ирина Владимировна Глазкова. - (Практикум для вузов). Ч. 1 Выходные данные : Москва: ВЛАДОС, 2005 Колич.характеристики :95 с.: ил. ISBN, Цена 5-691-00875-7: 20 грн 11 к. ББК : 22.151.34я73 Предметные рубрики: Природничі науки Навчальні видання Практикум Математика Нарисна геометрія Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 409. - [x]Вид документа : Однотомное издание Шифр издания : Б 63602 22.17/Т 33 Автор(ы) : Мынбаева, Гульшат Узакбаевна, Дмитриев, Иван Григорьевич, Борисов, Владимир Захарович, Саввин, Афанасий Семенович Заглавие : Теория вероятностей в примерах и задачах : учеб. пособие Выходные данные : Б.м., 2005 Колич.характеристики :433, с Примечания : Библиогр. : с. 434 ISBN, Цена 5-9502-0122-1: 90 грн 61к. ББК : 22.171я73 Предметные рубрики:Математика Теорія ймовірностей Випадкові величини Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 410. - [x]Вид документа : Однотомное издание Шифр издания : Б 63433 22.1/Л 33 Автор(ы) : Лебег, Анри Заглавие : Об измерении величин : пер. с фр. . -Изд. 3-е, стер. Выходные данные : Б.м., 2005 Колич.характеристики :203, с ISBN, Цена 5-484-00118-8: 36 грн 76 к. ББК : 22.151 Предметные рубрики: Природничі науки Математика Вимірювання-- величини Фігури-- площина Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 411. - [x]Вид документа : Однотомное издание Шифр издания : Б 63457 22.16/Л 22 Автор(ы) : Ландау, Эдмунд Георг Герман Заглавие : Введение в дифференциальное и интегральное исчисление . -Изд. 2-е, испр. Выходные данные : Б.м., 2005 Колич.характеристики :458 с ISBN, Цена 5-484-00092-0: 70 грн 99 к. ББК : 22.161.1 Предметные рубрики: Природничі науки Математика Диференційне числення Інтегральне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 412. - [x]Вид документа : Однотомное издание Шифр издания : Б 63445 22.15/К 78 Автор(ы) : Краснощёченко, Владимир Иванович, Крищенко, Александр Петрович Заглавие : Нелинейные системы : геометрические методы анализа и синтеза : Монография Выходные данные : Б.м., 2005 Колич.характеристики :519 с Примечания : Библиогр.: с. 509-516 ISBN, Цена 5-7038-2182-7: (в пер.) : 64 грн 58 к. ББК : 22.151я9 Предметные рубрики: Природничі науки Математика Геометричні методи аналізу Нелінійні автоматичні системи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 413. - [x]Вид документа : Однотомное издание Шифр издания : Б 31977 22.16/Л 47 Автор(ы) : Леонтьева, Татьяна Алексеевна , Панферов, Валерий Семенович, Серов, Валерий Сергеевич Заглавие : Задачи по теории функций комплексного переменного с решениями : учеб. пособие . -2-е изд., испр. и доп. Выходные данные : Б.м., 2005 Колич.характеристики :358, с.: ил. Примечания : Библиогр.: с. 354 ISBN, Цена 5-03-003692-Х: ББК : 22.161.55я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорії функцій комплексної змінної Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 414. - [x]Вид документа : Однотомное издание Шифр издания : В 6791 22.18/Л 50 Автор(ы) : Лесневский, Александр Станиславович Заглавие : Объектно-ориентированное программирование для начинающих Выходные данные : Б.м., 2005 Колич.характеристики :232 с Примечания : Библиогр.: с. 227 ISBN, Цена 5-94774-251-9: (в пер.) : 39 грн 40 к. ББК : 22.183.4 Предметные рубрики: Природничі науки Математика Математичне програмування Мови програмування Об'єктно-орієнтоване програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 415. - [x]Вид документа : Однотомное издание Шифр издания : В 6634 22.18/М 34 Автор(ы) : Глушик М. М. Заглавие : Математичне програмування : навч. посібник Выходные данные : Б.м., 2005 Колич.характеристики :212, с Примечания : Бібліогр.: с. 213 ISBN, Цена 966-7827-45-3: (в опр.) : 38 грн 10 к. ББК : 22.183.4я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне програмування Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 416. - [x]Вид документа : Однотомное издание Шифр издания : Б 62723 22.1/О-76 Автор(ы) : Острейковский, Владислав Алексеевич Заглавие : Анализ устойчивости и управляемости динамических систем методами теории катастроф : учеб. пособие для вузов Выходные данные : Б.м., 2005 Колич.характеристики :325, с.: ил. Примечания : Библиогр.: с. 322-323 ISBN, Цена 5-06-004610-9: (в пер.) : 76 грн 7 к. ББК : 22.18я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інформатика Теорія катастроф Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 417. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-1(1) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [Монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т.1, Гл.1-4, кн.1: Типы языков и машин Выходные данные : Киев: Феникс, 2005 Колич.характеристики :270, с Примечания : Библиогр.: с. 264-269 ISBN, Цена 966-651-251-3: (в пер.) : 21 грн 25 к. ББК : 22.1я9 Предметные рубрики: Природничі науки Математика Кібернетика Переробка інформації Перетворювачі Мови Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 418. - [x]Вид документа : Шифр издания : КД-2670 22.1/У 93 Автор(ы) : Ушаков Р. П. Заглавие : Открытая математика. Версия 2.6: Стереометрия : для общеобразоват. учреждений Выходные данные : Б.м., 2005 Колич.характеристики :1 электрон. опт. диск Цена : 45 грн 74 к. ББК : 22.151.02я7 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 419. - [x]Вид документа : Однотомное издание Шифр издания : Б 61802 22.1/Х-21 Автор(ы) : Харрис, Джо Заглавие : Алгебраическая геометрия. Начальный курс Выходные данные : Б.м., 2005 Колич.характеристики :400 с.: ил. Примечания : Парал. тит. л.: англ. -Библиогр. : с.383-385 ISBN, Цена 5-94057-084-4: (в пер.) : 112 грн 46к. ББК : 22.147 Предметные рубрики:Математика Алгебраїчна геометрія Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 420. - [x]Вид документа : Однотомное издание Шифр издания : Б 66892 22.1/Г 95 Автор(ы) : Гуринович, Константин Михайлович, Хамидулина, Любовь Константиновна Заглавие : Справочник по математике : Практ.пособие для шк.и абитуриентов Выходные данные : Б.м., 2005 Колич.характеристики :381, с.: ил. ISBN, Цена 985-6751-26-8: (в пер.): 26 грн 58к. УДК : 51(076.1)(075.4) ББК : 22.1я2 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 421. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-1(3) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т.1, Гл. 13-17, кн. 3: Алгебра и автоматы. II Выходные данные : Киев: Феникс, 2005 Колич.характеристики :261, с Примечания : Библиогр.: с. 256-261 ISBN, Цена 966-651-250-5: (в пер.) : 21 грн 25 к. ББК : 22.1я9 Предметные рубрики: Природничі науки Кібернетика Математика Лінійні рівняння Перетворювачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 422. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-1(2) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т.1, Гл.5-12, Кн.2: Алгебра и автоматы. I Выходные данные : Киев: Феникс, 2005 Колич.характеристики :265, с Примечания : Библиогр.: с. 260-263 ISBN, Цена 966-651-225-4: (в пер.) : 21 грн 25 к. ББК : 22.1я9 Предметные рубрики: Природничі науки Кібернетика Математика Алгебраїчна теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 423. - [x]Вид документа : Однотомное издание Шифр издания : Б 62030 22.161/Б 50 Автор(ы) : Берман, Георгий Николаевич Заглавие : Сборник задач по курсу математического анализа : Решение типичных и трудных задач : учеб. пособие Выходные данные : Б.м., 2005 Колич.характеристики :604 с.: ил. Серия: Учебники для вузов. Специальная литература ISBN, Цена 5-8114-0657-6: (в пер.): 69 грн 8 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Задачі-- завдання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 424. - [x]Вид документа : Однотомное издание Шифр издания : В 6520 22.1/Ш 24 Автор(ы) : Шапорев, Сергей Дмитриевич Заглавие : Математическая логика : курс лекций и практических занятий : учеб. пособие для вузов Выходные данные : Б.м., 2005 Колич.характеристики :410 с Примечания : Библиогр.: с. 405 ISBN, Цена 5-94157-702-8: (в пер.) : 42 грн 84 к. ББК : 22.122я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична логіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 425. - [x]Вид документа : Однотомное издание Шифр издания : Б 63424 22.1/Ф 53 Автор(ы) : Филипс Г. Заглавие : Дифференциальные уравнения : пер. с англ. . -4-е изд., стер. Выходные данные : Б.м., 2005 Колич.характеристики :104 с ISBN, Цена 5-484-00054-8: 24 грн 41к. ББК : 22.161.62 Предметные рубрики: Диференціальні рівняння Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 426. - [x]Вид документа : Однотомное издание Шифр издания : Б 67648 22.1/В 15 Автор(ы) : Валєєв, Кім Галямович, Джалладова, Ірада Агаверді Заглавие : Математика на ступних випробуваннях : Навч.посіб. Выходные данные : Б.м., 2006 Колич.характеристики :397 с.: іл. Примечания : Бібліогр.: с.393 ISBN, Цена 966-574-901-3: (в пер.): 51 грн 75к. УДК : 51(075.4) ББК : 22.1я729 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 427. - [x]Вид документа : Однотомное издание Шифр издания : Б 67685 22.1/В 15 Автор(ы) : Валєєв, Кім Галямович, Джалладова, Ірада Агаверді Заглавие : Збірник задач з теорії ймовірностей та математичної статистики : навч. посібник Выходные данные : Б.м., 2006 Колич.характеристики :351 с Примечания : Бібліогр.: с. 315-316 ISBN, Цена 966-574-855-6: 37 грн 50 к. ББК : 22.17 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 428. - [x]Вид документа : Однотомное издание Шифр издания : В 7079 22.1/К 43 Автор(ы) : Киркинский, Александр Сергеевич Заглавие : Математический анализ : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :526 с. Серия: Gaudeamus Примечания : Библиогр. : с. 525 ISBN, Цена 5-8291-0761-9: 120 грн. 96 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 429. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-2(4) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [Монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т.2, Кн. 4: Размеченные деревья Выходные данные : Киев: Феникс, 2006 Колич.характеристики :351 с Примечания : Библиогр.: с. 341-348 ISBN, Цена 966-651-393-5: (в пер.) : 15 грн. ББК : 22.1 Предметные рубрики: Природничі науки Математика Кібернетика Перетворювачі Схематология Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 430. - [x]Вид документа : Однотомное издание Шифр издания : Б48195 22.1/В 58 Автор(ы) : Власова, Елена Александровна Заглавие : Ряды : учебник для вузов . -Изд. 3-е, испр. Выходные данные : Б.м., 2006 Колич.характеристики :611 с Серия:Математика в техническом университете; вып. 9 Примечания : Библиогр.: с. 600-602 ISBN, Цена 5-7038-2884-8: (в пер.): 66 грн 76к. ББК : 22.161.3я73 Предметные рубрики: Природничі науки Навчальні видання Математика Ряди (мат) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 431. - [x]Вид документа : Однотомное издание Шифр издания : Б 66287 22.1/М 34 Заглавие : Математика : навч. посіб. Выходные данные : Б.м., 2006 Колич.характеристики :554 с.: іл. Примечания : Бібліогр.: с. 541-544 ISBN, Цена 966-7982-72-6: (в опр.) : 64 грн 70 к. ББК : 22.1я73 Предметные рубрики:Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 432. - [x]Вид документа : Однотомное издание Шифр издания : Б 66642 22.1/Л 13 Автор(ы) : Лавров, Игорь Андреевич Заглавие : Математическая логика : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :240 с Серия: Университетский учебник. Прикладная математика и информатика Примечания : Библиогр.: с. 230 ISBN, Цена 5-7695-2735-8: (в пер.) : 71 грн 43 к. ББК : 22.122я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична логіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 433. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-3(5) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [Монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т.3, Кн.5: Типы функций. Интерполяция. Фрактальность Выходные данные : Киев: Феникс, 2006 Колич.характеристики :318, с Примечания : Библиогр.: с. 307-317 ISBN, Цена 966-651-375-7: (в пер.) : 15 грн. ББК : 22.1 Предметные рубрики: Природничі науки Математика Кібернетика Перетворювачі Функції Інтерполяція Фрактальність Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 434. - [x]Вид документа : Однотомное издание Шифр издания : Б 65512 22.1/К 89 Автор(ы) : Кузьмин, Олег Викторович Заглавие : Комбинаторные методы решения логических задач : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :187, с Серия: Высшее педагогическое образование Примечания : Библиогр.: с. 188 ISBN, Цена 5-7107-8579-2: (в пер.) : 42 грн 37 к. ББК : 22.122я73 Предметные рубрики:Математика Логіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 435. - [x]Вид документа : Однотомное издание Шифр издания : Б 65165 22.1/К 78 Заглавие : Краткий справочник по математике : для студентов и инженеров Выходные данные : Б.м., 2006 Колич.характеристики :189, с.: ил. Примечания : Библиогр.: с. 191 ISBN, Цена 5-93037-151-2: 34 грн 20 к. ББК : 22.1я2 Предметные рубрики:Математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 436. - [x]Вид документа : Однотомное издание Шифр издания : Б 64822 22.1/Р 48 Автор(ы) : Ржевський, Сергій Володимирович Заглавие : Дослідження операцій : підручник Выходные данные : Б.м., 2006 Колич.характеристики :558 с. Серия: Альма-Матер Примечания : Библиогр. : с. 543-546 ISBN, Цена 966-8226-33-Х: 40 грн. 80 к. ББК : 22.183 я 73 Предметные рубрики:Математика Кібернетика Математичне программування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 437. - [x]Вид документа : Однотомное издание Шифр издания : Б 10742 22.15/Ф 33 Автор(ы) : Федорчук, Виталий Витальевич, Филиппов, Владимир Васильевич Заглавие : Общая топология.Основные конструкции : учеб. пособие для вузов . -2-е изд., испр. и доп. Выходные данные : Б.м., 2006 Колич.характеристики :332 с Серия: Классический университетский учебник Примечания : Библиогр.: с. 331-332. ISBN, Цена 5-9221-0618-Х: (в пер.) : 72 грн 83 к. ББК : 22.152.1я73 Предметные рубрики: Топологія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 438. - [x]Вид документа : Однотомное издание Шифр издания : Б 65403 22.1/Х-20 Автор(ы) : Харди, Годфри Гарольд Заглавие : Курс чистой математики . -2-е изд., стер. Выходные данные : Б.м., 2006 Колич.характеристики :512 с. Примечания : Бібліогр. в підрядк. прим. ISBN, Цена 5-484-00336-9: 73 грн 30к. ББК : 22.161 Предметные рубрики:Математика Математичні функції Прикладна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 439. - [x]Вид документа : Однотомное издание Шифр издания : Б 68725 22.1/Є 72 Автор(ы) : Єрмакова, Олена Анатоліївна Заглавие : Математика : навч. посіб. Выходные данные : Б.м., 2006 Колич.характеристики :177 с Примечания : Бібліогр.: с. 175 ISBN, Цена 966-388-061-9: 19 грн 90 к. ББК : 22.1я729 Предметные рубрики: Навчальні видання Математика Тригонометрія Математичний аналіз Векторна алгебра Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 440. - [x]Вид документа : Однотомное издание Шифр издания : Б 64806 22.1/Ш 63 Автор(ы) : Шипачев, Виктор Семенович Заглавие : Математический анализ. Теория и практика : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :349, с Серия: Высшее образование ISBN, Цена 5-7107-8950-X: (в пер.) : 64 грн 03 к. ББК : 22.161я73 Предметные рубрики:Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 441. - [x]Вид документа : Однотомное издание Шифр издания : Б 64451 22.1/Е 67 Автор(ы) : Епихин, Валерий Евгеньевич Заглавие : Алгебра и теория пределов : учеб. пособие Выходные данные : Б.м., 2006 Колич.характеристики :352 с.: ил. Серия: Элективный курс Примечания : Библиогр.: с. 340 ISBN, Цена 5-94774-449-Х: (в пер.): 49 грн 86 к. ББК : 22.1я73 Предметные рубрики: Навчальні видання Математика Алгебра Теорія множин Теорія меж Тригонометрія Функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 442. - [x]Вид документа : Однотомное издание Шифр издания : Б 64379 22.1/И 73 Автор(ы) : Антамошин А. Н. Заглавие : Интеллектуальные системы управления организационно-техническими системами Выходные данные : Б.м., 2006 Колич.характеристики :160 с.: ил. Примечания : Библиогр.:с. 146-154 ISBN, Цена 5-93517-289-5: (в пер.): 146 грн 52 к. ББК : 22.161.8+32.813 Предметные рубрики: Природничі науки Математика Системи управління Штучний інтелект Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 443. - [x]Вид документа : Однотомное издание Шифр издания : Б 64785 22.1/Г 24 Автор(ы) : Гачев, Георгий Дмитриевич Заглавие : Математика глазами гуманитария (дневник удивлений математике) : книга издается в авторской редакции Выходные данные : Б.м., 2006 Колич.характеристики :350 с. ISBN, Цена 5-8323-0409-9: 34 грн. 02 к. ББК : 22.1 Предметные рубрики:Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 444. - [x]Вид документа : Однотомное издание Шифр издания : Б 64218 22.1/Б 20 Автор(ы) : Балдин, Константин Васильевич , Башлыков, Виктор Николаевич , Рукосуев, Андрей Вадимович Заглавие : Математика : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :543 с.: ил. Примечания : Библиогр.в конце разд. ISBN, Цена 5-238-00980-1: (в пер.):122 грн 40к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Дискретна математика Функції Інтегральне числення Диференційні рівняння Математична статистика Теорія ймовірностей Економічно-математичні методи Лінійне програмування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 445. - [x]Вид документа : Однотомное издание Шифр издания : Б 64691 22.1/Ш 64 Автор(ы) : Ширяев, Альберт Николаевич Заглавие : Задачи по теории вероятностей : учеб. пособие Выходные данные : Б.м., 2006 Колич.характеристики :416 с Примечания : Библиогр.: с. 399-404 ISBN, Цена 5-94057-107-7: (в пер.) : 65 грн 59 к. ББК : 22.171я73 Предметные рубрики: Ймовірностей теорія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 446. - [x]Вид документа : Однотомное издание Шифр издания : Б 66518 22.1/М 80 Автор(ы) : Морозова, Инна Михайловна Заглавие : Математика : учеб.-тренир. тесты для подгот. к централиз. тестированию и экзамену . -2-е изд. Выходные данные : Б.м., 2006 Колич.характеристики :96 с ISBN, Цена 985-470-509-9: 16 грн. 56 к. ББК : 22.1я729 Предметные рубрики:Математика Тестування Тести-- математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 447. - [x]Вид документа : Однотомное издание Шифр издания : Б 66513 22.1/П 16 Автор(ы) : Пантелеев, Андрей Владимирович Заглавие : Вариационное исчисление в примерах и задачах : учеб. пособие для втузов Выходные данные : Б.м., 2006 Колич.характеристики :272 с Серия: Прикладная математика для ВТУзов Примечания : Библиогр.: с. 270-271 ISBN, Цена 5-06-005327-Х: (в пер.) : 76 грн 97 к. ББК : 22.161.8я73 Предметные рубрики: Природничі науки Навчальні видання Математика Варіаційне числення Приклади Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 448. - [x]Вид документа : Однотомное издание Шифр издания : Б 66646 22.1/К 72 Автор(ы) : Костомаров, Дмитрий Павлович, Фаворский, Антон Павлович Заглавие : Вводные лекции по численным методам : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :184 с Серия: Классический университетский учебник Примечания : Библиогр.: с. 184 ISBN, Цена 5-98704-160-0: (в пер.) : 60 грн 98 к. ББК : 22.193я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інформатика Алгебраїчні рівняння-- лінійні Методи вирішення Обчислювальні процеси Комп'ютери Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 449. - [x]Вид документа : Многотомное издание Шифр издания : Мб 1099-1 22.16/М 34 Автор(ы) : Заглавие : Математические основы теории автоматического управления: В 3 т. : учеб. пособие для вузов/ В.А.Иванов и др. . - Изд. 3-е, перераб. и доп. Т. 1 Выходные данные : Москва: Изд-во МГТУ им. Н. Э. Баумана, 2006 Колич.характеристики :550, с.: ил. Примечания : Библиогр.: с. 535-536 ISBN, Цена 5-7038-2808-2: (в пер.) : 43 грн 22 к. ББК : 22.161.6я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія автоматичного управління Математичні основи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 450. - [x]Вид документа : Однотомное издание Шифр издания : Б 67677 22.1/З-12 Автор(ы) : Заборовець, Марія Олександрівна , Левченко, Франко Андрійович, Охріменко, Михайло Гаврилович Заглавие : Сучасні методи роз'вязування систем лінійних алгебраїчних рівнянь : Навч.-метод. посіб. Выходные данные : Б.м., 2006 Колич.характеристики :75 с Коллективы : Київ. нац. екон. ун-т ім. В. Гетьмана Примечания : Бібліогр.: с. 74-75 ISBN, Цена 966-574-865-3: 12 грн 60 к. ББК : 22.143я73 Предметные рубрики: Природничі науки Навчальні видання Математика Алгебраїчні рівняння Рішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 451. - [x]Вид документа : Однотомное издание Шифр издания : Б 67068 22.1/Ф 76 Автор(ы) : Фоміна, Ася Олександрівна, Павлов, Геннадій Вікторович Заглавие : Типові розрахунки для спецкурсу з математики : метод. вказівки Выходные данные : Б.м., 2006 Колич.характеристики :54 с.: іл. Коллективы : Нац. ун-т корабледування ім. адм. Макарова Примечания : Бібліогр.: с. 53 Цена : 5 грн 90 к. ББК : 22.1р30 Предметные рубрики:Математика Розрахунки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 452. - [x]Вид документа : Однотомное издание Шифр издания : Б 65433 22.19/П 69 Автор(ы) : Москальков М. М., Риженко А. І. , Войцеховський С. О. Заглавие : Практикум з методів обчислень : метод вказівки та навч. завдання до практ. і лаб.робіт із чисельного розв'язання рівнянь і систем Выходные данные : Б.м., 2006 Колич.характеристики :77, с Примечания : Бібліогр.: с. 77-78 ISBN, Цена 966-608-504-6: 9 грн 55 к. ББК : 22.193я73 Предметные рубрики: Чисельні методи Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 453. - [x]Вид документа : Однотомное издание Шифр издания : В 7161 22/Л 40 Автор(ы) : Лежнев, Алексей Викторович Заглавие : Динамическое программирование в экономических задачах : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :174, с Примечания : Библиогр.: с. 175 ISBN, Цена 5-94774-344-2: (в пер.) :43 грн 53 к. ББК : 22.183.47я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне програмування Економічні задачі Вирішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 454. - [x]Вид документа : Однотомное издание Шифр издания : Б 18864 22.16/С 17 Автор(ы) : Самойленко, Анатолий Михайлович, Кривошея, Сергей Арсентьевич, Перестюк, Николай Алексеевич Заглавие : Дифференциальные уравнения : практический курс : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :383 с Серия: Учебник для высших учебных заведений. Математика Примечания : Библиогр.: с. 381 ISBN, Цена 5-06-005326-1: (в пер.) : 90 грн. ББК : 22.161.61я73 Предметные рубрики: Природничі науки Навчальні видання Математика Диференціальні рівняння Теорія рішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 455. - [x]Вид документа : Однотомное издание Шифр издания : Б 66832 22.1/Т 33 Автор(ы) : Печинкин, Александр Владимирович, Тескин, Олег Иванович, Цветкова, Галина Михайловна, Бочаров, Павел Петрович Заглавие : Теория вероятностей : учеб. для вузов . -4-е изд., стер. Выходные данные : Б.м., 2006 Колич.характеристики :455 с. Серия:Математика в техническом университете; вып. 16 Примечания : Библиогр. : с. 446-447 ISBN, Цена 5-7038-2485-0: 66 грн 76 к. ББК : 22.171я73 Предметные рубрики:Математика Теорія ймовірностей Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 456. - [x]Вид документа : Многотомное издание Шифр издания : Б54023-1 22.16/И 46 Автор(ы) : Ильин, Владимир Александрович и др. Заглавие : Математический анализ: Учебник : В 2 ч./ Садовничий Виктор Антонович, Сендов Благовест Христов. Ч.1 . -3-е изд., перераб. и доп. Выходные данные : Москва: Изд-во Моск. ун-та, 2006 - Колич.характеристики :660 с ISBN, Цена 5-482-00445-7: (в пер.): 54 грн. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 457. - [x]Вид документа : Однотомное издание Шифр издания : В 6624 22.1/Ч-81 Автор(ы) : Чубатюк, Валентин Миколайович Заглавие : Вища математика : навч. посіб. Выходные данные : Б.м., 2006 Колич.характеристики :422 с Коллективы : Вінниц. фін.-екон. ун-т Примечания : Бібліогр.: с. 422 ISBN, Цена 966-370-028-9: (в опр.) : 58 грн 65 к. ББК : 22.11я73 Предметные рубрики: Природничі науки Навчальні видання Математика Вища математика Лінійна алгебра Векторна алгебра Математичний аналіз Інтеграли Диференціальні рівняння Ряди Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 458. - [x]Вид документа : Многотомное издание Шифр издания : Б54023-2 22.16/И 46 Автор(ы) : Ильин, Владимир Александрович и др. Заглавие : Математический анализ: Учебник: В 2 ч./ Ильин Владимир Александрович, Садовничий Виктор Антонович, Сендов Благовест Христов. Ч. 2 . -2-е изд., перераб. и доп. Выходные данные : Москва: Изд-во Моск. ун-та, 2006 - Колич.характеристики :353, с ISBN, Цена 5-482-00444-9: (в пер.): 32 грн 40 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 459. - [x]Вид документа : Однотомное издание Шифр издания : Б57848 22.1/К 91 Автор(ы) : Кундышева, Елена Сергеевна Заглавие : Математика : учеб. пособие для экономистов Выходные данные : Москва: Дашков и К, 2006 Колич.характеристики :534 с. Примечания : Библиогр. : с. 525-526 ISBN, Цена 5-91131-149-6: 59 грн. 40 к. ББК : 22.1я73 Предметные рубрики:Математика Теорія ймовірностей Математична статистика Лінійна алгебра Лінійне програмування Аналітична геометрія Математичний аналіз Економіко-математичні методи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 460. - [x]Вид документа : Однотомное издание Шифр издания : Б 62945 22.16/К 82 Автор(ы) : Кристалинский, Роман Ефимович, Кристалинский, Владимир Романович Заглавие : Преобразование Фурье и Лапласа в системах компьютерной математики : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :216 с.: ил. Примечания : Библиогр.: с. 214-215 ISBN, Цена 5-93517-250-Х: ББК : 22.161.2я73 Предметные рубрики: Інтегральні перетворення Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 461. - [x]Вид документа : Однотомное издание Шифр издания : Б 62550 22.1/З-15 Автор(ы) : Репета В. К. Заглавие : Задачі з параметрами : Навч. посіб. для учнів загальноосвіт. навч. закл. Выходные данные : Б.м., 2006 Колич.характеристики :299, с.: іл. Примечания : Бібліогр.: с. 301 ISBN, Цена 966-642-306-5: (в опр.): 20 грн 50 к. ББК : 22.1я721 Предметные рубрики: Природничі науки Навчальні видання для середньої школи Математика Математичні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 462. - [x]Вид документа : Однотомное издание Шифр издания : Б 70123 65.05/О-65 Автор(ы) : Оревков, Владимир Павлович, Оревкова, Ольга Алексеевна Заглавие : Дискретная математика для гуманитариев : учеб. пособие Выходные данные : Б.м., 2006 Колич.характеристики :80 с Примечания : Библиогр.: с. 80 ISBN, Цена 5-288-04059-1: 33 грн 98 к. ББК : 65.050.214 Предметные рубрики: Природничі науки Навчальні видання Математика Дискретна математика Соціологія Інформаційні системи Бази даних-- упраління Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 463. - [x]Вид документа : Однотомное издание Шифр издания : Б49396 22.18/П 83 Автор(ы) : Протасов, Игорь Дмитриевич Заглавие : Теория игр и исследование операций : Учеб. пособие . -2-е изд. Выходные данные : Б.м., 2006 Колич.характеристики :368 с. Примечания : Библиогр. : с. 365-366 ISBN, Цена 5-85438-133-8: (в пер.) : 59 грн 47к. ББК : 22.183.2я7 Предметные рубрики:Математика Ігор теорія Прикладна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 464. - [x]Вид документа : Однотомное издание Шифр издания : Б 62003 22.1/Д 14 Автор(ы) : Дадаян, Александр Арсенович Заглавие : Математика для педагогических училищ : учеб. для сред. проф. образования Выходные данные : Б.м., 2006 Колич.характеристики :510 с Серия: Профессиональное образование ISBN, Цена 5-8199-0233-5: (в пер.) : 43 грн 16 к. ББК : 22.1я723 Предметные рубрики:Математика Визначення формули Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 465. - [x]Вид документа : Однотомное издание Шифр издания : Б 69958 22.1/К 89 Автор(ы) : Кузнецов, Альберт Николаевич Заглавие : Решебник задач по теме "Дифференциальные управления" : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :308, с Примечания : Библиогр.: с. 308 ISBN, Цена 978-966-321-061-2: (в пер.) : 18 грн. ББК : 22.161.61я73 Предметные рубрики: Диференціальні рівняння Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 466. - [x]Вид документа : Однотомное издание Шифр издания : Б 62115 22.1/Ш 57 Автор(ы) : Шикин, Евгений Викторович, Шикина, Гузель Евгеньевна Заглавие : Исследование операций : учебник для вузов Выходные данные : Б.м., 2006 Колич.характеристики :275, с Коллективы : МГУ им. М. В. Ломоносова Примечания : Библиогр.: с. 275 ISBN, Цена 5-482-00521-6: (в пер.) : 43 грн 20 к. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійне програмування Транспортні задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 467. - [x]Вид документа : Однотомное издание Шифр издания : Б 64213 22.1/Г 69 Автор(ы) : Горлач, Борис Алексеевич Заглавие : Математика : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :912 с. ISBN, Цена 5-238-01054-0: 108 грн. ББК : 22.1я73 Предметные рубрики:Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 468. - [x]Вид документа : Однотомное издание Шифр издания : Б 65008 22.1/П 32 Автор(ы) : Пилиди, Владимир Ставрович Заглавие : Курс математики для гуманитариев Выходные данные : Б.м., 2006 Колич.характеристики :195 с. Примечания : Библиогр. : с. 193 ISBN, Цена 5-9502-0148-5: 56 грн 63 к. ББК : 22.1я73 Предметные рубрики:Математика Математична логіка Теорія ймовірностей Математика для гуманітаріїв Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 469. - [x]Вид документа : Однотомное издание Шифр издания : В 6864 22.14/Д 21 Автор(ы) : Даугавет, Игорь Карлович Заглавие : Теория приближенных методов.Линейные уравнения : учеб. пособие . -2-е изд.,перераб.и доп. Выходные данные : Б.м., 2006 Колич.характеристики :272 с Примечания : Библиогр.: с.271-272 ISBN, Цена 5-94157-737-0: (в пер.) : 42 грн 84 к. ББК : 22.143я73+22.161.6я73 Предметные рубрики: Лінійні рівняння Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 470. - [x]Вид документа : Однотомное издание Шифр издания : Б 63906 22.18/М 36 Автор(ы) : Маховенко, Елена Борисовна Заглавие : Теоретико-числовые методы в криптографии : учеб. пособие для вузов Выходные данные : Б.м., 2006 Колич.характеристики :318, с ISBN, Цена 5-85438-143-5: 46 грн 73 к. ББК : 22.181я73 Предметные рубрики: Природничі науки Навчальні видання Математика Комп'ютерна безпека Безпека-- криптосистеми Алгебраїчні задачі Арифметика великих чисел-- алгоритми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 471. - [x]Вид документа : Однотомное издание Шифр издания : Б 67474 22.1/К 44 Автор(ы) : Кісілевич, Олександра Василівна Заглавие : Математика : Навч.-метод.посіб.для абітурієнтів Выходные данные : Б.м., 2006 Колич.характеристики :318 с Примечания : Бібліогр.: с.316-317 ISBN, Цена 966-418-013-0: 40 грн 80к. УДК : 51(072) ББК : 22.1я729 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 472. - [x]Вид документа : Однотомное издание Шифр издания : Б 63703 22.15/С 86 Автор(ы) : Стройк, Дирк Ян Заглавие : Очерк истории дифференциальной геометрии . -2-е изд., стер. Выходные данные : Б.м., 2006 Колич.характеристики :80, с Примечания : Библиогр.: с. 72-74 ISBN, Цена 5-484-00326-1: 30 грн 31 к. ББК : 22.151.6г Предметные рубрики: Природничі науки Математика Диференціальна геометрія Історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 473. - [x]Вид документа : Многотомное издание Шифр издания : Б 63089-1 22.1/П 22 Автор(ы) : Пащенко, Федор Федорович Заглавие : Введение в состоятельные методы моделирования систем: учеб. пособие : В 2ч./ Федор Федорович Пащенко. Ч. 1: Математические основы моделирования систем Выходные данные : Москва: Финансы и статистика, 2006 Колич.характеристики :327 с.: ил. Примечания : Бібліогр. в кінці гл. ISBN, Цена 5-279-02922-Х: 60 грн 48к. ББК : 22.181я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні основи Моделюввання систем Ідентифікація систем Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 474. - [x]Вид документа : Однотомное издание Шифр издания : Б 63731 22.12/Х-26 Автор(ы) : Хаусдорф, Феликс Заглавие : Теория множеств . -Изд. 3-е, стер. Выходные данные : Б.м., 2006 Колич.характеристики :302 с. Примечания : Библиогр. : с.291-295 ISBN, Цена 5-484-00296-6: 52 грн 13к. ББК : 22.126 Предметные рубрики:Математика Множин теорія Прикладна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 475. - [x]Вид документа : Однотомное издание Шифр издания : Б 69641 22.1/Т 33 Автор(ы) : Мхитарян, Владимир Сергеевич , Трошин, Лев Иванович, Астафьева, Екатерина Викторовна, Миронкина, Юлия Николаевна Заглавие : Теория вероятностей и математическая статистика : учеб. пособие Выходные данные : Б.м., 2007 Колич.характеристики :240 с.: ил. Серия: Университетская серия Примечания : Библиогр. : с. 223 ISBN, Цена 978-5-7958-0169-8: 33 грн 66к. ББК : 22.17я73 Предметные рубрики:Математика Математична статистика Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 476. - [x]Вид документа : Однотомное издание Шифр издания : Б 21403. 22.1/Б 74 Автор(ы) : Богомолов, Николай Васильевич Заглавие : Практические занятия по математике : учеб. пособие для сред. проф. учеб. заведений . -9-е изд., стер. Выходные данные : Москва: Высш. шк., 2007 Колич.характеристики :495 с. ISBN, Цена 5-06-005713-5: 144 грн. 61 к. ББК : 22.1я723 Предметные рубрики:Математика Обчислювальна математика Алгебра Геометрія Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 477. - [x]Вид документа : Шифр издания : Б 79563 22.1/Ц 83 Автор(ы) : Цубербиллер О. Н. Заглавие : Задачи и упражнения по аналитической геометрии : учеб. для вузов . -33-е изд., стер. Выходные данные : Б.м., 2007 Колич.характеристики :336 с. Серия: Учебники для вузов. Специальная литература ISBN, Цена 978-5-8114-0475-9: (в пер.) : 159 грн 1 к. ББК : 22.151.54я73 Предметные рубрики:Математика Алгебра Векторна алгебра Лінійна алгебра Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 478. - [x]Вид документа : Однотомное издание Шифр издания : Б 70568 22.18/Ш 23 Автор(ы) : Шапкин, Александр Сергеевич, Мазаева, Наталья Петровна Заглавие : Математические методы и модели исследования операций : учебник для вузов . -4-е изд. Выходные данные : Б.м., 2007 Колич.характеристики :395, с.: ил. Примечания : Библиогр.: с. 395-396 ISBN, Цена 5-91131-331-6 : (в пер.) : 64 грн 35 к. ББК : 22.183я73+65в661я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійне програмування Економіко-математичні методи Моделі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 479. - [x]Вид документа : Однотомное издание Шифр издания : Б 70586 22.18/М 51 Автор(ы) : Меньшиков, Иван Станиславович Заглавие : Лекции по теории игр и экономическому моделированию : учеб. пособие для вузов Выходные данные : Б.м., 2007 Колич.характеристики :207 с Серия: Естественные науки, математика, информатика Примечания : Библиогр.: с. 207 ISBN, Цена 5-94073-099-Х: 74 грн 25 к. ББК : 22.183.2 Предметные рубрики: Природничі науки Навчальні видання Математика Економічне моделювання Теорія ігор Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 480. - [x]Вид документа : Однотомное издание Шифр издания : Б 70826 22.1/М 54 Автор(ы) : Сокуренко Є. В. Заглавие : Методичні вказівки для самостійної роботи та виконання модульних завдань з курсу "Елементи теорії ймовірностей та математичної статистики" Выходные данные : Б.м., 2007 Колич.характеристики :51, с.: іл. Коллективы : Нац. ун-т кораблебудування ім. адмірала Макарова, Херсон. філ. НУК Примечания : Библиогр.: с. 52 Цена : 3 грн 40к. ББК : 22.17р30 Предметные рубрики: Природничі науки Навчальні видання Методичні вказівки Математика Математична статистика Теорія ймовірностей Модульні завдання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 481. - [x]Вид документа : Однотомное издание Шифр издания : Б 69897 22.17/К 72 Автор(ы) : Костюкова, Нина Ивановна Заглавие : Графы и их применение. Комбинаторные алгоритмы для программистов : учеб. пособие Выходные данные : Б.м., 2007 Колич.характеристики :311 с Серия: Основы информационных технологий Примечания : Библиогр.: с. 311 ISBN, Цена 978-5-94774-545-0: (в пер.) : 92 грн 70 к. ББК : 22.174.2я7+22.181я7 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія графов Програмування Комбінаторні методи обчислення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 482. - [x]Вид документа : Однотомное издание Шифр издания : Б 70302 22.19/Л 24 Автор(ы) : Лапчик, Михаил Павлович, Рагулина , Марина Ивановна, Хеннер, Евгений Карлович Заглавие : Элементы численных методов : учебник Выходные данные : Б.м., 2007 Колич.характеристики :223 с Серия: Среднее профессиональное образование. Информатика и вычислительная техника Примечания : Библиогр.: с. 221 ISBN, Цена 978-5-7695-2700-5: (в пер.) : 67 грн 32 к. ББК : 22.193я723 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні задачі Вирішення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 483. - [x]Вид документа : Однотомное издание Шифр издания : Б 70408 22.1/Б 91 Автор(ы) : Бурмистрова, Елена Борисовна, Лобанов, Сергей Григорьевич Заглавие : Линейная алгебра с элементами аналитической геометрии : учеб. пособие . -2-е изд., доп. Выходные данные : Б.м., 2007 Колич.характеристики :220 с Примечания : Библиогр.: с. 216 ISBN, Цена 978-5-7598-0308-9: 60 грн 89 к. ББК : 22.143я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійна алгебра Аналітична геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 484. - [x]Вид документа : Однотомное издание Шифр издания : В 7217 22.1/Ф 91 Автор(ы) : Фролов, Сергей Аркадьевич Заглавие : Начертательная геометрия : учеб. для вузов . -3-е изд., перераб. и доп. Выходные данные : Б.м., 2007 Колич.характеристики :285 с.: ил. Серия: Высшее образование Примечания : Библиогр.: с. 281 ISBN, Цена 5-16-001849-2: (в пер.) : 94 грн 60 к. ББК : 22.151.34я73 Предметные рубрики: Нарисна геометрія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 485. - [x]Вид документа : Однотомное издание Шифр издания : В 8191 22.1/З-48 Автор(ы) : Зеленський, Кирило Харитонович, Макаренко, Володимир Миколайович Заглавие : Теорія ймовірностей і математична статистика : Навч. посіб. для дистанц. навч. Выходные данные : Б.м., 2007 Колич.характеристики :201 с.: іл. Примечания : Бібліогр. : с. 198-199 ISBN, Цена 966-388-122-4: 30 грн ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Математична статистика Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 486. - [x]Вид документа : Однотомное издание Шифр издания : Б 49977. 22.11/В 19 Автор(ы) : Васильченко Іван Петрович Заглавие : Вища математика для економістів : підручник . -3-тє вид., випр. і допов. Выходные данные : Б.м., 2007 Колич.характеристики :454 с ISBN, Цена 966-346-226-4: 66 грн. УДК : 51:33-057.86](075.8) ББК : 22.11я73 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 487. - [x]Вид документа : Однотомное издание Шифр издания : А 6387 22.1/Ц 59 Автор(ы) : Цикунов А. Е. Заглавие : Сборник формул по математике . -Изд. 3-е Выходные данные : Б.м., 2007 Колич.характеристики :160 с. Серия: Карманный справочник ISBN, Цена 5-88782-281-3: 7 грн 20к. ББК : 22.1я2 Предметные рубрики:Математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 488. - [x]Вид документа : Однотомное издание Шифр издания : Б 7472. 22.1/К 89 Автор(ы) : Кузнецов, Олег Петрович Заглавие : Дискретная математика для инженера . -5-е изд. стериот. Выходные данные : Б.м., 2007 Колич.характеристики :395 с.: ил. Примечания : Библиогр.: с. 388-389 Цена : (в пер.) : 144 грн 99 к. ББК : 22.174 Предметные рубрики:Математика Дискретна математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 489. - [x]Вид документа : Однотомное издание Шифр издания : Б 68081 22.1/Т 41 Автор(ы) : Тимофеева, Ирина Леонидовна Заглавие : Математическая логика.Курс лекций : учеб. пособие для вузов . -2-е изд., перераб. Выходные данные : Б.м., 2007 Колич.характеристики :302, с Примечания : Библиогр. : с. 294 ISBN, Цена 978-5-98227-307-9: 93 грн 56к. ББК : 22.122я73 Предметные рубрики:Математика Математична логіка Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 490. - [x]Вид документа : Однотомное издание Шифр издания : Б 67951 22.17/К 78 Автор(ы) : Крамер, Дункан Заглавие : Математическая обработка данных в социальных науках: современные методы : учеб. пособие для вузов Выходные данные : Б.м., 2007 Колич.характеристики :287 с Примечания : Библиогр.: с. 285-286 ISBN, Цена 978-5-7695-2878-1: (в пер.) : 138 грн 78 к. ББК : 22.171я73+60в641я73 Предметные рубрики: Природничі науки Навчальні видання Математика математична обробка даних Соціальні науки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 491. - [x]Вид документа : Однотомное издание Шифр издания : Б 67952 22.1/С 72 Автор(ы) : Спирина, Марина Савельевна, Спирин, Павел Алексеевич Заглавие : Дискретная математика : учебник . -3-е изд., стер. Выходные данные : Б.м., 2007 Колич.характеристики :368 с Серия: Среднее профессиональное образование . Информатика и вычислительная техника Примечания : Библиогр.: с. 366 ISBN, Цена 978-5-7695-3785-1: (в пер.) : 127 грн 66 к. ББК : 22.174я723 Предметные рубрики: Природничі науки Навчальні видання Математика Дискретна математика Класична логіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 492. - [x]Вид документа : Однотомное издание Шифр издания : Б 69113 22.1/В 67 Автор(ы) : Волков, Владимир Тарасович, Ягола Анатолий Григорьевич Заглавие : Интегральные уравнения. Вариационное исчисление : методы решения задач : учеб. пособие для вузов Выходные данные : Б.м., 2007 Колич.характеристики :137, с Примечания : Библиогр.: с. 138 ISBN, Цена 978-5-98227-315-4: 49 грн 14 к. ББК : 22.161.67я73 + 22.161.8я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інтегральні рівняння Задачі-- розв'язування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 493. - [x]Вид документа : Однотомное издание Шифр издания : Б 101504 821.161.1/Л 38 Автор(ы) : Лёвшин, Владимир Артурович Заглавие : Магистр Рассеянных Наук : повести Выходные данные : Б.м., 2007 Колич.характеристики :318 с.: ил Серия: Научные развлечения ISBN, Цена 978-5-91045-040-4: (в пер.) : 30 грн. УДК : 821.161.1-311.3-93 Предметные рубрики: Художня література Російська література, 20 ст. Пригодницькі романи Дитяча література Математика Аннотация: Герои этой книги носятся по всему белому свету в поисках приключений: плывут по океану, катаются на льдине, гуляют по краю кратера вулкана. Они знают, как заставить Черепаху и Гепарда бежать, не обгоняя друг друга, почему пятиконечную звезду древние называли золотой, а также умеют подобрать математический ключ к любому фруктовому магазину. С помощью участников Клуба Рассеянного Магистра читатель мгновенно запоминает правила и парадоксы самой точной в мире науки. Для среднего школьного возраста. Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 494. - [x]Вид документа : Однотомное издание Шифр издания : Б 54795. 22.17/Б 44 Автор(ы) : Белько, Иван Васильевич, Свирид, Георгий Петрович Заглавие : Теория вероятностей и математическая статистика : примеры и задачи : учеб. пособие для вузов . -3-е изд., стер. Выходные данные : Б.м., 2007 Колич.характеристики :250 с Примечания : Библиогр.: с. 245 ISBN, Цена 978-985-475-289-1: 57 грн 19 к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 495. - [x]Вид документа : Однотомное издание Шифр издания : Б 66130 22.3/К 49 Автор(ы) : Климонтович, Юрий Львович Заглавие : Турбулентное движение и структура хаоса: Новый подход к статистической теории открытых систем . -3-е изд. Выходные данные : Б.м., 2007 Колич.характеристики :323 с Серия: Синергетика: от прошлого к будущему Примечания : Библиогр.: с 305-317 ISBN, Цена 978-5-484-00899-5: 77 грн 22 к. ББК : 22.311+22.317.2 Предметные рубрики:Математика Фізика Турбулентність Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 496. - [x]Вид документа : Однотомное издание Шифр издания : В 7476 22.1/З-12 Автор(ы) : Забара, Станіслав Сергійович, Зеленський, Кирило Харитонович Заглавие : Математичні основи інформаційної діяльності : навч. посіб. для дистанційного навч. Выходные данные : Б.м., 2007 Колич.характеристики :336 с Примечания : Библиогр.: с. 332-333 ISBN, Цена 966-388-115-1: 21 грн 60 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Інформаційна діяльність Математичні основи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 497. - [x]Вид документа : Однотомное издание Шифр издания : Б 70168 22.17/С 72 Автор(ы) : Спирина, Марина Савельевна, Спирин, Павел Алексеевич Заглавие : Теория вероятностей и математическая статистика : учеб. для сред. проф. образования Выходные данные : Б.м., 2007 Колич.характеристики :352 с.: ил. Серия: Среднее профессиональное образование. Информатика и вычислительная техника Примечания : Библиогр.: с. 350 ISBN, Цена 978-5-7695-2768-5: (в пер.) : 92 грн 7 к. ББК : 22.17я723 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 498. - [x]Вид документа : Однотомное издание Шифр издания : Б 4223. 22.17/Ч-68 Автор(ы) : Чистяков, Владимир Павлович Заглавие : Курс теории вероятностей : учебник для вузов Выходные данные : Б.м., 2007 Колич.характеристики :253 с Примечания : Библиогр.: с. 248 ISBN, Цена 978-5-358-03022-0: (в пер.) : 73 грн 56 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 499. - [x]Вид документа : Многотомное издание Шифр издания : Мб997-1 22.16/Ф 65 Автор(ы) : Фихтенгольц, Григорий Михайлович Заглавие : Курс дифференциального и интегрального исчисления: учеб. для вузов/ Григорий Михайлович Фихтенгольц. - 8-е изд. Т. 1 Выходные данные : Москва: Физматлит, 2007 Колич.характеристики :679 с.: ил. ISBN, Цена 978-5-9221-0436-4: (в пер.): 91 грн 94 к. ББК : 22.161.11я73+22.161.12я73 Предметные рубрики: Числення Математика Диферанціальне числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 500. - [x]Вид документа : Многотомное издание Шифр издания : Мв 240-3(6) 22.1/Л 63 Автор(ы) : Лисовик, Леонид Петрович Заглавие : Теория трансдьюсеров: [Монография]/ Леонид Петрович Лисовик. - (Математическое мышление). Т. 3, Кн. 6: Функции и апплекативные системы Выходные данные : Киев: Феникс, 2007 Колич.характеристики :284 с Примечания : Библиогр.: с. 273-284 ISBN, Цена 978-966-651-466-3: 35 грн 16 к. ББК : 22.1я9 Предметные рубрики: Природничі науки Кібернетика Математика Перетворювачі Функції Теорія Аплекативні системи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 501. - [x]Вид документа : Однотомное издание Шифр издания : Б51447 22.16/З-15 Автор(ы) : Бараненков Г. С. Заглавие : Задачи и упражнения по математическому анализу для втузов : учеб. пособие Выходные данные : Москва: АСТ : Астрель, 2007 Колич.характеристики :495 с.: ил. ISBN, Цена 5-17-002965-9: (в пер.): 86 грн 40 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Задачі та вправи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 502. - [x]Вид документа : Однотомное издание Шифр издания : Б 68640 22.11/М 34 Автор(ы) : Лавренчук, Володимир Петрович, Готинчан, Тетяна Іванівна, Дронь, Віталій Сільвестрович Заглавие : Математика для економістів: теорія та застосування : підручник Выходные данные : Б.м., 2007 Колич.характеристики :595 с Примечания : Бібліогр.: с. 568-571 ISBN, Цена 966-351-059-5: (в опр.) : 70 грн 50 к. ББК : 22.11я73 Предметные рубрики:Математика Лінійні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 503. - [x]Вид документа : Однотомное издание Шифр издания : Б 63686. Автор(ы) : Блехман, Илья Израилевич , Мышкис, Анатолий Дмитриевич, Пановко , Яков Гилелевич Заглавие : Прикладная математика : предмет, логика, особенности подходов : с примерами из механики . -Изд. 4-е Выходные данные : Б.м., 2007 Примечания : Библиогр.: с. 329-350 ISBN, Цена 978-5-382-00210-1: 99 грн 20 к. ББК : 22.2я73+22.11я73 Предметные рубрики: Природничі науки Математика Фізика Прикладна математика Математичні досліди Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 504. - [x]Вид документа : Однотомное издание Шифр издания : Б 67833 22.1/К 21 Автор(ы) : Карагодова О. О., Кігель Р. В., Рожок В. Д. Заглавие : Дослідження операцій : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2007 Колич.характеристики :254, с Коллективы : Київ. екон. ін-т менеджменту Примечания : Бібліогр.: с. 250-251 ISBN, Цена 978-966-364-446-2: (в опр.): 39 грн. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне моделювання Економічне моделювання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 505. - [x]Вид документа : Однотомное издание Шифр издания : А 7682 22.1/Б 48 Автор(ы) : Березина Н.А., Максина Е. Л. Заглавие : Математика : учеб. пособие для сред. учеб. заведений Выходные данные : Б.м., 2007 Колич.характеристики :175 с Серия: Профессиональное образование ISBN, Цена 5-369-00061-1: 20 грн 88 к. ББК : 22.1я723 Предметные рубрики: Природничі науки Навчальні видання Математика Множини Функції Похідна Тригонометрія Теорема Ферма Формула Маклорена Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 506. - [x]Вид документа : Однотомное издание Шифр издания : Б 75057 22.1/А 94 Автор(ы) : Афанасьев, Владимир Васильевич Заглавие : Теория вероятностей : учеб. пособие для вузов Выходные данные : Б.м., 2007 Колич.характеристики :350 с.: ил Серия: Учебники для вузов Примечания : Библиогр.: с. 349-350 ISBN, Цена 978-5-691-01525-0 : (в пер.) : 78 грн. 75 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Випадкові величини Содержание : Элементы комбинаторика ; Случайные события ; Случайные величины ; Энтропия и информация ; Математическая статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 507. - [x]Вид документа : Многотомное издание Шифр издания : Б 70315-1 22.1/К 88 Автор(ы) : Кудинов А. Т., Щепанский С. Б. Заглавие : Основы математики и информатики: учеб. пособие для юрид. вузов/ А. Т. Кудинов. Ч. 1 Выходные данные : : Элит Б.м., 2007 Колич.характеристики :224 с Примечания : Библиогр.: с. 221-222 ISBN, Цена 978-5-902405-19: 36 грн. ББК : 22.1я73 Предметные рубрики:Математика Інформатика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 508. - [x]Вид документа : Однотомное издание Шифр издания : Б 99186 22.1/М 54 Автор(ы) : Казарезов А. Я., Галь А. Ф., Фарионова Т. А., Чернов С. К. Заглавие : Методы оптимизации в принятии технических и экономических решений : учеб. пособие Выходные данные : Б.м., 2007 Колич.характеристики :145 с Примечания : Библиогр.: с. 141-143 ISBN, Цена 966-8147-94-4: 23 грн. ББК : 22.183.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичне програмування Проектування Суднобудування Рішення -- технічні-- економічні Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 509. - [x]Вид документа : Однотомное издание Шифр издания : В 4710. 22.19/Б 30 Автор(ы) : Бахвалов, Николай Сергеевич, Жидков, Николай Петрович, Кобельков, Георгий Михайлович Заглавие : Численные методы : учеб. пособие для вузов . -5-е издание Выходные данные : Б.м., 2007 Колич.характеристики :636 с Коллективы : МГУ им. М. А. Ломоносова Примечания : Библиогр.: с. 624-628 ISBN, Цена 5-94774-620-4: (в пер.) : 85 грн 50 к. ББК : 22.192я73 + 22.193я73 Предметные рубрики: Природничі науки Навчальні видання Математика Численні методи Содержание : Погрешность результата численного решения задачи ; Интерполяция и численное дифференцирование ; Численное интегрирование ; Приближение функций и смежные вопросы ; Многомерные задачи ; Численные методы алгебры ; Решение систем нелинейных уравнений и задач оптимизации ; Численные методы решения задачи Коши для обыкновенных дифференциальных уравнений ; Численные методы решения краевых задач для обыкновенных дифференциальных уравнений ; Методы решения уравнений в частных производных Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 510. - [x]Вид документа : Однотомное издание Шифр издания : Б 75565 22.1/Л 72 Автор(ы) : Лозовий, Богдан Леонтійович, Пушак, Ярослав Сильвестрович, Шабат, Орислава Євгенівна Заглавие : Практикум з вищої математики : навч. посіб. . -Вид. 2-ге, допов. і переробл. Выходные данные : Б.м., 2007 Колич.характеристики :284 с Серия: Вища освіта в Україні Примечания : Бібліогр.: с. 283-284 ISBN, Цена 966-2025-16-2 : (в опр.) : 48 грн. ББК : 22.11я73 Предметные рубрики:Математика Вища математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 511. - [x]Вид документа : Однотомное издание Шифр издания : Б 69804 22.1/К 24 Автор(ы) : Кармелюк, Ганна Іванівна Заглавие : Теорія йомівірностей та математична статистика : посіб. з розв'язування задач для вищ. навч. закл. Выходные данные : Київ: Центр учб. літ, 2007 Колич.характеристики :575 с.: іл. Примечания : Бібліогр.: с. 565-566 ISBN, Цена 978-966-364-495-0: 75 грн. ББК : 22.17я73 Предметные рубрики: Статистика Математика Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 512. - [x]Вид документа : Однотомное издание Шифр издания : В 7595 22.1/Ч-57 Автор(ы) : Чжун, Кай Лай, АитСахлиа , Фарид Заглавие : Элементарный курс теории вероятностей : стохастические процессы и финансовая математика Выходные данные : Б.м., 2007 Колич.характеристики :455 с Примечания : Библиогр.: с. 444-445 ISBN, Цена 5-94774-347-7: (в пер.): 89 грн 10 к. ББК : 22.171 Предметные рубрики: Природничі науки Математика Теорія ймовірностей Стохастичні процеси Фінансова математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 513. - [x]Вид документа : Однотомное издание Шифр издания : Б 58234. 22.1/Ш 23 Автор(ы) : Шапкин, Александр Сергеевич Заглавие : Задачи по высшей математике, теории вероятностей, математической статистике, математическому программированию с решениями : учеб. пособие для вузов . -5-е изд. Выходные данные : Б.м., 2007 Колич.характеристики :431 с Примечания : Библиогр.: с. 428 ISBN, Цена 978-5-91131-659-4: 89 грн 10 к. ББК : 22.11 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Математичне програмування Задачі-- розв'язування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 514. - [x]Вид документа : Однотомное издание Шифр издания : Б 68827 22.1/Г 85 Автор(ы) : Грисенко, Марина Віталіївна Заглавие : Математика для економістів. Методи й моделі, приклади й задачі : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2007 Колич.характеристики :719 с Примечания : Бібліогр.: с. 709 ISBN, Цена 978-966-06-0436-0: (в опр.): 54 грн. УДК : 51:33-057.86](075.8) ББК : 22.1я73 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 515. - [x]Вид документа : Однотомное издание Шифр издания : Б 72794 22.1/М 34 Заглавие : Математические методы и модели исследования операций : учебник для вузов Выходные данные : Б.м., 2008 Колич.характеристики :591 с.: ил. Примечания : Библиогр.: с. 588-589 ISBN, Цена 978-5-238-01325-1 (в пер.): 126 грн 36 к. ББК : 22.183.5я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні методи-- застосування-- економіка Задачі-- економічні Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 516. - [x]Вид документа : Однотомное издание Шифр издания : Б 68444 22.1/О-90 Автор(ы) : Оуэн Г. Заглавие : Теория игр . -3-е изд. Выходные данные : Б.м., 2008 Колич.характеристики :215 с.: ил. ISBN, Цена 978-5-9502-0330-5: 136 грн 62 к. ББК : 22.183.2 Предметные рубрики: Природничі науки Математика Теорія ігор Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 517. - [x]Вид документа : Однотомное издание Шифр издания : Б 75922 22.1/С 23 Заглавие : Сборник задач по геометрии : учеб. пособие . -Изд. 2-е, стер. Выходные данные : Б.м., 2008 Колич.характеристики :238 с Серия: Классическая учебная литература по математике Примечания : Библиогр.: с. 238 ISBN, Цена 978-5-8114-0815-3: (в пер.) : 139 грн 77 к. ББК : 22.151 Предметные рубрики: Природничі науки Навчальні видання Математика Геометрія Збірник задач Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 518. - [x]Вид документа : Однотомное издание Шифр издания : Б 75534 22.1/Ф 53 Автор(ы) : Филиппов, Алексей Федорович Заглавие : Сборник задач по дифференциальным уравнениям . -изд. 2-е Выходные данные : Б.м., 2008 Колич.характеристики :237 с Примечания : На обл. подзаг.: Более 1400 задач с ответами ISBN, Цена 978-5-382-00455-6: 68 грн 89 к. ББК : 22.161.61 Предметные рубрики: Диференціальні рівняння Гомологічна алгебра Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 519. - [x]Вид документа : Однотомное издание Шифр издания : Б 74521 22.1/К 68 Автор(ы) : Королев, Виктор Юрьевич Заглавие : Теория вероятностей и математическая статистика : учебник для вузов Выходные данные : Б.м., 2008 Колич.характеристики :160 с Коллективы : Моск. гос. ун-т им. М. В. Ломоносова Примечания : Библиогр.: с. 160 ISBN, Цена 978-5-482-01946-7: (в пер.) : 44 грн 55 к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 520. - [x]Вид документа : Однотомное издание Шифр издания : Б 75460 22.1/Б 85 Автор(ы) : Босс В. Заглавие : Интуиция и математика . -Изд. 3-е, испр. и доп. Выходные данные : Б.м., 2008 Колич.характеристики :212 с.: ил. ISBN, Цена 978-5-382-00797-7: 67 грн. 38 к. ББК : 22.1 Предметные рубрики: Фізико-математичні науки Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 521. - [x]Вид документа : Однотомное издание Шифр издания : Б 75012. 22.11/К 26 Автор(ы) : Карпук, Андрей Андреевич Заглавие : Высшая математика для технических университетов : Интеграл. исчисление функций одной переменной: техн. ун-тов Выходные данные : Минск: Харвест, 2008 Колич.характеристики :300 с.: ил. Примечания : Библиогр.: с. 295 ISBN, Цена 978-985-16-3975-1: 98 грн 28 к. ББК : 22.11 Предметные рубрики:Математика Інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 522. - [x]Вид документа : Однотомное издание Шифр издания : Б 76322 22.18/С 25 Автор(ы) : Святецький, Микола Володимирович Заглавие : Математичне програмування : метод вказівки і контр.завдання Выходные данные : Б.м., 2008 Колич.характеристики :41, с.: іл. Примечания : Бібліогр.: с.42 Цена : 2 грн 50 к. ББК : 22.183.4р30 Предметные рубрики:математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 523. - [x]Вид документа : Однотомное издание Шифр издания : Б 75833 22/М 17 Автор(ы) : Максина, Елена Леонидовна, Березина, Наталья Алексеевна, Лапухина, Татьяна Юрьевна Заглавие : Справочник по техническим дисциплинам : высшая математика, физика, химия Выходные данные : Б.м., 2008 Колич.характеристики :381 с.: ил. Серия: Справочник ISBN, Цена 978-5-222-12410-9: 45 грн. 41 к. ББК : 22я2+24я2 Предметные рубрики: Фізико-математичні науки Фізика Хімія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 524. - [x]Вид документа : Однотомное издание Шифр издания : Б58428 22.17/К 75 Автор(ы) : Кочетков, Евгений Семенович, Смерчинская, Светлана Олеговна Заглавие : Теория вероятностей в задачах и упражнениях : учеб. пособие . -2-е изд. Выходные данные : Б.м., 2008 Колич.характеристики :479 с.: ил. Примечания : Библиогр.: с. 478-479 ISBN, Цена 978-5-91134-181-7: (в пер.) : 87 грн 44 к. ББК : 22.171я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Задачі Вправи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 525. - [x]Вид документа : Многотомное издание Шифр издания : Б 76296-4 22.1/О-26 Автор(ы) : Обуховська, Ганна Ізидорівна Заглавие : Вища математика: метод вказівки для самост. роботи студ./ М. В. Святецький. Ч. 4: Диференціальні рівняння, ряди Выходные данные : : ППІ НУК Б.м., 2008 Колич.характеристики :62 с.: іл. Примечания : Бібліогр .: с. 61 Цена : 4 грн. ББК : 22.11р30 Предметные рубрики:Математика Диференціальні рівняння Функціональні ряди Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 526. - [x]Вид документа : Однотомное издание Шифр издания : Б58668 22.1/П 14 Автор(ы) : Палий, Ирина Абрамовна Заглавие : Прикладная статистика : учеб. пособие для вузов Выходные данные : Б.м., 2008 Колич.характеристики :222, с.: ил. Примечания : Библиогр.: с. 222-223 ISBN, Цена 978-5-91131-563-4: (в пер.) : 51 грн 98 к. ББК : 22.172я73+60.601я73 Предметные рубрики: Природничі науки Навчальні видання Математика Прикладна статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 527. - [x]Вид документа : Шифр издания : Б 80201 22.1/А 46 Автор(ы) : Александрова, Надежда Вячеславовна Заглавие : История математических терминов, понятий, обозначений : слов.-справ. . -Изд. 3-е, испр. Выходные данные : Москва: ЛКИ, 2008 Колич.характеристики :246 с. Примечания : Библиогр. : с. 225-231 ISBN, Цена 978-5-382-00839-4: 116 грн. 67 к. ББК : 22.1я2 Предметные рубрики:Математика Математичні терміни-- історія Довідкові видання-- математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 528. - [x]Вид документа : Однотомное издание Шифр издания : Б 58234, 22.1/Ш 23 Автор(ы) : Шапкин, Александр Сергеевич Заглавие : Задачи по высшей математике, теории вероятностей, математической статистике, математическому программированию с решениями : учеб. пособие . -6-е изд. Выходные данные : Б.м., 2008 Колич.характеристики :432 с Примечания : Библиогр.: с. 428 ISBN, Цена 798-5-91131659-4: 93 грн 56 к. ББК : 22.11 Предметные рубрики: Природничі науки Навчальні видання Математика Вища математика Теорія ймовірностей Математична статистика Математичне програмування Задачі-- розв'язування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 529. - [x]Вид документа : Однотомное издание Шифр издания : Б 51646. 22.161/Б 50 Автор(ы) : Бермант, Анисим Фёдорович, Араманович, Исаак Генрихович Заглавие : Краткий курс математического анализа : учеб. пособие для вузов . -Изд. 14-е, стер. Выходные данные : Б.м., 2008 Колич.характеристики :736 с Примечания : Библиогр.: с. 736 ISBN, Цена 978-5-8114-0499-5: (в пер.) : 227 грн 70 к. ББК : 22.161я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичний аналіз Функції Інтеграли Диференційні числення Ряди Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 530. - [x]Вид документа : Однотомное издание Шифр издания : Б 75229 22.1/Ш 37 Автор(ы) : Шевчук, Валерий Петрович Заглавие : Расчет динамических погрешностей интеллектуальных измерительных систем Выходные данные : Б.м., 2008 Колич.характеристики :283 с.: ил Серия:Математика. Прикладная математика Примечания : Библиогр.: с. 279-283 ISBN, Цена 978-5-9221-0915-4 : (в пер.) : 91 грн 39 к. ББК : 22.18 Предметные рубрики: Природничі науки Математика Інтелектуальні виміріючі системи Розрахунки Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 531. - [x]Вид документа : Однотомное издание Шифр издания : В5574 22.1/Н 73 Автор(ы) : Новиков, Федор Александрович Заглавие : Дискретная математика для программистов : учеб. пособие для вузов . -3-е изд. Выходные данные : Б.м., 2008 Колич.характеристики :383 с.: ил. Серия: Учебник для вузов Примечания : Библиогр.: с.368-369 ISBN, Цена 978-5-91180-759-7: (в пер.) : 79 грн 47 к. ББК : 22.174я73 Предметные рубрики: Дискретна математика Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 532. - [x]Вид документа : Однотомное издание Шифр издания : Б 74588 22.1/А 92 Автор(ы) : Аттетков, Александр Владимирович, Зарубин, Владимир Степанович, Канатников, Анатолий Николаевич Заглавие : Введение в методы оптимизации : учеб. пособие Выходные данные : Б.м., 2008 Колич.характеристики :269 с.: ил. Примечания : Библиогр.: с. 260-265 ISBN, Цена 978-5-279-03251-8 : 118 грн. 94 к. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія оптимізації Содержание : Задачи оптимизации ; Методы одномерной минимизации ; Многомерная безусловная минимизация ; Аналитические методы нелинейного программирования ; Численные методы нелинейного программирования ; Методы линейного программирования Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 533. - [x]Вид документа : Однотомное издание Шифр издания : Б 75165 22.1/А 95 Автор(ы) : Ахтямов, Азамат Мухтарович Заглавие : Математика для социологов и экономистов : учеб. пособие для вузов . -Изд. 2-е, испр. и доп. Выходные данные : Б.м., 2008 Колич.характеристики :464 с Примечания : Библиогр.: с. 456-457 ISBN, Цена 978-5-9221-0919-2: (в пер.) : 135 грн 72к. ББК : 22.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Соціально - економічні спеціальності Содержание : Введение в анализ ; Дифференциальное исчисление ; Интегральное исчисление ; Функции многих переменных ; Дифференциальные и разностные уравнения Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 534. - [x]Вид документа : Однотомное издание Шифр издания : Б 72179 22.1/М 34 Автор(ы) : Виноградов, Юрий Николаевич, Гомола, Александр Иванович, Потапов, Виктор Петрович, Соколова, Елена Владимировна Заглавие : Математика и информатика : учеб. для сред. профобразования Выходные данные : Б.м., 2008 Колич.характеристики :272 с Серия: Среднее профессиональное образование.Экономика и управление Примечания : Библиогр.: с. 270 ISBN, Цена 978-5-7695-4092-9 : (в пер.) : 76 грн 99 к. ББК : 22.161я723+32.973.2я723 Предметные рубрики:Математика Інформатика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 535. - [x]Вид документа : Однотомное издание Шифр издания : Б 72150 22.1/М 23 Автор(ы) : Манин, Юрий Иванович Заглавие : Математика как метафора Выходные данные : Б.м., 2008 Колич.характеристики :400 с. Примечания : Библиогр. в конце частей ISBN, Цена 978-5-94057-287-9: 98 грн. 96 к. ББК : 22.1г Предметные рубрики:Математика Наука -- філософські аспекти Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 536. - [x]Вид документа : Однотомное издание Шифр издания : Б 95253. 22.17/М 42 Автор(ы) : Медведєв М. Г., Пащенко І. О. Заглавие : Теорія ймовірностей та математична статистика : підручник Выходные данные : Київ: Ліра-К, 2008 Колич.характеристики :535 с.: іл. Примечания : Бібліогр.: с. 529-535 ISBN, Цена 978-966-96938-3-9: (в опр.) : 168 грн. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Математична статистика Функції випадкових величин Экземпляры : всего : ЗРФ(2) Свободны: ЗРФ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 537. - [x]Вид документа : Однотомное издание Шифр издания : Б 96715 74.26/С 23 Заглавие : Сборник материалов математических олимпиад : 906 самых интересных задач и примеров с решениями : рек. учащимся шк., лицеев, гимназий... Выходные данные : Б.м., 2008 Колич.характеристики :335 с ISBN, Цена 978-966-338-099-5: (в пер.) : 18 грн. ББК : 74.262.21 Предметные рубрики:Математика Олімпіади Математичні задачі Экземпляры : всего : ЗРФ(2) Свободны: ЗРФ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 538. - [x]Вид документа : Однотомное издание Шифр издания : А 8880 22.1/М 86 Автор(ы) : Мочалов, Леонид Петрович Заглавие : Головоломки Выходные данные : Б.м., 2008 Колич.характеристики :140, с.: ил. Серия: Мир вокруг нас ISBN, Цена 978-5-275-01838-7: 470 грн. ББК : 22.1я9 Предметные рубрики:Математика Популярні видання Математичні ігри Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 539. - [x]Вид документа : Однотомное издание Шифр издания : А 5988. 22.1/В 92 Автор(ы) : Выгодский Марк Яковлевич Заглавие : Справочник по высшей математике Выходные данные : Б.м., 2008 Колич.характеристики :991 с ISBN, Цена 978-5-17-012238-7: 114 грн 75 к. УДК : 51(035) ББК : 22.11я2 Предметные рубрики:Математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 540. - [x]Вид документа : Однотомное издание Шифр издания : Б 72209 22.1/Б 42 Автор(ы) : Беклемишев, Дмитрий Владимирович Заглавие : Дополнительные главы линейной алгебры : учеб. пособие . -Изд. 2-е, перераб. и доп. Выходные данные : Москва; Краснодар: Лань, 2008 Колич.характеристики :490 с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 481-484 ISBN, Цена 978-58114-0811-5 (в пер.): 201 грн 91 к. ББК : 22.143я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійна алгебра Содержание : Линейные отображения ; Функции от матриц ; Введение в численные методы ; Псевдорешения и псевдообратные матрицы ; Системы линейных неравенств Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 541. - [x]Вид документа : Шифр издания : Б 77753 22.1/П 14 Автор(ы) : Палий, Ирина Абрамовна Заглавие : Дискретная математика. Курс лекций : учеб. пособие Выходные данные : Б.м., 2008 Колич.характеристики :350 с Серия: Техническое образование Примечания : Библиогр.: с. 344 ISBN, Цена 978-5-699-27101-6: (в пер.) : 116 грн 97 к. ББК : 22.174 я73 Предметные рубрики: Природничі науки Навчальні видання Математика Дискретна математика Інформатика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 542. - [x]Вид документа : Шифр издания : Б 77969 74.2/З-17 Автор(ы) : Зайцева, Светлана Анатольевна, Румянцева, Ирина Борисовна, Целищева, Ирина Ивановна Заглавие : Методика обучения математике в начальной школе Выходные данные : Б.м., 2008 Колич.характеристики :192 с.: ил Серия: Библиотека учителя начальной школы Примечания : Библиогр.: с.190-192 ISBN, Цена 978-5-691-01635-6: 58 грн 90 к. ББК : 74.262-46 Предметные рубрики:Математика Методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 543. - [x]Вид документа : Однотомное издание Шифр издания : Б 72146 22/К 71 Автор(ы) : Косарев, Евгений Леонидович Заглавие : Методы обработки экспериментальных данных : учеб. пособие для вузов Выходные данные : Б.м., 2008 Колич.характеристики :207 с Серия: Физтеховский учебник Примечания : Библиогр.: с. 202-205 ISBN, Цена 978-5-9221-0608-5: (в пер.) : 113 грн 09 к. ББК : 22в641я73 Предметные рубрики: Природничі науки Навчальні видання Математика Фізика Експериментальні показники Обробка Методи Комп'ютерні програми Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 544. - [x]Вид документа : Однотомное издание Шифр издания : Б 97488 83.3(0)/В 76 Автор(ы) : Ильяшенко, Ольга Николаевна Заглавие : 800. Золотая коллекция рефератов : рус. и зарубеж. лит., химия, физика, астрономия, всемир. история, география, худож. культура мира, биология, валеология, соврем. компьютер. технологии, маематика и математ. статистика Выходные данные : Донецк: БАО, 2008 Колич.характеристики :607 с.: ил Примечания : Соответствует новой шк. программе.- На пер. подзаг.: Глубина разработки тем. Чёткость структуры. Грамотность. Детал. анализ. Достоверность информации. ISBN, Цена 978-966-338-618-8: 17 грн. ББК : 83.3(0)я721 + 92 Предметные рубрики:Математика Світова література Хімія Фізика Астрономія Всесвітня історія Географія Біологія Валеологія Навчальні видання для середньої школи Экземпляры : всего : ЗРФ(2) Свободны: ЗРФ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 545. - [x]Вид документа : Однотомное издание Шифр издания : Б 70879 22.17/К 89 Автор(ы) : Кузнецов, Альберт Миколайович, Зароський Р. І., Нєдєлько Є. Ю. Заглавие : Конспект лекцій з математичної статистики : навч. посіб. Выходные данные : Б.м., 2008 Колич.характеристики :110 с Примечания : Бібліогр.: с. 109 Цена : 8 грн. ББК : 22.172я73 Предметные рубрики:Математика Статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 546. - [x]Вид документа : Шифр издания : Б 76323 22.1/Щ 33 Автор(ы) : Щеглов, Олександр Олександрович Заглавие : Методичні вказівки до виконання типових розрахунків з вищої математики для студентів економічних спеціальностей Выходные данные : Б.м., 2008 Колич.характеристики :23, с.: іл. Примечания : Бібліогр.: с.24 Цена : 1 грн ББК : 22.11р30 Предметные рубрики:Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 547. - [x]Вид документа : Однотомное издание Шифр издания : Б 73923 22.12/Н 13 Автор(ы) : Набебин, Алексей Александрович, Кораблин, Юрий Прокофьевич Заглавие : Математическая логика и теория алгоритмов : учеб. пособие для обучающихся по направлению " Информатика и вичислит. техника" спец. " Программное обеспечение вычеслит. техники автоматезир. систем", а также спец. "Информ. системы и технологии" Выходные данные : Б.м., 2008 Колич.характеристики :343 с.: ил. Примечания : Библиогр.: с. 334 ISBN, Цена 978-5-91522-001-9: 164 грн 3 к. ББК : 22.12я73 Предметные рубрики:Математика Логіка алгоритмів Математична логіка Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 548. - [x]Вид документа : Однотомное издание Шифр издания : Б 73961 22.18/Н 73 Автор(ы) : Новиков, Дмитрий Александрович Заглавие : Математические модели формирования и функционирования команд : монография Выходные данные : Б.м., 2008 Колич.характеристики :186 с Коллективы : РАН ; Ин-т проблем упр. им. В. А. Трапезникова Примечания : Библиогр.: с. 176-186 ISBN, Цена 9875-94052-146-0: 32 грн 18 к. ББК : 22.181я9 Предметные рубрики:Математика Математичне моделювання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 549. - [x]Вид документа : Однотомное издание Шифр издания : Б 72622 22.1/Т 76 Автор(ы) : Трохимчук, Ростислав Миколайович Заглавие : Збірник задач і вправ з математичної логіки : навч. посіб. Выходные данные : Б.м., 2008 Колич.характеристики :114 с. Коллективы : Міжрегіон. акад. упр. персоналом Примечания : Бібліогр. : с. 113 ISBN, Цена 978-966-608-854-6: 17 грн. ББК : 22.122я73 Предметные рубрики: Математична логіка Логіка Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 550. - [x]Вид документа : Однотомное издание Шифр издания : Б 72865 22.1/М 15 Автор(ы) : Макаренко, Віта Олександрівна Заглавие : Вища математика для економістів : навч. посіб. Выходные данные : Київ: Знання, 2008 Колич.характеристики :517 с.: іл. Примечания : Бібліогр.: с. 515-517 ISBN, Цена 978-966-346-347-6: 95 грн 40 к. ББК : 22.11я73 Предметные рубрики: Вища математика Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 551. - [x]Вид документа : Однотомное издание Шифр издания : Б 27441 22.161/Е 14 Автор(ы) : Евграфов, Марат Андреевич Заглавие : Аналитические функции : учеб. пособие . -4-е изд., стер. Выходные данные : Б.м., 2008 Колич.характеристики :447 с Серия: Классическая учебная литература по математике Примечания : Библиогр.: с. 441-442 ISBN, Цена 978-5-8114-0809-2: (в пер.) : 170 грн 78 к. ББК : 22.161.55я73 Предметные рубрики: Аналітичні функції Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 552. - [x]Вид документа : Шифр издания : Б 81344 65.05/Д 79 Автор(ы) : Дубовой, Володимир Михайлович , Ковалюк, Олег Олександрович Заглавие : Моделі прийняття рішень в управлінні розподіленими динамічними системами : монографія Выходные данные : Б.м., 2008 Колич.характеристики :184 с.: іл. Коллективы : Вінниц. нац. техн. ун-т Примечания : Бібліогр.: с. 177-184 ISBN, Цена 978-966-641-251-8: 17 грн ББК : 65.050.030.4я9+22.183.1я9 Предметные рубрики: прийняття рішень -теорія математика кібернетика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 553. - [x]Вид документа : Однотомное издание Шифр издания : Б 72897 22.1/Б 20 Автор(ы) : Балдин, Константин Васильевич, Башлыков, Виктор Николаевич , Рукосуев, Андрей Вадимович Заглавие : Теория вероятностей и математическая статистика : учебник Выходные данные : Б.м., 2008 Колич.характеристики :472 с Примечания : Библиогр.: с. 433-434 ISBN, Цена 978-5-91131-633-4: ( в пер.) : 103 грн 95к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Ймовірностей теорія Випадкові величини Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 554. - [x]Вид документа : Однотомное издание Шифр издания : А 6387. 22.1/Ц 59 Автор(ы) : Цикунов А. Е. Заглавие : Сборник формул по математике . -Изд. 3-е Выходные данные : Б.м., 2008 Колич.характеристики :160 с.: ил. Серия: Карманный справочник ISBN, Цена 978-5-88782-281-5: 14 грн 44к. ББК : 22.1я2 Предметные рубрики:Математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 555. - [x]Вид документа : Однотомное издание Шифр издания : Б 71949 22.1/Ф 91 Автор(ы) : Фролов, Сергей Аркадьевич Заглавие : Начертательная геометрия : сб. задач: учеб. пособие для вузов . -3-е изд., испр. Выходные данные : Б.м., 2008 Колич.характеристики :170, с.: ил. Серия: Высшее образование ISBN, Цена 978-5-16-003273-3 : (в пер.) : 55 грн 84 к. ББК : 22.151.34я73 Предметные рубрики: Нарисна геометрія Математика Геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 556. - [x]Вид документа : Однотомное издание Шифр издания : В 7906 22.1/В 88 Автор(ы) : Вуколов, Эдуард Александрович Заглавие : Основы статистического анализа. : Практикум по статистическим методам и исследованию операций с использованием пакетов STATISTICA и EXCEL : учеб. пособие для вузов . -2-е изд., испр. и доп. Выходные данные : Б.м., 2008 Колич.характеристики :463 с Серия: Высшее образование Примечания : Библиогр.: с. 455-456 ISBN, Цена 978-5-91134-231-9: (в пер.) : 121 грн 45 к. ББК : 22.17я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ймовірностей Статистичні методи Статистичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 557. - [x]Вид документа : Однотомное издание Шифр издания : Б 76137 22.1/С 77 Автор(ы) : Старков, Сергей Николаевич Заглавие : Справочник по математическим формулам и графикам функций для студентов Выходные данные : Б.м., 2008 Колич.характеристики :234 с.: ил. Серия: Учебное пособие Примечания : Библиогр.: с. 229 - 230 ISBN, Цена 978-5-91180-830-3: 47 грн 70 к. ББК : 22.1я2 Предметные рубрики: Природничі науки Довідкові видання Математика Математичні формули Графіки функцій Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 558. - [x]Вид документа : Однотомное издание Шифр издания : Б 102822 514/П 27 Автор(ы) : Перельман, Яків Ісидорович Заглавие : Захоплююча геометрія Выходные данные : Тернопіль: Навч. кн. - Богдан, 2008 Колич.характеристики :286 с.: іл Серия: Класики популяризації науки . Країна Перельманія ISBN, Цена 978-966-408-365-9: 49 грн 81 к. УДК : 514(089.3) Предметные рубрики:Математика Геометричні форми Популярні видання Аннотация: «Захоплююча геометрія» Я.І. Перельмана — одна з найзахопливіших книжок про геометрію. У ній ця наука, за словами самого автора, «відривається від класної дошки», з якою тісно пов’язувалася у свідомості пересічного читача, й виводиться у вільний простір — у ліс, поле, до річки, на дорогу, а згодом — у далекі моря та захмарні глибини простору. Часто ці розповіді ілюструються уривками з творів класиків літератури та висловами великих учених. Прочитавши книгу, читач знатиме незмірно більше про застосування геометрії, ніж про це можна довідатися з десятків «шкільних» підручників та посібників. Книгу перекладено багатьма мовами світу. Українською виходить уперше. Видання зацікавить не тільки тих, хто вже любить і займається математикою, а й усіх, хто хотів би мати про неї правдиве уявлення, зокрема, знати про її справжнє місце у нашому житті. Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 559. - [x]Вид документа : Однотомное издание Шифр издания : Б 74291 22.16/Ш 18 Автор(ы) : Шалдырван, Валерий Анатольевич, Медведев, Кирилл Валериевич Заглавие : Дифференциальные уравнения. : учеб. пособие Выходные данные : Б.м., 2008 Колич.характеристики :356 с.: ил. Примечания : Библиогр.: с. 352-353 ISBN, Цена 978-5-9502-0317-6: 175 грн 23 к. ББК : 22.161.6я73 Предметные рубрики: Природничі науки Навчальні видання Математика Диференціальні рівнення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 560. - [x]Вид документа : Шифр издания : Б63668 22.1/Д 22 Автор(ы) : Двайт, Герберт Бристол Заглавие : Таблицы интегралов и другие математические формулы . -Изд.10-е,стер. Выходные данные : Б.м., 2009 Колич.характеристики :228 с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 227-228 ISBN, Цена 978-5-8114-0642-5: (в пер.) : 183 грн 33 к. ББК : 22.194.5я73 Предметные рубрики: Інтеграли Математика Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 561. - [x]Вид документа : Шифр издания : Б58162 74.26/Ф 88 Автор(ы) : Фридман, Лев Моисеевич Заглавие : Теоретические основы методики обучения математике : учеб.-метод.пособие . -Изд.3-е Выходные данные : Б.м., 2009 Колич.характеристики :244 с.: ил. Серия: Психология, педагогика, технология обучения Примечания : Библиогр.: с. 239-244 ISBN, Цена 978-5-397-00141-0: 130 грн 93 к. ББК : 74.262я7 Предметные рубрики:математика методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 562. - [x]Вид документа : Шифр издания : В 9053 22.17/Л 14 Автор(ы) : Лагутин, Михаил Борисович Заглавие : Наглядная математическая статистика : учеб. пособие . -2-е изд., испр. Выходные данные : Б.м., 2009 Колич.характеристики :466 с.: ил Примечания : Библиогр.: с. 456-459 ISBN, Цена 978-5-94774-996-0: (в пер.) : 141 грн 66 к. ББК : 22.172я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Приклади Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 563. - [x]Вид документа : Шифр издания : В 9072 22.1/П 23 Автор(ы) : Пегат, Анджей Заглавие : Нечёткое моделирование и управление Выходные данные : Б.м., 2009 Колич.характеристики :798 с.: ил. Серия: Адаптивные и интеллектуальные системы Примечания : Библиогр.: с. 767-785 ISBN, Цена 978-5-94774-353-1: (в пер.) : 318 грн 73 к. ББК : 22.181 Предметные рубрики: Природничі науки Математика Математичні методи Управління системами Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 564. - [x]Вид документа : Шифр издания : Б 80230 22.1/Ж 86 Автор(ы) : Жуков, Александр Владимирович Заглавие : Вездесущее число "Пи" . -Изд. 3-е Выходные данные : Б.м., 2009 Колич.характеристики :214 с.: ил Примечания : Библиогр.: с. 209-214 ISBN, Цена 978-5-397-00521-0: 116 грн 67 к. ББК : 22.131 Предметные рубрики: Природничі науки Математика Визначники (мат.) Число "Пи" Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 565. - [x]Вид документа : Шифр издания : Б 66393 22.1/Л 53 Автор(ы) : Лефевр, Владимир Александрович, Смолян, Георгий Львович Заглавие : Алгебра конфликта . -Изд. 3-е Выходные данные : Б.м., 2009 Колич.характеристики :63 с Примечания : Библиогр.: с. 62 ISBN, Цена 978-5-397-00547-0: 81 грн 41 к. ББК : 22.183.2 Предметные рубрики: Природничі науки Математика Алгебра Конфліктні ситуації Рефлексивні ігри Підходи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 566. - [x]Вид документа : Шифр издания : Б 80377 22.17/М 48 Автор(ы) : Мельников, Олег Исидорович Заглавие : Теория графов в занимательных задачах : учеб.-метод. пособие . -Изд. 3-е, испр. и доп. Выходные данные : Б.м., 2009 Колич.характеристики :231 с Примечания : Библиогр.: с. 226 ISBN, Цена 978-5-397-00033-8: 140 грн 10 к. ББК : 22.174.2я722 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія графов Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 567. - [x]Вид документа : Шифр издания : Б 80319 22.1/Ш 57 Автор(ы) : Шикин, Евгений Викторович Заглавие : Гуманитариям о математике : Математика. Пути знакомства. Основные понятия. Методы. Модели : учебник . -Изд 3-е. Выходные данные : Б.м., 2009 Колич.характеристики :267 с ISBN, Цена 978-5-397-00450-3: 140 грн 10 к. ББК : 22.1я73 Предметные рубрики: Природничі науки Математика Математичні моделі Математичні методи-- застосування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 568. - [x]Вид документа : Шифр издания : Б 80313 22.1/Ш 64 Автор(ы) : Ширяев, Владимир Иванович Заглавие : Финансовая математика: потоки платежей, производные финансовые инструменты : учеб. пособие для вузов . -изд. 2-е, испр. и доп. Выходные данные : Москва: ЛИБРОКОМ, 2009 Колич.характеристики :231 с.: ил Примечания : Библиогр. в конце глав ISBN, Цена 978-5-397-00357-5: 140 грн 10 к. ББК : 22.1я73+65в641я73 + 65в641я73 Предметные рубрики: Фінансова математика Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 569. - [x]Вид документа : Шифр издания : Б 78586 22.18/С 87 Автор(ы) : Струченков, Валерий Иванович Заглавие : Методы оптимизации в прикладных задачах Выходные данные : Б.м., 2009 Колич.характеристики :319 с.: ил. Серия: Библиотека профессионала Примечания : Библиогр.: с. 309-310 ISBN, Цена 978-5-91359-061-9: 155 грн 85 к. ББК : 22.183.4 Предметные рубрики: Природничі науки Математика Математичне програмування Оптимізація-- математична -- методи Задачі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 570. - [x]Вид документа : Шифр издания : Б 55475 22.1/Л 64 Автор(ы) : Литвин, Ірина Іванівна, Конончук, Оксана Миколаївна, Желізняк, Галина Олексиївна Заглавие : Вища математика : навч. посіб. для студ. вищ. навч. закл. Выходные данные : Київ: Центр учбової л-ри, 2009 Колич.характеристики :367 с Примечания : Бібліогр.: с. 367 ISBN, Цена 978-966-364-854-5: (в опр.) : 75 грн ББК : 22.11я73 Предметные рубрики:Математика Вища математика Система числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 571. - [x]Вид документа : Шифр издания : Б 54621 22.172/М 28 Автор(ы) : Мармоза, Анатолій Тимофійович Заглавие : Практикум з математичної статистики : навч. посіб. Выходные данные : Б.м., 2009 Колич.характеристики :257 с. Примечания : Бібліогр. : с. 253 ISBN, Цена 966-7982-30-0: 58 грн. 50 к. ББК : 22.172я73 Предметные рубрики:Математика Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 572. - [x]Вид документа : Шифр издания : Б 78584 22.18/П 23 Автор(ы) : Певзнер, Леонид Давидович, Чураков, Евгений Павлович Заглавие : Математические основы теории систем : учеб. пособие для вузов Выходные данные : Б.м., 2009 Колич.характеристики :503 с.: ил. Серия: Для высших учебных заведений Примечания : Библиогр. в конце глав ISBN, Цена 978-5-06-004860-5: (в пер.) : 388 грн 36 к. ББК : 22.181я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія систем Математичні основи Математичні моделі Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 573. - [x]Вид документа : Шифр издания : Б 78962 22.1/Е 30 Автор(ы) : Егоров, Иван Петрович Заглавие : Геометрия : учеб. пособие . -2-е изд. Выходные данные : Москва: ЛИБРОКОМ, 2009 Колич.характеристики :256 с.: ил Примечания : Библиогр.: с. 254 ISBN, Цена 978-5-397-00278-3: 152 грн 81к. ББК : 22.151я73 Предметные рубрики: Геометрія Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 574. - [x]Вид документа : Шифр издания : Б 79250 22.1/Ч 75 Автор(ы) : Чорней, Руслан Костянтинович Заглавие : Практикум по теории вероятностей и математической статистике : учеб. пособие Выходные данные : Б.м., 2009 Колич.характеристики :331 с Коллективы : МАУП Серия: МАУП Примечания : Библиогр.: с. 326-327 ISBN, Цена 978-966-608-929-1: 85 грн ББК : 22.17 я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 575. - [x]Вид документа : Шифр издания : В 6910. 22.1/О-62 Автор(ы) : Оппенгейм, А., Шафер Р. Заглавие : Цифровая обработка сигналов . -2-е изд., испр. Выходные данные : Б.м., 2009 Колич.характеристики :855 с Примечания : Библиогр.: с. 843-852 ISBN, Цена 978-5-94836-202-1: (в пер.) : 313 грн 84 к. ББК : 22.161.2+32.811.3 + 32.811.3 Предметные рубрики: Природничі науки Математика Комп'юерна математика Цифрова обробка сигналів Дискретні сигнали Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 576. - [x]Вид документа : Шифр издания : Б 4614 22.1/С 74 Заглавие : Справочник по математике для экономистов : учеб. пособие . -3-е изд., перераб. и доп. Выходные данные : Б.м., 2009 Колич.характеристики :456 с.: ил Серия: Высшее образование ISBN, Цена 978-5-16-003542-0: (в пер.) 148 грн 10 к. ББК : 22.1я2 Предметные рубрики: Природничі науки Довідкові видання Математика Економічні задачі Математичне програмування Статистичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 577. - [x]Вид документа : Шифр издания : В 8946 22.1/Н 64 Автор(ы) : Нікольський, Юрій Володимирович, Пасічник, Володимир Володимирович, Щербина, Юрій Миколайович Заглавие : Дискретна математика : підручник Выходные данные : Б.м., 2009 Колич.характеристики :431 с.: іл Серия: Комп"ютинг Примечания : Бібліогр.: с.430-431 ISBN, Цена 978-966-2025-76-7: (в опр.) : 120 грн. ББК : 22.174я73 Предметные рубрики: Комбінаторика Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 578. - [x]Вид документа : Шифр издания : Б 82921 22.1/В 46 Автор(ы) : Вильямс, Джон Д. Заглавие : Совершенный стратег, или Букварь по теории стратегических игр . -Изд. 2-е Выходные данные : Б.м., 2009 Колич.характеристики :269 с.: ил. Серия: Науку-Всем! Шедевры научно-популярной литературы. Математика Примечания : Парал. назв. : англ. - Библиогр. : с. 265 ISBN, Цена 978-5-397-00529-6: 131 грн 95 к. ББК : 22.183.2 Предметные рубрики: Ігри Математика Стратегії Математичні ігри Теорія ігор Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 579. - [x]Вид документа : Многотомное издание Шифр издания : Б 83959-2 22.1/Д 95 Автор(ы) : Дюженкова Любовь Ивановна Заглавие : Практикум по высшей математике: учеб. пособие : в 2 ч./ Ольга Юрьевна Дюженкова, Геннадий Александрович Михалин. - (Математика). Ч. 2. Выходные данные : Москва: Бином. Лаборатория знаний, 2009 Колич.характеристики :468 с Примечания : Библиогр.: с. 462 ISBN, Цена 978-5-94774-999-1: (в пер.): 140 грн 27 к. ББК : 22.11я73 Предметные рубрики:Математика Інтегральні системи Диференціальні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 580. - [x]Вид документа : Шифр издания : Б 82127 22.1/Ш 66 Автор(ы) : Шкільняк, Степан Степанович Заглавие : Математична логіка. Основи теорії алгоритмів : навч.посіб. Выходные данные : Б.м., 2009 Колич.характеристики :279 с Примечания : Бібліогр.: с. 268-269 в кінці ч. ISBN, Цена 978-966-608-979-6: (в опр.): 82 грн 80 к. УДК : 510.5./.6(075.8) ББК : 22.12я73 Предметные рубрики: логіка математика алгоритмічна теорія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 581. - [x]Вид документа : Шифр издания : Б 82391 22.1/К 19 Автор(ы) : Канель-Белов, Алексей Яковлевич, Ковальджи, Александр Кириллович Заглавие : Как решают нестандартные задачи . -5-е изд., испр. Выходные данные : Б.м., 2009 Колич.характеристики :94 с.: ил Серия: Школьные математические кружки Примечания : Библиогр.: с. 90-94 ISBN, Цена 978-5-94057-471-2: 37 грн 18 к. ББК : 22.1я721 Предметные рубрики: Природничі науки Навчальні видання Математика Олімпіади Задачі-- розв'язування Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 582. - [x]Вид документа : Многотомное издание Шифр издания : Б 80437-1 22.1/И 60 Заглавие : Индивидуальные задания по высшей математике: учеб. пособие : в 4 ч. Ч. 1: Линейная и векторная алгебра. Аналитическая геометрия. Дифференциальное исчисление функций одной переменной . -5-е изд. Выходные данные : Минск: Вышейш. шк., 2009 Колич.характеристики :303, с.: ил. Примечания : Библиогр.: с. 302 ISBN, Цена 978-985-06-1764-4 ( в пер.): 94 грн 83 к. ББК : 22.11я73 Предметные рубрики: Природничі науки Навчальні видання Математика Лінійна алгебра Аналітична геометрія Диференційні числення Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 583. - [x]Вид документа : Однотомное издание Шифр издания : Б 92239 22.1/М 19 Автор(ы) : Малинина Т.Б. Заглавие : Высшая математика для социологов : курс лекций Выходные данные : Б.м., 2009 Колич.характеристики :93 с. ISBN, Цена 978-5-288-21891-0: 119 грн. ББК : 22.11я73 Предметные рубрики: Вища математика Математика Лінійна алгебра Лінійні рівняння Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 584. - [x]Вид документа : Однотомное издание Шифр издания : 030/У-59 Автор(ы) : Иваница С. В., Завязкин О. В., Заведея Т. Л., Меженко Ю. С. Заглавие : Универсальный энциклопедический справочник в таблицах Выходные данные : Донецк: БАО, 2009 Колич.характеристики :1022 с ISBN, Цена 978-966-338-812-0: (в пер.) : 55 грн. УДК : 030(083.4) ББК : 92я721 Предметные рубрики: Природничі науки Інформатика Техніка Всесвітня історія Хімія Біологія Математика Географія Довідкові видання Содержание : Алгебра ; Геометрия ; Физика ; Химия ; Биология ; География ; Всемирная история ; История культуры ; Наука и техника ; Информатика ; Обработка и передача информации Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 585. - [x]Вид документа : Многотомное издание Шифр издания : Б 85417-1 22.1/Х-76 Автор(ы) : Хоменко, Євген Васильович Заглавие : Природа й математика/ Є. В. Хоменко. Ч.1: Математичне дерево Выходные данные : : "Отаман" Б.м., 2009 - Колич.характеристики :77 с.: іл. Примечания : Бібліогр. : с. 72 Цена : 22 грн. ББК : 22.1 Предметные рубрики:Математика Система числення Інтеграли Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 586. - [x]Вид документа : Однотомное издание Шифр издания : Б 75012. 22.1/К 26 Автор(ы) : Карпук, Андрей Андреевич Заглавие : Высшая математика для технических университетов : интеграл. исчисление функций многих переменных: для техн. ун-тов Выходные данные : Минск: Харвест, 2009 Колич.характеристики :267 с.: ил Примечания : Библиогр.: с. 264 ISBN, Цена 978-985-16-5759-5: 97 грн 72 к. ББК : 22.11я73 Предметные рубрики:Математика Інтеграли Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 587. - [x]Вид документа : Шифр издания : Б 81015 22.161/С 19 Автор(ы) : Саприкіна, Людмила Тихонівна, Майборода, Олександр Валерійович, Кондратенко, Євген Олександрович Заглавие : Вступ до математичного аналізу : метод. вказівки Выходные данные : Б.м., 2009 Колич.характеристики :77, с Коллективы : НУК ім. адм. Макарова, Каф. вищ. математики Примечания : Бібліогр.: с. 78 Цена : 9 грн. ББК : 22.161р30 Предметные рубрики: Природничі науки Методичні вказівки Математика Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 588. - [x]Вид документа : Многотомное издание Шифр издания : Б 80437-3 22.1/И 60 Заглавие : Индивидуальные задания по высшей математике: учеб. пособие : в 4 ч. Ч.3: Ряды. Кратные и криволинейные интегралы. Элементы теории поля . -5-е изд.,испр. Выходные данные : Минск: Вышейш.шк., 2009 Колич.характеристики :367 с Примечания : Библиогр.: с. 366 ISBN, Цена 978-985-06-1677-7: (в пер.): 139 грн 84 к. ББК : 22.11я73 Предметные рубрики: Природничі науки Навчальні видання Вища математика Математика Ряди (мат) Інтеграли Теорія поля Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 589. - [x]Вид документа : Многотомное издание Шифр издания : Б 80437-2 22.1/И 60 Заглавие : Индивидуальные задания по высшей математике: учеб. пособие : в 4 ч. Ч. 2: Комплексные числа. Неопределенные и определенные интегралы. Функции нескольких переменных. Обыкновенные дифференциальные уравнения . -4-е изд., испр. Выходные данные : Минск: Вышейш.шк., 2009 Колич.характеристики :395 с Примечания : Библиогр.: с. 394 Цена : 167.07 грн ББК : 22.11я73 Предметные рубрики: Природничі науки Навчальні видання Математика Комплексні числа Інтеграли Функції Диференційні рівняння Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 590. - [x]Вид документа : Шифр издания : Б 81036 22.1/К 56 Автор(ы) : Коваленко, Игорь Иванович, Фарионова, Татьяна Анатольевна, Приходько, Сергей Борисович Заглавие : Методы принятия решений : учеб. пособие Выходные данные : Б.м., 2009 Колич.характеристики :178 с Примечания : Библиогр.: с. 176 ISBN, Цена 978-966-321-118-3: 25 грн ББК : 22.183.1я73 Предметные рубрики: Прийняття рішень Математика Навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 591. - [x]Вид документа : Однотомное издание Шифр издания : Б 81175 22.16/Ц 29 Автор(ы) : Цеберна, Наталя Олександрівна, Юрченко, Тамара Акимівна Заглавие : Невизначений та визначений інтеграл : навч. посіб. Выходные данные : Б.м., 2009 Колич.характеристики :124 с.: іл Примечания : Бібліогр. : с. 123 Цена : 8 грн. ББК : 22.161.12я73 Предметные рубрики:Математика Інтеграли Математичний аналіз Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 592. - [x]Вид документа : Шифр издания : КД-2819 22.3/К 30 Автор(ы) : Кафтанова , Юлия Викторовна Заглавие : Специальные функции математической физики: в 3 ч./ Юлия Викторовна Кафтанова . - Прил. к одноим. кн. Ч. 3: Моделирование аномальных и экстраординарных природных и техногенных процессов Выходные данные : , 2009 Колич.характеристики :1 электрон. опт. диск (CD- ROM) Цена : 20 грн ББК : 22.311 Предметные рубрики: Фізика Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 593. - [x]Вид документа : Шифр издания : В 9358 22.1/М 59 Автор(ы) : Микони, Станислав Витальевич Заглавие : Многокритериальный выбор на конечном множестве альтернатив : учеб. пособие Выходные данные : Санкт-Петербург: Лань , 2009 Колич.характеристики :265 с Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 258-262 ISBN, Цена 978-5-8114-0984-6 : (в пер.) : 346 грн 98 к. ББК : 22.183.1я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія прийняття рішень Альтернативи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 594. - [x]Вид документа : Шифр издания : Б 81228 22.1кр/К 89 Автор(ы) : Кузнецов, Альберт Николаевич, Чорный, Александр Леонидович Заглавие : Практикум по неопределённым интегралам : учеб. пособие Выходные данные : Б.м., 2009 Колич.характеристики :227 с Примечания : Библиогр.: с. 226 Цена : 30 грн. ББК : 22.161.12я73кр Предметные рубрики: Невизначені інтеграли Математика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 595. - [x]Вид документа : Многотомное издание Шифр издания : Б 81309-1 22.3/К 30 Автор(ы) : Кафтанова , Юлия Викторовна Заглавие : Специальные функции математической физики: в 3 ч./ Юлия Викторовна Кафтанова . Ч. 1: Функции Бесселя и цилиндрические функции в элементарном изложении с программами вычислений Выходные данные : Х.: Новое слово, 2009 Колич.характеристики :178, +1 электрон. опт. диск (CD № 2818) ISBN, Цена 978-966-2046-62-5: 32 грн УДК : 531.01:51-7 ББК : 22.311 Предметные рубрики: математична фізика математика фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 596. - [x]Вид документа : Однотомное издание Шифр издания : В 9945 22.1/Б 72 Автор(ы) : Бобик, Омелян Іванович Заглавие : Рівняння математичної фізики : (практикум) : навч.посіб. Выходные данные : Б.м., 2010 Колич.характеристики :252 с Серия: КОМП'ЮТИНГ Примечания : Бібліогр.: с. 252 ISBN, Цена 978-966-418-122-5: 84 грн. УДК : 517.958(076) ББК : 22.161.68я73 Предметные рубрики:математика фізика Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 597. - [x]Вид документа : Однотомное издание Шифр издания : Б 86532 22.16/Б 93 Автор(ы) : Буфеев С. В. Заглавие : Функции и графики : учеб. пособие Выходные данные : Б.м., 2010 Колич.характеристики :110 с.: ил Коллективы : МГТУ им. Н. Э. Баумана Примечания : Библиогр.: с. 100-101 ISBN, Цена 978-5-7038-3411-4: 60 грн. ББК : 22.161.5я722 Предметные рубрики: Природничі науки Математика Математичні перетворення Функції-- графіки Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 598. - [x]Вид документа : Однотомное издание Шифр издания : В 9840 22.17/З-35 Автор(ы) : Зароський, Ромуальд Іванович Заглавие : Основи дискретної математики : підруч. для вищ. навч. закл. Выходные данные : Б.м., 2010 Колич.характеристики :310 с.: іл Примечания : Бібліогр.: с.306-307 ISBN, Цена 978-966-321-150-3: (В пер.) : 60 грн. УДК : 510.3/.6:519.1](075.8) ББК : 22.174я73 Предметные рубрики:математика комбінаторика навчальний посібник для вищої школи Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 599. - [x]Вид документа : Многотомное издание Шифр издания : Б 85479-1 22.1/Х-58 Автор(ы) : Хлопенко, Наталія Юріївна Заглавие : Дійсні числа і вирази: навч.посіб./ Наталія Юріївна Хлопенко. Ч. 1 Выходные данные : Миколаїв: Вид-во НУК, 2010 Колич.характеристики :47 с Цена : 20.00 грн. ББК : 22.161.0я73 Предметные рубрики: математичний аналіз математика варіаційне числення Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 600. - [x]Вид документа : Однотомное издание Шифр издания : В 10005 28/Б 87 Автор(ы) : Братусь, Александр Сергеевич, Новожилов А.С., Платонов А.П. Заглавие : Динамические системы и модели биологии Выходные данные : Б.м., 2010 Примечания : Библиогр.: с.390-393 ISBN, Цена 978-5-9221-1192-8: (в пер.) : 469 грн. ББК : 28в631.0 Предметные рубрики: біологія природничі науки математика моделювання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 601. - [x]Вид документа : Однотомное издание Шифр издания : В 10087 22.1/Б 95 Автор(ы) : Быльцов, Сергей Заглавие : Логические головоломки и задачи : занимат. математика для всей семьи Выходные данные : Б.м., 2010 Колич.характеристики :156, с.: ил. ISBN, Цена 978-5-49807-785-7 (в пер.): 97 грн. УДК : 51-8:793.7 ББК : 22.1я9 Предметные рубрики:математика логічні задачі популярні видання Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 602. - [x]Вид документа : Однотомное издание Шифр издания : Б 87931 74.26/С 67 Автор(ы) : Соуза, Дэвид Заглавие : Как мозг осваивает математику. Пратические советы учителю Выходные данные : Б.м., 2010 Колич.характеристики :238 с Серия: Школа завтра Примечания : Парал. тит. л.: англ. ISBN, Цена 978-5-91678-035-2: (в пер.) : 143 грн. ББК : 74.262 Предметные рубрики:Математика Методи навчання Экземпляры :ЗФ(1) Свободны: ЗФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 603. - [x]Вид документа : Однотомное издание Шифр издания : Б 86810 22.17/Ж 94 Автор(ы) : Жучок, Юрій Володимирович Заглавие : Дискретна математика : навч. посіб. Выходные данные : Б.м., 2010 Колич.характеристики :219 с Коллективы : Держ.закл."Луган.нац.ун-т імені Тараса Шевченка" Примечания : Бібліогр.: с. 209-210 ISBN, Цена 978-966-617-186-6: (в опр.): 127 грн. УДК : 510.21:512.5:519.156](075.8) ББК : 22.174я73 Предметные рубрики:Математика Комбінаторика Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 604. - [x]Вид документа : Однотомное издание Шифр издания : Б 87203 22.1/С 79 Автор(ы) : Стеклов, Владимир Андреевич Заглавие : Математика и ее значение для человечества . -Изд. 2-е Выходные данные : Б.м., 2010 Колич.характеристики :136, с Серия: Физико-математическое наследие. Математика (философия математики) ISBN, Цена 978-5-397-01400-7: 85 грн. ББК : 22.1в Предметные рубрики: Природничі науки Математика Розвиток Історія Філософський аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 605. - [x]Вид документа : Шифр издания : Б 58439. 22.1/Т 33 Заглавие : Терия вероятностей и математическая статистика : учеб. пособие Выходные данные : Б.м., 2010 Колич.характеристики :279 с. Серия: Высшее образование Примечания : Библиогр. : с. 276-279 ISBN, Цена 978-5-16-001561-3: (в пер.) : 92 грн 07 к. ББК : 22.17я73 Предметные рубрики:Математика Теорія ймовірностей Математична статистика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 606. - [x]Вид документа : Шифр издания : Б 82069 22.1/М 34 Заглавие : Математика ХХ века. Взгляд из Петербурга Выходные данные : Б.м., 2010 Колич.характеристики :184 с.: ил Примечания : Библиогр. в конце ст. ISBN, Цена 978-5-94057-586-3: 111 грн 15 к. ББК : 22.1г Предметные рубрики: Природничі науки Математика Досягнення XX століття Історичний аспект Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 607. - [x]Вид документа : Шифр издания : Б 79814 22.17/О-66 Автор(ы) : Орлов, Александр Иванович Заглавие : Вероятность и прикладная статистика : основные факты : справочник Выходные данные : Москва: КНОРУС, 2010 Колич.характеристики :190 с Примечания : Библиогр.: с. 168-190 ISBN, Цена 978-5-406-00173-8: 75 грн 12 к. ББК : 22.172я2 Предметные рубрики: Природничі науки Довідкові видання Математика Статистика Статистичні методи Теорія ймовірностей Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 608. - [x]Вид документа : Шифр издания : Б 80346 22.18/Х-47 Автор(ы) : Хинчин, Александр Яковлевич Заглавие : Работы по математической теории массового обслуживания . -Изд. 4-е Выходные данные : Б.м., 2010 Колич.характеристики :235 с. Серия: Физико-математическое наследие. Математика (теория вероятностей) Примечания : Библиогр. : с. 234-235 ISBN, Цена 978-5-397-01037-5: 130 грн 92 к. ББК : 22.183.5 Предметные рубрики:Математика Теорія масового обслуговування Математична кібернетика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 609. - [x]Вид документа : Шифр издания : Б 82858 22.1/С 12 Автор(ы) : Саати, Томас Л. Заглавие : Элементы теории массового обслуживания и ее приложения . -3-е изд. Выходные данные : Москва: Либроком, 2010 Колич.характеристики :520 с. Примечания : Библиогр. : с. 489- 509 ISBN, Цена 978-5-397-01263-6: 240 грн. 49 к. ББК : 22.183.5 Предметные рубрики: Масове обслуговування Математична кібернетика Математика Математичні методи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 610. - [x]Вид документа : Шифр издания : В 9642 22.16/М 26 Автор(ы) : Маркович, Богдан Михайлович Заглавие : Рівняння математичної фізики : навч. посібник Выходные данные : Львів: Львів. політехніка, 2010 Колич.характеристики :383 с. Примечания : Бібліогр. : с. 377-378 ISBN, Цена 978-966-553-891-2: 153 грн. ББК : 22.161.68я73 Предметные рубрики:Математика Фізика Математична фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 611. - [x]Вид документа : Шифр издания : Б 84312 22.1/Ц 32 Автор(ы) : Цейтен, Иероним Георг Заглавие : История математики в древности и в Средние века : пер. с фр. П. С. Юшкевича . -Изд. 3-е Выходные данные : Москва: ЛИБРОКОМ, 2010 Колич.характеристики :231 с. Серия: Физико-математическое наследие. Математика (история математики) ISBN, Цена 978-5-397-01169-3: 144 грн 01 к. ББК : 22.1г Предметные рубрики:Математика Історія математики Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 612. - [x]Вид документа : Шифр издания : Б 83066 22.17/Б 91 Автор(ы) : Буріменко, Юрій Іванович, Сінявський О. В. Заглавие : Елементи теорії ймовірностей, математичної статистики та випадкових процесів : навч. посіб. для вищ. навч. закл. Выходные данные : К. : Освіта України, 2010 Колич.характеристики :118 с.: іл Примечания : Бібліогр. : с. 103-104 ISBN, Цена 978-966-188-140-1: 69 грн. УДК : 519.2(075.8) ББК : 22.17я73 Предметные рубрики: ймовірностей теорія випадкові величини математика навчальні видання для вищої школи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 613. - [x]Вид документа : Шифр издания : В 9604 22.1/Б 95 Автор(ы) : Быльцов, Сергей Заглавие : Математические игры, пасьянсы и фокусы. Занимательная математика для всей семьи Выходные данные : СПб.: Питер, 2010 Колич.характеристики :158 с.: ил. ISBN, Цена 978-5-4237-0018-8: (в пер.): 82 грн 24 к. УДК : 51-8:793.7 ББК : 22.1я9 Предметные рубрики:Математика ігри дозвілля Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 614. - [x]Вид документа : Однотомное издание Шифр издания : Б 88405 65.26/Ф 59 Заглавие : Финансовая математика : учеб. пособие Выходные данные : Б.м., 2010 Колич.характеристики :224 с.: ил Серия: Для бакалавров Примечания : Библиогр.: с.221 ISBN, Цена 978-5-406-00574-3: (в пер.) : 114 грн. ББК : 65.26в631я73 Предметные рубрики: фінансова математика математика навчальні видання для вищої школи Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 615. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-4 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Плюс кмітливість/ Борис Анастасійович Кордемський ; пер. з рос. Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 4 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :60 с ISBN, Цена 978-966-10-0690-3: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Четверта книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та казочок, фантазій і просто задач. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 616. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-5 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Тринадцять захопливих диваків/ Борис Анастасійович Кордемський ; пер. з рос.: Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 5 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :36 с ISBN, Цена 978-966-10-0692-7: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: П'ята книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та казочок, фантазій і просто задач. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 617. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-1 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Усяка всячина/ Борис Анастасійович Кордемський ; пер. з рос. Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 1 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :60 с ISBN, Цена 978-966-10-0663-7: 8 грн 91 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Перша книга із серії “Математичні заманинки” майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та казочок, фантазій і просто задач. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 618. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-3 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Події та пригоди на стежинках математики/ Борис Анастасійович Кордемський ; пер. з рос. Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 3 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :48 с ISBN, Цена 978-966-10-0689-7: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Третя книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та казочок, фантазій і просто задач. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 619. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-6 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Маленькі таємниці чисел та фігур/ Борис Анастасійович Кордемський ; пер. з рос.: Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 6 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :72 с ISBN, Цена 978-966-10-0693-4: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Шоста книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та досліджень і просто задач, що увібрали в себе «жар холодних чисел» і красу геометричних абстракцій. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 620. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-8 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Незвичайне - у звичайному/ Борис Анастасійович Кордемський ; пер. з рос.: Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 8 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :48 с ISBN, Цена 978-966-10-0697-2: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв, фантазій і просто задач, дошукавшись розв’язку яких, можна відкрити «незвичайне — у звичайному». Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 621. - [x]Вид документа : Однотомное издание Шифр издания : Б 89377 22.18/М 13 Автор(ы) : Мазалов, Владимир Викторович Заглавие : Математическая теория игр и приложения : учеб. пособие Выходные данные : Санкт-Петербург; Москва: Лань, 2010 Колич.характеристики :446 с.: ил. Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 431-438 ISBN, Цена 978-5-8114-1025-5: 540 грн. ББК : 22.183.2я73 Предметные рубрики:Математика Теорія гри Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 622. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-7 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Віддалеки через віки/ Борис Анастасійович Кордемський ; пер. з рос.: Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). - (Математичні заманинки). Кн. 7 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :84 с ISBN, Цена 978-966-10-0694-1: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Сьома книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв з історичної тематики, задач-легенд, жартів, а також задач з алфаметики — «зашифрованої» арифметики. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 623. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-2 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Галерея казок і фантазій/ Борис Анастасійович Кордемський ; пер. з рос.: Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. - (Класики популяризації науки). Кн. 2 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :48 с ISBN, Цена 978-966-10-0684-2: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Друга книга із серії “Математичні заманинки” майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та казочок, фантазій і просто задач. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 624. - [x]Вид документа : Однотомное издание Шифр издания : Б 89021 22.1/Н 20 Автор(ы) : Наймарк, Марк Аронович Заглавие : Нормированные кольца . -3-е изд. Выходные данные : Б.м., 2010 Колич.характеристики :684 с. Серия: Классика и современность. Математика Примечания : Библиогр. : с. 612 ISBN, Цена 978-5-9221-1273-4: (в пер. ) : 477 грн. ББК : 22.162.5 Предметные рубрики: Топологія Математика Функціональний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 625. - [x]Вид документа : Однотомное издание Шифр издания : Б 89079 22.16/Л 33 Автор(ы) : Лебедев, Николай Николаевич Заглавие : Специальные функции и их приложение : учеб. пособие . -Изд. 3-е, стер. Выходные данные : Б.м., 2010 Колич.характеристики :358 с.: ил Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 355-358 ISBN, Цена 978-5-8114-1023-1: (в пер.) : 442 грн. ББК : 22.161.4я73 Предметные рубрики: Природничі науки Навчальні видання Математика Функції Інтеграли Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 626. - [x]Вид документа : Однотомное издание Шифр издания : Б 88878 22.1/Б 91 Автор(ы) : Бунин, Валентин Алексеевич Заглавие : Биоподобие техногенных систем : математический код метагармонии Выходные данные : Б.м., 2010 Колич.характеристики :92 с Серия: Relata Refero Примечания : Библиогр.: с. 86-92 ISBN, Цена 978-5-396-00105-3: 60 грн. ББК : 22.12 Предметные рубрики:Математика Технічні системи Математична логіка Гармонія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 627. - [x]Вид документа : Однотомное издание Шифр издания : Б 88893 22.1/Г 29 Автор(ы) : Гейтинг, Аренд Заглавие : Интуиционизм. Введение : монография . -Изд.2-е, испр. Выходные данные : Б.м., 2010 Колич.характеристики :164 с Коллективы : пер. с англ. В. В. Янкова ; под ред. А. А. Маркова Серия: Физико-математическое наследие. Математика (основания математики и логики) ISBN, Цена 978-5-397-01321-5: 139 грн. ББК : 22.12я9 Предметные рубрики:математика математична логіка Экземпляры :ЗФ(1) Свободны: ЗФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 628. - [x]Вид документа : Однотомное издание Шифр издания : Б 101460 51/П 27 Автор(ы) : Перельман, Яков Исидорович Заглавие : Научные фокусы и загадки Выходные данные : Москва: АСТ : Астрель, 2010 Колич.характеристики :155 с.: ил ISBN, Цена 978-5-17-056636-5: (в пер.) : 20 грн. УДК : 51:793.7 Предметные рубрики:Математика Загадки математичні Аннотация: "Научные фокусы и загадки" составлены известным мастером занимательного жанра Я.И. Перельманом. Это увлекательная коллекция хитрых вопросов, занимательных задач, интересных загадок, головоломок, фокусов и игр. Эта книга для веселых, находчивых и сообразительных читателей! Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 629. - [x]Вид документа : Однотомное издание Шифр издания : Б 102883 514/Д 96 Автор(ы) : Дюпен, Шарль П'єр Франсуа Заглавие : Геометрія мистецтв і ремесел Выходные данные : Тернопіль: Навч. кн. - Богдан , 2010 Колич.характеристики :334 с.: іл Серия: Класики популяризації науки ISBN, Цена 978-966-10-1433-5 : 58 грн 32 к. УДК : 514.01-8:001.92 Предметные рубрики: Геометрія Математика Геометричні фігури Аннотация: Книга написана майже 200 років тому відомим французьким математиком, інженером, економістом, громадським діячем та просвітителем Шарлем Дюпеном (1784–1873). Незважаючи на свій поважний вік, вона залишається цікавою і для сучасного читача як чудова збірка численних практичних застосувань геометрії та майстерний опис важливих ідей, що вплинули на розвиток геометрії у XIX – ХХ ст. Це тим більше цікаво, що багато із цих ідей належать самому автору. Книга добре ілюстрована, тому її з цікавістю прочитають навіть ті, хто «не дружить» з геометрією через переобтяженість абстрактними поняттями і доведеннями. А вчитель знайде у ній велику кількість прикладів практичного застосування геометрії, які так потрібні йому для зацікавлення учнів своїм предметом. У свій час книга видавалася багатьма європейськими мовами. Українською мовою друкується вперше. Для учнів та вчителів математики загальноосвітніх навчальних закладів різних профілів, студентів та викладачів математичних спеціальностей вищих навчальних закладів, а також для усіх, хто цікавиться точними науками та їхньою історією. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 630. - [x]Вид документа : Многотомное издание Шифр издания : А 10075-9 51/К 69 Автор(ы) : Кордемський, Борис Анастасійович Заглавие : Робимо відкриття/ пер. з рос. Андрія Кравчука, Володимира Дячуна ; за ред. В. К. Дячуна. Кн. 9 Выходные данные : Тернопіль: Навч. кн-Богдан, 2010 Колич.характеристики :48 с Серия: Класики популяризації науки . Математичні заманинки ISBN, Цена 978-966-10-0699-6: 8 грн 50 к. УДК : 51-8:001.92 Предметные рубрики:Математика Популярні видання Аннотация: Дев’ята книга із серії «Математичні заманинки» майстра науково-популярної літератури Бориса Анастасійовича Кордемського — збірник математичних мініатюр: різноманітних цікавих есеїв та заманливих задач, які приводять до незвичайних «відкриттів» у царині як чисел, так і фігур. Усі, хто захоплюється математикою, — незалежно від віку — матимуть можливість потренувати мислення, кмітливість та винахідливість. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 631. - [x]Вид документа : Однотомное издание Шифр издания : Б 102799 51/Р 15 Автор(ы) : Радемахер, Ганс Адольф, Тепліц, Отто Заглавие : Числа і фігури : висліди мат. мислення Выходные данные : Тернопіль: Навч. кн. - Богдан, 2010 Колич.характеристики :318 с.: іл Серия: Класики популяризації науки ISBN, Цена 978-966-10-0251-6: 53 грн 46 к. УДК : 510.2 Предметные рубрики:Математика Елементарна математика Математичне мислення Аннотация: Книга містить 27 невеликих нарисів (тем), присвячених різним питанням елементарної математики. Кожна з тем є взірцем витонченого й доступного навіть для початківців наукового дослідження. Тому читання книги не потребує жодної спеціальної математичної підготовки, що виходить за межі звичайної шкільної програми для неповної середньої школи. Цінність книги полягає у тому, що вона не тільки знайомить читача з окремими новими поняттями і фактами, а й демонструє в дії найважливіші методи наукового пізнання, що застосовуються у математичних дослідженнях. З цієї точки зору книга є унікальним явищем у світовій науково-популярній літературі. Вона й досі перевидається багатьма мовами світу. Українською мовою друкується вперше. Для широкого кола читачів, від учнів до вчителів та викладачів математики. Особливо буде корисною тим, хто мріє про наукову кар’єру та хотів би дізнатися про це більше і випробувати себе. Стане в пригоді учасникам математичних гуртків та членам Малих академій наук по секції «Математика». Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 632. - [x]Вид документа : Однотомное издание Шифр издания : Б 88567 22.12/Н 16 Автор(ы) : Нагель, Эрнест, Ньюмен, Джеймс Рой Заглавие : Теорема Гёделя . -Изд. 3-е Выходные данные : Москва: КРАСАНД, 2011 Колич.характеристики :118 с Серия: Наука - всем! Шедевры научно-популярной литературы ISBN, Цена 978-5-396-00321-7: 103 грн. ББК : 22.122г Предметные рубрики:Математика Математика - історія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 633. - [x]Вид документа : Шифр издания : Б 88746 22.1/К 17 Автор(ы) : Калошина, Инна Павловна Заглавие : Большая теорема Ферма и психология творчества : монография Выходные данные : Б.м., 2011 Колич.характеристики :319 с.: ил. Примечания : Библиогр.: с. 317 ISBN, Цена 978-5-238-02124-9: 355 грн. ББК : 22.132я9+88в631я9 Предметные рубрики:математика психологія-математичні методи чисел-теорія Экземпляры :ЗФ(1) Свободны: ЗФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 634. - [x]Вид документа : Однотомное издание Шифр издания : Б 89033 74.1/Ф 86 Автор(ы) : Фрейлах, Наталья Ивановна Заглавие : Методика математического развития : учеб.пособие для сред.проф.образования Выходные данные : Б.м., 2011 Колич.характеристики :207 с Серия: Профессиональное образование Примечания : Библиогр.: с. 198-201 ISBN, Цена 5-8199-0278-5: (в пер.): 71 грн. ББК : 74.102.13я723 Предметные рубрики:Математика Дошкільна підготовка Навчальні видання Педагогіка Математичний розвиток -- методика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 635. - [x]Вид документа : Однотомное издание Шифр издания : Б 88558 22.16/К 68 Автор(ы) : Корпусов , Максим Олегович Заглавие : Разрушение в неклассических нелокальных уравнениях Выходные данные : Б.м., 2011 Колич.характеристики :374 с Примечания : Библиогр.: с. 372-374 ISBN, Цена 978-5-397-01577-6: 246 грн ББК : 22.161.627 Предметные рубрики: Природничі науки Математика Диференційні рівняння-- нелінійні Початково-крайові задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 636. - [x]Вид документа : Однотомное издание Шифр издания : Б 88495 22.1/П 50 Автор(ы) : Полищук, Ефим Михайлович Заглавие : Софус Ли: 1842-1899 . -Изд. 2-е Выходные данные : Б.м., 2011 Колич.характеристики :213 с. Коллективы : отв. ред. Ю.Д.Бураго Серия: Физико-математическое наследие. Математика (история математики) Примечания : Библиогр. : с. 204-212 ISBN, Цена 978-5-397-01649-0: 151 грн. ББК : 22.1г Предметные рубрики:Математика Математика-історія Географич. рубрики: Норвежський математик Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 637. - [x]Вид документа : Многотомное издание Шифр издания : Мб 1271-1 22.1/С 20 Автор(ы) : Сарданашвили, Геннадий Александрович Заглавие : Современные методы теории поля/ Геннадий Александрович Сарданашвили. Т. 1: Геометрия и классические поля . -Изд. 2-е, испр. Выходные данные : Москва: ЛИБРОКОМ, 2011 Колич.характеристики :212 с Примечания : Библиогр.: с. 203-204 ISBN, Цена 978-5-397-01670-4: 234 грн. ББК : 22.152+22.315 + 22.315 Предметные рубрики: Природничі науки Математика Геометрія Квантова теорія поля Математичний апарат Диференційна геометрія Геометрична теорія поля Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 638. - [x]Вид документа : Однотомное издание Шифр издания : Б 93281 22г/П 18 Заглавие : Остап Парасюк. Слава української науки : пропам'ятна зб. Выходные данные : Б.м., 2011 Колич.характеристики :590 с. с. іл. Коллективы : Нац. акад. наук України ; Ін-т теорет. фізики ім. М. М. Боголюбова Примечания : Бібліогр. : с. 441-461 ISBN, Цена 978-966-02-6216-4: (в опр.) : 70 грн. ББК : 22г(4УКР) Предметные рубрики: Природничі науки Математика Механіка Фізика Квантова теорія поля Вчені українські Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 639. - [x]Вид документа : Однотомное издание Шифр издания : Б 90889 22.17/Т 68 Автор(ы) : Триумфгородских, Максим Валерьевич Заглавие : Дискретная математика и математическая логика для информатиков, экономистов и менеджеров : учеб. пособие для вузов Выходные данные : Б.м., 2011 Колич.характеристики :180 с.: ил. Примечания : Библиогр. : с. 177 ISBN, Цена 978-5-86404-238-0: 101 грн. ББК : 22.174я73 Предметные рубрики: Дискретна математика Математична логіка Математика Комбінаторика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 640. - [x]Вид документа : Однотомное издание Шифр издания : Б 89668 22.16/П 69 Автор(ы) : Кузнецов А. Н., Неделько Е. Ю., Емельянова Т. В., Цеберная Н. А. Заглавие : Практикум по теории рядов и их приложениям Выходные данные : Б.м., 2011 Колич.характеристики :244 с Примечания : Библиогр.: с. 243 ISBN, Цена 978-966-321-177-0: 58 грн. ББК : 22.161.3 Предметные рубрики: Ряди (мат) Функціональні ряди Числові ряди Математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 641. - [x]Вид документа : Однотомное издание Шифр издания : Б 49215. 22.11/В 44 Автор(ы) : Виленкин, Игорь Владимирович, Гробер, Владимир Михайлович Заглавие : Высшая математика. Линейная алгебра. Аналитическая геометрия. Дифференциальное и интегральное исчисление : учеб. пособие . -Изд. 6-е Выходные данные : Б.м., 2011 Колич.характеристики :415 с.: ил. Серия: Высшее образование Примечания : Библиогр. : с. 409 ISBN, Цена 978-5-222-18236-9: 136 грн. ББК : 22.11я73 Предметные рубрики: Вища математика Математика Лінійна алгебра Аналітична геометрія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 642. - [x]Вид документа : Однотомное издание Шифр издания : Б 88380 65.26/К 82 Автор(ы) : Криничанский, Константин Владимирович Заглавие : Финансовая математика : учеб. пособие . -2-е изд., перераб. и доп. Выходные данные : Б.м., 2011 Колич.характеристики :331 с.: ил Примечания : Библиогр.: с.322-324 ISBN, Цена 978-5-8118-0520-7: (в пер.) : 162 грн. ББК : 65.26в631я73 Предметные рубрики: фінансова математика математика навчальні видання для вищої школи Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 643. - [x]Вид документа : Однотомное издание Шифр издания : Б 86339 22.16/П 16 Автор(ы) : Панюкова, Татьяна Анатольевна Заглавие : Основы теории дифференциальных уравнений для экономистов : учеб. пособие для вузов Выходные данные : Б.м., 2011 Колич.характеристики :254 с.: ил Примечания : Библиогр.: с. 254 ISBN, Цена 978-5-397-01671-1: 143 грн. ББК : 22.161.61я73 + 65в631я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математичні методи Економічна освіта Теорія дифференційних рівнянь Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 644. - [x]Вид документа : Однотомное издание Шифр издания : Б 86846 22.11/В 15 Автор(ы) : Валєєв, Кім Галямович, Джалладова, Ірада Агаверді, Дегтяр С.В. Заглавие : Вища математика для економістів : навч.посіб. Выходные данные : Б.м., 2011 Колич.характеристики :287 с Примечания : Бібліогр.: с. 287 ISBN, Цена 978-966-346-646-0: (в опр.): 129 грн. УДК : 51:33-057.86](075.8) ББК : 22.11я73 Предметные рубрики:математика лінійна алгебра аналітична геометрія Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 645. - [x]Вид документа : Однотомное издание Шифр издания : Б 86225 22.1/Д 26 Автор(ы) : Деза, Елена Ивановна, Модель Дмитрий Лазаревич Заглавие : Основы дискретной математики : учеб. пособие для вузов . -Изд. 2-е, испр. и доп. Выходные данные : Б.м., 2011 Колич.характеристики :218 с Примечания : Библиогр.: с. 218 ISBN, Цена 978-5-397-01588-2: 151 грн. ББК : 22.174я73 Предметные рубрики: Дискретна математика Математика Комбінаторика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 646. - [x]Вид документа : Однотомное издание Шифр издания : Б 85474 22.1/С 19 Автор(ы) : Саприкіна, Людмила Тихонівна, Цеберна, Наталя Олександрівна, Пєтков , Ігор Васильович Заглавие : Аналітична геометрія : навч. посібник Выходные данные : Б.м., 2011 Колич.характеристики :116 с.: іл Примечания : Бібліогр.: с. 115 Цена : 35 грн. ББК : 22.151.54 Предметные рубрики: Природничі науки Навчальні видання Математика Аналітична геометрія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 647. - [x]Вид документа : Однотомное издание Шифр издания : Б 85761 22.17/К 56 Автор(ы) : Коваленко, Іван Іванович Заглавие : Сучасні методи статистичного аналізу даних Выходные данные : Б.м., 2011 Колич.характеристики :189 с Примечания : Бібліогр.: с.178-182 ISBN, Цена 978-966-321-184-8: 40 грн. УДК : 519.24(075.8) ББК : 22.172я73 Предметные рубрики:математика математична статистика статистика Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 648. - [x]Вид документа : Однотомное издание Шифр издания : Б 87993 22.1/В 76 Автор(ы) : Воскобойников Юрий Евгеньевич Заглавие : Регрессионный анализ данных в пакете MATHCAD : учеб. пособие Выходные данные : Б.м., 2011 Колич.характеристики :224 с Серия: Учебник для вузов. Специальная литература Примечания : Библиогр.: с. 220 ISBN, Цена 978-5-8114-1096-5: ( в пер. ) : 320 грн. ББК : 22.172я73 Предметные рубрики: Природничі науки Навчальні видання Математика Математична статистика Програми MATHCAD Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 649. - [x]Вид документа : Однотомное издание Шифр издания : Б 88152 22.17/П 58 Автор(ы) : Попов, Александр Михайлович, Сотников, Валерий Николаевич Заглавие : Теория вероятностей и математическая статистика : учеб. для бакалавров Выходные данные : Б.м., 2011 Колич.характеристики :440 с. Серия: Бакалавр Примечания : Библиогр. : с. 382-383 ISBN, Цена 978-5-9916-1290-6 (в пер.): 193 грн. ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Математична статистика Математика Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 650. - [x]Вид документа : Однотомное издание Шифр издания : Б 87370 22.19/С 60 Автор(ы) : Соловьев, Игорь Алексеевич, Червяков, Александр Викторович, Репин Андрей Юрьевич Заглавие : Вычислительная математика на смартфонах, коммуникаторах и ноутбуках с использованием программных сред Python : учеб. пособие Выходные данные : Б.м., 2011 Колич.характеристики :265 с.: ил Серия: Учебники для вузов. Специальная литература Примечания : Библиогр.: с. 261-262 ISBN, Цена 978-5-8114-1120-7: (в пер.) : 340 грн. ББК : 22.19с51я73 Предметные рубрики: Природничі науки Навчальні видання Математика Обчислювальна математикка Смартфони Ноутбуки Програмне забезпечення Python, мова Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 651. - [x]Вид документа : Однотомное издание Шифр издания : Б 87127 22.1/П 50 Автор(ы) : Полищук, Ефим Михайлович Заглавие : Эмиль Борель: 1871-1956 . -Изд. 2-е Выходные данные : Б.м., 2011 Колич.характеристики :167, с Коллективы : отв. ред. Ф. А. Медведев Серия: Физико-математическое наследие. Математика (история математики) Примечания : Библиогр.: с. 165 и в подстроч. примеч. ISBN, Цена 978-5-397-01646-9: 112 грн. ББК : 22.1г Предметные рубрики:математика - історія математика Экземпляры :ЗФ(1) Свободны: ЗФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 652. - [x]Вид документа : Многотомное издание Шифр издания : В 10904 -1 22.1/М 34 Заглавие : Математична лінгвістика: навч. посіб. - (Комп'ютинг). Кн. 1: Квантитативна лінгвістика Выходные данные : Львів: Новий Світ-2000, 2012 Колич.характеристики :358 с.: ілюстр. Примечания : Бібліогр.: с. 351-358 ISBN, Цена 978-966-418-191-1: (в опр.) : 136 грн. УДК : 519.76(075.8) ББК : 22.17я73+81.1я73 Предметные рубрики:Математика Прикладне мовознавство Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 653. - [x]Вид документа : Однотомное издание Шифр издания : Б 92778 22.17/М 77 Автор(ы) : Монсик, Владислав Борисович, Скрынников, Андрей Александрович Заглавие : Вероятность и статистика : учеб. пособие Выходные данные : Б.м., 2012 Колич.характеристики :381 с.: ил. Примечания : Библиогр. : с. 376 ISBN, Цена 978-5-9963-0637-4: 182 грн. ББК : 22.17я73 Предметные рубрики:Математика Теорія ймовірностей Математична статистика Випадкові величини Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 654. - [x]Вид документа : Однотомное издание Шифр издания : Б 92698 22.1/П 75 Автор(ы) : Копич І. М. , Копитко Б. І., Сороківський В. М., Бабенко В. В., Стефаняк В. І. Заглавие : Прикладна математична статистика для економістів : навч. посіб. Выходные данные : Б.м., 2012 Колич.характеристики :407 с.: ілюстр Коллективы : Серия: Вища освіта в Україні Примечания : Бібліогр.: с. 398-402 ISBN, Цена 978-966-418-214-7: 114 грн. ББК : 22.172 я73 Предметные рубрики:Математика Статистика Навчальні видання для вищої школи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 655. - [x]Вид документа : Однотомное издание Шифр издания : Б 92594 22.1/Т 33 Заглавие : Теорія ймовірностей та математична статистика : навч. посіб. Выходные данные : Б.м., 2012 Колич.характеристики :394 с.: ілюстр Коллективы : Х. Т. Дрогомирецька [та ін.] Примечания : Бібліогр.: с.379-380 ISBN, Цена 978-617-607-260-7: (в опр.) : 130 грн. УДК : 519.2(075.8) ББК : 22.17я73 Предметные рубрики: ймовірностей теорія математика навчальні видання для вищої школи Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 656. - [x]Вид документа : Однотомное издание Шифр издания : Б 92721 74.26/Щ 92 Автор(ы) : Щоголєва, Людмила Олександрівна Заглавие : Розв'язування рівнянь і нерівностей з параметрами : (метод посіб. для вчителів) Выходные данные : Б.м., 2012 Колич.характеристики :164 с Коллективы : Волин. ін-т післядипл.пед.освіти Примечания : Бібліогр.: с. 159-160 ISBN, Цена 978-617-517-101-1: 59 грн. ББК : 74.262.21 Предметные рубрики:Математика Методика викладання Методичні посібники Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 657. - [x]Вид документа : Однотомное издание Шифр издания : Б 93058 22.1/Х-58 Автор(ы) : Хлопенко, Наталія Юріївна, Фатєєва, Тетяна Василівна Заглавие : Ірраціональні рівняння, нерівності та їх системи : навч. посіб. Выходные данные : Б.м., 2012 Колич.характеристики :51 с Примечания : Бібліогр. : с. 50 Цена : 10 грн. ББК : 22.161.61я73 Предметные рубрики:Математика Рівняння Навчальні видання Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 658. - [x]Вид документа : Однотомное издание Шифр издания : Б 94144 32.97/М 33 Автор(ы) : Матвієнко, Микола Павлович Заглавие : Комп'ютерна логіка : навч. посіб. для вищ. навч. закл. Выходные данные : Київ: Ліра-К, 2012 Колич.характеристики :286 с Примечания : Бібліогр.: с. 285-286. ISBN, Цена 966-2609-09-7: 60, 70 грн, грн. ББК : 32.973-047.20я73 Предметные рубрики: Комп'ютери Логіка Математична логіка Математика Навчальні видання для вищої школи Экземпляры :ВПНД(1), ДЕП(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 659. - [x]Вид документа : Однотомное издание Шифр издания : Б 93073 22.17/З-35 Автор(ы) : Зароський, Ромуальд Іванович, Кошкін К. В., Ніконова Л. М. Заглавие : Практикум з дискретної математики : навч. посіб. Выходные данные : Б.м., 2012 Колич.характеристики :266 с.: ілюстр. Примечания : Бібліогр. : с. 262-264 ISBN, Цена 978-966-321-224-1: (в опр.) : 50 грн. УДК : 519.854(076.5) ББК : 22.174я73 Предметные рубрики:математика комбінаторика навчальні видання для вищої школи Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 660. - [x]Вид документа : Однотомное издание Шифр издания : Б 92841 22/П 27 Автор(ы) : Перельман, Яков Исидорович Заглавие : Большая книга занимательных наук : алгебра, геометрия, физика, головоломки, задачи, опыты : сборник Выходные данные : Москва: АСТ : Астрель, 2012 Колич.характеристики : 542 с.: ил. ISBN, Цена 978-5-17-055460-7: (в пер.) : 168 грн. УДК : 51/53-8:793.7 ББК : 22я92 Предметные рубрики: фізікоматематичні науки фізика математика популярні видання Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 661. - [x]Вид документа : Однотомное издание Шифр издания : Б 92557 22.17/Р 83 Автор(ы) : Руденко, Володимир Миколайович Заглавие : Математична статистика : навч. посіб. для вузів Выходные данные : Б.м., 2012 Колич.характеристики :303 с.: іл. Примечания : Бібліогр. : с. 298-303 ISBN, Цена 978-611-01-0277-3: 70 грн. УДК : 519.22(075.8) ББК : 22.172я73 Предметные рубрики:Математика Статистика Математична статистика Прикладна математика Экземпляры :ЗРФ(2) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 662. - [x]Вид документа : Однотомное издание Шифр издания : Б 91310 22.1/О-77 Автор(ы) : Островский, Геннадий Макович, Зиятдинов, Надир Низамович, Лаптева, Татьяна Владимировна Заглавие : Оптимизация технических систем : учеб. пособие для вузов Выходные данные : Б.м., 2012 Колич.характеристики :421 с.: ил. Примечания : Библиогр.: с. 404-411 ISBN, Цена 978-5-406-01094-5: (в пер.) : 336 грн. ББК : 22.183я 73 Предметные рубрики: Природничі науки Математика Навчальні видання Технічні системи Оптимізація математична Програмування Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 663. - [x]Вид документа : Однотомное издание Шифр издания : Б 23546 22.11/Ш 63 Автор(ы) : Шипачев, Виктор Семенович Заглавие : Высшая математика : учеб. пособие для бакалавров . -8-е изд., перераб. и доп. Выходные данные : Б.м., 2012 Колич.характеристики :447 с Коллективы : под ред. А. Н. Тихонова Серия: Бакалавр. Базовый курс ISBN, Цена 978-5-9916-2031-4: (в пер.) : 238 грн. ББК : 22.11я73 Предметные рубрики: Вища математика Математика Аналітична геометрія Диференціальне числення Інтегральне числення Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 664. - [x]Вид документа : Однотомное издание Шифр издания : Б 91068 22.11/П 58 Автор(ы) : Попов, Александр Михайлович, Сотников, Валерий Николаевич Заглавие : Высшая математика для эконономистов : учеб. для бакалавров Выходные данные : Б.м., 2012 Колич.характеристики :564 с.: табл. Коллективы : под ред. А.М. Попова Серия: Бакалавр Примечания : Библиогр. : с. 563-564 ISBN, Цена 978-5-9916-1383-5: (в пер.) : 223 грн. ББК : 22.11я73 Предметные рубрики: Вища математика Математика Лінійна алгебра Математичний аналіз Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 665. - [x]Вид документа : Однотомное издание Шифр издания : Б 103806 51/Б 24 Автор(ы) : Баран, Олег Іванович Заглавие : Математичне асорті Выходные данные : Київ: Ленвіт, 2012 Колич.характеристики :295 с.: іл Примечания : Бібліогр.: с. 291-295 ISBN, Цена 978-966-8995-64-4: 15 грн. УДК : 51-8(07) Предметные рубрики:Математика Популярні видання Аннотация: Книга спрямована на активізацію пізнавальної діяльності учнів і студентів, які вивчають математику, на поглиблення здобутих в межах шкільної програми знань і підвищення зацікавленості предметом, розширення загальної культури і світогляду. Розділи книги складаються з мініатюр переважно розважального характеру. Математичні розваги, математичні кросворди і чайнворди, афоризми будуть цікаві всім читачам і стануть у нагоді вчителям-практикам як при організації навчального процесу, так і в позакласній роботі: при проведенні засідань математичних гуртків і розробці завдань для вікторин, при створені сценаріїв математичних вечорів, КВВ тощо. Матеріали книги розраховані на увагу учнів, студентів, вчителів і справжніх любителів математики. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 666. - [x]Вид документа : Однотомное издание Шифр издания : Б 90769 22.18/С 86 Автор(ы) : Стронгин, Роман Григорьевич Заглавие : Исследование операций. Модели экономического поведения : учебник Выходные данные : Б.м., 2012 Колич.характеристики :207 с.: ил. Серия: Основы информационных технологий Примечания : Библиогр. в подстроч. примеч. ISBN, Цена 978-5-94774-547-4 : (в пер.) : 109 грн. ББК : 22.183я73 Предметные рубрики: Природничі науки Навчальні видання Математика Теорія ігор Моделі операцій Экземпляры :ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 667. - [x]Вид документа : Однотомное издание Шифр издания : В 12963 517/Л 64 Автор(ы) : Личковський, Едуард Іванович, Свердан, Петро Леонович Заглавие : Вища математика. Теорія наукових досліджень у фармації та медицині : підручник Выходные данные : Київ: Знання, 2012 Колич.характеристики :476 с.: іл ISBN, Цена 978-966-346-999-7: 154 грн. УДК : 517(075.8) Предметные рубрики: Вища математика Математика Наукова діяльність Математичний аналіз Навчальні видання для вищої школи Аннотация: У підручнику викладено теорію і практику наукових досліджень: математичне моделювання, планування експерименту, оцінювання та статистичну перевірку гіпотез, дисперсійний, кореляційний та регресійний аналіз, аналіз рядів динаміки, лінійне програмування, теорію систем масового обслуговування, аналіз експертних оцінок. Значна увага приділена практиці розв’язання типових задач, аналізу та графічному представленню результатів. Подано завдання для самоконтролю, вправи для аудиторної та домашньої роботи, а також довідкові матеріали. Для студентів, аспірантів, науковців у галузі фармації, медицини, біології, хімії, соціології, комерції та промислових технологій. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 668. - [x]Вид документа : Однотомное издание Шифр издания : Б 92229 65.26/З-15 Заглавие : Задачи по финансовой математике : учеб. пособие для студентов Выходные данные : М.: КНОРУС, 2012 Колич.характеристики : 272 с.: ил. Коллективы : П.П. Брусов, Н.П.Орехова, С.В.Скородулина Серия: Для бакалавров ISBN, Цена 978-5-406-01554-4: (в пер.) : 202 грн. ББК : 65.26в631я73 Предметные рубрики: фінансова математика математика Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 669. - [x]Вид документа : Однотомное издание Шифр издания : Б 92232 22.1/Н 40 Автор(ы) : Невежин, Виктор Павлович Заглавие : Теория игр : примеры и задачи : учеб. пособие Выходные данные : Б.м., 2012 Колич.характеристики :127 с. Серия: Высшее образование Примечания : Библиогр. : с. 110 ISBN, Цена 978-5-91134-645-4: 87 грн. ББК : 22.183.2я73 Предметные рубрики: Ігор теорія Математика Теорія навчальні видання для вищої школи Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 670. - [x]Вид документа : Однотомное издание Шифр издания : Б 91955 22.13/О-66 Автор(ы) : Орлов, Петр Макарович Заглавие : Новые методики в арифметике целых чисел Выходные данные : Б.м., 2012 Колич.характеристики :104 с Серия: Relata Refero Примечания : Библиогр.: с. 104 ISBN, Цена 978-5-397-02782-3: 83 грн. ББК : 22.130 Предметные рубрики: Природничі науки Математика Арифметика Рівняння Теорема Ферма Числа Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 671. - [x]Вид документа : Однотомное издание Шифр издания : Б 91961 22.1/М 48 Автор(ы) : Мельников, Олег Исидорович Заглавие : Незнайка в стране графов . -Изд. 5-е Выходные данные : Б.м., 2012 Колич.характеристики :160 с.: ил Примечания : Библиогр.: с. 158 ISBN, Цена 978-5-397-02364-1: 129 грн. ББК : 22.174.2 я 9 Предметные рубрики: Природничі науки Математика Математичні терміни Теорія графов Задачі Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 672. - [x]Вид документа : Однотомное издание Шифр издания : Б 97575 20/И 19 Автор(ы) : Иваница, Сергей Васильевич, Матвеева, Марина Олеговна Заглавие : Современный справочник школьника и абитуриента. 5-11 классы : в табл., схемах, рис. Выходные данные : Донецк: БАО, 2013 Колич.характеристики :559 с.: ил Примечания : Содерж.: Математика. Геометрия. Химия. Алгебра. Физика ISBN, Цена 978-966-481-827-5: (в пер.) : 50 грн. УДК : 51/54-053.5(035) ББК : 20я721 Предметные рубрики: Природні науки Фізико-математичні науки Хімія Фізика Алгебра Математика Довідкові видання Экземпляры : всего : ДЕТ(1), ЗРФ(1) Свободны: ДЕТ(1), ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 673. - [x]Вид документа : Однотомное издание Шифр издания : Б 94253 22.3/Я 88 Автор(ы) : Яу, Шинтан, Надис, Стив Заглавие : Теория струн и скрытые измерения Вселенной Выходные данные : Б.м., 2013 Колич.характеристики :399 с.: ил. Серия: Династия ISBN, Цена 978-5-496-00247-9: 109 грн ББК : 22.315 Предметные рубрики: Фізика Математика Всесвіт Теоретична механіка релятивістська космологія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 674. - [x]Вид документа : Однотомное издание Шифр издания : Б 93546 20/І-19 Автор(ы) : Іваниця, Сергій Васильович, Матвєєва, Марина Олегівна Заглавие : Сучасний довідник школяра та абітурієнта. 5-11 класи Выходные данные : Б.м., 2013 Колич.характеристики :559 с.: іл. ISBN, Цена 978-966-481-826-8: 41 грн. ББК : 20.я721 Предметные рубрики: Фізика Природничі науки Геометрія Хімія Математика Алгебра Довідкові видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 675. - [x]Вид документа : Однотомное издание Шифр издания : Б 21403 22.1/Б 74 Автор(ы) : Богомолов, Николай Васильевич Заглавие : Практические занятия по математике : учеб. пособие для бакалавров . -11-е изд., перераб. и доп. Выходные данные : Б.м., 2013 Колич.характеристики :495 с. Серия: Бакалавр. Базовый курс ISBN, Цена 978-5-9916-2215-8: 252 грн. ББК : 22.1я73 Предметные рубрики: Фізико-математичні науки Математика Обчислювальна математика Алгебра Геометрія Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 676. - [x]Вид документа : Однотомное издание Шифр издания : КД-4896 22.17/Н 42 Автор(ы) : Неделько, Евгений Юрьевич Заглавие : Лабораторные работы по математической статистике : учеб. пособие Выходные данные : Николаев: НУК , 2013 Колич.характеристики :1 электрон. опт. диск Коллективы : НУК им. адмирала Макарова Цена : 20 грн. ББК : 22.172я73 Предметные рубрики:Математика Статистика Навчальні видання для вищої школи Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 677. - [x]Вид документа : Однотомное издание Шифр издания : Б 95201 22.17/М 33 Автор(ы) : Матвієнко, Микола Павлович Заглавие : Дискретна математика. ХХІ століття : навч. посіб. для вищ. навч. закл. Выходные данные : Київ: Ліра-К, 2014 Колич.характеристики :347 с.: іл. ISBN, Цена 978-966-2609-32-5: 94, 136, грн. УДК : 510.6:519.156](075.8) ББК : 22.174я73 Предметные рубрики: Чисел теорія Математика Комбінаторика Навчальні видання для вищої школи Множення теорія Автоматів теорія Аннотация: У навчальному посібнику в логічний послідовності викладено основні поняття дискретної математики згідно галузевого стандарту вищої освіти України по комп'ютерним і другим наукам. Теоретичний матеріал книги проілюстровано значною кількістю вправ і задач для набуття читачем практичного досвіду. За змістом та обсягом навчальний посібник відповідає навчалним планам дисципліни "Дискретна математика" для студентів різних спеціальностей вищих навчальних закладів, які вивчають дану дисціпліну, аспірантів і спеціалістів, які використовують відповідні математичні і комп'ютерні методи, а також окремі розділи навчального посібника можуть бути використані відповідними технічними навчальними закладами і коледжами. Экземпляры :ДЕП(1), ДЕТ(1) Свободны: ДЕП(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 678. - [x]Вид документа : Однотомное издание Шифр издания : Б 93763 22.1/Т 76 Автор(ы) : Трохимчук, Петро Павлович Заглавие : Математичні основи знань. Поліметричний підхід : монографія . -Вид. 2-ге, випр. та допов. Выходные данные : Б.м., 2014 Колич.характеристики :622 с.: ілюстр. Коллективы : Східноєвроп. нац. ун-т ім. Л. Українки Примечания : Бібліогр. : с. 532-564 ISBN, Цена 978-966-2750-01-7: (в опр.) : 57 грн. УДК : 510.2+53+007+001](0.064) ББК : 22.121я9 Предметные рубрики:математика математичні моделі фізико-математичні дисципліни фізико-математичні науки оптика Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 679. - [x]Вид документа : Однотомное издание Шифр издания : В 43-1 22.1/К 93 Автор(ы) : Кураса , Александр Александрович Заглавие : Цифровая теория чисел: в 2 ч./ Александр Александрович Кураса . Ч.1: Структура универсального множества натуральных чисел в системе предельных сумм Выходные данные : Николаев: НУК, 2014 Колич.характеристики :138 с Примечания : Библиогр.: с 138 Цена : 35 грн. УДК : 511(075.8) ББК : 22.13 Предметные рубрики:математика теорія аналітичні функції Экземпляры :ВПНД(1) Свободны: ВПНД(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 680. - [x]Вид документа : Однотомное издание Шифр издания : Б 98708 74.26/Х-58 Автор(ы) : Хлопенко, Наталія Юріївна Заглавие : Текстові задачі : навч. посіб. Выходные данные : Б.м., 2014 Колич.характеристики :59 с Примечания : Бібліогр. : с. 58 Цена : 26 грн. ББК : 74.262.21я73 Предметные рубрики:Математика Практичні методи навчання Математичні задачі навчальні видання для абітурієнтів Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 681. - [x]Вид документа : Однотомное издание Шифр издания : В 11495 22.1/С 86 Автор(ы) : Строгац, Стивен Заглавие : Удовольствие от х : увлекательное путешествие в мир математики от одного из лучших преподавателей в мире : пер. с англ. Выходные данные : Б.м., 2014 Колич.характеристики :292 с.: ил ISBN, Цена 978-500057-008-1: (в пер.) : 346 грн. ББК : 22.1я9 Предметные рубрики: Природничі науки Математика Геометрія Числа Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 682. - [x]Вид документа : Однотомное издание Шифр издания : 030/І-49 Заглавие : Ілюстрований довідник учня початкової школи Выходные данные : Б.м., 2015 Колич.характеристики :256 с ISBN, Цена 978-966-431-876-8: 45 грн. УДК : 030(035.053.2) Предметные рубрики: Українська мова Природознавство Математика Безпека життєдіяльності Довідкові видання Дитяча література Аннотация: Довідник містить основний матеріал курсу початкової школи з предметів: "Українська мова", " Літературне читання ", " Математика", "Я у світі", " Природознавство" та "Основи здоров'я" згідно з чинною програмою та Державним стандартом початкової загальної освіти. Робота з довідником дозволить систематизувати і доповнити знання учнів. Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 683. - [x]Вид документа : Однотомное издание Шифр издания : КД-4911 22.1/К 89 Автор(ы) : Кузнецов, Альберт Миколайович, Романчук Н. О. Заглавие : Курс лекцій з математичного аналізу : навч. посіб. для студентів, які навчаються за скороченим терміном Выходные данные : Б.м., 2015 Колич.характеристики :1 електрон. опт. диск Цена : 20 грн. ББК : 22.161я73 Предметные рубрики: Аналіз Математика Інтеграли функціональний аналіз Навчальні видання для вищої школи Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 684. - [x]Вид документа : Однотомное издание Шифр издания : Б 54621. 519/М 26 Автор(ы) : Мармоза, Анатолій Тимофійович Заглавие : Практикум з математичної статистики : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2015 Колич.характеристики :264 с Примечания : Бібліогр.: с.253. ISBN, Цена 966-7982-30-0: (в опр.) : 144 грн. УДК : 519.22/.25(075.8) Предметные рубрики:Математика Статистика Навчальні видання для вищої школи Аннотация: Практикум написаний відповідно до діючої програми курсу "Статистика" для студентів економічних спеціальностей вищих навчальних закладів. У ньому розглядаються статистичні розподіли та їх характеристики, статистична оцінка параметрів розподілу, перевірка статистичних гіпотез, дисперсійний і кореляційний аналіз тощо. Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 685. - [x]Вид документа : Однотомное издание Шифр издания : Б 72606. 519.8/В 19 Автор(ы) : Васильченко, Іван Петрович, Васильченко, Зоя Миколаївна Заглавие : Фінансова математика : навч. посіб. для вищ. навч. закл. . -Вид. 2-ге, допов. Выходные данные : Б.м., 2015 Колич.характеристики :250 с ISBN, Цена 978-966-351-350-356-0 : 144 грн. УДК : 519.8:330.4](075.8) Предметные рубрики:Математика Фінанси Математичне моделювання Навчальні ивдання для вищої школи Аннотация: Посібник присвячений вивченню основних елементів фінансової математики та деяких питань методології економіко-математичного моделювання, зокрема моделюванню економічної безпеки банку, математичній економіці запасів та імітаційному моделюванню. Посібник розрахований на студентів економічних спеціальностей вищих навчальних закладів, осіб, які навчаються за програмами підготовки бакалаврів, спеціалістів, магістрів. Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 686. - [x]Вид документа : Однотомное издание Шифр издания : Б 95253 22.17/М 42 Автор(ы) : Медведєв, Микола Георгійович, Пащенко І. О. Заглавие : Теорія ймовірностей та математична статистика : підручник Выходные данные : Київ: Ліра-К, 2015 Колич.характеристики :535 с.: іл Примечания : Бібліогр.: с. 529-535 ISBN, Цена 978-966-96938-3-9: (в опр.) : 90, 146, грн. УДК : 519.2(075.8) ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Випадкові величини Математика Статистика Навчальні видання для вищої школи Аннотация: У підручнику розглядається основний теоретичний матеріал з курсу “Теорія ймовірностей та математична статистика” за кредитно – модульною системою згідно вимог Болонської конференції. Особлива увага приділена доступності викладання матеріалу, який супроводжується великою кількістю прикладів. На електронному носії знаходяться задачі для самостійної роботи та модульного контролю студентів, біографічний довідник. Для студентів, викладачів та всіх, хто вивчає курс “Теорія ймовірностей та математична статистика”. Экземпляры :ДЕП(1), ДЕТ(1) Свободны: ДЕП(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 687. - [x]Вид документа : Однотомное издание Шифр издания : Б 95014 22.1/М 33 Автор(ы) : Матвієнко, Микола Павлович, Шаповалов, Сергій Павлович Заглавие : Математична логіка та теорія алгоритмів : навч. посіб. для вищ. навч. закл. Выходные данные : Київ: Ліра-К, 2015 Колич.характеристики :211 с Коллективы : Сум. держ. ун-т Примечания : Бібліогр.: с. 210-211. ISBN, Цена 978-966-2609-74-5: 65, 110 грн., 157, грн. ББК : 22.122я73 + 22.127я73 Предметные рубрики: Логіка Математика Алгоритмів теорія Навчальні видання для вищої школи Экземпляры :ЗРФ(4) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 688. - [x]Вид документа : Однотомное издание Шифр издания : Б 97592 22.1/А 65 Автор(ы) : Андреєва, Юлія Юріївна Заглавие : Унікальний спосіб вивчити таблицю множення і навчитися швидко рахувати : для дітей і дорослих Выходные данные : Б.м., 2015 Колич.характеристики :29, с ISBN, Цена 978-966-481-544-1: 12 грн. ББК : 22.194.2 Предметные рубрики: Математичні дії Математика Экземпляры : всего : ЗРФ(2) Свободны: ЗРФ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 689. - [x]Вид документа : Однотомное издание Шифр издания : Б 94144. 32.97/М 33 Автор(ы) : Матвієнко, Микола Павлович Заглавие : Комп'ютерна логіка : навч. посіб. Выходные данные : Київ: Ліра-К, 2015 Коллективы : М-во освіти і науки, молоді та спорту Укр. Примечания : Бібліогр.: с. 285-286. ISBN, Цена 966-2609-09-7: 145 , 145 грн., грн. УДК : 004.38:510.6](075.8) ББК : 32.97-047.20я73 Предметные рубрики: Комп'ютерна схемотехніка Комп'ютери Логіка Математика Навчальні видання для вищої школи Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 690. - [x]Вид документа : Однотомное издание Шифр издания : Б 97655 22.3/К84 Автор(ы) : Крутиков, Виктор Сергеевич Заглавие : Волны в областях с подвижными проницаемыми границами. Вопросы управления : монография Выходные данные : Б.м., 2015 Колич.характеристики :390 с.: ил Примечания : Парал. тит. л. и ч. текста : англ. - Библиогр.: с.341-370. ISBN, Цена 978-966-321-304-0: (в пер.) ББК : 22.311я9 Предметные рубрики: Математична фізика Математика Фізика Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 691. - [x]Вид документа : Однотомное издание Шифр издания : Б 99357 22.1/Б 25 Автор(ы) : Барковський, Віктор Володимирович , Барковська Н. В. Заглавие : Вища математика для економістів : навч. посіб. . -5-те вид. Выходные данные : Б.м., 2016 Колич.характеристики :445 с ISBN, Цена 978-966-364-991-7: 194 грн. УДК : 51:33-057.86](075.8) ББК : 22.11я73 Предметные рубрики:Математика Математична логіка Математичний аналіз Фінансова математика Алгебраїчні рівняння Аналітична геометрія Навчальні видання для вищої школи Аннотация: Навчальний посібник «Вища математика для економістів» містить теоретичні відомості всіх традиційних розділів курсу вищої математики, рекомендованих типовою навчальною програмою Міністерства освіти України для економічних спеціальностей, а також основні поняття математичної логіки, комбінаторики, теорії графів, опуклих множин, різницевих рівнянь, математики в фінансах та обліку. Посібник містить достатню кількість задач економічного змісту та таблиці, що використовуються для їх розв’язання. Для студентів економічних спеціальностей. Посібник може бути корисним викладачам ліцеїв, коледжів, а також фінансистам, бізнесменам, соціологам, фахівцям менеджменту та обліку. Экземпляры :ДЕТ(2) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 692. - [x]Вид документа : Однотомное издание Шифр издания : Б 50395. 22.17/Б 25 Автор(ы) : Барковський, Віктор Володимирович, Барковстка Н. В., Лопатін О. К. Заглавие : Теорія імовірностей та математична статистика : навч. посіб. . -5-те вид. Выходные данные : Б.м., 2016 Колич.характеристики :424 с Примечания : Бібліогр. : с. 421-422. ISBN, Цена 978-966-364-992-4: 262 грн. УДК : 519.2(075.8) ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Математика Статистика Навчальні видання для вищої школи Аннотация: Навчальний посібник написаний відповідно до навчальної програми дисципліни “Ви ща математика для економістів” для підготовки бакалаврів, спеціалістів та магістрів з економіки. Посібник містить основні поняття, методи, теореми та формули, багато розв’язаних типових задач, необхідні таблиці, зразки контрольних робіт, тести для перевірки залишкових знань студентів. Навчальний посібник може бути корисним фінансистам, актуаріям, керівникам різного рівня, бізнесменам, соціологам та політологам як довідник для повсякденного користування. Экземпляры :ДЕТ(1), ЗРФ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 693. - [x]Вид документа : Однотомное издание Шифр издания : Б 99748 22.11/Т 89 Автор(ы) : Турчанінова, Людмила Іванівна, Доля, Олена Вікторівна Заглавие : Вища математика в прикладах і задачах : навч. посіб. Выходные данные : Б.м., 2016 Колич.характеристики :347 с Коллективы : Київ. нац. ун-т будівництва і архітектури Примечания : Бібліогр. : с. 345-347 ISBN, Цена 978-617-7320-80-6: 200, 185, грн. УДК : 51(076.1)(075.8) ББК : 20.11я73 Предметные рубрики:Математика Лінійна алгебра Аналітична геометрія Інтеграли Навчальні видання для вищої школи Аннотация: Навчальний посібник є результатом узагальнення досвіду викладання курсу вищої математики для студентів денної форми навчання архітектурних, природничих, інженерних та інших спеціальностей КНУБА. Матеріал посібника розділено на шістнадцять тем, які традиційно вивчаються на І курсі вищих навчальних закладів і відповідають логічно завершеним змістовним модулям. Кожна з тем містить стислі теоретичні відомості і практичну частину, в якій наведені приклади розв’язання типових вправ і задачі для аудиторної та самостійної роботи. Для активізації пізнавальної діяльності студентів і індивідуалізації навчання в посібнику спеціально наведені практикуми з основних розділів курсу і контрольні тестові завдання. Посібник розрахований на студентів всіх форм навчання, особливо дистанційної, аспірантів, слухачів мережі ФПК. Він стане в нагоді і викладачам для проведення практичних, самостійних, а також контрольних робіт. Экземпляры : всего : ДЕТ(3) Свободны: ДЕТ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 694. - [x]Вид документа : Однотомное издание Шифр издания : Б 60278. 22.1/К 48 Автор(ы) : Клепко, Віктор Юхимович, Голець, Валентина Леонтіївна Заглавие : Вища математика в прикладах і задачах : навч. посіб. для вищ. навч. закл. . -2-ге вид. Выходные данные : Б.м., 2016 Колич.характеристики :592 с Коллективы : Київ. екон. ін-т менеджменту (Екомен) Примечания : Бібліогр. : с. 581-582. ISBN, Цена 978-966-364-928-3: 194 грн. УДК : 517(076.1)(075.8) ББК : 22.11я73 Предметные рубрики:Математика Диференційні рівняння Математичні задачі Лінійна алгебра Векторна алгебра Аналітична геометрія Математичний аналіз Диференційне числення Інтергаційне числення Навчальні видання для вищої школи Аннотация: Навчальний посібник містить задачі та приклади до всіх розділів вищої математики відповідно до програми загального курсу вищої математики для студентів економічних спеціальностей. Наведено необхідний довідковий матеріал, розв’язування типових прикладів і задач, набори прикладів і задач для практичних занять та самостійної роботи студентів. Экземпляры :ДЕТ(2) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 695. - [x]Вид документа : Однотомное издание Шифр издания : Б 99020 22.3/К 84 Автор(ы) : Крутиков, Віктор Сергійович Заглавие : Хвилі в областях з рухливими проникними границями. Питання керування : монографія Выходные данные : Б.м., 2016 Колич.характеристики :421 с Коллективы : Міжнар. технол. ун-т "Миколаїв. політехніка" Примечания : Бібліогр. : с. 356-382 ISBN, Цена 978-617-7240-02-16: ( в опр.) : 35 грн. УДК : 534.1:517.9](0.064) ББК : 22.311я9 Предметные рубрики: Математична фізика Математика Фізика Хвилі Рівняння Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 696. - [x]Вид документа : Однотомное издание Шифр издания : Б 99003-1 22.11/П 69 Заглавие : Практикум з вищої математики. Комп'ютерна система для дистанційного навчання: навч. посіб. : в 2 ч. Ч. І Выходные данные : : МНАУ Б.м., 2016 Колич.характеристики :236 с Примечания : Бібліогр.: с.231-232 ISBN, Цена 978-617-7149-13-1: (в опр.) : 60 грн. УДК : 512.64:514.12:517.53](076.5) ББК : 22.11я73 Предметные рубрики:Математика Лінійна алгебра Аналітична геометрія Навчальні видання для вищої школи Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 697. - [x]Вид документа : Однотомное издание Шифр издания : Б 55475. 22.1/Л 64 Автор(ы) : Литвин, Ірина Іванівна, Конончук, Оксана Миколаївна, Желізняк, Галина Олексіївна Заглавие : Вища математика : навч. посіб. для вищ. навч. закл. . -2-ге вид. Выходные данные : Б.м., 2016 Колич.характеристики :368 с Коллективы : Держ. ком. зв'язку та інформатизації України ; Львів. коледж держ. ун-ту інформ.-комунікац. технологій Примечания : Бібліогр. : с. 367. ISBN, Цена 978-966-364-5: 231 грн. УДК : 517(076.1)(075.8) ББК : 22.11я73 Предметные рубрики:Математика Числення Диференційні рівняння Ймовірностей теорія Лінійна алгебра Векторна алгебра функції (мат.) Навчальні видання для вищої школи Аннотация: Посібник містить виклад матеріалу курсу Вища математика в обсязі діючої програ ми для вищих закладів освіти першого та другого рівня акредитації, затвердженої управлінням кадрів та навчальних закладів Міністерства зв’язку України в 1998 році. Посібник призначений для студентів різних форм навчання та викладачів технікумів, ліцеїв, коледжів та гуманітарних інститутів. Достатьо широка система вправ дозво ляє використовувати посібник як задачник. Экземпляры :ДЕТ(2) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 698. - [x]Вид документа : Однотомное издание Шифр издания : Б 101619 519/К 84 Автор(ы) : Крикун, Іван Григорович Заглавие : Теорія ймовірностей та математична статистика : навч. посіб. Выходные данные : Миколаїв: НУК, 2017 Колич.характеристики :148 с ISBN, Цена 978-966-321-337-8: (в опр.) : 40 грн. УДК : 519.2(075.8) Предметные рубрики: Ймовірностей теорія Теорія Випадкові величини Математика Навчальні видання для вищої школи Аннотация: Подано основи теорії ймовірностей та математичної статистики. Кожен розділ містить теоретичні відомості та результати, приклади розв'язання типових задач,питання та задачі для самоперевірки, а також таблиці математичної статистики, які будукть корисними при розв'язанні теоретичних та практичних задач. Призначено для студентів економічних та технічних спеціальностей усіх форм навчання, а такожможе бути використаний за основу студентами інших спеціальностей та викладачами, які застосовують теорію ймовірностей та математичну статистику у власних курсах та дослідженнях. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 699. - [x]Вид документа : Однотомное издание Шифр издания : Б 103354 336/Г 13 Автор(ы) : Гадецька, Світлана Вікторівна, Савченко, Ганна Олександрівна Заглавие : Фінансова математика : навч. посіб. Выходные данные : Львів: Новий Світ-2000, 2017 Колич.характеристики :212 с Серия: Вища освіта в Україні Примечания : Бібліогр.: с. 209-210 ISBN, Цена 978-966-418-277-2: 200 грн. УДК : 336.11:51-7](075.8) Предметные рубрики:Математика Фінанси Навчальні видання для вищої школи Аннотация: Мета посібника - надати фахівцям економічної сфери необхідну допомогу з вивчення перевірених практикою методів кількісного аналізу фінансово-кредитних операцій. Головна увага приділяється теорії простих і складних відсотків та їх застосуванню, рентам, аналізу інвестиційних проектів, оцінкам фінансових інструментів та дохідності цінних паперів. Посібник призначений для студентів економічних спеціальностей, осіб - користувачів фінансових фінансових послуг, а також для тих, що застосовують фінансові розрахунки у професійній діяльності. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 700. - [x]Вид документа : Однотомное издание Шифр издания : Б 95377. 512/З-17 Автор(ы) : Зайцев, Євгеній Павлович Заглавие : Вища математика : лінійна та векторна алгебра, аналіт. геометрія, вступ до математ. аналізу : навч. посіб. . -2-ге вид., стер. Выходные данные : Б.м., 2017 Колич.характеристики :573 с.: іл Примечания : Бібліогр.: с. 549-552 ISBN, Цена 978-617-566-231-1 : 175 грн. УДК : 512.64+514.7+517](075.8) Предметные рубрики:Математика Лінійна алгебра Векторна алгебра Математичний аналіз Навчальні видання для вищої школи Аннотация: Запропонований навчальний посібник призначений для початкового ознайомлення з основами вищої математики і розвитку навичок розв’язання практичних задач для студентів нематематичних спеціальностей вищих навчальних закладів. У навчальному посібнику розглядаються елементи теорії множин, лінійна та векторна алгебра, аналітична геометрія, теорія границь, диференціальне числення та його застосування. Теоретичний матеріал супроводжується великою кількістю прикладів і задач. Крім того, у кінці кожного розділу наведено питання для самоперевірки, 30 комплексних індивідуальних завдань, а також розв’язок варіанта індивідуального завдання. До книги включено розширений довідковий розділ з елементарної математики. У додатку наведено числові таблиці та 120 графіків раціональних, дробово-раціональних, ірраціональних і трансцендентних функцій. Посібник може одночасно виконувати роль підручника, довідника, задачника і методичних вказівок щодо виконання індивідуальних завдань. Призначається для студентів технікумів і ВНЗ, аспірантів, осіб, що займаються самоосвітою, та викладачів, які організовують навчальний процес за кредитно-модульною системою. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 701. - [x]Вид документа : Однотомное издание Шифр издания : Б 104676 51/Ф 82 Автор(ы) : Фрай, Ханна Заглавие : Математика кохання : стереотипи, докази і пошук остаточного рішення Выходные данные : Б.м., 2017 Колич.характеристики :128 с.: іл. Серия: Ted Books Примечания : Бібліогр.: с.118-120. - Парал. тит. арк. англ. ISBN, Цена 978-966-942-122-7: 63 грн 96 к. УДК : 51-9:177.61 Предметные рубрики:Математика Почуття Любов Популярні видання Экземпляры :ОК/ДГН(1) Свободны: ОК/ДГН(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 702. - [x]Вид документа : Однотомное издание Шифр издания : Б 95014. 22.1/М 33 Автор(ы) : Матвієнко, Микола Павлович, Шаповалов, Сергій Павлович Заглавие : Математична логіка та теорія алгоритмів : навч. посіб. для вищ. навч. закл. Выходные данные : Б.м., 2017 Коллективы : Сум. держ. ун-т Примечания : Бібліогр.: с.210-211. ISBN, Цена 978-966-2609-74-5: 89, 121, 132, грн. УДК : 510.6:519.254](075.8) ББК : 22.122я73 + 22.127я73 Предметные рубрики: Логіка Математика Алгоритмів теорія Автоматів теорія Навчальні видання для вищої школи Аннотация: У навчальному посібнику наведено основні поняття і методи математичної логіки, а також основи теорії алгоритмів. Теоретичний матеріал проілюстровано вправами та задачами для набуття читачем практичного досвіду. Призначений для студентів, аспірантів і спеціалістів, які використовують відповінді методи класичної і некласичної математичної логіки та теорії алгоритмів. Экземпляры :ДЕТ(1), ЗРФ(3) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 703. - [x]Вид документа : Однотомное издание Шифр издания : Б 95253. 22.1/М 42 Автор(ы) : Медведєв, Микола Георгійович, Пащенко І. О. Заглавие : Теорія ймовірностей та математична статистика : підручник Выходные данные : Б.м., 2017 Колич.характеристики :535 с Примечания : Бібліогр.: с.529-535. - До кн. дод. 1 електрон. опт. диск. (КД-5214) ISBN, Цена 978-966-96938-3-9: (в опр.) : 165, 145, 165, грн. УДК : 519.2(075.8) ББК : 22.17я73 Предметные рубрики: Ймовірностей теорія Випадкові величини Математика Статистика Навчальні видання для вищої школи Аннотация: У підручнику розглядається основний теоретичний матеріал з курсу «Теорія ймовірностей та математична статистика» за кредитно-модульною системою згідно вимог Болонської конвенції. Особлива увага приділена доступності викладення матеріалу, який супроводжується великою кількістю прикладів. На електронному носії знаходяться задачі для самостійної роботи та модульного контролю студентів, біографічний довідник. Для студентів, викладачів та всіх, хто вивчає курс «Теорія ймовірностей та математична статистика». Вміст підручника є за повнотою викладу прийнятним, за орфографією стерпним (технічний суржик з орфографічними та стилістичними помилками), за оформленням жахливим (усі можливі проблеми використання Microsoft Word, включно з рисунками за допомогою автофігур). Экземпляры : всего : ЗРФ(2), ДЕТ(2) Свободны: ЗРФ(2), ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 704. - [x]Вид документа : Однотомное издание Шифр издания : Б 95201. 519/М 31 Автор(ы) : Матвієнко, Микола Павлович Заглавие : Дискретна математика ХХІ століття : підручник . -Вид. 2-ге перероб. і допов. Выходные данные : Б.м., 2017 Колич.характеристики :323 с Коллективы : Конотоп. ін-т Сум. держ. ун-ту Примечания : Бібліогр.: с. 321-323 ISBN, Цена 978-966-2609-32-5 : 228, 215, грн. УДК : 519.854"20"(075.8) Предметные рубрики:Математика Комбінаторика Дисретне програмування Навчальні видання для вищої школи Аннотация: У підручнику у логічний послідовності викладено основні поняття дискретної математики згідно галузевого стандарту вищої освіти України з комп’ютерних і других наук. Теоретичний матеріал книги проілюстровано значною кількістю вправ і задач для набуття читачем практичного досвіду. За змістом та обсягом підручник відповідає навчалним планам дисципліни «Дискретна математика» для студентів різних спеціальностей вищих навчальних закладів, які вивчають дану дисціпліну, аспірантів і спеціалістів, які застовують відповідні математичні й комп’ютерні методи, а також окремі розділи підручника можуть бути використані відповідними технічними навчальними закладами і коледжами. Экземпляры :ДЕТ(1), ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 705. - [x]Вид документа : Однотомное издание Шифр издания : Б 103365 519.8/Р 74 Автор(ы) : Романюк В. В. Заглавие : Теорія антагоністичних ігор : навч. посіб. для вищ. шк. Выходные данные : Б.м., 2018 Колич.характеристики :294 с Серия: Вища освіта в Україні Примечания : Бібліогр.: с.286-288. ISBN, Цена 978-966-418-170-6: 180 грн. УДК : 519.832(075.8) Предметные рубрики: Ігор теорія Математика Теорія Навчальні видання для вищої школи Аннотация: У посібнику висвітлюються питання теорії антагоністичних ігор як самостійної галузі наук і навчальної дисципліни, що формує теоретично-практичні знання, навики та уміння будувати та досліджувати математичні моделі конфліктних ситуацій, явищ, процесів. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 706. - [x]Вид документа : Однотомное издание Шифр издания : Б 103772 519/П 76 Автор(ы) : Шебанін О. В. Заглавие : Прикладна математика : навч. посіб. Выходные данные : Б.м., 2018 Колич.характеристики :164 с ISBN, Цена 978-617-7140-36-0: 35 грн. УДК : 519.677(075.8) Предметные рубрики:Математика Математичні задачі Обчислювальна математика Навчальні видання для вищої школи Аннотация: У навчальному посібнику розкрито суть алгоритми методів наближеного рішення задач різних галузей знань. Подано завдання для закріплених отриманих знань та контрольні питання. Містить навчальні матеріали з основних тем курсу прикладної математики, що передбачені освітньо-професійними програми підготовки ЗВО зі спеціальностей 073 "Менеджмент", 141 "Електроенергетика, електротехніка та електромеханіка", 208 "Агроінженерія". Призначено для студентів та викладачів закладів вищої освіти. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 707. - [x]Вид документа : Однотомное издание Шифр издания : Б 104185 51/Б 46 Автор(ы) : Бенджамін, Артур Заглавие : Магія математики. Як знайти х і навіщо це потрібно Выходные данные : Київ: КМ-БУКС, 2018 Колич.характеристики :350 с.: іл ISBN, Цена 978-617-7498-94-9: 148 грн 50 к. УДК : 51-8(076) Предметные рубрики:Математика Дидактичні ігри Математичні задачі Аннотация: Магія математики - це підручник з математики, який ви хотіли б мати в школі. Використовуючи чудовий асортимент методів - від прикладів з кульками морозива до вимірювання гір і створення чарівних квадратів, ця книга охоплює основні математичні сфери, включаючи арифметику, алгебру, геометрію та обчислення, плюс числа Фібоначчі, нескінченність і, звичайно, математичні магічні трюки. Артур Бенджамін, відомий в усьому світі як "математичний чарівник", поєднує математику та магію, щоб зробити предмет веселим, привабливим і однаково легким у розумінні як для фанатів математики, так і для тих, кого вона лякає. "Час від часу математичні та інші наукові періодичні видання проводять серед своїх читачів опитування на тему вибору найкрасивішого рівняння. І щоразу лідером виявляється це дивовижне рівняння, яке називають тотожністю Ейлера" "Позитивно радісна розвідка математики", - Publishers Weekly. "Кожен [трюк] більш сліпучий, ніж попередній", - Physics World. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 708. - [x]Вид документа : Однотомное издание Шифр издания : Б 104948 004/Л 63 Автор(ы) : Лиходєєва, Ганна Володимирівна Заглавие : Комп'ютерний практикум з математичної статистики : навч. посіб. Выходные данные : Київ: ЦУЛ, 2018 Колич.характеристики :97 с.: іл Коллективы : Бердян. держ. пед. ун-т Примечания : Бібліогр.: с. 96-97 ISBN, Цена 978-617-673-751-3: 87 грн. УДК : 004:519.22](075.8) Предметные рубрики: Комп'ютери Математика Статистика Комп'ютерні технології Навчальні видання для вищої школи Аннотация: У навчальному посібнику містяться лабораторні роботи, короткі теоретичні відомості з математичної статистики та методичні рекомендації для проведення комп'ютерного практикуму з математичної статистики, метою навчання якого є формування у студентів знань і умінь щодо опрацювання статистичних даних засобами комп'ютерних технологій, досвіду дослідницької діяльності. Навчальний посібник призначено для студентів педагогічних університетів. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 709. - [x]Вид документа : Однотомное издание Шифр издания : Б 102822. 514/П 27 Автор(ы) : Перельман, Яків Ісидорович Заглавие : Захоплююча геометрія Выходные данные : Тернопіль: Навч. кн. - Богдан, 2018 Колич.характеристики :286 с.: іл Серия: Класики популяризації науки . Країна Перельманія ISBN, Цена 978-966-408-365-9: 164 грн 25 к. УДК : 514(089.3) Предметные рубрики:Математика Геометричні форми Популярні видання Аннотация: «Захоплююча геометрія» Я.І. Перельмана — одна з найзахопливіших книжок про геометрію. У ній ця наука, за словами самого автора, «відривається від класної дошки», з якою тісно пов’язувалася у свідомості пересічного читача, й виводиться у вільний простір — у ліс, поле, до річки, на дорогу, а згодом — у далекі моря та захмарні глибини простору. Часто ці розповіді ілюструються уривками з творів класиків літератури та висловами великих учених. Прочитавши книгу, читач знатиме незмірно більше про застосування геометрії, ніж про це можна довідатися з десятків «шкільних» підручників та посібників. Книгу перекладено багатьма мовами світу. Українською виходить уперше. Видання зацікавить не тільки тих, хто вже любить і займається математикою, а й усіх, хто хотів би мати про неї правдиве уявлення, зокрема, знати про її справжнє місце у нашому житті. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 710. - [x]Вид документа : Однотомное издание Шифр издания : Б 104203 517/З-17 Автор(ы) : Зайцев, Євгеній Павлович Заглавие : Вища математика : інтегральне числення функцій однієї та багатьох змінних, звичайні дифференціальні рівняння, ряди : навч. посіб. Выходные данные : Київ: Алерта, 2018 Колич.характеристики :608 с.: іл ISBN, Цена 978-617-566-487-2: 290, 340 грн., грн. УДК : 517.3+517.91+517.5](075.8) Предметные рубрики:Математика Диференціальні числення Функціональні ряди Навчальні видання для вищої школи Аннотация: Запропонований навчальний посібник є продовженням першої книги з вищої математики і призначений для подальшого вивчення цієї дисципліни, а також розвитку навичок розв’язання практичних задач. Незважаючи на стислість викладення, теоретичний матеріал подається по можливості строго і доступно та ілюструється великою кількістю прикладів і задач. У кінці кожного розділу наведено питання для самоперевірки, 30 комплексних індивідуальних завдань, а також розв’язок варіанта індивідуального завдання. Посібник може одночасно виконувати роль підручника, довідника, задачника і методичних вказівок щодо виконання індивідуальних завдань. Буде корисним для студентів ВНЗ, аспірантів, осіб, що займаються самоосвітою, та викладачів, які організовують навчальний процес за STEM-технологіями. Экземпляры :ОК/ДЕТ(1), ЗРФ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 711. - [x]Вид документа : Однотомное издание Шифр издания : Б 103388 519/Т 33 Автор(ы) : Копич, Іван Михайлович, Сороківський, Василь Михайлович, Кісілевич, Олександра Василівна, Пенцак, Оксана Степанівна Заглавие : Теорія ймовірностей та математична статистика : навч. посіб. Выходные данные : Львів: Новий Світ - 2000, 2018 Колич.характеристики :381 с.: іл Серия: Вища освіта в Україні Примечания : Бібліогр.: с. 375-376 ISBN, Цена 978-966-1537-08-7: 200 грн. УДК : 519.2(075.8) Предметные рубрики: Ймовірностей теорія Математика Статистика Навчальні видання для вищої школи Випадкові величини Аннотация: Навчальний посібник містить основні розділи курсу "Теорія ймовірностей та математична статистика". Виклад теоретичного матьеріалу ведеться від інтуітивного введення і обгрунтування понять теорії ймовірностей до рекомендацій щодо практичного застосування методів математичної статистики. Кожна тема супроводжується прикладами, контрольними запитаннями та практичними завданнями і відповідями до них. Для студентів, аспірантів та викладачів економічних навчальних закладів. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 712. - [x]Вид документа : Однотомное издание Шифр издания : Б 72606. 519/В 19 Автор(ы) : Васильченко, Іван Петрович, Васильченко, Зоя Миколаївна Заглавие : Фінансова математика : навч. посіб. для вищ. навч. закл. . -Вид. 2-ге, допов. Выходные данные : Київ: Кондор, 2018 Колич.характеристики :248 с.: іл Примечания : Бібліогр. в підрядк. прим. ISBN, Цена 978-966-351-350-356-0: 200 грн. УДК : 519.8:330.4](075.4) Предметные рубрики:Математика Фінансовий аналіз Навчальні видання для вищої школи Аннотация: Посібник присвячений вивченню основних елементів фінансової математики та деяких питань методології економіко-математичного моделювання, зокрема моделюванню економічної безпеки банку, математичній економіці запасів та імітаційному моделюванню. Посібник розрахований на студентів економічних спеціальностей вищих навчальних закладів, осіб, які навчаються за програмами підготовки бакалаврів, спеціалістів, магістрів. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 713. - [x]Вид документа : Однотомное издание Шифр издания : Б 54621. 519/М 26 Автор(ы) : Мармоза , Анатолій Тимофійович Заглавие : Практикум з математичної статистики : навч. посіб. для вищ. навч. закл. Выходные данные : Київ: Кондор, 2018 Колич.характеристики :257 с Примечания : Бібліогр.: с. 253 ISBN, Цена 966-7982-30-0: 200 грн. УДК : 519.22/.25(076.5+075.8) Предметные рубрики:Математика Статистика Навчальні видання для вищої школи Аннотация: У посібнику розглядаються статистичні розподіли та їх характеристики, статистична оцінка параметрів розподілу, перевірка статистичних гіпотез, дисперсійний і кореляційний аналіз тощо, а також основні методи і прийоми обробки результатів спостережень. До кожного з п'яти розділів посібника подані короткі теоретичні відомості і приклади розв'язування типових задач, показано використання методів математичної статистики для розрахунку конкретних показників. Посібник призначений для студентів економічних спеціальностей вищих навчальних закладів, аспірантів, викладачів, працівників народного господарства, які бажають поглибити свої знання з математичної статистики. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 714. - [x]Вид документа : Однотомное издание Шифр издания : Б 103323 517/Б 72 Автор(ы) : Бобик, Омелян Іванович, Бобик, Ігор Омелянович, Литвин, Василь Володимирович Заглавие : Рівняння математичної фізики : (практикум) : навч. посіб. Выходные данные : Львів: Новий Світ-2000, 2018 Колич.характеристики :252 с.: іл Серия: Комп'ютинг Примечания : Бібліогр.: с. 252 ISBN, Цена 978-966-418-122-5: 200 грн. УДК : 517.958(076.5) Предметные рубрики:Математика Фізика Рівняння Навчальні видання для вищої школи Аннотация: У навчальному посібнику викладені основні поняття математичної фізики і методи розв'язування типових задач : метод характеристик, метод Фур'є, а також методи функції точкового джерела, потенціалів та інтегральних перетворень. Пропонований навчальний посібник рекомендується студентам вищих навчальних закладів, що навчаються за програмами, які передбачають поглиблене вивчення вищої математики, де дисципліна "Рівняння математичної фізики" вивчається окремим курсом або входить в інші її розділи. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 715. - [x]Вид документа : Однотомное издание Шифр издания : Б 92698. 519/П 76 Автор(ы) : Копич, Іван Михайлович, Копитко, Богдан Іванович, Сороківський, Василь Михайлович Заглавие : Прикладна математична статистика для економістів : навч. посіб. Выходные данные : Львів: Новий Світ - 2000, 2018 Колич.характеристики :407 с.: іл Серия: Вища освіта в Україні Примечания : Бібліогр.: с. 398-402 ISBN, Цена 978-966-418-214-7: 250 грн. УДК : 519.22:33](075.8) Предметные рубрики:Математика Статистика Навчальні видання для вищої школи Аннотация: У навчальному посібнику викладено теоретичні засади математичної статистики в супроводі з практичними застосуваннями в STATISTICA 6.0. Запропоновані численні завдання з відповідями для аудиторної та самостійної роботи студентів економічних напрямів підготовки, що допоможе їм оволодіти сучасними методами статистичного моделювання економічних процесів та явищ. Навчальний посібник може бути корисним також студентам старших курсів, аспірантам, викладачам та науковим співробітникам, які використовують статистичні методи у своїх дослідженнях. Экземпляры :ДЕТ(1) Свободны: ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 716. - [x]Вид документа : Однотомное издание Шифр издания : Б 106032 51/С 89 Автор(ы) : Строґац, Стівен Заглавие : Екскурсія математикою. Як через готелі, риб, камінці і пасажирів зрозуміти цю науку Выходные данные : Б.м., 2019 Колич.характеристики :256 с.: іл Примечания : Парал. тит. арк. англ. ISBN, Цена 978-617-7682-56-0: 146 грн 25 к. УДК : 51-028.31(02.062) Предметные рубрики:Математика Елементарна математика Популярні видання Аннотация:Математика — це не прісний набір формул. Вона скрізь, якщо знати, куди поглянути. Як розгледіти синусоїди у смужках зебри? Що таке «математика матраців»? Як кидки в баскетболі пояснюють основи лічби? У цій книжці автор проведе математичну екскурсію: від рівня садочка до університетського, даючи можливість будь-кому розібратися в цьому предметі й полюбити його. Книжка для найширшого кола читачів, всіх, хто любить математику або навпаки не розуміє її, для батьків, викладачів. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 717. - [x]Вид документа : Однотомное издание Шифр издания : Б 106071 512/П 27 Автор(ы) : Перельман, Яків Ісидорович Заглавие : Захоплююча алгебра Выходные данные : Київ: Навч. кн.-Богдан, 2019 Колич.характеристики :334 с.: іл Серия: Класики популяризації науки . Країна Перельманія ISBN, Цена 978-966-10-2341-1: 186 грн 75 к. УДК : 512(089.3) Предметные рубрики:Математика Популярні видання Алгебраїчні рівняння Аннотация: «Захоплююча алгебра» Я. І. Перельмана — одна з найзахопливіших книжок про алгебру. У ній ця наука ніби відокремлюється від класної дошки, з якою вона тісно «зрослася» у свідомості більшості читачів, і переноситься у вільний простір, де знаходить найрізноманітніші застосування. Основне своє завдання автор вбачає у тому, аби прищепити читачеві інтерес до занять алгеброю та викликати у нього бажання самостійно усунути прогалини своєї математичної підготовки. Прочитавши книгу, читач знатиме незмірно більше про основні ідеї алгебри та про її застосування, ніж про це можна довідатися з десятків «шкільних» підручників та посібників. Книга перекладена багатьма мовами світу. Повний переклад українською мовою виходить уперше. Книга зацікавить усіх, хто хоче мати правдиве уявлення про математику, зокрема, — переконливі свідчення про можливості її практичного застосування. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 718. - [x]Вид документа : Однотомное издание Шифр издания : А 10359 51/Я 60 Автор(ы) : Янґ, Трейсі, Ґ'юветт, Кейт Заглавие : Крута Математика : 50 супер. фактів для дітей - малих і великих Выходные данные : Б.м., 2019 Колич.характеристики :112 с.: іл Серия: Круті знання для пізнання ISBN, Цена 978-966-97730-2-9: 198 грн 75 к. УДК : 51(089.3) Предметные рубрики:Математика Аннотация: Читання, письмо й арифметика — три кити всіх освітніх систем на світі. Однак ці предмети завжди чомусь вважають нуднуватими. Останнім часом Дж. К. Ролінґ доклала чималих зусиль, щоб читання вважалося крутим і веселим заняттям, і навіть надихнула багатьох дівчат і хлопців спробувати себе в ролі письменниць і письменників. І це супер! Але як там поживає бідолашна стара добра арифметика? Ну, по-перше, тепер її ніхто так більше не називає. Арифметика стала математикою. А британці навіть повну назву не вимовляють, а кажуть скорочено: maths (а американці ще коротше —math). Тим, хто не палає бажанням по-справжньому зрозуміти математику, вона здається нічим не особливою, тоді як інші вважають її цариною, де поміж числами й рівняннями, теоріями й формулами панують чари і творяться чудеса. Математика здатна оживити буденні речі й перетворити повсякденну ситуацію на безкінечні, неможливі, неймовірні веселощі. Так-так, тут справді слова МАТЕМАТИКА й ВЕСЕЛОЩІ в одному абзаці. Але ж як їх поєднати? А ось як… Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 719. - [x]Вид документа : Однотомное издание Шифр издания : Б 105565 519/В 41 Автор(ы) : Вигоднер І. В., Білоусова Т. П., Ляхович Т. П. Заглавие : Теорія ймовірностей та математична статистика : навч. посіб. Выходные данные : Б.м., 2019 Колич.характеристики :336 с Коллективы : Херсон. нац. техн. ун-т Примечания : Бібліогр.: с.332 ISBN, Цена 978-966-289-284-0: 240 грн. УДК : 519.2(075.8) Предметные рубрики: Ймовірностей теорія Теорія Випадкові величини Гармонічний аналіз Математика Статистика Навчальні видання для вищої школи Аннотация: Навчальний посібник побудовано відповідно до діючих навчальної та робочої програми підготовки студентів з дисципліни "Теорія ймовірностей та математична статистика". Посібник містить теоретичний матеріал з усіх розділів даного курсу, приклади розв’язання типових задач. Для організації самостійної роботи студентів кожна тема супроводжується варіантами індивідуальних завдань, контрольними запитаннями для самоперевірки та списком літератури. Індивідуальні завдання складаються з 30 різних варіантів. Варіанти завдань подані у вигляді, який зручний для використання в якості дидактичних матеріалів. Посібник має додатки, які містять довідниковий матеріал, необхідний при розв'язанні задач. Даний посібник рекомендується для студентів інженерних, технологічних та економічних спеціальностей. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 720. - [x]Вид документа : Однотомное издание Шифр издания : Б 105765 51/Б 24 Автор(ы) : Баран, Олег Іванович, Васильєва, Лариса Яківна Заглавие : 884 задачі Миколаївських математичних олімпіад (1998-2019) Выходные данные : Миколаїв: Іліон, 2019 Колич.характеристики :290 с.: іл Примечания : Бібліогр.: с. 285-290 ISBN, Цена 978-617-534-537-5: 150, 180, грн. УДК : 51-8(477.73)"1998/2019"(076.3) Предметные рубрики:Математика Математичні задачі Навчальні видання Географич. рубрики: Миколаїв Аннотация: Представлено 884 задачі Миколаївських математичних олімпіад 1998-2019 років, переважно другого (районного і міського) етапів, і задачі Інтернет-олімпіад для школярів п'ятих класів. Для всіх задач наведені розв'язки або вказівки і відповіді. Для зручності використання посібника у навчальній роботі наводиться тематичний покажчик задач. Представлені рекомендації учасникам олімпіад, критерії оцінювання задач, методичні вказівки до окремих розділів. Наводиться список інформаційних джерел. Для учнів 5-11 класів середніх навчальних закладів усіх типів, які цікавляться проблемами підготовки до математичних змагань, а також для вчителів математики і керівників математичних гуртків. Экземпляры : всего : ДПК(5) Свободны: ДПК(3) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 721. - [x]Вид документа : Однотомное издание Шифр издания : Б 75565. 51/Л 75 Автор(ы) : Лозовий, Богдан Леонтійович, Пушак, Ярослав Сильвестрович, Шабат, Орислава Євгенівна Заглавие : Практикум з вищої математики : навч. посіб. . -Вид. 3-тє, допов. і перероб. Выходные данные : Б.м., 2019 Колич.характеристики :285 с Серия: Вища освіта в Україні Примечания : Бібліогр.: с.283-284 ISBN, Цена 966-2025-16-2 : 276 грн 88 к. УДК : 51(075.8+076.6) Предметные рубрики:Математика Лінійна алгебра Векторна алгебра Аналітична геометрія Математичний аналіз Диференціальні рівняння Ймовірностей теорія Навчальні видання для вищої школи Аннотация: Книга написана у відповідності з програмою курсу вищої математики для технічних вищих навчальних закладів. Не зважаючи на стислість викладу матеріал подано послідовно і по можливості доступно. Велика увага приділена застосуванні теоретичних положень до розв’язування практичних завдань. До кожного розділу програми дано розв’язки типових задач, які можуть бути використані, як взірці при виконанні студентами самостійних робіт. Приведено понад 1000 завдань для виконання індивідуальних домашніх робіт. У 3-му виданні посібника майже всі розділи доповненні більшою кількістю розв’язків типових задач, а також великою кількістю довідкових матеріалів (формули, таблиці по всіх розділах програми). Книга може бути використана студентами стаціонарнорго, а особливо заочного навчання при воконанні ними контрольних робіт. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 722. - [x]Вид документа : Однотомное издание Шифр издания : В 13655 51/Ф 80 Автор(ы) : Фортуна, Василь Васильович Заглавие : Вища та прикладна математика : навч. посіб. для студ. ден. форми навч. екон. спец. . -2-ге вид., стер. Выходные данные : Б.м., 2019 Колич.характеристики :647 с Коллективы : Донец. нац. ун-т економіки і торгівлі ім. М. Туган-Барановського, Каф. вищ. і приклад. математики Серия: Вища освіта в Україні Примечания : Бібліогр.: с.523 ISBN, Цена 978-617-574-071-2: 136 грн 88 к. УДК : 51(075.8) Предметные рубрики:Математика Математична кібернетика Математичне програмування Навчальні видання для вищої школи Аннотация: При підготовці даного курсу вищої і прикладної математики автори керувалися принципом підвищення рівня фундаментальної математичної підготовки майбутніх спеціалістів. Разом зтим, автори намагалися не переобтяжувати посібник складними викладками доведеннями, тому іноді доведення скорочено або зовсім опущено. Всюди, де це можливо, дається геометричний і математичний зміст економічних понять. Розглядається багато економічних застосувань методів вищої математики. Більшість наведених застосувань не вимагають від студентів першого курсу додаткової економічної інформації. Можна сказати, що в даному посібнику автори намагалися поєднати в одній книжці теоретичний матеріал із практичним - розв’язуванням задач як математичного змісту, так і економічного. Усі теми викладаються за єдиним математичним принципом. Спочатку подається основні означення, формулюються і доводяться теореми (іноді доведення опускаються), наводяться ілюстровані приклади розв’язання задач, розглядаються відомі економіко-математичні моделі та прикладні задачі економічного змісту. Краще розібратися в теоретичному матеріалі і оцінити його засвоєння допоможуть контрольні запитання, завдання для самостійного розв’язування з відповідями і тестові завдання, наведені до кожного розділу. Наприкінці, для кожної теми даються понад 30 варіантів завдань для індивідуального розв’язання, що дозволить перевіряти практичні навички студентів з ключових питань дисципліни та здійснювати контроль знань за модульно-рейтинговою системою. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 723. - [x]Вид документа : Однотомное издание Шифр издания : Б 105329 519/В 49 Автор(ы) : Висоцька, Вікторія Анатоліївна, Литвин, Василь Володимирович, Лозинська, Ольга Володимирівна Заглавие : Дискретна математика : практикум : (зб. задач з дискретної математики) Выходные данные : Львів: Новий Світ - 2000, 2019 Колич.характеристики :575 с.: іл Коллективы : Нац. ун-ит "Львів. політехніка" Серия: Комп'ютинг Примечания : Бібліогр.: с. 540-551 ISBN, Цена 978-617-7519-29-3 : 300 грн. УДК : 519.1+510.6](075.8+076.5) Предметные рубрики: Дискретна математика Математика Комбінаторика Навчальні видання для вищої школи Аннотация: Навчальний посібник містить матеріал для вивчення основних теоретичних засад, функціональних можливостей та практичного застосування дискретної математики, основи математичної логіки і теорії множин, елементи комбінаторного аналізу, основи теорії відношень, основи теорії графів та дерев, основи теорії кодування, булеві функції, мови, граматики та автомати, основи теорії алгоритмів, основи теорії кодування. Посібник призначений для здобуття студентами практичних навичок застосування класичних алгоритмів дискретної математики та сучасних алгоритмів штучного інтелекту, що використовуються при побудові комп’ютерних програм. Теоретичний та практичний матеріал викладено у доступній формі. Викладення матеріалу супроводжується значною кількістю прикладів, що полегшує його сприйняття та засвоєння. Навчальний посібник призначається для студентів, що навчаються за спеціальностями 122 «Комп’ютерні науки», 124 «Системний аналіз», 126 «Інформаційні системи та технології» та споріднених спеціальностей, які пов’язані з інформатикою та інформаційними технологіями. В результаті освоєння матеріалу студент буде вміти застосовувати відомі методи та алгоритми дискретної математики для дослідження предметної області та побудови математичного опису прикладних проблем. Посібник може бути використаний аспірантами як підґрунтя для наукових досліджень та викладачами як дидактичний матеріал, а також для самостійного вивчення. Книга призначена для спеціалістів із проектування, розроблення та впровадження інтелектуальних систем опрацювання інформаційних ресурсів, науковців у галузі глобальних інформаційних системи, систем штучного інтелекту, Інтернет-технологій, фахівців з електронної комерції, Інтернет-маркетингу та Інтернет-реклами, менеджерів комплексних Web-проектів, а також для здобувачів 3-ого (освітньо-наукового) рівня вищої освіти в галузі знань 12 «Інформаційні технології». Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 724. - [x]Вид документа : Однотомное издание Шифр издания : Б 106752 519/Б 64 Автор(ы) : Білоусова, Тетяна Петрівна, Вигоднер, Інна Валентинівна, Ляхович, Тетяна Павлівна Заглавие : Прикладна математика : навч. посіб. Выходные данные : Херсон: Олді-плюс, 2019 Колич.характеристики :158 с Коллективы : Херсон. нац. техн. ун-т Примечания : Бібліогр.: с. 158 ISBN, Цена 978-966-289-278-9: 170 грн. УДК : 519.6(075.8) Предметные рубрики: Обчислювальна математика Математичний аналіз Математика Чисельні методи Рівняння Навчальні видання для вищої школи Аннотация: Навчальний посібник містить лабораторні роботи з дисципліни "Прикладна математика". Наведено основні теоретичні відомості. Особливістю викладення матеріалу є залучення зразків розв’язання не тільки типових прикладів, але й окремих випадків, які виникають при розв’язанні задач. Це робить доцільним використання посібника для організації самостійної роботи студентів. Кожна лабораторна робота супроводжується докладним зразком її виконання та варіантами індивідуальних завдань для студентів. Посібник розрахований на використання викладачами та студентами, які ведуть або, відповідно, вивчають дисципліну "Прикладна математика". Також він може бути корисний при написанні курсових та дипломних робіт з фахових дисциплін студентам інженерних спеціальностей. Экземпляры :ОК(1) Свободны: ОК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 725. - [x]Вид документа : Однотомное издание Шифр издания : Б 105331 510/Б 90 Автор(ы) : Бубняк, Тарас Іванович Заглавие : Вища математика : навч. посіб. Выходные данные : Львів: Новий Світ - 2000, 2019 Колич.характеристики :436 с Примечания : Бібліогр.: с.389-390 ISBN, Цена 966-7827-41-0: 350 грн. УДК : 510:330.4](075.8) Предметные рубрики:Математика Економіка Навчальні видання для вищої школи Аннотация: До навчального посібника увійшли основні розділи вищої математики, рекомендовані освітньо-професійною програмою вищої освіти для студентів економічних, природничих, технологічних та інженерних спеціальностей. Теоретичний матеріал супроводжується рисунками, розв'язаними прикладами та вправами для самостійної роботи. Основу посібника складає курс лекцій який читався автором студентам першого та другого курсів Львівського державного аграрного університету на економічному та механічному факультетах. Посібник буде корисним для студентів-заочників з дистанційною формою навчання, а також викладачам для проведення практичних занять. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 726. - [x]Вид документа : Однотомное издание Шифр издания : Б 107595 519/Н 71 Автор(ы) : Нікольський, Юрій Володимирович, Пасічник, Володимир Володимирович, Щербина, Юрій Миколайович Заглавие : Дискретна математика : підручник . -Вид. 5-те, випр. та доп. Выходные данные : Б.м., 2020 Колич.характеристики :432 с Серия: Комп'ютинг Примечания : Бібліогр.: с. 430-431 ISBN, Цена 978-966-2025-76-7: 410 грн. УДК : 519.1(075.8) Предметные рубрики:Математика Комбінаторика Навчальні видання для вищої школи Аннотация: У підручнику розкрито основні поняття дискретної математики. В ньому представлено велика кількість розібраних прикладів, кожен розділ закінчується задачами для самостійного розв'язування. Экземпляры :ОК(1) Свободны: ОК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 727. - [x]Вид документа : Однотомное издание Шифр издания : Б 72671. 510/Б 83 Автор(ы) : Борисенко, Олексій Андрійович Заглавие : Дискретна математика : підруч. для вищ. навч. закл. Выходные данные : Суми: Унів. кн., 2020 Колич.характеристики :255 с Примечания : Бібліогр.: с.253-254 ISBN, Цена 978-966-680-376-7: 205 грн. УДК : 510.6:519.156](075.8) Предметные рубрики: Комбінаторика Вища математика Прикладна математика Математика Навчальні видання для вищої школи Аннотация: У підручнику, який складається з чотирьох частин і додатків, послідовно викладені елементарні питання теорії множин, логіки, комбінаторики та систем числення. Матеріал викладається конепективно за темами лекцій з великою кількістю прикладів. У першій частині посібника викладені основні поняття теорії множин і операції з ними. У другій частині розглядаються елементи математичної логіки. Особлива увага звертається на диз'юнкгивні і кон'юнкгивні нормальні форми логічних функцій та методи їх мінімізації. Третя частина містить елементарні комбінаторні конфігурації і біном Ньютона. У четвертій частині подається теорія і практика сучасних пози- ційних систем числення. Для студентів за напрямками підготовки «Електроніка», «Інформатика», «Автоматика». Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 728. - [x]Вид документа : Однотомное издание Шифр издания : Б 59983. 51/В 51 Автор(ы) : Лиман Ф. М. [та ін.] Заглавие : Вища математика : навч. посіб. для вищ. навч. закл. : у 2 ч. Выходные данные : Б.м., 2020 Колич.характеристики :614 с Примечания : Бібліогр.: с.604-605 ISBN, Цена 978-966-680-230-9: 460 грн. УДК : 51(075.8) Предметные рубрики: Природничі науки Математика Лінійна алгебра Векторна алгебра Аналітична геометрія Комплексні числа Інтеграли Навчальні видання для вищої школи Аннотация: У посібнику поєднано теоретичну і практичну складові курсу вищої математики. Повністю забезпечено першу частину курсу, що включає елементи теорії визначників, матриць, систем лінійних рівнянь, векторної алгебри, аналітичної геометрії, теорії многочленів, та другу частину курсу, що включає елементи диференціального та інтегрального числення, теорії диференціальних рівнянь, теорії рядів, теорії ймовірностей та математичної статистики. Для студентів вищих навчальних закладів, які вивчають вищу математику як загальноосвітній предмет. Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 729. - [x]Вид документа : Однотомное издание Шифр издания : Б 95253... 519/М 40 Автор(ы) : Медведєв, Микола Георгійович, Пащенко І. О. Заглавие : Теорія ймовірностей та математична статистика : підручник Выходные данные : Б.м., 2020 Колич.характеристики :536 с Примечания : Бібліогр.: с.529-535 ISBN, Цена 978-966-96938-3-9: 180 грн. УДК : 519.2(075.8) Предметные рубрики: Ймовірностей теорія Гармонічний аналіз Математика Статистика Навчальні видання для вищої школи Аннотация: У підручнику розглядається основний теоретичний матеріал з курсу «Теорія ймовірностей та математична статистика» за кредитно-модульною системою згідно вимог Болонської конвенції. Особлива увага приділена доступності викладення матеріалу, який супроводжується великою кількістю прикладів. На електронному носії знаходяться задачі для самостійної роботи та модульного контролю студентів, біографічний довідник. Экземпляры :ОК(1) Свободны: ОК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 730. - [x]Вид документа : Однотомное издание Шифр издания : Б 107444 373/П 12 Автор(ы) : Пагута Т. І. Заглавие : Методика формування елементарних математичних уявлень у дошкільників : навч.-метод. посіб. для спец. "Дошк. освіта" Выходные данные : Львів: Новий Світ-2000, 2020 Колич.характеристики :299 с.: іл Коллективы : ПВНЗ "Міжнар. економіко-гуманіт. ун-т ім. академіка С. Дем'янчука" Примечания : Бібліогр.: с. 289-299 ISBN, Цена 978-617-7519-25-5: 300 грн. УДК : 373.2.016:511-028.31](072) Предметные рубрики: Дошкільне навчання Дошкільна педагогіка Дошкільна освіта Математика Навчальні видання Аннотация: У навчально-методичному посібнику представлені теоретичні основи й методичні підходи до розвитку математичних уявлень у дітей дошкільного віку. Розкриті дидактичні основи математичного розвитку, питання формування кількісних, просторових, часових, геометричних уявлень у дітей дошкільного віку. Запропоновано завдання для самоперевірки та тестові завдання для підсумкового контролю. Рекомендовано для студентів спеціальності «Дошкільна освіта», викладачів закладів вищої освіти, вихователів. Экземпляры :ОК(1) Свободны: ОК(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 731. - [x]Вид документа : Однотомное издание Шифр издания : 37/Р 74 Автор(ы) : Романишин, Романа, Лесів, Андрій Заглавие : Числа = Numbers : вчимося рахувати Выходные данные : Київ: Укр. PEN, 2023 Колич.характеристики :46 іл. Примечания : Текст парал англ. ISBN, Цена 978-1-915068-26-6: 58 грн. УДК : 37.018.1:511.11](02.025.2.053.2) Предметные рубрики: Сімейне виховання Числа Математика Ілюстровані видання Література для дітей Экземпляры : всего : ЗРФ(2), ЦМП(1) Свободны: ЗРФ(2) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 732. - [x]Вид документа : Однотомное издание Шифр издания : 51/С 89 Автор(ы) : Строгац, Стівен Заглавие : Екскурсія математикою. Як через готелі. риб, камінці і пасажирів зрозуміти цю науку . -4-те вид. Выходные данные : Київ: Наш Формат, 2023 Колич.характеристики :256 с Примечания : Парал. тит. арк. англ. ISBN, Цена 978-617-7682-56-0: 350 грн. УДК : 51-028.31(02.062) Предметные рубрики:Математика Елементарна математика Популярні видання Аннотация:Математика - це не прісний набір формул. Вона скрізь, якщо знати, куди поглянути. Як розгледіти синусоїди у смужках зебри? Що таке «математика матраців»? Як кидки в баскетболі пояснюють основи лічби? У цій книжці автор проведе математичну екскурсію: від рівня садочка до університетського, даючи можливість будь-кому розібратися в цьому предметі й полюбити його. Книжка для найширшого кола читачів, всіх, хто любить математику або навпаки не розуміє її, для батьків, викладачів. Навіть якщо маєте страх перед математикою, ви просто не зможете відірватися від книжки та зрештою зрозумієте цей предмет. Книжку перекладено 15-ма мовами. Экземпляры :ОК/ДЕТ(1) Свободны: ОК/ДЕТ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 733. - [x]Вид документа : Однотомное издание Шифр издания : Б 92219 22.1/П 34 Автор(ы) : Писаревский, Борис Меерович, Харин, Виталий Тимофеевич Заглавие : О математике, математиках и не только . -2-е изд., испр. и доп. Выходные данные : Б.м., 212 Колич.характеристики :301 с.: ил. Примечания : Библиогр. : с. 295-298 ISBN, Цена 978-5-9963-0631-2: (в пер.) : 160 грн. ББК : 22.1 Предметные рубрики:Математика Фізико- математичні науки Математики Экземпляры :ЗРФ(1) Свободны: ЗРФ(1) Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 734. - [x]Вид документа : Однотомное издание Шифр издания : 51/М 34 Заглавие : Математический отдел "Научного образования" Выходные данные : Б.м., Приплетено: 1. Гаусс Общие исследования о кривых поверхностях/ Карл Фридрих Гаусс. - С. 2-4 с. 2. Эйлер О неопределенных уравнениях/ Леонардо Эйлер. - С. 4-12 с. 3. Гаусс Общие исследования о кривых поверхностях/ Карл Фридрих Гаусс ; пер. с прим. М. М. Филиппова. - С. 2-8 с. 4. Гаусс Общие исследования о крвых поверхностях/ Карл Фридрих Гаусс ; пер. с прим. М. М. Филиппова. - С. 2-12 с. 5. Протоколы заседаний С.-Петербургского Математического Общества 18 93/94 год: заседание 18 сентября 1893 г. - С. 2-20 с. - //Прил. к журн. "Научное обозрение" 6. Гаусс Общие исследования о кривых поверхностях/ Карла Фридрих Гаусс ; пер. с прим. М. М. Филиппова. - С. 22-48 с. 7. Вычисление небесных треугольников. - С. 2-16 с. 8. Протоколы заседаний С.-Петербургского Математического Общества 18 93/94 г.: о значениях знаков : =, -, +, :, . ; T, U, K, S, V в теории квартерионов. - С. 2-23 с. Цена : Б.ц. УДК : 51+517.518.18](082) Предметные рубрики:Математика Поверхні (мат.) Математичні терміни Аналітична геометрія Астрономія Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 735. - [x]Вид документа : Однотомное издание Шифр издания : 512/К 12 Автор(ы) : Каган В. Ф. Заглавие : Что такое алгебра? Выходные данные : Б.м., Колич.характеристики :72 с Цена : Б.ц. Предметные рубрики:математика фізико-математичні науки методика викладання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 736. - [x]Вид документа : Однотомное издание Шифр издания : 512/Л 33 Автор(ы) : Лебединцев К. Ф. Заглавие : Учение о простейших функциях, их графиках и теория пределов : доп. к 1-й ч. "Концентр. рук. алгебры для сред. учеб. заведений"... Выходные данные : Б.м., [1916?] Колич.характеристики :140 с., 73 черт. на 11 табл. Цена : Б.ц. Предметные рубрики:математика алгебра функції (мат.) елементарні функції навчальні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 737. - [x]Вид документа : Однотомное издание Шифр издания : 512/П 91 Автор(ы) : Пфейффер Г. В. Заглавие : Представление областей особенных точек алгебраических поверхностей рядами, расположенными по целым положительным степеням двух параметров : (разрежение особ. точек алгебр. поверхностей) Выходные данные : Б.м., [191?] Колич.характеристики :134 с., 15 л. черт. Примечания : Отт. из журн. Унив. Изв. - 1910 Цена : Б.ц. Предметные рубрики: алгебра математика функції Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") 738. - [x]Вид документа : Однотомное издание Шифр издания : 511/П 27 Автор(ы) : Перельман, Яков Исидорович Заглавие : Занимательная арифметика. Загадки и диковинки в мире чисел . -Изд. 4-е, вновь просм. и доп. Выходные данные : Б.м., Колич.характеристики :207, с.: ил. Цена : (В пер.) : 4 р. Предметные рубрики:математика числа популярні видання Найти похожие "Поиск с ранжированием по запросу, составленному из ключевых слов записи") СтандартныйРасширенныйПрофессиональныйРаспределенныйТематический навигатор © Международная Ассоциация пользователей и разработчиков электронных библиотек и новых информационных технологий (Ассоциация ЭБНИТ)
1787	https://www.youtube.com/watch?v=dgz_zg4mnD8	Read and speak any number up to 10 million and beyond! The Maths Man 6090 subscribers 934 likes Description 48847 views Posted: 13 Sep 2020 Follow me on instagram: This explains how breaking numbers down in families helps to read large numbers. This video was made to help children who are studying maths as part of the Year 6 English national curriculum for the objective "read and write numbers up to 10 million". Transcript: hey everyone welcome to today's youtube video in this video today you are going to learn how to say any number up to 10 million and maybe even past 10 million let's learn it let's learn it now [Music] so here i have a five digit number one two three four five and here i have a six digit number one two three four five six so i'm going to show you how to say these numbers correctly okay the first thing that you need to understand though is that the comma is actually there for a reason in some big numbers you'll see a comma and it's to separate the units from the thousands okay so the units is a family of three the thousands is also a family of three and if i wanted to say this number you start with the number on the left just like you would with any other number and you say the numbers that are in the thousand family so i have 36 500 and twelve okay so thirty six thousand five hundred and twelve can you see how i said the numbers in their families okay rather than saying three six five hundred and twelve that wouldn't make sense thirty six thousand five hundred five hundred and twelve so let's take a look at this six digit number now so i'm going to break the numbers down in my head okay this is what i'm doing in my head okay i'm gonna break the numbers down into their different families so these are the units and these are the thousands so all i need to do is i need to read the number that is in the thousands 250 000 one hundred and five two hundred and fifty thousand one hundred and five let's take a look now at some larger numbers that go over a million so the two examples i have in front of me here are both million numbers okay so this is one two three four five six seven digits any number that has seven digits in it is over a million this has one two three four five six seven eight digits in it now the same as i did before i need to break this number down into families okay so these are the units these are the thousands and this time i have a number that is part of the million family million and the same as i did before i need to say these numbers in their families to say the full uh to say the full number so 2 million 708 000 500 and thirteen two million seven hundred and eight thousand five hundred and thirteen let's take a look at the last example so i'm gonna again break this down into units thousands and the minion family so just say exactly what you see 17 million 68 109 17 million and 17 million 68 hundred and nine seventeen million sixty eight thousand one hundred and nine if you did find this video helpful today then please smash that like button down below and if you would like to see more maths videos then please feel free to subscribe to the mathsman youtube channel catch you in the next video
1788	https://www.free-online-calculator-use.com/decimal-to-octal-converter.html	Decimal to Octal Converter with Built-in Conversion Chart Creator Menu Favorites Ad-Free VersionLOGIN Scroll To Decimal to Octal Converter Calc Calc Calculator Calculator Steps Steps Steps Instructions Terms PCalc Data? HomeWhat's New?About MeContact MeSite MapMy Dashboard Love the calculators but not the ads? Take a sneak peek at the Ad-Free Design being enjoyed by hundreds of members for less than 25¢ a week. Full details here! Discover more Calculators Calculator Financial Calculators financial calculators Financial calculator Even Spacing Calculator calculator calculators Home » Conversion Calcs » Decimal to Octal Converter Decimal to Octal Converter to Convert Base 10 to Base 8 This calculator will convert decimal numbers into octal numbers and display an interactive conversion chart to show how it arrived at the result. Tapping any number within the generated conversion chart will display an explanation of how the calculator arrived at the number located within the tapped cell. Be sure to check out my other decimal conversion calculators for converting: Octal to Base 10 Base 10 to Binary Base 10 to Hexadecimal Read more ... Also on this page: What is an octal number? How to convert decimal to octal, with example. Discover more Calculators Even Spacing Calculator calculator Financial Calculators Financial calculator Calculator calculators financial calculators Calculator Preferences (Click to change width of calculator) Adjust Calculator Width % Show/Hide Keypads Stick/Unstick Tools Adjust Calculator Width: 56 Show Hide Stick Unstick Decimal to Octal Converter Convert decimal to octal and create the conversion chart. Special Instructions Save your entries: To save your entries in between visits, click the Data tab in the right-hand column. Learn More Selected Data Record: A Data Record is a set of calculator entries that are stored in your web browser's Local Storage. If a Data Record is currently selected in the "Data" tab, this line will list the name you gave to that data record. If no data record is selected, or you have no entries stored for this calculator, the line will display "None". Data Data record Data record Selected data record: None Learn More Decimal:Decimal:Decimal number (base 10):Decimal number (base 10): #### Decimal number (base 10): Enter the decimal (base 10) number you would like to convert into an octal number. Note that the entered number must be positive and may only consist of digits 0-9, a single decimal point, and the leading number must not be a zero. For numbers containing a decimal point, the converter will only convert out to the last digit entered. Decimal number No text Convert Decimal to Octal Learn More Octal:Octal:Equivalent octal (base 8):Equivalent octal (base 8): Octal (base 8) equivalent: This is the octal equivalent to the entered decimal number. Note that after clicking the Convert Decimal to Octal button the converter will display a result summary and an interactive conversion chart below this line. the Data tab to save this set of entries to your current web browser so you won't have to start over from scratch on your next visit. Subscribe to the Ad-Free Cloud Version to save your entries to a secure online database, allowing you to access your saved entries from any device or web browser. Take a Sneak Peek at the Cool, Ad-Free Design! Reset If you received value from this calculator, please pay it forward with a Share, Like, Tweet, Pin, or Link. Thank you! -Dan 0 Save Feedback Appreciated Feedback Greatly Appreciated Your Feedback Would Be Greatly Appreciated Your Feedback Is Greatly Appreciated This Form cannot be submitted until the missing fields (labelled below in red) have been filled in Feedback Please note that all fields preceded by a red asterisk must be filled in. Learn More Letter grade: Letter grade: Please give the calculator a letter grade. If you grade the calculator less than A, please tell me what I would need to do to the calculator to get an A. If the calculator didn't work at all, please try downloading the latest version of Google Chrome or Firefox. Chances are, if the calculator is not working at all, you may be missing out on other content on the web due to an outdated or non-conforming web browser. What letter grade would you give the calculator on this page? 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Back to Calculator Show Help and Tools Help and Tools Hide ?Instructions Terms Data PCalc Click the ? tab for Help & Tools instructions. Global Instructions Instructions How to use the Decimal to Octal Converter IMPORTANT: Numeric entry fields must not contain dollar signs, percent signs, commas, spaces, etc. (only digits 0-9 and decimal points are allowed). Click the Terms tab above for a more detailed description of each entry. Step #1: Enter the decimal number you want to convert to octal. Step #2: Click the "Convert Decimal to Octal" button, which will display the result of the conversion and generate an interactive conversion chart with tap-enabled help cells. Glossary Fields, Terms, and Definitions. Octal number system: Number system that uses the base 8 system of expressing values, which consists of the digits 0, 1, 2, 3, 4, 5, 6, and 7. Base 8 numbers are usually written with a subscripted 8 behind them (175 8). Reset button: Clicking the "Reset" button will restore the calculator to its default settings. Save Entries and Notes Pocket Calculator Discover more Financial calculator financial calculators calculator Even Spacing Calculator Calculators Calculator calculators Financial Calculators Other Related Articles Learn: How to read and write Roman numerals. — Roman Numeral Converter Learn: How to convert binary to hexadecimal. — Binary to Hexadecimal Converter Tools: Common kitchen measurement conversions and 2 mini cooking measurement converters. — Recipe Conversion Calculator Learn What an octal number is and how to convert it to a decimal number. What is an Octal Number? The easiest way to understand what an octal number is is to compare it to something you already know -- a decimal number. As you know, a decimal number uses the base 10 system for counting and expressing value. It's called "base 10" because it uses ten numeric characters (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to count and express values. The octal number system, on the other hand, uses the base-8 method for counting and expressing value. It's called "base 8" because it uses 8 numeric characters to count and express value. The 8 octal numeric characters are 0, 1, 2, 3, 4, 5, 6, and 7. Now, since we are looking to convert a base 10 number into a base 8 number, let's compare the base 10 place values to the place values in a base 8 number system: Place Values of Decimal Vs. Octal Systems ← swipe left and right →← swipe left and right → Power of 10:10 3 10 2 10 1 10 0.10-1 10-2 10-3 Base 10 Place value:1000 100 10 1.1/10 1/100 1/1000 Power of 8:8 3 8 2 8 1 8 0.8-1 8-2 8-3 Base 8 Place value:512 64 8 1.1/8 1/64 1/512 As you can see, since each place value in a base 10 number is different than the corresponding place value in a base 8 system, we need a method for converting 0-9 base 10 place values into 0-7 base 8 place values. How to Convert Decimal to Octal To convert a base 10 number into a base 8 number, the first step is to find the first base 8 place value that is greater than or equal to the decimal number you are converting -- starting at the 8 0 place and working your way to the left. For example, suppose you want to convert the decimal number 125 into an octal number. In that case, you would find the first base 8 place value that is greater than or equal to 125, which would be 512. Step# 1: First Base 8 Value Greater than 125 10 Power of 8:8 3 8 2 8 1 8 0 Place value:51264 8 1 Once you have located your base 8 place value starting point, the next step is to create a conversion chart like the one shown below. Step #2: Create Decimal to Octal Conversion Chart ← swipe left and right → A Power of 8:8 3 8 2 8 1 8 0 B Remainder of Division:125 C Place value(A result):512 64 8 1 D Octal digit (B÷C): Notice that the decimal you want to convert is placed in the left-most cell of Row B, just above the base 8 place value row (C). Step #3: Complete Conversion Chart for Converting 125 10 to Octal Next, attempt to divide the amount in row B into the amount in row C. If the amount in row C is greater than the amount in row B, enter a "0" in row D and move the amount in row B one cell to the right. Otherwise, if the amount in row C is less than the amount in row B, enter the number of times row C goes into row B in row D and enter the remainder in the next open cell in row B. Then simply repeat this process for each subsequent column. Here is how the completed conversion chart would look: ← swipe left and right →← swipe left and right → A Power of 8:8 3 8 2 8 1 8 0 B Remainder of Division:125 125 61 5 C Place value(A result):512 64 8 1 D Octal digit (B÷C):0 1 7 5 Step #4: Combine Numbers In Last Row of Chart Combining the numbers in the last row of the chart we can see that the base 10 number 125 converts to the base 8 number 175 (164 + 78 + 51 = 125). Note that the leading zeros are dropped since they represent no value (just like the base 10 system). As you can see, converting a decimal number to an octal number is a simple process of identifying the first base 8 place value greater than or equal to the base 10 number you are converting, and then dividing each place value into the remainder of previous division. Back to Calculator Back to Calculator Back to Decimal to Octal Converter Back up to Decimal to Octal Converter Free Online Calculators » 2. 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1789	https://study.com/skill/learn/how-to-find-the-range-of-a-quadratic-function-explanation.html	How to Find the Range of a Quadratic Function \| Algebra \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up How to Find the Range of a Quadratic Function Algebra 2 Skills Practice Click for sound 3:14 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:05 introduction Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Robert Vittoe, Amy McKenney Instructors Robert Vittoe Robert has taught high school chemistry, college astronomy, physical science, and physics. He has an MS in Space Studies/Aerospace Science from APU, an MS in Education from IU, and a BS in Physics from Purdue. He holds a Missouri educator licenses for chemistry and physics. View bio Amy McKenney Amy has taught high school mathematics for over 14 years. She has a master's degree in education from Plymouth State University and her undergraduate degree in mathematics. She is certified to teach grades 7-12 mathematics. View bio Example SolutionsPractice Questions Learn How to Find the Range of a Quadratic Function in Vertex Form In this lesson, we are going to learn how to determine the range of a quadratic function that is in vertex form. There are three steps we must go through when solving for the distance equation. Step 1: Determine if the function has a maximum or a minimum. Step 2: Determine the vertex of the function and identify the y coordinate. Step 3: Write the answer in set notation. Learn How to Find the Range of a Quadratic Function in Vertex Form Equations and Vocabulary The range of a function is a set of values that are the output of a function. Real numbers include both rational and irrational numbers. A maximum is a point where a function's value is greater than all the other values of the function. A minimum is a point where a function's value is less than all the other values of the function. The vertex of a parabola is the point where the parabola and its line of symmetry meet. It will be located at the maximum or minimum of the parabola. Example Problem 1: Finding the Range of a Quadratic Function in Vertex Form We will use the following quadratic equation for our first example. f(x)=−2(x−4)2+3 Step 1: Determine if the function has a maximum or a minimum. To determine if we are looking for a maximum or minimum, we look to see if the a value of our quadratic equation is positive or negative. The a is the coefficient of the (x - h) squared term. y=a(x−h)2+k If a<0 we are looking for a maximum. The graph of the function opens down If a>0 we are looking for a minimum. The graph of the function opens up Examining our quadratic equation, we can determine the a coefficient. y=a(x−h)2+k y=−2(x−4)2+3 We can see that a is equal to -2, which is less than zero. Therefore, we will be looking for a maximum, and the graph of the function will open down. Step 2: Determine the vertex of the function and identify the y coordinate. By examining our quadratic equation again, we can determine the vertex. y=a(x−h)2+k y=−2(x−4)2+3 vertex=(h,k)=(4,3) This tells us that the maximum value of the function is located at the vertex (4, 3). The y value of our vertex is 3. This is the maximum value of the output for our quadratic function. All the other values in the range will be less than or equal to 3. Step 3: Write the answer in set notation. In set notation, our range is: R:{y\|y⩽3,y∈R} Example Problem 2: Finding the Range of a Quadratic Function in Vertex Form We will use the following quadratic equation for our first example. f(x)=5(x−7)2+5 Step 1: Determine if the function has a maximum or a minimum. To determine if we are looking for a maximum or minimum, we look to see if the a value of our quadratic equation is positive or negative. The a is the coefficient of the (x - h) squared term. y=a(x−h)2+k If a<0 we are looking for a maximum. The graph of the function opens down If a>0 we are looking for a minimum. The graph of the function opens up Examining our quadratic equation, we can determine the a coefficient. y=5(x−7)2+5 We can see that a is equal to 5, which is greater than zero. Therefore, we will be looking for a minimum and the graph of the function will open up. Step 2: Determine the vertex of the function and identify the y coordinate. By examining our quadratic equation again, we can determine the vertex. y=5(x−7)2+5 vertex=(7,5) This tells us that the minimum value of the function is located at the vertex (7, 5). The y value of our vertex is 5. This is the minimum value of the output for our quadratic function. All the other values in the range will be greater than or equal to 5. Step 3: Write the answer in set notation. In set notation, our range is: R:{y\|y⩾5,y∈R} Get access to thousands of practice questions and explanations! Create an account Table of Contents Learn in Vertex Form Learn in Vertex Form Equations and Vocabulary Example Problem 1 Example Problem 2 Test your current knowledge Practice Finding the Range of a Quadratic Function Related Courses Algebra I: High School Remedial Algebra I Math 104: Calculus Math 101: College Algebra High School Algebra II: Tutoring Solution Related Lessons Rational Function \| Formula, Properties & Examples Exponentials, Logarithms & the Natural Log Inverse Functions \| Definition, Methods & Calculation The Quadratic Formula: Definition & Example Step Function \| Definition, Equation & Graph Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources Solving Problems Involving Systems of Equations Ethnic Groups in Indonesia \| Demographics & People Gods of the Winter Solstice Holes by Louis Sachar \| Themes, Quotes & Analysis Aurangzeb \| Empire, Achievements & Failures Libya Ethnic Groups \| Demographics, Population & Cultures Cold War Lesson for Kids: Facts & Timeline The Chronicles of Narnia Series by C.S. 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1790	https://www.gffbrokers.com/speculators-vs-hedgers-a-simple-guide/	The Global Update Speculators vs Hedgers – A Simple Guide Just about every type of futures account application will ask whether the account holder will be hedging or speculating. Chances are that if the customer is wondering which box to check off, they are almost certainly a speculator. Even if the trader is planning to be use futures contracts to hedge a position, or somehow planning on adopting a hedge-based trading strategy, they most likely still do not qualify as a hedger. This can be confusing, as there’s nothing explaining the context of the question. Speculating and hedging are two types of trading roles that deal with futures contracts. Beyond this, the motives and strategies are quite different. So, let’s break it down in plain language. SPECULATORS –Who are they? Think of speculators as the risk-takers of the futures market. They don’t want the physical commodity itself. Instead, they just care about betting on the price movements. What’s their objective in the market? Speculators aim to profit from price fluctuations. They’ll buy a futures contract if they think the price will rise, and sell if they believe it will fall. How will they deal with market risk? Speculators can only generate returns by taking market risks. So, in a way, they seek risk in order to take advantage of market opportunities. For example: Let’s imagine you’re a speculator who believes the price of gold will rise in the next three months. You buy a futures contract. If the price of gold does increase in that time frame, you can sell the contract for a profit. If it decreases, you’ll face a loss. HEDGERS – Who are they? Hedgers are the safety-seekers of the futures market. They’re typically the “commercial” participants who grow, produce, profit, or make use of the actual commodity. What’s their objective in the market? Hedgers use futures to “lock-in” prices. This way, they can protect themselves against unfavorable price fluctuations. In other words, their main goal isn’t to make a profit (like speculators) but rather, to manage their risk. How will they deal with market risk? Hedgers don’t like price risk. They’re using futures to stabilize the cost of the commodity that is central to their business. That’s why they’re main goal is to lock-in a commodity’s price. For example: Let’s say you’re a farmer who grows wheat. If you’re worried that the price of wheat might drop by the time you harvest, you can sell a futures contract now at the current price. This way, even if the market price drops in the future, meaning, you’ll be selling your wheat at a lower cost, the “profit” you make from your futures will compensate for your loss in selling the physical product. A quick and simple comparison Speculators are risk-takers looking to make a profit from price fluctuations. Hedgers are safety-seekers aiming to lock in prices to protect their businesses from unexpected price changes. The bottom line Speculators and hedgers both use futures, but their strategies and objectives differ greatly. For more information on common futures trading terms, Click Here to View the GFF Brokers Futures Glossary. Please be aware that the content of this blog is based upon the opinions and research of GFF Brokers and its staff and should not be treated as trade recommendations. There is a substantial risk of loss in trading futures, options and forex. Past performance is not necessarily indicative of future results. We use cookies to improve the performance of our site, to analyze the traffic to our site, and to personalize your experience of the site. You can control cookies through your browser settings. Please find more information on the cookies used on our site here.ACCEPTDECLINECookies Policy
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1792	https://www.sciencedirect.com/science/article/abs/pii/S0083672905720045	Skip to article My account Sign in Access through your organization Purchase PDF Patient Access Article preview Abstract Introduction Section snippets References (201) Cited by (45) Vitamins & Hormones Volume 72, 2005, Pages 111-154 Plant Peroxisomes Author links open overlay panel, rights and content Peroxisomes, one of single membranebound organelles, are present ubiquitously in eukaryotic cells. They were originally identified as organelles for production of hydrogen peroxide, the degradation of its hydrogen peroxide, and metabolism of fatty acids, which are functions common to almost all the organisms. Meanwhile, photorespiration and assimilation of symbiotically induced nitrogen are plantspecific functions. Recent postgenetic approaches such as transcriptome and proteome showed that plant peroxisomes are differentiated in various tissues, and revealed that peroxisomes have more important roles in various metabolic processes including biosynthesis of plant hormones than we speculated. All peroxisomal proteins, including metabolic enzymes in the matrix, membrane proteins, and factors responsible for peroxisome biogenesis, are nuclear encoded, and are provided from the outside of peroxisomes. Peroxisome biogenesis, such as protein transport, division, and enlargement, requires various complicated steps and is one of the most intriguing topics. Analyses using peroxisome biogenesis mutants and the wholescale sequencing projects among several organisms revealed the existence of essential factors responsible for peroxisome biogenesis such as peroxins. This review addresses a comprehensive issue relating to function and biogenesis of plant peroxisomes and Arabidopsis mutants that have been accelerating our understanding of peroxisomes in planta. Introduction Peroxisomes are single membranebound organelles that are found ubiquitously in eukaryotic cells including mammalian, yeasts, and plant cells. Peroxisomes are involved in various metabolic reactions in cells, such as fatty acid Î²oxidation, photorespiration, scavenging of hydrogen peroxide, and biosynthesis of plant hormones, jasmonic acid (JA), and auxin. However, the functions differ with the species, developmental stage, and organ, even in the same organism. Based on their functions, plant peroxisomes are subcategorized into three groups, glyoxysomes, leaf peroxisomes, and unspecialized peroxisomes. Interestingly, each peroxisome can transform directly to another type of peroxisome. In addition to cDNA cloning and characterization of peroxisomal genes, genetic analyses using mutants with defects in peroxisome function of various organisms have led to understanding the importance of correct maintenance of peroxisomes within cells. For example, the functional consequence of human peroxisome biogenesis disorders causes diseases such as Zellweger syndrome, neonatal adrenoleukodystrophy, infantile Refsum disease, and rhizomelic chondrodysplasia in mammals. Peroxisomedeficient mutants of yeasts are specifically unable to grow on fatty acids, methanol, or oleate as a sole carbon source. In higher plants, Arabidopsis mutants with defects in peroxisomal function or peroxisome biogenesis are unable to grow normally such as being unable to germinate or being sterile. In this review, we briefly summarize the discovery, structure, and function of plant peroxisome, then review our current knowledge about peroxisome biogenesis and how they are regulated, and finally introduce the results that make use of the complete sequences of the Arabidopsis genome. Section snippets Discovery, Structure, and Localization in Cells In the early 1960s, peroxisomes were discovered as distinctive organelles about 0.21.5 Î¼m in diameter, bounded by a single membrane, and with an amorphous or granular interior (Fig. 1). These organelles were initially given the name microbody based on the observation of animal tissues (Rhodin, 1958). Therefore, some textbooks use the word microbody. However, they are now mainly called peroxisomes based on one of their important functions, peroxidase activity. In the early stage of Peroxisome Functions in Higher Plants Peroxisomes have a variety of functions, such as lipid metabolism, photorespiration, scavenging of hydrogen peroxide, biosynthesis of plant hormones, catabolism of branched chain amino acids, assimilation of symbiotically induced nitrogen, and so on. Among them, there are speciesspecific peroxisomal functions such as synthesis of cholesterol, bile acids, and plasmalogens in mammals; methanol oxidation in yeasts; or photorespiration, the biosynthesis of plant hormones, and the metabolism of Transition of Peroxisome Function Functional transition of peroxisomes is the most interesting phenomenon in higher plants. Glyoxysomes, which have a role in lipid metabolism in etiolated tissues of germinating seeds, are transformed directly into leaf peroxisomes that are involved in photorespiration in photosynthetic tissues (Nishimura 1986, Titus 1985). In the process of the transition from glyoxysomes to leaf peroxisomes, activities of glyoxysomal enzymes such as malate synthase and isocitrate lyase decreased, whereas Peroxisome Biogenesis Since peroxisomes do not possess a genome and proteinsynthesizing machinery, all the proteins constituting peroxisomes are nuclear encoded, synthesized in the cytosol, and then imported to peroxisomes. Several pathways of protein transport to peroxisomes have been reported in higher plants (Johnson 2001, Mullen 2001), as in mammals and yeasts. Peroxisomes proliferate by division of preexisting peroxisomes (Miyagishima et al., 1999). The analysis using peroxisomedefective mutants as described Challenge of Comprehensive Approach in the Postgenomic Era The largescale sequencing projects of the whole plant genome, such as Arabidopsis (Arabidopsis Genome Initiative, 2000) and rice (Goff 2002, Yu 2002), and the large number of deposits of EST clones in the public databases provide us an enormous amount of biological information. Moreover, the pools of transgenic Arabidopsis plants, which contain transposonable elements or TDNA insertion in their genome, have been rapidly expanding so that it is easier to identify gene disruption mutants. By Few Final Considerations and Perspectives Considerable progress has been made during the past decades in the research on plant peroxisomes. The peroxisome has emerged as an essential organelle with a variety of functions in the cell. These various peroxisomal functions are indispensable for maintenance of normal plant growth. However, much remains to be elucidated, although not described here because of limited space. An example is our limited knowledge on the origin of peroxisomes. It is apparent that mitochondria and plastids evolved References (201) AlbertiniM. et al. ### Pex14p, a peroxisomal membrane protein binding both receptors of the two PTSdependent import pathways ### Cell (1997) BrocardC. et al. ### The tetratricopeptide repeatdomain of the PAS10 protein of Saccharomyces cerevisiae is essential for binding the peroxisomal targeting signal SKL ### Biochim. Biophys. Acta (1994) CooperT.G. et al. ### Î²Oxidation in glyoxysomes from castor bean endosperm ### J. Biol. Chem. (1969) CornahJ.E. et al. ### Lipid utilization, gluconeogenesis and seedling growth in Arabidopsis mutants lacking the glyoxylate cycle enzyme malate synthase ### J. Biol. Chem. (2004) CorpasF.J. et al. ### Peroxisomes as a source of reactive oxygen species and nitric oxide signal molecules in plant cells ### Trends Plant Sci. 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(2002) - FukaoY. et al. ### Developmental analysis of a putative ATP/ADP carrier protein localized on glyoxysomal membranes during the peroxisome transition in pumpkin cotyledons ### Plant Cell Physiol. (2001) Cited by (45) Abiotic environmental stress induced changes in the Arabidopsis thaliana chloroplast, mitochondria and peroxisome proteomes 2009, Journal of Proteomics Exposure to adverse abiotic environmental conditions causes oxidative stress in plants, leading to debilitation and death or to response and tolerance. The subcellular energy organelles (chloroplast, mitochondria and peroxisomes) in plants are responsible for major metabolic processes including photosynthesis, photorespiration, oxidative phosphorylation, Î²-oxidation and the tricarboxylic acid cycle. Here we analyze data and review a collection of both whole tissue and organellar proteomic studies that have investigated the effects of environmental stress in the model plant Arabidopsis thaliana. We assess these data from an organellar perspective to begin to build an understanding of the changes in protein abundance within these organelles during environmental stresses. We found 279 claims of proteins that change in abundance that could be assigned to protein components of the energy organelles. These could be placed into eight different functional categories and nearly 80% of the specific protein isoforms detected were only reported to change in a single environmental stress. We propose primary and secondary mechanisms in organelles by which the protein changes observed could be mediated in order to begin developing an integrated and mechanistic understanding of environmental stress response. ### Plant peroxisomes as a source of signalling molecules 2006, Biochimica Et Biophysica Acta Molecular Cell Research Peroxisomes are pleoimorphic, metabolically plastic organelles. Their essentially oxidative function led to the adoption of the name peroxisome. The dynamic and diverse nature of peroxisome metabolism has led to the realisation that peroxisomes are an important source of signalling molecules that can function to integrate cellular activity and multicellular development. In plants defence against predators and a hostile environment is of necessity a metabolic and developmental responsea plant has no place to hide. Mutant screens are implicating peroxisomes in disease resistance and signalling in response to light. Characterisation of mutants disrupted in peroxisomal Î²-oxidation has led to a growing appreciation of the importance of this pathway in the production of jasmonic acid, conversion of indole butyric acid to indole acetic acid and possibly in the production of other signalling molecules. Likewise the role of peroxisomes in the production and detoxification of reactive oxygen, and possibly reactive nitrogen species and changes in redox status, suggests considerable scope for peroxisomes to contribute to perception and response to a wide range of biotic and abiotic stresses. Whereas the peroxisome is the sole site of Î²-oxidation in plants, the production and detoxification of ROS in many cell compartments makes the specific contribution of the peroxisome much more difficult to establish. However progress in identifying peroxisome specific isoforms of enzymes associated with ROS metabolism should allow a more definitive assessment of these contributions in the future. ### Redox homeostasis via gene families of ascorbate-glutathione pathway 2015, Frontiers in Environmental Science ### Plant organelle proteomics: Collaborating for optimal cell function 2011, Mass Spectrometry Reviews ### Peroxisomal malate dehydrogenase is not essential for photorespiration in arabidopsis but its absence causes an increase in the stoichiometry of photorespiratory CO2 release 2008, Plant Physiology ### Arabidopsis peroxin11c-e, fission1b, and dynamin-related protein3A cooperate in cell cycle-associated replication of peroxisomes 2008, Plant Cell View all citing articles on Scopus View full text Copyright © 2005 Elsevier Inc. All rights reserved.
1793	https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:radians/e/degrees_to_radians	Radians & degrees (practice) \| Trigonometry \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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1794	https://math.colgate.edu/~integers/y34/y34.pdf	A34 INTEGERS 24 (2024) EFFECTIVE ESTIMATES FOR SOME FUNCTIONS DEFINED OVER PRIMES Christian Axler Heinrich Heine University D¨ usseldorf, Faculty of Mathematics and Natural Sciences, Mathematical Institut, D¨ usseldorf, Germany christian.axler@hhu.de Received: 5/12/22, Revised: 7/21/23, Accepted: 3/15/24, Published: 4/8/24 Abstract In this paper we give eﬀective estimates for some classical arithmetic functions deﬁned over prime numbers. First we ﬁnd the smallest real number x0 so that some inequality involving Chebyshev’s ϑ-function holds for every x ≥x0. Then we give some new results concerning the existence of prime numbers in short intervals. Also we derive new upper and lower bounds for some functions deﬁned over prime numbers, for instance the prime counting function π(x), which improve current best estimates of similar shape. 1. Introduction First, we consider Chebyshev’s ϑ-function ϑ(x) = P p≤x log p, where p runs over all primes not exceeding x. Since there are inﬁnitely many primes, we have ϑ(x) →∞ as x →∞. Hadamard and de la Vall´ ee-Poussin independently proved a result concerning the asymptotic behavior for ϑ(x), namely ϑ(x) ∼x (x →∞), (1) which is known as the Prime Number Theorem. In a later paper , where the existence of a zero-free region for the Riemann zeta function to the left of the line Re(s) = 1 was proved, de la Vall´ ee-Poussin also estimated the error term in the Prime Number Theorem by showing that ϑ(x) = x + O(xe−c0 √log x) (x →∞), (2) where c0 is a positive absolute constant. The current best explicit version of this result is due to Fiori, Kadiri, and Swidinsky [34, Corollary 14]. They found that \|ϑ(x) −x\| ≤121.0961 log x R 3/2 exp −2 r log x R ! (3) DOI: 10.5281/zenodo.10943995 INTEGERS: 24 (2024) 2 for every x ≥2, where R = 5.5666305. The work of Korobov and Vinogradov implies the current asymptotically strongest error term in (1), namely ϑ(x) = x + O x exp −c1 log3/5 x(log log x)−1/5 (x →∞), (4) where c1 is a positive absolute constant. An explicit version of (4) was recently given by Johnston and Yang [40, Theorem 1.4]. Now, (2)–(4) each imply that for every positive integer k and every positive real number ηk there is real number x1 = x1(k, ηk) > 1 so that for every x ≥x1, we have \|ϑ(x) −x\| < ηkx logk x . (5) In the case where k = 3 and η3 = 0.024334, Broadbent et al. found that \|ϑ(x) −x\| < 0.024334x log3 x (x ≥e29). (6) In our ﬁrst result, we compute the smallest positive integer N so that (6) holds for every x ≥N. Proposition 1. The inequality (6) holds for every x ≥1, 757, 126, 630, 797 = p64,707,865,143. Estimates for ϑ(x) of the form (5) can be used to specify short intervals containing at least one prime number. Here, we ﬁnd the following result. Theorem 1. Let a = (1.42969 × 1012 −1)−1 and b = (1.59753 × 1012 −1)−1. Further, let n be a positive integer with 1 ≤n ≤5. Then there is a prime number p such that x < p ≤x 1 + an logn x for every x ≥Xn, where an and Xn are given as in Table 1. n an Xn 1 43a = 3.00 . . . × 10−11 952, 527, 672, 606, 693 2 46nb = 1.32 . . . × 10−9 684, 943, 746, 324, 434 3 46nb = 6.09 . . . × 10−8 543, 684, 371, 469, 081 4 46nb = 2.802 . . . × 10−6 336, 149, 866, 771, 577 5 46nb = 1.289 . . . × 10−4 246, 782, 656, 239, 427 Table 1: Explicit values for an and Xn. INTEGERS: 24 (2024) 3 Remark 1. Note that the values of Xn, where 1 ≤n ≤5, are the smallest positive integers so that there is always a prime number in the interval (x, x(1+an/ logn x)]. For n ≥6, we are only able to ﬁnd X(1) n and X(2) n , so that analogous results are valid for all x ∈R with X(1) n ≤x ≤X(2) n . To prove these results for every x > X(2) n as well, we would need estimates of the form ϑ(x) > x −ηnx/ logn x with n ≥6. Let π(x) denote the number of primes not exceeding x. Chebyshev’s ϑ-function and the prime counting function π(x) are connected by the identity π(x) = ϑ(x) log x + Z x 2 ϑ(t) t log2 t dt, (7) which holds for every x ≥2 (see [2, Theorem 4.3]). If we combine (3) and (7), we see that π(x) = li(x) + O(xe−c2 √log x) (x →∞), (8) where c2 is a positive absolute constant. Here, the integral logarithm li(x) is deﬁned for every x ≥0 as li(x) = Z x 0 dt log t = lim ε→0+ Z 1−ε 0 dt log t + Z x 1+ε dt log t and plays an important role in this paper. The current best explicit version of (8) is due to Johnston and Yang [40, Corollary 1.3]. Again, the work of Korobov and Vinogradov implies the current asymptotically strongest error term for the diﬀerence π(x) −li(x), namely π(x) = li(x) + O x exp −c3(log x)3/5(log log x)−1/5 (x →∞), (9) where c3 is a positive absolute constant. Ford [36, p. 2] has found that the constant c3 in (9) can be chosen to be equal to 0.2098. Johnston and Yang [40, Theorem 1.4] used explicit zero-free regions and zero-density estimates for the Riemann zeta-function to show that the inequality \|π(x) −li(x)\| ≤0.028x(log x)0.801 exp −0.1853(log x)3/5(log log x)−1/5 (10) holds for every x ≥71. Panaitopol [53, p. 55] gave another completely diﬀerent asymptotic formula for the prime counting function by showing that for every pos-itive integer m, one has π(x) = x log x −1 − k1 log x − k2 log2 x −. . . − km logm x + O x logm+2 x (x →∞), (11) where the positive integers k1, . . . , km are deﬁned by the recurrence formula km + 1!km−1 + 2!km−2 + . . . + (m −1)!k1 = m · m!. INTEGERS: 24 (2024) 4 For instance, we have k1 = 1, k2 = 3, k3 = 13, k4 = 71, k5 = 461, and k6 = 3441. The computation of the prime counting function π(x) for large values of x is a diﬃcult problem (the latest record is due to Baugh and Walisch and was π(1028) = 157, 589, 269, 275, 973, 410, 412, 739, 598). Also the asymptotic formula (8) (or (11)) is not very meaningful with regard to the computation of π(x) for some ﬁxed x. Hence we are interested in ﬁnding new eﬀective estimates for the prime counting function π(x) which correspond to the ﬁrst terms of (11). For instance, those estimates for the prime counting function are used to get eﬀective estimates for 1/π(x) (see ) or the nth prime number (see ). In this paper, we use Proposition 1 to establish the following upper bound for π(x) which corresponds to the ﬁrst terms of the asymptotic formula (11). Theorem 2. Let a5 = 461.364417856444 and a6 = 4331.1. Then for every x ≥48, we have π(x) < x log x −1 − 1 log x −3.024334 log2 x −12.975666 log3 x −71.048668 log4 x − a5 log5 x − a6 log6 x . (12) For all suﬃciently large values of x, Theorem 2 is a consequence of (10). On the other hand, we get the following lower bound for the π(x) which corresponds to the ﬁrst terms of (11). Theorem 3. Let b5 = 460.634397856444 and b6 = 3444.031844143556. Then for every x ≥1, 751, 189, 194, 177 = p64,497,259,289, we have π(x) > x log x −1 − 1 log x −2.975666 log2 x −13.024334 log3 x −70.951332 log4 x − b5 log5 x − b6 log6 x . (13) Again, for all suﬃciently large values of x, Theorem 3 follows directly from (10). The asymptotic expansion (11) implies that the slightly sharper inequality π(x) > x log x −1 − 1 log x − 3 log2 x (14) holds for all suﬃciently large values of x. In [6, Theorem 1], the present author was able to prove that the inequality (14) holds for every x such that 65 405 887 ≤x ≤ 2.7358·1040 and every x ≥4.8447·1019377. Under the assumption that the Riemann hypothesis is true, the present author[6, Proposition 2] showed that the inequality (14) holds for every x ≥65, 405, 887. In the following theorem we ﬁnally see that the inequality (14) holds for every x ≥65, 405, 887 even without the assumption that the Riemann hypothesis is true. Theorem 4. The inequality (14) holds unconditionally for every x ≥65, 405, 887. Our next goal is to establish new explicit estimates for the functions X p≤x 1 p and X p≤x log p p , INTEGERS: 24 (2024) 5 where p runs over primes not exceeding x, respectively. Euler proved that the sum of the reciprocals of all prime numbers diverges. Mertens [49, p. 52] found that log log x is the right order of magnitude for this sum by showing X p≤x 1 p = log log x + B + O 1 log x . (15) Here B denotes the Mertens’ constant and is deﬁned by B = γ + X p log 1 −1 p + 1 p = 0.26149 . . . , (16) where γ = 0.577215 . . . denotes the Euler-Mascheroni constant. In Section 6, we apply Proposition 1 to some identity obtained by Rosser and Schoenfeld and derive the following result which improves all other results of this form. Theorem 5. For every x ≥1, 757, 126, 630, 797, we have X p≤x 1 p −log log x −B ≤0.024334 3 log3 x 1 + 15 4 log x . (17) In 1874, Mertens showed that X p≤x log p p = log x + O(1). (18) Landau [45, §55] improved (18) by ﬁnding X p≤x log p p = log x + E + O(exp(− 14 p log x)), where E is a constant deﬁned by E = −γ − X p log p p(p −1) = −1.3325 . . . . (19) Similar to Theorem 5, we establish the following explicit estimates for P p≤x log(p)/p which improve [5, Proposition 8]. Theorem 6. For every x ≥1, 757, 126, 630, 797, we have X p≤x log p p −log x −E ≤0.024334 2 log2 x 1 + 2 log x . Remark 2. Note that the positive integer N0 = 1, 757, 126, 630, 797 in Theorem 6 might not be the smallest positive integer N so that the inequality given in Theorem 6 holds for every x ≥N. INTEGERS: 24 (2024) 6 2. Proof of Proposition 1 In the following proof of Proposition 1, we ﬁrst utilize an identity investigated by Rosser and Schoenfeld to express Chebyshev’s ϑ-function in terms of the diﬀerence π(x) −li(x). Then we apply Walisch’s primecount C++ code to ﬁnd a lower bound for π(x) −li(x) in a certain restricted interval. Proof of Proposition 1. By (6) and [12, Corollary 11.1], it suﬃces to check that the inequality ϑ(x) > x −0.024334x log3 x (20) holds for every x satisfying 1, 757, 126, 630, 797 ≤x ≤e29. Using [63, (2.26)] with f(x) = log x, we get ϑ(x) = x −2 + li(2) log 2 + (π(x) −li(x)) log x − Z x 2 π(t) −li(t) t dt (21) for every x ≥2. Now we can use [54, Corollary 1] to see that −2 + li(2) log 2 − Z x 2 π(t) −li(t) t dt ≥−2 + li(2) log 2 − Z 9 2 π(t) −li(t) t dt ≥0.129 (22) for every x with 9 ≤x ≤e29. Applying (23) to (21), we get ϑ(x) > x + (π(x) −li(x)) log x (23) for every x so that 9 ≤x ≤e29. Now we use Walisch’s primecount C++ code to get π(x) −li(x) ≥−0.024334x log4 x (24) for every x with 1, 760, 505, 892, 241 ≤x ≤2, 342, 911, 050, 819 and every x with 2, 346, 094, 807, 193 ≤x ≤4 × 1012. If we combine (24) with (23), we get (20) for every x satisfying 1, 760, 505, 892, 241 ≤x ≤2, 342, 911, 050, 819 and every x with 2, 346, 094, 807, 193 ≤x ≤e29 ≤4 × 1012. In order to verify the re-quired inequality (20) in the case where x satisﬁes 1, 757, 126, 630, 797 ≤x < 1, 760, 505, 892, 241, we can check with a computer that ϑ(pn) > g(pn+1) for every integer n such that π(1, 757, 126, 630, 797) ≤n ≤π(1, 760, 505, 892, 241). Finally, a direct computer check shows that the inequality (20) also holds for every x such that 2, 342, 911, 050, 819 ≤x ≤2, 346, 094, 807, 193. The present author [5, Theorem 1, Proposition 1, and Equations (4.4) and (4.5)] INTEGERS: 24 (2024) 7 utilized [30, Table 1 and Corollary 4.5] to show that \|ϑ(x) −x\| < 0.043x log3 x (x ≥e40), (25) \|ϑ(x) −x\| < 0.15x log3 x (e35 ≤x < e5000), (26) \|ϑ(x) −x\| < 99.07x log4 x (x ≥e25), (27) \|ϑ(x) −x\| < 100x log4 x (x ≥70, 111). (28) Broadbent et al. [12, p. 2299] pointed out that the main theorem of is incor-rect and thus bounds claimed in are likely aﬀected, in particular [30, Table 1] for bounds for ψ(x), and consequently the inequalities (25)–(28). Except for the corresponding line for the value b = 2500, all other explicit values in [30, Table 1] were conﬁrmed and even improved by Broadbent et al. [12, Table 8] while the corresponding line for the value b = 2500 was recently conﬁrmed and even improved by Fiori, Kadiri, and Swidinsky [33, Table 5]. Hence, we can recover [30, Table 1]. Proposition 2. The explicit values for ε given in [30, Table 1] are correct. To show that the inequalities (25)–(28) still hold, it suﬃces to note that [30, Proposition 4.4] combined with [12, Proposition 4] yield the correctness of [30, Corollary 4.5]. Hence, we get Proposition 3. The inequalities (25)–(28) for Chebyshev’s ϑ-function are correct. Remark 3. Note that Proposition 1 already provides the correctness of the in-equalities (25) and (26). Remark 4. To ﬁnd other explicit estimates for ϑ(x) in the restricted interval [2, 1020], one can also apply the method used by Dusart in . Let π0(x) = lim ε→0 π(x −ε) + π(x + ε) 2 = ( π(x) −1/2, if x is prime, π(x), otherwise. Riemann published the formula π0(x) = ∞ X n=1 µ(n) n f(x1/n), (29) where µ(n) is the M¨ obius function, and f(x) is the Riemann prime counting function f(x) = li(x) − X ρ li(xρ) + Z ∞ x dt t(t2 −1) log t −log 2. INTEGERS: 24 (2024) 8 Here the sum means limT →∞ P \|ρ\|≤T li(xρ), and the ρ’s are the nontrivial zeros of the Riemann zeta function. A ﬁrst proof of (29) was given by von Mangoldt in 1895. Now let R(x) = ∞ X n=1 µ(n) n li(x1/n) = 1 + ∞ X k=1 logk x k!kζ(k + 1). (30) The latter series for it is known as Gram series. Since log x < x for every real x > 0, this series converges for all positive x by comparison with the series for ex. In , Riesel and G¨ ohl showed that the function g(x) = R(x) − 1 log x + 1 π arctan π log x is a quite good approximation to π0(x). The diﬀerence between g(x) and π0(x) heuristically oscillates with an amplitude of about √x/ log x. So we deﬁne ∆(x) = π0(x) −R(x) + 1 log x −1 π arctan π log x log x √x , (31) the function which represents the ﬂuctuations of the distribution of primes. We can use (30) and (31) to get π(x) −li(x) ≤1 2 + f2(x) + √x log x × ∆(x) − 1 log x + 1 π arctan π log x, (32) where fk(x) = ∞ X n=k µ(n) n li(x1/n). Since µ(4) = 0 and f5(x) is strictly decreasing on (1, ∞), the inequality (32) implies that π(x) −li(x) ≤−li(√x) 2 −li(x1/3) 3 + √x log x × ∆(x) (33) for every x ≥2, 000. Similarly, we see that π(x) −li(x) ≥ 5 X n=2 µ(n) n li(x1/n) + √x log x × ∆(x) (34) for every x ≥10, 326. Applying (33) and (34) to (21), we get ϑ(x) > x + (∆(x) −1)√x − max 2000≤t≤x ∆(t) × li(√x) − 3 √x −li( 5 √x) log x 5 + c1 for every x ≥10, 326, where c1 is a constant. Analogously, we see that the inequality ϑ(x) < x + (∆(x) −1)√x − min 10,236≤t≤x ∆(t) × li(√x) − 3 √x + li( 5 √x) log x 5 − 5 √x + c2 INTEGERS: 24 (2024) 9 holds for every x ≥10, 326, where c2 is a constant. Now one can use the exten-sive table of the minimum and maximum values of ∆(x) in to obtain explicit estimates for ϑ(x) in the restricted interval [2, 1020]. Remark 5. Under the assumption that the Riemann hypothesis is true, von Koch deduced the asymptotic formula ϑ(x) = x + O(√x log2 x). An explicit version was given by Schoenfeld [66, Theorem 10]. Under the assumption that the Riemann hypothesis is true, Schoenfeld has found that \|ϑ(x) −x\| < √x 8π log2 x (35) for every x ≥599. Recently, Schoenfeld’s result was slightly improved by Dusart [31, Proposition 2.5]. In 2016, B¨ uthe [14, Theorem 2] investigated a method to show that the inequality (35) holds unconditionally for every x such that 599 ≤x ≤ 1.4 × 1025. B¨ uthe’s result was improved by Platt and Trudgian [55, Corollary 1]. They proved that the inequality (35) holds unconditionally for every x satisfying 599 ≤x ≤2.169 × 1025. Recently, Johnston [39, Corollary 3.3] extended the last result by showing that the inequality (35) holds unconditionally for every x with 599 ≤x ≤1.101 × 1026. 3. Proof of Theorem 1 Bertrand’s postulate states that for each positive integer n there is a prime number p with n < p ≤2n. It was proved, for instance, by Chebyshev . In the following, we note some improvements of Bertrand’s postulate. The ﬁrst result is due to Schoenfeld [66, Theorem 12]. He discovered that for every x ≥2, 010, 759.9 there is a prime number p with x < p < x(1 + 1/16, 597). Ramar´ e and Saouter [58, Theorem 3] proved that for every x ≥10, 726, 905, 041 there is a prime number p so that x < p ≤x(1 + 1/28, 313, 999). Further, they gave a table of sharper results which hold for large x, see [58, Table 1]. Kadiri and Lumley [41, Table 2] obtained a series of improvements. For instance, they showed that for every x ≥4 × 1018 there is a prime number p such that x < p < x(1 + 1/36, 082, 898). Recently, Cully-Hugill and Lee [22, Theorem 1] improved the results of Kadiri and Lumley. In particular, they found that for every x ≥4×1018 there is a prime number p so that x < p ≤x(1+1/1, 429, 689, 999, 999). Dusart [28, Th´ eoreme 1] proved that for every x ≥3, 275 there exists a prime number p such that x < p ≤x(1 + 1/(2 log2 x)) and then reduced the interval himself [29, Proposition 6.8] by showing that for every x ≥396, 738 there is a prime number p satisfying x < p ≤x(1 + 1/(25 log2 x)). Trudgian [70, Corollary 2] proved that for every x ≥2, 898, 242 there exists a prime number p with x < p ≤x 1 + 1 111 log2 x . (36) INTEGERS: 24 (2024) 10 In [3, Theorem 1.26], it is shown that for every x ≥58, 833 there is a prime number p such that x < p ≤x 1 + 1.274 log3 x . This was improved in [4, Theorem 1.5] by showing that for every x ≥58, 837 there is a prime number p such that x < p ≤x(1 + 1.1817/ log3 x). Dusart [30, p. 243] used (the recovered) Table 1 of (cf. Proposition 2) to show the inequality \|ϑ(x) −x\| < 0.499x log3 x (x ≥4 × 1018). (37) Alternatively, the inequality (37) follows directly from Proposition 1. Then, he [30, Proposition 5.4] utilized the inequality (37) to see that for every x ≥89, 693 there exists a prime number p such that x < p ≤x 1 + 1 log3 x (38) and concluded from this that for every x ≥468, 991, 632 there exists a prime number p such that x < p ≤x 1 + 1 5, 000 log2 x (39) which improves Trudgian’s result (36). In [5, Theorem 4], the present author com-bined (39) and the (recovered) inequality (25) (cf. Proposition 3) to obtain that for every x ≥6, 034, 256 there exists a prime number p such that x < p ≤x 1 + 0.087 log3 x . (40) Further, the present author [5, Theorem 4] used the (recovered) inequality (27) (cf. Proposition 3) and (39) to ﬁnd found that for every x > 1 there is a prime number p with x < p ≤x 1 + 198.2 log4 x . (41) Now we give a proof of Theorem 1 where we give improvements of (38)–(41) by decreasing the coeﬃcient of the term 1/ logn x and on the other hand by increasing the exponent of the log x term. Proof of Theorem 1. In order to prove that there is a prime number p with x < p ≤ x(1+a1/ log x) for every x ≥X1 = 952, 527, 672, 606, 693, we ﬁrst consider the case where x ≥exp(4, 000). Here, we can use [12, Table 15] to get that ϑ x 1 + a1 log x −ϑ(x) > x log x a1 −2ε −a1ε log x ≥0, INTEGERS: 24 (2024) 11 where ε = 5.741 × 10−13, which implies that for every x ≥exp(4, 000) there is a prime number p satisfying x < p ≤x(1 + a1/ log x). For every x with 4 × 1018 ≤ x < exp(4, 000), the claim follows directly from [22, Theorem 1]. So it suﬃces to consider the case where 952, 527, 672, 606, 693 ≤x < 4 × 1018. Let n be an integer so that 1, 721, 649, 982, 233, 847 ≤pn ≤π(4 × 1018) and let x be a real number satisfying pn ≤x < pn+1. Then, we can utilize [51, Table 8] to see that pn+1 −pn ≤1, 476 < a1pn log pn ≤a1x log x. This implies that for every x with 1, 721, 649, 982, 233, 847 ≤x < 4 × 1018 there is a prime number in the interval (x, x(1+a1/ log x)]. Similar, we can see that for every x satisfying 1, 041, 648, 882, 338, 903 ≤x < 1, 721, 649, 982, 233, 847 there is always a prime in the interval (x, x(1 + a1/ log x)]. Next, we can use Walisch’s primesieve program to obtain that pn+1 −pn ≤860 < a1pn log pn ≤a1x log x for every integer n satisfying 9.88 × 1014 ≤pn < 1.042 × 1015 and every x with pn ≤x < pn+1. So there exists a prime number p with x < p ≤x(1 + a1/ log x) for every x so that 9.88 × 1014 ≤x ≤1, 041, 648, 882, 338, 903. If n is an integer with 9.53 × 1014 ≤pn < 9.88 × 1014 and x satisﬁes pn ≤x < pn+1, we can use Walisch’s primesieve program to see that pn+1 −pn ≤802 < a1pn/ log pn ≤a1x/ log x. This provides that for every x with 9.53 × 1014 ≤x < 9.88 × 1014 there is always a prime number in the interval (x, x(1 + a1/ log x)]. For every integer n satisfying 952, 527, 672, 607, 523 ≤pn < 9.53×1014 and every x with pn ≤x < pn+1, we apply Walisch’s primesieve program to obtain that pn+1 −pn ≤708 < a1pn/ log pn ≤ a1x/ log x and it turns out that for every x with 952, 527, 672, 607, 523 ≤x < 9.53 × 1014 there is a prime number in the interval (x, x(1 + a1/ log x)]. Finally, it suﬃces to consider the case where x belongs to the interval [a, b) where a = 952, 527, 672, 606, 693 and b = 952, 527, 672, 607, 523. In this situation, we have π(x(1 + a1/ log x)) −π(x) ≥1 as desired. The proof of the remaining assertions is similar to the above proof and we leave the details to the reader. Remark 6. Beginning with Hoheisel , many authors have found shorter inter-vals of the form [x −xδ, x] that must contain a prime number for all suﬃciently large values of x. The most recent result is due to Baker, Harman, and Pintz . They found the value δ = 0.525. Under the assumption that the Riemann hypoth-esis is true, much better results are known. For more details, see, for instance, Ramar´ e and Saouter , Dudek , Dudek, Greni´ e, and Molteni , Carneiro, Milinovich, and Soundararajan , Cully-Hugill and Dudek , Cully-Hugill and Johnston , and Cully-Hugill and Dudek . INTEGERS: 24 (2024) 12 4. Proof of Theorem 2 First, we note some well known estimates for the prime counting function π(x). A classic method of ﬁnding explicit estimates for π(x) is the following. Let k be a positive integer and ηk a positive real number. By (5), there is a real number x1 = x1(k, ηk) > 1 so that \|ϑ(x) −x\| < ηkx logk x for every x ≥x1. In order to prove Theorem 2, we deﬁne the auxiliary function Jk;ηk;x1(x) = π(x1) −ϑ(x1) log x1 + x log x + ηkx logk+1 x + Z x x1 1 log2 t + ηk logk+2 t dt (42) and note the following both inequalities involving the prime counting function π(x). Lemma 1. For every x ≥x1, we have Jk;−ηk;x1(x) ≤π(x) ≤Jk;ηk;x1(x). Proof. The claim follows directly form (7) and (5). One of the ﬁrst estimates for π(x) is due to Gauss. In 1793, he computed that π(x) ≤li(x) (43) holds for every x with 2 ≤x ≤3, 000, 000 and conjectured that the inequality (43) holds for every x ≥2. This conjecture was disproven by Littlewood . More precisely, he proved that the function π(x) −li(x) changes the sign inﬁnitely many times. Unfortunetely, Littlewood’s proof is nonconstructive and there is still no example of x such that π(x) > li(x). Skewes proved the existence of a number x0 with x0 < exp(exp(exp(exp(7.705)))) such that π(x0) > li(x0). Lehman improved this last upper bound considerably by showing that exists a number x0 with x0 < 1.65 × 101165 such that π(x0) > li(x0). After some further improvements (see, for instance, te Riele , Bays and Hudson , Chao and Plymen , Saouter and Demichel , Stoll and Demichel ), the current best upper bound was found by Saouter, Trudgian, and Demichel . They proved that there exists a number x0 with x0 < exp(727.951335621) such that π(x0) > li(x0). All these upper bounds have been proved by using computer calculations of zeros of the Riemann zeta function. The ﬁrst lower bound for a number x0 with π(x0) > li(x0) was given by the calculation of Gauss, namely x0 > 3, 000, 000. This lower bound was improved in a series of papers. For details, see Rosser and Schoenfeld , Brent , Kotnik , Platt and Trudgian , and Stoll and Demichel . For our further inverstigation, we use the following improvement. Lemma 2 (B¨ uthe ). For every x with 2 ≤x ≤1019, we have π(x) ≤li(x). Remark 7. Recently. Dusart [31, Lemma 2.2] showed that π(x) ≤li(x) for every x with 2 ≤x ≤1020. INTEGERS: 24 (2024) 13 Now we use Proposition 1 and the Lemmata 1 and 2 to give a proof of Theorem 2. Proof of Theorem 2. First, we combine Lemma 1 with Proposition 1 to see that J3;−0.024334;x1(x) ≤π(x) ≤J3;0.024334;x1(x) (44) for every x ≥x1, where x1 ≥1, 757, 126, 630, 797. Now, let x2 = 1018 and let f(x) be given by the right-hand side of (12). We consider the function g(x) = f(x) − J3,0.024334,x2(x). By , we have ϑ(x2) ≥999, 999, 999, 144, 115, 634. Further, π(x2) = 24, 739, 954, 287, 740, 860 and so we compute g(x2) ≥2 × 108. Since the derivative of g is positive for every x ≥x2, we get f(x) −J3,0.024334,x2(x) > 0 for every x ≥x1, and we conclude from (44) that the inequality (12) holds for every x ≥x1. Comparing f(x) with the integral logarithm li(x), we see that f(x) > li(x) for every x ≥121, 141, 948. Now we can utilize Lemma 2 to see that the desired inequality also holds for every x such that 121, 141, 948 ≤x < 1018. A computer check for smaller values of x completes the proof. Under the assumption that the Riemann hypothesis is true, von Koch de-duced that π(x) = li(x) + O(√x log x) as x →∞. Actually, one can show that the asymptotic formula π(x) = li(x)+O(√x log x) as x →∞is even a suﬃcient criterion for the truth of the Riemann hypothesis. An explicit version of von Koch’s result is due to Schoenfeld [66, Corollary 1]. Under the assumption that the Riemann hypothesis is true, Schoenfeld found that the inequality \|π(x) −li(x)\| < 1 8π √x log x (45) holds for every x ≥2, 657. In 2014, B¨ uthe [14, p. 2,495] proved that the inequality (45) holds unconditionally for every x such that 2, 657 ≤x ≤1.4 × 1025. Platt and Trudgian [55, Corollary 1] improved B¨ uthe’s result by showing that the inequality (45) holds unconditionally for every x satisfying 2, 657 ≤x ≤2.169×1025. Johnston [39, Corollary 3.3] extended the last result by showing the following Lemma 3 (Johnston). The inequality (45) holds unconditionally for every x sat-isfying 2, 657 ≤x ≤1.101 × 1026. Now we can use Theorem 2 and the Lemmata 2 and (3) to ﬁnd the following weaker but more compact upper bounds for the prime counting function π(x) of the form π(x) < x log x −a0 − a1 log x −· · · − am logm x (x ≥x0), where m is a integer with 0 ≤m ≤5 and a0, . . . , am are suitable positive real numbers. INTEGERS: 24 (2024) 14 Corollary 1. We have π(x) < x log x −a0 − a1 log x − a2 log2 x for every x ≥x0, where a0, a1, a2, and x0 are given as in Table 2. a0 1.0343 1 1 a1 0 1.109 1 a2 0 0 3.48 x0 106, 640, 139, 304, 611 81, 250, 795, 096, 339 145, 413, 088, 724, 077 Table 2: Explicit values for a0, a1, a2, and x0. Proof. Theorem 2 implies that the inequality π(x) < x log x −1.0343 (46) holds for every x ≥108, 943, 258, 198, 427. If we compare the right-hand side of (46) with li(x), we can use Lemma 2 to see that the required inequality (46) holds for every x with 106, 910, 668, 441, 596 ≤x ≤108, 943, 258, 198, 427. Finally, we use Walisch’s primecount program to obtain that the inequality (46) is also valid for every x satisfying 106, 640, 139, 304, 611 ≤x ≤106, 910, 668, 441, 596. The proof of each of the next three inequalities is similar to the proof of (46) and we leave the details to the reader. Next, we show that the inequality π(x) < x log x −1 − 1 log x −3.024334 log2 x −12.975666 log3 x −79.962 log4 x (47) holds for every x ≥22. First, we can use Theorem 2 to obtain the inequality (47) for every x ≥1.101 × 1026. Let f(x) denote the right-hand side of (46). We get that f(x) ≥li(x)+√x log(x)/(8π) for every x with 22, 066, 689, 219, 741, 110 ≤x ≤ 1.101 × 1026. Now we can apply Lemma 3 to see that the required inequality (47) also holds for every x satisfying 22, 066, 689, 219, 741, 110 ≤x ≤1.101 × 1026. A comparison with li(x) shows that f(x) > li(x) for every x ≥259, 576, 712, 645 and Lemma 2 yields the desired inequality (47) for every x with 259, 576, 712, 645 ≤ x ≤22, 066, 689, 219, 741, 110. Finally, it suﬃces to apply Walisch’s primecount program to see that the inequality (47) also holds for every x satisfying 22 ≤ x ≤259, 576, 712, 645. Again, the proof of the remaining inequality is similar to the proof of (47) and we leave the details to the reader. Remark 8. In the appendix at the end of this paper, we give lots of other weaker upper bounds in the case where m ∈{0, 1, 2}. INTEGERS: 24 (2024) 15 Corollary 2. We have π(x) < x log x −1 − 1 log x −3.024334 log2 x − a3 log3 x − a4 log4 x − a5 log5 x for every x ≥x0, where a3, a4, a5, and x0 are given as in Table 3. a3 14.893 12.975666 12.975666 a4 0 79.962 71.048668 a5 0 0 533.594 x0 142, 464, 507, 937, 911 22 32 Table 3: Explicit values for a3, a4, a5, and x0. Proof. Since the proof is similar to the proof of Corollary 1, we leave the details to the reader. Using an estimate for Chebyshev’s ϑ-function found by Broadbent et al. [12, Table 15], we get the following upper bound for π(x) which improves the inequality (12) for all suﬃciently large values of x. Proposition 4. For every x ≥29.53, we have π(x) < x log x −1 − 1 log x − 3 log2 x −70.935 log3 x . Proof. The proof is similar to the proof of Theorem 2 and we leave the details to the reader. We denote the right-hand side of (4) by f(x) and let x1 = 1018. We combine Lemma 1 with [12, Table 15] to see that π(x) ≤J4,57.184,x1(x) for every x ≥x1. So it suﬃces to compare f(x) with J4,57.184,x1(x) to get that f(x) > π(x) for every x ≥1018. Since f(x) > li(x) for every x such that 1, 098 ≤x < 1018, we can apply Lemma 2 to obtain that (4) also holds for every x such that 1, 098 ≤x < 1018. A direct computation for smaller values of x completes the proof. Integration by parts in (8) implies that for every positive integer m, one has π(x) = x log x + x log2 x + 2x log3 x + 6x log4 x + 24x log5 x + . . . + (m −1)!x logm x + O x logm+1 x (48) as x →∞. In this direction, we get the following upper bound for π(x). Proposition 5. For every x > 1, we have π(x) < x log x + x log2 x + 2x log3 x + 6.024334x log4 x + 24.024334x log5 x + 120.12167x log6 x + 720.73002x log7 x + 6098x log8 x . INTEGERS: 24 (2024) 16 Proof. We set x1 = 1018. Further, let f(x) be the right-hand side of the required inequality. We have f(x) > J3,0.024334,x1(x) for every x ≥x1. So, we can use (44) to get f(x) > π(x) for every x ≥x1. Since f(x) > li(x) for every x ≥204, 182, 829, we can apply Lemma 2 to obtain f(x) > π(x) for every x such that 204, 182, 829 ≤ x ≤x1. A direct computation for smaller values of x completes the proof. Proposition 5 yields the following weaker but more compact upper bounds for the prime counting function π(x). Corollary 3. For every x ≥x0, we have π(x) < x log x + x log2 x + (2 + ε)x log3 x , where ε and x0 are given as in Table 4. ε 0.21 0.215 0.22 x0 160, 930, 932, 942, 272 83, 016, 503, 500, 865 43, 999, 690, 220, 699 ε 0.225 0.23 0.24 x0 23, 824, 649, 646, 672 13, 279, 102, 022, 111 4, 511, 700, 549, 332 ε 0.25 0.26 0.2651 x0 1, 615, 202, 653, 795 643, 809, 266, 445 406, 742, 886, 708 ε 0.27 0.28 0.29 x0 265, 248, 130, 170 117, 997, 473, 286 57, 720, 805, 589 Table 4: Explicit values for ε and x0. Proof. Let x0 = 160, 930, 932, 942, 272 and f(x) = x/ log x+x/ log2 x+2.21x/ log3 x. Proposition 5 implies that π(x) < f(x) for every x ≥180, 250, 881, 352, 396. If we compare f(x) with the integral logarithm li(x), we get by Lemma 2 that π(x) < f(x) for every x ≥162, 791, 795, 110, 834. Next, we use a computer to verify the inequality π(x) < f(x) for every x with x0 ≤x ≤162, 791, 795, 110, 834. The remaining inequalities can be proved in the same way. 5. Proof of Theorem 3 In order to give a proof of Theorem 3, we use (44) and a numerical calculation that veriﬁes the desired inequality for smaller values of x. 1This inequality was already known to be true for every x ≥8 × 1011 (see [50, Proposition 3.3] INTEGERS: 24 (2024) 17 Proof of Theorem 3. Let x1 = 1, 757, 126, 630, 797. Further, let g(x) be the right-hand side of (13). We can compute that J3,−0.024334,x1(x1) −g(x1) > 6 × 103. In addition we have J′ 3,−0.024334,x1(x) > g′(x) for every x ≥44.42. Therefore, we get J3,−0.024334,x1(x) > g(x) for every x ≥x1. Using (44), we get the required inequality for every x ≥x1. For smaller values of x we use a computer. Remark 9. Let x1 = 1, 751, 189, 194, 177. Then the inequality (13) does not hold for x = x1 −0.1. Remark 10. Theorem 3 improves the lower bound for π(x) obtained in [5, Theorem 3]. In the next corollary, we establish some weaker lower bounds for the prime count-ing function. Corollary 4. We have π(x) > x log x −1 − 1 log x −2.975666 log2 x − a3 log3 x − a4 log4 x − a5 log5 x for every x ≥x0, where a3, a4, a5, and x0 are given as in Table 5. a3 13.024334 13.024334 13.024334 0 a4 70.951332 70.951332 0 0 a5 460.634397856444 0 0 0 x0 1, 035, 745, 443, 241 153, 887, 581, 621 7, 713, 187, 213 54, 941, 209 Table 5: Explicit values for a3, a4, a5, and x0. Proof. From Theorem 3, it follows that each required inequality holds for every x ≥1, 751, 189, 194, 177. For smaller values of x we use a computer. Let n be a positive integer. Then (48) provides the inequality π(x) > x log x + x log2 x + 2x log3 x + 6x log4 x + 24x log5 x + . . . + (n −1)!x logn x for all suﬃciently large values of x. In the following proposition, we describe a method to ﬁnd lower bounds for π(x) in the direction of (5) by using lower bounds for π(x) in the direction of (11). Proposition 6. Let n be a positive integer and let a0 > 0 and a1, . . . , an be neg-ative real numbers. Suppose that there is a positive real number x0 such that the inequalities a0 log x + a1 + a2 log x + . . . + an logn−1 x > 0 (49) INTEGERS: 24 (2024) 18 and π(x) > x a0 log x + a1 + a2 log x + . . . + an logn−1 x (50) hold simultaneously for every x ≥x0. Then we have π(x) > b0x log x + b1x log2 x + . . . + bnx logn+1 x for every x ≥x0, where b0, . . . , bn are real numbers recursively deﬁned by b0 = 1/a0, and bk = −1 a0 k X i=1 aibk−1 (1 ≤k ≤n). (51) Proof. For y > 0, we deﬁne R(y) = Pn k=0 ai/yi and S(y) = Pn i=0 bi/yi. For i ∈{1, . . . , 2n}, we set a′ i = ( ai, if i ∈{1, . . . , n}, 0, otherwise and b′ i = ( bi, if i ∈{1, . . . , n}, 0, otherwise. Using (51) together with b′ n+1 = 0, we can see that R(y)S(y) = 1 + 2n X k=n+1 k X i=1 a′ ib′ k−i yk . Since a′ ib′ k−i ≤0 for every i with 1 ≤i ≤2n and every k satisfying n + 1 ≤k ≤2n, we get R(y)S(y) ≤1. By (49), we have R(log x) > 0 for every x ≥x0. Now we can use (50) to get π(x) > x/(R(x) log x) ≥xS(log x)/ log x for every x ≥x0. The best explicit result in the direction of (50) was found in [5, Proposition 5]. The following reﬁnements of it are a consequence of Proposition 6, Theorem 3, and Corollary 4. Corollary 5. We have π(x) > x log x + x log2 x + 2x log3 x + 5.975666x log4 x + b5x log5 x + b6x log6 x + b7x log7 x + b8x log8 x for every x ≥x0, where b5, b6, b7, b8, and x0 are given as in Table 6. b5 b6 b7 b8 x0 23.975666 119.87833 719.26998 5034.88986 1, 681, 111, 802, 141 23.975666 119.87833 719.26998 0 721, 733, 241, 667 23.975666 119.87833 0 0 110, 838, 719, 141 23.975666 0 0 0 1, 331, 691, 853 0 0 0 0 10, 383, 799 Table 6: Explicit values for a2 and x2. INTEGERS: 24 (2024) 19 Proof. In order to prove the ﬁrst inequality, we combine Proposition 6 and Theorem 3 to see that this inequality holds for every x ≥1, 751, 189, 194, 177. For smaller values of x, we use a computer. Further, we use Proposition 6, Corollary 4, and a direct computation for smaller values of x to verify the remaining inequalities. 6. Proof of Theorem 4 In order to prove Theorem 4, we set R = 5.5666305 and, similar to [56, p. 879], we deﬁne the function a : R>0 →R by a(x) log6 x =                          2 −log 2 2 if 2 ≤x < 599, log2 x 8π√x if 599 ≤x < 1.101 × 1026, r 8 17π log x 6.455 1/4 exp − r log x 6.455 ! if 1.101 × 1026 ≤x < e673, 121.0961 log x R 3/2 exp −2 r log x R ! if x ≥e673. Then we get the following result concerning Chebyshev’s ϑ-function. Lemma 4. For every x ≥2, we have \|ϑ(x) −x\| ≤a(x)x log6 x. Proof. In the case where x satisﬁes 2 ≤x < 599, then the given bound is trivial. For second bound, see Johnston [39, Corollary 3.3]. The third bound was given by Trudgian [70, Theorem 1] and the last bound was recently established by Fiori, Kadiri, and Swidinsky [12, Corollary 14] (cf. (3)). We also need the following result on our function a. Lemma 5. Let x1 be a real number with x1 ≥e673. Then an(x) ≤an(x1) for every x ≥x1. Proof. By a straightforward calculation of the derivative, we see that a′(x) < 0 for every x ≥e673. Now we use Theorem 3 and Lemmata 4-5 to give the following proof of Theorem 4. Proof of Theorem 4. In [6, Theorem 1], the inequality was already proved for every x with 65, 405, 887 ≤x ≤2.7358×1040. If we utilize Theorem 3, it turns out that the INTEGERS: 24 (2024) 20 inequality (14) holds unconditionally for every x such that 65, 405, 887 ≤x ≤e540. Now, let f(x) denote the right-hand side of (14). In order to verify the required inequality for every x with e540 ≤x ≤e1680, we set c0 = 1 −1.6341 × 10−12. By [34, Table 3], we have ϑ(x) ≥c0x for every x > e500. Applying this inequality to (7), we get π(x) > g0(x) (52) for every x ≥e500, where g0(x) = c0(li(x) −li(e500) + e500/500). If we show that g0(x) > f(x) for every x satisfying e540 ≤x ≤e1680, we can use (52) to see that the required inequality (14) holds for every x with e540 ≤x ≤e1680. Since g′ 0(x) > f ′(x) for every x so that 9 ≤x ≤e1680, it remains to show that g0(x0) > f(x0), where x0 = e540. First, we note that t P6 k=1(k−1)!/ logk t < li(t) < 1.003t/ log t, where the left-hand side inequality holds for every t ≥565 and the right-hand side inequality is valid for every t ≥e500. Therefore, g0(x0) −f(x0) x0 > c0 6 X k=1 (k −1)! 540k −0.003c0 e40 −f(x0) x0 . Since the right-hand side of the last inequality is positive and we conclude that the required inequality holds for every x with x0 ≤x ≤e1680. The ﬁnal step of the proof consists in the veriﬁcation of the required inequality for every x ≥x1, where x1 = e1680. If we combine (7) with Lemma 4, we can see that π(x) ≥ x log x −xa(x) log7 x + Z x 2 dt log2 t − Z x 2 a(t) log8 t dt. (53) Integration by parts in (53) provides that π(x) ≥C + x 5 X k=0 k! logk+1 x + (720 −a(x))x log7 x + Z x x1 5040 −a(t) log8 t dt, where C = Z x1 2 5040 −a(t) log8 t dt −2 6 X k=1 k! logk+1 2 . Since 0 < a(t) ≤a(x1) for every t ≥x1 (cf. Lemma 5), it turns out that π(x) ≥C + x 5 X k=0 k! logk+1 x + (720 −a(x1))x log7 x + (5040 −a(x1)) Z x x1 dt log8 t. Note that 5040 −a(x1) < 0. Hence π(x) > C + x 5 X k=0 k! logk+1 x + (720 −a(x1))x log7 x + (5040 −a(x1))E(x1), INTEGERS: 24 (2024) 21 where E(x) = 1 5040 li(x) − 7 X k=1 (k −1)!x logk x ! . Since C + (5040 −a(x1))E(x1) < 0, we obtain that π(x) > x 5 X k=0 k! logk+1 x + x log7 x 720 −a(x1) + (5040 −a(x1))E(x1) × log7 x1 x1 . Now we use a computer to get that π(x) > x 5 X k=0 k! logk+1 x −6918930x log7 x for every x ≥x1. Now we set H(y) = 5 X k=0 k! yk+1 −6918930 y7 − 1 y −1 −1 y −3 y2 . It is easy to see that H′(y) < 0 for every y ≥859. Together with limy→∞H(y) = 0, it turns out that H(log x) ≥0 for every x ≥e859. If we combine the last inequality with (6), we get that π(x) > f(x) for every x ≥e1680 and we arrive at the end of the proof. Remark 11. The method employed in the proof of Theorem 4 can also be used to ﬁnd further lower bounds for π(x) given by truncating the asymptotic expansion (11) at later terms (with logn x in the denominator, where n ≥3). However, these bounds will only hold when x is exceptionally large. For instance, (11) provides the even sharper inequality π(x) > x log x −1 − 1 log x − 3 log2 x − 13 log3 x for all suﬃciently large values of x. Similar to the proof of Theorem 4, we get that this inequality holds for every x satisfying 11, 471, 757, 461 ≤x ≤e57.820987 and every x ≥e3661.424. Corollary 6. For every x ≥10, 384, 261, we have π(x) > x log x + x log2 x + 2x log3 x + 6x log4 x. Proof. It suﬃces to combine Proposition 6, Theorem 4, and [6, Theorem 2]. INTEGERS: 24 (2024) 22 7. Proof of Theorem 5 In this section, we want to ﬁnd unrestricted eﬀective estimates for the sum of the reciprocals of all prime numbers not exceeding x For this purpose, we use the method investigated by Rosser and Schoenfeld [63, p. 74]. They derived a remarkable identity which connects the sum of the reciprocals of all prime numbers not exceeding x with Chebyshev’s ϑ-function by showing that A1(x) = ϑ(x) −x x log x − Z ∞ x (ϑ(y) −y)(1 + log y) y2 log2 y dy, (54) where A1(x) = X p≤x 1 p −log log x −B. Here, the constant B is deﬁned as in (16). Applying (2) to (54), Rosser and Schoenfeld [63, p. 68] reﬁned the error term in Mertens’ result (15) by giving A1(x) = O(exp(−a√log x)) as x →∞, where a is an absolute positive constant. Then [63, Theorem 5] they used explicit estimates for Chebyshev’s ϑ-function to show that − 1 2 log2 x < A1(x) < 1 2 log2 x, where the left-hand side inequality is valid for every x > 1 and the right-hand side inequality holds for every x ≥286. Meanwhile there are several improvements of (7) (see, for instance, [30, Theorem 5.6] and [5, Proposition 7]). In Theorem 5, we give the current best unconditionally eﬀective estimates for A1(x). The proof is now rather simple. Proof of Theorem 5. It suﬃces to combine (54) with Proposition 1. Remark 12. Note that the positive integer N0 = 1, 757, 126, 630, 797 might not be the smallest positive integer N so that the inequality given in Theorem 5 holds for every x ≥N. Remark 13. Rosser and Schoenfeld [63, Theorem 20] used the calculation in to see that A1(x) > 0 for every 1 < x ≤108 and raised the question whether this inequality hold for every x > 1. Robin [62, Th´ eor eme 2] proved that the function A1(x) changes the sign inﬁnitely often, which leads to a negative answer to the obove question. By adapting a method for bounding Skewes’ number, B¨ uthe [13, Theorem 1.1] found that there exists an x0 ∈[exp(495.702833109), exp(495.702833165)] such that A1(x) is negative for every x ∈[x0 −exp(239.046541), x0]. Remark 14. Under the assumption that the Riemann hypothesis is true, Schoen-feld [66, Corollary 2] found some better estimate for the sum of the reciprocals of all prime numbers not exceeding x. This result was recently improved by Dusart [31, Theorem 4.1]. INTEGERS: 24 (2024) 23 Using the deﬁnition (16) of B, we get eγ log x Y p≤x 1 −1 p = e−S(x)−A1(x), (55) where S(x) = X p>x log 1 −1 p + 1 p = − ∞ X n=2 1 n X p>x 1 pn . (56) By Rosser and Schoenfeld [63, p. 87], we have − 1.02 (x −1) log x < S(x) < 0 (57) for every x > 1. Hence, the asymptotic formula (15) gives A2(x) = O(1/ log2 x) as x →∞, where A2(x) = e−γ log x − Y p≤x 1 −1 p . In [63, Theorem 7], Rosser and Schoenfeld found that e−γ log x 1 − 1 2 log2 x < Y p≤x 1 −1 p < e−γ log x 1 + 1 2 log2 x , where the left-hand side inequality is valid for every x ≥285 and the right-hand side inequality holds for every x > 1. We use (55) combined with Theorem 5 to obtain the following reﬁnement of [5, Proposition 9]. Proposition 7. For every x ≥1, 757, 126, 630, 797, we have e−γ log x exp(−f(x)) < Y p≤x 1 −1 p < e−γ log x exp f(x) + 1.02 (x −1) log x , where f(x) denotes the right-hand side of (17). Proof. First, we apply the left-hand side inequality of Theorem 5 to (55) and see that Y p≤x 1 −1 p < e−γ log x exp(−S(x) + f(x)) (58) for every x > 1, 757, 126, 630, 797. Now it suﬃces to apply the right-hand side inequality of (57) to (58) and we get the required right-hand side inequality. One the other hand, we have S(x) < 0 by (57). Applying this and (17) to (55), we arrive at the end of the proof. INTEGERS: 24 (2024) 24 Remark 15. Note that the positive integer N0 = 1, 757, 126, 630, 797 in Proposition 7 might not be the smallest positive integer N so that the inequality given holds for every x ≥N. Remark 16. Under the assumption that the Riemann hypothesis is true, Schoen-feld [66, Corollary 3] found that the inequality \|A2(x)\| < 3 log x + 5 8πeγ√x log x holds for every x ≥8. This was slightly improved by Dusart [31, Theorem 4.4] in 2018. Remark 17. Rosser and Schoenfeld [63, Theorem 23] found that A2(x) > 0 for every 0 < x ≤108 and stated [63, p. 73] the question whether this inequality also hold for every x > 108. In [62, Proposition 1], Robin answered this by showing that the function A2(x) changes the sign inﬁnitely often. Now we can use Proposition 7 to derive the following eﬀective estimates for Y p≤x 1 + 1 p , where p runs over primes not exceeding x. Corollary 7. For every x ≥1, 757, 126, 630, 797, one has 6eγ π2 exp −f(x) − 1.02 (x −1) log x log x < Y p≤x 1 + 1 p < 6eγ π2 1 + 1 x exp(f(x)) log x, where f(x) denotes the right-hand side of (17). Proof. Since 1 + 1/p = (1 −1/p2)/(1 −1/p) and ζ(2) = π2/6, it suﬃces to combine Proposition 7 and [31, Lemma 4.3]. Remark 18. Note that the positive integer N0 = 1, 757, 126, 630, 797 might not be the smallest positive integer N so that the inequality given in Corollary 7 holds for every x ≥N. Let us brieﬂy study S(x), deﬁned as in (56), in more detail. In the proof of the left-hand side inequality in (57), Rosser and Schoenfeld used the inequality ϑ(x) < 1.02x which is valid for every x > 0 (see [63, Theorem 9]). If we use approximations for ϑ(x) of the form (5), we get the following result. INTEGERS: 24 (2024) 25 Proposition 8. Let k be a positive integer and let ηk and x0 = x0(k) be positive real numbers with x0 > 1 so that \|ϑ(x) −x\| < ηkx/ logk x for every x ≥x0. Then, we have S(x) − ∞ X n=1 li(x−n) n + 1 < ηk logk+1 x (x + 1) log x x −1 −1 for every x ≥x0. In order to prove this proposition, we ﬁrst establish the following lemma. Lemma 6. Let n be a positive integer with n ≥2. Under the assumptions of Proposition 8, we have li(x1−n) + X p>x 1 pn < ηk xn−1 logk+1 x 1 + n n −1 for every x ≥x0. Proof. By [63, p. 87], we have X p>x 1 pn = − ϑ(x) xn log x + Z ∞ x (1 + n log y)ϑ(y) yn+1 log2 y dy. (59) Since we have assumed that \|ϑ(x) −x\| < ηkx/ logk x for every x ≥x0, we see that X p>x 1 pn ≤−li(x1−n) + ηk xn−1 logk+1 x + ηk Z ∞ x 1 + n log y yn logk+2 y dy (60) for every x ≥x0. Analogous to [63, Lemma 9], we get that Z ∞ x 1 + n log y yn logk+2 y dy ≤ n (n −1)xn−1 logk+1 x . Applying this inequality to (60), we see that the required upper bound holds for every x ≥x0. The proof of the required lower bound is quite similar and we leave the details to the reader. Now we can combine the deﬁnition (56) with Lemma 6 to get the following proof of Proposition 8. Proof of Proposition 8. If we apply Lemma 6 to (56), it turns out that S(x) − ∞ X n=1 li(x−n) n + 1 < ηk logk+1 x ∞ X n=2 1 + n n −1 1 nxn−1 INTEGERS: 24 (2024) 26 for every x ≥x0. Now, it suﬃces to apply the identity ∞ X n=2 1 + n n −1 1 nxn−1 = (x + 1) log x x −1 −1 to complete the proof. If we combine (35) and (59), we ﬁnd the following new necessary condition for the Riemann hypothesis including the sum given in Lemma 6. Proposition 9. Let n be a positive integer with n ≥2. Under the assumption that the Riemann hypothesis is true, we have li(x1−n) + X p>x 1 pn < 1 8πxn−1/2 1 + 2n 2n −1 log x + 2 2n −1 for every x ≥599. Proof. Instead of the assumption (5), we now use (35) in the proof of Lemma 6. 8. Proof of Theorem 6 Here we give the following proof of Theorem 6. Proof of Theorem 6. Let the constant E be deﬁned as in (19) and let A3(x) = X p≤x log p p −log x −E. By Rosser and Schoenfeld [63, p. 74], we have A3(x) = ϑ(x) −x x − Z ∞ x ϑ(y) −y y2 dy. (61) Similarly to the proof of Theorem 5, we may combine (61) and Proposition 1 to get that the desired both inequalities hold for every x ≥1, 757, 126, 630, 797. Remark 19. Under the assumption that the Riemann hypothesis is true, Schoen-feld [66, Corollary 2] found a better upper bound for \|A3(x)\|. This result was later improved by Dusart [31, Theorem 4.2]. Remark 20. Rosser and Schoenfeld [63, Theorem 21] also found that A3(x) > 0 for every 0 < x ≤108. Again, they asked whether this inequality also holds for every x > 108. Robin [62, Proposition 1] showed that the function A3(x) changes the sign inﬁnitely often, which leads again to a negative answer to the above question. Unfortunately, until today no x0 is known so that A3(x0) < 0. INTEGERS: 24 (2024) 27 Acknowledgements. The author would like to express his great appreciation to Kim Walisch, Tom´ as Oliveira e Silva, and Thomas Lessmann for the support in writing the C++ codes used in this paper. Furthermore the author thanks Samuel Broadbent, Habiba Kadiri, Allysa Lumley, Nathan Ng, and Kirsten Wilk, whose paper has motivated him to deal with the present topic again. Moreover, the author would also like to thank the two beautiful souls R. and O. for the never ending inspiration. Finally, the author thanks the anonymous reviewer for the useful comments and suggestions to improve the quality of this paper. References K. I. Appel and J. B. Rosser, Tables for estimating functions of primes, Comm. Res. Div. Tech. Rep. 4 (1961). T. Apostol, Introduction to Analytic Number Theory, Springer, New York-Heidelberg, 1976. C. Axler, ¨ Uber die Primzahl-Z¨ ahlfunktion, die n-te Primzahl und verallgemeinerte Ramanujan-Primzahlen, PhD Thesis, Heinrich Heine University, D¨ usseldorf (Germany), 2013. 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J. te Riele, On the sign of the diﬀerence π(x) −li(x), Math. Comp. 48 (1987), 323-328. T. S. Trudgian, Updating the error term in the prime number theorem, Ramanujan J. 39 (2016), 225-234. I. M. Vinogradov, A new estimate of the function ζ(1+it), Izv. Akad. Nauk SSSR. Ser. Mat. 22 (1958), 161-164. H. von Koch, Sur la distribution des nombres premiers, Acta Math. 24 (1901), 159-182. H. von Mangoldt, Zu Riemanns Abhandlung ”Ueber die Anzahl der Primzahlen unter einer gegebenen Gr¨ osse”, J. Reine Angew. Math. 114 (1895), 255-305. K. Walisch, primecount, version 7.4. Available at github.com/kimwalisch/primecount. K. Walisch, primesieve, version 8.0. Available at github.com/kimwalisch/primesieve. Appendix Here we use Corollary 1 and Walisch’s primecount program to note more weaker upper bounds for π(x) of the form (4), where m is an integer with 0 ≤m ≤2 and a0, . . . , am are suitable positive real numbers. We start with the case where m = 0. INTEGERS: 24 (2024) 31 Proposition 10. One has π(x) < x log x −a0 for every x ≥x0, where a0 and x0 are given as in Table 7 and Table 8. a0 1.0344 1.0345 1.0346 x0 98, 011, 218, 006, 714 90, 093, 726, 828, 053 82, 972, 765, 680, 514 a0 1.0347 1.0348 1.0349 x0 76, 292, 362, 570, 940 70, 363, 470, 737, 452 64, 716, 191, 738, 353 a0 1.035 1.036 1.037 x0 59, 667, 044, 596, 151 27, 086, 141, 056, 455 12, 806, 615, 320, 917 a0 1.038 1.039 1.04 x0 6, 317, 261, 904, 937 3, 231, 501, 496, 562 1, 697, 021, 254, 855 a0 1.041 1.042 1.043 x0 924, 640, 658, 874 519, 205, 451, 664 296, 735, 291, 225 a0 1.044 1.045 1.046 x0 175, 758, 684, 156 105, 640, 136, 371 65, 431, 161, 562 a0 1.047 1.048 1.049 x0 41, 022, 022, 044 25, 724, 702, 310 17, 231, 171, 472 a0 1.05 1.051 1.052 x0 11, 207, 440, 881 7, 538, 561, 672 5, 047, 295, 951 a0 1.053 1.054 1.055 x0 3, 745, 835, 388 2, 605, 443, 747 1, 810, 796, 757 a0 1.056 1.057 1.058 x0 1, 220, 594, 340 876, 542, 559 673, 828, 570 a0 1.059 1.06 1.061 x0 501, 155, 566 383, 446, 375 269, 585, 283 a0 1.062 1.063 1.064 x0 196, 894, 353 180, 220, 137 116, 749, 925 a0 1.065 1.066 1.067 x0 110, 166, 540 76, 223, 058 53, 431, 171 a0 1.068 1.069 1.07 x0 46, 097, 944 39, 706, 453 31, 027, 247 Table 7: Explicit values for a0 and x0. INTEGERS: 24 (2024) 32 a0 1.071 1.072 1.073 1.074 1.075 x0 22, 078, 017 18, 339, 738 13, 026, 859 12, 895, 928 8, 832, 927 a0 1.076 1.077 1.078 1.079 1.08 x0 7, 299, 254 7, 117, 256 5, 465, 656 4, 994, 010 3, 462, 478 a0 1.081 1.082 1.083 1.08366 1.084 x0 3, 455, 648 2, 279, 177 1, 529, 630 1, 526, 671 1, 525, 432 a0 1.085 1.086 1.087 1.088 1.089 x0 1, 515, 074 1, 200, 014 1, 195, 296 624, 878 618, 726 a0 1.09 1.091 1.092 1.093 1.094 x0 618, 058 445, 112 359, 804 356203 355, 990 a0 1.095 1.096 1.097 1.098 1.099 x0 355, 177 155, 935 155, 907 60, 297 60, 224 Table 8: Explicit values for a0 and x0. Proof. Let f(x) = x/(log x −1.0344). Corollary 1 implies that π(x) < x log x −1.0344 for every x ≥106, 640, 139, 304, 611. If we compare the right-hand side of (8) with the integral logarithm li(x), we can use Lemma 2 to see that the inequality (8) also holds for every x with 98, 269, 667, 551, 459 ≤x ≤106, 640, 139, 304, 611. We conclude by direct computation. Remark 21. The real number a0 = 1.08366 in Proposition 10 is mostly only of historical value. On the basis of his study of a limited table of primes, Legendre stated 1808 (see [46, p. 394]) that π(x) = x/(log x −A(x)), where limx→∞A(x) = 1.08366. Clearly Legendre’s conjecture is equivalent to (8). However, from (11), it follows that the best value of limx→∞A(x) is 1. At this point it should be mentioned that Panaitopol claimed to have proved the inequality π(x) < x log x −1.08366 (62) for every x > 106. In Proposition 10, it could be shown that N = 1, 526, 671 is the smallest possible positive integer so that the inequality (62) holds for every x ≥N. Next, we obtain the following eﬀective estimates for π(x) for the case where m = 1. The proof is similar to the proof of Proposition 10 and is left to the reader. INTEGERS: 24 (2024) 33 Proposition 11. We have π(x) < x log x −1 − a1 log x for every x ≥x1, where a1 and x1 are given as in Table 9. a1 1.11 1.1105 1.111 x1 62, 998, 850, 942, 976 55, 193, 608, 062, 217 49, 246, 036, 992, 716 a1 1.112 1.113 1.114 x1 38, 472, 138, 880, 411 30, 658, 643, 813, 468 23, 767, 640, 743, 883 a1 1.115 1.116 1.117 x1 19, 278, 513, 358, 342 15, 142, 627, 022, 527 12, 279, 648, 138, 508 a1 1.118 1.119 1.12 x1 9, 684, 114, 630, 824 7, 981, 446, 192, 206 6, 323, 967, 140, 812 a1 1.121 1.122 1.123 x1 5, 273, 225, 700, 761 4, 170, 462, 893, 841 3, 458, 549, 136, 539 a1 1.124 1.125 1.126 x1 2, 825, 539, 807, 244 2, 292, 448, 124, 593 1, 903, 596, 231, 542 a1 1.127 1.128 1.129 x1 1, 573, 767, 234, 188 1, 290, 096, 268, 844 1, 073, 403, 839, 693 a1 1.13 1.131 1.132 x1 889, 377, 392, 161 782, 989, 678, 664 608, 408, 258, 090 a1 1.133 1.134 1.135 x1 540, 050, 850, 157 452, 875, 824, 702 373, 479, 021, 700 a1 1.136 1.137 1.138 x1 335, 562, 521, 091 263, 728, 502, 964 242, 118, 904, 367 a1 1.139 1.14 1.141 x1 201, 924, 836, 111 161, 054, 192, 492 149, 061, 190, 565 a1 1.142 1.143 1.144 x1 125, 233, 112, 846 105, 053, 836, 224 86, 061, 321, 374 a1 1.145 1.146 1.147 x1 77, 278, 924, 451 61, 344, 524, 412 57, 720, 831, 343 a1 1.148 1.149 1.15 x1 46, 039, 922, 948 42, 575, 222, 481 38, 284, 442, 297 Table 9: Explicit values for a1 and x1. Finally, we consider the case where m = 2 and ﬁnd the following explicit estimates INTEGERS: 24 (2024) 34 for π(x). Again, the proof is quite similar to the proof of Proposition 10 and we leave the details to the reader. Proposition 12. We have π(x) < x log x −1 − 1 log x − a2 log2 x for every x ≥x2, where a2 and x2 are given as in Table 10. a2 3.49 3.495 3.5 x2 83, 027, 761, 686, 134 63, 024, 307, 127, 421 50, 794, 512, 296, 846 a2 3.51 3.52 3.53 x2 30, 594, 003, 254, 258 17, 348, 455, 129, 950 11, 655, 963, 556, 138 a2 3.54 3.55 3.56 x2 5, 539, 984, 798, 515 4, 489, 052, 430, 063 2, 180, 930, 569, 481 a2 3.57 3.58 3.59 x2 1, 464, 200, 206, 021 882, 055, 689, 961 584, 256, 118, 105 a2 3.6 3.61 3.62 x2 437, 882, 804, 654 332, 203, 763, 508 201, 890, 631, 296 a2 3.63 3.64 3.65 x2 148, 632, 348, 138 102, 965, 110, 268 55, 102, 251, 180 a2 3.66 3.67 3.68 x2 38, 278, 086, 931 24, 178, 954, 639 21, 729, 109, 565 Table 10: Explicit values for a2 and x2.
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Marco Nicolosi, C. Francesco on Google Scholar Tina, G. Marco Nicolosi, C. Francesco on PubMed Tina, G. Marco Nicolosi, C. Francesco /ajax/scifeed/subscribe Article Views 2383 Citations 3 Table of Contents Abstract Introduction Methodologies Results Discussion Conclusions Author Contributions Funding Institutional Review Board Statement Informed Consent Statement Data Availability Statement Conflicts of Interest References Altmetricshare Shareannouncement Helpformat_quote Citequestion_answer Discuss in SciProfiles Need Help? Support Find support for a specific problem in the support section of our website. Get Support Feedback Please let us know what you think of our products and services. Give Feedback Information Visit our dedicated information section to learn more about MDPI. Get Information clear JSmol Viewer clear first_page Download PDF settings Order Article Reprints Font Type: Arial Georgia Verdana Font Size: Aa Aa Aa Line Spacing:    Column Width:    Background: Open Access Article A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses by Giuseppe Marco Tina Giuseppe Marco Tina SciProfilesScilitPreprints.orgGoogle Scholar and Claudio Francesco Nicolosi Claudio Francesco Nicolosi SciProfilesScilitPreprints.orgGoogle Scholar Department of Electrical, Electronic and Computer Engineering (DIEEI), University of Catania, Viale Andrea Doria n.6, 95125 Catania, Italy Author to whom correspondence should be addressed. Appl. Sci.2023, 13(3), 1986; Submission received: 20 December 2022 / Revised: 31 January 2023 / Accepted: 31 January 2023 / Published: 3 February 2023 (This article belongs to the Section Energy Science and Technology) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Browse Figures Versions Notes Abstract The increasing presence of non-programmable renewable energy plants increases the intermittency of the electricity supply and thus threatens the adequacy of a power system. Hydropower can solve this problem due to its flexibility. This paper applies statistical approaches to assess water availability in the context of hydropower generation and adequacy analysis on a seasonal basis for one site in Sicily and the other in Sardinia, where major hydroelectric plants are present. First, an empirical relationship between soil moisture content (SMC) and potential evapotranspiration (ET0) is evaluated through linear regression analysis. Then, precipitation trends over the last twenty years are analyzed to determine any effects of global warming on water availability. Finally, Monte Carlo algorithms are used for the stochastic generation of hourly precipitation, direct runoff profiles, and daily SMC profiles. Strong positive and negative correlations between ET 0 and SMC (p< 0.05), and R 2 ≥ 0.5 are found for both sites, except for summer, and R 2 ≥ 0.5 is obtained. The cumulative pH-historical precipitation shows changes in seasonal trends, with evidence of a decrease at the annual level. The algorithms used to synthetically generate hourly precipitation and direct runoff profiles, as well as daily SMC profiles, effectively simulate the statistical variability of the historical profiles of these physical quantities. Keywords: climate change; hydropower; adequacy; Monte Carlo; water availability; direct runoff; linear regression; soil physics 1. Introduction 1.1. European Energy Sector Emissions and EU Actions to Counter Them In climate science, the climate system is defined as the whole of the atmosphere, the hydrosphere, the cryosphere, the biosphere, and the land surface ; thus, the climate is the state of the climate system . Furthermore, climate change can be defined as a statistically significant variation in climate from its mean state or its variability. The large variations in the current climate are likely due to human activities, which by causing an increase in the greenhouse gas (GHGs) concentrations in the atmosphere, trigger a change in climate . Specifically, different human sectors emit significant percentages of carbon dioxide into the atmosphere, and to set a figure, referring to the energy sector in 2021 global energy-related CO 2 emissions were 33.0 Gt . It is worth dwelling on some data pertaining to the European Union (EU). In the EU the two sectors that experience the highest CO 2 emissions are energy supply and transport . However, defining the CO 2 emission intensity as the ratio between the CO 2 emissions due to electricity generation and gross electricity generation , since 1990 the intensity of GHGs emissions from electricity generation has decreased markedly, reaching, an amount of CO 2 emitted by half of that emitted in 1990 for 1 kilowatt hour generated in 2020. This sharp decrease is mainly due to the high number of investments and installations of renewable energy systems. Indeed, in the decade 2009–2019, solar photovoltaics (PV), wind power, and solid biofuels, have driven the growth of renewable energy in the EU . Although the installed capacity of hydropower had only a slight increase in the decade 2009–2019 , wind and hydropower made up two-thirds of the total electricity generated from renewable energy sources (RES) in 2019 ; the remaining one-third was covered by PV (13%), solid biofuels (8%) and other RES (9%). Although the 13% (125.7 TWh) share of generation by PV as of 2019 may seem small, it must be considered that this generation technology started from a 1% (7.4 TWh) share in 2008; this sharp percentage increase over a decade makes PV technology the one with the fastest growth rate among all renewables in EU. The data cited above testify to a trend in the penetration of renewables, in the generation mixes of EU countries’ electric power systems (EPSs), that is set to increase further to meet the environmental targets of net zero GHGs emissions by 2050. Indeed, in 2019 the EU released the European Green Deal , an agreement among the member countries to achieve sustainable development goals, such as GHGs emission reductions of 55% below 1990 levels by 2030, and climate neutrality by 2050. To achieve these challenging goals, the EU plans to focus heavily on PV and wind power installations. It is worth mentioning a recent energy plan, also released by the EU, namely the REPowerEU Plan , which aims to increase the installation rates of wind and PV power plants already envisioned by . Energy savings, heat pumps, renewable hydrogen, biomethane, and electrification of industrial sectors currently fossil-fueled, are the other linchpins of the European strategy. An ever-increasing amount of renewable generation sources in the power systems of EU countries (and beyond) poses risks to system adequacy, and therefore the flexibility of other generation sources already present in the EPSs of the respective countries, along with storage, is crucial to the proper functioning of present and future power systems, as is illustrated in the following subsections. 1.2. Interaction between Climate Change and Power Systems The increasing presence of RES in the generation mixes of European and non-European electric power systems is, on the one hand, reducing the impact of power generation in terms of CO 2 CO 2 released into the atmosphere, but on the other hand, it is jeopardizing the adequacy of power systems . This is happening because the energy produced by nonprogrammable renewable sources (PV, wind power) [13,14,15], along with their efficiencies , depends on the dynamics of weather variables, which are inherently stochastic (i.e., it is not possible to predict a priori, instant by instant, the values assumed by weather variables). In turn, global warming is and will increasingly threaten the adequacy and safety of power systems. It follows from the above considerations that it is becoming increasingly pressing to analyze the impacts of current and future climate on current and future power systems. The most characteristic aspect of these analyses, however, is uncertainty: climate models’ projections of future climate are affected by uncertainty due to climate dynamics; the impact of future power generation technologies on GHGs concentrations in the atmosphere, cannot be known a priori ; the impact of future climate on future EPS is characterized by the double uncertainty given by future climate change projections and the energy policies of different countries . These uncertainties occur on both a spatial and temporal basis. The inherent presence of uncertainty in the above analyses suggests how statistical approaches can reliably model the impacts of climate on energy production and demand. Climate change is already increasing the frequency, severity, and persistence of extreme weather events, thereby impacting the resilience and reliability of electric power systems . Furthermore, climate change is characterized by marked spatial variability, on a regional scale; for example, the Mediterranean region will be one of the most affected by the effects of climate change in the coming decades, with temperature increases of 20% above the global average, and a marked reduction in the amount of annual precipitation . This spatial variability in the effects of climate change has immediate consequences for electricity production and demand: the European continent will experience polarized global warming effects on thermoelectric and hydroelectric generation between northern and southern Europe . Going into the specifics of individual countries, Italy would experience a general decrease in available PV and wind generation, and a sharp decrease in thermoelectric generation and efficiency [13,19]. In this framework, climate model projections can be useful in evaluating the level of production and efficiency of future generation plants, as well as future electricity demand. In this regard, one paper shows that a +2 °C increase in average temperatures would impact future European electricity generation and demand more markedly than a +1.5 °C increase . In this paper, climate models are not used; rather, statistical techniques are applied to simulate time series of meteorological and hydrological variables that directly influence the amount of water available for hydropower production, whether from run-of-the-river or reservoir systems. 1.3. The Role of Hydropower on the Adequacy of Power Systems Since adequacy is the ability of an EPS to meet energy demand at any given time and market area in a given country, and since nonprogrammable sources are stochastically dependent on trends in weather variables, today’s power systems are experiencing increasing precariousness in deterministically meeting load demand. Exacerbating this picture is the energy crisis sweeping the EU in the wake of geopolitical unrest in Eastern Europe. This crisis is not only about the commodity prices of electricity and gas in their respective markets, but also regards the terms of the availability of fuels for power generation and the consequent compliance with the adequacy levels of European power systems . Different solutions and different approaches can be devised to deal with the above issues and challenges, but hydropower can already be considered one of the main reliefs to the risks of the inadequacy of an EPS . In fact, hydropower is not only an important renewable energy source for several countries around the world, capable of covering both base loads and peak demand, but can be a useful source of relief for grid imbalances between generation and demand. In cases of grid imbalance, hydropower plants are among the fastest in terms of start-up time when there is a generation shortfall in demand coverage . This implies that flexible hydropower management can facilitate the integration of non-programmable renewable energy into EPSs. Going into national detail, a study conducted on the Spanish power system, and future energy scenarios in this country, shows how flexible hydropower management can play an important role in generation fleet adequacy, reducing power shortages . Taking advantage of the flexibility of hydropower technology, however, involves meeting both operational constraints (on the security of supply) and governmental and environmental constraints (on reservoir levels and river flows). Thus, modeling regarding hydropower involves consideration of all aspects underlying the interaction between hydropower and the power systems . Among these aspects is the modeling of water availability, and thus water inflow to hydropower plants, which has seasonal variability. 1.4. Literature Review (Analysis) of the Impact of Climate Variables on Hydropower Generation Despite the importance of hydropower generation to the adequacy of a power system, there is, to the author’s knowledge, little literature that addresses modelling the impact of climate variables on hydropower generation (both run-of-river and storage) with respect to the adequacy of a power system . The most important aspect of adequacy analyses is that they are carried out using statistical methods based on probability theory and stochastic modelling of generation and demand profiles. In this context, there is a lack of work in the literature that models synthetic profiles of meteorological and hydrological variables for the purpose of adequacy analyses. The study and theorising of these types of models are crucial today but will become even more important in the future as trends in weather variables continue to change due to climate change. This paper attempts to fill the gap in the literature. First, linear regression is used to search for empirical relationships linking soil moisture (expressed as soil water content saturation fraction) and potential evapotranspiration, which is evaluated for a reference crop for specific geographical locations. The purpose of this evaluation is that daily time series of potential evapotranspiration can be derived from the soil moisture data. Since precipitation is the main driver of the hydrological cycle, hourly precipitation trends are analysed next. This can be useful because while the direction of the impact of global warming on air and ocean temperature trends is well known, the direction of the impact on precipitation is still a subject of scientific debate [27,28]. Furthermore, the analysis of precipitation trends allows conclusions to be drawn about the development of water availability for hydropower plants. This is because the main source of water availability for a power plant is direct runoff, which in turn results from precipitation. For this reason, two models from the literature are presented here that can be used to assess direct runoff: one model considers only the precipitation intensity (mm/h), while the other includes the daily precipitation amount (mm). Since adequacy analyses are carried out with statistical approaches , this work also uses the Monte Carlo method for the stochastic generation of hourly precipitation and direct runoff profiles as well as daily soil moisture profiles. In this way, synthetic profiles are obtained that form the building blocks on which simulations of hydropower generation profiles can be built for future research. Finally, to complement the analyses just described, additional physical considerations are added to the assumptions and considerations. All simulations are performed in MATLAB. The paper is organized as follows: Section 2 explains in detail the methodologies implemented, the results of which are in Section 3. Section 4 provides final considerations on the methodologies and the results obtained. 2. Methodologies This section presents, in addition to an introductory subsection of the physics of a catchment, the methodologies and procedures applied for: the evaluation of empirical relationships between potential evapotranspiration and SMC through linear regression; the analysis of seasonal and annual precipitation trends over the past twenty years; application of Monte Carlo approaches for generating synthetic profiles of precipitation, direct runoff, and SMC. Satellite data of hourly precipitation (mm/h) and of daily (hourly values are not currently available) soil moisture content (SMC), expressed as soil saturation fraction (dimensionless), are used; these data are taken from the MERRA-2 product, one of the NASA’s databases having a spatial resolution of 0.5° × 0.625° (latitude and longitude, respectively) . 2.1. The Hydrology of a Catchment, Basic Concepts The water balance of a catchment area is influenced by the amount and intensity of precipitation, evapotranspiration, and runoff, as well as the change in moisture stored in the soil . Precipitation is one of the main sources of water for the catchment; depending on its intensity (mm/h) and its amount (mm), variates the fraction of water that will seep into the ground or runoff into the river or into the basin. However, not all of the atmospheric precipitation reaches the ground surface; in fact, depending on the type of crop and vegetation canopy, some of the precipitation is intercepted and evaporates without ever having reached the ground . Precipitation that does not remain intercepted in the foliage and reaches the soil surface will infiltrate, evaporate, or runoff, depending on climatic conditions (net radiation, air temperature, wind speed, and relative humidity) and soil types and conditions. In fact, depending on the soil texture, the physical properties of the soil change; these include field capacity, saturation capacity, permanent wilting point, and available water capacity . The saturation capacity is reached when all pores are filled with water; however, in this condition, gravity drains some of the water into the lower layers of soil. The amount of water that remains after this drainage is the field capacity of the soil. However, the field capacity does not correspond to the total amount of water available in the soil (available water capacity). In fact, a fraction of the field capacity is bound to solid soil particles by capillary forces and is available neither for evapotranspiration nor for plant nutrition; this amount of water is called the permanent wilting point . Finally, taking away the wilting point, the remaining water percentage of the field capacity is water that is free to move, that is, available water . The texture of the soil, together with the amount of moisture (water) in the soil, characterize the infiltration capacity of the soil. When atmospheric precipitation phenomena occur, if the soil is saturated or if the intensity of precipitation is greater than the infiltration rate, the amount of precipitation that reaches the soil will constitute direct runoff that will feed the river or basin near the catchment. In hydrology, the soil is usually divided into two zones, one called unsaturated, and the other called saturated, separated by an imaginary line called the water table. The water stored in the unsaturated zone is called soil water , and this zone is further divided into two zones: the rooting zone and the intermediate zone. The saturated zone, on the other hand, lies below the unsaturated zone and is bounded by two different layers: an upper layer that is the capillary fringe (located at the water table level) and an impermeable bottom that is the bedrock . Precipitation contributes to runoff when its intensity or duration is such that it exceeds the infiltration capacity of the soil and thus surface runoff, also called direct runoff, results. Another water contribution to a catchment also comes from indirect runoff, that is, some of the water that infiltrates into the soil will percolate into the lower layers of the soil until it reaches the saturated zone, going to form groundwater. Some of this water will go into the catchment, constituting indirect runoff. In this work, the rooting zone is considered, since it constitutes a large part of the unsaturated zone, and as such allows for modelling and consideration of fundamental aspects of soil physics that influence evapotranspiration and direct runoff . The SMC data for the root zone are taken from , where they refer to a soil depth ranging from the surface to 100 cm. The saturated soil zone, and thus the indirect runoff are ignored in this study. This, however, is a conservative consideration, being that indirect runoff would constitute more water available for hydropower generation. The meteorological and hydrological quantities, discussed so far, are linked together through the water balance equation, referred to as an atmosphere-land surface system. Indeed, considering that for a given time frame the hydrologic cycle as a closed system, the physical principles of conservation of energy and mass apply. In the case of water, the conservation of mass and energy posits the following expression : P±ET±R=±Δ S P±ET±R=±Δ S (1) where P P denotes precipitation, ET ET evapotranspiration, R R runoff, and Δ S Δ S is the change in water stored in the soil (deep groundwater is included). The symbol ± is used in the first member of (1) to indicate how the physical quantities can be a gain or a loss for the system, depending on the reference: for the atmosphere, the ET ET is a gain, while for soil hydrology, it is a loss. Since the hydrological point of view is applied in this paper, (1) leads to the following: P−ET−R=±Δ S P−ET−R=±Δ S (2) The first-member terms are the main constituents of a water availability study, and therefore will be discussed in the next subsections. 2.2. Seasonal Empirical Relationships between Potential Evapotranspiration and Soil Moisture Content The amount of potential evapotranspiration (ET 0 ET 0) depends on net radiation, air temperature, relative humidity (RH), and wind speed. For example, with the same wind speed and RH, the rate of ET 0 ET 0 will increase as net radiation and air temperature increase. This in turn leads to a change in the amount of water (moisture) contained in the soil. Indeed, potential evapotranspiration ET 0 ET 0 (mm/day) and SMC (dimensionless) depend on the same climate variables (net insolation, air temperature, wind speed, relative humidity). These physical considerations then lead to an attempt to find empirical relationships that link ET 0 ET 0 to soil moisture content, for the two geographic case studies and time frame (daily). The approach used is that of linear regression and is conducted on a seasonal basis, for each of the selected years of historical data. The purpose is to see if the coefficients of the fitting curves have consistency of magnitude and sign between the two different geographical locations chosen, over the four seasons considered, so that through fitting curve expressions, ET 0 ET 0 rates (mm/day) can be directly estimated from soil moisture data (known from ). A first-degree polynomial is used as fitting curve. ET 0 ET 0 values are not available in and therefore must be derived. The rate of evapotranspiration depends not only on atmospheric factors, but also on the type of vegetation covering the surface. Thus, several works in literature evaluate evapotranspiration depending on the type of crop considered; some papers evaluate the evapotranspiration in case of a reference crop, with the following characteristics : a fixed surface resistance of 69 s/m s/m, a fixed height of 0.12 m, and an albedo of 0.23. Furthermore, it is an extensive surface cover that completely shadows the ground . In this paper, the reference crop approach is used since it makes the results found comparable with other studies using reference crop, and since ET 0 ET 0 for a specific crop can be assessed through crop coefficients available in the literature. In the literature, among the models used to assess reference crop ET 0 ET 0, there is one recognized by the Food and Agriculture Organization (FAO): the Penman-Monteith formula (FAO56) , that allows ET 0 ET 0 to be evaluated on a daily basis. However, this formula requires as inputs environmental data that are not present in common in situ weather stations or satellite measurements, thus making it difficult to implement the FAO56 in several geographical regions, especially in developing countries. Another model allows daily ET 0 ET 0 estimates to be obtained close to those of FAO56 is that of Hargreaves-Samani (HS) , as described in [39,40,41]. The equation for this model requires maximum and minimum air temperature as the only climate data. The HS equation is reported below : ET 0=C 1·R ext·(T+17.8)·Δ T C 2(mm day)ET 0=C 1·R ext·T+17.8·Δ T C 2 mm day (3) where ET 0 ET 0 (mm/day mm/day) is the reference potential evapotranspiration, C 1 C 1 is a constant equal to 0.0023, C 2 C 2 is another constant equal to 0.5, and Δ T Δ T (°C) is the difference between the monthly average daily maximum and minimum temperatures; the sum of these two temperature values, divided by two, yields T (°C), which is thus an average temperature; R ext R ext (mm/day) is the extraterrestrial radiation expressed in equivalent water evaporation; finally, 17.8 is an empirical factor introduced to account for the temperature units used in the original formulations. In (3) in addition to explicitly appearing atmospheric quantities on which ET 0 ET 0 depends, ET 0 ET 0 binding to RH is also expressed, implicitly, since Δ T Δ T depends linearly on relative humidity . A specific formula must be used to evaluate R ext R ext : R ext=C 3·d r·(ω s sin ϕ sin δ+cos ϕ cos δ sin ω s)(mm day)R ext=C 3·d r·ω s sin ϕ sin δ+cos ϕ cos δ sin ω s mm day (4) where C 3 C 3 is a constant equal to 15.392; d r d r (dimensionless) is the relative distance between the Sun and the Earth; ϕ ϕ (radians) is the latitude of the site; δ δ (radians) is the solar declination; ω s ω s (radians) is the sunset hour angle. The d r d r, ω s ω s, and δ δ are expressed by : d r=1+0.033·cos(2 π 365 J)(dimensionless)d r=1+0.033·cos 2 π 365 J dimensionless (5) where J is the Julian day number of the year. δ=0.4093·sin(2 π 365 J−1.405)(radians)δ=0.4093·sin 2 π 365 J−1.405 radians (6) ω s=arccos(−tan ϕ·tan δ)(radians)ω s=arccos−tan ϕ·tan δ radians (7) 2.3. Precipitation Data Assessment and Models for Water Availability in Hydropower Plants Since precipitation is the prime driver of a catchment’s hydrologic cycle and water availability for hydropower production, it is useful to conduct an analysis of historical hourly precipitation profiles for the geographic locations of interest. The months constituting the seasons are aggregated as usually done in the literature : December-January-February (DJF) for winter, March-April-May (MAM) for spring, June-July-August (JJA) for summer, September-October-November (SON) for autumn. This analysis aims to understand whether the trend of hourly rainfall in recent years may positively or negatively impact the availability of water for hydropower generation. In addition, it may be interesting to observe what the effects of climate change may be on trends in extreme precipitation events. There is no univocal definition of extreme precipitation in the literature, and this paper considers extreme precipitation to be precipitation intensity greater than 3.6 mm/h, as defined in . Direct runoff is proportional to the amount of water that precipitates. Thus, analysis of precipitation trends over the past twenty years also allows conclusions to be drawn about the amount of direct runoff. There are two main methods for evaluating direct runoff from precipitation: the Rational method, and the Natural Resources Conservation Service (NRCS) method . The Rational method considers the intensity of precipitation (mm/h), the recurrence time of precipitation with certain intensities, and the specific catchment area considered (km 2 km 2), as shown by the following equation : Rf=c·Pi·A(mm·km 2 h)Rf=c·Pi·A mm·km 2 h (8) where Rf Rf is the direct runoff; c is called the runoff coefficient and varies according to the type of land cover and slope of the land (dimensionless); Pi Pi is the intensity of precipitation (mm/h); and A A is the extent of the drainage area (km 2 km 2). To obtain the direct runoff in Equation (8) with units equal to m 3/s m 3/s, one must multiply (8) by 0.278, which is thus a conversion factor. Regarding the runoff coefficient, the values of c referring to grassland (land cover assumed for ET 0 ET 0 evaluation) and for different land slopes, are given in the following Table 1 : Table 1. Runoff coefficient values in the case of grassland. The main limitation of the Rational method is that the considered drainage area should not exceed 0.8 km 2 km 2, so this value is here set for A. The NRCS method estimates, not the flow rate, but the amount of water available as runoff (mm) . This method, unlike the Rational one, uses the cumulative amount of precipitation (mm) as input. The equation that ‘converts’ this amount to runoff is as follows : Rf′=(Pr−I a)2 Pr−I a+S(mm)Rf′=Pr−I a 2 Pr−I a+S mm (9) where Pr Pr (mm) is the cumulative precipitation of a day; I a I a (mm), known as initial abstraction, is the loss of water to runoff due to infiltration, interception, and storage of water in surface depressions; S S (mm) is the maximum potential retention after runoff begins. I a I a and S S are related by the following empirical relationship : I a=0.2·S(mm)I a=0.2·S mm (10) S, in turn, is evaluated by the following formula : S=(1000 CN−10)·25.4(mm)S=1000 CN−10·25.4 mm (11) where CN CN (dimensionless) is the runoff curve number, that is, a numerical coefficient that depends on the total precipitation (both duration and intensity), soil moisture status prior to precipitation, soil type, land cover type, and temperature. These meteorological and hydrological variables that influence the values of CN CN, and thus the resulting runoff, are called Antecedent Runoff Conditions (ARC) and are divided into three classes: ARC I (dry conditions), ARC II (average conditions), and ARC III (wetter conditions) [45,46]. In this paper, it is assumed the ARC II class is valid. The values of CN CN in the case of grassland and different types of soil texture take the following values : CN=39 CN=39 for sands, CN=61 CN=61 for loams, CN=74 CN=74 for clay loams, and CN=80 CN=80 for clays. 2.4. Monte Carlo Approaches for Synthetic Generation of Hourly Profiles of Precipitation and Runoff, and Daily Profiles of Soil Moisture Content Analyses of power generation from both programmable and non-programmable renewable sources requires stochastic modeling. In the case of hydropower generation, the generation of synthetic time series of meteorological and hydrological variables, for stochastic water availability assessments, go in this direction. In this paper, the Monte Carlo method is used to generate hourly synthetic profiles of precipitation and runoff, along with daily SMC profiles; the latter have a daily resolution since no hourly resolution is available for them in . The Monte Carlo approach is chosen since it lends itself well to the stochastic simulations applied in this work, in which the number of variables makes the computational weight acceptable. In addition, ENTSO-E and several transmission system operators (including Italy’s Terna) apply adequacy analyses using the Monte Carlo approach; however, they use the Monte Carlo method for the stochastic generation of availability profiles, whereas in this work Monte Carlo is used for the stochastic generation of time series of meteorological and hydrological quantities. Synthetic hourly precipitation profiles (mm/h) are evaluated by fitting a probability distribution over the twenty years of available data; specifically, a family of Pearson distributions is chosen, from which a single distribution identified through the mean, standard deviation, kurtosis, and skewness of the historical data series is selected. Random numbers are then extracted according to the distributions thus identified, and synthetic hourly precipitation series are obtained as output. This procedure is repeated using each year of historical data as a basis for each season; in this way, the inter-seasonal variability of precipitation trends is considered. Specifically, to simulate the inter-annual variability of precipitation trends, the algorithm implemented in this paper randomly chooses a single year from the twenty years of data, and the just described fitting and extraction procedure is applied to it. Once the synthetic hourly precipitation hourly profiles are obtained, at each Monte Carlo simulation, they are put as inputs into Equations (8) and (9), thus also obtaining synthetic direct runoff profiles. Regarding SMC data, they are expressed in saturation fractions (range from zero to one), with one indicating a soil completely saturated. Compared to the stochastic generation of hourly precipitation profiles, a different Monte Carlo algorithm is used for the generation of synthetic daily profiles of SMC. In fact, SMC synthetic profiles are generated through inverse and bootstrap methods together. Specifically, the implemented algorithm starts by deriving an empirical cumulative distribution function (ECDF) of the historical SMC data; then, numbers belonging to the interval (0,1) are randomly generated according to a uniform distribution (U(0,1)). This set of random numbers is then placed as input to the inverse of the ECDF before being obtained, thus obtaining synthetic SMC profiles as outputs. Finally, from this set of synthetic SMC values, a number of elements equal to the number of days of the different seasons are sampled with replacement. As with synthetic precipitation profiles, in the case of SMC profiles, the algorithm randomly extracts one of twenty years of historical data in each simulation run; fifty Monte Carlo simulations are repeated for SMC profiles as well. To assess the goodness of the approaches used, two error indices expressed by the following formulas are evaluated: e mean=\|m syn−m his m syn+m his\|(dimensionless)e mean=m syn−m his m syn+m his dimensionless (12) e STD=\|s syn−s his s syn+s his\|(dimensionless)e STD=s syn−s his s syn+s his dimensionless (13) where in (12) m his m his (mm/h) is the sample mean of the historical seasonal hourly profile of the randomly chosen year and m syn m syn (mm/h) is the sample mean of the corresponding synthetic seasonal hourly profile; in (13) s his s his (mm/h) is the sample standard deviation (sample std) of the historical seasonal hourly profile and s syn s syn (mm/h) is the sample std of the corresponding synthetic seasonal hourly profile. Finally, fifty Monte Carlo simulations are carried out, for each of which the indices (12) and (13) are calculated. 2.5. Geographical Locations Case Studies The search for geographic locations to use for the case study is conducted by considering: areas where hydropower plants are present; areas afferent to the most important rivers in a given region; areas whose crop type is grassland, with sparsely forested areas; and geographic areas where snowfall is rare, since snow and its effects on the water balance of a catchment are here ignored. The methods described in the subsections above are applied to two geographical locations in the Italian regions of Sicily and Sardinia, respectively, shown in Figure 1. Figure 1. Map showing the geographical locations of the two Italian case studies Paternò (Sicily) and Busachi (Sardinia). The two sites selected are near Paternò (Sicily) and near Busachi (Sardinia), whose geographical position is 37°53′47″ N longitude 14°86′14″ E and 40°01′97″ N and 8°84′85″ E, respectively. Both sites are located in catchments of the main rivers of the two islands (in terms of catchment size), namely the Simeto (Sicily) and the Tirso (Sardinia). In addition, the sites are located near hydroelectric power plants (dams), which are among the largest hydroelectric power plants on the two islands in terms of installed capacity. The soil type in both localities is loam, and the vegetation cover type is pasture, with rare and small woodland clusters. This implies that the interception of precipitation by foliage does not decisively impact the relationship between precipitation and runoff. Satellite data reveal that the geographic site chosen for Sicily has slightly higher soil moisture saturation rates than the site in Sardinia. This is probably due to the amounts of annual precipitation, which are greater for the Sicilian site. The sites are affected by a Mediterranean climate where snowfall is very rare. This complies with the modelling of water availability, as snow and its impact on the hydrological cycle are neglected in this work. 2.6. Further Physical Considerations of General Utility The amount of water that the soil can hold in the unsaturated zone depends on the porosity of the soil, which in turn depends on the soil texture. In fact, porosity is defined as the ratio of pore volume to total soil volume ; thus, once the texture of the soil is known, its porosity value can also be known . It follows from the above definition that porosity is related to the amount of water contained in the soil. In fact, the soil is saturated when all the water fills the pore space, so denoting θ ST θ ST as the volumetric water content (VWC) of the soil at saturation (unsaturated zone), θ ST θ ST coincides with porosity. As a consequence, field capacity and the permanent wilting point are specific fractions of θ ST θ ST, and thus of porosity [33,47]; the value of this fraction, depends on the soil texture. The ET ET at first member of (2) (Section 2.1) is actual evapotranspiration, that is, it is also related to the amount of water in the soil. In fact, given the type of land cover and soil texture, evapotranspiration increases (decreases) as SMC increases (decreases), all other atmospheric variables being equal. However, the evapotranspiration evaluated through the equations in Section 2.2 does not consider the amount of water in the soil. Due to this, it is therefore necessary to estimate the actual evapotranspiration, ET 0,ACT ET 0,ACT, by tying potential ET 0 ET 0 to the VMC. For this purpose, the following function is introduced here: ET 0,ACT ET 0=min{max[θ−θ WP θ FC−θ WP,0],1}(dimensionless)ET 0,ACT ET 0=min max θ−θ WP θ FC−θ WP,0,1(dimensionless) (14) where the first member is the ratio of the actual (ET 0,ACT ET 0,ACT) to potential evapotranspiration (ET 0 ET 0), and at second member θ θ is the VMC; θ FC θ FC is the VMC corresponding to the field capacity and θ WP θ WP is that corresponding to the wilting point. Knowing ET 0 ET 0 from Equation (3) and evaluating the ratio in (14), for daily values of θ θ, it is possible to obtain the values of ET 0,ACT ET 0,ACT. Some hydrological considerations are worth drawing attention to: If the hydrological cycle was annual (rather than seasonal), then it would be possible to estimate approximately ET (2) as the difference between precipitation and runoff, being that SMC has annual cycles; From the perspective of hydropower generation, field capacity is useful water, being that more water causes both surface runoff and groundwater recharge; Since ET 0 is here calculated for grassland, the stomatal control of leaves does not come into play in the relationship between ET 0 and actual evapotranspiration . This physical aspect of grassland implies that the relationship between ET 0,ATC and ET 0 is largely given by the water content of the soil; In applying the methodologies described in this section some assumptions were implicitly made: over the several years of simulations conducted, human land use does not change, and no mining of groundwater is carried out. 3. Results Twenty years of historical data (from December 2001 to November 2021) of hourly precipitation and daily SMC time series are taken from , for the two sites of interest. Hourly values of SMC are currently (as of 2022) not available in . The following subsections illustrate the results obtained through the methods described in Section 2. 3.1. Linear Regression A polynomial of the first degree is chosen as the fitting curve, with the coefficient of determination R 2 R 2 evaluated for each season and year of the data for both sites. In addition, Spearman’s correlation coefficient (S) is used to quantify the magnitude and sign of the relationship between the ET 0 ET 0 calculated through the HS formula and the SMC data. Figure 2a,b shows two examples of the results obtained for the Paternò and Busachi case studies, respectively (95% prediction intervals are also shown). Instead, the Table 2a,b shows the mean, median, and standard deviation (STD) of the statistical quantities of interest for the two selected case studies. Figure 2. Fitting lines data points of soil moisture values versus potential evapotranspiration values for spring 2006 in Paternò (a) and autumn 2018 in Busachi (b). Additionally, plotted in purple are the 95% prediction intervals. Table 2. (a) Mean, median, and standard deviation (STD) of the correlation coefficients (Corr), coefficients of determination R 2, and the slope of the fitting line, and its intercept, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the correlation coefficients (Corr), coefficients of determination R 2, and the slope of the fitting line, and its intercept, for the Busachi case study. From the above tables, the seasonal results of the correlation coefficient statistics for the Paternò locality are consistent in magnitude and sign with those for the Busachi locality, except for the summer: in winter, ET 0 and SMC are related by a strong positive correlation (p< 0.05), while in spring and autumn, a strong anticorrelation is found (p< 0.05); in summer, a strong variability is found, going from positive to negative values (p< 0.05 or p> 0.05 depending on the year). Finally, the anticorrelation values in autumn are lower in modulus than in spring, probably due to the approaching winter. Regarding R 2 values, high and consistent values are obtained between the two locations, especially in spring and autumn: in these seasons, more than 60% of the variability of ET 0 is explained by the variability of SMC values. In winter this percentage drops to 50% but is still significant. Low R 2 values are obtained in summer, confirming the poor fitting that is obtained with a straight line for this season. The coefficients of the fitting lines show the consistency of magnitude and sign in the respective seasons over the twenty years of data, in the same case study and between the two case studies (except in summer). Moreover, the intervals of the values of the slopes and intercepts of the fitting lines of the two case studies have values in common. The statistical significance of the above results suggests that through the fitting lines found here, ET 0 values close to those of the HS formula can be obtained from the SMC data (satellite or in situ). 3.2. Precipitation Trends Among the various precipitation trends, along with their intensities, analyzed in this work (as described in Section 2.3.), significant ones are shown here, which may give indications of what precipitation trends might look similar in the years to come. First, trends related to the Paternò locality are shown, and then later those related to Busachi. Figure 3 shows the annual cumulative precipitation trends for the Sicilian location. Figure 3. Trend of cumulative annual precipitation (mm) over the past twenty years (2002–2021) for the Paternò case study. From Figure 3 it would appear that although the trend falls within a given statistical variability over the last twenty years, starting in 2014 there seem to be more frequent years when precipitation is low. It is useful then to analyze seasonal trends to understand whether precipitation trends have seasonal sensitivity, hence Figure 4a,b are shown, for the Paternò case study. Figure 4. Seasonal cumulative precipitation trends: (a) in winter and autumn, (b) in spring and summer, for the Paternò case study. From the above figures, the cumulative amount of winter precipitation has experienced a marked reduction, portending dry spells in future years this season that could impact energy production from hydroelectric plants afferent to the Paternò catchment. Cumulative autumn precipitation has experienced an increase over the past twenty years, although with a different slope than in winter. Regarding the spring trend, a reduction can be observed from 2011 onwards, but without a marked trend as observed in winter and autumn. Finally, a decreasing trend is noted in summer, with 2018 experiencing a particularly rainy summer (probably an outlier). In addition to the cumulative amount of seasonal and annual precipitation, it is useful to analyze trends in hourly intense precipitation events across seasons and years. Figure 5 shows annual, winter, and autumn trends of heavy precipitation. The summer trend is not shown being that in twenty years there has been no intense hourly rainfall, while the spring trend is not shown because the amount of data available does not allow inferring any trend for this season. Figure 5. Trends in the number of hours (events) with intense precipitation (greater than 3.6 mm/h), on an annual basis and on a seasonal basis (winter and autumn), for the Paternò case study. From Figure 5, in the last twenty years, the number of annual events with heavy precipitation seems to represent an increasing trend. Specifically, looking at the data from 2002 to 2010, there have been three years in which no intense precipitation occurred in any hour, while from 2010 until 2021 only one year (2016) recorded zero hourly intense precipitation. On the other hand, looking at the seasonal trend of heavy hourly precipitation, winter turns out to be the season with the highest density of hours with heavy precipitation, which from 2010 onward seems to have decreased compared to previous years. In contrast, the autumn pattern shows an increasing trend, which is likely to maintain this slope in the next years. A peak in the number of intense precipitation occurred in 2003: above-average air temperatures in the summer of 2003 probably led to an increase in the amount of water the atmosphere can hold, and thus to more frequent intense precipitation in the autumn. The procedure carried out for Paternò is repeated for Busachi; Figure 6 shows the annual cumulative precipitation trend. Figure 6. Trend of cumulative annual precipitation (mm) over the past twenty years (2002–2021) for the Busachi case study. As in Paternò, there seems to be a reduction in precipitation patterns for Busachi: starting in 2014 a lack of smoothness in the trend of interannual cumulative precipitation is observed. Now, it is useful to look at seasonal trends to see what the contribution of individual seasons might be to the overall annual contribution. Figure 7a,b shows the cumulative seasonal precipitation trends. Figure 7. Seasonal cumulative precipitation trends: (a) in winter and autumn, (b) in spring and summer, for the Busachi case study. Figure 7a shows how Busachi has experienced a reduction in winter cumulative precipitation over the past twenty years, while in autumn an increase has been experienced. Although the cumulative precipitation values for these two seasons are slightly lower than those of Paternò, the same trends are observed for both Busachi and Paternò. Turning then to Figure 7b, the spring and summer trends both show decreasing trends, as seen for Paternò. Indeed, 2021 was a year characterized by an outsized amount of precipitation, compared to the previous nineteen years. Regarding trends in intense precipitation, annual, spring, and autumn trends are shown in Figure 8, being that in winter no seasonal trends are observed, while in the summers of the past twenty years, there have been no heavy precipitation events. Figure 8. Trends in the number of hours (events) with intense precipitation (greater than 3.6 mm/h), on an annual basis and on a seasonal basis (spring and autumn), for the Busachi case study. From Figure 8, there seems to be an increasing annual trend for the frequency of intense precipitation events. On the other hand, regarding seasonal trends, the frequency of intense precipitation seems to increase in autumn, and this partly explains the increase in cumulative autumn precipitation across the years. In spring there seems to be a crescendo of extreme precipitation events with a cadence of a few years. 3.3. Monte Carlo Approaches In this subsection, results of stochastic time series generation of precipitation, runoff, and SMC are presented. Fifty Monte Carlo simulations are carried out for each of these physical quantities, and for each of these simulations, the errors in the mean and standard deviation are calculated. This yields samples of the errors whose mean, median, and standard deviation are calculated and reported in the Tables below. Table 3a,b shows the error statistics on the mean and standard deviation for Paternò and Busachi, respectively. Table 3. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation e STD) evaluated in the case of synthetic hourly precipitation profiles, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic hourly precipitation profiles, for the Busachi case study. The narrow values of error statistics show that the approach based on Pearson’s family of distributions appropriately simulates the statistical and physical variability of historical precipitation values, for both locations. Once the hourly synthetic precipitation profiles are obtained, they can be used as inputs to the equations for direct runoff assessments. Equation (8) is a linear combination of the precipitation values, so it is expected that the error statistics on the means and standard deviations would give results very close to those of precipitation. Regarding the direct runoff evaluated by (9), this equation binds the direct runoff nonlinearly with the precipitation, therefore, it could be expected that in this case, the results of the error statistics of the synthetic runoff profiles could deviate from the error statistics results of synthetic precipitation profiles. Table 4a,b shows the results obtained for the direct runoff with both the Rational and the NRCS methods, for the Paternò and Busachi case studies, respectively. Table 4. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic runoff profiles, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic runoff profiles, for the Busachi case study. The synthetic profiles of the direct runoff evaluated by the Rational method show error statistics very close in magnitude to those obtained for precipitation, with two exceptions for Busachi, i.e., the STD of the error on the mean and standard deviation in spring. In contrast, in the case of the NRCS method, the synthetic direct runoff error statistics provided better results on the mean than the synthetic precipitation, in both case studies. However, the error statistics on standard deviation give results that are an order of magnitude greater than the respective statistics of the synthetic precipitation profiles, and then of the synthetic runoff profiles (Rational method). This fact could be due to the nonlinear nature of the link between precipitation and direct runoff expressed by the NRCS equation. Regarding the synthetic generation of daily SMC profiles, the historical daily values of the specific seasons of the specific year are considered as data base. This implies that the synthetic generation algorithm is applied on significantly less data than hourly precipitation data. Table 5a,b shows the results obtained. Table 5. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and the standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Paternò case study, using data of individual seasons of individual years. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Busachi case study, using data of individual seasons of individual years. The error statistics on the mean reveal little deviation between the historical and simulated averages, for both locations, comparable even with the statistics obtained in the case of precipitation; on the other hand, the error statistics on the standard deviation are higher than those on the mean, even by an order of magnitude, as in summer in Paternò, and winter, spring, autumn in Busachi. These statistics of the error on the standard deviation may or not be considered acceptable depending on the objectives of a given research work: for the purposes of this paper, the percentage values of these statistics are considered acceptable. Furthermore, it is desired to understand whether, by considering more data for generating synthetic profiles of SMC, results in lower percentage errors on the standard deviation than in the previous case. For this purpose, for each season, historical data from all twenty years of data for that season are considered, thus creating four databases, one for each season. Table 6a,b shows the results obtained. Table 6. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Paternò case study, using twenty years of data for each individual season. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Busachi case study, using twenty years of data for each individual season. Looking at the errors statistics on the mean in Table 6a, it can be seen that a slight improvement in percentage values is observed for autumn, while a slight increase in percentage values is observed for the other seasons. On the other hand, percentage values of the error statistics on the standard deviation show marked improvements for all seasons, especially for winter and spring. For the Busachi case study (Table 6b), there are no significant changes in the error statistics on the mean compared to when there were fewer data in the database. On the other hand, the error statistics on the standard deviation show important reductions in the percentage values for winter and spring (as in Paternò), with improvements in summer as well; autumn turns out to be the only season to show slight worsening, the only worsening outcome between the two localities, compared to Table 5a,b. 4. Discussion There are only a few contributions in the literature on modelling hydropower generation as a function of meteorological and hydrological variables in the context of adequacy analyses. This paper attempts to fill this gap in the literature by providing tools for modelling water availability for power generation using statistical methods. The results obtained suggest that the Monte Carlo approaches applied may be suitable for simulating hourly values of water availability for hydropower generation. Indeed, by creating synthetic precipitation profiles, hourly synthetic profiles of direct runoff can be obtained, considering either the intensity of the precipitation or its accumulated amount. These simulated hourly profiles can thus be used as input for conducting adequacy analyses. In fact, simulations of hourly profiles of water availability can be used as input to hydropower generation models, thereby simulating hourly profiles of hydropower generation for adequacy analyses by researchers and transmission system operators. In addition, this approach can be used to simulate several years of hydropower plant operation, so that different possible scenarios can be considered in each simulated year with an acceptable computational effort. In this way, both researchers and transmission system operators can understand the limits within which a given hydropower plant can elastically provide its availability to meet energy demand in a given region characterized by a specific climate, thus improving the adequacy level of the market area in which the power plants are located. Furthermore, the method proposed here can be applied to hydropower plants with reservoirs of any size and flow rate. For plants that belong to low- extent catchments, the Rational Method is well-suited, while for plants that belong to extensive catchments, the NRCS method can be chosen. Regarding the results of the linear regressions presented in Section 3, the strong positive correlation found for both sites in winter can be explained by the fact that winter is modelled for the months of December, January, and February and the greater amount of precipitation that accumulates in the soil in December and January meets higher radiation intensity and temperatures in February. Indeed, more water is available to the soil for evapotranspiration due to the late winter precipitation, and more ET 0 ET 0 is available due to the higher irradiance and temperatures. The strong anti-correlation found for spring could be justified by the fact that irradiance and air temperature increase during this season, decreasing SMC on the one hand and increasing ET 0 ET 0 on the other. A strong anti-correlation was also found for autumn, but this is probably due to different physical reasons than in the case of spring. In fact, the values for irradiance and ambient temperature decrease during this season, leading in parallel to a decrease in ET 0 ET 0 and an increase in the residence time of water in the soil. The results for the summer season, however, deserve separate consideration. The different statistical character of the results for this season could be due to the fact that in summer most SMC daily values correspond to the wilting point, the amount of water not available for evaporation and transpiration, which introduces a bias into the data thus rendering meaningless the attempt to obtain statistically significant results. Indeed, inconsistency in the sign and modulus of the correlation coefficients was found between years and between the two sites. Finally, the fact that the coefficients of the fitting lines in the respective seasons (for all twenty years of data) for the two sites showed consistency in modulus and sign suggests that the fitting line found for the Paternò case study could also fit the Busachi case study and vice versa, although there is a difference in latitude of almost 3° between these two sites. This fact can be justified by the fact that the two places as well as the islands have the same climate. This makes it possible to claim that through the empirical relationships derived here, diurnal profiles of ET 0 ET 0 can be estimated from SMC diurnal data for other regions that share the same climate with the places analyzed here and whose soil textures are typical of Mediterranean flora, albeit at somewhat different latitudes than those of the case studies in this paper. The added advantage of these empirical relationships is that ET 0 ET 0 profiles can be obtained for a given weather-simulated season or year using daily SMC values as input, without having to use formulae that consider meteorological data that are not always available from satellite or in situ measurements. Analyses of seasonal and annual hourly rainfall trends, as well as evaluation of the number of extreme rainfall events, suggest that there is less precipitation in Busachi than in Paternò. This fact has implications for the statistical significance of Busachi’s historical data in terms of intense rainfall events, which are rarer in Busachi and require a longer time span than the twenty years considered (although all hourly rainfall data available in were used). Finally, the historical hourly precipitation profiles show that the cumulative precipitation amounts tend to decrease on an annual and seasonal basis at both locations studied (with the exception of autumn, where a slight increase is observed). These observations are consistent with several scientific papers that use climate model projections to define the Mediterranean as a hotspot of global warming, with a significant increase in air temperature values and increasingly frequent and prolonged droughts . As the amount of precipitation in the Mediterranean decreases, hydropower generation is increasingly constrained, leading to an increased risk of inadequate power systems in the various Mediterranean countries. Future research must therefore address the analysis of historical precipitation trends in Mediterranean countries and future climate projections for precipitation patterns so that power system operators can plan effective responses to this problem in the medium and long term. As for the Monte Carlo simulations, the synthetic hourly precipitation profiles reproduced well the statistical properties of the historical hourly profiles for both sites. Since the approach used for the synthetic generation of hourly profiles relies on statistics calculated from historical data, it can be concluded that statistically significant results can also be obtained for sites other than those considered in this paper. For the synthetic hourly profiles of direct runoff, the results obtained with the NRCS method show an order of magnitude larger standard deviation than the standard deviation of the profiles obtained with the Rational method. This can be justified by the fact that in the Rational method, the relationship between precipitation and direct runoff is linear, whereas in the NRCS method, it is non-linear. Regarding the synthetic generation of SMC daily profiles, it can be noted that overall significant improvements in the percentage values of standard deviation errors are obtained when all twenty years of available data for each season are used, compared to using data from only one year for the generation of SMC synthetic profiles. Future research should use the results of the synthetic profiles of the meteorological and hydrological variables obtained here as input for simulations of the hourly profiles of the active power generated by hydroelectric plants at the sites selected as case studies. 5. Conclusions In this paper statistical approaches to assess water availability in the context of hydropower generation and adequacy analysis on a seasonal basis are applied, two Italian locations are considered for the numerical analysis, one in Sicily and the other in Sardinia, where major hydroelectric plants are present. First, an empirical relationship between soil moisture content (SMC) and potential evapotranspiration (ET0) is evaluated through linear regression analysis. Linear regression carried out with first-degree polynomials yielded fitting lines whose coefficients have the consistency of magnitude and sign in both case studies. This fact, in addition to the high values obtained from Spearman’s correlation coefficient and the R 2 R 2 coefficient, suggests that the fitting lines evaluated in this work can be used to estimate ET 0 ET 0 from SMC data (publicly available), for geographic areas affected by the Mediterranean climate. Analyses of precipitation trends showed that at both locations there has been a reduction in cumulative precipitation over the past twenty years, with decreasing trends in winter and summer. It is expected that these trends will be confirmed in the years to come, and prolonged periods of drought in critical seasons for energy demand such as winter and summer may lead to major reductions in water availability, and thus hydropower production, with consequent adequacy risks for the Sicily and Sardinia market zones. Finally, Monte Carlo algorithms, applied to generate synthetic profiles of precipitation, direct runoff, and SMC, have proven successful in capturing the statistical variability of historical data of these meteorological quantities. Thus, through the synthetic profiles of water availability obtained in this work, stochastic analyses of hydropower production can be conducted for the purpose of the assessment of the adequacy of an electric power system. Future research should continue in this direction, including modeling snow as a natural reservoir of water availability for hydropower production, and conducting adequacy analyses on a seasonal basis, since weather quantities have marked inter-seasonal variability that impacts electricity generation and demand in different ways depending on the season. Author Contributions Conceptualization, G.M.T. and C.F.N.; methodology, G.M.T. and C.F.N.; software, G.M.T.; validation G.M.T. and C.F.N.; formal analysis, G.M.T. and C.F.N.; investigation, C.F.N. and G.M.T.; data curation, C.F.N.; writing—original draft preparation, C.F.N. and G.M.T.; writing—review and editing, G.M.T.; visualization, C.F.N.; supervision, G.M.T. All authors have read and agreed to the published version of the manuscript. Funding This research received no external funding. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. 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Trend of cumulative annual precipitation (mm) over the past twenty years (2002–2021) for the Paternò case study. Figure 4. Seasonal cumulative precipitation trends: (a) in winter and autumn, (b) in spring and summer, for the Paternò case study. Figure 5. Trends in the number of hours (events) with intense precipitation (greater than 3.6 mm/h), on an annual basis and on a seasonal basis (winter and autumn), for the Paternò case study. Figure 6. Trend of cumulative annual precipitation (mm) over the past twenty years (2002–2021) for the Busachi case study. Figure 7. Seasonal cumulative precipitation trends: (a) in winter and autumn, (b) in spring and summer, for the Busachi case study. Figure 8. Trends in the number of hours (events) with intense precipitation (greater than 3.6 mm/h), on an annual basis and on a seasonal basis (spring and autumn), for the Busachi case study. Table 1. Runoff coefficient values in the case of grassland. \| Slope Grass Cover (%) \| c Range \| :---: \| \| 0 ÷ 5 \| 0.10 ÷ 0.40 \| \| 5 ÷ 10 \| 0.16 ÷ 0.55 \| \| 10 ÷ 30 \| 0.22 ÷ 0.60 \| Table 2. (a) Mean, median, and standard deviation (STD) of the correlation coefficients (Corr), coefficients of determination R 2, and the slope of the fitting line, and its intercept, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the correlation coefficients (Corr), coefficients of determination R 2, and the slope of the fitting line, and its intercept, for the Busachi case study. (a) WinterSpring CorrR 2SlopeInterceptCorrR 2SlopeIntercept Mean 0.684 0.561 4.404−1.689−0.928 0.855−16.169 14.203 Median 0.735 0.545 3.255−0.983−0.943 0.866−14.808 13.813 STD 0.188 0.244 2.848 2.004 0.070 0.089 5.744 2.750 SummerAutumn CorrR 2SlopeInterceptCorrR 2SlopeIntercept Mean 0.006 0.120−1.106 6.518−0.850 0.715−18.518 12.245 Median−0.073 0.060−1.853 6.742−0.875 0.749−15.841 11.007 STD 0.352 0.149 5.078 2.242 0.106 0.155 8.610 3.964 (b) WinterSpring CorrR 2SlopeInterceptCorrR 2SlopeIntercept Mean 0.655 0.478 6.997−2.589−0.783 0.654−25.414 17.405 Median 0.719 0.468 6.905−2.451−0.861 0.718−24.403 18.279 STD 0.164 0.233 3.084 1.595 0.197 0.264 6.635 3.172 SummerAutumn CorrR 2SlopeInterceptCorrR 2SlopeIntercept Mean 0.352 0.196 3.221 4.420−0.758 0.612−21.693 11.880 Median 0.412 0.120 2.408 4.842−0.761 0.585−20.700 11.647 STD 0.367 0.203 5.218 2.169 0.153 0.186 7.289 2.719 Table 3. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation e STD) evaluated in the case of synthetic hourly precipitation profiles, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic hourly precipitation profiles, for the Busachi case study. (a) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.4 3.9 4.2 2.5 2.8 4.1 4.2 2.7 Median 1.6 3.3 3.3 2.3 2.1 3.4 3.3 2.3 STD 2.2 2.8 3.4 1.7 2.7 3.1 3.5 2.1 (b) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.7 3.4 6.0 3.3 3.6 3.3 5.9 3.5 Median 2.3 2.7 4.9 2.8 2.8 2.8 4.4 3.1 STD 2.0 2.5 4.3 2.6 3.1 2.6 5.0 2.9 Table 4. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic runoff profiles, for the Paternò case study. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic runoff profiles, for the Busachi case study. (a) Rational method e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.4 3.9 4.3 2.6 2.9 4.1 4.2 2.7 Median 1.5 3.2 3.3 2.3 2.1 3.5 3.3 2.3 STD 2.3 2.8 3.3 1.7 2.6 3.1 3.5 2.1 NRCS method e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 1.6 0.5 0.2 2.3 41.2 42.6 47.8 38.1 Median 0.8 0.4 0.1 1.9 42.7 43.3 49.2 38.5 STD 1.8 0.4 0.2 1.5 7.3 7.5 6.8 6.3 (b) Rational method e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.6 4.7 5.8 3.2 3.5 5.6 5.8 3.3 Median 2.2 2.5 4.5 2.8 2.7 3.0 4.2 3.1 STD 2.2 8.3 4.3 2.4 3.0 13.9 5.1 2.7 NRCS method e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 0.6 1.1 0.7 1.3 44.4 44.8 47.8 42.6 Median 0.5 0.6 0.1 1.1 44.8 44.3 47.3 42.5 STD 0.4 2.0 2.9 0.8 5.3 10.7 10.3 5.5 Table 5. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and the standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Paternò case study, using data of individual seasons of individual years. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Busachi case study, using data of individual seasons of individual years. (a) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 3.1 4.1 1.2 3.7 19.2 11.4 33.3 17.1 Median 2.8 4.0 1.2 3.6 10.3 7.6 20.4 13.9 STD 1.8 1.1 0.9 1.7 23.8 11.9 32.0 16.4 (b) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.6 3.2 2.2 2.9 31.8 15.9 28.0 33.7 Median 2.4 3.0 2.2 3.0 15.2 13.8 20.6 19.4 STD 1.5 1.0 1.2 1.4 34.7 12.6 26.2 31.5 Table 6. (a) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Paternò case study, using twenty years of data for each individual season. (b) Mean, median, and standard deviation (STD) of the errors on the mean (e mean) and standard deviation (e STD) evaluated in the case of synthetic soil moisture daily profiles, for the Busachi case study, using twenty years of data for each individual season. (a) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 5.9 6.4 1.6 2.5 8.1 7.7 30.5 16.6 Median 5.7 6.3 1.4 2.5 6.3 7.0 24.1 15.6 STD 1.6 1.2 1.0 1.2 6.7 5.3 26.7 9.9 (b) e mean(%)e STD(%) WinSprSumAutWinSprSumAut Mean 2.4 3.2 2.2 3.1 21.3 12.3 26.4 36.2 Median 2.3 3.1 2.3 2.8 13.8 7.4 24.6 28.4 STD 1.4 1.0 1.1 1.6 24.1 12.5 21.7 30.1 Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( Share and Cite MDPI and ACS Style Tina, G.M.; Nicolosi, C.F. A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Appl. Sci.2023, 13, 1986. AMA Style Tina GM, Nicolosi CF. A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Applied Sciences. 2023; 13(3):1986. Chicago/Turabian Style Tina, Giuseppe Marco, and Claudio Francesco Nicolosi. 2023. "A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses" Applied Sciences 13, no. 3: 1986. APA Style Tina, G. M., & Nicolosi, C. F. (2023). A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Applied Sciences, 13(3), 1986. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. Article Metrics No No Article Access Statistics For more information on the journal statistics, click here. Multiple requests from the same IP address are counted as one view. Zoom\|Orient\|As Lines\|As Sticks\|As Cartoon\|As Surface\|Previous Scene\|Next Scene Cite Export citation file: BibTeX) MDPI and ACS Style Tina, G.M.; Nicolosi, C.F. A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Appl. Sci.2023, 13, 1986. AMA Style Tina GM, Nicolosi CF. A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Applied Sciences. 2023; 13(3):1986. Chicago/Turabian Style Tina, Giuseppe Marco, and Claudio Francesco Nicolosi. 2023. "A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses" Applied Sciences 13, no. 3: 1986. APA Style Tina, G. M., & Nicolosi, C. F. (2023). A Statistical Assessment of Water Availability for Hydropower Generation in the Context of Adequacy Analyses. Applied Sciences, 13(3), 1986. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. clear Appl. 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1796	https://pro.arcgis.com/en/pro-app/3.4/tool-reference/3d-analyst/intersect-3d-lines.htm	Intersect 3D Lines (3D Analyst)—ArcGIS Pro \| Documentation Products ARCGIS ArcGIS Overview Esri's enterprise geospatial platform ArcGIS Online Complete SaaS mapping platform ArcGIS Pro The world's leading GIS software ArcGIS Enterprise Foundational system for GIS & mapping Developer Technology Build mapping & spatial analysis applications All products CAPABILITIES Mapping See & understand data spatially Analytics Bring location to analytics Data Management Manage, enhance & share your GIS data All capabilities BUY ARCGIS User Types Role-based access to ArcGIS Esri Store ArcGIS products from Esri How to Buy Esri products, partner products & developer subscriptions FEATURED SOLUTIONS ArcGIS for operational intelligenceArcGIS for operational intelligence is geospatial software that helps analysts move from strategic to tactical, transforming operations and intelligence data into actionable, real-time insight.Explore ArcGIS for operational intelligence Industries INDUSTRIES Architecture, Engineering & Construction Business Conservation Education Energy Utilities Facilities Management Health & Human Services National Government Natural Resources Nonprofit Public Safety Science State & Local Government Sustainable Development Telecommunications Transportation Water All industries Featured event Esri European National Security SummitRegister now to learn from national security experts and network with industry peers at the Esri European National Security Summit, November 11–13, 2025, in Berlin, Germany.Explore the event page Support & Services Support & Services Professional Services Technical Support Training SELF-SERVICE Path to Geospatial Excellence Esri Community ArcGIS Blog Documentation My Esri CONTACT US Contact Support FEATURED TRAINING New Training \| ArcGIS Experience Builder: Advanced TechniquesLevel up your GIS-infused web applications with powerful widgets, actions, and other features—without writing code.Explore class details and dates Stories ESRI STORIES WhereNext Magazine Executive-level news and insights Esri Blog Real-world, global GIS innovation Esri & The Science of Where Podcast Voices of business and technology leaders ArcUser Practical, technical resource for ArcGIS users ArcNews Industry news and ArcGIS updates ArcWatch Geospatial news, views, and trends All stories Featured STORY Historic film town gets a digital twin to promote revitalizationMagna, Utah, uses a digital twin to visualize plans and secure public buy-in for development. 3D models show existing buildings alongside design options to maintain historic charm.Read the story on the Esri Blog About About Esri About Esri Esri Programs & Initiatives Events Partners Careers Media & Analyst Relations Contact us About GIS What is GIS? 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Usage The tool works on a pairwise basis. If more than two lines intersect, the number of output points will be the mathematical combination of the intersecting lines. For example, three intersecting lines will produce three output points—one for the first and second intersection, another for the first and third intersection, and another for the second and third intersection. Likewise, if four input lines intersect, six output points will be produced—one each for the first and second, the first and third, the first and fourth, the second and third, the second and fourth, and the third and fourth. Input features must have a spatial index. Use the Add Spatial Index tool to create a new index (particularly for shapefiles) or rebuild an existing index if there is any doubt about it being correct. If 2D lines are provided as input, all vertices are assumed to be at elevation zero. When two lines with different z-values intersect because of the allowed vertical tolerance for detecting intersections, the output point will use the higher of the two elevations. Similarly, when two line segments with different z-values intersect, the output line feature will be based on the segments with the higher z-values. Some output lines may consist of multiple parts due to two input lines with multiple, discontinuous overlapping sections or lines getting closer and farther apart in the z-direction. When attributes are transferred to the output, the values placed into the first set of fields are from the first input line, and the values in the second set of fields are from the second input line. If two input line feature layers are specified, the first and second input lines are from the first and second input line feature layers, respectively. If only one input line feature layer is specified, the rule is the same, except that the columns appear to be repeated. The output point features will include the following attribute fields: I3L_TYPE—A value of 0 indicates the point marks the end point of the overlapping line segment, whereas a value of 1 indicates the point is an intersection of two lines. I3L_GROUP—A unique value that identifies all points that share the same position in horizontal space. I3L_OID1—The feature ID of the first intersecting line feature. I3L_OID2—The feature ID of the second intersecting line feature. I3L_LINE1Z—The Z value of the intersection point on the first line. I3L_LINE2Z—The Z value of the intersection point on the second line. I3L_Z_DIFF—The difference in Z between the two input lines at the XY location of the intersection. Parameters DialogPython Label Explanation Data Type Input Features The line features that will be evaluated for intersections. The input can consist of either one or two line feature layers or classes. If one input is specified, each feature will be compared with all other features in that feature class. No feature will be compared to itself.Feature Layer Maximum Z Difference (Optional)The maximum vertical distance between line segments that intersect.Linear Unit Attributes To Join (Optional)Specifies the attributes from the input features that will be transferred to the output feature class. All attributes—All the attributes from the input features will be transferred to the output feature class. This is the default. All attributes except feature IDs—All the attributes except the FID attribute from the input features will be transferred to the output feature class. Only feature IDs—Only the FID attribute from the input features will be transferred to the output feature class.String Output Point Feature Class (Optional)The output points representing the locations where the input lines intersect, including locations where overlapping line segments begin and end.Feature Class Output Line Feature Class (Optional)The output lines representing the overlapping sections that exist between the input lines.Feature Class Derived Output Label Explanation Data Type Output Intersection Count The number of intersecting locations in the input lines.Long arcpy.ddd.Intersect3DLines(in_lines, {max_z_diff}, {join_attributes}, {out_point_fc}, {out_line_fc}) Name Explanation Data Type in_lines [in_lines,...]The line features that will be evaluated for intersections. The input can consist of either one or two line feature layers or classes. If one input is specified, each feature will be compared with all other features in that feature class. No feature will be compared to itself.Feature Layer max_z_diff (Optional)The maximum vertical distance between line segments that intersect.Linear Unit join_attributes (Optional)Specifies the attributes from the input features that will be transferred to the output feature class. ALL—All the attributes from the input features will be transferred to the output feature class. This is the default. NO_FID—All the attributes except the FID attribute from the input features will be transferred to the output feature class. ONLY_FID—Only the FID attribute from the input features will be transferred to the output feature class.String out_point_fc (Optional)The output points representing the locations where the input lines intersect, including locations where overlapping line segments begin and end.Feature Class out_line_fc (Optional)The output lines representing the overlapping sections that exist between the input lines.Feature Class Derived Output Name Explanation Data Type out_intersection_count The number of intersecting locations in the input lines.Long Code sample Intersect3DLines example (Python window) The following sample demonstrates the use of this tool in the Python window: import arcpy arcpy.env.workspace = 'C:/data' arcpy.ddd.Intersect3DLines(['floor_centerlines.shp', 'stairs.shp'], '2 Meters', 'ONLY_FID', 'intersection_pts.shp') Environments Current Workspace, Scratch Workspace, Parallel Processing Factor, Extent, Output Coordinate System, Geographic Transformations, XY Resolution, XY Tolerance, Z Resolution, Z Tolerance Licensing information Basic: Requires 3D Analyst or ArcGIS Location Referencing Standard: Requires 3D Analyst or ArcGIS Location Referencing Advanced: Requires 3D Analyst or ArcGIS Location Referencing Related topics An overview of the 3D Intersections toolset Find a geoprocessing tool Feedback on this topic? 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1797	https://msmt.gov.cz/file/57691_1_1/	AP Microeconomics Course and Exam Description AP Economics Program The AP Program offers two courses in economics: AP Macroeconomics and AP Microeconomics. Each course corresponds to one semester of a typical introductory college course in economics. AP Macroeconomics focuses on the principles that apply to an economic system as a whole. AP Microeconomics focuses on the principles of economics that apply to the functions of individual decision -makers, both consumers and producers, within the economic system. AP M icroeconomics Course Overview AP Microeconomics is a college -level course that introduces students to the principles of economics that apply to the functions of individual economic decision -makers. The course also develops students’ familiarity with the operation of product and factor markets, distributions of income, market failure, and the role of government in promoting greater efficiency and equity in the economy. Students learn to use graphs, charts, and data to analyze, describe , and explain economic concepts. AP Microeconomics is equivalent to a one -semester introductory college course in economics . RECOMMENDED PREREQUISITES There are no prerequisites for AP Macroeconomics. AP Microeconomics Course Content Unit 1: Basic Economic Concepts ■ Unit 2: Supply and Demand ■ Unit 3: Production, Cost, and the Perfect Competition Model ■ Unit 4: Imperfect Competition ■ Unit 5: Factor Markets ■ Unit 6: Market Failure and the Role of Government Big Ideas Big Idea 1: Scarcity a nd Markets ■ Big Idea 2: Costs, Benefits, and Marginal Analysis ■ Big Idea 3: Production Choices and Behavior ■ Big Idea 4: Market Inefficiency and Public Policy Course Skills The following skill categories describe what skills students should develop during the course : ♦ Define economic principles and models. ■ Interpretation . ♦ Explain given economic outcomes. ■ Manipulation ♦ Determine outcomes of specific economic situations. ■ Graphing and Visuals ♦ Model economic situations using graphs or visual representations. AP Microeconomics course and exam UNIT 1 BASIC ECONOMIC CONCEPTS 1.1 Scarcity 1.2 Resource Allocation and Economic Systems 1.3 Production Possibilities Curve 1.4 Comparative Advantage and Trade 1.5 Cost -Benefit Analysis 1.6 Marginal Analysis and Consumer Choice UNIT 2 SUPPLY AND DEMAND 2.1 Demand 2.2 Supply 2.3 Price Elasticity of Demand 2.4 Price Elasticity of Supply 2.5 Other Elasticities 2.6 Market Equilibrium and Consumer and Producer Surplus 2.7 Market Disequilibrium and Changes in Equilibrium 2.8 The Effects of Government Intervention in Markets 2.9 International Trade and Public Policy UNIT 3 PRODUCTION, COST, AND THE PERFECT COMPETITION MODEL 3.1 The Production Function 3.2 Short -Run Production Costs 3.3 Long -Run Production Costs 3.4 Types of Profit 3.5 Profit Maximization 3.6 Firms´ Short -Run Decision to Produce and Long -Run Decisions to Enter or Exit a Market 3.7 Perfect Competition UNIT 4 IMPERFECT COMPETITION 4.1 Introduction to Imperfectly Competitive Markets 4.2 Monopoly 4.3 Price Discrimination 4.4. Monopolistic Competition 4.5 Oligopoly and Game Theory UNIT 5 FACTOR MARKETS 5.1 Introduction to Factor Markets 5.2 Changes in Factor Deman d and Factor Supply 5.3 Profit -Maximizing Behavior in Perfectly Competitive Factor Markets 5.4 Monopsonistic Markets UNIT 6 MARKET FAILURE AND THE ROLE OF GOVERNMENT 6.1 Socially Efficient and Inefficient Market Outcomes 6.2 Externalities 6.3 Public and Private Goods 6.4 The Effects of Government Intervention in Different Market Structures 6.5 Inequality AP Microeconomics EXAM: 2 Hours 10 minutes The AP M icroeconomics Exam assesses student understanding of the skills and learning objectives outlined in the course framework. The exam is 2 hours and 10 minutes long and includes 60 multiple -choice questions and 3 free -response questions. Další informace: AP Microeconomics Course Overview – 2 stránky -microeconomics -course - overview.pdf?course=ap -microeconomics AP Microeconomics Course at a glance – 3 strany -microeconomics -course -a-glance.pdf?course=ap - microeconomics AP Microeconomics Course and Exam Description – 18 2 stran -microeconomics -course -and -exa m- description.pdf?course=ap -microeconomics Příklady zkouškových otázek -microeconomics/exam/past -exam - questions?course=ap -microeconomics CTM, duben 2022
1798	https://byjus.com/chemistry/atomic-mass-of-elements/	The atomic mass of elements is measured with the help of unified atomic mass units. One unified atomic mass unit can be quantified as the weight of one-twelfth of the mass of a carbon-12 atom considering that it is at rest. Since protons and neutrons account for almost all of the mass of the given atom, the atomic mass of a given element is almost equal to its mass number. What is the Atomic Mass of Elements? The atomic mass of an element can be described as the total mass of one atom of the given element. Its unit is called the unified atomic mass unit and is denoted by the symbol ‘u’. Standard atomic weight is used to give the value of the mean of the atomic masses in a mixture of isotopes in a given sample of an element. Learn more ⇒Interactive Periodic Table Atomic Mass of First 30 Elements Given below is a table that lists the first 30 elements based on atomic number and their corresponding atomic mass. ATOMIC NUMBER ELEMENT ATOMIC MASS 1 Hydrogen 1.008 2 Helium 4.0026 3 Lithium 6.94 4 Beryllium 9.0122 5 Boron 10.81 6 Carbon 12.011 7 Nitrogen 14.007 8 Oxygen 15.999 9 Fluorine 18.998 10 Neon 20.180 11 Sodium 22.990 12 Magnesium 24.305 13 Aluminium 26.982 14 Silicon 28.085 15 Phosphorus 30.974 16 Sulfur 32.06 17 Chlorine 35.45 18 Argon 39.948 19 Potassium 39.098 20 Calcium 40.078 21 Scandium 44.956 22 Titanium 47.867 23 Vanadium 50.942 24 Chromium 51.996 25 Manganese 54.938 26 Iron 55.845 27 Cobalt 58.933 28 Nickel 58.693 29 Copper 63.546 30 Zinc 65.38 The Molecular mass of an element can be calculated by adding the atomic masses of each of its constituents. There are many ways to find the atomic mass of an element, but the easiest way is to look it up on the periodic table of elements. Test your Knowledge on Atomic Mass of Elements! Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Start Quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted View Quiz Answers and Analysis X Login To View Results Mobile Number Send OTP Did not receive OTP? Request OTP on Voice Call Login To View Results Name Email-id Grade Grade 1 2 3 4 5 6 7 8 9 10 11 12 City View Result Recommended Videos Periodic Table of Elements Atoms and Molecules Size of an Atom or Ion Atoms, Molecules and Ions One Mole of a substance is defined as the total number of atoms in 12 grams of Carbon-12 isotope. It is denoted by ‘mol’ and is also referred to as gram molecular weight. Frequently Asked Questions – FAQsQ1 What is atomic mass? Atomic mass is the average mass of the protons, neutrons, and electrons in an atom. Its unit is a unified atomic mass and is denoted by the symbol ‘u’. Q2 What is one amu? An Atomic mass unit or one amu is the mass unit equivalent to the one-twelfth mass of one atom of the carbon-12 isotope. Q3 Why was carbon-12 selected as the reference element for calculating atomic mass? Carbon-12 was chosen as a reference element for calculating atomic mass because it is naturally occurring and is present in abundance. Q4 Is the one amu SI unit of atomic mass? An Atomic mass unit or one amu is the physical constant accepted in the SI measurement system. Q5 What is the full form of amu? The atomic mass unit is the full form of amu. It is equivalent to the one-twelfth mass of one atom of the carbon-12 isotope. ATOMIC NUMBER \| ELEMENT \| ATOMIC MASS 1 \| Hydrogen \| 1.008 2 \| Helium \| 4.0026 3 \| Lithium \| 6.94 4 \| Beryllium \| 9.0122 5 \| Boron \| 10.81 6 \| Carbon \| 12.011 7 \| Nitrogen \| 14.007 8 \| Oxygen \| 15.999 9 \| Fluorine \| 18.998 10 \| Neon \| 20.180 11 \| Sodium \| 22.990 12 \| Magnesium \| 24.305 13 \| Aluminium \| 26.982 14 \| Silicon \| 28.085 15 \| Phosphorus \| 30.974 16 \| Sulfur \| 32.06 17 \| Chlorine \| 35.45 18 \| Argon \| 39.948 19 \| Potassium \| 39.098 20 \| Calcium \| 40.078 21 \| Scandium \| 44.956 22 \| Titanium \| 47.867 23 \| Vanadium \| 50.942 24 \| Chromium \| 51.996 25 \| Manganese \| 54.938 26 \| Iron \| 55.845 27 \| Cobalt \| 58.933 28 \| Nickel \| 58.693 29 \| Copper \| 63.546 30 \| Zinc \| 65.38 The Molecular mass of an element can be calculated by adding the atomic masses of each of its constituents. There are many ways to find the atomic mass of an element, but the easiest way is to look it up on the periodic table of elements. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below Request OTP on Voice Call Recommended Videos Periodic Table of Elements Atoms and Molecules Size of an Atom or Ion Atoms, Molecules and Ions One Mole of a substance is defined as the total number of atoms in 12 grams of Carbon-12 isotope. It is denoted by ‘mol’ and is also referred to as gram molecular weight. Frequently Asked Questions – FAQs What is atomic mass? Atomic mass is the average mass of the protons, neutrons, and electrons in an atom. Its unit is a unified atomic mass and is denoted by the symbol ‘u’. What is one amu? An Atomic mass unit or one amu is the mass unit equivalent to the one-twelfth mass of one atom of the carbon-12 isotope. Why was carbon-12 selected as the reference element for calculating atomic mass? Carbon-12 was chosen as a reference element for calculating atomic mass because it is naturally occurring and is present in abundance. Is the one amu SI unit of atomic mass? An Atomic mass unit or one amu is the physical constant accepted in the SI measurement system. What is the full form of amu? The atomic mass unit is the full form of amu. It is equivalent to the one-twelfth mass of one atom of the carbon-12 isotope. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Chemistry related queries and study materials Your result is as below Request OTP on Voice Call Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Request OTP on Voice Call Website Post My Comment Byjus is very good aap it’s amazing app This is good app to understand some concepts and this working websites helps to make notes for all the classes excellent explaination Thanks Thank u some much for th great explanation 😊 thanks Thank you 😊 Great explanation! Therefore,I have choosen BYJU’S as my learning partner. I advice u to choose BYJU’S as ur learning partner. Thanks 😊I am satisfied with this app so much thank u byjus to make study more amd more easy Thank you,Sir. thank u so much to keep my study easy thank u so much to keep my studies easy this was a very helpful aap Great app 👍👍 I am very happy with the explanation I liked this app and fell very easy in studies Great app, I’m very happy with the explanation. It is well defined. A very good explanation, find it easy to study. I love learning with byju’s. 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1799	https://pdfcoffee.com/heat-transfer-a-practical-approach-yunus-a-cengel-pdf-free.html	Heat Transfer - A Practical Approach - Yunus a. Cengel - PDFCOFFEE.COM Email: info@pdfcoffee.com Login Register English Deutsch Español Français Português Home Top Categories CAREER & MONEY PERSONAL GROWTH POLITICS & CURRENT AFFAIRS SCIENCE & TECH HEALTH & FITNESS LIFESTYLE ENTERTAINMENT BIOGRAPHIES & HISTORY FICTION Top stories Best stories Add Story My Stories Home Heat Transfer - A Practical Approach - Yunus a. Cengel Heat Transfer - A Practical Approach - Yunus a. Cengel Author / Uploaded vinodkonatham703 cen58933_fm.qxd 9/11/2002 10:56 AM Page vii CONTENTS Preface xviii Nomenclature xxvi CHAPTER TWO HEAT CONDUCTION E Views 7,600 Downloads 1,106 File size 12MB Report DMCA / Copyright DOWNLOAD FILE Recommend Stories ###### Heat-transfer-yunus-a-cengel- HEAT TRANSFER A Practical Approach YUNUS A. CENGEL SECOND EDITION cen58933_fm.qxd 9/11/2002 10:56 AM Page vii CO 7,120 6,099 12MB Read more ###### Heat Transfer Yunus a. Cengel 2nd Edition HEAT TRANSFER A Practical Approach YUNUS A. CENGEL SECOND EDITION cen58933_fm.qxd 9/11/2002 10:56 AM Page vii CO 641 82 12MB Read more ###### Heat Transfer Yunus A. Cengel 2nd Edition HEAT TRANSFER A Practical Approach YUNUS A. CENGEL SECOND EDITION cen58933_fm.qxd 9/11/2002 10:56 AM Page vii CO 807 131 12MB Read more ###### Yunus A. Cengel-Heat and Mass Transfer (SI Units) A Practical Approach, 3rd edition 3,246 1,243 43MB Read more ###### Yunus Cengel Heat and Mass Transfer 4th 1-1 Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications Fourth Edition Yunus A. Cengel & Afshin J 0 0 35MB Read more ###### Heat and Mass Transfer Yunus a Cengel Appendix cen58933_ap1.qxd 7/25/2002 1:56 PM Page 855 APPENDIX P R O P E R T Y TA B L E S A N D CHARTS (SI UNITS) 1 855 c 821 102 150KB Read more ###### SECOND EDITION SECOND EDITION SECOND EDITION HEAT TRANSFER HEAT TRANSFER HEAT TRANSFER A Practical Approach A Practical Approach A Practical Approach HEAT TRANSFER A Practical Approach YUNUS A. CENGEL www.mepcafe.com SECOND EDITION cen58933_fm.qxd 9/11/2002 10:56 9,284 6,357 11MB Read more ###### Yunus A. Çengel Heat and mass transfer HEAT AND MASS TRANSFER FUNDAMENTALS & APPLICATIONS Quotes on Ethics Without ethics, everything happens as if we were a 2,451 1,953 17MB Read more ###### Tables Heat Transfer Cengel cen29305_chAP1.qxd 11/30/05 3:13 PM Page 841 APPENDIX P R O P E R T Y TA B L E S A N D CHARTS (SI UNITS) Table A–1 321 18 299KB Read more Citation preview cen58933_fm.qxd 9/11/2002 10:56 AM Page vii CONTENTS Preface xviii Nomenclature xxvi CHAPTER TWO HEAT CONDUCTION EQUATION 61 2-1 CHAPTER ONE Steady versus Transient Heat Transfer 63 Multidimensional Heat Transfer 64 Heat Generation 66 BASICS OF HEAT TRANSFER 1 1-1 Thermodynamics and Heat Transfer 2 2-2 Application Areas of Heat Transfer 3 Historical Background 3 1-2 Engineering Heat Transfer 4 Heat and Other Forms of Energy 6 2-3 The First Law of Thermodynamics 11 Energy Balance for Closed Systems (Fixed Mass) 12 Energy Balance for Steady-Flow Systems 12 Surface Energy Balance 13 1-5 Heat Transfer Mechanisms 17 1-6 Conduction 17 2-4 Boundary and Initial Conditions 77 1 2 3 4 5 6 Thermal Conductivity 19 Thermal Diffusivity 23 2-5 1-7 Convection 25 1-8 Radiation 27 2-6 2-7 1-9 Simultaneous Heat Transfer Mechanisms 30 Specified Temperature Boundary Condition 78 Specified Heat Flux Boundary Condition 79 Convection Boundary Condition 81 Radiation Boundary Condition 82 Interface Boundary Conditions 83 Generalized Boundary Conditions 84 Solution of Steady One-Dimensional Heat Conduction Problems 86 Heat Generation in a Solid 97 Variable Thermal Conductivity, k(T) 104 Topic of Special Interest: A Brief Review of Differential Equations 107 Summary 111 References and Suggested Reading 112 Problems 113 1-10 Problem-Solving Technique 35 A Remark on Significant Digits 37 Engineering Software Packages 38 Engineering Equation Solver (EES) 39 Heat Transfer Tools (HTT) 39 Topic of Special Interest: Thermal Comfort 40 Summary 46 References and Suggested Reading 47 Problems 47 General Heat Conduction Equation 74 Rectangular Coordinates 74 Cylindrical Coordinates 75 Spherical Coordinates 76 Specific Heats of Gases, Liquids, and Solids 7 Energy Transfer 9 1-4 One-Dimensional Heat Conduction Equation 68 Heat Conduction Equation in a Large Plane Wall 68 Heat Conduction Equation in a Long Cylinder 69 Heat Conduction Equation in a Sphere 71 Combined One-Dimensional Heat Conduction Equation 72 Modeling in Heat Transfer 5 1-3 Introduction 62 CHAPTER THREE STEADY HEAT CONDUCTION 127 3-1 Steady Heat Conduction in Plane Walls 128 The Thermal Resistance Concept 129 vii cen58933_fm.qxd 9/11/2002 10:56 AM Page viii viii CONTENTS Thermal Resistance Network 131 Multilayer Plane Walls 133 3-2 3-3 3-4 Thermal Contact Resistance 138 Generalized Thermal Resistance Networks 143 Heat Conduction in Cylinders and Spheres 146 4 Complications 268 5 Human Nature 268 5-2 5-3 Boundary Conditions 274 Multilayered Cylinders and Spheres 148 3-5 3-6 Critical Radius of Insulation 153 Heat Transfer from Finned Surfaces 156 Fin Equation 157 Fin Efficiency 160 Fin Effectiveness 163 Proper Length of a Fin 165 3-7 5-4 5-5 FOUR TRANSIENT HEAT CONDUCTION 209 Lumped System Analysis 210 Criteria for Lumped System Analysis 211 Some Remarks on Heat Transfer in Lumped Systems 213 4-2 4-3 4-4 Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects 216 Transient Heat Conduction in Semi-Infinite Solids 228 Transient Heat Conduction in Multidimensional Systems 231 Topic of Special Interest: Refrigeration and Freezing of Foods 239 Summary 250 References and Suggested Reading 251 Problems 252 CHAPTER CHAPTER SIX FUNDAMENTALS OF CONVECTION 333 6-1 Physical Mechanism on Convection 334 Nusselt Number 336 6-2 Classification of Fluid Flows 337 Viscous versus Inviscid Flow 337 Internal versus External Flow 337 Compressible versus Incompressible Flow Laminar versus Turbulent Flow 338 Natural (or Unforced) versus Forced Flow Steady versus Unsteady (Transient) Flow One-, Two-, and Three-Dimensional Flows 6-3 337 338 338 338 Velocity Boundary Layer 339 Surface Shear Stress 340 6-4 Thermal Boundary Layer 341 6-5 Laminar and Turbulent Flows 342 Prandtl Number 341 FIVE NUMERICAL METHODS IN HEAT CONDUCTION 265 Reynolds Number 343 6-6 Why Numerical Methods? 266 Heat and Momentum Transfer in Turbulent Flow 343 Derivation of Differential Convection Equations 345 1 Limitations 267 2 Better Modeling 267 3 Flexibility 268 Conservation of Mass Equation 345 Conservation of Momentum Equations 346 Conservation of Energy Equation 348 6-7 5-1 Transient Heat Conduction 291 Transient Heat Conduction in a Plane Wall 293 Two-Dimensional Transient Heat Conduction 304 Topic of Special Interest: Controlling Numerical Error 309 Summary 312 References and Suggested Reading 314 Problems 314 Topic of Special Interest: Heat Transfer Through Walls and Roofs 175 Summary 185 References and Suggested Reading 186 Problems 187 4-1 Two-Dimensional Steady Heat Conduction 282 Boundary Nodes 283 Irregular Boundaries 287 Heat Transfer in Common Configurations 169 CHAPTER Finite Difference Formulation of Differential Equations 269 One-Dimensional Steady Heat Conduction 272 cen58933_fm.qxd 9/11/2002 10:56 AM Page ix ix CONTENTS 6-8 Solutions of Convection Equations for a Flat Plate 352 8-4 Constant Surface Heat Flux (q·s constant) 427 Constant Surface Temperature (Ts constant) 428 The Energy Equation 354 6-9 Nondimensionalized Convection Equations and Similarity 356 6-10 Functional Forms of Friction and Convection Coefficients 357 6-11 Analogies between Momentum and Heat Transfer 358 Summary 361 References and Suggested Reading 362 Problems 362 CHAPTER 8-5 8-6 7-2 7-3 Parallel Flow over Flat Plates 371 9-1 Friction Coefficient 372 Heat Transfer Coefficient 373 Flat Plate with Unheated Starting Length 375 Uniform Heat Flux 375 9-2 9-3 Laminar and Turbulent Flow in Tubes 422 The Entrance Region 423 Entry Lengths 425 9-4 Natural Convection from Finned Surfaces and PCBs 473 Natural Convection Cooling of Finned Surfaces (Ts constant) 473 Natural Convection Cooling of Vertical PCBs (q·s constant) 474 Mass Flow Rate through the Space between Plates 475 EIGHT Introduction 420 Mean Velocity and Mean Temperature 420 Natural Convection over Surfaces 466 Vertical Plates (Ts constant) 467 Vertical Plates (q·s constant) 467 Vertical Cylinders 467 Inclined Plates 467 Horizontal Plates 469 Horizontal Cylinders and Spheres 469 Flow across Tube Banks 389 INTERNAL FORCED CONVECTION 419 Physical Mechanism of Natural Convection 460 Equation of Motion and the Grashof Number 463 The Grashof Number 465 Flow across Cylinders and Spheres 380 CHAPTER 8-3 NINE NATURAL CONVECTION 459 Pressure Drop 392 Topic of Special Interest: Reducing Heat Transfer through Surfaces 395 Summary 406 References and Suggested Reading 407 Problems 408 8-1 8-2 CHAPTER Friction and Pressure Drag 368 Heat Transfer 370 Effect of Surface Roughness 382 Heat Transfer Coefficient 384 7-4 Turbulent Flow in Tubes 441 Rough Surfaces 442 Developing Turbulent Flow in the Entrance Region 443 Turbulent Flow in Noncircular Tubes 443 Flow through Tube Annulus 444 Heat Transfer Enhancement 444 Summary 449 References and Suggested Reading 450 Problems 452 SEVEN Drag Force and Heat Transfer in External Flow 368 Laminar Flow in Tubes 431 Pressure Drop 433 Temperature Profile and the Nusselt Number 434 Constant Surface Heat Flux 435 Constant Surface Temperature 436 Laminar Flow in Noncircular Tubes 436 Developing Laminar Flow in the Entrance Region 436 EXTERNAL FORCED CONVECTION 367 7-1 General Thermal Analysis 426 9-5 Natural Convection inside Enclosures 477 Effective Thermal Conductivity 478 Horizontal Rectangular Enclosures 479 Inclined Rectangular Enclosures 479 Vertical Rectangular Enclosures 480 Concentric Cylinders 480 Concentric Spheres 481 Combined Natural Convection and Radiation 481 cen58933_fm.qxd 9/11/2002 10:56 AM Page x x CONTENTS 9-6 Combined Natural and Forced Convection 486 Topic of Special Interest: Heat Transfer through Windows 489 Summary 499 References and Suggested Reading 500 Problems 501 11-6 Atmospheric and Solar Radiation 586 Topic of Special Interest: Solar Heat Gain through Windows 590 Summary 597 References and Suggested Reading 599 Problems 599 CHAPTER TEN C H A P T E R T W E LV E BOILING AND CONDENSATION 515 RADIATION HEAT TRANSFER 605 10-1 Boiling Heat Transfer 516 10-2 Pool Boiling 518 12-1 The View Factor 606 12-2 View Factor Relations 609 Boiling Regimes and the Boiling Curve 518 Heat Transfer Correlations in Pool Boiling 522 Enhancement of Heat Transfer in Pool Boiling 526 10-3 Flow Boiling 530 10-4 Condensation Heat Transfer 532 10-5 Film Condensation 532 Flow Regimes 534 Heat Transfer Correlations for Film Condensation 535 10-6 Film Condensation Inside Horizontal Tubes 545 10-7 Dropwise Condensation 545 Topic of Special Interest: Heat Pipes 546 Summary 551 References and Suggested Reading 553 Problems 553 1 The Reciprocity Relation 610 2 The Summation Rule 613 3 The Superposition Rule 615 4 The Symmetry Rule 616 View Factors between Infinitely Long Surfaces: The Crossed-Strings Method 618 12-3 Radiation Heat Transfer: Black Surfaces 620 12-4 Radiation Heat Transfer: Diffuse, Gray Surfaces 623 Radiosity 623 Net Radiation Heat Transfer to or from a Surface 623 Net Radiation Heat Transfer between Any Two Surfaces 625 Methods of Solving Radiation Problems 626 Radiation Heat Transfer in Two-Surface Enclosures 627 Radiation Heat Transfer in Three-Surface Enclosures 629 12-5 Radiation Shields and the Radiation Effect 635 Radiation Effect on Temperature Measurements 637 CHAPTER ELEVEN FUNDAMENTALS OF THERMAL RADIATION 561 11-1 11-2 11-3 11-4 Introduction 562 Thermal Radiation 563 Blackbody Radiation 565 Radiation Intensity 571 Solid Angle 572 Intensity of Emitted Radiation 573 Incident Radiation 574 Radiosity 575 Spectral Quantities 575 11-5 Radiative Properties 577 Emissivity 578 Absorptivity, Reflectivity, and Transmissivity 582 Kirchhoff’s Law 584 The Greenhouse Effect 585 12-6 Radiation Exchange with Emitting and Absorbing Gases 639 Radiation Properties of a Participating Medium 640 Emissivity and Absorptivity of Gases and Gas Mixtures 642 Topic of Special Interest: Heat Transfer from the Human Body 649 Summary 653 References and Suggested Reading 655 Problems 655 CHAPTER THIRTEEN HEAT EXCHANGERS 667 13-1 Types of Heat Exchangers 668 13-2 The Overall Heat Transfer Coefficient 671 Fouling Factor 674 13-3 Analysis of Heat Exchangers 678 cen58933_fm.qxd 9/11/2002 10:56 AM Page xi xi CONTENTS 13-4 The Log Mean Temperature Difference Method 680 Counter-Flow Heat Exchangers 682 Multipass and Cross-Flow Heat Exchangers: Use of a Correction Factor 683 13-5 The Effectiveness–NTU Method 690 13-6 Selection of Heat Exchangers 700 Heat Transfer Rate 700 Cost 700 Pumping Power 701 Size and Weight 701 Type 701 Materials 701 Other Considerations 702 Summary 703 References and Suggested Reading 704 Problems 705 CHAPTER FOURTEEN MASS TRANSFER 717 14-1 Introduction 718 14-2 Analogy between Heat and Mass Transfer 719 Temperature 720 Conduction 720 Heat Generation 720 Convection 721 14-3 Mass Diffusion 721 1 Mass Basis 722 2 Mole Basis 722 Special Case: Ideal Gas Mixtures 723 Fick’s Law of Diffusion: Stationary Medium Consisting of Two Species 723 14-4 14-5 14-6 14-7 14-8 Boundary Conditions 727 Steady Mass Diffusion through a Wall 732 Water Vapor Migration in Buildings 736 Transient Mass Diffusion 740 Diffusion in a Moving Medium 743 Special Case: Gas Mixtures at Constant Pressure and Temperature 747 Diffusion of Vapor through a Stationary Gas: Stefan Flow 748 Equimolar Counterdiffusion 750 14-10 Simultaneous Heat and Mass Transfer 763 Summary 769 References and Suggested Reading 771 Problems 772 CHAPTER COOLING OF ELECTRONIC EQUIPMENT 785 15-1 Introduction and History 786 15-2 Manufacturing of Electronic Equipment 787 The Chip Carrier 787 Printed Circuit Boards 789 The Enclosure 791 15-3 Cooling Load of Electronic Equipment 793 15-4 Thermal Environment 794 15-5 Electronics Cooling in Different Applications 795 15-6 Conduction Cooling 797 Conduction in Chip Carriers 798 Conduction in Printed Circuit Boards 803 Heat Frames 805 The Thermal Conduction Module (TCM) 810 15-7 Air Cooling: Natural Convection and Radiation 812 15-8 Air Cooling: Forced Convection 820 Fan Selection 823 Cooling Personal Computers 826 15-9 Liquid Cooling 833 15-10 Immersion Cooling 836 Summary 841 References and Suggested Reading 842 Problems 842 APPENDIX 1 PROPERTY TABLES AND CHARTS (SI UNITS) 855 Table A-1 Table A-2 14-9 Mass Convection 754 Analogy between Friction, Heat Transfer, and Mass Transfer Coefficients 758 Limitation on the Heat–Mass Convection Analogy 760 Mass Convection Relations 760 FIFTEEN Table A-3 Table A-4 Table A-5 Molar Mass, Gas Constant, and Critical-Point Properties 856 Boiling- and Freezing-Point Properties 857 Properties of Solid Metals 858 Properties of Solid Nonmetals 861 Properties of Building Materials 862 cen58933_fm.qxd 9/11/2002 10:56 AM Page xii xii CONTENTS Table A-6 Table A-7 Table A-8 Properties of Insulating Materials 864 Properties of Common Foods 865 Properties of Miscellaneous Materials 867 Table A-9 Properties of Saturated Water 868 Table A-10 Properties of Saturated Refrigerant-134a 869 Table A-11 Properties of Saturated Ammonia 870 Table A-12 Properties of Saturated Propane 871 Table A-13 Properties of Liquids 872 Table A-14 Properties of Liquid Metals 873 Table A-15 Properties of Air at 1 atm Pressure 874 Table A-16 Properties of Gases at 1 atm Pressure 875 Table A-17 Properties of the Atmosphere at High Altitude 877 Table A-18 Emissivities of Surfaces 878 Table A-19 Solar Radiative Properties of Materials 880 Figure A-20 The Moody Chart for the Friction Factor for Fully Developed Flow in Circular Tubes 881 APPENDIX 2 PROPERTY TABLES AND CHARTS (ENGLISH UNITS) 883 Table A-1E Molar Mass, Gas Constant, and Critical-Point Properties 884 Table A-2E Table A-3E Table A-4E Table A-5E Table A-6E Table A-7E Table A-8E Table A-9E Table A-10E Table A-11E Table A-12E Table A-13E Table A-14E Table A-15E Table A-16E Table A-17E Boiling- and Freezing-Point Properties 885 Properties of Solid Metals 886 Properties of Solid Nonmetals 889 Properties of Building Materials 890 Properties of Insulating Materials 892 Properties of Common Foods 893 Properties of Miscellaneous Materials 895 Properties of Saturated Water 896 Properties of Saturated Refrigerant-134a 897 Properties of Saturated Ammonia 898 Properties of Saturated Propane 899 Properties of Liquids 900 Properties of Liquid Metals 901 Properties of Air at 1 atm Pressure 902 Properties of Gases at 1 atm Pressure 903 Properties of the Atmosphere at High Altitude 905 APPENDIX 3 INTRODUCTION TO EES 907 INDEX 921 cen58933_fm.qxd 9/11/2002 10:56 AM Page xiii TA B L E O F E X A M P L E S CHAPTER ONE BASICS OF HEAT TRANSFER 1 Example 1-1 Example 1-2 Example 1-3 Example 1-4 Example 1-5 Example 1-6 Example 1-7 Example 1-8 Example 1-9 Example 1-10 Example 1-11 Example 1-12 Example 1-13 Example 1-14 Heating of a Copper Ball 10 Heating of Water in an Electric Teapot 14 Heat Loss from Heating Ducts in a Basement 15 Electric Heating of a House at High Elevation 16 The Cost of Heat Loss through a Roof 19 Measuring the Thermal Conductivity of a Material 23 Conversion between SI and English Units 24 Measuring Convection Heat Transfer Coefficient 26 Radiation Effect on Thermal Comfort 29 Heat Loss from a Person 31 Heat Transfer between Two Isothermal Plates 32 Heat Transfer in Conventional and Microwave Ovens 33 Heating of a Plate by Solar Energy 34 Solving a System of Equations with EES 39 CHAPTER TWO Example 2-2 Heat Generation in a Hair Dryer 67 Example 2-3 Heat Conduction through the Bottom of a Pan 72 Example 2-4 Heat Conduction in a Resistance Heater 72 Example 2-5 Cooling of a Hot Metal Ball in Air 73 Example 2-6 Heat Conduction in a Short Cylinder 76 Example 2-7 Heat Flux Boundary Condition 80 Example 2-8 Convection and Insulation Boundary Conditions 82 Example 2-9 Combined Convection and Radiation Condition 84 Example 2-10 Combined Convection, Radiation, and Heat Flux 85 Example 2-11 Heat Conduction in a Plane Wall 86 Example 2-12 A Wall with Various Sets of Boundary Conditions 88 Example 2-13 Heat Conduction in the Base Plate of an Iron 90 Example 2-14 Heat Conduction in a Solar Heated Wall 92 Example 2-15 Heat Loss through a Steam Pipe 94 Example 2-16 Heat Conduction through a Spherical Shell 96 Example 2-17 Centerline Temperature of a Resistance Heater 100 Example 2-18 Variation of Temperature in a Resistance Heater 100 Example 2-19 Heat Conduction in a Two-Layer Medium 102 HEAT CONDUCTION EQUATION 61 Example 2-1 Heat Gain by a Refrigerator 67 xiii cen58933_fm.qxd 9/11/2002 10:56 AM Page xiv xiv CONTENTS Example 2-20 Example 2-21 Variation of Temperature in a Wall with k(T) 105 Heat Conduction through a Wall with k(T) 106 CHAPTER THREE STEADY HEAT CONDUCTION 127 Example 3-1 Example 3-2 Example 3-3 Example 3-4 Example 3-5 Example 3-6 Example 3-7 Example 3-8 Example 3-9 Example 3-10 Example 3-11 Example 3-12 Example 3-13 Example 3-14 Example 3-15 Example 3-16 Example 3-17 Example 3-18 Example 3-19 Heat Loss through a Wall 134 Heat Loss through a Single-Pane Window 135 Heat Loss through Double-Pane Windows 136 Equivalent Thickness for Contact Resistance 140 Contact Resistance of Transistors 141 Heat Loss through a Composite Wall 144 Heat Transfer to a Spherical Container 149 Heat Loss through an Insulated Steam Pipe 151 Heat Loss from an Insulated Electric Wire 154 Maximum Power Dissipation of a Transistor 166 Selecting a Heat Sink for a Transistor 167 Effect of Fins on Heat Transfer from Steam Pipes 168 Heat Loss from Buried Steam Pipes 170 Heat Transfer between Hot and Cold Water Pipes 173 Cost of Heat Loss through Walls in Winter 174 The R-Value of a Wood Frame Wall 179 The R-Value of a Wall with Rigid Foam 180 The R-Value of a Masonry Wall 181 The R-Value of a Pitched Roof 182 CHAPTER FOUR TRANSIENT HEAT CONDUCTION 209 Example 4-1 Example 4-2 Example 4-3 Example 4-4 Example 4-5 Example 4-6 Example 4-7 Example 4-8 Example 4-9 Example 4-10 Example 4-11 Temperature Measurement by Thermocouples 214 Predicting the Time of Death 215 Boiling Eggs 224 Heating of Large Brass Plates in an Oven 225 Cooling of a Long Stainless Steel Cylindrical Shaft 226 Minimum Burial Depth of Water Pipes to Avoid Freezing 230 Cooling of a Short Brass Cylinder 234 Heat Transfer from a Short Cylinder 235 Cooling of a Long Cylinder by Water 236 Refrigerating Steaks while Avoiding Frostbite 238 Chilling of Beef Carcasses in a Meat Plant 248 CHAPTER FIVE NUMERICAL METHODS IN HEAT CONDUCTION 265 Example 5-1 Example 5-2 Example 5-3 Example 5-4 Example 5-5 Example 5-6 Example 5-7 Steady Heat Conduction in a Large Uranium Plate 277 Heat Transfer from Triangular Fins 279 Steady Two-Dimensional Heat Conduction in L-Bars 284 Heat Loss through Chimneys 287 Transient Heat Conduction in a Large Uranium Plate 296 Solar Energy Storage in Trombe Walls 300 Transient Two-Dimensional Heat Conduction in L-Bars 305 cen58933_fm.qxd 9/11/2002 10:56 AM Page xv xv CONTENTS CHAPTER SIX Example 8-6 FUNDAMENTALS OF CONVECTION 333 Example 6-1 Example 6-2 Temperature Rise of Oil in a Journal Bearing 350 Finding Convection Coefficient from Drag Measurement 360 CHAPTER SEVEN EXTERNAL FORCED CONVECTION 367 Example 7-1 Example 7-2 Example 7-3 Example 7-4 Example 7-5 Example 7-6 Example 7-7 Example 7-8 Example 7-9 Flow of Hot Oil over a Flat Plate 376 Cooling of a Hot Block by Forced Air at High Elevation 377 Cooling of Plastic Sheets by Forced Air 378 Drag Force Acting on a Pipe in a River 383 Heat Loss from a Steam Pipe in Windy Air 386 Cooling of a Steel Ball by Forced Air 387 Preheating Air by Geothermal Water in a Tube Bank 393 Effect of Insulation on Surface Temperature 402 Optimum Thickness of Insulation 403 Example 9-2 Example 9-3 Example 9-4 Example 9-5 Example 9-6 Example 9-7 Example 9-8 Example 9-9 BOILING AND CONDENSATION 515 Example 10-1 EIGHT INTERNAL FORCED CONVECTION 419 Example 10-3 Example 8-1 Example 10-4 Example 8-2 Example 8-3 Example 8-4 Example 8-5 Heating of Water in a Tube by Steam 430 Pressure Drop in a Pipe 438 Flow of Oil in a Pipeline through a Lake 439 Pressure Drop in a Water Pipe 445 Heating of Water by Resistance Heaters in a Tube 446 Heat Loss from Hot Water Pipes 470 Cooling of a Plate in Different Orientations 471 Optimum Fin Spacing of a Heat Sink 476 Heat Loss through a Double-Pane Window 482 Heat Transfer through a Spherical Enclosure 483 Heating Water in a Tube by Solar Energy 484 U-Factor for Center-of-Glass Section of Windows 496 Heat Loss through Aluminum Framed Windows 497 U-Factor of a Double-Door Window 498 CHAPTER TEN Example 10-2 CHAPTER NINE NATURAL CONVECTION 459 Example 9-1 CHAPTER Heat Loss from the Ducts of a Heating System 448 Example 10-5 Example 10-6 Example 10-7 Nucleate Boiling Water in a Pan 526 Peak Heat Flux in Nucleate Boiling 528 Film Boiling of Water on a Heating Element 529 Condensation of Steam on a Vertical Plate 541 Condensation of Steam on a Tilted Plate 542 Condensation of Steam on Horizontal Tubes 543 Condensation of Steam on Horizontal Tube Banks 544 cen58933_fm.qxd 9/11/2002 10:56 AM Page xvi xvi CONTENTS Example 10-8 Replacing a Heat Pipe by a Copper Rod 550 Example 12-12 Example 12-13 CHAPTER ELEVEN Example 12-14 FUNDAMENTALS OF THERMAL RADIATION 561 Example 12-15 Example 11-1 Example 11-2 Example 11-3 Example 11-4 Example 11-5 Example 11-6 Radiation Emission from a Black Ball 568 Emission of Radiation from a Lightbulb 571 Radiation Incident on a Small Surface 576 Emissivity of a Surface and Emissive Power 581 Selective Absorber and Reflective Surfaces 589 Installing Reflective Films on Windows 596 C H A P T E R T W E LV E RADIATION HEAT TRANSFER 605 Example 12-1 Example 12-2 Example 12-3 Example 12-4 Example 12-5 Example 12-6 Example 12-7 Example 12-8 Example 12-9 Example 12-10 Example 12-11 View Factors Associated with Two Concentric Spheres 614 Fraction of Radiation Leaving through an Opening 615 View Factors Associated with a Tetragon 617 View Factors Associated with a Triangular Duct 617 The Crossed-Strings Method for View Factors 619 Radiation Heat Transfer in a Black Furnace 621 Radiation Heat Transfer between Parallel Plates 627 Radiation Heat Transfer in a Cylindrical Furnace 630 Radiation Heat Transfer in a Triangular Furnace 631 Heat Transfer through a Tubular Solar Collector 632 Radiation Shields 638 Radiation Effect on Temperature Measurements 639 Effective Emissivity of Combustion Gases 646 Radiation Heat Transfer in a Cylindrical Furnace 647 Effect of Clothing on Thermal Comfort 652 CHAPTER THIRTEEN HEAT EXCHANGERS 667 Example 13-1 Example 13-2 Example 13-3 Example 13-4 Example 13-5 Example 13-6 Example 13-7 Example 13-8 Example 13-9 Example 13-10 Overall Heat Transfer Coefficient of a Heat Exchanger 675 Effect of Fouling on the Overall Heat Transfer Coefficient 677 The Condensation of Steam in a Condenser 685 Heating Water in a Counter-Flow Heat Exchanger 686 Heating of Glycerin in a Multipass Heat Exchanger 687 Cooling of an Automotive Radiator 688 Upper Limit for Heat Transfer in a Heat Exchanger 691 Using the Effectiveness– NTU Method 697 Cooling Hot Oil by Water in a Multipass Heat Exchanger 698 Installing a Heat Exchanger to Save Energy and Money 702 CHAPTER FOURTEEN MASS TRANSFER 717 Example 14-1 Example 14-2 Example 14-3 Example 14-4 Determining Mass Fractions from Mole Fractions 727 Mole Fraction of Water Vapor at the Surface of a Lake 728 Mole Fraction of Dissolved Air in Water 730 Diffusion of Hydrogen Gas into a Nickel Plate 732 cen58933_fm.qxd 9/11/2002 10:56 AM Page xvii xvii CONTENTS Example 14-5 Example 14-6 Example 14-7 Example 14-8 Example 14-9 Example 14-10 Example 14-11 Example 14-12 Example 14-13 Diffusion of Hydrogen through a Spherical Container 735 Condensation and Freezing of Moisture in the Walls 738 Hardening of Steel by the Diffusion of Carbon 742 Venting of Helium in the Atmosphere by Diffusion 751 Measuring Diffusion Coefficient by the Stefan Tube 752 Mass Convection inside a Circular Pipe 761 Analogy between Heat and Mass Transfer 762 Evaporative Cooling of a Canned Drink 765 Heat Loss from Uncovered Hot Water Baths 766 Example 15-5 Example 15-6 Example 15-7 Example 15-8 Example 15-9 Example 15-10 Example 15-11 Example 15-12 Example 15-13 Example 15-14 CHAPTER FIFTEEN COOLING OF ELECTRONIC EQUIPMENT 785 Example 15-1 Example 15-2 Example 15-3 Example 15-4 Predicting the Junction Temperature of a Transistor 788 Determining the Junction-to-Case Thermal Resistance 789 Analysis of Heat Conduction in a Chip 799 Predicting the Junction Temperature of a Device 802 Example 15-15 Example 15-16 Example 15-17 Example 15-18 Example 15-19 Heat Conduction along a PCB with Copper Cladding 804 Thermal Resistance of an Epoxy Glass Board 805 Planting Cylindrical Copper Fillings in an Epoxy Board 806 Conduction Cooling of PCBs by a Heat Frame 807 Cooling of Chips by the Thermal Conduction Module 812 Cooling of a Sealed Electronic Box 816 Cooling of a Component by Natural Convection 817 Cooling of a PCB in a Box by Natural Convection 818 Forced-Air Cooling of a Hollow-Core PCB 826 Forced-Air Cooling of a Transistor Mounted on a PCB 828 Choosing a Fan to Cool a Computer 830 Cooling of a Computer by a Fan 831 Cooling of Power Transistors on a Cold Plate by Water 835 Immersion Cooling of a Logic Chip 840 Cooling of a Chip by Boiling 840 cen58933_fm.qxd 9/11/2002 10:56 AM Page xviii PREFACE OBJECTIVES eat transfer is a basic science that deals with the rate of transfer of thermal energy. This introductory text is intended for use in a first course in heat transfer for undergraduate engineering students, and as a reference book for practicing engineers. The objectives of this text are H • To cover the basic principles of heat transfer. • To present a wealth of real-world engineering applications to give students a feel for engineering practice. • To develop an intuitive understanding of the subject matter by emphasizing the physics and physical arguments. Students are assumed to have completed their basic physics and calculus sequence. The completion of first courses in thermodynamics, fluid mechanics, and differential equations prior to taking heat transfer is desirable. The relevant concepts from these topics are introduced and reviewed as needed. In engineering practice, an understanding of the mechanisms of heat transfer is becoming increasingly important since heat transfer plays a crucial role in the design of vehicles, power plants, refrigerators, electronic devices, buildings, and bridges, among other things. Even a chef needs to have an intuitive understanding of the heat transfer mechanism in order to cook the food “right” by adjusting the rate of heat transfer. We may not be aware of it, but we already use the principles of heat transfer when seeking thermal comfort. We insulate our bodies by putting on heavy coats in winter, and we minimize heat gain by radiation by staying in shady places in summer. We speed up the cooling of hot food by blowing on it and keep warm in cold weather by cuddling up and thus minimizing the exposed surface area. That is, we already use heat transfer whether we realize it or not. GENERAL APPROACH This text is the outcome of an attempt to have a textbook for a practically oriented heat transfer course for engineering students. The text covers the standard topics of heat transfer with an emphasis on physics and real-world applications, while de-emphasizing intimidating heavy mathematical aspects. This approach is more in line with students’ intuition and makes learning the subject matter much easier. The philosophy that contributed to the warm reception of the first edition of this book has remained unchanged. The goal throughout this project has been to offer an engineering textbook that xviii cen58933_fm.qxd 9/11/2002 10:56 AM Page xix xix PREFACE • Talks directly to the minds of tomorrow’s engineers in a simple yet precise manner. • Encourages creative thinking and development of a deeper understanding of the subject matter. • Is read by students with interest and enthusiasm rather than being used as just an aid to solve problems. Special effort has been made to appeal to readers’ natural curiosity and to help students explore the various facets of the exciting subject area of heat transfer. The enthusiastic response we received from the users of the first edition all over the world indicates that our objectives have largely been achieved. Yesterday’s engineers spent a major portion of their time substituting values into the formulas and obtaining numerical results. However, now formula manipulations and number crunching are being left to computers. Tomorrow’s engineer will have to have a clear understanding and a firm grasp of the basic principles so that he or she can understand even the most complex problems, formulate them, and interpret the results. A conscious effort is made to emphasize these basic principles while also providing students with a look at how modern tools are used in engineering practice. NEW IN THIS EDITION All the popular features of the previous edition are retained while new ones are added. The main body of the text remains largely unchanged except that the coverage of forced convection is expanded to three chapters and the coverage of radiation to two chapters. Of the three applications chapters, only the Cooling of Electronic Equipment is retained, and the other two are deleted to keep the book at a reasonable size. The most significant changes in this edition are highlighted next. EXPANDED COVERAGE OF CONVECTION Forced convection is now covered in three chapters instead of one. In Chapter 6, the basic concepts of convection and the theoretical aspects are introduced. Chapter 7 deals with the practical analysis of external convection while Chapter 8 deals with the practical aspects of internal convection. See the Content Changes and Reorganization section for more details. ADDITIONAL CHAPTER ON RADIATION Radiation is now covered in two chapters instead of one. The basic concepts associated with thermal radiation, including radiation intensity and spectral quantities, are covered in Chapter 11. View factors and radiation exchange between surfaces through participating and nonparticipating media are covered in Chapter 12. See the Content Changes and Reorganization section for more details. TOPICS OF SPECIAL INTEREST Most chapters now contain a new end-of-chapter optional section called “Topic of Special Interest” where interesting applications of heat transfer are discussed. Some existing sections such as A Brief Review of Differential Equations in Chapter 2, Thermal Insulation in Chapter 7, and Controlling Numerical Error in Chapter 5 are moved to these sections as topics of special cen58933_fm.qxd 9/11/2002 10:56 AM Page xx xx PREFACE interest. Some sections from the two deleted chapters such as the Refrigeration and Freezing of Foods, Solar Heat Gain through Windows, and Heat Transfer through the Walls and Roofs are moved to the relevant chapters as special topics. Most topics selected for these sections provide real-world applications of heat transfer, but they can be ignored if desired without a loss in continuity. COMPREHENSIVE PROBLEMS WITH PARAMETRIC STUDIES A distinctive feature of this edition is the incorporation of about 130 comprehensive problems that require conducting extensive parametric studies, using the enclosed EES (or other suitable) software. Students are asked to study the effects of certain variables in the problems on some quantities of interest, to plot the results, and to draw conclusions from the results obtained. These problems are designated by computer-EES and EES-CD icons for easy recognition, and can be ignored if desired. Solutions of these problems are given in the Instructor’s Solutions Manual. CONTENT CHANGES AND REORGANIZATION With the exception of the changes already mentioned, the main body of the text remains largely unchanged. This edition involves over 500 new or revised problems. The noteworthy changes in various chapters are summarized here for those who are familiar with the previous edition. • In Chapter 1, surface energy balance is added to Section 1-4. In a new section Problem-Solving Technique, the problem-solving technique is introduced, the engineering software packages are discussed, and overviews of EES (Engineering Equation Solver) and HTT (Heat Transfer Tools) are given. The optional Topic of Special Interest in this chapter is Thermal Comfort. • In Chapter 2, the section A Brief Review of Differential Equations is moved to the end of chapter as the Topic of Special Interest. • In Chapter 3, the section on Thermal Insulation is moved to Chapter 7, External Forced Convection, as a special topic. The optional Topic of Special Interest in this chapter is Heat Transfer through Walls and Roofs. • Chapter 4 remains mostly unchanged. The Topic of Special Interest in this chapter is Refrigeration and Freezing of Foods. • In Chapter 5, the section Solutions Methods for Systems of Algebraic Equations and the FORTRAN programs in the margin are deleted, and the section Controlling Numerical Error is designated as the Topic of Special Interest. • Chapter 6, Forced Convection, is now replaced by three chapters: Chapter 6 Fundamentals of Convection, where the basic concepts of convection are introduced and the fundamental convection equations and relations (such as the differential momentum and energy equations and the Reynolds analogy) are developed; Chapter 7 External Forced Convection, where drag and heat transfer for flow over surfaces, including flow over tube banks, are discussed; and Chapter 8 Internal Forced Convection, where pressure drop and heat transfer for flow in tubes are cen58933_fm.qxd 9/11/2002 10:56 AM Page xxi xxi PREFACE • • • • • presented. Reducing Heat Transfer through Surfaces is added to Chapter 7 as the Topic of Special Interest. Chapter 7 (now Chapter 9) Natural Convection is completely rewritten. The Grashof number is derived from a momentum balance on a differential volume element, some Nusselt number relations (especially those for rectangular enclosures) are updated, and the section Natural Convection from Finned Surfaces is expanded to include heat transfer from PCBs. The optional Topic of Special Interest in this chapter is Heat Transfer through Windows. Chapter 8 (now Chapter 10) Boiling and Condensation remained largely unchanged. The Topic of Special Interest in this chapter is Heat Pipes. Chapter 9 is split in two chapters: Chapter 11 Fundamentals of Thermal Radiation, where the basic concepts associated with thermal radiation, including radiation intensity and spectral quantities, are introduced, and Chapter 12 Radiation Heat Transfer, where the view factors and radiation exchange between surfaces through participating and nonparticipating media are discussed. The Topic of Special Interest are Solar Heat Gain through Windows in Chapter 11, and Heat Transfer from the Human Body in Chapter 12. There are no significant changes in the remaining three chapters of Heat Exchangers, Mass Transfer, and Cooling of Electronic Equipment. In the appendices, the values of the physical constants are updated; new tables for the properties of saturated ammonia, refrigerant-134a, and propane are added; and the tables on the properties of air, gases, and liquids (including liquid metals) are replaced by those obtained using EES. Therefore, property values in tables for air, other ideal gases, ammonia, refrigerant-134a, propane, and liquids are identical to those obtained from EES. LEARNING TOOLS EMPHASIS ON PHYSICS A distinctive feature of this book is its emphasis on the physical aspects of subject matter rather than mathematical representations and manipulations. The author believes that the emphasis in undergraduate education should remain on developing a sense of underlying physical mechanism and a mastery of solving practical problems an engineer is likely to face in the real world. Developing an intuitive understanding should also make the course a more motivating and worthwhile experience for the students. EFFECTIVE USE OF ASSOCIATION An observant mind should have no difficulty understanding engineering sciences. After all, the principles of engineering sciences are based on our everyday experiences and experimental observations. A more physical, intuitive approach is used throughout this text. Frequently parallels are drawn between the subject matter and students’ everyday experiences so that they can relate the subject matter to what they already know. The process of cooking, for example, serves as an excellent vehicle to demonstrate the basic principles of heat transfer. cen58933_fm.qxd 9/11/2002 10:56 AM Page xxii xxii PREFACE SELF-INSTRUCTING The material in the text is introduced at a level that an average student can follow comfortably. It speaks to students, not over students. In fact, it is selfinstructive. Noting that the principles of sciences are based on experimental observations, the derivations in this text are based on physical arguments, and thus they are easy to follow and understand. EXTENSIVE USE OF ARTWORK Figures are important learning tools that help the students “get the picture.” The text makes effective use of graphics. It contains more figures and illustrations than any other book in this category. Figures attract attention and stimulate curiosity and interest. Some of the figures in this text are intended to serve as a means of emphasizing some key concepts that would otherwise go unnoticed; some serve as paragraph summaries. CHAPTER OPENERS AND SUMMARIES Each chapter begins with an overview of the material to be covered and its relation to other chapters. A summary is included at the end of each chapter for a quick review of basic concepts and important relations. NUMEROUS WORKED-OUT EXAMPLES Each chapter contains several worked-out examples that clarify the material and illustrate the use of the basic principles. An intuitive and systematic approach is used in the solution of the example problems, with particular attention to the proper use of units. A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMS The end-of-chapter problems are grouped under specific topics in the order they are covered to make problem selection easier for both instructors and students. The problems within each group start with concept questions, indicated by “C,” to check the students’ level of understanding of basic concepts. The problems under Review Problems are more comprehensive in nature and are not directly tied to any specific section of a chapter. The problems under the Design and Essay Problems title are intended to encourage students to make engineering judgments, to conduct independent exploration of topics of interest, and to communicate their findings in a professional manner. Several economics- and safety-related problems are incorporated throughout to enhance cost and safety awareness among engineering students. Answers to selected problems are listed immediately following the problem for convenience to students. A SYSTEMATIC SOLUTION PROCEDURE A well-structured approach is used in problem solving while maintaining an informal conversational style. The problem is first stated and the objectives are identified, and the assumptions made are stated together with their justifications. The properties needed to solve the problem are listed separately. Numerical values are used together with their units to emphasize that numbers without units are meaningless, and unit manipulations are as important as manipulating the numerical values with a calculator. The significance of the findings is discussed following the solutions. This approach is also used consistently in the solutions presented in the Instructor’s Solutions Manual. cen58933_fm.qxd 9/11/2002 10:56 AM Page xxiii xxiii PREFACE A CHOICE OF SI ALONE OR SI / ENGLISH UNITS In recognition of the fact that English units are still widely used in some industries, both SI and English units are used in this text, with an emphasis on SI. The material in this text can be covered using combined SI/English units or SI units alone, depending on the preference of the instructor. The property tables and charts in the appendices are presented in both units, except the ones that involve dimensionless quantities. Problems, tables, and charts in English units are designated by “E” after the number for easy recognition, and they can be ignored easily by the SI users. CONVERSION FACTORS Frequently used conversion factors and the physical constants are listed on the inner cover pages of the text for easy reference. SUPPLEMENTS These supplements are available to the adopters of the book. COSMOS SOLUTIONS MANUAL Available to instructors only. The detailed solutions for all text problems will be delivered in our new electronic Complete Online Solution Manual Organization System (COSMOS). COSMOS is a database management tool geared towards assembling homework assignments, tests and quizzes. No longer do instructors need to wade through thick solutions manuals and huge Word files. COSMOS helps you to quickly find solutions and also keeps a record of problems assigned to avoid duplication in subsequent semesters. Instructors can contact their McGraw-Hill sales representative at to obtain a copy of the COSMOS solutions manual. EES SOFTWARE Developed by Sanford Klein and William Beckman from the University of Wisconsin–Madison, this software program allows students to solve problems, especially design problems, and to ask “what if” questions. EES (pronounced “ease”) is an acronym for Engineering Equation Solver. EES is very easy to master since equations can be entered in any form and in any order. The combination of equation-solving capability and engineering property data makes EES an extremely powerful tool for students. EES can do optimization, parametric analysis, and linear and nonlinear regression and provides publication-quality plotting capability. Equations can be entered in any form and in any order. EES automatically rearranges the equations to solve them in the most efficient manner. EES is particularly useful for heat transfer problems since most of the property data needed for solving such problems are provided in the program. For example, the steam tables are implemented such that any thermodynamic property can be obtained from a built-in function call in terms of any two properties. Similar capability is provided for many organic refrigerants, ammonia, methane, carbon dioxide, and many other fluids. Air tables are built-in, as are psychrometric functions and JANAF table data for many common gases. Transport properties are also provided for all substances. EES also allows the user to enter property data or functional relationships with look-up tables, with internal functions written cen58933_fm.qxd 9/11/2002 10:56 AM Page xxiv xxiv PREFACE with EES, or with externally compiled functions written in Pascal, C, C, or FORTRAN. The Student Resources CD that accompanies the text contains the Limited Academic Version of the EES program and the scripted EES solutions of about 30 homework problems (indicated by the “EES-CD” logo in the text). Each EES solution provides detailed comments and on-line help, and can easily be modified to solve similar problems. These solutions should help students master the important concepts without the calculational burden that has been previously required. HEAT TRANSFER TOOLS (HTT) One software package specifically designed to help bridge the gap between the textbook fundamentals and commercial software packages is Heat Transfer Tools, which can be ordered “bundled” with this text (Robert J. Ribando, ISBN 0-07-246328-7). While it does not have the power and functionality of the professional, commercial packages, HTT uses research-grade numerical algorithms behind the scenes and modern graphical user interfaces. Each module is custom designed and applicable to a single, fundamental topic in heat transfer. BOOK-SPECIFIC WEBSITE The book website can be found at www.mhhe.com/cengel/. Visit this site for book and supplement information, author information, and resources for further study or reference. At this site you will also find PowerPoints of selected text figures. ACKNOWLEDGMENTS I would like to acknowledge with appreciation the numerous and valuable comments, suggestions, criticisms, and praise of these academic evaluators: Sanjeev Chandra University of Toronto, Canada Fan-Bill Cheung The Pennsylvania State University Nicole DeJong San Jose State University David M. Doner West Virginia University Institute of Technology Mark J. Holowach The Pennsylvania State University Mehmet Kanoglu Gaziantep University, Turkey Francis A. Kulacki University of Minnesota Sai C. Lau Texas A&M University Joseph Majdalani Marquette University Jed E. Marquart Ohio Northern University Robert J. Ribando University of Virginia Jay M. Ochterbeck Clemson University James R. Thomas Virginia Polytechnic Institute and State University John D. Wellin Rochester Institute of Technology cen58933_fm.qxd 9/11/2002 10:56 AM Page xxv xxv PREFACE Their suggestions have greatly helped to improve the quality of this text. I also would like to thank my students who provided plenty of feedback from their perspectives. Finally, I would like to express my appreciation to my wife Zehra and my children for their continued patience, understanding, and support throughout the preparation of this text. Yunus A. Çengel cen58933_ch01.qxd 9/10/2002 8:29 AM Page 1 CHAPTER B A S I C S O F H E AT T R A N S F E R he science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and makes no reference to how long the process will take. But in engineering, we are often interested in the rate of heat transfer, which is the topic of the science of heat transfer. We start this chapter with a review of the fundamental concepts of thermodynamics that form the framework for heat transfer. We first present the relation of heat to other forms of energy and review the first law of thermodynamics. We then present the three basic mechanisms of heat transfer, which are conduction, convection, and radiation, and discuss thermal conductivity. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent, less energetic ones as a result of interactions between the particles. Convection is the mode of heat transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. We close this chapter with a discussion of simultaneous heat transfer. T 1 CONTENTS 1–1 Thermodynamics and Heat Transfer 2 1–2 Engineering Heat Transfer 4 1–3 Heat and Other Forms of Energy 6 1–4 The First Law of Thermodynamics 11 1–5 Heat Transfer Mechanisms 17 1–6 Conduction 17 1–7 Convection 25 1–8 Radiation 27 1–9 Simultaneous Heat Transfer Mechanism 30 1–10 Problem-Solving Technique 35 Topic of Special Interest: Thermal Comfort 40 1 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 2 2 HEAT TRANSFER 1–1 Thermos bottle Hot coffee Insulation FIGURE 1–1 We are normally interested in how long it takes for the hot coffee in a thermos to cool to a certain temperature, which cannot be determined from a thermodynamic analysis alone. Hot coffee 70°C Cool environment 20°C Heat FIGURE 1–2 Heat flows in the direction of decreasing temperature. ■ THERMODYNAMICS AND HEAT TRANSFER We all know from experience that a cold canned drink left in a room warms up and a warm canned drink left in a refrigerator cools down. This is accomplished by the transfer of energy from the warm medium to the cold one. The energy transfer is always from the higher temperature medium to the lower temperature one, and the energy transfer stops when the two mediums reach the same temperature. You will recall from thermodynamics that energy exists in various forms. In this text we are primarily interested in heat, which is the form of energy that can be transferred from one system to another as a result of temperature difference. The science that deals with the determination of the rates of such energy transfers is heat transfer. You may be wondering why we need to undertake a detailed study on heat transfer. After all, we can determine the amount of heat transfer for any system undergoing any process using a thermodynamic analysis alone. The reason is that thermodynamics is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and it gives no indication about how long the process will take. A thermodynamic analysis simply tells us how much heat must be transferred to realize a specified change of state to satisfy the conservation of energy principle. In practice we are more concerned about the rate of heat transfer (heat transfer per unit time) than we are with the amount of it. For example, we can determine the amount of heat transferred from a thermos bottle as the hot coffee inside cools from 90°C to 80°C by a thermodynamic analysis alone. But a typical user or designer of a thermos is primarily interested in how long it will be before the hot coffee inside cools to 80°C, and a thermodynamic analysis cannot answer this question. Determining the rates of heat transfer to or from a system and thus the times of cooling or heating, as well as the variation of the temperature, is the subject of heat transfer (Fig. 1–1). Thermodynamics deals with equilibrium states and changes from one equilibrium state to another. Heat transfer, on the other hand, deals with systems that lack thermal equilibrium, and thus it is a nonequilibrium phenomenon. Therefore, the study of heat transfer cannot be based on the principles of thermodynamics alone. However, the laws of thermodynamics lay the framework for the science of heat transfer. The first law requires that the rate of energy transfer into a system be equal to the rate of increase of the energy of that system. The second law requires that heat be transferred in the direction of decreasing temperature (Fig. 1–2). This is like a car parked on an inclined road that must go downhill in the direction of decreasing elevation when its brakes are released. It is also analogous to the electric current flowing in the direction of decreasing voltage or the fluid flowing in the direction of decreasing total pressure. The basic requirement for heat transfer is the presence of a temperature difference. There can be no net heat transfer between two mediums that are at the same temperature. The temperature difference is the driving force for heat transfer, just as the voltage difference is the driving force for electric current flow and pressure difference is the driving force for fluid flow. The rate of heat transfer in a certain direction depends on the magnitude of the temperature gradient (the temperature difference per unit length or the rate of change of cen58933_ch01.qxd 9/10/2002 8:29 AM Page 3 3 CHAPTER 1 temperature) in that direction. The larger the temperature gradient, the higher the rate of heat transfer. Application Areas of Heat Transfer Heat transfer is commonly encountered in engineering systems and other aspects of life, and one does not need to go very far to see some application areas of heat transfer. In fact, one does not need to go anywhere. The human body is constantly rejecting heat to its surroundings, and human comfort is closely tied to the rate of this heat rejection. We try to control this heat transfer rate by adjusting our clothing to the environmental conditions. Many ordinary household appliances are designed, in whole or in part, by using the principles of heat transfer. Some examples include the electric or gas range, the heating and air-conditioning system, the refrigerator and freezer, the water heater, the iron, and even the computer, the TV, and the VCR. Of course, energy-efficient homes are designed on the basis of minimizing heat loss in winter and heat gain in summer. Heat transfer plays a major role in the design of many other devices, such as car radiators, solar collectors, various components of power plants, and even spacecraft. The optimal insulation thickness in the walls and roofs of the houses, on hot water or steam pipes, or on water heaters is again determined on the basis of a heat transfer analysis with economic consideration (Fig. 1–3). Historical Background Heat has always been perceived to be something that produces in us a sensation of warmth, and one would think that the nature of heat is one of the first things understood by mankind. But it was only in the middle of the nineteenth The human body Air-conditioning systems Circuit boards Water in Water out Car radiators Power plants Refrigeration systems FIGURE 1–3 Some application areas of heat transfer. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 4 4 HEAT TRANSFER Contact surface Hot body Cold body Caloric FIGURE 1–4 In the early nineteenth century, heat was thought to be an invisible fluid called the caloric that flowed from warmer bodies to the cooler ones. century that we had a true physical understanding of the nature of heat, thanks to the development at that time of the kinetic theory, which treats molecules as tiny balls that are in motion and thus possess kinetic energy. Heat is then defined as the energy associated with the random motion of atoms and molecules. Although it was suggested in the eighteenth and early nineteenth centuries that heat is the manifestation of motion at the molecular level (called the live force), the prevailing view of heat until the middle of the nineteenth century was based on the caloric theory proposed by the French chemist Antoine Lavoisier (1743–1794) in 1789. The caloric theory asserts that heat is a fluidlike substance called the caloric that is a massless, colorless, odorless, and tasteless substance that can be poured from one body into another (Fig. 1–4). When caloric was added to a body, its temperature increased; and when caloric was removed from a body, its temperature decreased. When a body could not contain any more caloric, much the same way as when a glass of water could not dissolve any more salt or sugar, the body was said to be saturated with caloric. This interpretation gave rise to the terms saturated liquid and saturated vapor that are still in use today. The caloric theory came under attack soon after its introduction. It maintained that heat is a substance that could not be created or destroyed. Yet it was known that heat can be generated indefinitely by rubbing one’s hands together or rubbing two pieces of wood together. In 1798, the American Benjamin Thompson (Count Rumford) (1753–1814) showed in his papers that heat can be generated continuously through friction. The validity of the caloric theory was also challenged by several others. But it was the careful experiments of the Englishman James P. Joule (1818–1889) published in 1843 that finally convinced the skeptics that heat was not a substance after all, and thus put the caloric theory to rest. Although the caloric theory was totally abandoned in the middle of the nineteenth century, it contributed greatly to the development of thermodynamics and heat transfer. 1–2 ■ ENGINEERING HEAT TRANSFER Heat transfer equipment such as heat exchangers, boilers, condensers, radiators, heaters, furnaces, refrigerators, and solar collectors are designed primarily on the basis of heat transfer analysis. The heat transfer problems encountered in practice can be considered in two groups: (1) rating and (2) sizing problems. The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference. A heat transfer process or equipment can be studied either experimentally (testing and taking measurements) or analytically (by analysis or calculations). The experimental approach has the advantage that we deal with the actual physical system, and the desired quantity is determined by measurement, within the limits of experimental error. However, this approach is expensive, time-consuming, and often impractical. Besides, the system we are analyzing may not even exist. For example, the size of a heating system of a building must usually be determined before the building is actually built on the basis of the dimensions and specifications given. The analytical approach (including numerical approach) has the advantage that it is fast and cen58933_ch01.qxd 9/10/2002 8:29 AM Page 5 5 CHAPTER 1 inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. In heat transfer studies, often a good compromise is reached by reducing the choices to just a few by analysis, and then verifying the findings experimentally. Modeling in Heat Transfer The descriptions of most scientific problems involve expressions that relate the changes in some key variables to each other. Usually the smaller the increment chosen in the changing variables, the more general and accurate the description. In the limiting case of infinitesimal or differential changes in variables, we obtain differential equations that provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. Therefore, differential equations are used to investigate a wide variety of problems in sciences and engineering, including heat transfer. However, most heat transfer problems encountered in practice can be solved without resorting to differential equations and the complications associated with them. The study of physical phenomena involves two important steps. In the first step, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. The equation itself is very instructive as it shows the degree of dependence of some variables on others, and the relative importance of various terms. In the second step, the problem is solved using an appropriate approach, and the results are interpreted. Many processes that seem to occur in nature randomly and without any order are, in fact, being governed by some visible or not-so-visible physical laws. Whether we notice them or not, these laws are there, governing consistently and predictably what seem to be ordinary events. Most of these laws are well defined and well understood by scientists. This makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and timeconsuming experiments. This is where the power of analysis lies. Very accurate results to meaningful practical problems can be obtained with relatively little effort by using a suitable and realistic mathematical model. The preparation of such models requires an adequate knowledge of the natural phenomena involved and the relevant laws, as well as a sound judgment. An unrealistic model will obviously give inaccurate and thus unacceptable results. An analyst working on an engineering problem often finds himself or herself in a position to make a choice between a very accurate but complex model, and a simple but not-so-accurate model. The right choice depends on the situation at hand. The right choice is usually the simplest model that yields adequate results. For example, the process of baking potatoes or roasting a round chunk of beef in an oven can be studied analytically in a simple way by modeling the potato or the roast as a spherical solid ball that has the properties of water (Fig. 1–5). The model is quite simple, but the results obtained are sufficiently accurate for most practical purposes. As another example, when we analyze the heat losses from a building in order to select the right size for a heater, we determine the heat losses under anticipated worst conditions and select a furnace that will provide sufficient heat to make up for those losses. Oven Potato Actual 175°C Water Ideal FIGURE 1–5 Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some accuracy. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 6 6 HEAT TRANSFER Often we tend to choose a larger furnace in anticipation of some future expansion, or just to provide a factor of safety. A very simple analysis will be adequate in this case. When selecting heat transfer equipment, it is important to consider the actual operating conditions. For example, when purchasing a heat exchanger that will handle hard water, we must consider that some calcium deposits will form on the heat transfer surfaces over time, causing fouling and thus a gradual decline in performance. The heat exchanger must be selected on the basis of operation under these adverse conditions instead of under new conditions. Preparing very accurate but complex models is usually not so difficult. But such models are not much use to an analyst if they are very difficult and timeconsuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents. There are many significant realworld problems that can be analyzed with a simple model. But it should always be kept in mind that the results obtained from an analysis are as accurate as the assumptions made in simplifying the problem. Therefore, the solution obtained should not be applied to situations for which the original assumptions do not hold. A solution that is not quite consistent with the observed nature of the problem indicates that the mathematical model used is too crude. In that case, a more realistic model should be prepared by eliminating one or more of the questionable assumptions. This will result in a more complex problem that, of course, is more difficult to solve. Thus any solution to a problem should be interpreted within the context of its formulation. 1–3 ■ HEAT AND OTHER FORMS OF ENERGY Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear, and their sum constitutes the total energy E (or e on a unit mass basis) of a system. The forms of energy related to the molecular structure of a system and the degree of the molecular activity are referred to as the microscopic energy. The sum of all microscopic forms of energy is called the internal energy of a system, and is denoted by U (or u on a unit mass basis). The international unit of energy is joule (J) or kilojoule (1 kJ 1000 J). In the English system, the unit of energy is the British thermal unit (Btu), which is defined as the energy needed to raise the temperature of 1 lbm of water at 60°F by 1°F. The magnitudes of kJ and Btu are almost identical (1 Btu 1.055056 kJ). Another well-known unit of energy is the calorie (1 cal 4.1868 J), which is defined as the energy needed to raise the temperature of 1 gram of water at 14.5°C by 1°C. Internal energy may be viewed as the sum of the kinetic and potential energies of the molecules. The portion of the internal energy of a system associated with the kinetic energy of the molecules is called sensible energy or sensible heat. The average velocity and the degree of activity of the molecules are proportional to the temperature. Thus, at higher temperatures the molecules will possess higher kinetic energy, and as a result, the system will have a higher internal energy. The internal energy is also associated with the intermolecular forces between the molecules of a system. These are the forces that bind the molecules cen58933_ch01.qxd 9/10/2002 8:29 AM Page 7 7 CHAPTER 1 to each other, and, as one would expect, they are strongest in solids and weakest in gases. If sufficient energy is added to the molecules of a solid or liquid, they will overcome these molecular forces and simply break away, turning the system to a gas. This is a phase change process and because of this added energy, a system in the gas phase is at a higher internal energy level than it is in the solid or the liquid phase. The internal energy associated with the phase of a system is called latent energy or latent heat. The changes mentioned above can occur without a change in the chemical composition of a system. Most heat transfer problems fall into this category, and one does not need to pay any attention to the forces binding the atoms in a molecule together. The internal energy associated with the atomic bonds in a molecule is called chemical (or bond) energy, whereas the internal energy associated with the bonds within the nucleus of the atom itself is called nuclear energy. The chemical and nuclear energies are absorbed or released during chemical or nuclear reactions, respectively. In the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and Pv. For the sake of simplicity and convenience, this combination is defined as enthalpy h. That is, h u Pv where the term Pv represents the flow energy of the fluid (also called the flow work), which is the energy needed to push a fluid and to maintain flow. In the energy analysis of flowing fluids, it is convenient to treat the flow energy as part of the energy of the fluid and to represent the microscopic energy of a fluid stream by enthalpy h (Fig. 1–6). Flowing fluid Stationary fluid Energy = h Energy = u FIGURE 1–6 The internal energy u represents the microscopic energy of a nonflowing fluid, whereas enthalpy h represents the microscopic energy of a flowing fluid. Specific Heats of Gases, Liquids, and Solids You may recall that an ideal gas is defined as a gas that obeys the relation Pv RT or P RT (1-1) where P is the absolute pressure, v is the specific volume, T is the absolute temperature, is the density, and R is the gas constant. It has been experimentally observed that the ideal gas relation given above closely approximates the P-v-T behavior of real gases at low densities. At low pressures and high temperatures, the density of a gas decreases and the gas behaves like an ideal gas. In the range of practical interest, many familiar gases such as air, nitrogen, oxygen, hydrogen, helium, argon, neon, and krypton and even heavier gases such as carbon dioxide can be treated as ideal gases with negligible error (often less than one percent). Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, however, should not always be treated as ideal gases since they usually exist at a state near saturation. You may also recall that specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree (Fig. 1–7). In general, this energy depends on how the process is executed. In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume Cv and specific heat at constant pressure Cp. The specific heat at constant volume Cv can be viewed as the energy required to raise the temperature of a unit mass of a substance by one degree as the volume is held constant. The energy required to do the same as the pressure is held constant is the specific heat at constant pressure Cp. The specific heat at constant m = 1 kg ∆T = 1°C Specific heat = 5 kJ/kg·°C 5 kJ FIGURE 1–7 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 8 8 THERMODYNAMICS Air m = 1 kg 300 → 301 K Air m = 1 kg 1000 → 1001 K 0.718 kJ 0.855 kJ FIGURE 1–8 The specific heat of a substance changes with temperature. pressure Cp is greater than Cv because at constant pressure the system is allowed to expand and the energy for this expansion work must also be supplied to the system. For ideal gases, these two specific heats are related to each other by Cp Cv R. A common unit for specific heats is kJ/kg · °C or kJ/kg · K. Notice that these two units are identical since ∆T(°C) ∆T(K), and 1°C change in temperature is equivalent to a change of 1 K. Also, 1 kJ/kg · °C 1 J/g · °C 1 kJ/kg · K 1 J/g · K The specific heats of a substance, in general, depend on two independent properties such as temperature and pressure. For an ideal gas, however, they depend on temperature only (Fig. 1–8). At low pressures all real gases approach ideal gas behavior, and therefore their specific heats depend on temperature only. The differential changes in the internal energy u and enthalpy h of an ideal gas can be expressed in terms of the specific heats as du Cv dT and dh Cp dT (1-2) The finite changes in the internal energy and enthalpy of an ideal gas during a process can be expressed approximately by using specific heat values at the average temperature as u Cv, aveT and h Cp, aveT (J/g) (1-3) or U mCv, aveT IRON 25°C C = Cv = Cp = 0.45 kJ/kg·°C FIGURE 1–9 The Cv and Cp values of incompressible substances are identical and are denoted by C. and H mCp, aveT (J) (1-4) where m is the mass of the system. A substance whose specific volume (or density) does not change with temperature or pressure is called an incompressible substance. The specific volumes of solids and liquids essentially remain constant during a process, and thus they can be approximated as incompressible substances without sacrificing much in accuracy. The constant-volume and constant-pressure specific heats are identical for incompressible substances (Fig. 1–9). Therefore, for solids and liquids the subscripts on Cv and Cp can be dropped and both specific heats can be represented by a single symbol, C. That is, Cp Cv C. This result could also be deduced from the physical definitions of constant-volume and constantpressure specific heats. Specific heats of several common gases, liquids, and solids are given in the Appendix. The specific heats of incompressible substances depend on temperature only. Therefore, the change in the internal energy of solids and liquids can be expressed as U mCaveT (J) (1-5) cen58933_ch01.qxd 9/10/2002 8:29 AM Page 9 9 CHAPTER 1 where Cave is the average specific heat evaluated at the average temperature. Note that the internal energy change of the systems that remain in a single phase (liquid, solid, or gas) during the process can be determined very easily using average specific heats. Energy Transfer Energy can be transferred to or from a given mass by two mechanisms: heat Q and work W. An energy interaction is heat transfer if its driving force is a temperature difference. Otherwise, it is work. A rising piston, a rotating shaft, and an electrical wire crossing the system boundaries are all associated with work interactions. Work done per unit time is called power, and is denoted · by W. The unit of power is W or hp (1 hp 746 W). Car engines and hydraulic, steam, and gas turbines produce work; compressors, pumps, and mixers consume work. Notice that the energy of a system decreases as it does work, and increases as work is done on it. In daily life, we frequently refer to the sensible and latent forms of internal energy as heat, and we talk about the heat content of bodies (Fig. 1–10). In thermodynamics, however, those forms of energy are usually referred to as thermal energy to prevent any confusion with heat transfer. The term heat and the associated phrases such as heat flow, heat addition, heat rejection, heat absorption, heat gain, heat loss, heat storage, heat generation, electrical heating, latent heat, body heat, and heat source are in common use today, and the attempt to replace heat in these phrases by thermal energy had only limited success. These phrases are deeply rooted in our vocabulary and they are used by both the ordinary people and scientists without causing any misunderstanding. For example, the phrase body heat is understood to mean the thermal energy content of a body. Likewise, heat flow is understood to mean the transfer of thermal energy, not the flow of a fluid-like substance called heat, although the latter incorrect interpretation, based on the caloric theory, is the origin of this phrase. Also, the transfer of heat into a system is frequently referred to as heat addition and the transfer of heat out of a system as heat rejection. Keeping in line with current practice, we will refer to the thermal energy as heat and the transfer of thermal energy as heat transfer. The amount of heat transferred during the process is denoted by Q. The amount of heat transferred · per unit time is called heat transfer rate, and is denoted by Q . The overdot · stands for the time derivative, or “per unit time.” The heat transfer rate Q has the unit J/s, which is equivalent to W. · When the rate of heat transfer Q is available, then the total amount of heat transfer Q during a time interval t can be determined from Q t · Q dt (J) (1-6) 0 · provided that the variation of Q with time is known. For the special case of · Q constant, the equation above reduces to · Q Q t (J) (1-7) Vapor 80°C Liquid 80°C Heat transfer 25°C FIGURE 1–10 The sensible and latent forms of internal energy can be transferred as a result of a temperature difference, and they are referred to as heat or thermal energy. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 10 10 HEAT TRANSFER . The rate of heat transfer per unit area normal to the direction of heat transfer is called heat flux, and the average heat flux is expressed as (Fig. 1–11) Q = 24 W = const. 3m A = 6 m2 Q· q· A (W/m2) (1-8) where A is the heat transfer area. The unit of heat flux in English units is Btu/h · ft2. Note that heat flux may vary with time as well as position on a surface. 2m . . Q 24 W q = — = –—— = 4 W/m2 A 6 m2 FIGURE 1–11 Heat flux is heat transfer per unit time and per unit area, and is equal · · to q· Q /A when Q is uniform over the area A. EXAMPLE 1–1 Heating of a Copper Ball A 10-cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes (Fig. 1–12). Taking the average density and specific heat of copper in this temperature range to be 8950 kg/m3 and Cp 0.395 kJ/kg · °C, respectively, determine (a) the total amount of heat transfer to the copper ball, (b) the average rate of heat transfer to the ball, and (c) the average heat flux. T2 = 150°C SOLUTION The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined. T1 = 100°C A = πD 2 Q FIGURE 1–12 Schematic for Example 1–1. Assumptions Constant properties can be used for copper at the average temperature. Properties The average density and specific heat of copper are given to be 8950 kg/m3 and Cp 0.395 kJ/kg · °C. Analysis (a) The amount of heat transferred to the copper ball is simply the change in its internal energy, and is determined from Energy transfer to the system Energy increase of the system Q U mCave (T2 T1) where m V 3 D (8950 kg/m3)(0.1 m)3 4.69 kg 6 6 Substituting, Q (4.69 kg)(0.395 kJ/kg · °C)(150 100)°C 92.6 kJ Therefore, 92.6 kJ of heat needs to be transferred to the copper ball to heat it from 100°C to 150°C. (b) The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Therefore, Q 92.6 kJ · Q ave 0.0514 kJ/s 51.4 W t 1800 s cen58933_ch01.qxd 9/10/2002 8:29 AM Page 11 11 CHAPTER 1 (c) Heat flux is defined as the heat transfer per unit time per unit area, or the rate of heat transfer per unit area. Therefore, the average heat flux in this case is Q· ave Q· ave 51.4 W q·ave 1636 W/m2 A D2 (0.1 m)2 Discussion Note that heat flux may vary with location on a surface. The value calculated above is the average heat flux over the entire surface of the ball. 1–4 ■ THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics, also known as the conservation of energy principle, states that energy can neither be created nor destroyed; it can only change forms. Therefore, every bit of energy must be accounted for during a process. The conservation of energy principle (or the energy balance) for any system undergoing any process may be expressed as follows: The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process. That is, Total energy Total energy Change in the entering the leaving the total energy of system system the system (1-9) Noting that energy can be transferred to or from a system by heat, work, and mass flow, and that the total energy of a simple compressible system consists of internal, kinetic, and potential energies, the energy balance for any system undergoing any process can be expressed as Ein Eout 1424 3 Net energy transfer by heat, work, and mass Esystem 123 (J) (1-10) (W) (1-11) Change in internal, kinetic, potential, etc., energies or, in the rate form, as · · Ein Eout 1424 3 Rate of net energy transfer by heat, work, and mass dEsystem/dt 1424 3 Rate of change in internal kinetic, potential, etc., energies Energy is a property, and the value of a property does not change unless the state of the system changes. Therefore, the energy change of a system is zero (Esystem 0) if the state of the system does not change during the process, that is, the process is steady. The energy balance in this case reduces to (Fig. 1–13) Steady, rate form: · Ein 123 Rate of net energy transfer in by heat, work, and mass · Eout 123 (1-12) Rate of net energy transfer out by heat, work, and mass In the absence of significant electric, magnetic, motion, gravity, and surface tension effects (i.e., for stationary simple compressible systems), the change · Ein · Eout Heat Work Heat Steady system Mass Work Mass · · Ein = Eout FIGURE 1–13 In steady operation, the rate of energy transfer to a system is equal to the rate of energy transfer from the system. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 12 12 HEAT TRANSFER in the total energy of a system during a process is simply the change in its internal energy. That is, Esystem Usystem. In heat transfer analysis, we are usually interested only in the forms of energy that can be transferred as a result of a temperature difference, that is, heat or thermal energy. In such cases it is convenient to write a heat balance and to treat the conversion of nuclear, chemical, and electrical energies into thermal energy as heat generation. The energy balance in that case can be expressed as Qin Qout Egen Ethermal, system 1424 3 123 1442443 Net heat transfer Heat generation (J) (1-13) Change in thermal energy of the system Energy Balance for Closed Systems (Fixed Mass) Specific heat = Cv Mass = m Initial temp = T1 Final temp = T2 Q = mCv (T1 – T2 ) FIGURE 1–14 In the absence of any work interactions, the change in the energy content of a closed system is equal to the net heat transfer. A closed system consists of a fixed mass. The total energy E for most systems encountered in practice consists of the internal energy U. This is especially the case for stationary systems since they don’t involve any changes in their velocity or elevation during a process. The energy balance relation in that case reduces to Stationary closed system: Ein Eout U mCvT (J) (1-14) where we expressed the internal energy change in terms of mass m, the specific heat at constant volume Cv, and the temperature change T of the system. When the system involves heat transfer only and no work interactions across its boundary, the energy balance relation further reduces to (Fig. 1–14) Stationary closed system, no work: Q mCvT (J) (1-15) where Q is the net amount of heat transfer to or from the system. This is the form of the energy balance relation we will use most often when dealing with a fixed mass. Energy Balance for Steady-Flow Systems A large number of engineering devices such as water heaters and car radiators involve mass flow in and out of a system, and are modeled as control volumes. Most control volumes are analyzed under steady operating conditions. The term steady means no change with time at a specified location. The opposite of steady is unsteady or transient. Also, the term uniform implies no change with position throughout a surface or region at a specified time. These meanings are consistent with their everyday usage (steady girlfriend, uniform distribution, etc.). The total energy content of a control volume during a steady-flow process remains constant (ECV constant). That is, the change in the total energy of the control volume during such a process is zero (ECV 0). Thus the amount of energy entering a control volume in all forms (heat, work, mass transfer) for a steady-flow process must be equal to the amount of energy leaving it. The amount of mass flowing through a cross section of a flow device per unit time is called the mass flow rate, and is denoted by m· . A fluid may flow in and out of a control volume through pipes or ducts. The mass flow rate of a fluid flowing in a pipe or duct is proportional to the cross-sectional area Ac of cen58933_ch01.qxd 9/10/2002 8:29 AM Page 13 13 CHAPTER 1 the pipe or duct, the density , and the velocity of the fluid. The mass flow rate through a differential area dAc can be expressed as δm· n dAc where n is the velocity component normal to dAc. The mass flow rate through the entire cross-sectional area is obtained by integration over Ac. The flow of a fluid through a pipe or duct can often be approximated to be one-dimensional. That is, the properties can be assumed to vary in one direction only (the direction of flow). As a result, all properties are assumed to be uniform at any cross section normal to the flow direction, and the properties are assumed to have bulk average values over the entire cross section. Under the one-dimensional flow approximation, the mass flow rate of a fluid flowing in a pipe or duct can be expressed as (Fig. 1–15) m· Ac (kg/s) (1-16) where is the fluid density, is the average fluid velocity in the flow direction, and Ac is the cross-sectional area of the pipe or duct. The volume of a fluid flowing through a pipe or duct per unit time is called · the volume flow rate V, and is expressed as · m· V Ac (m3/s) (kJ/s) (1-18) · where Q is the rate of net heat transfer into or out of the control volume. This is the form of the energy balance relation that we will use most often for steady-flow systems. Surface Energy Balance As mentioned in the chapter opener, heat is transferred by the mechanisms of conduction, convection, and radiation, and heat often changes vehicles as it is transferred from one medium to another. For example, the heat conducted to the outer surface of the wall of a house in winter is convected away by the cold outdoor air while being radiated to the cold surroundings. In such cases, it may be necessary to keep track of the energy interactions at the surface, and this is done by applying the conservation of energy principle to the surface. A surface contains no volume or mass, and thus no energy. Thereore, a surface can be viewed as a fictitious system whose energy content remains constant during a process (just like a steady-state or steady-flow system). Then the energy balance for a surface can be expressed as Surface energy balance: · · Ein Eout Ac = π D 2/4 for a circular pipe m· = ρAc FIGURE 1–15 The mass flow rate of a fluid at a cross section is equal to the product of the fluid density, average fluid velocity, and the cross-sectional area. (1-17) Note that the mass flow rate of a fluid through a pipe or duct remains constant during steady flow. This is not the case for the volume flow rate, however, unless the density of the fluid remains constant. For a steady-flow system with one inlet and one exit, the rate of mass flow into the control volume must be equal to the rate of mass flow out of it. That is, m· in m· out m· . When the changes in kinetic and potential energies are negligible, which is usually the case, and there is no work interaction, the energy balance for such a steady-flow system reduces to (Fig. 1–16) · Q m· h m· CpT (1-19) Control volume m· T1 m· T2 · · (T – T ) Etransfer = mC p 2 1 FIGURE 1–16 Under steady conditions, the net rate of energy transfer to a fluid in a control volume is equal to the rate of increase in the energy of the fluid stream flowing through the control volume. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 14 14 HEAT TRANSFER WALL Control surface This relation is valid for both steady and transient conditions, and the surface energy balance does not involve heat generation since a surface does not have a volume. The energy balance for the outer surface of the wall in Fig. 1–17, for example, can be expressed as radiation · · · Q1 Q2 Q3 . Q3 conduction . Q1 . Q2 convection FIGURE 1–17 Energy interactions at the outer wall surface of a house. (1-20) · · where Q 1 is conduction through the wall to the surface, Q 2 is convection from · the surface to the outdoor air, and Q 3 is net radiation from the surface to the surroundings. When the directions of interactions are not known, all energy interactions can be assumed to be towards the surface, and the surface energy balance can · be expressed as E in 0. Note that the interactions in opposite direction will end up having negative values, and balance this equation. EXAMPLE 1–2 Heating of Water in an Electric Teapot 1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipped with a 1200-W electric heating element inside (Fig. 1–18). The teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated. Electric heating element Water 15°C 1200 W FIGURE 1–18 Schematic for Example 1–2. SOLUTION Liquid water is to be heated in an electric teapot. The heating time is to be determined. Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water. Properties The average specific heats are given to be 0.7 kJ/kg · °C for the teapot and 4.18 kJ/kg · °C for water. Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in this case can be expressed as Ein Eout Esystem Ein Usystem Uwater Uteapot Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is Ein (mCT )water (mCT )teapot (1.2 kg)(4.18 kJ/kg · °C)(95 15)°C (0.5 kg)(0.7 kJ/kg · °C) (95 15)°C 429.3 kJ The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 429.3 kJ of heat is determined from t Total energy transferred Ein 429.3 kJ · 358 s 6.0 min Rate of energy transfer 1.2 kJ/s E transfer cen58933_ch01.qxd 9/10/2002 8:29 AM Page 15 15 CHAPTER 1 Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable during heating. EXAMPLE 1–3 Heat Loss from Heating Ducts in a Basement A 5-m-long section of an air heating system of a house passes through an unheated space in the basement (Fig. 1–19). The cross section of the rectangular duct of the heating system is 20 cm 25 cm. Hot air enters the duct at 100 kPa and 60°C at an average velocity of 5 m/s. The temperature of the air in the duct drops to 54°C as a result of heat loss to the cool space in the basement. Determine the rate of heat loss from the air in the duct to the basement under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace that has an efficiency of 80 percent, and the cost of the natural gas in that area is $0.60/therm (1 therm 100,000 Btu 105,500 kJ). SOLUTION The temperature of the air in the heating duct of a house drops as a result of heat loss to the cool space in the basement. The rate of heat loss from the hot air and its cost are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air can be treated as an ideal gas with constant properties at room temperature. Properties The constant pressure specific heat of air at the average temperature of (54 60)/2 57°C is 1.007 kJ/kg · °C (Table A-15). Analysis We take the basement section of the heating system as our system, which is a steady-flow system. The rate of heat loss from the air in the duct can be determined from · Q m· Cp T where m· is the mass flow rate and T is the temperature drop. The density of air at the inlet conditions is 100 kPa P 1.046 kg/m3 RT (0.287 kPa · m3/kg · K)(60 273)K The cross-sectional area of the duct is Ac (0.20 m)(0.25 m) 0.05 m2 Then the mass flow rate of air through the duct and the rate of heat loss become m· Ac (1.046 kg/m3)(5 m/s)(0.05 m2) 0.2615 kg/s and · Q loss m· Cp(Tin Tout) (0.2615 kg/s)(1.007 kJ/kg · °C)(60 54)°C 1.580 kJ/s 5m 20 cm Hot air 100 kPa 60°C 5 m/s 54°C 25 cm · Qloss FIGURE 1–19 Schematic for Example 1–3. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16 16 HEAT TRANSFER or 5688 kJ/h. The cost of this heat loss to the home owner is (Rate of heat loss)(Unit cost of energy input) Furnace efficiency (5688 kJ/h)($0.60/therm) 1 therm 0.80 105,500 kJ Cost of heat loss $0.040/h Discussion The heat loss from the heating ducts in the basement is costing the home owner 4 cents per hour. Assuming the heater operates 2000 hours during a heating season, the annual cost of this heat loss adds up to $80. Most of this money can be saved by insulating the heating ducts in the unheated areas. EXAMPLE 1–4 Patm = 12.2 psia 9 ft 70°F 50°F 40 ft 50 ft FIGURE 1–20 Schematic for Example 1–4. Electric Heating of a House at High Elevation Consider a house that has a floor space of 2000 ft2 and an average height of 9 ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia (Fig. 1–20). Initially the house is at a uniform temperature of 50°F. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 70°F. Determine the amount of energy transferred to the air assuming (a) the house is air-tight and thus no air escapes during the heating process and (b) some air escapes through the cracks as the heated air in the house expands at constant pressure. Also determine the cost of this heat for each case if the cost of electricity in that area is $0.075/kWh. SOLUTION The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constant-volume and constant-pressure cases. Assumptions 1 Air can be treated as an ideal gas with constant properties at room temperature. 2 Heat loss from the house during heating is negligible. 3 The volume occupied by the furniture and other things is negligible. Properties The specific heats of air at the average temperature of (50 70)/2 60°F are Cp 0.240 Btu/lbm · °F and Cv Cp R 0.171 Btu/lbm · °F (Tables A-1E and A-15E). Analysis The volume and the mass of the air in the house are V (Floor area)(Height) (2000 ft2)(9 ft) 18,000 ft3 (12.2 psia)(18,000 ft3) PV m 1162 lbm RT (0.3704 psia · ft3/lbm · R)(50 460)R (a) The amount of energy transferred to air at constant volume is simply the change in its internal energy, and is determined from Ein Eout Esystem Ein, constant volume Uair mCv T (1162 lbm)(0.171 Btu/lbm · °F)(70 50)°F 3974 Btu At a unit cost of $0.075/kWh, the total cost of this energy is cen58933_ch01.qxd 9/10/2002 8:29 AM Page 17 17 CHAPTER 1 Cost of energy (Amount of energy)(Unit cost of energy) 1 kWh (3974 Btu)($0.075/kWh) 3412 Btu $0.087 (b) The amount of energy transferred to air at constant pressure is the change in its enthalpy, and is determined from Ein, constant pressure Hair mCpT (1162 lbm)(0.240 Btu/lbm · °F)(70 50)°F 5578 Btu At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy (Amount of energy)(Unit cost of energy) 1 kWh (5578 Btu)($0.075/kWh) 3412 Btu $0.123 Discussion It will cost about 12 cents to raise the temperature of the air in this house from 50°F to 70°F. The second answer is more realistic since every house has cracks, especially around the doors and windows, and the pressure in the house remains essentially constant during a heating process. Therefore, the second approach is used in practice. This conservative approach somewhat overpredicts the amount of energy used, however, since some of the air will escape through the cracks before it is heated to 70°F. 1–5 ■ HEAT TRANSFER MECHANISMS In Section 1–1 we defined heat as the form of energy that can be transferred from one system to another as a result of temperature difference. A thermodynamic analysis is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. The science that deals with the determination of the rates of such energy transfers is the heat transfer. The transfer of energy as heat is always from the higher-temperature medium to the lower-temperature one, and heat transfer stops when the two mediums reach the same temperature. Heat can be transferred in three different modes: conduction, convection, and radiation. All modes of heat transfer require the existence of a temperature difference, and all modes are from the high-temperature medium to a lower-temperature one. Below we give a brief description of each mode. A detailed study of these modes is given in later chapters of this text. 1–6 ■ CONDUCTION Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Conduction can take place in solids, liquids, or gases. In gases and liquids, conduction is due to the collisions and diffusion of the cen58933_ch01.qxd 9/10/2002 8:29 AM Page 18 18 HEAT TRANSFER T1 T2 . Q A A ∆x 0 x FIGURE 1–21 Heat conduction through a large plane wall of thickness x and area A. molecules during their random motion. In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. A cold canned drink in a warm room, for example, eventually warms up to the room temperature as a result of heat transfer from the room to the drink through the aluminum can by conduction. The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, and the material of the medium, as well as the temperature difference across the medium. We know that wrapping a hot water tank with glass wool (an insulating material) reduces the rate of heat loss from the tank. The thicker the insulation, the smaller the heat loss. We also know that a hot water tank will lose heat at a higher rate when the temperature of the room housing the tank is lowered. Further, the larger the tank, the larger the surface area and thus the rate of heat loss. Consider steady heat conduction through a large plane wall of thickness x L and area A, as shown in Fig. 1–21. The temperature difference across the wall is T T2 T1. Experiments have shown that the rate of heat trans· fer Q through the wall is doubled when the temperature difference T across the wall or the area A normal to the direction of heat transfer is doubled, but is halved when the wall thickness L is doubled. Thus we conclude that the rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, but is inversely proportional to the thickness of the layer. That is, Rate of heat conduction (Area)(Temperature difference) Thickness or, T1 T2 · T Q cond kA kA x x 30°C 20°C . q = 4010 W/m 2 1m (a) Copper (k = 401 W/m·°C) 30°C 20°C . q = 1480 W/m 2 1m (b) Silicon (k = 148 W/m·°C) FIGURE 1–22 The rate of heat conduction through a solid is directly proportional to its thermal conductivity. (W) (1-21) where the constant of proportionality k is the thermal conductivity of the material, which is a measure of the ability of a material to conduct heat (Fig. 1–22). In the limiting case of x → 0, the equation above reduces to the differential form dT · Q cond kA dx (W) (1-22) which is called Fourier’s law of heat conduction after J. Fourier, who expressed it first in his heat transfer text in 1822. Here dT/dx is the temperature gradient, which is the slope of the temperature curve on a T-x diagram (the rate of change of T with x), at location x. The relation above indicates that the rate of heat conduction in a direction is proportional to the temperature gradient in that direction. Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. The negative sign in Eq. 1–22 ensures that heat transfer in the positive x direction is a positive quantity. The heat transfer area A is always normal to the direction of heat transfer. For heat loss through a 5-m-long, 3-m-high, and 25-cm-thick wall, for example, the heat transfer area is A 15 m2. Note that the thickness of the wall has no effect on A (Fig. 1–23). cen58933_ch01.qxd 9/10/2002 8:29 AM Page 19 19 CHAPTER 1 EXAMPLE 1–5 The Cost of Heat Loss through a Roof The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k 0.8 W/m · °C (Fig. 1–24). The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine (a) the rate of heat loss through the roof that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh. SOLUTION The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined. Assumptions 1 Steady operating conditions exist during the entire night since the surface temperatures of the roof remain constant at the specified values. 2 Constant properties can be used for the roof. Properties The thermal conductivity of the roof is given to be k 0.8 W/m · °C. Analysis (a) Noting that heat transfer through the roof is by conduction and the area of the roof is A 6 m 8 m 48 m2, the steady rate of heat transfer through the roof is determined to be T1 T2 (15 4)°C · Q kA (0.8 W/m · °C)(48 m2) 1690 W 1.69 kW L 0.25 m H A=W×H · Q W L FIGURE 1–23 In heat conduction analysis, A represents the area normal to the direction of heat transfer. Concrete roof 6m 0.25 m 8m 4°C 15°C (b) The amount of heat lost through the roof during a 10-hour period and its cost are determined from · Q Q t (1.69 kW)(10 h) 16.9 kWh Cost (Amount of energy)(Unit cost of energy) (16.9 kWh)($0.08/kWh) $1.35 Discussion The cost to the home owner of the heat loss through the roof that night was $1.35. The total heating bill of the house will be much larger since the heat losses through the walls are not considered in these calculations. Thermal Conductivity We have seen that different materials store heat differently, and we have defined the property specific heat Cp as a measure of a material’s ability to store thermal energy. For example, Cp 4.18 kJ/kg · °C for water and Cp 0.45 kJ/kg · °C for iron at room temperature, which indicates that water can store almost 10 times the energy that iron can per unit mass. Likewise, the thermal conductivity k is a measure of a material’s ability to conduct heat. For example, k 0.608 W/m · °C for water and k 80.2 W/m · °C for iron at room temperature, which indicates that iron conducts heat more than 100 times faster than water can. Thus we say that water is a poor heat conductor relative to iron, although water is an excellent medium to store thermal energy. Equation 1–22 for the rate of conduction heat transfer under steady conditions can also be viewed as the defining equation for thermal conductivity. Thus the thermal conductivity of a material can be defined as the rate of FIGURE 1–24 Schematic for Example 1–5. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 20 20 HEAT TRANSFER TABLE 1–1 The thermal conductivities of some materials at room temperature Material k, W/m · °C Diamond Silver Copper Gold Aluminum Iron Mercury (l) Glass Brick Water (l) Human skin Wood (oak) Helium (g) Soft rubber Glass fiber Air (g) Urethane, rigid foam 2300 429 401 317 237 80.2 8.54 0.78 0.72 0.613 0.37 0.17 0.152 0.13 0.043 0.026 0.026 Multiply by 0.5778 to convert to Btu/h · ft · °F. Electric heater Insulation Insulation T1 Sample material k . . T2 Q = We A . L We Insulation . L Q k = ———— A(T1 – T2) FIGURE 1–25 A simple experimental setup to determine the thermal conductivity of a material. heat transfer through a unit thickness of the material per unit area per unit temperature difference. The thermal conductivity of a material is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator. The thermal conductivities of some common materials at room temperature are given in Table 1–1. The thermal conductivity of pure copper at room temperature is k 401 W/m · °C, which indicates that a 1-m-thick copper wall will conduct heat at a rate of 401 W per m2 area per °C temperature difference across the wall. Note that materials such as copper and silver that are good electric conductors are also good heat conductors, and have high values of thermal conductivity. Materials such as rubber, wood, and styrofoam are poor conductors of heat and have low conductivity values. A layer of material of known thickness and area can be heated from one side by an electric resistance heater of known output. If the outer surfaces of the heater are well insulated, all the heat generated by the resistance heater will be transferred through the material whose conductivity is to be determined. Then measuring the two surface temperatures of the material when steady heat transfer is reached and substituting them into Eq. 1–22 together with other known quantities give the thermal conductivity (Fig. 1–25). The thermal conductivities of materials vary over a wide range, as shown in Fig. 1–26. The thermal conductivities of gases such as air vary by a factor of 104 from those of pure metals such as copper. Note that pure crystals and metals have the highest thermal conductivities, and gases and insulating materials the lowest. Temperature is a measure of the kinetic energies of the particles such as the molecules or atoms of a substance. In a liquid or gas, the kinetic energy of the molecules is due to their random translational motion as well as their vibrational and rotational motions. When two molecules possessing different kinetic energies collide, part of the kinetic energy of the more energetic (higher-temperature) molecule is transferred to the less energetic (lowertemperature) molecule, much the same as when two elastic balls of the same mass at different velocities collide, part of the kinetic energy of the faster ball is transferred to the slower one. The higher the temperature, the faster the molecules move and the higher the number of such collisions, and the better the heat transfer. The kinetic theory of gases predicts and the experiments confirm that the thermal conductivity of gases is proportional to the square root of the absolute temperature T, and inversely proportional to the square root of the molar mass M. Therefore, the thermal conductivity of a gas increases with increasing temperature and decreasing molar mass. So it is not surprising that the thermal conductivity of helium (M 4) is much higher than those of air (M 29) and argon (M 40). The thermal conductivities of gases at 1 atm pressure are listed in Table A-16. However, they can also be used at pressures other than 1 atm, since the thermal conductivity of gases is independent of pressure in a wide range of pressures encountered in practice. The mechanism of heat conduction in a liquid is complicated by the fact that the molecules are more closely spaced, and they exert a stronger intermolecular force field. The thermal conductivities of liquids usually lie between those cen58933_ch01.qxd 9/10/2002 8:29 AM Page 21 21 CHAPTER 1 NONMETALLIC CRYSTALS 1000 PURE METALS k, W/m·°C METAL ALLOYS Silver Copper Aluminum NONMETALLIC alloys Iron SOLIDS 100 Oxides 10 Bronze Steel Nichrome Diamond Graphite Silicon carbide Beryllium oxide Manganese Quartz LIQUIDS Mercury Rock 1 INSULATORS Water Food Fibers 0.1 GASES Hydrogen Helium Wood Air Foams Rubber Oils Carbon dioxide 0.01 of solids and gases. The thermal conductivity of a substance is normally highest in the solid phase and lowest in the gas phase. Unlike gases, the thermal conductivities of most liquids decrease with increasing temperature, with water being a notable exception. Like gases, the conductivity of liquids decreases with increasing molar mass. Liquid metals such as mercury and sodium have high thermal conductivities and are very suitable for use in applications where a high heat transfer rate to a liquid is desired, as in nuclear power plants. In solids, heat conduction is due to two effects: the lattice vibrational waves induced by the vibrational motions of the molecules positioned at relatively fixed positions in a periodic manner called a lattice, and the energy transported via the free flow of electrons in the solid (Fig. 1–27). The thermal conductivity of a solid is obtained by adding the lattice and electronic components. The relatively high thermal conductivities of pure metals are primarily due to the electronic component. The lattice component of thermal conductivity strongly depends on the way the molecules are arranged. For example, diamond, which is a highly ordered crystalline solid, has the highest known thermal conductivity at room temperature. Unlike metals, which are good electrical and heat conductors, crystalline solids such as diamond and semiconductors such as silicon are good heat conductors but poor electrical conductors. As a result, such materials find widespread use in the electronics industry. Despite their higher price, diamond heat sinks are used in the cooling of sensitive electronic components because of the FIGURE 1–26 The range of thermal conductivity of various materials at room temperature. GAS Molecular collisions Molecular diffusion LIQUID Molecular collisions Molecular diffusion electrons SOLID Lattice vibrations Flow of free electrons FIGURE 1–27 The mechanisms of heat conduction in different phases of a substance. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 22 22 HEAT TRANSFER TABLE 1–2 The thermal conductivity of an alloy is usually much lower than the thermal conductivity of either metal of which it is composed Pure metal or alloy k, W/m · °C, at 300 K Copper Nickel Constantan (55% Cu, 45% Ni) 401 91 Copper Aluminum Commercial bronze (90% Cu, 10% Al) 401 237 23 52 TABLE 1–3 Thermal conductivities of materials vary with temperature T, K Copper Aluminum 100 200 300 400 600 800 482 413 401 393 379 366 302 237 237 240 231 218 excellent thermal conductivity of diamond. Silicon oils and gaskets are commonly used in the packaging of electronic components because they provide both good thermal contact and good electrical insulation. Pure metals have high thermal conductivities, and one would think that metal alloys should also have high conductivities. One would expect an alloy made of two metals of thermal conductivities k1 and k2 to have a conductivity k between k1 and k2. But this turns out not to be the case. The thermal conductivity of an alloy of two metals is usually much lower than that of either metal, as shown in Table 1–2. Even small amounts in a pure metal of “foreign” molecules that are good conductors themselves seriously disrupt the flow of heat in that metal. For example, the thermal conductivity of steel containing just 1 percent of chrome is 62 W/m · °C, while the thermal conductivities of iron and chromium are 83 and 95 W/m · °C, respectively. The thermal conductivities of materials vary with temperature (Table 1–3). The variation of thermal conductivity over certain temperature ranges is negligible for some materials, but significant for others, as shown in Fig. 1–28. The thermal conductivities of certain solids exhibit dramatic increases at temperatures near absolute zero, when these solids become superconductors. For example, the conductivity of copper reaches a maximum value of about 20,000 W/m · °C at 20 K, which is about 50 times the conductivity at room temperature. The thermal conductivities and other thermal properties of various materials are given in Tables A-3 to A-16. 10,000 k, W/m·°C Solids Liquids Gases Diamonds 1000 Type IIa Type IIb Type I Silver Copper 100 Gold Aluminum Tungsten Platinum Iron 10 Aluminum oxide Pyroceram glass Clear fused quartz 1 Water 0.1 Helium Carbon tetrachloride FIGURE 1–28 The variation of the thermal conductivity of various solids, liquids, and gases with temperature (from White, Ref. 10). Air Steam Argon 0.01 200 400 600 800 T, K 1000 1200 1400 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 23 23 CHAPTER 1 The temperature dependence of thermal conductivity causes considerable complexity in conduction analysis. Therefore, it is common practice to evaluate the thermal conductivity k at the average temperature and treat it as a constant in calculations. In heat transfer analysis, a material is normally assumed to be isotropic; that is, to have uniform properties in all directions. This assumption is realistic for most materials, except those that exhibit different structural characteristics in different directions, such as laminated composite materials and wood. The thermal conductivity of wood across the grain, for example, is different than that parallel to the grain. Thermal Diffusivity The product Cp, which is frequently encountered in heat transfer analysis, is called the heat capacity of a material. Both the specific heat Cp and the heat capacity Cp represent the heat storage capability of a material. But Cp expresses it per unit mass whereas Cp expresses it per unit volume, as can be noticed from their units J/kg · °C and J/m3 · °C, respectively. Another material property that appears in the transient heat conduction analysis is the thermal diffusivity, which represents how fast heat diffuses through a material and is defined as k Heat conducted Cp Heat stored (m2/s) (1-23) Note that the thermal conductivity k represents how well a material conducts heat, and the heat capacity Cp represents how much energy a material stores per unit volume. Therefore, the thermal diffusivity of a material can be viewed as the ratio of the heat conducted through the material to the heat stored per unit volume. A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat into the medium. A small value of thermal diffusivity means that heat is mostly absorbed by the material and a small amount of heat will be conducted further. The thermal diffusivities of some common materials at 20°C are given in Table 1–4. Note that the thermal diffusivity ranges from 0.14 106 m2/s for water to 174 106 m2/s for silver, which is a difference of more than a thousand times. Also note that the thermal diffusivities of beef and water are the same. This is not surprising, since meat as well as fresh vegetables and fruits are mostly water, and thus they possess the thermal properties of water. EXAMPLE 1–6 Measuring the Thermal Conductivity of a Material A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material, as shown in Fig. 1–29. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one-dimensional. Two thermocouples are embedded into each sample some distance L apart, and a TABLE 1–4 The thermal diffusivities of some materials at room temperature , m2/s Material Silver Gold Copper Aluminum Iron Mercury (l) Marble Ice Concrete Brick Heavy soil (dry) Glass Glass wool Water (l) Beef Wood (oak) 149 127 113 97.5 22.8 4.7 1.2 1.2 0.75 0.52 0.52 0.34 0.23 0.14 0.14 0.13 106 106 106 106 106 106 106 106 106 106 106 106 106 106 106 106 Multiply by 10.76 to convert to ft2/s. Cooling fluid Sample Insulation Thermocouple L ∆T1 a Resistance heater Sample a L ∆T1 Cooling fluid FIGURE 1–29 Apparatus to measure the thermal conductivity of a material using two identical samples and a thin resistance heater (Example 1–6). cen58933_ch01.qxd 9/10/2002 8:29 AM Page 24 24 HEAT TRANSFER differential thermometer reads the temperature drop T across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter 5 cm and length 10 cm are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15°C. Determine the thermal conductivity of the sample. SOLUTION The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis The electrical power consumed by the resistance heater and converted to heat is · W e VI (110 V)(0.4 A) 44 W The rate of heat flow through each sample is · · Q 21 W e 21 (44 W) 22 W since only half of the heat generated will flow through each sample because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possesses thermal symmetry. The heat transfer area is the area normal to the direction of heat flow, which is the cross-sectional area of the cylinder in this case: A 14 D2 14 (0.05 m)2 0.00196 m2 Noting that the temperature drops by 15°C within 3 cm in the direction of heat flow, the thermal conductivity of the sample is determined to be Q· L (22 W)(0.03 m) · T Q kA → k 22.4 W/m · °C L A T (0.00196 m2)(15°C) Discussion Perhaps you are wondering if we really need to use two samples in the apparatus, since the measurements on the second sample do not give any additional information. It seems like we can replace the second sample by insulation. Indeed, we do not need the second sample; however, it enables us to verify the temperature measurements on the first sample and provides thermal symmetry, which reduces experimental error. EXAMPLE 1–7 Conversion between SI and English Units An engineer who is working on the heat transfer analysis of a brick building in English units needs the thermal conductivity of brick. But the only value he can cen58933_ch01.qxd 9/10/2002 8:29 AM Page 25 25 CHAPTER 1 find from his handbooks is 0.72 W/m · °C, which is in SI units. To make matters worse, the engineer does not have a direct conversion factor between the two unit systems for thermal conductivity. Can you help him out? SOLUTION The situation this engineer is facing is not unique, and most engineers often find themselves in a similar position. A person must be very careful during unit conversion not to fall into some common pitfalls and to avoid some costly mistakes. Although unit conversion is a simple process, it requires utmost care and careful reasoning. The conversion factors for W and m are straightforward and are given in conversion tables to be 1 W 3.41214 Btu/h 1 m 3.2808 ft But the conversion of °C into °F is not so simple, and it can be a source of error if one is not careful. Perhaps the first thought that comes to mind is to replace °C by (°F 32)/1.8 since T(°C) [T(°F) 32]/1.8. But this will be wrong since the °C in the unit W/m · °C represents per °C change in temperature. Noting that 1°C change in temperature corresponds to 1.8°F, the proper conversion factor to be used is 1°C 1.8°F Substituting, we get 1 W/m · °C 3.41214 Btu/h 0.5778 Btu/h · ft · °F (3.2808 ft)(1.8°F) k = 0.72 W/m·°C = 0.42 Btu/h·ft·°F which is the desired conversion factor. Therefore, the thermal conductivity of the brick in English units is kbrick 0.72 W/m · °C 0.72 (0.5778 Btu/h · ft · °F) 0.42 Btu/h · ft · °F Discussion Note that the thermal conductivity value of a material in English units is about half that in SI units (Fig. 1–30). Also note that we rounded the result to two significant digits (the same number in the original value) since expressing the result in more significant digits (such as 0.4160 instead of 0.42) would falsely imply a more accurate value than the original one. 1–7 ■ CONVECTION Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction. The presence of bulk motion of the fluid enhances the heat transfer between the solid surface and the fluid, but it also complicates the determination of heat transfer rates. FIGURE 1–30 The thermal conductivity value in English units is obtained by multiplying the value in SI units by 0.5778. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 26 26 HEAT TRANSFER Velocity variation of air Air flow As T T Temperature variation of air · Qconv Ts Hot Block FIGURE 1–31 Heat transfer from a hot surface to air by convection. Forced convection Natural convection Air Air hot egg hot egg FIGURE 1–32 Consider the cooling of a hot block by blowing cool air over its top surface (Fig. 1–31). Energy is first transferred to the air layer adjacent to the block by conduction. This energy is then carried away from the surface by convection, that is, by the combined effects of conduction within the air that is due to random motion of air molecules and the bulk or macroscopic motion of the air that removes the heated air near the surface and replaces it by the cooler air. Convection is called forced convection if the fluid is forced to flow over the surface by external means such as a fan, pump, or the wind. In contrast, convection is called natural (or free) convection if the fluid motion is caused by buoyancy forces that are induced by density differences due to the variation of temperature in the fluid (Fig. 1–32). For example, in the absence of a fan, heat transfer from the surface of the hot block in Fig. 1–31 will be by natural convection since any motion in the air in this case will be due to the rise of the warmer (and thus lighter) air near the surface and the fall of the cooler (and thus heavier) air to fill its place. Heat transfer between the block and the surrounding air will be by conduction if the temperature difference between the air and the block is not large enough to overcome the resistance of air to movement and thus to initiate natural convection currents. Heat transfer processes that involve change of phase of a fluid are also considered to be convection because of the fluid motion induced during the process, such as the rise of the vapor bubbles during boiling or the fall of the liquid droplets during condensation. Despite the complexity of convection, the rate of convection heat transfer is observed to be proportional to the temperature difference, and is conveniently expressed by Newton’s law of cooling as The cooling of a boiled egg by forced and natural convection. TABLE 1–5 Typical values of convection heat transfer coefficient Type of convection Free convection of gases Free convection of liquids Forced convection of gases Forced convection of liquids Boiling and condensation h, W/m2 · °C 2–25 10–1000 25–250 50–20,000 · Q conv hAs (Ts T ) (W) (1-24) where h is the convection heat transfer coefficient in W/m2 · °C or Btu/h · ft2 · °F, As is the surface area through which convection heat transfer takes place, Ts is the surface temperature, and T is the temperature of the fluid sufficiently far from the surface. Note that at the surface, the fluid temperature equals the surface temperature of the solid. The convection heat transfer coefficient h is not a property of the fluid. It is an experimentally determined parameter whose value depends on all the variables influencing convection such as the surface geometry, the nature of fluid motion, the properties of the fluid, and the bulk fluid velocity. Typical values of h are given in Table 1–5. Some people do not consider convection to be a fundamental mechanism of heat transfer since it is essentially heat conduction in the presence of fluid motion. But we still need to give this combined phenomenon a name, unless we are willing to keep referring to it as “conduction with fluid motion.” Thus, it is practical to recognize convection as a separate heat transfer mechanism despite the valid arguments to the contrary. 2500–100,000 Multiply by 0.176 to convert to Btu/h · ft2 · °F. EXAMPLE 1–8 Measuring Convection Heat Transfer Coefficient A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as shown in Fig. 1–33. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152°C in steady cen58933_ch01.qxd 9/10/2002 8:29 AM Page 27 27 CHAPTER 1 operation. Also, the voltage drop and electric current through the wire are measured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. T• = 15°C 152°C 1.5 A 60 V FIGURE 1–33 SOLUTION The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible. Analysis When steady operating conditions are reached, the rate of heat loss from the wire will equal the rate of heat generation in the wire as a result of resistance heating. That is, · · Q Egenerated VI (60 V)(1.5 A) 90 W The surface area of the wire is As DL (0.003 m)(2 m) 0.01885 m2 Newton’s law of cooling for convection heat transfer is expressed as · Q conv hAs (Ts T ) Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficient is determined to be h · Q conv 90 W 34.9 W/m2 · °C As(Ts T ) (0.01885 m2)(152 15)°C Discussion Note that the simple setup described above can be used to determine the average heat transfer coefficients from a variety of surfaces in air. Also, heat transfer by radiation can be eliminated by keeping the surrounding surfaces at the temperature of the wire. 1–8 ■ RADIATION Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. Unlike conduction and convection, the transfer of energy by radiation does not require the presence of an intervening medium. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. This is how the energy of the sun reaches the earth. In heat transfer studies we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. It differs from other forms of electromagnetic radiation such as x-rays, gamma rays, microwaves, radio waves, and television waves that are not related to temperature. All bodies at a temperature above absolute zero emit thermal radiation. Radiation is a volumetric phenomenon, and all solids, liquids, and gases emit, absorb, or transmit radiation to varying degrees. However, radiation is Schematic for Example 1–8. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 28 28 HEAT TRANSFER q· emit, max = σ Ts4 = 1452 W/ m 2 Ts = 400 K Blackbody (ε = 1) FIGURE 1–34 Blackbody radiation represents the maximum amount of radiation that can be emitted from a surface at a specified temperature. TABLE 1–6 Emissivities of some materials at 300 K Material Emissivity Aluminum foil Anodized aluminum Polished copper Polished gold Polished silver Polished stainless steel Black paint White paint White paper Asphalt pavement Red brick Human skin Wood Soil Water Vegetation 0.07 0.82 0.03 0.03 0.02 0.17 0.98 0.90 0.92–0.97 0.85–0.93 0.93–0.96 0.95 0.82–0.92 0.93–0.96 0.96 0.92–0.96 usually considered to be a surface phenomenon for solids that are opaque to thermal radiation such as metals, wood, and rocks since the radiation emitted by the interior regions of such material can never reach the surface, and the radiation incident on such bodies is usually absorbed within a few microns from the surface. The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts (in K or R) is given by the Stefan–Boltzmann law as · Q emit, max AsT s4 · · Qref = (1 – α ) Qincident · · Qabs = α Qincident FIGURE 1–35 The absorption of radiation incident on an opaque surface of absorptivity . (1-25) where 5.67 108 W/m2 · K4 or 0.1714 108 Btu/h · ft2 · R4 is the Stefan–Boltzmann constant. The idealized surface that emits radiation at this maximum rate is called a blackbody, and the radiation emitted by a blackbody is called blackbody radiation (Fig. 1–34). The radiation emitted by all real surfaces is less than the radiation emitted by a blackbody at the same temperature, and is expressed as · Q emit AsT s4 (W) (1-26) where is the emissivity of the surface. The property emissivity, whose value is in the range 0 1, is a measure of how closely a surface approximates a blackbody for which 1. The emissivities of some surfaces are given in Table 1–6. Another important radiation property of a surface is its absorptivity , which is the fraction of the radiation energy incident on a surface that is absorbed by the surface. Like emissivity, its value is in the range 0 1. A blackbody absorbs the entire radiation incident on it. That is, a blackbody is a perfect absorber ( 1) as it is a perfect emitter. In general, both and of a surface depend on the temperature and the wavelength of the radiation. Kirchhoff’s law of radiation states that the emissivity and the absorptivity of a surface at a given temperature and wavelength are equal. In many practical applications, the surface temperature and the temperature of the source of incident radiation are of the same order of magnitude, and the average absorptivity of a surface is taken to be equal to its average emissivity. The rate at which a surface absorbs radiation is determined from (Fig. 1–35) · · Q absorbed Q incident · Qincident (W) (W) (1-27) · where Q incident is the rate at which radiation is incident on the surface and is the absorptivity of the surface. For opaque (nontransparent) surfaces, the portion of incident radiation not absorbed by the surface is reflected back. The difference between the rates of radiation emitted by the surface and the radiation absorbed is the net radiation heat transfer. If the rate of radiation absorption is greater than the rate of radiation emission, the surface is said to be gaining energy by radiation. Otherwise, the surface is said to be losing energy by radiation. In general, the determination of the net rate of heat transfer by radiation between two surfaces is a complicated matter since it depends on the properties of the surfaces, their orientation relative to each other, and the interaction of the medium between the surfaces with radiation. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 29 29 CHAPTER 1 When a surface of emissivity and surface area As at an absolute temperature Ts is completely enclosed by a much larger (or black) surface at absolute temperature Tsurr separated by a gas (such as air) that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by (Fig. 1–36) · 4 Q rad As (Ts4 Tsurr ) Air (W) (W) · Qemitted (1-28) In this special case, the emissivity and the surface area of the surrounding surface do not have any effect on the net radiation heat transfer. Radiation heat transfer to or from a surface surrounded by a gas such as air occurs parallel to conduction (or convection, if there is bulk gas motion) between the surface and the gas. Thus the total heat transfer is determined by adding the contributions of both heat transfer mechanisms. For simplicity and convenience, this is often done by defining a combined heat transfer coefficient hcombined that includes the effects of both convection and radiation. Then the total heat transfer rate to or from a surface by convection and radiation is expressed as · Q total hcombined As (Ts T ) Surrounding surfaces at Tsurr · Qincident ε , As, Ts · Qrad = εσ As (T 4s – T 4surr ) FIGURE 1–36 Radiation heat transfer between a surface and the surfaces surrounding it. (1-29) Note that the combined heat transfer coefficient is essentially a convection heat transfer coefficient modified to include the effects of radiation. Radiation is usually significant relative to conduction or natural convection, but negligible relative to forced convection. Thus radiation in forced convection applications is usually disregarded, especially when the surfaces involved have low emissivities and low to moderate temperatures. EXAMPLE 1–9 Radiation Effect on Thermal Comfort It is a common experience to feel “chilly” in winter and “warm” in summer in our homes even when the thermostat setting is kept the same. This is due to the so called “radiation effect” resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling. Consider a person standing in a room maintained at 22°C at all times. The inner surfaces of the walls, floors, and the ceiling of the house are observed to be at an average temperature of 10°C in winter and 25°C in summer. Determine the rate of radiation heat transfer between this person and the surrounding surfaces if the exposed surface area and the average outer surface temperature of the person are 1.4 m2 and 30°C, respectively (Fig. 1–37). SOLUTION The rates of radiation heat transfer between a person and the surrounding surfaces at specified temperatures are to be determined in summer and winter. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is not considered. 3 The person is completely surrounded by the interior surfaces of the room. 4 The surrounding surfaces are at a uniform temperature. Properties The emissivity of a person is 0.95 (Table 1–6). Analysis The net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and floor in winter and summer are Room Tsurr 30°C 1.4 m2 · Qrad FIGURE 1–37 Schematic for Example 1–9. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 30 30 HEAT TRANSFER · 4 Q rad, winter As (Ts4 Tsurr, winter) (0.95)(5.67 108 W/m2 · K4)(1.4 m2) [(30 273)4 (10 273)4] K4 152 W and · 4 Q rad, summer As (Ts4 Tsurr, summer) (0.95)(5.67 108 W/m2 · K4)(1.4 m2) [(30 273)4 (25 273)4] K4 40.9 W Discussion Note that we must use absolute temperatures in radiation calculations. Also note that the rate of heat loss from the person by radiation is almost four times as large in winter than it is in summer, which explains the “chill” we feel in winter even if the thermostat setting is kept the same. 1–9 T1 OPAQUE SOLID T2 1 mode Conduction T1 GAS T2 Radiation 2 modes Conduction or convection T1 VACUUM Radiation T2 1 mode FIGURE 1–38 Although there are three mechanisms of heat transfer, a medium may involve only two of them simultaneously. ■ SIMULTANEOUS HEAT TRANSFER MECHANISMS We mentioned that there are three mechanisms of heat transfer, but not all three can exist simultaneously in a medium. For example, heat transfer is only by conduction in opaque solids, but by conduction and radiation in semitransparent solids. Thus, a solid may involve conduction and radiation but not convection. However, a solid may involve heat transfer by convection and/or radiation on its surfaces exposed to a fluid or other surfaces. For example, the outer surfaces of a cold piece of rock will warm up in a warmer environment as a result of heat gain by convection (from the air) and radiation (from the sun or the warmer surrounding surfaces). But the inner parts of the rock will warm up as this heat is transferred to the inner region of the rock by conduction. Heat transfer is by conduction and possibly by radiation in a still fluid (no bulk fluid motion) and by convection and radiation in a flowing fluid. In the absence of radiation, heat transfer through a fluid is either by conduction or convection, depending on the presence of any bulk fluid motion. Convection can be viewed as combined conduction and fluid motion, and conduction in a fluid can be viewed as a special case of convection in the absence of any fluid motion (Fig. 1–38). Thus, when we deal with heat transfer through a fluid, we have either conduction or convection, but not both. Also, gases are practically transparent to radiation, except that some gases are known to absorb radiation strongly at certain wavelengths. Ozone, for example, strongly absorbs ultraviolet radiation. But in most cases, a gas between two solid surfaces does not interfere with radiation and acts effectively as a vacuum. Liquids, on the other hand, are usually strong absorbers of radiation. Finally, heat transfer through a vacuum is by radiation only since conduction or convection requires the presence of a material medium. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 31 31 CHAPTER 1 EXAMPLE 1–10 Heat Loss from a Person Room air 20°C Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m2 · °C (Fig. 1–39). · Qconv 29°C · Qrad SOLUTION The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined. Assumptions 1 Steady operating conditions exist. 2 The person is completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the room. 4 Heat conduction to the floor through the feet is negligible. Properties The emissivity of a person is 0.95 (Table 1–6). Analysis The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m2) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is · Q conv hAs (Ts T ) (6 W/m2 · °C)(1.6 m2)(29 20)°C 86.4 W The person will also lose heat by radiation to the surrounding wall surfaces. We take the temperature of the surfaces of the walls, ceiling, and floor to be equal to the air temperature in this case for simplicity, but we recognize that this does not need to be the case. These surfaces may be at a higher or lower temperature than the average temperature of the room air, depending on the outdoor conditions and the structure of the walls. Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces, the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and floor is · 4 Q rad As (Ts4 Tsurr ) (0.95)(5.67 108 W/m2 · K4)(1.6 m2) [(29 273)4 (20 273)4] K4 81.7 W Note that we must use absolute temperatures in radiation calculations. Also note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature. Then the rate of total heat transfer from the body is determined by adding these two quantities: · · · Q total Q conv Q rad (86.4 81.7) W 168.1 W · Qcond FIGURE 1–39 Heat transfer from the person described in Example 1–10. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 32 32 HEAT TRANSFER Discussion The heat transfer would be much higher if the person were not dressed since the exposed surface temperature would be higher. Thus, an important function of the clothes is to serve as a barrier against heat transfer. In these calculations, heat transfer through the feet to the floor by conduction, which is usually very small, is neglected. Heat transfer from the skin by perspiration, which is the dominant mode of heat transfer in hot environments, is not considered here. T1 = 300 K · Q T2 = 200 K L = 1 cm ε=1 FIGURE 1–40 EXAMPLE 1–11 Heat Transfer between Two Isothermal Plates Consider steady heat transfer between two large parallel plates at constant temperatures of T1 300 K and T2 200 K that are L 1 cm apart, as shown in Fig. 1–40. Assuming the surfaces to be black (emissivity 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with urethane insulation, and (d) filled with superinsulation that has an apparent thermal conductivity of 0.00002 W/m · °C. Schematic for Example 1–11. SOLUTION The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. Assumptions 1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus 1. Properties The thermal conductivity at the average temperature of 250 K is k 0.0219 W/m · °C for air (Table A-11), 0.026 W/m · °C for urethane insulation (Table A-6), and 0.00002 W/m · °C for the superinsulation. Analysis (a) The rates of conduction and radiation heat transfer between the plates through the air layer are T1 T2 (300 200)°C · Q cond kA (0.0219 W/m · °C)(1 m2) 219 W L 0.01 m and · Q rad A(T 14 T 24) (1)(5.67 108 W/m2 · K4)(1 m2)[(300 K)4 (200 K)4] 368 W Therefore, · · · Q total Q cond Q rad 219 368 587 W The heat transfer rate in reality will be higher because of the natural convection currents that are likely to occur in the air space between the plates. (b) When the air space between the plates is evacuated, there will be no conduction or convection, and the only heat transfer between the plates will be by radiation. Therefore, · · Q total Q rad 368 W (c) An opaque solid material placed between two plates blocks direct radiation heat transfer between the plates. Also, the thermal conductivity of an insulating material accounts for the radiation heat transfer that may be occurring through cen58933_ch01.qxd 9/10/2002 8:30 AM Page 33 33 CHAPTER 1 300 K 200 K 300 K 200 K 300 K 200 K 300 K 200 K · Q = 587 W · Q = 368 W · Q = 260 W · Q = 0.2 W 1 cm 1 cm 1 cm 1 cm (a) Air space (b) Vacuum (c) Insulation (d ) Superinsulation FIGURE 1–41 Different ways of reducing heat transfer between two isothermal plates, and their effectiveness. the voids in the insulating material. The rate of heat transfer through the urethane insulation is T1 T2 (300 200)°C · · Q total Q cond kA (0.026 W/m · °C)(1 m2) 260 W L 0.01 m Note that heat transfer through the urethane material is less than the heat transfer through the air determined in (a), although the thermal conductivity of the insulation is higher than that of air. This is because the insulation blocks the radiation whereas air transmits it. (d ) The layers of the superinsulation prevent any direct radiation heat transfer between the plates. However, radiation heat transfer between the sheets of superinsulation does occur, and the apparent thermal conductivity of the superinsulation accounts for this effect. Therefore, T1 T2 (300 200)°C · Q total kA (0.00002 W/m · °C)(1 m2) 0.2 W L 0.01 m 1 of the heat transfer through the vacuum. The results of this exwhich is 1840 ample are summarized in Fig. 1–41 to put them into perspective. Discussion This example demonstrates the effectiveness of superinsulations, which are discussed in the next chapter, and explains why they are the insulation of choice in critical applications despite their high cost. EXAMPLE 1–12 Heat Transfer in Conventional and Microwave Ovens The fast and efficient cooking of microwave ovens made them one of the essential appliances in modern kitchens (Fig. 1–42). Discuss the heat transfer mechanisms associated with the cooking of a chicken in microwave and conventional ovens, and explain why cooking in a microwave oven is more efficient. FIGURE 1–42 SOLUTION Food is cooked in a microwave oven by absorbing the electromagnetic radiation energy generated by the microwave tube, called the magnetron. A chicken being cooked in a microwave oven (Example 1–12). cen58933_ch01.qxd 9/10/2002 8:30 AM Page 34 34 HEAT TRANSFER The radiation emitted by the magnetron is not thermal radiation, since its emission is not due to the temperature of the magnetron; rather, it is due to the conversion of electrical energy into electromagnetic radiation at a specified wavelength. The wavelength of the microwave radiation is such that it is reflected by metal surfaces; transmitted by the cookware made of glass, ceramic, or plastic; and absorbed and converted to internal energy by food (especially the water, sugar, and fat) molecules. In a microwave oven, the radiation that strikes the chicken is absorbed by the skin of the chicken and the outer parts. As a result, the temperature of the chicken at and near the skin rises. Heat is then conducted toward the inner parts of the chicken from its outer parts. Of course, some of the heat absorbed by the outer surface of the chicken is lost to the air in the oven by convection. In a conventional oven, the air in the oven is first heated to the desired temperature by the electric or gas heating element. This preheating may take several minutes. The heat is then transferred from the air to the skin of the chicken by natural convection in most ovens or by forced convection in the newer convection ovens that utilize a fan. The air motion in convection ovens increases the convection heat transfer coefficient and thus decreases the cooking time. Heat is then conducted toward the inner parts of the chicken from its outer parts as in microwave ovens. Microwave ovens replace the slow convection heat transfer process in conventional ovens by the instantaneous radiation heat transfer. As a result, microwave ovens transfer energy to the food at full capacity the moment they are turned on, and thus they cook faster while consuming less energy. EXAMPLE 1–13 Heating of a Plate by Solar Energy A thin metal plate is insulated on the back and exposed to solar radiation at the front surface (Fig. 1–43). The exposed surface of the plate has an absorptivity of 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of 700 W/m2 and the surrounding air temperature is 25°C, determine the surface temperature of the plate when the heat loss by convection and radiation equals the solar energy absorbed by the plate. Assume the combined convection and radiation heat transfer coefficient to be 50 W/m2 · °C. 700 W/ m2 α = 0.6 25°C FIGURE 1–43 Schematic for Example 1–13. SOLUTION The back side of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient remains constant. Properties The solar absorptivity of the plate is given to be 0.6. Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar radiation incident on the plate will be absorbed continuously. As a result, the temperature of the plate will rise, and the temperature difference between the plate and the surroundings will increase. This increasing temperature difference will cause the rate of heat loss from the plate to the surroundings to increase. At some point, the rate of heat loss from the plate will equal the rate of solar cen58933_ch01.qxd 9/10/2002 8:30 AM Page 35 35 CHAPTER 1 energy absorbed, and the temperature of the plate will no longer change. The temperature of the plate when steady operation is established is determined from · · Egained Elost or As q·incident, solar hcombined As (Ts T ) Solving for Ts and substituting, the plate surface temperature is determined to be Ts T q· incident, solar 0.6 (700 W/m2) 25°C 33.4°C hcombined 50 W/m2 · °C Discussion Note that the heat losses will prevent the plate temperature from rising above 33.4°C. Also, the combined heat transfer coefficient accounts for the effects of both convection and radiation, and thus it is very convenient to use in heat transfer calculations when its value is known with reasonable accuracy. SOLUTION PROBLEM-SOLVING TECHNIQUE The first step in learning any science is to grasp the fundamentals, and to gain a sound knowledge of it. The next step is to master the fundamentals by putting this knowledge to test. This is done by solving significant real-world problems. Solving such problems, especially complicated ones, requires a systematic approach. By using a step-by-step approach, an engineer can reduce the solution of a complicated problem into the solution of a series of simple problems (Fig. 1–44). When solving a problem, we recommend that you use the following steps zealously as applicable. This will help you avoid some of the common pitfalls associated with problem solving. Step 1: Problem Statement In your own words, briefly state the problem, the key information given, and the quantities to be found. This is to make sure that you understand the problem and the objectives before you attempt to solve the problem. Step 2: Schematic Draw a realistic sketch of the physical system involved, and list the relevant information on the figure. The sketch does not have to be something elaborate, but it should resemble the actual system and show the key features. Indicate any energy and mass interactions with the surroundings. Listing the given information on the sketch helps one to see the entire problem at once. Also, check for properties that remain constant during a process (such as temperature during an isothermal process), and indicate them on the sketch. Step 3: Assumptions State any appropriate assumptions made to simplify the problem to make it possible to obtain a solution. Justify the questionable assumptions. Assume reasonable values for missing quantities that are necessary. For example, in the absence of specific data for atmospheric pressure, it can be taken to be AY SY W HARD WAY 1–10 ■ EA PROBLEM FIGURE 1–44 A step-by-step approach can greatly simplify problem solving. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 36 36 HEAT TRANSFER Given: Air temperature in Denver 1 atm. However, it should be noted in the analysis that the atmospheric pressure decreases with increasing elevation. For example, it drops to 0.83 atm in Denver (elevation 1610 m) (Fig. 1–45). To be found: Density of air Missing information: Atmospheric pressure Assumption #1: Take P = 1 atm (Inappropriate. Ignores effect of altitude. Will cause more than 15% error.) Assumption #2: Take P = 0.83 atm (Appropriate. Ignores only minor effects such as weather.) Step 4: Physical Laws Apply all the relevant basic physical laws and principles (such as the conservation of energy), and reduce them to their simplest form by utilizing the assumptions made. However, the region to which a physical law is applied must be clearly identified first. For example, the heating or cooling of a canned drink is usually analyzed by applying the conservation of energy principle to the entire can. Step 5: Properties FIGURE 1–45 The assumptions made while solving an engineering problem must be reasonable and justifiable. Determine the unknown properties at known states necessary to solve the problem from property relations or tables. List the properties separately, and indicate their source, if applicable. Step 6: Calculations Energy use: $80/yr Substitute the known quantities into the simplified relations and perform the calculations to determine the unknowns. Pay particular attention to the units and unit cancellations, and remember that a dimensional quantity without a unit is meaningless. Also, don’t give a false implication of high accuracy by copying all the digits from the screen of the calculator—round the results to an appropriate number of significant digits. Energy saved by insulation: $200/yr Step 7: Reasoning, Verification, and Discussion IMPOSSIBLE! FIGURE 1–46 The results obtained from an engineering analysis must be checked for reasonableness. Check to make sure that the results obtained are reasonable and intuitive, and verify the validity of the questionable assumptions. Repeat the calculations that resulted in unreasonable values. For example, insulating a water heater that uses $80 worth of natural gas a year cannot result in savings of $200 a year (Fig. 1–46). Also, point out the significance of the results, and discuss their implications. State the conclusions that can be drawn from the results, and any recommendations that can be made from them. Emphasize the limitations under which the results are applicable, and caution against any possible misunderstandings and using the results in situations where the underlying assumptions do not apply. For example, if you determined that wrapping a water heater with a $20 insulation jacket will reduce the energy cost by $30 a year, indicate that the insulation will pay for itself from the energy it saves in less than a year. However, also indicate that the analysis does not consider labor costs, and that this will be the case if you install the insulation yourself. Keep in mind that you present the solutions to your instructors, and any engineering analysis presented to others is a form of communication. Therefore neatness, organization, completeness, and visual appearance are of utmost importance for maximum effectiveness. Besides, neatness also serves as a great checking tool since it is very easy to spot errors and inconsistencies in a neat work. Carelessness and skipping steps to save time often ends up costing more time and unnecessary anxiety. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 37 37 CHAPTER 1 The approach just described is used in the solved example problems without explicitly stating each step, as well as in the Solutions Manual of this text. For some problems, some of the steps may not be applicable or necessary. However, we cannot overemphasize the importance of a logical and orderly approach to problem solving. Most difficulties encountered while solving a problem are not due to a lack of knowledge; rather, they are due to a lack of coordination. You are strongly encouraged to follow these steps in problem solving until you develop your own approach that works best for you. A Remark on Significant Digits In engineering calculations, the information given is not known to more than a certain number of significant digits, usually three digits. Consequently, the results obtained cannot possibly be accurate to more significant digits. Reporting results in more significant digits implies greater accuracy than exists, and it should be avoided. For example, consider a 3.75-L container filled with gasoline whose density is 0.845 kg/L, and try to determine its mass. Probably the first thought that comes to your mind is to multiply the volume and density to obtain 3.16875 kg for the mass, which falsely implies that the mass determined is accurate to six significant digits. In reality, however, the mass cannot be more accurate than three significant digits since both the volume and the density are accurate to three significant digits only. Therefore, the result should be rounded to three significant digits, and the mass should be reported to be 3.17 kg instead of what appears in the screen of the calculator. The result 3.16875 kg would be correct only if the volume and density were given to be 3.75000 L and 0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly confident that the volume is accurate within 0.01 L, and it cannot be 3.74 or 3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all round to 3.75 L (Fig. 1–47). It is more appropriate to retain all the digits during intermediate calculations, and to do the rounding in the final step since this is what a computer will normally do. When solving problems, we will assume the given information to be accurate to at least three significant digits. Therefore, if the length of a pipe is given to be 40 m, we will assume it to be 40.0 m in order to justify using three significant digits in the final results. You should also keep in mind that all experimentally determined values are subject to measurement errors, and such errors will reflect in the results obtained. For example, if the density of a substance has an uncertainty of 2 percent, then the mass determined using this density value will also have an uncertainty of 2 percent. You should also be aware that we sometimes knowingly introduce small errors in order to avoid the trouble of searching for more accurate data. For example, when dealing with liquid water, we just use the value of 1000 kg/m3 for density, which is the density value of pure water at 0C. Using this value at 75C will result in an error of 2.5 percent since the density at this temperature is 975 kg/m3. The minerals and impurities in the water will introduce additional error. This being the case, you should have no reservation in rounding the final results to a reasonable number of significant digits. Besides, having a few percent uncertainty in the results of engineering analysis is usually the norm, not the exception. Given: Volume: Density: V = 3.75 L ρ = 0.845 kg/L (3 significant digits) Also, Find: 3.75 × 0.845 = 3.16875 Mass: m = ρV = 3.16875 kg Rounding to 3 significant digits: m = 3.17 kg FIGURE 1–47 A result with more significant digits than that of given data falsely implies more accuracy. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 38 38 HEAT TRANSFER Engineering Software Packages FIGURE 1–48 An excellent word-processing program does not make a person a good writer; it simply makes a good writer a better and more efficient writer. Perhaps you are wondering why we are about to undertake a painstaking study of the fundamentals of heat transfer. After all, almost all such problems we are likely to encounter in practice can be solved using one of several sophisticated software packages readily available in the market today. These software packages not only give the desired numerical results, but also supply the outputs in colorful graphical form for impressive presentations. It is unthinkable to practice engineering today without using some of these packages. This tremendous computing power available to us at the touch of a button is both a blessing and a curse. It certainly enables engineers to solve problems easily and quickly, but it also opens the door for abuses and misinformation. In the hands of poorly educated people, these software packages are as dangerous as sophisticated powerful weapons in the hands of poorly trained soldiers. Thinking that a person who can use the engineering software packages without proper training on fundamentals can practice engineering is like thinking that a person who can use a wrench can work as a car mechanic. If it were true that the engineering students do not need all these fundamental courses they are taking because practically everything can be done by computers quickly and easily, then it would also be true that the employers would no longer need high-salaried engineers since any person who knows how to use a word-processing program can also learn how to use those software packages. However, the statistics show that the need for engineers is on the rise, not on the decline, despite the availability of these powerful packages. We should always remember that all the computing power and the engineering software packages available today are just tools, and tools have meaning only in the hands of masters. Having the best word-processing program does not make a person a good writer, but it certainly makes the job of a good writer much easier and makes the writer more productive (Fig. 1–48). Hand calculators did not eliminate the need to teach our children how to add or subtract, and the sophisticated medical software packages did not take the place of medical school training. Neither will engineering software packages replace the traditional engineering education. They will simply cause a shift in emphasis in the courses from mathematics to physics. That is, more time will be spent in the classroom discussing the physical aspects of the problems in greater detail, and less time on the mechanics of solution procedures. All these marvelous and powerful tools available today put an extra burden on today’s engineers. They must still have a thorough understanding of the fundamentals, develop a “feel” of the physical phenomena, be able to put the data into proper perspective, and make sound engineering judgments, just like their predecessors. However, they must do it much better, and much faster, using more realistic models because of the powerful tools available today. The engineers in the past had to rely on hand calculations, slide rules, and later hand calculators and computers. Today they rely on software packages. The easy access to such power and the possibility of a simple misunderstanding or misinterpretation causing great damage make it more important today than ever to have a solid training in the fundamentals of engineering. In this text we make an extra effort to put the emphasis on developing an intuitive and physical understanding of natural phenomena instead of on the mathematical details of solution procedures. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 39 39 CHAPTER 1 Engineering Equation Solver (EES) EES is a program that solves systems of linear or nonlinear algebraic or differential equations numerically. It has a large library of built-in thermodynamic property functions as well as mathematical functions, and allows the user to supply additional property data. Unlike some software packages, EES does not solve thermodynamic problems; it only solves the equations supplied by the user. Therefore, the user must understand the problem and formulate it by applying any relevant physical laws and relations. EES saves the user considerable time and effort by simply solving the resulting mathematical equations. This makes it possible to attempt significant engineering problems not suitable for hand calculations, and to conduct parametric studies quickly and conveniently. EES is a very powerful yet intuitive program that is very easy to use, as shown in the examples below. The use and capabilities of EES are explained in Appendix 3. Heat Transfer Tools (HTT) One software package specifically designed to help bridge the gap between the textbook fundamentals and these powerful software packages is Heat Transfer Tools, which may be ordered “bundled” with this text. The software included in that package was developed for instructional use only and thus is applicable only to fundamental problems in heat transfer. While it does not have the power and functionality of the professional, commercial packages, HTT uses research-grade numerical algorithms behind the scenes and modern graphical user interfaces. Each module is custom designed and applicable to a single, fundamental topic in heat transfer to ensure that almost all time at the computer is spent learning heat transfer. Nomenclature and all inputs and outputs are consistent with those used in this and most other textbooks in the field. In addition, with the capability of testing parameters so readily available, one can quickly gain a physical feel for the effects of all the nondimensional numbers that arise in heat transfer. EXAMPLE 1–14 Solving a System of Equations with EES The difference of two numbers is 4, and the sum of the squares of these two numbers is equal to the sum of the numbers plus 20. Determine these two numbers. SOLUTION Relations are given for the difference and the sum of the squares of two numbers. They are to be determined. Analysis We start the EES program by double-clicking on its icon, open a new file, and type the following on the blank screen that appears: x-y=4 x^2+y^2=x+y+20 which is an exact mathematical expression of the problem statement with x and y denoting the unknown numbers. The solution to this system of two cen58933_ch01.qxd 9/10/2002 8:30 AM Page 40 40 HEAT TRANSFER nonlinear equations with two unknowns is obtained by a single click on the “calculator” symbol on the taskbar. It gives x=5 and y=1 Discussion Note that all we did is formulate the problem as we would on paper; EES took care of all the mathematical details of solution. Also note that equations can be linear or nonlinear, and they can be entered in any order with unknowns on either side. Friendly equation solvers such as EES allow the user to concentrate on the physics of the problem without worrying about the mathematical complexities associated with the solution of the resulting system of equations. Throughout the text, problems that are unsuitable for hand calculations and are intended to be solved using EES are indicated by a computer icon. TOPIC OF SPECIAL INTEREST Thermal Comfort Baby Bird Fox FIGURE 1–49 Most animals come into this world with built-in insulation, but human beings come with a delicate skin. Unlike animals such as a fox or a bear that are born with built-in furs, human beings come into this world with little protection against the harsh environmental conditions (Fig. 1–49). Therefore, we can claim that the search for thermal comfort dates back to the beginning of human history. It is believed that early human beings lived in caves that provided shelter as well as protection from extreme thermal conditions. Probably the first form of heating system used was open fire, followed by fire in dwellings through the use of a chimney to vent out the combustion gases. The concept of central heating dates back to the times of the Romans, who heated homes by utilizing double-floor construction techniques and passing the fire’s fumes through the opening between the two floor layers. The Romans were also the first to use transparent windows made of mica or glass to keep the wind and rain out while letting the light in. Wood and coal were the primary energy sources for heating, and oil and candles were used for lighting. The ruins of south-facing houses indicate that the value of solar heating was recognized early in the history. The term air-conditioning is usually used in a restricted sense to imply cooling, but in its broad sense it means to condition the air to the desired level by heating, cooling, humidifying, dehumidifying, cleaning, and deodorizing. The purpose of the air-conditioning system of a building is to provide complete thermal comfort for its occupants. Therefore, we need to understand the thermal aspects of the human body in order to design an effective air-conditioning system. The building blocks of living organisms are cells, which resemble miniature factories performing various functions necessary for the survival of organisms. The human body contains about 100 trillion cells with an average diameter of 0.01 mm. In a typical cell, thousands of chemical reactions This section can be skipped without a loss in continuity. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 41 41 CHAPTER 1 occur every second during which some molecules are broken down and energy is released and some new molecules are formed. The high level of chemical activity in the cells that maintain the human body temperature at a temperature of 37.0°C (98.6°F) while performing the necessary bodily functions is called the metabolism. In simple terms, metabolism refers to the burning of foods such as carbohydrates, fat, and protein. The metabolizable energy content of foods is usually expressed by nutritionists in terms of the capitalized Calorie. One Calorie is equivalent to 1 Cal 1 kcal 4.1868 kJ. The rate of metabolism at the resting state is called the basal metabolic rate, which is the rate of metabolism required to keep a body performing the necessary bodily functions such as breathing and blood circulation at zero external activity level. The metabolic rate can also be interpreted as the energy consumption rate for a body. For an average man (30 years old, 70 kg, 1.73 m high, 1.8 m2 surface area), the basal metabolic rate is 84 W. That is, the body is converting chemical energy of the food (or of the body fat if the person had not eaten) into heat at a rate of 84 J/s, which is then dissipated to the surroundings. The metabolic rate increases with the level of activity, and it may exceed 10 times the basal metabolic rate when someone is doing strenuous exercise. That is, two people doing heavy exercising in a room may be supplying more energy to the room than a 1-kW resistance heater (Fig. 1–50). An average man generates heat at a rate of 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position. The maximum metabolic rate of an average man is 1250 W at age 20 and 730 at age 70. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. Metabolic rates during various activities are given in Table 1–7 per unit body surface area. The surface area of a nude body was given by D. DuBois in 1916 as As 0.202m0.425 h0.725 (m2) (1-30) where m is the mass of the body in kg and h is the height in m. Clothing increases the exposed surface area of a person by up to about 50 percent. The metabolic rates given in the table are sufficiently accurate for most purposes, but there is considerable uncertainty at high activity levels. More accurate values can be determined by measuring the rate of respiratory oxygen consumption, which ranges from about 0.25 L/min for an average resting man to more than 2 L/min during extremely heavy work. The entire energy released during metabolism can be assumed to be released as heat (in sensible or latent forms) since the external mechanical work done by the muscles is very small. Besides, the work done during most activities such as walking or riding an exercise bicycle is eventually converted to heat through friction. The comfort of the human body depends primarily on three environmental factors: the temperature, relative humidity, and air motion. The temperature of the environment is the single most important index of comfort. Extensive research is done on human subjects to determine the “thermal comfort zone” and to identify the conditions under which the body feels 1.2 kJ/s 1 kJ/s FIGURE1–50 Two fast-dancing people supply more heat to a room than a 1-kW resistance heater. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 42 42 HEAT TRANSFER TABLE 1–7 Metabolic rates during various activities (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 8, Table 4). Metabolic rate Activity W/m2 Resting: Sleeping Reclining Seated, quiet Standing, relaxed 40 45 60 70 Walking (on the level): 2 mph (0.89 m/s) 3 mph (1.34 m/s) 4 mph (1.79 m/s) 115 150 220 Office Activities: Reading, seated Writing Typing Filing, seated Filing, standing Walking about Lifting/packing 55 60 65 70 80 100 120 Driving/Flying: Car Aircraft, routine Heavy vehicle 60–115 70 185 Miscellaneous Occupational Activities: Cooking Cleaning house Machine work: Light Heavy Handling 50-kg bags Pick and shovel work 95–115 115–140 115–140 235 235 235–280 Miscellaneous Leisure Activities: Dancing, social Calisthenics/exercise Tennis, singles Basketball Wrestling, competitive 140–255 175–235 210–270 290–440 410–505 Multiply by 1.8 m2 to obtain metabolic rates for an average man. Multiply by 0.3171 to convert to Btu/h · ft2. comfortable in an environment. It has been observed that most normally clothed people resting or doing light work feel comfortable in the operative temperature (roughly, the average temperature of air and surrounding surfaces) range of 23°C to 27°C or 73°C to 80°F (Fig. 1–51). For unclothed people, this range is 29°C to 31°C. Relative humidity also has a considerable effect on comfort since it is a measure of air’s ability to absorb moisture and thus it affects the amount of heat a body can dissipate by evaporation. High relative humidity slows down heat rejection by evaporation, especially at high temperatures, and low relative humidity speeds it up. The desirable level of relative humidity is the broad range of 30 to 70 percent, with 50 percent being the most desirable level. Most people at these conditions feel neither hot nor cold, and the body does not need to activate any of the defense mechanisms to maintain the normal body temperature (Fig. 1–52). Another factor that has a major effect on thermal comfort is excessive air motion or draft, which causes undesired local cooling of the human body. Draft is identified by many as a most annoying factor in work places, automobiles, and airplanes. Experiencing discomfort by draft is most common among people wearing indoor clothing and doing light sedentary work, and least common among people with high activity levels. The air velocity should be kept below 9 m/min (30 ft/min) in winter and 15 m/min (50 ft/min) in summer to minimize discomfort by draft, especially when the air is cool. A low level of air motion is desirable as it removes the warm, moist air that builds around the body and replaces it with fresh air. Therefore, air motion should be strong enough to remove heat and moisture from the vicinity of the body, but gentle enough to be unnoticed. High speed air motion causes discomfort outdoors as well. For example, an environment at 10°C (50°F) with 48 km/h winds feels as cold as an environment at 7°C (20°F) with 3 km/h winds because of the chilling effect of the air motion (the wind-chill factor). A comfort system should provide uniform conditions throughout the living space to avoid discomfort caused by nonuniformities such as drafts, asymmetric thermal radiation, hot or cold floors, and vertical temperature stratification. Asymmetric thermal radiation is caused by the cold surfaces of large windows, uninsulated walls, or cold products and the warm surfaces of gas or electric radiant heating panels on the walls or ceiling, solar-heated masonry walls or ceilings, and warm machinery. Asymmetric radiation causes discomfort by exposing different sides of the body to surfaces at different temperatures and thus to different heat loss or gain by radiation. A person whose left side is exposed to a cold window, for example, will feel like heat is being drained from that side of his or her body (Fig. 1–53). For thermal comfort, the radiant temperature asymmetry should not exceed 5°C in the vertical direction and 10°C in the horizontal direction. The unpleasant effect of radiation asymmetry can be minimized by properly sizing and installing heating panels, using double-pane windows, and providing generous insulation at the walls and the roof. Direct contact with cold or hot floor surfaces also causes localized discomfort in the feet. The temperature of the floor depends on the way it is constructed (being directly on the ground or on top of a heated room, being made of wood or concrete, the use of insulation, etc.) as well as the floor cen58933_ch01.qxd 9/10/2002 8:30 AM Page 43 43 CHAPTER 1 °C 2.0 Clothing insulation (clo) covering used such as pads, carpets, rugs, and linoleum. A floor temperature of 23 to 25°C is found to be comfortable to most people. The floor asymmetry loses its significance for people with footwear. An effective and economical way of raising the floor temperature is to use radiant heating panels instead of turning the thermostat up. Another nonuniform condition that causes discomfort is temperature stratification in a room that exposes the head and the feet to different temperatures. For thermal comfort, the temperature difference between the head and foot levels should not exceed 3°C. This effect can be minimized by using destratification fans. It should be noted that no thermal environment will please everyone. No matter what we do, some people will express some discomfort. The thermal comfort zone is based on a 90 percent acceptance rate. That is, an environment is deemed comfortable if only 10 percent of the people are dissatisfied with it. Metabolism decreases somewhat with age, but it has no effect on the comfort zone. Research indicates that there is no appreciable difference between the environments preferred by old and young people. Experiments also show that men and women prefer almost the same environment. The metabolism rate of women is somewhat lower, but this is compensated by their slightly lower skin temperature and evaporative loss. Also, there is no significant variation in the comfort zone from one part of the world to another and from winter to summer. Therefore, the same thermal comfort conditions can be used throughout the world in any season. Also, people cannot acclimatize themselves to prefer different comfort conditions. In a cold environment, the rate of heat loss from the body may exceed the rate of metabolic heat generation. Average specific heat of the human body is 3.49 kJ/kg · °C, and thus each 1°C drop in body temperature corresponds to a deficit of 244 kJ in body heat content for an average 70-kg man. A drop of 0.5°C in mean body temperature causes noticeable but acceptable discomfort. A drop of 2.6°C causes extreme discomfort. A sleeping person will wake up when his or her mean body temperature drops by 1.3°C (which normally shows up as a 0.5°C drop in the deep body and 3°C in the skin area). The drop of deep body temperature below 35°C may damage the body temperature regulation mechanism, while a drop below 28°C may be fatal. Sedentary people reported to feel comfortable at a mean skin temperature of 33.3°C, uncomfortably cold at 31°C, shivering cold at 30°C, and extremely cold at 29°C. People doing heavy work reported to feel comfortable at much lower temperatures, which shows that the activity level affects human performance and comfort. The extremities of the body such as hands and feet are most easily affected by cold weather, and their temperature is a better indication of comfort and performance. A hand-skin temperature of 20°C is perceived to be uncomfortably cold, 15°C to be extremely cold, and 5°C to be painfully cold. Useful work can be performed by hands without difficulty as long as the skin temperature of fingers remains above 16°C (ASHRAE Handbook of Fundamentals, Ref. 1, Chapter 8). The first line of defense of the body against excessive heat loss in a cold environment is to reduce the skin temperature and thus the rate of heat loss from the skin by constricting the veins and decreasing the blood flow to the skin. This measure decreases the temperature of the tissues subjacent to the skin, but maintains the inner body temperature. The next preventive 20 25 Sedentary 50% RH ≤ 30 fpm (0.15 m/s) 1.5 30 Heavy clothing 1.0 Winter clothing 0.5 Summer clothing 0 64 68 72 76 80 °F Operative temperature 84 Upper acceptability limit Optimum Lower acceptability limit FIGURE 1–51 The effect of clothing on the environment temperature that feels comfortable (1 clo 0.155 m2 · °C/W 0.880 ft2 · °F · h/Btu) (from ASHRAE Standard 55-1981). 23°C RH = 50% Air motion 5 m/min FIGURE 1–52 A thermally comfortable environment. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 44 44 HEAT TRANSFER Cold window Warm wall Radiation Radiation FIGURE 1–53 Cold surfaces cause excessive heat loss from the body by radiation, and thus discomfort on that side of the body. Brrr! Shivering FIGURE 1–54 The rate of metabolic heat generation may go up by six times the resting level during total body shivering in cold weather. measure is increasing the rate of metabolic heat generation in the body by shivering, unless the person does it voluntarily by increasing his or her level of activity or puts on additional clothing. Shivering begins slowly in small muscle groups and may double the rate of metabolic heat production of the body at its initial stages. In the extreme case of total body shivering, the rate of heat production may reach six times the resting levels (Fig. 1–54). If this measure also proves inadequate, the deep body temperature starts falling. Body parts furthest away from the core such as the hands and feet are at greatest danger for tissue damage. In hot environments, the rate of heat loss from the body may drop below the metabolic heat generation rate. This time the body activates the opposite mechanisms. First the body increases the blood flow and thus heat transport to the skin, causing the temperature of the skin and the subjacent tissues to rise and approach the deep body temperature. Under extreme heat conditions, the heart rate may reach 180 beats per minute in order to maintain adequate blood supply to the brain and the skin. At higher heart rates, the volumetric efficiency of the heart drops because of the very short time between the beats to fill the heart with blood, and the blood supply to the skin and more importantly to the brain drops. This causes the person to faint as a result of heat exhaustion. Dehydration makes the problem worse. A similar thing happens when a person working very hard for a long time stops suddenly. The blood that has flooded the skin has difficulty returning to the heart in this case since the relaxed muscles no longer force the blood back to the heart, and thus there is less blood available for pumping to the brain. The next line of defense is releasing water from sweat glands and resorting to evaporative cooling, unless the person removes some clothing and reduces the activity level (Fig. 1–55). The body can maintain its core temperature at 37°C in this evaporative cooling mode indefinitely, even in environments at higher temperatures (as high as 200°C during military endurance tests), if the person drinks plenty of liquids to replenish his or her water reserves and the ambient air is sufficiently dry to allow the sweat to evaporate instead of rolling down the skin. If this measure proves inadequate, the body will have to start absorbing the metabolic heat and the deep body temperature will rise. A person can tolerate a temperature rise of 1.4°C without major discomfort but may collapse when the temperature rise reaches 2.8°C. People feel sluggish and their efficiency drops considerably when the core body temperature rises above 39°C. A core temperature above 41°C may damage hypothalamic proteins, resulting in cessation cen58933_ch01.qxd 9/10/2002 8:30 AM Page 45 45 CHAPTER 1 of sweating, increased heat production by shivering, and a heat stroke with irreversible and life-threatening damage. Death can occur above 43°C. A surface temperature of 46°C causes pain on the skin. Therefore, direct contact with a metal block at this temperature or above is painful. However, a person can stay in a room at 100°C for up to 30 min without any damage or pain on the skin because of the convective resistance at the skin surface and evaporative cooling. We can even put our hands into an oven at 200°C for a short time without getting burned. Another factor that affects thermal comfort, health, and productivity is ventilation. Fresh outdoor air can be provided to a building naturally by doing nothing, or forcefully by a mechanical ventilation system. In the first case, which is the norm in residential buildings, the necessary ventilation is provided by infiltration through cracks and leaks in the living space and by the opening of the windows and doors. The additional ventilation needed in the bathrooms and kitchens is provided by air vents with dampers or exhaust fans. With this kind of uncontrolled ventilation, however, the fresh air supply will be either too high, wasting energy, or too low, causing poor indoor air quality. But the current practice is not likely to change for residential buildings since there is not a public outcry for energy waste or air quality, and thus it is difficult to justify the cost and complexity of mechanical ventilation systems. Mechanical ventilation systems are part of any heating and air conditioning system in commercial buildings, providing the necessary amount of fresh outdoor air and distributing it uniformly throughout the building. This is not surprising since many rooms in large commercial buildings have no windows and thus rely on mechanical ventilation. Even the rooms with windows are in the same situation since the windows are tightly sealed and cannot be opened in most buildings. It is not a good idea to oversize the ventilation system just to be on the “safe side” since exhausting the heated or cooled indoor air wastes energy. On the other hand, reducing the ventilation rates below the required minimum to conserve energy should also be avoided so that the indoor air quality can be maintained at the required levels. The minimum fresh air ventilation requirements are listed in Table 1–8. The values are based on controlling the CO2 and other contaminants with an adequate margin of safety, which requires each person be supplied with at least 7.5 L/s (15 ft3/min) of fresh air. Another function of the mechanical ventilation system is to clean the air by filtering it as it enters the building. Various types of filters are available for this purpose, depending on the cleanliness requirements and the allowable pressure drop. Evaporation FIGURE 1–55 In hot environments, a body can dissipate a large amount of metabolic heat by sweating since the sweat absorbs the body heat and evaporates. TABLE 1–8 Minimum fresh air requirements in buildings (from ASHRAE Standard 62-1989) Requirement (per person) L/s ft3/min Classrooms, libraries, supermarkets 8 15 Dining rooms, conference rooms, offices 10 20 Hospital rooms 13 25 Application Hotel rooms Smoking lounges 15 30 (per room) (per room) 30 Retail stores 1.0–1.5 (per m2) 60 0.2–0.3 (per ft2) Residential 0.35 air change per buildings hour, but not less than 7.5 L/s (or 15 ft3/min) per person cen58933_ch01.qxd 9/10/2002 8:30 AM Page 46 46 HEAT TRANSFER SUMMARY In this chapter, the basics of heat transfer are introduced and discussed. The science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, whereas the science of heat transfer deals with the rate of heat transfer, which is the main quantity of interest in the design and evaluation of heat transfer equipment. The sum of all forms of energy of a system is called total energy, and it includes the internal, kinetic, and potential energies. The internal energy represents the molecular energy of a system, and it consists of sensible, latent, chemical, and nuclear forms. The sensible and latent forms of internal energy can be transferred from one medium to another as a result of a temperature difference, and are referred to as heat or thermal energy. Thus, heat transfer is the exchange of the sensible and latent forms of internal energy between two mediums as a result of a temperature difference. The amount of heat transferred · per unit time is called heat transfer rate and is denoted by Q . · The rate of heat transfer per unit area is called heat flux, q. A system of fixed mass is called a closed system and a system that involves mass transfer across its boundaries is called an open system or control volume. The first law of thermodynamics or the energy balance for any system undergoing any process can be expressed as Ein Eout Esystem When a stationary closed system involves heat transfer only and no work interactions across its boundary, the energy balance relation reduces to Q mCvT where Q is the amount of net heat transfer to or from the sys· tem. When heat is transferred at a constant rate of Q , the amount of heat transfer during a time interval t can be deter· mined from Q Q t. Under steady conditions and in the absence of any work interactions, the conservation of energy relation for a control volume with one inlet and one exit with negligible changes in kinetic and potential energies can be expressed as · Q m· CpT · where m· Ac is the mass flow rate and Q is the rate of net heat transfer into or out of the control volume. Heat can be transferred in three different modes: conduction, convection, and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles, and is expressed by Fourier’s law of heat conduction as · dT Q cond kA dx where k is the thermal conductivity of the material, A is the area normal to the direction of heat transfer, and dT/dx is the temperature gradient. The magnitude of the rate of heat conduction across a plane layer of thickness L is given by · T Q cond kA L where T is the temperature difference across the layer. Convection is the mode of heat transfer between a solid surface and the adjacent liquid or gas that is in motion, and involves the combined effects of conduction and fluid motion. The rate of convection heat transfer is expressed by Newton’s law of cooling as · Q convection hAs (Ts T∞) where h is the convection heat transfer coefficient in W/m2 · °C or Btu/h · ft2 · °F, As is the surface area through which convection heat transfer takes place, Ts is the surface temperature, and T∞ is the temperature of the fluid sufficiently far from the surface. Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts is given by the Stefan–Boltzmann · law as Q emit, max AsTs4, where 5.67 108 W/m2 · K4 or 0.1714 108 Btu/h · ft2 · R4 is the Stefan–Boltzmann constant. When a surface of emissivity and surface area As at an absolute temperature Ts is completely enclosed by a much larger (or black) surface at absolute temperature Tsurr separated by a gas (such as air) that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by · 4 ) Q rad As (Ts4 Tsurr In this case, the emissivity and the surface area of the surrounding surface do not have any effect on the net radiation heat transfer. The rate at which a surface absorbs radiation is determined · · · from Q absorbed Q incident where Q incident is the rate at which radiation is incident on the surface and is the absorptivity of the surface. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 47 47 CHAPTER 1 REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and AirConditioning Engineers, Handbook of Fundamentals. Atlanta: ASHRAE, 1993. 2. Y. A. Çengel and R. H. Turner. Fundamentals of ThermalFluid Sciences. New York: McGraw-Hill, 2001. 3. Y. A. Çengel and M. A. Boles. Thermodynamics—An Engineering Approach. 4th ed. New York: McGraw-Hill, 2002. 4. J. P. Holman. Heat Transfer. 9th ed. New York: McGrawHill, 2002. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. 7. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed. Upper Saddle River, NJ: Prentice-Hall, 1999. 8. M. N. Ozisik. Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. 9. Robert J. Ribando. Heat Transfer Tools. New York: McGraw-Hill, 2002. 10. F. M. White. Heat and Mass Transfer. Reading, MA: Addison-Wesley, 1988. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. PROBLEMS 1–1C How does the science of heat transfer differ from the science of thermodynamics? 1–9C What are the mechanisms of energy transfer to a closed system? How is heat transfer distinguished from the other forms of energy transfer? 1–2C What is the driving force for (a) heat transfer, (b) electric current flow, and (c) fluid flow? 1–10C How are heat, internal energy, and thermal energy related to each other? 1–3C What is the caloric theory? When and why was it abandoned? 1–11C An ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? Thermodynamics and Heat Transfer 1–4C How do rating problems in heat transfer differ from the sizing problems? 1–5C What is the difference between the analytical and experimental approach to heat transfer? Discuss the advantages and disadvantages of each approach. 1–6C What is the importance of modeling in engineering? How are the mathematical models for engineering processes prepared? 1–7C When modeling an engineering process, how is the right choice made between a simple but crude and a complex but accurate model? Is the complex model necessarily a better choice since it is more accurate? Heat and Other Forms of Energy 1–8C What is heat flux? How is it related to the heat transfer rate? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 1–12 A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces. 1–13E A logic chip used in a computer dissipates 3 W of power in an environment at 120°F, and has a heat transfer surface area of 0.08 in2. Assuming the heat transfer from the surface to be uniform, determine (a) the amount of heat this chip dissipates during an eight-hour work day, in kWh, and (b) the heat flux on the surface of the chip, in W/in2. 1–14 Consider a 150-W incandescent lamp. The filament of the lamp is 5 cm long and has a diameter of 0.5 mm. The diameter of the glass bulb of the lamp is 8 cm. Determine the heat flux, in W/m2, (a) on the surface of the filament and (b) on the surface of the glass bulb, and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is $0.08/kWh. Answers: (a) 1.91 106 W/m2, (b) 7500 W/m2, (c) $35.04/yr 1–15 A 1200-W iron is left on the ironing board with its base exposed to the air. About 90 percent of the heat generated in the iron is dissipated through its base whose surface area is 150 cm2, and the remaining 10 percent through other surfaces. Assuming the heat transfer from the surface to be uniform, cen58933_ch01.qxd 9/10/2002 8:30 AM Page 48 48 HEAT TRANSFER D = 8 cm Filament d = 0.5 mm L = 5 cm FIGURE P1–14 determine (a) the amount of heat the iron dissipates during a 2-hour period, in kWh, (b) the heat flux on the surface of the iron base, in W/m2, and (c) the total cost of the electrical energy consumed during this 2-hour period. Take the unit cost of electricity to be $0.07/kWh. 1–16 A 15-cm 20-cm circuit board houses on its surface 120 closely spaced logic chips, each dissipating 0.12 W. If the heat transfer from the back surface of the board is negligible, determine (a) the amount of heat this circuit board dissipates during a 10-hour period, in kWh, and (b) the heat flux on the surface of the circuit board, in W/m2. Chips 15 cm 1–18 The average specific heat of the human body is 3.6 kJ/kg · °C. If the body temperature of a 70-kg man rises from 37°C to 39°C during strenuous exercise, determine the increase in the thermal energy content of the body as a result of this rise in body temperature. 1–19 Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of 200 m2 and an average height of 3 m at 1000 m elevation, where the standard atmospheric pressure is 89.6 kPa. The house is maintained at a temperature of 22°C, and the infiltration losses are estimated to amount to 0.7 ACH. Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is 5°C. Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $0.082/kWh. Answers: 53.8 kWh/day, $4.41/day 1–20 Consider a house with a floor space of 200 m2 and an average height of 3 m at sea level, where the standard atmospheric pressure is 101.3 kPa. Initially the house is at a uniform temperature of 10°C. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 22°C. Determine how much heat is absorbed by the air assuming some air escapes through the cracks as the heated air in the house expands at constant pressure. Also, determine the cost of this heat if the unit cost of electricity in that area is $0.075/kWh. 1–21E Consider a 60-gallon water heater that is initially filled with water at 45°F. Determine how much energy needs to be transferred to the water to raise its temperature to 140°F. Take the density and specific heat of water to be 62 lbm/ft3 and 1.0 Btu/lbm · °F, respectively. The First Law of Thermodynamics 20 cm FIGURE P1–16 1–17 A 15-cm-diameter aluminum ball is to be heated from 80°C to an average temperature of 200°C. Taking the average density and specific heat of aluminum in this temperature range to be 2700 kg/m3 and Cp 0.90 kJ/kg · °C, respectively, determine the amount of energy that needs to be transAnswer: 515 kJ ferred to the aluminum ball. 1–22C On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will his room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed. 1–23C Consider two identical rooms, one with a refrigerator in it and the other without one. If all the doors and windows are closed, will the room that contains the refrigerator be cooler or warmer than the other room? Why? 1–24C Define mass and volume flow rates. How are they related to each other? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 49 49 CHAPTER 1 1–25 Two 800-kg cars moving at a velocity of 90 km/h have a head-on collision on a road. Both cars come to a complete rest after the crash. Assuming all the kinetic energy of cars is converted to thermal energy, determine the average temperature rise of the remains of the cars immediately after the crash. Take the average specific heat of the cars to be 0.45 kJ/kg · °C. 1–26 A classroom that normally contains 40 people is to be air-conditioned using window air-conditioning units of 5-kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of 360 kJ/h. There are 10 lightbulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window airAnswer: two units conditioning units required. 1–27E A rigid tank contains 20 lbm of air at 50 psia and 80°F. The air is now heated until its pressure is doubled. Determine (a) the volume of the tank and (b) the amount of heat Answers: (a) 80 ft3, (b) 2035 Btu transfer. 1–28 A 1-m3 rigid tank contains hydrogen at 250 kPa and 420 K. The gas is now cooled until its temperature drops to 300 K. Determine (a) the final pressure in the tank and (b) the amount of heat transfer from the tank. 1–29 A 4-m 5-m 6-m room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 7°C to 25°C within 15 minutes. Assuming no heat losses from the room and an atmospheric pressure of 100 kPa, determine the required power rating of the resistance heater. Assume constant specific heats at room temperature. Answer: 3.01 kW 1–30 A 4-m 5-m 7-m room is heated by the radiator of a steam heating system. The steam radiator transfers heat at a rate of 10,000 kJ/h and a 100-W fan is used to distribute the warm air in the room. The heat losses from the room are estimated to be at a rate of about 5000 kJ/h. If the initial temperature of the room air is 10°C, determine how long it will take for the air temperature to rise to 20°C. Assume constant specific heats at room temperature. 5000 kJ/h Room 4m×6m×6m Fan FIGURE P1–31 1–31 A student living in a 4-m 6-m 6-m dormitory room turns his 150-W fan on before she leaves her room on a summer day hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 hours later. Use specific heat values at room temperature and assume the room to be at 100 kPa and 15°C in the morning when she leaves. Answer: 58.1°C 1–32E A 10-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Neglecting the energy stored in the paddle wheel, determine the work done by the paddle wheel. 1–33 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 7000 kJ/h, it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in kW. 1–34 A 50-kg mass of copper at 70°C is dropped into an insulated tank containing 80 kg of water at 25°C. Determine the final equilibrium temperature in the tank. 1–35 A 20-kg mass of iron at 100°C is brought into contact with 20 kg of aluminum at 200°C in an insulated enclosure. Determine the final equilibrium temperature of the combined system. Answer: 168°C 1–36 An unknown mass of iron at 90°C is dropped into an insulated tank that contains 80 L of water at 20°C. At the same Room 4m×5m×7m Steam · Wpw FIGURE P1–30 Water Iron Wpw 10,000 kJ/ h FIGURE P1–36 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 50 50 HEAT TRANSFER time, a paddle wheel driven by a 200-W motor is activated to stir the water. Thermal equilibrium is established after 25 minutes with a final temperature of 27°C. Determine the mass of the iron. Neglect the energy stored in the paddle wheel, and Answer: 72.1 kg take the density of water to be 1000 kg/m3. 1–37E A 90-lbm mass of copper at 160°F and a 50-lbm mass of iron at 200°F are dropped into a tank containing 180 lbm of water at 70°F. If 600 Btu of heat is lost to the surroundings during the process, determine the final equilibrium temperature. 1–38 A 5-m 6-m 8-m room is to be heated by an electrical resistance heater placed in a short duct in the room. Initially, the room is at 15°C, and the local atmospheric pressure is 98 kPa. The room is losing heat steadily to the outside at a rate of 200 kJ/min. A 200-W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of 50 kg/min. The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 15 minutes for the room air to reach an average temperature of 25°C, find (a) the power rating of the electric heater and (b) the temperature rise that the air experiences each time it passes through the heater. 1–39 A house has an electric heating system that consists of a 300-W fan and an electric resistance heating element placed in a duct. Air flows steadily through the duct at a rate of 0.6 kg/s and experiences a temperature rise of 5°C. The rate of heat loss from the air in the duct is estimated to be 250 W. Determine the power rating of the electric resistance heating element. 1–40 A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors where it is heated. Air enters a 1200-W hair dryer at 100 kPa and 22°C, and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer, determine (a) the volume flow rate of air at the inlet and (b) the velocity of the air at the Answers: (a) 0.0404 m3/s, (b) 7.30 m/s exit. T2 = 47°C A 2 = 60 cm 2 P1 = 100 kPa T1 = 22°C · We = 1200 W FIGURE P1–40 1–41 The ducts of an air heating system pass through an unheated area. As a result of heat losses, the temperature of the air in the duct drops by 3°C. If the mass flow rate of air is 120 kg/min, determine the rate of heat loss from the air to the cold environment. 1–42E Air enters the duct of an air-conditioning system at 15 psia and 50°F at a volume flow rate of 450 ft3/min. The diameter of the duct is 10 inches and heat is transferred to the air in the duct from the surroundings at a rate of 2 Btu/s. Determine (a) the velocity of the air at the duct inlet and (b) the temperaAnswers: (a) 825 ft/min, (b) 64°F ture of the air at the exit. 1–43 Water is heated in an insulated, constant diameter tube by a 7-kW electric resistance heater. If the water enters the heater steadily at 15°C and leaves at 70°C, determine the mass flow rate of water. 70°C Water 15°C Resistance heater, 7 kW FIGURE P1–43 Heat Transfer Mechanisms 1–44C Define thermal conductivity and explain its significance in heat transfer. 1–45C What are the mechanisms of heat transfer? How are they distinguished from each other? 1–46C What is the physical mechanism of heat conduction in a solid, a liquid, and a gas? 1–47C Consider heat transfer through a windowless wall of a house in a winter day. Discuss the parameters that affect the rate of heat conduction through the wall. 1–48C Write down the expressions for the physical laws that govern each mode of heat transfer, and identify the variables involved in each relation. 1–49C How does heat conduction differ from convection? 1–50C Does any of the energy of the sun reach the earth by conduction or convection? 1–51C How does forced convection differ from natural convection? 1–52C Define emissivity and absorptivity. What is Kirchhoff’s law of radiation? 1–53C What is a blackbody? How do real bodies differ from blackbodies? 1–54C Judging from its unit W/m · °C, can we define thermal conductivity of a material as the rate of heat transfer through the material per unit thickness per unit temperature difference? Explain. 1–55C Consider heat loss through the two walls of a house on a winter night. The walls are identical, except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain. 1–56C Which is a better heat conductor, diamond or silver? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 51 51 CHAPTER 1 1–57C Consider two walls of a house that are identical except that one is made of 10-cm-thick wood, while the other is made of 25-cm-thick brick. Through which wall will the house lose more heat in winter? 1–58C How do the thermal conductivity of gases and liquids vary with temperature? 105°C 1–59C Why is the thermal conductivity of superinsulation orders of magnitude lower than the thermal conductivity of ordinary insulation? 1–60C Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of the ordinary thermal conductivity? 1–61C Consider an alloy of two metals whose thermal conductivities are k1 and k2. Will the thermal conductivity of the alloy be less than k1, greater than k2, or between k1 and k2? 1–62 The inner and outer surfaces of a 5-m 6-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m · °C are maintained at temperatures of 20°C and 5°C, respectively. Determine the rate of heat transfer through the wall, in W. Answer: 1035 W 800 W FIGURE P1–65 1–66E The north wall of an electrically heated home is 20 ft long, 10 ft high, and 1 ft thick, and is made of brick whose thermal conductivity is k 0.42 Btu/h · ft · °F. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about 62°F and 25°F, respectively, for a period of 8 hours. Determine (a) the rate of heat loss through the wall that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.07/kWh. 1–67 In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used (see Fig. 1–29). The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.6 A at 110 V, and both differential thermometers read a temperature difference of 10°C. Determine the thermal conductivity of the Answer: 78.8 W/m · °C sample. Brick wall 20°C 0.4 cm 5°C 30 cm FIGURE P1–62 1–63 The inner and outer surfaces of a 0.5-cm-thick 2-m 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m · °C, determine the amount of heat loss, in kJ, through the glass over a period of 5 hours. What would your answer be if the glass were Answers: 78,624 kJ, 39,312 kJ 1 cm thick? Reconsider Problem 1–63. Using EES (or other) software, plot the amount of heat loss through the glass as a function of the window glass thickness in the range of 0.1 cm to 1.0 cm. Discuss the results. 1–68 One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two 0.5-cm-thick samples 10 cm 10 cm in size are used. When steady operation is reached, the heater is observed to draw 35 W of electric power, and the temperature of each sample is observed to drop from 82°C at the inner surface to 74°C at the outer surface. Determine the thermal conductivity of the material at the average temperature. Samples Insulation Wattmeter 1–64 1–65 An aluminum pan whose thermal conductivity is 237 W/m · °C has a flat bottom with diameter 20 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 800 W. If the inner surface of the bottom of the pan is at 105°C, determine the temperature of the outer surface of the bottom of the pan. Insulation ~ Source 0.5 cm Resistance heater FIGURE P1–68 1–69 Repeat Problem 1–68 for an electric power consumption of 28 W. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 52 52 HEAT TRANSFER 1–70 A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m2 through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of Answer: 0.0938 W/m · °C the refrigerator door. 1–77 A 50-cm-long, 800-W electric resistance heating element with diameter 0.5 cm and surface temperature 120°C is immersed in 60 kg of water initially at 20°C. Determine how long it will take for this heater to raise the water temperature to 80°C. Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1–71 Consider a person standing in a room maintained at 20°C at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of 12°C in winter and 23°C in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person are 1.6 m2, 0.95, and 32°C, respectively. 1–78 A 5-cm-external-diameter, 10-m-long hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/m2 · °C. Determine the rate of heat loss from the pipe by natural conAnswer: 2945 W vection, in W. 1–72 Reconsider Problem 1–71. Using EES (or other) software, plot the rate of radiation heat transfer in winter as a function of the temperature of the inner surface of the room in the range of 8°C to 18°C. Discuss the results. 1–79 A hollow spherical iron container with outer diameter 20 cm and thickness 0.4 cm is filled with iced water at 0°C. If the outer surface temperature is 5°C, determine the approximate rate of heat loss from the sphere, in kW, and the rate at which ice melts in the container. The heat from fusion of water is 333.7 kJ/kg. 5°C 1–73 For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine the rate of heat loss from this man by convection in an environment at Answer: 336 W 20°C. Iced water 0.4 cm 1–74 Hot air at 80°C is blown over a 2-m 4-m flat surface at 30°C. If the average convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of heat transfer from the air to Answer: 22 kW the plate, in kW. 1–75 Reconsider Problem 1–74. Using EES (or other) software, plot the rate of heat transfer as a function of the heat transfer coefficient in the range of 20 W/m2 · °C to 100 W/m2 · °C. Discuss the results. 1–76 The heat generated in the circuitry on the surface of a silicon chip (k 130 W/m · °C) is conducted to the ceramic substrate to which it is attached. The chip is 6 mm 6 mm in size and 0.5 mm thick and dissipates 3 W of power. Disregarding any heat transfer through the 0.5-mm-high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation. Silicon chip 0.5 mm Ceramic substrate FIGURE P1–76 6 6 mm m m 3W FIGURE P1–79 1–80 Reconsider Problem 1–79. Using EES (or other) software, plot the rate at which ice melts as a function of the container thickness in the range of 0.2 cm to 2.0 cm. Discuss the results. 1–81E The inner and outer glasses of a 6-ft 6-ft doublepane window are at 60°F and 42°F, respectively. If the 0.25-in. space between the two glasses is filled with still air, determine the rate of heat transfer through the window. Answer: 439 Btu/h 1–82 Two surfaces of a 2-cm-thick plate are maintained at 0°C and 80°C, respectively. If it is determined that heat is transferred through the plate at a rate of 500 W/m2, determine its thermal conductivity. 1–83 Four power transistors, each dissipating 15 W, are mounted on a thin vertical aluminum plate 22 cm 22 cm in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at 25°C, which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is 25 W/m2 · °C, determine the temperature of the aluminum plate. Disregard any radiation effects. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 53 53 CHAPTER 1 1–84 An ice chest whose outer dimensions are 30 cm 40 cm 40 cm is made of 3-cm-thick Styrofoam (k 0.033 W/m · °C). Initially, the chest is filled with 40 kg of ice at 0°C, and the inner surface temperature of the ice chest can be taken to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7 kJ/kg, and the surrounding ambient air is at 30°C. Disregarding any heat transfer from the 40-cm 40-cm base of the ice chest, determine how long it will take for the ice in the chest to melt completely if the outer surfaces of the ice chest are at 8°C. 1–87E A 200-ft-long section of a steam pipe whose outer diameter is 4 inches passes through an open space at 50°F. The average temperature of the outer surface of the pipe is measured to be 280°F, and the average heat transfer coefficient on that surface is determined to be 6 Btu/h · ft2 · °F. Determine (a) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $0.58/therm (1 therm 100,000 Btu). Answers: (a) 289,000 Btu/h, (b) $17,074/yr Answer: 32.7 days Tair = 30°C Ice chest 0°C 3 cm 0°C Styrofoam FIGURE P1–84 1–85 A transistor with a height of 0.4 cm and a diameter of 0.6 cm is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of 30 W/m2 · °C. If the air temperature is 55°C and the transistor case temperature is not to exceed 70°C, determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base. 1–88 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm) is 196°C. Therefore, nitrogen is commonly used in low temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm. Consider a 4-m-diameter spherical tank initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to 20°C ambient air with a heat transfer coefficient of 25 W/m2 · °C. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air. N2 vapor Tair = 20°C Air 55°C 1 atm Liquid N2 –196°C Power transistor Ts ≤ 70°C 0.6 cm 0.4 cm FIGURE P1–85 1–86 Reconsider Problem 1–85. Using EES (or other) software, plot the amount of power the transistor can dissipate safely as a function of the maximum case temperature in the range of 60°C to 90°C. Discuss the results. FIGURE P1–88 1–89 Repeat Problem 1–88 for liquid oxygen, which has a boiling temperature of 183°C, a heat of vaporization of 213 kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure. 1–90 Reconsider Problem 1–88. Using EES (or other) software, plot the rate of evaporation of liquid nitrogen as a function of the ambient air temperature in the range of 0°C to 35°C. Discuss the results. 1–91 Consider a person whose exposed surface area is 1.7 m2, emissivity is 0.7, and surface temperature is 32°C. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 54 54 HEAT TRANSFER Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of (a) 300 K and Answers: (a) 37.4 W, (b) 169.2 W (b) 280 K. 1–92 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. The board is impregnated with copper fillings and has an effective thermal conductivity of 16 W/m · °C. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between Answer: 0.042°C the two sides of the circuit board. 1–93 Consider a sealed 20-cm-high electronic box whose base dimensions are 40 cm 40 cm placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 55°C, determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible. 40 cm 40 cm 100 W ε = 0.95 Ts = 60°C Electronic box 20 cm Stand FIGURE P1–93 1–94 Using the conversion factors between W and Btu/h, m and ft, and K and R, express the Stefan–Boltzmann constant 5.67 108 W/m2 · K4 in the English unit Btu/h · ft2 · R4. 1–95 An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is 20 W/m2 · °C, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between W and Btu/h, m and ft, and °C and °F, express the given convection heat transfer coefficient in Btu/h · ft2 · °F. Answer: 3.52 Btu/h · ft2 · °F Simultaneous Heat Transfer Mechanisms 1–96C Can all three modes of heat transfer occur simultaneously (in parallel) in a medium? 1–97C Can a medium involve (a) conduction and convection, (b) conduction and radiation, or (c) convection and radiation simultaneously? Give examples for the “yes” answers. 1–98C The deep human body temperature of a healthy person remains constant at 37°C while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter. 1–99C We often turn the fan on in summer to help us cool. Explain how a fan makes us feel cooler in the summer. Also explain why some people use ceiling fans also in winter. 1–100 Consider a person standing in a room at 23°C. Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are 1.7 m2 and 32°C, respectively, and the convection heat transfer coefficient is 5 W/m2 · °C. Take the emissivity of the skin and the clothes to be 0.9, and assume the temperature of the inner surfaces of the room to be the same as the air temperature. Answer: 161 W 1–101 Consider steady heat transfer between two large parallel plates at constant temperatures of T1 290 K and T2 150 K that are L 2 cm apart. Assuming the surfaces to be black (emissivity 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with fiberglass insulation, and (d) filled with superinsulation having an apparent thermal conductivity of 0.00015 W/m · °C. 1–102 A 1.4-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at 20°C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 240°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 110 V and 3 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: 170.5 W/m2 · °C Room 20°C 240°C Electric resistance heater FIGURE P1–102 1–103 Reconsider Problem 1–102. Using EES (or other) software, plot the convection heat transfer coefficient as a function of the wire surface temperature in the range of 100°C to 300°C. Discuss the results. 1–104E A 2-in-diameter spherical ball whose surface is maintained at a temperature of 170°F is suspended in the middle of a room at 70°F. If the convection heat transfer coefficient is 12 Btu/h · ft2 · °F and the emissivity of the surface is 0.8, determine the total rate of heat transfer from the ball. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 55 55 CHAPTER 1 1–105 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The convection heat transfer coefficient between the base surface and the surrounding air is 35 W/m2 · °C. If the base has an emissivity of 0.6 and a surface area of 0.02 m2, determine the temperature of Answer: 674°C the base of the iron. Iron 1000 W of the collector is 100°F. The emissivity of the exposed surface of the collector is 0.9. Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is 70°F and the effective sky temperature for radiation exchange is 50°F. Take the convection heat transfer coefficient on the exposed surface to be 2.5 Btu/h · ft2 · °F. 20°C Tsky = 50°F Solar collector 70°F FIGURE P1–105 1–106 The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed. 1–107 A 3-m-internal-diameter spherical tank made of 1-cmthick stainless steel is used to store iced water at 0°C. The tank is located outdoors at 25°C. Assuming the entire steel tank to be at 0°C and thus the thermal resistance of the tank to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-hour period. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. The emissivity of the outer surface of the tank is 0.6, and the convection heat transfer coefficient on the outer surface can be taken to be 30 W/m2 · °C. Assume the average surrounding surface temperature for radiation exAnswer: 5898 kg change to be 15°C. The roof of a house consists of a 15-cm-thick concrete slab (k 2 W/m · °C) that is 15 m wide and 20 m long. The emissivity of the outer surface of the roof is 0.9, and the convection heat transfer coefficient on that surface is estimated to be 15 W/m2 · °C. The inner surface of the roof is maintained at 15°C. On a clear winter night, the ambient air is reported to be at 10°C while the night sky temperature for radiation heat transfer is 255 K. Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is $0.60/therm (1 therm 105,500 kJ of energy content), determine the money lost through the roof that night during a 14-hour period. FIGURE P1–109E Problem Solving Technique and EES 1–110C What is the value of the engineering software packages in (a) engineering education and (b) engineering practice? 1–111 Determine a positive real root of the following equation using EES: 2x3 10x0.5 3x 3 1–112 Solve the following system of two equations with two unknowns using EES: x3 y2 7.75 3xy y 3.5 1–108 1–109E Consider a flat plate solar collector placed horizontally on the flat roof of a house. The collector is 5 ft wide and 15 ft long, and the average temperature of the exposed surface 1–113 Solve the following system of three equations with three unknowns using EES: 2x y z 5 3x2 2y z 2 xy 2z 8 1–114 Solve the following system of three equations with three unknowns using EES: x2y z 1 x 3y0.5 xz 2 xyz2 Special Topic: Thermal Comfort 1–115C What is metabolism? What is the range of metabolic rate for an average man? Why are we interested in metabolic cen58933_ch01.qxd 9/10/2002 8:30 AM Page 56 56 HEAT TRANSFER rate of the occupants of a building when we deal with heating and air conditioning? 1–116C Why is the metabolic rate of women, in general, lower than that of men? What is the effect of clothing on the environmental temperature that feels comfortable? 1–117C What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room? 1–118C How do (a) draft and (b) cold floor surfaces cause discomfort for a room’s occupants? Resistance heater 1–119C What is stratification? Is it likely to occur at places with low or high ceilings? How does it cause thermal discomfort for a room’s occupants? How can stratification be prevented? 1–120C Why is it necessary to ventilate buildings? What is the effect of ventilation on energy consumption for heating in winter and for cooling in summer? Is it a good idea to keep the bathroom fans on all the time? Explain. Review Problems 1–121 2.5 kg of liquid water initially at 18°C is to be heated to 96°C in a teapot equipped with a 1200-W electric heating element inside. The teapot is 0.8 kg and has an average specific heat of 0.6 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated. 1–122 A 4-m-long section of an air heating system of a house passes through an unheated space in the attic. The inner diameter of the circular duct of the heating system is 20 cm. Hot air enters the duct at 100 kPa and 65°C at an average velocity of 3 m/s. The temperature of the air in the duct drops to 60°C as a result of heat loss to the cool space in the attic. Determine the rate of heat loss from the air in the duct to the attic under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace having an efficiency of 82 percent, and the cost of the natural gas in that area is $0.58/therm (1 therm 105,500 kJ). Answers: 0.488 kJ/s, $0.012/h 4m 65°C 3 m/s Hot air 60°C FIGURE P1–122 1–123 Reconsider Problem 1–122. Using EES (or other) software, plot the cost of the heat loss per hour as a function of the average air velocity in the range of 1 m/s to 10 m/s. Discuss the results. 1–124 Water flows through a shower head steadily at a rate of 10 L/min. An electric resistance heater placed in the water pipe heats the water from 16°C to 43°C. Taking the density of FIGURE P1–124 water to be 1 kg/L, determine the electric power input to the heater, in kW. In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of 39°C through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case. If the price of the electric energy is 8.5 ¢/kWh, determine how much money is saved during a 10-minute shower as a result of installing this heat exchanger. Answers: 18.8 kW, 10.8 kW, $0.0113 1–125 It is proposed to have a water heater that consists of an insulated pipe of 5 cm diameter and an electrical resistor inside. Cold water at 15°C enters the heating section steadily at a rate of 18 L/min. If water is to be heated to 50°C, determine (a) the power rating of the resistance heater and (b) the average velocity of the water in the pipe. 1–126 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/h is maintained at 22°C at all times during a winter night for 10 hours. The house is to be heated by 50 glass containers each containing 20 L of water heated to 80°C during the day by absorbing solar energy. A thermostat-controlled 15-kW back-up electric resistance heater turns on whenever necessary to keep the house at 22°C. (a) How long did the electric heating system run that night? (b) How long would the electric heater have run that night if the house incorporated no solar heating? Answers: (a) 4.77 h, (b) 9.26 h 1–127 It is well known that wind makes the cold air feel much colder as a result of the windchill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The windchill effect is usually expressed in terms of the windchill factor, which is the difference between the actual air temperature and the equivalent calm-air cen58933_ch01.qxd 9/10/2002 8:30 AM Page 57 57 CHAPTER 1 50,000 kJ/h and the surrounding air temperature is 10°C, determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be 30 W/m2 · °C, and disregard any heat loss by radiation. 1–129 A 4-m 5-m 6-m room is to be heated by one ton (1000 kg) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of 10,000 kJ/h. The room is initially at 20°C and 100 kPa, and is maintained at an average temperature of 20°C at all times. If the hot water is to meet the heating requirements of this room for a 24-hour period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. 22°C Water Answer: 77.4°C 80°C FIGURE P1–126 temperature. For example, a windchill factor of 20°C for an actual air temperature of 5°C means that the windy air at 5°C feels as cold as the still air at 15°C. In other words, a person will lose as much heat to air at 5°C with a windchill factor of 20°C as he or she would in calm air at 15°C. For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine the rate of heat loss from this man by convection in still air at 20°C. What would your answer be if the convection heat transfer coefficient is increased to 50 W/m2 · °C as a result of winds? What is the windAnswers: 336 W, 1120 W, 32.7°C chill factor in this case? 1–130 Consider a 3-m 3-m 3-m cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of 1200 K. The base surface has an emissivity of 0.7, and is maintained at 800 K. Determine the net rate of radiation heat transfer to the base surface from the top and Answer: 594,400 W side surfaces. 1–131 Consider a refrigerator whose dimensions are 1.8 m 1.2 m 0.8 m and whose walls are 3 cm thick. The refrigerator consumes 600 W of power when operating and has a COP of 2.5. It is observed that the motor of the refrigerator remains on for 5 minutes and then is off for 15 minutes periodically. If the average temperatures at the inner and outer surfaces of the refrigerator are 6°C and 17°C, respectively, determine the average thermal conductivity of the refrigerator walls. Also, determine the annual cost of operating this refrigerator if the unit cost of electricity is $0.08/kWh. 1–128 A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of 0.7 for solar radiation. If solar radiation is incident on the plate at a rate of 700 W/m2 17°C 6°C 700 W/ m2 α = 0.7 10°C FIGURE P1–128 FIGURE P1–131 1–132 A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at 0°C. Also, determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 58 58 HEAT TRANSFER Ice, 0°C Water 0.2 L 20°C FIGURE P1–132 0.6 cm and a thermal conductivity of 0.7 W/m · C. Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of 10 W/m2 · °C and an ambient temperature of 15°C. Determine the fraction of heat lost from the glass cover by radiation. 1–138 The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the U-factor. The value of the U-factor ranges from about 1.25 W/m2 · °C (or 0.22 Btu/h · ft2 · °F) for low-e coated, argon-filled, quadruple-pane windows to 6.25 W/m2 · °C (or 1.1 Btu/h · ft2 · °F) for a singlepane window with aluminum frames. Determine the range for the rate of heat loss through a 1.2-m 1.8-m window of a house that is maintained at 20°C when the outdoor air temperature is 8°C. 1.2 m 1–133 Reconsider Problem 1–132. Using EES (or other) software, plot the amount of ice that needs to be added to the water as a function of the ice temperature in the range of 24°C to 0°C. Discuss the results. Indoors 20°C 1–134E In order to cool 1 short ton (2000 lbm) of water at 70°F in a tank, a person pours 160 lbm of ice at 25°F into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are 32°F and 143.5 Btu/lbm, respectively. . Q Answer: 56.3°F 1–135 Engine valves (Cp 440 J/kg · °C and 7840 kg/m3) are to be heated from 40°C to 800°C in 5 minutes in the heat treatment section of a valve manufacturing facility. The valves have a cylindrical stem with a diameter of 8 mm and a length of 10 cm. The valve head and the stem may be assumed to be of equal surface area, with a total mass of 0.0788 kg. For a single valve, determine (a) the amount of heat transfer, (b) the average rate of heat transfer, and (c) the average heat flux, (d) the number of valves that can be heat treated per day if the heating section can hold 25 valves, and it is used 10 hours per day. 1–136 The hot water needs of a household are met by an electric 60-L hot water tank equipped with a 1.6-kW heating element. The tank is initially filled with hot water at 80°C, and the cold water temperature is 20°C. Someone takes a shower by mixing constant flow rates of hot and cold waters. After a showering period of 8 minutes, the average water temperature in the tank is measured to be 60°C. The heater is kept on during the shower and hot water is replaced by cold water. If the cold water is mixed with the hot water stream at a rate of 0.06 kg/s, determine the flow rate of hot water and the average temperature of mixed water used during the shower. 1–137 Consider a flat plate solar collector placed at the roof of a house. The temperatures at the inner and outer surfaces of glass cover are measured to be 28°C and 25°C, respectively. The glass cover has a surface area of 2.2. m2 and a thickness of Outdoors –8°C 1.8 m FIGURE P1–138 1–139 Reconsider Problem 1–138. Using EES (or other) software, plot the rate of heat loss through the window as a function of the U-factor. Discuss the results. Design and Essay Problems 1–140 Write an essay on how microwave ovens work, and explain how they cook much faster than conventional ovens. Discuss whether conventional electric or microwave ovens consume more electricity for the same task. 1–141 Using information from the utility bills for the coldest month last year, estimate the average rate of heat loss from your house for that month. In your analysis, consider the contribution of the internal heat sources such as people, lights, and appliances. Identify the primary sources of heat loss from your house and propose ways of improving the energy efficiency of your house. 1–142 Design a 1200-W electric hair dryer such that the air temperature and velocity in the dryer will not exceed 50°C and 3/ms, respectively. 1–143 Design an electric hot water heater for a family of four in your area. The maximum water temperature in the tank cen58933_ch01.qxd 9/10/2002 8:30 AM Page 59 59 CHAPTER 1 and the power consumption are not to exceed 60°C and 4 kW, respectively. There are two showers in the house, and the flow rate of water through each of the shower heads is about 10 L/min. Each family member takes a 5-minute shower every morning. Explain why a hot water tank is necessary, and determine the proper size of the tank for this family. 1–144 Conduct this experiment to determine the heat transfer coefficient between an incandescent lightbulb and the surrounding air using a 60-W lightbulb. You will need an indoor– outdoor thermometer, which can be purchased for about $10 in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light, calculate the heat transfer coefficient from Newton’s law of cooling. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 60 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 61 CHAPTER H E AT C O N D U C T I O N E Q U AT I O N eat transfer has direction as well as magnitude. The rate of heat conduction in a specified direction is proportional to the temperature gradient, which is the change in temperature per unit length in that direction. Heat conduction in a medium, in general, is three-dimensional and time dependent. That is, T T(x, y, z, t) and the temperature in a medium varies with position as well as time. Heat conduction in a medium is said to be steady when the temperature does not vary with time, and unsteady or transient when it does. Heat conduction in a medium is said to be one-dimensional when conduction is significant in one dimension only and negligible in the other two dimensions, two-dimensional when conduction in the third dimension is negligible, and three-dimensional when conduction in all dimensions is significant. We start this chapter with a description of steady, unsteady, and multidimensional heat conduction. Then we derive the differential equation that governs heat conduction in a large plane wall, a long cylinder, and a sphere, and generalize the results to three-dimensional cases in rectangular, cylindrical, and spherical coordinates. Following a discussion of the boundary conditions, we present the formulation of heat conduction problems and their solutions. Finally, we consider heat conduction problems with variable thermal conductivity. This chapter deals with the theoretical and mathematical aspects of heat conduction, and it can be covered selectively, if desired, without causing a significant loss in continuity. The more practical aspects of heat conduction are covered in the following two chapters. H 2 CONTENTS 2–1 Introduction 62 2–2 One-Dimensional Heat Conduction Equation 68 2–3 General Heat Conduction Equation 74 2–4 Boundary and Initial Conditions 77 2–5 Solution of Steady One-Dimensional Heat Conduction Problems 86 2–6 Heat Generation in a Solid 97 2–7 Variable Thermal Conductivity k (T ) 104 Topic of Special Interest: A Brief Review of Differential Equations 107 61 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 62 62 HEAT TRANSFER Magnitude of temperature at a point A (no direction) 50°C Hot baked potato 80 W/ m2 A Magnitude and direction of heat flux at the same point FIGURE 2–1 Heat transfer has direction as well as magnitude, and thus it is a vector quantity. · Q = 500 W Hot medium 0 Cold medium x L · Q = –500 W Cold medium 0 Hot medium L x FIGURE 2–2 Indicating direction for heat transfer (positive in the positive direction; negative in the negative direction). 2–1 ■ INTRODUCTION In Chapter 1 heat conduction was defined as the transfer of thermal energy from the more energetic particles of a medium to the adjacent less energetic ones. It was stated that conduction can take place in liquids and gases as well as solids provided that there is no bulk motion involved. Although heat transfer and temperature are closely related, they are of a different nature. Unlike temperature, heat transfer has direction as well as magnitude, and thus it is a vector quantity (Fig. 2–1). Therefore, we must specify both direction and magnitude in order to describe heat transfer completely at a point. For example, saying that the temperature on the inner surface of a wall is 18°C describes the temperature at that location fully. But saying that the heat flux on that surface is 50 W/m2 immediately prompts the question “in what direction?” We can answer this question by saying that heat conduction is toward the inside (indicating heat gain) or toward the outside (indicating heat loss). To avoid such questions, we can work with a coordinate system and indicate direction with plus or minus signs. The generally accepted convention is that heat transfer in the positive direction of a coordinate axis is positive and in the opposite direction it is negative. Therefore, a positive quantity indicates heat transfer in the positive direction and a negative quantity indicates heat transfer in the negative direction (Fig. 2–2). The driving force for any form of heat transfer is the temperature difference, and the larger the temperature difference, the larger the rate of heat transfer. Some heat transfer problems in engineering require the determination of the temperature distribution (the variation of temperature) throughout the medium in order to calculate some quantities of interest such as the local heat transfer rate, thermal expansion, and thermal stress at some critical locations at specified times. The specification of the temperature at a point in a medium first requires the specification of the location of that point. This can be done by choosing a suitable coordinate system such as the rectangular, cylindrical, or spherical coordinates, depending on the geometry involved, and a convenient reference point (the origin). The location of a point is specified as (x, y, z) in rectangular coordinates, as (r, , z) in cylindrical coordinates, and as (r, , ) in spherical coordinates, where the distances x, y, z, and r and the angles and are as shown in Figure 2–3. Then the temperature at a point (x, y, z) at time t in rectangular coordinates is expressed as T(x, y, z, t). The best coordinate system for a given geometry is the one that describes the surfaces of the geometry best. For example, a parallelepiped is best described in rectangular coordinates since each surface can be described by a constant value of the x-, y-, or z-coordinates. A cylinder is best suited for cylindrical coordinates since its lateral surface can be described by a constant value of the radius. Similarly, the entire outer surface of a spherical body can best be described by a constant value of the radius in spherical coordinates. For an arbitrarily shaped body, we normally use rectangular coordinates since it is easier to deal with distances than with angles. The notation just described is also used to identify the variables involved in a heat transfer problem. For example, the notation T(x, y, z, t) implies that the temperature varies with the space variables x, y, and z as well as time. The cen58933_ch02.qxd 9/10/2002 8:46 AM Page 63 63 CHAPTER 2 z z z P(x, y, z) P(r, φ , z) z φ z x y φ r y P(r, φ , θ ) θ r y y x x (a) Rectangular coordinates x (b) Cylindrical coordinates (c) Spherical coordinates FIGURE 2–3 The various distances and angles involved when describing the location of a point in different coordinate systems. notation T(x), on the other hand, indicates that the temperature varies in the x-direction only and there is no variation with the other two space coordinates or time. Steady versus Transient Heat Transfer Heat transfer problems are often classified as being steady (also called steadystate) or transient (also called unsteady). The term steady implies no change with time at any point within the medium, while transient implies variation with time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer through a medium at any location, although both quantities may vary from one location to another (Fig. 2–4). For example, heat transfer through the walls of a house will be steady when the conditions inside the house and the outdoors remain constant for several hours. But even in this case, the temperatures on the inner and outer surfaces of the wall will be different unless the temperatures inside and outside the house are the same. The cooling of an apple in a refrigerator, on the other hand, is a transient heat transfer process since the temperature at any fixed point within the apple will change with time during cooling. During transient heat transfer, the temperature normally varies with time as well as position. In the special case of variation with time but not with position, the temperature of the medium changes uniformly with time. Such heat transfer systems are called lumped systems. A small metal object such as a thermocouple junction or a thin copper wire, for example, can be analyzed as a lumped system during a heating or cooling process. Most heat transfer problems encountered in practice are transient in nature, but they are usually analyzed under some presumed steady conditions since steady processes are easier to analyze, and they provide the answers to our questions. For example, heat transfer through the walls and ceiling of a typical house is never steady since the outdoor conditions such as the temperature, the speed and direction of the wind, the location of the sun, and so on, change constantly. The conditions in a typical house are not so steady either. Therefore, it is almost impossible to perform a heat transfer analysis of a house accurately. But then, do we really need an in-depth heat transfer analysis? If the Time = 2 PM 15°C Time = 5 PM 7°C 12°C 5°C · Q1 · · Q2 ≠ Q1 (a) Transient 15°C 7°C 15°C · Q1 7°C · · Q2 = Q1 (b) Steady-state FIGURE 2–4 Steady and transient heat conduction in a plane wall. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 64 64 HEAT TRANSFER purpose of a heat transfer analysis of a house is to determine the proper size of a heater, which is usually the case, we need to know the maximum rate of heat loss from the house, which is determined by considering the heat loss from the house under worst conditions for an extended period of time, that is, during steady operation under worst conditions. Therefore, we can get the answer to our question by doing a heat transfer analysis under steady conditions. If the heater is large enough to keep the house warm under the presumed worst conditions, it is large enough for all conditions. The approach described above is a common practice in engineering. Multidimensional Heat Transfer 80°C 65°C T(x, y) · Qy 70°C 65°C 80°C · Qx 70°C 65°C 80°C 70°C z y x FIGURE 2–5 Two-dimensional heat transfer in a long rectangular bar. Negligible heat transfer · Q Primary direction of heat transfer FIGURE 2–6 Heat transfer through the window of a house can be taken to be one-dimensional. Heat transfer problems are also classified as being one-dimensional, twodimensional, or three-dimensional, depending on the relative magnitudes of heat transfer rates in different directions and the level of accuracy desired. In the most general case, heat transfer through a medium is three-dimensional. That is, the temperature varies along all three primary directions within the medium during the heat transfer process. The temperature distribution throughout the medium at a specified time as well as the heat transfer rate at any location in this general case can be described by a set of three coordinates such as the x, y, and z in the rectangular (or Cartesian) coordinate system; the r, , and z in the cylindrical coordinate system; and the r, , and in the spherical (or polar) coordinate system. The temperature distribution in this case is expressed as T(x, y, z, t), T(r, , z, t), and T(r, , , t) in the respective coordinate systems. The temperature in a medium, in some cases, varies mainly in two primary directions, and the variation of temperature in the third direction (and thus heat transfer in that direction) is negligible. A heat transfer problem in that case is said to be two-dimensional. For example, the steady temperature distribution in a long bar of rectangular cross section can be expressed as T(x, y) if the temperature variation in the z-direction (along the bar) is negligible and there is no change with time (Fig. 2–5). A heat transfer problem is said to be one-dimensional if the temperature in the medium varies in one direction only and thus heat is transferred in one direction, and the variation of temperature and thus heat transfer in other directions are negligible or zero. For example, heat transfer through the glass of a window can be considered to be one-dimensional since heat transfer through the glass will occur predominantly in one direction (the direction normal to the surface of the glass) and heat transfer in other directions (from one side edge to the other and from the top edge to the bottom) is negligible (Fig. 2–6). Likewise, heat transfer through a hot water pipe can be considered to be onedimensional since heat transfer through the pipe occurs predominantly in the radial direction from the hot water to the ambient, and heat transfer along the pipe and along the circumference of a cross section (z- and -directions) is typically negligible. Heat transfer to an egg dropped into boiling water is also nearly one-dimensional because of symmetry. Heat will be transferred to the egg in this case in the radial direction, that is, along straight lines passing through the midpoint of the egg. We also mentioned in Chapter 1 that the rate of heat conduction through a medium in a specified direction (say, in the x-direction) is proportional to the temperature difference across the medium and the area normal to the direction cen58933_ch02.qxd 9/10/2002 8:46 AM Page 65 65 CHAPTER 2 of heat transfer, but is inversely proportional to the distance in that direction. This was expressed in the differential form by Fourier’s law of heat conduction for one-dimensional heat conduction as · dT Q cond kA dx (W) dT slope — < 0 dx (2-1) where k is the thermal conductivity of the material, which is a measure of the ability of a material to conduct heat, and dT/dx is the temperature gradient, which is the slope of the temperature curve on a T-x diagram (Fig. 2–7). The thermal conductivity of a material, in general, varies with temperature. But sufficiently accurate results can be obtained by using a constant value for thermal conductivity at the average temperature. Heat is conducted in the direction of decreasing temperature, and thus the temperature gradient is negative when heat is conducted in the positive x-direction. The negative sign in Eq. 2–1 ensures that heat transfer in the positive x-direction is a positive quantity. To obtain a general relation for Fourier’s law of heat conduction, consider a medium in which the temperature distribution is three-dimensional. Figure 2–8 shows an isothermal surface in that medium. The heat flux vector at a point P on this surface must be perpendicular to the surface, and it must point in the direction of decreasing temperature. If n is the normal of the isothermal surface at point P, the rate of heat conduction at that point can be expressed by Fourier’s law as · T Q n kA n T (W) T(x) · Q>0 Heat flow x FIGURE 2–7 The temperature gradient dT/dx is simply the slope of the temperature curve on a T-x diagram. (2-2) In rectangular coordinates, the heat conduction vector can be expressed in terms of its components as → · · → · → · → Qn Qx i Qy j Qz k (2-3) → → → · · · where i, j, and k are the unit vectors, and Q x, Q y, and Q z are the magnitudes of the heat transfer rates in the x-, y-, and z-directions, which again can be determined from Fourier’s law as · T Q x kAx , x · T Q y kAy , y and · T Q z kAz z (2-4) Here Ax, Ay and Az are heat conduction areas normal to the x-, y-, and z-directions, respectively (Fig. 2–8). Most engineering materials are isotropic in nature, and thus they have the same properties in all directions. For such materials we do not need to be concerned about the variation of properties with direction. But in anisotropic materials such as the fibrous or composite materials, the properties may change with direction. For example, some of the properties of wood along the grain are different than those in the direction normal to the grain. In such cases the thermal conductivity may need to be expressed as a tensor quantity to account for the variation with direction. The treatment of such advanced topics is beyond the scope of this text, and we will assume the thermal conductivity of a material to be independent of direction. z Az n Ay · Qn · Qz Ax · Qy P y · Qx An isotherm x FIGURE 2–8 The heat transfer vector is always normal to an isothermal surface and can be resolved into its components like any other vector. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 66 66 HEAT TRANSFER Heat Generation FIGURE 2–9 Heat is generated in the heating coils of an electric range as a result of the conversion of electrical energy to heat. Sun Solar radiation q· s x Solar energy absorbed by water Water · = q· g(x) s, absorbed(x) A medium through which heat is conducted may involve the conversion of electrical, nuclear, or chemical energy into heat (or thermal) energy. In heat conduction analysis, such conversion processes are characterized as heat generation. For example, the temperature of a resistance wire rises rapidly when electric current passes through it as a result of the electrical energy being converted to heat at a rate of I2R, where I is the current and R is the electrical resistance of the wire (Fig. 2–9). The safe and effective removal of this heat away from the sites of heat generation (the electronic circuits) is the subject of electronics cooling, which is one of the modern application areas of heat transfer. Likewise, a large amount of heat is generated in the fuel elements of nuclear reactors as a result of nuclear fission that serves as the heat source for the nuclear power plants. The natural disintegration of radioactive elements in nuclear waste or other radioactive material also results in the generation of heat throughout the body. The heat generated in the sun as a result of the fusion of hydrogen into helium makes the sun a large nuclear reactor that supplies heat to the earth. Another source of heat generation in a medium is exothermic chemical reactions that may occur throughout the medium. The chemical reaction in this case serves as a heat source for the medium. In the case of endothermic reactions, however, heat is absorbed instead of being released during reaction, and thus the chemical reaction serves as a heat sink. The heat generation term becomes a negative quantity in this case. Often it is also convenient to model the absorption of radiation such as solar energy or gamma rays as heat generation when these rays penetrate deep into the body while being absorbed gradually. For example, the absorption of solar energy in large bodies of water can be treated as heat generation throughout the water at a rate equal to the rate of absorption, which varies with depth (Fig. 2–10). But the absorption of solar energy by an opaque body occurs within a few microns of the surface, and the solar energy that penetrates into the medium in this case can be treated as specified heat flux on the surface. Note that heat generation is a volumetric phenomenon. That is, it occurs throughout the body of a medium. Therefore, the rate of heat generation in a medium is usually specified per unit volume and is denoted by g·, whose unit is W/m3 or Btu/h · ft3. The rate of heat generation in a medium may vary with time as well as position within the medium. When the variation of heat generation with position is known, the total rate of heat generation in a medium of volume V can be determined from · G FIGURE 2–10 The absorption of solar radiation by water can be treated as heat generation. g·dV (W) (2-5) V In the special case of uniform heat generation, as in the case of electric resistance heating throughout a homogeneous material, the relation in Eq. 2–5 · reduces to G g·V, where g· is the constant rate of heat generation per unit volume. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 67 67 CHAPTER 2 Heat transfer EXAMPLE 2–1 Heat Gain by a Refrigerator In order to size the compressor of a new refrigerator, it is desired to determine the rate of heat transfer from the kitchen air into the refrigerated space through the walls, door, and the top and bottom section of the refrigerator (Fig. 2–11). In your analysis, would you treat this as a transient or steady-state heat transfer problem? Also, would you consider the heat transfer to be one-dimensional or multidimensional? Explain. SOLUTION The heat transfer process from the kitchen air to the refrigerated space is transient in nature since the thermal conditions in the kitchen and the refrigerator, in general, change with time. However, we would analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the lowest thermostat setting for the refrigerated space, and the anticipated highest temperature in the kitchen (the so-called design conditions). If the compressor is large enough to keep the refrigerated space at the desired temperature setting under the presumed worst conditions, then it is large enough to do so under all conditions by cycling on and off. Heat transfer into the refrigerated space is three-dimensional in nature since heat will be entering through all six sides of the refrigerator. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer to be onedimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated values of heat transfer at each surface. EXAMPLE 2–2 FIGURE 2–11 Schematic for Example 2–1. Heat Generation in a Hair Dryer The resistance wire of a 1200-W hair dryer is 80 cm long and has a diameter of D 0.3 cm (Fig. 2–12). Determine the rate of heat generation in the wire per unit volume, in W/cm3, and the heat flux on the outer surface of the wire as a result of this heat generation. SOLUTION The power consumed by the resistance wire of a hair dryer is given. The heat generation and the heat flux are to be determined. Assumptions Heat is generated uniformly in the resistance wire. Analysis A 1200-W hair dryer will convert electrical energy into heat in the wire at a rate of 1200 W. Therefore, the rate of heat generation in a resistance wire is equal to the power consumption of a resistance heater. Then the rate of heat generation in the wire per unit volume is determined by dividing the total rate of heat generation by the volume of the wire, G· G· 1200 W g· 212 W/cm3 Vwire (D2/4)L (0.3 cm)2/4 Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate of heat generation by the surface area of the wire, G· G· 1200 W q· 15.9 W/cm2 Awire DL (0.3 cm)(80 cm) Hair dryer 1200 W FIGURE 2–12 Schematic for Example 2–2. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 68 68 HEAT TRANSFER Discussion Note that heat generation is expressed per unit volume in W/cm3 or Btu/h · ft3, whereas heat flux is expressed per unit surface area in W/cm2 or Btu/h · ft2. 2–2 · G Volume element A · Qx L Consider heat conduction through a large plane wall such as the wall of a house, the glass of a single pane window, the metal plate at the bottom of a pressing iron, a cast iron steam pipe, a cylindrical nuclear fuel element, an electrical resistance wire, the wall of a spherical container, or a spherical metal ball that is being quenched or tempered. Heat conduction in these and many other geometries can be approximated as being one-dimensional since heat conduction through these geometries will be dominant in one direction and negligible in other directions. Below we will develop the onedimensional heat conduction equation in rectangular, cylindrical, and spherical coordinates. Consider a thin element of thickness x in a large plane wall, as shown in Figure 2–13. Assume the density of the wall is , the specific heat is C, and the area of the wall normal to the direction of heat transfer is A. An energy balance on this thin element during a small time interval t can be expressed as 0 x + ∆x ONE-DIMENSIONAL HEAT CONDUCTION EQUATION Heat Conduction Equation in a Large Plane Wall · Qx + ∆ x x ■ x Ax = Ax + ∆ x = A FIGURE 2–13 One-dimensional heat conduction through a volume element in a large plane wall. Rate of heat Rate of heat conduction conduction at x at x x Rate of heat generation inside the element Rate of change of the energy content of the element or Eelement · · · Q x Q x x Gelement t (2-6) But the change in the energy content of the element and the rate of heat generation within the element can be expressed as Eelement Et t Et mC(Tt t Tt) CA x(Tt t Tt) · Gelement g·Velement g·A x (2-7) (2-8) Substituting into Equation 2–6, we get Tt t Tt · · Q x Q x x g·A x CA x t (2-9) Dividing by A x gives · · Tt t Tt 1 Q x x Q x g· C A x t (2-10) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 69 69 CHAPTER 2 Taking the limit as x → 0 and t → 0 yields T T 1 kA g· C x t A x (2-11) since, from the definition of the derivative and Fourier’s law of heat conduction, lim x → 0 Q· x x Q· x Q· T kA x x x x (2-12) Noting that the area A is constant for a plane wall, the one-dimensional transient heat conduction equation in a plane wall becomes (2-13) The thermal conductivity k of a material, in general, depends on the temperature T (and therefore x), and thus it cannot be taken out of the derivative. However, the thermal conductivity in most practical applications can be assumed to remain constant at some average value. The equation above in that case reduces to Constant conductivity: 2T g· 1 T x2 k t (2-14) where the property k/ C is the thermal diffusivity of the material and represents how fast heat propagates through a material. It reduces to the following forms under specified conditions (Fig. 2–14): (1) Steady-state: (/t 0) d 2T g· 0 k dx2 (2-15) (2) Transient, no heat generation: (g· 0) 2T 1 T x2 t (2-16) (3) Steady-state, no heat generation: (/t 0 and g· 0) d 2T 0 dx2 (2-17) General, one dimensional: No Steadygeneration state 0 0 2T g· 1 T t x2 k  → T T k g· C x x t  → Variable conductivity: Steady, one-dimensional: d2T 0 dx2 FIGURE 2–14 The simplification of the onedimensional heat conduction equation in a plane wall for the case of constant conductivity for steady conduction with no heat generation. L Note that we replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case since the partial and ordinary derivatives of a function are identical when the function depends on a single variable only [T T(x) in this case]. · G Heat Conduction Equation in a Long Cylinder Now consider a thin cylindrical shell element of thickness r in a long cylinder, as shown in Figure 2–15. Assume the density of the cylinder is , the specific heat is C, and the length is L. The area of the cylinder normal to the direction of heat transfer at any location is A 2rL where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. An energy balance on this thin cylindrical shell element during a small time interval t can be expressed as 0 · Qr r · Qr + ∆r r + ∆r r Volume element FIGURE 2–15 One-dimensional heat conduction through a volume element in a long cylinder. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 70 70 HEAT TRANSFER Rate of heat Rate of heat conduction conduction at r at r r Rate of heat generation inside the element Rate of change of the energy content of the element or Eelement · · · Q r Q r r Gelement t (2-18) The change in the energy content of the element and the rate of heat generation within the element can be expressed as Eelement Et t Et mC(Tt t Tt) CA r(Tt t Tt) · Gelement g·Velement g·A r (2-19) (2-20) Substituting into Eq. 2–18, we get Tt t Tt · · Q r Q r r g·A r CA r t (2-21) where A 2rL. You may be tempted to express the area at the middle of the element using the average radius as A 2(r r/2)L. But there is nothing we can gain from this complication since later in the analysis we will take the limit as r → 0 and thus the term r/2 will drop out. Now dividing the equation above by A r gives · · Tt t Tt 1 Q r r Q r g· C A r t (2-22) Taking the limit as r → 0 and t → 0 yields T T 1 g· C kA r t A r (2-23) since, from the definition of the derivative and Fourier’s law of heat conduction, lim r → 0 Q· r r Q· r Q· T kA r r r r (2-24) Noting that the heat transfer area in this case is A 2rL, the onedimensional transient heat conduction equation in a cylinder becomes Variable conductivity: T T 1 · r r rk r g C t (2-25) For the case of constant thermal conductivity, the equation above reduces to Constant conductivity: g· 1 T T 1 r r r r k t (2-26) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 71 71 CHAPTER 2 where again the property k/ C is the thermal diffusivity of the material. Equation 2–26 reduces to the following forms under specified conditions (Fig. 2–16): g· dT 1 d r dr r dr k 0 T 1 1 T r r r r t dT d r 0 dr dr (1) Steady-state: (/t 0) (2) Transient, no heat generation: (g· 0) (3) Steady-state, no heat generation: (/t 0 and g· 0) (2-27) (2-28) (2-29) Note that we again replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case since the partial and ordinary derivatives of a function are identical when the function depends on a single variable only [T T(r) in this case]. Now consider a sphere with density , specific heat C, and outer radius R. The area of the sphere normal to the direction of heat transfer at any location is A 4r2, where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case also, and thus it varies with location. By considering a thin spherical shell element of thickness r and repeating the approach described above for the cylinder by using A 4r2 instead of A 2rL, the one-dimensional transient heat conduction equation for a sphere is determined to be (Fig. 2–17) T 1 2 T r k g· C r t r 2 r (2-30) which, in the case of constant thermal conductivity, reduces to g· 1 T 1 2 T r 2 r r t k r Constant conductivity: (2-31) where again the property k/ C is the thermal diffusivity of the material. It reduces to the following forms under specified conditions: (1) Steady-state: (/t 0) g· 1 d 2 dT r 0 dr k r 2 dr (2) Transient, no heat generation: (g· 0) (3) Steady-state, no heat generation: (/t 0 and g· 0) 1 2 T 1 T r r t r 2 r (2-32) (2-33) d 2 dT r 0 dr dr or r dT d 2T 2 0 dr dr 2 dT d r 0 dr dr (b) The equivalent alternative form r d 2T dT 0 dr 2 dr FIGURE 2–16 Two equivalent forms of the differential equation for the onedimensional steady heat conduction in a cylinder with no heat generation. · G Heat Conduction Equation in a Sphere Variable conductivity: (a) The form that is ready to integrate (2-34) where again we replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case. · Qr + ∆r · Qr r + ∆r 0 r R r Volume element FIGURE 2–17 One-dimensional heat conduction through a volume element in a sphere. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 72 72 HEAT TRANSFER Combined One-Dimensional Heat Conduction Equation An examination of the one-dimensional transient heat conduction equations for the plane wall, cylinder, and sphere reveals that all three equations can be expressed in a compact form as T 1 n T g· C r k r n r r t (2-35) where n 0 for a plane wall, n 1 for a cylinder, and n 2 for a sphere. In the case of a plane wall, it is customary to replace the variable r by x. This equation can be simplified for steady-state or no heat generation cases as described before. EXAMPLE 2–3 Heat Conduction through the Bottom of a Pan Consider a steel pan placed on top of an electric range to cook spaghetti (Fig. 2–18). The bottom section of the pan is L 0.4 cm thick and has a diameter of D 18 cm. The electric heating unit on the range top consumes 800 W of power during cooking, and 80 percent of the heat generated in the heating element is transferred uniformly to the pan. Assuming constant thermal conductivity, obtain the differential equation that describes the variation of the temperature in the bottom section of the pan during steady operation. 800 W FIGURE 2–18 Schematic for Example 2–3. SOLUTION The bottom section of the pan has a large surface area relative to its thickness and can be approximated as a large plane wall. Heat flux is applied to the bottom surface of the pan uniformly, and the conditions on the inner surface are also uniform. Therefore, we expect the heat transfer through the bottom section of the pan to be from the bottom surface toward the top, and heat transfer in this case can reasonably be approximated as being onedimensional. Taking the direction normal to the bottom surface of the pan to be the x-axis, we will have T T (x) during steady operation since the temperature in this case will depend on x only. The thermal conductivity is given to be constant, and there is no heat generation in the medium (within the bottom section of the pan). Therefore, the differential equation governing the variation of temperature in the bottom section of the pan in this case is simply Eq. 2–17, d 2T 0 dx2 which is the steady one-dimensional heat conduction equation in rectangular coordinates under the conditions of constant thermal conductivity and no heat generation. Note that the conditions at the surface of the medium have no effect on the differential equation. EXAMPLE 2–4 Heat Conduction in a Resistance Heater A 2-kW resistance heater wire with thermal conductivity k 15 W/m · °C, diameter D 0.4 cm, and length L 50 cm is used to boil water by immersing cen58933_ch02.qxd 9/10/2002 8:46 AM Page 73 73 CHAPTER 2 it in water (Fig. 2–19). Assuming the variation of the thermal conductivity of the wire with temperature to be negligible, obtain the differential equation that describes the variation of the temperature in the wire during steady operation. SOLUTION The resistance wire can be considered to be a very long cylinder since its length is more than 100 times its diameter. Also, heat is generated uniformly in the wire and the conditions on the outer surface of the wire are uniform. Therefore, it is reasonable to expect the temperature in the wire to vary in the radial r direction only and thus the heat transfer to be one-dimensional. Then we will have T T (r ) during steady operation since the temperature in this case will depend on r only. The rate of heat generation in the wire per unit volume can be determined from Water Resistance heater FIGURE 2–19 Schematic for Example 2–4. G· G· 2000 W g· 0.318 109 W/m3 Vwire (D2/4)L (0.004 m)2/4 Noting that the thermal conductivity is given to be constant, the differential equation that governs the variation of temperature in the wire is simply Eq. 2–27, g· dT 1 d r dr r dr k 0 which is the steady one-dimensional heat conduction equation in cylindrical coordinates for the case of constant thermal conductivity. Note again that the conditions at the surface of the wire have no effect on the differential equation. EXAMPLE 2–5 Cooling of a Hot Metal Ball in Air A spherical metal ball of radius R is heated in an oven to a temperature of 600°F throughout and is then taken out of the oven and allowed to cool in ambient air at T 75°F by convection and radiation (Fig. 2–20). The thermal conductivity of the ball material is known to vary linearly with temperature. Assuming the ball is cooled uniformly from the entire outer surface, obtain the differential equation that describes the variation of the temperature in the ball during cooling. SOLUTION The ball is initially at a uniform temperature and is cooled uniformly from the entire outer surface. Also, the temperature at any point in the ball will change with time during cooling. Therefore, this is a one-dimensional transient heat conduction problem since the temperature within the ball will change with the radial distance r and the time t. That is, T T (r, t ). The thermal conductivity is given to be variable, and there is no heat generation in the ball. Therefore, the differential equation that governs the variation of temperature in the ball in this case is obtained from Eq. 2–30 by setting the heat generation term equal to zero. We obtain T 1 2 T C r k r t r 2 r 75°F Metal ball 600°F · Q FIGURE 2–20 Schematic for Example 2–5. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 74 74 HEAT TRANSFER which is the one-dimensional transient heat conduction equation in spherical coordinates under the conditions of variable thermal conductivity and no heat generation. Note again that the conditions at the outer surface of the ball have no effect on the differential equation. 2–3 ■ GENERAL HEAT CONDUCTION EQUATION In the last section we considered one-dimensional heat conduction and assumed heat conduction in other directions to be negligible. Most heat transfer problems encountered in practice can be approximated as being onedimensional, and we will mostly deal with such problems in this text. However, this is not always the case, and sometimes we need to consider heat transfer in other directions as well. In such cases heat conduction is said to be multidimensional, and in this section we will develop the governing differential equation in such systems in rectangular, cylindrical, and spherical coordinate systems. · Qz + ∆ z Rectangular Coordinates · Qy + ∆y Volume element · Qx ∆z · x∆y∆ z g∆ · Qx + ∆ x · Qy z ∆y ∆x y x · Qz Consider a small rectangular element of length x, width y, and height z, as shown in Figure 2–21. Assume the density of the body is and the specific heat is C. An energy balance on this element during a small time interval t can be expressed as Rate of heat conduction at x, y, and z Rate of heat conduction at x x, y y, and z z Rate of heat generation inside the element Rate of change of the energy content of the element or FIGURE 2–21 Three-dimensional heat conduction through a rectangular volume element. Eelement · · · · · · · Q x Q y Q z Q x x Q y y Q z z Gelement t (2-36) Noting that the volume of the element is Velement x y z, the change in the energy content of the element and the rate of heat generation within the element can be expressed as Eelement Et t Et mC(Tt t Tt) C x y z(Tt t Tt) · Gelement g·Velement g· x y z Substituting into Eq. 2–36, we get Tt t Tt · · · · · · Q x Q y Q z Q x x Q y y Q z z g· x y z C x y z t Dividing by x y z gives · · · · · · Tt t Tt 1 Q x x Q x 1 Q y y Q y 1 Q z z Q z · g C y z x x z y x y z t (2-37) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 75 75 CHAPTER 2 Noting that the heat transfer areas of the element for heat conduction in the x, y, and z directions are Ax y z, Ay x z, and Az x y, respectively, and taking the limit as x, y, z and t → 0 yields T T T T k k k g· C x x y y z z t (2-38) since, from the definition of the derivative and Fourier’s law of heat conduction, · · T T 1 Q x x Q x 1 Qx 1 k y z k x → 0 y z x x x x y z x y z x Q· y y Q· y T T 1 Qy 1 lim 1 k x z k y → 0 x z y y y y x z y x z y · · T T 1 Q z z Q z 1 Qz 1 lim k x y k z → 0 x y z z z z x y z x y z lim Equation 2–38 is the general heat conduction equation in rectangular coordinates. In the case of constant thermal conductivity, it reduces to 2T 2T 2T g· 1 T x2 y2 z2 k t 2T 2T 2T g· 0 x2 y2 z2 k (2-39) 2T 2T 2T 1 T x2 y2 z2 t where the property k/ C is again the thermal diffusivity of the material. Equation 2–39 is known as the Fourier-Biot equation, and it reduces to these forms under specified conditions: (1) Steady-state: (called the Poisson equation) (2) Transient, no heat generation: (called the diffusion equation) (3) Steady-state, no heat generation: (called the Laplace equation) 2T 2T 2T g· 0 x2 y2 z2 k 2T 2T 2T 1 T x2 y2 z2 t 2T 2T 2T 0 x2 y2 z2 2T 2T 2T 0 x2 y2 z2 (2-40) FIGURE 2–22 The three-dimensional heat conduction equations reduce to the one-dimensional ones when the temperature varies in one dimension only. (2-41) (2-42) Note that in the special case of one-dimensional heat transfer in the x-direction, the derivatives with respect to y and z drop out and the equations above reduce to the ones developed in the previous section for a plane wall (Fig. 2–22). z dz r dr z Cylindrical Coordinates The general heat conduction equation in cylindrical coordinates can be obtained from an energy balance on a volume element in cylindrical coordinates, shown in Figure 2–23, by following the steps just outlined. It can also be obtained directly from Eq. 2–38 by coordinate transformation using the following relations between the coordinates of a point in rectangular and cylindrical coordinate systems: x r cos , y r sin , and zz y x φ dφ FIGURE 2–23 A differential volume element in cylindrical coordinates. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 76 76 HEAT TRANSFER After lengthy manipulations, we obtain z T T T T 1 1 · r r kr r r 2 kr z k z g C t dr θ (2-43) Spherical Coordinates r dθ dφ φ y x FIGURE 2–24 A differential volume element in spherical coordinates. The general heat conduction equations in spherical coordinates can be obtained from an energy balance on a volume element in spherical coordinates, shown in Figure 2–24, by following the steps outlined above. It can also be obtained directly from Eq. 2–38 by coordinate transformation using the following relations between the coordinates of a point in rectangular and spherical coordinate systems: x r cos sin , y r sin sin , and z cos Again after lengthy manipulations, we obtain T T T T 1 1 1 kr 2 k k sin 2 2 2 g· C r t r sin r sin r 2 r (2-44) Obtaining analytical solutions to these differential equations requires a knowledge of the solution techniques of partial differential equations, which is beyond the scope of this introductory text. Here we limit our consideration to one-dimensional steady-state cases or lumped systems, since they result in ordinary differential equations. Heat loss 600°F Metal billet EXAMPLE 2–6 T = 65°F z Heat Conduction in a Short Cylinder A short cylindrical metal billet of radius R and height h is heated in an oven to a temperature of 600°F throughout and is then taken out of the oven and allowed to cool in ambient air at T 65°F by convection and radiation. Assuming the billet is cooled uniformly from all outer surfaces and the variation of the thermal conductivity of the material with temperature is negligible, obtain the differential equation that describes the variation of the temperature in the billet during this cooling process. r R φ FIGURE 2–25 Schematic for Example 2–6. SOLUTION The billet shown in Figure 2–25 is initially at a uniform temperature and is cooled uniformly from the top and bottom surfaces in the z-direction as well as the lateral surface in the radial r-direction. Also, the temperature at any point in the ball will change with time during cooling. Therefore, this is a two-dimensional transient heat conduction problem since the temperature within the billet will change with the radial and axial distances r and z and with time t. That is, T T (r, z, t ). The thermal conductivity is given to be constant, and there is no heat generation in the billet. Therefore, the differential equation that governs the variation of temperature in the billet in this case is obtained from Eq. 2–43 by setting the heat generation term and the derivatives with respect to equal to zero. We obtain T T T 1 r r kr r z k z C t cen58933_ch02.qxd 9/10/2002 8:46 AM Page 77 77 CHAPTER 2 In the case of constant thermal conductivity, it reduces to 2T T 1 1 T r r r r z2 t which is the desired equation. ■ BOUNDARY AND INITIAL CONDITIONS The heat conduction equations above were developed using an energy balance on a differential element inside the medium, and they remain the same regardless of the thermal conditions on the surfaces of the medium. That is, the differential equations do not incorporate any information related to the conditions on the surfaces such as the surface temperature or a specified heat flux. Yet we know that the heat flux and the temperature distribution in a medium depend on the conditions at the surfaces, and the description of a heat transfer problem in a medium is not complete without a full description of the thermal conditions at the bounding surfaces of the medium. The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions. From a mathematical point of view, solving a differential equation is essentially a process of removing derivatives, or an integration process, and thus the solution of a differential equation typically involves arbitrary constants (Fig. 2–26). It follows that to obtain a unique solution to a problem, we need to specify more than just the governing differential equation. We need to specify some conditions (such as the value of the function or its derivatives at some value of the independent variable) so that forcing the solution to satisfy these conditions at specified points will result in unique values for the arbitrary constants and thus a unique solution. But since the differential equation has no place for the additional information or conditions, we need to supply them separately in the form of boundary or initial conditions. Consider the variation of temperature along the wall of a brick house in winter. The temperature at any point in the wall depends on, among other things, the conditions at the two surfaces of the wall such as the air temperature of the house, the velocity and direction of the winds, and the solar energy incident on the outer surface. That is, the temperature distribution in a medium depends on the conditions at the boundaries of the medium as well as the heat transfer mechanism inside the medium. To describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant (Fig. 2–27). Therefore, we need to specify two boundary conditions for one-dimensional problems, four boundary conditions for two-dimensional problems, and six boundary conditions for three-dimensional problems. In the case of the wall of a house, for example, we need to specify the conditions at two locations (the inner and the outer surfaces) of the wall since heat transfer in this case is one-dimensional. But in the case of a parallelepiped, we need to specify six boundary conditions (one at each face) when heat transfer in all three dimensions is significant. The differential equation: d 2T 0 dx2 General solution: T(x) C1x C2 →  → 2–4 Arbitrary constants Some specific solutions: T(x) 2x 5 T(x) x 12 T(x) 3 T(x) 6.2x M FIGURE 2–26 The general solution of a typical differential equation involves arbitrary constants, and thus an infinite number of solutions. T Some solutions of 2T d–— =0 dx 2 50°C 0 15°C The only solution L x that satisfies the conditions T(0) = 50°C and T(L) = 15°C. FIGURE 2–27 To describe a heat transfer problem completely, two boundary conditions must be given for each direction along which heat transfer is significant. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 78 78 HEAT TRANSFER The physical argument presented above is consistent with the mathematical nature of the problem since the heat conduction equation is second order (i.e., involves second derivatives with respect to the space variables) in all directions along which heat conduction is significant, and the general solution of a second-order linear differential equation involves two arbitrary constants for each direction. That is, the number of boundary conditions that needs to be specified in a direction is equal to the order of the differential equation in that direction. Reconsider the brick wall already discussed. The temperature at any point on the wall at a specified time also depends on the condition of the wall at the beginning of the heat conduction process. Such a condition, which is usually specified at time t 0, is called the initial condition, which is a mathematical expression for the temperature distribution of the medium initially. Note that we need only one initial condition for a heat conduction problem regardless of the dimension since the conduction equation is first order in time (it involves the first derivative of temperature with respect to time). In rectangular coordinates, the initial condition can be specified in the general form as T(x, y, z, 0) f(x, y, z) (2-45) where the function f(x, y, z) represents the temperature distribution throughout the medium at time t 0. When the medium is initially at a uniform temperature of Ti, the initial condition of Eq. 2–45 can be expressed as T(x, y, z, 0) Ti. Note that under steady conditions, the heat conduction equation does not involve any time derivatives, and thus we do not need to specify an initial condition. The heat conduction equation is first order in time, and thus the initial condition cannot involve any derivatives (it is limited to a specified temperature). However, the heat conduction equation is second order in space coordinates, and thus a boundary condition may involve first derivatives at the boundaries as well as specified values of temperature. Boundary conditions most commonly encountered in practice are the specified temperature, specified heat flux, convection, and radiation boundary conditions. 150°C T(x, t) 70°C 1 Specified Temperature Boundary Condition 0 L x T(0, t) = 150°C T(L, t) = 70°C FIGURE 2–28 Specified temperature boundary conditions on both surfaces of a plane wall. The temperature of an exposed surface can usually be measured directly and easily. Therefore, one of the easiest ways to specify the thermal conditions on a surface is to specify the temperature. For one-dimensional heat transfer through a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed as (Fig. 2–28) T(0, t) T1 T(L, t) T2 (2-46) where T1 and T2 are the specified temperatures at surfaces at x 0 and x L, respectively. The specified temperatures can be constant, which is the case for steady heat conduction, or may vary with time. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 79 79 CHAPTER 2 2 Specified Heat Flux Boundary Condition When there is sufficient information about energy interactions at a surface, it may be possible to determine the rate of heat transfer and thus the heat flux q· (heat transfer rate per unit surface area, W/m2) on that surface, and this information can be used as one of the boundary conditions. The heat flux in the positive x-direction anywhere in the medium, including the boundaries, can be expressed by Fourier’s law of heat conduction as T Heat flux in the q· k x positive x-direction (W/m2) (2-47) Then the boundary condition at a boundary is obtained by setting the specified heat flux equal to k(T/x) at that boundary. The sign of the specified heat flux is determined by inspection: positive if the heat flux is in the positive direction of the coordinate axis, and negative if it is in the opposite direction. Note that it is extremely important to have the correct sign for the specified heat flux since the wrong sign will invert the direction of heat transfer and cause the heat gain to be interpreted as heat loss (Fig. 2–29). For a plate of thickness L subjected to heat flux of 50 W/m2 into the medium from both sides, for example, the specified heat flux boundary conditions can be expressed as k T(0, t) 50 x and k T(L, t) 50 x Heat flux Conduction ∂T(0, t) q0 = – k ——— ∂x Heat flux Conduction ∂T(L, t) – k ——— = qL ∂x 0 x L FIGURE 2–29 Specified heat flux boundary conditions on both surfaces of a plane wall. (2-48) Note that the heat flux at the surface at x L is in the negative x-direction, and thus it is 50 W/m2. Special Case: Insulated Boundary Some surfaces are commonly insulated in practice in order to minimize heat loss (or heat gain) through them. Insulation reduces heat transfer but does not totally eliminate it unless its thickness is infinity. However, heat transfer through a properly insulated surface can be taken to be zero since adequate insulation reduces heat transfer through a surface to negligible levels. Therefore, a well-insulated surface can be modeled as a surface with a specified heat flux of zero. Then the boundary condition on a perfectly insulated surface (at x 0, for example) can be expressed as (Fig. 2–30) k T(0, t) 0 x or T(0, t) 0 x (2-49) That is, on an insulated surface, the first derivative of temperature with respect to the space variable (the temperature gradient) in the direction normal to the insulated surface is zero. This also means that the temperature function must be perpendicular to an insulated surface since the slope of temperature at the surface must be zero. Another Special Case: Thermal Symmetry Some heat transfer problems possess thermal symmetry as a result of the symmetry in imposed thermal conditions. For example, the two surfaces of a large hot plate of thickness L suspended vertically in air will be subjected to Insulation T(x, t) 0 60°C L x ∂T(0, t) ——— =0 ∂x T(L, t) = 60°C FIGURE 2–30 A plane wall with insulation and specified temperature boundary conditions. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 80 80 HEAT TRANSFER Center plane Zero slope Temperature distribution (symmetric about center plane) 0 L — 2 ∂T(L /2, t) ———— =0 ∂x L the same thermal conditions, and thus the temperature distribution in one half of the plate will be the same as that in the other half. That is, the heat transfer problem in this plate will possess thermal symmetry about the center plane at x L/2. Also, the direction of heat flow at any point in the plate will be toward the surface closer to the point, and there will be no heat flow across the center plane. Therefore, the center plane can be viewed as an insulated surface, and the thermal condition at this plane of symmetry can be expressed as (Fig. 2–31) T(L/2, t) 0 x x FIGURE 2–31 Thermal symmetry boundary condition at the center plane of a plane wall. which resembles the insulation or zero heat flux boundary condition. This result can also be deduced from a plot of temperature distribution with a maximum, and thus zero slope, at the center plane. In the case of cylindrical (or spherical) bodies having thermal symmetry about the center line (or midpoint), the thermal symmetry boundary condition requires that the first derivative of temperature with respect to r (the radial variable) be zero at the centerline (or the midpoint). EXAMPLE 2–7 Water x 110°C L 0 q· 0 FIGURE 2–32 Schematic for Example 2–7. (2-50) Heat Flux Boundary Condition Consider an aluminum pan used to cook beef stew on top of an electric range. The bottom section of the pan is L 0.3 cm thick and has a diameter of D 20 cm. The electric heating unit on the range top consumes 800 W of power during cooking, and 90 percent of the heat generated in the heating element is transferred to the pan. During steady operation, the temperature of the inner surface of the pan is measured to be 110°C. Express the boundary conditions for the bottom section of the pan during this cooking process. SOLUTION The heat transfer through the bottom section of the pan is from the bottom surface toward the top and can reasonably be approximated as being one-dimensional. We take the direction normal to the bottom surfaces of the pan as the x axis with the origin at the outer surface, as shown in Figure 2–32. Then the inner and outer surfaces of the bottom section of the pan can be represented by x 0 and x L, respectively. During steady operation, the temperature will depend on x only and thus T T (x). The boundary condition on the outer surface of the bottom of the pan at x 0 can be approximated as being specified heat flux since it is stated that 90 percent of the 800 W (i.e., 720 W) is transferred to the pan at that surface. Therefore, k dT(0) q·0 dx where 0.720 kW Heat transfer rate q·0 22.9 kW/m2 Bottom surface area (0.1 m)2 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 81 81 CHAPTER 2 The temperature at the inner surface of the bottom of the pan is specified to be 110°C. Then the boundary condition on this surface can be expressed as T(L) 110°C where L 0.003 m. Note that the determination of the boundary conditions may require some reasoning and approximations. 3 Convection Boundary Condition Convection is probably the most common boundary condition encountered in practice since most heat transfer surfaces are exposed to an environment at a specified temperature. The convection boundary condition is based on a surface energy balance expressed as Heat conduction Heat convection at the surface in a at the surface in selected direction the same direction For one-dimensional heat transfer in the x-direction in a plate of thickness L, the convection boundary conditions on both surfaces can be expressed as T(0, t) k h1[T 1 T(0, t)] x (2-51a) Convection Conduction ∂T(0, t) h1[T 1 – T(0, t)] = – k ——— ∂x Conduction Convection h1 T 1 and T(L, t) k h2[T(L, t) T 2] x (2-51b) where h1 and h2 are the convection heat transfer coefficients and T 1 and T 2 are the temperatures of the surrounding mediums on the two sides of the plate, as shown in Figure 2–33. In writing Eqs. 2–51 for convection boundary conditions, we have selected the direction of heat transfer to be the positive x-direction at both surfaces. But those expressions are equally applicable when heat transfer is in the opposite direction at one or both surfaces since reversing the direction of heat transfer at a surface simply reverses the signs of both conduction and convection terms at that surface. This is equivalent to multiplying an equation by 1, which has no effect on the equality (Fig. 2–34). Being able to select either direction as the direction of heat transfer is certainly a relief since often we do not know the surface temperature and thus the direction of heat transfer at a surface in advance. This argument is also valid for other boundary conditions such as the radiation and combined boundary conditions discussed shortly. Note that a surface has zero thickness and thus no mass, and it cannot store any energy. Therefore, the entire net heat entering the surface from one side must leave the surface from the other side. The convection boundary condition simply states that heat continues to flow from a body to the surrounding medium at the same rate, and it just changes vehicles at the surface from conduction to convection (or vice versa in the other direction). This is analogous to people traveling on buses on land and transferring to the ships at the shore. h2 T 2 ∂T(L, t) – k ——— = h2[T(L, t) – T 2] ∂x 0 L x FIGURE 2–33 Convection boundary conditions on the two surfaces of a plane wall. Convection Conduction ∂T(0, t) h1[T 1 – T(0, t)] = – k ——— ∂x h1, T 1 Convection Conduction ∂T(0, t) h1[T(0, t) – T 1] = k ——— ∂x 0 L x FIGURE 2–34 The assumed direction of heat transfer at a boundary has no effect on the boundary condition expression. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 82 82 HEAT TRANSFER If the passengers are not allowed to wander around at the shore, then the rate at which the people are unloaded at the shore from the buses must equal the rate at which they board the ships. We may call this the conservation of “people” principle. Also note that the surface temperatures T(0, t) and T(L, t) are not known (if they were known, we would simply use them as the specified temperature boundary condition and not bother with convection). But a surface temperature can be determined once the solution T(x, t) is obtained by substituting the value of x at that surface into the solution. EXAMPLE 2–8 Steam flows through a pipe shown in Figure 2–35 at an average temperature of T 200°C. The inner and outer radii of the pipe are r1 8 cm and r2 8.5 cm, respectively, and the outer surface of the pipe is heavily insulated. If the convection heat transfer coefficient on the inner surface of the pipe is h 65 W/m2 · °C, express the boundary conditions on the inner and outer surfaces of the pipe during transient periods. Insulation h1 T Convection and Insulation Boundary Conditions r2 r1 SOLUTION During initial transient periods, heat transfer through the pipe maFIGURE 2–35 Schematic for Example 2–8. terial will predominantly be in the radial direction, and thus can be approximated as being one-dimensional. Then the temperature within the pipe material will change with the radial distance r and the time t. That is, T T (r, t ). It is stated that heat transfer between the steam and the pipe at the inner surface is by convection. Then taking the direction of heat transfer to be the positive r direction, the boundary condition on that surface can be expressed as k T(r1, t) h[T T(r1)] r The pipe is said to be well insulated on the outside, and thus heat loss through the outer surface of the pipe can be assumed to be negligible. Then the boundary condition at the outer surface can be expressed as T(r2, t) 0 r That is, the temperature gradient must be zero on the outer surface of the pipe at all times. 4 Radiation Boundary Condition In some cases, such as those encountered in space and cryogenic applications, a heat transfer surface is surrounded by an evacuated space and thus there is no convection heat transfer between a surface and the surrounding medium. In such cases, radiation becomes the only mechanism of heat transfer between the surface under consideration and the surroundings. Using an energy balance, the radiation boundary condition on a surface can be expressed as Heat conduction Radiation exchange at the surface in a at the surface in selected direction the same direction cen58933_ch02.qxd 9/10/2002 8:46 AM Page 83 83 CHAPTER 2 For one-dimensional heat transfer in the x-direction in a plate of thickness L, the radiation boundary conditions on both surfaces can be expressed as (Fig. 2–36) T(0, t) 4 4 k 1[T surr, 1 T(0, t) ] x (2-52a) Radiation Conduction ∂T(0, t) 4 4 ——— ε1σ [Tsurr, 1 – T(0, t) ] = – k ∂x ε1 Tsurr, 1 and ε2 Tsurr, 2 Conduction Radiation k T(L, t) 4 2[T(L, t)4 T surr, 2] x ∂T(L, t) 4 – k ——— = ε 2σ [T(L, t)4 – Tsurr, 2] ∂x (2-52b) 0 where 1 and 2 are the emissivities of the boundary surfaces, 5.67 108 W/m2 · K4 is the Stefan–Boltzmann constant, and Tsurr, 1 and Tsurr, 2 are the average temperatures of the surfaces surrounding the two sides of the plate, respectively. Note that the temperatures in radiation calculations must be expressed in K or R (not in °C or °F). The radiation boundary condition involves the fourth power of temperature, and thus it is a nonlinear condition. As a result, the application of this boundary condition results in powers of the unknown coefficients, which makes it difficult to determine them. Therefore, it is tempting to ignore radiation exchange at a surface during a heat transfer analysis in order to avoid the complications associated with nonlinearity. This is especially the case when heat transfer at the surface is dominated by convection, and the role of radiation is minor. L x FIGURE 2–36 Radiation boundary conditions on both surfaces of a plane wall. Interface Material A 5 Interface Boundary Conditions TA(x0, t) = TB(x0, t) Some bodies are made up of layers of different materials, and the solution of a heat transfer problem in such a medium requires the solution of the heat transfer problem in each layer. This, in turn, requires the specification of the boundary conditions at each interface. The boundary conditions at an interface are based on the requirements that (1) two bodies in contact must have the same temperature at the area of contact and (2) an interface (which is a surface) cannot store any energy, and thus the heat flux on the two sides of an interface must be the same. The boundary conditions at the interface of two bodies A and B in perfect contact at x x0 can be expressed as (Fig. 2–37) TA(x0, t) TB(x0, t) (2-53) and kA TA(x0, t) TB(x0, t) kB x x Material B (2-54) where kA and kB are the thermal conductivities of the layers A and B, respectively. The case of imperfect contact results in thermal contact resistance, which is considered in the next chapter. TA(x, t) TB(x, t) Conduction Conduction 0 ∂TA(x0, t) ∂TB(x0, t) – kA ———— = – kB ———— ∂x ∂x x0 L x FIGURE 2–37 Boundary conditions at the interface of two bodies in perfect contact. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 84 84 HEAT TRANSFER 6 Generalized Boundary Conditions So far we have considered surfaces subjected to single mode heat transfer, such as the specified heat flux, convection, or radiation for simplicity. In general, however, a surface may involve convection, radiation, and specified heat flux simultaneously. The boundary condition in such cases is again obtained from a surface energy balance, expressed as Heat transfer Heat transfer to the surface from the surface in all modes in all modes (2-55) This is illustrated in Examples 2–9 and 2–10. EXAMPLE 2–9 Radia tion Tsurr = 525 R n Convection o cti u nd Metal ball Co 0 T = 78°F r0 Ti = 600°F FIGURE 2–38 Schematic for Example 2–9. Combined Convection and Radiation Condition A spherical metal ball of radius r0 is heated in an oven to a temperature of 600°F throughout and is then taken out of the oven and allowed to cool in ambient air at T 78°F, as shown in Figure 2–38. The thermal conductivity of the ball material is k 8.3 Btu/h · ft · °F, and the average convection heat transfer coefficient on the outer surface of the ball is evaluated to be h 4.5 Btu/h · ft2 · °F. The emissivity of the outer surface of the ball is 0.6, and the average temperature of the surrounding surfaces is Tsurr 525 R. Assuming the ball is cooled uniformly from the entire outer surface, express the initial and boundary conditions for the cooling process of the ball. r SOLUTION The ball is initially at a uniform temperature and is cooled uniformly from the entire outer surface. Therefore, this is a one-dimensional transient heat transfer problem since the temperature within the ball will change with the radial distance r and the time t. That is, T T (r, t ). Taking the moment the ball is removed from the oven to be t 0, the initial condition can be expressed as T(r, 0) Ti 600°F The problem possesses symmetry about the midpoint (r 0) since the isotherms in this case will be concentric spheres, and thus no heat will be crossing the midpoint of the ball. Then the boundary condition at the midpoint can be expressed as T(0, t) 0 r The heat conducted to the outer surface of the ball is lost to the environment by convection and radiation. Then taking the direction of heat transfer to be the positive r direction, the boundary condition on the outer surface can be expressed as k T(r0, t) 4 h[T(r0) T ] [T(r0)4 T surr ] r cen58933_ch02.qxd 9/10/2002 8:46 AM Page 85 85 CHAPTER 2 All the quantities in the above relations are known except the temperatures and their derivatives at r 0 and r0. Also, the radiation part of the boundary condition is often ignored for simplicity by modifying the convection heat transfer coefficient to account for the contribution of radiation. The convection coefficient h in that case becomes the combined heat transfer coefficient. EXAMPLE 2–10 Combined Convection, Radiation, and Heat Flux Consider the south wall of a house that is L 0.2 m thick. The outer surface of the wall is exposed to solar radiation and has an absorptivity of 0.5 for solar energy. The interior of the house is maintained at T 1 20°C, while the ambient air temperature outside remains at T 2 5°C. The sky, the ground, and the surfaces of the surrounding structures at this location can be modeled as a surface at an effective temperature of Tsky 255 K for radiation exchange on the outer surface. The radiation exchange between the inner surface of the wall and the surfaces of the walls, floor, and ceiling it faces is negligible. The convection heat transfer coefficients on the inner and the outer surfaces of the wall are h1 6 W/m2 · °C and h2 25 W/m2 · °C, respectively. The thermal conductivity of the wall material is k 0.7 W/m · °C, and the emissivity of the outer surface is 2 0.9. Assuming the heat transfer through the wall to be steady and one-dimensional, express the boundary conditions on the inner and the outer surfaces of the wall. SOLUTION We take the direction normal to the wall surfaces as the x-axis with the origin at the inner surface of the wall, as shown in Figure 2–39. The heat transfer through the wall is given to be steady and one-dimensional, and thus the temperature depends on x only and not on time. That is, T T (x). The boundary condition on the inner surface of the wall at x 0 is a typical convection condition since it does not involve any radiation or specified heat flux. Taking the direction of heat transfer to be the positive x-direction, the boundary condition on the inner surface can be expressed as dT(0) k h1[T 1 T(0)] dx The boundary condition on the outer surface at x 0 is quite general as it involves conduction, convection, radiation, and specified heat flux. Again taking the direction of heat transfer to be the positive x-direction, the boundary condition on the outer surface can be expressed as Tsky Sun South wall Inner surface R Conduction Solar Co h1 T 1 nv ec tio where q· solar is the incident solar heat flux. Assuming the opposite direction for heat transfer would give the same result multiplied by 1, which is equivalent to the relation here. All the quantities in these relations are known except the temperatures and their derivatives at the two boundaries. n h2 T 2 Convection Conduction dT(L) 4 k h2[T(L) T 2] 2[T(L)4 T sky ] q·solar dx n tio ia ad 0 Outer surface L x FIGURE 2–39 Schematic for Example 2–10. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 86 86 HEAT TRANSFER Note that a heat transfer problem may involve different kinds of boundary conditions on different surfaces. For example, a plate may be subject to heat flux on one surface while losing or gaining heat by convection from the other surface. Also, the two boundary conditions in a direction may be specified at the same boundary, while no condition is imposed on the other boundary. For example, specifying the temperature and heat flux at x 0 of a plate of thickness L will result in a unique solution for the one-dimensional steady temperature distribution in the plate, including the value of temperature at the surface x L. Although not necessary, there is nothing wrong with specifying more than two boundary conditions in a specified direction, provided that there is no contradiction. The extra conditions in this case can be used to verify the results. 2–5 Heat transfer problem Mathematical formulation (Differential equation and boundary conditions) General solution of differential equation Application of boundary conditions Solution of the problem FIGURE 2–40 Basic steps involved in the solution of heat transfer problems. ■ SOLUTION OF STEADY ONE-DIMENSIONAL HEAT CONDUCTION PROBLEMS So far we have derived the differential equations for heat conduction in various coordinate systems and discussed the possible boundary conditions. A heat conduction problem can be formulated by specifying the applicable differential equation and a set of proper boundary conditions. In this section we will solve a wide range of heat conduction problems in rectangular, cylindrical, and spherical geometries. We will limit our attention to problems that result in ordinary differential equations such as the steady one-dimensional heat conduction problems. We will also assume constant thermal conductivity, but will consider variable conductivity later in this chapter. If you feel rusty on differential equations or haven’t taken differential equations yet, no need to panic. Simple integration is all you need to solve the steady one-dimensional heat conduction problems. The solution procedure for solving heat conduction problems can be summarized as (1) formulate the problem by obtaining the applicable differential equation in its simplest form and specifying the boundary conditions, (2) obtain the general solution of the differential equation, and (3) apply the boundary conditions and determine the arbitrary constants in the general solution (Fig. 2–40). This is demonstrated below with examples. EXAMPLE 2–11 Plane wall T1 T2 0 L x FIGURE 2–41 Schematic for Example 2–11. Heat Conduction in a Plane Wall Consider a large plane wall of thickness L 0.2 m, thermal conductivity k 1.2 W/m · °C, and surface area A 15 m2. The two sides of the wall are maintained at constant temperatures of T1 120°C and T2 50°C, respectively, as shown in Figure 2–41. Determine (a) the variation of temperature within the wall and the value of temperature at x 0.1 m and (b) the rate of heat conduction through the wall under steady conditions. SOLUTION A plane wall with specified surface temperatures is given. The variation of temperature and the rate of heat transfer are to be determined. Assumptions 1 Heat conduction is steady. 2 Heat conduction is onedimensional since the wall is large relative to its thickness and the thermal cen58933_ch02.qxd 9/10/2002 8:46 AM Page 87 87 CHAPTER 2 conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation. Properties The thermal conductivity is given to be k 1.2 W/m · °C. Analysis (a) Taking the direction normal to the surface of the wall to be the x-direction, the differential equation for this problem can be expressed as d 2T 0 dx2 Differential equation: T(0) T1 120°C T(L) T2 50°C The differential equation is linear and second order, and a quick inspection of it reveals that it has a single term involving derivatives and no terms involving the unknown function T as a factor. Thus, it can be solved by direct integration. Noting that an integration reduces the order of a derivative by one, the general solution of the differential equation above can be obtained by two simple successive integrations, each of which introduces an integration constant. Integrating the differential equation once with respect to x yields dT C1 dx where C1 is an arbitrary constant. Notice that the order of the derivative went down by one as a result of integration. As a check, if we take the derivative of this equation, we will obtain the original differential equation. This equation is not the solution yet since it involves a derivative. Integrating one more time, we obtain d 2T 0 dx2 Integrate: dT C1 dx Integrate again: T(x) C1x C2 →  General T(0) C1 0 C2 → C2 T1  Arbitrary constants solution FIGURE 2–42 Obtaining the general solution of a simple second order differential equation by integration. Boundary condition: T(x) C1x C2 T(0) T1 General solution: T(x) C1x C2 Applying the boundary condition: T(x) C1x C2 ↑ ↑ 0 0 { T1 Substituting: which is the general solution of the differential equation (Fig. 2–42). The general solution in this case resembles the general formula of a straight line whose slope is C1 and whose value at x 0 is C2. This is not surprising since the second derivative represents the change in the slope of a function, and a zero second derivative indicates that the slope of the function remains constant. Therefore, any straight line is a solution of this differential equation. The general solution contains two unknown constants C1 and C2, and thus we need two equations to determine them uniquely and obtain the specific solution. These equations are obtained by forcing the general solution to satisfy the specified boundary conditions. The application of each condition yields one equation, and thus we need to specify two conditions to determine the constants C1 and C2. When applying a boundary condition to an equation, all occurrences of the dependent and independent variables and any derivatives are replaced by the specified values. Thus the only unknowns in the resulting equations are the arbitrary constants. The first boundary condition can be interpreted as in the general solution, replace all the x’s by zero and T (x ) by T1. That is (Fig. 2–43), → →  with boundary conditions T1 C1 0 C2 → C2 T1 It cannot involve x or T(x) after the boundary condition is applied. FIGURE 2–43 When applying a boundary condition to the general solution at a specified point, all occurrences of the dependent and independent variables should be replaced by their specified values at that point. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 88 88 HEAT TRANSFER The second boundary condition can be interpreted as in the general solution, replace all the x’s by L and T (x ) by T2. That is, T(L) C1L C2 → T2 C1L T1 → C1 T2 T 1 L Substituting the C1 and C2 expressions into the general solution, we obtain T(x) T2 T1 x T1 L (2-56) which is the desired solution since it satisfies not only the differential equation but also the two specified boundary conditions. That is, differentiating Eq. 2–56 with respect to x twice will give d 2T /dx 2, which is the given differential equation, and substituting x 0 and x L into Eq. 2–56 gives T (0) T1 and T (L) T2, respectively, which are the specified conditions at the boundaries. Substituting the given information, the value of the temperature at x 0.1 m is determined to be T(0.1 m) (50 120)°C (0.1 m) 120°C 85°C 0.2 m (b) The rate of heat conduction anywhere in the wall is determined from Fourier’s law to be T1 T2 T2 T1 · dT Q wall kA kAC1 kA kA L L dx (2-57) The numerical value of the rate of heat conduction through the wall is determined by substituting the given values to be T1 T2 (120 50)°C · Q kA (1.2 W/m · °C)(15 m2) 6300 W L 0.2 m Discussion Note that under steady conditions, the rate of heat conduction through a plane wall is constant. EXAMPLE 2–12 A Wall with Various Sets of Boundary Conditions Consider steady one-dimensional heat conduction in a large plane wall of thickness L and constant thermal conductivity k with no heat generation. Obtain expressions for the variation of temperature within the wall for the following pairs of boundary conditions (Fig. 2–44): dT(0) q·0 40 W/cm2 dx dT(0) (b) k q·0 40 W/cm2 dx dT(0) (c) k q·0 40 W/cm2 dx (a) k and and and T(0) T0 15°C dT(L) q·L 25 W/cm2 dx dT(L) k q·0 40 W/cm2 dx k cen58933_ch02.qxd 9/10/2002 8:46 AM Page 89 89 CHAPTER 2 15°C Plane wall 40 W/cm2 Plane wall 40 W/cm2 T(x) Plane wall 40 W/cm2 T(x) T(x) 25 W/cm2 0 L 0 x (a) L 0 x (b) 40 W/cm2 L x (c) FIGURE 2–44 Schematic for Example 2–12. SOLUTION This is a steady one-dimensional heat conduction problem with constant thermal conductivity and no heat generation in the medium, and the heat conduction equation in this case can be expressed as (Eq. 2–17) d 2T 0 dx2 whose general solution was determined in the previous example by direct integration to be T(x) C1x C2 where C1 and C2 are two arbitrary integration constants. The specific solutions corresponding to each specified pair of boundary conditions are determined as follows. (a) In this case, both boundary conditions are specified at the same boundary at x 0, and no boundary condition is specified at the other boundary at x L. Noting that dT C1 dx the application of the boundary conditions gives k dT(0) q·0 dx → kC1 q·0 q·0 → C1 k and T(0) T0 → T0 C1 0 C2 → C2 T0 Substituting, the specific solution in this case is determined to be q·0 T(x) T0 k Therefore, the two boundary conditions can be specified at the same boundary, and it is not necessary to specify them at different locations. In fact, the fundamental theorem of linear ordinary differential equations guarantees that a cen58933_ch02.qxd 9/10/2002 8:46 AM Page 90 90 HEAT TRANSFER unique solution exists when both conditions are specified at the same location. But no such guarantee exists when the two conditions are specified at different boundaries, as you will see below. Differential equation: T (x) 0 (b) In this case different heat fluxes are specified at the two boundaries. The application of the boundary conditions gives General solution: T(x) C1x C2 (a) Unique solution: · kT(0) q·0 T(x) q 0 x T 0 T(0) T0 k (b) No solution: kT(0) q·0 kT(L) q·L FIGURE 2–45 A boundary-value problem may have a unique solution, infinitely many solutions, or no solutions at all. dT(0) q·0 dx → kC1 q·0 → C1 q·0 k k dT(L) q·L dx → kC1 q·L → C1 q·L k and T(x) None (c) Multiple solutions: · kT(0) q·0 T(x) q 0 x C 2 kT(L) q·0 k ↑ Arbitrary k Since q· 0 q· L and the constant C1 cannot be equal to two different things at the same time, there is no solution in this case. This is not surprising since this case corresponds to supplying heat to the plane wall from both sides and expecting the temperature of the wall to remain steady (not to change with time). This is impossible. (c) In this case, the same values for heat flux are specified at the two boundaries. The application of the boundary conditions gives k dT(0) q·0 dx → kC1 q·0 q·0 → C1 k k dT(L) q·0 dx → kC1 q·0 q·0 → C1 k and Thus, both conditions result in the same value for the constant C1, but no value for C2. Substituting, the specific solution in this case is determined to be q·0 T(x) x C2 k Resistance heater 1200 W T = 20°C which is not a unique solution since C2 is arbitrary. This solution represents a family of straight lines whose slope is q· 0/k. Physically, this problem corresponds to requiring the rate of heat supplied to the wall at x 0 be equal to the rate of heat removal from the other side of the wall at x L. But this is a consequence of the heat conduction through the wall being steady, and thus the second boundary condition does not provide any new information. So it is not surprising that the solution of this problem is not unique. The three cases discussed above are summarized in Figure 2–45. x EXAMPLE 2–13 Base plate Insulation 300 cm2 h L FIGURE 2–46 Schematic for Example 2–13. Heat Conduction in the Base Plate of an Iron Consider the base plate of a 1200-W household iron that has a thickness of L 0.5 cm, base area of A 300 cm2, and thermal conductivity of k 15 W/m · °C. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside, and the outer surface loses heat to the surroundings at T 20°C by convection, as shown in Figure 2–46. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 91 91 CHAPTER 2 Taking the convection heat transfer coefficient to be h 80 W/m2 · °C and disregarding heat loss by radiation, obtain an expression for the variation of temperature in the base plate, and evaluate the temperatures at the inner and the outer surfaces. SOLUTION The base plate of an iron is considered. The variation of temperature in the plate and the surface temperatures are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since the surface area of the base plate is large relative to its thickness, and the thermal conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation in the medium. 5 Heat transfer by radiation is negligible. 6 The upper part of the iron is well insulated so that the entire heat generated in the resistance wires is transferred to the base plate through its inner surface. Properties The thermal conductivity is given to be k 15 W/m · °C. Analysis The inner surface of the base plate is subjected to uniform heat flux at a rate of Q· 0 1200 W q·0 40,000 W/m2 Abase 0.03 m2 The outer side of the plate is subjected to the convection condition. Taking the direction normal to the surface of the wall as the x-direction with its origin on the inner surface, the differential equation for this problem can be expressed as (Fig. 2–47) d 2T 0 dx2 with the boundary conditions k dT(0) q·0 40,000 W/m2 dx k dT(L) h[T(L) T ] dx The general solution of the differential equation is again obtained by two successive integrations to be dT C1 dx and T(x) C1x C2 (a) where C1 and C2 are arbitrary constants. Applying the first boundary condition, k dT(0) q·0 dx → kC1 q·0 q·0 → C1 k Noting that dT /dx C1 and T (L) C1L C2, the application of the second boundary condition gives Base plate Heat flux Conduction h T dT(0) ·q = – k –—— 0 dx Conduction Convection dT(L) – k –—— = h[T(L) – T ] dx 0 L x FIGURE 2–47 The boundary conditions on the base plate of the iron discussed in Example 2–13. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 92 92 HEAT TRANSFER k dT(L) h[T(L) T ] → kC1 h[(C1L C2) T ] dx Substituting C1 q· 0/k and solving for C2, we obtain C2 T q·0 q·0 L h k Now substituting C1 and C2 into the general solution (a) gives Lx 1 T(x) T q·0 k h (b) which is the solution for the variation of the temperature in the plate. The temperatures at the inner and outer surfaces of the plate are determined by substituting x 0 and x L, respectively, into the relation (b): L 1 T(0) T q·0 k h m 1 150.005 W/m · °C 80 W/m 20°C (40,000 W/m2) 2 · °C 533°C and 40,000 W/m2 1 T(L) T q·0 0 20°C 520°C h 80 W/m2 · °C Discussion Note that the temperature of the inner surface of the base plate will be 13°C higher than the temperature of the outer surface when steady operating conditions are reached. Also note that this heat transfer analysis enables us to calculate the temperatures of surfaces that we cannot even reach. This example demonstrates how the heat flux and convection boundary conditions are applied to heat transfer problems. EXAMPLE 2–14 So l Plane wall ar Sun Conduction tio a di Ra n T1 ε α 0 L Space x FIGURE 2–48 Schematic for Example 2–14. Heat Conduction in a Solar Heated Wall Consider a large plane wall of thickness L 0.06 m and thermal conductivity k 1.2 W/m · °C in space. The wall is covered with white porcelain tiles that have an emissivity of 0.85 and a solar absorptivity of 0.26, as shown in Figure 2–48. The inner surface of the wall is maintained at T1 300 K at all times, while the outer surface is exposed to solar radiation that is incident at a rate of q· solar 800 W/m2. The outer surface is also losing heat by radiation to deep space at 0 K. Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached. What would your response be if no solar radiation was incident on the surface? SOLUTION A plane wall in space is subjected to specified temperature on one side and solar radiation on the other side. The outer surface temperature and the rate of heat transfer are to be determined. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 93 93 CHAPTER 2 Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness, and the thermal conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation. Properties The thermal conductivity is given to be k 1.2 W/m · °C. Analysis Taking the direction normal to the surface of the wall as the x-direction with its origin on the inner surface, the differential equation for this problem can be expressed as d 2T 0 dx2 with boundary conditions T(0) T1 300 K dT(L) 4 ] q·solar k [T(L)4 T space dx where Tspace 0. The general solution of the differential equation is again obtained by two successive integrations to be T(x) C1x C2 (a) where C1 and C2 are arbitrary constants. Applying the first boundary condition yields T(0) C1 0 C2 → C2 T1 Noting that dT /dx C1 and T (L) C1L C2 C1L T1, the application of the second boundary conditions gives k dT(L) T(L)4 q·solar dx → kC1 (C1L T1)4 q·solar Although C1 is the only unknown in this equation, we cannot get an explicit expression for it because the equation is nonlinear, and thus we cannot get a closed-form expression for the temperature distribution. This should explain why we do our best to avoid nonlinearities in the analysis, such as those associated with radiation. Let us back up a little and denote the outer surface temperature by T (L) TL instead of T (L) C1L T1. The application of the second boundary condition in this case gives k dT(L) T(L)4 q·solar dx → kC1 TL4 q·solar Solving for C1 gives C1 q·solar TL4 k (b) Now substituting C1 and C2 into the general solution (a), we obtain T(x) q·solar TL4 x T1 k (c) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 94 94 HEAT TRANSFER (1) Rearrange the equation to be solved: 100 TL 310.4 0.240975 TL 4 The equation is in the proper form since the left side consists of TL only. (2) Guess the value of TL, say 300 K, and substitute into the right side of the equation. It gives TL 290.2 K (3) Now substitute this value of TL into the right side of the equation and get which is the solution for the variation of the temperature in the wall in terms of the unknown outer surface temperature TL. At x L it becomes TL TL 292.6 K TL 292.7 K TL 0.26 (800 W/m2) 0.85 (5.67 108 W/m2 · K4) TL4 (0.06 m) 300 K 1.2 W/m · K which simplifies to FIGURE 2–49 A simple method of solving a nonlinear equation is to arrange the equation such that the unknown is alone on the left side while everything else is on the right side, and to iterate after an initial guess until convergence. 4 TL 310.4 0.240975 TL 100 This equation can be solved by one of the several nonlinear equation solvers available (or by the old fashioned trial-and-error method) to give (Fig. 2–49) TL 292.7 K TL 292.7 K Therefore, the solution is TL 292.7 K. The result is independent of the initial guess. (d ) which is an implicit relation for the outer surface temperature TL. Substituting the given values, we get TL 293.1 K (4) Repeat step (3) until convergence to desired accuracy is achieved. The subsequent iterations give q·solar TL4 L T1 k Knowing the outer surface temperature and knowing that it must remain constant under steady conditions, the temperature distribution in the wall can be determined by substituting the TL value above into Eq. (c): T(x) 0.26 (800 W/m 2) 0.85 (5.67 10 8 W/m 2 · K 4)(292.7 K)4 x 300 K 1.2 W/m · K which simplifies to T(x) (121.5 K/m)x 300 K Note that the outer surface temperature turned out to be lower than the inner surface temperature. Therefore, the heat transfer through the wall will be toward the outside despite the absorption of solar radiation by the outer surface. Knowing both the inner and outer surface temperatures of the wall, the steady rate of heat conduction through the wall can be determined from T0 TL (300 292.7) K q· k (1.2 W/m · K) 146 W/m2 L 0.06 m Discussion In the case of no incident solar radiation, the outer surface temperature, determined from Eq. (d ) by setting q· solar 0, will be TL 284.3 K. It is interesting to note that the solar energy incident on the surface causes the surface temperature to increase by about 8 K only when the inner surface temperature of the wall is maintained at 300 K. L T2 T1 0 r1 EXAMPLE 2–15 r2 r FIGURE 2–50 Schematic for Example 2–15. Heat Loss through a Steam Pipe Consider a steam pipe of length L 20 m, inner radius r1 6 cm, outer radius r2 8 cm, and thermal conductivity k 20 W/m · °C, as shown in Figure 2–50. The inner and outer surfaces of the pipe are maintained at average temperatures of T1 150°C and T2 60°C, respectively. Obtain a general relation cen58933_ch02.qxd 9/10/2002 8:46 AM Page 95 95 CHAPTER 2 for the temperature distribution inside the pipe under steady conditions, and determine the rate of heat loss from the steam through the pipe. SOLUTION A steam pipe is subjected to specified temperatures on its surfaces. The variation of temperature and the rate of heat transfer are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction, and thus T T (r ). 3 Thermal conductivity is constant. 4 There is no heat generation. Properties The thermal conductivity is given to be k 20 W/m · °C. Analysis The mathematical formulation of this problem can be expressed as dT d 0 r dr dr with boundary conditions T(r1) T1 150°C T(r2) T2 60°C Integrating the differential equation once with respect to r gives r dT C1 dr where C1 is an arbitrary constant. We now divide both sides of this equation by r to bring it to a readily integrable form, dT C1 r dr Again integrating with respect to r gives (Fig. 2–51) Differential equation: T(r) C1 ln r C2 We now apply both boundary conditions by replacing all occurrences of r and T (r ) in Eq. (a) with the specified values at the boundaries. We get T(r1) T1 T(r2) T2 → C1 ln r1 C2 T1 → C1 ln r2 C2 T2 T2 T1 ln(r2/r1) C2 T1 and ln(r /r (T T ) T ln(r/r1) 2 1 2 1 1 r dT C1 dr dT C1 r dr Integrate again: T(r) C1 ln r C2 T2 T1 ln r1 ln(r2/r1) which is the general solution. Substituting them into Eq. (a) and rearranging, the variation of temperature within the pipe is determined to be T(r) Integrate: Divide by r (r 0): which are two equations in two unknowns, C1 and C2. Solving them simultaneously gives C1 dT d r 0 dr dr (a) (2-58) The rate of heat loss from the steam is simply the total rate of heat conduction through the pipe, and is determined from Fourier’s law to be FIGURE 2–51 Basic steps involved in the solution of the steady one-dimensional heat conduction equation in cylindrical coordinates. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 96 96 HEAT TRANSFER C1 T1 T2 · dT Q cylinder kA k(2rL) r 2kLC1 2kL dr ln(r2/r1) (2-59) The numerical value of the rate of heat conduction through the pipe is determined by substituting the given values (150 60)°C · Q 2(20 W/m · °C)(20 m) 786 kW ln(0.08/0.06) DISCUSSION Note that the total rate of heat transfer through a pipe is constant, but the heat flux is not since it decreases in the direction of heat trans· fer with increasing radius since q· Q /(2rL). T2 EXAMPLE 2–16 T1 0 r1 r2 r FIGURE 2–52 Schematic for Example 2–16. Heat Conduction through a Spherical Shell Consider a spherical container of inner radius r1 8 cm, outer radius r2 10 cm, and thermal conductivity k 45 W/m · °C, as shown in Figure 2–52. The inner and outer surfaces of the container are maintained at constant temperatures of T1 200°C and T2 80°C, respectively, as a result of some chemical reactions occurring inside. Obtain a general relation for the temperature distribution inside the shell under steady conditions, and determine the rate of heat loss from the container. SOLUTION A spherical container is subjected to specified temperatures on its surfaces. The variation of temperature and the rate of heat transfer are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint, and thus T T (r ). 3 Thermal conductivity is constant. 4 There is no heat generation. Properties The thermal conductivity is given to be k 45 W/m · °C. Analysis The mathematical formulation of this problem can be expressed as d 2 dT 0 r dr dr with boundary conditions T(r1) T1 200°C T(r2) T2 80°C Integrating the differential equation once with respect to r yields r2 dT C1 dr where C1 is an arbitrary constant. We now divide both sides of this equation by r 2 to bring it to a readily integrable form, dT C1 dr r 2 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 97 97 CHAPTER 2 Again integrating with respect to r gives C1 T(r) r C2 (a) We now apply both boundary conditions by replacing all occurrences of r and T (r ) in the relation above by the specified values at the boundaries. We get T(r1) T1 T(r2) T2 C1 → r C2 T1 1 C1 → r C2 T2 2 · · Q2 = Q1 which are two equations in two unknowns, C1 and C2 . Solving them simultaneously gives r1r2 C1 r r (T1 T2) 2 1 and C2 · Q1 r2T2 r1T1 r2 r1 Substituting into Eq. (a), the variation of temperature within the spherical shell is determined to be T(r) r2T2 r1T1 r1r2 (T T2) r r 2 1 r (r2 r1) 1 C1 T1 T2 · dT Q sphere kA k(4r 2) 2 4kC1 4kr1r2 r r 2 1 dr r (2-61) The numerical value of the rate of heat conduction through the wall is determined by substituting the given values to be (200 80)°C · Q 4(45 W/m · °C)(0.08 m)(0.10 m) 27,140 W (0.10 0.08) m Discussion Note that the total rate of heat transfer through a spherical shell is · constant, but the heat flux, q· Q /4r 2, is not since it decreases in the direction of heat transfer with increasing radius as shown in Figure 2–53. r2 r q· 1 (2-60) The rate of heat loss from the container is simply the total rate of heat conduction through the container wall and is determined from Fourier’s law r1 0 q· 2 < q· 1 · Q 27.14 kW q· 1 = —1 = —————2 = 337.5 kW/m2 A1 4π (0.08 m) · Q 27.14 kW q· 2 = —2 = —————2 = 216.0 kW/m2 A2 4π (0.10 m) FIGURE 2–53 During steady one-dimensional heat conduction in a spherical (or cylindrical) container, the total rate of heat transfer remains constant, but the heat flux decreases with increasing radius. Chemical reactions 2–6 HEAT GENERATION IN A SOLID Many practical heat transfer applications involve the conversion of some form of energy into thermal energy in the medium. Such mediums are said to involve internal heat generation, which manifests itself as a rise in temperature throughout the medium. Some examples of heat generation are resistance heating in wires, exothermic chemical reactions in a solid, and nuclear reactions in nuclear fuel rods where electrical, chemical, and nuclear energies are converted to heat, respectively (Fig. 2–54). The absorption of radiation throughout the volume of a semitransparent medium such as water can also be considered as heat generation within the medium, as explained earlier. Nuclear fuel rods Electric resistance wires FIGURE 2–54 Heat generation in solids is commonly encountered in practice. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 98 98 HEAT TRANSFER Heat generation is usually expressed per unit volume of the medium, and is denoted by g·, whose unit is W/m3. For example, heat generation in an electrical wire of outer radius r0 and length L can be expressed as g· h, T Ts V k · · Q = Egen Heat generation · · Egen = gV FIGURE 2–55 At steady conditions, the entire heat generated in a solid must leave the solid through its outer surface. E· g.electric I 2 Re 2 Vwire ro L (W/m3) (2-62) where I is the electric current and Re is the electrical resistance of the wire. The temperature of a medium rises during heat generation as a result of the absorption of the generated heat by the medium during transient start-up period. As the temperature of the medium increases, so does the heat transfer from the medium to its surroundings. This continues until steady operating conditions are reached and the rate of heat generation equals the rate of heat transfer to the surroundings. Once steady operation has been established, the temperature of the medium at any point no longer changes. The maximum temperature Tmax in a solid that involves uniform heat generation will occur at a location farthest away from the outer surface when the outer surface of the solid is maintained at a constant temperature Ts. For example, the maximum temperature occurs at the midplane in a plane wall, at the centerline in a long cylinder, and at the midpoint in a sphere. The temperature distribution within the solid in these cases will be symmetrical about the center of symmetry. The quantities of major interest in a medium with heat generation are the surface temperature Ts and the maximum temperature Tmax that occurs in the medium in steady operation. Below we develop expressions for these two quantities for common geometries for the case of uniform heat generation (g· constant) within the medium. Consider a solid medium of surface area As, volume V, and constant thermal conductivity k, where heat is generated at a constant rate of g· per unit volume. Heat is transferred from the solid to the surrounding medium at T , with a constant heat transfer coefficient of h. All the surfaces of the solid are maintained at a common temperature Ts. Under steady conditions, the energy balance for this solid can be expressed as (Fig. 2–55) Rate of Rate of heat transfer energy generation from the solid within the solid (2-63) or · Q g·V (W) (2-64) Disregarding radiation (or incorporating it in the heat transfer coefficient h), the heat transfer rate can also be expressed from Newton’s law of cooling as · Q hAs (Ts T ) (W) (2-65) Combining Eqs. 2–64 and 2–65 and solving for the surface temperature Ts gives Ts T g· V hAs (2-66) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 99 99 CHAPTER 2 For a large plane wall of thickness 2L (As 2Awall and V 2LAwall), a long solid cylinder of radius ro (As 2ro L and V r o2 L), and a solid sphere of radius r0 (As 4r o2 and V 43 r o3), Eq. 2–66 reduces to g· L h g· ro Ts, cylinder T 2h g· ro Ts, sphere T 3h Ts, plane wall T (2-67) (2-68) (2-69) Note that the rise in surface temperature Ts is due to heat generation in the solid. Reconsider heat transfer from a long solid cylinder with heat generation. We mentioned above that, under steady conditions, the entire heat generated within the medium is conducted through the outer surface of the cylinder. Now consider an imaginary inner cylinder of radius r within the cylinder (Fig. 2–56). Again the heat generated within this inner cylinder must be equal to the heat conducted through the outer surface of this inner cylinder. That is, from Fourier’s law of heat conduction, kAr dT g·Vr dr · · Q = Egen Ar Vr r ro · · Egen = gV r FIGURE 2–56 Heat conducted through a cylindrical shell of radius r is equal to the heat generated within a shell. (2-70) where Ar 2rL and Vr r 2 L at any location r. Substituting these expressions into Eq. 2–70 and separating the variables, we get k(2rL) g· dT g·(r 2 L) → dT rdr dr 2k To = Tmax Integrating from r 0 where T(0) T0 to r ro where T(ro) Ts yields g·ro2 Tmax, cylinder To Ts 4k (2-71) (2-72) The approach outlined above can also be used to determine the maximum temperature rise in a plane wall of thickness 2L and a solid sphere of radius r0, with these results: g·L2 2k g·ro2 Tmax, sphere 6k Tmax, plane wall ∆Tmax Ts T T Heat generation where To is the centerline temperature of the cylinder, which is the maximum temperature, and Tmax is the difference between the centerline and the surface temperatures of the cylinder, which is the maximum temperature rise in the cylinder above the surface temperature. Once Tmax is available, the centerline temperature can easily be determined from (Fig. 2–57) Tcenter To Ts Tmax Ts (2-73) (2-74) Symmetry line FIGURE 2–57 The maximum temperature in a symmetrical solid with uniform heat generation occurs at its center. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 100 100 HEAT TRANSFER Again the maximum temperature at the center can be determined from Eq. 2–72 by adding the maximum temperature rise to the surface temperature of the solid. EXAMPLE 2–17 · Q Water q· A 2-kW resistance heater wire whose thermal conductivity is k 15 W/m · °C has a diameter of D 4 mm and a length of L 0.5 m, and is used to boil water (Fig. 2–58). If the outer surface temperature of the resistance wire is Ts 105°C, determine the temperature at the center of the wire. D = 4 m m Ts = 105°C Centerline Temperature of a Resistance Heater To FIGURE 2–58 Schematic for Example 2–17. SOLUTION The surface temperature of a resistance heater submerged in water is to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform. Properties The thermal conductivity is given to be k 15 W/m · °C. Analysis The 2-kW resistance heater converts electric energy into heat at a rate of 2 kW. The heat generation per unit volume of the wire is g· Q· gen Q· gen 2000 W 0.318 109 W/m3 Vwire ro2L (0.002 m)2(0.5 m) Then the center temperature of the wire is determined from Eq. 2–71 to be To Ts g·ro2 (0.318 109 W/m3)(0.002 m)2 105°C 126°C 4k 4 (15 W/m · °C) Discussion Note that the temperature difference between the center and the surface of the wire is 21°C. Water 226°F 0 r0 r g· FIGURE 2–59 Schematic for Example 2–18. We have developed these relations using the intuitive energy balance approach. However, we could have obtained the same relations by setting up the appropriate differential equations and solving them, as illustrated in Examples 2–18 and 2–19. EXAMPLE 2–18 Variation of Temperature in a Resistance Heater A long homogeneous resistance wire of radius r0 0.2 in. and thermal conductivity k 7.8 Btu/h · ft · °F is being used to boil water at atmospheric pressure by the passage of electric current, as shown in Figure 2–59. Heat is generated in the wire uniformly as a result of resistance heating at a rate of g· 2400 Btu/h · in3. If the outer surface temperature of the wire is measured to be Ts 226°F, obtain a relation for the temperature distribution, and determine the temperature at the centerline of the wire when steady operating conditions are reached. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 101 101 CHAPTER 2 SOLUTION This heat transfer problem is similar to the problem in Example 2–17, except that we need to obtain a relation for the variation of temperature within the wire with r. Differential equations are well suited for this purpose. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is no thermal symmetry about the centerline and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties The thermal conductivity is given to be k 7.8 Btu/h · ft · °F. Analysis The differential equation which governs the variation of temperature in the wire is simply Eq. 2–27, g· dT 1 d 0 r r dr dr k This is a second-order linear ordinary differential equation, and thus its general solution will contain two arbitrary constants. The determination of these constants requires the specification of two boundary conditions, which can be taken to be dT(0) –—— =0 dr T(r0) Ts 226°F T T(r) and dT(0) 0 dr 0 The first boundary condition simply states that the temperature of the outer surface of the wire is 226°F. The second boundary condition is the symmetry condition at the centerline, and states that the maximum temperature in the wire will occur at the centerline, and thus the slope of the temperature at r 0 must be zero (Fig. 2–60). This completes the mathematical formulation of the problem. Although not immediately obvious, the differential equation is in a form that can be solved by direct integration. Multiplying both sides of the equation by r and rearranging, we obtain g· dT d r r dr dr k Integrating with respect to r gives r g· r 2 dT C1 dr k 2 (a) since the heat generation is constant, and the integral of a derivative of a function is the function itself. That is, integration removes a derivative. It is convenient at this point to apply the second boundary condition, since it is related to the first derivative of the temperature, by replacing all occurrences of r and dT /dr in Eq. (a) by zero. It yields 0 g· dT(0) 0 C1 dr 2k → C1 0 r0 r g· FIGURE 2–60 The thermal symmetry condition at the centerline of a wire in which heat is generated uniformly. cen58933_ch02.qxd 9/10/2002 8:47 AM Page 102 102 HEAT TRANSFER Thus C1 cancels from the solution. We now divide Eq. (a) by r to bring it to a readily integrable form, g· dT r dr 2k Again integrating with respect to r gives T(r) g· 2 r C2 4k (b) We now apply the first boundary condition by replacing all occurrences of r by r0 and all occurrences of T by Ts. We get Ts g· 2 r C2 4k 0 → C2 Ts g· 2 r 4k 0 Substituting this C2 relation into Eq. (b) and rearranging give T(r) Ts g· (r 2 r 2) 4k 0 (c) which is the desired solution for the temperature distribution in the wire as a function of r. The temperature at the centerline (r 0) is obtained by replacing r in Eq. (c) by zero and substituting the known quantities, T(0) Ts g· 2 2400 Btu/h · in3 12 in. (0.2 in.)2 263°F r0 226°F 4k 4 (7.8 Btu/h · ft · °F) 1 ft Discussion The temperature of the centerline will be 37°F above the temperature of the outer surface of the wire. Note that the expression above for the centerline temperature is identical to Eq. 2–71, which was obtained using an energy balance on a control volume. EXAMPLE 2–19 Heat Conduction in a Two-Layer Medium Consider a long resistance wire of radius r1 0.2 cm and thermal conductivity kwire 15 W/m · °C in which heat is generated uniformly as a result of resistance heating at a constant rate of g· 50 W/cm3 (Fig. 2–61). The wire is embedded in a 0.5-cm-thick layer of ceramic whose thermal conductivity is kceramic 1.2 W/m · °C. If the outer surface temperature of the ceramic layer is measured to be Ts 45°C, determine the temperatures at the center of the resistance wire and the interface of the wire and the ceramic layer under steady conditions. Interface Wire r1 r2 Ts = 45°C r Ceramic layer FIGURE 2–61 Schematic for Example 2–19. SOLUTION The surface and interface temperatures of a resistance wire covered with a ceramic layer are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since this two-layer heat transfer problem possesses symmetry about the centerline and involves no change in the axial direction, and thus T T (r ). 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform. Properties It is given that kwire 15 W/m · °C and kceramic 1.2 W/m · ° C. cen58933_ch02.qxd 9/10/2002 8:47 AM Page 103 103 CHAPTER 2 Analysis Letting TI denote the unknown interface temperature, the heat transfer problem in the wire can be formulated as dTwire g· 1 d r dr r dr k 0 with Twire(r1) TI dTwire(0) 0 dr This problem was solved in Example 2–18, and its solution was determined to be Twire(r) TI g· (r 2 r 2) 4kwire 1 (a) Noting that the ceramic layer does not involve any heat generation and its outer surface temperature is specified, the heat conduction problem in that layer can be expressed as dTceramic d 0 r dr dr with Tceramic (r1) TI Tceramic (r2) Ts 45°C This problem was solved in Example 2–15, and its solution was determined to be Tceramic (r) ln(r/r1) (T TI) TI ln(r2/r1) s (b) We have already utilized the first interface condition by setting the wire and ceramic layer temperatures equal to TI at the interface r r1. The interface temperature TI is determined from the second interface condition that the heat flux in the wire and the ceramic layer at r r1 must be the same: kwire dTwire (r1) Ts TI 1 dTceramic (r1) g·r1 kceramic → kceramic 2 dr dr ln(r2/r1) r1 Solving for TI and substituting the given values, the interface temperature is determined to be g·r12 r2 ln r Ts 1 2kceramic (50 106 W/m3)(0.002 m)2 0.007 m ln 45° C 149.4°C 2(1.2 W/m · °C) 0.002 m TI Knowing the interface temperature, the temperature at the centerline (r 0) is obtained by substituting the known quantities into Eq. (a), Twire (0) TI g·r12 (50 106 W/m3)(0.002 m)2 149.4°C 152.7°C 4kwire 4 (15 W/m · °C) cen58933_ch02.qxd 9/10/2002 8:47 AM Page 104 104 HEAT TRANSFER Thus the temperature of the centerline will be slightly above the interface temperature. Discussion This example demonstrates how steady one-dimensional heat conduction problems in composite media can be solved. We could also solve this problem by determining the heat flux at the interface by dividing the total heat generated in the wire by the surface area of the wire, and then using this value as the specifed heat flux boundary condition for both the wire and the ceramic layer. This way the two problems are decoupled and can be solved separately. 2–7 500 400 300 Silver Copper Gold Aluminum Thermal conductivity (W/ m·K) 200 100 Tungsten Platinum 50 Iron 20 10 Stainless steel, AISI 304 Aluminum oxide 5 ■ VARIABLE THERMAL CONDUCTIVITY, k (T ) You will recall from Chapter 1 that the thermal conductivity of a material, in general, varies with temperature (Fig. 2–62). However, this variation is mild for many materials in the range of practical interest and can be disregarded. In such cases, we can use an average value for the thermal conductivity and treat it as a constant, as we have been doing so far. This is also common practice for other temperature-dependent properties such as the density and specific heat. When the variation of thermal conductivity with temperature in a specified temperature interval is large, however, it may be necessary to account for this variation to minimize the error. Accounting for the variation of the thermal conductivity with temperature, in general, complicates the analysis. But in the case of simple one-dimensional cases, we can obtain heat transfer relations in a straightforward manner. When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between T1 and T2 can be determined from Pyroceram kave 2 Fused quartz 1 100 300 500 1000 2000 4000 Temperature (K) FIGURE 2–62 Variation of the thermal conductivity of some solids with temperature. T2 k(T)dT T1 (2-75) T2 T1 This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity kave equals the rate of heat transfer through the same medium with variable conductivity k(T). Note that in the case of constant thermal conductivity k(T) k, Eq. 2–75 reduces to kave k, as expected. Then the rate of steady heat transfer through a plane wall, cylindrical layer, or spherical layer for the case of variable thermal conductivity can be determined by replacing the constant thermal conductivity k in Eqs. 2–57, 2–59, and 2–61 by the kave expression (or value) from Eq. 2–75: T1 T2 A · Q plane wall kave A L L T1 T2 T1 T2 · 2L Q cylinder 2kave L ln(r2/r1) ln(r2/r1) T1 T2 4r1r2 · Q sphere 4kaver1r2 r r r r 2 1 2 1 k(T)dT T1 (2-76) k(T)dT (2-77) k(T)dT (2-78) T2 T1 T2 cen58933_ch02.qxd 9/10/2002 8:47 AM Page 105 105 CHAPTER 2 The variation in thermal conductivity of a material with temperature in the temperature range of interest can often be approximated as a linear function and expressed as Plane wall k(T) k0(1 T) (2-79) where is called the temperature coefficient of thermal conductivity. The average value of thermal conductivity in the temperature range T1 to T2 in this case can be determined from kave T2 T1 k0(1 T)dT T2 T1 T2 T 1 k0 1 k(Tave) 2 k(T) = k0(1 + β T) β>0 β=0 T1 T2 β >D and z > 1.5D) (2) Vertical isothermal cylinder of length L buried in a semi-infinite medium (L >>D) T2 T2 T1 2π L S = ———– ln (4z /D) L 2π L S = ———– ln (4 L /D) z T1 D D L (3) Two parallel isothermal cylinders placed in an infinite medium (L >>D1 , D2 , z) T1 (4) A row of equally spaced parallel isothermal cylinders buried in a semi-infinite medium (L >>D, z and w > 1.5D) T2 T2 D1 2π L S = —————–———— 2 2 2 4z – D 1 – D 2 cosh–1 –——————— 2D1D2 D2 2π L S = —————–— 2w sinh —— 2π z ln —— πD w L T1 z D L (per cylinder) w z (5) Circular isothermal cylinder of length L in the midplane of an infinite wall (z > 0.5D) w (6) Circular isothermal cylinder of length L at the center of a square solid bar of the same length T2 T2 T1 2π L S = ———– ln (8z /πD) w z D z L 2π L S = —————– ln (1.08 w/D) T1 T2 L D w (7) Eccentric circular isothermal cylinder of length L in a cylinder of the same length (L > D2) T (8) Large plane wall T2 1 D1 2π L S = —————–———— D 21 + D 22 – 4z 2 cosh–1 –——————— 2D1D2 T1 T2 A S = —– L z D2 L L A (continued) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 172 172 HEAT TRANSFER TABLE 3-5 (CONCLUDED) (9) A long cylindrical layer (10) A square flow passage T2 T2 (a) For a / b > 1.4, 2π L S = ———–— ln (D2 / D1) 2π L S = ———–———– 0.93 ln (0.948a /b) D1 (b) For a / b < 1.41, D2 T1 T1 L 2π L S = ———–———– 0.785 ln (a /b) (11) A spherical layer L b a (12) Disk buried parallel to the surface in a semi-infinite medium (z >> D) 2π D1D2 S = ————– D2 – D1 D2 D1 T2 S = 4D z T1 (S = 2 D when z = 0) D T1 T2 (13) The edge of two adjoining walls of equal thickness (14) Corner of three walls of equal thickness T2 S = 0.54 w S = 0.15L L T1 (inside) L T1 w (15) Isothermal sphere buried in a semi-infinite medium T2 2π D S = ————— 1 – 0.25D/z (16) Isothermal sphere buried in a semi-infinite medium at T2 whose surface is insulated T1 z 2π D S = ————— 1 + 0.25D/z D T2 (outside) L T2 L L Insulated z T2 (medium) T1 D cen58933_ch03.qxd 9/10/2002 8:59 AM Page 173 173 CHAPTER 3 Then the steady rate of heat transfer from the pipe becomes · Q Sk(T1 T2) (62.9 m)(0.9 W/m · °C)(80 10)°C 3963 W Discussion Note that this heat is conducted from the pipe surface to the surface of the earth through the soil and then transferred to the atmosphere by convection and radiation. Heat Transfer between Hot and Cold Water Pipes SOLUTION Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is twodimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. Properties The thermal conductivity of concrete is given to be k 0.75 W/m · °C. Analysis The shape factor for this configuration is given in Table 3–5 to be S 2 L 4z2 D21 D22 2D1D2 cosh1 where z is the distance between the centerlines of the pipes and L is their length. Substituting, S 2 (5 m) cosh1 4 0.32 0.052 0.052 2 0.05 0.05 6.34 m Then the steady rate of heat transfer between the pipes becomes · Q Sk(T1 T2) (6.34 m)(0.75 W/m · °C)(70 15°)C 262 W Discussion We can reduce this heat loss by placing the hot and cold water pipes further away from each other. It is well known that insulation reduces heat transfer and saves energy and money. Decisions on the right amount of insulation are based on a heat transfer analysis, followed by an economic analysis to determine the “monetary value” of energy loss. This is illustrated with Example 3–15. T2 = 15°C cm 5m 2 D L= =5 cm =5 1 A 5-m-long section of hot and cold water pipes run parallel to each other in a thick concrete layer, as shown in Figure 3–50. The diameters of both pipes are 5 cm, and the distance between the centerline of the pipes is 30 cm. The surface temperatures of the hot and cold pipes are 70°C and 15°C, respectively. Taking the thermal conductivity of the concrete to be k 0.75 W/m · °C, determine the rate of heat transfer between the pipes. T1 = 70°C D EXAMPLE 3–14 z = 30 cm FIGURE 3–50 Schematic for Example 3-14. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 174 174 HEAT TRANSFER EXAMPLE 3–15 Cost of Heat Loss through Walls in Winter Consider an electrically heated house whose walls are 9 ft high and have an R-value of insulation of 13 (i.e., a thickness-to-thermal conductivity ratio of L/k 13 h · ft2 · °F/Btu). Two of the walls of the house are 40 ft long and the others are 30 ft long. The house is maintained at 75°F at all times, while the temperature of the outdoors varies. Determine the amount of heat lost through the walls of the house on a certain day during which the average temperature of the outdoors is 45°F. Also, determine the cost of this heat loss to the homeowner if the unit cost of electricity is $0.075/kWh. For combined convection and radiation heat transfer coefficients, use the ASHRAE (American Society of Heating, Refrigeration, and Air Conditioning Engineers) recommended values of hi 1.46 Btu/h · ft2 · °F for the inner surface of the walls and ho 4.0 Btu/h · ft2 · °F for the outer surface of the walls under 15 mph wind conditions in winter. SOLUTION An electrically heated house with R-13 insulation is considered. Wall, R=13 75°F The amount of heat lost through the walls and its cost are to be determined. Assumptions 1 The indoor and outdoor air temperatures have remained at the given values for the entire day so that heat transfer through the walls is steady. 2 Heat transfer through the walls is one-dimensional since any significant temperature gradients in this case will exist in the direction from the indoors to the outdoors. 3 The radiation effects are accounted for in the heat transfer coefficients. Analysis This problem involves conduction through the wall and convection at its surfaces and can best be handled by making use of the thermal resistance concept and drawing the thermal resistance network, as shown in Fig. 3–51. The heat transfer area of the walls is A Circumference Height (2 30 ft 2 40 ft)(9 ft) 1260 ft2 T1 T2 45°F Then the individual resistances are evaluated from their definitions to be 1 1 0.00054 h · °F/Btu hi A (1.46 Btu/h · ft2 · °F)(1260 ft2) R-value 13 h · ft2 · °F/Btu L 0.01032 h · °F/Btu A kA 1260 ft2 Ri Rconv, i Rwall Ro Rconv, o 1 1 0.00020 h · °F/Btu hc A (4.0 Btu/h · ft2 · °F)(1260 ft2) Noting that all three resistances are in series, the total resistance is Ri Rwall Ro T1 T2 T1 T2 FIGURE 3–51 Schematic for Example 3–15. Rtotal Ri Rwall Ro 0.00054 0.01032 0.00020 0.01106 h · °F/Btu Then the steady rate of heat transfer through the walls of the house becomes T1 T2 (75 45)°F · Q 2712 Btu/h Rtotal 0.01106 h · °F/Btu Finally, the total amount of heat lost through the walls during a 24-h period and its cost to the home owner are · Q Q t (2712 Btu/h)(24-h/day) 65,099 Btu/day 19.1 kWh/day cen58933_ch03.qxd 9/10/2002 8:59 AM Page 175 175 CHAPTER 3 since 1 kWh 3412 Btu, and Heating cost (Energy lost)(Cost of energy) (19.1 kWh/day)($0.075/kWh) $1.43/day Discussion The heat losses through the walls of the house that day will cost the home owner $1.43 worth of electricity. TOPIC OF SPECIAL INTEREST Heat Transfer Through Walls and Roofs Under steady conditions, the rate of heat transfer through any section of a building wall or roof can be determined from A(Ti To) · Q UA(Ti To) R (3-80) where Ti and To are the indoor and outdoor air temperatures, A is the heat transfer area, U is the overall heat transfer coefficient (the U-factor), and R 1/U is the overall unit thermal resistance (the R-value). Walls and roofs of buildings consist of various layers of materials, and the structure and operating conditions of the walls and the roofs may differ significantly from one building to another. Therefore, it is not practical to list the R-values (or U-factors) of different kinds of walls or roofs under different conditions. Instead, the overall R-value is determined from the thermal resistances of the individual components using the thermal resistance network. The overall thermal resistance of a structure can be determined most accurately in a lab by actually assembling the unit and testing it as a whole, but this approach is usually very time consuming and expensive. The analytical approach described here is fast and straightforward, and the results are usually in good agreement with the experimental values. The unit thermal resistance of a plane layer of thickness L and thermal conductivity k can be determined from R L/k. The thermal conductivity and other properties of common building materials are given in the appendix. The unit thermal resistances of various components used in building structures are listed in Table 3–6 for convenience. Heat transfer through a wall or roof section is also affected by the convection and radiation heat transfer coefficients at the exposed surfaces. The effects of convection and radiation on the inner and outer surfaces of walls and roofs are usually combined into the combined convection and radiation heat transfer coefficients (also called surface conductances) hi and ho, respectively, whose values are given in Table 3–7 for ordinary surfaces ( 0.9) and reflective surfaces ( 0.2 or 0.05). Note that surfaces having a low emittance also have a low surface conductance due to the reduction in radiation heat transfer. The values in the table are based on a surface This section can be skipped without a loss of continuity. TABLE 3–7 Combined convection and radiation heat transfer coefficients at window, wall, or roof surfaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 1). Position Direction of Heat Flow h, W/m2 · °C Surface Emittance, 0.90 0.20 0.05 Still air (both indoors and Horiz. Up ↑ 9.26 Horiz. Down ↓ 6.13 45° slope Up ↑ 9.09 45° slope Down ↓ 7.50 Vertical Horiz. → 8.29 outdoors) 5.17 4.32 2.10 1.25 5.00 4.15 3.41 2.56 4.20 3.35 Moving air (any position, any direction) Winter condition (winds at 15 mph or 24 km/h) 34.0 — — Summer condition (winds at 7.5 mph or 12 km/h) 22.7 — — Multiply by 0.176 to convert to Btu/h · ft2 · °F. Surface resistance can be obtained from R 1/h. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 176 176 HEAT TRANSFER TABLE 3–6 Unit thermal resistance (the R-value) of common components used in buildings R-value Component R-value 2 m · °C/W Outside surface (winter) Outside surface (summer) Inside surface, still air Plane air space, vertical, ordinary 13 mm (12 in.) 20 mm (43 in.) 40 mm (1.5 in.) 90 mm (3.5 in.) Insulation, 25 mm (1 in.) Glass fiber Mineral fiber batt Urethane rigid foam Stucco, 25 mm (1 in.) Face brick, 100 mm (4 in.) Common brick, 100 mm (4 in.) Steel siding Slag, 13 mm (12 in.) Wood, 25 mm (1 in.) Wood stud, nominal 2 in. 4 in. (3.5 in. or 90 mm wide) 2 ft · h · °F/Btu 0.030 0.17 0.044 0.25 0.12 0.68 surfaces (eff 0.82): 0.16 0.90 0.17 0.94 0.16 0.90 0.16 0.91 0.70 0.66 0.98 0.037 0.075 0.12 0.00 0.067 0.22 4.00 3.73 5.56 0.21 0.43 0.79 0.00 0.38 1.25 0.63 3.58 Component m2 · °C/W Wood stud, nominal 2 in. 6 in. (5.5 in. or 140 mm wide) Clay tile, 100 mm (4 in.) Acoustic tile Asphalt shingle roofing Building paper Concrete block, 100 mm (4 in.): Lightweight Heavyweight Plaster or gypsum board, 13 mm (21 in.) Wood fiberboard, 13 mm (21 in.) Plywood, 13 mm (12 in.) Concrete, 200 mm (8 in.): Lightweight Heavyweight Cement mortar, 13 mm (1/2 in.) Wood bevel lapped siding, 13 mm 200 mm (1/2 in. 8 in.) ft2 · h · °F/Btu 0.98 0.18 0.32 0.077 0.011 5.56 1.01 1.79 0.44 0.06 0.27 0.13 1.51 0.71 0.079 0.23 0.11 0.45 1.31 0.62 1.17 0.12 0.018 6.67 0.67 0.10 0.14 0.81 temperature of 21°C (72°F) and a surface–air temperature difference of 5.5°C (10°F). Also, the equivalent surface temperature of the environment is assumed to be equal to the ambient air temperature. Despite the convenience it offers, this assumption is not quite accurate because of the additional radiation heat loss from the surface to the clear sky. The effect of sky radiation can be accounted for approximately by taking the outside temperature to be the average of the outdoor air and sky temperatures. The inner surface heat transfer coefficient hi remains fairly constant throughout the year, but the value of ho varies considerably because of its dependence on the orientation and wind speed, which can vary from less than 1 km/h in calm weather to over 40 km/h during storms. The commonly used values of hi and ho for peak load calculations are hi 8.29 W/m2 · °C 1.46 Btu/h · ft2 · °F 34.0 W/m2 · °C 6.0 Btu/h · ft2 · °F ho 22.7 W/m2 · °C 4.0 Btu/h · ft2 · °F (winter and summer) (winter) (summer) which correspond to design wind conditions of 24 km/h (15 mph) for winter and 12 km/h (7.5 mph) for summer. The corresponding surface thermal resistances (R-values) are determined from Ri 1/hi and Ro 1/ho. The surface conductance values under still air conditions can be used for interior surfaces as well as exterior surfaces in calm weather. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 177 177 CHAPTER 3 Building components often involve trapped air spaces between various layers. Thermal resistances of such air spaces depend on the thickness of the layer, the temperature difference across the layer, the mean air temperature, the emissivity of each surface, the orientation of the air layer, and the direction of heat transfer. The emissivities of surfaces commonly encountered in buildings are given in Table 3–8. The effective emissivity of a plane-parallel air space is given by 1 1 1 effective 1 2 1 (3-81) where 1 and 2 are the emissivities of the surfaces of the air space. Table 3–8 also lists the effective emissivities of air spaces for the cases where (1) the emissivity of one surface of the air space is while the emissivity of the other surface is 0.9 (a building material) and (2) the emissivity of both surfaces is . Note that the effective emissivity of an air space between building materials is 0.82/0.03 27 times that of an air space between surfaces covered with aluminum foil. For specified surface temperatures, radiation heat transfer through an air space is proportional to effective emissivity, and thus the rate of radiation heat transfer in the ordinary surface case is 27 times that of the reflective surface case. Table 3–9 lists the thermal resistances of 20-mm-, 40-mm-, and 90-mm(0.75-in., 1.5-in., and 3.5-in.) thick air spaces under various conditions. The thermal resistance values in the table are applicable to air spaces of uniform thickness bounded by plane, smooth, parallel surfaces with no air leakage. Thermal resistances for other temperatures, emissivities, and air spaces can be obtained by interpolation and moderate extrapolation. Note that the presence of a low-emissivity surface reduces radiation heat transfer across an air space and thus significantly increases the thermal resistance. The thermal effectiveness of a low-emissivity surface will decline, however, if the condition of the surface changes as a result of some effects such as condensation, surface oxidation, and dust accumulation. The R-value of a wall or roof structure that involves layers of uniform thickness is determined easily by simply adding up the unit thermal resistances of the layers that are in series. But when a structure involves components such as wood studs and metal connectors, then the thermal resistance network involves parallel connections and possible twodimensional effects. The overall R-value in this case can be determined by assuming (1) parallel heat flow paths through areas of different construction or (2) isothermal planes normal to the direction of heat transfer. The first approach usually overpredicts the overall thermal resistance, whereas the second approach usually underpredicts it. The parallel heat flow path approach is more suitable for wood frame walls and roofs, whereas the isothermal planes approach is more suitable for masonry or metal frame walls. The thermal contact resistance between different components of building structures ranges between 0.01 and 0.1 m2 · °C/W, which is negligible in most cases. However, it may be significant for metal building components such as steel framing members. TABLE 3–8 Emissivities of various surfaces and the effective emissivity of air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 3). Effective Emissivity of Air Space Surface 1 1 2 0.9 2 Aluminum foil, bright 0.05 0.05 Aluminum sheet 0.12 0.12 Aluminumcoated paper, polished 0.20 0.20 Steel, galvanized, bright 0.25 0.24 Aluminum paint 0.50 0.47 Building materials: Wood, paper, masonry, nonmetallic paints 0.90 0.82 Ordinary glass 0.84 0.77 0.03 0.06 0.11 0.15 0.35 0.82 0.72 Surface emissivity of aluminum foil increases to 0.30 with barely visible condensation, and to 0.70 with clearly visible condensation. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 178 TABLE 3–9 Unit thermal resistances (R-values) of well-sealed plane air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 2) (a) SI units (in m2 · °C/W) Position of Air Space Direction of Heat Flow Mean Temp., °C Temp. Diff., °C 32.2 5.6 10.0 16.7 Horizontal Up ↑ 10.0 5.6 17.8 11.1 32.2 5.6 10.0 16.7 45° slope Up ↑ 10.0 5.6 17.8 11.1 32.2 5.6 10.0 16.7 Vertical Horizontal → 10.0 5.6 17.8 11.1 32.2 5.6 10.0 16.7 45° slope Down ↓ 10.0 5.6 17.8 11.1 32.2 5.6 10.0 16.7 Horizontal Down ↓ 10.0 5.6 17.8 11.1 (b) English units (in h · ft2 · °F/Btu) Position of Air Space Direction of Heat Flow Horizontal Up ↑ 45° slope Up ↑ Vertical Horizontal → 45° slope Down ↓ Horizontal Down ↓ 178 Mean Temp. Temp., Diff., °F °F 90 50 50 0 90 50 50 0 90 50 50 0 90 50 50 0 90 50 50 0 10 30 10 20 10 30 10 20 10 30 10 20 10 30 10 20 10 30 10 20 20-mm Air Space 40-mm Air Space 90-mm Air Space Effective Emissivity, eff Effective Emissivity, eff Effective Emissivity, eff 0.03 0.05 0.41 0.30 0.40 0.32 0.52 0.35 0.51 0.37 0.62 0.51 0.65 0.55 0.62 0.60 0.67 0.66 0.62 0.66 0.68 0.74 0.39 0.29 0.39 0.32 0.49 0.34 0.48 0.36 0.57 0.49 0.61 0.53 0.58 0.57 0.63 0.63 0.58 0.62 0.63 0.70 0.5 0.18 0.17 0.20 0.20 0.20 0.19 0.23 0.23 0.21 0.23 0.25 0.28 0.21 0.24 0.26 0.30 0.21 0.25 0.26 0.32 0.82 0.03 0.05 0.13 0.14 0.15 0.16 0.14 0.14 0.17 0.18 0.15 0.17 0.18 0.21 0.15 0.17 0.18 0.22 0.15 0.18 0.18 0.23 0.45 0.33 0.44 0.35 0.51 0.38 0.51 0.40 0.70 0.45 0.67 0.49 0.89 0.63 0.90 0.68 1.07 1.10 1.16 1.24 0.42 0.32 0.42 0.34 0.48 0.36 0.48 0.39 0.64 0.43 0.62 0.47 0.80 0.59 0.82 0.64 0.94 0.99 1.04 1.13 0.5 0.19 0.18 0.21 0.22 0.20 0.20 0.23 0.24 0.22 0.22 0.26 0.26 0.24 0.25 0.28 0.31 0.25 0.30 0.30 0.39 0.82 0.03 0.05 0.14 0.14 0.16 0.17 0.14 0.15 0.17 0.18 0.15 0.16 0.18 0.20 0.16 0.18 0.19 0.22 0.17 0.20 0.20 0.26 0.50 0.27 0.49 0.40 0.56 0.40 0.55 0.43 0.65 0.47 0.64 0.51 0.85 0.62 0.83 0.67 1.77 1.69 1.96 1.92 0.47 0.35 0.47 0.38 0.52 0.38 0.52 0.41 0.60 0.45 0.60 0.49 0.76 0.58 0.77 0.64 1.44 1.44 1.63 1.68 0.5 0.82 0.20 0.19 0.23 0.23 0.21 0.20 0.24 0.24 0.22 0.22 0.25 0.27 0.24 0.25 0.28 0.31 0.28 0.33 0.34 0.43 0.14 0.15 0.16 0.18 0.14 0.15 0.17 0.19 0.15 0.16 0.18 0.20 0.16 0.18 0.19 0.22 0.18 0.21 0.22 0.29 0.75-in. Air Space 1.5-in. Air Space 3.5-in. Air Space Effective Emissivity, eff Effective Emissivity, eff Effective Emissivity, eff 0.03 0.05 2.34 1.71 2.30 1.83 2.96 1.99 2.90 2.13 3.50 2.91 3.70 3.14 3.53 3.43 3.81 3.75 3.55 3.77 3.84 4.18 2.22 1.66 2.21 1.79 2.78 1.92 2.75 2.07 3.24 2.77 3.46 3.02 3.27 3.23 3.57 3.57 3.29 3.52 3.59 3.96 0.5 1.04 0.99 1.16 1.16 1.15 1.08 1.29 1.28 1.22 1.30 1.43 1.58 1.22 1.39 1.45 1.72 1.22 1.44 1.45 1.81 0.82 0.03 0.05 0.75 0.77 0.87 0.93 0.81 0.82 0.94 1.00 0.84 0.94 1.01 1.18 0.84 0.99 1.02 1.26 0.85 1.02 1.02 1.30 2.55 1.87 2.50 2.01 2.92 2.14 2.88 2.30 3.99 2.58 3.79 2.76 5.07 3.58 5.10 3.85 6.09 6.27 6.61 7.03 2.41 1.81 2.40 1.95 2.73 2.06 2.74 2.23 3.66 2.46 3.55 2.66 4.55 3.36 4.66 3.66 5.35 5.63 5.90 6.43 0.5 1.08 1.04 1.21 1.23 1.14 1.12 1.29 1.34 1.27 1.23 1.45 1.48 1.36 1.42 1.60 1.74 1.43 1.70 1.73 2.19 0.82 0.03 0.05 0.77 0.80 0.89 0.97 0.80 0.84 0.94 1.04 0.87 0.90 1.02 1.12 0.91 1.00 1.09 1.27 0.94 1.14 1.15 1.49 2.84 2.09 2.80 2.25 3.18 2.26 3.12 2.42 3.69 2.67 3.63 2.88 4.81 3.51 4.74 3.81 10.07 9.60 11.15 10.90 2.66 2.01 2.66 2.18 2.96 2.17 2.95 2.35 3.40 2.55 3.40 2.78 4.33 3.30 4.36 3.63 8.19 8.17 9.27 9.52 0.5 0.82 1.13 1.10 1.28 1.32 1.18 1.15 1.34 1.38 1.24 1.25 1.42 1.51 1.34 1.40 1.57 1.74 1.57 1.88 1.93 2.47 0.80 0.84 0.93 1.03 0.82 0.86 0.96 1.06 0.85 0.91 1.01 1.14 0.90 1.00 1.08 1.27 1.00 1.22 1.24 1.62 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 179 179 CHAPTER 3 EXAMPLE 3–16 The R-Value of a Wood Frame Wall Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a wood frame wall that is built around 38-mm 90-mm (2 4 nominal) wood studs with a center-to-center distance of 400 mm. The 90-mm-wide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13-mm gypsum wallboard and the outside with 13-mm wood fiberboard and 13-mm 200-mm wood bevel lapped siding. The insulated cavity constitutes 75 percent of the heat transmission area while the studs, plates, and sills constitute 21 percent. The headers constitute 4 percent of the area, and they can be treated as studs. Also, determine the rate of heat loss through the walls of a house whose perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose winter design temperature is 2°C. Take the indoor design temperature to be 22°C and assume 20 percent of the wall area is occupied by glazing. SOLUTION The R-value and the U-factor of a wood frame wall as well as the rate of heat loss through such a wall in Las Vegas are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant. Properties The R-values of different materials are given in Table 3–6. Analysis The schematic of the wall as well as the different elements used in its construction are shown here. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the U-factors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from Roverall 1/Uoverall where Uoverall (U farea)insulation (U farea)stud and the value of the area fraction farea is 0.75 for the insulation section and 0.25 for the stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available R-values from Table 3–6 and calculating others, the total R-values for each section can be determined in a systematic manner in the table in this sample. We conclude that the overall unit thermal resistance of the wall is 2.23 m2 · °C/W, and this value accounts for the effects of the studs and headers. It corresponds to an R-value of 2.23 5.68 12.7 (or nearly R-13) in English units. Note that if there were no wood studs and headers in the wall, the overall thermal resistance would be 3.05 m2 · °C/W, which is 37 percent greater than 2.23 m2 · °C/W. Therefore, the wood studs and headers in this case serve as thermal bridges in wood frame walls, and their effect must be considered in the thermal analysis of buildings. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 180 180 HEAT TRANSFER Schematic R-value, m2 · °C/W Construction 4b At Studs 0.030 0.030 0.14 0.14 0.23 0.23 2.45 — 1. 1 Outside surface, 24 km/h wind 2. Wood bevel lapped siding 3. Wood fiberboard sheeting, 13 mm 4a. Glass fiber insulation, 90 mm 4b. Wood stud, 38 mm 6 90 mm 5 4a 5. Gypsum wallboard, 3 13 mm 2 6. Inside surface, still air Between Studs — 0.079 0.12 0.63 0.079 0.12 Total unit thermal resistance of each section, R (in m2 · °C/W) 3.05 1.23 The U-factor of each section, U 1/R, in W/m2 · °C 0.328 0.813 Area fraction of each section, farea 0.75 0.25 Overall U-factor: U farea, i Ui 0.75 0.328 0.25 0.813 0.449 W/m2 · °C Overall unit thermal resistance: R 1/U 2.23 m2 · °C/W The perimeter of the building is 50 m and the height of the walls is 2.5 m. Noting that glazing constitutes 20 percent of the walls, the total wall area is Awall 0.80(Perimeter)(Height) 0.80(50 m)(2.5 m) 100 m2 Then the rate of heat loss through the walls under design conditions becomes · Q wall (UA)wall (Ti To) (0.449 W/m2 · °C)(100 m2)[22 (2)°C] 1078 W Discussion Note that a 1-kW resistance heater in this house will make up almost all the heat lost through the walls, except through the doors and windows, when the outdoor air temperature drops to 2°C. EXAMPLE 3–17 The R-Value of a Wall with Rigid Foam The 13-mm-thick wood fiberboard sheathing of the wood stud wall discussed in the previous example is replaced by a 25-mm-thick rigid foam insulation. Determine the percent increase in the R-value of the wall as a result. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 181 181 CHAPTER 3 SOLUTION The overall R-value of the existing wall was determined in Example 3–16 to be 2.23 m2 · °C/W. Noting that the R-values of the fiberboard and the foam insulation are 0.23 m2 · °C/W and 0.98 m2 · °C/W, respectively, and the added and removed thermal resistances are in series, the overall R-value of the wall after modification becomes Rnew Rold Rremoved Radded 2.23 0.23 0.98 2.98 m2 · °C/W This represents an increase of (2.98 2.23)/2.23 0.34 or 34 percent in the R-value of the wall. This example demonstrated how to evaluate the new R-value of a structure when some structural members are added or removed. EXAMPLE 3–18 The R-Value of a Masonry Wall Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a masonry cavity wall that is built around 6-in.-thick concrete blocks made of lightweight aggregate with 3 cores filled with perlite (R 4.2 h · ft2 · °F/Btu). The outside is finished with 4-in. face brick with 21 -in. cement mortar between the bricks and concrete blocks. The inside finish consists of 12 in. gypsum wallboard separated from the concrete block by 43 -in.-thick (1-in. 3-in. nominal) vertical furring (R 4.2 h · ft2 · °F/Btu) whose center-to-center distance is 16 in. Both sides of the 34 -in.-thick air space between the concrete block and the gypsum board are coated with reflective aluminum foil ( 0.05) so that the effective emissivity of the air space is 0.03. For a mean temperature of 50°F and a temperature difference of 30°F, the R-value of the air space is 2.91 h · ft2 · °F/Btu. The reflective air space constitutes 80 percent of the heat transmission area, while the vertical furring constitutes 20 percent. SOLUTION The R-value and the U-factor of a masonry cavity wall are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant. Properties The R-values of different materials are given in Table 3–6. Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Following the approach described here and using the available R-values from Table 3–6, the overall R-value of the wall is determined in this table. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 182 182 HEAT TRANSFER Schematic R-value, h · ft2 · °F/Btu Construction Between Furring 1. 5b 2 1 3 Outside surface, 15 mph wind 2. Face brick, 4 in. 3. Cement mortar, 0.5 in. 4. Concrete block, 6 in. 5a. Reflective air space, 3 in. 4 5b. Nominal 1 3 7 vertical furring 6 5a 6. Gypsum wallboard, 4 0.5 in. 7. Inside surface, still air At Furring 0.17 0.43 0.17 0.43 0.10 4.20 0.10 4.20 2.91 — — 0.94 0.45 0.45 0.68 0.68 Total unit thermal resistance of each section, R 8.94 6.97 0.112 0.143 The U-factor of each section, U 1/R, in Btu/h · ft2 · °F Area fraction of each section, farea 0.80 0.20 Overall U-factor: U farea, i Ui 0.80 0.112 0.20 0.143 0.118 Btu/h · ft2 · °F Overall unit thermal resistance: R 1/U 8.46 h · ft2 · °F/Btu Therefore, the overall unit thermal resistance of the wall is 8.46 h · ft2 · °F/Btu and the overall U-factor is 0.118 Btu/h · ft2 · °F. These values account for the effects of the vertical furring. EXAMPLE 3–19 The R-Value of a Pitched Roof Determine the overall unit thermal resistance (the R-value) and the overall heat transfer coefficient (the U-factor) of a 45° pitched roof built around nominal 2-in. 4-in. wood studs with a center-to-center distance of 16 in. The 3.5-in.wide air space between the studs does not have any reflective surface and thus its effective emissivity is 0.84. For a mean temperature of 90°F and a temperature difference of 30°F, the R-value of the air space is 0.86 h · ft2 · °F/Btu. The lower part of the roof is finished with 21 -in. gypsum wallboard and the upper part with 58 -in. plywood, building paper, and asphalt shingle roofing. The air space constitutes 75 percent of the heat transmission area, while the studs and headers constitute 25 percent. SOLUTION The R-value and the U-factor of a 45° pitched roof are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the roof is one-dimensional. 3 Thermal properties of the roof and the heat transfer coefficients are constant. Properties The R-values of different materials are given in Table 3–6. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 183 183 CHAPTER 3 Analysis The schematic of the pitched roof as well as the different elements used in its construction are shown below. Following the approach described above and using the available R-values from Table 3–6, the overall R-value of the roof can be determined in the table here. Schematic R-value, h · ft2 · °F/Btu Construction Outside surface, 15 mph wind 2. Asphalt shingle roofing 45° 3. Building paper 4. Plywood deck, 58 in. 5a. Nonreflective air 1 2 3 4 5a 5b 6 7 space, 3.5 in. 5b. Wood stud, 2 in. by 4 in. 6. Gypsum wallboard, 0.5 in. 7. Inside surface, 45° slope, still air Between At Studs Studs 1. 0.17 0.17 0.44 0.10 0.78 0.44 0.10 0.78 0.86 — 0.45 — 3.58 0.45 0.63 0.63 Total unit thermal resistance of each section, R 3.43 6.15 0.292 0.163 The U-factor of each section, U 1/R, in Btu/h · ft2 · °F Area fraction of each section, farea 0.75 0.25 Overall U-factor: U farea, i Ui 0.75 0.292 0.25 0.163 0.260 Btu/h · ft2 · °F Overall unit thermal resistance: R 1/U 3.85 h · ft2 · °F/Btu Therefore, the overall unit thermal resistance of this pitched roof is 3.85 h · ft2 · °F/Btu and the overall U-factor is 0.260 Btu/h · ft2 · °F. Note that the wood studs offer much larger thermal resistance to heat flow than the air space between the studs. The construction of wood frame flat ceilings typically involve 2-in. 6-in. joists on 400-mm (16-in.) or 600-mm (24-in.) centers. The fraction of framing is usually taken to be 0.10 for joists on 400-mm centers and 0.07 for joists on 600-mm centers. Most buildings have a combination of a ceiling and a roof with an attic space in between, and the determination of the R-value of the roof–attic– ceiling combination depends on whether the attic is vented or not. For adequately ventilated attics, the attic air temperature is practically the same as the outdoor air temperature, and thus heat transfer through the roof is governed by the R-value of the ceiling only. However, heat is also transferred between the roof and the ceiling by radiation, and it needs to be considered (Fig. 3–52). The major function of the roof in this case is to serve as a radiation shield by blocking off solar radiation. Effectively ventilating the attic in summer should not lead one to believe that heat gain to the building through the attic is greatly reduced. This is because most of the heat transfer through the attic is by radiation. Air exhaust 6 in. 3 in. Air intake Radiant barrier 3 in. Air intake FIGURE 3–52 Ventilation paths for a naturally ventilated attic and the appropriate size of the flow areas around the radiant barrier for proper air circulation (from DOE/CE-0335P, U.S. Dept. of Energy). cen58933_ch03.qxd 9/10/2002 9:00 AM Page 184 184 HEAT TRANSFER Roof decking Rafter Radiant barrier Joist Air space Insulation (a) Under the roof deck Rafter Radiant barrier Joist Roof decking Roof decking Insulation (b) At the bottom of rafters Rafter Radiant barrier Joist Insulation (c) On top of attic floor insulation FIGURE 3–53 Three possible locations for an attic radiant barrier (from DOE/CE-0335P, U.S. Dept. of Energy). To Rroof Shingles Rafter Tattic Attic Deck Aroof Aceiling Rceiling Ceiling joist Ti FIGURE 3–54 Thermal resistance network for a pitched roof–attic–ceiling combination for the case of an unvented attic. Radiation heat transfer between the ceiling and the roof can be minimized by covering at least one side of the attic (the roof or the ceiling side) by a reflective material, called radiant barrier, such as aluminum foil or aluminum-coated paper. Tests on houses with R-19 attic floor insulation have shown that radiant barriers can reduce summer ceiling heat gains by 16 to 42 percent compared to an attic with the same insulation level and no radiant barrier. Considering that the ceiling heat gain represents about 15 to 25 percent of the total cooling load of a house, radiant barriers will reduce the air conditioning costs by 2 to 10 percent. Radiant barriers also reduce the heat loss in winter through the ceiling, but tests have shown that the percentage reduction in heat losses is less. As a result, the percentage reduction in heating costs will be less than the reduction in the airconditioning costs. Also, the values given are for new and undusted radiant barrier installations, and percentages will be lower for aged or dusty radiant barriers. Some possible locations for attic radiant barriers are given in Figure 3–53. In whole house tests on houses with R-19 attic floor insulation, radiant barriers have reduced the ceiling heat gain by an average of 35 percent when the radiant barrier is installed on the attic floor, and by 24 percent when it is attached to the bottom of roof rafters. Test cell tests also demonstrated that the best location for radiant barriers is the attic floor, provided that the attic is not used as a storage area and is kept clean. For unvented attics, any heat transfer must occur through (1) the ceiling, (2) the attic space, and (3) the roof (Fig. 3–54). Therefore, the overall R-value of the roof–ceiling combination with an unvented attic depends on the combined effects of the R-value of the ceiling and the R-value of the roof as well as the thermal resistance of the attic space. The attic space can be treated as an air layer in the analysis. But a more practical way of accounting for its effect is to consider surface resistances on the roof and ceiling surfaces facing each other. In this case, the R-values of the ceiling and the roof are first determined separately (by using convection resistances for the still-air case for the attic surfaces). Then it can be shown that the overall R-value of the ceiling–roof combination per unit area of the ceiling can be expressed as cen58933_ch03.qxd 9/10/2002 9:00 AM Page 185 185 CHAPTER 3 R Rceiling Rroof A Aceiling (3-82) roof where Aceiling and Aroof are the ceiling and roof areas, respectively. The area ratio is equal to 1 for flat roofs and is less than 1 for pitched roofs. For a 45° pitched roof, the area ratio is Aceiling/Aroof 1/ 2 0.707. Note that the pitched roof has a greater area for heat transfer than the flat ceiling, and the area ratio accounts for the reduction in the unit R-value of the roof when expressed per unit area of the ceiling. Also, the direction of heat flow is up in winter (heat loss through the roof) and down in summer (heat gain through the roof). The R-value of a structure determined by analysis assumes that the materials used and the quality of workmanship meet the standards. Poor workmanship and substandard materials used during construction may result in R-values that deviate from predicted values. Therefore, some engineers use a safety factor in their designs based on experience in critical applications. SUMMARY One-dimensional heat transfer through a simple or composite body exposed to convection from both sides to mediums at temperatures T1 and T2 can be expressed as T1 T2 · Q Rtotal (W) where Rtotal is the total thermal resistance between the two mediums. For a plane wall exposed to convection on both sides, the total resistance is expressed as Rtotal Rconv, 1 Rwall Rconv, 2 1 L 1 h1 A kA h2 A This relation can be extended to plane walls that consist of two or more layers by adding an additional resistance for each additional layer. The elementary thermal resistance relations can be expressed as follows: L Conduction resistance (plane wall): Rwall kA ln(r2 /r1) Conduction resistance (cylinder): Rcyl 2 Lk r2 r1 Conduction resistance (sphere): Rsph 4 r1r2 k 1 Convection resistance: Rconv hA Interface resistance: Radiation resistance: Rc 1 Rinterface hc A A 1 Rrad hrad A where hc is the thermal contact conductance, Rc is the thermal contact resistance, and the radiation heat transfer coefficient is defined as 2 hrad (Ts2 Tsurr )(Ts Tsurr) Once the rate of heat transfer is available, the temperature drop across any layer can be determined from · T Q R The thermal resistance concept can also be used to solve steady heat transfer problems involving parallel layers or combined series-parallel arrangements. Adding insulation to a cylindrical pipe or a spherical shell will increase the rate of heat transfer if the outer radius of the insulation is less than the critical radius of insulation, defined as kins h 2kins rcr, sphere h rcr, cylinder The effectiveness of an insulation is often given in terms of its R-value, the thermal resistance of the material per unit surface area, expressed as R-value L k (flat insulation) where L is the thickness and k is the thermal conductivity of the material. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 186 186 HEAT TRANSFER Finned surfaces are commonly used in practice to enhance heat transfer. Fins enhance heat transfer from a surface by exposing a larger surface area to convection. The temperature distribution along the fin for very long fins and for fins with negligible heat transfer at the fin are given by Very long fin: Adiabatic fin tip: T(x) T exhp/kAc Tb T T(x) T cosh a(L x) Tb T cosh aL where a hp/kAc, p is the perimeter, and Ac is the cross sectional area of the fin. The rates of heat transfer for both cases are given to be Very dT · long Q long fin kAc dx fin: Adiabatic dT · fin Q insulated tip kAc dx tip: x0 x0 hpkAc (Tb T) hpkAc (Tb T) tanh aL Fins exposed to convection at their tips can be treated as fins with insulated tips by using the corrected length Lc L Ac/p instead of the actual fin length. The temperature of a fin drops along the fin, and thus the heat transfer from the fin will be less because of the decreasing temperature difference toward the fin tip. To account for the effect of this decrease in temperature on heat transfer, we define fin efficiency as Q· fin Actual heat transfer rate from the fin fin · Ideal heat transfer rate from the fin if Q fin, max the entire fin were at base temperature The performance of the fins is judged on the basis of the enhancement in heat transfer relative to the no-fin case and is expressed in terms of the fin effectiveness fin, defined as Heat transfer rate from the fin of base area Ab Q· fin Q· fin fin · hA (T T ) Heat transfer rate from Q no fin b b the surface of area Ab Here, Ab is the cross-sectional area of the fin at the base and · Q no fin represents the rate of heat transfer from this area if no fins are attached to the surface. The overall effectiveness for a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same surface if there were no fins, Q· total, fin h(Aunfin fin Afin)(Tb T) fin, overall · hAno fin (Tb T) Q total, no fin Fin efficiency and fin effectiveness are related to each other by fin Afin Ab fin Certain multidimensional heat transfer problems involve two surfaces maintained at constant temperatures T1 and T2. The steady rate of heat transfer between these two surfaces is expressed as · Q Sk(T1 T2) where S is the conduction shape factor that has the dimension of length and k is the thermal conductivity of the medium between the surfaces. When the fin efficiency is available, the rate of heat transfer from a fin can be determined from · · Q fin finQ fin, max finhAfin (Tb T) REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers. Handbook of Fundamentals. Atlanta: ASHRAE, 1993. E. Fried. “Thermal Conduction Contribution to Heat Transfer at Contacts.” Thermal Conductivity, vol. 2, ed. R. P. Tye. London: Academic Press, 1969. R. V. Andrews. “Solving Conductive Heat Transfer Problems with Electrical-Analogue Shape Factors.” Chemical Engineering Progress 5 (1955), p. 67. K. A. Gardner. “Efficiency of Extended Surfaces.” Trans. ASME 67 (1945), pp. 621–31. R. Barron. Cryogenic Systems. New York: McGraw-Hill, 1967. 4. L. S. Fletcher. “Recent Developments in Contact Conductance Heat Transfer.” Journal of Heat Transfer 110, no. 4B (1988), pp. 1059–79. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. 8. D. Q. Kern and A. D. Kraus. Extended Surface Heat Transfer. New York: McGraw-Hill, 1972. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 187 187 CHAPTER 3 M. N. Özis,ik. Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. 10. G. P. Peterson. “Thermal Contact Resistance in Waste Heat Recovery Systems.” Proceedings of the 18th ASME/ETCE Hydrocarbon Processing Symposium. Dallas, TX, 1987, pp. 45–51. J. E. Sunderland and K. R. Johnson. “Shape Factors for Heat Conduction through Bodies with Isothermal or Convective Boundary Conditions,” Trans. ASME 10 (1964), pp. 237–41. 13. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West Publishing, 1995. S. Song, M. M. Yovanovich, and F. O. Goodman. “Thermal Gap Conductance of Conforming Surfaces in Contact.” Journal of Heat Transfer 115 (1993), p. 533. PROBLEMS Steady Heat Conduction in Plane Walls 3–1C Consider one-dimensional heat conduction through a cylindrical rod of diameter D and length L. What is the heat transfer area of the rod if (a) the lateral surfaces of the rod are insulated and (b) the top and bottom surfaces of the rod are insulated? 3–2C Consider heat conduction through a plane wall. Does the energy content of the wall change during steady heat conduction? How about during transient conduction? Explain. 3–3C Consider heat conduction through a wall of thickness L and area A. Under what conditions will the temperature distributions in the wall be a straight line? 3–4C What does the thermal resistance of a medium represent? 3–5C How is the combined heat transfer coefficient defined? What convenience does it offer in heat transfer calculations? 3–6C Can we define the convection resistance per unit surface area as the inverse of the convection heat transfer coefficient? 3–7C Why are the convection and the radiation resistances at a surface in parallel instead of being in series? 3–8C Consider a surface of area A at which the convection and radiation heat transfer coefficients are hconv and hrad, respectively. Explain how you would determine (a) the single equivalent heat transfer coefficient, and (b) the equivalent thermal resistance. Assume the medium and the surrounding surfaces are at the same temperature. 3–9C How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 3–10C Consider steady one-dimensional heat transfer · through a multilayer medium. If the rate of heat transfer Q is known, explain how you would determine the temperature drop across each layer. 3–11C Consider steady one-dimensional heat transfer through a plane wall exposed to convection from both sides to environments at known temperatures T1 and T2 with known heat transfer coefficients h1 and h2. Once the rate of heat trans· fer Q has been evaluated, explain how you would determine the temperature of each surface. 3–12C Someone comments that a microwave oven can be viewed as a conventional oven with zero convection resistance at the surface of the food. Is this an accurate statement? 3–13C Consider a window glass consisting of two 4-mmthick glass sheets pressed tightly against each other. Compare the heat transfer rate through this window with that of one consisting of a single 8-mm-thick glass sheet under identical conditions. 3–14C Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surface of the wall is three times that of the inner surface as a result of the winds. On which surface of the wall do you think the temperature will be closer to the surrounding air temperature? Explain. 3–15C The bottom of a pan is made of a 4-mm-thick aluminum layer. In order to increase the rate of heat transfer through the bottom of the pan, someone proposes a design for the bottom that consists of a 3-mm-thick copper layer sandwiched between two 2-mm-thick aluminum layers. Will the new design conduct heat better? Explain. Assume perfect contact between the layers. 3–16C Consider two cold canned drinks, one wrapped in a blanket and the other placed on a table in the same room. Which drink will warm up faster? 3–17 Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick wall whose thermal conductivity is k 0.8 W/m · °C . On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 14°C and 6°C, respectively. Determine the rate of heat loss through the wall on that day. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 188 188 HEAT TRANSFER surfaces of the window to be h1 10 W/m2 · °C and h2 25 W/m2 · °C, and disregard any heat transfer by radiation. Answers: 114 W, 19.2°C 3–20 Repeat Problem 3–19, assuming the space between the two glass layers is evacuated. 3–21 2 mm 3 mm 2 mm Aluminum Copper FIGURE P3–15C 3–18 Consider a 1.2-m-high and 2-m-wide glass window whose thickness is 6 mm and thermal conductivity is k 0.78 W/m · °C. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is 5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 10 W/m2 · °C and h2 25 W/m2 · °C, and disregard any heat transfer by radiation. 3–19 Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k 0.78 W/m · °C) separated by a 12-mm-wide stagnant air space (k 0.026 W/m · °C). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is 5°C. Take the convection heat transfer coefficients on the inner and outer Reconsider Problem 3–19. Using EES (or other) software, plot the rate of heat transfer through the window as a function of the width of air space in the range of 2 mm to 20 mm, assuming pure conduction through the air. Discuss the results. 3–22E Consider an electrically heated brick house (k 0.40 Btu/h · ft · °F) whose walls are 9 ft high and 1 ft thick. Two of the walls of the house are 40 ft long and the others are 30 ft long. The house is maintained at 70°F at all times while the temperature of the outdoors varies. On a certain day, the temperature of the inner surface of the walls is measured to be at 55°F while the average temperature of the outer surface is observed to remain at 45°F during the day for 10 h and at 35°F at night for 14 h. Determine the amount of heat lost from the house that day. Also determine the cost of that heat loss to the homeowner for an electricity price of $0.09/kWh. 9 ft Tin = 70°F 40 ft 30 ft FIGURE P3–22E 3–23 A cylindrical resistor element on a circuit board dissipates 0.15 W of power in an environment at 40°C. The resistor is 1.2 cm long, and has a diameter of 0.3 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-h period, (b) the heat flux on the surface of the resistor, in W/m2, and (c) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of 9 W/m2 · °C. 3–24 Consider a power transistor that dissipates 0.2 W of power in an environment at 30°C. The transistor is 0.4 cm long and has a diameter of 0.5 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-h period, in kWh; (b) the heat flux on the surface of the transistor, in W/m2; and (c) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of 18 W/m2 · °C. Glass 3 12 3 mm Frame FIGURE P3–19 3–25 A 12-cm 18-cm circuit board houses on its surface 100 closely spaced logic chips, each dissipating 0.07 W in an environment at 40°C. The heat transfer from the back surface of the board is negligible. If the heat transfer coefficient on the cen58933_ch03.qxd 9/10/2002 9:00 AM Page 189 189 CHAPTER 3 is filled with fiberglass insulation (k 0.020 Btu/h · ft · °F). Determine (a) the thermal resistance of the wall, and (b) its R-value of insulation in English units. 30°C Power transistor 0.2 W 0.5 cm 0.4 cm FIGURE P3–24 surface of the board is 10 W/m2 · °C, determine (a) the heat flux on the surface of the circuit board, in W/m2; (b) the surface temperature of the chips; and (c) the thermal resistance between the surface of the circuit board and the cooling medium, in °C/W. 3–26 Consider a person standing in a room at 20°C with an exposed surface area of 1.7 m2. The deep body temperature of the human body is 37°C, and the thermal conductivity of the human tissue near the skin is about 0.3 W/m · °C. The body is losing heat at a rate of 150 W by natural convection and radiation to the surroundings. Taking the body temperature 0.5 cm beneath the skin to be 37°C, determine the skin temperature of Answer: 35.5° C the person. 3–27 Water is boiling in a 25-cm-diameter aluminum pan (k 237 W/m · °C) at 95°C. Heat is transferred steadily to the boiling water in the pan through its 0.5-cm-thick flat bottom at a rate of 800 W. If the inner surface temperature of the bottom of the pan is 108°C, determine (a) the boiling heat transfer coefficient on the inner surface of the pan, and (b) the outer surface temperature of the bottom of the pan. 3–28E A wall is constructed of two layers of 0.5-in-thick sheetrock (k 0.10 Btu/h · ft · °F), which is a plasterboard made of two layers of heavy paper separated by a layer of gypsum, placed 5 in. apart. The space between the sheetrocks Fiberglass insulation Sheetrock 0.5 in. 5 in. FIGURE P3–28E 0.5 in. 3–29 The roof of a house consists of a 3–cm-thick concrete slab (k 2 W/m · °C) that is 15 m wide and 20 m long. The convection heat transfer coefficients on the inner and outer surfaces of the roof are 5 and 12 W/m2 · °C, respectively. On a clear winter night, the ambient air is reported to be at 10°C, while the night sky temperature is 100 K. The house and the interior surfaces of the wall are maintained at a constant temperature of 20°C. The emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfers, determine the rate of heat transfer through the roof, and the inner surface temperature of the roof. If the house is heated by a furnace burning natural gas with an efficiency of 80 percent, and the price of natural gas is $0.60/therm (1 therm 105,500 kJ of energy content), determine the money lost through the roof that night during a 14-h period. Tsky = 100 K Tair = 10°C Concrete roof 15 cm 20 m 15 m Tin = 20°C FIGURE P3–29 3–30 A 2-m 1.5-m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the outer surface of this section is measured to be 80°C. The temperature of the furnace room is 30°C, and the combined convection and radiation heat transfer coefficient at the surface of the outer furnace is 10 W/m2 · °C. It is proposed to insulate this section of the furnace wall with glass wool insulation (k 0.038 W/m · °C) in order to reduce the heat loss by 90 percent. Assuming the outer surface temperature of the metal section still remains at about 80°C, determine the thickness of the insulation that needs to be used. The furnace operates continuously and has an efficiency of 78 percent. The price of the natural gas is $0.55/therm (1 therm 105,500 kJ of energy content). If the installation of the insulation will cost $250 for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves. 3–31 Repeat Problem. 3–30 for expanded perlite insulation assuming conductivity is k 0.052 W/m · °C. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 190 190 HEAT TRANSFER 3–32 Reconsider Problem 3–30. Using EES (or other) software, investigate the effect of thermal conductivity on the required insulation thickness. Plot the thickness of insulation as a function of the thermal conductivity of the insulation in the range of 0.02 W/m · °C to 0.08 W/m · °C, and discuss the results. Sheet metal Kitchen air 25°C 3–33E Consider a house whose walls are 12 ft high and 40 ft long. Two of the walls of the house have no windows, while each of the other two walls has four windows made of 0.25-in.thick glass (k 0.45 Btu/h · ft · °F), 3 ft 5 ft in size. The walls are certified to have an R-value of 19 (i.e., an L/k value of 19 h · ft2 · °F/Btu). Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 2 and 4 Btu/h · ft2 · °F, respectively, determine the ratio of the heat transfer through the walls with and without windows. Attic space 12 ft Refrigerated space 3°C Insulation 10°C 1 mm L 1 mm FIGURE P3–35 4 W/m2 · °C and 9 W/m2 · °C, respectively. The kitchen temperature averages 25°C. It is observed that condensation occurs on the outer surfaces of the refrigerator when the temperature of the outer surface drops to 20°C. Determine the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces. 3–36 40 ft 40 ft Windows FIGURE P3–33E 3–34 Consider a house that has a 10-m 20-m base and a 4-m-high wall. All four walls of the house have an R-value of 2.31 m2 · °C/W. The two 10-m 4-m walls have no windows. The third wall has five windows made of 0.5-cm-thick glass (k 0.78 W/m · °C), 1.2 m 1.8 m in size. The fourth wall has the same size and number of windows, but they are doublepaned with a 1.5-cm-thick stagnant air space (k 0.026 W/m · °C) enclosed between two 0.5-cm-thick glass layers. The thermostat in the house is set at 22°C and the average temperature outside at that location is 8°C during the seven-monthlong heating season. Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 7 and 15 W/m2 · °C, respectively, determine the average rate of heat transfer through each wall. If the house is electrically heated and the price of electricity is $0.08/kWh, determine the amount of money this household will save per heating season by converting the single-pane windows to double-pane windows. 3–35 The wall of a refrigerator is constructed of fiberglass insulation (k 0.035 W/m · °C) sandwiched between two layers of 1-mm-thick sheet metal (k 15.1 W/m · °C). The refrigerated space is maintained at 3°C, and the average heat transfer coefficients at the inner and outer surfaces of the wall are Reconsider Problem 3–35. Using EES (or other) software, investigate the effects of the thermal conductivities of the insulation material and the sheet metal on the thickness of the insulation. Let the thermal conductivity vary from 0.02 W/m · °C to 0.08 W/m · °C for insulation and 10 W/m · °C to 400 W/m · °C for sheet metal. Plot the thickness of the insulation as the functions of the thermal conductivities of the insulation and the sheet metal, and discuss the results. 3–37 Heat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy layers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the copper (k 386 W/m · °C) and epoxy (k 0.26 W/m · °C) layers. Also determine the effective thermal conductivity of the board. Answers: 0.8 percent, 99.2 percent, and 29.9 W/m · °C 3–38E A 0.03-in-thick copper plate (k 223 Btu/h · ft · °F) is sandwiched between two 0.1-in.-thick epoxy boards (k 0.15 Btu/h · ft · °F) that are 7 in. 9 in. in size. Determine the effective thermal conductivity of the board along its 9-in.-long side. What fraction of the heat conducted along that side is conducted through copper? Thermal Contact Resistance 3–39C What is thermal contact resistance? How is it related to thermal contact conductance? 3–40C Will the thermal contact resistance be greater for smooth or rough plain surfaces? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 191 191 CHAPTER 3 Copper plate Transistor Plexiglas cover 9 in. Epoxy boards Copper plate 85°C 15°C 7 in. 0.03 in. FIGURE P3–38E 1.2 cm 3–41C A wall consists of two layers of insulation pressed against each other. Do we need to be concerned about the thermal contact resistance at the interface in a heat transfer analysis or can we just ignore it? 3–42C A plate consists of two thin metal layers pressed against each other. Do we need to be concerned about the thermal contact resistance at the interface in a heat transfer analysis or can we just ignore it? 3–43C Consider two surfaces pressed against each other. Now the air at the interface is evacuated. Will the thermal contact resistance at the interface increase or decrease as a result? 3–44C Explain how the thermal contact resistance can be minimized. 3–45 The thermal contact conductance at the interface of two 1-cm-thick copper plates is measured to be 18,000 W/m2 · °C. Determine the thickness of the copper plate whose thermal resistance is equal to the thermal resistance of the interface between the plates. 3–46 Six identical power transistors with aluminum casing are attached on one side of a 1.2-cm-thick 20-cm 30-cm copper plate (k 386 W/m · °C) by screws that exert an average pressure of 10 MPa. The base area of each transistor is 9 cm2, and each transistor is placed at the center of a 10-cm 10-cm section of the plate. The interface roughness is estimated to be about 1.4 m. All transistors are covered by a thick Plexiglas layer, which is a poor conductor of heat, and thus all the heat generated at the junction of the transistor must be dissipated to the ambient at 15°C through the back surface of the copper plate. The combined convection/radiation heat transfer coefficient at the back surface can be taken to be 30 W/m2 · °C. If the case temperature of the transistor is not to exceed 85°C, determine the maximum power each transistor can dissipate safely, and the temperature jump at the case-plate interface. FIGURE P3–46 3–47 Two 5-cm-diameter, 15–cm-long aluminum bars (k 176 W/m · °C) with ground surfaces are pressed against each other with a pressure of 20 atm. The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the two-bar system are maintained at temperatures of 150°C and 20°C, respectively, determine (a) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop Answers: (a) 142.4 W, (b) 6.4°C at the interface. 3–48 A 1-mm-thick copper plate (k 386 W/m · °C) is sandwiched between two 5-mm-thick epoxy boards (k 0.26 W/m · °C) that are 15 cm 20 cm in size. If the thermal contact conductance on both sides of the copper plate is estimated to be 6000 W/m · °C, determine the error involved in the total thermal resistance of the plate if the thermal contact conductances are ignored. Copper Epoxy Epoxy hc Heat flow 5 mm 5 mm FIGURE P3–48 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 192 192 HEAT TRANSFER Generalized Thermal Resistance Networks 3–49C When plotting the thermal resistance network associated with a heat transfer problem, explain when two resistances are in series and when they are in parallel. 3–50C The thermal resistance networks can also be used approximately for multidimensional problems. For what kind of multidimensional problems will the thermal resistance approach give adequate results? 3–51C What are the two approaches used in the development of the thermal resistance network for two-dimensional problems? 3–52 A 4-m-high and 6-m-wide wall consists of a long 18-cm 30-cm cross section of horizontal bricks (k 0.72 W/m · °C) separated by 3-cm-thick plaster layers (k 0.22 W/m · °C). There are also 2-cm-thick plaster layers on each side of the wall, and a 2-cm-thick rigid foam (k 0.026 W/m · °C) on the inner side of the wall. The indoor and the outdoor temperatures are 22°C and 4°C, and the convection heat transfer coefficients on the inner and the outer sides are h1 10 W/m2 · °C and h2 20 W/m2 · °C, respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall. Foam Plaster 1.5 cm Brick 30 cm instead. The manganese steel nails (k 50 W/m · °C) are 10 cm long and have a diameter of 0.4 cm. A total of 50 nails are used to connect the two studs, which are mounted to the wall such that the nails cross the wall. The temperature difference between the inner and outer surfaces of the wall is 8°C. Assuming the thermal contact resistance between the two layers to be negligible, determine the rate of heat transfer (a) through a solid stud and (b) through a stud pair of equal length and width nailed to each other. (c) Also determine the effective conductivity of the nailed stud pair. 3–55 A 12-m-long and 5-m-high wall is constructed of two layers of 1-cm-thick sheetrock (k 0.17 W/m · °C) spaced 12 cm by wood studs (k 0.11 W/m · °C) whose cross section is 12 cm 5 cm. The studs are placed vertically 60 cm apart, and the space between them is filled with fiberglass insulation (k 0.034 W/m · °C). The house is maintained at 20°C and the ambient temperature outside is 5°C. Taking the heat transfer coefficients at the inner and outer surfaces of the house to be 8.3 and 34 W/m2 · °C, respectively, determine (a) the thermal resistance of the wall considering a representative section of it and (b) the rate of heat transfer through the wall. 3–56E A 10-in.-thick, 30-ft-long, and 10-ft-high wall is to be constructed using 9-in.-long solid bricks (k 0.40 Btu/h · ft · °F) of cross section 7 in. 7 in., or identical size bricks with nine square air holes (k 0.015 Btu/h · ft · °F) that are 9 in. long and have a cross section of 1.5 in. 1.5 in. There is a 0.5-in.-thick plaster layer (k 0.10 Btu/h · ft · °F) between two adjacent bricks on all four sides and on both sides of the wall. The house is maintained at 80°F and the ambient temperature outside is 30°F. Taking the heat transfer coefficients at the inner and outer surfaces of the wall to be 1.5 and 4 Btu/h · ft2 · °F, respectively, determine the rate of heat transfer through the wall constructed of (a) solid bricks and (b) bricks with air holes. 1.5 cm 2 2 18 cm Air channels 1.5 in. × 1.5 in. × 9 in. 2 FIGURE P3–52 0.5 in. 7 in. 3–53 Reconsider Problem 3–52. Using EES (or other) software, plot the rate of heat transfer through the wall as a function of the thickness of the rigid foam in the range of 1 cm to 10 cm. Discuss the results. 3–54 A 10-cm-thick wall is to be constructed with 2.5-mlong wood studs (k 0.11 W/m · °C) that have a cross section of 10 cm 10 cm. At some point the builder ran out of those studs and started using pairs of 2.5-m-long wood studs that have a cross section of 5 cm 10 cm nailed to each other 0.5 in. Plaster Brick 0.5 in. 9 in. FIGURE P3–56E 0.5 in. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 193 193 CHAPTER 3 3–57 Consider a 5-m-high, 8-m-long, and 0.22-m-thick wall whose representative cross section is as given in the figure. The thermal conductivities of various materials used, in W/m · °C, are kA kF 2, kB 8, kC 20, kD 15, and kE 35. The left and right surfaces of the wall are maintained at uniform temperatures of 300°C and 100°C, respectively. Assuming heat transfer through the wall to be one-dimensional, determine (a) the rate of heat transfer through the wall; (b) the temperature at the point where the sections B, D, and E meet; and (c) the temperature drop across the section F. Disregard any contact resistances at the interfaces. 100°C 300°C D C A 4 cm F 6 cm B 4 cm 1 cm What would your response be if the jacket is made of a single layer of 0.5-mm-thick synthetic fabric? What should be the thickness of a wool fabric (k 0.035 W/m · °C) if the person is to achieve the same level of thermal comfort wearing a thick wool coat instead of a five-layer ski jacket? 3–60 Repeat Problem 3–59 assuming the layers of the jacket are made of cotton fabric (k 0.06 W/m · °C). 3–61 A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20-cm-thick concrete walls and ceiling (k 0.9 W/m · °C). The kiln is maintained at 40°C by injecting hot steam into it. The two ends of the kiln, 4 m 5 m in size, are made of a 3-mm-thick sheet metal covered with 2-cm-thick Styrofoam (k 0.033 W/m · °C). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are 3000 W/m2 · °C and 25 W/m2 · °C, respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at 4°C. · Q Tout = – 4°C E C 4 cm 6 cm 5 cm 10 cm 40 m 6 cm 8m FIGURE P3–57 4m Tin = 40°C 20 cm 3–58 Repeat Problem 3–57 assuming that the thermal contact resistance at the interfaces D-F and E-F is 0.00012 m2 · °C/W. 3–59 Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made of five layers of 0.1-mm-thick synthetic fabric (k 0.13 W/m · °C) with 1.5-mm-thick air space (k 0.026 W/m · °C) between the layers. Assuming the inner surface temperature of the jacket to be 28°C and the surface area to be 1.1 m2, determine the rate of heat loss through the jacket when the temperature of the outdoors is 5°C and the heat transfer coefficient at the outer surface is 25 W/m2 · °C. Multilayered ski jacket FIGURE P3–59 5m FIGURE P3–61 3–62 Reconsider Problem 3–61. Using EES (or other) software, investigate the effects of the thickness of the wall and the convection heat transfer coefficient on the outer surface of the rate of heat loss from the kiln. Let the thickness vary from 10 cm to 30 cm and the convection heat transfer coefficient from 5 W/m2 · °C to 50 W/m2 · °C. Plot the rate of heat transfer as functions of wall thickness and the convection heat transfer coefficient, and discuss the results. 3–63E Consider a 6-in. 8-in. epoxy glass laminate (k 0.10 Btu/h · ft · °F) whose thickness is 0.05 in. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k 223 Btu/h · ft · °F) of 0.02 in. diameter are to be planted throughout the board, with a center-to-center distance of 0.06 in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification. Answer: 0.00064 h · °F/Btu cen58933_ch03.qxd 9/10/2002 9:00 AM Page 194 194 HEAT TRANSFER 0.02 in. transfer coefficients at the inner and the outer surfaces of the tank are 80 W/m2 · °C and 10 W/m2 · °C, respectively. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. 0.06 in. Copper filling 3–68 Steam at 320°C flows in a stainless steel pipe (k 15 W/m · °C) whose inner and outer diameters are 5 cm and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k 0.038 W/m · °C). Heat is lost to the surroundings at 5°C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of 15 W/m2 · °C. Taking the heat transfer coefficient inside the pipe to be 80 W/m2 · °C, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation. Epoxy board FIGURE P3–63E 3–69 Heat Conduction in Cylinders and Spheres 3–64C What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not? 3–65C Consider a short cylinder whose top and bottom surfaces are insulated. The cylinder is initially at a uniform temperature Ti and is subjected to convection from its side surface to a medium at temperature T, with a heat transfer coefficient of h. Is the heat transfer in this short cylinder one- or twodimensional? Explain. 3–66C Can the thermal resistance concept be used for a solid cylinder or sphere in steady operation? Explain. 3–67 A 5-m-internal-diameter spherical tank made of 1.5-cm-thick stainless steel (k 15 W/m · °C) is used to store iced water at 0°C. The tank is located in a room whose temperature is 30°C. The walls of the room are also at 30°C. The outer surface of the tank is black (emissivity 1), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat Reconsider Problem 3–68. Using EES (or other) software, investigate the effect of the thickness of the insulation on the rate of heat loss from the steam and the temperature drop across the insulation layer. Let the insulation thickness vary from 1 cm to 10 cm. Plot the rate of heat loss and the temperature drop as a function of insulation thickness, and discuss the results. 3–70 A 50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open space at 15°C. The average temperature of the outer surface of the pipe is measured to be 150°C. If the combined heat transfer coefficient on the outer surface of the pipe is 20 W/m2 · °C, determine (a) the rate of heat loss from the steam pipe, (b) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $0.52/therm (1 therm 105,500 kJ), and (c) the thickness of fiberglass insulation (k 0.035 W/m · °C) needed in order to save 90 percent of the heat lost. Assume the pipe temperature to remain constant at 150°C. Tair = 15°C 150°C Troom = 30°C Iced water Steam 50 m Di = 5 m Tin = 0°C FIGURE P3–67 1.5 cm Fiberglass insulation FIGURE P3–70 3–71 Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and maintains the hot water at 55°C. The tank is located in a small room whose average temperature is cen58933_ch03.qxd 9/10/2002 9:00 AM Page 195 195 CHAPTER 3 3°C 3 cm 40 cm 12.5 cm 27°C 2m Tw = 55°C Tair = 25°C Foam insulation Water heater FIGURE P3–71 27°C, and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and 12 W/m2 · °C, respectively. The tank is placed in another 46-cm-diameter sheet metal tank of negligible thickness, and the space between the two tanks is filled with foam insulation (k 0.03 W/m · °C). The thermal resistances of the water tank and the outer thin sheet metal shell are very small and can be neglected. The price of electricity is $0.08/kWh, and the home owner pays $280 a year for water heating. Determine the fraction of the hot water energy cost of this household that is due to the heat loss from the tank. Hot water tank insulation kits consisting of 3-cm-thick fiberglass insulation (k 0.035 W/m · °C) large enough to wrap the entire tank are available in the market for about $30. If such an insulation is installed on this water tank by the home owner himself, how long will it take for this additional insulation to Answers: 17.5 percent, 1.5 years pay for itself? 3–72 Reconsider Problem 3–71. Using EES (or other) software, plot the fraction of energy cost of hot water due to the heat loss from the tank as a function of the hot water temperature in the range of 40°C to 90°C. Discuss the results. 3–73 Consider a cold aluminum canned drink that is initially at a uniform temperature of 3°C. The can is 12.5 cm high and has a diameter of 6 cm. If the combined convection/radiation heat transfer coefficient between the can and the surrounding air at 25°C is 10 W/m2 · °C, determine how long it will take for the average temperature of the drink to rise to 10°C. In an effort to slow down the warming of the cold drink, a person puts the can in a perfectly fitting 1-cm-thick cylindrical rubber insulation (k 0.13 W/m · °C). Now how long will it take for the average temperature of the drink to rise to 10°C? Assume the top of the can is not covered. 6 cm FIGURE P3–73 3–74 Repeat Problem 3–73, assuming a thermal contact resistance of 0.00008 m2 · °C/W between the can and the insulation. 3–75E Steam at 450°F is flowing through a steel pipe (k 8.7 Btu/h · ft · °F) whose inner and outer diameters are 3.5 in. and 4.0 in., respectively, in an environment at 55°F. The pipe is insulated with 2-in.-thick fiberglass insulation (k 0.020 Btu/h · ft · °F). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and 5 Btu/h · ft2 · °F, respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations? Steel pipe Steam 450°F Insulation FIGURE P3–75E 3–76 Hot water at an average temperature of 90°C is flowing through a 15-m section of a cast iron pipe (k 52 W/m · °C) whose inner and outer diameters are 4 cm and 4.6 cm, respectively. The outer surface of the pipe, whose emissivity is 0.7, is exposed to the cold air at 10°C in the basement, with a heat transfer coefficient of 15 W/m2 · °C. The heat transfer coefficient at the inner surface of the pipe is 120 W/m2 · °C. Taking the walls of the basement to be at 10°C also, determine the rate of heat loss from the hot water. Also, determine the average cen58933_ch03.qxd 9/10/2002 9:00 AM Page 196 196 HEAT TRANSFER velocity of the water in the pipe if the temperature of the water drops by 3°C as it passes through the basement. 3–77 Repeat Problem 3–76 for a pipe made of copper (k 386 W/m · °C) instead of cast iron. 3–78E Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper pipes (k 223 Btu/h · ft · °F) of inner diameter 0.4 in. and outer diameter 0.6 in. at an average temperature of 70°F. The heat of vaporization of water at 100°F is 1037 Btu/lbm. The heat transfer coefficients are 1500 Btu/h · ft2 · °F on the steam side and 35 Btu/h · ft2 · °F on the water side. Determine the length of the tube required to conAnswer: 1148 ft dense steam at a rate of 120 lbm/h. N2 vapor Tair = 15°C 1 atm Liquid N2 –196°C Insulation FIGURE P3–81 Steam, 100°F 120 lbm/h Cooling water Liquid water FIGURE P3–78E Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to ambient air at 15°C, with a combined convection and radiation heat transfer coefficient of 35 W/m2 · °C. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is (a) not insulated, (b) insulated with 5-cm-thick fiberglass insulation (k 0.035 W/m · °C), and (c) insulated with 2-cm-thick superinsulation which has an effective thermal conductivity of 0.00005 W/m · °C. 3–82 Repeat Problem 3–81 for liquid oxygen, which has a boiling temperature of 183°C, a heat of vaporization of 213 kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure. 3–79E Repeat Problem 3–78E, assuming that a 0.01-in.-thick layer of mineral deposit (k 0.5 Btu/h · ft · °F) has formed on the inner surface of the pipe. Critical Radius of Insulation 3–80 3–84C A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken off. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature? Reconsider Problem 3–78E. Using EES (or other) software, investigate the effects of the thermal conductivity of the pipe material and the outer diameter of the pipe on the length of the tube required. Let the thermal conductivity vary from 10 Btu/h · ft · °F to 400 Btu/h · ft · °F and the outer diameter from 0.5 in. to 1.0 in. Plot the length of the tube as functions of pipe conductivity and the outer pipe diameter, and discuss the results. 3–81 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm pressure) is 196°C. Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm. 3–83C What is the critical radius of insulation? How is it defined for a cylindrical layer? 3–85C A pipe is insulated to reduce the heat loss from it. However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right? 3–86C Consider a pipe at a constant temperature whose radius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid? 3–87C Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 197 197 CHAPTER 3 3–88 A 2-mm-diameter and 10-m-long electric wire is tightly wrapped with a 1-mm-thick plastic cover whose thermal conductivity is k 0.15 W/m · °C. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T 30°C with a heat transfer coefficient of h 24 W/m2 · °C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature. Tair = 30°C Electrical wire Insulation 10 m FIGURE P3–88 3–89E A 0.083-in.-diameter electrical wire at 115°F is covered by 0.02-in.-thick plastic insulation (k 0.075 Btu/h · ft · °F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h · ft2 · °F. Determine if the plastic insulation on the wire will increase or decrease heat transfer from the wire. Answer: It helps 3–90E Repeat Problem 3–89E, assuming a thermal contact resistance of 0.001 h · ft2 · °F/Btu at the interface of the wire and the insulation. 3–91 A 5-mm-diameter spherical ball at 50°C is covered by a 1-mm-thick plastic insulation (k 0.13 W/m · °C). The ball is exposed to a medium at 15°C, with a combined convection and radiation heat transfer coefficient of 20 W/m2 · °C. Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball. Plastic insulation 5 mm 1 mm FIGURE P3–91 3–94C What is the difference between the fin effectiveness and the fin efficiency? 3–95C The fins attached to a surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins? 3–96C Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface. 3–97C How does the overall effectiveness of a finned surface differ from the effectiveness of a single fin? 3–98C Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? 3–99C Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes? 3–100C Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain. 3–101C The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip? 3–102C Does the (a) efficiency and (b) effectiveness of a fin increase or decrease as the fin length is increased? 3–103C Two pin fins are identical, except that the diameter of one of them is twice the diameter of the other. For which fin will the (a) fin effectiveness and (b) fin efficiency be higher? Explain. 3–104C Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin will the (a) fin effectiveness and (b) fin efficiency be higher? Explain. 3–92 3–105C Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface will the (a) fin effectiveness and (b) fin efficiency be higher? Explain. Heat Transfer from Finned Surfaces 3–106 Obtain a relation for the fin efficiency for a fin of constant cross-sectional area Ac, perimeter p, length L, and thermal conductivity k exposed to convection to a medium at T with a heat transfer coefficient h. Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly T. Take the temperature of the fin at the base to be Tb and neglect heat Reconsider Problem 3–91. Using EES (or other) software, plot the rate of heat transfer from the ball as a function of the plastic insulation thickness in the range of 0.5 mm to 20 mm. Discuss the results. 3–93C What is the reason for the widespread use of fins on surfaces? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 198 198 HEAT TRANSFER 2.5 cm h, T Tb k T = 25°C 3 cm D Ab = Ac 180°C p = πD, Ac = πD2/4 1 mm FIGURE P3–106 3 mm transfer from the fin tips. Simplify the relation for (a) a circular fin of diameter D and (b) rectangular fins of thickness t. 3–107 The case-to-ambient thermal resistance of a power transistor that has a maximum power rating of 15 W is given to be 25°C/W. If the case temperature of the transistor is not to exceed 80°C, determine the power at which this transistor can be operated safely in an environment at 40°C. 3–108 A 40-W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table 3–4. Select a heat sink that will allow the case temperature of the transistor not to exceed 90° in the ambient air at 20°. FIGURE P3–110 surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is 3 Btu/h · ft2 · °F, determine the temperature difference across the exposed surface of Answer: 124.6°F the spoon handle. State your assumptions. Spoon Tair = 20°C Tair = 75°F 90°C 7 in. 40 W Boiling water 200°F FIGURE P3–111E FIGURE P3–108 3–109 A 30-W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table 3–4. Select a heat sink that will allow the case temperature of the transistor not to exceed 80°C in the ambient air at 35°C. 3–110 Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 180°C. Circular aluminum alloy 2024-T6 fins (k 186 W/m · °C) of outer diameter 6 cm and constant thickness 1 mm are attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at T 25°C, with a heat transfer coefficient of 40 W/m2 · °C. Determine the increase in heat transfer from the tube per meter of its length as Answer: 2639 W a result of adding fins. 3–111E Consider a stainless steel spoon (k 8.7 Btu/h · ft · °F) partially immersed in boiling water at 200°F in a kitchen at 75°F. The handle of the spoon has a cross section of 0.08 in. 0.5 in., and extends 7 in. in the air from the free 3–112E Repeat Problem 3–111 for a silver spoon (k 247 Btu/h · ft · °F). 3–113E Reconsider Problem 3–111E. Using EES (or other) software, investigate the effects of the thermal conductivity of the spoon material and the length of its extension in the air on the temperature difference across the exposed surface of the spoon handle. Let the thermal conductivity vary from 5 Btu/h · ft · °F to 225 Btu/h · ft · °F and the length from 5 in. to 12 in. Plot the temperature difference as the functions of thermal conductivity and length, and discuss the results. 3–114 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.04 W. The board is impregnated with copper fillings and has an effective thermal conductivity of 20 W/m · °C. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at 40°C, with a heat transfer coefficient of 50 W/m2 · °C. (a) Determine the temperatures on the two sides of the circuit board. (b) Now a 0.2-cm-thick, 12-cm-high, and cen58933_ch03.qxd 9/10/2002 9:00 AM Page 199 199 CHAPTER 3 18-cm-long aluminum plate (k 237 W/m · °C) with 864 2-cm-long aluminum pin fins of diameter 0.25 cm is attached to the back side of the circuit board with a 0.02-cm-thick epoxy adhesive (k 1.8 W/m · °C). Determine the new temperatures on the two sides of the circuit board. 10 cm 9.2 cm Tair = 8°C 3–115 Repeat Problem 3–114 using a copper plate with copper fins (k 386 W/m · °C) instead of aluminum ones. 1 cm 1 cm 3–116 A hot surface at 100°C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter aluminum pin fins (k 237 W/m · °C) to it, with a center-to-center distance of 0.6 cm. The temperature of the surrounding medium is 30°C, and the heat transfer coefficient on the surfaces is 35 W/m2 · °C. Determine the rate of heat transfer from the surface for a 1-m 1-m section of the plate. Also determine the overall effectiveness of the fins. 20 cm Steam 200°C FIGURE P3–119 3 cm 0.6 cm 0.25 cm Heat Transfer in Common Configurations 3-120C What is a conduction shape factor? How is it related to the thermal resistance? 3-121C What is the value of conduction shape factors in engineering? FIGURE P3–116 3-122 A 20-m-long and 8-cm-diameter hot water pipe of a district heating system is buried in the soil 80 cm below the ground surface. The outer surface temperature of the pipe is 60°C. Taking the surface temperature of the earth to be 5°C and the thermal conductivity of the soil at that location to be 0.9 W/m · °C, determine the rate of heat loss from the pipe. 3–117 Repeat Problem 3–116 using copper fins (k 386 W/m · °C) instead of aluminum ones. 5°C 3–118 Reconsider Problem 3–116. Using EES (or other) software, investigate the effect of the center-to-center distance of the fins on the rate of heat transfer from the surface and the overall effectiveness of the fins. Let the center-to-center distance vary from 0.4 cm to 2.0 cm. Plot the rate of heat transfer and the overall effectiveness as a function of the center-to-center distance, and discuss the results. 3–119 Two 3-m-long and 0.4-cm-thick cast iron (k 52 W/m · °C) steam pipes of outer diameter 10 cm are connected to each other through two 1-cm-thick flanges of outer diameter 20 cm. The steam flows inside the pipe at an average temperature of 200°C with a heat transfer coefficient of 180 W/m2 · °C. The outer surface of the pipe is exposed to an ambient at 12°C, with a heat transfer coefficient of 25 W/m2 · °C. (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes? 80 cm 60°C D = 8 cm 20 m FIGURE P3–122 3–123 Reconsider Problem 3–122. Using EES (or other) software, plot the rate of heat loss from the pipe as a function of the burial depth in the range of 20 cm to 2.0 m. Discuss the results. 3-124 Hot and cold water pipes 8 m long run parallel to each other in a thick concrete layer. The diameters of both pipes are 5 cm, and the distance between the centerlines of the pipes is 40 cm. The surface temperatures of the hot and cold pipes are 60°C and 15°C, respectively. Taking the thermal conductivity of the concrete to be k 0.75 W/m · °C, determine the rate of heat transfer between the pipes. Answer: 306 W cen58933_ch03.qxd 9/10/2002 9:00 AM Page 200 200 HEAT TRANSFER 3–125 Reconsider Problem 3–124. Using EES (or other) software, plot the rate of heat transfer between the pipes as a function of the distance between the centerlines of the pipes in the range of 10 cm to 1.0 m. Discuss the results. 3-126E A row of 3-ft-long and 1-in.-diameter used uranium fuel rods that are still radioactive are buried in the ground parallel to each other with a center-to-center distance of 8 in. at a depth 15 ft from the ground surface at a location where the thermal conductivity of the soil is 0.6 Btu/h · ft · °F. If the surface temperature of the rods and the ground are 350°F and 60°F, respectively, determine the rate of heat transfer from the fuel rods to the atmosphere through the soil. 60°F 350°F 3f t 1i n. 15 ft 8 in. 8 in. 8 in. FIGURE P3–126 3-127 Hot water at an average temperature of 60°C and an average velocity of 0.6 m/s is flowing through a 5-m section of a thin-walled hot water pipe that has an outer diameter of 2.5 cm. The pipe passes through the center of a 14-cm-thick wall filled with fiberglass insulation (k 0.035 W/m · °C). If the surfaces of the wall are at 18°C, determine (a) the rate of heat transfer from the pipe to the air in the rooms and (b) the temperature drop of the hot water as it flows through this Answers: 23.5 W, 0.02°C 5-m-long section of the wall. Hot water pipe 8°C 0°C 3m 80°C 20 m FIGURE P3–128 (k 1.5 W/m · °C) vertically for 3 m, and continues horizontally at this depth for 20 m more before it enters the next building. The first section of the pipe is exposed to the ambient air at 8°C, with a heat transfer coefficient of 22 W/m2 · °C. If the surface of the ground is covered with snow at 0°C, determine (a) the total rate of heat loss from the hot water and (b) the temperature drop of the hot water as it flows through this 25-m-long section of the pipe. 3-129 Consider a house with a flat roof whose outer dimensions are 12 m 12 m. The outer walls of the house are 6 m high. The walls and the roof of the house are made of 20-cmthick concrete (k 0.75 W/m · °C). The temperatures of the inner and outer surfaces of the house are 15°C and 3°C, respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a 12 m 12 m surface and the walls as 6 m 12 m surfaces for simplicity? 3-130 Consider a 10-m-long thick-walled concrete duct (k 0.75 W/m · °C) of square cross-section. The outer dimensions of the duct are 20 cm 20 cm, and the thickness of the duct wall is 2 cm. If the inner and outer surfaces of the duct are at 100°C and 15°C, respectively, determine the rate of heat transAnswer: 22.9 kW fer through the walls of the duct. 15°C Hot water 18°C 5m 2.5 cm 60°C Wall 100°C 10 m 16 cm 20 cm FIGURE P3–127 3-128 Hot water at an average temperature of 80°C and an average velocity of 1.5 m/s is flowing through a 25-m section of a pipe that has an outer diameter of 5 cm. The pipe extends 2 m in the ambient air above the ground, dips into the ground FIGURE P3–130 3-131 A 3-m-diameter spherical tank containing some radioactive material is buried in the ground (k 1.4 W/m · °C). The distance between the top surface of the tank and the ground surface is 4 m. If the surface temperatures of the tank and the cen58933_ch03.qxd 9/10/2002 9:00 AM Page 201 201 CHAPTER 3 ground are 140°C and 15°C, respectively, determine the rate of heat transfer from the tank. 4b 3–132 Reconsider Problem 3–131. Using EES (or other) software, plot the rate of heat transfer from the tank as a function of the tank diameter in the range of 0.5 m to 5.0 m. Discuss the results. 3-133 Hot water at an average temperature of 85°C passes through a row of eight parallel pipes that are 4 m long and have an outer diameter of 3 cm, located vertically in the middle of a concrete wall (k 0.75 W/m · °C) that is 4 m high, 8 m long, and 15 cm thick. If the surfaces of the concrete walls are exposed to a medium at 32°C, with a heat transfer coefficient of 12 W/m2 · °C, determine the rate of heat loss from the hot water and the surface temperature of the wall. Special Topics: Heat Transfer through the Walls and Roofs 3–134C What is the R-value of a wall? How does it differ from the unit thermal resistance of the wall? How is it related to the U-factor? 3–135C What is effective emissivity for a plane-parallel air space? How is it determined? How is radiation heat transfer through the air space determined when the effective emissivity is known? 3–136C The unit thermal resistances (R-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3–9 to be 0.22 m2 · °C/W, which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain. 6 3 4a 5 2 1 FIGURE P3–139 3–141E Determine the winter R-value and the U-factor of a masonry cavity wall that is built around 4-in.-thick concrete blocks made of lightweight aggregate. The outside is finished with 4-in. face brick with 21 -in. cement mortar between the bricks and concrete blocks. The inside finish consists of 21 -in. gypsum wallboard separated from the concrete block by 43 -in.thick (1-in. by 3-in. nominal) vertical furring whose center-tocenter distance is 16 in. Neither side of the 34 -in.-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the R-value of the air space, the temperature difference across it can be taken to be 30°F with a mean air temperature of 50°F. The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent. 3–137C What is a radiant barrier? What kind of materials are suitable for use as radiant barriers? Is it worthwhile to use radiant barriers in the attics of homes? 5b 3–138C Consider a house whose attic space is ventilated effectively so that the air temperature in the attic is the same as the ambient air temperature at all times. Will the roof still have any effect on heat transfer through the ceiling? Explain. 3–139 Determine the summer R-value and the U-factor of a wood frame wall that is built around 38-mm 140-mm wood studs with a center-to-center distance of 400 mm. The 140mm-wide cavity between the studs is filled with mineral fiber batt insulation. The inside is finished with 13-mm gypsum wallboard and the outside with 13-mm wood fiberboard and 13-mm 200-mm wood bevel lapped siding. The insulated cavity constitutes 80 percent of the heat transmission area, while the studs, headers, plates, and sills constitute 20 percent. Answers: 3.213 m2 · °C/W, 0.311 W/m2 · °C 3–140 The 13-mm-thick wood fiberboard sheathing of the wood stud wall in Problem 3–139 is replaced by a 25-mmthick rigid foam insulation. Determine the percent increase in the R-value of the wall as a result. 4 2 5a 6 7 3 1 FIGURE P3–141E 3–142 Consider a flat ceiling that is built around 38-mm 90-mm wood studs with a center-to-center distance of 400 mm. The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor (R 0.166 m2 · °C/W), a 13-mm plywood, a layer of felt (R 0.011 m2 · °C/W), and linoleum (R 0.009 m2 · °C/W). Both cen58933_ch03.qxd 9/10/2002 9:00 AM Page 202 202 HEAT TRANSFER while the vertical furring and similar structures constitute 16 percent. Answers: 1.02 m2 · °C/W, 0.978 W/m2 · °C 3–144 Repeat Problem 3–143 assuming one side of both air spaces is coated with a reflective film of 0.05. 3–145 Determine the winter R-value and the U-factor of a masonry wall that consists of the following layers: 100-mm face bricks, 100-mm common bricks, 25-mm urethane rigid foam insulation, and 13-mm gypsum wallboard. Answers: 1.404 m2 · °C/W, 0.712 W/m2 · °C 3–146 The overall heat transfer coefficient (the U-value) of a wall under winter design conditions is U 1.55 W/m2 · °C. Determine the U-value of the wall under summer design conditions. 1 2 3 4 5 6 7 8 FIGURE P3–142 sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter R-value and the U-factor of the ceiling assuming the 90-mmwide air space between the studs (a) does not have any reflective surface, (b) has a reflective surface with 0.05 on one side, and (c) has reflective surfaces with 0.05 on both sides. Assume a mean temperature of 10°C and a temperature difference of 5.6°C for the air space. 3–143 Determine the winter R-value and the U-factor of a masonry cavity wall that consists of 100-mm common bricks, a 90-mm air space, 100-mm concrete blocks made of lightweight aggregate, 20-mm air space, and 13-mm gypsum wallboard separated from the concrete block by 20-mm-thick (1-in. 3-in. nominal) vertical furring whose center-to-center distance is 400 mm. Neither side of the two air spaces is coated with any reflective films. When determining the R-value of the air spaces, the temperature difference across them can be taken to be 16.7°C with a mean air temperature of 10°C. The air space constitutes 84 percent of the heat transmission area, 3–147 The overall heat transfer coefficient (the U-value) of a wall under winter design conditions is U 2.25 W/m2 · °C. Now a layer of 100-mm face brick is added to the outside, leaving a 20-mm air space between the wall and the bricks. Determine the new U-value of the wall. Also, determine the rate of heat transfer through a 3-m-high, 7-m-long section of the wall after modification when the indoor and outdoor temperatures are 22°C and 5°C, respectively. Face brick Existing wall FIGURE P3–147 3–148 Determine the summer and winter R-values, in m2 · °C/W, of a masonry wall that consists of 100-mm face bricks, 13-mm of cement mortar, 100-mm lightweight concrete block, 40-mm air space, and 20-mm plasterboard. Answers: 0.809 and 0.795 m2 · °C/W 3–149E The overall heat transfer coefficient of a wall is determined to be U 0.09 Btu/h · ft2 · °F under the conditions of still air inside and winds of 7.5 mph outside. What will the U-factor be when the wind velocity outside is doubled? 4 3 2 1 FIGURE P3–143 5 6 7 Answer: 0.0907 Btu/h · ft2 · °F 3–150 Two homes are identical, except that the walls of one house consist of 200-mm lightweight concrete blocks, 20-mm air space, and 20-mm plasterboard, while the walls of the other house involve the standard R-2.4 m2 · °C/W frame wall construction. Which house do you think is more energy efficient? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 203 203 CHAPTER 3 3–151 Determine the R-value of a ceiling that consists of a layer of 19-mm acoustical tiles whose top surface is covered with a highly reflective aluminum foil for winter conditions. Assume still air below and above the tiles. Highly reflective foil 19 mm Acoustical tiles FIGURE P3–151 propane at 1 atm is 425 kJ/kg. The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at 42°C through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the propane tank to empty if the tank is (a) not insulated and (b) insulated with 7.5-cm-thick glass wool insulation (k 0.038 W/m · °C). 3–155 Hot water is flowing at an average velocity of 1.5 m/s through a cast iron pipe (k 52 W/m · °C) whose inner and outer diameters are 3 cm and 3.5 cm, respectively. The pipe passes through a 15-m-long section of a basement whose temperature is 15°C. If the temperature of the water drops from 70°C to 67°C as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is 400 W/m2 · °C, determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. Answer: 272.5 W/m2 · °C Review Problems 3–152E Steam is produced in the copper tubes (k 223 Btu/h · ft · °F) of a heat exchanger at a temperature of 250°F by another fluid condensing on the outside surfaces of the tubes at 350°F. The inner and outer diameters of the tube are 1 in. and 1.3 in., respectively. When the heat exchanger was new, the rate of heat transfer per foot length of the tube was 2 104 Btu/h. Determine the rate of heat transfer per foot length of the tube when a 0.01-in.-thick layer of limestone (k 1.7 Btu/h · ft · °F) has formed on the inner surface of the tube after extended use. 3–153E Repeat Problem 3–152E, assuming that a 0.01-in.thick limestone layer has formed on both the inner and outer surfaces of the tube. 3–154 A 1.2-m-diameter and 6-m-long cylindrical propane tank is initially filled with liquid propane whose density is 581 kg/m3. The tank is exposed to the ambient air at 30°C, with a heat transfer coefficient of 25 W/m2 · °C. Now a crack develops at the top of the tank and the pressure inside drops to 1 atm while the temperature drops to 42°C, which is the boiling temperature of propane at 1 atm. The heat of vaporization of 3–156 Newly formed concrete pipes are usually cured first overnight by steam in a curing kiln maintained at a temperature of 45°C before the pipes are cured for several days outside. The heat and moisture to the kiln is provided by steam flowing in a pipe whose outer diameter is 12 cm. During a plant inspection, it was noticed that the pipe passes through a 10-m section that is completely exposed to the ambient air before it reaches the kiln. The temperature measurements indicate that the average temperature of the outer surface of the steam pipe is 82°C when the ambient temperature is 8°C. The combined convection and radiation heat transfer coefficient at the outer surface of the pipe is estimated to be 25 W/m2 · °C. Determine the amount of heat lost from the steam during a 10-h curing process that night. Steam is supplied by a gas-fired steam generator that has an efficiency of 80 percent, and the plant pays $0.60/therm of natural gas (1 therm 105,500 kJ). If the pipe is insulated and 90 percent of the heat loss is saved as a result, determine the amount of money this facility will save a year as a result of insulating the steam pipes. Assume that the concrete pipes are cured 110 nights a year. State your assumptions. Tair = 8°C Propane vapor Furnace Kiln 82°C Tair = 30°C Steam 12 cm Steam pipe PROPANE TANK 1.2 m T = – 42°C P = 1 atm 6m FIGURE P3–154 10 m FIGURE P3–156 3–157 Consider an 18-cm 18-cm multilayer circuit board dissipating 27 W of heat. The board consists of four layers of 0.2-mm-thick copper (k 386 W/m · °C) and three layers of cen58933_ch03.qxd 9/10/2002 9:00 AM Page 204 204 HEAT TRANSFER heat transfer coefficient and the same surface temperature, determine how long it will take for the potato to experience the same temperature drop if it is wrapped completely in a 0.12-in.-thick towel (k 0.035 Btu/h · ft · °F). You may use the properties of water for potato. Copper 3–161E Repeat Problem 3–160E assuming there is a 0.02in.-thick air space (k 0.015 Btu/h · ft · °F) between the potato and the towel. Epoxy glass 18 cm 18 cm FIGURE P3–157 1.5-mm-thick epoxy glass (k 0.26 W/m · °C) sandwiched together, as shown in the figure. The circuit board is attached to a heat sink from both ends, and the temperature of the board at those ends is 35°C. Heat is considered to be uniformly generated in the epoxy layers of the board at a rate of 0.5 W per 1-cm 18-cm epoxy laminate strip (or 1.5 W per 1-cm 18-cm strip of the board). Considering only a portion of the board because of symmetry, determine the magnitude and location of the maximum temperature that occurs in the board. Assume heat transfer from the top and bottom faces of the board to be negligible. 3–158 The plumbing system of a house involves a 0.5-m section of a plastic pipe (k 0.16 W/m · °C) of inner diameter 2 cm and outer diameter 2.4 cm exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about 5°C for a period of 14 h. The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be 40 W/m2 · °C, and the heat of fusion of water is 333.7 kJ/kg. Assuming the pipe to contain stationary water initially at 0°C, determine if the water in that section of the pipe will completely freeze that night. Exposed water pipe 3–162 An ice chest whose outer dimensions are 30 cm 40 cm 50 cm is made of 3-cm-thick Styrofoam (k 0.033 W/m · °C). Initially, the chest is filled with 45 kg of ice at 0°C, and the inner surface temperature of the ice chest can be taken to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7 kJ/kg, and the heat transfer coefficient between the outer surface of the ice chest and surrounding air at 35°C is 18 W/m2 · °C. Disregarding any heat transfer from the 40-cm 50-cm base of the ice chest, determine how long it will take for the ice in the chest to melt completely. Tair = 35°C Ice chest 0°C 0°C Styrofoam FIGURE P3–162 3–163 A 4-m-high and 6-m-long wall is constructed of two large 2-cm-thick steel plates (k 15 W/m · °C) separated by 1-cm-thick and 20-cm-wide steel bars placed 99 cm apart. The Steel plates Tair = –5°C 2.4 cm 3 cm AIR Fiberglass insulation Water SOIL FIGURE P3–158 99 cm 3–159 Repeat Problem 3–158 for the case of a heat transfer coefficient of 10 W/m2 · °C on the outer surface as a result of putting a fence around the pipe that blocks the wind. 3–160E The surface temperature of a 3-in.-diameter baked potato is observed to drop from 300°F to 200°F in 5 minutes in an environment at 70°F. Determine the average heat transfer coefficient between the potato and its surroundings. Using this 1 cm 2 cm 20 cm FIGURE P3–163 2 cm cen58933_ch03.qxd 9/10/2002 9:00 AM Page 205 205 CHAPTER 3 remaining space between the steel plates is filled with fiberglass insulation (k 0.035 W/m · °C). If the temperature difference between the inner and the outer surfaces of the walls is 22°C, determine the rate of heat transfer through the wall. Can we ignore the steel bars between the plates in heat transfer analysis since they occupy only 1 percent of the heat transfer surface area? 3–164 A 0.2-cm-thick, 10-cm-high, and 15-cm-long circuit board houses electronic components on one side that dissipate a total of 15 W of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of 12 W/m · °C. All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at 37°C, with a heat transfer coefficient of 45 W/m2 · °C. (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15-cm-long aluminum plate (k 237 W/m · °C) with 20 0.2-cm-thick, 2-cmlong, and 15-cm-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a 0.03–cmthick epoxy adhesive (k 1.8 W/m · °C). Determine the new temperatures on the two sides of the circuit board. Electronic components Fin 15 cm 0.3 cm 10 cm 0.2 cm 20 fins 2 cm 1 mm 2 mm FIGURE P3–164 3–165 Repeat Problem 3–164 using a copper plate with copper fins (k 386 W/m · °C) instead of aluminum ones. 3-166 A row of 10 parallel pipes that are 5 m long and have an outer diameter of 6 cm are used to transport steam at 150°C through the concrete floor (k 0.75 W/m · °C) of a 10-m 5-m room that is maintained at 25°C. The combined convection and radiation heat transfer coefficient at the floor is 12 W/m2 · °C. If the surface temperature of the concrete floor is not to exceed 40°C, determine how deep the steam pipes should be buried below the surface of the concrete floor. Room 25°C 10 m 40°C D = 6 cm Steam pipes Concrete floor FIGURE P3–166 3–167 Consider two identical people each generating 60 W of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1-mm-thick leather (k 0.159 W/m · °C) that covers half of the body while the second one is wearing clothes made of 1-mm-thick synthetic fabric (k 0.13 W/m · °C) that covers the body completely. The ambient air is at 30°C, the heat transfer coefficient at the outer surface is 15 W/m2 · °C, and the inner surface temperature of the clothes can be taken to be 32°C. Treating the body of each person as a 25-cm-diameter 1.7-m-long cylinder, determine the fractions of heat lost from each person by perspiration. 3–168 A 6-m-wide 2.8-m-high wall is constructed of one layer of common brick (k 0.72 W/m · °C) of thickness 20 cm, one inside layer of light-weight plaster (k 0.36 W/m · °C) of thickness 1 cm, and one outside layer of cement based covering (k 1.40 W/m · °C) of thickness 2 cm. The inner surface of the wall is maintained at 23°C while the outer surface is exposed to outdoors at 8°C with a combined convection and radiation heat transfer coefficient of 17 W/m2 · °C. Determine the rate of heat transfer through the wall and temperature drops across the plaster, brick, covering, and surfaceambient air. 3–169 Reconsider Problem 3–168. It is desired to insulate the wall in order to decrease the heat loss by 85 percent. For the same inner surface temperature, determine the thickness of insulation and the outer surface temperature if the wall is insulated with (a) polyurethane foam (k 0.025 W/m · °C) and (b) glass fiber (k 0.036 W/m · °C). 3–170 Cold conditioned air at 12°C is flowing inside a 1.5-cm-thick square aluminum (k 237 W/m · °C) duct of inner cross section 22 cm 22 cm at a mass flow rate of 0.8 kg/s. The duct is exposed to air at 33°C with a combined convection-radiation heat transfer coefficient of 8 W/m2 · °C. The convection heat transfer coefficient at the inner surface is 75 W/m2 · °C. If the air temperature in the duct should not increase by more than 1°C determine the maximum length of the duct. 3–171 When analyzing heat transfer through windows, it is important to consider the frame as well as the glass area. Consider a 2-m-wide 1.5-m-high wood-framed window with cen58933_ch03.qxd 9/10/2002 9:00 AM Page 206 206 HEAT TRANSFER 85 percent of the area covered by 3-mm-thick single-pane glass (k 0.7 W/m · °C). The frame is 5 cm thick, and is made of pine wood (k 0.12 W/m · °C). The heat transfer coefficient is 7 W/m2 · °C inside and 13 W/m2 · °C outside. The room is maintained at 24°C, and the temperature outdoors is 40°C. Determine the percent error involved in heat transfer when the window is assumed to consist of glass only. 3–172 Steam at 235°C is flowing inside a steel pipe (k 61 W/m · °C) whose inner and outer diameters are 10 cm and 12 cm, respectively, in an environment at 20°C. The heat transfer coefficients inside and outside the pipe are 105 W/m2 · °C and 14 W/m2 · °C, respectively. Determine (a) the thickness of the insulation (k 0.038 W/m · °C) needed to reduce the heat loss by 95 percent and (b) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to 40°C for safety reasons. 3–173 When the transportation of natural gas in a pipeline is not feasible for economic or other reasons, it is first liquefied at about 160°C, and then transported in specially insulated tanks placed in marine ships. Consider a 6-m-diameter spherical tank that is filled with liquefied natural gas (LNG) at 160°C. The tank is exposed to ambient air at 18°C with a heat transfer coefficient of 22 W/m2 · °C. The tank is thinshelled and its temperature can be taken to be the same as the LNG temperature. The tank is insulated with 5-cm-thick super insulation that has an effective thermal conductivity of 0.00008 W/m · °C. Taking the density and the specific heat of LNG to be 425 kg/m3 and 3.475 kJ/kg · °C, respectively, estimate how long it will take for the LNG temperature to rise to 150°C. 3–174 A 15-cm 20-cm hot surface at 85°C is to be cooled by attaching 4-cm-long aluminum (k 237 W/m · °C) fins of 2-mm 2-mm square cross section. The temperature of surrounding medium is 25°C and the heat transfer coefficient on the surfaces can be taken to be 20 W/m2 · °C. If it is desired to triple the rate of heat transfer from the bare hot surface, determine the number of fins that needs to be attached. 3–175 Reconsider Problem 3–174. Using EES (or other) software, plot the number of fins as a function of the increase in the heat loss by fins relative to no fin case (i.e., overall effectiveness of the fins) in the range of 1.5 to 5. Discuss the results. Is it realistic to assume the heat transfer coefficient to remain constant? 3–176 A 1.4-m-diameter spherical steel tank filled with iced water at 0°C is buried underground at a location where the thermal conductivity of the soil is k 0.55 W/m · °C. The distance between the tank center and the ground surface is 2.4 m. For ground surface temperature of 18°C, determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were 18°C and the ground surface were insulated? 3–177 A 0.6-m-diameter 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at 160°C is placed at the center of a 1.9-m-long 1.4-m 1.4-m square solid bar made of an insulating material with k 0.0006 W/m · °C. If the outer surface temperature of the bar is 20°C, determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be 425 kg/m3 and 3.475 kJ/kg · °C, respectively. Design and Essay Problems 3–178 The temperature in deep space is close to absolute zero, which presents thermal challenges for the astronauts who do space walks. Propose a design for the clothing of the astronauts that will be most suitable for the thermal environment in space. Defend the selections in your design. 3–179 In the design of electronic components, it is very desirable to attach the electronic circuitry to a substrate material that is a very good thermal conductor but also a very effective electrical insulator. If the high cost is not a major concern, what material would you propose for the substrate? 3–180 Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known. 3–181 Find out about the wall construction of the cabins of large commercial airplanes, the range of ambient conditions under which they operate, typical heat transfer coefficients on the inner and outer surfaces of the wall, and the heat generation rates inside. Determine the size of the heating and airconditioning system that will be able to maintain the cabin at 20°C at all times for an airplane capable of carrying 400 people. 3–182 Repeat Problem 3–181 for a submarine with a crew of 60 people. 3–183 A house with 200-m2 floor space is to be heated with geothermal water flowing through pipes laid in the ground under the floor. The walls of the house are 4 m high, and there are 10 single-paned windows in the house that are 1.2 m wide and 1.8 m high. The house has R-19 (in h · ft2 · °F/Btu) insulation in the walls and R-30 on the ceiling. The floor temperature is not to exceed 40°C. Hot geothermal water is available at 90°C, and the inner and outer diameter of the pipes to be used are 2.4 cm and 3.0 cm. Design such a heating system for this house in your area. 3–184 Using a timer (or watch) and a thermometer, conduct this experiment to determine the rate of heat gain of your refrigerator. First, make sure that the door of the refrigerator is not opened for at least a few hours to make sure that steady operating conditions are established. Start the timer when the refrigerator stops running and measure the time t1 it stays off cen58933_ch03.qxd 9/10/2002 9:00 AM Page 207 207 CHAPTER 3 before it kicks in. Then measure the time t2 it stays on. Noting that the heat removed during t2 is equal to the heat gain of the refrigerator during t1 t2 and using the power consumed by the refrigerator when it is running, determine the average rate of heat gain for your refrigerator, in watts. Take the COP (coefficient of performance) of your refrigerator to be 1.3 if it is not available. Now, clean the condenser coils of the refrigerator and remove any obstacles on the way of airflow through the coils By replacing these measurements, determine the improvement in the COP of the refrigerator. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 208 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 209 CHAPTER T R A N S I E N T H E AT CONDUCTION he temperature of a body, in general, varies with time as well as position. In rectangular coordinates, this variation is expressed as T(x, y, z, t), where (x, y, z) indicates variation in the x, y, and z directions, respectively, and t indicates variation with time. In the preceding chapter, we considered heat conduction under steady conditions, for which the temperature of a body at any point does not change with time. This certainly simplified the analysis, especially when the temperature varied in one direction only, and we were able to obtain analytical solutions. In this chapter, we consider the variation of temperature with time as well as position in one- and multidimensional systems. We start this chapter with the analysis of lumped systems in which the temperature of a solid varies with time but remains uniform throughout the solid at any time. Then we consider the variation of temperature with time as well as position for one-dimensional heat conduction problems such as those associated with a large plane wall, a long cylinder, a sphere, and a semi-infinite medium using transient temperature charts and analytical solutions. Finally, we consider transient heat conduction in multidimensional systems by utilizing the product solution. T 4 CONTENTS 4–1 Lumped Systems Analysis 210 4–2 Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects 216 4–3 Transient Heat Conduction in Semi-Infinite Solids 228 4–4 Transient Heat Conduction in Multidimensional Systems 231 Topic of Special Interest: Refrigeration and Freezing of Foods 239 209 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 210 210 HEAT TRANSFER 4–1 70°C 70°C 70°C 70°C 70°C (a) Copper ball 110°C 90°C 40°C (b) Roast beef FIGURE 4–1 A small copper ball can be modeled as a lumped system, but a roast beef cannot. As SOLID BODY h T m = mass V = volume ρ = density Ti = initial temperature T = T(t) · Q = h As[T – T(t)] FIGURE 4–2 The geometry and parameters involved in the lumped system analysis. ■ LUMPED SYSTEM ANALYSIS In heat transfer analysis, some bodies are observed to behave like a “lump” whose interior temperature remains essentially uniform at all times during a heat transfer process. The temperature of such bodies can be taken to be a function of time only, T(t). Heat transfer analysis that utilizes this idealization is known as lumped system analysis, which provides great simplification in certain classes of heat transfer problems without much sacrifice from accuracy. Consider a small hot copper ball coming out of an oven (Fig. 4–1). Measurements indicate that the temperature of the copper ball changes with time, but it does not change much with position at any given time. Thus the temperature of the ball remains uniform at all times, and we can talk about the temperature of the ball with no reference to a specific location. Now let us go to the other extreme and consider a large roast in an oven. If you have done any roasting, you must have noticed that the temperature distribution within the roast is not even close to being uniform. You can easily verify this by taking the roast out before it is completely done and cutting it in half. You will see that the outer parts of the roast are well done while the center part is barely warm. Thus, lumped system analysis is not applicable in this case. Before presenting a criterion about applicability of lumped system analysis, we develop the formulation associated with it. Consider a body of arbitrary shape of mass m, volume V, surface area As, density , and specific heat Cp initially at a uniform temperature Ti (Fig. 4–2). At time t 0, the body is placed into a medium at temperature T, and heat transfer takes place between the body and its environment, with a heat transfer coefficient h. For the sake of discussion, we will assume that T Ti, but the analysis is equally valid for the opposite case. We assume lumped system analysis to be applicable, so that the temperature remains uniform within the body at all times and changes with time only, T T(t). During a differential time interval dt, the temperature of the body rises by a differential amount dT. An energy balance of the solid for the time interval dt can be expressed as The increase in the Heat transfer into the body energy of the body during dt during dt or hAs(T T) dt mCp dT (4-1) Noting that m V and dT d(T T) since T constant, Eq. 4–1 can be rearranged as hAs d(T T) dt T T VCp (4-2) Integrating from t 0, at which T Ti, to any time t, at which T T(t), gives ln hAs T(t) T t Ti T VCp (4-3) cen58933_ch04.qxd 9/10/2002 9:12 AM Page 211 211 CHAPTER 4 Taking the exponential of both sides and rearranging, we obtain T(t) T ebt Ti T T(t) (4-4) T b3 b2 where b hAs VCp (1/s) (4-5) is a positive quantity whose dimension is (time)1. The reciprocal of b has time unit (usually s), and is called the time constant. Equation 4–4 is plotted in Fig. 4–3 for different values of b. There are two observations that can be made from this figure and the relation above: 1. Equation 4–4 enables us to determine the temperature T(t) of a body at time t, or alternatively, the time t required for the temperature to reach a specified value T(t). 2. The temperature of a body approaches the ambient temperature T exponentially. The temperature of the body changes rapidly at the beginning, but rather slowly later on. A large value of b indicates that the body will approach the environment temperature in a short time. The larger the value of the exponent b, the higher the rate of decay in temperature. Note that b is proportional to the surface area, but inversely proportional to the mass and the specific heat of the body. This is not surprising since it takes longer to heat or cool a larger mass, especially when it has a large specific heat. b1 b3 > b2 > b1 Ti t FIGURE 4–3 The temperature of a lumped system approaches the environment temperature as time gets larger. Once the temperature T(t) at time t is available from Eq. 4–4, the rate of convection heat transfer between the body and its environment at that time can be determined from Newton’s law of cooling as · Q (t) hAs[T(t) T] (W) (4-6) The total amount of heat transfer between the body and the surrounding medium over the time interval t 0 to t is simply the change in the energy content of the body: Q mCp[T(t) Ti] (kJ) (4-7) The amount of heat transfer reaches its upper limit when the body reaches the surrounding temperature T. Therefore, the maximum heat transfer between the body and its surroundings is (Fig. 4–4) Qmax mCp(T Ti) (kJ) (4-8) Ti Ti Criteria for Lumped System Analysis The lumped system analysis certainly provides great convenience in heat transfer analysis, and naturally we would like to know when it is appropriate Ti Ti Ti We could also obtain this equation by substituting the T(t) relation from Eq. · 4–4 into the Q (t) relation in Eq. 4–6 and integrating it from t 0 to t → . h T t=0 t→ T Ti Ti T T T T T T Q = Qmax = mCp (Ti – T) FIGURE 4–4 Heat transfer to or from a body reaches its maximum value when the body reaches the environment temperature. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 212 212 HEAT TRANSFER to use it. The first step in establishing a criterion for the applicability of the lumped system analysis is to define a characteristic length as Convection Conduction h T SOLID BODY Lc V As Bi hLc k and a Biot number Bi as (4-9) It can also be expressed as (Fig. 4–5) heat convection Bi = ———————– heat conduction FIGURE 4–5 The Biot number can be viewed as the ratio of the convection at the surface to conduction within the body. Bi h T Convection at the surface of the body k /Lc T Conduction within the body or Bi Conduction resistance within the body Lc /k 1/h Convection resistance at the surface of the body When a solid body is being heated by the hotter fluid surrounding it (such as a potato being baked in an oven), heat is first convected to the body and subsequently conducted within the body. The Biot number is the ratio of the internal resistance of a body to heat conduction to its external resistance to heat convection. Therefore, a small Biot number represents small resistance to heat conduction, and thus small temperature gradients within the body. Lumped system analysis assumes a uniform temperature distribution throughout the body, which will be the case only when the thermal resistance of the body to heat conduction (the conduction resistance) is zero. Thus, lumped system analysis is exact when Bi 0 and approximate when Bi 0. Of course, the smaller the Bi number, the more accurate the lumped system analysis. Then the question we must answer is, How much accuracy are we willing to sacrifice for the convenience of the lumped system analysis? Before answering this question, we should mention that a 20 percent uncertainty in the convection heat transfer coefficient h in most cases is considered “normal” and “expected.” Assuming h to be constant and uniform is also an approximation of questionable validity, especially for irregular geometries. Therefore, in the absence of sufficient experimental data for the specific geometry under consideration, we cannot claim our results to be better than 20 percent, even when Bi 0. This being the case, introducing another source of uncertainty in the problem will hardly have any effect on the overall uncertainty, provided that it is minor. It is generally accepted that lumped system analysis is applicable if Bi 0.1 When this criterion is satisfied, the temperatures within the body relative to the surroundings (i.e., T T) remain within 5 percent of each other even for well-rounded geometries such as a spherical ball. Thus, when Bi 0.1, the variation of temperature with location within the body will be slight and can reasonably be approximated as being uniform. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 213 213 CHAPTER 4 The first step in the application of lumped system analysis is the calculation of the Biot number, and the assessment of the applicability of this approach. One may still wish to use lumped system analysis even when the criterion Bi 0.1 is not satisfied, if high accuracy is not a major concern. Note that the Biot number is the ratio of the convection at the surface to conduction within the body, and this number should be as small as possible for lumped system analysis to be applicable. Therefore, small bodies with high thermal conductivity are good candidates for lumped system analysis, especially when they are in a medium that is a poor conductor of heat (such as air or another gas) and motionless. Thus, the hot small copper ball placed in quiescent air, discussed earlier, is most likely to satisfy the criterion for lumped system analysis (Fig. 4–6). Some Remarks on Heat Transfer in Lumped Systems To understand the heat transfer mechanism during the heating or cooling of a solid by the fluid surrounding it, and the criterion for lumped system analysis, consider this analogy (Fig. 4–7). People from the mainland are to go by boat to an island whose entire shore is a harbor, and from the harbor to their destinations on the island by bus. The overcrowding of people at the harbor depends on the boat traffic to the island and the ground transportation system on the island. If there is an excellent ground transportation system with plenty of buses, there will be no overcrowding at the harbor, especially when the boat traffic is light. But when the opposite is true, there will be a huge overcrowding at the harbor, creating a large difference between the populations at the harbor and inland. The chance of overcrowding is much lower in a small island with plenty of fast buses. In heat transfer, a poor ground transportation system corresponds to poor heat conduction in a body, and overcrowding at the harbor to the accumulation of heat and the subsequent rise in temperature near the surface of the body relative to its inner parts. Lumped system analysis is obviously not applicable when there is overcrowding at the surface. Of course, we have disregarded radiation in this analogy and thus the air traffic to the island. Like passengers at the harbor, heat changes vehicles at the surface from convection to conduction. Noting that a surface has zero thickness and thus cannot store any energy, heat reaching the surface of a body by convection must continue its journey within the body by conduction. Consider heat transfer from a hot body to its cooler surroundings. Heat will be transferred from the body to the surrounding fluid as a result of a temperature difference. But this energy will come from the region near the surface, and thus the temperature of the body near the surface will drop. This creates a temperature gradient between the inner and outer regions of the body and initiates heat flow by conduction from the interior of the body toward the outer surface. When the convection heat transfer coefficient h and thus convection heat transfer from the body are high, the temperature of the body near the surface will drop quickly (Fig. 4–8). This will create a larger temperature difference between the inner and outer regions unless the body is able to transfer heat from the inner to the outer regions just as fast. Thus, the magnitude of the maximum temperature difference within the body depends strongly on the ability of a body to conduct heat toward its surface relative to the ability of h = 15 W/m2 ·°C Spherical copper ball k = 401 W/ m·°C D = 12 cm 1– πD 3 1 V = —— 6 = – D = 0.02 m Lc = — As π D 2 6 hL 15 × 0.02 Bi = —–c = ———— = 0.00075 < 0.1 k 401 FIGURE 4–6 Small bodies with high thermal conductivities and low convection coefficients are most likely to satisfy the criterion for lumped system analysis. Boat Bus ISLAND FIGURE 4–7 Analogy between heat transfer to a solid and passenger traffic to an island. T = 20°C 50°C 70°C 85°C 110°C 130°C Convection h = 2000 W/ m2 ·°C FIGURE 4–8 When the convection coefficient h is high and k is low, large temperature differences occur between the inner and outer regions of a large solid. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 214 214 HEAT TRANSFER the surrounding medium to convect this heat away from the surface. The Biot number is a measure of the relative magnitudes of these two competing effects. Recall that heat conduction in a specified direction n per unit surface area is expressed as q· k T/ n, where T/ n is the temperature gradient and k is the thermal conductivity of the solid. Thus, the temperature distribution in the body will be uniform only when its thermal conductivity is infinite, and no such material is known to exist. Therefore, temperature gradients and thus temperature differences must exist within the body, no matter how small, in order for heat conduction to take place. Of course, the temperature gradient and the thermal conductivity are inversely proportional for a given heat flux. Therefore, the larger the thermal conductivity, the smaller the temperature gradient. EXAMPLE 4–1 Thermocouple wire Gas T, h Junction D = 1 mm T(t) FIGURE 4–9 Schematic for Example 4–1. Temperature Measurement by Thermocouples The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1-mm-diameter sphere, as shown in Fig. 4–9. The properties of the junction are k 35 W/m · °C, 8500 kg/m3, and Cp 320 J/kg · °C, and the convection heat transfer coefficient between the junction and the gas is h 210 W/m2 · °C. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference. SOLUTION The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the initial T is to be determined. Assumptions 1 The junction is spherical in shape with a diameter of D 0.001 m. 2 The thermal properties of the junction and the heat transfer coefficient are constant. 3 Radiation effects are negligible. Properties The properties of the junction are given in the problem statement. Analysis The characteristic length of the junction is Lc 1 3 V 6 D 1 1 D (0.001 m) 1.67 104 m As 6 6 D 2 Then the Biot number becomes Bi hLc (210 W/m2 · °C)(1.67 104 m) 0.001 0.1 k 35 W/m · °C Therefore, lumped system analysis is applicable, and the error involved in this approximation is negligible. In order to read 99 percent of the initial temperature difference Ti T between the junction and the gas, we must have T (t ) T 0.01 Ti T For example, when Ti 0°C and T 100°C, a thermocouple is considered to have read 99 percent of this applied temperature difference when its reading indicates T (t ) 99°C. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 215 215 CHAPTER 4 The value of the exponent b is b hAs 210 W/m2 · °C h 0.462 s1 CpV Cp Lc (8500 kg/m3)(320 J/kg · °C)(1.67 104 m) We now substitute these values into Eq. 4–4 and obtain T (t ) T ebt → Ti T 0.01 e(0.462 s 1 )t which yields t 10 s Therefore, we must wait at least 10 s for the temperature of the thermocouple junction to approach within 1 percent of the initial junction-gas temperature difference. Discussion Note that conduction through the wires and radiation exchange with the surrounding surfaces will affect the result, and should be considered in a more refined analysis. EXAMPLE 4–2 Predicting the Time of Death A person is found dead at 5 PM in a room whose temperature is 20°C. The temperature of the body is measured to be 25°C when found, and the heat transfer coefficient is estimated to be h 8 W/m2 · °C. Modeling the body as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that person (Fig. 4–10). SOLUTION A body is found while still warm. The time of death is to be estimated. Assumptions 1 The body can be modeled as a 30-cm-diameter, 1.70-m-long cylinder. 2 The thermal properties of the body and the heat transfer coefficient are constant. 3 The radiation effects are negligible. 4 The person was healthy(!) when he or she died with a body temperature of 37°C. Properties The average human body is 72 percent water by mass, and thus we can assume the body to have the properties of water at the average temperature of (37 25)/2 31°C; k 0.617 W/m · °C, 996 kg/m3, and Cp 4178 J/kg · °C (Table A-9). Analysis The characteristic length of the body is Lc ro2 L (0.15 m)2(1.7 m) V 0.0689 m 2 As 2 ro L 2 ro 2 (0.15 m)(1.7 m) 2 (0.15 m)2 Then the Biot number becomes Bi hLc (8 W/m2 · °C)(0.0689 m) 0.89 0.1 k 0.617 W/m · °C FIGURE 4–10 Schematic for Example 4–2. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 216 216 HEAT TRANSFER Therefore, lumped system analysis is not applicable. However, we can still use it to get a “rough” estimate of the time of death. The exponent b in this case is b hAs 8 W/m2 · °C h 3 CpV Cp Lc (996 kg/m )(4178 J/kg · °C)(0.0689 m) 2.79 105 s1 We now substitute these values into Eq. 4–4, T (t ) T ebt → Ti T 25 20 5 1 e(2.79 10 s )t 37 20 which yields t 43,860 s 12.2 h Therefore, as a rough estimate, the person died about 12 h before the body was found, and thus the time of death is 5 AM. This example demonstrates how to obtain “ball park” values using a simple analysis. 4–2 ■ TRANSIENT HEAT CONDUCTION IN LARGE PLANE WALLS, LONG CYLINDERS, AND SPHERES WITH SPATIAL EFFECTS In Section, 4–1, we considered bodies in which the variation of temperature within the body was negligible; that is, bodies that remain nearly isothermal during a process. Relatively small bodies of highly conductive materials approximate this behavior. In general, however, the temperature within a body will change from point to point as well as with time. In this section, we consider the variation of temperature with time and position in one-dimensional problems such as those associated with a large plane wall, a long cylinder, and a sphere. Consider a plane wall of thickness 2L, a long cylinder of radius ro, and a sphere of radius ro initially at a uniform temperature Ti, as shown in Fig. 4–11. At time t 0, each geometry is placed in a large medium that is at a constant temperature T and kept in that medium for t 0. Heat transfer takes place between these bodies and their environments by convection with a uniform and constant heat transfer coefficient h. Note that all three cases possess geometric and thermal symmetry: the plane wall is symmetric about its center plane (x 0), the cylinder is symmetric about its centerline (r 0), and the sphere is symmetric about its center point (r 0). We neglect radiation heat transfer between these bodies and their surrounding surfaces, or incorporate the radiation effect into the convection heat transfer coefficient h. The variation of the temperature profile with time in the plane wall is illustrated in Fig. 4–12. When the wall is first exposed to the surrounding medium at T Ti at t 0, the entire wall is at its initial temperature Ti. But the wall temperature at and near the surfaces starts to drop as a result of heat transfer from the wall to the surrounding medium. This creates a temperature cen58933_ch04.qxd 9/10/2002 9:12 AM Page 217 217 CHAPTER 4 T h Initially T = Ti 0 T h T h L x (a) A large plane wall Initially T = Ti 0 T h T Initially T = Ti 0 r h ro ro r (b) A long cylinder (c) A sphere gradient in the wall and initiates heat conduction from the inner parts of the wall toward its outer surfaces. Note that the temperature at the center of the wall remains at Ti until t t2, and that the temperature profile within the wall remains symmetric at all times about the center plane. The temperature profile gets flatter and flatter as time passes as a result of heat transfer, and eventually becomes uniform at T T. That is, the wall reaches thermal equilibrium with its surroundings. At that point, the heat transfer stops since there is no longer a temperature difference. Similar discussions can be given for the long cylinder or sphere. The formulation of the problems for the determination of the onedimensional transient temperature distribution T(x, t) in a wall results in a partial differential equation, which can be solved using advanced mathematical techniques. The solution, however, normally involves infinite series, which are inconvenient and time-consuming to evaluate. Therefore, there is clear motivation to present the solution in tabular or graphical form. However, the solution involves the parameters x, L, t, k, , h, Ti, and T, which are too many to make any graphical presentation of the results practical. In order to reduce the number of parameters, we nondimensionalize the problem by defining the following dimensionless quantities: Dimensionless temperature: (x, t) Dimensionless distance from the center: X Dimensionless heat transfer coefficient: Bi Dimensionless time: T(x, t) T Ti T x L hL k t 2 L (Biot number) (Fourier number) The nondimensionalization enables us to present the temperature in terms of three parameters only: X, Bi, and . This makes it practical to present the solution in graphical form. The dimensionless quantities defined above for a plane wall can also be used for a cylinder or sphere by replacing the space variable x by r and the half-thickness L by the outer radius ro. Note that the characteristic length in the definition of the Biot number is taken to be the FIGURE 4–11 Schematic of the simple geometries in which heat transfer is one-dimensional. Ti t = t1 t=0 t = t2 t = t3 T 0 h Initially T = Ti t→ L x T h FIGURE 4–12 Transient temperature profiles in a plane wall exposed to convection from its surfaces for Ti T. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 218 218 HEAT TRANSFER half-thickness L for the plane wall, and the radius ro for the long cylinder and sphere instead of V/A used in lumped system analysis. The one-dimensional transient heat conduction problem just described can be solved exactly for any of the three geometries, but the solution involves infinite series, which are difficult to deal with. However, the terms in the solutions converge rapidly with increasing time, and for 0.2, keeping the first term and neglecting all the remaining terms in the series results in an error under 2 percent. We are usually interested in the solution for times with 0.2, and thus it is very convenient to express the solution using this oneterm approximation, given as Plane wall: Cylinder: Sphere: T(x, t) T 2 A1e1 cos (1x/L), 0.2 Ti T T(r, t) T 2 (r, t)cyl A1e1 J0(1r/ro), 0.2 Ti T T(r, t) T 2 sin(1r/ro) (r, t)sph A1e1 , 0.2 Ti T 1r/ro (x, t)wall (4-10) (4-11) (4-12) where the constants A1 and 1 are functions of the Bi number only, and their values are listed in Table 4–1 against the Bi number for all three geometries. The function J0 is the zeroth-order Bessel function of the first kind, whose value can be determined from Table 4–2. Noting that cos (0) J0(0) 1 and the limit of (sin x)/x is also 1, these relations simplify to the next ones at the center of a plane wall, cylinder, or sphere: Center of plane wall (x 0): Center of cylinder (r 0): Center of sphere (r 0): To T 2 A1e1 Ti T To T 2 0, cyl A1e1 Ti T To T 2 0, sph A1e1 Ti T 0, wall (4-13) (4-14) (4-15) Once the Bi number is known, the above relations can be used to determine the temperature anywhere in the medium. The determination of the constants A1 and 1 usually requires interpolation. For those who prefer reading charts to interpolating, the relations above are plotted and the one-term approximation solutions are presented in graphical form, known as the transient temperature charts. Note that the charts are sometimes difficult to read, and they are subject to reading errors. Therefore, the relations above should be preferred to the charts. The transient temperature charts in Figs. 4–13, 4–14, and 4–15 for a large plane wall, long cylinder, and sphere were presented by M. P. Heisler in 1947 and are called Heisler charts. They were supplemented in 1961 with transient heat transfer charts by H. Gröber. There are three charts associated with each geometry: the first chart is to determine the temperature To at the center of the geometry at a given time t. The second chart is to determine the temperature at other locations at the same time in terms of To. The third chart is to determine the total amount of heat transfer up to the time t. These plots are valid for 0.2. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 219 219 CHAPTER 4 TABLE 4–1 TABLE 4–2 Coefficients used in the one-term approximate solution of transient onedimensional heat conduction in plane walls, cylinders, and spheres (Bi hL/k for a plane wall of thickness 2L, and Bi hro /k for a cylinder or sphere of radius ro ) The zeroth- and first-order Bessel functions of the first kind Bi 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 20.0 30.0 40.0 50.0 100.0 Plane Wall 1 A1 0.0998 0.1410 0.1987 0.2425 0.2791 0.3111 0.4328 0.5218 0.5932 0.6533 0.7051 0.7506 0.7910 0.8274 0.8603 1.0769 1.1925 1.2646 1.3138 1.3496 1.3766 1.3978 1.4149 1.4289 1.4961 1.5202 1.5325 1.5400 1.5552 1.5708 1.0017 1.0033 1.0066 1.0098 1.0130 1.0161 1.0311 1.0450 1.0580 1.0701 1.0814 1.0918 1.1016 1.1107 1.1191 1.1785 1.2102 1.2287 1.2403 1.2479 1.2532 1.2570 1.2598 1.2620 1.2699 1.2717 1.2723 1.2727 1.2731 1.2732 Cylinder Sphere 1 A1 1 A1 0.1412 0.1995 0.2814 0.3438 0.3960 0.4417 0.6170 0.7465 0.8516 0.9408 1.0184 1.0873 1.1490 1.2048 1.2558 1.5995 1.7887 1.9081 1.9898 2.0490 2.0937 2.1286 2.1566 2.1795 2.2880 2.3261 2.3455 2.3572 2.3809 2.4048 1.0025 1.0050 1.0099 1.0148 1.0197 1.0246 1.0483 1.0712 1.0931 1.1143 1.1345 1.1539 1.1724 1.1902 1.2071 1.3384 1.4191 1.4698 1.5029 1.5253 1.5411 1.5526 1.5611 1.5677 1.5919 1.5973 1.5993 1.6002 1.6015 1.6021 0.1730 0.2445 0.3450 0.4217 0.4860 0.5423 0.7593 0.9208 1.0528 1.1656 1.2644 1.3525 1.4320 1.5044 1.5708 2.0288 2.2889 2.4556 2.5704 2.6537 2.7165 2.7654 2.8044 2.8363 2.9857 3.0372 3.0632 3.0788 3.1102 3.1416 1.0030 1.0060 1.0120 1.0179 1.0239 1.0298 1.0592 1.0880 1.1164 1.1441 1.1713 1.1978 1.2236 1.2488 1.2732 1.4793 1.6227 1.7202 1.7870 1.8338 1.8673 1.8920 1.9106 1.9249 1.9781 1.9898 1.9942 1.9962 1.9990 2.0000 Note that the case 1/Bi k/hL 0 corresponds to h → , which corresponds to the case of specified surface temperature T. That is, the case in which the surfaces of the body are suddenly brought to the temperature T at t 0 and kept at T at all times can be handled by setting h to infinity (Fig. 4–16). The temperature of the body changes from the initial temperature Ti to the temperature of the surroundings T at the end of the transient heat conduction process. Thus, the maximum amount of heat that a body can gain (or lose if Ti T) is simply the change in the energy content of the body. That is, Qmax mCp(T Ti ) VCp(T Ti ) (kJ) (4-16) Jo() J1() 0.0 0.1 0.2 0.3 0.4 1.0000 0.9975 0.9900 0.9776 0.9604 0.0000 0.0499 0.0995 0.1483 0.1960 0.5 0.6 0.7 0.8 0.9 0.9385 0.9120 0.8812 0.8463 0.8075 0.2423 0.2867 0.3290 0.3688 0.4059 1.0 1.1 1.2 1.3 1.4 0.7652 0.7196 0.6711 0.6201 0.5669 0.4400 0.4709 0.4983 0.5220 0.5419 1.5 1.6 1.7 1.8 1.9 0.5118 0.4554 0.3980 0.3400 0.2818 0.5579 0.5699 0.5778 0.5815 0.5812 2.0 2.1 2.2 2.3 2.4 0.2239 0.1666 0.1104 0.0555 0.0025 0.5767 0.5683 0.5560 0.5399 0.5202 2.6 2.8 3.0 3.2 0.0968 0.1850 0.2601 0.3202 0.4708 0.4097 0.3391 0.2613 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 220 220 HEAT TRANSFER To – T Ti – T 1.0 0.7 0.5 0.4 0.3 0.2 θo = k hL = 1 Bi = 0.6 0.4 0.7 0.5 0.3 35 7 6 25 30 16 3 2 1.8 1.6 1.4 1.2 0.05 2.5 0 2 50 40 20 18 5 4 0.2 0.1 1 45 9 8 8 0 12 10 0. 0.01 0.007 0.005 0.004 0.003 0.002 3 4 6 8 10 14 18 22 26 30 50 τ = αt/L2 70 100 120 150 T h (a) Midplane temperature (from M. P. Heisler) 300 Initially T = Ti 0 T – T To – T x/L = 0.2 1.0 Q Qmax 1.0 0.9 0.9 θ= 0.4 Bi = hL/k 0.4 0.8 50 10 5 2 1 0.5 0.05 0.1 0.2 0.00 5 0.01 0.02 0.3 0.9 0.1 1.0 0 0.01 0.1 0.00 1 0.00 2 0.5 0.2 x L 2L Bi = 0.6 0.5 0.3 T h 0.7 0.6 0.6 0.4 600 700 0.8 0.8 0.7 400 500 20 0.001 100 80 90 60 70 14 1.0 0.1 0.07 0.05 0.04 0.03 0.02 Plate 0.2 Plate 1.0 10 100 0.1 0 10–5 Plate 10– 4 10–3 10–2 1 k = Bi hL (b) Temperature distribution (from M. P. Heisler) 10–1 1 Bi 2 τ = h2 α t/k 2 10 102 103 104 (c) Heat transfer (from H. Gröber et al.) FIGURE 4–13 Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature Ti subjected to convection from both sides to an environment at temperature T with a convection coefficient of h. where m is the mass, V is the volume, is the density, and Cp is the specific heat of the body. Thus, Qmax represents the amount of heat transfer for t → . The amount of heat transfer Q at a finite time t will obviously be less than this cen58933_ch04.qxd 9/10/2002 9:12 AM Page 221 221 CHAPTER 4 θo = To – T Ti – T 1.0 0.7 Cylinder 0.5 0.4 0.3 5 0.2 0.1 k 3 5 8 1. 16 90 18 70 14 12 1.6 10 0 80 60 9 1.2 50 10 7 0.8 0.6 8 45 35 30 0.3 0.1 0 0.5 6 40 0.4 0.2 0.01 0.007 0.005 0.004 0.003 25 20 2 1 .4 1.0 0.02 = 1 Bi = o 4 0.1 0.07 0.05 0.04 0.03 hr 0.002 0.001 0 1 2 3 4 6 8 10 14 18 22 26 τ = αt /ro2 30 50 70 100 120 (a) Centerline temperature (from M. P. Heisler) 140 150 Q Qmax 1.0 0.9 0.9 0.4 0.4 0.8 50 20 10 5 2 0.3 0.2 0.9 0.1 1.0 0 0.1 0.01 1 0.5 0.4 0.2 0.00 1 0.00 2 0.00 5 0.01 0.02 0.6 0.5 0.6 0.05 0.1 0.2 0.7 0.5 0.3 Bi = hro /k 0.8 0.7 0.6 ro r Bi = 0.8 350 T Initially T h T = Ti h 0 T – T θ= To – T 1.0 r/ro = 0.2 250 Cylinder 1.0 10 100 1 k = Bi hro (b) Temperature distribution (from M. P. Heisler) 0.1 0 10–5 Cylinder 10– 4 10–3 10–2 10–1 Bi 2 τ = 1 10 102 103 104 h2 α t/k 2 (c) Heat transfer (from H. Gröber et al.) FIGURE 4–14 Transient temperature and heat transfer charts for a long cylinder of radius ro initially at a uniform temperature Ti subjected to convection from all sides to an environment at temperature T with a convection coefficient of h. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 222 222 HEAT TRANSFER To – T Ti – T 1.0 0.7 0.5 0.4 0.3 0.2 12 14 2. 0.02 0.5 1.0 1.5 2 2.5 3 4 5 6 7 8 9 10 20 τ = αt/ro2 30 40 50 100 150 (a) Midpoint temperature (from M. P. Heisler) T – T T h 1.0 0.9 0.9 0.4 0 ro r Bi = hro /k 0.3 0.8 0.3 0.2 0.9 0.2 0.1 1.0 1.0 10 50 20 10 0.1 Sphere 0.1 0.5 1 0.4 0.05 0.1 0.2 0.5 0.4 0.00 1 0.00 2 0.6 0.6 0.00 5 0.01 0.02 0.7 0.5 0 0.01 T h 0.8 0.7 0.6 Initially T = Ti Q Qmax To – T r/ro = 0.2 1.0 0.8 250 Bi = θ= 200 5 0 0.75 0.5 0.35 0.2 0.1 .05 0 0 0.01 0.007 0.005 0.004 0.003 0.002 4 3 .5 2.0 2.2 8 1.6 1. .2 1.4 1 1.0 6 2.8 2. 4 50 40 45 0 35 3 25 20 18 16 10 9 8 7 6 5 3.0 0.1 0.07 0.05 0.04 0.03 0.001 100 80 90 60 70 Sphere k hr = 1 o Bi = 2 θo = 100 0 10–5 Sphere 10– 4 10–3 10–2 1 = k Bi hro (b) Temperature distribution (from M. P. Heisler) 10–1 1 Bi 2 τ = h2 α t/k 2 10 102 103 104 (c) Heat transfer (from H. Gröber et al.) FIGURE 4–15 Transient temperature and heat transfer charts for a sphere of radius ro initially at a uniform temperature Ti subjected to convection from all sides to an environment at temperature T with a convection coefficient of h. maximum. The ratio Q/Qmax is plotted in Figures 4–13c, 4–14c, and 4–15c against the variables Bi and h2t/k2 for the large plane wall, long cylinder, and cen58933_ch04.qxd 9/10/2002 9:12 AM Page 223 223 CHAPTER 4 sphere, respectively. Note that once the fraction of heat transfer Q/Qmax has been determined from these charts for the given t, the actual amount of heat transfer by that time can be evaluated by multiplying this fraction by Qmax. A negative sign for Qmax indicates that heat is leaving the body (Fig. 4–17). The fraction of heat transfer can also be determined from these relations, which are based on the one-term approximations already discussed: Plane wall: Q Q max wall Cylinder: Q Q Q Q max cyl Sphere: max sph 1 0, wall sin 1 1 (4-17) 1 20, cyl J1(1) 1 (4-18) 1 30, sph sin 1 1 cos 1 31 (4-19) Ts Ts ≠ T h h T (a) Finite convection coefficient The use of the Heisler/Gröber charts and the one-term solutions already discussed is limited to the conditions specified at the beginning of this section: the body is initially at a uniform temperature, the temperature of the medium surrounding the body and the convection heat transfer coefficient are constant and uniform, and there is no energy generation in the body. We discussed the physical significance of the Biot number earlier and indicated that it is a measure of the relative magnitudes of the two heat transfer mechanisms: convection at the surface and conduction through the solid. A small value of Bi indicates that the inner resistance of the body to heat conduction is small relative to the resistance to convection between the surface and the fluid. As a result, the temperature distribution within the solid becomes fairly uniform, and lumped system analysis becomes applicable. Recall that when Bi 0.1, the error in assuming the temperature within the body to be uniform is negligible. To understand the physical significance of the Fourier number , we express it as (Fig. 4–18) The rate at which heat is conducted across L of a body of volume L3 t kL2 (1/L) T 2 The rate at which heat is stored L Cp L3/t T in a body of volume L3 Ts T (4-20) Therefore, the Fourier number is a measure of heat conducted through a body relative to heat stored. Thus, a large value of the Fourier number indicates faster propagation of heat through a body. Perhaps you are wondering about what constitutes an infinitely large plate or an infinitely long cylinder. After all, nothing in this world is infinite. A plate whose thickness is small relative to the other dimensions can be modeled as an infinitely large plate, except very near the outer edges. But the edge effects on large bodies are usually negligible, and thus a large plane wall such as the wall of a house can be modeled as an infinitely large wall for heat transfer purposes. Similarly, a long cylinder whose diameter is small relative to its length can be analyzed as an infinitely long cylinder. The use of the transient temperature charts and the one-term solutions is illustrated in the following examples. T h→ Ts Ts h→ T Ts = T (b) Infinite convection coefficient FIGURE 4–16 The specified surface temperature corresponds to the case of convection to an environment at T with a convection coefficient h that is infinite. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 224 224 HEAT TRANSFER . Qmax t=0 EXAMPLE 4–3 T = Ti T = T m, Cp h T (a) Maximum heat transfer (t → ) . An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig. 4–19). The egg is initially at a uniform temperature of 5°C and is dropped into boiling water at 95°C. Taking the convection heat transfer coefficient to be h 1200 W/m2 · °C, determine how long it will take for the center of the egg to reach 70°C. SOLUTION An egg is cooked in boiling water. The cooking time of the egg is to be determined. Q t=0 T = Ti T = T(r, t) m, Cp h T Bi = . . . 2αt h—— = Bi2τ = . . . k2 Q —— = . . . Qmax (Gröber chart) (b) Actual heat transfer for time t FIGURE 4–17 The fraction of total heat transfer Q/Qmax up to a specified time t is determined using the Gröber charts. L L Boiling Eggs L Assumptions 1 The egg is spherical in shape with a radius of r0 2.5 cm. 2 Heat conduction in the egg is one-dimensional because of thermal symmetry about the midpoint. 3 The thermal properties of the egg and the heat transfer coefficient are constant. 4 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The water content of eggs is about 74 percent, and thus the thermal conductivity and diffusivity of eggs can be approximated by those of water at the average temperature of (5 70)/2 37.5°C; k 0.627 W/m · °C and k/Cp 0.151 106 m2/s (Table A-9). Analysis The temperature within the egg varies with radial distance as well as time, and the temperature at a specified location at a given time can be determined from the Heisler charts or the one-term solutions. Here we will use the latter to demonstrate their use. The Biot number for this problem is Bi which is much greater than 0.1, and thus the lumped system analysis is not applicable. The coefficients 1 and A1 for a sphere corresponding to this Bi are, from Table 4–1, 1 3.0753, · Qconducted · Q hr0 (1200 W/m2 · °C)(0.025 m) 47.8 k 0.627 W/m · °C A1 1.9958 Substituting these and other values into Eq. 4–15 and solving for gives · Qstored · conducted αt = Q Fourier number: τ = —– ———— · 2 L Qstored FIGURE 4–18 Fourier number at time t can be viewed as the ratio of the rate of heat conducted to the rate of heat stored at that time. To T 2 A1e1 Ti T → 70 95 2 1.9958e(3.0753) 5 95 → 0.209 which is greater than 0.2, and thus the one-term solution is applicable with an error of less than 2 percent. Then the cooking time is determined from the definition of the Fourier number to be ro2 (0.209)(0.025 m)2 t 865 s 14.4 min 0.151 106 m2/s Therefore, it will take about 15 min for the center of the egg to be heated from 5°C to 70°C. Discussion Note that the Biot number in lumped system analysis was defined differently as Bi hLc /k h(r /3)/k. However, either definition can be used in determining the applicability of the lumped system analysis unless Bi 0.1. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 225 225 CHAPTER 4 EXAMPLE 4–4 Heating of Large Brass Plates in an Oven In a production facility, large brass plates of 4 cm thickness that are initially at a uniform temperature of 20°C are heated by passing them through an oven that is maintained at 500°C (Fig. 4–20). The plates remain in the oven for a period of 7 min. Taking the combined convection and radiation heat transfer coefficient to be h 120 W/m2 · °C, determine the surface temperature of the plates when they come out of the oven. SOLUTION Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined. Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 2 The thermal properties of the plate and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The properties of brass at room temperature are k 110 W/m · °C, 8530 kg/m3, Cp 380 J/kg · °C, and 33.9 106 m2/s (Table A-3). More accurate results are obtained by using properties at average temperature. Analysis The temperature at a specified location at a given time can be determined from the Heisler charts or one-term solutions. Here we will use the charts to demonstrate their use. Noting that the half-thickness of the plate is L 0.02 m, from Fig. 4–13 we have 100 W/m · °C k 1 45.8 Bi hL (120 W/m2 · °C)(0.02 m) 6 2 t (33.9 10 m /s)(7 60 s) 2 35.6 (0.02 m)2 L Also, k 1 45.8 Bi hL x L 1 L L To T 0.46 Ti T T T 0.99 To T Therefore, T T To T T T 0.46 0.99 0.455 Ti T To T Ti T and T T 0.455(Ti T) 500 0.455(20 500) 282°C Therefore, the surface temperature of the plates will be 282°C when they leave the oven. Discussion We notice that the Biot number in this case is Bi 1/45.8 0.022, which is much less than 0.1. Therefore, we expect the lumped system analysis to be applicable. This is also evident from (T T)/(To T) 0.99, which indicates that the temperatures at the center and the surface of the plate relative to the surrounding temperature are within 1 percent of each other. Egg Ti = 5°C h = 1200 W/ m 2·°C T = 95°C FIGURE 4–19 Schematic for Example 4–3. T = 500°C h = 120 W/m 2·°C 2 L = 4 cm Brass plate Ti = 20°C FIGURE 4–20 Schematic for Example 4–4. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 226 226 HEAT TRANSFER Noting that the error involved in reading the Heisler charts is typically at least a few percent, the lumped system analysis in this case may yield just as accurate results with less effort. The heat transfer surface area of the plate is 2A, where A is the face area of the plate (the plate transfers heat through both of its surfaces), and the volume of the plate is V (2L)A, where L is the half-thickness of the plate. The exponent b used in the lumped system analysis is determined to be hAs h(2A) h CpV Cp (2LA) Cp L 120 W/m2 · °C 0.00185 s1 (8530 kg/m3)(380 J/kg · °C)(0.02 m) b Then the temperature of the plate at t 7 min 420 s is determined from T (t ) T ebt → Ti T T (t ) 500 1 e(0.00185 s )(420 s) 20 500 It yields T (t ) 279°C which is practically identical to the result obtained above using the Heisler charts. Therefore, we can use lumped system analysis with confidence when the Biot number is sufficiently small. EXAMPLE 4–5 T = 200°C h = 80 W/ m2 ·°C Stainless steel shaft Ti = 600°C D = 20 cm FIGURE 4–21 Schematic for Example 4–5. Cooling of a Long Stainless Steel Cylindrical Shaft A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out of an oven at a uniform temperature of 600°C (Fig. 4–21). The shaft is then allowed to cool slowly in an environment chamber at 200°C with an average heat transfer coefficient of h 80 W/m2 · °C. Determine the temperature at the center of the shaft 45 min after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. SOLUTION A long cylindrical shaft at 600°C is allowed to cool slowly. The center temperature and the heat transfer per unit length are to be determined. Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the centerline. 2 The thermal properties of the shaft and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The properties of stainless steel 304 at room temperature are k 14.9 W/m · °C, 7900 kg/m3, Cp 477 J/kg · °C, and 3.95 106 m2/s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis The temperature within the shaft may vary with the radial distance r as well as time, and the temperature at a specified location at a given time can cen58933_ch04.qxd 9/10/2002 9:12 AM Page 227 227 CHAPTER 4 be determined from the Heisler charts. Noting that the radius of the shaft is ro 0.1 m, from Fig. 4–14 we have 14.9 W/m · °C k 1 1.86 Bi hro (80 W/m2 · °C)(0.1 m) (3.95 106 m2/s)(45 60 s) t2 1.07 (0.1 m)2 ro To T 0.40 Ti T and To T 0.4(Ti T) 200 0.4(600 200) 360°C Therefore, the center temperature of the shaft will drop from 600°C to 360°C in 45 min. To determine the actual heat transfer, we first need to calculate the maximum heat that can be transferred from the cylinder, which is the sensible energy of the cylinder relative to its environment. Taking L 1 m, m V ro2 L (7900 kg/m3) (0.1 m)2(1 m) 248.2 kg Qmax mCp(T Ti) (248.2 kg)(0.477 kJ/kg · °C)(600 200)°C 47,354 kJ The dimensionless heat transfer ratio is determined from Fig. 4–14c for a long cylinder to be Bi 1 1 0.537 1/Bi 1.86 h 2 t Bi2 (0.537)2(1.07) 0.309 k2 Q 0.62 Qmax Therefore, Q 0.62Qmax 0.62 (47,354 kJ) 29,360 kJ which is the total heat transfer from the shaft during the first 45 min of the cooling. ALTERNATIVE SOLUTION We could also solve this problem using the one-term solution relation instead of the transient charts. First we find the Biot number Bi hro (80 W/m2 · °C)(0.1 m) 0.537 14.9 W/m · °C k The coefficients 1 and A1 for a cylinder corresponding to this Bi are determined from Table 4–1 to be 1 0.970, A1 1.122 Substituting these values into Eq. 4–14 gives 0 To T 2 2 A1e1 1.122e(0.970) (1.07) 0.41 Ti T cen58933_ch04.qxd 9/10/2002 9:12 AM Page 228 228 HEAT TRANSFER and thus To T 0.41(Ti T) 200 0.41(600 200) 364°C The value of J1(1) for 1 0.970 is determined from Table 4–2 to be 0.430. Then the fractional heat transfer is determined from Eq. 4–18 to be J1(1) Q 0.430 1 20 1 2 0.41 0.636 Qmax 1 0.970 and thus Q 0.636Qmax 0.636 (47,354 kJ) 30,120 kJ Discussion The slight difference between the two results is due to the reading error of the charts. 4–3 Plane surface T h x 0 FIGURE 4–22 Schematic of a semi-infinite body. ■ TRANSIENT HEAT CONDUCTION IN SEMI-INFINITE SOLIDS A semi-infinite solid is an idealized body that has a single plane surface and extends to infinity in all directions, as shown in Fig. 4–22. This idealized body is used to indicate that the temperature change in the part of the body in which we are interested (the region close to the surface) is due to the thermal conditions on a single surface. The earth, for example, can be considered to be a semi-infinite medium in determining the variation of temperature near its surface. Also, a thick wall can be modeled as a semi-infinite medium if all we are interested in is the variation of temperature in the region near one of the surfaces, and the other surface is too far to have any impact on the region of interest during the time of observation. Consider a semi-infinite solid that is at a uniform temperature Ti. At time t 0, the surface of the solid at x 0 is exposed to convection by a fluid at a constant temperature T, with a heat transfer coefficient h. This problem can be formulated as a partial differential equation, which can be solved analytically for the transient temperature distribution T(x, t). The solution obtained is presented in Fig. 4–23 graphically for the nondimensionalized temperature defined as 1 (x, t) 1 T(x, t) T T(x, t ) Ti Ti T T Ti (4-21) against the dimensionless variable x/(2 t) for various values of the parameter h t/k. Note that the values on the vertical axis correspond to x 0, and thus represent the surface temperature. The curve h t/k corresponds to h → , which corresponds to the case of specified temperature T at the surface at x 0. That is, the case in which the surface of the semi-infinite body is suddenly brought to temperature T at t 0 and kept at T at all times can be handled by setting h to infinity. The specified surface temperature case is closely cen58933_ch04.qxd 9/10/2002 9:12 AM Page 229 229 CHAPTER 4 1.0 Ambient T(x, t) T, h 0.5 0.4 T(x, t) – T 1 – ————— = 1 – θ(x, t) Ti – T 0.3 3 2 0.2 x 1 0.5 0.4 0.3 0.2 0.1 0.1 0.05 0.04 h α t k = 0.0 5 0.03 0.02 0.01 0 0.25 0.5 0.75 1.0 1.25 1.5 x ξ = ——– 2 αt FIGURE 4–23 Variation of temperature with position and time in a semi-infinite solid initially at Ti subjected to convection to an environment at T with a convection heat transfer coefficient of h (from P. J. Schneider, Ref. 10). approximated in practice when condensation or boiling takes place on the surface. For a finite heat transfer coefficient h, the surface temperature approaches the fluid temperature T as the time t approaches infinity. The exact solution of the transient one-dimensional heat conduction problem in a semi-infinite medium that is initially at a uniform temperature of Ti and is suddenly subjected to convection at time t 0 has been obtained, and is expressed as T(x, t) Ti h t x hx h2t x erfc exp 2 erfc T Ti k k k 2 t 2 t (4-22) where the quantity erfc ( ) is the complementary error function, defined as erfc ( ) 1 2 e u2 du (4-23) 0 Despite its simple appearance, the integral that appears in the above relation cannot be performed analytically. Therefore, it is evaluated numerically for different values of , and the results are listed in Table 4–3. For the special case of h → , the surface temperature Ts becomes equal to the fluid temperature T, and Eq. 4–22 reduces to T(x, t) Ti x erfc Ts Ti 2 t (4-24) cen58933_ch04.qxd 9/10/2002 9:12 AM Page 230 230 HEAT TRANSFER TABLE 4–3 The complementary error function erfc () erfc () erfc () erfc () erfc () erfc () 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 1.00000 0.9774 0.9549 0.9324 0.9099 0.8875 0.8652 0.8431 0.8210 0.7991 0.7773 0.7557 0.7343 0.7131 0.6921 0.6714 0.6509 0.6306 0.6107 0.38 0.40 0.42 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.5910 0.5716 0.5525 0.5338 0.5153 0.4973 0.4795 0.4621 0.4451 0.4284 0.4121 0.3961 0.3806 0.3654 0.3506 0.3362 0.3222 0.3086 0.2953 0.76 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98 1.00 1.02 1.04 1.06 1.08 1.10 1.12 0.2825 0.2700 0.2579 0.2462 0.2349 0.2239 0.2133 0.2031 0.1932 0.1837 0.1746 0.1658 0.1573 0.1492 0.1413 0.1339 0.1267 0.1198 0.1132 1.14 1.16 1.18 1.20 1.22 1.24 1.26 1.28 1.30 1.32 1.34 1.36 1.38 1.40 1.42 1.44 1.46 1.48 1.50 0.1069 0.10090 0.09516 0.08969 0.08447 0.07950 0.07476 0.07027 0.06599 0.06194 0.05809 0.05444 0.05098 0.04772 0.04462 0.04170 0.03895 0.03635 0.03390 1.52 1.54 1.56 1.58 1.60 1.62 1.64 1.66 1.68 1.70 1.72 1.74 1.76 1.78 1.80 1.82 1.84 1.86 1.88 0.03159 0.02941 0.02737 0.02545 0.02365 0.02196 0.02038 0.01890 0.01751 0.01612 0.01500 0.01387 0.01281 0.01183 0.01091 0.01006 0.00926 0.00853 0.00784 1.90 1.92 1.94 1.96 1.98 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 3.20 3.40 3.60 0.00721 0.00662 0.00608 0.00557 0.00511 0.00468 0.00298 0.00186 0.00114 0.00069 0.00041 0.00024 0.00013 0.00008 0.00004 0.00002 0.00001 0.00000 0.00000 This solution corresponds to the case when the temperature of the exposed surface of the medium is suddenly raised (or lowered) to Ts at t 0 and is maintained at that value at all times. Although the graphical solution given in Fig. 4–23 is a plot of the exact analytical solution given by Eq. 4–23, it is subject to reading errors, and thus is of limited accuracy. EXAMPLE 4–6 Ts = –10°C Soil x Water pipe Ti = 15°C FIGURE 4–24 Schematic for Example 4–6. Minimum Burial Depth of Water Pipes to Avoid Freezing In areas where the air temperature remains below 0°C for prolonged periods of time, the freezing of water in underground pipes is a major concern. Fortunately, the soil remains relatively warm during those periods, and it takes weeks for the subfreezing temperatures to reach the water mains in the ground. Thus, the soil effectively serves as an insulation to protect the water from subfreezing temperatures in winter. The ground at a particular location is covered with snow pack at 10°C for a continuous period of three months, and the average soil properties at that location are k 0.4 W/m · °C and 0.15 106 m2/s (Fig. 4–24). Assuming an initial uniform temperature of 15°C for the ground, determine the minimum burial depth to prevent the water pipes from freezing. SOLUTION The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined. Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semiinfinite medium with a specified surface temperature of 10°C. 2 The thermal properties of the soil are constant. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 231 231 CHAPTER 4 Properties The properties of the soil are as given in the problem statement. Analysis The temperature of the soil surrounding the pipes will be 0°C after three months in the case of minimum burial depth. Therefore, from Fig. 4–23, we have h t k (since h → ) T (x, t ) T 0 (10) 1 0.6 1 Ti T 15 (10) x 0.36 2 t We note that t (90 days)(24 h/day)(3600 s/h) 7.78 106 s and thus x 2 t 2 0.36 (0.15 106 m2/s)(7.78 106 s) 0.77 m Therefore, the water pipes must be buried to a depth of at least 77 cm to avoid freezing under the specified harsh winter conditions. ALTERNATIVE SOLUTION The solution of this problem could also be determined from Eq. 4–24: T (x, t ) Ti erfc Ts Ti x 2 t → 0 15 x erfc 0.60 10 15 2 t The argument that corresponds to this value of the complementary error function is determined from Table 4–3 to be 0.37. Therefore, T h T h T(r, t) x 2 t 2 0.37 (0.15 106 m2/s)(7.78 106 s) 0.80 m Heat transfer Again, the slight difference is due to the reading error of the chart. (a) Long cylinder 4–4 ■ TRANSIENT HEAT CONDUCTION IN MULTIDIMENSIONAL SYSTEMS The transient temperature charts presented earlier can be used to determine the temperature distribution and heat transfer in one-dimensional heat conduction problems associated with a large plane wall, a long cylinder, a sphere, and a semi-infinite medium. Using a superposition approach called the product solution, these charts can also be used to construct solutions for the twodimensional transient heat conduction problems encountered in geometries such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or plate, and even three-dimensional problems associated with geometries such as a rectangular prism or a semi-infinite rectangular bar, provided that all surfaces of the solid are subjected to convection to the same fluid at temperature T h T(r, x, t) Heat transfer (b) Short cylinder (two-dimensional) FIGURE 4–25 The temperature in a short cylinder exposed to convection from all surfaces varies in both the radial and axial directions, and thus heat is transferred in both directions. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 232 232 HEAT TRANSFER T h Plane wall a ro Long cylinder FIGURE 4–26 A short cylinder of radius ro and height a is the intersection of a long cylinder of radius ro and a plane wall of thickness a. Plane wall T h T, with the same heat transfer coefficient h, and the body involves no heat generation (Fig. 4–25). The solution in such multidimensional geometries can be expressed as the product of the solutions for the one-dimensional geometries whose intersection is the multidimensional geometry. Consider a short cylinder of height a and radius ro initially at a uniform temperature Ti. There is no heat generation in the cylinder. At time t 0, the cylinder is subjected to convection from all surfaces to a medium at temperature T with a heat transfer coefficient h. The temperature within the cylinder will change with x as well as r and time t since heat transfer will occur from the top and bottom of the cylinder as well as its side surfaces. That is, T T(r, x, t) and thus this is a two-dimensional transient heat conduction problem. When the properties are assumed to be constant, it can be shown that the solution of this two-dimensional problem can be expressed as T(r, x, t) T Ti T short cylinder T(x, t) T Ti T plane wall infinite cylinder (4-25) That is, the solution for the two-dimensional short cylinder of height a and radius ro is equal to the product of the nondimensionalized solutions for the one-dimensional plane wall of thickness a and the long cylinder of radius ro, which are the two geometries whose intersection is the short cylinder, as shown in Fig. 4–26. We generalize this as follows: the solution for a multidimensional geometry is the product of the solutions of the one-dimensional geometries whose intersection is the multidimensional body. For convenience, the one-dimensional solutions are denoted by T(x, t) T T T T(r, t) T (r, t) T T T(x, t) T (x, t) T T wall(x, t) Plane wall semi-inf a i i i cyl b FIGURE 4–27 A long solid bar of rectangular profile a b is the intersection of two plane walls of thicknesses a and b. T(r, t) T Ti T plane wall infinite cylinder semi-infinite solid (4-26) For example, the solution for a long solid bar whose cross section is an a b rectangle is the intersection of the two infinite plane walls of thicknesses a and b, as shown in Fig. 4–27, and thus the transient temperature distribution for this rectangular bar can be expressed as T(x, y, t) T Ti T rectangular bar wall(x, t)wall(y, t) (4-27) The proper forms of the product solutions for some other geometries are given in Table 4–4. It is important to note that the x-coordinate is measured from the surface in a semi-infinite solid, and from the midplane in a plane wall. The radial distance r is always measured from the centerline. Note that the solution of a two-dimensional problem involves the product of two one-dimensional solutions, whereas the solution of a three-dimensional problem involves the product of three one-dimensional solutions. A modified form of the product solution can also be used to determine the total transient heat transfer to or from a multidimensional geometry by using the one-dimensional values, as shown by L. S. Langston in 1982. The cen58933_ch04.qxd 9/10/2002 9:13 AM Page 233 233 CHAPTER 4 TABLE 4–4 Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a uniform temperature Ti and exposed to convection from all surfaces to a medium at T x 0 ro r r x θ(r, t) = θcyl(r, t) Infinite cylinder r θ (x,r, t) = θcyl (r, t) θwall (x,t) Short cylinder θ (x,r, t) = θcyl (r, t) θsemi-inf (x, t) Semi-infinite cylinder y x x y z θ (x, t) = θsemi-inf (x, t) Semi-infinite medium θ (x,y,t) = θsemi-inf (x, t) θsemi-inf (y, t) Quarter-infinite medium x θ (x, y, z, t) = θsemi-inf (x, t) θsemi-inf (y, t) θsemi-inf (z,t) Corner region of a large medium 2L 2L y 0 x L x y z θ(x, t) = θwall(x, t) Infinite plate (or plane wall) θ(x, y, t) = θwall (x, t) θsemi-inf (y, t) Semi-infinite plate x θ (x,y,z, t) = θwall (x, t) θsemi-inf (y, t) θsemi-inf (z , t) Quarter-infinite plate y x z y x z y x θ(x, y, t) = θwall (x, t) θwall ( y, t) Infinite rectangular bar θ (x,y,z, t) = θwall (x, t) θwall (y, t) θsemi-inf (z , t) Semi-infinite rectangular bar θ(x,y,z, t) = θwall (x, t) θwall (y, t) θwall (z , t) Rectangular parallelepiped cen58933_ch04.qxd 9/10/2002 9:13 AM Page 234 234 HEAT TRANSFER transient heat transfer for a two-dimensional geometry formed by the intersection of two one-dimensional geometries 1 and 2 is Q Q max total, 2D Q Q 1-Q Q Q Q max 1 max 2 max 1 (4-28) Transient heat transfer for a three-dimensional body formed by the intersection of three one-dimensional bodies 1, 2, and 3 is given by Q Q max total, 3D Q Q 1-Q Q Q Q 11Q Q Q Q Q Q max 1 max 2 max 1 max 3 max 1 max (4-29) 2 The use of the product solution in transient two- and three-dimensional heat conduction problems is illustrated in the following examples. EXAMPLE 4–7 Cooling of a Short Brass Cylinder A short brass cylinder of diameter D 10 cm and height H 12 cm is initially at a uniform temperature Ti 120°C. The cylinder is now placed in atmospheric air at 25°C, where heat transfer takes place by convection, with a heat transfer coefficient of h 60 W/m2 · °C. Calculate the temperature at (a) the center of the cylinder and (b) the center of the top surface of the cylinder 15 min after the start of the cooling. SOLUTION A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The properties of brass at room temperature are k 110 W/m · °C and 33.9 106 m2/s (Table A-3). More accurate results can be obtained by using properties at average temperature. T = 25°C h = 60 W/ m2 ·°C x 0 L r Ti = 120°C ro Analysis (a) This short cylinder can physically be formed by the intersection of a long cylinder of radius ro 5 cm and a plane wall of thickness 2L 12 cm, as shown in Fig. 4–28. The dimensionless temperature at the center of the plane wall is determined from Figure 4–13a to be L FIGURE 4–28 Schematic for Example 4–7. t (3.39 105 m2/s)(900 s) 8.48 L2 (0.06 m)2 1 k 110 W/m · °C Bi hL (60 W/m2 · °C)(0.06 m) 30.6 wall(0, t ) T (0, t ) T 0.8 Ti T cen58933_ch04.qxd 9/10/2002 9:13 AM Page 235 235 CHAPTER 4 Similarly, at the center of the cylinder, we have t (3.39 105 m2/s)(900 s) 12.2 ro2 (0.05 m)2 1 k 110 W/m · °C 36.7 Bi hro (60 W/m2 · °C)(0.05 m) cyl(0, t ) T (0, t ) T 0.5 Ti T Therefore, T (0, 0, t ) T Ti T short cylinder wall(0, t ) cyl(0, t ) 0.8 0.5 0.4 and T (0, 0, t ) T 0.4(Ti T) 25 0.4(120 25) 63°C This is the temperature at the center of the short cylinder, which is also the center of both the long cylinder and the plate. (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r 0), but at the outer surface of the plane wall (x L). Therefore, we first need to find the surface temperature of the wall. Noting that x L 0.06 m, x 0.06 m 1 L 0.06 m 1 k 110 W/m · °C Bi hL (60 W/m2 · °C)(0.06 m) 30.6 T (L, t ) T 0.98 To T Then wall(L, t ) T (L, t ) T To T T (L, t ) T 0.98 0.8 0.784 Ti T To T Ti T Therefore, T (L, 0, t ) T Ti T short cylinder wall(L, t )cyl(0, t ) 0.784 0.5 0.392 and T(L, 0, t ) T 0.392(Ti T) 25 0.392(120 25) 62.2°C which is the temperature at the center of the top surface of the cylinder. EXAMPLE 4–8 Heat Transfer from a Short Cylinder Determine the total heat transfer from the short brass cylinder ( 8530 kg/m3, Cp 0.380 kJ/kg · °C) discussed in Example 4–7. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 236 236 HEAT TRANSFER SOLUTION We first determine the maximum heat that can be transferred from the cylinder, which is the sensible energy content of the cylinder relative to its environment: m V ro2 L (8530 kg/m3) (0.05 m)2(0.06 m) 4.02 kg Qmax mCp(Ti T) (4.02 kg)(0.380 kJ/kg · °C)(120 25)°C 145.1 kJ Then we determine the dimensionless heat transfer ratios for both geometries. For the plane wall, it is determined from Fig. 4–13c to be Bi 1 1 0.0327 1/Bi 30.6 h 2t Bi2 (0.0327)2(8.48) 0.0091 k2 Similarly, for the cylinder, we have Bi 1 1 0.0272 1/Bi 36.7 h 2t Bi2 (0.0272)2(12.2) 0.0090 k2 Q Q 0.23 plane wall max Q Q max infinite cylinder 0.47 Then the heat transfer ratio for the short cylinder is, from Eq. 4–28, Q Q max short cyl Q Q 1 Q Q Q Q max 1 max 2 max 1 0.23 0.47(1 0.23) 0.592 Therefore, the total heat transfer from the cylinder during the first 15 min of cooling is Q 0.592Qmax 0.592 (145.1 kJ) 85.9 kJ EXAMPLE 4–9 Cooling of a Long Cylinder by Water A semi-infinite aluminum cylinder of diameter D 20 cm is initially at a uniform temperature Ti 200°C. The cylinder is now placed in water at 15°C where heat transfer takes place by convection, with a heat transfer coefficient of h 120 W/m2 · °C. Determine the temperature at the center of the cylinder 15 cm from the end surface 5 min after the start of the cooling. SOLUTION A semi-infinite aluminum cylinder is cooled by water. The temperature at the center of the cylinder 15 cm from the end surface is to be determined. Assumptions 1 Heat conduction in the semi-infinite cylinder is twodimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 237 237 CHAPTER 4 Properties The properties of aluminum at room temperature are k 237 W/m · °C and 9.71 106 m2/s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis This semi-infinite cylinder can physically be formed by the intersection of an infinite cylinder of radius ro 10 cm and a semi-infinite medium, as shown in Fig. 4–29. We will solve this problem using the one-term solution relation for the cylinder and the analytic solution for the semi-infinite medium. First we consider the infinitely long cylinder and evaluate the Biot number: Bi 2 hro (120 W/m · °C)(0.1 m) 0.05 237 W/m · °C k The coefficients 1 and A1 for a cylinder corresponding to this Bi are determined from Table 4–1 to be 1 0.3126 and A1 1.0124. The Fourier number in this case is t (9.71 105 m2/s)(5 60 s) 2.91 0.2 ro2 (0.1 m)2 and thus the one-term approximation is applicable. Substituting these values into Eq. 4–14 gives 0 cyl(0, t ) A1e1 1.0124e(0.3126) (2.91) 0.762 2 2 The solution for the semi-infinite solid can be determined from 1 semi-inf(x, t ) erfc h t x x 2 t exp hxk hk t erfc 2 t k 2 2 First we determine the various quantities in parentheses: 0.15 m x 0.44 2 t 2 (9.71 105 m2/s)(5 60 s) h t (120 W/m2 · °C)(9.71 105 m2/s)(300 s) 0.086 237 W/m · °C k hx (120 W/m2 · °C)(0.15 m) 0.0759 237 W/m · °C k h t h 2t k k2 2 (0.086) 0.0074 2 Substituting and evaluating the complementary error functions from Table 4–3, semi-inf(x, t ) 1 erfc (0.44) exp (0.0759 0.0074) erfc (0.44 0.086) 1 0.5338 exp (0.0833) 0.457 0.963 Now we apply the product solution to get T (x, 0, t ) T Ti T semi-infinite cylinder semi-inf(x, t )cyl(0, t ) 0.963 0.762 0.734 T = 15°C h = 120 W/ m2 ·°C Ti = 200°C D = 20 cm x x = 15 cm 0r FIGURE 4–29 Schematic for Example 4–9. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 238 238 HEAT TRANSFER and T (x, 0, t ) T 0.734(Ti T) 15 0.734(200 15) 151°C which is the temperature at the center of the cylinder 15 cm from the exposed bottom surface. EXAMPLE 4–10 5°F 35°F Steak 1 in. FIGURE 4–30 Schematic for Example 4–10. Refrigerating Steaks while Avoiding Frostbite In a meat processing plant, 1-in.-thick steaks initially at 75°F are to be cooled in the racks of a large refrigerator that is maintained at 5°F (Fig. 4–30). The steaks are placed close to each other, so that heat transfer from the 1-in.-thick edges is negligible. The entire steak is to be cooled below 45°F, but its temperature is not to drop below 35°F at any point during refrigeration to avoid “frostbite.” The convection heat transfer coefficient and thus the rate of heat transfer from the steak can be controlled by varying the speed of a circulating fan inside. Determine the heat transfer coefficient h that will enable us to meet both temperature constraints while keeping the refrigeration time to a minimum. The steak can be treated as a homogeneous layer having the properties 74.9 lbm/ft3, Cp 0.98 Btu/lbm · °F, k 0.26 Btu/h · ft · °F, and 0.0035 ft2/h. SOLUTION Steaks are to be cooled in a refrigerator maintained at 5°F. The heat transfer coefficient that will allow cooling the steaks below 45°F while avoiding frostbite is to be determined. Assumptions 1 Heat conduction through the steaks is one-dimensional since the steaks form a large layer relative to their thickness and there is thermal symmetry about the center plane. 2 The thermal properties of the steaks and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The properties of the steaks are as given in the problem statement. Analysis The lowest temperature in the steak will occur at the surfaces and the highest temperature at the center at a given time, since the inner part will be the last place to be cooled. In the limiting case, the surface temperature at x L 0.5 in. from the center will be 35°F, while the midplane temperature is 45°F in an environment at 5°F. Then, from Fig. 4–13b, we obtain x 0.5 in. 1 L 0.5 in. T (L, t ) T 35 5 0.75 To T 45 5 k 1 1.5 Bi hL which gives h 1 k 0.26 Btu/h · ft · °F 4.16 Btu/h · ft2 · °F 1.5 L 1.5(0.5/12 ft) Discussion The convection heat transfer coefficient should be kept below this value to satisfy the constraints on the temperature of the steak during refrigeration. We can also meet the constraints by using a lower heat transfer coefficient, but doing so would extend the refrigeration time unnecessarily. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 239 239 CHAPTER 4 The restrictions that are inherent in the use of Heisler charts and the oneterm solutions (or any other analytical solutions) can be lifted by using the numerical methods discussed in Chapter 5. TOPIC OF SPECIAL INTEREST Refrigeration and Freezing of Foods Control of Microorganisms in Foods Microorganisms such as bacteria, yeasts, molds, and viruses are widely encountered in air, water, soil, living organisms, and unprocessed food items, and cause off-flavors and odors, slime production, changes in the texture and appearances, and the eventual spoilage of foods. Holding perishable foods at warm temperatures is the primary cause of spoilage, and the prevention of food spoilage and the premature degradation of quality due to microorganisms is the largest application area of refrigeration. The first step in controlling microorganisms is to understand what they are and the factors that affect their transmission, growth, and destruction. Of the various kinds of microorganisms, bacteria are the prime cause for the spoilage of foods, especially moist foods. Dry and acidic foods create an undesirable environment for the growth of bacteria, but not for the growth of yeasts and molds. Molds are also encountered on moist surfaces, cheese, and spoiled foods. Specific viruses are encountered in certain animals and humans, and poor sanitation practices such as keeping processed foods in the same area as the uncooked ones and being careless about handwashing can cause the contamination of food products. When contamination occurs, the microorganisms start to adapt to the new environmental conditions. This initial slow or no-growth period is called the lag phase, and the shelf life of a food item is directly proportional to the length of this phase (Fig. 4–31). The adaptation period is followed by an exponential growth period during which the population of microorganisms can double two or more times every hour under favorable conditions unless drastic sanitation measures are taken. The depletion of nutrients and the accumulation of toxins slow down the growth and start the death period. The rate of growth of microorganisms in a food item depends on the characteristics of the food itself such as the chemical structure, pH level, presence of inhibitors and competing microorganisms, and water activity as well as the environmental conditions such as the temperature and relative humidity of the environment and the air motion (Fig. 4–32). Microorganisms need food to grow and multiply, and their nutritional needs are readily provided by the carbohydrates, proteins, minerals, and vitamins in a food. Different types of microorganisms have different nutritional needs, and the types of nutrients in a food determine the types of microorganisms that may dwell on them. The preservatives added to the This section can be skipped without a loss of continuity. Microorganism population Lag Exponential growth Death Time FIGURE 4–31 Typical growth curve of microorganisms. 50 ENVIRONMENT 0 100 % Temperature Oxygen level Relative humidity Air motion FOOD Water content Chemical composition Contamination level The use of inhibitors pH level FIGURE 4–32 The factors that affect the rate of growth of microorganisms. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 240 240 HEAT TRANSFER Rate of growth Temperature FIGURE 4–33 The rate of growth of microorganisms in a food product increases exponentially with increasing environmental temperature. food may also inhibit the growth of certain microorganisms. Different kinds of microorganisms that exist compete for the same food supply, and thus the composition of microorganisms in a food at any time depends on the initial make-up of the microorganisms. All living organisms need water to grow, and microorganisms cannot grow in foods that are not sufficiently moist. Microbiological growth in refrigerated foods such as fresh fruits, vegetables, and meats starts at the exposed surfaces where contamination is most likely to occur. Fresh meat in a package left in a room will spoil quickly, as you may have noticed. A meat carcass hung in a controlled environment, on the other hand, will age healthily as a result of dehydration on the outer surface, which inhibits microbiological growth there and protects the carcass. Microorganism growth in a food item is governed by the combined effects of the characteristics of the food and the environmental factors. We cannot do much about the characteristics of the food, but we certainly can alter the environmental conditions to more desirable levels through heating, cooling, ventilating, humidification, dehumidification, and control of the oxygen levels. The growth rate of microorganisms in foods is a strong function of temperature, and temperature control is the single most effective mechanism for controlling the growth rate. Microorganisms grow best at “warm” temperatures, usually between 20 and 60°C. The growth rate declines at high temperatures, and death occurs at still higher temperatures, usually above 70°C for most microorganisms. Cooling is an effective and practical way of reducing the growth rate of microorganisms and thus extending the shelf life of perishable foods. A temperature of 4°C or lower is considered to be a safe refrigeration temperature. Sometimes a small increase in refrigeration temperature may cause a large increase in the growth rate, and thus a considerable decrease in shelf life of the food (Fig. 4–33). The growth rate of some microorganisms, for example, doubles for each 3°C rise in temperature. Another factor that affects microbiological growth and transmission is the relative humidity of the environment, which is a measure of the water content of the air. High humidity in cold rooms should be avoided since condensation that forms on the walls and ceiling creates the proper environment for mold growth and buildups. The drip of contaminated condensate onto food products in the room poses a potential health hazard. Different microorganisms react differently to the presence of oxygen in the environment. Some microorganisms such as molds require oxygen for growth, while some others cannot grow in the presence of oxygen. Some grow best in low-oxygen environments, while others grow in environments regardless of the amount of oxygen. Therefore, the growth of certain microorganisms can be controlled by controlling the amount of oxygen in the environment. For example, vacuum packaging inhibits the growth of microorganisms that require oxygen. Also, the storage life of some fruits can be extended by reducing the oxygen level in the storage room. Microorganisms in food products can be controlled by (1) preventing contamination by following strict sanitation practices, (2) inhibiting growth by altering the environmental conditions, and (3) destroying the organisms by heat treatment or chemicals. The best way to minimize contamination cen58933_ch04.qxd 9/10/2002 9:13 AM Page 241 241 CHAPTER 4 in food processing areas is to use fine air filters in ventilation systems to capture the dust particles that transport the bacteria in the air. Of course, the filters must remain dry since microorganisms can grow in wet filters. Also, the ventilation system must maintain a positive pressure in the food processing areas to prevent any airborne contaminants from entering inside by infiltration. The elimination of condensation on the walls and the ceiling of the facility and the diversion of plumbing condensation drip pans of refrigerators to the drain system are two other preventive measures against contamination. Drip systems must be cleaned regularly to prevent microbiological growth in them. Also, any contact between raw and cooked food products should be minimized, and cooked products must be stored in rooms with positive pressures. Frozen foods must be kept at 18°C or below, and utmost care should be exercised when food products are packaged after they are frozen to avoid contamination during packaging. The growth of microorganisms is best controlled by keeping the temperature and relative humidity of the environment in the desirable range. Keeping the relative humidity below 60 percent, for example, prevents the growth of all microorganisms on the surfaces. Microorganisms can be destroyed by heating the food product to high temperatures (usually above 70°C), by treating them with chemicals, or by exposing them to ultraviolet light or solar radiation. Distinction should be made between survival and growth of microorganisms. A particular microorganism that may not grow at some low temperature may be able to survive at that temperature for a very long time (Fig. 4–34). Therefore, freezing is not an effective way of killing microorganisms. In fact, some microorganism cultures are preserved by freezing them at very low temperatures. The rate of freezing is also an important consideration in the refrigeration of foods since some microorganisms adapt to low temperatures and grow at those temperatures when the cooling rate is very low. Refrigeration and Freezing of Foods The storage life of fresh perishable foods such as meats, fish, vegetables, and fruits can be extended by several days by storing them at temperatures just above freezing, usually between 1 and 4°C. The storage life of foods can be extended by several months by freezing and storing them at subfreezing temperatures, usually between 18 and 35°C, depending on the particular food (Fig. 4–35). Refrigeration slows down the chemical and biological processes in foods, and the accompanying deterioration and loss of quality and nutrients. Sweet corn, for example, may lose half of its initial sugar content in one day at 21°C, but only 5 percent of it at 0°C. Fresh asparagus may lose 50 percent of its vitamin C content in one day at 20°C, but in 12 days at 0°C. Refrigeration also extends the shelf life of products. The first appearance of unsightly yellowing of broccoli, for example, may be delayed by three or more days by refrigeration. Early attempts to freeze food items resulted in poor-quality products because of the large ice crystals that formed. It was determined that the rate of freezing has a major effect on the size of ice crystals and the quality, texture, and nutritional and sensory properties of many foods. During slow Z Z Z Microorganisms Frozen food FIGURE 4–34 Freezing may stop the growth of microorganisms, but it may not necessarily kill them. Freezer –18 to –35°C Refrigerator 1 to 4°C Frozen foods Fresh foods FIGURE 4–35 Recommended refrigeration and freezing temperatures for most perishable foods. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 242 242 HEAT TRANSFER Temperature Cooling (above freezing) Beginning of freezing Freezing End of freezing Cooling (below freezing) Time FIGURE 4–36 Typical freezing curve of a food item. TABLE 4–5 Thermal properties of beef Quantity Typical value Average density 1070 kg/m3 Specific heat: Above freezing 3.14 kJ/kg · °C Below freezing 1.70 kJ/kg · °C Freezing point –2.7°C Latent heat of fusion 249 kJ/kg Thermal 0.41 W/m · °C conductivity (at 6°C) freezing, ice crystals can grow to a large size, whereas during fast freezing a large number of ice crystals start forming at once and are much smaller in size. Large ice crystals are not desirable since they can puncture the walls of the cells, causing a degradation of texture and a loss of natural juices during thawing. A crust forms rapidly on the outer layer of the product and seals in the juices, aromatics, and flavoring agents. The product quality is also affected adversely by temperature fluctuations of the storage room. The ordinary refrigeration of foods involves cooling only without any phase change. The freezing of foods, on the other hand, involves three stages: cooling to the freezing point (removing the sensible heat), freezing (removing the latent heat), and further cooling to the desired subfreezing temperature (removing the sensible heat of frozen food), as shown in Figure 4–36. Beef Products Meat carcasses in slaughterhouses should be cooled as fast as possible to a uniform temperature of about 1.7°C to reduce the growth rate of microorganisms that may be present on carcass surfaces, and thus minimize spoilage. The right level of temperature, humidity, and air motion should be selected to prevent excessive shrinkage, toughening, and discoloration. The deep body temperature of an animal is about 39°C, but this temperature tends to rise a couple of degrees in the midsections after slaughter as a result of the heat generated during the biological reactions that occur in the cells. The temperature of the exposed surfaces, on the other hand, tends to drop as a result of heat losses. The thickest part of the carcass is the round, and the center of the round is the last place to cool during chilling. Therefore, the cooling of the carcass can best be monitored by inserting a thermometer deep into the central part of the round. About 70 percent of the beef carcass is water, and the carcass is cooled mostly by evaporative cooling as a result of moisture migration toward the surface where evaporation occurs. But this shrinking translates into a loss of salable mass that can amount to 2 percent of the total mass during an overnight chilling. To prevent excessive loss of mass, carcasses are usually washed or sprayed with water prior to cooling. With adequate care, spray chilling can eliminate carcass cooling shrinkage almost entirely. The average total mass of dressed beef, which is normally split into two sides, is about 300 kg, and the average specific heat of the carcass is about 3.14 kJ/kg · °C (Table 4–5). The chilling room must have a capacity equal to the daily kill of the slaughterhouse, which may be several hundred. A beef carcass is washed before it enters the chilling room and absorbs a large amount of water (about 3.6 kg) at its surface during the washing process. This does not represent a net mass gain, however, since it is lost by dripping or evaporation in the chilling room during cooling. Ideally, the carcass does not lose or gain any net weight as it is cooled in the chilling room. However, it does lose about 0.5 percent of the total mass in the holding room as it continues to cool. The actual product loss is determined by first weighing the dry carcass before washing and then weighing it again after it is cooled. The refrigerated air temperature in the chilling room of beef carcasses must be sufficiently high to avoid freezing and discoloration on the outer cen58933_ch04.qxd 9/10/2002 9:13 AM Page 243 243 CHAPTER 4 40 Maximum carcass temp (deep round) Temperature, °C 30 Avg. carcass temp (all parts) 20 Minimum carcass temp (chuck surface) 10 0 Room air temp. –10 0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 Time from start of chill, hours surfaces of the carcass. This means a long residence time for the massive beef carcasses in the chilling room to cool to the desired temperature. Beef carcasses are only partially cooled at the end of an overnight stay in the chilling room. The temperature of a beef carcass drops to 1.7 to 7°C at the surface and to about 15°C in mid parts of the round in 10 h. It takes another day or two in the holding room maintained at 1 to 2°C to complete chilling and temperature equalization. But hog carcasses are fully chilled during that period because of their smaller size. The air circulation in the holding room is kept at minimum levels to avoid excessive moisture loss and discoloration. The refrigeration load of the holding room is much smaller than that of the chilling room, and thus it requires a smaller refrigeration system. Beef carcasses intended for distant markets are shipped the day after slaughter in refrigerated trucks, where the rest of the cooling is done. This practice makes it possible to deliver fresh meat long distances in a timely manner. The variation in temperature of the beef carcass during cooling is given in Figure 4–37. Initially, the cooling process is dominated by sensible heat transfer. Note that the average temperature of the carcass is reduced by about 28°C (from 36 to 8°C) in 20 h. The cooling rate of the carcass could be increased by lowering the refrigerated air temperature and increasing the air velocity, but such measures also increase the risk of surface freezing. Most meats are judged on their tenderness, and the preservation of tenderness is an important consideration in the refrigeration and freezing of meats. Meat consists primarily of bundles of tiny muscle fibers bundled together inside long strings of connective tissues that hold it together. The tenderness of a certain cut of beef depends on the location of the cut, the age, and the activity level of the animal. Cuts from the relatively inactive mid-backbone section of the animal such as short loins, sirloin, and prime ribs are more tender than the cuts from the active parts such as the legs and the neck (Fig. 4–38). The more active the animal, the more the connective tissue, and the tougher the meat. The meat of an older animal is more flavorful, however, and is preferred for stewing since the toughness of the meat does not pose a problem for moist-heat cooking such as boiling. The FIGURE 4–37 Typical cooling curve of a beef carcass in the chilling and holding rooms at an average temperature of 0°C (from ASHRAE, Handbook: Refrigeration, Ref. 3, Chap. 11, Fig. 2). Chuck Rib Short loin Brisket Flank Foreshank Short plate Sirloin Round FIGURE 4–38 Various cuts of beef (from National Livestock and Meat Board). cen58933_ch04.qxd 9/10/2002 9:13 AM Page 244 244 HEAT TRANSFER Tenderness scale 10 5 0 5 Time in days 10 FIGURE 4–39 Variation of tenderness of meat stored at 2°C with time after slaughter. Meat freezer Air – 40 to –30°C 2.5 to 5 m/s Meat FIGURE 4–40 The freezing time of meat can be reduced considerably by using low temperature air at high velocity. TABLE 4–6 Storage life of frozen meat products at different storage temperatures (from ASHRAE Handbook: Refrigeration, Chap. 10, Table 7) Storage Life, Months Temperature Product Beef Lamb Veal Pork Chopped beef Cooked foods 12°C 18°C 23°C 4–12 3–8 3–4 2–6 3–4 2–3 6–18 12–24 6–16 12–18 4–14 8 4–12 8–15 4–6 8 2–4 protein collagen, which is the main component of the connective tissue, softens and dissolves in hot and moist environments and gradually transforms into gelatin, and tenderizes the meat. The old saying “one should either cook an animal immediately after slaughter or wait at least two days” has a lot of truth in it. The biomechanical reactions in the muscle continue after the slaughter until the energy supplied to the muscle to do work diminishes. The muscle then stiffens and goes into rigor mortis. This process begins several hours after the animal is slaughtered and continues for 12 to 36 h until an enzymatic action sets in and tenderizes the connective tissue, as shown in Figure 4–39. It takes about seven days to complete tenderization naturally in storage facilities maintained at 2°C. Electrical stimulation also causes the meat to be tender. To avoid toughness, fresh meat should not be frozen before rigor mortis has passed. You have probably noticed that steaks are tender and rather tasty when they are hot but toughen as they cool. This is because the gelatin that formed during cooking thickens as it cools, and meat loses its tenderness. So it is no surprise that first-class restaurants serve their steak on hot thick plates that keep the steaks warm for a long time. Also, cooking softens the connective tissue but toughens the tender muscle fibers. Therefore, barbecuing on low heat for a long time results in a tough steak. Variety meats intended for long-term storage must be frozen rapidly to reduce spoilage and preserve quality. Perhaps the first thought that comes to mind to freeze meat is to place the meat packages into the freezer and wait. But the freezing time is too long in this case, especially for large boxes. For example, the core temperature of a 4–cm-deep box containing 32 kg of variety meat can be as high as 16°C 24 h after it is placed into a 30°C freezer. The freezing time of large boxes can be shortened considerably by adding some dry ice into it. A more effective method of freezing, called quick chilling, involves the use of lower air temperatures, 40 to 30°C, with higher velocities of 2.5 m/s to 5 m/s over the product (Fig. 4–40). The internal temperature should be lowered to 4°C for products to be transferred to a storage freezer and to 18°C for products to be shipped immediately. The rate of freezing depends on the package material and its insulating properties, the thickness of the largest box, the type of meat, and the capacity of the refrigeration system. Note that the air temperature will rise excessively during initial stages of freezing and increase the freezing time if the capacity of the system is inadequate. A smaller refrigeration system will be adequate if dry ice is to be used in packages. Shrinkage during freezing varies from about 0.5 to 1 percent. Although the average freezing point of lean meat can be taken to be 2°C with a latent heat of 249 kJ/kg, it should be remembered that freezing occurs over a temperature range, with most freezing occurring between 1 and 4°C. Therefore, cooling the meat through this temperature range and removing the latent heat takes the most time during freezing. Meat can be kept at an internal temperature of 2 to 1°C for local use and storage for under a week. Meat must be frozen and stored at much lower temperatures for long-term storage. The lower the storage temperature, the longer the storage life of meat products, as shown in Table 4–6. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 245 245 CHAPTER 4 The internal temperature of carcasses entering the cooling sections varies from 38 to 41°C for hogs and from 37 to 39°C for lambs and calves. It takes about 15 h to cool the hogs and calves to the recommended temperature of 3 to 4°C. The cooling-room temperature is maintained at 1 to 0°C and the temperature difference between the refrigerant and the cooling air is kept at about 6°C. Air is circulated at a rate of about 7 to 12 air changes per hour. Lamb carcasses are cooled to an internal temperature of 1 to 2°C, which takes about 12 to 14 h, and are held at that temperature with 85 to 90 percent relative humidity until shipped or processed. The recommended rate of air circulation is 50 to 60 air changes per hour during the first 4 to 6 h, which is reduced to 10 to 12 changes per hour afterward. Freezing does not seem to affect the flavor of meat much, but it affects the quality in several ways. The rate and temperature of freezing may influence color, tenderness, and drip. Rapid freezing increases tenderness and reduces the tissue damage and the amount of drip after thawing. Storage at low freezing temperatures causes significant changes in animal fat. Frozen pork experiences more undesirable changes during storage because of its fat structure, and thus its acceptable storage period is shorter than that of beef, veal, or lamb. Meat storage facilities usually have a refrigerated shipping dock where the orders are assembled and shipped out. Such docks save valuable storage space from being used for shipping purposes and provide a more acceptable working environment for the employees. Packing plants that ship whole or half carcasses in bulk quantities may not need a shipping dock; a load-out door is often adequate for such cases. A refrigerated shipping dock, as shown in Figure 4–41, reduces the refrigeration load of freezers or coolers and prevents temperature fluctuations in the storage area. It is often adequate to maintain the shipping docks at 4 to 7°C for the coolers and about 1.5°C for the freezers. The dew point of the dock air should be below the product temperature to avoid condensation on the surface of the products and loss of quality. The rate of airflow through the loading doors and other openings is proportional to the square root of the temperature difference, and thus reducing the temperature difference at the opening by half by keeping the shipping dock at the average temperature reduces the rate of airflow into the dock and thus into the freezer by 1 0.5 0.3, or 30 percent. Also, the air that flows into the freezer is already cooled to about 1.5°C by the refrigeration unit of the dock, which represents about 50 percent of the cooling load of the incoming air. Thus, the net effect of the refrigerated shipping dock is a reduction of the infiltration load of the freezer by about 65 percent since 1 0.7 0.5 0.65. The net gain is equal to the difference between the reduction of the infiltration load of the freezer and the refrigeration load of the shipping dock. Note that the dock refrigerators operate at much higher temperatures (1.5°C instead of about 23°C), and thus they consume much less power for the same amount of cooling. Poultry Products Poultry products can be preserved by ice-chilling to 1 to 2°C or deep chilling to about 2°C for short-term storage, or by freezing them to 18°C or Freezer –23°C Refrigerated dock 1.5°C Sliding door Refrigerated truck FIGURE 4–41 A refrigerated truck dock for loading frozen items to a refrigerated truck. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 246 246 HEAT TRANSFER Air chilling H 2O 1000 g 980 g Immersion chilling H 2O 1000 g 1050 g FIGURE 4–42 Air chilling causes dehydration and thus weight loss for poultry, whereas immersion chilling causes a weight gain as a result of water absorption. below for long-term storage. Poultry processing plants are completely automated, and the small size of the birds makes continuous conveyor line operation feasible. The birds are first electrically stunned before cutting to prevent struggling. Following a 90- to 120-s bleeding time, the birds are scalded by immersing them into a tank of warm water, usually at 51 to 55°C, for up to 120 s to loosen the feathers. Then the feathers are removed by featherpicking machines, and the eviscerated carcass is washed thoroughly before chilling. The internal temperature of the birds ranges from 24 to 35°C after washing, depending on the temperatures of the ambient air and the washing water as well as the extent of washing. To control the microbial growth, the USDA regulations require that poultry be chilled to 4°C or below in less than 4 h for carcasses of less than 1.8 kg, in less than 6 h for carcasses of 1.8 to 3.6 kg. and in less than 8 h for carcasses more than 3.6 kg. Meeting these requirements today is not difficult since the slow air chilling is largely replaced by the rapid immersion chilling in tanks of slush ice. Immersion chilling has the added benefit that it not only prevents dehydration, but it causes a net absorption of water and thus increases the mass of salable product. Cool air chilling of unpacked poultry can cause a moisture loss of 1 to 2 percent, while water immersion chilling can cause a moisture absorption of 4 to 15 percent (Fig. 4–42). Water spray chilling can cause a moisture absorption of up to 4 percent. Most water absorbed is held between the flesh and the skin and the connective tissues in the skin. In immersion chilling, some soluble solids are lost from the carcass to the water, but the loss has no significant effect on flavor. Many slush ice tank chillers today are replaced by continuous flow-type immersion slush ice chillers. Continuous slush ice-chillers can reduce the internal temperature of poultry from 32 to 4°C in about 30 minutes at a rate up to 10, 000 birds per hour. Ice requirements depend on the inlet and exit temperatures of the carcass and the water, but 0.25 kg of ice per kg of carcass is usually adequate. However, bacterial contamination such as salmonella remains a concern with this method, and it may be necessary to chloride the water to control contamination. Tenderness is an important consideration for poultry products just as it is for red meat, and preserving tenderness is an important consideration in the cooling and freezing of poultry. Birds cooked or frozen before passing through rigor mortis remain very tough. Natural tenderization begins soon after slaughter and is completed within 24 h when birds are held at 4°C. Tenderization is rapid during the first three hours and slows down thereafter. Immersion in hot water and cutting into the muscle adversely affect tenderization. Increasing the scalding temperature or the scalding time has been observed to increase toughness, and decreasing the scalding time has been observed to increase tenderness. The beating action of mechanical feather-picking machines causes considerable toughening. Therefore, it is recommended that any cutting be done after tenderization. Cutting up the bird into pieces before natural tenderization is completed reduces tenderness considerably. Therefore, it is recommended that any cutting be done after tenderization. Rapid chilling of poultry can also have a toughening cen58933_ch04.qxd 9/10/2002 9:13 AM Page 247 247 CHAPTER 4 Storage life (days) 12 10 8 6 4 2 0 –2 0 5 10 15 20 25 Storage temperature, °C FIGURE 4–43 The storage life of fresh poultry decreases exponentially with increasing storage temperature. 9 8 7 Freezing time, hours effect. It is found that the tenderization process can be speeded up considerably by a patented electrical stunning process. Poultry products are highly perishable, and thus they should be kept at the lowest possible temperature to maximize their shelf life. Studies have shown that the populations of certain bacteria double every 36 h at 2°C, 14 h at 0°C, 7 h at 5°C, and less than 1 h at 25°C (Fig. 4–43). Studies have also shown that the total bacterial counts on birds held at 2°C for 14 days are equivalent to those held at 10°C for 5 days or 24°C for 1 day. It has also been found that birds held at 1°C had 8 days of additional shelf life over those held at 4°C. The growth of microorganisms on the surfaces of the poultry causes the development of an off-odor and bacterial slime. The higher the initial amount of bacterial contamination, the faster the sliming occurs. Therefore, good sanitation practices during processing such as cleaning the equipment frequently and washing the carcasses are as important as the storage temperature in extending shelf life. Poultry must be frozen rapidly to ensure a light, pleasing appearance. Poultry that is frozen slowly appears dark and develops large ice crystals that damage the tissue. The ice crystals formed during rapid freezing are small. Delaying freezing of poultry causes the ice crystals to become larger. Rapid freezing can be accomplished by forced air at temperatures of 23 to 40°C and velocities of 1.5 to 5 m/s in air-blast tunnel freezers. Most poultry is frozen this way. Also, the packaged birds freeze much faster on open shelves than they do in boxes. If poultry packages must be frozen in boxes, then it is very desirable to leave the boxes open or to cut holes on the boxes in the direction of airflow during freezing. For best results, the blast tunnel should be fully loaded across its cross-section with even spacing between the products to assure uniform airflow around all sides of the packages. The freezing time of poultry as a function of refrigerated air temperature is given in Figure 4–44. Thermal properties of poultry are given in Table 4–7. Other freezing methods for poultry include sandwiching between cold plates, immersion into a refrigerated liquid such as glycol or calcium chloride brine, and cryogenic cooling with liquid nitrogen. Poultry can be frozen in several hours by cold plates. Very high freezing rates can be obtained by immersing the packaged birds into a low-temperature brine. The freezing time of birds in 29°C brine can be as low as 20 min, depending on the size of the bird (Fig. 4–45). Also, immersion freezing produces a very appealing light appearance, and the high rates of heat transfer make continuous line operation feasible. It also has lower initial and maintenance costs than forced air, but leaks into the packages through some small holes or cracks remain a concern. The convection heat transfer coefficient is 17 W/m2 · °C for air at 29°C and 2.5 m/s whereas it is 170 W/m2 · °C for sodium chloride brine at 18°C and a velocity of 0.02 m/s. Sometimes liquid nitrogen is used to crust freeze the poultry products to 73°C. The freezing is then completed with air in a holding room at 23°C. Properly packaged poultry products can be stored frozen for up to about a year at temperatures of 18°C or lower. The storage life drops considerably at higher (but still below-freezing) temperatures. Significant changes Giblets Inside surface 13 mm depth Under skin 6 5 4 3 2 1 0 – 84 –73 –62 –51 –40 –29 –18 Air temperature, degrees Celsius –7 Note: Freezing time is the time required for temperature to fall from 0 to –4°C. The values are for 2.3 to 3.6 kg chickens with initial temperature of 0 to 2°C and with air velocity of 2.3 to 2.8 m/s. FIGURE 4–44 The variation of freezing time of poultry with air temperature (from van der Berg and Lentz, Ref. 11). cen58933_ch04.qxd 9/10/2002 9:13 AM Page 248 248 HEAT TRANSFER 5 0 Giblets Temperature, °C –5 FIGURE 4–45 The variation of temperature of the breast of 6.8-kg turkeys initially at 1°C with depth during immersion cooling at 29°C (from van der Berg and Lentz, Ref. 11). TABLE 4–7 Thermal properties of poultry Quantity Typical value Average density: Muscle 1070 kg/m3 Skin 1030 kg/m3 Specific heat: Above freezing 2.94 kJ/kg · °C Below freezing 1.55 kJ/kg · °C Freezing point 2.8°C Latent heat of fusion 247 kJ/kg Thermal conductivity: (in W/m · °C) Breast muscle 0.502 at 20°C 1.384 at 20°C 1.506 at 40°C Dark muscle 1.557 at 40°C –10 –15 Inside surface –20 38 mm depth 25 mm depth 13 mm depth 6.5 mm depth Under skin Skin surface –25 –30 –35 0 25 50 75 100 150 125 Time, min. 175 200 225 250 occur in flavor and juiciness when poultry is frozen for too long, and a stale rancid odor develops. Frozen poultry may become dehydrated and experience freezer burn, which may reduce the eye appeal of the product and cause toughening of the affected area. Dehydration and thus freezer burn can be controlled by humidification, lowering the storage temperature, and packaging the product in essentially impermeable film. The storage life can be extended by packing the poultry in an oxygen-free environment. The bacterial counts in precooked frozen products must be kept at safe levels since bacteria may not be destroyed completely during the reheating process at home. Frozen poultry can be thawed in ambient air, water, refrigerator, or oven without any significant difference in taste. Big birds like turkey should be thawed safely by holding it in a refrigerator at 2 to 4°C for two to four days, depending on the size of the bird. They can also be thawed by immersing them into cool water in a large container for 4 to 6 h, or holding them in a paper bag. Care must be exercised to keep the bird’s surface cool to minimize microbiological growth when thawing in air or water. EXAMPLE 4–5 Chilling of Beef Carcasses in a Meat Plant The chilling room of a meat plant is 18 m 20 m 5.5 m in size and has a capacity of 450 beef carcasses. The power consumed by the fans and the lights of the chilling room are 26 and 3 kW, respectively, and the room gains heat through its envelope at a rate of 13 kW. The average mass of beef carcasses is 285 kg. The carcasses enter the chilling room at 36°C after they are washed to facilitate evaporative cooling and are cooled to 15°C in 10 h. The water is expected to evaporate at a rate of 0.080 kg/s. The air enters the evaporator section of the refrigeration system at 0.7°C and leaves at 2°C. The air side of the evaporator is heavily finned, and the overall heat transfer coefficient of the evaporator based on the air side is 20 W/m2 · °C. Also, the average temperature difference between the air and the refrigerant in the evaporator is 5.5°C. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 249 249 CHAPTER 4 Determine (a) the refrigeration load of the chilling room, (b) the volume flow rate of air, and (c) the heat transfer surface area of the evaporator on the air side, assuming all the vapor and the fog in the air freezes in the evaporator. SOLUTION The chilling room of a meat plant with a capacity of 450 beef carcasses is considered. The cooling load, the airflow rate, and the heat transfer area of the evaporator are to be determined. Assumptions 1 Water evaporates at a rate of 0.080 kg/s. 2 All the moisture in the air freezes in the evaporator. Properties The heat of fusion and the heat of vaporization of water at 0°C are 333.7 kJ/kg and 2501 kJ/kg (Table A-9). The density and specific heat of air at 0°C are 1.292 kg/m3 and 1.006 kJ/kg · °C (Table A-15). Also, the specific heat of beef carcass is determined from the relation in Table A-7b to be Cp 1.68 2.51 (water content) 1.68 2.51 0.58 3.14 kJ/kg · °C Analysis (a) A sketch of the chilling room is given in Figure 4–46. The amount of beef mass that needs to be cooled per unit time is mbeef (Total beef mass cooled)/(Cooling time) (450 carcasses)(285 kg/carcass)/(10 3600 s) 3.56 kg/s Lights, 3 kW 13 kW Beef carcass The product refrigeration load can be viewed as the energy that needs to be removed from the beef carcass as it is cooled from 36 to 15°C at a rate of 3.56 kg/s and is determined to be Evaporation 0.080 kg/s 36°C 285 kg Refrigerated air · Q beef (mC T)beef (3.56 kg/s)(3.14 kJ/kg · °C)(36 15)°C 235 kW Fans, 26 kW Then the total refrigeration load of the chilling room becomes · · · · · Q total, chillroom Q beef Q fan Q lights Q heat gain 235 26 3 13 277 kW 0.7°C Evaporator –2°C The amount of carcass cooling due to evaporative cooling of water is · Q beef, evaporative (mhfg)water (0.080 kg/s)(2490 kJ/kg) 199 kW which is 199/235 85 percent of the total product cooling load. The remaining 15 percent of the heat is transferred by convection and radiation. (b) Heat is transferred to air at the rate determined above, and the temperature of the air rises from 2°C to 0.7°C as a result. Therefore, the mass flow rate of air is m· air Q· air 277 kW 102.0 kg/s (Cp Tair) (1.006 kJ/kg · °C)[0.7 (2)°C] Then the volume flow rate of air becomes m· air · Vair air 102 kg/s 78.9 m3/s 1.292 kg/m3 (c) Normally the heat transfer load of the evaporator is the same as the refrigeration load. But in this case the water that enters the evaporator as a liquid is · Qevap FIGURE 4–46 Schematic for Example 4–5. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 250 250 HEAT TRANSFER frozen as the temperature drops to 2°C, and the evaporator must also remove the latent heat of freezing, which is determined from · Q freezing (m· hlatent)water (0.080 kg/s)(334 kJ/kg) 27 kW Therefore, the total rate of heat removal at the evaporator is · · · Q evaporator Q total, chill room Q freezing 277 27 304 kW Then the heat transfer surface area of the evaporator on the air side is deter· mined from Q evaporator (UA)airside T, A Q· evaporator U T 304,000 W 2764 m2 (20 W/m2 · °C)(5.5°C) Obviously, a finned surface must be used to provide such a large surface area on the air side. SUMMARY In this chapter we considered the variation of temperature with time as well as position in one- or multidimensional systems. We first considered the lumped systems in which the temperature varies with time but remains uniform throughout the system at any time. The temperature of a lumped body of arbitrary shape of mass m, volume V, surface area As, density , and specific heat Cp initially at a uniform temperature Ti that is exposed to convection at time t 0 in a medium at temperature T with a heat transfer coefficient h is expressed as Q mCp[T(t) Ti] The amount of heat transfer reaches its upper limit when the body reaches the surrounding temperature T. Therefore, the maximum heat transfer between the body and its surroundings is Qmax mCp (T Ti ) Bi where hAs h Cp V Cp Lc (kJ) The error involved in lumped system analysis is negligible when T(t) T ebt Ti T b (kJ) (1/s) is a positive quantity whose dimension is (time)1. This relation can be used to determine the temperature T(t) of a body at time t or, alternately, the time t required for the temperature to reach a specified value T(t). Once the temperature T(t) at time t is available, the rate of convection heat transfer between the body and its environment at that time can be determined from Newton’s law of cooling as · (W) Q (t) hAs [T(t) T] The total amount of heat transfer between the body and the surrounding medium over the time interval t 0 to t is simply the change in the energy content of the body, hLc 0.1 k where Bi is the Biot number and Lc V/As is the characteristic length. When the lumped system analysis is not applicable, the variation of temperature with position as well as time can be determined using the transient temperature charts given in Figs. 4–13, 4–14, 4–15, and 4–23 for a large plane wall, a long cylinder, a sphere, and a semi-infinite medium, respectively. These charts are applicable for one-dimensional heat transfer in those geometries. Therefore, their use is limited to situations in which the body is initially at a uniform temperature, all surfaces are subjected to the same thermal conditions, and the body does not involve any heat generation. These charts can also be used to determine the total heat transfer from the body up to a specified time t. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 251 251 CHAPTER 4 Using a one-term approximation, the solutions of onedimensional transient heat conduction problems are expressed analytically as Plane wall: Cylinder: Sphere: T(x, t) T Ti T 2 A1e1 cos (1x/L), 0.2 T(r, t) T (r, t)cyl Ti T 2 A1e1 J0(1r/ro), 0.2 T(r, t) T (r, t)sph Ti T 2 sin(1r/ro) , 0.2 A1e1 1r/ro (x, t)wall where the constants A1 and 1 are functions of the Bi number only, and their values are listed in Table 4–1 against the Bi number for all three geometries. The error involved in oneterm solutions is less than 2 percent when 0.2. Using the one-term solutions, the fractional heat transfers in different geometries are expressed as Plane wall: Cylinder: Sphere: sin 1 1 max wall J1(1) 1 20, cyl 1 max cyl Q sin 1 1 cos 1 1 30, sph Qmax sph 31 Q Q Q Q 1 0, wall The analytic solution for one-dimensional transient heat conduction in a semi-infinite solid subjected to convection is given by T(x, t) Ti x hx h2t erfc exp 2 T Ti k k 2 t h t x erfc k 2 t where the quantity erfc () is the complementary error function. For the special case of h → , the surface temperature Ts becomes equal to the fluid temperature T, and the above equation reduces to T(x, t) Ti x erfc Ts Ti 2 t (Ts constant) Using a clever superposition principle called the product solution these charts can also be used to construct solutions for the two-dimensional transient heat conduction problems encountered in geometries such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or plate, and even three-dimensional problems associated with geometries such as a rectangular prism or a semi-infinite rectangular bar, provided that all surfaces of the solid are subjected to convection to the same fluid at temperature T, with the same convection heat transfer coefficient h, and the body involves no heat generation. The solution in such multidimensional geometries can be expressed as the product of the solutions for the one-dimensional geometries whose intersection is the multidimensional geometry. The total heat transfer to or from a multidimensional geometry can also be determined by using the one-dimensional values. The transient heat transfer for a two-dimensional geometry formed by the intersection of two one-dimensional geometries 1 and 2 is Q Q max total, 2D Q Q 1-Q Q Q Q max 1 max 2 max 1 Transient heat transfer for a three-dimensional body formed by the intersection of three one-dimensional bodies 1, 2, and 3 is given by Q Q max total, 3D Q Q 1-Q Q Q Q 11Q Q Q Q Q Q max 1 max 2 max 1 max 3 max 1 max 2 REFERENCES AND SUGGESTED READING 1. ASHRAE. Handbook of Fundamentals. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, Inc., 1993. 2. ASHRAE. Handbook of Fundamentals. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, Inc., 1994. 3. H. S. Carslaw and J. C. Jaeger. Conduction of Heat in Solids. 2nd ed. London: Oxford University Press, 1959. H. Gröber, S. Erk, and U. Grigull. Fundamentals of Heat Transfer. New York: McGraw-Hill, 1961. 5. M. P. Heisler. “Temperature Charts for Induction and Constant Temperature Heating.” ASME Transactions 69 (1947), pp. 227–36. 6. H. Hillman. Kitchen Science. Mount Vernon, NY: Consumers Union, 1981. 7. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 252 252 HEAT TRANSFER L. S. Langston. “Heat Transfer from Multidimensional Objects Using One-Dimensional Solutions for Heat Loss.” International Journal of Heat and Mass Transfer 25 (1982), pp. 149–50. 9. M. N. Özis,ik, Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. P. J. Schneider. Conduction Heat Transfer. Reading, MA: Addison-Wesley, 1955. 11. L. van der Berg and C. P. Lentz. “Factors Affecting Freezing Rate and Appearance of Eviscerated Poultry Frozen in Air.” Food Technology 12 (1958). PROBLEMS Lumped System Analysis 4–1C What is lumped system analysis? When is it applicable? 4–2C Consider heat transfer between two identical hot solid bodies and the air surrounding them. The first solid is being cooled by a fan while the second one is allowed to cool naturally. For which solid is the lumped system analysis more likely to be applicable? Why? 4–3C Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air. For which solid is the lumped system analysis more likely to be applicable? Why? 4–4C Consider a hot baked potato on a plate. The temperature of the potato is observed to drop by 4°C during the first minute. Will the temperature drop during the second minute be less than, equal to, or more than 4°C? Why? Cool air Hot baked potato FIGURE P4–4C 4–5C Consider a potato being baked in an oven that is maintained at a constant temperature. The temperature of the potato is observed to rise by 5°C during the first minute. Will the temperature rise during the second minute be less than, equal to, or more than 5°C? Why? 4–6C What is the physical significance of the Biot number? Is the Biot number more likely to be larger for highly conducting solids or poorly conducting ones? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 4–7C Consider two identical 4–kg pieces of roast beef. The first piece is baked as a whole, while the second is baked after being cut into two equal pieces in the same oven. Will there be any difference between the cooking times of the whole and cut roasts? Why? 4–8C Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temperature and are exposed to convection in the same environment. Which do you think will cool faster, the cylinder or the sphere? Why? 4–9C In what medium is the lumped system analysis more likely to be applicable: in water or in air? Why? 4–10C For which solid is the lumped system analysis more likely to be applicable: an actual apple or a golden apple of the same size? Why? 4–11C For which kind of bodies made of the same material is the lumped system analysis more likely to be applicable: slender ones or well-rounded ones of the same volume? Why? 4–12 Obtain relations for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder of radius ro, and a sphere of radius ro. 4–13 Obtain a relation for the time required for a lumped system to reach the average temperature 12 (Ti T), where Ti is the initial temperature and T is the temperature of the environment. 4–14 The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.2-mm-diameter sphere. The properties of the junction are k 35 W/m · °C, 8500 kg/m3, and Cp 320 J/kg · °C, and the heat transfer coefficient between the junction and the gas is h 65 W/m2 · °C. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature Answer: 38.5 s difference. 4–15E In a manufacturing facility, 2-in.-diameter brass balls (k 64.1 Btu/h · ft · °F, 532 lbm/ft3, and Cp 0.092 Btu/lbm · °F) initially at 250°F are quenched in a water bath at 120°F for a period of 2 min at a rate of 120 balls per minute. If the convection heat transfer coefficient is 42 Btu/h · ft2 · °F, determine (a) the temperature of the balls after quenching and (b) the rate at which heat needs to be removed from the water in order to keep its temperature constant at 120°F. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 253 253 CHAPTER 4 250°F Brass balls 120°F Water bath FIGURE P4–15E 4–16E 4–20 Consider a 1000-W iron whose base plate is made of 0.5-cm-thick aluminum alloy 2024-T6 ( 2770 kg/m3, Cp 875 J/kg · °C, 7.3 105 m2/s). The base plate has a surface area of 0.03 m2. Initially, the iron is in thermal equilibrium with the ambient air at 22°C. Taking the heat transfer coefficient at the surface of the base plate to be 12 W/m2 · °C and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach 140°C. Is it realistic to assume the plate temperature to be uniform at all times? Repeat Problem 4–15E for aluminum balls. Air 22°C 4–17 To warm up some milk for a baby, a mother pours milk into a thin-walled glass whose diameter is 6 cm. The height of the milk in the glass is 7 cm. She then places the glass into a large pan filled with hot water at 60°C. The milk is stirred constantly, so that its temperature is uniform at all times. If the heat transfer coefficient between the water and the glass is 120 W/m2 · °C, determine how long it will take for the milk to warm up from 3°C to 38°C. Take the properties of the milk to be the same as those of water. Can the milk in this case be Answer: 5.8 min treated as a lumped system? Why? 4–18 Repeat Problem 4–17 for the case of water also being stirred, so that the heat transfer coefficient is doubled to 240 W/m2 · °C. 4–19E During a picnic on a hot summer day, all the cold drinks disappeared quickly, and the only available drinks were those at the ambient temperature of 80°F. In an effort to cool a 12-fluid-oz drink in a can, which is 5 in. high and has a diameter of 2.5 in., a person grabs the can and starts shaking it in the iced water of the chest at 32°F. The temperature of the drink can be assumed to be uniform at all times, and the heat transfer coefficient between the iced water and the aluminum can is 30 Btu/h · ft2 · °F. Using the properties of water for the drink, estimate how long it will take for the canned drink to cool to 45°F. 1000 W iron FIGURE P4–20 4–21 Reconsider Problem 4–20. Using EES (or other) software, investigate the effects of the heat transfer coefficient and the final plate temperature on the time it will take for the plate to reach this temperature. Let the heat transfer coefficient vary from 5 W/m2 · °C to 25 W/m2 · °C and the temperature from 30°C to 200°C. Plot the time as functions of the heat transfer coefficient and the temperature, and discuss the results. 4–22 Stainless steel ball bearings ( 8085 kg/m3, k 15.1 W/m · °C, Cp 0.480 kJ/kg · °C, and 3.91 106 m2/s) having a diameter of 1.2 cm are to be quenched in water. The balls leave the oven at a uniform temperature of 900°C and are exposed to air at 30°C for a while before they are dropped into the water. If the temperature of the balls is not to fall below 850°C prior to quenching and the heat transfer coefficient in the air is 125 W/m2 · °C, determine how long they can stand in Answer: 3.7 s the air before being dropped into the water. 4–23 Carbon steel balls ( 7833 kg/m3, k 54 W/m · °C, Cp 0.465 kJ/kg · °C, and 1.474 106 m2/s) 8 mm in Air, 35°C Furnace 900°C FIGURE P4–19E FIGURE P4–23 Steel ball 100°C cen58933_ch04.qxd 9/10/2002 9:13 AM Page 254 254 HEAT TRANSFER diameter are annealed by heating them first to 900°C in a furnace and then allowing them to cool slowly to 100°C in ambient air at 35°C. If the average heat transfer coefficient is 75 W/m2 · °C, determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air. 4–24 Reconsider Problem 4–23. Using EES (or other) software, investigate the effect of the initial temperature of the balls on the annealing time and the total rate of heat transfer. Let the temperature vary from 500°C to 1000°C. Plot the time and the total rate of heat transfer as a function of the initial temperature, and discuss the results. 4–25 An electronic device dissipating 30 W has a mass of 20 g, a specific heat of 850 J/kg · °C, and a surface area of 5 cm2. The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of 25°C. Taking the heat transfer coefficient to be 12 W/m2 · °C, determine the temperature of the device at the end of the 5-min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of 200 g and a surface area of 80 cm2? Assume the device and the heat sink to be nearly isothermal. Would you use lumped system analysis or the transient temperature charts when determining the midpoint temperature of the sphere? Why? 4–33 A student calculates that the total heat transfer from a spherical copper ball of diameter 15 cm initially at 200°C and its environment at a constant temperature of 25°C during the first 20 min of cooling is 4520 kJ. Is this result reasonable? Why? 4–34 An ordinary egg can be approximated as a 5.5-cmdiameter sphere whose properties are roughly k 0.6 W/m · °C and 0.14 106 m2/s. The egg is initially at a uniform temperature of 8°C and is dropped into boiling water at 97°C. Taking the convection heat transfer coefficient to be h 1400 W/m2 · °C, determine how long it will take for the center of the egg to reach 70°C. Boiling water Egg Ti = 8°C Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects 4–26C What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not? For example, is it proper to use this model when finding the temperatures near the bottom or top surfaces of a cylinder? Explain. 4–27C Can the transient temperature charts in Fig. 4–13 for a plane wall exposed to convection on both sides be used for a plane wall with one side exposed to convection while the other side is insulated? Explain. 4–28C Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time? 4–29C What is the physical significance of the Fourier number? Will the Fourier number for a specified heat transfer problem double when the time is doubled? 97°C FIGURE P4–34 4–35 Reconsider Problem 4–34. Using EES (or other) software, investigate the effect of the final center temperature of the egg on the time it will take for the center to reach this temperature. Let the temperature vary from 50°C to 95°C. Plot the time versus the temperature, and discuss the results. 4–36 In a production facility, 3-cm-thick large brass plates (k 110 W/m · °C, 8530 kg/m3, Cp 380 J/kg · °C, and 33.9 106 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 80 W/m2 · °C, determine the surface temperature of the plates when they come out of the oven. Furnace, 700°C 4–30C How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient? 4–31C A body at an initial temperature of Ti is brought into a medium at a constant temperature of T. How can you determine the maximum possible amount of heat transfer between the body and the surrounding medium? 4–32C The Biot number during a heat transfer process between a sphere and its surroundings is determined to be 0.02. 3 cm Brass plate 25°C FIGURE P4–36 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 255 255 CHAPTER 4 4–37 Reconsider Problem 4–36. Using EES (or other) software, investigate the effects of the temperature of the oven and the heating time on the final surface temperature of the plates. Let the oven temperature vary from 500°C to 900°C and the time from 2 min to 30 min. Plot the surface temperature as the functions of the oven temperature and the time, and discuss the results. 4–38 A long 35-cm-diameter cylindrical shaft made of stainless steel 304 (k 14.9 W/m · °C, 7900 kg/m3, Cp 477 J/kg · °C, and 3.95 106 m2/s) comes out of an oven at a uniform temperature of 400°C. The shaft is then allowed to cool slowly in a chamber at 150°C with an average convection heat transfer coefficient of h 60 W/m2 · °C. Determine the temperature at the center of the shaft 20 min after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Answers: 390°C, 16,015 kJ/m 4–39 Reconsider Problem 4–38. Using EES (or other) software, investigate the effect of the cooling time on the final center temperature of the shaft and the amount of heat transfer. Let the time vary from 5 min to 60 min. Plot the center temperature and the heat transfer as a function of the time, and discuss the results. 4–40E Long cylindrical AISI stainless steel rods (k 7.74 Btu/h · ft · °F and 0.135 ft2/h) of 4-in. diameter are heattreated by drawing them at a velocity of 10 ft/min through a 30-ft-long oven maintained at 1700°F. The heat transfer coefficient in the oven is 20 Btu/h · ft2 · °F. If the rods enter the oven at 85°F, determine their centerline temperature when they leave. 500°C in a fireplace with a heat transfer coefficient of 13.6 W/m2 · °C on the surface. If the ignition temperature of the wood is 420°C, determine how long it will be before the log ignites. 4–43 In Betty Crocker’s Cookbook, it is stated that it takes 2 h 45 min to roast a 3.2-kg rib initially at 4.5°C “rare” in an oven maintained at 163°C. It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers 60°C. The rib can be treated as a homogeneous spherical object with the properties 1200 kg/m3, Cp 4.1 kJ/kg · °C, k 0.45 W/m · °C, and 0.91 107 m2/s. Determine (a) the heat transfer coefficient at the surface of the rib, (b) the temperature of the outer surface of the rib when it is done, and (c) the amount of heat transferred to the rib. (d) Using the values obtained, predict how long it will take to roast this rib to “medium” level, which occurs when the innermost temperature of the rib reaches 71°C. Compare your result to the listed value of 3 h 20 min. If the roast rib is to be set on the counter for about 15 min before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about 4°C below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Answers: (a) 156.9 W/m2 · °C, (b) 159.5°C, (c) 1629 kJ, (d) 3.0 h Oven, 163°C Oven 1700°F 10 ft/min 30 ft Stainless steel 85°F FIGURE P4–40E 4–41 In a meat processing plant, 2-cm-thick steaks (k 0.45 W/m · °C and 0.91 107 m2/s) that are initially at 25°C are to be cooled by passing them through a refrigeration room at 11°C. The heat transfer coefficient on both sides of the steaks is 9 W/m2 · °C. If both surfaces of the steaks are to be cooled to 2°C, determine how long the steaks should be kept in the refrigeration room. 4–42 A long cylindrical wood log (k 0.17 W/m · °C and 1.28 107 m2/s) is 10 cm in diameter and is initially at a uniform temperature of 10°C. It is exposed to hot gases at Rib Ti = 4.5°C FIGURE P4–43 4–44 Repeat Problem 4–43 for a roast rib that is to be “welldone” instead of “rare.” A rib is considered to be well-done when its center temperature reaches 77°C, and the roasting in this case takes about 4 h 15 min. 4–45 For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that is initially at 8°C is dropped into the boiling water at 100°C. The heat transfer coefficient at the surface of the egg is estimated to be 800 W/m2 · °C. If the egg is considered cooked when its center temperature reaches 60°C, determine how long the egg should be kept in the boiling water. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 256 256 HEAT TRANSFER 4–46 Repeat Problem 4–45 for a location at 1610-m elevation such as Denver, Colorado, where the boiling temperature of water is 94.4°C. 4–47 The author and his 6-year-old son have conducted the following experiment to determine the thermal conductivity of a hot dog. They first boiled water in a large pan and measured the temperature of the boiling water to be 94°C, which is not surprising, since they live at an elevation of about 1650 m in Reno, Nevada. They then took a hot dog that is 12.5 cm long and 2.2 cm in diameter and inserted a thermocouple into the midpoint of the hot dog and another thermocouple just under the skin. They waited until both thermocouples read 20°C, which is the ambient temperature. They then dropped the hot dog into boiling water and observed the changes in both temperatures. Exactly 2 min after the hot dog was dropped into the boiling water, they recorded the center and the surface temperatures to be 59°C and 88°C, respectively. The density of the hot dog can be taken to be 980 kg/m3, which is slightly less than the density of water, since the hot dog was observed to be floating in water while being almost completely immersed. The specific heat of a hot dog can be taken to be 3900 J/kg · °C, which is slightly less than that of water, since a hot dog is mostly water. Using transient temperature charts, determine (a) the thermal diffusivity of the hot dog, (b) the thermal conductivity of the hot dog, and (c) the convection heat transfer coefficient. Answers: (a) 2.02 107 m2/s, (b) 0.771 W/m · °C, (c) 467 W/m2 · °C. Boiling water 94°C Refrigerator 5°F Chicken Ti = 72°F FIGURE P4–49E minimum. The chicken can be treated as a homogeneous spherical object having the properties 74.9 lbm/ft3, Cp 0.98 Btu/lbm · °F, k 0.26 Btu/h · ft · °F, and 0.0035 ft2/h. 4–50 A person puts a few apples into the freezer at 15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2 · °C. Treating the apples as 9-cm-diameter spheres and taking their properties to be 840 kg/m3, Cp 3.81 kJ/kg · °C, k 0.418 W/m · °C, and 1.3 107 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. 4–51 Tsurface HOT DOG Tcenter FIGURE P4–47 4–48 Using the data and the answers given in Problem 4–47, determine the center and the surface temperatures of the hot dog 4 min after the start of the cooking. Also determine the amount of heat transferred to the hot dog. 4–49E In a chicken processing plant, whole chickens averaging 5 lb each and initially at 72°F are to be cooled in the racks of a large refrigerator that is maintained at 5°F. The entire chicken is to be cooled below 45°F, but the temperature of the chicken is not to drop below 35°F at any point during refrigeration. The convection heat transfer coefficient and thus the rate of heat transfer from the chicken can be controlled by varying the speed of a circulating fan inside. Determine the heat transfer coefficient that will enable us to meet both temperature constraints while keeping the refrigeration time to a Reconsider Problem 4–50. Using EES (or other) software, investigate the effect of the initial temperature of the apples on the final center and surface temperatures and the amount of heat transfer. Let the initial temperature vary from 2°C to 30°C. Plot the center temperature, the surface temperature, and the amount of heat transfer as a function of the initial temperature, and discuss the results. 4–52 Citrus fruits are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy them. Consider an 8-cm-diameter orange that is initially at Ambient air –15°C Orange Ti = 15°C FIGURE P4–52 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 257 257 CHAPTER 4 15°C. A cold front moves in one night, and the ambient temperature suddenly drops to 6°C, with a heat transfer coefficient of 15 W/m2 · °C. Using the properties of water for the orange and assuming the ambient conditions to remain constant for 4 h before the cold front moves out, determine if any part of the orange will freeze that night. 4–53 An 8-cm-diameter potato ( 1100 kg/m3, Cp 3900 J/kg · °C, k 0.6 W/m · °C, and 1.4 107 m2/s) that is initially at a uniform temperature of 25°C is baked in an oven at 170°C until a temperature sensor inserted to the center of the potato indicates a reading of 70°C. The potato is then taken out of the oven and wrapped in thick towels so that almost no heat is lost from the baked potato. Assuming the heat transfer coefficient in the oven to be 25 W/m2 · °C, determine (a) how long the potato is baked in the oven and (b) the final equilibrium temperature of the potato after it is wrapped. 4–54 White potatoes (k 0.50 W/m · °C and 0.13 106 m2/s) that are initially at a uniform temperature of 25°C and have an average diameter of 6 cm are to be cooled by refrigerated air at 2°C flowing at a velocity of 4 m/s. The average heat transfer coefficient between the potatoes and the air is experimentally determined to be 19 W/m2 · °C. Determine how long it will take for the center temperature of the potatoes to drop to 6°C. Also, determine if any part of the potatoes will experience chilling injury during this process. Air 2°C 4 m/s Potato Ti = 25°C FIGURE P4–54 4–55E Oranges of 2.5-in. diameter (k 0.26 Btu/h · ft · °F and 1.4 106 ft2/s) initially at a uniform temperature of 78°F are to be cooled by refrigerated air at 25°F flowing at a velocity of 1 ft/s. The average heat transfer coefficient between the oranges and the air is experimentally determined to be 4.6 Btu/h · ft2 · °F. Determine how long it will take for the center temperature of the oranges to drop to 40°F. Also, determine if any part of the oranges will freeze during this process. 4–56 A 65-kg beef carcass (k 0.47 W/m · °C and 0.13 106 m2/s) initially at a uniform temperature of 37°C is to be cooled by refrigerated air at 6°C flowing at a velocity of 1.8 m/s. The average heat transfer coefficient between the carcass and the air is 22 W/m2 · °C. Treating the carcass as a cylinder of diameter 24 cm and height 1.4 m and disregarding heat transfer from the base and top surfaces, determine how long it will take for the center temperature of the carcass to drop to 4°C. Also, determine if any part of the carcass will Answer: 14.0 h freeze during this process. Air – 6°C 1.8 m/s Beef 37°C FIGURE P4–56 4–57 Layers of 23-cm-thick meat slabs (k 0.47 W/m · °C and 0.13 106 m2/s) initially at a uniform temperature of 7°C are to be frozen by refrigerated air at 30°C flowing at a velocity of 1.4 m/s. The average heat transfer coefficient between the meat and the air is 20 W/m2 · °C. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to 18°C. Also, determine the surface temperature of the meat slab at that time. 4–58E Layers of 6-in.-thick meat slabs (k 0.26 Btu/h · ft · °F and 1.4 106 ft2/s) initially at a uniform temperature of 50°F are cooled by refrigerated air at 23°F to a temperature of 36°F at their center in 12 h. Estimate the average heat transfer coefficient during this cooling process. Answer: 1.5 Btu/h · ft2 · °F 4–59 Chickens with an average mass of 1.7 kg (k 0.45 W/m · °C and 0.13 106 m2/s) initially at a uniform temperature of 15°C are to be chilled in agitated brine at 10°C. The average heat transfer coefficient between the chicken and the brine is determined experimentally to be 440 W/m2 · °C. Taking the average density of the chicken to be 0.95 g/cm3 and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in 2 h and 30 min. Also, determine if any part of the chicken will freeze during this process. Chicken 1.7 kg Brine –10°C FIGURE P4–59 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 258 258 HEAT TRANSFER Transient Heat Conduction in Semi-Infinite Solids 4–60C What is a semi-infinite medium? Give examples of solid bodies that can be treated as semi-infinite mediums for heat transfer purposes. 4–61C Under what conditions can a plane wall be treated as a semi-infinite medium? 4–62C Consider a hot semi-infinite solid at an initial temperature of Ti that is exposed to convection to a cooler medium at a constant temperature of T, with a heat transfer coefficient of h. Explain how you can determine the total amount of heat transfer from the solid up to a specified time to. 4–63 In areas where the air temperature remains below 0°C for prolonged periods of time, the freezing of water in underground pipes is a major concern. Fortunately, the soil remains relatively warm during those periods, and it takes weeks for the subfreezing temperatures to reach the water mains in the ground. Thus, the soil effectively serves as an insulation to protect the water from the freezing atmospheric temperatures in winter. The ground at a particular location is covered with snow pack at 8°C for a continuous period of 60 days, and the average soil properties at that location are k 0.35 W/m · °C and 0.15 806 m2/s. Assuming an initial uniform temperature of 8°C for the ground, determine the minimum burial depth to prevent the water pipes from freezing. 4–64 The soil temperature in the upper layers of the earth varies with the variations in the atmospheric conditions. Before a cold front moves in, the earth at a location is initially at a uniform temperature of 10°C. Then the area is subjected to a temperature of 10°C and high winds that resulted in a convection heat transfer coefficient of 40 W/m2 · °C on the earth’s surface for a period of 10 h. Taking the properties of the soil at that location to be k 0.9 W/m · °C and 1.6 105 m2/s, determine the soil temperature at distances 0, 10, 20, and 50 cm from the earth’s surface at the end of this 10-h period. Winds, –10°C furnace and the surrounding air are in thermal equilibrium at 70°F. The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at 1800°F with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to 70.1°F. Answer: 181 min 4–67 A thick wood slab (k 0.17 W/m · °C and 1.28 107 m2/s) that is initially at a uniform temperature of 25°C is exposed to hot gases at 550°C for a period of 5 minutes. The heat transfer coefficient between the gases and the wood slab is 35 W/m2 · °C. If the ignition temperature of the wood is 450°C, determine if the wood will ignite. 4–68 A large cast iron container (k 52 W/m · °C and 1.70 105 m2/s) with 5-cm-thick walls is initially at a uniform temperature of 0°C and is filled with ice at 0°C. Now the outer surfaces of the container are exposed to hot water at 60°C with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be 250 W/m2 · °C, determine the rate of heat transfer to the ice through a 1.2-m-wide and 2-m-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to 0.1°C. Hot water 60°C Cast iron chest Ice 0°C 5 cm FIGURE P4–68 Transient Heat Conduction in Multidimensional Systems Soil 10°C FIGURE P4–64 4–65 Reconsider Problem 4–64. Using EES (or other) software, plot the soil temperature as a function of the distance from the earth’s surface as the distance varies from 0 m to 1m, and discuss the results. 4–66E The walls of a furnace are made of 1.5-ft-thick concrete (k 0.64 Btu/h · ft · °F and 0.023 ft2/h). Initially, the 4–69C What is the product solution method? How is it used to determine the transient temperature distribution in a twodimensional system? 4–70C How is the product solution used to determine the variation of temperature with time and position in threedimensional systems? 4–71C A short cylinder initially at a uniform temperature Ti is subjected to convection from all of its surfaces to a medium at temperature T. Explain how you can determine the temperature of the midpoint of the cylinder at a specified time t. 4–72C Consider a short cylinder whose top and bottom surfaces are insulated. The cylinder is initially at a uniform temperature Ti and is subjected to convection from its side surface cen58933_ch04.qxd 9/10/2002 9:13 AM Page 259 259 CHAPTER 4 to a medium at temperature T with a heat transfer coefficient of h. Is the heat transfer in this short cylinder one- or twodimensional? Explain. 4–77E Repeat Problem 4–76E for a location at 5300 ft elevation such as Denver, Colorado, where the boiling temperature of water is 202°F. 4–73 A short brass cylinder ( 8530 kg/m3, Cp 0.389 kJ/kg · °C, k 110 W/m · °C, and 3.39 105 m2/s) of diameter D 8 cm and height H 15 cm is initially at a uniform temperature of Ti 150°C. The cylinder is now placed in atmospheric air at 20°C, where heat transfer takes place by convection with a heat transfer coefficient of h 40 W/m2 · °C. Calculate (a) the center temperature of the cylinder, (b) the center temperature of the top surface of the cylinder, and (c) the total heat transfer from the cylinder 15 min after the start of the cooling. 4–78 A 5-cm-high rectangular ice block (k 2.22 W/m · °C and 0.124 107 m2/s) initially at 20°C is placed on a table on its square base 4 cm 4 cm in size in a room at 18°C. The heat transfer coefficient on the exposed surfaces of the ice block is 12 W/m2 · °C. Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear? Brass cylinder Room air 18°C Ambient air 20°C Ice block –20°C 15 cm 8 cm Ti = 150°C FIGURE P4–78 FIGURE P4–73 4–74 Reconsider Problem 4–73. Using EES (or other) software, investigate the effect of the cooling time on the center temperature of the cylinder, the center temperature of the top surface of the cylinder, and the total heat transfer. Let the time vary from 5 min to 60 min. Plot the center temperature of the cylinder, the center temperature of the top surface, and the total heat transfer as a function of the time, and discuss the results. 4–75 A semi-infinite aluminum cylinder (k 237 W/m · °C, 9.71 105 m2/s) of diameter D 15 cm is initially at a uniform temperature of Ti 150°C. The cylinder is now placed in water at 10°C, where heat transfer takes place by convection with a heat transfer coefficient of h 140 W/m2 · °C. Determine the temperature at the center of the cylinder 5 cm from the end surface 8 min after the start of cooling. 4–76E A hot dog can be considered to be a cylinder 5 in. long and 0.8 in. in diameter whose properties are 61.2 lbm/ft3, Cp 0.93 Btu/lbm · °F, k 0.44 Btu/h · ft · °F, and 0.0077 ft2/h. A hot dog initially at 40°F is dropped into boiling water at 212°F. If the heat transfer coefficient at the surface of the hot dog is estimated to be 120 Btu/h · ft2 · °F, determine the center temperature of the hot dog after 5, 10, and 15 min by treating the hot dog as (a) a finite cylinder and (b) an infinitely long cylinder. 4–79 Reconsider Problem 4–78. Using EES (or other) software, investigate the effect of the initial temperature of the ice block on the time period before the ice block starts melting. Let the initial temperature vary from 26°C to 4°C. Plot the time versus the initial temperature, and discuss the results. 4–80 A 2-cm-high cylindrical ice block (k 2.22 W/m · °C and 0.124 107 m2/s) is placed on a table on its base of diameter 2 cm in a room at 20°C. The heat transfer coefficient on the exposed surfaces of the ice block is 13 W/m2 · °C, and heat transfer from the base of the ice block to the table is negligible. If the ice block is not to start melting at any point for at least 2 h, determine what the initial temperature of the ice block should be. 4–81 Consider a cubic block whose sides are 5 cm long and a cylindrical block whose height and diameter are also 5 cm. Both blocks are initially at 20°C and are made of granite (k 2.5 W/m · °C and 1.15 106 m2/s). Now both blocks are exposed to hot gases at 500°C in a furnace on all of their surfaces with a heat transfer coefficient of 40 W/m2 · °C. Determine the center temperature of each geometry after 10, 20, and 60 min. 4–82 Repeat Problem 4–81 with the heat transfer coefficient at the top and the bottom surfaces of each block being doubled to 80 W/m2 · °C. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 260 260 HEAT TRANSFER To reduce the cooling time, it is proposed to cool the carcass with refrigerated air at –10°C. How would you evaluate this proposal? 5 cm 5 cm 5 cm Ti = 20°C 5 cm 5 cm Ti = 20°C Hot gases, 500°C FIGURE P4–81 4–83 A 20-cm-long cylindrical aluminum block ( 2702 kg/m3, Cp 0.896 kJ/kg · °C, k 236 W/m · °C, and 9.75 105 m2/s), 15 cm in diameter, is initially at a uniform temperature of 20°C. The block is to be heated in a furnace at 1200°C until its center temperature rises to 300°C. If the heat transfer coefficient on all surfaces of the block is 80 W/m2 · °C, determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to 20°C throughout. 4–84 Repeat Problem 4–83 for the case where the aluminum block is inserted into the furnace on a low-conductivity material so that the heat transfer to or from the bottom surface of the block is negligible. 4–85 Reconsider Problem 4–83. Using EES (or other) software, investigate the effect of the final center temperature of the block on the heating time and the amount of heat transfer. Let the final center temperature vary from 50°C to 1000°C. Plot the time and the heat transfer as a function of the final center temperature, and discuss the results. Special Topic: Refrigeration and Freezing of Foods 4–86C What are the common kinds of microorganisms? What undesirable changes do microorganisms cause in foods? 4–87C How does refrigeration prevent or delay the spoilage of foods? Why does freezing extend the storage life of foods for months? 4–88C What are the environmental factors that affect the growth rate of microorganisms in foods? 4–89C What is the effect of cooking on the microorganisms in foods? Why is it important that the internal temperature of a roast in an oven be raised above 70°C? 4–90C How can the contamination of foods with microorganisms be prevented or minimized? How can the growth of microorganisms in foods be retarded? How can the microorganisms in foods be destroyed? 4–93C Consider the freezing of packaged meat in boxes with refrigerated air. How do (a) the temperature of air, (b) the velocity of air, (c) the capacity of the refrigeration system, and (d) the size of the meat boxes affect the freezing time? 4–94C How does the rate of freezing affect the tenderness, color, and the drip of meat during thawing? 4–95C It is claimed that beef can be stored for up to two years at –23°C but no more than one year at –12°C. Is this claim reasonable? Explain. 4–96C What is a refrigerated shipping dock? How does it reduce the refrigeration load of the cold storage rooms? 4–97C How does immersion chilling of poultry compare to forced-air chilling with respect to (a) cooling time, (b) moisture loss of poultry, and (c) microbial growth. 4–98C What is the proper storage temperature of frozen poultry? What are the primary methods of freezing for poultry? 4–99C What are the factors that affect the quality of frozen fish? 4–100 The chilling room of a meat plant is 15 m 18 m 5.5 m in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and 2 kW, respectively, and the room gains heat through its envelope at a rate of 11 kW. The average mass of beef carcasses is 280 kg. The carcasses enter the chilling room at 35°C, after they are washed to facilitate evaporative cooling, and are cooled to 16°C in 12 h. The air enters the chilling room at 2.2°C and leaves at 0.5°C. Determine (a) the refrigeration load of the chilling room and (b) the volume flow rate of air. The average specific heats of beef carcasses and air are 3.14 and 1.0 kJ/kg · °C, respectively, and the density of air can be taken to be 1.28 kg/m3. 4–101 Turkeys with a water content of 64 percent that are initially at 1°C and have a mass of about 7 kg are to be frozen by submerging them into brine at 29°C. Using Figure 4–45, determine how long it will take to reduce the temperature of the turkey breast at a depth of 3.8 cm to 18°C. If the temperature at a depth of 3.8 cm in the breast represents the average Brine –29°C Turkey 7 kg 1°C 4–91C How does (a) the air motion and (b) the relative humidity of the environment affect the growth of microorganisms in foods? 4–92C The cooling of a beef carcass from 37°C to 5°C with refrigerated air at 0°C in a chilling room takes about 48 h. FIGURE P4–101 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 261 261 CHAPTER 4 temperature of the turkey, determine the amount of heat transfer per turkey assuming (a) the entire water content of the turkey is frozen and (b) only 90 percent of the water content of the turkey is frozen at 18°C. Take the specific heats of turkey to be 2.98 and 1.65 kJ/kg · °C above and below the freezing point of 2.8°C, respectively, and the latent heat of fusion of Answers: (a) 1753 kJ, (b) 1617 kJ turkey to be 214 kJ/kg. 4–102E Chickens with a water content of 74 percent, an initial temperature of 32°F, and a mass of about 6 lbm are to be frozen by refrigerated air at 40°F. Using Figure 4–44, determine how long it will take to reduce the inner surface temperature of chickens to 25°F. What would your answer be if the air temperature were 80°F? 4–103 Chickens with an average mass of 2.2 kg and average specific heat of 3.54 kJ/kg · °C are to be cooled by chilled water that enters a continuous-flow-type immersion chiller at 0.5°C. Chickens are dropped into the chiller at a uniform temperature of 15°C at a rate of 500 chickens per hour and are cooled to an average temperature of 3°C before they are taken out. The chiller gains heat from the surroundings at a rate of 210 kJ/min. Determine (a) the rate of heat removal from the chicken, in kW, and (b) the mass flow rate of water, in kg/s, if the temperature rise of water is not to exceed 2°C. 4–104 In a meat processing plant, 10-cm-thick beef slabs (ρ 1090 kg/m3, Cp 3.54 kJ/kg · °C, k 0.47 W/m · °C, and α 0.13 10–6 m2/s) initially at 15°C are to be cooled in the racks of a large freezer that is maintained at 12°C. The meat slabs are placed close to each other so that heat transfer from the 10-cm-thick edges is negligible. The entire slab is to be cooled below 5°C, but the temperature of the steak is not to drop below 1°C anywhere during refrigeration to avoid “frost bite.” The convection heat transfer coefficient and thus the rate of heat transfer from the steak can be controlled by varying the speed of a circulating fan inside. Determine the heat transfer coefficient h that will enable us to meet both temperature constraints while keeping the refrigeration time to a Answer: 9.9 W/m2 · °C. minimum. Air –12°C Meat 10 cm FIGURE P4–104 between the two plates has bonded them together. In an effort to melt the ice between the plates and separate them, the worker takes a large hairdryer and blows hot air at 50°C all over the exposed surface of the plate on the top. The convection heat transfer coefficient at the top surface is estimated to be 40 W/m2 · °C. Determine how long the worker must keep blowing hot air before the two plates separate. Answer: 482 s 4–106 Consider a curing kiln whose walls are made of 30-cm-thick concrete whose properties are k 0.9 W/m · °C and 0.23 105 m2/s. Initially, the kiln and its walls are in equilibrium with the surroundings at 2°C. Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to 42°C and is maintained at that level for 3 h. The curing kiln is then opened and exposed to the atmospheric air after the stream flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for 3 h only, or would it make no difference? Base your answer on calculations. 2°C 42°C 30 cm FIGURE P4–106 4–107 The water main in the cities must be placed at sufficient depth below the earth’s surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at 15°C and the earth’s surface temperature under the worst conditions is expected to remain at 10°C for a period of 75 days. Take the properties of soil at that location to be k 0.7 W/m · °C and Answer: 7.05 m 1.4 105 m2/s. 4–108 A hot dog can be considered to be a 12-cm-long cylinder whose diameter is 2 cm and whose properties are 980 kg/m3, Cp 3.9 kJ/kg · °C, k 0.76 W/m · °C, and Review Problems 4–105 Consider two 2-cm-thick large steel plates (k 43 W/m · °C and 1.17 105 m2/s) that were put on top of each other while wet and left outside during a cold winter night at 15°C. The next day, a worker needs one of the plates, but the plates are stuck together because the freezing of the water Hot dog Water, 100°C FIGURE P4–108 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 262 262 HEAT TRANSFER 2 107 m2/s. A hot dog initially at 5°C is dropped into boiling water at 100°C. The heat transfer coefficient at the surface of the hot dog is estimated to be 600 W/m2 · °C. If the hot dog is considered cooked when its center temperature reaches 80°C, determine how long it will take to cook it in the boiling water. 4–109 A long roll of 2-m-wide and 0.5-cm-thick 1-Mn manganese steel plate coming off a furnace at 820°C is to be quenched in an oil bath (Cp 2.0 kJ/kg · °C) at 45°C. The metal sheet is moving at a steady velocity of 10 m/min, and the oil bath is 5 m long. Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 · °C, determine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at 45°C. During a fire, the trunks of some dry oak trees (k 0.17 W/m · °C and 1.28 107 m2/s) that are initially at a uniform temperature of 30°C are exposed to hot gases at 520°C for a period of 5 h, with a heat transfer coefficient of 65 W/m2 · °C on the surface. The ignition temperature of the trees is 410°C. Treating the trunks of the trees as long cylindrical rods of diameter 20 cm, determine if these dry trees will ignite as the fire sweeps through them. Hot gases 520°C 30°C 20 cm Furnace 10 m/min 4–111 Steel plate FIGURE P4–111 Oil bath, 45°C FIGURE P4–109 4–110E In Betty Crocker’s Cookbook, it is stated that it takes 5 h to roast a 14-lb stuffed turkey initially at 40°F in an oven maintained at 325°F. It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers 185°F. The turkey can be treated as a homogeneous spherical object with the properties 75 lbm/ft3, Cp 0.98 Btu/lbm · °F, k 0.26 Btu/h · ft · °F, and 0.0035 ft2/h. Assuming the tip of the thermometer is at one-third radial distance from the center of the turkey, determine (a) the average heat transfer coefficient at the surface of the turkey, (b) the temperature of the skin of the turkey when it is done, and (c) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than 185°F 5 min after the turkey is taken out of the oven? 4–112 We often cut a watermelon in half and put it into the freezer to cool it quickly. But usually we forget to check on it and end up having a watermelon with a frozen layer on the top. To avoid this potential problem a person wants to set the timer such that it will go off when the temperature of the exposed surface of the watermelon drops to 3°C. Consider a 30-cm-diameter spherical watermelon that is cut into two equal parts and put into a freezer at 12°C. Initially, the entire watermelon is at a uniform temperature of 25°C, and the heat transfer coefficient on the surfaces is 30 W/m2 · °C. Assuming the watermelon to have the properties of water, determine how long it will take for the center of the exposed cut surfaces of the watermelon to drop to 3°C. Freezer –12°C Watermelon, 25°C FIGURE P4–112 FIGURE P4–110 4–113 The thermal conductivity of a solid whose density and specific heat are known can be determined from the relation k /Cp after evaluating the thermal diffusivity . Consider a 2-cm-diameter cylindrical rod made of a sample material whose density and specific heat are 3700 kg/m3 and 920 J/kg · °C, respectively. The sample is initially at a uniform temperature of 25°C. In order to measure the temperatures of cen58933_ch04.qxd 9/10/2002 9:13 AM Page 263 263 CHAPTER 4 Tsurface Thermocouples Rod Tcenter Boiling water 100°C FIGURE P4–113 the sample at its surface and its center, a thermocouple is inserted to the center of the sample along the centerline, and another thermocouple is welded into a small hole drilled on the surface. The sample is dropped into boiling water at 100°C. After 3 min, the surface and the center temperatures are recorded to be 93°C and 75°C, respectively. Determine the thermal diffusivity and the thermal conductivity of the material. 4–114 In desert climates, rainfall is not a common occurrence since the rain droplets formed in the upper layer of the atmosphere often evaporate before they reach the ground. Consider a raindrop that is initially at a temperature of 5°C and has a diameter of 5 mm. Determine how long it will take for the diameter of the raindrop to reduce to 3 mm as it falls through ambient air at 18°C with a heat transfer coefficient of 400 W/m2 · °C. The water temperature can be assumed to remain constant and uniform at 5°C at all times. 4–115E Consider a plate of thickness 1 in., a long cylinder of diameter 1 in., and a sphere of diameter 1 in., all initially at 400°F and all made of bronze (k 15.0 Btu/h · ft · °F and 0.333 ft2/h). Now all three of these geometries are exposed to cool air at 75°F on all of their surfaces, with a heat transfer coefficient of 7 Btu/h · ft2 · °F. Determine the center temperature of each geometry after 5, 10, and 30 min. Explain why the center temperature of the sphere is always the lowest. Plate 1 in. Sphere Cylinder 1 in. 1 in. FIGURE P4–115 4–116E Repeat Problem 4–115E for cast iron geometries (k 29 Btu/h · ft · °F and 0.61 ft2/h). 4–117E Reconsider Problem 4–115E. Using EES (or other) software, plot the center temperature of each geometry as a function of the cooling time as the time varies fom 5 min to 60 min, and discuss the results. 4–118 Engine valves (k 48 W/m · °C, Cp 440 J/kg · °C, and 7840 kg/m3) are heated to 800°C in the heat treatment section of a valve manufacturing facility. The valves are then quenched in a large oil bath at an average temperature of 45°C. The heat transfer coefficient in the oil bath is 650 W/m2 · °C. The valves have a cylindrical stem with a diameter of 8 mm and a length of 10 cm. The valve head and the stem may be assumed to be of equal surface area, and the volume of the valve head can be taken to be 80 percent of the volume of steam. Determine how long will it take for the valve temperature to drop to (a) 400°C, (b) 200°C, and (c) 46°C and (d) the maximum heat transfer from a single valve. 4–119 A watermelon initially at 35°C is to be cooled by dropping it into a lake at 15°C. After 4 h and 40 min of cooling, the center temperature of watermelon is measured to be 20°C. Treating the watermelon as a 20-cm-diameter sphere and using the properties k 0.618 W/m · °C, 0.15 106 m2/s, 995 kg/m3, and Cp 4.18 kJ/kg · °C, determine the average heat transfer coefficient and the surface temperature of watermelon at the end of the cooling period. 4–120 10-cm-thick large food slabs tightly wrapped by thin paper are to be cooled in a refrigeration room maintained at 0°C. The heat transfer coefficient on the box surfaces is 25 W/m2 · °C and the boxes are to be kept in the refrigeration room for a period of 6 h. If the initial temperature of the boxes is 30°C determine the center temperature of the boxes if the boxes contain (a) margarine (k 0.233 W/m · °C and 0.11 106 m2/s), (b) white cake (k 0.082 W/m · °C and 0.10 106 m2/s), and (c) chocolate cake (k 0.106 W/m · °C and 0.12 106 m2/s). 4–121 A 30-cm-diameter, 3.5-m-high cylindrical column of a house made of concrete (k 0.79 W/m · °C, 5.94 107 m2/s, 1600 kg/m3, and Cp 0.84 kJ/kg · °C) cooled to 16°C during a cold night is heated again during the day by being exposed to ambient air at an average temperature of 28°C with an average heat transfer coefficient of 14 W/m2 · °C. Determine (a) how long it will take for the column surface temperature to rise to 27°C, (b) the amount of heat transfer until the center temperature reaches to 28°C, and (c) the amount of heat transfer until the surface temperature reaches to 27°C. 4–122 Long aluminum wires of diameter 3 mm ( 2702 kg/m3, Cp 0.896 kJ/kg · °C, k 236 W/m · °C, and 9.75 105 m2/s) are extruded at a temperature of 350°C and exposed to atmospheric air at 30°C with a heat transfer coefficient of 35 W/m2 · °C. (a) Determine how long it will take for the wire temperature to drop to 50°C. (b) If the wire is extruded at a velocity of 10 m/min, determine how far the wire travels after extrusion by the time its temperature drops to 50°C. What change in the cooling process would you propose to shorten this distance? (c) Assuming the aluminum wire leaves the extrusion room at 50°C, determine the rate of heat transfer from the wire to the extrusion room. Answers: (a) 144 s, (b) 24 m, (c) 856 W cen58933_ch04.qxd 9/10/2002 9:13 AM Page 264 264 HEAT TRANSFER 350°C body as 28-cm diameter, 1.80-m-long cylinder, estimate how long it has been since he died. Take the properties of the body to be k 0.62 W/m · °C and 0.15 106 m2/s, and assume the initial temperature of the body to be 36°C. Tair = 30°C 10 m/min Aluminum wire FIGURE P4–122 4–123 Repeat Problem 4–122 for a copper wire ( 8950 kg/m3, Cp 0.383 kJ/kg · °C, k 386 W/m · °C, and 1.13 104 m2/s). 4–124 Consider a brick house (k 0.72 W/m · °C and 0.45 106 m2/s) whose walls are 10 m long, 3 m high, and 0.3 m thick. The heater of the house broke down one night, and the entire house, including its walls, was observed to be 5°C throughout in the morning. The outdoors warmed up as the day progressed, but no change was felt in the house, which was tightly sealed. Assuming the outer surface temperature of the house to remain constant at 15°C, determine how long it would take for the temperature of the inner surfaces of the walls to rise to 5.1°C. 15°C 5°C FIGURE P4–124 4–125 A 40-cm-thick brick wall (k 0.72 W/m · °C, and 1.6 107 m2/s) is heated to an average temperature of 18°C by the heating system and the solar radiation incident on it during the day. During the night, the outer surface of the wall is exposed to cold air at 2°C with an average heat transfer coefficient of 20 W/m2 · °C, determine the wall temperatures at distances 15, 30, and 40 cm from the outer surface for a period of 2 hours. 4–126 Consider the engine block of a car made of cast iron (k 52 W/m · °C and 1.7 105 m2/s). The engine can be considered to be a rectangular block whose sides are 80 cm, 40 cm, and 40 cm. The engine is at a temperature of 150°C when it is turned off. The engine is then exposed to atmospheric air at 17°C with a heat transfer coefficient of 6 W/m2 · °C. Determine (a) the center temperature of the top surface whose sides are 80 cm and 40 cm and (b) the corner temperature after 45 min of cooling. 4–127 A man is found dead in a room at 16°C. The surface temperature on his waist is measured to be 23°C and the heat transfer coefficient is estimated to be 9 W/m2 · °C. Modeling the Computer, Design, and Essay Problems 4–128 Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You may measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every 5 min for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above. 4–129 Repeat Problem 4–128 using a banana instead of an apple. The thermal properties of bananas are practically the same as those of apples. 4–130 Conduct the following experiment to determine the time constant for a can of soda and then predict the temperature of the soda at different times. Leave the soda in the refrigerator overnight. Measure the air temperature in the kitchen and the temperature of the soda while it is still in the refrigerator by taping the sensor of the thermometer to the outer surface of the can. Then take the soda out and measure its temperature again in 5 min. Using these values, calculate the exponent b. Using this b-value, predict the temperatures of the soda in 10, 15, 20, 30, and 60 min and compare the results with the actual temperature measurements. Do you think the lumped system analysis is valid in this case? 4–131 Citrus trees are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy the crop. In order to protect the trees from occasional cold fronts with subfreezing temperatures, tree growers in Florida usually install water sprinklers on the trees. When the temperature drops below a certain level, the sprinklers spray water on the trees and their fruits to protect them against the damage the subfreezing temperatures can cause. Explain the basic mechanism behind this protection measure and write an essay on how the system works in practice. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 265 CHAPTER NUMERICAL METHODS I N H E AT C O N D U C T I O N o far we have mostly considered relatively simple heat conduction problems involving simple geometries with simple boundary conditions because only such simple problems can be solved analytically. But many problems encountered in practice involve complicated geometries with complex boundary conditions or variable properties and cannot be solved analytically. In such cases, sufficiently accurate approximate solutions can be obtained by computers using a numerical method. Analytical solution methods such as those presented in Chapter 2 are based on solving the governing differential equation together with the boundary conditions. They result in solution functions for the temperature at every point in the medium. Numerical methods, on the other hand, are based on replacing the differential equation by a set of n algebraic equations for the unknown temperatures at n selected points in the medium, and the simultaneous solution of these equations results in the temperature values at those discrete points. There are several ways of obtaining the numerical formulation of a heat conduction problem, such as the finite difference method, the finite element method, the boundary element method, and the energy balance (or control volume) method. Each method has its own advantages and disadvantages, and each is used in practice. In this chapter we will use primarily the energy balance approach since it is based on the familiar energy balances on control volumes instead of heavy mathematical formulations, and thus it gives a better physical feel for the problem. Besides, it results in the same set of algebraic equations as the finite difference method. In this chapter, the numerical formulation and solution of heat conduction problems are demonstrated for both steady and transient cases in various geometries. S 5 CONTENTS 5–1 Why Numerical Methods 266 5–2 Finite Difference Formulation of Differential Equations 269 5–3 One-Dimensional Steady Heat Conduction 272 5–4 Two-Dimensional Steady Heat Conduction 282 5–5 Transient Heat Conduction 291 Topic of Special Interest: Controlling the Numerical Error 309 265 cen58933_ch05.qxd 9/4/2002 11:41 AM Page 266 266 HEAT TRANSFER 5–1 g· 0 1 — d 2 dT — — — =0 r + r 2 dr dr k ( ) r0 Sphere 0 dT(0) ——– =0 dr T(r0) = T1 Solution: g· T(r) = T1 + —0 (r02 – r 2) 6k 4 π g· 0r 3 · dT Q(r) = –kA — = ——— dr 3 FIGURE 5–1 The analytical solution of a problem requires solving the governing differential equation and applying the boundary conditions. ■ WHY NUMERICAL METHODS? The ready availability of high-speed computers and easy-to-use powerful software packages has had a major impact on engineering education and practice in recent years. Engineers in the past had to rely on analytical skills to solve significant engineering problems, and thus they had to undergo a rigorous training in mathematics. Today’s engineers, on the other hand, have access to a tremendous amount of computation power under their fingertips, and they mostly need to understand the physical nature of the problem and interpret the results. But they also need to understand how calculations are performed by the computers to develop an awareness of the processes involved and the limitations, while avoiding any possible pitfalls. In Chapter 2 we solved various heat conduction problems in various geometries in a systematic but highly mathematical manner by (1) deriving the governing differential equation by performing an energy balance on a differential volume element, (2) expressing the boundary conditions in the proper mathematical form, and (3) solving the differential equation and applying the boundary conditions to determine the integration constants. This resulted in a solution function for the temperature distribution in the medium, and the solution obtained in this manner is called the analytical solution of the problem. For example, the mathematical formulation of one-dimensional steady heat conduction in a sphere of radius r0 whose outer surface is maintained at a uniform temperature of T1 with uniform heat generation at a rate of g·0 was expressed as (Fig. 5–1) g·0 1 d 2 dT r 0 2 dr k dr r dT(0) 0 and T(r0) T1 dr (5-1) whose (analytical) solution is T(r) T1 g·0 2 (r r 2) 6k 0 (5-2) This is certainly a very desirable form of solution since the temperature at any point within the sphere can be determined simply by substituting the r-coordinate of the point into the analytical solution function above. The analytical solution of a problem is also referred to as the exact solution since it satisfies the differential equation and the boundary conditions. This can be verified by substituting the solution function into the differential equation and the boundary conditions. Further, the rate of heat flow at any location within the sphere or its surface can be determined by taking the derivative of the solution function T(r) and substituting it into Fourier’s law as g·0r 4g·0r 3 dT · Q (r) kA k(4r 2) 3 3k dr (5-3) The analysis above did not require any mathematical sophistication beyond the level of simple integration, and you are probably wondering why anyone would ask for something else. After all, the solutions obtained are exact and cen58933_ch05.qxd 9/4/2002 11:41 AM Page 267 267 CHAPTER 5 easy to use. Besides, they are instructive since they show clearly the functional dependence of temperature and heat transfer on the independent variable r. Well, there are several reasons for searching for alternative solution methods. T, h k = constant No radiation T, h h, T h = constant T = constant FIGURE 5–2 Analytical solution methods are limited to simplified problems in simple geometries. An oval-shaped body 2 Better Modeling We mentioned earlier that analytical solutions are exact solutions since they do not involve any approximations. But this statement needs some clarification. Distinction should be made between an actual real-world problem and the mathematical model that is an idealized representation of it. The solutions we get are the solutions of mathematical models, and the degree of applicability of these solutions to the actual physical problems depends on the accuracy of the model. An “approximate” solution of a realistic model of a physical problem is usually more accurate than the “exact” solution of a crude mathematical model (Fig. 5–3). When attempting to get an analytical solution to a physical problem, there is always the tendency to oversimplify the problem to make the mathematical model sufficiently simple to warrant an analytical solution. Therefore, it is common practice to ignore any effects that cause mathematical complications such as nonlinearities in the differential equation or the boundary conditions. So it comes as no surprise that nonlinearities such as temperature dependence of thermal conductivity and the radiation boundary conditions are seldom considered in analytical solutions. A mathematical model intended for a numerical solution is likely to represent the actual problem better. Therefore, the numerical solution of engineering problems has now become the norm rather than the exception even when analytical solutions are available. No radiation Long cylinder 1 Limitations Analytical solution methods are limited to highly simplified problems in simple geometries (Fig. 5–2). The geometry must be such that its entire surface can be described mathematically in a coordinate system by setting the variables equal to constants. That is, it must fit into a coordinate system perfectly with nothing sticking out or in. In the case of one-dimensional heat conduction in a solid sphere of radius r0, for example, the entire outer surface can be described by r r0. Likewise, the surfaces of a finite solid cylinder of radius r0 and height H can be described by r r0 for the side surface and z 0 and z H for the bottom and top surfaces, respectively. Even minor complications in geometry can make an analytical solution impossible. For example, a spherical object with an extrusion like a handle at some location is impossible to handle analytically since the boundary conditions in this case cannot be expressed in any familiar coordinate system. Even in simple geometries, heat transfer problems cannot be solved analytically if the thermal conditions are not sufficiently simple. For example, the consideration of the variation of thermal conductivity with temperature, the variation of the heat transfer coefficient over the surface, or the radiation heat transfer on the surfaces can make it impossible to obtain an analytical solution. Therefore, analytical solutions are limited to problems that are simple or can be simplified with reasonable approximations. h, T Simplified model Realistic model A sphere Exact (analytical) solution of model, but crude solution of actual problem Approximate (numerical) solution of model, but accurate solution of actual problem FIGURE 5–3 The approximate numerical solution of a real-world problem may be more accurate than the exact (analytical) solution of an oversimplified model of that problem. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 268 268 HEAT TRANSFER z 3 Flexibility L T T(r, z) 0 r0 T0 Engineering problems often require extensive parametric studies to understand the influence of some variables on the solution in order to choose the right set of variables and to answer some “what-if ” questions. This is an iterative process that is extremely tedious and time-consuming if done by hand. Computers and numerical methods are ideally suited for such calculations, and a wide range of related problems can be solved by minor modifications in the code or input variables. Today it is almost unthinkable to perform any significant optimization studies in engineering without the power and flexibility of computers and numerical methods. 4 Complications r Analytical solution: T(r, z) – T J0(λnr) sinh λn(L – z) ————– = ∑ ———— –————— λ J ( λ r ) sinh (λnL) T0 – T n=1 n 1 n 0 where λn’s are roots of J0(λnr0) = 0 FIGURE 5–4 Some analytical solutions are very complex and difficult to use. Some problems can be solved analytically, but the solution procedure is so complex and the resulting solution expressions so complicated that it is not worth all that effort. With the exception of steady one-dimensional or transient lumped system problems, all heat conduction problems result in partial differential equations. Solving such equations usually requires mathematical sophistication beyond that acquired at the undergraduate level, such as orthogonality, eigenvalues, Fourier and Laplace transforms, Bessel and Legendre functions, and infinite series. In such cases, the evaluation of the solution, which often involves double or triple summations of infinite series at a specified point, is a challenge in itself (Fig. 5–4). Therefore, even when the solutions are available in some handbooks, they are intimidating enough to scare prospective users away. 5 Human Nature FIGURE 5–5 The ready availability of highpowered computers with sophisticated software packages has made numerical solution the norm rather than the exception. As human beings, we like to sit back and make wishes, and we like our wishes to come true without much effort. The invention of TV remote controls made us feel like kings in our homes since the commands we give in our comfortable chairs by pressing buttons are immediately carried out by the obedient TV sets. After all, what good is cable TV without a remote control. We certainly would love to continue being the king in our little cubicle in the engineering office by solving problems at the press of a button on a computer (until they invent a remote control for the computers, of course). Well, this might have been a fantasy yesterday, but it is a reality today. Practically all engineering offices today are equipped with high-powered computers with sophisticated software packages, with impressive presentation-style colorful output in graphical and tabular form (Fig. 5–5). Besides, the results are as accurate as the analytical results for all practical purposes. The computers have certainly changed the way engineering is practiced. The discussions above should not lead you to believe that analytical solutions are unnecessary and that they should be discarded from the engineering curriculum. On the contrary, insight to the physical phenomena and engineering wisdom is gained primarily through analysis. The “feel” that engineers develop during the analysis of simple but fundamental problems serves as an invaluable tool when interpreting a huge pile of results obtained from a computer when solving a complex problem. A simple analysis by hand for a limiting case can be used to check if the results are in the proper range. Also, cen58933_ch05.qxd 9/4/2002 11:41 AM Page 269 269 CHAPTER 5 nothing can take the place of getting “ball park” results on a piece of paper during preliminary discussions. The calculators made the basic arithmetic operations by hand a thing of the past, but they did not eliminate the need for instructing grade school children how to add or multiply. In this chapter, you will learn how to formulate and solve heat transfer problems numerically using one or more approaches. In your professional life, you will probably solve the heat transfer problems you come across using a professional software package, and you are highly unlikely to write your own programs to solve such problems. (Besides, people will be highly skeptical of the results obtained using your own program instead of using a wellestablished commercial software package that has stood the test of time.) The insight you will gain in this chapter by formulating and solving some heat transfer problems will help you better understand the available software packages and be an informed and responsible user. 5–2 ■ FINITE DIFFERENCE FORMULATION OF DIFFERENTIAL EQUATIONS The numerical methods for solving differential equations are based on replacing the differential equations by algebraic equations. In the case of the popular finite difference method, this is done by replacing the derivatives by differences. Below we will demonstrate this with both first- and second-order derivatives. But first we give a motivational example. Consider a man who deposits his money in the amount of A0 $100 in a savings account at an annual interest rate of 18 percent, and let us try to determine the amount of money he will have after one year if interest is compounded continuously (or instantaneously). In the case of simple interest, the money will earn $18 interest, and the man will have 100 100 0.18 $118.00 in his account after one year. But in the case of compounding, the interest earned during a compounding period will also earn interest for the remaining part of the year, and the year-end balance will be greater than $118. For example, if the money is compounded twice a year, the balance will be 100 100 (0.18/2) $109 after six months, and 109 109 (0.18/2) $118.81 at the end of the year. We could also determine the balance A directly from A A0(1 i)n ($100)(1 0.09)2 $118.81 (5-4) where i is the interest rate for the compounding period and n is the number of periods. Using the same formula, the year-end balance is determined for monthly, daily, hourly, minutely, and even secondly compounding, and the results are given in Table 5–1. Note that in the case of daily compounding, the year-end balance will be $119.72, which is $1.72 more than the simple interest case. (So it is no wonder that the credit card companies usually charge interest compounded daily when determining the balance.) Also note that compounding at smaller time intervals, even at the end of each second, does not change the result, and we suspect that instantaneous compounding using “differential” time intervals dt will give the same result. This suspicion is confirmed by obtaining the differential TABLE 5–1 Year-end balance of a $100 account earning interest at an annual rate of 18 percent for various compounding periods Compounding Period Number of Year-End Periods, n Balance 1 year 1 $118.00 6 months 2 118.81 1 month 12 119.56 1 week 52 119.68 1 day 365 119.72 1 hour 8760 119.72 1 minute 525,600 119.72 1 second 31,536,000 119.72 Instantaneous 119.72 cen58933_ch05.qxd 9/4/2002 11:41 AM Page 270 270 HEAT TRANSFER equation dA/dt iA for the balance A, whose solution is A A0 exp(it). Substitution yields A ($100)exp(0.18 1) $119.72 which is identical to the result for daily compounding. Therefore, replacing a differential time interval dt by a finite time interval of t 1 day gave the same result, which leads us into believing that reasonably accurate results can be obtained by replacing differential quantities by sufficiently small differences. Next, we develop the finite difference formulation of heat conduction problems by replacing the derivatives in the differential equations by differences. In the following section we will do it using the energy balance method, which does not require any knowledge of differential equations. Derivatives are the building blocks of differential equations, and thus we first give a brief review of derivatives. Consider a function f that depends on x, as shown in Figure 5–6. The first derivative of f(x) at a point is equivalent to the slope of a line tangent to the curve at that point and is defined as f (x) f f(x x) f(x) df(x) xlim xlim → 0 → 0 x x dx f (x + ∆ x) (5-5) ∆f which is the ratio of the increment f of the function to the increment x of the independent variable as x → 0. If we don’t take the indicated limit, we will have the following approximate relation for the derivative: f (x) ∆x Tangent line x x + ∆x x FIGURE 5–6 The derivative of a function at a point represents the slope of the function at that point. T (x) Tm + 1 Tm Tm – 1 01 2 L m m+1… M M –1 m – 1– m + 1– 2 2 … m –1 x FIGURE 5–7 Schematic of the nodes and the nodal temperatures used in the development of the finite difference formulation of heat transfer in a plane wall. (5-6) This approximate expression of the derivative in terms of differences is the finite difference form of the first derivative. The equation above can also be obtained by writing the Taylor series expansion of the function f about the point x, f(x x) f(x) x Plane wall 0 df(x) f(x x) f(x) x dx df(x) 1 d 2f(x) x2 ··· 2 dx dx2 (5-7) and neglecting all the terms in the expansion except the first two. The first term neglected is proportional to x2, and thus the error involved in each step of this approximation is also proportional to x2. However, the commutative error involved after M steps in the direction of length L is proportional to x since Mx2 (L/x)x2 Lx. Therefore, the smaller the x, the smaller the error, and thus the more accurate the approximation. Now consider steady one-dimensional heat transfer in a plane wall of thickness L with heat generation. The wall is subdivided into M sections of equal thickness x L/M in the x-direction, separated by planes passing through M 1 points 0, 1, 2, . . . , m 1, m, m 1, . . . , M called nodes or nodal points, as shown in Figure 5–7. The x-coordinate of any point m is simply xm mx, and the temperature at that point is simply T(xm) Tm. The heat conduction equation involves the second derivatives of temperature with respect to the space variables, such as d 2T/dx2, and the finite difference formulation is based on replacing the second derivatives by appropriate cen58933_ch05.qxd 9/4/2002 11:41 AM Page 271 271 CHAPTER 5 differences. But we need to start the process with first derivatives. Using Eq. 5–6, the first derivative of temperature dT/dx at the midpoints m 21 and m 12 of the sections surrounding the node m can be expressed as dT dx m 1 2 Tm Tm 1 x and dT dx m 1 2 Tm 1 Tm x (5-8) Noting that the second derivative is simply the derivative of the first derivative, the second derivative of temperature at node m can be expressed as 2 d T dx2 m dT dx m 1 2 x dT dx m 1 2 Tm1 Tm Tm Tm1 x x x Tm 1 2Tm Tm 1 x2 (5-9) which is the finite difference representation of the second derivative at a general internal node m. Note that the second derivative of temperature at a node m is expressed in terms of the temperatures at node m and its two neighboring nodes. Then the differential equation d 2T g· 0 k dx 2 (5-10) which is the governing equation for steady one-dimensional heat transfer in a plane wall with heat generation and constant thermal conductivity, can be expressed in the finite difference form as (Fig. 5–8) Tm 1 2Tm Tm 1 g·m 0, k x2 m 1, 2, 3, . . . , M 1 (5-11) where g·m is the rate of heat generation per unit volume at node m. If the surface temperatures T0 and TM are specified, the application of this equation to each of the M 1 interior nodes results in M 1 equations for the determination of M 1 unknown temperatures at the interior nodes. Solving these equations simultaneously gives the temperature values at the nodes. If the temperatures at the outer surfaces are not known, then we need to obtain two more equations in a similar manner using the specified boundary conditions. Then the unknown temperatures at M 1 nodes are determined by solving the resulting system of M 1 equations in M 1 unknowns simultaneously. Note that the boundary conditions have no effect on the finite difference formulation of interior nodes of the medium. This is not surprising since the control volume used in the development of the formulation does not involve any part of the boundary. You may recall that the boundary conditions had no effect on the differential equation of heat conduction in the medium either. The finite difference formulation above can easily be extended to two- or three-dimensional heat transfer problems by replacing each second derivative by a difference equation in that direction. For example, the finite difference formulation for steady two-dimensional heat conduction in a region with Plane wall Differential equation: 2T g· d—– + — =0 2 k dx Valid at every point Finite difference equation: Tm – 1 – 2Tm + Tm + 1 g· m —–—————— — +— =0 k ∆x2 Valid at discrete points ∆x FIGURE 5–8 The differential equation is valid at every point of a medium, whereas the finite difference equation is valid at discrete points (the nodes) only. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 272 272 HEAT TRANSFER heat generation and constant thermal conductivity can be expressed in rectangular coordinates as (Fig. 5–9) Tm 1, n 2Tm, n Tm 1, n m, n + 1 n+1 ∆y m – 1, n ∆y n n–1 x2 m, n m–1 m m+1 FIGURE 5–9 Finite difference mesh for twodimensional conduction in rectangular coordinates. 5–3 Plane wall · Qcond, left Volume element of node m g· m · Qcond, right A general interior node 0 0 1 2 y2 g·m, n 0 k (5-12) for m 1, 2, 3, . . . , M 1 and n 1, 2, 3, . . . , N 1 at any interior node (m, n). Note that a rectangular region that is divided into M equal subregions in the x-direction and N equal subregions in the y-direction has a total of (M 1)(N 1) nodes, and Eq. 5–12 can be used to obtain the finite difference equations at (M 1)(N 1) of these nodes (i.e., all nodes except those at the boundaries). The finite difference formulation is given above to demonstrate how difference equations are obtained from differential equations. However, we will use the energy balance approach in the following sections to obtain the numerical formulation because it is more intuitive and can handle boundary conditions more easily. Besides, the energy balance approach does not require having the differential equation before the analysis. ∆x ∆x x Tm, n 1 2Tm, n Tm, n 1 m + 1, n m, n – 1 y m–1 m m+1 ∆x ∆x ∆x FIGURE 5–10 The nodal points and volume elements for the finite difference formulation of one-dimensional conduction in a plane wall. L M x ■ ONE-DIMENSIONAL STEADY HEAT CONDUCTION In this section we will develop the finite difference formulation of heat conduction in a plane wall using the energy balance approach and discuss how to solve the resulting equations. The energy balance method is based on subdividing the medium into a sufficient number of volume elements and then applying an energy balance on each element. This is done by first selecting the nodal points (or nodes) at which the temperatures are to be determined and then forming elements (or control volumes) over the nodes by drawing lines through the midpoints between the nodes. This way, the interior nodes remain at the middle of the elements, and the properties at the node such as the temperature and the rate of heat generation represent the average properties of the element. Sometimes it is convenient to think of temperature as varying linearly between the nodes, especially when expressing heat conduction between the elements using Fourier’s law. To demonstrate the approach, again consider steady one-dimensional heat transfer in a plane wall of thickness L with heat generation g·(x) and constant conductivity k. The wall is now subdivided into M equal regions of thickness x L/M in the x-direction, and the divisions between the regions are selected as the nodes. Therefore, we have M 1 nodes labeled 0, 1, 2, . . . , m 1, m, m 1, . . . , M, as shown in Figure 5–10. The x-coordinate of any node m is simply xm mx, and the temperature at that point is T(xm) Tm. Elements are formed by drawing vertical lines through the midpoints between the nodes. Note that all interior elements represented by interior nodes are full-size elements (they have a thickness of x), whereas the two elements at the boundaries are half-sized. To obtain a general difference equation for the interior nodes, consider the element represented by node m and the two neighboring nodes m 1 and m 1. Assuming the heat conduction to be into the element on all surfaces, an energy balance on the element can be expressed as cen58933_ch05.qxd 9/4/2002 11:41 AM Page 273 273 CHAPTER 5 Rate of heat Rate of heat Rate of heat Rate of change conduction conduction generation of the energy at the left at the right inside the content of surface surface element the element or Eelement · · · Q cond, left Q cond, right Gelement 0 t (5-13) since the energy content of a medium (or any part of it) does not change under steady conditions and thus Eelement 0. The rate of heat generation within the element can be expressed as · Gelement g·mVelement g·m Ax (5-14) where g·m is the rate of heat generation per unit volume in W/m3 evaluated at node m and treated as a constant for the entire element, and A is heat transfer area, which is simply the inner (or outer) surface area of the wall. Recall that when temperature varies linearly, the steady rate of heat conduction across a plane wall of thickness L can be expressed as · T Q cond kA L (5-15) where T is the temperature change across the wall and the direction of heat transfer is from the high temperature side to the low temperature. In the case of a plane wall with heat generation, the variation of temperature is not linear and thus the relation above is not applicable. However, the variation of temperature between the nodes can be approximated as being linear in the determination of heat conduction across a thin layer of thickness x between two nodes (Fig. 5–11). Obviously the smaller the distance x between two nodes, the more accurate is this approximation. (In fact, such approximations are the reason for classifying the numerical methods as approximate solution methods. In the limiting case of x approaching zero, the formulation becomes exact and we obtain a differential equation.) Noting that the direction of heat transfer on both surfaces of the element is assumed to be toward the node m, the rate of heat conduction at the left and right surfaces can be expressed as Tm1 Tm · Q cond, left kA x and Tm1 Tm · Q cond, right kA x (5-16) Tm1 Tm Tm1 Tm kA g·m Ax 0 x x (5-17) which simplifies to Tm1 2Tm Tm1 g·m 0, k x2 m 1, 2, 3, . . . , M 1 (5-18) Volume element k Tm + 1 Tm Linear Linear ∆x m–1 Tm – 1 – Tm k A ————– ∆x A Substituting Eqs. 5–14 and 5–16 into Eq. 5–13 gives kA Tm – 1 ∆x m m+1 Tm + 1 – Tm k A ————– ∆x A FIGURE 5–11 In finite difference formulation, the temperature is assumed to vary linearly between the nodes. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 274 274 HEAT TRANSFER g· 2 A∆ x T1 – T2 k A ——— ∆x T2 – T3 k A ——— ∆x 1 2 3 Volume element of node 2 T2 – T3 · T1 – T2 – k A ——— + g2 A∆ x = 0 k A ——— ∆x ∆x or T1 – 2T2 + T3 + g· 2 A∆ x 2 / k = 0 (a) Assuming heat transfer to be out of the volume element at the right surface. g· 2 A∆ x T1 – T2 k A ——— ∆x T3 – T2 k A ——— ∆x 1 2 which is identical to the difference equation (Eq. 5–11) obtained earlier. Again, this equation is applicable to each of the M 1 interior nodes, and its application gives M 1 equations for the determination of temperatures at M 1 nodes. The two additional equations needed to solve for the M 1 unknown nodal temperatures are obtained by applying the energy balance on the two elements at the boundaries (unless, of course, the boundary temperatures are specified). You are probably thinking that if heat is conducted into the element from both sides, as assumed in the formulation, the temperature of the medium will have to rise and thus heat conduction cannot be steady. Perhaps a more realistic approach would be to assume the heat conduction to be into the element on the left side and out of the element on the right side. If you repeat the formulation using this assumption, you will again obtain the same result since the heat conduction term on the right side in this case will involve Tm Tm 1 instead of Tm 1 Tm, which is subtracted instead of being added. Therefore, the assumed direction of heat conduction at the surfaces of the volume elements has no effect on the formulation, as shown in Figure 5–12. (Besides, the actual direction of heat transfer is usually not known.) However, it is convenient to assume heat conduction to be into the element at all surfaces and not worry about the sign of the conduction terms. Then all temperature differences in conduction relations are expressed as the temperature of the neighboring node minus the temperature of the node under consideration, and all conduction terms are added. 3 Volume element of node 2 T3 – T2 · T1 – T2 + k A ——— + g2 A∆ x = 0 k A ——— ∆x ∆x or T1 – 2T2 + T3 + g· 2 A∆ x 2 / k = 0 (b) Assuming heat transfer to be into the volume element at all surfaces. FIGURE 5–12 The assumed direction of heat transfer at surfaces of a volume element has no effect on the finite difference formulation. Boundary Conditions Above we have developed a general relation for obtaining the finite difference equation for each interior node of a plane wall. This relation is not applicable to the nodes on the boundaries, however, since it requires the presence of nodes on both sides of the node under consideration, and a boundary node does not have a neighboring node on at least one side. Therefore, we need to obtain the finite difference equations of boundary nodes separately. This is best done by applying an energy balance on the volume elements of boundary nodes. Boundary conditions most commonly encountered in practice are the specified temperature, specified heat flux, convection, and radiation boundary conditions, and here we develop the finite difference formulations for them for the case of steady one-dimensional heat conduction in a plane wall of thickness L as an example. The node number at the left surface at x 0 is 0, and at the right surface at x L it is M. Note that the width of the volume element for either boundary node is x/2. The specified temperature boundary condition is the simplest boundary condition to deal with. For one-dimensional heat transfer through a plane wall of thickness L, the specified temperature boundary conditions on both the left and right surfaces can be expressed as (Fig. 5–13) T(0) T0 Specified value T(L) TM Specified value (5-19) where T0 and Tm are the specified temperatures at surfaces at x 0 and x L, respectively. Therefore, the specified temperature boundary conditions are cen58933_ch05.qxd 9/4/2002 11:41 AM Page 275 275 CHAPTER 5 incorporated by simply assigning the given surface temperatures to the boundary nodes. We do not need to write an energy balance in this case unless we decide to determine the rate of heat transfer into or out of the medium after the temperatures at the interior nodes are determined. When other boundary conditions such as the specified heat flux, convection, radiation, or combined convection and radiation conditions are specified at a boundary, the finite difference equation for the node at that boundary is obtained by writing an energy balance on the volume element at that boundary. The energy balance is again expressed as · · Q Gelement 0 Plane wall 35°C 0 82°C 0 1 for heat transfer under steady conditions. Again we assume all heat transfer to be into the volume element from all surfaces for convenience in formulation, except for specified heat flux since its direction is already specified. Specified heat flux is taken to be a positive quantity if into the medium and a negative quantity if out of the medium. Then the finite difference formulation at the node m 0 (at the left boundary where x 0) of a plane wall of thickness L during steady one-dimensional heat conduction can be expressed as (Fig. 5–14) (5-22) Special case: Insulated Boundary (q·0 0) T1 T0 g·0(Ax/2) 0 x (5-23) Convection Boundary Condition hA(T T0) kA T1 T0 g·0(Ax/2) 0 x FIGURE 5–13 Finite difference formulation of specified temperature boundary conditions on both surfaces of a plane wall. ∆ x– — 2 g· 0 Specified Heat Flux Boundary Condition kA M T0 = 35°C TM = 82°C (5-21) where Ax/2 is the volume of the volume element (note that the boundary element has half thickness), g·0 is the rate of heat generation per unit volume (in W/m3) at x 0, and A is the heat transfer area, which is constant for a plane wall. Note that we have x in the denominator of the second term instead of x/2. This is because the ratio in that term involves the temperature difference between nodes 0 and 1, and thus we must use the distance between those two nodes, which is x. The finite difference form of various boundary conditions can be obtained · from Eq. 5–21 by replacing Q left surface by a suitable expression. Next this is done for various boundary conditions at the left boundary. T1 T0 q·0A kA g·0(Ax/2) 0 x L … (5-20) all sides T1 T0 · Q left surface kA g·0(Ax/2) 0 x 2 (5-24) Volume element of node 0 T1 – T0 k A ——— ∆x · Qleft surface 0 0 1 ∆x 2 … L x ∆x T1 – T0 · ∆ x · Qleft surface + k A ——— + g0 A —– = 0 ∆x 2 FIGURE 5–14 Schematic for the finite difference formulation of the left boundary node of a plane wall. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 276 276 HEAT TRANSFER Radiation Boundary Condition Tsurr 4 T04) kA A(Tsurr ∆ x– — 2 ε T1 – T0 k A ——— ∆x hA(T – T0) 0 0 1 2 ∆x A … 4 T04) kA hA(T T0) A(Tsurr L x hcombined A(T T0) kA A Medium B kB kAA Tm + 1 – Tm kB A ————– ∆x m ∆x x m+1 ∆x ∆ x —– ∆x —– 2 2 (5-27) (5-28) Interface Boundary Condition Two different solid media A and B are assumed to be in perfect contact, and thus at the same temperature at the interface at node m (Fig. 5–16). Subscripts A and B indicate properties of media A and B, respectively. Interface m–1 T1 T0 g·0(Ax/2) 0 x T1 T0 4 T04) kA q·0A hA(T T0) A(Tsurr g·0(Ax/2) 0 x FIGURE 5–15 Schematic for the finite difference formulation of combined convection and radiation on the left boundary of a plane wall. Tm – 1 – Tm kA A ————– ∆x (5-26) Combined Convection, Radiation, and Heat Flux Boundary Condition T1 – T0 · ∆ x + g0 A —– = 0 + kA ——— ∆x 2 g·A,m g·B,m T1 T0 g·0(Ax/2) 0 x or ∆x hA(T – T0) + εσ A(T 4surr – T 40 ) Medium A kA (5-25) Combined Convection and Radiation Boundary Condition (Fig. 5–15) g· 0 εσ A(T 4surr – T 40 ) T1 T0 g·0(Ax/2) 0 x A Tm + 1 – Tm Tm – 1 – Tm kA A ————– + kB A ————– ∆x ∆x ∆ x ∆ x · · + gA, m A —– + gB, m A —– = 0 2 2 FIGURE 5–16 Schematic for the finite difference formulation of the interface boundary condition for two mediums A and B that are in perfect thermal contact. Tm1 Tm Tm1 Tm kBA g·A, m(Ax/2) g·B, m(Ax/2) 0 x x (5-29) In these relations, q·0 is the specified heat flux in W/m2, h is the convection coefficient, hcombined is the combined convection and radiation coefficient, T is the temperature of the surrounding medium, Tsurr is the temperature of the surrounding surfaces, is the emissivity of the surface, and is the Stefan– Boltzman constant. The relations above can also be used for node M on the right boundary by replacing the subscript “0” by “M” and the subscript “1” by “M 1”. Note that absolute temperatures must be used in radiation heat transfer calculations, and all temperatures should be expressed in K or R when a boundary condition involves radiation to avoid mistakes. We usually try to avoid the radiation boundary condition even in numerical solutions since it causes the finite difference equations to be nonlinear, which are more difficult to solve. Treating Insulated Boundary Nodes as Interior Nodes: The Mirror Image Concept One way of obtaining the finite difference formulation of a node on an insulated boundary is to treat insulation as “zero” heat flux and to write an energy balance, as done in Eq. 5–23. Another and more practical way is to treat the node on an insulated boundary as an interior node. Conceptually this is done cen58933_ch05.qxd 9/4/2002 11:41 AM Page 277 277 CHAPTER 5 by replacing the insulation on the boundary by a mirror and considering the reflection of the medium as its extension (Fig. 5–17). This way the node next to the boundary node appears on both sides of the boundary node because of symmetry, converting it into an interior node. Then using the general formula (Eq. 5–18) for an interior node, which involves the sum of the temperatures of the adjoining nodes minus twice the node temperature, the finite difference formulation of a node m 0 on an insulated boundary of a plane wall can be expressed as Tm1 2Tm Tm1 g·m 0 k x2 → T1 2T0 T1 g·0 0 k x2 EXAMPLE 5–1 0 1 x 2 Mirror Equivalent interior node Mirror image (5-30) which is equivalent to Eq. 5–23 obtained by the energy balance approach. The mirror image approach can also be used for problems that possess thermal symmetry by replacing the plane of symmetry by a mirror. Alternately, we can replace the plane of symmetry by insulation and consider only half of the medium in the solution. The solution in the other half of the medium is simply the mirror image of the solution obtained. Insulated boundary node Insulation x 2 1 0 1 x 2 FIGURE 5–17 A node on an insulated boundary can be treated as an interior node by replacing the insulation by a mirror. Steady Heat Conduction in a Large Uranium Plate Consider a large uranium plate of thickness L 4 cm and thermal conductivity k 28 W/m · °C in which heat is generated uniformly at a constant rate of g· 5 106 W/m3. One side of the plate is maintained at 0°C by iced water while the other side is subjected to convection to an environment at T 30°C with a heat transfer coefficient of h 45 W/m2 · °C, as shown in Figure 5–18. Considering a total of three equally spaced nodes in the medium, two at the boundaries and one at the middle, estimate the exposed surface temperature of the plate under steady conditions using the finite difference approach. SOLUTION A uranium plate is subjected to specified temperature on one side and convection on the other. The unknown surface temperature of the plate is to be determined numerically using three equally spaced nodes. Assumptions 1 Heat transfer through the wall is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible. Properties The thermal conductivity is given to be k 28 W/m · °C. Analysis The number of nodes is specified to be M 3, and they are chosen to be at the two surfaces of the plate and the midpoint, as shown in the figure. Then the nodal spacing x becomes x 0.04 m L 0.02 m M1 31 We number the nodes 0, 1, and 2. The temperature at node 0 is given to be T0 0°C, and the temperatures at nodes 1 and 2 are to be determined. This problem involves only two unknown nodal temperatures, and thus we need to have only two equations to determine them uniquely. These equations are obtained by applying the finite difference method to nodes 1 and 2. Uranium plate 0°C 0 h T k = 28 W/m·°C g· = 5 × 106 W/m3 L 0 1 2 x FIGURE 5–18 Schematic for Example 5–1. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 278 278 HEAT TRANSFER Node 1 is an interior node, and the finite difference formulation at that node is obtained directly from Eq. 5–18 by setting m 1: T0 2T1 T2 g·1 0 k x2 → 0 2T1 T2 g·1 0 k x2 → 2T1 T2 g·1x2 k (1) Node 2 is a boundary node subjected to convection, and the finite difference formulation at that node is obtained by writing an energy balance on the volume element of thickness x/2 at that boundary by assuming heat transfer to be into the medium at all sides: hA(T T2) kA T1 T2 g·2(Ax/2) 0 x Canceling the heat transfer area A and rearranging give T1 1 g·2x hx hx T2 T k k 2k 2 (2) Equations (1) and (2) form a system of two equations in two unknowns T1 and T2. Substituting the given quantities and simplifying gives 2T1 T2 71.43 T1 1.032T2 36.68 (in °C) (in °C) This is a system of two algebraic equations in two unknowns and can be solved easily by the elimination method. Solving the first equation for T1 and substituting into the second equation result in an equation in T2 whose solution is T2 136.1°C This is the temperature of the surface exposed to convection, which is the desired result. Substitution of this result into the first equation gives T1 103.8°C, which is the temperature at the middle of the plate. h T Plate 0 1 2 cm 2 x 2 cm Finite difference solution: T2 = 136.1°C Exact solution: T2 = 136.0°C FIGURE 5–19 Despite being approximate in nature, highly accurate results can be obtained by numerical methods. Discussion The purpose of this example is to demonstrate the use of the finite difference method with minimal calculations, and the accuracy of the result was not a major concern. But you might still be wondering how accurate the result obtained above is. After all, we used a mesh of only three nodes for the entire plate, which seems to be rather crude. This problem can be solved analytically as described in Chapter 2, and the analytical (exact) solution can be shown to be T(x) 0.5g·hL2/k g·L Th g·x2 x hL k 2k Substituting the given quantities, the temperature of the exposed surface of the plate at x L 0.04 m is determined to be 136.0°C, which is almost identical to the result obtained here with the approximate finite difference method (Fig. 5–19). Therefore, highly accurate results can be obtained with numerical methods by using a limited number of nodes. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 279 279 CHAPTER 5 EXAMPLE 5–2 Heat Transfer from Triangular Fins Consider an aluminum alloy fin (k 180 W/m · °C) of triangular cross section with length L 5 cm, base thickness b 1 cm, and very large width w in the direction normal to the plane of paper, as shown in Figure 5–20. The base of the fin is maintained at a temperature of T0 200°C. The fin is losing heat to the surrounding medium at T 25°C with a heat transfer coefficient of h 15 W/m2 · °C. Using the finite difference method with six equally spaced nodes along the fin in the x-direction, determine (a) the temperatures at the nodes, (b) the rate of heat transfer from the fin for w 1 m, and (c) the fin efficiency. h, T T0 b 0 SOLUTION A long triangular fin attached to a surface is considered. The nodal temperatures, the rate of heat transfer, and the fin efficiency are to be determined numerically using six equally spaced nodes. Triangular fin w b/ 2 tan θ = —– L θ 1 2 3 ∆x 4 5 Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 The temperature along the fin varies in the x direction only. 3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible. ∆x ——– cos θ [L – (m + 1– )∆ x]tan θ 2 θ Properties The thermal conductivity is given to be k 180 W/m · °C. Analysis (a) The number of nodes in the fin is specified to be M 6, and their location is as shown in the figure. Then the nodal spacing x becomes x 0.05 m L 0.01 m M1 61 The temperature at node 0 is given to be T0 200°C, and the temperatures at the remaining five nodes are to be determined. Therefore, we need to have five equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and the finite difference formulation for a general interior node m is obtained by applying an energy balance on the volume element of this node. Noting that heat transfer is steady and there is no heat generation in the fin and assuming heat transfer to be into the medium at all sides, the energy balance can be expressed as · Q 0 all sides → kAleft Tm1 Tm Tm1 Tm kAright hAconv(T Tm) 0 x x Note that heat transfer areas are different for each node in this case, and using geometrical relations, they can be expressed as Aleft (Height Width)@m 12 2w[L (m 1/2)x]tan Aright (Height Width)@m 12 2w[L (m 1/2)x]tan Aconv 2 Length Width 2w(x/cos ) Substituting, Tm1 Tm x Tm1 Tm 2wx 2kw[L (m 12)x]tan h (T Tm) 0 cos x 2kw[L (m 12)x]tan x L m–1 m m+1 ∆x L – (m – 1– )∆ x 2 [L – (m – 1– )∆ x]tan θ 2 FIGURE 5–20 Schematic for Example 5–2 and the volume element of a general interior node of the fin. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 280 280 HEAT TRANSFER Dividing each term by 2kwL tan /x gives 1 (m 1 ) 2 x x (Tm 1 Tm) 1 (m 12) (Tm 1 Tm) L L h(x)2 (T Tm) 0 kL sin Note that tan b/2 0.5 cm 0.1 L 5 cm → tan10.1 5.71° Also, sin 5.71° 0.0995. Then the substitution of known quantities gives (5.5 m)Tm 1 (10.00838 2m)Tm (4.5 m)Tm 1 0.209 Now substituting 1, 2, 3, and 4 for m results in these finite difference equations for the interior nodes: m 1: m 2: m 3: m 4: kAleft θ 4 5 ∆x —– 2 ∆x —– 2 FIGURE 5–21 Schematic of the volume element of node 5 at the tip of a triangular fin. (1) (2) (3) (4) The finite difference equation for the boundary node 5 is obtained by writing an energy balance on the volume element of length x/2 at that boundary, again by assuming heat transfer to be into the medium at all sides (Fig. 5–21): ∆x/2 —–— cos θ ∆x —– tan θ 2 8.00838T1 3.5T2 900.209 3.5T1 6.00838T2 2.5T3 0.209 2.5T2 4.00838T3 1.5T4 0.209 1.5T3 2.00838T4 0.5T5 0.209 T4 T5 hAconv (T T5) 0 x where Aleft 2w x tan 2 and Aconv 2w x/2 cos Canceling w in all terms and substituting the known quantities gives T4 1.00838T5 0.209 (5) Equations (1) through (5) form a linear system of five algebraic equations in five unknowns. Solving them simultaneously using an equation solver gives T1 198.6°C, T4 194.3°C, T2 197.1°C, T5 192.9°C T3 195.7°C, which is the desired solution for the nodal temperatures. (b) The total rate of heat transfer from the fin is simply the sum of the heat transfer from each volume element to the ambient, and for w 1 m it is determined from cen58933_ch05.qxd 9/4/2002 11:41 AM Page 281 281 CHAPTER 5 · Q fin 5 Q m0 · element, m 5 hA conv, m(Tm T) m0 Noting that the heat transfer surface area is wx/cos for the boundary nodes 0 and 5, and twice as large for the interior nodes 1, 2, 3, and 4, we have · wx Q fin h [(T T) 2(T1 T) 2(T2 T) 2(T3 T) cos 0 2(T4 T) (T5 T)] wx [T 2(T1 T2 T3 T4) T5 10T] cos 0 (1 m)(0.01 m) (15 W/m2 · °C) [200 2 785.7 192.9 10 25] cos 5.71° 258.4 W h (c) If the entire fin were at the base temperature of T0 200°C, the total rate of heat transfer from the fin for w 1 m would be · Q max hAfin, total (T0 T) h(2wL/cos )(T0 T) (15 W/m2 · °C)2(1 m)(0.05 m)/cos5.71°°C 263.8 W Then the fin efficiency is determined from Q· fin 258.4 W fin · 0.98 Q max 263.8 W which is less than 1, as expected. We could also determine the fin efficiency in this case from the proper fin efficiency curve in Chapter 3, which is based on the analytical solution. We would read 0.98 for the fin efficiency, which is identical to the value determined above numerically. The finite difference formulation of steady heat conduction problems usually results in a system of N algebraic equations in N unknown nodal temperatures that need to be solved simultaneously. When N is small (such as 2 or 3), we can use the elementary elimination method to eliminate all unknowns except one and then solve for that unknown (see Example 5–1). The other unknowns are then determined by back substitution. When N is large, which is usually the case, the elimination method is not practical and we need to use a more systematic approach that can be adapted to computers. There are numerous systematic approaches available in the literature, and they are broadly classified as direct and iterative methods. The direct methods are based on a fixed number of well-defined steps that result in the solution in a systematic manner. The iterative methods, on the other hand, are based on an initial guess for the solution that is refined by iteration until a specified convergence criterion is satisfied (Fig. 5–22). The direct methods usually require a large amount of computer memory and computation time, Direct methods: Solve in a systematic manner following a series of well-defined steps. Iterative methods: Start with an initial guess for the solution, and iterate until solution converges. FIGURE 5–22 Two general categories of solution methods for solving systems of algebraic equations. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 282 282 HEAT TRANSFER and they are more suitable for systems with a relatively small number of equations. The computer memory requirements for iterative methods are minimal, and thus they are usually preferred for large systems. The convergence of iterative methods to the desired solution, however, may pose a problem. 5–4 y … N ∆y ∆y Node (m, n) … n+1 n n–1 2 1 0 ∆x ∆x … 0 1 2 … m m–1 m+1 M x FIGURE 5–23 The nodal network for the finite difference formulation of twodimensional conduction in rectangular coordinates. m, n + 1 n+1 ∆y Volume element m – 1, n n g· m, n m, n m + 1, n ∆x m–1 ∆x m Rate of heat conduction Rate of heat Rate of change of at the left, top, right, generation inside the energy content and bottom surfaces the element of the element m, n – 1 y TWO-DIMENSIONAL STEADY HEAT CONDUCTION In Section 5–3 we considered one-dimensional heat conduction and assumed heat conduction in other directions to be negligible. Many heat transfer problems encountered in practice can be approximated as being one-dimensional, but this is not always the case. Sometimes we need to consider heat transfer in other directions as well when the variation of temperature in other directions is significant. In this section we will consider the numerical formulation and solution of two-dimensional steady heat conduction in rectangular coordinates using the finite difference method. The approach presented below can be extended to three-dimensional cases. Consider a rectangular region in which heat conduction is significant in the x- and y-directions. Now divide the x-y plane of the region into a rectangular mesh of nodal points spaced x and y apart in the x- and y-directions, respectively, as shown in Figure 5–23, and consider a unit depth of z 1 in the z-direction. Our goal is to determine the temperatures at the nodes, and it is convenient to number the nodes and describe their position by the numbers instead of actual coordinates. A logical numbering scheme for two-dimensional problems is the double subscript notation (m, n) where m 0, 1, 2, . . . , M is the node count in the x-direction and n 0, 1, 2, . . . , N is the node count in the y-direction. The coordinates of the node (m, n) are simply x mx and y ny, and the temperature at the node (m, n) is denoted by Tm, n. Now consider a volume element of size x y 1 centered about a general interior node (m, n) in a region in which heat is generated at a rate of g· and the thermal conductivity k is constant, as shown in Figure 5–24. Again assuming the direction of heat conduction to be toward the node under consideration at all surfaces, the energy balance on the volume element can be expressed as ∆y n–1 ■ or m+1 x FIGURE 5–24 The volume element of a general interior node (m, n) for twodimensional conduction in rectangular coordinates. Eelement · · · · · Q cond, left Q cond, top Q cond, right Q cond, bottom Gelement 0 t (5-31) for the steady case. Again assuming the temperatures between the adjacent nodes to vary linearly and noting that the heat transfer area is Ax y 1 y in the x-direction and Ay x 1 x in the y-direction, the energy balance relation above becomes cen58933_ch05.qxd 9/4/2002 11:41 AM Page 283 283 CHAPTER 5 ky Tm1, n Tm, n Tm, n1 Tm, n Tm1, n Tm, n kx ky x y x Tm, n1 Tm, n kx g·m, n x y 0 y (5-32) Dividing each term by x y and simplifying gives Tm1, n 2Tm, n Tm1, n x 2 Tm, n1 2Tm, n Tm, n1 y 2 g·m, n 0 k (5-33) for m 1, 2, 3, . . . , M 1 and n 1, 2, 3, . . . , N 1. This equation is identical to Eq. 5–12 obtained earlier by replacing the derivatives in the differential equation by differences for an interior node (m, n). Again a rectangular region M equally spaced nodes in the x-direction and N equally spaced nodes in the y-direction has a total of (M 1)(N 1) nodes, and Eq. 5–33 can be used to obtain the finite difference equations at all interior nodes. In finite difference analysis, usually a square mesh is used for simplicity (except when the magnitudes of temperature gradients in the x- and y-directions are very different), and thus x and y are taken to be the same. Then x y l, and the relation above simplifies to Tm 1, n Tm 1, n Tm, n 1 Tm, n 1 4Tm, n g·m, nl 2 0 k (5-34) That is, the finite difference formulation of an interior node is obtained by adding the temperatures of the four nearest neighbors of the node, subtracting four times the temperature of the node itself, and adding the heat generation term. It can also be expressed in this form, which is easy to remember: Tleft Ttop Tright Tbottom 4Tnode g· nodel 2 0 k (5-35) When there is no heat generation in the medium, the finite difference equation for an interior node further simplifies to Tnode (Tleft Ttop Tright Tbottom)/4, which has the interesting interpretation that the temperature of each interior node is the arithmetic average of the temperatures of the four neighboring nodes. This statement is also true for the three-dimensional problems except that the interior nodes in that case will have six neighboring nodes instead of four. Boundary Nodes The development of finite difference formulation of boundary nodes in two(or three-) dimensional problems is similar to the development in the onedimensional case discussed earlier. Again, the region is partitioned between the nodes by forming volume elements around the nodes, and an energy balance is written for each boundary node. Various boundary conditions can be handled as discussed for a plane wall, except that the volume elements in the two-dimensional case involve heat transfer in the y-direction as well as the x-direction. Insulated surfaces can still be viewed as “mirrors, ” and the cen58933_ch05.qxd 9/4/2002 11:41 AM Page 284 284 HEAT TRANSFER Volume element of node 2 h, T 1 · Qtop 2 3 · Qleft ∆y mirror image concept can be used to treat nodes on insulated boundaries as interior nodes. For heat transfer under steady conditions, the basic equation to keep in mind when writing an energy balance on a volume element is (Fig. 5–25) Boundary subjected to convection · Qright whether the problem is one-, two-, or three-dimensional. Again we assume, for convenience in formulation, all heat transfer to be into the volume element from all surfaces except for specified heat flux, whose direction is already specified. This is demonstrated in Example 5–3 for various boundary conditions. 4 g· 2V2 · · · · Qleft + Qtop + Qright + Qbottom + —— =0 k FIGURE 5–25 The finite difference formulation of a boundary node is obtained by writing an energy balance on its volume element. EXAMPLE 5–3 Convection h, T y 1 2 3 4 5 6 7 8 10 11 12 13 14 15 ∆ x = ∆y = l ∆y · 9 qR ∆y x ∆x ∆x 90°C ∆x ∆x h, T h, T 2 1 2 3 5 4 (a) Node 1 Steady Two-Dimensional Heat Conduction in L-Bars Consider steady heat transfer in an L-shaped solid body whose cross section is given in Figure 5–26. Heat transfer in the direction normal to the plane of the paper is negligible, and thus heat transfer in the body is two-dimensional. The thermal conductivity of the body is k 15 W/m · °C, and heat is generated in the body at a rate of g· 2 106 W/m3. The left surface of the body is insulated, and the bottom surface is maintained at a uniform temperature of 90°C. The entire top surface is subjected to convection to ambient air at T 25°C with a convection coefficient of h 80 W/m2 · °C, and the right surface is subjected to heat flux at a uniform rate of q· R 5000 W/m2. The nodal network of the problem consists of 15 equally spaced nodes with x y 1.2 cm, as shown in the figure. Five of the nodes are at the bottom surface, and thus their temperatures are known. Obtain the finite difference equations at the remaining nine nodes and determine the nodal temperatures by solving them. ∆x FIGURE 5–26 Schematic for Example 5–3 and the nodal network (the boundaries of volume elements of the nodes are indicated by dashed lines). 1 (5-36) all sides · Qbottom ∆x · Q g·Velement 0 (b) Node 2 FIGURE 5–27 Schematics for energy balances on the volume elements of nodes 1 and 2. SOLUTION Heat transfer in a long L-shaped solid bar with specified boundary conditions is considered. The nine unknown nodal temperatures are to be determined with the finite difference method. Assumptions 1 Heat transfer is steady and two-dimensional, as stated. 2 Thermal conductivity is constant. 3 Heat generation is uniform. 4 Radiation heat transfer is negligible. Properties The thermal conductivity is given to be k 15 W/m · °C. Analysis We observe that all nodes are boundary nodes except node 5, which is an interior node. Therefore, we will have to rely on energy balances to obtain the finite difference equations. But first we form the volume elements by partitioning the region among the nodes equitably by drawing dashed lines between the nodes. If we consider the volume element represented by an interior node to be full size (i.e., x y 1), then the element represented by a regular boundary node such as node 2 becomes half size (i.e., x y/2 1), and a corner node such as node 1 is quarter size (i.e., x/2 y/2 1). Keeping Eq. 5–36 in mind for the energy balance, the finite difference equations for each of the nine nodes are obtained as follows: (a) Node 1. The volume element of this corner node is insulated on the left and subjected to convection at the top and to conduction at the right and bottom surfaces. An energy balance on this element gives [Fig. 5–27a] cen58933_ch05.qxd 9/4/2002 11:41 AM Page 285 285 CHAPTER 5 0h y T2 T1 x T4 T1 x x y (T T1) k k g·1 0 2 2 2 2 2 x y Taking x y l, it simplifies to – 2 g·1l 2 hl hl T1 T2 T4 T k k 2k (b) Node 2. The volume element of this boundary node is subjected to convection at the top and to conduction at the right, bottom, and left surfaces. An energy balance on this element gives [Fig. 5–27b] hx(T T2) k T5 T2 y y T3 T2 y T1 T2 kx k g·2x 0 2 2 2 x y x Taking x y l, it simplifies to T1 4 g·2l 2hl 2hl T2 T3 2T5 T k k k 2 (c) Node 3. The volume element of this corner node is subjected to convection at the top and right surfaces and to conduction at the bottom and left surfaces. An energy balance on this element gives [Fig. 5–28a] h y x2 2 (T T3) k y T2 T3 x T6 T3 x y k g·3 0 2 2 2 2 y x 2 1 Mirror h, T 3 (5) 4 5 h, T Taking x y l, it simplifies to T2 2 g·3l 2 2hl 2hl T3 T6 T k k 2k (d ) Node 4. This node is on the insulated boundary and can be treated as an interior node by replacing the insulation by a mirror. This puts a reflected image of node 5 to the left of node 4. Noting that x y l, the general interior node relation for the steady two-dimensional case (Eq. 5–35) gives [Fig. 5–28b] T5 T1 T5 T10 4T4 (a) Node 3 (b) Node 4 FIGURE 5–28 Schematics for energy balances on the volume elements of nodes 3 and 4. g·4l 2 0 k 3 2 or, noting that T10 90° C, T1 4T4 2T5 90 g·4l 2 k (e) Node 5. This is an interior node, and noting that x y l, the finite difference formulation of this node is obtained directly from Eq. 5–35 to be [Fig. 5–29a] T4 T2 T6 T11 4T5 10 6 g·5l 2 0 k 4 5 6 h, T 7 6 5 12 11 (a) Node 5 (b) Node 6 FIGURE 5–29 Schematics for energy balances on the volume elements of nodes 5 and 6. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 286 286 HEAT TRANSFER or, noting that T11 90°C, T2 T4 4T5 T6 90 g·5l 2 k (f ) Node 6. The volume element of this inner corner node is subjected to convection at the L-shaped exposed surface and to conduction at other surfaces. An energy balance on this element gives [Fig. 5–29b] h y x2 2 (T ky T6) k T12 T6 y T7 T6 kx 2 x y T5 T6 3xy x T3 T6 0 k g·6 2 4 x y Taking x y l and noting that T12 90°C, it simplifies to T3 2T5 6 6 7 8 T13 T7 y T8 T7 kx 2 x y y y T6 T7 k g·7x 0 2 2 x 9 hx(T T7) k q· R 13 2 (g) Node 7. The volume element of this boundary node is subjected to convection at the top and to conduction at the right, bottom, and left surfaces. An energy balance on this element gives [Fig. 5–30a] h, T h, T 3g·6l 2hl 2hl T6 T7 180 T k k 2k 15 FIGURE 5–30 Schematics for energy balances on the volume elements of nodes 7 and 9. Taking x y l and noting that T13 90°C, it simplifies to T6 4 g·7l 2 2hl 2hl T7 T8 180 T k k k (h) Node 8. This node is identical to Node 7, and the finite difference formulation of this node can be obtained from that of Node 7 by shifting the node numbers by 1 (i.e., replacing subscript m by m 1). It gives T7 4 g·8l 2hl 2hl T8 T9 180 T k k k 2 (i ) Node 9. The volume element of this corner node is subjected to convection at the top surface, to heat flux at the right surface, and to conduction at the bottom and left surfaces. An energy balance on this element gives [Fig. 5–30b] h y y T8 T9 x T15 T9 x x y (T T9) q·R k k g·9 0 2 2 2 2 2 2 y x Taking x y l and noting that T15 90°C, it simplifies to T8 2 g·9l 2 q·Rl hl hl T9 90 T k k k 2k cen58933_ch05.qxd 9/4/2002 11:41 AM Page 287 287 CHAPTER 5 This completes the development of finite difference formulation for this problem. Substituting the given quantities, the system of nine equations for the determination of nine unknown nodal temperatures becomes –2.064T1 T2 T4 11.2 T1 4.128T2 T3 2T5 22.4 T2 2.128T3 T6 12.8 T1 4T4 2T5 109.2 T2 T4 4T5 T6 109.2 T3 2T5 6.128T6 T7 212.0 T6 4.128T7 T8 202.4 T7 4.128T8 T9 202.4 T8 2.064T9 105.2 which is a system of nine algebraic equations with nine unknowns. Using an equation solver, its solution is determined to be T1 112.1°C T4 109.4°C T7 97.3°C T2 110.8°C T5 108.1°C T8 96.3°C T3 106.6°C T6 103.2°C T9 97.6°C Note that the temperature is the highest at node 1 and the lowest at node 8. This is consistent with our expectations since node 1 is the farthest away from the bottom surface, which is maintained at 90°C and has one side insulated, and node 8 has the largest exposed area relative to its volume while being close to the surface at 90°C. Irregular Boundaries In problems with simple geometries, we can fill the entire region using simple volume elements such as strips for a plane wall and rectangular elements for two-dimensional conduction in a rectangular region. We can also use cylindrical or spherical shell elements to cover the cylindrical and spherical bodies entirely. However, many geometries encountered in practice such as turbine blades or engine blocks do not have simple shapes, and it is difficult to fill such geometries having irregular boundaries with simple volume elements. A practical way of dealing with such geometries is to replace the irregular geometry by a series of simple volume elements, as shown in Figure 5–31. This simple approach is often satisfactory for practical purposes, especially when the nodes are closely spaced near the boundary. More sophisticated approaches are available for handling irregular boundaries, and they are commonly incorporated into the commercial software packages. EXAMPLE 5–4 Actual boundary Approximation Heat Loss through Chimneys Hot combustion gases of a furnace are flowing through a square chimney made of concrete (k 1.4 W/m · °C). The flow section of the chimney is 20 cm 20 cm, and the thickness of the wall is 20 cm. The average temperature of the FIGURE 5–31 Approximating an irregular boundary with a rectangular mesh. cen58933_ch05.qxd 9/4/2002 11:41 AM Page 288 288 HEAT TRANSFER hot gases in the chimney is Ti 300°C, and the average convection heat transfer coefficient inside the chimney is hi 70 W/m2 · °C. The chimney is losing heat from its outer surface to the ambient air at To 20°C by convection with a heat transfer coefficient of ho 21 W/m2 · °C and to the sky by radiation. The emissivity of the outer surface of the wall is 0.9, and the effective sky temperature is estimated to be 260 K. Using the finite difference method with x y 10 cm and taking full advantage of symmetry, determine the temperatures at the nodal points of a cross section and the rate of heat loss for a 1-m-long section of the chimney. SOLUTION Heat transfer through a square chimney is considered. The nodal temperatures and the rate of heat loss per unit length are to be determined with the finite difference method. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the chimney is two-dimensional since the height of the chimney is large relative to its cross section, and thus heat conduction through the chimney in the axial direction is negligible. It is tempting to simplify the problem further by considering heat transfer in each wall to be one-dimensional, which would be the case if the walls were thin and thus the corner effects were negligible. This assumption cannot be justified in this case since the walls are very thick and the corner sections constitute a considerable portion of the chimney structure. 3 Thermal conductivity is constant. Symmetry lines (Equivalent to insulation) Properties 0.9. h1 T1 1 2 3 4 5 6 7 8 h0, T0 Tsky 9 Representative section of chimney FIGURE 5–32 Schematic of the chimney discussed in Example 5–4 and the nodal network for a representative section. h, T 1 h, T 2 1 2 The properties of chimney are given to be k 1.4 W/m · °C and Analysis The cross section of the chimney is given in Figure 5–32. The most striking aspect of this problem is the apparent symmetry about the horizontal and vertical lines passing through the midpoint of the chimney as well as the diagonal axes, as indicated on the figure. Therefore, we need to consider only one-eighth of the geometry in the solution whose nodal network consists of nine equally spaced nodes. No heat can cross a symmetry line, and thus symmetry lines can be treated as insulated surfaces and thus “mirrors” in the finite difference formulation. Then the nodes in the middle of the symmetry lines can be treated as interior nodes by using mirror images. Six of the nodes are boundary nodes, so we will have to write energy balances to obtain their finite difference formulations. First we partition the region among the nodes equitably by drawing dashed lines between the nodes through the middle. Then the region around a node surrounded by the boundary or the dashed lines represents the volume element of the node. Considering a unit depth and using the energy balance approach for the boundary nodes (again assuming all heat transfer into the volume element for convenience) and the formula for the interior nodes, the finite difference equations for the nine nodes are determined as follows: (a) Node 1. On the inner boundary, subjected to convection, Figure 5–33a 0 hi 3 (a) Node 1 4 (b) Node 2 FIGURE 5–33 Schematics for energy balances on the volume elements of nodes 1 and 2. y T2 T1 x T3 T1 x (T T1) k k 00 2 i 2 2 x y Taking x y l, it simplifies to – 2 hi l hi l T1 T2 T3 T k k i cen58933_ch05.qxd 9/4/2002 11:42 AM Page 289 289 CHAPTER 5 (b) Node 2. On the inner boundary, subjected to convection, Figure 5–33b k y T1 T2 T4 T2 x hi (T T2) 0 kx 0 2 2 i x y Taking x y l, it simplifies to T1 3 (4) 1 2 (4) Mirror images 3 4 5 (8) 6 7 8 9 Mirror image hi l hi l T2 2T4 T k k i (c) Nodes 3, 4, and 5. (Interior nodes, Fig. 5–34) Node 3: T4 T1 T4 T6 4T3 0 Node 4: T3 T2 T5 T7 4T4 0 Node 5: T4 T4 T8 T8 4T5 0 Mirror Mirror FIGURE 5–34 Converting the boundary nodes 3 and 5 on symmetry lines to interior nodes by using mirror images. (d ) Node 6. (On the outer boundary, subjected to convection and radiation) 0k ho y T7 T6 x T3 T6 k 2 2 y x x x 4 (T T6) (T T64) 0 2 o 2 sky Taking x y l, it simplifies to T2 T3 2 ho l ho l l 4 T6 T (Tsky T64) k k o k (e) Node 7. (On the outer boundary, subjected to convection and radiation, Fig. 5–35) k Insulation y T6 T7 y T8 T7 T4 T 7 kx k 2 2 x y x 4 4 hox(To T7) x(Tsky T74) 0 Taking x y l, it simplifies to 2ho l 2ho l 2 l 4 2T4 T6 4 T7 T8 T (Tsky T74) k k o k (f ) Node 8. Same as Node 7, except shift the node numbers up by 1 (replace 4 by 5, 6 by 7, 7 by 8, and 8 by 9 in the last relation) 2T5 T7 4 2ho l 2ho l 2 l 4 T8 T9 T (Tsky T84) k k o k (g) Node 9. (On the outer boundary, subjected to convection and radiation, Fig. 5–35) k y T8 T9 x x 4 0 ho (T T9) (T T94) 0 2 2 o 2 sky x 6 7 8 9 h, T Tsky FIGURE 5–35 Schematics for energy balances on the volume elements of nodes 7 and 9. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 290 290 HEAT TRANSFER Taking x y l, it simplifies to T8 1 ho l ho l l 4 T9 T (Tsky T94) k k o k This problem involves radiation, which requires the use of absolute temperature, and thus all temperatures should be expressed in Kelvin. Alternately, we could use °C for all temperatures provided that the four temperatures in the radiation terms are expressed in the form (T 273)4. Substituting the given quantities, the system of nine equations for the determination of nine unknown nodal temperatures in a form suitable for use with the Gauss-Seidel iteration method becomes T1 (T2 T3 2865)/7 T2 (T1 2T4 2865)/8 T3 (T1 2T4 T6)/4 T4 (T2 T3 T5 T7)/4 T5 (2T4 2T8)/4 T6 (T2 T3 456.2 0.3645 109 T64)/3.5 T7 (2T4 T6 T8 912.4 0.729 109 T74)/7 T8 (2T5 T7 T9 912.4 0.729 109 T84)/7 T9 (T8 456.2 0.3645 109 T94)/2.5 which is a system of nonlinear equations. Using an equation solver, its solution is determined to be 23 40 55 60 55 40 23 T1 545.7 K 272.6°C T4 411.2 K 138.0°C T7 328.1 K 54.9°C 40 Temperature, °C 55 60 55 40 89 152 89 138 152 256 273 273 256 138 273 152 138 256 273 256 138 89 138 152 138 89 40 55 60 55 40 FIGURE 5–36 The variation of temperature in the chimney. T2 529.2 K 256.1°C T5 362.1 K 89.0°C T8 313.1 K 39.9°C T3 425.2 K 152.1°C T6 332.9 K 59.7°C T9 296.5 K 23.4°C 23 40 55 The variation of temperature in the chimney is shown in Figure 5–36. Note that the temperatures are highest at the inner wall (but less than 300°C) and lowest at the outer wall (but more that 260 K), as expected. The average temperature at the outer surface of the chimney weighed by the surface area is 60 Twall, out 55 40 (0.5T6 T7 T8 0.5T9) (0.5 1 1 0.5) 0.5 332.9 328.1 313.1 0.5 296.5 318.6 K 3 23 Then the rate of heat loss through the 1-m-long section of the chimney can be determined approximately from cen58933_ch05.qxd 9/4/2002 11:42 AM Page 291 291 CHAPTER 5 · 4 4 Q chimney ho Ao (Twall, out To) Ao (Twall, out Tsky) 2 (21 W/m · K)4 (0.6 m)(1 m)K 0.9(5.67 108 W/m2 · K4) 4 (0.6 m)(1 m)4 (260 K)4] 1291 702 1993 W We could also determine the heat transfer by finding the average temperature of the inner wall, which is (272.6 256.1)/2 264.4°C, and applying Newton’s law of cooling at that surface: · Q chimney hi Ai (Ti Twall, in) (70 W/m2 · K)4 (0.2 m)(1 m)°C 1994 W The difference between the two results is due to the approximate nature of the numerical analysis. Discussion We used a relatively crude numerical model to solve this problem to keep the complexities at a manageable level. The accuracy of the solution obtained can be improved by using a finer mesh and thus a greater number of nodes. Also, when radiation is involved, it is more accurate (but more laborious) to determine the heat losses for each node and add them up instead of using the average temperature. 5–5 ■ TRANSIENT HEAT CONDUCTION So far in this chapter we have applied the finite difference method to steady heat transfer problems. In this section we extend the method to solve transient problems. We applied the finite difference method to steady problems by discretizing the problem in the space variables and solving for temperatures at discrete points called the nodes. The solution obtained is valid for any time since under steady conditions the temperatures do not change with time. In transient problems, however, the temperatures change with time as well as position, and thus the finite difference solution of transient problems requires discretization in time in addition to discretization in space, as shown in Figure 5–37. This is done by selecting a suitable time step t and solving for the unknown nodal temperatures repeatedly for each t until the solution at the desired time is obtained. For example, consider a hot metal object that is taken out of the oven at an initial temperature of Ti at time t 0 and is allowed to cool in ambient air. If a time step of t 5 min is chosen, the determination of the temperature distribution in the metal piece after 3 h requires the determination of the temperatures 3 60/5 36 times, or in 36 time steps. Therefore, the computation time of this problem will be 36 times that of a steady problem. Choosing a smaller t will increase the accuracy of the solution, but it will also increase the computation time. In transient problems, the superscript i is used as the index or counter of time steps, with i 0 corresponding to the specified initial condition. In the case of the hot metal piece discussed above, i 1 corresponds to t 1 t 5 min, i 2 corresponds to t 2 t 10 min, and a general t i +1 T i +1 Tmi +1 – 1 Tm m +1 i+1 ∆t Tmi – 1 Tmi i 1 0 ∆t ∆x 0 1 ∆x m–1 Tmi +1 ∆x m m+1 x FIGURE 5–37 Finite difference formulation of timedependent problems involves discrete points in time as well as space. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 292 292 HEAT TRANSFER time step i corresponds to ti it. The notation Tmi is used to represent the temperature at the node m at time step i. The formulation of transient heat conduction problems differs from that of steady ones in that the transient problems involve an additional term representing the change in the energy content of the medium with time. This additional term appears as a first derivative of temperature with respect to time in the differential equation, and as a change in the internal energy content during t in the energy balance formulation. The nodes and the volume elements in transient problems are selected as they are in the steady case, and, again assuming all heat transfer is into the element for convenience, the energy balance on a volume element during a time interval t can be expressed as Heat transferred into Heat generated The change in the the volume element within the energy content of from all of its surfaces volume element the volume element during t during t during t or t · · Q t G element Eelement (5-37) All sides · where the rate of heat transfer Q normally consists of conduction terms for interior nodes, but may involve convection, heat flux, and radiation for boundary nodes. Noting that Eelement mCT Velement CT, where is density and C is the specific heat of the element, dividing the earlier relation by t gives Eelement · · T Q G element Velement C t t All sides (5-38) or, for any node m in the medium and its volume element, All sides Volume element (can be any shape) Node m ρ = density V = volume ρV = mass C = specific heat ∆ T = temperature change ∆U = ρVC∆T = ρVC(Tmi+ 1 – Tmi ) FIGURE 5–38 The change in the energy content of the volume element of a node during a time interval t. Tmi t i1 Tm · · Q G element Velement C (5-39) where Tmi and Tmi1 are the temperatures of node m at times ti it and ti 1 (i 1)t, respectively, and Tmi1 Tmi represents the temperature change of the node during the time interval t between the time steps i and i 1 (Fig. 5–38). Note that the ratio (Tmi1 Tmi )/t is simply the finite difference approximation of the partial derivative T/t that appears in the differential equations of transient problems. Therefore, we would obtain the same result for the finite difference formulation if we followed a strict mathematical approach instead of the energy balance approach used above. Also note that the finite difference formulations of steady and transient problems differ by the single term on the right side of the equal sign, and the format of that term remains the same in all coordinate systems regardless of whether heat transfer is one-, two-, or three-dimensional. For the special case of Tmi1 Tmi (i.e., no change in temperature with time), the formulation reduces to that of steady case, as expected. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 293 293 CHAPTER 5 The nodal temperatures in transient problems normally change during each time step, and you may be wondering whether to use temperatures at the previous time step i or the new time step i 1 for the terms on the left side of Eq. 5–39. Well, both are reasonable approaches and both are used in practice. The finite difference approach is called the explicit method in the first case and the implicit method in the second case, and they are expressed in the general form as (Fig. 5–39) Explicit method: Tm · i1 Q i 1 G· element Velement C All sides · Tmi t · ∑ Q + Gelement All sides Tmi+ 1 – Tmi = ρVelementC ————– ∆t If expressed at i: Explicit method i1 Tm · Q i G· ielement Velement C All sides Implicit method: Tmi t If expressed at i + 1: Implicit method FIGURE 5–39 The formulation of explicit and implicit methods differs at the time step (previous or new) at which the heat transfer and heat generation terms are expressed. (5-40) i1 (5-41) It appears that the time derivative is expressed in forward difference form in the explicit case and backward difference form in the implicit case. Of course, it is also possible to mix the two fundamental formulations of Eqs. 5–40 and 5–41 and come up with more elaborate formulations, but such formulations offer little insight and are beyond the scope of this text. Note that both formulations are simply expressions between the nodal temperatures before and after a time interval and are based on determining the new temperatures Tmi1 using the previous temperatures Tmi . The explicit and implicit formulations given here are quite general and can be used in any coordinate system regardless of the dimension of heat transfer. The volume elements in multidimensional cases simply have more surfaces and thus involve more terms in the summation. The explicit and implicit methods have their advantages and disadvantages, and one method is not necessarily better than the other one. Next you will see that the explicit method is easy to implement but imposes a limit on the allowable time step to avoid instabilities in the solution, and the implicit method requires the nodal temperatures to be solved simultaneously for each time step but imposes no limit on the magnitude of the time step. We will limit the discussion to one- and two-dimensional cases to keep the complexities at a manageable level, but the analysis can readily be extended to three-dimensional cases and other coordinate systems. A Plane wall g· m Volume element of node m Tmi+ 1 Tmi kA Transient Heat Conduction in a Plane Wall Consider transient one-dimensional heat conduction in a plane wall of thickness L with heat generation g·(x, t) that may vary with time and position and constant conductivity k with a mesh size of x L/M and nodes 0, 1, 2, . . . , M in the x-direction, as shown in Figure 5–40. Noting that the volume element of a general interior node m involves heat conduction from two sides and the volume of the element is Velement Ax, the transient finite difference formulation for an interior node can be expressed on the basis of Eq. 5–39 as kA Tmi1 Tmi Tm1 Tm Tm1 Tm kA g·m Ax AxC x x t (5-42) Tmi – 1 – Tmi ————– Tmi + 1 – Tmi kA ————– ∆x ∆x ∆x 0 1 2 ∆x m –1 m m +1 ∆x M–1 M x FIGURE 5–40 The nodal points and volume elements for the transient finite difference formulation of one-dimensional conduction in a plane wall. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 294 294 HEAT TRANSFER Canceling the surface area A and multiplying by x/k, it simplifies to Tm 1 2Tm Tm 1 g·m x2 x2 i1 (T Tmi ) k t m (5-43) where k/ C is the thermal diffusivity of the wall material. We now define a dimensionless mesh Fourier number as αt x2 (5-44) Then Eq. 5–43 reduces to Tm 1 2Tm Tm 1 g·mx2 Tmi1 Tmi k (5-45) Note that the left side of this equation is simply the finite difference formulation of the problem for the steady case. This is not surprising since the formulation must reduce to the steady case for Tmi1 Tmi . Also, we are still not committed to explicit or implicit formulation since we did not indicate the time step on the left side of the equation. We now obtain the explicit finite difference formulation by expressing the left side at time step i as i i Tm1 2Tmi Tm1 g·mi x2 Tmi1 Tmi k (explicit) (5-46) This equation can be solved explicitly for the new temperature Tmi1 (and thus the name explicit method) to give i i Tm1 ) (1 2) Tmi Tmi1 (Tm1 g·mi x2 k (5-47) for all interior nodes m 1, 2, 3, . . . , M 1 in a plane wall. Expressing the left side of Eq. 5–45 at time step i 1 instead of i would give the implicit finite difference formulation as ∆ –x — 2 A hA(T – T0i ) ∆x ρ A —– C 2 i1 i1 Tm1 2Tmi1 Tm1 T0i + 1 – T0i ————– (implicit) (5-48) ∆t which can be rearranged as g· 0 i1 i1 Tm1 (1 2) Tmi1 Tm1 T1i – T0i kA ——— ∆x ∆x ∆x 0 g·mi1x2 Tmi1 Tmi k 1 2 … FIGURE 5–41 Schematic for the explicit finite difference formulation of the convection condition at the left boundary of a plane wall. L x g·mi1x2 Tmi 0 k (5-49) The application of either the explicit or the implicit formulation to each of the M 1 interior nodes gives M 1 equations. The remaining two equations are obtained by applying the same method to the two boundary nodes unless, of course, the boundary temperatures are specified as constants (invariant with time). For example, the formulation of the convection boundary condition at the left boundary (node 0) for the explicit case can be expressed as (Fig. 5–41) hA(T T0i) kA i1 i T1i T0i x T0 T0 x g·i0 A A C 2 2 x t (5-50) cen58933_ch05.qxd 9/4/2002 11:42 AM Page 295 295 CHAPTER 5 which simplifies to T0i1 1 2 2 g· i0 x 2 hx i hx T0 2T1i 2 T k k k (5-51) Note that in the case of no heat generation and 0.5, the explicit finite difference formulation for a general interior node reduces to Tmi1 i i (Tm1 Tm1 )/2, which has the interesting interpretation that the temperature of an interior node at the new time step is simply the average of the temperatures of its neighboring nodes at the previous time step. Once the formulation (explicit or implicit) is complete and the initial condition is specified, the solution of a transient problem is obtained by marching in time using a step size of t as follows: select a suitable time step t and determine the nodal temperatures from the initial condition. Taking the initial temperatures as the previous solution Tmi at t 0, obtain the new solution Tmi1 at all nodes at time t t using the transient finite difference relations. Now using the solution just obtained at t t as the previous solution Tmi , obtain the new solution Tmi1 at t 2t using the same relations. Repeat the process until the solution at the desired time is obtained. Stability Criterion for Explicit Method: Limitation on t The explicit method is easy to use, but it suffers from an undesirable feature that severely restricts its utility: the explicit method is not unconditionally stable, and the largest permissible value of the time step t is limited by the stability criterion. If the time step t is not sufficiently small, the solutions obtained by the explicit method may oscillate wildly and diverge from the actual solution. To avoid such divergent oscillations in nodal temperatures, the value of t must be maintained below a certain upper limit established by the stability criterion. It can be shown mathematically or by a physical argument based on the second law of thermodynamics that the stability criterion is satisfied if the coefficients of all Tmi in the Tmi1 expressions (called the primary coefficients) are greater than or equal to zero for all nodes m (Fig. 5–42). Of course, all the terms involving Tmi for a particular node must be grouped together before this criterion is applied. Different equations for different nodes may result in different restrictions on the size of the time step t, and the criterion that is most restrictive should be used in the solution of the problem. A practical approach is to identify the equation with the smallest primary coefficient since it is the most restrictive and to determine the allowable values of t by applying the stability criterion to that equation only. A t value obtained this way will also satisfy the stability criterion for all other equations in the system. For example, in the case of transient one-dimensional heat conduction in a plane wall with specified surface temperatures, the explicit finite difference equations for all the nodes (which are interior nodes) are obtained from Eq. 5–47. The coefficient of Tmi in the Tmi1 expression is 1 2, which is independent of the node number m, and thus the stability criterion for all nodes in this case is 1 2 0 or t 1 x2 2 nodes, one-dimensional heat interior transfer in rectangular coordinates (5-52) Explicit formulation: T0i1 a0T0i · · · T1i1 a1T1i · · · M Tmi1 amTmi · · · M TMi1 aMTMi · · · Stability criterion: am 0, m 0, 1, 2, . . . m, . . . M FIGURE 5–42 The stability criterion of the explicit method requires all primary coefficients to be positive or zero. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 296 296 HEAT TRANSFER When the material of the medium and thus its thermal diffusivity is known and the value of the mesh size x is specified, the largest allowable value of the time step t can be determined from this relation. For example, in the case of a brick wall ( 0.45 106 m2/s) with a mesh size of x 0.01 m, the upper limit of the time step is t (0.01 m)2 1 x2 111s 1.85 min 2 2(0.45 106 m2/s) The boundary nodes involving convection and/or radiation are more restrictive than the interior nodes and thus require smaller time steps. Therefore, the most restrictive boundary node should be used in the determination of the maximum allowable time step t when a transient problem is solved with the explicit method. To gain a better understanding of the stability criterion, consider the explicit finite difference formulation for an interior node of a plane wall (Eq. 5–47) for the case of no heat generation, i i Tmi1 (Tm1 Tm1 ) (1 2)Tmi 80°C 50°C 50°C 20°C m–1 m m–1 m+1 m m+1 Time step: i + 1 Time step: i FIGURE 5–43 The violation of the stability criterion in the explicit method may result in the violation of the second law of thermodynamics and thus divergence of solution. i i Assume that at some time step i the temperatures Tm1 and Tm1 are equal but i i i i less than Tm (say, Tm1 Tm1 50°C and Tm 80°C). At the next time step, we expect the temperature of node m to be between the two values (say, 70°C). However, if the value of exceeds 0.5 (say, 1), the temperature of node m at the next time step will be less than the temperature of the neighboring nodes (it will be 20°C), which is physically impossible and violates the second law of thermodynamics (Fig. 5–43). Requiring the new temperature of node m to remain above the temperature of the neighboring nodes is equivalent to requiring the value of to remain below 0.5. The implicit method is unconditionally stable, and thus we can use any time step we please with that method (of course, the smaller the time step, the better the accuracy of the solution). The disadvantage of the implicit method is that it results in a set of equations that must be solved simultaneously for each time step. Both methods are used in practice. EXAMPLE 5–5 Uranium plate 0°C k = 28 W/m·°C g· = 5 × 106 W/m3 h T α = 12.5 × 10 – 6 m2/s ∆x ∆x 0 0 1 L 2 Tinitial = 200°C FIGURE 5–44 Schematic for Example 5–5. x Transient Heat Conduction in a Large Uranium Plate Consider a large uranium plate of thickness L 4 cm, thermal conductivity k 28 W/m · °C, and thermal diffusivity 12.5 106 m2/s that is initially at a uniform temperature of 200°C. Heat is generated uniformly in the plate at a constant rate of g· 5 106 W/m3. At time t 0, one side of the plate is brought into contact with iced water and is maintained at 0°C at all times, while the other side is subjected to convection to an environment at T 30°C with a heat transfer coefficient of h 45 W/m2 · °C, as shown in Figure 5–44. Considering a total of three equally spaced nodes in the medium, two at the boundaries and one at the middle, estimate the exposed surface temperature of the plate 2.5 min after the start of cooling using (a) the explicit method and (b) the implicit method. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 297 297 CHAPTER 5 SOLUTION We have solved this problem in Example 5–1 for the steady case, and here we repeat it for the transient case to demonstrate the application of the transient finite difference methods. Again we assume one-dimensional heat transfer in rectangular coordinates and constant thermal conductivity. The number of nodes is specified to be M 3, and they are chosen to be at the two surfaces of the plate and at the middle, as shown in the figure. Then the nodal spacing x becomes 0.04 m L 0.02 m M1 31 x We number the nodes as 0, 1, and 2. The temperature at node 0 is given to be T0 0°C at all times, and the temperatures at nodes 1 and 2 are to be determined. This problem involves only two unknown nodal temperatures, and thus we need to have only two equations to determine them uniquely. These equations are obtained by applying the finite difference method to nodes 1 and 2. (a) Node 1 is an interior node, and the explicit finite difference formulation at that node is obtained directly from Eq. 5–47 by setting m 1: T1i1 (T0 T2i) (1 2) T1i g·1x2 k (1) Node 2 is a boundary node subjected to convection, and the finite difference formulation at that node is obtained by writing an energy balance on the volume element of thickness x/2 at that boundary by assuming heat transfer to be into the medium at all sides (Fig. 5–45): hA(T T2i) kA i1 i T1i T2i x T2 T2 x g·2 A A C 2 2 x x T1i – T2i kA ——— ∆x Dividing by kA/2x and using the definitions of thermal diffusivity k/ C and the dimensionless mesh Fourier number t/(x)2 gives g·2x2 T2i1 T2i 2hx (T T2i) 2(T1i T2i) k k which can be solved for T2i1 to give T2i1 1 2 2 g·2x2 hx i hx T2 2T1i 2 T k k k (2) Note that we did not use the superscript i for quantities that do not change with time. Next we need to determine the upper limit of the time step t from the stability criterion, which requires the coefficient of T1i in Equation 1 and the coefficient of T2i in the second equation to be greater than or equal to zero. The coefficient of T2i is smaller in this case, and thus the stability criterion for this problem can be expressed as 1 2 2 hx 0 k A → x2 1 → t 2(1 hx/k) 2(1 hx/k) g· 2 Volume element of node 2 hA(T – T2i ) T2i + 1 T2i 0 0 1 ∆ –x — 2 2 x FIGURE 5–45 Schematic for the explicit finite difference formulation of the convection condition at the right boundary of a plane wall. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 298 298 HEAT TRANSFER since t/(x)2. Substituting the given quantities, the maximum allowable value of the time step is determined to be t (0.02 m)2 15.5 s 2(12.5 106 m2/s)[1 (45 W/m2 · °C)(0.02 m)/28 W/m · °C] Therefore, any time step less than 15.5 s can be used to solve this problem. For convenience, let us choose the time step to be t 15 s. Then the mesh Fourier number becomes (12.5 106 m2/s)(15 s) αt 0.46875 (for t 15 s) 2 (x) (0.02 m)2 Substituting this value of and other given quantities, the explicit finite difference equations (1) and (2) developed here reduce to T1i1 0.0625T1i 0.46875T2i 33.482 T2i1 0.9375T1i 0.032366T2i 34.386 The initial temperature of the medium at t 0 and i 0 is given to be 200°C throughout, and thus T10 T20 200°C. Then the nodal temperatures at T11 and T21 at t t 15 s are determined from these equations to be T11 0.0625T10 0.46875T20 33.482 0.0625 200 0.46875 200 33.482 139.7°C T21 0.9375T10 0.032366T20 34.386 0.9375 200 0.032366 200 34.386 228.4°C Similarly, the nodal temperatures T12 and T22 at t 2t 2 15 30 s are determined to be TABLE 5–2 The variation of the nodal temperatures in Example 5–5 with time obtained by the explicit method Time Step, i 0 1 2 3 4 5 6 7 8 9 10 20 30 40 Time, s 0 15 30 45 60 75 90 105 120 135 150 300 450 600 Node Temperature, °C T1i 200.0 139.7 149.3 123.8 125.6 114.6 114.3 109.5 108.9 106.7 106.3 103.8 103.7 103.7 T2i 200.0 228.4 172.8 179.9 156.3 157.1 146.9 146.3 141.8 141.1 139.0 136.1 136.0 136.0 T12 0.0625T11 0.46875T21 33.482 0.0625 139.7 0.46875 228.4 33.482 149.3°C T22 0.9375T11 0.032366T21 34.386 0.9375 139.7 0.032366 228.4 34.386 172.8°C Continuing in the same manner, the temperatures at nodes 1 and 2 are determined for i 1, 2, 3, 4, 5, . . . , 50 and are given in Table 5–2. Therefore, the temperature at the exposed boundary surface 2.5 min after the start of cooling is TL2.5 min T210 139.0°C (b) Node 1 is an interior node, and the implicit finite difference formulation at that node is obtained directly from Eq. 5–49 by setting m 1: T0 (1 2) T1i1 T2i1 g·0 x2 T1i 0 k (3) Node 2 is a boundary node subjected to convection, and the implicit finite difference formulation at that node can be obtained from this formulation by expressing the left side of the equation at time step i 1 instead of i as cen58933_ch05.qxd 9/4/2002 11:42 AM Page 299 299 CHAPTER 5 g·2 x2 T2i1 T2i 2hx (T T2i1) 2(T1i1 T2i1) k k which can be rearranged as 2T1i1 1 2 2 g·2 x2 hx i1 hx T T2i 0 T2 2 k k k (4) Again we did not use the superscript i or i 1 for quantities that do not change with time. The implicit method imposes no limit on the time step, and thus we can choose any value we want. However, we will again choose t 15 s, and thus 0.46875, to make a comparison with part (a) possible. Substituting this value of and other given quantities, the two implicit finite difference equations developed here reduce to –1.9375T1i1 0.46875T2i1 T1i 33.482 0 0.9375T1i1 1.9676T2i1 T2i 34.386 0 Again T10 T20 200°C at t 0 and i 0 because of the initial condition, and for i 0, these two equations reduce to 1.9375T11 0.46875T21 200 33.482 0 0.9375T11 1.9676T21 200 34.386 0 The unknown nodal temperatures T11 and T21 at t t 15 s are determined by solving these two equations simultaneously to be T11 168.8°C and T21 199.6°C Similarly, for i 1, these equations reduce to TABLE 5–3 1.9375T12 0.46875T22 168.8 33.482 0 0.9375T12 1.9676T22 199.6 34.386 0 The unknown nodal temperatures T12 and T22 at t t 2 15 30 s are determined by solving these two equations simultaneously to be T12 150.5°C and T22 190.6°C Continuing in this manner, the temperatures at nodes 1 and 2 are determined for i 2, 3, 4, 5, . . . , 40 and are listed in Table 5–3, and the temperature at the exposed boundary surface (node 2) 2.5 min after the start of cooling is obtained to be TL2.5 min T210 143.9°C which is close to the result obtained by the explicit method. Note that either method could be used to obtain satisfactory results to transient problems, except, perhaps, for the first few time steps. The implicit method is preferred when it is desirable to use large time steps, and the explicit method is preferred when one wishes to avoid the simultaneous solution of a system of algebraic equations. The variation of the nodal temperatures in Example 5–5 with time obtained by the implicit method Time Step, i Time, s 0 1 2 3 4 5 6 7 8 9 10 20 30 40 0 15 30 45 60 75 90 105 120 135 150 300 450 600 Node Temperature, °C T1i T2i 200.0 168.8 150.5 138.6 130.3 124.1 119.5 115.9 113.2 111.0 109.4 104.2 103.8 103.8 200.0 199.6 190.6 180.4 171.2 163.6 157.6 152.8 149.0 146.1 143.9 136.7 136.1 136.1 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 300 300 HEAT TRANSFER EXAMPLE 5–6 South Warm air Sun’s rays Trombe wall Heat loss Heat gain Vent Cool air Glazing FIGURE 5–46 Schematic of a Trombe wall (Example 5–6). TABLE 5–4 The hourly variation of monthly average ambient temperature and solar heat flux incident on a vertical surface for January in Reno, Nevada Time of Day 7 10 1 4 7 10 1 4 Ambient Solar Temperature, Radiation, °F Btu/h · ft2 AM–10 AM AM–1 PM PM–4 PM PM–7 PM PM–10 PM PM–1 AM AM–4 AM AM–7 AM 33 43 45 37 32 27 26 25 114 242 178 0 0 0 0 0 Solar Energy Storage in Trombe Walls Dark painted thick masonry walls called Trombe walls are commonly used on south sides of passive solar homes to absorb solar energy, store it during the day, and release it to the house during the night (Fig. 5–46). The idea was proposed by E. L. Morse of Massachusetts in 1881 and is named after Professor Felix Trombe of France, who used it extensively in his designs in the 1970s. Usually a single or double layer of glazing is placed outside the wall and transmits most of the solar energy while blocking heat losses from the exposed surface of the wall to the outside. Also, air vents are commonly installed at the bottom and top of the Trombe walls so that the house air enters the parallel flow channel between the Trombe wall and the glazing, rises as it is heated, and enters the room through the top vent. Consider a house in Reno, Nevada, whose south wall consists of a 1-ft-thick Trombe wall whose thermal conductivity is k 0.40 Btu/h · ft · °F and whose thermal diffusivity is 4.78 106 ft2/s. The variation of the ambient temperature Tout and the solar heat flux q· solar incident on a south-facing vertical surface throughout the day for a typical day in January is given in Table 5–4 in 3-h intervals. The Trombe wall has single glazing with an absorptivity-transmissivity product of 0.77 (that is, 77 percent of the solar energy incident is absorbed by the exposed surface of the Trombe wall), and the average combined heat transfer coefficient for heat loss from the Trombe wall to the ambient is determined to be hout 0.7 Btu/h · ft2 · °F. The interior of the house is maintained at Tin 70°F at all times, and the heat transfer coefficient at the interior surface of the Trombe wall is hin 1.8 Btu/h · ft2 · °F. Also, the vents on the Trombe wall are kept closed, and thus the only heat transfer between the air in the house and the Trombe wall is through the interior surface of the wall. Assuming the temperature of the Trombe wall to vary linearly between 70°F at the interior surface and 30°F at the exterior surface at 7 AM and using the explicit finite difference method with a uniform nodal spacing of x 0.2 ft, determine the temperature distribution along the thickness of the Trombe wall after 12, 24, 36, and 48 h. Also, determine the net amount of heat transferred to the house from the Trombe wall during the first day and the second day. Assume the wall is 10 ft high and 25 ft long. SOLUTION The passive solar heating of a house through a Trombe wall is considered. The temperature distribution in the wall in 12-h intervals and the amount of heat transfer during the first and second days are to be determined. Assumptions 1 Heat transfer is one-dimensional since the exposed surface of the wall is large relative to its thickness. 2 Thermal conductivity is constant. 3 The heat transfer coefficients are constant. Properties The wall properties are given to be k 0.40 Btu/h · ft · °F, 4.78 106 ft2/s, and 0.77. Analysis The nodal spacing is given to be x 0.2 ft, and thus the total number of nodes along the Trombe wall is M 1 ft L 1 16 0.2 ft x We number the nodes as 0, 1, 2, 3, 4, and 5, with node 0 on the interior surface of the Trombe wall and node 5 on the exterior surface, as shown in Figure 5–47. Nodes 1 through 4 are interior nodes, and the explicit finite difference formulations of these nodes are obtained directly from Eq. 5–47 to be cen58933_ch05.qxd 9/4/2002 11:42 AM Page 301 301 CHAPTER 5 Node 1 (m 1): Node 2 (m 2): Node 3 (m 3): Node 4 (m 4): T1i1 (T0i T2i) (1 2)T1i T2i1 (T1i T3i) (1 2)T2i T3i1 (T2i T4i) (1 2)T3i T4i1 (T3i T5i) (1 2)T4i Trombe wall (2) k = 0.40 Btu/h·ft·°F α = 4.78 × 10 – 6 ft2/s (3) (4) The interior surface is subjected to convection, and thus the explicit formulation of node 0 can be obtained directly from Eq. 5–51 to be T0i1 0 (5) The exterior surface of the Trombe wall is subjected to convection as well as to heat flux. The explicit finite difference formulation at that boundary is obtained by writing an energy balance on the volume element represented by node 5, i1 i T4i T5i x T5 T5 i hout A(Tout T5i) Aq·isolar kA A C 2 x t (5-53) which simplifies to T5i1 1 2 2 i q·solar x hout x i hout x i (5-54) T5 2T4i 2 Tout 2 k k k where t/x2 is the dimensionless mesh Fourier number. Note that we kept the superscript i for quantities that vary with time. Substituting the quantities hout, x, k, and , which do not change with time, into this equation gives i 0.770q·isolar) T5i1 (1 2.70) T5i (2T4i 0.70Tout (6) where the unit of q· isolar is Btu/h · ft2. Next we need to determine the upper limit of the time step t from the stability criterion since we are using the explicit method. This requires the identification of the smallest primary coefficient in the system. We know that the boundary nodes are more restrictive than the interior nodes, and thus we examine the formulations of the boundary nodes 0 and 5 only. The smallest and thus the most restrictive primary coefficient in this case is the coefficient of T0i in the formulation of node 0 since 1 3.8 1 2.7, and thus the stability criterion for this problem can be expressed as 1 3.80 0 → αx 1 x2 3.80 Substituting the given quantities, the maximum allowable value of the time step is determined to be t (0.2 ft)2 x2 2202 s 3.80α 3.80 (4.78 106 ft2/s) hout, Tout ∆ x = 0.2 ft Substituting the quantities hin, x, k, and Tin, which do not change with time, into this equation gives T0i1 (1 3.80) T0i (2T1i 126.0) Initial temperature distribution at 7 AM (t = 0) 70°F hin, Tin hin x i hin x T0 2T1i 2 1 2 2 Tin k k q· solar (1) 0 1 2 30°F 3 4 5 L x FIGURE 5–47 The nodal network for the Trombe wall discussed in Example 5–6. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 302 302 HEAT TRANSFER Therefore, any time step less than 2202 s can be used to solve this problem. For convenience, let us choose the time step to be t 900 s 15 min. Then the mesh Fourier number becomes (4.78 106 ft2/s)(900 s) αt 0.10755 2 (x) (0.2 ft)2 (for t 15 min) Initially (at 7 AM or t 0), the temperature of the wall is said to vary linearly between 70°F at node 0 and 30°F at node 5. Noting that there are five nodal spacings of equal length, the temperature change between two neighboring nodes is (70 30)°F/5 8°F. Therefore, the initial nodal temperatures are T00 70°F, T30 46°F, T10 62°F, T40 38°F, T20 54°F, T50 30°F Then the nodal temperatures at t t 15 min (at 7:15 from these equations to be °F 170 1st day 2nd day 130 7 PM 110 1 AM 90 1 PM 70 50 Initial temperature 30 0 are determined T01 (1 3.80) T00 (2T10 126.0) (1 3.80 0.10755) 70 0.10755(2 62 126.0) 68.3° F 1 T1 (T00 T20) (1 2) T10 0.10755(70 54) (1 2 0.10755)62 62°F 1 T2 (T10 T30) (1 2) T20 0.10755(62 46) (1 2 0.10755)54 54°F 1 T3 (T20 T40) (1 2) T30 0.10755(54 38) (1 2 0.10755)46 46°F 1 T4 (T30 T50) (1 2) T40 0.10755(46 30) (1 2 0.10755)38 38°F 1 0 T5 (1 2.70) T50 (2T40 0.70Tout 0.770q·0solar) (1 2.70 0.10755)30 0.10755(2 38 0.70 33 0.770 114) 41.4°F Temperature 150 AM) 7 AM 0.2 0.4 0.6 0.8 Distance along the Trombe wall FIGURE 5–48 The variation of temperatures in the Trombe wall discussed in Example 5–6. 1 ft Note that the inner surface temperature of the Trombe wall dropped by 1.7°F and the outer surface temperature rose by 11.4°F during the first time step while the temperatures at the interior nodes remained the same. This is typical of transient problems in mediums that involve no heat generation. The nodal temperatures at the following time steps are determined similarly with the help of a computer. Note that the data for ambient temperature and the incident solar radiation change every 3 hours, which corresponds to 12 time steps, and this must be reflected in the computer program. For example, the value of q· isolar must be taken to be q· isolar 75 for i 1–12, q· isolar 242 for i 13–24, q· isolar 178 for i 25–36, and q· isolar 0 for i 37–96. The results after 6, 12, 18, 24, 30, 36, 42, and 48 h are given in Table 5–5 and are plotted in Figure 5–48 for the first day. Note that the interior temperature of the Trombe wall drops in early morning hours, but then rises as the solar energy absorbed by the exterior surface diffuses through the wall. The exterior surface temperature of the Trombe wall rises from 30 to 142°F in just 6 h because of the solar energy absorbed, but then drops to 53°F by next morning as a result of heat loss at night. Therefore, it may be worthwhile to cover the outer surface at night to minimize the heat losses. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 303 303 CHAPTER 5 TABLE 5–5 The temperatures at the nodes of a Trombe wall at various times Time 0 6 12 18 24 30 36 42 48 h h h h h h h h h (7 (1 (7 (1 (7 (1 (7 (1 (7 AM) PM) PM) AM) AM) PM) PM) AM) AM) Nodal Temperatures, °F Time Step, i T0 T1 T2 T3 T4 T5 0 24 48 72 96 120 144 168 192 70.0 65.3 71.6 73.3 71.2 70.3 75.4 75.8 73.0 62.0 61.7 74.2 75.9 71.9 71.1 81.1 80.7 75.1 54.0 61.5 80.4 77.4 70.9 74.3 89.4 83.5 72.2 46.0 69.7 88.4 76.3 67.7 84.2 98.2 83.0 66.0 38.0 94.1 91.7 71.2 61.7 108.3 101.0 77.4 66.0 30.0 142.0 82.4 61.2 53.0 153.2 89.7 66.2 56.3 The rate of heat transfer from the Trombe wall to the interior of the house during each time step is determined from Newton’s law using the average temperature at the inner surface of the wall (node 0) as · QiTrombe wall Q iTrombe wall t hin A(T0i Tin) t hin A[(T0i T0i1)/2 Tin]t Therefore, the amount of heat transfer during the first time step (i 1) or during the first 15-min period is Q1Trombe wall hin A[(T01 T00)/2 Tin] t (1.8 Btu/h · ft2 · °F)(10 25 ft2)(68.3 70)/2 70°F 95.6 Btu The negative sign indicates that heat is transferred to the Trombe wall from the air in the house, which represents a heat loss. Then the total heat transfer during a specified time period is determined by adding the heat transfer amounts for each time step as I QTrombe wall i1 · Q iTrombe wall I h in A[(T0i T0i1)/2 Tin] t (5-55) i1 where I is the total number of time intervals in the specified time period. In this case I 48 for 12 h, 96 for 24 h, and so on. Following the approach described here using a computer, the amount of heat transfer between the Trombe wall and the interior of the house is determined to be QTrombe wall 17, 048 Btu after 12 h QTrombe wall 2483 Btu after 24 h QTrombe wall 5610 Btu after 36 h QTrombe wall 34, 400 Btu after 48 h (17, 078 Btu during the first 12 h) (14, 565 Btu during the second 12 h) (8093 Btu during the third 12 h) (28, 790 Btu during the fourth 12 h) Therefore, the house loses 2483 Btu through the Trombe wall the first day as a result of the low start-up temperature but delivers a total of 36,883 Btu of heat to the house the second day. It can be shown that the Trombe wall will deliver even more heat to the house during the third day since it will start the day at a higher average temperature. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 304 304 HEAT TRANSFER Two-Dimensional Transient Heat Conduction m, n + 1 n+1 ∆y Volume element m – 1, n n g· m, n m, n m + 1, n ∆y m, n – 1 n–1 ∆x y m–1 ∆x m m+1 x FIGURE 5–49 The volume element of a general interior node (m, n) for twodimensional transient conduction in rectangular coordinates. Consider a rectangular region in which heat conduction is significant in the x- and y-directions, and consider a unit depth of z 1 in the z-direction. Heat may be generated in the medium at a rate of g·(x, y, t), which may vary with time and position, with the thermal conductivity k of the medium assumed to be constant. Now divide the x-y-plane of the region into a rectangular mesh of nodal points spaced x and y apart in the x- and y-directions, respectively, and consider a general interior node (m, n) whose coordinates are x mx and y ny, as shown in Figure 5–49. Noting that the volume element centered about the general interior node (m, n) involves heat conduction from four sides (right, left, top, and bottom) and the volume of the element is Velement x y 1 xy, the transient finite difference formulation for a general interior node can be expressed on the basis of Eq. 5–39 as Tm1, n Tm, n Tm, n1 Tm, n Tm1, n Tm, n kx ky x y x i1 Tm, n1 Tm, n Tm Tmi kx g·m, n xy xy C y t ky (5-56) Taking a square mesh (x y l ) and dividing each term by k gives after simplifying, Tm 1, n Tm 1, n Tm, n 1 Tm, n 1 4Tm, n g·m, nl 2 Tmi1 Tmi k (5-57) where again k/ C is the thermal diffusivity of the material and t/l 2 is the dimensionless mesh Fourier number. It can also be expressed in terms of the temperatures at the neighboring nodes in the following easy-to-remember form: Tleft Ttop Tright Tbottom 4Tnode i1 i g·nodel 2 Tnode Tnode k (5-58) Again the left side of this equation is simply the finite difference formulation of the problem for the steady case, as expected. Also, we are still not committed to explicit or implicit formulation since we did not indicate the time step on the left side of the equation. We now obtain the explicit finite difference formulation by expressing the left side at time step i as i i i i i Ttop Tright Tbottom 4Tnode Tleft i i1 i l 2 Tnode Tnode g·node k (5-59) Expressing the left side at time step i 1 instead of i would give the implicit formulation. This equation can be solved explicitly for the new temperature i1 Tnode to give i1 i i i i i Tnode (Tleft Ttop Tright Tbottom ) (1 4) Tnode i g·node l2 k (5-60) for all interior nodes (m, n) where m 1, 2, 3, . . . , M 1 and n 1, 2, 3, . . . , N 1 in the medium. In the case of no heat generation and 14, the cen58933_ch05.qxd 9/4/2002 11:42 AM Page 305 305 CHAPTER 5 explicit finite difference formulation for a general interior node reduces to i1 i i i i Tnode Ttop Tright Tbottom (Tleft )/4, which has the interpretation that the temperature of an interior node at the new time step is simply the average of the temperatures of its neighboring nodes at the previous time step (Fig. 5–50). The stability criterion that requires the coefficient of Tmi in the Tmi1 expression to be greater than or equal to zero for all nodes is equally valid for twoor three-dimensional cases and severely limits the size of the time step t that can be used with the explicit method. In the case of transient two-dimensional heat transfer in rectangular coordinates, the coefficient of Tmi in the Tmi1 expression is 1 4, and thus the stability criterion for all interior nodes in this case is 1 4 0, or t 1 4 l2 (interior nodes, two-dimensional heat transfer in rectangular coordinates) Time step i: 30°C Tmi 20°C 40°C Node m 10°C Time step i + 1: (5-61) Tmi+ 1 25°C Node m where x y l. When the material of the medium and thus its thermal diffusivity are known and the value of the mesh size l is specified, the largest allowable value of the time step t can be determined from the relation above. Again the boundary nodes involving convection and/or radiation are more restrictive than the interior nodes and thus require smaller time steps. Therefore, the most restrictive boundary node should be used in the determination of the maximum allowable time step t when a transient problem is solved with the explicit method. The application of Eq. 5–60 to each of the (M 1) (N 1) interior nodes gives (M 1) (N 1) equations. The remaining equations are obtained by applying the method to the boundary nodes unless, of course, the boundary temperatures are specified as being constant. The development of the transient finite difference formulation of boundary nodes in two- (or three-) dimensional problems is similar to the development in the one-dimensional case discussed earlier. Again the region is partitioned between the nodes by forming volume elements around the nodes, and an energy balance is written for each boundary node on the basis of Eq. 5–39. This is illustrated in Example 5–7. EXAMPLE 5–7 FIGURE 5–50 In the case of no heat generation and 14, the temperature of an interior node at the new time step is the average of the temperatures of its neighboring nodes at the previous time step. Transient Two-Dimensional Heat Conduction in L-Bars Consider two-dimensional transient heat transfer in an L-shaped solid body that is initially at a uniform temperature of 90°C and whose cross section is given in Figure 5–51. The thermal conductivity and diffusivity of the body are k 15 W/m · °C and 3.2 106 m2/s, respectively, and heat is generated in the body at a rate of g· 2 106 W/m3. The left surface of the body is insulated, and the bottom surface is maintained at a uniform temperature of 90°C at all times. At time t 0, the entire top surface is subjected to convection to ambient air at T 25°C with a convection coefficient of h 80 W/m2 · °C, and the right surface is subjected to heat flux at a uniform rate of q· R 5000 W/m2. The nodal network of the problem consists of 15 equally spaced nodes with x y 1.2 cm, as shown in the figure. Five of the nodes are at the bottom surface, and thus their temperatures are known. Using the explicit method, determine the temperature at the top corner (node 3) of the body after 1, 3, 5, 10, and 60 min. Convection h, T y 1 2 3 4 5 6 7 8 10 11 12 13 14 15 ∆ x = ∆y = l ∆y · 9 qR ∆y x ∆x ∆x 90°C ∆x ∆x ∆x FIGURE 5–51 Schematic and nodal network for Example 5–7. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 306 306 HEAT TRANSFER h, T 1 SOLUTION This is a transient two-dimensional heat transfer problem in rectangular coordinates, and it was solved in Example 5–3 for the steady case. Therefore, the solution of this transient problem should approach the solution for the steady case when the time is sufficiently large. The thermal conductivity and heat generation rate are given to be constants. We observe that all nodes are boundary nodes except node 5, which is an interior node. Therefore, we will have to rely on energy balances to obtain the finite difference equations. The region is partitioned among the nodes equitably as shown in the figure, and the explicit finite difference equations are determined on the basis of the energy balance for the transient case expressed as h, T 2 1 2 3 All sides 5 4 (a) Node 1 (b) Node 2 FIGURE 5–52 Schematics for energy balances on the volume elements of nodes 1 and 2. Tmi t i1 Tm · · Q i G ielement Velement C The quantities h, T, g· , and q· R do not change with time, and thus we do not need to use the superscript i for them. Also, the energy balance expressions are simplified using the definitions of thermal diffusivity k/ C and the dimensionless mesh Fourier number t/l2, where x y l. (a) Node 1. (Boundary node subjected to convection and insulation, Fig. 5–52a) h i i y T2i T1i x T4 T1 x (T T1i) k k 2 2 2 x y i1 i x y x y T1 T1 C g·1 2 2 2 2 t Dividing by k/4 and simplifying, g·1l 2 T1i1 T1i 2hl (T T1i) 2(T2i T1i) 2(T4i T1i) k k which can be solved for T1i1 to give T1i1 1 4 2 g·1l 2 hl hl T1i 2 T2i T4i T k k 2k (b) Node 2. (Boundary node subjected to convection, Fig. 5–52b) T5i T2i y T3i T2i kx 2 x y y y T1i T2i y T2i1 T2i k g·2 x x C 2 2 2 x t hx(T T2i) k Dividing by k/2, simplifying, and solving for T2i1 gives T2i1 1 4 2 g·2l 2 hl 2hl T2i T1i T3i 2T5i T k k k cen58933_ch05.qxd 9/4/2002 11:42 AM Page 307 307 CHAPTER 5 (c) Node 3. (Boundary node subjected to convection on two sides, Fig. 5–53a) h i i x T6 T3 x y (T T3i) k 2 2 2 y k 2 3 5 10 (a) Node 3 g·3l 2 hl hl T3i 2 T4i T6i 2 T k k 2k 4 6 Dividing by k/4, simplifying, and solving for T3i1 gives (5) h, T i1 i y T2i T3i x y x y T3 T3 g·3 2 2 2 2 2 x t T3i1 1 4 4 1 Mirror h, T (b) Node 4 FIGURE 5–53 Schematics for energy balances on the volume elements of nodes 3 and 4. (d ) Node 4. (On the insulated boundary, and can be treated as an interior node, Fig. 5–53b). Noting that T10 90°C, Eq. 5–60 gives T4i1 (1 4) T4i T1i 2T5i 90 g·4l 2 k (e) Node 5. (Interior node, Fig. 5–54a). Noting that T11 90°C, Eq. 5–60 gives T5i1 (1 4) T5i T2i T4i T6i 90 g·5l 2 k (f ) Node 6. (Boundary node subjected to convection on two sides, Fig. 5–54b) h kx ky x y i i 3xy 3xy T6i1 T6i x T3 T6 g·6 C 2 4 4 y t y x2 2 (T T6i) k y 2 T7i T6i T12i T6i T5i x T6i 3 2 4 5 6 12 11 FIGURE 5–54 Schematics for energy balances on the volume elements of nodes 5 and 6. g·6l 2 hl 2T3i 4T5i 2T7i 4 90 4 T 3 k 3 k (g) Node 7. (Boundary node subjected to convection, Fig. 5–55) hx(T T7i) k Dividing by k/2, simplifying, and solving for T7i1 gives g·7l 2 2hl hl T7i T6i T8i 2 90 T k k k h, T h, T T13i T7i y T8i T7i kx 2 x y i i y y T6 T7 y T7i1 T7i k g·7x x C 2 2 2 x t T7i1 1 4 2 (b) Node 6 (a) Node 5 hl Ti 3k 3 7 6 5 Dividing by 3k/4, simplifying, and solving for T6i1 gives T6i1 1 4 4 h, T 6 7 8 9 q· R 13 15 FIGURE 5–55 Schematics for energy balances on the volume elements of nodes 7 and 9. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 308 308 HEAT TRANSFER (h) Node 8. This node is identical to node 7, and the finite difference formulation of this node can be obtained from that of node 7 by shifting the node numbers by 1 (i.e., replacing subscript m by subscript m 1). It gives T8i1 1 4 2 g·8l 2 2hl hl T8i T7i T9i 2 90 T k k k (i ) Node 9. (Boundary node subjected to convection on two sides, Fig. 5–55) y x T15 T9 x (T T9i) q·R k 2 2 2 y i1 i ky T8i T9i x y x y T9 T9 g·9 C 2 2 2 2 2 x t i h i Dividing by k/4, simplifying, and solving for T9i1 gives T9i1 1 4 2 g·9l 2 q·R l hl hl T T9i 2 T8i 90 k k 2k k This completes the finite difference formulation of the problem. Next we need to determine the upper limit of the time step t from the stability criterion, which requires the coefficient of Tmi in the Tmi1 expression (the primary coefficient) to be greater than or equal to zero for all nodes. The smallest primary coefficient in the nine equations here is the coefficient of T3i in the expression, and thus the stability criterion for this problem can be expressed as 1 4 4 hl 0 k → l2 1 → t 4(1 hl/k) 4(1 hl/k) since t /l 2. Substituting the given quantities, the maximum allowable value of the time step is determined to be t 4(3.2 10 6 (0.012 m)2 10.6 s m /s)[1 (80 W/m2 · °C)(0.012 m)/(15 W/m · °C)] 2 Therefore, any time step less than 10.6 s can be used to solve this problem. For convenience, let us choose the time step to be t 10 s. Then the mesh Fourier number becomes t (3.2 106 m2/s)(10 s) 0.222 l2 (0.012 m)2 (for t 10 s) Substituting this value of and other given quantities, the developed transient finite difference equations simplify to T1i1 0.0836T1i 0.444(T2i T4i 11.2) T2i1 0.0836T2i 0.222(T1i T3i 2T5i 22.4) T3i1 0.0552T3i 0.444(T2i T6i 12.8) T4i1 0.112T4i 0.222(T1i 2T5i 109.2) T5i1 0.112T1i 0.222(T2i T4i T6i 109.2) cen58933_ch05.qxd 9/4/2002 11:42 AM Page 309 309 CHAPTER 5 T6i1 0.0931T6i 0.074(2T3i 4T5i 2T7i 424) T7i1 0.0836T7i 0.222(T6i T8i 202.4) T8i1 0.0836T8i 0.222(T7i T9i 202.4) T9i1 0.0836T9i 0.444(T8i 105.2) Using the specified initial condition as the solution at time t 0 (for i 0), sweeping through these nine equations will give the solution at intervals of 10 s. The solution at the upper corner node (node 3) is determined to be 100.2, 105.9, 106.5, 106.6, and 106.6°C at 1, 3, 5, 10, and 60 min, respectively. Note that the last three solutions are practically identical to the solution for the steady case obtained in Example 5–3. This indicates that steady conditions are reached in the medium after about 5 min. TOPIC OF SPECIAL INTEREST Controlling the Numerical Error A comparison of the numerical results with the exact results for temperature distribution in a cylinder would show that the results obtained by a numerical method are approximate, and they may or may not be sufficiently close to the exact (true) solution values. The difference between a numerical solution and the exact solution is the error involved in the numerical solution, and it is primarily due to two sources: • The discretization error (also called the truncation or formulation error), which is caused by the approximations used in the formulation of the numerical method. • The round-off error, which is caused by the computer’s use of a limited number of significant digits and continuously rounding (or chopping) off the digits it cannot retain. Below we discuss both types of errors. T(xm, t) Discretization Error The discretization error involved in numerical methods is due to replacing the derivatives by differences in each step, or the actual temperature distribution between two adjacent nodes by a straight line segment. Consider the variation of the solution of a transient heat transfer problem with time at a specified nodal point. Both the numerical and actual (exact) solutions coincide at the beginning of the first time step, as expected, but the numerical solution deviates from the exact solution as the time t increases. The difference between the two solutions at t t is due to the approximation at the first time step only and is called the local discretization error. One would expect the situation to get worse with each step since the second step uses the erroneous result of the first step as its starting point and adds a second local discretization error on top of it, as shown in Figure 5–56. The accumulation of the local discretization errors continues with the increasing number of time steps, and the total discretization error at any Local error Actual solution T(x0, t) Global error T3 T2 T0 T1 Step 1 t0 Numerical solution Step 2 t1 Step 3 t2 t3 Time FIGURE 5–56 The local and global discretization errors of the finite difference method at the third time step at a specified nodal point. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 310 310 HEAT TRANSFER step is called the global or accumulated discretization error. Note that the local and global discretization errors are identical for the first time step. The global discretization error usually increases with the increasing number of steps, but the opposite may occur when the solution function changes direction frequently, giving rise to local discretization errors of opposite signs, which tend to cancel each other. To have an idea about the magnitude of the local discretization error, consider the Taylor series expansion of the temperature at a specified nodal point m about time ti, T(xm, ti t) T(xm, ti) t T(xm, ti) 1 2 2T(xm, ti) t ··· t 2 t2 (5-62) The finite difference formulation of the time derivative at the same nodal point is expressed as T(xm, ti) T(xm, ti t) T(xm, ti) Tmi1 Tmi t t t (5-63) or T(xm, ti t) T(xm, ti) t T(xm, ti) t (5-64) which resembles the Taylor series expansion terminated after the first two terms. Therefore, the third and later terms in the Taylor series expansion represent the error involved in the finite difference approximation. For a sufficiently small time step, these terms decay rapidly as the order of derivative increases, and their contributions become smaller and smaller. The first term neglected in the Taylor series expansion is proportional to t 2, and thus the local discretization error of this approximation, which is the error involved in each step, is also proportional to t 2. The local discretization error is the formulation error associated with a single step and gives an idea about the accuracy of the method used. However, the solution results obtained at every step except the first one involve the accumulated error up to that point, and the local error alone does not have much significance. What we really need to know is the global discretization error. At the worst case, the accumulated discretization error after I time steps during a time period t0 is i(t)2 (t0/t)(t)2 t0t, which is proportional to t. Thus, we conclude that the local discretization error is proportional to the square of the step size t 2 while the global discretization error is proportional to the step size t itself. Therefore, the smaller the mesh size (or the size of the time step in transient problems), the smaller the error, and thus the more accurate is the approximation. For example, halving the step size will reduce the global discretization error by half. It should be clear from the discussions above that the discretization error can be minimized by decreasing the step size in space or time as much as possible. The discretization error approaches zero as the difference quantities such as x and t approach the differential quantities such as dx and dt. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 311 311 CHAPTER 5 Round-off Error If we had a computer that could retain an infinite number of digits for all numbers, the difference between the exact solution and the approximate (numerical) solution at any point would entirely be due to discretization error. But we know that every computer (or calculator) represents numbers using a finite number of significant digits. The default value of the number of significant digits for many computers is 7, which is referred to as single precision. But the user may perform the calculations using 15 significant digits for the numbers, if he or she wishes, which is referred to as double precision. Of course, performing calculations in double precision will require more computer memory and a longer execution time. In single precision mode with seven significant digits, a computer will register the number 44444.666666 as 44444.67 or 44444.66, depending on the method of rounding the computer uses. In the first case, the excess digits are said to be rounded to the closest integer, whereas in the second case they are said to be chopped off. Therefore, the numbers a 44444.12345 and b 44444.12032 are equivalent for a computer that performs calculations using seven significant digits. Such a computer would give a b 0 instead of the true value 0.00313. The error due to retaining a limited number of digits during calculations is called the round-off error. This error is random in nature and there is no easy and systematic way of predicting it. It depends on the number of calculations, the method of rounding off, the type of computer, and even the sequence of calculations. In algebra you learned that a b c a c b, which seems quite reasonable. But this is not necessarily true for calculations performed with a computer, as demonstrated in Figure 5–57. Note that changing the sequence of calculations results in an error of 30.8 percent in just two operations. Considering that any significant problem involves thousands or even millions of such operations performed in sequence, we realize that the accumulated round-off error has the potential to cause serious error without giving any warning signs. Experienced programmers are very much aware of this danger, and they structure their programs to prevent any buildup of the round-off error. For example, it is much safer to multiply a number by 10 than to add it 10 times. Also, it is much safer to start any addition process with the smallest numbers and continue with larger numbers. This rule is particularly important when evaluating series with a large number of terms with alternating signs. The round-off error is proportional to the number of computations performed during the solution. In the finite difference method, the number of calculations increases as the mesh size or the time step size decreases. Halving the mesh or time step size, for example, will double the number of calculations and thus the accumulated round-off error. Controlling the Error in Numerical Methods The total error in any result obtained by a numerical method is the sum of the discretization error, which decreases with decreasing step size, and the round-off error, which increases with decreasing step size, as shown in Figure 5–58. Therefore, decreasing the step size too much in order to get more Given: a 7777777 b 7777776 c 0.4444432 Find: Dabc Eacb Solution: D 7777777 7777776 0.4444432 1 0.4444432 1.444443 (Correct result) E 7777777 0.4444432 7777776 7777777 7777776 1.000000 (In error by 30.8%) FIGURE 5–57 A simple arithmetic operation performed with a computer in single precision using seven significant digits, which results in 30.8 percent error when the order of operation is reversed. Error Total error Discretization error Round-off error Optimum step size Step size FIGURE 5–58 As the mesh or time step size decreases, the discretization error decreases but the round-off error increases. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 312 312 HEAT TRANSFER accurate results may actually backfire and give less accurate results because of a faster increase in the round-off error. We should be careful not to let round-off error get out of control by avoiding a large number of computations with very small numbers. In practice, we will not know the exact solution of the problem, and thus we will not be able to determine the magnitude of the error involved in the numerical method. Knowing that the global discretization error is proportional to the step size is not much help either since there is no easy way of determining the value of the proportionality constant. Besides, the global discretization error alone is meaningless without a true estimate of the round-off error. Therefore, we recommend the following practical procedures to assess the accuracy of the results obtained by a numerical method. • Start the calculations with a reasonable mesh size x (and time step size t for transient problems) based on experience. Then repeat the calculations using a mesh size of x/2. If the results obtained by halving the mesh size do not differ significantly from the results obtained with the full mesh size, we conclude that the discretization error is at an acceptable level. But if the difference is larger than we can accept, then we have to repeat the calculations using a mesh size x/4 or even a smaller one at regions of high temperature gradients. We continue in this manner until halving the mesh size does not cause any significant change in the results, which indicates that the discretization error is reduced to an acceptable level. • Repeat the calculations using double precision holding the mesh size (and the size of the time step in transient problems) constant. If the changes are not significant, we conclude that the round-off error is not a problem. But if the changes are too large to accept, then we may try reducing the total number of calculations by increasing the mesh size or changing the order of computations. But if the increased mesh size gives unacceptable discretization errors, then we may have to find a reasonable compromise. It should always be kept in mind that the results obtained by any numerical method may not reflect any trouble spots in certain problems that require special consideration such as hot spots or areas of high temperature gradients. The results that seem quite reasonable overall may be in considerable error at certain locations. This is another reason for always repeating the calculations at least twice with different mesh sizes before accepting them as the solution of the problem. Most commercial software packages have built-in routines that vary the mesh size as necessary to obtain highly accurate solutions. But it is a good engineering practice to be aware of any potential pitfalls of numerical methods and to examine the results obtained with a critical eye. SUMMARY Analytical solution methods are limited to highly simplified problems in simple geometries, and it is often necessary to use a numerical method to solve real world problems with complicated geometries or nonuniform thermal conditions. The cen58933_ch05.qxd 9/4/2002 11:42 AM Page 313 313 CHAPTER 5 numerical finite difference method is based on replacing derivatives by differences, and the finite difference formulation of a heat transfer problem is obtained by selecting a sufficient number of points in the region, called the nodal points or nodes, and writing energy balances on the volume elements centered about the nodes. For steady heat transfer, the energy balance on a volume element can be expressed in general as · Q g·Velement 0 also be used to solve a system of equations simultaneously at the press of a button. The finite difference formulation of transient heat conduction problems is based on an energy balance that also accounts for the variation of the energy content of the volume element during a time interval t. The heat transfer and heat generation terms are expressed at the previous time step i in the explicit method, and at the new time step i 1 in the implicit method. For a general node m, the finite difference formulations are expressed as All sides whether the problem is one-, two-, or three-dimensional. For convenience in formulation, we always assume all heat transfer to be into the volume element from all surfaces toward the node under consideration, except for specified heat flux whose direction is already specified. The finite difference formulations for a general interior node under steady conditions are expressed for some geometries as follows: One-dimensional Tm1 2Tm Tm1 g·m steady conduction 0 k (x)2 in a plane wall: Twodimensional g·nodel 2 steady 0 Tleft Ttop Tright Tbottom 4Tnode conduction k in rectangular coordinates: where x is the nodal spacing for the plane wall and x y l is the nodal spacing for the two-dimensional case. Insulated boundaries can be viewed as mirrors in formulation, and thus the nodes on insulated boundaries can be treated as interior nodes by using mirror images. The finite difference formulation at node 0 at the left boundary of a plane wall for steady one-dimensional heat conduction can be expressed as T1 T 0 · Q left surface kA g·0(Ax/2) 0 x where Ax/2 is the volume of the volume, g·0 is the rate of heat generation per unit volume at x 0, and A is the heat transfer area. The form of the first term depends on the boundary condition at x 0 (convection, radiation, specified heat flux, etc.). The finite difference formulation of heat conduction problems usually results in a system of N algebraic equations in N unknown nodal temperatures that need to be solved simultaneously. There are numerous systematic approaches available in the literature. Several widely available equation solvers can Explicit method: Implicit method: Tmi t i1 Tm · · Q i G ielement Velement C Tm · i1 · Velement C Q i1 G element All sides Tmi t i1 All sides where Tmi and Tmi1 are the temperatures of node m at times ti it and ti1 (i 1)t, respectively, and Tmi1 Tmi represents the temperature change of the node during the time interval t between the time steps i and i 1. The explicit and implicit formulations given here are quite general and can be used in any coordinate system regardless of heat transfer being one-, two-, or three-dimensional. The explicit formulation of a general interior node for oneand two-dimensional heat transfer in rectangular coordinates can be expressed as Oneg·mi x2 i i Tmi1 (Tm1 dimen Tm1 ) (1 2) Tmi k sional case: Twoi i i1 i i Tnode (Tleft Ttop Tright Tbottom ) dimeni 2 g·nodel sional i (1 4) Tnode case: k where t x2 is the dimensionless mesh Fourier number and k/ C is the thermal diffusivity of the medium. The implicit method is inherently stable, and any value of t can be used with that method as the time step. The largest value of the time step t in the explicit method is limited by the stability criterion, expressed as: the coefficients of all Tmi in the Tmi1 expressions (called the primary coefficients) must be greater than or equal to zero for all nodes m. The maximum value of t is determined by applying the stability criterion to the equation with the smallest primary coefficient since it is the cen58933_ch05.qxd 9/4/2002 11:42 AM Page 314 314 HEAT TRANSFER most restrictive. For problems with specified temperatures or heat fluxes at all the boundaries, the stability criterion can be expressed as 12 for one-dimensional problems and 14 for the two-dimensional problems in rectangular coordinates. REFERENCES AND SUGGESTED READING 1. D. A. Anderson, J. C. Tannehill, and R. H. Pletcher. Computational Fluid Mechanics and Heat Transfer. New York: Hemisphere, 1984. W. J. Minkowycz, E. M. Sparrow, G. E. Schneider, and R. H. Pletcher. Handbook of Numerical Heat Transfer. New York: John Wiley & Sons, 1988. C. A. Brebbia. The Boundary Element Method for Engineers. New York: Halsted Press, 1978. G. E. Myers. Analytical Methods in Conduction Heat Transfer. New York: McGraw-Hill, 1971. G. E. Forsythe and W. R. Wasow. Finite Difference Methods for Partial Differential Equations. New York: John Wiley & Sons, 1960. D. H. Norrie and G. DeVries. An Introduction to Finite Element Analysis. New York: Academic Press, 1978. B. Gebhart. Heat Conduction and Mass Diffusion. New York: McGraw-Hill, 1993. 5. K. H. Huebner and E. A. Thornton. The Finite Element Method for Engineers. 2nd ed. New York: John Wiley & Sons, 1982. M. N. Özi¸sik. Finite Difference Methods in Heat Transfer. Boca Raton, FL: CRC Press, 1994. 11. S. V. Patankhar. Numerical Heat Transfer and Fluid Flow. New York: Hemisphere, 1980. 12. T. M. Shih. Numerical Heat Transfer. New York: Hemisphere, 1984. Y. Jaluria and K. E. Torrance. Computational Heat Transfer. New York: Hemisphere, 1986. PROBLEMS Why Numerical Methods? 5–1C What are the limitations of the analytical solution methods? 5–2C How do numerical solution methods differ from analytical ones? What are the advantages and disadvantages of numerical and analytical methods? 5–3C What is the basis of the energy balance method? How does it differ from the formal finite difference method? For a specified nodal network, will these two methods result in the same or a different set of equations? 5–4C Consider a heat conduction problem that can be solved both analytically, by solving the governing differential equation and applying the boundary conditions, and numerically, by a software package available on your computer. Which Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. approach would you use to solve this problem? Explain your reasoning. 5–5C Two engineers are to solve an actual heat transfer problem in a manufacturing facility. Engineer A makes the necessary simplifying assumptions and solves the problem analytically, while engineer B solves it numerically using a powerful software package. Engineer A claims he solved the problem exactly and thus his results are better, while engineer B claims that he used a more realistic model and thus his results are better. To resolve the dispute, you are asked to solve the problem experimentally in a lab. Which engineer do you think the experiments will prove right? Explain. Finite Difference Formulation of Differential Equations 5–6C Define these terms used in the finite difference formulation: node, nodal network, volume element, nodal spacing, and difference equation. 5–7 Consider three consecutive nodes n 1, n, and n 1 in a plane wall. Using the finite difference form of the first derivative at the midpoints, show that the finite difference form of the second derivative can be expressed as Tn1 2Tn Tn1 0 x2 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 315 315 CHAPTER 5 nodes 0, 1, 2, 3, 4, and 5 with a uniform nodal spacing of x. Using the finite difference form of the first derivative (not the energy balance approach), obtain the finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0) and radiation at the right boundary (node 5) with an emissivity of and surrounding temperature of Tsurr. T(x) Tn + 1 Tn One-Dimensional Steady Heat Conduction Tn – 1 5–11C Explain how the finite difference form of a heat conduction problem is obtained by the energy balance method. 5–12C In the energy balance formulation of the finite difference method, it is recommended that all heat transfer at the boundaries of the volume element be assumed to be into the volume element even for steady heat conduction. Is this a valid recommendation even though it seems to violate the conservation of energy principle? ∆x ∆x n–1 n n+1 x FIGURE P5–7 5–8 The finite difference formulation of steady twodimensional heat conduction in a medium with heat generation and constant thermal conductivity is given by Tm1, n 2Tm, n Tm1, n x 2 Tm, n1 2Tm, n Tm, n1 y2 g·m, n 0 k in rectangular coordinates. Modify this relation for the threedimensional case. 5–9 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, and 4 with a uniform nodal spacing of x. Using the finite difference form of the first derivative (not the energy balance approach), obtain the finite difference formulation of the boundary nodes for the case of uniform heat flux q·0 at the left boundary (node 0) and convection at the right boundary (node 4) with a convection coefficient of h and an ambient temperature of T. 5–10 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of 5–13C How is an insulated boundary handled in the finite difference formulation of a problem? How does a symmetry line differ from an insulated boundary in the finite difference formulation? 5–14C How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain. 5–15C Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as Tm1 2Tm Tm1 g·m 0 k x2 (a) (b) (c) (d) (e) Is heat transfer in this medium steady or transient? Is heat transfer one-, two-, or three-dimensional? Is there heat generation in the medium? Is the nodal spacing constant or variable? Is the thermal conductivity of the medium constant or variable? 5–16 Consider steady heat conduction in a plane wall whose left surface (node 0) is maintained at 30°C while the right surface (node 8) is subjected to a heat flux of 800 W/m2. Express the finite difference formulation of the boundary nodes 0 and 8 30°C Insulation Tsurr ∆x 0 1 2 FIGURE P5–10 W 800 —–2 m No heat generation · g(x) 3 4 Radiation ∆x ε 0 1 2 3 5 FIGURE P5–16 4 5 6 7 8 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 316 316 HEAT TRANSFER for the case of no heat generation. Also obtain the finite difference formulation for the rate of heat transfer at the left boundary. 5–17 Consider steady heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, and 4 with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of the boundary nodes for the case of uniform heat flux q·0 at the left boundary (node 0) and convection at the right boundary (node 4) with a convection coefficient of h and an ambient temperature of T. 5–18 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, 4, and 5 with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0) and radiation at the right boundary (node 5) with an emissivity of and surrounding temperature of Tsurr. 5–19 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, 4, and 5 with a uniform nodal spacing of x. The temperature at the right boundary (node 5) is specified. Using the energy balance approach, obtain the finite difference formulation of the boundary node 0 on the left boundary for the case of combined convection, radiation, and heat flux at the left boundary with an emissivity of , convection coefficient of h, ambient temperature of T, surrounding temperature of Tsurr, and uniform heat flux of q·0. Also, obtain the finite difference formulation for the rate of heat transfer at the right boundary. q· 0 Tsurr ε Convection T5 · g(x) Radiation ∆x 0 1 2 3 4 5 boundary (node 0) and radiation at the right boundary (node 2) with an emissivity of and surrounding temperature of Tsurr. 5–21 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and variable thermal conductivity. The nodal network of the medium consists of nodes 0, 1, and 2 with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of specified heat flux q·0 to the wall and convection at the left boundary (node 0) with a convection coefficient of h and ambient temperature of T, and radiation at the right boundary (node 2) with an emissivity of and surrounding surface temperature of Tsurr. Tsurr q· 0 · g(x) k(T) Convection h T Radiation ε ∆x 0 1 2 FIGURE P5–21 5–22 Consider steady one-dimensional heat conduction in a pin fin of constant diameter D with constant thermal conductivity. The fin is losing heat by convection to the ambient air at T with a heat transfer coefficient of h. The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of this problem to determine T1 and T2 for the case of specified temperature at the fin base and negligible heat transfer at the fin tip. All temperatures are in °C. 5–23 Consider steady one-dimensional heat conduction in a pin fin of constant diameter D with constant thermal conductivity. The fin is losing heat by convection to the ambient air at T with a convection coefficient of h, and by radiation to the surrounding surfaces at an average temperature of Tsurr. h, T Tsurr FIGURE P5–19 5–20 Consider steady one-dimensional heat conduction in a composite plane wall consisting of two layers A and B in perfect contact at the interface. The wall involves no heat generation. The nodal network of the medium consists of nodes 0, 1 (at the interface), and 2 with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of insulation at the left T0 Radiation ε 0 ∆x 1 h, T Convection FIGURE P5–23 D 2 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 317 317 CHAPTER 5 The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of this problem to determine T1 and T2 for the case of specified temperature at the fin base and negligible heat transfer at the fin tip. All temperatures are in °C. 5–24 Consider a large uranium plate of thickness 5 cm and thermal conductivity k 28 W/m · °C in which heat is generated uniformly at a constant rate of g· 6 105 W/m3. One side of the plate is insulated while the other side is subjected to convection to an environment at 30°C with a heat transfer coefficient of h 60 W/m2 · °C. Considering six equally spaced nodes with a nodal spacing of 1 cm, (a) obtain the finite difference formulation of this problem and (b) determine the nodal temperatures under steady conditions by solving those equations. 5–25 Consider an aluminum alloy fin (k 180 W/m · °C) of triangular cross section whose length is L 5 cm, base thickness is b 1 cm, and width w in the direction normal to the plane of paper is very large. The base of the fin is maintained at a temperature of T0 180°C. The fin is losing heat by convection to the ambient air at T 25°C with a heat transfer coefficient of h 25 W/m2 · °C and by radiation to the surrounding surfaces at an average temperature of Tsurr 290 K. Using the finite difference method with six equally spaced nodes along the fin in the x-direction, determine (a) the temperatures at the nodes and (b) the rate of heat transfer from the fin for w 1 m. Take the emissivity of the fin surface to be 0.9 and assume steady one-dimensional heat transfer in the fin. from 100°C to 200°C. Plot the fin tip temperature and the rate of heat transfer as a function of the fin base temperature, and discuss the results. 5–27 Consider a large plane wall of thickness L 0.4 m, thermal conductivity k 2.3 W/m · °C, and surface area A 20 m2. The left side of the wall is maintained at a constant temperature of 80°C, while the right side loses heat by convection to the surrounding air at T 15°C with a heat transfer coefficient of h 24 W/m2 · °C. Assuming steady onedimensional heat transfer and taking the nodal spacing to be 10 cm, (a) obtain the finite difference formulation for all nodes, (b) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall. 5–28 Consider the base plate of a 800-W household iron having a thickness of L 0.6 cm, base area of A 160 cm2, and thermal conductivity of k 20 W/m · °C. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 85°C. Disregarding any heat loss through the upper part of the iron and taking the nodal spacing to be 0.2 cm, (a) obtain the finite difference formulation for the nodes and (b) determine the inner surface temperature of the plate by Answer: (b) 100°C solving those equations. Insulation Resistance heater, 800 W ∆ x = 0.2 cm Triangular fin w 0 h, T 1 2 3 x 160 cm2 T0 FIGURE P5–28 b 0 b/2 tan θ = —– L θ 1 2 3 ∆x 4 5 x L FIGURE P5–25 5–26 85°C Base plate Reconsider Problem 5–25. Using EES (or other) software, investigate the effect of the fin base temperature on the fin tip temperature and the rate of heat transfer from the fin. Let the temperature at the fin base vary 5–29 Consider a large plane wall of thickness L 0.3 m, thermal conductivity k 2.5 W/m · °C, and surface area A 12 m2. The left side of the wall is subjected to a heat flux of q·0 700 W/m2 while the temperature at that surface is measured to be T0 60°C. Assuming steady one-dimensional heat transfer and taking the nodal spacing to be 6 cm, (a) obtain the finite difference formulation for the six nodes and (b) determine the temperature of the other surface of the wall by solving those equations. 5–30E A large steel plate having a thickness of L 5 in., thermal conductivity of k 7.2 Btu/h · ft · °F, and an emissivity of 0.6 is lying on the ground. The exposed surface of cen58933_ch05.qxd 9/4/2002 11:42 AM Page 318 318 HEAT TRANSFER Tsurr = 295 K Tsky Radiation ε 6 Convection h, T 0 1 2 Plate 3 4 5 T = 25°C h = 13 W/m2·°C 1 in. 5 4 18 cm 3 2 1 3 cm 0 0.6 ft 6 Soil 95° 7 8 FIGURE P5–32 9 10 5–34 x FIGURE P5–30E the plate exchanges heat by convection with the ambient air at T 80°F with an average heat transfer coefficient of h 3.5 Btu/h · ft2 · °F as well as by radiation with the open sky at an equivalent sky temperature of Tsky 510 R. The ground temperature below a certain depth (say, 3 ft) is not affected by the weather conditions outside and remains fairly constant at 50°F at that location. The thermal conductivity of the soil can be taken to be ksoil 0.49 Btu/h · ft · °F, and the steel plate can be assumed to be in perfect contact with the ground. Assuming steady one-dimensional heat transfer and taking the nodal spacings to be 1 in. in the plate and 0.6 ft in the ground, (a) obtain the finite difference formulation for all 11 nodes shown in Figure P5–30E and (b) determine the top and bottom surface temperatures of the plate by solving those equations. 5–31E Repeat Problem 5–30E by disregarding radiation heat transfer from the upper surface. Answers: (b) 78.7°F, 78.4°F 5–32 Consider a stainless steel spoon (k 15.1 W/m · C, 0.6) that is partially immersed in boiling water at 95°C in a kitchen at 25°C. The handle of the spoon has a cross section of about 0.2 cm 1 cm and extends 18 cm in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of h 13 W/m2 · °C as well as by radiation to the surrounding surfaces at an average temperature of Tsurr 295 K. Assuming steady one-dimensional heat transfer along the spoon and taking the nodal spacing to be 3 cm, (a) obtain the finite difference formulation for all nodes, (b) determine the temperature of the tip of the spoon by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the spoon. 5–33 Repeat Problem 5–32 using a nodal spacing of 1.5 cm. Reconsider Problem 5–33. Using EES (or other) software, investigate the effects of the thermal conductivity and the emissivity of the spoon material on the temperature at the spoon tip and the rate of heat transfer from the exposed surfaces of the spoon. Let the thermal conductivity vary from 10 W/m · °C to 400 W/m · °C, and the emissivity from 0.1 to 1.0. Plot the spoon tip temperature and the heat transfer rate as functions of thermal conductivity and emissivity, and discuss the results. 5–35 One side of a 2-m-high and 3-m-wide vertical plate at 130°C is to be cooled by attaching aluminum fins (k 237 W/m · °C) of rectangular profile in an environment at 35°C. The fins are 2 cm long, 0.3 cm thick, and 0.4 cm apart. The heat transfer coefficient between the fins and the surrounding air for combined convection and radiation is estimated to be 30 W/m2 · °C. Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be 0.5 cm, determine (a) the finite difference formulation of this problem, (b) the nodal temperatures along the fin by solving these equations, (c) the rate of heat transfer from a single fin, 130°C T = 35°C 0.4 cm 0.5 cm 0.3 cm 0 1 2 3 x 4 3m 2 cm FIGURE P5–35 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 319 319 CHAPTER 5 and (d) the rate of heat transfer from the entire finned surface of the plate. 10 cm 9.2 cm 5–36 A hot surface at 100°C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter aluminum pin fins (k 237 W/m · °C) with a center-to-center distance of 0.6 cm. The temperature of the surrounding medium is 30°C, and the combined heat transfer coefficient on the surfaces is 35 W/m2 · °C. Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be 0.5 cm, determine (a) the finite difference formulation of this problem, (b) the nodal temperatures along the fin by solving these equations, (c) the rate of heat transfer from a single fin, and (d) the rate of heat transfer from a 1-m 1-m section of the plate. Tair = 8°C 1 cm 1 cm 20 cm Steam 200°C 3 cm FIGURE P5–38 0.6 cm 0.25 cm 100°C 1 2 3 Reconsider Problem 5–38. Using EES (or other) software, investigate the effects of the steam temperature and the outer heat transfer coefficient on the flange tip temperature and the rate of heat transfer from the exposed surfaces of the flange. Let the steam temperature vary from 150°C to 300°C and the heat transfer coefficient from 15 W/m2 · °C to 60 W/m2 · °C. Plot the flange tip temperature and the heat transfer rate as functions of steam temperature and heat transfer coefficient, and discuss the results. 5–40 0.5 cm 0 5–39 4 5 6 x (a) FIGURE P5–36 5–37 Repeat Problem 5–36 using copper fins (k 386 W/m · °C) instead of aluminum ones. (b) Answers: (b) 98.6°C, 97.5°C, 96.7°C, 96.0°C, 95.7°C, 95.5°C 5–38 Two 3-m-long and 0.4-cm-thick cast iron (k 52 W/m · °C, 0.8) steam pipes of outer diameter 10 cm are connected to each other through two 1-cm-thick flanges of outer diameter 20 cm, as shown in the figure. The steam flows inside the pipe at an average temperature of 200°C with a heat transfer coefficient of 180 W/m2 · °C. The outer surface of the pipe is exposed to convection with ambient air at 8°C with a heat transfer coefficient of 25 W/m2 · °C as well as radiation with the surrounding surfaces at an average temperature of Tsurr 290 K. Assuming steady one-dimensional heat conduction along the flanges and taking the nodal spacing to be 1 cm along the flange (a) obtain the finite difference formulation for all nodes, (b) determine the temperature at the tip of the flange by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the flange. Using EES (or other) software, solve these systems of algebraic equations. 3x1 x2 3x3 0 x1 2x2 x3 3 2x1 x2 x3 2 4x1 2x22 0.5x3 2 x31 x2 x3 11.964 x1 x2 x3 3 Answers: (a) x1 2, x2 3, x3 1, (b) x1 2.33, x2 2.29, x3 1.62 5–41 (a) (b) Using EES (or other) software, solve these systems of algebraic equations. 3x1 2x2 x3 x4 6 x1 2x2 x4 3 2x1 x2 3x3 x4 2 3x2 x3 4x4 6 3x1 x22 2x3 8 x21 3x2 2x3 6.293 2x1 x42 4x3 12 cen58933_ch05.qxd 9/4/2002 11:42 AM Page 320 320 HEAT TRANSFER 5–42 (a) (b) Using EES (or other) software, solve these systems of algebraic equations. 4x1 x2 2x3 x4 6 x1 3x2 x3 4x4 1 x1 2x2 5x4 5 2x2 4x3 3x4 5 Insulation g· = 6 × 106 W/ m3 2x1 x42 2x3 x4 1 x21 4x2 2x23 2x4 3 x1 x42 5x3 10 3x1 x23 8x4 15 5 cm 5 cm 5–43C Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as (a) (b) (c) (d) (e) g·nodel 2 0 k Is heat transfer in this medium steady or transient? Is heat transfer one-, two-, or three-dimensional? Is there heat generation in the medium? Is the nodal spacing constant or variable? Is the thermal conductivity of the medium constant or variable? 5–44C Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as Tnode (Tleft Ttop Tright Tbottom)/4 (a) (b) (c) (d) (e) 305 3 290 240 1 350°C 2 325 Convection T = 20°C, h = 50 W/ m2 · °C Two-Dimensional Steady Heat Conduction Tleft Ttop Tright Tbottom 4Tnode 260 200°C Is heat transfer in this medium steady or transient? Is heat transfer one-, two-, or three-dimensional? Is there heat generation in the medium? Is the nodal spacing constant or variable? Is the thermal conductivity of the medium constant or variable? 5–45C What is an irregular boundary? What is a practical way of handling irregular boundary surfaces with the finite difference method? 5–46 Consider steady two-dimensional heat transfer in a long solid body whose cross section is given in the figure. The temperatures at the selected nodes and the thermal conditions at the boundaries are as shown. The thermal conductivity of the body is k 45 W/m · °C, and heat is generated in the body uniformly at a rate of g· 6 106 W/m3. Using the finite difference method with a mesh size of x y 5.0 cm, determine (a) the temperatures at nodes 1, 2, and 3 and (b) the rate of heat loss from the bottom surface through a 1-m-long section of the body. FIGURE P5–46 5–47 Consider steady two-dimensional heat transfer in a long solid body whose cross section is given in the figure. The measured temperatures at selected points of the outer surfaces are as shown. The thermal conductivity of the body is k 45 W/m · °C, and there is no heat generation. Using the finite difference method with a mesh size of x y 2.0 cm, determine the temperatures at the indicated points in the medium. Hint: Take advantage of symmetry. 150 180 200 1 180 2 150°C 3 180 180 4 5 6 200 200 7 8 9 180 180 2 cm 2 cm 150 180 200 180 150 FIGURE P5–47 5–48 Consider steady two-dimensional heat transfer in a long solid bar whose cross section is given in the figure. The measured temperatures at selected points of the outer surfaces are as shown. The thermal conductivity of the body is k 20 W/m · °C, and there is no heat generation. Using the finite difference method with a mesh size of x y 1.0 cm, determine the temperatures at the indicated points in the medium. Answers: T1 185°C, T2 T3 T4 190°C 5–49 Starting with an energy balance on a volume element, obtain the steady two-dimensional finite difference equation for a general interior node in rectangular coordinates for T(x, y) for the case of variable thermal conductivity and uniform heat generation. cen58933_ch05.qxd 9/4/2002 11:42 AM Page 321 321 CHAPTER 5 W/m3. Plot the temperatures at nodes 1 and 3, and the rate of heat loss as functions of thermal conductivity and heat generation rate, and discuss the results. 200°C 1 2 3 4 180 5–52 Consider steady two-dimensional heat transfer in a long solid bar whose cross section is given in the figure. The measured temperatures at selected points on the outer surfaces are as shown. The thermal conductivity of the body is k 20 W/m · °C, and there is no heat generation. Using the finite difference method with a mesh size of x y 1.0 cm, determine the temperatures at the indicated points in the medium. Hint: Take advantage of symmetry. 200 Insulation (a) 120 100 Answers: (b) T1 T4 143°C, T2 T3 136°C 120 1 100°C 100°C 2 120 x 120 y 150 3 140 140 1 200 FIGURE P5–48 100 100 4 250 5 6 300 Insulation (a) Insulation 100 100 100 1 2 100 3 100°C 4 1 cm 1 cm 200 100°C 200 200 200 200°C (b) g· = 107 W/ m3 FIGURE P5–52 1 2 120 120 3 4 150 150 0.1 m 0.1 m Convection 200 200 1 Reconsider Problem 5–50. Using EES (or other) software, investigate the effects of the thermal conductivity and the heat generation rate on the temperatures at nodes 1 and 3, and the rate of heat loss from the top surface. Let the thermal conductivity vary from 10 W/m · °C to 400 W/m · °C and the heat generation rate from 105 W/m3 to 108 q· L 2 4 h, T 3 5 6 1.5 cm 1.5 cm 120°C FIGURE P5–53 7 8 Insulation 200 5–53 Consider steady two-dimensional heat transfer in an L-shaped solid body whose cross section is given in the figure. The thermal conductivity of the body is k 45 W/m · °C, and heat is generated in the body at a rate of g· 5 106 W/m3. FIGURE P5–50 5–51 3 1 cm 300 5–50 Consider steady two-dimensional heat transfer in a long solid body whose cross section is given in the figure. The temperatures at the selected nodes and the thermal conditions on the boundaries are as shown. The thermal conductivity of the body is k 180 W/m · °C, and heat is generated in the body uniformly at a rate of g· 107 W/m3. Using the finite difference method with a mesh size of x y 10 cm, determine (a) the temperatures at nodes 1, 2, 3, and 4 and (b) the rate of heat loss from the top surface through a 1-m-long section of the body. 200 2 250 (b) 200 1 cm Insulation 100 150 4 Insulation 180 Insulation 150 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 322 322 HEAT TRANSFER The right surface of the body is insulated, and the bottom surface is maintained at a uniform temperature of 120°C. The entire top surface is subjected to convection with ambient air at T 30°C with a heat transfer coefficient of h 55 W/m2 · °C, and the left surface is subjected to heat flux at a uniform rate of q·L 8000 W/m2. The nodal network of the problem consists of 13 equally spaced nodes with x y 1.5 cm. Five of the nodes are at the bottom surface and thus their temperatures are known. (a) Obtain the finite difference equations at the remaining eight nodes and (b) determine the nodal temperatures by solving those equations. 5–54E Consider steady two-dimensional heat transfer in a long solid bar of square cross section in which heat is generated uniformly at a rate of g· 0.19 105 Btu/h · ft3. The cross section of the bar is 0.4 ft 0.4 ft in size, and its thermal conductivity is k 16 Btu/h · ft · °F. All four sides of the bar are subjected to convection with the ambient air at T 70°F with a heat transfer coefficient of h 7.9 Btu/h · ft2 · °F. Using the finite difference method with a mesh size of x y 0.2 ft, determine (a) the temperatures at the nine nodes and (b) the rate of heat loss from the bar through a 1-ft-long section. Answer: (b) 3040 Btu/h h, T 1 2 3 g· h, T 4 5 7 8 6 h, T 9 h, T FIGURE P5–54E 5–55 Hot combustion gases of a furnace are flowing through a concrete chimney (k 1.4 W/m · °C) of rectangular cross Tsky section. The flow section of the chimney is 20 cm 40 cm, and the thickness of the wall is 10 cm. The average temperature of the hot gases in the chimney is Ti 280°C, and the average convection heat transfer coefficient inside the chimney is hi 75 W/m2 · °C. The chimney is losing heat from its outer surface to the ambient air at To 15°C by convection with a heat transfer coefficient of ho 18 W/m2 · °C and to the sky by radiation. The emissivity of the outer surface of the wall is 0.9, and the effective sky temperature is estimated to be 250 K. Using the finite difference method with x y 10 cm and taking full advantage of symmetry, (a) obtain the finite difference formulation of this problem for steady twodimensional heat transfer, (b) determine the temperatures at the nodal points of a cross section, and (c) evaluate the rate of heat loss for a 1-m-long section of the chimney. 5–56 Repeat Problem 5–55 by disregarding radiation heat transfer from the outer surfaces of the chimney. 5–57 Reconsider Problem 5–55. Using EES (or other) software, investigate the effects of hot-gas temperature and the outer surface emissivity on the temperatures at the outer corner of the wall and the middle of the inner surface of the right wall, and the rate of heat loss. Let the temperature of the hot gases vary from 200°C to 400°C and the emissivity from 0.1 to 1.0. Plot the temperatures and the rate of heat loss as functions of the temperature of the hot gases and the emissivity, and discuss the results. Consider a long concrete dam (k 0.6 W/m · °C, s 0.7 m2/s) of triangular cross section whose exposed surface is subjected to solar heat flux of q·s 800 W/m2 and to convection and radiation to the environment at 25°C with a combined heat transfer coefficient of 30 W/m2 · °C. The 2-m-high vertical section of the dam is subjected to convection by water at 15°C with a heat transfer coefficient of 150 W/m2 · °C, and heat transfer through the 2-m-long base is considered to be negligible. Using the finite difference method with a mesh size of x y 1 m and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surAnswers: 21.3°C, 43.2°C, 43.6°C face of the dam. 5–58 ε To, ho Chimney 1 q· s 10 cm h T Hot gases 20 cm Ti, hi 1m 2 3 Water 10 cm 1m 4 10 cm FIGURE P5–55 40 cm 1m 5 1m 6 10 cm FIGURE P5–58 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 323 323 CHAPTER 5 5–59E Consider steady two-dimensional heat transfer in a V-grooved solid body whose cross section is given in the figure. The top surfaces of the groove are maintained at 32°F while the bottom surface is maintained at 212°F. The side surfaces of the groove are insulated. Using the finite difference method with a mesh size of x y 1 ft and taking advantage of symmetry, determine the temperatures at the middle of the insulated surfaces. 32°F 2 4 3 5 6 9 7 10 8 11 Insulation Insulation 1 1 ft 1 ft dimensional heat transfer and (b) determine the unknown nodal temperatures by solving those equations. Answers: (b) 85.7°C, 86.4°C, 87.6°C 5–62 Consider a 5-m-long constantan block (k 23 W/m · °C) 30 cm high and 50 cm wide. The block is completely submerged in iced water at 0°C that is well stirred, and the heat transfer coefficient is so high that the temperatures on both sides of the block can be taken to be 0°C. The bottom surface of the bar is covered with a low-conductivity material so that heat transfer through the bottom surface is negligible. The top surface of the block is heated uniformly by a 6-kW resistance heater. Using the finite difference method with a mesh size of x y 10 cm and taking advantage of symmetry, (a) obtain the finite difference formulation of this problem for steady two-dimensional heat transfer, (b) determine the unknown nodal temperatures by solving those equations, and (c) determine the rate of heat transfer from the block to the iced water. 6 kW heater Insulation 212°F FIGURE P5–59E 0°C Reconsider Problem 5–59E. Using EES (or other) software, investigate the effects of the temperatures at the top and bottom surfaces on the temperature in the middle of the insulated surface. Let the temperatures at the top and bottom surfaces vary from 32°F to 212°F. Plot the temperature in the middle of the insulated surface as functions of the temperatures at the top and bottom surfaces, and discuss the results. 0°C 5–60 5–61 Consider a long solid bar whose thermal conductivity is k 12 W/m · °C and whose cross section is given in the figure. The top surface of the bar is maintained at 50°C while the bottom surface is maintained at 120°C. The left surface is insulated and the remaining three surfaces are subjected to convection with ambient air at T 25°C with a heat transfer coefficient of h 30 W/m2 · °C. Using the finite difference method with a mesh size of x y 10 cm, (a) obtain the finite difference formulation of this problem for steady two50°C Insulation Convection h, T 10 cm 10 cm Insulation FIGURE P5–62 Transient Heat Conduction 5–63C How does the finite difference formulation of a transient heat conduction problem differ from that of a steady heat conduction problem? What does the term AxC(Tmi1 Tmi )/t represent in the transient finite difference formulation? 5–64C What are the two basic methods of solution of transient problems based on finite differencing? How do heat transfer terms in the energy balance formulation differ in the two methods? 5–65C The explicit finite difference formulation of a general interior node for transient heat conduction in a plane wall is given by i i Tm1 2Tmi Tm1 120°C FIGURE P5–61 10 cm 10 cm i g· im x2 T i1 m Tm k Obtain the finite difference formulation for the steady case by simplifying the relation above. cen58933_ch05.qxd 9/4/2002 11:43 AM Page 324 324 HEAT TRANSFER 5–66C The explicit finite difference formulation of a general interior node for transient two-dimensional heat conduction is given by · t) g(x, q· 0 i1 i i i i Tnode (Tleft Ttop Tright Tbottom ) i 2 ·node g l i (1 4)Tnode k 5–67C Is there any limitation on the size of the time step t in the solution of transient heat conduction problems using (a) the explicit method and (b) the implicit method? 5–68C Express the general stability criterion for the explicit method of solution of transient heat conduction problems. 5–69C Consider transient one-dimensional heat conduction in a plane wall that is to be solved by the explicit method. If both sides of the wall are at specified temperatures, express the stability criterion for this problem in its simplest form. 5–70C Consider transient one-dimensional heat conduction in a plane wall that is to be solved by the explicit method. If both sides of the wall are subjected to specified heat flux, express the stability criterion for this problem in its simplest form. 5–71C Consider transient two-dimensional heat conduction in a rectangular region that is to be solved by the explicit method. If all boundaries of the region are either insulated or at specified temperatures, express the stability criterion for this problem in its simplest form. 5–72C The implicit method is unconditionally stable and thus any value of time step t can be used in the solution of transient heat conduction problems. To minimize the computation time, someone suggests using a very large value of t since there is no danger of instability. Do you agree with this suggestion? Explain. 5–73 Consider transient heat conduction in a plane wall whose left surface (node 0) is maintained at 50°C while the right surface (node 6) is subjected to a solar heat flux of 600 W/m2. The wall is initially at a uniform temperature of 50°C. Express the explicit finite difference formulation of the boundary nodes 0 and 6 for the case of no heat generation. Also, obtain the finite difference formulation for the total amount of heat transfer at the left boundary during the first three time steps. 5–74 Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, and 4 with a uniform nodal spacing of x. The wall is initially at a specified temperature. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary nodes for the case of uniform heat flux q·0 at the left boundary ∆x 0 Obtain the finite difference formulation for the steady case by simplifying the relation above. h, T 1 2 3 x 4 FIGURE P5–74 (node 0) and convection at the right boundary (node 4) with a convection coefficient of h and an ambient temperature of T. Do not simplify. 5–75 tion. Repeat Problem 5–74 for the case of implicit formula- 5–76 Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, 4, and 5 with a uniform nodal spacing of x. The wall is initially at a specified temperature. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0) and radiation at the right boundary (node 5) with an emissivity of and surrounding temperature of Tsurr. 5–77 Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, 3, and 4 with a uniform nodal spacing of x. The wall is initially at a specified temperature. The temperature at the right boundary (node 4) is specified. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary node 0 for the case of combined convection, radiation, and heat flux at the left boundary with an emissivity of , convection coefficient of h, ambient temperature of T, surrounding temperature of Tsurr, and uniform heat flux of q·0 toward the wall. Also, obtain the finite difference formulation for the total amount of heat transfer at the right boundary for the first 20 time steps. Tsurr Radiation TL · t) g(x, ∆x q· 0 0 h, T Convection FIGURE P5–77 L 1 2 3 4 x cen58933_ch05.qxd 9/4/2002 11:43 AM Page 325 325 CHAPTER 5 5–78 Starting with an energy balance on a volume element, obtain the two-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for T(x, y, t) for the case of constant thermal conductivity and no heat generation. 5–79 Starting with an energy balance on a volume element, obtain the two-dimensional transient implicit finite difference equation for a general interior node in rectangular coordinates for T(x, y, t) for the case of constant thermal conductivity and no heat generation. 5–80 Starting with an energy balance on a disk volume element, derive the one-dimensional transient explicit finite difference equation for a general interior node for T(z, t) in a cylinder whose side surface is insulated for the case of constant thermal conductivity with uniform heat generation. 5–81 Consider one-dimensional transient heat conduction in a composite plane wall that consists of two layers A and B with perfect contact at the interface. The wall involves no heat generation and initially is at a specified temperature. The nodal network of the medium consists of nodes 0, 1 (at the interface), and 2 with a uniform nodal spacing of x. Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of insulation at the left boundary (node 0) and radiation at the right boundary (node 2) with an emissivity of and surrounding temperature of Tsurr. Tsurr Insulation A ε B Radiation ∆x 0 ∆x 1 2 5–84 Consider a large uranium plate of thickness L 8 cm, thermal conductivity k 28 W/m · °C, and thermal diffusivity 12.5 106 m2/s that is initially at a uniform temperature of 100°C. Heat is generated uniformly in the plate at a constant rate of g· 106 W/m3. At time t 0, the left side of the plate is insulated while the other side is subjected to convection with an environment at T 20°C with a heat transfer coefficient of h 35 W/m2 · °C. Using the explicit finite difference approach with a uniform nodal spacing of x 2 cm, determine (a) the temperature distribution in the plate after 5 min and (b) how long it will take for steady conditions to be reached in the plate. 5–85 Reconsider Problem 5–84. Using EES (or other) software, investigate the effect of the cooling time on the temperatures of the left and right sides of the plate. Let the time vary from 5 min to 60 min. Plot the temperatures at the left and right surfaces as a function of time, and discuss the results. 5–86 Consider a house whose south wall consists of a 30-cmthick Trombe wall whose thermal conductivity is k 0.70 W/m · °C and whose thermal diffusivity is 0.44 106 m2/s. The variations of the ambient temperature Tout and the solar heat flux q·solar incident on a south-facing vertical surface throughout the day for a typical day in February are given in the table in 3-h intervals. The Trombe wall has single glazing with an absorptivity-transmissivity product of 0.76 (that is, 76 percent of the solar energy incident is absorbed by the exposed surface of the Trombe wall), and the average combined heat transfer coefficient for heat loss from the Trombe wall to the ambient is determined to be hout 3.4 W/m2 · °C. The interior of the house is maintained at Tin 20°C at all times, and the heat transfer coefficient at the interior surface of the Trombe wall is hin 9.1 W/m2 · °C. Also, the vents on the Trombe wall are kept closed, and thus the only heat transfer between the air in the house and the Trombe wall is through the Interface FIGURE P5–81 5–82 Consider transient one-dimensional heat conduction in a pin fin of constant diameter D with constant thermal conductivity. The fin is losing heat by convection to the ambient air at T with a heat transfer coefficient of h and by radiation to the surrounding surfaces at an average temperature of Tsurr. The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of x. Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of a specified temperature at the fin base and negligible heat transfer at the fin tip. 5–83 Repeat Problem 5–82 for the case of implicit formulation. Sun’s rays Trombe wall Heat gain Heat loss Glazing FIGURE P5–86 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 326 326 HEAT TRANSFER 5–88 TABLE P5–86 The hourly variations of the monthly average ambient temperature and solar heat flux incident on a vertical surface Time of Day 7 10 1 4 7 10 1 4 Ambient Temperature, °C Solar Insolation, W/m2 0 4 6 1 2 3 4 4 375 750 580 95 0 0 0 0 AM–10 AM AM–1 PM PM–4 PM PM–7 PM PM–10 PM PM–1 AM AM–4 AM AM–7 AM Reconsider Problem 5–87. Using EES (or other) software, plot the temperature at the top corner as a function of heating time varies from 2 min to 30 min, and discuss the results. 5–89 Consider a long solid bar (k 28 W/m · °C and 12 106 m2/s) of square cross section that is initially at a uniform temperature of 20°C. The cross section of the bar is 20 cm 20 cm in size, and heat is generated in it uniformly at a rate of g· 8 105 W/m3. All four sides of the bar are subjected to convection to the ambient air at T 30°C with a heat transfer coefficient of h 45 W/m2 · °C. Using the explicit finite difference method with a mesh size of x y 10 cm, determine the centerline temperature of the bar (a) after 10 min and (b) after steady conditions are established. h, T 1 interior surface of the wall. Assuming the temperature of the Trombe wall to vary linearly between 20°C at the interior surface and 0°C at the exterior surface at 7 AM and using the explicit finite difference method with a uniform nodal spacing of x 5 cm, determine the temperature distribution along the thickness of the Trombe wall after 6, 12, 18, 24, 30, 36, 42, and 48 hours and plot the results. Also, determine the net amount of heat transferred to the house from the Trombe wall during the first day if the wall is 2.8 m high and 7 m long. 5–87 Consider two-dimensional transient heat transfer in an L-shaped solid bar that is initially at a uniform temperature of 140°C and whose cross section is given in the figure. The thermal conductivity and diffusivity of the body are k 15 W/m · °C and 3.2 106 m2/s, respectively, and heat is generated in the body at a rate of g· 2 107 W/m3. The right surface of the body is insulated, and the bottom surface is maintained at a uniform temperature of 140°C at all times. At time t 0, the entire top surface is subjected to convection with ambient air at T 25°C with a heat transfer coefficient of h 80 W/m2 · °C, and the left surface is subjected to uniform heat flux at a rate of q·L 8000 W/m2. The nodal network of the problem consists of 13 equally spaced nodes with x y 1.5 cm. Using the explicit method, determine the temperature at the top corner (node 3) of the body after 2, 5, and 30 min. Convection q· L 2 4 h, T 3 5 6 1.5 cm 1.5 cm 7 5 10 cm 10 cm 7 8 6 h, T 9 h, T FIGURE P5–89 5–90E Consider a house whose windows are made of 0.375-in.-thick glass (k 0.48 Btu/h · ft · °F and 4.2 106 ft2/s). Initially, the entire house, including the walls and the windows, is at the outdoor temperature of To 35°F. It is observed that the windows are fogged because the indoor temperature is below the dew-point temperature of 54°F. Now the heater is turned on and the air temperature in the house is raised to Ti 72°F at a rate of 2°F rise per minute. The heat transfer coefficients at the inner and outer surfaces of the wall can be taken to be hi 1.2 and ho 2.6 Btu/h · ft2 · °F, respectively, and the outdoor temperature can be assumed to remain constant. Using the explicit finite difference method with a mesh size of x 0.125 in., determine how long it will take Window glass ∆x 1 2 House Fog 140°C FIGURE P5–87 4 h, T hi 8 3 g· Ti Insulation 1 2 FIGURE P5–90E 3 4 0.375 in To ho Outdoors cen58933_ch05.qxd 9/4/2002 11:43 AM Page 327 327 CHAPTER 5 for the fog on the windows to clear up (i.e., for the inner surface temperature of the window glass to reach 54°F). 5–91 A common annoyance in cars in winter months is the formation of fog on the glass surfaces that blocks the view. A practical way of solving this problem is to blow hot air or to attach electric resistance heaters to the inner surfaces. Consider the rear window of a car that consists of a 0.4-cm-thick glass (k 0.84 W/m · °C and 0.39 106 m2/s). Strip heater wires of negligible thickness are attached to the inner surface of the glass, 4 cm apart. Each wire generates heat at a rate of 10 W/m length. Initially the entire car, including its windows, is at the outdoor temperature of To 3°C. The heat transfer coefficients at the inner and outer surfaces of the glass can be taken to be hi 6 and ho 20 W/m2 · °C, respectively. Using the explicit finite difference method with a mesh size of x 0.2 cm along the thickness and y 1 cm in the direction normal to the heater wires, determine the temperature distribution throughout the glass 15 min after the strip heaters are turned on. Also, determine the temperature distribution when steady conditions are reached. Thermal symmetry line Inner surface Outer surface Glass Heater 10 W/ m 0.2 cm 1 cm Thermal symmetry line Tsky Radiation Convection ho, To ε 15 cm ε Convection hi, Ti Radiation Ti FIGURE P5–93 are maintained at a constant temperature of 20°C during the night, and the emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfers and using the explicit finite difference method with a time step of t 5 min and a mesh size of x 3 cm, determine the temperatures of the inner and outer surfaces of the roof at 6 AM. Also, determine the average rate of heat transfer through the roof during that night. 5–94 Consider a refrigerator whose outer dimensions are 1.80 m 0.8 m 0.7 m. The walls of the refrigerator are constructed of 3-cm-thick urethane insulation (k 0.026 W/m · ° C and 0.36 106 m2/s) sandwiched between two layers of sheet metal with negligible thickness. The refrigerated space is maintained at 3°C and the average heat transfer coefficients at the inner and outer surfaces of the wall are 6 W/m2 · °C and 9 W/m2 · °C, respectively. Heat transfer through the bottom surface of the refrigerator is negligible. The kitchen temperature remains constant at about 25°C. Initially, the refrigerator contains 15 kg of food items at an average specific heat of 3.6 kJ/kg · °C. Now a malfunction occurs and the refrigerator stops running for 6 h as a result. Assuming the FIGURE P5–91 5–92 Refrigerator Air Repeat Problem 5–91 using the implicit method with a time step of 1 min. 5–93 The roof of a house consists of a 15-cm-thick concrete slab (k 1.4 W/m · °C and 0.69 106 m2/s) that is 20 m wide and 20 m long. One evening at 6 PM, the slab is observed to be at a uniform temperature of 18°C. The average ambient air and the night sky temperatures for the entire night are predicted to be 6°C and 260 K, respectively. The convection heat transfer coefficients at the inner and outer surfaces of the roof can be taken to be hi 5 and ho 12 W/m2 · °C, respectively. The house and the interior surfaces of the walls and the floor Concrete roof Heat 3°C 25°C ho hi Food 15 kg FIGURE P5–94 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 328 328 HEAT TRANSFER temperature of the contents of the refrigerator, including the air inside, rises uniformly during this period, predict the temperature inside the refrigerator after 6 h when the repairman arrives. Use the explicit finite difference method with a time step of t 1 min and a mesh size of x 1 cm and disregard corner effects (i.e., assume one-dimensional heat transfer in the walls). 5–95 Reconsider Problem 5–94. Using EES (or other) software, plot the temperature inside the refrigerator as a function of heating time as time varies from 1 h to 10 h, and discuss the results. Special Topic: Controlling the Numerical Error coordinates for T(x, y, z, t) for the case of constant thermal conductivity and no heat generation. 5–108 Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes 0, 1, 2, and 3 with a uniform nodal spacing of x. The temperature at the left boundary (node 0) is specified. Using the energy balance approach, obtain the finite difference formulation of boundary node 3 at the right boundary for the case of combined convection and radiation with an emissivity of , convection coefficient of h, ambient temperature of T, and surrounding temperature of Tsurr. Also, obtain the finite difference formulation for the rate of heat transfer at the left boundary. 5–96C Why do the results obtained using a numerical method differ from the exact results obtained analytically? What are the causes of this difference? 5–97C What is the cause of the discretization error? How does the global discretization error differ from the local discretization error? Tsurr Radiation 5–98C Can the global (accumulated) discretization error be less than the local error during a step? Explain. 5–99C How is the finite difference formulation for the first derivative related to the Taylor series expansion of the solution function? 5–100C Explain why the local discretization error of the finite difference method is proportional to the square of the step size. Also explain why the global discretization error is proportional to the step size itself. 5–101C What causes the round-off error? What kind of calculations are most susceptible to round-off error? 5–102C What happens to the discretization and the roundoff errors as the step size is decreased? Suggest some practical ways of reducing the round-off error. 5–103C 5–104C What is a practical way of checking if the round-off error has been significant in calculations? 5–105C What is a practical way of checking if the discretization error has been significant in calculations? ε · g(x) T0 h, T Convection ∆x 0 1 2 3 FIGURE P5–108 5–109 Consider one-dimensional transient heat conduction in a plane wall with variable heat generation and variable thermal conductivity. The nodal network of the medium consists of nodes 0, 1, and 2 with a uniform nodal spacing of x. Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of specified heat flux q·0 and convection at the left boundary (node 0) with a convection coefficient of h and ambient temperature of T, and radiation at the right boundary (node 2) with an emissivity of and surrounding temperature of Tsurr. 5–110 Repeat Problem 5–109 for the case of implicit formulation. 5–111 Consider steady one-dimensional heat conduction in a pin fin of constant diameter D with constant thermal conductivity. The fin is losing heat by convection with the ambient air Review Problems 5–106 Starting with an energy balance on the volume element, obtain the steady three-dimensional finite difference equation for a general interior node in rectangular coordinates for T(x, y, z) for the case of constant thermal conductivity and uniform heat generation. 5–107 Starting with an energy balance on the volume element, obtain the three-dimensional transient explicit finite difference equation for a general interior node in rectangular Convection h, T ε 0 1 ∆x FIGURE P5–111 Tsurr Radiation 2 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 329 329 CHAPTER 5 at T (in °C) with a convection coefficient of h, and by radiation to the surrounding surfaces at an average temperature of Tsurr (in K). The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of x. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of a specified temperature at the fin base and convection and radiation heat transfer at the fin tip. 5–112 Starting with an energy balance on the volume element, obtain the two-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for T(x, y, t) for the case of constant thermal conductivity and uniform heat generation. 5–113 Starting with an energy balance on a disk volume element, derive the one-dimensional transient implicit finite difference equation for a general interior node for T(z, t) in a cylinder whose side surface is subjected to convection with a convection coefficient of h and an ambient temperature of T for the case of constant thermal conductivity with uniform heat generation. 5–114E The roof of a house consists of a 5-in.-thick concrete slab (k 0.81 Btu/h · ft · °F and 7.4 106 ft2/s) that is 45 ft wide and 55 ft long. One evening at 6 PM, the slab is observed to be at a uniform temperature of 70°F. The ambient air temperature is predicted to be at about 50°F from 6 PM to 10 PM, 42°F from 10 PM to 2 AM, and 38°F from 2 AM to 6 AM, while the night sky temperature is expected to be about 445 R for the entire night. The convection heat transfer coefficients at the inner and outer surfaces of the roof can be taken to be hi 0.9 and ho 2.1 Btu/h · ft2 · °F, respectively. The house and the interior surfaces of the walls and the floor are maintained at a constant temperature of 70°F during the night, and the emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfers and using the explicit finite difference method with a mesh size of x 1 in. and a time step of t 5 min, determine the temperatures of the inner and outer surfaces of the roof at 6 AM. Also, determine the average rate of heat transfer through the roof during that night. 5–115 Solar radiation incident on a large body of clean water (k 0.61 W/m · °C and 0.15 106 m2/s) such as a lake, a river, or a pond is mostly absorbed by water, and the amount of absorption varies with depth. For solar radiation incident at a 45° angle on a 1-m-deep large pond whose bottom surface is black (zero reflectivity), for example, 2.8 percent of the solar energy is reflected back to the atmosphere, 37.9 percent is absorbed by the bottom surface, and the remaining 59.3 percent is absorbed by the water body. If the pond is considered to be four layers of equal thickness (0.25 m in this case), it can be shown that 47.3 percent of the incident solar energy is absorbed by the top layer, 6.1 percent by the upper mid layer, 3.6 percent by the lower mid layer, and 2.4 percent by the bottom layer [for more information see Çengel and Özi¸sik, Solar Energy, 33, no. 6 (1984), pp. 581–591]. The radiation absorbed by the water can be treated conveniently as heat generation in the heat transfer analysis of the pond. Consider a large 1-m-deep pond that is initially at a uniform temperature of 15°C throughout. Solar energy is incident on the pond surface at 45° at an average rate of 500 W/m2 for a period of 4 h. Assuming no convection currents in the water and using the explicit finite difference method with a mesh size of x 0.25 m and a time step of t 15 min, determine the temperature distribution in the pond under the most favorable conditions (i.e., no heat losses from the top or bottom surfaces of the pond). The solar energy absorbed by the bottom surface of the pond can be treated as a heat flux to the water at that surface in this case. Sun Tsky Radiation Convection ho, To ε Solar radiation Concrete roof 45° 0 1 ε Radiation Convection hi, Ti 2 3 Ti 4 Top layer Solar pond Upper mid layer Lower mid layer Bottom layer x FIGURE P5–114E q· s, W/ m2 FIGURE P5–115 Black cen58933_ch05.qxd 9/4/2002 11:43 AM Page 330 330 HEAT TRANSFER 5–116 Reconsider Problem 5–115. The absorption of solar radiation in that case can be expressed more accurately as a fourth-degree polynomial as g·(x) q·s(0.859 3.415x 6.704x2 6.339x3 2.278x4), W/m3 where q·s is the solar flux incident on the surface of the pond in W/m2 and x is the distance from the free surface of the pond in m. Solve Problem 5–115 using this relation for the absorption of solar radiation. 5–117 A hot surface at 120°C is to be cooled by attaching 8 cm long, 0.8 cm in diameter aluminum pin fins (k 237 W/m · °C and 97.1 106 m2/s) to it with a center-tocenter distance of 1.6 cm. The temperature of the surrounding medium is 15°C, and the heat transfer coefficient on the surfaces is 35 W/m2 · °C. Initially, the fins are at a uniform temperature of 30°C, and at time t 0, the temperature of the hot surface is raised to 120°C. Assuming one-dimensional heat conduction along the fin and taking the nodal spacing to be x 2 cm and a time step to be t 0.5 s, determine the nodal temperatures after 5 min by using the explicit finite difference method. Also, determine how long it will take for steady conditions to be reached. T0 Convection h, T 2 cm 0 1 2 3 4 FIGURE P5–117 5–118E Consider a large plane wall of thickness L 0.3 ft and thermal conductivity k 1.2 Btu/h · ft · °F in space. The wall is covered with a material having an emissivity of 0.80 and a solar absorptivity of s 0.45. The inner surface of the wall is maintained at 520 R at all times, while the outer surface is exposed to solar radiation that is incident at a rate of q·s 300 Btu/h · ft2. The outer surface is also losing heat αs T0 by radiation to deep space at 0 R. Using a uniform nodal spacing of x 0.1 ft, (a) obtain the finite difference formulation for steady one-dimensional heat conduction and (b) determine the nodal temperatures by solving those equations. Answers: (b) 522 R, 525 R, 527 R 5–119 Frozen food items can be defrosted by simply leaving them on the counter, but it takes too long. The process can be speeded up considerably for flat items such as steaks by placing them on a large piece of highly conducting metal, called the defrosting plate, which serves as a fin. The increased surface area enhances heat transfer and thus reduces the defrosting time. Consider two 1.5-cm-thick frozen steaks at 18°C that resemble a 15-cm-diameter circular object when placed next to each other. The steaks are now placed on a 1-cm-thick blackanodized circular aluminum defrosting plate (k 237 W/m · °C, 97.1 106 m2/s, and 0.90) whose outer diameter is 30 cm. The properties of the frozen steaks are 970 kg/m3, Cp 1.55 kJ/kg · °C, k 1.40 W/m · °C, 0.93 106 m2/s, and 0.95, and the heat of fusion is hif 187 kJ/kg. The steaks can be considered to be defrosted when their average temperature is 0°C and all of the ice in the steaks is melted. Initially, the defrosting plate is at the room temperature of 20°C, and the wooden countertop it is placed on can be treated as insulation. Also, the surrounding surfaces can be taken to be at the same temperature as the ambient air, and the convection heat transfer coefficient for all exposed surfaces can be taken to be 12 W/m2 · °C. Heat transfer from the lateral surfaces of the steaks and the defrosting plate can be neglected. Assuming one-dimensional heat conduction in both the steaks and the defrosting plate and using the explicit finite difference method, determine how long it will take to defrost the steaks. Use four nodes with a nodal spacing of x 0.5 cm for the steaks, and three nodes with a nodal spacing of r 3.75 cm for the exposed portion of the defrosting plate. Also, use a time step of t 5 s. Hint: First, determine the total amount of heat transfer needed to defrost the steaks, and then determine how long it will take to transfer that much heat. Symmetry line 20°C Heat transfer q· s 6 ∆x 0 1 2 3 ε 5 1 2 3 4 Tsurr Defrosting plate Radiation FIGURE P5–118E Steaks FIGURE P5–119 cen58933_ch05.qxd 9/4/2002 11:43 AM Page 331 331 CHAPTER 5 5–120 Repeat Problem 5–119 for a copper defrosting plate using a time step of t 3 s. Design and Essay Problems 5–121 Write a two-page essay on the finite element method, and explain why it is used in most commercial engineering software packages. Also explain how it compares to the finite difference method. 5–122 Numerous professional software packages are available in the market for performing heat transfer analysis, and they are widely advertised in professional magazines such as the Mechanical Engineering magazine published by the American Society of Mechanical Engineers (ASME). Your company decides to purchase such a software package and asks you to prepare a report on the available packages, their costs, capabilities, ease of use, and compatibility with the available hardware, and other software as well as the reputation of the software company, their history, financial health, customer support, training, and future prospects, among other things. After a preliminary investigation, select the top three packages and prepare a full report on them. 5–123 Design a defrosting plate to speed up defrosting of flat food items such as frozen steaks and packaged vegetables and evaluate its performance using the finite difference method (see Prob. 5–119). Compare your design to the defrosting plates currently available on the market. The plate must perform well, and it must be suitable for purchase and use as a household utensil, durable, easy to clean, easy to manufacture, and affordable. The frozen food is expected to be at an initial temperature of 18°C at the beginning of the thawing process and 0°C at the end with all the ice melted. Specify the material, shape, size, and thickness of the proposed plate. Justify your recommendations by calculations. Take the ambient and surrounding surface temperatures to be 20°C and the convection heat transfer coefficient to be 15 W/m2 · °C in your analysis. For a typical case, determine the defrosting time with and without the plate. 5–124 Design a fire-resistant safety box whose outer dimensions are 0.5 m 0.5 m 0.5 m that will protect its combustible contents from fire which may last up to 2 h. Assume the box will be exposed to an environment at an average temperature of 700°C with a combined heat transfer coefficient of 70 W/m2 · °C and the temperature inside the box must be below 150°C at the end of 2 h. The cavity of the box must be as large as possible while meeting the design constraints, and the insulation material selected must withstand the high temperatures to which it will be exposed. Cost, durability, and strength are also important considerations in the selection of insulation materials. cen58933_ch05.qxd 9/4/2002 11:43 AM Page 332 cen58933_ch06.qxd 9/4/2002 12:05 PM Page 333 CHAPTER F U N D A M E N TA L S O F CONVECTION o far, we have considered conduction, which is the mechanism of heat transfer through a solid or a quiescent fluid. We now consider convection, which is the mechanism of heat transfer through a fluid in the presence of bulk fluid motion. Convection is classified as natural (or free) and forced convection, depending on how the fluid motion is initiated. In forced convection, the fluid is forced to flow over a surface or in a pipe by external means such as a pump or a fan. In natural convection, any fluid motion is caused by natural means such as the buoyancy effect, which manifests itself as the rise of warmer fluid and the fall of the cooler fluid. Convection is also classified as external and internal, depending on whether the fluid is forced to flow over a surface or in a channel. We start this chapter with a general physical description of the convection mechanism. We then discuss the velocity and thermal boundary layers, and laminar and turbulent flows. We continue with the discussion of the dimensionless Reynolds, Prandtl, and Nusselt numbers, and their physical significance. Next we derive the convection equations of on the basis of mass, momentum, and energy conservation, and obtain solutions for flow over a flat plate. We then nondimensionalize the convection equations, and obtain functional forms of friction and convection coefficients. Finally, we present analogies between momentum and heat transfer. S 6 CONTENTS 6–1 Physical Mechanism on Convection 334 6–2 Classification of Fluid Flows 337 6–3 Velocity Boundary Layer 339 6–4 Thermal Boundary Layer 341 6–5 Laminar and Turbulent Flows 342 6–6 Heat and Momentum Transfer in Turbulent Flow 343 6–7 Derivation of Differential Convection Equations 345 6–8 Solutions of Convection Equations for a Flat Plate 352 6–9 Nondimensionalized Convection Equations and Similarity 356 6–10 Functional Forms of Friction and Convection Coefficients 357 6–11 Analogies between Momentum and Heat Transfer 358 333 cen58933_ch06.qxd 9/4/2002 12:05 PM Page 334 334 HEAT TRANSFER 6–1 20°C 5 m/s AIR · Q 50°C (a) Forced convection · Q AIR Warmer air rising (b) Free convection · No convection Q currents AIR (c) Conduction FIGURE 6–1 Heat transfer from a hot surface to the surrounding fluid by convection and conduction. Hot plate, 110°C Fluid · Q Heat transfer through the fluid Cold plate, 30°C FIGURE 6–2 Heat transfer through a fluid sandwiched between two parallel plates. ■ PHYSICAL MECHANISM OF CONVECTION We mentioned earlier that there are three basic mechanisms of heat transfer: conduction, convection, and radiation. Conduction and convection are similar in that both mechanisms require the presence of a material medium. But they are different in that convection requires the presence of fluid motion. Heat transfer through a solid is always by conduction, since the molecules of a solid remain at relatively fixed positions. Heat transfer through a liquid or gas, however, can be by conduction or convection, depending on the presence of any bulk fluid motion. Heat transfer through a fluid is by convection in the presence of bulk fluid motion and by conduction in the absence of it. Therefore, conduction in a fluid can be viewed as the limiting case of convection, corresponding to the case of quiescent fluid (Fig. 6–1). Convection heat transfer is complicated by the fact that it involves fluid motion as well as heat conduction. The fluid motion enhances heat transfer, since it brings hotter and cooler chunks of fluid into contact, initiating higher rates of conduction at a greater number of sites in a fluid. Therefore, the rate of heat transfer through a fluid is much higher by convection than it is by conduction. In fact, the higher the fluid velocity, the higher the rate of heat transfer. To clarify this point further, consider steady heat transfer through a fluid contained between two parallel plates maintained at different temperatures, as shown in Figure 6–2. The temperatures of the fluid and the plate will be the same at the points of contact because of the continuity of temperature. Assuming no fluid motion, the energy of the hotter fluid molecules near the hot plate will be transferred to the adjacent cooler fluid molecules. This energy will then be transferred to the next layer of the cooler fluid molecules. This energy will then be transferred to the next layer of the cooler fluid, and so on, until it is finally transferred to the other plate. This is what happens during conduction through a fluid. Now let us use a syringe to draw some fluid near the hot plate and inject it near the cold plate repeatedly. You can imagine that this will speed up the heat transfer process considerably, since some energy is carried to the other side as a result of fluid motion. Consider the cooling of a hot iron block with a fan blowing air over its top surface, as shown in Figure 6–3. We know that heat will be transferred from the hot block to the surrounding cooler air, and the block will eventually cool. We also know that the block will cool faster if the fan is switched to a higher speed. Replacing air by water will enhance the convection heat transfer even more. Experience shows that convection heat transfer strongly depends on the fluid properties dynamic viscosity , thermal conductivity k, density , and specific heat Cp, as well as the fluid velocity . It also depends on the geometry and the roughness of the solid surface, in addition to the type of fluid flow (such as being streamlined or turbulent). Thus, we expect the convection heat transfer relations to be rather complex because of the dependence of convection on so many variables. This is not surprising, since convection is the most complex mechanism of heat transfer. Despite the complexity of convection, the rate of convection heat transfer is observed to be proportional to the temperature difference and is conveniently expressed by Newton’s law of cooling as cen58933_ch06.qxd 9/4/2002 12:05 PM Page 335 335 CHAPTER 6 q·conv h(Ts T) (W/m2) (6-1) (W) (6-2) Velocity profile T = 15°C = 3 m/s or · Q conv hAs(Ts T) · Qconv where · Qcond h convection heat transfer coefficient, W/m2 ˚C As heat transfer surface area, m2 Ts temperature of the surface, ˚C T temperature of the fluid sufficiently far from the surface, ˚C Hot iron block 400°C Judging from its units, the convection heat transfer coefficient h can be defined as the rate of heat transfer between a solid surface and a fluid per unit surface area per unit temperature difference. You should not be deceived by the simple appearance of this relation, because the convection heat transfer coefficient h depends on the several of the mentioned variables, and thus is difficult to determine. When a fluid is forced to flow over a solid surface that is nonporous (i.e., impermeable to the fluid), it is observed that the fluid in motion comes to a complete stop at the surface and assumes a zero velocity relative to the surface. That is, the fluid layer in direct contact with a solid surface “sticks” to the surface and there is no slip. In fluid flow, this phenomenon is known as the no-slip condition, and it is due to the viscosity of the fluid (Fig. 6–4). The no-slip condition is responsible for the development of the velocity profile for flow. Because of the friction between the fluid layers, the layer that sticks to the wall slows the adjacent fluid layer, which slows the next layer, and so on. A consequence of the no-slip condition is that all velocity profiles must have zero values at the points of contact between a fluid and a solid. The only exception to the no-slip condition occurs in extremely rarified gases. A similar phenomenon occurs for the temperature. When two bodies at different temperatures are brought into contact, heat transfer occurs until both bodies assume the same temperature at the point of contact. Therefore, a fluid and a solid surface will have the same temperature at the point of contact. This is known as no-temperature-jump condition. An implication of the no-slip and the no-temperature jump conditions is that heat transfer from the solid surface to the fluid layer adjacent to the surface is by pure conduction, since the fluid layer is motionless, and can be expressed as T ` q·conv q·cond kfluid y y0 (W/m2) (6-3) where T represents the temperature distribution in the fluid and (T/y)y0 is the temperature gradient at the surface. This heat is then convected away from the surface as a result of fluid motion. Note that convection heat transfer from a solid surface to a fluid is merely the conduction heat transfer from the solid surface to the fluid layer adjacent to the surface. Therefore, we can equate Eqs. 6-1 and 6-3 for the heat flux to obtain FIGURE 6–3 The cooling of a hot block by forced convection. Uniform approach velocity, Relative velocities of fluid layers Zero velocity at the surface Solid block FIGURE 6–4 A fluid flowing over a stationary surface comes to a complete stop at the surface because of the no-slip condition. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 336 336 HEAT TRANSFER h kfluid(T/y)y0 Ts T (W/m2 ˚C) (6-4) for the determination of the convection heat transfer coefficient when the temperature distribution within the fluid is known. The convection heat transfer coefficient, in general, varies along the flow (or x-) direction. The average or mean convection heat transfer coefficient for a surface in such cases is determined by properly averaging the local convection heat transfer coefficients over the entire surface. Nusselt Number In convection studies, it is common practice to nondimensionalize the governing equations and combine the variables, which group together into dimensionless numbers in order to reduce the number of total variables. It is also common practice to nondimensionalize the heat transfer coefficient h with the Nusselt number, defined as Nu T2 Fluid layer · Q L T1 ∆T = T2 – T1 FIGURE 6–5 Heat transfer through a fluid layer of thickness L and temperature difference T. hLc k (6-5) where k is the thermal conductivity of the fluid and Lc is the characteristic length. The Nusselt number is named after Wilhelm Nusselt, who made significant contributions to convective heat transfer in the first half of the twentieth century, and it is viewed as the dimensionless convection heat transfer coefficient. To understand the physical significance of the Nusselt number, consider a fluid layer of thickness L and temperature difference T T2 T1, as shown in Fig. 6–5. Heat transfer through the fluid layer will be by convection when the fluid involves some motion and by conduction when the fluid layer is motionless. Heat flux (the rate of heat transfer per unit time per unit surface area) in either case will be q·conv h T (6-6) T q·cond k L (6-7) q˙conv hL h T Nu q˙cond k T/L k (6-8) and Blowing on food FIGURE 6–6 We resort to forced convection whenever we need to increase the rate of heat transfer. Taking their ratio gives which is the Nusselt number. Therefore, the Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer. The larger the Nusselt number, the more effective the convection. A Nusselt number of Nu 1 for a fluid layer represents heat transfer across the layer by pure conduction. We use forced convection in daily life more often than you might think (Fig. 6–6). We resort to forced convection whenever we want to increase the cen58933_ch06.qxd 9/4/2002 12:05 PM Page 337 337 CHAPTER 6 rate of heat transfer from a hot object. For example, we turn on the fan on hot summer days to help our body cool more effectively. The higher the fan speed, the better we feel. We stir our soup and blow on a hot slice of pizza to make them cool faster. The air on windy winter days feels much colder than it actually is. The simplest solution to heating problems in electronics packaging is to use a large enough fan. 6–2 ■ CLASSIFICATION OF FLUID FLOWS Convection heat transfer is closely tied with fluid mechanics, which is the science that deals with the behavior of fluids at rest or in motion, and the interaction of fluids with solids or other fluids at the boundaries. There are a wide variety of fluid flow problems encountered in practice, and it is usually convenient to classify them on the basis of some common characteristics to make it feasible to study them in groups. There are many ways to classify the fluid flow problems, and below we present some general categories. Viscous versus Inviscid Flow When two fluid layers move relative to each other, a friction force develops between them and the slower layer tries to slow down the faster layer. This internal resistance to flow is called the viscosity, which is a measure of internal stickiness of the fluid. Viscosity is caused by cohesive forces between the molecules in liquids, and by the molecular collisions in gases. There is no fluid with zero viscosity, and thus all fluid flows involve viscous effects to some degree. Flows in which the effects of viscosity are significant are called viscous flows. The effects of viscosity are very small in some flows, and neglecting those effects greatly simplifies the analysis without much loss in accuracy. Such idealized flows of zero-viscosity fluids are called frictionless or inviscid flows. Internal versus External Flow A fluid flow is classified as being internal and external, depending on whether the fluid is forced to flow in a confined channel or over a surface. The flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe is external flow. The flow in a pipe or duct is internal flow if the fluid is completely bounded by solid surfaces. Water flow in a pipe, for example, is internal flow, and air flow over an exposed pipe during a windy day is external flow (Fig. 6–7). The flow of liquids in a pipe is called open-channel flow if the pipe is partially filled with the liquid and there is a free surface. The flow of water in rivers and irrigation ditches are examples of such flows. Compressible versus Incompressible Flow A fluid flow is classified as being compressible or incompressible, depending on the density variation of the fluid during flow. The densities of liquids are essentially constant, and thus the flow of liquids is typically incompressible. Therefore, liquids are usually classified as incompressible substances. A pressure of 210 atm, for example, will cause the density of liquid water at 1 atm to change by just 1 percent. Gases, on the other hand, are highly compressible. A External flow Water Air Internal flow FIGURE 6–7 Internal flow of water in a pipe and the external flow of air over the same pipe. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 338 338 HEAT TRANSFER pressure change of just 0.01 atm, for example, will cause a change of 1 percent in the density of atmospheric air. However, gas flows can be treated as incompressible if the density changes are under about 5 percent, which is usually the case when the flow velocity is less than 30 percent of the velocity of sound in that gas (i.e., the Mach number of flow is less than 0.3). The velocity of sound in air at room temperature is 346 m/s. Therefore, the compressibility effects of air can be neglected at speeds under 100 m/s. Note that the flow of a gas is not necessarily a compressible flow. Laminar versus Turbulent Flow Some flows are smooth and orderly while others are rather chaotic. The highly ordered fluid motion characterized by smooth streamlines is called laminar. The flow of high-viscosity fluids such as oils at low velocities is typically laminar. The highly disordered fluid motion that typically occurs at high velocities characterized by velocity fluctuations is called turbulent. The flow of low-viscosity fluids such as air at high velocities is typically turbulent. The flow regime greatly influences the heat transfer rates and the required power for pumping. Natural (or Unforced) versus Forced Flow Cold water Hot water Light hot water rising Hot water Hot water storage tank (above the top of collectors) Steady versus Unsteady (Transient) Flow The terms steady and uniform are used frequently in engineering, and thus it is important to have a clear understanding of their meanings. The term steady implies no change with time. The opposite of steady is unsteady, or transient. The term uniform, however, implies no change with location over a specified region. Many devices such as turbines, compressors, boilers, condensers, and heat exchangers operate for long periods of time under the same conditions, and they are classified as steady-flow devices. During steady flow, the fluid properties can change from point to point within a device, but at any fixed point they remain constant. So lar co ll ect So lar ors rad iat i on Cold water A fluid flow is said to be natural or forced, depending on how the fluid motion is initiated. In forced flow, a fluid is forced to flow over a surface or in a pipe by external means such as a pump or a fan. In natural flows, any fluid motion is due to a natural means such as the buoyancy effect, which manifests itself as the rise of the warmer (and thus lighter) fluid and the fall of cooler (and thus denser) fluid. This thermosiphoning effect is commonly used to replace pumps in solar water heating systems by placing the water tank sufficiently above the solar collectors (Fig. 6–8). One-, Two-, and Three-Dimensional Flows Dense cold water sinking A flow field is best characterized by the velocity distribution, and thus a flow is said to be one-, two-, or three-dimensional if the flow velocity varies in one, two, or three primary dimensions, respectively. A typical fluid flow involves a three-dimensional geometry and the velocity may vary in all three dimensions rendering the flow three-dimensional [ (x, y, z) in rectangular or (r, , z) in cylindrical coordinates]. However, the variation of velocity in FIGURE 6–8 Natural circulation of water in a solar water heater by thermosiphoning. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 339 339 CHAPTER 6 certain direction can be small relative to the variation in other directions, and can be ignored with negligible error. In such cases, the flow can be modeled conveniently as being one- or two-dimensional, which is easier to analyze. When the entrance effects are disregarded, fluid flow in a circular pipe is one-dimensional since the velocity varies in the radial r direction but not in the angular - or axial z-directions (Fig. 6–9). That is, the velocity profile is the same at any axial z-location, and it is symmetric about the axis of the pipe. Note that even in this simplest flow, the velocity cannot be uniform across the cross section of the pipe because of the no-slip condition. However, for convenience in calculations, the velocity can be assumed to be constant and thus uniform at a cross section. Fluid flow in a pipe usually approximated as onedimensional uniform flow. 6–3 ■ velocity profile (remains unchanged) r (r) z FIGURE 6–9 One-dimensional flow in a circular pipe. VELOCITY BOUNDARY LAYER Consider the parallel flow of a fluid over a flat plate, as shown in Fig. 6–10. Surfaces that are slightly contoured such as turbine blades can also be approximated as flat plates with reasonable accuracy. The x-coordinate is measured along the plate surface from the leading edge of the plate in the direction of the flow, and y is measured from the surface in the normal direction. The fluid approaches the plate in the x-direction with a uniform upstream velocity of , which is practically identical to the free-stream velocity u over the plate away from the surface (this would not be the case for cross flow over blunt bodies such as a cylinder). For the sake of discussion, we can consider the fluid to consist of adjacent layers piled on top of each other. The velocity of the particles in the first fluid layer adjacent to the plate becomes zero because of the no-slip condition. This motionless layer slows down the particles of the neighboring fluid layer as a result of friction between the particles of these two adjoining fluid layers at different velocities. This fluid layer then slows down the molecules of the next layer, and so on. Thus, the presence of the plate is felt up to some normal distance from the plate beyond which the free-stream velocity u remains essentially unchanged. As a result, the x-component of the fluid velocity, u, will vary from 0 at y 0 to nearly u at y (Fig. 6–11). Laminar boundary layer Turbulent boundary layer u u y 0 Transition region Turbulent layer Buffer layer Laminar sublayer x xcr Boundary-layer thickness, δ FIGURE 6–10 The development of the boundary layer for flow over a flat plate, and the different flow regimes. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 340 340 HEAT TRANSFER Relative velocities of fluid layers u 0.99 δ Zero velocity at the surface FIGURE 6–11 The development of a boundary layer on a surface is due to the no-slip condition. Viscosity The region of the flow above the plate bounded by in which the effects of the viscous shearing forces caused by fluid viscosity are felt is called the velocity boundary layer. The boundary layer thickness, , is typically defined as the distance y from the surface at which u 0.99u. The hypothetical line of u 0.99u divides the flow over a plate into two regions: the boundary layer region, in which the viscous effects and the velocity changes are significant, and the inviscid flow region, in which the frictional effects are negligible and the velocity remains essentially constant. Surface Shear Stress Consider the flow of a fluid over the surface of a plate. The fluid layer in contact with the surface will try to drag the plate along via friction, exerting a friction force on it. Likewise, a faster fluid layer will try to drag the adjacent slower layer and exert a friction force because of the friction between the two layers. Friction force per unit area is called shear stress, and is denoted by . Experimental studies indicate that the shear stress for most fluids is proportional to the velocity gradient, and the shear stress at the wall surface is as s Liquids Gases Temperature FIGURE 6–12 The viscosity of liquids decreases and the viscosity of gases increases with temperature. u ` y y0 (N/m2) (6-9) where the constant of proportionality is called the dynamic viscosity of the fluid, whose unit is kg/m s (or equivalently, N s/m2, or Pa s, or poise 0.1 Pa s). The fluids that that obey the linear relationship above are called Newtonian fluids, after Sir Isaac Newton who expressed it first in 1687. Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Blood and liquid plastics are examples of non-Newtonian fluids. In this text we will consider Newtonian fluids only. In fluid flow and heat transfer studies, the ratio of dynamic viscosity to density appears frequently. For convenience, this ratio is given the name kinematic viscosity and is expressed as /. Two common units of kinematic viscosity are m2/s and stoke (1 stoke 1 cm2/s 0.0001 m2/s). The viscosity of a fluid is a measure of its resistance to flow, and it is a strong function of temperature. The viscosities of liquids decrease with temperature, whereas the viscosities of gases increase with temperature (Fig. 6–12). The viscosities of some fluids at 20˚C are listed in Table 6–1. Note that the viscosities of different fluids differ by several orders of magnitude. The determination of the surface shear stress s from Eq. 6-9 is not practical since it requires a knowledge of the flow velocity profile. A more practical approach in external flow is to relate s to the upstream velocity as s Cf 2 2 (N/m2) (6-10) where Cf is the dimensionless friction coefficient, whose value in most cases is determined experimentally, and is the density of the fluid. Note that the friction coefficient, in general, will vary with location along the surface. Once the average friction coefficient over a given surface is available, the friction force over the entire surface is determined from cen58933_ch06.qxd 9/4/2002 12:05 PM Page 341 341 CHAPTER 6 Ff Cf As 2 2 (N) (6-11) where As is the surface area. The friction coefficient is an important parameter in heat transfer studies since it is directly related to the heat transfer coefficient and the power requirements of the pump or fan. 6–4 ■ THERMAL BOUNDARY LAYER We have seen that a velocity boundary layer develops when a fluid flows over a surface as a result of the fluid layer adjacent to the surface assuming the surface velocity (i.e., zero velocity relative to the surface). Also, we defined the velocity boundary layer as the region in which the fluid velocity varies from zero to 0.99u. Likewise, a thermal boundary layer develops when a fluid at a specified temperature flows over a surface that is at a different temperature, as shown in Fig. 6–13. Consider the flow of a fluid at a uniform temperature of T over an isothermal flat plate at temperature Ts. The fluid particles in the layer adjacent to the surface will reach thermal equilibrium with the plate and assume the surface temperature Ts. These fluid particles will then exchange energy with the particles in the adjoining-fluid layer, and so on. As a result, a temperature profile will develop in the flow field that ranges from Ts at the surface to T sufficiently far from the surface. The flow region over the surface in which the temperature variation in the direction normal to the surface is significant is the thermal boundary layer. The thickness of the thermal boundary layer t at any location along the surface is defined as the distance from the surface at which the temperature difference T Ts equals 0.99(T Ts). Note that for the special case of Ts 0, we have T 0.99T at the outer edge of the thermal boundary layer, which is analogous to u 0.99u for the velocity boundary layer. The thickness of the thermal boundary layer increases in the flow direction, since the effects of heat transfer are felt at greater distances from the surface further down stream. The convection heat transfer rate anywhere along the surface is directly related to the temperature gradient at that location. Therefore, the shape of the temperature profile in the thermal boundary layer dictates the convection heat transfer between a solid surface and the fluid flowing over it. In flow over a heated (or cooled) surface, both velocity and thermal boundary layers will develop simultaneously. Noting that the fluid velocity will have a strong influence on the temperature profile, the development of the velocity boundary layer relative to the thermal boundary layer will have a strong effect on the convection heat transfer. Prandtl Number The relative thickness of the velocity and the thermal boundary layers is best described by the dimensionless parameter Prandtl number, defined as Pr Molecular diffusivity of momentum Cp Molecular diffusivity of heat k (6-12) TABLE 6–1 Dynamic viscosities of some fluids at 1 atm and 20˚C (unless otherwise stated) Dynamic viscosity , kg/m s Fluid Glycerin: 20˚C 0˚C 20˚C 40˚C Engine oil: SAE 10W SAE 10W30 SAE 30 SAE 50 Mercury Ethyl alcohol Water: 0˚C 20˚C 100˚C (liquid) 100˚C (vapor) Blood, 37˚C Gasoline Ammonia Air Hydrogen, 0˚C T 134.0 12.1 1.49 0.27 0.10 0.17 0.29 0.86 0.0015 0.0012 0.0018 0.0010 0.0003 0.000013 0.0004 0.00029 0.00022 0.000018 0.000009 Free-stream T T x δt Thermal boundary Ts layer Ts + 0.99(T – Ts ) FIGURE 6–13 Thermal boundary layer on a flat plate (the fluid is hotter than the plate surface). cen58933_ch06.qxd 9/4/2002 12:05 PM Page 342 342 HEAT TRANSFER TABLE 6–2 Typical ranges of Prandtl numbers for common fluids Fluid Liquid metals Gases Water Light organic fluids Oils Glycerin Pr 0.004–0.030 0.7–1.0 1.7–13.7 5–50 50–100,000 2000–100,000 It is named after Ludwig Prandtl, who introduced the concept of boundary layer in 1904 and made significant contributions to boundary layer theory. The Prandtl numbers of fluids range from less than 0.01 for liquid metals to more than 100,000 for heavy oils (Table 6–2). Note that the Prandtl number is in the order of 10 for water. The Prandtl numbers of gases are about 1, which indicates that both momentum and heat dissipate through the fluid at about the same rate. Heat diffuses very quickly in liquid metals (Pr 1) and very slowly in oils (Pr 1) relative to momentum. Consequently the thermal boundary layer is much thicker for liquid metals and much thinner for oils relative to the velocity boundary layer. 6–5 Smoke Turbulent flow Laminar flow FIGURE 6–14 Laminar and turbulent flow regimes of cigarette smoke. (a) Laminar flow Die trace (b) Turbulent flow FIGURE 6–15 The behavior of colored fluid injected into the flow in laminar and turbulent flows in a tube. ■ LAMINAR AND TURBULENT FLOWS If you have been around smokers, you probably noticed that the cigarette smoke rises in a smooth plume for the first few centimeters and then starts fluctuating randomly in all directions as it continues its journey toward the lungs of others (Fig. 6–14). Likewise, a careful inspection of flow in a pipe reveals that the fluid flow is streamlined at low velocities but turns chaotic as the velocity is increased above a critical value, as shown in Figure 6–15. The flow regime in the first case is said to be laminar, characterized by smooth streamlines and highly-ordered motion, and turbulent in the second case, where it is characterized by velocity fluctuations and highly-disordered motion. The transition from laminar to turbulent flow does not occur suddenly; rather, it occurs over some region in which the flow fluctuates between laminar and turbulent flows before it becomes fully turbulent. We can verify the existence of these laminar, transition, and turbulent flow regimes by injecting some dye streak into the flow in a glass tube, as the British scientist Osborn Reynolds (1842–1912) did over a century ago. We will observe that the dye streak will form a straight and smooth line at low velocities when the flow is laminar (we may see some blurring because of molecular diffusion), will have bursts of fluctuations in the transition regime, and will zigzag rapidly and randomly when the flow becomes fully turbulent. These zigzags and the dispersion of the dye are indicative of the fluctuations in the main flow and the rapid mixing of fluid particles from adjacent layers. Typical velocity profiles in laminar and turbulent flow are also given in Figure 6–10. Note that the velocity profile is approximately parabolic in laminar flow and becomes flatter in turbulent flow, with a sharp drop near the surface. The turbulent boundary layer can be considered to consist of three layers. The very thin layer next to the wall where the viscous effects are dominant is the laminar sublayer. The velocity profile in this layer is nearly linear, and the flow is streamlined. Next to the laminar sublayer is the buffer layer, in which the turbulent effects are significant but not dominant of the diffusion effects, and next to it is the turbulent layer, in which the turbulent effects dominate. The intense mixing of the fluid in turbulent flow as a result of rapid fluctuations enhances heat and momentum transfer between fluid particles, which increases the friction force on the surface and the convection heat transfer rate. It also causes the boundary layer to enlarge. Both the friction and heat transfer coefficients reach maximum values when the flow becomes fully turbulent. So it will come as no surprise that a special effort is made in the design cen58933_ch06.qxd 9/4/2002 12:05 PM Page 343 343 CHAPTER 6 of heat transfer coefficients associated with turbulent flow. The enhancement in heat transfer in turbulent flow does not come for free, however. It may be necessary to use a larger pump to overcome the larger friction forces accompanying the higher heat transfer rate. Reynolds Number The transition from laminar to turbulent flow depends on the surface geometry, surface roughness, free-stream velocity, surface temperature, and type of fluid, among other things. After exhaustive experiments in the 1880s, Osborn Reynolds discovered that the flow regime depends mainly on the ratio of the inertia forces to viscous forces in the fluid. This ratio is called the Reynolds number, which is a dimensionless quantity, and is expressed for external flow as (Fig. 6–16) Re Inertia forces Lc Lc Viscous (6-13) where is the upstream velocity (equivalent to the free-stream velocity u for a flat plate), Lc is the characteristic length of the geometry, and / is the kinematic viscosity of the fluid. For a flat plate, the characteristic length is the distance x from the leading edge. Note that kinematic viscosity has the unit m2/s, which is identical to the unit of thermal diffusivity, and can be viewed as viscous diffusivity or diffusivity for momentum. At large Reynolds numbers, the inertia forces, which are proportional to the density and the velocity of the fluid, are large relative to the viscous forces, and thus the viscous forces cannot prevent the random and rapid fluctuations of the fluid. At small Reynolds numbers, however, the viscous forces are large enough to overcome the inertia forces and to keep the fluid “in line.” Thus the flow is turbulent in the first case and laminar in the second. The Reynolds number at which the flow becomes turbulent is called the critical Reynolds number. The value of the critical Reynolds number is different for different geometries. For flow over a flat plate, the generally accepted value of the critical Reynolds number is Recr xcr/ uxcr/ 5 105, where xcr is the distance from the leading edge of the plate at which transition from laminar to turbulent flow occurs. The value of Recr may change substantially, however, depending on the level of turbulence in the free stream. 6–6 ■ Inertia forces Re = –––––––––––– Viscous forces HEAT AND MOMENTUM TRANSFER IN TURBULENT FLOW Most flows encountered in engineering practice are turbulent, and thus it is important to understand how turbulence affects wall shear stress and heat transfer. Turbulent flow is characterized by random and rapid fluctuations of groups of fluid particles, called eddies, throughout the boundary layer. These fluctuations provide an additional mechanism for momentum and heat transfer. In laminar flow, fluid particles flow in an orderly manner along streamlines, and both momentum and heat are transferred across streamlines by molecular diffusion. In turbulent flow, the transverse motion of eddies transport momentum and heat to other regions of flow before they mix with the rest of the fluid and lose their identity, greatly enhancing momentum and heat L ρ 2/L = –––––– µ/L2 ρ L = –––– µ L = ––– ν FIGURE 6–16 The Reynolds number can be viewed as the ratio of the inertia forces to viscous forces acting on a fluid volume element. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 344 344 HEAT TRANSFER 2 2 2 2 2 5 5 5 5 5 7 7 7 7 7 12 12 12 12 12 (a) Before turbulence 12 2 5 7 5 2 5 7 2 12 7 12 7 5 12 2 7 5 12 2 (b) After turbulence FIGURE 6–17 The intense mixing in turbulent flow brings fluid particles at different temperatures into close contact, and thus enhances heat transfer. u u– u' Time, t FIGURE 6–18 Fluctuations of the velocity component u with time at a specified location in turbulent flow. transfer. As a result, turbulent flow is associated with much higher values of friction and heat transfer coefficients (Fig. 6–17). Even when the mean flow is steady, the eddying motion in turbulent flow causes significant fluctuations in the values of velocity, temperature, pressure, and even density (in compressible flow). Figure 6–18 shows the variation of the instantaneous velocity component u with time at a specified location, as can be measured with a hot-wire anemometer probe or other sensitive device. We observe that the instantaneous values of the velocity fluctuate about a mean value, _which suggests that the velocity can be expressed as the sum of a mean value u and a fluctuating component u, _ u u u (6-14) This is also the case for other_ properties such as the velocity _ _ component v in the y direction, and thus v v v, P P P, and T T T. The mean value of a property at some location is determined by averaging it over a time interval that is sufficiently large so that the net effect of fluctuations _is zero. Therefore, the time average of fluctuating components is zero, e.g., u 0. _ The magnitude of u is usually just a few percent of u , but the high frequencies of eddies (in the order of a thousand per second) makes them very effective for the transport of momentum and thermal energy. In steady turbulent flow, the mean values of properties (indicated by an overbar) are independent of time. Consider the upward eddy motion of a fluid during flow over a surface. The mass flow rate of fluid per unit area normal to flow is v. Noting that h CpT represents the energy of the fluid and T is the eddy temperature relative to the mean value, the rate of thermal energy transport by turbulent eddies is q· t CpvT. By a similar argument on momentum transfer, the turbulent shear stress can be shown to be t uv. Note that uv 0 even though u 0 and v 0, and experimental results show that uv is a negative quantity. Terms such as uv are called Reynolds stresses. The random eddy motion of groups of particles resembles the random motion of molecules in a gas—colliding with each other after traveling a certain distance and exchanging momentum and heat in the process. Therefore, momentum and heat transport by eddies in turbulent boundary layers is analogous to the molecular momentum and heat diffusion. Then turbulent wall shear stress and turbulent heat transfer can be expressed in an analogous manner as _ t uv t u y and q˙t Cp vT kt T y (6-15) where t is called the turbulent viscosity, which accounts for momentum transport by turbulent eddies, and kt is called the turbulent thermal conductivity, which accounts for thermal energy transport by turbulent eddies. Then the total shear stress and total heat flux can be expressed conveniently as _ _ u u ( M) y y (6-16) T T Cp( H) y y (6-17) total ( t) and q˙total (k kt) cen58933_ch06.qxd 9/4/2002 12:05 PM Page 345 345 CHAPTER 6 where M t/ is the eddy diffusivity of momentum and H kt/Cp is the eddy diffusivity of heat. Eddy motion and thus eddy diffusivities are much larger than their molecular counterparts in the core region of a turbulent boundary layer. The eddy motion loses its intensity close to the wall, and diminishes at the wall because of the no-slip condition. Therefore, the velocity and temperature profiles are nearly uniform in the core region of a turbulent boundary layer, but very steep in the thin layer adjacent to the wall, resulting in large velocity and temperature gradients at the wall surface. So it is no surprise that the wall shear stress and wall heat flux are much larger in turbulent flow than they are in laminar flow (Fig. 6–19). Note that molecular diffusivities and (as well as and k) are fluid properties, and their values can be found listed in fluid handbooks. Eddy diffusivities M and H (as well as t and kt), however are not fluid properties and their values depend on flow conditions. Eddy diffusivities M and H decrease towards the wall, becoming zero at the wall. 6–7 ■ DERIVATION OF DIFFERENTIAL CONVECTION EQUATIONS T or u ∂u ∂y ∂y y0 y0 Laminar T or u ∂u– ∂y ∂T or ∂y y0 y0 Turbulent In this section we derive the governing equations of fluid flow in the boundary layers. To keep the analysis at a manageable level, we assume the flow to be steady and two-dimensional, and the fluid to be Newtonian with constant properties (density, viscosity, thermal conductivity, etc.). Consider the parallel flow of a fluid over a surface. We take the flow direction along the surface to be x and the direction normal to the surface to be y, and we choose a differential volume element of length dx, height dy, and unit depth in the z-direction (normal to the paper) for analysis (Fig. 6–20). The fluid flows over the surface with a uniform free-stream velocity u, but the velocity within boundary layer is two-dimensional: the x-component of the velocity is u, and the y-component is v. Note that u u(x, y) and v v(x, y) in steady two-dimensional flow. Next we apply three fundamental laws to this fluid element: Conservation of mass, conservation of momentum, and conservation of energy to obtain the continuity, momentum, and energy equations for laminar flow in boundary layers. FIGURE 6–19 The velocity and temperature gradients at the wall, and thus the wall shear stress and heat transfer rate, are much larger for turbulent flow than they are for laminar flow (T is shown relative to Ts). T u y x Velocity boundary layer dy dx v Conservation of Mass Equation The conservation of mass principle is simply a statement that mass cannot be created or destroyed, and all the mass must be accounted for during an analysis. In steady flow, the amount of mass within the control volume remains constant, and thus the conservation of mass can be expressed as of mass flow Rate of mass flow intoRate the control volume out of the control volume ∂T or This and the upcoming sections of this chapter deal with theoretical aspects of convection, and can be skipped and be used as a reference if desired without a loss in continuity. ∂y dy dy u u x, y (6-18) ∂v ∂u ∂x dx dx v FIGURE 6–20 Differential control volume used in the derivation of mass balance in velocity boundary layer in two-dimensional flow over a surface. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 346 346 HEAT TRANSFER Noting that mass flow rate is equal to the product of density, mean velocity, and cross-sectional area normal to flow, the rate at which fluid enters the control volume from the left surface is u(dy 1). The rate at which the fluid leaves the control volume from the right surface can be expressed as u u dx (dy 1) x (6-19) Repeating this for the y direction and substituting the results into Eq. 6-18, we obtain u(dy 1) v(dx 1) u v u dx (dy 1) dy (dx 1) (6-20) x y Simplifying and dividing by dx dy 1 gives u v 0 x y (6-21) This is the conservation of mass relation, also known as the continuity equation, or mass balance for steady two-dimensional flow of a fluid with constant density. Conservation of Momentum Equations The differential forms of the equations of motion in the velocity boundary layer are obtained by applying Newton’s second law of motion to a differential control volume element in the boundary layer. Newton’s second law is an expression for the conservation of momentum, and can be stated as the net force acting on the control volume is equal to the mass times the acceleration of the fluid element within the control volume, which is also equal to the net rate of momentum outflow from the control volume. The forces acting on the control volume consist of body forces that act throughout the entire body of the control volume (such as gravity, electric, and magnetic forces) and are proportional to the volume of the body, and surface forces that act on the control surface (such as the pressure forces due to hydrostatic pressure and shear stresses due to viscous effects) and are proportional to the surface area. The surface forces appear as the control volume is isolated from its surroundings for analysis, and the effect of the detached body is replaced by a force at that location. Note that pressure represents the compressive force applied on the fluid element by the surrounding fluid, and is always directed to the surface. We express Newton’s second law of motion for the control volume as (Mass) Acceleration Net force (body and surface) in a specified direction acting in that direction (6-22) or m ax Fsurface, x Fbody, x (6-23) where the mass of the fluid element within the control volume is m (dx dy 1) (6-24) cen58933_ch06.qxd 9/4/2002 12:05 PM Page 347 347 CHAPTER 6 Noting that flow is steady and two-dimensional and thus u u(x, y), the total differential of u is du u u dx dy x y (6-25) Then the acceleration of the fluid element in the x direction becomes ax du u dx u dy u u u v dt x dt y dt x y (6-26) You may be tempted to think that acceleration is zero in steady flow since acceleration is the rate of change of velocity with time, and in steady flow there is no change with time. Well, a garden hose nozzle will tell us that this understanding is not correct. Even in steady flow and thus constant mass flow rate, water will accelerate through the nozzle (Fig. 6–21). Steady simply means no change with time at a specified location (and thus u/t 0), but the value of a quantity may change from one location to another (and thus u/x and u/y may be different from zero). In the case of a nozzle, the velocity of water remains constant at a specified point, but it changes from inlet to the exit (water accelerates along the nozzle, which is the reason for attaching a nozzle to the garden hose in the first place). The forces acting on a surface are due to pressure and viscous effects. In two-dimensional flow, the viscous stress at any point on an imaginary surface within the fluid can be resolved into two perpendicular components: one normal to the surface called normal stress (which should not be confused with pressure) and another along the surface called shear stress. The normal stress is related to the velocity gradients u/x and v/y, that are much smaller than u/y, to which shear stress is related. Neglecting the normal stresses for simplicity, the surface forces acting on the control volume in the x-direction will be as shown in Fig. 6–22. Then the net surface force acting in the x-direction becomes P dx (dy 1) (dx dy 1) y dy(dx 1) P x y x u P (dx dy 1) x y Fsurface, x Garden hose nozzle Water FIGURE 6–21 During steady flow, a fluid may not accelerate in time at a fixed point, but it may accelerate in space. τ (6-27) since (u/y). Substituting Eqs. 6-21, 6-23, and 6-24 into Eq. 6-20 and dividing by dx dy 1 gives u u u u P v 2 x y x y 2 (6-28) This is the relation for the conservation of momentum in the x-direction, and is known as the x-momentum equation. Note that we would obtain the same result if we used momentum flow rates for the left-hand side of this equation instead of mass times acceleration. If there is a body force acting in the x-direction, it can be added to the right side of the equation provided that it is expressed per unit volume of the fluid. In a boundary layer, the velocity component in the flow direction is much larger than that in the normal direction, and thus u v, and v/x and v/y are ∂y x, y τ dy dy Differential control volume P 2 2 ∂τ P ∂P ∂x dx dx FIGURE 6–22 Differential control volume used in the derivation of x-momentum equation in velocity boundary layer in twodimensional flow over a surface. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 348 348 HEAT TRANSFER T u y x 1) 2) 3) v u Velocity components: vu Velocity grandients: v v 0, 0 x y u u x y Temperature gradients: T T x y FIGURE 6–23 Boundary layer approximations. negligible. Also, u varies greatly with y in the normal direction from zero at the wall surface to nearly the free-stream value across the relatively thin boundary layer, while the variation of u with x along the flow is typically small. Therefore, u/y u/x. Similarly, if the fluid and the wall are at different temperatures and the fluid is heated or cooled during flow, heat conduction will occur primarily in the direction normal to the surface, and thus T/y T/x. That is, the velocity and temperature gradients normal to the surface are much greater than those along the surface. These simplifications are known as the boundary layer approximations. These approximations greatly simplify the analysis usually with little loss in accuracy, and make it possible to obtain analytical solutions for certain types of flow problems (Fig. 6–23). When gravity effects and other body forces are negligible and the boundary layer approximations are valid, applying Newton’s second law of motion on the volume element in the y-direction gives the y-momentum equation to be P 0 y (6-29) That is, the variation of pressure in the direction normal to the surface is negligible, and thus P P(x) and P/x dP/dx. Then it follows that for a given x, the pressure in the boundary layer is equal to the pressure in the free stream, and the pressure determined by a separate analysis of fluid flow in the free stream (which is typically easier because of the absence of viscous effects) can readily be used in the boundary layer analysis. The velocity components in the free stream region of a flat plate are u u constant and v 0. Substituting these into the x-momentum equations (Eq. 6-28) gives P/x 0. Therefore, for flow over a flat plate, the pressure remains constant over the entire plate (both inside and outside the boundary layer). Conservation of Energy Equation The energy balance for any system undergoing any process is expressed as Ein Eout Esystem, which states that the change in the energy content of a system during a process is equal to the difference between the energy input and the energy output. During a steady-flow process, the total energy content of a control volume remains constant (and thus Esystem 0), and the amount of energy entering a control volume in all forms must be equal to the amount of energy leaving it. Then the rate form of the general energy equation reduces · · for a steady-flow process to E in E out 0. Noting that energy can be transferred by heat, work, and mass only, the energy balance for a steady-flow control volume can be written explicitly as · · · · · · (Ein Eout)by heat (Ein Eout)by work (Ein Eout )by mass 0 (6-30) The total energy of a flowing fluid stream per unit mass is estream h ke pe where h is the enthalpy (which is the sum of internal energy and flow energy), pe gz is the potential energy, and ke 2/2 (u2 v2)/2 is the kinetic energy of the fluid per unit mass. The kinetic and potential energies are usually very small relative to enthalpy, and therefore it is common practice to neglect them (besides, it can be shown that if kinetic energy is included in the analysis below, all the terms due to this inclusion cancel each other). We cen58933_ch06.qxd 9/4/2002 12:05 PM Page 349 349 CHAPTER 6 assume the density , specific heat Cp, viscosity , and the thermal conductivity k of the fluid to be constant. Then the energy of the fluid per unit mass can be expressed as estream h CpT. Energy is a scalar quantity, and thus energy interactions in all directions can be combined in one equation. Noting that mass flow rate of the fluid entering the control volume from the left is u(dy 1), the rate of energy transfer to the control volume by mass in the x-direction is, from Fig. 6–24, ˙ stream)x ( me ˙ stream)x (E˙in E˙ out)by mass, x ( me ˙ stream)x (me dx x [u(dy 1)Cp T ] u T dx Cp u T dx dy (6-31) x x x Repeating this for the y-direction and adding the results, the net rate of energy transfer to the control volume by mass is determined to be T T v dx dy C u x y v u T T T T dx dy Cp v dx dy (E˙ in E˙ out) by mass Cp u x x y y (6-32) p since u/x v/y 0 from the continuity equation. The net rate of heat conduction to the volume element in the x-direction is Q˙ x x dx T T dx dy k(dy 1) dx k x x x (E˙ in E˙ out ) by heat, x Q˙ x Q˙ x 2 2 (6-33) Repeating this for the y-direction and adding the results, the net rate of energy transfer to the control volume by heat conduction becomes 2T 2T 2T 2T (E˙ in E˙ out ) by heat k 2 dx dy k 2 dx dy k dx dy x y x2 y2 (6-34) Another mechanism of energy transfer to and from the fluid in the control volume is the work done by the body and surface forces. The work done by a body force is determined by multiplying this force by the velocity in the direction of the force and the volume of the fluid element, and this work needs to be considered only in the presence of significant gravitational, electric, or magnetic effects. The surface forces consist of the forces due to fluid pressure and the viscous shear stresses. The work done by pressure (the flow work) is already accounted for in the analysis above by using enthalpy for the microscopic energy of the fluid instead of internal energy. The shear stresses that result from viscous effects are usually very small, and can be neglected in many cases. This is especially the case for applications that involve low or moderate velocities. Then the energy equation for the steady two-dimensional flow of a fluid with constant properties and negligible shear stresses is obtained by substituting Eqs. 6-32 and 6-34 into 6-30 to be Cp u 2T 2T T T v k x y x2 y2 (6-35) Eheat. out. y Emass. out. y dv Eheat in, x Eheat out, x Emass in, x Emass out, y dx Eheat in, y Emass in, y FIGURE 6–24 The energy transfers by heat and mass flow associated with a differential control volume in the thermal boundary layer in steady twodimensional flow. cen58933_ch06.qxd 9/4/2002 12:05 PM Page 350 350 HEAT TRANSFER which states that the net energy convected by the fluid out of the control volume is equal to the net energy transferred into the control volume by heat conduction. When the viscous shear stresses are not negligible, their effect is accounted for by expressing the energy equation as Cp u 2T 2T T T v k 2 2 x y x y (6-36) where the viscous dissipation function is obtained after a lengthy analysis (see an advanced book such as the one by Schlichting (Ref. 9) for details) to be 2 ux vy uy vx 2 2 2 (6-37) Viscous dissipation may play a dominant role in high-speed flows, especially when the viscosity of the fluid is high (like the flow of oil in journal bearings). This manifests itself as a significant rise in fluid temperature due to the conversion of the kinetic energy of the fluid to thermal energy. Viscous dissipation is also significant for high-speed flights of aircraft. For the special case of a stationary fluid, u v 0 and the energy equation reduces, as expected, to the steady two-dimensional heat conduction equation, 2T 2T 0 x2 y2 EXAMPLE 6–1 = 12 m/s L u(y) 0 Temperature Rise of Oil in a Journal Bearing The flow of oil in a journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. Such flows are known as Couette flow. Consider two large isothermal plates separated by 2-mm-thick oil film. The upper plates moves at a constant velocity of 12 m/s, while the lower plate is stationary. Both plates are maintained at 20˚C. (a) Obtain relations for the velocity and temperature distributions in the oil. (b) Determine the maximum temperature in the oil and the heat flux from the oil to each plate (Fig. 6–25). Moving plate y (6-38) x Stationary plate FIGURE 6–25 Schematic for Example 6–1. SOLUTION Parallel flow of oil between two plates is considered. The velocity and temperature distributions, the maximum temperature, and the total heat transfer rate are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. 4 The plates are large so that there is no variation in the z direction. Properties The properties of oil at 20˚C are (Table A-10): k 0.145 W/m K and 0.800 kg/m s 0.800 N s/m2 Analysis (a) We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v 0. Then the continuity equation (Eq. 6-21) reduces to Continuity: u u v 0 → 0 x y x → u u( y) cen58933_ch06.qxd 9/4/2002 12:05 PM Page 351 351 CHAPTER 6 Therefore, the x-component of velocity does not change in the flow direction (i.e., the velocity profile remains unchanged). Noting that u u(y), v 0, and P/x 0 (flow is maintained by the motion of the upper plate rather than the pressure gradient), the x-momentum equation (Eq. 6-28) reduces to x-momentum: u u u 2u P v 2 x y x y d 2u 0 dy 2 → This is a second-order ordinary differential equation, and integrating it twice gives u(y) C1y C2 The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no-slip condition. Therefore, the boundary conditions are u(0) 0 and u(L) , and applying them gives the velocity distribution to be u(y) y L Frictional heating due to viscous dissipation in this case is significant because of the high viscosity of oil and the large plate velocity. The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T T(y). Also, u u(y) and v 0. Then the energy equation with dissipation (Eqs. 6-36 and 6-37) reduce to Energy: 0k u 2T y y 2 2 d 2T k 2 L dy → 2 since u/y /L. Dividing both sides by k and integrating twice give y T( y) 2k L C yC 2 3 4 Applying the boundary conditions T(0) T0 and T(L) T0 gives the temperature distribution to be T( y) T0 2 y y2 2k L L2 (b) The temperature gradient is determined by differentiating T ( y ) with respect to y, 2 y dT 12 L dy 2kL The location of maximum temperature is determined by setting dT/dy 0 and solving for y, 2 y dT 12 0 L dy 2kL → y L 2 Therefore, maximum temperature will occur at mid plane, which is not surprising since both plates are maintained at the same temperature. The maximum temperature is the value of temperature at y L/2, L / 2 (L / 2) T L2 T 2k L L 8k (0.8 N s/m )(12m/s) 1W 20 119°C 8(0.145 W/m °C) 1 N m /s Tmax T 2 2 2 2 2 0 2 0 cen58933_ch06.qxd 9/4/2002 12:05 PM Page 352 352 HEAT TRANSFER Heat flux at the plates is determined from the definition of heat flux, dT 2 2 k (1 0) dy y0 2kL 2L (0.8 N s/m 2 )(12 m/s) 2 1W 28,800 W/m2 2(0.002 m) 1N m/s 2 2 dT q˙0 28,800 W/m2 q˙L k k (1 2) 2L dy yL 2kL q˙0 k Therefore, heat fluxes at the two plates are equal in magnitude but opposite in sign. Discussion A temperature rise of 99˚C confirms our suspicion that viscous dissipation is very significant. Also, the heat flux is equivalent to the rate of mechanical energy dissipation. Therefore, mechanical energy is being converted to thermal energy at a rate of 57.2 kW/m2 of plate area to overcome friction in the oil. Finally, calculations are done using oil properties at 20˚C, but the oil temperature turned out to be much higher. Therefore, knowing the strong dependence of viscosity on temperature, calculations should be repeated using properties at the average temperature of 70˚C to improve accuracy. 6–8 ■ SOLUTIONS OF CONVECTION EQUATIONS FOR A FLAT PLATE Consider laminar flow of a fluid over a flat plate, as shown in Fig. 6–19. Surfaces that are slightly contoured such as turbine blades can also be approximated as flat plates with reasonable accuracy. The x-coordinate is measured along the plate surface from the leading edge of the plate in the direction of the flow, and y is measured from the surface in the normal direction. The fluid approaches the plate in the x-direction with a uniform upstream velocity, which is equivalent to the free stream velocity u. When viscous dissipation is negligible, the continuity, momentum, and energy equations (Eqs. 6-21, 6-28, and 6-35) reduce for steady, incompressible, laminar flow of a fluid with constant properties over a flat plate to Continuity: Momentum: T u Energy: T u y x u (x, 0) 0 v (x, 0) 0 T (x, 0) Ts FIGURE 6–26 Boundary conditions for flow over a flat plate. u v 0 x y u 2u u u v v 2 x y y T 2T T v 2 u x y y (6-39) (6-40) (6-41) with the boundary conditions (Fig. 6–26) At x 0: At y 0: As y → : u(0, y) u, u(x, 0) 0, u(x, ) u, T(0, y) T v(x, 0) 0, T(x, 0) Ts T(x, ) T (6-42) When fluid properties are assumed to be constant and thus independent of temperature, the first two equations can be solved separately for the velocity components u and v. Once the velocity distribution is available, we can determine cen58933_ch06.qxd 9/4/2002 12:05 PM Page 353 353 CHAPTER 6 the friction coefficient and the boundary layer thickness using their definitions. Also, knowing u and v, the temperature becomes the only unknown in the last equation, and it can be solved for temperature distribution. The continuity and momentum equations were first solved in 1908 by the German engineer H. Blasius, a student of L. Prandtl. This was done by transforming the two partial differential equations into a single ordinary differential equation by introducing a new independent variable, called the similarity variable. The finding of such a variable, assuming it exists, is more of an art than science, and it requires to have a good insight of the problem. Noticing that the general shape of the velocity profile remains the same along the plate, Blasius reasoned that the nondimensional velocity profile u/u should remain unchanged when plotted against the nondimensional distance y/ , where is the thickness of the local velocity boundary layer at a given x. That is, although both and u at a given y vary with x, the velocity u at a fixed y/ remains constant. Blasius was also aware from the work of Stokes that is proportional to vx/u, and thus he defined a dimensionless similarity variable as u y vx (6-43) and thus u/u function(). He then introduced a stream function (x, y) as u y v x and (6-44) so that the continuity equation (Eq. 6-39) is automatically satisfied and thus eliminated (this can be verified easily by direct substitution). He then defined a function f() as the dependent variable as f() u vx/u (6-45) Then the velocity components become u vx df u df u u y y u d vx d (6-46) vx f u v 1 uv df u f f x u x 2 u x 2 x d (6-47) By differentiating these u and v relations, the derivatives of the velocity components can be shown to be u d 2f u u 2 , y vx d u d 2f u 2, x 2x d 2 u u 2 d 3f y 2 vx d3 (6-48) Substituting these relations into the momentum equation and simplifying, we obtain 2 d 3f d 2f f 0 d3 d 2 (6-49) which is a third-order nonlinear differential equation. Therefore, the system of two partial differential equations is transformed into a single ordinary cen58933_ch06.qxd 9/4/2002 12:05 PM Page 354 354 HEAT TRANSFER differential equation by the use of a similarity variable. Using the definitions of f and , the boundary conditions in terms of the similarity variables can be expressed as f(0) 0, TABLE 6–3 Similarity function f and its derivatives for laminar boundary layer along a flat plate. f df u d u d 2f d 2 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 0 0.042 0.166 0.370 0.650 0.996 1.397 1.838 2.306 2.790 3.283 3.781 4.280 0 0.166 0.330 0.487 0.630 0.751 0.846 0.913 0.956 0.980 0.992 0.997 0.999 1 0.332 0.331 0.323 0.303 0.267 0.217 0.161 0.108 0.064 0.034 0.016 0.007 0.002 0 df 0, ` d 0 and df 1 ` d (6-50) The transformed equation with its associated boundary conditions cannot be solved analytically, and thus an alternative solution method is necessary. The problem was first solved by Blasius in 1908 using a power series expansion approach, and this original solution is known as the Blasius solution. The problem is later solved more accurately using different numerical approaches, and results from such a solution are given in Table 6–3. The nondimensional velocity profile can be obtained by plotting u/u against . The results obtained by this simplified analysis are in excellent agreement with experimental results. Recall that we defined the boundary layer thickness as the distance from the surface for which u/u 0.99. We observe from Table 6–3 that the value of corresponding to u/u 0.992 is 5.0. Substituting 5.0 and y into the definition of the similarity variable (Eq. 6-43) gives 5.0 u /vx. Then the velocity boundary layer thickness becomes 5.0 5.0x u / vx Rex (6-51) since Rex ux/ , where x is the distance from the leading edge of the plate. Note that the boundary layer thickness increases with increasing kinematic viscosity and with increasing distance from the leading edge x, but it decreases with increasing free-stream velocity u. Therefore, a large free-stream velocity will suppress the boundary layer and cause it to be thinner. The shear stress on the wall can be determined from its definition and the u/y relation in Eq. 6-48: w u d 2f u u vx 2 y y0 d 0 (6-52) Substituting the value of the second derivative of f at 0 from Table 6–3 gives w 0.332u u 0.332u 2 x Rex (6-53) Then the local skin friction coefficient becomes Cf, x w w 0.664 Re 1/2 x 2 /2 u 2 / 2 (6-54) Note that unlike the boundary layer thickness, wall shear stress and the skin friction coefficient decrease along the plate as x1/2. The Energy Equation Knowing the velocity profile, we are now ready to solve the energy equation for temperature distribution for the case of constant wall temperature Ts. First we introduce the dimensionless temperature as cen58933_ch06.qxd 9/4/2002 12:05 PM Page 355 355 CHAPTER 6 (x, y) T(x, y) Ts T Ts (6-55) Noting that both Ts and T are constant, substitution into the energy equation gives u 2 v 2 x y y (6-56) Temperature profiles for flow over an isothermal flat plate are similar, just like the velocity profiles, and thus we expect a similarity solution for temperature to exist. Further, the thickness of the thermal boundary layer is proportional to vx/u , just like the thickness of the velocity boundary layer, and thus the similarity variable is also , and (). Using the chain rule and substituting the u and v expressions into the energy equation gives u df d 1 uv df d d 2 f 2 d d x 2 x d d y d y 2 (6-57) Simplifying and noting that Pr / give 2 d 2 d Pr f 0 d d 2 (6-58) with the boundary conditions (0) 0 and () 1. Obtaining an equation for as a function of alone confirms that the temperature profiles are similar, and thus a similarity solution exists. Again a closed-form solution cannot be obtained for this boundary value problem, and it must be solved numerically. It is interesting to note that for Pr 1, this equation reduces to Eq. 6-49 when is replaced by df/d, which is equivalent to u/u (see Eq. 6-46). The boundary conditions for and df/d are also identical. Thus we conclude that the velocity and thermal boundary layers coincide, and the nondimensional velocity and temperature profiles (u/u and ) are identical for steady, incompressible, laminar flow of a fluid with constant properties and Pr 1 over an isothermal flat plate (Fig. 6–27). The value of the temperature gradient at the surface (y 0 or 0) in this case is, from Table 6–3, d /d d 2f/d2 0.332. Equation 6-58 is solved for numerous values of Prandtl numbers. For Pr 0.6, the nondimensional temperature gradient at the surface is found to be proportional to Pr 1/3, and is expressed as d 0.332 Pr 1/3 ` d 0 (6-59) The temperature gradient at the surface is d T (T Ts) (T Ts) y y0 y y0 d 0 y y0 u 0.332 Pr1/3(T Ts) vx (6-60) Then the local convection coefficient and Nusselt number become hx qs k(T/y) y0 u 0.332 Pr1/3k Ts T Ts T vx (6-61) T u Pr 1 u/u or y Velocity or thermal boundary layer x FIGURE 6–27 When Pr 1, the velocity and thermal boundary layers coincide, and the nondimensional velocity and temperature profiles are identical for steady, incompressible, laminar flow over a flat plate. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 356 356 HEAT TRANSFER and Nux hx x 0.332 Pr1/3 Re1/2 x k Pr 0.6 (6-62) The Nux values obtained from this relation agree well with measured values. Solving Eq. 6-58 numerically for the temperature profile for different Prandtl numbers, and using the definition of the thermal boundary layer, it is determined that / t Pr 1/3. Then the thermal boundary layer thickness becomes t 5.0x Pr1/3 Pr1/3 Rex (6-63) Note that these relations are valid only for laminar flow over an isothermal flat plate. Also, the effect of variable properties can be accounted for by evaluating all such properties at the film temperature defined as Tf (Ts T)/2. The Blasius solution gives important insights, but its value is largely historical because of the limitations it involves. Nowadays both laminar and turbulent flows over surfaces are routinely analyzed using numerical methods. 6–9 ■ NONDIMENSIONALIZED CONVECTION EQUATIONS AND SIMILARITY When viscous dissipation is negligible, the continuity, momentum, and energy equations for steady, incompressible, laminar flow of a fluid with constant properties are given by Eqs. 6-21, 6-28, and 6-35. These equations and the boundary conditions can be nondimensionalized by dividing all dependent and independent variables by relevant and meaningful constant quantities: all lengths by a characteristic length L (which is the length for a plate), all velocities by a reference velocity (which is the free stream velocity for a plate), pressure by 2 (which is twice the free stream dynamic pressure for a plate), and temperature by a suitable temperature difference (which is T Ts for a plate). We get x x , L y y , L u u , v v , P P , 2 and T T Ts T Ts where the asterisks are used to denote nondimensional variables. Introducing these variables into Eqs. 6-21, 6-28, and 6-35 and simplifying give Continuity: Momentum: Energy: u v 0 x y u u 1 2u dP v u 2 Re L y x y dx 2 T T 1 T v u x y ReL Pr y 2 (6-64) (6-65) (6-66) with the boundary conditions u(0, y) 1, T(0, y) 1, u(x, 0) 0, u(x, ) 1, v(x, 0) 0, T(x, 0) 0, T(x, ) 1 (6-67) cen58933_ch06.qxd 9/4/2002 12:06 PM Page 357 357 CHAPTER 6 where ReL L/ is the dimensionless Reynolds number and Pr / is the Prandtl number. For a given type of geometry, the solutions of problems with the same Re and Nu numbers are similar, and thus Re and Nu numbers serve as similarity parameters. Two physical phenomena are similar if they have the same dimensionless forms of governing differential equations and boundary conditions (Fig. 6–28). A major advantage of nondimensionalizing is the significant reduction in the number of parameters. The original problem involves 6 parameters (L, , T, Ts, , ), but the nondimensionalized problem involves just 2 parameters (ReL and Pr). For a given geometry, problems that have the same values for the similarity parameters have identical solutions. For example, determining the convection heat transfer coefficient for flow over a given surface will require numerical solutions or experimental investigations for several fluids, with several sets of velocities, surface lengths, wall temperatures, and free stream temperatures. The same information can be obtained with far fewer investigations by grouping data into the dimensionless Re and Pr numbers. Another advantage of similarity parameters is that they enable us to group the results of a large number of experiments and to report them conveniently in terms of such parameters (Fig. 6–29). 6–10 ■ FUNCTIONAL FORMS OF FRICTION AND CONVECTION COEFFICIENTS u u f (x , ReL) y y0 L y y 0 L 2 (6-68) (6-69) Substituting into its definition gives the local friction coefficient, Cf, x /L s 2 f (x , ReL ) f (x , Rel ) f3(x , ReL ) 2 ReL 2 /2 2 /2 2 L1 Re2 2 Air L2 If Re1 Re2, then Cƒ 1 Cƒ 2 FIGURE 6–28 Two geometrically similar bodies have the same value of friction coefficient at the same Reynolds number. L, , T , Ts , v, Then the shear stress at the surface becomes s Re1 Parameters before nondimensionalizing The three nondimensionalized boundary layer equations (Eqs. 6-64, 6-65, and 6-66) involve three unknown functions u, v, and T, two independent variables x and y, and two parameters ReL and Pr. The pressure P(x) depends on the geometry involved (it is constant for a flat plate), and it has the same value inside and outside the boundary layer at a specified x. Therefore, it can be determined separately from the free stream conditions, and dP/dx in Eq. 6-65 can be treated as a known function of x. Note that the boundary conditions do not introduce any new parameters. For a given geometry, the solution for u can be expressed as u f1(x, y, ReL) 1 Water (6-70) Thus we conclude that the friction coefficient for a given geometry can be expressed in terms of the Reynolds number Re and the dimensionless space variable x alone (instead of being expressed in terms of x, L, , , and ). This is a very significant finding, and shows the value of nondimensionalized equations. Similarly, the solution of Eq. 6-66 for the dimensionless temperature T for a given geometry can be expressed as Parameters after nondimensionalizing: Re, Pr FIGURE 6–29 The number of parameters is reduced greatly by nondimensionalizing the convection equations. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 358 358 HEAT TRANSFER T g1(x, y, ReL, Pr) (6-71) Using the definition of T, the convection heat transfer coefficient becomes h k T k(T/y) y0 k(T Ts) T Ts T L(Ts T) y y 0 L y y 0 (6-72) Substituting this into the Nusselt number relation gives [or alternately, we can rearrange the relation above in dimensionless form as hL/k (T/y)\|y0 and define the dimensionless group hL/k as the Nusselt number] Nux T ∂T y ∂y x Nu y0 Laminar FIGURE 6–30 The Nusselt number is equivalent to the dimensionless temperature gradient at the surface. Local Nusselt number: Nux function (x, ReL, Pr) Average Nusselt number: Nu function (ReL, Pr) A common form of Nusselt number: Nu = C Re Lm Pr n FIGURE 6–31 For a given geometry, the average Nusselt number is a function of Reynolds and Prandtl numbers. hL T ` g2(x , Re L, Pr) k y y0 (6-73) Note that the Nusselt number is equivalent to the dimensionless temperature gradient at the surface, and thus it is properly referred to as the dimensionless heat transfer coefficient (Fig. 6–30). Also, the Nusselt number for a given geometry can be expressed in terms of the Reynolds number Re, the Prandtl number Pr, and the space variable x, and such a relation can be used for different fluids flowing at different velocities over similar geometries of different lengths. The average friction and heat transfer coefficients are determined by integrating Cf,x and Nux over the surface of the given body with respect to x from 0 to 1. Integration will remove the dependence on x, and the average friction coefficient and Nusselt number can be expressed as Cf f4(ReL) and Nu g3(ReL , Pr) (6-74) These relations are extremely valuable as they state that for a given geometry, the friction coefficient can be expressed as a function of Reynolds number alone, and the Nusselt number as a function of Reynolds and Prandtl numbers alone (Fig. 6–31). Therefore, experimentalists can study a problem with a minimum number of experiments, and report their friction and heat transfer coefficient measurements conveniently in terms of Reynolds and Prandtl numbers. For example, a friction coefficient relation obtained with air for a given surface can also be used for water at the same Reynolds number. But it should be kept in mind that the validity of these relations is limited by the limitations on the boundary layer equations used in the analysis. The experimental data for heat transfer is often represented with reasonable accuracy by a simple power-law relation of the form Nu C RemL Pr n (6-75) where m and n are constant exponents (usually between 0 and 1), and the value of the constant C depends on geometry. Sometimes more complex relations are used for better accuracy. 6–11 ■ ANALOGIES BETWEEN MOMENTUM AND HEAT TRANSFER In forced convection analysis, we are primarily interested in the determination of the quantities Cf (to calculate shear stress at the wall) and Nu (to calculate heat transfer rates). Therefore, it is very desirable to have a relation between cen58933_ch06.qxd 9/4/2002 12:06 PM Page 359 359 CHAPTER 6 Cf and Nu so that we can calculate one when the other is available. Such relations are developed on the basis of the similarity between momentum and heat transfers in boundary layers, and are known as Reynolds analogy and Chilton–Colburn analogy. Reconsider the nondimensionalized momentum and energy equations for steady, incompressible, laminar flow of a fluid with constant properties and negligible viscous dissipation (Eqs. 6-65 and 6-66). When Pr 1 (which is approximately the case for gases) and P/x 0 (which is the case when, u u constant in the free stream, as in flow over a flat plate), these equations simplify to Momentum: Energy: u u 1 2u v x y ReL y 2 T T 1 2T u v x y ReL y 2 u (6-76) (6-77) which are exactly of the same form for the dimensionless velocity u and temperature T. The boundary conditions for u and T are also identical. Therefore, the functions u and T must be identical, and thus the first derivatives of u and T at the surface must be equal to each other, u T y y 0 y y 0 (6-78) Then from Eqs. 6-69, 6-70, and 6-73 we have Cf, x ReL Nux 2 (Pr 1) (6-79) which is known as the Reynolds analogy (Fig. 6–32). This is an important analogy since it allows us to determine the heat transfer coefficient for fluids with Pr 1 from a knowledge of friction coefficient which is easier to measure. Reynolds analogy is also expressed alternately as Cf, x Stx 2 (Pr 1) (6-80) Profiles: u T Gradients: u T y 0 y y y 0 Analogy: Cf, x ReL Nux 2 where St Nu h Cp ReLPr (6-81) is the Stanton number, which is also a dimensionless heat transfer coefficient. Reynolds analogy is of limited use because of the restrictions Pr 1 and P/x 0 on it, and it is desirable to have an analogy that is applicable over a wide range of Pr. This is done by adding a Prandtl number correction. The friction coefficient and Nusselt number for a flat plate are determined in Section 6-8 to be Cf, x 0.664 Re1/2 x and Nux 0.332 Pr1/3 Re 1/2 x (6-82) Taking their ratio and rearranging give the desired relation, known as the modified Reynolds analogy or Chilton–Colburn analogy, FIGURE 6–32 When Pr 1 and P/x 0, the nondimensional velocity and temperature profiles become identical, and Nu is related to Cf by Reynolds analogy. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 360 360 HEAT TRANSFER ReL Nux Pr1/3 Cf, x 2 or Cf, x hx Pr2/3 jH 2 C (6-83) p for 0.6 Pr 60. Here jH is called the Colburn j-factor. Although this relation is developed using relations for laminar flow over a flat plate (for which P/x 0), experimental studies show that it is also applicable approximately for turbulent flow over a surface, even in the presence of pressure gradients. For laminar flow, however, the analogy is not applicable unless P/x 0. Therefore, it does not apply to laminar flow in a pipe. Analogies between Cf and Nu that are more accurate are also developed, but they are more complex and beyond the scope of this book. The analogies given above can be used for both local and average quantities. EXAMPLE 6–2 Air 20°C, 7 m/s Finding Convection Coefficient from Drag Measurement A 2-m 3-m flat plate is suspended in a room, and is subjected to air flow parallel to its surfaces along its 3-m-long side. The free stream temperature and velocity of air are 20˚C and 7 m/s. The total drag force acting on the plate is measured to be 0.86 N. Determine the average convection heat transfer coefficient for the plate (Fig. 6–33). SOLUTION A flat plate is subjected to air flow, and the drag force acting on it is measured. The average convection coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 The edge effects are negligible. 3 The local atmospheric pressure is 1 atm. Properties The properties of air at 20˚C and 1 atm are (Table A-15): 1.204 kg/m3, L=3m Cp 1.007 kJ/kg K, Pr 0.7309 Analysis The flow is along the 3-m side of the plate, and thus the characteristic length is L 3 m. Both sides of the plate are exposed to air flow, and thus the total surface area is As 2WL 2(2 m)(3 m) 12 m2 FIGURE 6–33 Schematic for Example 6–2. For flat plates, the drag force is equivalent to friction force. The average friction coefficient Cf can be determined from Eq. 6-11, Ff C f As 2 2 Solving for Cf and substituting, Cf Ff As2/2 1kg m/s2 0.86 N 0.00243 1N (1.204 kg/m3)(12 m2)(7 m/s)2/2 Then the average heat transfer coefficient can be determined from the modified Reynolds analogy (Eq. 6-83) to be h Cf Cp 0.00243 (1.204 kg/m3)(7 m/s)(1007 J/kg °C) 12.7 W/m2 · °C 2 Pr2/3 2 0.73092/3 Discussion This example shows the great utility of momentum-heat transfer analogies in that the convection heat transfer coefficient can be determined from a knowledge of friction coefficient, which is easier to determine. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 361 361 CHAPTER 6 SUMMARY Convection heat transfer is expressed by Newton’s law of cooling as · Q conv hAs(Ts T) where h is the convection heat transfer coefficient, Ts is the surface temperature, and T is the free-stream temperature. The convection coefficient is also expressed as h kfluid(T/y)y0 Ts T tance from the surface at which the temperature difference T Ts equals 0.99(T Ts). The relative thickness of the velocity and the thermal boundary layers is best described by the dimensionless Prandtl number, defined as Pr Molecular diffusivity of momentum v Cp Molecular diffusivity of heat k For external flow, the dimensionless Reynolds number is expressed as Re The Nusselt number, which is the dimensionless heat transfer coefficient, is defined as Nu hLc k where k is the thermal conductivity of the fluid and Lc is the characteristic length. The highly ordered fluid motion characterized by smooth streamlines is called laminar. The highly disordered fluid motion that typically occurs at high velocities is characterized by velocity fluctuations is called turbulent. The random and rapid fluctuations of groups of fluid particles, called eddies, provide an additional mechanism for momentum and heat transfer. The region of the flow above the plate bounded by in which the effects of the viscous shearing forces caused by fluid viscosity are felt is called the velocity boundary layer. The boundary layer thickness, , is defined as the distance from the surface at which u 0.99u. The hypothetical line of u 0.99u divides the flow over a plate into the boundary layer region in which the viscous effects and the velocity changes are significant, and the inviscid flow region, in which the frictional effects are negligible. The friction force per unit area is called shear stress, and the shear stress at the wall surface is expressed as u s ` y y0 or 2 s Cf 2 where is the dynamic viscosity, is the upstream velocity, and Cf is the dimensionless friction coefficient. The property / is the kinematic viscosity. The friction force over the entire surface is determined from Ff C f A s 2 2 The flow region over the surface in which the temperature variation in the direction normal to the surface is significant is the thermal boundary layer. The thickness of the thermal boundary layer t at any location along the surface is the dis- Lc Lc Inertia forces v Viscous forces For a flat plate, the characteristic length is the distance x from the leading edge. The Reynolds number at which the flow becomes turbulent is called the critical Reynolds number. For flow over a flat plate, its value is taken to be Recr xcr/v 5 10 5. The continuity, momentum, and energy equations for steady two-dimensional incompressible flow with constant properties are determined from mass, momentum, and energy balances to be Continuity: u v 0 x y x-momentum: u Energy: Cp u u u 2u P v 2 x y x y 2T 2T T T v k 2 2 x y x y where the viscous dissipation function is 2 ux vy uy vx 2 2 2 Using the boundary layer approximations and a similarity variable, these equations can be solved for parallel steady incompressible flow over a flat plate, with the following results: Velocity boundary layer thickness: Local friction coefficient: Local Nusselt number: Cf, x Nu x Thermal boundary layer thickness: 5.0 5.0x /vx Rex w 0.664 Re 1/2 x 2 /2 hx x 0.332 Pr1/3 Re 1/2 x k 5.0x t 1/3 1/3 Pr Pr Rex The average friction coefficient and Nusselt number are expressed in functional form as cen58933_ch06.qxd 9/4/2002 12:06 PM Page 362 362 HEAT TRANSFER Cf f4(ReL) Nu g3(ReL, Pr) and The Nusselt number can be expressed by a simple power-law relation of the form Nu C Re Lm Pr is the Stanton number. The analogy is extended to other Prandtl numbers by the modified Reynolds analogy or Chilton– Colburn analogy, expressed as Cf, x n or where m and n are constant exponents, and the value of the constant C depends on geometry. The Reynolds analogy relates the convection coefficient to the friction coefficient for fluids with Pr 1, and is expressed as Cf, x ReL Nux 2 or Cf, x Stx 2 ReL NuxPr1/3 2 Cf, x hx Pr2/3 jH (0.6 Pr 60) 2 Cp These analogies are also applicable approximately for turbulent flow over a surface, even in the presence of pressure gradients. where St Nu h Cp ReL Pr REFERENCES AND SUGGESTED READING 1. H. Blasius. “The Boundary Layers in Fluids with Little Friction (in German).” Z. Math. Phys., 56, 1 (1908); pp. 1–37; English translation in National Advisory Committee for Aeronautics Technical Memo No. 1256, February 1950. 2. R. W. Fox and A. T. McDonald. Introduction to Fluid Mechanics. 5th. ed. New York, Wiley, 1999. 3. J. P. Holman. Heat Transfer. 9th ed. New York: McGrawHill, 2002. O. Reynolds. “On the Experimental Investigation of the Circumstances Which Determine Whether the Motion of Water Shall Be Direct or Sinuous, and the Law of Resistance in Parallel Channels.” Philosophical Transactions of the Royal Society of London 174 (1883), pp. 935–82. 9. H. Schlichting. Boundary Layer Theory. 7th ed. New York: McGraw-Hill, 1979. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. G. G. Stokes. “On the Effect of the Internal Friction of Fluids on the Motion of Pendulums.” Cambridge Philosophical Transactions, IX, 8, 1851. W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West, 1995. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. F. M. White. Heat and Mass Transfer. Reading, MA: Addison-Wesley, 1988. M. N. Özisik. Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. PROBLEMS Mechanism and Types of Convection 6–1C What is forced convection? How does it differ from natural convection? Is convection caused by winds forced or natural convection? 6–2C What is external forced convection? How does it differ from internal forced convection? Can a heat transfer system involve both internal and external convection at the same time? Give an example. Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 363 363 CHAPTER 6 6–3C In which mode of heat transfer is the convection heat transfer coefficient usually higher, natural convection or forced convection? Why? Air 5°C 1 atm 6–4C Consider a hot baked potato. Will the potato cool faster or slower when we blow the warm air coming from our lungs on it instead of letting it cool naturally in the cooler air in the room? Explain. 6–5C What is the physical significance of the Nusselt number? How is it defined? 6–6C When is heat transfer through a fluid conduction and when is it convection? For what case is the rate of heat transfer higher? How does the convection heat transfer coefficient differ from the thermal conductivity of a fluid? 6–7C Define incompressible flow and incompressible fluid. Must the flow of a compressible fluid necessarily be treated as compressible? 6–8 During air cooling of potatoes, the heat transfer coefficient for combined convection, radiation, and evaporation is determined experimentally to be as shown: Air Velocity, m/s Heat Transfer Coefficient, W/m2 ˚C 0.66 1.00 1.36 1.73 14.0 19.1 20.2 24.4 Consider a 10-cm-diameter potato initially at 20˚C with a thermal conductivity of 0.49 W/m ˚C. Potatoes are cooled by refrigerated air at 5˚C at a velocity of 1 m/s. Determine the initial rate of heat transfer from a potato, and the initial value of the temperature gradient in the potato at the surface. Answers: 9.0 W, 585°C/m 6–9 An average man has a body surface area of 1.8 m2 and a skin temperature of 33˚C. The convection heat transfer coefficient for a clothed person walking in still air is expressed as h 8.60.53 for 0.5 2 m/s, where is the walking velocity in m/s. Assuming the average surface temperature of the clothed person to be 30˚C, determine the rate of heat loss from an average man walking in still air at 10˚C by convection at a walking velocity of (a) 0.5 m/s, (b) 1.0 m/s, (c) 1.5 m/s, and (d) 2.0 m/s. Orange FIGURE P6–11 is determined experimentally and is expressed as h 5.05 kairRe1/3/D, where the diameter D is the characteristic length. Oranges are cooled by refrigerated air at 5˚C and 1 atm at a velocity of 0.5 m/s. Determine (a) the initial rate of heat transfer from a 7-cm-diameter orange initially at 15˚C with a thermal conductivity of 0.50 W/m ˚C, (b) the value of the initial temperature gradient inside the orange at the surface, and (c) the value of the Nusselt number. Velocity and Thermal Boundary Layers 6–12C What is viscosity? What causes viscosity in liquids and in gases? Is dynamic viscosity typically higher for a liquid or for a gas? 6–13C fluid? What is Newtonian fluid? Is water a Newtonian 6–14C What is the no-slip condition? What causes it? 6–15C Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why? 6–16C How does the dynamic viscosity of (a) liquids and (b) gases vary with temperature? 6–17C What fluid property is responsible for the development of the velocity boundary layer? For what kind of fluids will there be no velocity boundary layer on a flat plate? 6–18C What is the physical significance of the Prandtl number? Does the value of the Prandtl number depend on the type of flow or the flow geometry? Does the Prandtl number of air change with pressure? Does it change with temperature? 6–19C Will a thermal boundary layer develop in flow over a surface even if both the fluid and the surface are at the same temperature? 6–10 The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 14.80.69 for 0.15 1.5 m/s, where is the air velocity. For a person with a body surface area of 1.7 m2 and an average surface temperature of 29˚C, determine the rate of heat loss from the person in windy air at 10˚C by convection for air velocities of (a) 0.5 m/s, (b) 1.0 m/s, and (c) 1.5 m/s. Laminar and Turbulent Flows 6–11 During air cooling of oranges, grapefruit, and tangelos, the heat transfer coefficient for combined convection, radiation, and evaporation for air velocities of 0.11 0.33 m/s 6–22C What does the friction coefficient represent in flow over a flat plate? How is it related to the drag force acting on the plate? 6–20C How does turbulent flow differ from laminar flow? For which flow is the heat transfer coefficient higher? 6–21C What is the physical significance of the Reynolds number? How is it defined for external flow over a plate of length L? cen58933_ch06.qxd 9/4/2002 12:06 PM Page 364 364 HEAT TRANSFER 6–23C What is the physical mechanism that causes the friction factor to be higher in turbulent flow? 6–24C 12 m/s What is turbulent viscosity? What is it caused by? 6–25C What is turbulent thermal conductivity? What is it caused by? Convection Equations and Similarity Solutions 6–26C Under what conditions can a curved surface be treated as a flat plate in fluid flow and convection analysis? 6–27C Express continuity equation for steady twodimensional flow with constant properties, and explain what each term represents. 6–28C Is the acceleration of a fluid particle necessarily zero in steady flow? Explain. 6–29C For steady two-dimensional flow, what are the boundary layer approximations? 6–30C For what types of fluids and flows is the viscous dissipation term in the energy equation likely to be significant? u(y) FIGURE P6–37 6–38 Repeat Problem 6–37 for a spacing of 0.4 mm. 6–39 A 6-cm-diameter shaft rotates at 3000 rpm in a 20-cmlong bearing with a uniform clearance of 0.2 mm. At steady operating conditions, both the bearing and the shaft in the vicinity of the oil gap are at 50˚C, and the viscosity and thermal conductivity of lubricating oil are 0.05 N s/m2 and 0.17 W/m K. By simplifying and solving the continuity, momentum, and energy equations, determine (a) the maximum temperature of oil, (b) the rates of heat transfer to the bearing and the shaft, and (c) the mechanical power wasted by the viscous dissipation Answers: (a) 53.3°C, (b) 419 W, (c) 838 W in the oil. 3000 rpm 6–31C For steady two-dimensional flow over an isothermal flat plate in the x-direction, express the boundary conditions for the velocity components u and v, and the temperature T at the plate surface and at the edge of the boundary layer. 6–32C What is a similarity variable, and what is it used for? For what kinds of functions can we expect a similarity solution for a set of partial differential equations to exist? 6 cm 20 cm FIGURE P6-39 6–33C Consider steady, laminar, two-dimensional flow over an isothermal plate. Does the thickness of the velocity boundary layer increase or decrease with (a) distance from the leading edge, (b) free-stream velocity, and (c) kinematic viscosity? 6–40 Repeat Problem 6–39 by assuming the shaft to have reached peak temperature and thus heat transfer to the shaft to be negligible, and the bearing surface still to be maintained at 50˚C. 6–34C Consider steady, laminar, two-dimensional flow over an isothermal plate. Does the wall shear stress increase, decrease, or remain constant with distance from the leading edge? 6–41 Reconsider Problem 6–39. Using EES (or other) software, investigate the effect of shaft velocity on the mechanical power wasted by viscous dissipation. Let the shaft rotation vary from 0 rpm to 5000 rpm. Plot the power wasted versus the shaft rpm, and discuss the results. 6–35C What are the advantages of nondimensionalizing the convection equations? 6–36C Consider steady, laminar, two-dimensional, incompressible flow with constant properties and a Prandtl number of unity. For a given geometry, is it correct to say that both the average friction and heat transfer coefficients depend on the Reynolds number only? 6–37 Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates with one plate moving at a constant velocity of 12 m/s and the other stationary. Consider such a flow with a uniform spacing of 0.7 mm between the plates. The temperatures of the upper and lower plates are 40˚C and 15˚C, respectively. By simplifying and solving the continuity, momentum, and energy equations, determine (a) the velocity and temperature distributions in the oil, (b) the maximum temperature and where it occurs, and (c) the heat flux from the oil to each plate. 6–42 Consider a 5-cm-diameter shaft rotating at 2500 rpm in a 10-cm-long bearing with a clearance of 0.5 mm. Determine the power required to rotate the shaft if the fluid in the gap is (a) air, (b) water, and (c) oil at 40˚C and 1 atm. 6–43 Consider the flow of fluid between two large parallel isothermal plates separated by a distance L. The upper plate is moving at a constant velocity of and maintained at temperature T0 while the lower plate is stationary and insulated. By simplifying and solving the continuity, momentum, and energy equations, obtain relations for the maximum temperature of fluid, the location where it occurs, and heat flux at the upper plate. 6–44 Reconsider Problem 6–43. Using the results of this problem, obtain a relation for the volumetric heat generation rate g·, in W/m3. Then express the convection problem as an equivalent cen58933_ch06.qxd 9/4/2002 12:06 PM Page 365 365 CHAPTER 6 conduction problem in the oil layer. Verify your model by solving the conduction problem and obtaining a relation for the maximum temperature, which should be identical to the one obtained in the convection analysis. 6–45 A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k 70 W/m K) with a uniform clearance of 0.6 mm filled with lubricating oil ( 0.03 N s/m2 and k 0.14 W/m K). The bearing is cooled externally by a liquid, and its outer surface is maintained at 40˚C. Disregarding heat conduction through the shaft and assuming one-dimensional heat transfer, determine (a) the rate of heat transfer to the coolant, (b) the surface temperature of the shaft, and (c) the mechanical power wasted by the viscous dissipation in oil. 4500 rpm 5 cm 8 cm 15 cm FIGURE P6–45 6–50 A metallic airfoil of elliptical cross section has a mass of 50 kg, surface area of 12 m2, and a specific heat of 0.50 kJ/kg ˚C). The airfoil is subjected to air flow at 1 atm, 25˚C, and 8 m/s along its 3-m-long side. The average temperature of the airfoil is observed to drop from 160˚C to 150˚C within 2 min of cooling. Assuming the surface temperature of the airfoil to be equal to its average temperature and using momentum-heat transfer analogy, determine the average friction coefficient of Answer: 0.000227 the airfoil surface. 6–51 Repeat Problem 6–50 for an air-flow velocity of 12 m/s. 6–52 The electrically heated 0.6-m-high and 1.8-m-long windshield of a car is subjected to parallel winds at 1 atm, 0˚C, and 80 km/h. The electric power consumption is observed to be 50 W when the exposed surface temperature of the windshield is 4˚C. Disregarding radiation and heat transfer from the inner surface and using the momentum-heat transfer analogy, determine drag force the wind exerts on the windshield. 6–53 Consider an airplane cruising at an altitude of 10 km where standard atmospheric conditions are 50˚C and 26.5 kPa at a speed of 800 km/h. Each wing of the airplane can be modeled as a 25-m 3-m flat plate, and the friction coefficient of the wings is 0.0016. Using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions. Answer: 89.6 W/m2 · °C 6–46 Repeat Problem 6–45 for a clearance of 1 mm. Design and Essay Problems Momentum and Heat Transfer Analogies 6–54 Design an experiment to measure the viscosity of liquids using a vertical funnel with a cylindrical reservoir of height h and a narrow flow section of diameter D and length L. Making appropriate assumptions, obtain a relation for viscosity in terms of easily measurable quantities such as density and volume flow rate. 6–47C How is Reynolds analogy expressed? What is the value of it? What are its limitations? 6–48C How is the modified Reynolds analogy expressed? What is the value of it? What are its limitations? A 4-m 4-m flat plate maintained at a constant temperature of 80˚C is subjected to parallel flow of air at 1 atm, 20˚C, and 10 m/s. The total drag force acting on the upper surface of the plate is measured to be 2.4 N. Using momentum-heat transfer analogy, determine the average convection heat transfer coefficient, and the rate of heat transfer between the upper surface of the plate and the air. 6–49 6–55 A facility is equipped with a wind tunnel, and can measure the friction coefficient for flat surfaces and airfoils. Design an experiment to determine the mean heat transfer coefficient for a surface using friction coefficient data. cen58933_ch06.qxd 9/4/2002 12:06 PM Page 366 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 367 CHAPTER EXTERNAL FORCED CONVECTION n Chapter 6 we considered the general and theoretical aspects of forced convection, with emphasis on differential formulation and analytical solutions. In this chapter we consider the practical aspects of forced convection to or from flat or curved surfaces subjected to external flow, characterized by the freely growing boundary layers surrounded by a free flow region that involves no velocity and temperature gradients. We start this chapter with an overview of external flow, with emphasis on friction and pressure drag, flow separation, and the evaluation of average drag and convection coefficients. We continue with parallel flow over flat plates. In Chapter 6, we solved the boundary layer equations for steady, laminar, parallel flow over a flat plate, and obtained relations for the local friction coefficient and the Nusselt number. Using these relations as the starting point, we determine the average friction coefficient and Nusselt number. We then extend the analysis to turbulent flow over flat plates with and without an unheated starting length. Next we consider cross flow over cylinders and spheres, and present graphs and empirical correlations for the drag coefficients and the Nusselt numbers, and discuss their significance. Finally, we consider cross flow over tube banks in aligned and staggered configurations, and present correlations for the pressure drop and the average Nusselt number for both configurations. I 7 CONTENTS 7–1 Drag and Heat Transfer in External Flow 368 7–2 Parallel Flow over Flat Plates 371 7–3 Flow across Cylinders and Spheres 380 7–4 Flow across Tube Banks 389 Topic of Special Interest: Reducing Heat Transfer through Surfaces: Thermal Insulation 395 367 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 368 368 HEAT TRANSFER 7–1 ■ DRAG AND HEAT TRANSFER IN EXTERNAL FLOW Fluid flow over solid bodies frequently occurs in practice, and it is responsible for numerous physical phenomena such as the drag force acting on the automobiles, power lines, trees, and underwater pipelines; the lift developed by airplane wings; upward draft of rain, snow, hail, and dust particles in high winds; and the cooling of metal or plastic sheets, steam and hot water pipes, and extruded wires. Therefore, developing a good understanding of external flow and external forced convection is important in the mechanical and thermal design of many engineering systems such as aircraft, automobiles, buildings, electronic components, and turbine blades. The flow fields and geometries for most external flow problems are too complicated to be solved analytically, and thus we have to rely on correlations based on experimental data. The availability of high-speed computers has made it possible to conduct series of “numerical experimentations” quickly by solving the governing equations numerically, and to resort to the expensive and time-consuming testing and experimentation only in the final stages of design. In this chapter we will mostly rely on relations developed experimentally. The velocity of the fluid relative to an immersed solid body sufficiently far from the body (outside the boundary layer) is called the free-stream velocity, and is denoted by u. It is usually taken to be equal to the upstream velocity also called the approach velocity, which is the velocity of the approaching fluid far ahead of the body. This idealization is nearly exact for very thin bodies, such as a flat plate parallel to flow, but approximate for blunt bodies such as a large cylinder. The fluid velocity ranges from zero at the surface (the noslip condition) to the free-stream value away from the surface, and the subscript “infinity” serves as a reminder that this is the value at a distance where the presence of the body is not felt. The upstream velocity, in general, may vary with location and time (e.g., the wind blowing past a building). But in the design and analysis, the upstream velocity is usually assumed to be uniform and steady for convenience, and this is what we will do in this chapter. Friction and Pressure Drag Wind tunnel 60 mph FD FIGURE 7–1 Schematic for measuring the drag force acting on a car in a wind tunnel. You may have seen high winds knocking down trees, power lines, and even trailers, and have felt the strong “push” the wind exerts on your body. You experience the same feeling when you extend your arm out of the window of a moving car. The force a flowing fluid exerts on a body in the flow direction is called drag (Fig. 7–1) A stationary fluid exerts only normal pressure forces on the surface of a body immersed in it. A moving fluid, however, also exerts tangential shear forces on the surface because of the no-slip condition caused by viscous effects. Both of these forces, in general, have components in the direction of flow, and thus the drag force is due to the combined effects of pressure and wall shear forces in the flow direction. The components of the pressure and wall shear forces in the normal direction to flow tend to move the body in that direction, and their sum is called lift. In general, both the skin friction (wall shear) and pressure contribute to the drag and the lift. In the special case of a thin flat plate aligned parallel to the flow direction, the drag force depends on the wall shear only and is cen58933_ch07.qxd 9/4/2002 12:12 PM Page 369 369 CHAPTER 7 independent of pressure. When the flat plate is placed normal to the flow direction, however, the drag force depends on the pressure only and is independent of the wall shear since the shear stress in this case acts in the direction normal to flow (Fig. 7–2). For slender bodies such as wings, the shear force acts nearly parallel to the flow direction. The drag force for such slender bodies is mostly due to shear forces (the skin friction). The drag force FD depends on the density of the fluid, the upstream velocity , and the size, shape, and orientation of the body, among other things. The drag characteristics of a body is represented by the dimensionless drag coefficient CD defined as Drag coefficient: FD 2A 2 CD 1 (7-1) where A is the frontal area (the area projected on a plane normal to the direction of flow) for blunt bodies—bodies that tends to block the flow. The frontal area of a cylinder of diameter D and length L, for example, is A LD. For parallel flow over flat plates or thin airfoils, A is the surface area. The drag coefficient is primarily a function of the shape of the body, but it may also depend on the Reynolds number and the surface roughness. The drag force is the net force exerted by a fluid on a body in the direction of flow due to the combined effects of wall shear and pressure forces. The part of drag that is due directly to wall shear stress w is called the skin friction drag (or just friction drag) since it is caused by frictional effects, and the part that is due directly to pressure P is called the pressure drag (also called the form drag because of its strong dependence on the form or shape of the body). When the friction and pressure drag coefficients are available, the total drag coefficient is determined by simply adding them, CD CD, friction CD, pressure (7-2) The friction drag is the component of the wall shear force in the direction of flow, and thus it depends on the orientation of the body as well as the magnitude of the wall shear stress w. The friction drag is zero for a surface normal to flow, and maximum for a surface parallel to flow since the friction drag in this case equals the total shear force on the surface. Therefore, for parallel flow over a flat plate, the drag coefficient is equal to the friction drag coefficient, or simply the friction coefficient (Fig. 7–3). That is, Flat plate: CD CD, friction Cf (7-3) Once the average friction coefficient Cf is available, the drag (or friction) force over the surface can be determined from Eq. 7-1. In this case A is the surface area of the plate exposed to fluid flow. When both sides of a thin plate are subjected to flow, A becomes the total area of the top and bottom surfaces. Note that the friction coefficient, in general, will vary with location along the surface. Friction drag is a strong function of viscosity, and an “idealized” fluid with zero viscosity would produce zero friction drag since the wall shear stress would be zero (Fig. 7–4). The pressure drag would also be zero in this case during steady flow regardless of the shape of the body since there will be no pressure losses. For flow in the horizontal direction, for example, the pressure along a horizontal line will be constant (just like stationary fluids) since the High pressure Low pressure + + + + + + + + – – – – – – – – Wall shear FIGURE 7–2 Drag force acting on a flat plate normal to flow depends on the pressure only and is independent of the wall shear, which acts normal to flow. CD, pressure = 0 CD = CD, friction = Cf FD, pressure = 0 FD = FD, friction = Ff = Cf A ρ2 2 FIGURE 7–3 For parallel flow over a flat plate, the pressure drag is zero, and thus the drag coefficient is equal to the friction coefficient and the drag force is equal to the friction force. FD = 0 if µ = 0 FIGURE 7–4 For the flow of an “idealized” fluid with zero viscosity past a body, both the friction drag and pressure drag are zero regardless of the shape of the body. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 370 370 HEAT TRANSFER Separation point Reattachment point upstream velocity is constant, and thus there will be no net pressure force acting on the body in the horizontal direction. Therefore, the total drag is zero for the case of ideal inviscid fluid flow. At low Reynolds numbers, most drag is due to friction drag. This is especially the case for highly streamlined bodies such as airfoils. The friction drag is also proportional to the surface area. Therefore, bodies with a larger surface area will experience a larger friction drag. Large commercial airplanes, for example, reduce their total surface area and thus drag by retracting their wing extensions when they reach the cruising altitudes to save fuel. The friction drag coefficient is independent of surface roughness in laminar flow, but is a strong function of surface roughness in turbulent flow due to surface roughness elements protruding further into the highly viscous laminar sublayer. The pressure drag is proportional to the difference between the pressures acting on the front and back of the immersed body, and the frontal area. Therefore, the pressure drag is usually dominant for blunt bodies, negligible for streamlined bodies such as airfoils, and zero for thin flat plates parallel to the flow. When a fluid is forced to flow over a curved surface at sufficiently high velocities, it will detach itself from the surface of the body. The low-pressure region behind the body where recirculating and back flows occur is called the separation region. The larger the separation area is, the larger the pressure drag will be. The effects of flow separation are felt far downstream in the form of reduced velocity (relative to the upstream velocity). The region of flow trailing the body where the effect of the body on velocity is felt is called the wake (Fig. 7–5). The separated region comes to an end when the two flow streams reattach, but the wake keeps growing behind the body until the fluid in the wake region regains its velocity. The viscous effects are the most significant in the boundary layer, the separated region, and the wake. The flow outside these regions can be considered to be inviscid. Heat Transfer Separation region Wake region (within dashed line) FIGURE 7–5 Separation and reattachment during flow over a cylinder, and the wake region. The phenomena that affect drag force also affect heat transfer, and this effect appears in the Nusselt number. By nondimensionalizing the boundary layer equations, it was shown in Chapter 6 that the local and average Nusselt numbers have the functional form Nux f1(x, Rex, Pr) and Nu f2(ReL, Pr) (7-4a, b) The experimental data for heat transfer is often represented conveniently with reasonable accuracy by a simple power-law relation of the form Nu C Re Lm Pr n (7-5) where m and n are constant exponents, and the value of the constant C depends on geometry and flow. The fluid temperature in the thermal boundary layer varies from Ts at the surface to about T at the outer edge of the boundary. The fluid properties also vary with temperature, and thus with position across the boundary layer. In order to account for the variation of the properties with temperature, the fluid properties are usually evaluated at the so-called film temperature, defined as cen58933_ch07.qxd 9/4/2002 12:12 PM Page 371 371 CHAPTER 7 Tf Ts T 2 (7-6) which is the arithmetic average of the surface and the free-stream temperatures. The fluid properties are then assumed to remain constant at those values during the entire flow. An alternative way of accounting for the variation of properties with temperature is to evaluate all properties at the free stream temperature and to multiply the Nusselt number relation in Eq. 7-5 by (Pr/Prs)r or (/s)r. The local drag and convection coefficients vary along the surface as a result of the changes in the velocity boundary layers in the flow direction. We are usually interested in the drag force and the heat transfer rate for the entire surface, which can be determined using the average friction and convection coefficient. Therefore, we present correlations for both local (identified with the subscript x) and average friction and convection coefficients. When relations for local friction and convection coefficients are available, the average friction and convection coefficients for the entire surface can be determined by integration from CD 1 L C h 1 L h dx L 0 D, x dx (7-7) and L 0 x (7-8) When the average drag and convection coefficients are available, the drag force can be determined from Eq. 7-1 and the rate of heat transfer to or from an isothermal surface can be determined from Q˙ h As(Ts T) (7-9) where As is the surface area. 7–2 ■ PARALLEL FLOW OVER FLAT PLATES Consider the parallel flow of a fluid over a flat plate of length L in the flow direction, as shown in Figure 7–6. The x-coordinate is measured along the plate surface from the leading edge in the direction of the flow. The fluid approaches the plate in the x-direction with uniform upstream velocity and temperature T. The flow in the velocity boundary layer starts out as laminar, but if the plate is sufficiently long, the flow will become turbulent at a distance xcr from the leading edge where the Reynolds number reaches its critical value for transition. The transition from laminar to turbulent flow depends on the surface geometry, surface roughness, upstream velocity, surface temperature, and the type of fluid, among other things, and is best characterized by the Reynolds number. The Reynolds number at a distance x from the leading edge of a flat plate is expressed as T Turbulent Laminar y x xcr Ts L FIGURE 7–6 Laminar and turbulent regions of the boundary layer during flow over a flat plate. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 372 372 HEAT TRANSFER x x Rex v (7-10) Note that the value of the Reynolds number varies for a flat plate along the flow, reaching ReL L /v at the end of the plate. For flow over a flat plate, transition from laminar to turbulent is usually taken to occur at the critical Reynolds number of Recr xcr 5 5 10 (7-11) The value of the critical Reynolds number for a flat plate may vary from 105 to 3 106, depending on the surface roughness and the turbulence level of the free stream. Friction Coefficient Based on analysis, the boundary layer thickness and the local friction coefficient at location x for laminar flow over a flat plate were determined in Chapter 6 to be 1 LC dx Cf = –– L 0 f, x 1 L ––––– 0.664 dx = –– L 0 Re1/2 x 0.664 = ––––– L x –––– ν ( ) ( ) ( ) L 0 0.664 –––– = ––––– ν L Laminar: × 0.664 L = 2––––––– ν L 5x Re1/2 x Cf, x and 0.664 , Re1/2 x Rex 5 10 5 (7-12a, b) The corresponding relations for turbulent flow are –1/2 dx –1/2 v, x L x1/2 –––– 1 –– 2 0 –1/2 1.328 = ––––– 1/2 Re L FIGURE 7–7 The average friction coefficient over a surface is determined by integrating the local friction coefficient over the entire surface. Turbulent: v, x 0.382x and Re1/5 x Cf, x 0.0592 , Re1/5 x 5 10 5 Rex 107 (7-13a, b) where x is the distance from the leading edge of the plate and Rex x/v is the and thus Reynolds number at location x. Note that Cf, x. is proportional to Re1/2 x to x1/2 for laminar flow. Therefore, Cf, x is supposedly infinite at the leading edge (x 0) and decreases by a factor of x1/2 in the flow direction. The local friction coefficients are higher in turbulent flow than they are in laminar flow because of the intense mixing that occurs in the turbulent boundary layer. Note that Cf, x reaches its highest values when the flow becomes fully turbulent, and then decreases by a factor of x1/5 in the flow direction. The average friction coefficient over the entire plate is detennined by substituting the relations above into Eq. 7-7 and performing the integrations (Fig.7–7). We get Laminar: Turbulent: Cf Cf 1.328 Re 1/2 L 0.074 Re 1/5 L ReL 5 10 5 (7-14) 5 10 5 ReL 10 7 (7-15) The first relation gives the average friction coefficient for the entire plate when the flow is laminar over the entire plate. The second relation gives the average friction coefficient for the entire plate only when the flow is turbulent over the entire plate, or when the laminar flow region of the plate is too small relative to the turbulent flow region (that is, xcr L where the length of the plate xcr over which the flow is laminar can be determined from Recr 5 105 xcr/v). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 373 373 CHAPTER 7 In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to disregard the laminar flow region. In such cases, the average friction coefficient over the entire plate is determined by performing the integration in Eq. 7-7 over two parts: the laminar region 0 x xcr and the turbulent region xcr x L as Cf 1 L xcr 0 Cf, x laminar dx C L xcr f, x, turbulent dx (7-16) Note that we included the transition region with the turbulent region. Again taking the critical Reynolds number to be Recr 5 105 and performing the integrations of Eq. 7-16 after substituting the indicated expressions, the average friction coefficient over the entire plate is determined to be Cf 0.074 1742 ReL Re 1/5 L 5 10 5 ReL 107 (7-17) The constants in this relation will be different for different critical Reynolds numbers. Also, the surfaces are assumed to be smooth, and the free stream to be turbulent free. For laminar flow, the friction coefficient depends on only the Reynolds number, and the surface roughness has no effect. For turbulent flow, however, surface roughness causes the friction coefficient to increase severalfold, to the point that in fully turbulent regime the friction coefficient is a function of surface roughness alone, and independent of the Reynolds number (Fig. 7–8). A curve fit of experimental data for the average friction coefficient in this regime is given by Schlichting as Rough surface, turbulent: Cf 1.89 1.62 log L 2.5 (7-18) were is the surface roughness, and L is the length of the plate in the flow direction. In the absence of a better relation, the relation above can be used for turbulent flow on rough surfaces for Re l06, especially when L 104. Relative roughness, L 0.0 1 105 1 104 1 103 FIGURE 7–8 For turbulent flow, surface roughness may cause the friction coefficient to increase severalfold. h Cf The local Nusselt number at a location x for laminar flow over a flat plate was determined in Chapter 6 by solving the differential energy equation to be Nu x hx x 1/3 0.332 Re0.5 x Pr k Pr 0.60 0.0029 0.0032 0.0049 0.0084 Smooth surface for Re = 107. Others calculated from Eq. 7-18. Heat Transfer Coefficient Laminar: Friction coefficient Cf hx or Cf, x (7-19) δx The corresponding relation for turbulent flow is Turbulent: hx x 1/3 0.0296 Re 0.8 Nu x x Pr k 0.6 Pr 60 5 105 Rex 10 7 T (7-20) 0.5 Note that hx is proportional to Re 0.5 for laminar flow. Therex and thus to x fore, hx is infinite at the leading edge (x 0) and decreases by a factor of x0.5 in the flow direction. The variation of the boundary layer thickness and the friction and heat transfer coefficients along an isothermal flat plate are shown in Figure 7–9. The local friction and heat transfer coefficients are higher in Laminar Transition Turbulent x FIGURE 7–9 The variation of the local friction and heat transfer coefficients for flow over a flat plate. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 374 374 HEAT TRANSFER turbulent flow than they are in laminar flow. Also, hx reaches its highest values when the flow becomes fully turbulent, and then decreases by a factor of x0.2 in the flow direction, as shown in the figure. The average Nusselt number over the entire plate is determined by substituting the relations above into Eq. 7-8 and performing the integrations. We get Nu Laminar: Turbulent: Nu hL 1/3 0.664 Re 0.5 L Pr k hL 1/3 0.037 Re 0.8 L Pr k ReL 5 10 5 (7-21) 0.6 Pr 60 5 10 5 ReL 10 7 (7-22) The first relation gives the average heat transfer coefficient for the entire plate when the flow is laminar over the entire plate. The second relation gives the average heat transfer coefficient for the entire plate only when the flow is turbulent over the entire plate, or when the laminar flow region of the plate is too small relative to the turbulent flow region. In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to disregard the laminar flow region. In such cases, the average heat transfer coefficient over the entire plate is determined by performing the integration in Eq. 7-8 over two parts as h 1 L xcr 0 hx, laminar dx h L xcr x, trubulent dx (7-23) Again taking the critical Reynolds number to be Recr 5 105 and performing the integrations in Eq. 7-23 after substituting the indicated expressions, the average Nusselt number over the entire plate is determined to be (Fig. 7–10) h hx,turbulent Nu h average hx, laminar Laminar 0 xcr Turbulent L x FIGURE 7–10 Graphical representation of the average heat transfer coefficient for a flat plate with combined laminar and turbulent flow. hL 1/3 (0.037 Re0.8 L 871)Pr k 0.6 Pr 60 5 10 5 ReL 10 7 (7-24) The constants in this relation will be different for different critical Reynolds numbers. Liquid metals such as mercury have high thermal conductivities, and are commonly used in applications that require high heat transfer rates. However, they have very small Prandtl numbers, and thus the thermal boundary layer develops much faster than the velocity boundary layer. Then we can assume the velocity in the thermal boundary layer to be constant at the free stream value and solve the energy equation. It gives Nu x 0.565(Re x Pr) 1/2 Pr 0.05 (7-25) It is desirable to have a single correlation that applies to all fluids, including liquid metals. By curve-fitting existing data, Churchill and Ozoe (Ref. 3) proposed the following relation which is applicable for all Prandtl numbers and is claimed to be accurate to 1%, Nu x hx x 0.3387 Pr 1/3 Re1/2 x k [1 (0.0468/Pr)2/3]1/4 (7-26) These relations have been obtained for the case of isothermal surfaces but could also be used approximately for the case of nonisothermal surfaces by assuming the surface temperature to be constant at some average value. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 375 375 CHAPTER 7 Also, the surfaces are assumed to be smooth, and the free stream to be turbulent free. The effect of variable properties can be accounted for by evaluating all properties at the film temperature. Flat Plate with Unheated Starting Length So far we have limited our consideration to situations for which the entire plate is heated from the leading edge. But many practical applications involve surfaces with an unheated starting section of length , shown in Figure 7–11, and thus there is no heat transfer for 0 x . In such cases, the velocity boundary layer starts to develop at the leading edge (x 0), but the thermal boundary layer starts to develop where heating starts (x = ). Consider a flat plate whose heated section is maintained at a constant temperature (T Ts constant for x ). Using integral solution methods (see Kays and Crawford, 1994), the local Nusselt numbers for both laminar and turbulent flows are determined to be Laminar: Turbulent: 1/3 Nux (for 0) 0.332 Re0.5 x Pr Nux 3/4 1/3 3/4 1/3 [1 (/x) ] [1 (/x) ] 1/3 Nux (for 0) 0.0296 Re0.8 x Pr Nux 9/10 1/9 9/10 1/9 [1 ( /x) ] [1 ( /x) ] Velocity boundary layer Ts ξ (7-28) x FIGURE 7–11 Laminar: h 2[1 ( /x) 3/4 ] h xL 1 /L (7-29) Turbulent: h 5[1 ( /x)9/10] h xL 4(1 /L) (7-30) The first relation gives the average convection coefficient for the entire heated section of the plate when the flow is laminar over the entire plate. Note that for 0 it reduces to hL 2hx L , as expected. The second relation gives the average convection coefficient for the case of turbulent flow over the entire plate or when the laminar flow region is small relative to the turbulent region. Uniform Heat Flux When a flat plate is subjected to uniform heat flux instead of uniform temperature, the local Nusselt number is given by Turbulent: Thermal boundary layer (7-27) for x . Note that for 0, these Nux relations reduce to Nux (for 0), which is the Nusselt number relation for a flat plate without an unheated starting length. Therefore, the terms in brackets in the denominator serve as correction factors for plates with unheated starting lengths. The determination of the average Nusselt number for the heated section of a plate requires the integration of the local Nusselt number relations above, which cannot be done analytically. Therefore, integrations must be done numerically. The results of numerical integrations have been correlated for the average convection coefficients [Thomas, (1977) Ref. 11] as Laminar: T 1/3 Nu x 0.453 Re 0.5 x Pr (7-31) Nu x 0.0308 Re x0.8 Pr1/3 (7-32) Flow over a flat plate with an unheated starting length. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 376 376 HEAT TRANSFER These relations give values that are 36 percent higher for laminar flow and 4 percent higher for turbulent flow relative to the isothermal plate case. When the plate involves an unheated starting length, the relations developed for the uniform surface temperature case can still be used provided that Eqs. 7-31 and 7-32 are used for Nux(for 0) in Eqs. 7-27 and 7-28, respectively. When heat flux q·s is prescribed, the rate of heat transfer to or from the plate and the surface temperature at a distance x are determined from Q˙ q˙s A s (7-33) and q˙s h x [Ts( x) T] → Ts(x) T q˙s hx (7-34) where As is the heat transfer surface area. EXAMPLE 7–1 Flow of Hot Oil over a Flat Plate Engine oil at 60°C flows over the upper surface of a 5-m-long flat plate whose temperature is 20°C with a velocity of 2 m/s (Fig. 7–12). Determine the total drag force and the rate of heat transfer per unit width of the entire plate. T = 60°C = 2 m/s . Oil Q Ts = 20°C A L=5m FIGURE 7–12 Schematic for Example 7-1. SOLUTION Engine oil flows over a flat plate. The total drag force and the rate of heat transfer per unit width of the plate are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Recr 5 105. Properties The properties of engine oil at the film temperature of Tf (Ts T)/ 2 (20 60)/2 40°C are (Table A–14). 876 kg/m3 k 0.144 W/m °C Pr 2870 242 106 m2/s Analysis Noting that L 5 m, the Reynolds number at the end of the plate is (2 m/s)(5 m) L ReL 4.13 104 0.242 105 m 2/s which is less than the critical Reynolds number. Thus we have laminar flow over the entire plate, and the average friction coefficient is Cf 1.328 Re L0.5 1.328 (4.13 103)0.5 0.0207 Noting that the pressure drag is zero and thus CD Cf for a flat plate, the drag force acting on the plate per unit width becomes (876 kg/m3)(2 m/s) 2 2 1N 0.0207 (5 1 m2) 2 2 1 kg · m/s2 181 N FD Cf As The total drag force acting on the entire plate can be determined by multiplying the value obtained above by the width of the plate. This force per unit width corresponds to the weight of a mass of about 18 kg. Therefore, a person who applies an equal and opposite force to the plate to keep cen58933_ch07.qxd 9/4/2002 12:12 PM Page 377 377 CHAPTER 7 it from moving will feel like he or she is using as much force as is necessary to hold a 18-kg mass from dropping. Similarly, the Nusselt number is determined using the laminar flow relations for a flat plate, Nu hL 0.664 ReL0.5 Pr1/3 0.664 (4.13 104)0.5 28701/3 1918 k Then, h 0.144 W/m · °C k Nu (1918) 55.2 W/m2 · °C L 5m and · Q hAs(T Ts) (55.2 W/m 2 · °C)(5 1 m2)(60 20)°C 11,040 W Discussion Note that heat transfer is always from the higher-temperature medium to the lower-temperature one. In this case, it is from the oil to the plate. The heat transfer rate is per m width of the plate. The heat transfer for the entire plate can be obtained by multiplying the value obtained by the actual width of the plate. EXAMPLE 7–2 Cooling of a Hot Block by Forced Air at High Elevation The local atmospheric pressure in Denver, Colorado (elevation 1610 m), is 83.4 kPa. Air at this pressure and 20°C flows with a velocity of 8 m/s over a 1.5 m 6 m flat plate whose temperature is 140°C (Fig. 7-13). Determine the rate of heat transfer from the plate if the air flows parallel to the (a) 6-m-long side and (b) the 1.5-m side. k 0.02953 W/m · °C @ 1 atm 2.097 105 m2/s Pr 0.7154 The atmospheric pressure in Denver is P (83.4 kPa)/(101.325 kPa/atm) 0.823 atm. Then the kinematic viscosity of air in Denver becomes @ 1 atm /P (2.097 105 m2/s)/0.823 2.548 105 m2/s Analysis (a ) When air flow is parallel to the long side, we have L 6 m, and the Reynolds number at the end of the plate becomes L (8 m/s)(6 m) ReL 1.884 106 2.548 105 m2/s Ts = 140°C T = 20°C = 8 m/s · Q m Air 1.5 SOLUTION The top surface of a hot block is to be cooled by forced air. The rate of heat transfer is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr 5 105. 3 Radiation effects are negligible. 4 Air is an ideal gas. Properties The properties k, , Cp, and Pr of ideal gases are independent of pressure, while the properties and are inversely proportional to density and thus pressure. The properties of air at the film temperature of Tf (Ts T)/2 (140 20)/2 80°C and 1 atm pressure are (Table A–15) Patm = 83.4 kPa 6m FIGURE 7–13 Schematic for Example 7-2. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 378 378 HEAT TRANSFER which is greater than the critical Reynolds number. Thus, we have combined laminar and turbulent flow, and the average Nusselt number for the entire plate is determined to be hL (0.037 ReL0.8 871)Pr1/3 k [0.037(1.884 106)0.8 871]0.71541/3 2687 Nu Then 0.02953 W/m · °C k Nu (2687) 13.2 W/m2 · °C L 6m As wL (1.5 m)(6 m) 9 m2 h and · Q hAs(Ts T) (13.2 W/m2 · °C)(9 m2)(140 20)°C 1.43 104 W 20°C 8 m/s 140°C Air Note that if we disregarded the laminar region and assumed turbulent flow over the entire plate, we would get Nu 3466 from Eq. 7–22, which is 29 percent higher than the value calculated above. (b) When air flow is along the short side, we have L 1.5 m, and the Reynolds number at the end of the plate becomes L (8 m/s)(1.5 m) 4.71 10 5 ReL 2.548 105 m2/s 1.5 m · Qconv = 14,300 W 6m which is less than the critical Reynolds number. Thus we have laminar flow over the entire plate, and the average Nusselt number is (a) Flow along the long side Air 20°C 8 m/s Nu 140°C 6m (b) Flow along the short side FIGURE 7–14 The direction of fluid flow can have a significant effect on convection heat transfer. Then 0.02953 W/m · °C k Nu (408) 8.03 W/m2 · °C L 1.5 m m h 1.5 · Qconv = 8670 W hL 0.664 ReL0.5 Pr1/3 0.664 (4.71 105)0.5 0.71541/3 408 k and · Q hAs(Ts T) (8.03 W/m2 · °C)(9 m2)(140 20)°C 8670 W which is considerably less than the heat transfer rate determined in case (a). Discussion Note that the direction of fluid flow can have a significant effect on convection heat transfer to or from a surface (Fig. 7-14). In this case, we can increase the heat transfer rate by 65 percent by simply blowing the air along the long side of the rectangular plate instead of the short side. EXAMPLE 7–3 Cooling of Plastic Sheets by Forced Air The forming section of a plastics plant puts out a continuous sheet of plastic that is 4 ft wide and 0.04 in. thick at a velocity of 30 ft/min. The temperature of the plastic sheet is 200°F when it is exposed to the surrounding air, and a 2-ft-long section of the plastic sheet is subjected to air flow at 80°F at a velocity of 10 ft/s on both sides along its surfaces normal to the direction of motion cen58933_ch07.qxd 9/4/2002 12:12 PM Page 379 379 CHAPTER 7 of the sheet, as shown in Figure 7–15. Determine (a) the rate of heat transfer from the plastic sheet to air by forced convection and radiation and (b) the temperature of the plastic sheet at the end of the cooling section. Take the density, specific heat, and emissivity of the plastic sheet to be 75 lbm/ft3, Cp 0.4 Btu/lbm · °F, and 0.9. SOLUTION Plastic sheets are cooled as they leave the forming section of a plastics plant. The rate of heat loss from the plastic sheet by convection and radiation and the exit temperature of the plastic sheet are to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr 5 105. 3 Air is an ideal gas. 4 The local atmospheric pressure is 1 atm. 5 The surrounding surfaces are at the temperature of the room air. Properties The properties of the plastic sheet are given in the problem statement. The properties of air at the film temperature of Tf (Ts T)/2 (200 80)/2 140°F and 1 atm pressure are (Table A–15E) k 0.01623 Btu/h · ft · °F Pr 0.7202 0.7344 ft2/h 0.204 103 ft2/s Analysis (a) We expect the temperature of the plastic sheet to drop somewhat as it flows through the 2-ft-long cooling section, but at this point we do not know the magnitude of that drop. Therefore, we assume the plastic sheet to be isothermal at 200°F to get started. We will repeat the calculations if necessary to account for the temperature drop of the plastic sheet. Noting that L 4 ft, the Reynolds number at the end of the air flow across the plastic sheet is (10 ft /s)(4 ft) L ReL 1.961 10 5 0.204 103 ft 2/s which is less than the critical Reynolds number. Thus, we have laminar flow over the entire sheet, and the Nusselt number is determined from the laminar flow relations for a flat plate to be Nu hL 0.664 ReL0.5 Pr1/3 0.664 (1.961 105)0.5 (0.7202)1/3 263.6 k Then, 0.01623 Btu/ h ft °F k Nu (263.6) 1.07 Btu/h · ft2 · °F L 4 ft As (2 ft)(4 ft)(2 sides) 16 ft2 h and · Q conv hAs(Ts T ) (1.07 Btu/h · ft2 · °F)(16 ft2)(200 80)°F 2054 Btu/h · 4 ) Q rad As(Ts4 Tsurr (0.9)(0.1714 108 Btu/h · ft2 · R4)(16 ft2)[(660 R)4 (540 R)4] 2584 Btu/h 2 ft 200°F Plastic sheet Air 80°F, 10 ft/s 4 ft 0.04 in 30 ft/min FIGURE 7–15 Schematic for Example 7–3. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 380 380 HEAT TRANSFER Therefore, the rate of cooling of the plastic sheet by combined convection and radiation is · · · Qtotal Q conv Q rad 2054 2584 4638 Btu/h (b) To find the temperature of the plastic sheet at the end of the cooling section, we need to know the mass of the plastic rolling out per unit time (or the mass flow rate), which is determined from 4 0.04 3 30 ft m· Acplastic (75 lbm/ft3) ft /s 0.5 lbm/s 12 60 Then, an energy balance on the cooled section of the plastic sheet yields · Q m· Cp(T2 T1) · Q → T2 T1 · m Cp · Noting that Q is a negative quantity (heat loss) for the plastic sheet and substituting, the temperature of the plastic sheet as it leaves the cooling section is determined to be T2 200°F 4638 Btu/h 1h 193.6°F (0.5 lbm/s)(0.4 Btu/lbm · °F) 3600 s Discussion The average temperature of the plastic sheet drops by about 6.4°F as it passes through the cooling section. The calculations now can be repeated by taking the average temperature of the plastic sheet to be 196.8°F instead of 200°F for better accuracy, but the change in the results will be insignificant because of the small change in temperature. 7–3 Midplane θ D Wake Separation Stagnation point point Boundary layer FIGURE 7–16 Typical flow patterns in cross flow over a cylinder. ■ FLOW ACROSS CYLINDERS AND SPHERES Flow across cylinders and spheres is frequently encountered in practice. For example, the tubes in a shell-and-tube heat exchanger involve both internal flow through the tubes and external flow over the tubes, and both flows must be considered in the analysis of the heat exchanger. Also, many sports such as soccer, tennis, and golf involve flow over spherical balls. The characteristic length for a circular cylinder or sphere is taken to be the external diameter D. Thus, the Reynolds number is defined as Re D/ where is the uniform velocity of the fluid as it approaches the cylinder or sphere. The critical Reynolds number for flow across a circular cylinder or sphere is about Recr 2 105. That is, the boundary layer remains laminar for about Re 2 105 and becomes turbulent for Re 2 105. Cross flow over a cylinder exhibits complex flow patterns, as shown in Figure 7–16. The fluid approaching the cylinder branches out and encircles the cylinder, forming a boundary layer that wraps around the cylinder. The fluid particles on the midplane strike the cylinder at the stagnation point, bringing the fluid to a complete stop and thus raising the pressure at that point. The pressure decreases in the flow direction while the fluid velocity increases. At very low upstream velocities (Re 1), the fluid completely wraps around the cylinder and the two arms of the fluid meet on the rear side of the cylinder cen58933_ch07.qxd 9/4/2002 12:12 PM Page 381 381 CHAPTER 7 400 200 100 60 40 CD 20 10 6 4 2 Smooth cylinder 1 0.6 0.4 Sphere 0.2 0.1 0.06 10–1 100 101 102 103 104 105 106 Re FIGURE 7–17 Average drag coefficient for cross flow over a smooth circular cylinder and a smooth sphere (from Schlichting, Ref. 10). in an orderly manner. Thus, the fluid follows the curvature of the cylinder. At higher velocities, the fluid still hugs the cylinder on the frontal side, but it is too fast to remain attached to the surface as it approaches the top of the cylinder. As a result, the boundary layer detaches from the surface, forming a separation region behind the cylinder. Flow in the wake region is characterized by random vortex formation and pressures much lower than the stagnation point pressure. The nature of the flow across a cylinder or sphere strongly affects the total drag coefficient CD. Both the friction drag and the pressure drag can be significant. The high pressure in the vicinity of the stagnation point and the low pressure on the opposite side in the wake produce a net force on the body in the direction of flow. The drag force is primarily due to friction drag at low Reynolds numbers (Re 10) and to pressure drag at high Reynolds numbers (Re 5000). Both effects are significant at intermediate Reynolds numbers. The average drag coefficients CD for cross flow over a smooth single circular cylinder and a sphere are given in Figure 7–17. The curves exhibit different behaviors in different ranges of Reynolds numbers: • For Re 1, we have creeping flow, and the drag coefficient decreases with increasing Reynolds number. For a sphere, it is CD 24/Re. There is no flow separation in this regime. • At about Re 10, separation starts occurring on the rear of the body with vortex shedding starting at about Re 90. The region of separation increases with increasing Reynolds number up to about Re 103. At this point, the drag is mostly (about 95 percent) due to pressure drag. The drag coefficient continues to decrease with increasing Reynolds number in this range of 10 Re 103. (A decrease in the drag coefficient does not necessarily indicate a decrease in drag. The drag force is proportional to the square of the velocity, and the increase in velocity at higher Reynolds numbers usually more than offsets the decrease in the drag coefficient.) cen58933_ch07.qxd 9/4/2002 12:12 PM Page 382 382 HEAT TRANSFER • In the moderate range of 103 Re 105, the drag coefficient remains relatively constant. This behavior is characteristic of blunt bodies. The flow in the boundary layer is laminar in this range, but the flow in the separated region past the cylinder or sphere is highly turbulent with a wide turbulent wake. • There is a sudden drop in the drag coefficient somewhere in the range of 105 Re 106 (usually, at about 2 105). This large reduction in CD is due to the flow in the boundary layer becoming turbulent, which moves the separation point further on the rear of the body, reducing the size of the wake and thus the magnitude of the pressure drag. This is in contrast to streamlined bodies, which experience an increase in the drag coefficient (mostly due to friction drag) when the boundary layer becomes turbulent. Flow separation occurs at about 80 (measured from the stagnation point) when the boundary layer is laminar and at about 140 when it is turbulent (Fig. 7–18). The delay of separation in turbulent flow is caused by the rapid fluctuations of the fluid in the transverse direction, which enables the turbulent boundary layer to travel further along the surface before separation occurs, resulting in a narrower wake and a smaller pressure drag. In the range of Reynolds numbers where the flow changes from laminar to turbulent, even the drag force FD decreases as the velocity (and thus Reynolds number) increases. This results in a sudden decrease in drag of a flying body and instabilities in flight. Laminar boundary layer Separation (a) Laminar flow (Re < 2 × 10 5) Laminar boundary layer Transition Effect of Surface Roughness Turbulent boundary layer Separation (b) Turbulence occurs (Re > 2 × 10 5) FIGURE 7–18 Turbulence delays flow separation. We mentioned earlier that surface roughness, in general, increases the drag coefficient in turbulent flow. This is especially the case for streamlined bodies. For blunt bodies such as a circular cylinder or sphere, however, an increase in the surface roughness may actually decrease the drag coefficient, as shown in Figure 7–19 for a sphere. This is done by tripping the flow into turbulence at a lower Reynolds number, and thus causing the fluid to close in behind the body, narrowing the wake and reducing pressure drag considerably. This results in a much smaller drag coefficient and thus drag force for a roughsurfaced cylinder or sphere in a certain range of Reynolds number compared to a smooth one of identical size at the same velocity. At Re 105, for example, CD 0.1 for a rough sphere with /D 0.0015, whereas CD 0.5 for a smooth one. Therefore, the drag coefficient in this case is reduced by a factor of 5 by simply roughening the surface. Note, however, that at Re 106, CD 0.4 for the rough sphere while CD 0.1 for the smooth one. Obviously, roughening the sphere in this case will increase the drag by a factor of 4 (Fig. 7–20). The discussion above shows that roughening the surface can be used to great advantage in reducing drag, but it can also backfire on us if we are not careful—specifically, if we do not operate in the right range of Reynolds number. With this consideration, golf balls are intentionally roughened to induce turbulence at a lower Reynolds number to take advantage of the sharp drop in the drag coefficient at the onset of turbulence in the boundary layer (the typical velocity range of golf balls is 15 to 150 m/s, and the Reynolds number is less than 4 105). The critical Reynolds number of dimpled golf balls is cen58933_ch07.qxd 9/4/2002 12:12 PM Page 383 383 CHAPTER 7 0.6 ε = relative roughness D CD = FD 1 ρ 2 π D2 2 4 ( ( 0.5 0.4 Golf ball 0.3 0.2 0.1 0 ε = 1.25×10–2 D ε = 5×10–3 D ε = 1.5×10–3 D 4×104 ε = 0 (smooth) D 105 4×105 106 4×106 D Re = v FIGURE 7–19 The effect of surface roughness on the drag coefficient of a sphere (from Blevins, Ref. 1). about 4 104. The occurrence of turbulent flow at this Reynolds number reduces the drag coefficient of a golf ball by half, as shown in Figure 7–19. For a given hit, this means a longer distance for the ball. Experienced golfers also give the ball a spin during the hit, which helps the rough ball develop a lift and thus travel higher and further. A similar argument can be given for a tennis ball. For a table tennis ball, however, the distances are very short, and the balls never reach the speeds in the turbulent range. Therefore, the surfaces of table tennis balls are made smooth. Once the drag coefficient is available, the drag force acting on a body in cross flow can be determined from Eq. 7-1 where A is the frontal area (A LD for a cylinder of length L and A D2/4 for a sphere). It should be kept in mind that the free-stream turbulence and disturbances by other bodies in flow (such as flow over tube bundles) may affect the drag coefficients significantly. EXAMPLE 7–4 Drag Force Acting on a Pipe in a River A 2.2-cm-outer-diameter pipe is to cross a river at a 30-m-wide section while being completely immersed in water (Fig. 7–21). The average flow velocity of water is 4 m/s and the water temperature is 15C. Determine the drag force exerted on the pipe by the river. SOLUTION A pipe is crossing a river. The drag force that acts on the pipe is to be determined. Assumptions 1 The outer surface of the pipe is smooth so that Figure 7–17 can be used to determine the drag coefficient. 2 Water flow in the river is steady. 3 The direction of water flow is normal to the pipe. 4 Turbulence in river flow is not considered. CD Re Smooth surface Rough surface, /D 0.0015 105 106 0.5 0.1 0.1 0.4 FIGURE 7–20 Surface roughness may increase or decrease the drag coefficient of a spherical object, depending on the value of the Reynolds number. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 384 384 HEAT TRANSFER River Pipe FIGURE 7–21 Schematic for Example 7–4. 30 m Properties The density and dynamic viscosity of water at 15C are 999.1 kg/m3 and 1.138 103 kg/m · s (Table A-9). Analysis Noting that D 0.022 m, the Reynolds number for flow over the pipe is 3 D D (999.1 kg/m )(4 m/s)(0.022 m) Re 7.73 104 1.138 103 kg/m · s The drag coefficient corresponding to this value is, from Figure 7-17, CD 1.0. Also, the frontal area for flow past a cylinder is A LD. Then the drag force acting on the pipe becomes 800 FD CD A θ 700 2 (999.1 kg/m3)(4 m/s)2 1N 1.0(30 0.022 m2) 2 2 1 kg · m/s2 5275 N D Discussion Note that this force is equivalent to the weight of a mass over 500 kg. Therefore, the drag force the river exerts on the pipe is equivalent to hanging a total of over 500 kg in mass on the pipe supported at its ends 30 m apart. The necessary precautions should be taken if the pipe cannot support this force. 600 Nuθ Re = 219 ,0 186,00 00 0 500 170,000 140,0 00 400 300 101,300 Heat Transfer Coefficient 70,800 Flows across cylinders and spheres, in general, involve flow separation, which is difficult to handle analytically. Therefore, such flows must be studied experimentally or numerically. Indeed, flow across cylinders and spheres has been studied experimentally by numerous investigators, and several empirical correlations have been developed for the heat transfer coefficient. The complicated flow pattern across a cylinder greatly influences heat transfer. The variation of the local Nusselt number Nu around the periphery of a cylinder subjected to cross flow of air is given in Figure 7–22. Note that, for all cases, the value of Nu starts out relatively high at the stagnation point ( 0°) but decreases with increasing as a result of the thickening of the laminar boundary layer. On the two curves at the bottom corresponding to Re 70,800 and 101,300, Nu reaches a minimum at 80°, which is the separation point in laminar flow. Then Nu increases with increasing as a result of the intense mixing in the separated flow region (the wake). The curves 200 100 0 0° 40° 80° 120° θ from stagnation point 160° FIGURE 7–22 Variation of the local heat transfer coefficient along the circumference of a circular cylinder in cross flow of air (from Giedt, Ref. 5). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 385 385 CHAPTER 7 at the top corresponding to Re 140,000 to 219,000 differ from the first two curves in that they have two minima for Nu. The sharp increase in Nu at about 90° is due to the transition from laminar to turbulent flow. The later decrease in Nu is again due to the thickening of the boundary layer. Nu reaches its second minimum at about 140°, which is the flow separation point in turbulent flow, and increases with as a result of the intense mixing in the turbulent wake region. The discussions above on the local heat transfer coefficients are insightful; however, they are of little value in heat transfer calculations since the calculation of heat transfer requires the average heat transfer coefficient over the entire surface. Of the several such relations available in the literature for the average Nusselt number for cross flow over a cylinder, we present the one proposed by Churchill and Bernstein: Nucyl 0.62 Re1/2 Pr1/3 Re hD 0.3 1 [1 (0.4/Pr)2/3]1/4 282,000 k 5/8 4/5 (7-35) This relation is quite comprehensive in that it correlates available data well for Re Pr 0.2. The fluid properties are evaluated at the film temperature Tf 12 (T Ts), which is the average of the free-stream and surface temperatures. For flow over a sphere, Whitaker recommends the following comprehensive correlation: Nusph hD 2 [0.4 Re1/2 0.06 Re2/3] Pr0.4 s k 1/4 (7-36) which is valid for 3.5 Re 80,000 and 0.7 Pr 380. The fluid properties in this case are evaluated at the free-stream temperature T, except for s, which is evaluated at the surface temperature Ts. Although the two relations above are considered to be quite accurate, the results obtained from them can be off by as much as 30 percent. The average Nusselt number for flow across cylinders can be expressed compactly as Nucyl hD C Rem Prn k (7-37) where n 13 and the experimentally determined constants C and m are given in Table 7–1 for circular as well as various noncircular cylinders. The characteristic length D for use in the calculation of the Reynolds and the Nusselt numbers for different geometries is as indicated on the figure. All fluid properties are evaluated at the film temperature. The relations for cylinders above are for single cylinders or cylinders oriented such that the flow over them is not affected by the presence of others. Also, they are applicable to smooth surfaces. Surface roughness and the freestream turbulence may affect the drag and heat transfer coefficients significantly. Eq. 7-37 provides a simpler alternative to Eq. 7-35 for flow over cylinders. However, Eq. 7-35 is more accurate, and thus should be preferred in calculations whenever possible. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 386 386 HEAT TRANSFER TABLE 7-1 Empirical correlations for the average Nusselt number for forced convection over circular and noncircular cylinders in cross flow (from Zukauskas, Ref. 14, and Jakob, Ref. 6) Cross-section of the cylinder Fluid Circle D Square Range of Re Nusselt number 0.989Re0.330 0.911Re0.385 0.683Re0.466 0.193Re0.618 0.027Re0.805 Pr1/3 Pr1/3 Pr1/3 Pr1/3 Pr1/3 Gas or liquid 0.4–4 4–40 40–4000 4000–40,000 40,000–400,000 Nu Nu Nu Nu Nu Gas 5000–100,000 Nu 0.102Re0.675 Pr1/3 Gas 5000–100,000 Nu 0.246Re0.588 Pr1/3 Gas 5000–100,000 Nu 0.153Re0.638 Pr1/3 Gas 5000–19,500 19,500–100,000 Nu 0.160Re0.638 Pr1/3 Nu 0.0385Re0.782 Pr1/3 Gas 4000–15,000 Nu 0.228Re0.731 Pr1/3 Gas 2500–15,000 Nu 0.248Re0.612 Pr1/3 D Square (tilted 45°) D Hexagon D Hexagon (tilted 45°) Vertical plate D D Ellipse Ts = 110°C D Wind = 8 m/s T = 10°C .1 D =0 m EXAMPLE 7–5 FIGURE 7–23 Schematic for Example 7–5. Heat Loss from a Steam Pipe in Windy Air A long 10-cm-diameter steam pipe whose external surface temperature is 110°C passes through some open area that is not protected against the winds (Fig. 7–23). Determine the rate of heat loss from the pipe per unit of its length cen58933_ch07.qxd 9/4/2002 12:12 PM Page 387 387 CHAPTER 7 when the air is at 1 atm pressure and 10°C and the wind is blowing across the pipe at a velocity of 8 m/s. SOLUTION A steam pipe is exposed to windy air. The rate of heat loss from the steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas. Properties The properties of air at the average film temperature of Tf (Ts T)/2 (110 10)/2 60°C and 1 atm pressure are (Table A-15) k 0.02808 W/m · °C 1.896 105 m2/s Pr 0.7202 Analysis The Reynolds number is Re D (8 m/s)(0.1 m) 4.219 104 1.896 105 m2/s The Nusselt number can be determined from Nu 0.3 0.62(4.219 10 ) (0.7202) [1 (0.4/0.7202)2/3]1/4 4 1/2 1/3 5/8 4/5 0.62 Re1/2 Pr1/3 Re hD 0.3 1 [1 (0.4/Pr)2/3]1/4 282,000 k 4.219 10 4 282,000 1 5/8 4/5 124 and h 0.02808 W/m · °C k Nu (124) 34.8 W/m2 · °C D 0.1 m Then the rate of heat transfer from the pipe per unit of its length becomes As pL DL (0.1 m)(1 m) 0.314 m2 · Q hAs(Ts T) (34.8 W/m2 · C)(0.314 m2)(110 10)°C 1093 W The rate of heat loss from the entire pipe can be obtained by multiplying the value above by the length of the pipe in m. Discussion The simpler Nusselt number relation in Table 7–1 in this case would give Nu 128, which is 3 percent higher than the value obtained above using Eq. 7-35. EXAMPLE 7–6 Cooling of a Steel Ball by Forced Air A 25-cm-diameter stainless steel ball ( 8055 kg/m , Cp 480 J/kg · °C) is removed from the oven at a uniform temperature of 300°C (Fig. 7–24). The ball is then subjected to the flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. The surface temperature of the ball eventually drops to 200°C. Determine the average convection heat transfer coefficient during this cooling process and estimate how long the process will take. Air 3 T = 25°C = 3 m/s Steel ball 300°C FIGURE 7–24 Schematic for Example 7–6. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 388 388 HEAT TRANSFER SOLUTION A hot stainless steel ball is cooled by forced air. The average convection heat transfer coefficient and the cooling time are to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas. 4 The outer surface temperature of the ball is uniform at all times. 5 The surface temperature of the ball during cooling is changing. Therefore, the convection heat transfer coefficient between the ball and the air will also change. To avoid this complexity, we take the surface temperature of the ball to be constant at the average temperature of (300 200)/2 250°C in the evaluation of the heat transfer coefficient and use the value obtained for the entire cooling process. Properties The dynamic viscosity of air at the average surface temperature is s @ 250°C 2.76 105 kg/m · s. The properties of air at the free-stream temperature of 25°C and 1 atm are (Table A-15) k 0.02551 W/m · °C 1.849 105 kg/m · s 1.562 105 m2/s Pr 0.7296 Analysis The Reynolds number is determined from (3 m/s)(0.25 m) D Re 4.802 104 1.562 105 m2/s The Nusselt number is Nu hD 2 [0.4 Re1/2 0.06 Re2/3] Pr0.4 s k 1/4 2 0.4(4.802 104)1/2 0.06(4.802 104)2/30.4 1.849 105 1/4 2.76 10 5 135 Then the average convection heat transfer coefficient becomes h 0.02551 W/m · °C k Nu (135) 13.8 W/m2 °C D 0.25 m In order to estimate the time of cooling of the ball from 300°C to 200°C, we determine the average rate of heat transfer from Newton’s law of cooling by using the average surface temperature. That is, As D 2 (0.25 m)2 0.1963 m2 · Q ave hAs(Ts, ave T) (13.8 W/m2 · °C)(0.1963 m2)(250 25)°C 610 W Next we determine the total heat transferred from the ball, which is simply the change in the energy of the ball as it cools from 300°C to 200°C: m V 16D 3 (8055 kg/m3) 16(0.25 m)3 65.9 kg Qtotal mCp(T2 T1) (65.9 kg)(480 J/kg · °C)(300 200)°C 3,163,000 J In this calculation, we assumed that the entire ball is at 200°C, which is not necessarily true. The inner region of the ball will probably be at a higher temperature than its surface. With this assumption, the time of cooling is determined to be cen58933_ch07.qxd 9/4/2002 12:12 PM Page 389 389 CHAPTER 7 Q t · Q ave 3,163,000 J 5185 s 1 h 26 min 610 J/s Discussion The time of cooling could also be determined more accurately using the transient temperature charts or relations introduced in Chapter 4. But the simplifying assumptions we made above can be justified if all we need is a ballpark value. It will be naive to expect the time of cooling to be exactly 1 h 26 min, but, using our engineering judgment, it is realistic to expect the time of cooling to be somewhere between one and two hours. Flow direction ↑  7–4 ■ FLOW ACROSS TUBE BANKS Cross-flow over tube banks is commonly encountered in practice in heat transfer equipment such as the condensers and evaporators of power plants, refrigerators, and air conditioners. In such equipment, one fluid moves through the tubes while the other moves over the tubes in a perpendicular direction. In a heat exchanger that involves a tube bank, the tubes are usually placed in a shell (and thus the name shell-and-tube heat exchanger), especially when the fluid is a liquid, and the fluid flows through the space between the tubes and the shell. There are numerous types of shell-and-tube heat exchangers, some of which are considered in Chap. 13. In this section we will consider the general aspects of flow over a tube bank, and try to develop a better and more intuitive understanding of the performance of heat exchangers involving a tube bank. Flow through the tubes can be analyzed by considering flow through a single tube, and multiplying the results by the number of tubes. This is not the case for flow over the tubes, however, since the tubes affect the flow pattern and turbulence level downstream, and thus heat transfer to or from them, as shown in Figure 7–25. Therefore, when analyzing heat transfer from a tube bank in cross flow, we must consider all the tubes in the bundle at once. The tubes in a tube bank are usually arranged either in-line or staggered in the direction of flow, as shown in Figure 7–26. The outer tube diameter D is taken as the characteristic length. The arrangement of the tubes in the tube bank is characterized by the transverse pitch ST, longitudinal pitch SL , and the diagonal pitch SD between tube centers. The diagonal pitch is determined from SD S 2L (S T /2) 2 (7-38) As the fluid enters the tube bank, the flow area decreases from A1 STL to AT (ST D)L between the tubes, and thus flow velocity increases. In staggered arrangement, the velocity may increase further in the diagonal region if the tube rows are very close to each other. In tube banks, the flow characteristics are dominated by the maximum velocitiy max that occurs within the tube bank rather than the approach velocity . Therefore, the Reynolds number is defined on the basis of maximum velocity as ReD max D max D (7-39) FIGURE 7-25 Flow patterns for staggered and in-line tube banks (photos by R. D. Willis, Ref 12). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 390 390 HEAT TRANSFER The maximum velocity is determined from the conservation of mass requirement for steady incompressible flow. For in-line arrangement, the maximum velocity occurs at the minimum flow area between the tubes, and the conservation of mass can be expressed as (see Fig. 7-26a) A1 maxAT or ST max(ST D). Then the maximum velocity becomes SL , T1 ST D A1 AT max 1st row 2nd row 3rd row (a) In-line SL , T1 SD ST D AD A1 AT AD A1 = ST L AT = (ST D)L AD = (SD D)L (b) Staggered FIGURE 7–26 Arrangement of the tubes in in-line and staggered tube banks (A1, AT, and AD are flow areas at indicated locations, and L is the length of the tubes). ST ST D (7-40) In staggered arrangement, the fluid approaching through area A1 in Figure 7–26b passes through area AT and then through area 2AD as it wraps around the pipe in the next row. If 2AD AT, maximum velocity will still occur at AT between the tubes, and thus the max relation Eq. 7-40 can also be used for staggered tube banks. But if 2AD [or, if 2(SD D) (ST D)], maximum velocity will occur at the diagonal cross sections, and the maximum velocity in this case becomes Staggered and SD (ST D)/2: max ST 2(SD D) (7-41) since A1 max(2AD) or ST 2max(SD D). The nature of flow around a tube in the first row resembles flow over a single tube discussed in section 7–3, especially when the tubes are not too close to each other. Therefore, each tube in a tube bank that consists of a single transverse row can be treated as a single tube in cross-flow. The nature of flow around a tube in the second and subsequent rows is very different, however, because of wakes formed and the turbulence caused by the tubes upstream. The level of turbulence, and thus the heat transfer coefficient, increases with row number because of the combined effects of upstream rows. But there is no significant change in turbulence level after the first few rows, and thus the heat transfer coefficient remains constant. Flow through tube banks is studied experimentally since it is too complex to be treated analytically. We are primarily interested in the average heat transfer coefficient for the entire tube bank, which depends on the number of tube rows along the flow as well as the arrangement and the size of the tubes. Several correlations, all based on experimental data, have been proposed for the average Nusselt number for cross flow over tube banks. More recently, Zukauskas has proposed correlations whose general form is Nu D hD C Re mD Pr n(Pr/Prs) 0.25 k (7-42) where the values of the constants C, m, and n depend on value Reynolds number. Such correlations are given in Table 7–2 explicitly for 0.7 Pr 500 and 0 ReD 2 106. The uncertainty in the values of Nusselt number obtained from these relations is 15 percent. Note that all properties except Prs are to be evaluated at the arithmetic mean temperature of the fluid determined from Tm Ti Te 2 (7-43) where Ti and Te are the fluid temperatures at the inlet and the exit of the tube bank, respectively. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 391 391 CHAPTER 7 TABLE 7–2 Nusselt number correlations for cross flow over tube banks for N 16 and 0.7 Pr 500 (from Zukauskas, Ref. 15, 1987) Arrangement Range of ReD Correlation NuD 0.9 Re D0.4Pr 0.36(Pr/Prs )0.25 0–100 NuD 0.52 Re D0.5Pr 0.36(Pr/Prs )0.25 100–1000 In-line 1000–2 10 5 NuD 0.27 Re D0.63Pr 0.36(Pr/Prs )0.25 2 105–2 106 NuD 0.033 Re D0.8Pr 0.4(Pr/Prs )0.25 NuD 1.04 Re D0.4Pr 0.36(Pr/Prs )0.25 0–500 NuD 0.71 Re D0.5Pr 0.36(Pr/Prs )0.25 500–1000 Staggered 1000–2 10 NuD 0.35(ST /SL)0.2 Re D0.6Pr 0.36(Pr/Prs )0.25 5 2 105–2 106 NuD 0.031(ST /SL)0.2 Re D0.8Pr 0.36(Pr/Prs )0.25 All properties except Prs are to be evaluated at the arithmetic mean of the inlet and outlet temperatures of the fluid (Prs is to be evaluated at Ts ). The average Nusselt number relations in Table 7–2 are for tube banks with 16 or more rows. Those relations can also be used for tube banks with NL provided that they are modified as NuD, NL F NuD (7-44) where F is a correction factor F whose values are given in Table 7–3. For ReD 1000, the correction factor is independent of Reynolds number. Once the Nusselt number and thus the average heat transfer coefficient for the entire tube bank is known, the heat transfer rate can be determined from Newton’s law of cooling using a suitable temperature difference T. The first thought that comes to mind is to use T Ts Tm Ts (Ti Te)/2. But this will, in general, over predict the heat transfer rate. We will show in the next chapter that the proper temperature difference for internal flow (flow over tube banks is still internal flow through the shell) is the 1ogarithmic mean temperature difference Tln defined as Tln (Ts Te) (Ts Ti) Te Ti ln[(Ts Te)/(Ts Ti)] ln(Te /Ti ) (7-45) We will also show that the exit temperature of the fluid Te can be determined from TABLE 7–3 Correction factor F to be used in NuD, NL, = FNuD for NL 16 and ReD 1000 (from Zukauskas, Ref 15, 1987). NL 1 2 3 4 5 7 10 13 In-line 0.70 0.80 0.86 0.90 0.93 0.96 0.98 0.99 Staggered 0.64 0.76 0.84 0.89 0.93 0.96 0.98 0.99 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 392 392 HEAT TRANSFER As h Te Ts (Ts Ti) exp · m Cp (7-46) where As NDL is the heat transfer surface area and m· (N T S T L) is the mass flow rate of the fluid. Here N is the total number of tubes in the bank, NT is the number of tubes in a transverse plane, L is the length of the tubes, and is the velocity of the fluid just before entering the tube bank. Then the heat transfer rate can be determined from ˙ p(Te Ti) Q˙ h As Tln mC (7-47) The second relation is usually more convenient to use since it does not require the calculation of Tln. Pressure Drop Another quantity of interest associated with tube banks is the pressure drop P, which is the difference between the pressures at the inlet and the exit of the tube bank. It is a measure of the resistance the tubes offer to flow over them, and is expressed as P NL f 2 max 2 (7-48) where f is the friction factor and is the correction factor, both plotted in Figures 7–27a and 7–27b against the Reynolds number based on the maximum velocity max. The friction factor in Figure 7–27a is for a square in-line tube bank (ST SL), and the correction factor given in the insert is used to account for the effects of deviation of rectangular in-line arrangements from square arrangement. Similarly, the friction factor in Figure 7–27b is for an equilateral staggered tube bank (ST SD), and the correction factor is to account for the effects of deviation from equilateral arrangement. Note that 1 for both square and equilateral triangle arrangements. Also, pressure drop occurs in the flow direction, and thus we used NL (the number of rows) in the P relation. The power required to move a fluid through a tube bank is proportional to the pressure drop, and when the pressure drop is available, the pumping power required can be determined from ˙ mP ˙ W˙ pump VP (7-49) where V˙ (NT S T L) is the volume flow rate and m˙ V˙ (NT S T L) is the mass flow rate of the fluid through the tube bank. Note that the power required to keep a fluid flowing through the tube bank (and thus the operating cost) is proportional to the pressure drop. Therefore, the benefits of enhancing heat transfer in a tube bank via rearrangement should be weighed against the cost of additional power requirements. In this section we limited our consideration to tube banks with base surfaces (no fins). Tube banks with finned surfaces are also commonly used in practice, especially when the fluid is a gas, and heat transfer and pressure drop correlations can be found in the literature for tube banks with pin fins, plate fins, strip fins, etc. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 393 393 CHAPTER 7 60 40 SL PL SL/D 20 10 PT ST /D 1.5 8 6 4 ST 10 6 PT PL 103 104 2 1 0.6 PL 1.25 ReD,max 105 106 0.2 Friction factor, ƒ 2 0.1 0.2 2.0 1 0.8 0.6 0.4 0.6 1 2 6 10 (PT 1)/(PL 1) 0.2 2.5 3.0 0.1 8 6 4 6 8 101 2 4 6 8 102 2 4 6 8 103 2 4 6 8104 2 4 6 8 105 2 4 6 8 106 ReD,max (a) In-line arrangement 80 60 40 1.6 SD ReD,max 102 20 1.5 10 1.2 104 104 103 4 Friction factor, ƒ 103 SD ST 8 6 1.0 2 PT 1.25 2.0 1 0.8 0.6 0.4 105 102 0.4 0.6 0.8 1 2 4 PT /PL 2.5 3.5 0.2 0.1 105 1.4 2 4 6 8101 2 4 6 8 102 2 4 6 8103 2 4 6 8 104 2 4 6 8105 2 4 6 8 106 ReD,max (b) Staggered arrangement EXAMPLE 7–7 Preheating Air by Geothermal Water in a Tube Bank In an industrial facility, air is to be preheated before entering a furnace by geothermal water at 120ºC flowing through the tubes of a tube bank located in a duct. Air enters the duct at 20ºC and 1 atm with a mean velocity of 4.5 m/s, and flows over the tubes in normal direction. The outer diameter of the tubes is 1.5 cm, and the tubes are arranged in-line with longitudinal and transverse pitches of SL ST 5 cm. There are 6 rows in the flow direction with 10 tubes in each row, as shown in Figure 7–28. Determine the rate of heat transfer per unit length of the tubes, and the pressure drop across the tube bank. SOLUTION Air is heated by geothermal water in a tube bank. The rate of heat transfer to air and the pressure drop of air are to be determined. 2 FIGURE 7–27 Friction factor f and correction factor for tube banks (from Zukauskas, Ref. 16, 1985). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 394 394 HEAT TRANSFER AIR = 4.5 m/s T1 = 20°C Ts = 120°C Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the temperature of geothermal water. Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the assumed mean temperature of 60ºC (will be checked later) and 1 atm are Table A–15): k 0.02808 W/m K, 1.06 kg/m3 Cp 1.007 kJ/kg K, 2.008 10 5 Pr 0.7202 kg/m s Prs Pr@Ts 0.7073 Also, the density of air at the inlet temperature of 20ºC (for use in the mass flow rate calculation at the inlet) is 1 1.204 kg/m3 Analysis It is given that D 0.015 m, SL ST 0.05 m, and 4.5 m/s. Then the maximum velocity and the Reynolds number based on the maximum velocity become ST SL = ST = 5 cm D = 1.5 cm FIGURE 7–28 Schematic for Example 7–7. ST 0.05 (4.5 m /s) 6.43 m /s ST D 0.05 0.015 max D (1.06 kg /m 3)(6.43 m /s)(0.015 m) ReD 5091 2.008 105 kg /m s max The average Nusselt number is determined using the proper relation from Table 7–2 to be 0.36 NuD 0.27 Re 0.63 (Pr/Prs) 0.25 D Pr 0.27(5091)0.63(0.7202)0.36(0.7202/0.7073)0.25 52.2 This Nusselt number is applicable to tube banks with NL 16. In our case, the number of rows is NL 6, and the corresponding correction factor from Table 7–3 is F 0.945. Then the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, NL FNu D (0.945)(52.2) 49.3 NuD, NLk 49.3(0.02808 W/m ºC) 92.2 W/m2 ºC h D 0.015 m The total number of tubes is N NL NT 6 10 60. For a unit tube length (L 1 m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As NDL 60(0.015 m)(1 m) 2.827 m2 ˙ 1 1(NT S T L) m˙ m (1.204 kg/m3)(4.5 m/s)(10)(0.05 m)(1 m) 2.709 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become Ash ˙ p mC (2.827 m 2)(92.2 W/m 2 ºC) 120 (120 20) exp 29.11ºC (2.709 kg/s)(1007 J/kg ºC) Te Ts (Ts Ti) exp cen58933_ch07.qxd 9/4/2002 12:12 PM Page 395 395 CHAPTER 7 (Ts Te) (Ts Ti) (120 29.11) (120 20) 95.4ºC ln[(Ts Te)/(Ts Ti)] ln[(120 29.11)/(120 20)] Q˙ hAsTln (92.2 W/m2 ºC)(2.827 m2)(95.4ºC) 2.49 104 W Tln The rate of heat transfer can also be determined in a simpler way from · Q hAsTin m· Cp(Te Ti) (2.709 kg/s)(1007 J/kg · °C)(29. 11 20)°C 2.49 104 W For this square in-line tube bank, the friction coefficient corresponding to ReD 5088 and SL/D 5/1.5 3.33 is, from Fig. 7–27a, f 0.16. Also, 1 for the square arrangements. Then the pressure drop across the tube bank becomes V 2max 2 (1.06 kg/m3)(6.43 m/s)3 1N 6(0.16)(1) 21 Pa 2 1 kg m /s 2 P NL f Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 (20 110.9)/2 65.4ºC, which is fairly close to the assumed value of 60°C. Therefore, there is no need to repeat calculations by reevaluating the properties at 65.4°C (it can be shown that doing so would change the results by less than 1 percent, which is much less than the uncertainty in the equations and the charts used). TOPIC OF SPECIAL INTEREST Reducing Heat Transfer through Surfaces: Thermal Insulation Thermal insulations are materials or combinations of materials that are used primarily to provide resistance to heat flow (Fig. 7–29). You are probably familiar with several kinds of insulation available in the market. Most insulations are heterogeneous materials made of low thermal conductivity materials, and they involve air pockets. This is not surprising since air has one of the lowest thermal conductivities and is readily available. The Styrofoam commonly used as a packaging material for TVs, VCRs, computers, and just about anything because of its light weight is also an excellent insulator. Temperature difference is the driving force for heat flow, and the greater the temperature difference, the larger the rate of heat transfer. We can slow down the heat flow between two mediums at different temperatures by putting “barriers” on the path of heat flow. Thermal insulations serve as such barriers, and they play a major role in the design and manufacture of all energy-efficient devices or systems, and they are usually the cornerstone of energy conservation projects. A 1991 Drexel University study of the energy-intensive U.S. industries revealed that insulation saves the U.S. This section can be skipped without a loss in continuity. Insulation Heat loss Heat FIGURE 7–29 Thermal insulation retards heat transfer by acting as a barrier in the path of heat flow. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 396 396 HEAT TRANSFER Combustion gases FIGURE 7–30 Insulation also helps the environment by reducing the amount of fuel burned and the air pollutants released. FIGURE 7–31 In cold weather, we minimize heat loss from our bodies by putting on thick layers of insulation (coats or furs). industry nearly 2 billion barrels of oil per year, valued at $60 billion a year in energy costs, and more can be saved by practicing better insulation techniques and retrofitting the older industrial facilities. Heat is generated in furnaces or heaters by burning a fuel such as coal, oil, or natural gas or by passing electric current through a resistance heater. Electricity is rarely used for heating purposes since its unit cost is much higher. The heat generated is absorbed by the medium in the furnace and its surfaces, causing a temperature rise above the ambient temperature. This temperature difference drives heat transfer from the hot medium to the ambient, and insulation reduces the amount of heat loss and thus saves fuel and money. Therefore, insulation pays for itself from the energy it saves. Insulating properly requires a one-time capital investment, but its effects are dramatic and long term. The payback period of insulation is often less than one year. That is, the money insulation saves during the first year is usually greater than its initial material and installation costs. On a broader perspective, insulation also helps the environment and fights air pollution and the greenhouse effect by reducing the amount of fuel burned and thus the amount of CO2 and other gases released into the atmosphere (Fig. 7–30). Saving energy with insulation is not limited to hot surfaces. We can also save energy and money by insulating cold surfaces (surfaces whose temperature is below the ambient temperature) such as chilled water lines, cryogenic storage tanks, refrigerated trucks, and air-conditioning ducts. The source of “coldness” is refrigeration, which requires energy input, usually electricity. In this case, heat is transferred from the surroundings to the cold surfaces, and the refrigeration unit must now work harder and longer to make up for this heat gain and thus it must consume more electrical energy. A cold canned drink can be kept cold much longer by wrapping it in a blanket. A refrigerator with well-insulated walls will consume much less electricity than a similar refrigerator with little or no insulation. Insulating a house will result in reduced cooling load, and thus reduced electricity consumption for air-conditioning. Whether we realize it or not, we have an intuitive understanding and appreciation of thermal insulation. As babies we feel much better in our blankies, and as children we know we should wear a sweater or coat when going outside in cold weather (Fig. 7–31). When getting out of a pool after swimming on a windy day, we quickly wrap in a towel to stop shivering. Similarly, early man used animal furs to keep warm and built shelters using mud bricks and wood. Cork was used as a roof covering for centuries. The need for effective thermal insulation became evident with the development of mechanical refrigeration later in the nineteenth century, and a great deal of work was done at universities and government and private laboratories in the 1910s and 1920s to identify and characterize thermal insulation. Thermal insulation in the form of mud, clay, straw, rags, and wood strips was first used in the eighteenth century on steam engines to keep workmen from being burned by hot surfaces. As a result, boiler room temperatures dropped and it was noticed that fuel consumption was also reduced. The realization of improved engine efficiency and energy savings prompted the search for materials with improved thermal efficiency. One of the first such materials was mineral wool insulation, which, like many materials, was cen58933_ch07.qxd 9/4/2002 12:12 PM Page 397 397 CHAPTER 7 discovered by accident. About 1840, an iron producer in Wales aimed a stream of high-pressure steam at the slag flowing from a blast furnace, and manufactured mineral wool was born. In the early 1860s, this slag wool was a by-product of manufacturing cannons for the Civil War and quickly found its way into many industrial uses. By 1880, builders began installing mineral wool in houses, with one of the most notable applications being General Grant’s house. The insulation of this house was described in an article: “it keeps the house cool in summer and warm in winter; it prevents the spread of fire; and it deadens the sound between floors” [Edmunds (1989), Ref. 4]. An article published in 1887 in Scientific American detailing the benefits of insulating the entire house gave a major boost to the use of insulation in residential buildings. The energy crisis of the 1970s had a tremendous impact on the public awareness of energy and limited energy reserves and brought an emphasis on energy conservation. We have also seen the development of new and more effective insulation materials since then, and a considerable increase in the use of insulation. Thermal insulation is used in more places than you may be aware of. The walls of your house are probably filled with some kind of insulation, and the roof is likely to have a thick layer of insulation. The “thickness” of the walls of your refrigerator is due to the insulation layer sandwiched between two layers of sheet metal (Fig. 7–32). The walls of your range are also insulated to conserve energy, and your hot water tank contains less water than you think because of the 2- to 4-cm-thick insulation in the walls of the tank. Also, your hot water pipe may look much thicker than the cold water pipe because of insulation. Insulation Heat transfer Reasons for Insulating If you examine the engine compartment of your car, you will notice that the firewall between the engine and the passenger compartment as well as the inner surface of the hood are insulated. The reason for insulating the hood is not to conserve the waste heat from the engine but to protect people from burning themselves by touching the hood surface, which will be too hot if not insulated. As this example shows, the use of insulation is not limited to energy conservation. Various reasons for using insulation can be summarized as follows: • Energy Conservation Conserving energy by reducing the rate of heat flow is the primary reason for insulating surfaces. Insulation materials that will perform satisfactorily in the temperature range of 268°C to 1000°C (450°F to 1800°F) are widely available. • Personnel Protection and Comfort A surface that is too hot poses a danger to people who are working in that area of accidentally touching the hot surface and burning themselves (Fig. 7–33). To prevent this danger and to comply with the OSHA (Occupational Safety and Health Administration) standards, the temperatures of hot surfaces should be reduced to below 60°C (140°F) by insulating them. Also, the excessive heat coming off the hot surfaces creates an unpleasant environment in which to work, which adversely affects the performance or productivity of the workers, especially in summer months. FIGURE 7–32 The insulation layers in the walls of a refrigerator reduce the amount of heat flow into the refrigerator and thus the running time of the refrigerator, saving electricity. Insulation FIGURE 7–33 The hood of the engine compartment of a car is insulated to reduce its temperature and to protect people from burning themselves. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 398 398 HEAT TRANSFER Insulation z z z FIGURE 7–34 Insulation materials absorb vibration and sound waves, and are used to minimize sound transmission. • Maintaining Process Temperature Some processes in the chemical industry are temperature-sensitive, and it may become necessary to insulate the process tanks and flow sections heavily to maintain the same temperature throughout. • Reducing Temperature Variation and Fluctuations The temperature in an enclosure may vary greatly between the midsection and the edges if the enclosure is not insulated. For example, the temperature near the walls of a poorly insulated house is much lower than the temperature at the midsections. Also, the temperature in an uninsulated enclosure will follow the temperature changes in the environment closely and fluctuate. Insulation minimizes temperature nonuniformity in an enclosure and slows down fluctuations. • Condensation and Corrosion Prevention Water vapor in the air condenses on surfaces whose temperature is below the dew point, and the outer surfaces of the tanks or pipes that contain a cold fluid frequently fall below the dew-point temperature unless they have adequate insulation. The liquid water on exposed surfaces of the metal tanks or pipes may promote corrosion as well as algae growth. • Fire Protection Damage during a fire may be minimized by keeping valuable combustibles in a safety box that is well insulated. Insulation may lower the rate of heat flow to such levels that the temperature in the box never rises to unsafe levels during fire. • Freezing Protection Prolonged exposure to subfreezing temperatures may cause water in pipes or storage vessels to freeze and burst as a result of heat transfer from the water to the cold ambient. The bursting of pipes as a result of freezing can cause considerable damage. Adequate insulation will slow down the heat loss from the water and prevent freezing during limited exposure to subfreezing temperatures. For example, covering vegetables during a cold night will protect them from freezing, and burying water pipes in the ground at a sufficient depth will keep them from freezing during the entire winter. Wearing thick gloves will protect the fingers from possible frostbite. Also, a molten metal or plastic in a container will solidify on the inner surface if the container is not properly insulated. • Reducing Noise and Vibration An added benefit of thermal insulation is its ability to dampen noise and vibrations (Fig. 7–34). The insulation materials differ in their ability to reduce noise and vibration, and the proper kind can be selected if noise reduction is an important consideration. There are a wide variety of insulation materials available in the market, but most are primarily made of fiberglass, mineral wool, polyethylene, foam, or calcium silicate. They come in various trade names such as Ethafoam Polyethylene Foam Sheeting, Solimide Polimide Foam Sheets, FPC Fiberglass Reinforced Silicone Foam Sheeting, Silicone Sponge Rubber Sheets, fiberglass/mineral wool insulation blankets, wire-reinforced cen58933_ch07.qxd 9/4/2002 12:12 PM Page 399 399 CHAPTER 7 mineral wool insulation, Reflect-All Insulation, granulated bulk mineral wool insulation, cork insulation sheets, foil-faced fiberglass insulation, blended sponge rubber sheeting, and numerous others. Today various forms of fiberglass insulation are widely used in process industries and heating and air-conditioning applications because of their low cost, light weight, resiliency, and versatility. But they are not suitable for some applications because of their low resistance to moisture and fire and their limited maximum service temperature. Fiberglass insulations come in various forms such as unfaced fiberglass insulation, vinyl-faced fiberglass insulation, foil-faced fiberglass insulation, and fiberglass insulation sheets. The reflective foil-faced fiberglass insulation resists vapor penetration and retards radiation because of the aluminum foil on it and is suitable for use on pipes, ducts, and other surfaces. Mineral wool is resilient, lightweight, fibrous, wool-like, thermally efficient, fire resistant up to 1100°C (2000°F), and forms a sound barrier. Mineral wool insulation comes in the form of blankets, rolls, or blocks. Calcium silicate is a solid material that is suitable for use at high temperatures, but it is more expensive. Also, it needs to be cut with a saw during installation, and thus it takes longer to install and there is more waste. Superinsulators You may be tempted to think that the most effective way to reduce heat transfer is to use insulating materials that are known to have very low thermal conductivities such as urethane or rigid foam (k 0.026 W/m · °C) or fiberglass (k 0.035 W/m · °C). After all, they are widely available, inexpensive, and easy to install. Looking at the thermal conductivities of materials, you may also notice that the thermal conductivity of air at room temperature is 0.026 W/m · °C, which is lower than the conductivities of practically all of the ordinary insulating materials. Thus you may think that a layer of enclosed air space is as effective as any of the common insulating materials of the same thickness. Of course, heat transfer through the air will probably be higher than what a pure conduction analysis alone would indicate because of the natural convection currents that are likely to occur in the air layer. Besides, air is transparent to radiation, and thus heat will also be transferred by radiation. The thermal conductivity of air is practically independent of pressure unless the pressure is extremely high or extremely low. Therefore, we can reduce the thermal conductivity of air and thus the conduction heat transfer through the air by evacuating the air space. In the limiting case of absolute vacuum, the thermal conductivity will be zero since there will be no particles in this case to “conduct” heat from one surface to the other, and thus the conduction heat transfer will be zero. Noting that the thermal conductivity cannot be negative, an absolute vacuum must be the ultimate insulator, right? Well, not quite. The purpose of insulation is to reduce “total” heat transfer from a surface, not just conduction. A vacuum totally eliminates conduction but offers zero resistance to radiation, whose magnitude can be comparable to conduction or natural convection in gases (Fig. 7–35). Thus, a vacuum is T1 T2 Vacuum Radiation FIGURE 7–35 Evacuating the space between two surfaces completely eliminates heat transfer by conduction or convection but leaves the door wide open for radiation. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 400 400 HEAT TRANSFER Thin metal sheets T1 T2 Vacuum FIGURE 7–36 Superinsulators are built by closely packing layers of highly reflective thin metal sheets and evacuating the space between them. k, Btu/h·ft·°F Insulation L, ft L R-value = — k FIGURE 7–37 The R-value of an insulating material is simply the ratio of the thickness of the material to its thermal conductivity in proper units. no more effective in reducing heat transfer than sealing off one of the lanes of a two-lane road is in reducing the flow of traffic on a one-way road. Insulation against radiation heat transfer between two surfaces is achieved by placing “barriers” between the two surfaces, which are highly reflective thin metal sheets. Radiation heat transfer between two surfaces is inversely proportional to the number of such sheets placed between the surfaces. Very effective insulations are obtained by using closely packed layers of highly reflective thin metal sheets such as aluminum foil (usually 25 sheets per cm) separated by fibers made of insulating material such as glass fiber (Fig. 7–36). Further, the space between the layers is evacuated to form a vacuum under 0.000001 atm pressure to minimize conduction or convection heat transfer through the air space between the layers. The result is an insulating material whose apparent thermal conductivity is below 2 105 W/m · °C, which is one thousand times less than the conductivity of air or any common insulating material. These specially built insulators are called superinsulators, and they are commonly used in space applications and cryogenics, which is the branch of heat transfer dealing with temperatures below 100 K (173°C) such as those encountered in the liquefaction, storage, and transportation of gases, with helium, hydrogen, nitrogen, and oxygen being the most common ones. The R-value of Insulation The effectiveness of insulation materials is given by some manufacturers in terms of their R-value, which is the thermal resistance of the material per unit surface area. For flat insulation the R-value is obtained by simply dividing the thickness of the insulation by its thermal conductivity. That is, R-value L k (flat insulation) (7-50) where L is the thickness and k is the thermal conductivity of the material. Note that doubling the thickness L doubles the R-value of flat insulation. For pipe insulation, the R-value is determined using the thermal resistance relation from r2 r2 R-value ln r (pipe insulation) (7-51) 1 k where r1 is the inside radius of insulation and r2 is the outside radius of insulation. Once the R-value is available, the rate of heat transfer through the insulation can be determined from · Q T Area R-value (7-52) where T is the temperature difference across the insulation and Area is the outer surface area for a cylinder. In the United States, the R-values of insulation are expressed without any units, such as R-19 and R-30. These R-values are obtained by dividing the thickness of the material in feet by its thermal conductivity in the unit Btu/h · ft · °F so that the R-values actually have the unit h · ft2 · °F/Btu. For example, the R-value of 6-in.-thick glass fiber insulation whose thermal conductivity is 0.025 Btu/h · ft · °F is (Fig. 7–37) cen58933_ch07.qxd 9/4/2002 12:13 PM Page 401 401 CHAPTER 7 R-value 0.5 ft L 20 h · ft2 · °F/Btu k 0.025 Btu/h · ft · °F Thus, this 6-in.-thick glass fiber insulation would be referred to as R-20 insulation by the builders. The unit of R-value is m2 · °C/W in SI units, with the conversion relation 1 m2 · °C/W 5.678 h · ft2 · °F/Btu. Therefore, a small R-value in SI corresponds to a large R-value in English units. Optimum Thickness of Insulation It should be realized that insulation does not eliminate heat transfer; it merely reduces it. The thicker the insulation, the lower the rate of heat transfer but also the higher the cost of insulation. Therefore, there should be an optimum thickness of insulation that corresponds to a minimum combined cost of insulation and heat lost. The determination of the optimum thickness of insulation is illustrated in Figure 7–38. Notice that the cost of insulation increases roughly linearly with thickness while the cost of heat loss decreases exponentially. The total cost, which is the sum of the insulation cost and the lost heat cost, decreases first, reaches a minimum, and then increases. The thickness corresponding to the minimum total cost is the optimum thickness of insulation, and this is the recommended thickness of insulation to be installed. If you are mathematically inclined, you can determine the optimum thickness by obtaining an expression for the total cost, which is the sum of the expressions for the lost heat cost and insulation cost as a function of thickness; differentiating the total cost expression with respect to the thickness; and setting it equal to zero. The thickness value satisfying the resulting equation is the optimum thickness. The cost values can be determined from an annualized lifetime analysis or simply from the requirement that the insulation pay for itself within two or three years. Note that the optimum thickness of insulation depends on the fuel cost, and the higher the fuel cost, the larger the optimum thickness of insulation. Considering that insulation will be in service for many years and the fuel prices are likely to escalate, a reasonable increase in fuel prices must be assumed in calculations. Otherwise, what is optimum insulation today will be inadequate insulation in the years to come, and we may have to face the possibility of costly retrofitting projects. This is what happened in the 1970s and 1980s to insulations installed in the 1960s. The discussion above on optimum thickness is valid when the type and manufacturer of insulation are already selected, and the only thing to be determined is the most economical thickness. But often there are several suitable insulations for a job, and the selection process can be rather confusing since each insulation can have a different thermal conductivity, different installation cost, and different service life. In such cases, a selection can be made by preparing an annualized cost versus thickness chart like Figure 7–39 for each insulation, and determining the one having the lowest minimum cost. The insulation with the lowest annual cost is obviously the most economical insulation, and the insulation thickness corresponding to the minimum total cost is the optimum thickness. When the optimum thickness falls between two commercially available thicknesses, it is a good practice to be conservative and choose the thicker insulation. The Cost per year Total cost Insu st n co latio Lost heat cost Minimum total cost Optimum thickness 0 Insulation thickness FIGURE 7–38 Determination of the optimum thickness of insulation on the basis of minimum total cost. Cost per year Total cost curves for different insulation A B C Optimum thickness for D 0 D Minimum cost Insulation thickness FIGURE 7–39 Determination of the most economical type of insulation and its optimum thickness. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 402 402 HEAT TRANSFER TABLE 7–4 Recommended insulation thicknesses for flat hot surfaces as a function of surface temperature (from TIMA Energy Savings Guide) Surface temperature Insulation thickness 150°F 250°F 350°F 550°F 750°F 950°F 2! (5.1 cm) 3! (7.6 cm) 4! (10.2 cm) 6! (15.2 cm) 9! (22.9 cm) 10! (25.44 cm) (66°C) (121°C) (177°C) (288°C) (400°C) (510°C) extra thickness will provide a little safety cushion for any possible decline in performance over time and will help the environment by reducing the production of greenhouse gases such as CO2. The determination of the optimum thickness of insulation requires a heat transfer and economic analysis, which can be tedious and time-consuming. But a selection can be made in a few minutes using the tables and charts prepared by TIMA (Thermal Insulation Manufacturers Association) and member companies. The primary inputs required for using these tables or charts are the operating and ambient temperatures, pipe diameter (in the case of pipe insulation), and the unit fuel cost. Recommended insulation thicknesses for hot surfaces at specified temperatures are given in Table 7–4. Recommended thicknesses of pipe insulations as a function of service temperatures are 0.5 to 1 in. for 150°F, 1 to 2 in. for 250°F, 1.5 to 3 in. for 350°F, 2 to 4.5 in. for 450°F, 2.5 to 5.5 in. for 550°F, and 3 to 6 in. for 650°F for nominal pipe diameters of 0.5 to 36 in. The lower recommended insulation thicknesses are for pipes with small diameters, and the larger ones are for pipes with large diameters. EXAMPLE 7–8 Effect of Insulation on Surface Temperature Hot water at Ti 120°C flows in a stainless steel pipe (k 15 W/m · °C) whose inner diameter is 1.6 cm and thickness is 0.2 cm. The pipe is to be covered with adequate insulation so that the temperature of the outer surface of the insulation does not exceed 40°C when the ambient temperature is To 25°C. Taking the heat transfer coefficients inside and outside the pipe to be hi 70 W/m2 · °C and ho 20 W/m2 · °C, respectively, determine the thickness of fiberglass insulation (k 0.038 W/m · °C) that needs to be installed on the pipe. SOLUTION A steam pipe is to be covered with enough insulation to reduce the exposed surface temperature. The thickness of insulation that needs to be installed is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. To ho r1 r2 Ti hi Steam Properties The thermal conductivities are given to be k 15 W/m · °C for the steel pipe and k 0.038 W/m · °C for fiberglass insulation. 40°C r3 Insulation Ti T1 Ri T2 R1 T3 R2 FIGURE 7–40 Schematic for Example 7–8. To Ro Analysis The thermal resistance network for this problem involves four resistances in series and is given in Figure 7–40. The inner radius of the pipe is r1 0.8 cm and the outer radius of the pipe and thus the inner radius of the insulation is r2 1.0 cm. Letting r3 represent the outer radius of the insulation, the areas of the surfaces exposed to convection for an L 1-m-long section of the pipe become A1 2r1L 2(0.008 m)(1 m) 0.0503 m2 A3 2r3L 2r3 (1 m) 6.28r3 m2 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 403 403 CHAPTER 7 Then the individual thermal resistances are determined to be Ri Rconv, 1 R1 Rpipe 1 1 0.284°C/ W hi A1 (70 W/m2 · °C)(0.0503 m2) ln(r2 /r1) ln(0.01/0.008) 0.0024°C/ W 2k1 L 2(15 W/m · °C)(1 m) R2 Rinsulation ln(r3 /r2) ln(r3 /0.01) 2(0.038 W/m · °C)(1 m) 2k2 L 4.188 ln(r3 /0.01)°C/ W 1 1 1 °C/ W Ro Rconv, 2 ho A3 (20 W/m2 · °C)(6.28r3 m2) 125.6r3 Noting that all resistances are in series, the total resistance is determined to be Rtotal Ri R1 R2 R0 [0.284 0.0024 4.188 ln(r3/0.01) 1/125.6r3]°C/ W Then the steady rate of heat loss from the steam becomes Ti To (120 125)°C · Q Rtotal [0.284 0.0024 4.188 ln(r3 /0.01) 1/125.6r3 ]°C/ W Noting that the outer surface temperature of insulation is specified to be 40°C, the rate of heat loss can also be expressed as T3 To (40 25)°C · Q 1884r3 Ro (1/125.6r3)°C/ W Setting the two relations above equal to each other and solving for r3 gives r3 0.0170 m. Then the minimum thickness of fiberglass insulation required is t r3 r2 0.0170 0.0100 0.0070 m 0.70 cm Discussion Insulating the pipe with at least 0.70-cm-thick fiberglass insulation will ensure that the outer surface temperature of the pipe will be at 40°C or below. EXAMPLE 7–9 Optimum Thickness of Insulation During a plant visit, you notice that the outer surface of a cylindrical curing oven is very hot, and your measurements indicate that the average temperature of the exposed surface of the oven is 180°F when the surrounding air temperature is 75°F. You suggest to the plant manager that the oven should be insulated, but the manager does not think it is worth the expense. Then you propose to the manager to pay for the insulation yourself if he lets you keep the savings from the fuel bill for one year. That is, if the fuel bill is $5000/yr before insulation and drops to $2000/yr after insulation, you will get paid $3000. The manager agrees since he has nothing to lose, and a lot to gain. Is this a smart bet on your part? The oven is 12 ft long and 8 ft in diameter, as shown in Figure 7–41. The plant operates 16 h a day 365 days a year, and thus 5840 h/yr. The insulation 180°F T = 75°F Curing oven 8 ft 12 ft FIGURE 7–41 Schematic for Example 7–9. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 404 404 HEAT TRANSFER to be used is fiberglass (kins 0.024 Btu/h · ft · °F), whose cost is $0.70/ft2 per inch of thickness for materials, plus $2.00/ft2 for labor regardless of thickness. The combined heat transfer coefficient on the outer surface is estimated to be ho 3.5 Btu/h · ft2 · °F. The oven uses natural gas, whose unit cost is $0.75/therm input (1 therm 100,000 Btu), and the efficiency of the oven is 80 percent. Disregarding any inflation or interest, determine how much money you will make out of this venture, if any, and the thickness of insulation (in whole inches) that will maximize your earnings. SOLUTION A cylindrical oven is to be insulated to reduce heat losses. The optimum thickness of insulation and the potential earnings are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulation is one-dimensional. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 The surfaces of the cylindrical oven can be treated as plain surfaces since its diameter is greater than 3 ft. Properties The thermal conductivity of insulation is given to be k 0.024 Btu/ h · ft · °F. Analysis The exposed surface area of the oven is As 2Abase Aside 2r 2 2rL 2(4 ft)2 2(4 ft)(12 ft) 402 ft2 The rate of heat loss from the oven before the insulation is installed is determined from · Q ho As(Ts T) (3.5 Btu/h · ft2 · °F)(402 ft2)(180 75)°F 147,700 Btu/h Noting that the plant operates 5840 h/yr, the total amount of heat loss from the oven per year is · Q Q t (147,700 Btu/h)(5840 h/yr) 0.863 109 Btu/yr The efficiency of the oven is given to be 80 percent. Therefore, to generate this much heat, the oven must consume energy (in the form of natural gas) at a rate of Qin Q/"oven (0.863 109 Btu/yr)/0.80 1.079 109 Btu/yr 10,790 therms since 1 therm 100,000 Btu. Then the annual fuel cost of this oven before insulation becomes Annual cost Qin Unit cost (10,790 therm/yr)($0.75/therm) $8093/yr That is, the heat losses from the exposed surfaces of the oven are currently costing the plant over $8000/yr. When insulation is installed, the rate of heat transfer from the oven can be determined from Ts T Ts T Ts T · Q ins As t Rtotal Rins Rconv ins 1 kins ho We expect the surface temperature of the oven to increase and the heat transfer coefficient to decrease somewhat when insulation is installed. We assume cen58933_ch07.qxd 9/4/2002 12:13 PM Page 405 405 CHAPTER 7 these two effects to counteract each other. Then the relation above for 1-in.thick insulation gives the rate of heat loss to be As(Ts T) (402 ft2)(180 75)°F · Q ins t 1/12 ft 1 ins 1 0.024 Btu/h · ft · °F 3.5 Btu/h · ft2 · °F kins ho 11,230 Btu/h Also, the total amount of heat loss from the oven per year and the amount and cost of energy consumption of the oven become · Qins Q ins t (11,230 Btu/h)(5840 h/yr) 0.6558 108 Btu/yr Qin, ins Qins /"oven (0.6558 108 Btu/yr)/0.80 0.820 108 Btu/yr 820 therms Annual cost Qin, ins Unit cost (820 therm/yr)($0.75/therm) $615/yr Therefore, insulating the oven by 1-in.-thick fiberglass insulation will reduce the fuel bill by $8093 $615 $7362 per year. The unit cost of insulation is given to be $2.70/ft2. Then the installation cost of insulation becomes Insulation cost (Unit cost)(Surface area) ($2.70/ft2)(402 ft2) $1085 The sum of the insulation and heat loss costs is Total cost Insulation cost Heat loss cost $1085 $615 $1700 Then the net earnings will be Earnings Income Expenses $8093 $1700 $6393 To determine the thickness of insulation that maximizes your earnings, we repeat the calculations above for 2-, 3-, 4-, and 5-in.-thick insulations, and list the results in Table 7–5. Note that the total cost of insulation decreases first with increasing insulation thickness, reaches a minimum, and then starts to increase. TABLE 7–5 The variation of total insulation cost with insulation thickness Insulation thickness 1 2 3 4 5 in. in. in. in. in. Heat loss, Btu/h 11,230 5838 3944 2978 2392 Lost fuel, therms/yr 820 426 288 217 175 Lost fuel cost, $/yr 615 320 216 163 131 Insulation cost, $ 1085 1367 1648 1930 2211 Total cost, $ 1700 1687 1864 2093 2342 We observe that the total insulation cost is a minimum at $1687 for the case of 2-in.-thick insulation. The earnings in this case are Maximum earnings Income Minimum expenses $8093 $1687 $6406 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 406 406 HEAT TRANSFER which is not bad for a day’s worth of work. The plant manager is also a big winner in this venture since the heat losses will cost him only $320/yr during the second and consequent years instead of $8093/yr. A thicker insulation could probably be justified in this case if the cost of insulation is annualized over the lifetime of insulation, say 20 years. Several energy conservation measures are being marketed as explained above by several power companies and private firms. SUMMARY The force a flowing fluid exerts on a body in the flow direction is called drag. The part of drag that is due directly to wall shear stress w is called the skin friction drag since it is caused by frictional effects, and the part that is due directly to pressure is called the pressure drag or form drag because of its strong dependence on the form or shape of the body. The drag coefficient CD is a dimensionless number that represents the drag characteristics of a body, and is defined as FD 2 A 2 Cf Laminar: 1.328 Re 1/2 L 0.074 5 10 5 Re L 10 7 Re 1/5 L 0.074 1742 Cf 5 10 5 ReL 10 7 ReL Re1/5 L Cf Turbulent: Combined: Rough surface, turbulent: CD 1 where A is the frontal area for blunt bodies, and surface area for parallel flow over flat plates or thin airfoils. For flow over a flat plate, the Reynolds number is x x Rex xcr 5 5 10 For parallel flow over a flat plate, the local friction and convection coefficients are 0.664 Re x 5 10 5 Re 1/2 x hx x 1/3 Nu x Pr 0.6 0.332 Re0.5 x Pr k 0.0592 , 5 10 5 Re x 10 7 Turbulent: Cf, x Re 1/5 x hx x 0.6 Pr 60 1/3 Nu x 0.0296 Re 0.8 x Pr k 5 10 5 Re x 10 7 Laminar: C f, x The average friction coefficient relations for flow over a flat plate are: Cf 1.89 1.62 log L 2.5 The average Nusselt number relations for flow over a flat plate are: Nu Laminar: hL 1/3 0.664 Re 0.5 L Pr k ReL 5 10 5 Turbulent: Nu Transition from laminar to turbulent occurs at the critical Reynolds number of Re x, cr Re L 5 10 5 0.6 Pr 60 hL 1/3 0.037 Re0.8 L Pr k 5 10 5 ReL 10 7 Combined: 0.6 Pr 60 hL 1/3 Nu (0.037 Re0.8 L 871) Pr , k 5 10 5 ReL 10 7 For isothermal surfaces with an unheated starting section of length , the local Nusselt number and the average convection coefficient relations are Laminar: Turbulent: Laminar: Turbulent: 1/3 Nux (for 0) 0.332 Re0.5 x Pr 3/4 1/3 3/4 1/3 [1 ( /x) ] [1 ( /x) ] 1/3 Nu x (for 0) 0.0296 Re0.8 x Pr Nu x 9/10 1/ 9 9/10 1/9 [1 (/x) ] [1 (/x) ] 2[1 ( /x) 3/4] hxL h 1 /L 9/10 5[1 ( /x) hxL h (1 / L) Nux cen58933_ch07.qxd 9/4/2002 12:13 PM Page 407 407 CHAPTER 7 These relations are for the case of isothermal surfaces. When a flat plate is subjected to uniform heat flux, the local Nusselt number is given by Laminar: Turbulent: 1/3 Nux 0.453 Re 0.5 x Pr 0.8 Nux 0.0308 Re x P r 1/3 The average Nusselt numbers for cross flow over a cylinder and sphere are Nucyl 0.62 Re1/2 Pr1/3 Re hD 0.3 1 k [1 (0.4/ Pr)2/3]1/4 282,000 hD 2 [0.4 Re1/2 0.06 Re 2/3]Pr 0.4 s k 1/4 which is valid for 3.5 Re 80,000 and 0.7 Pr 380. The fluid properties are evaluated at the film temperature Tf (T Ts)/2 in the case of a cylinder, and at the freestream temperature T (except for s , which is evaluated at the surface temperature Ts) in the case of a sphere. In tube banks, the Reynolds number is based on the maximum velocity max that is related to the approach velocity as In-line and Staggered with SD (ST D)/2: ST max ST D Staggered with SD (ST D)/2: ST max 2(SD D) where ST the transverse pitch and SD is the diagonal pitch. The average Nusselt number for cross flow over tube banks is expressed as hD C Re Dm Pr n(Pr/Prs) 0.25 k where the values of the constants C, m, and n depend on value Reynolds number. Such correlations are given in Table 7–2. All properties except Prs are to be evaluated at the arithmetic mean of the inlet and outlet temperatures of the fluid defined as Tm (Ti Te)/2. The average Nusselt number for tube banks with less than 16 rows is expressed as 5/8 4/5 which is valid for Re Pr 0.2, and Nu sph Nu D Nu D, NL F NuD where F is the correction factor whose values are given in Table 7-3. The heat transfer rate to or from a tube bank is determined from ˙ p (Te Ti) Q˙ h A s Tln mC where Tln is the logarithmic mean temperature difference defined as Tln (Ts Te ) (Ts Ti ) Te Ti ln[(Ts Te )/(Ts Ti )] ln(Te /Ti) and the exit temperature of the fluid Te is Te Ts (Ts Ti ) exp mC ˙ Ash p where As NDL is the heat transfer surface area and m· (NT ST L) is the mass flow rate of the fluid. The pressure drop P for a tube bank is expressed as P NL f 2max 2 where f is the friction factor and is the correction factor, both given in Figs. 7–27. REFERENCES AND SUGGESTED READING 1. R. D. Blevin. Applied Fluid Dynamics Handbook. New York: Van Nostrand Reinhold, 1984. 2. S. W. Churchill and M. Bernstein. “A Correlating Equation for Forced Convection from Gases and Liquids to a Circular Cylinder in Cross Flow.” Journal of Heat Transfer 99 (1977), pp. 300–306. W. H. Giedt. “Investigation of Variation of Point UnitHeat Transfer Coefficient around a Cylinder Normal to an Air Stream.” Transactions of the ASME 71 (1949), pp. 375–381. 6. M. Jakob. Heat Transfer. Vol. l. New York: John Wiley & Sons, 1949. S. W. Churchill and H. Ozoe. “Correlations for Laminar Forced Convection in Flow over an Isothermal Flat Plate and in Developing and Fully Developed Flow in an Isothermal Tube.” Journal of Heat Transfer 95 (Feb. 1973), pp. 78–84. W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993. W. M. Edmunds. “Residential Insulation.” ASTM Standardization News (Jan. 1989), pp. 36–39. H. Schlichting. Boundary Layer Theory, 7th ed. New York, McGraw-Hill, 1979. F. Kreith and M. S. Bohn. Principles of Heat Transfer, 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 408 408 HEAT TRANSFER N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West, 1995. 11. W. C. Thomas. “Note on the Heat Transfer Equation for Forced Convection Flow over a Flat Plate with and Unheated Starting Length.” Mechanical Engineering News, 9, no.1 (1977), p. 361. 12. R. D. Willis. “Photographic Study of Fluid Flow Between Banks of Tubes.” Engineering (1934), pp. 423–425. 13. A. Zukauskas, “Convection Heat Transfer in Cross Flow.” In Advances in Heat Transfer, J. P. Hartnett and T. F. Irvine, Jr., Eds. New York: Academic Press, 1972, Vol. 8, pp. 93–106. A. Zukauskas. “Heat Transfer from Tubes in Cross Flow.” In Advances in Heat Transfer, ed. J. P. Hartnett and T. F. Irvine, Jr. Vol. 8. New York: Academic Press, 1972. 15. A. Zukauskas. “Heat Transfer from Tubes in Cross Flow.” In Handbook of Single Phase Convective Heat Transfer, Eds. S. Kakac, R. K. Shah, and Win Aung. New York: Wiley Interscience, 1987. 16. A. Zukauskas and R. Ulinskas, “Efficiency Parameters for Heat Transfer in Tube Banks.” Heat Transfer Engineering no. 2 (1985), pp. 19–25. PROBLEMS Drag Force and Heat Transfer in External Flow 7–1C What is the difference between the upstream velocity and the free-stream velocity? For what types of flow are these two velocities equal to each other? 7–2C What is the difference between streamlined and blunt bodies? Is a tennis ball a streamlined or blunt body? 7–3C What is drag? What causes it? Why do we usually try to minimize it? 7–4C What is lift? What causes it? Does wall shear contribute to the lift? 7–5C During flow over a given body, the drag force, the upstream velocity, and the fluid density are measured. Explain how you would detennine the drag coefficient. What area would you use in calculations? 7–6C Define frontal area of a body subjected to external flow. When is it appropriate to use the frontal area in drag and lift calculations? 7–7C What is the difference between skin friction drag and pressure drag? Which is usually more significant for slender bodies such as airfoils? 7–8C What is the effect of surface roughness on the friction drag coefficient in laminar and turbulent flows? 7–9C What is the effect of streamlining on (a) friction drag and (b) pressure drag? Does the total drag acting on a body necessarily decrease as a result of streamlining? Explain. Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 7–10C What is flow separation? What causes it? What is the effect of flow separation on the drag coefficient? Flow Over Flat Plates 7–11C What does the friction coefficient represent in flow over a flat plate? How is it related to the drag force acting on the plate? 7–12C Consider laminar flow over a flat plate. Will the friction coefficient change with distance from the leading edge? How about the heat transfer coefficient? 7–13C How are the average friction and heat transfer coefficients determined in flow over a flat plate? 7–14 Engine oil at 80°C flows over a 6-m-long flat plate whose temperature is 30°C with a velocity of 3 m/s. Determine the total drag force and the rate of heat transfer over the entire plate per unit width. 7–15 The local atmospheric pressure in Denver, Colorado (elevation 1610 m), is 83.4 kPa. Air at this pressure and at 30°C flows with a velocity of 6 m/s over a 2.5-m 8-m flat plate whose temperature is 120°C. Determine the rate of heat transfer from the plate if the air flows parallel to the (a) 8-m-long side and (b) the 2.5-m side. 7–16 During a cold winter day, wind at 55 km/h is blowing parallel to a 4-m-high and 10-m-long wall of a house. If the air outside is at 5°C and the surface temperature of the wall is Attic space Air 5°C 55 km/h FIGURE P7–16 4m 10 m 12°C cen58933_ch07.qxd 9/4/2002 12:13 PM Page 409 409 CHAPTER 7 12°C, determine the rate of heat loss from that wall by convection. What would your answer be if the wind velocity was douAnswers: 9081 W, 16,200 W bled? 7–17 Reconsider Problem 7–16. Using EES (or other) software, investigate the effects of wind velocity and outside air temperature on the rate of heat loss from the wall by convection. Let the wind velocity vary from 10 km/h to 80 km/h and the outside air temperature from 0ºC to 10ºC. Plot the rate of heat loss as a function of the wind velocity and of the outside temperature, and discuss the results. 7–18E Air at 60°F flows over a 10-ft-long flat plate at 7 ft/s. Determine the local friction and heat transfer coefficients at intervals of 1 ft, and plot the results against the distance from the leading edge. 7–19E Reconsider Problem 7–18. Using EES (or other) software, evaluate the local friction and heat transfer coefficients along the plate at intervals of 0.1 ft, and plot them against the distance from the leading edge. 7–20 Consider a hot automotive engine, which can be approximated as a 0.5-m-high, 0.40-m-wide, and 0.8-m-long rectangular block. The bottom surface of the block is at a temperature of 80°C and has an emissivity of 0.95. The ambient air is at 20°C, and the road surface is at 25°C. Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of 80 km/h. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block. 7–21 The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a rate of 15 m/min. The temperature of the plastic sheet is 90°C when it is exposed to the surrounding air, and the sheet is subjected to air flow at 30°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in 2 s. Determine the rate of heat transfer from the plastic sheet to the air. Plastic sheet 15 m/min FIGURE P7–21 Air 30°C 200 W/ m2 70 km/h FIGURE P7–22 7–23 Reconsider Problem 7–22. Using EES (or other) software, investigate the effects of the train velocity and the rate of absorption of solar radiation on the equilibrium temperature of the top surface of the car. Let the train velocity vary from 10 km/h to 120 km/h and the rate of solar absorption from 100 W/m2 to 500 W/m2. Plot the equilibrium temperature as functions of train velocity and solar radiation absorption rate, and discuss the results. 7–24 A 15-cm 15-cm circuit board dissipating 15 W of power uniformly is cooled by air, which approaches the circuit board at 20°C with a velocity of 5 m/s. Disregarding any heat transfer from the back surface of the board, determine the surface temperature of the electronic components (a) at the leading edge and (b) at the end of the board. Assume the flow to be turbulent since the electronic components are expected to act as turbulators. 7–25 Consider laminar flow of a fluid over a flat plate maintained at a constant temperature. Now the free-stream velocity of the fluid is doubled. Determine the change in the drag force on the plate and rate of heat transfer between the fluid and the plate. Assume the flow to remain laminar. Air 30°C, 3 m/s 90°C 7–22 The top surface of the passenger car of a train moving at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top surface is absorbing solar radiation at a rate of 200 W/m2, and the temperature of the ambient air is 30°C. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface Answer: 35.1°C of the car. 7–26E Consider a refrigeration truck traveling at 55 mph at a location where the air temperature is 80°F. The refrigerated compartment of the truck can be considered to be a 9-ft-wide, 8-ft-high, and 20-ft-long rectangular box. The refrigeration system of the truck can provide 3 tons of refrigeration (i.e., it can remove heat at a rate of 600 Btu/min). The outer surface of the truck is coated with a low-emissivity material, and thus radiation heat transfer is very small. Determine the average temperature of the outer surface of the refrigeration compartment of the truck if the refrigeration system is observed to be cen58933_ch07.qxd 9/4/2002 12:13 PM Page 410 410 HEAT TRANSFER passages between the fins. The heat sink is to dissipate 20 W of heat and the base temperature of the heat sink is not to exceed 60°C. Assuming the fins and the base plate to be nearly isothermal and the radiation heat transfer to be negligible, determine the minimum free-stream velocity the fan needs to supply to avoid overheating. Air, 80°F = 55 mph 20 ft 8 ft Refrigeration truck Air 25°C FIGURE P7–26E 60°C operating at half the capacity. Assume the air flow over the entire outer surface to be turbulent and the heat transfer coefficient at the front and rear surfaces to be equal to that on side surfaces. 7–27 Solar radiation is incident on the glass cover of a solar collector at a rate of 700 W/m2. The glass transmits 88 percent of the incident radiation and has an emissivity of 0.90. The entire hot water needs of a family in summer can be met by two collectors 1.2 m high and 1 m wide. The two collectors are attached to each other on one side so that they appear like a single collector 1.2 m 2 m in size. The temperature of the glass cover is measured to be 35°C on a day when the surrounding air temperature is 25°C and the wind is blowing at 30 km/h. The effective sky temperature for radiation exchange between the glass cover and the open sky is 40°C. Water enters the tubes attached to the absorber plate at a rate of 1 kg/min. Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine (a) the total rate of heat loss from the collector, (b) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and (c) the temperature rise of water as it flows through the collector. Tsky = – 40°C Air 25°C Solar collector 35°C Fins 0.5 cm 10 cm 7–28 A transformer that is 10 cm long, 6.2 cm wide, and 5 cm high is to be cooled by attaching a 10 cm 6.2 cm wide polished aluminum heat sink (emissivity 0.03) to its top surface. The heat sink has seven fins, which are 5 mm high, 2 mm thick, and 10 cm long. A fan blows air at 25°C parallel to the 5 cm 6.2 cm FIGURE P7–28 7–29 Repeat Problem 7–28 assuming the heat sink to be black-anodized and thus to have an effective emissivity of 0.90. Note that in radiation calculations the base area (10 cm 6.2 cm) is to be used, not the total surface area. 7–30 An array of power transistors, dissipating 6 W of power each, are to be cooled by mounting them on a 25-cm 25-cm square aluminum plate and blowing air at 35°C over the plate with a fan at a velocity of 4 m/s. The average temperature of the plate is not to exceed 65°C. Assuming the heat transfer from the back side of the plate to be negligible and disregarding radiation, determine the number of transistors that can be placed on this plate. Solar radiation FIGURE P7–27 Transformer 20 W Aluminum plate Power transistor, 6 W 35°C Air 4 m/s 25 cm 65°C 25 cm FIGURE P7–30 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 411 411 CHAPTER 7 7–31 Repeat Problem 7–30 for a location at an elevation of 1610 m where the atmospheric pressure is 83.4 kPa. Answer: 4 7–32 Air at 25°C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge of the plate where the flow becomes turbulent, and the thickness of the boundary layer at that location. 7–33 Repeat Problem 7–32 for water. 7–34 The weight of a thin flat plate 50 cm 50 cm in size is balanced by a counterweight that has a mass of 2 kg, as shown in the figure. Now a fan is turned on, and air at 1 atm and 25°C flows downward over both surfaces of the plate with a freestream velocity of 10 m/s. Determine the mass of the counterweight that needs to be added in order to balance the plate in this case. of air at 1 atm pressure and 30°C with a velocity of 6 m/s. The surface temperature of the ball eventually drops to 250°C. Determine the average convection heat transfer coefficient during this cooling process and estimate how long this process has taken. 7–41 Reconsider Problem 7–40. Using EES (or other) software, investigate the effect of air velocity on the average convection heat transfer coefficient and the cooling time. Let the air velocity vary from 1 m/s to 10 m/s. Plot the heat transfer coefficient and the cooling time as a function of air velocity, and discuss the results. 7–42E A person extends his uncovered arms into the windy air outside at 54°F and 20 mph in order to feel nature closely. Initially, the skin temperature of the arm is 86°F. Treating the arm as a 2-ft-long and 3-in.-diameter cylinder, determine the rate of heat loss from the arm. Air 25°C, 10 m /s Air 54°F, 20 mph 86°F Plate 50 cm 50 cm FIGURE P7–34 Flow across Cylinders and Spheres 7–35C Consider laminar flow of air across a hot circular cylinder. At what point on the cylinder will the heat transfer be highest? What would your answer be if the flow were turbulent? 7–36C In flow over cylinders, why does the drag coefficient suddenly drop when the flow becomes turbulent? Isn’t turbulence supposed to increase the drag coefficient instead of decreasing it? 7–37C In flow over blunt bodies such as a cylinder, how does the pressure drag differ from the friction drag? 7–38C Why is flow separation in flow over cylinders delayed in turbulent flow? 7–39 A long 8-cm-diameter steam pipe whose external surface temperature is 90°C passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 7°C and the wind is blowing across the pipe at a velocity of 50 km/h. 7–40 A stainless steel ball ( 8055 kg/m3, Cp 480 J/kg · °C) of diameter D 15 cm is removed from the oven at a uniform temperature of 350°C. The ball is then subjected to the flow FIGURE P7–42E 7–43E Reconsider Problem 7–42E. Using EES (or other) software, investigate the effects of air temperature and wind velocity on the rate of heat loss from the arm. Let the air temperature vary from 20°F to 80°F and the wind velocity from 10 mph to 40 mph. Plot the rate of heat loss as a function of air temperature and of wind velocity, and discuss the results. 7–44 An average person generates heat at a rate of 84 W while resting. Assuming one-quarter of this heat is lost from the head and disregarding radiation, determine the average surface temperature of the head when it is not covered and is subjected to winds at 10°C and 35 km/h. The head can be approximated as a 30-cm-diameter sphere. Answer: 12.7°C 7–45 Consider the flow of a fluid across a cylinder maintained at a constant temperature. Now the free-stream velocity of the fluid is doubled. Determine the change in the drag force on the cylinder and the rate of heat transfer between the fluid and the cylinder. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 412 412 HEAT TRANSFER 7–46 A 6-mm-diameter electrical transmission line carries an electric current of 50 A and has a resistance of 0.002 ohm per meter length. Determine the surface temperature of the wire during a windy day when the air temperature is 10°C and the wind is blowing across the transmission line at 40 km/h. the person. The average human body can be treated as a 1-ftdiameter cylinder with an exposed surface area of 18 ft2. Disregard any heat transfer by radiation. What would your answer Answers: 95.1°F, 91.6°F be if the air velocity were doubled? Wind, 40 km/h 10°C 85°F 6 ft /s 300 Btu/h Transmission lines FIGURE P7–46 7–47 Reconsider Problem 7–46. Using EES (or other) software, investigate the effect of the wind velocity on the surface temperature of the wire. Let the wind velocity vary from 10 km/h to 80 km/h. Plot the surface temperature as a function of wind velocity, and discuss the results. 7–48 A heating system is to be designed to keep the wings of an aircraft cruising at a velocity of 900 km/h above freezing temperatures during flight at 12,200-m altitude where the standard atmospheric conditions are 55.4°C and 18.8 kPa. Approximating the wing as a cylinder of elliptical cross section whose minor axis is 30 cm and disregarding radiation, determine the average convection heat transfer coefficient on the wing surface and the average rate of heat transfer per unit surface area. 7–49 A long aluminum wire of diameter 3 mm is extruded at a temperature of 370°C. The wire is subjected to cross air flow at 30°C at a velocity of 6 m/s. Determine the rate of heat transfer from the wire to the air per meter length when it is first exposed to the air. 370°C 30°C 6 m/s FIGURE P7–50E 7–51 An incandescent lightbulb is an inexpensive but highly inefficient device that converts electrical energy into light. It converts about 10 percent of the electrical energy it consumes into light while converting the remaining 90 percent into heat. (A fluorescent lightbulb will give the same amount of light while consuming only one-fourth of the electrical energy, and it will last 10 times longer than an incandescent lightbulb.) The glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by convection and radiation. Consider a 10-cm-diameter 100-W lightbulb cooled by a fan that blows air at 25°C to the bulb at a velocity of 2 m/s. The surrounding surfaces are also at 25°C, and the emissivity of the glass is 0.9. Assuming 10 percent of the energy passes through the glass bulb as light with negligible absorption and the rest of the energy is absorbed and dissipated by the bulb itself, determine the equilibrium temperature of the glass bulb. 3 mm Aluminum wire FIGURE P7–49 7–50E Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his entire body to air flow. The air temperature is 85°F and the fan is blowing air at a velocity of 6 ft/s. If the person is doing light work and generating sensible heat at a rate of 300 Btu/h, determine the average temperature of the outer surface (skin or clothing) of Air 25°C 2 m/s 100 W ε = 0.9 Light, 10 W FIGURE P7–51 7–52 During a plant visit, it was noticed that a 12-m-long section of a 10-cm-diameter steam pipe is completely exposed to the ambient air. The temperature measurements indicate that cen58933_ch07.qxd 9/4/2002 12:13 PM Page 413 413 CHAPTER 7 the average temperature of the outer surface of the steam pipe is 75°C when the ambient temperature is 5°C. There are also light winds in the area at 10 km/h. The emissivity of the outer surface of the pipe is 0.8, and the average temperature of the surfaces surrounding the pipe, including the sky, is estimated to be 0°C. Determine the amount of heat lost from the steam during a 10-h-long work day. Steam is supplied by a gas-fired steam generator that has an efficiency of 80 percent, and the plant pays $0.54/therm of natural gas (1 therm 105,500 kJ). If the pipe is insulated and 90 percent of the heat loss is saved, determine the amount of money this facility will save a year as a result of insulating the steam pipes. Assume the plant operates every day of the year for 10 h. State your assumptions. at 30°C flowing over the duct with a velocity of 200 m/min. If the surface temperature of the duct is not to exceed 65°C, determine the total power rating of the electronic devices that can Answer: 640 W be mounted into the duct. Electronic components inside 30°C 200 m/min Air 1.5 m Tsurr = 0°C ε = 0.8 75°C 10 cm 65°C 20 cm FIGURE P7–55 Steam pipe 7–56 Repeat Problem 7–55 for a location at 4000-m altitude where the atmospheric pressure is 61.66 kPa. 5°C 10 km/h FIGURE P7–52 7–53 Reconsider Problem 7–52. There seems to be some uncertainty about the average temperature of the surfaces surrounding the pipe used in radiation calculations, and you are asked to determine if it makes any significant difference in overall heat transfer. Repeat the calculations for average surrounding and surface temperatures of 20°C and 25°C, respectively, and determine the change in the values obtained. 7–54E A 12-ft-long, 1.5-kW electrical resistance wire is made of 0.1-in.-diameter stainless steel (k 8.7 Btu/h · ft · °F). The resistance wire operates in an environment at 85°F. Determine the surface temperature of the wire if it is cooled by a fan blowing air at a velocity of 20 ft/s. 85°F 20 ft/s 1.5 kW resistance heater FIGURE P7–54E 7–55 The components of an electronic system are located in a 1.5-m-long horizontal duct whose cross section is 20 cm 20 cm. The components in the duct are not allowed to come into direct contact with cooling air, and thus are cooled by air 7–57 A 0.4-W cylindrical electronic component with diameter 0.3 cm and length 1.8 cm and mounted on a circuit board is cooled by air flowing across it at a velocity of 150 m/min. If the air temperature is 40°C, determine the surface temperature of the component. 7–58 Consider a 50-cm-diameter and 95-cm-long hot water tank. The tank is placed on the roof of a house. The water inside the tank is heated to 80ºC by a flat-plate solar collector during the day. The tank is then exposed to windy air at 18ºC with an average velocity of 40 km/h during the night. Estimate the temperature of the tank after a 45-mm period. Assume the tank surface to be at the same temperature as the water inside, and the heat transfer coefficient on the top and bottom surfaces to be the same as that on the side surface. 7–59 Reconsider Problem 7–58. Using EES (or other) software, plot the temperature of the tank as a function of the cooling time as the time varies from 30 mm to 5 h, and discuss the results. 7–60 A 1.8-m-diameter spherical tank of negligible thickness contains iced water at 0ºC. Air at 25ºC flows over the tank with a velocity of 7 m/s. Determine the rate of heat transfer to the tank and the rate at which ice melts. The heat of fusion of water at 0ºC is 333.7 kJ/kg. 7-61 A 10-cm-diameter, 30-cm-high cylindrical bottle contains cold water at 3ºC. The bottle is placed in windy air at 27ºC. The water temperature is measured to be 11ºC after 45 minutes of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 414 414 HEAT TRANSFER Flow across Tube Banks Water 15°C 0.8 m/s 7–62C In flow across tube banks, why is the Reynolds number based on the maximum velocity instead of the uniform approach velocity? SL = 4 cm D = 1 cm Ts = 90°C ST = 3 cm 7–63C In flow across tube banks, how does the heat transfer coefficient vary with the row number in the flow direction? How does it vary with in the transverse direction for a given row number? 7–64 Combustion air in a manufacturing facility is to be preheated before entering a furnace by hot water at 90ºC flowing through the tubes of a tube bank located in a duct. Air enters the duct at 15ºC and 1 atm with a mean velocity of 3.8 m/s, and flows over the tubes in normal direction. The outer diameter of the tubes is 2.1 cm, and the tubes are arranged in-line with longitudinal and transverse pitches of SL ST 5 cm. There are eight rows in the flow direction with eight tubes in each row. Determine the rate of heat transfer per unit length of the tubes, and the pressure drop across the tube bank. 7–65 Repeat Problem 7–64 for staggered arrangement with SL ST 5 cm. 7–66 Air is to be heated by passing it over a bank of 3-m-long tubes inside which steam is condensing at 100ºC. Air approaches the tube bank in the normal direction at 20ºC and 1 atm with a mean velocity of 5.2 m/s. The outer diameter of the tubes is 1.6 cm, and the tubes are arranged staggered with longitudinal and transverse pitches of SL ST 4 cm. There are 20 rows in the flow direction with 10 tubes in each row. Determine (a) the rate of heat transfer, (b) and pressure drop across the tube bank, and (c) the rate of condensation of steam inside the tubes. FIGURE P7–69 7–70 Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.4-m-long tubes inside which the refrigerant is evaporating at 20ºC. Air approaches the tube bank in the normal direction at 0ºC and 1 atm with a mean velocity of 4 m/s. The tubes are arranged in-line with longitudinal and transverse pitches of SL ST 1.5 cm. There are 30 rows in the flow direction with 15 tubes in each row. Determine (a) the refrigeration capacity of this system and (b) and pressure drop across the tube bank. 0°C 1 atm 4 m/s Air 0.4 m ST = 1.5 cm 0.8 cm SL = 1.5 cm 7–67 Repeat Problem 7–66 for in-line arrangement with SL ST 5 cm. 7–68 Exhaust gases at 1 atm and 300ºC are used to preheat water in an industrial facility by passing them over a bank of tubes through which water is flowing at a rate of 6 kg/s. The mean tube wall temperature is 80ºC. Exhaust gases approach the tube bank in normal direction at 4.5 m/s. The outer diameter of the tubes is 2.1 cm, and the tubes are arranged in-line with longitudinal and transverse pitches of SL ST 8 cm. There are 16 rows in the flow direction with eight tubes in each row. Using the properties of air for exhaust gases, determine (a) the rate of heat transfer per unit length of tubes, (b) and pressure drop across the tube bank, and (c) the temperature rise of water flowing through the tubes per unit length of tubes. 7–69 Water at 15ºC is to be heated to 65ºC by passing it over a bundle of 4-m-long 1-cm-diameter resistance heater rods maintained at 90ºC. Water approaches the heater rod bundle in normal direction at a mean velocity of 0.8 m/s. The rods arc arranged in-line with longitudinal and transverse pitches of SL 4 cm and ST 3 cm. Determine the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise. Refrigerant, 20°C FIGURE P7–70 7–71 Repeat Problem 7–70 by solving it for staggered arrangement with SL ST 1.5 cm, and compare the performance of the evaporator for the in-line and staggered arrangements. 7–72 A tube bank consists of 300 tubes at a distance of 6 cm between the centerlines of any two adjacent tubes. Air approaches the tube bank in the normal direction at 40ºC and 1 atm with a mean velocity of 7 m/s. There are 20 rows in the flow direction with 15 tubes in each row with an average surface temperature of 140ºC. For an outer tube diameter of 2 cm, determine the average heat transfer coefficient. Special Topic: Thermal Insulation 7–73C What is thermal insulation? How does a thermal insulator differ in purpose from an electrical insulator and from a sound insulator? 7–74C Does insulating cold surfaces save energy? Explain. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 415 415 CHAPTER 7 7–75C What is the R-value of insulation? How is it determined? Will doubling the thickness of flat insulation double its R-value? 7–76C How does the R-value of an insulation differ from its thermal resistance? 7–77C Why is the thermal conductivity of superinsulation orders of magnitude lower than the thermal conductivities of ordinary insulations? 7–78C Someone suggests that one function of hair is to insulate the head. Do you agree with this suggestion? 7–79C Name five different reasons for using insulation in industrial facilities. 7–80C What is optimum thickness of insulation? How is it determined? 7–81 What is the thickness of flat R-8 (in SI units) insulation whose thermal conductivity is 0.04 W/m · °C? 7–82E What is the thickness of flat R-20 (in English units) insulation whose thermal conductivity is 0.02 Btu/h · ft · °F? 7–83 Hot water at 110°C flows in a cast iron pipe (k 52 W/m · °C) whose inner radius is 2.0 cm and thickness is 0.3 cm. The pipe is to be covered with adequate insulation so that the temperature of the outer surface of the insulation does not exceed 30°C when the ambient temperature is 22°C. Taking the heat transfer coefficients inside and outside the pipe to be hi 80 W/m2 · °C and ho 22 W/m2 · °C, respectively, determine the thickness of fiber glass insulation (k 0.038 W/m · °C) that needs to be installed on the pipe. Answer: 1.32 cm 7–84 Reconsider Problem 7–83. Using EES (or other) software, plot the thickness of the insulation as a function of the maximum temperature of the outer surface of insulation in the range of 24ºC to 48ºC. Discuss the results. 7–85 Consider a furnace whose average outer surface temperature is measured to be 90°C when the average surrounding air temperature is 27°C. The furnace is 6 m long and 3 m in diameter. The plant operates 80 h per week for 52 weeks per year. You are to insulate the furnace using fiberglass insulation (kins 0.038 W/m · °C) whose cost is $10/m2 per cm of thickness for materials, plus $30/m2 for labor regardless of thickness. The combined heat transfer coefficient on the outer surface is estimated to be ho 30 W/m2 · °C. The furnace uses natural gas whose unit cost is $0.50/therm input (1 therm 105,500 kJ), and the efficiency of the furnace is 78 percent. The management is willing to authorize the installation of the thickest insulation (in whole cm) that will pay for itself (materials and labor) in one year. That is, the total cost of insulation should be roughly equal to the drop in the fuel cost of the furnace for one year. Determine the thickness of insulation to be used and the money saved per year. Assume the surface temperature of the furnace and the heat transfer coefficient are to remain constant. Answer: 14 cm 7–85 Repeat Problem 7–85 for an outer surface temperature of 75°C for the furnace. 7–87E Steam at 400°F is flowing through a steel pipe (k 8.7 Btu/h · ft · °F) whose inner and outer diameters are 3.5 in. and 4.0 in., respectively, in an environment at 60°F. The pipe is insulated with 1-in.-thick fiberglass insulation (k 0.020 Btu/h · ft · °F), and the heat transfer coefficients on the inside and the outside of the pipe are 30 Btu/h · ft2 · °F and 5 Btu/h · ft2 · °F, respectively. It is proposed to add another 1-in.-thick layer of fiberglass insulation on top of the existing one to reduce the heat losses further and to save energy and money. The total cost of new insulation is $7 per ft length of the pipe, and the net fuel cost of energy in the steam is $0.01 per 1000 Btu (therefore, each 1000 Btu reduction in the heat loss will save the plant $0.01). The policy of the plant is to implement energy conservation measures that pay for themselves within two years. Assuming continuous operation (8760 h/year), determine if the proposed additional insulation is justified. 7–88 The plumbing system of a plant involves a section of a plastic pipe (k 0.16 W/m · °C) of inner diameter 6 cm and outer diameter 6.6 cm exposed to the ambient air. You are to insulate the pipe with adequate weather-jacketed fiberglass insulation (k 0.035 W/m · °C) to prevent freezing of water in the pipe. The plant is closed for the weekends for a period of 60 h, and the water in the pipe remains still during that period. The ambient temperature in the area gets as low as 10°C in winter, and the high winds can cause heat transfer coefficients as high as 30 W/m2 · °C. Also, the water temperature in the pipe can be as cold as 15°C, and water starts freezing when its temperature drops to 0°C. Disregarding the convection resistance inside the pipe, determine the thickness of insulation that will protect the water from freezing under worst conditions. 7–89 Repeat Problem 7–88 assuming 20 percent of the water in the pipe is allowed to freeze without jeopardizing safety. Answer: 27.9 cm Review Problems 7–90 Consider a house that is maintained at 22°C at all times. The walls of the house have R-3.38 insulation in SI units (i.e., an L/k value or a thermal resistance of 3.38 m2 · °C/W). During a cold winter night, the outside air temperature is 4°C and wind at 50 km/h is blowing parallel to a 3-m-high and 8-m-long wall of the house. If the heat transfer coefficient on the interior surface of the wall is 8 W/m2 · °C, determine the rate of heat loss from that wall of the house. Draw the thermal resistance network and disregard radiation heat transfer. Answer: 122 W 7–91 An automotive engine can be approximated as a 0.4-mhigh, 0.60-m-wide, and 0.7-m-long rectangular block. The bottom surface of the block is at a temperature of 75°C and has an emissivity of 0.92. The ambient air is at 5°C, and the road surface is at 10°C. Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation cen58933_ch07.qxd 9/4/2002 12:13 PM Page 416 416 HEAT TRANSFER as the car travels at a velocity of 60 km/h. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block. How will the heat transfer be affected when a 2-mm-thick gunk (k 3 W/m · °C) has formed at the bottom surface as a result of the dirt and oil collected at that surface over time? Assume the metal temperature under the gunk still to be 75°C. Engine block Air 60 km/h 5°C 75°C Gunk ε = 0.92 2 mm Road 10°C FIGURE P7–91 7–92E The passenger compartment of a minivan traveling at 60 mph can be modeled as a 3.2-ft-high, 6-ft-wide, and 11-ftlong rectangular box whose walls have an insulating value of R-3 (i.e., a wall thickness–to–thermal conductivity ratio of 3 h · ft2 · °F/Btu). The interior of a minivan is maintained at an average temperature of 70°F during a trip at night while the outside air temperature is 90°F. The average heat transfer coefficient on the interior surfaces of the van is 1.2 Btu/h · ft2 · °F. The air flow over the exterior surfaces can be assumed to be turbulent because of the intense vibrations involved, and the heat transfer coefficient on the front and back surfaces can be taken to be equal to that on the top surface. Disregarding any heat gain or loss by radiation, determine the rate of heat transfer from the ambient air to the van. 7–94 Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his body to air flow. The air temperature is 32°C, and the fan is blowing air at a velocity of 5 m/s. The surrounding surfaces are at 40°C, and the emissivity of the person can be taken to be 0.9. If the person is doing light work and generating sensible heat at a rate of 90 W, determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 30-cm-diameter cylinder with an Answer: 36.2°C exposed surface area of 1.7 m2. 7–95 Four power transistors, each dissipating 12 W, are mounted on a thin vertical aluminum plate (k 237 W/m · °C) 22 cm 22 cm in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at 20°C, which is blown over the plate by a fan at a velocity of 250 m/min. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. Determine the temperature of the aluminum plate. 7–96 A 3-m-internal-diameter spherical tank made of 1-cmthick stainless steel (k 15 W/m · °C) is used to store iced water at 0°C. The tank is located outdoors at 30°C and is subjected to winds at 25 km/h. Assuming the entire steel tank to be at 0°C and thus its thermal resistance to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. Disregard any heat transfer by radiation. Troom = 30°C 25 km/h Iced Water Di = 3 m Air 60 mph 90°F 1 cm Tin = 0°C FIGURE P7–96 FIGURE P7–92E 7–93 Consider a house that is maintained at a constant temperature of 22°C. One of the walls of the house has three single-pane glass windows that are 1.5 m high and 1.2 m long. The glass (k 0.78 W/m · °C) is 0.5 cm thick, and the heat transfer coefficient on the inner surface of the glass is 8 W/m2 · C. Now winds at 60 km/h start to blow parallel to the surface of this wall. If the air temperature outside is 2°C, determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. 7–97 Repeat Problem 7–96, assuming the inner surface of the tank to be at 0°C but by taking the thermal resistance of the tank and heat transfer by radiation into consideration. Assume the average surrounding surface temperature for radiation exchange to be 15°C and the outer surface of the tank to have an emissivity of 0.9. Answers: (a) 9630 W, (b) 2493 kg 7–98E A transistor with a height of 0.25 in. and a diameter of 0.22 in. is mounted on a circuit board. The transistor is cooled by air flowing over it at a velocity of 500 ft/min. If the air temperature is 120°F and the transistor case temperature is not to exceed 180°F, determine the amount of power this transistor can dissipate safely. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 417 417 CHAPTER 7 Air, 500 ft/min 120°F Power transistor Ts ≤ 180°F 0.22 in. 0.25 in. FIGURE P7–98E 7–99 The roof of a house consists of a 15-cm-thick concrete slab (k 2 W/m · °C) 15 m wide and 20 m long. The convection heat transfer coefficient on the inner surface of the roof is 5 W/m2 · °C. On a clear winter night, the ambient air is reported to be at 10°C, while the night sky temperature is 100 K. The house and the interior surfaces of the wall are maintained at a constant temperature of 20°C. The emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfer, determine the rate of heat transfer through the roof when wind at 60 km/h is blowing over the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the price of natural gas is $0.60/therm (1 therm 105,500 kJ of energy content), determine the money lost through the roof that night during a Answers: 28 kW, $9.44 14-h period. 7–101 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm pressure) is 196°C. Therefore, nitrogen is commonly used in low-temperature scientific studies, since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm. Consider a 4-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to 20°C ambient air and 40 km/h winds. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of heat transfer from the ambient air if the tank is (a) not insulated, (b) insulated with 5-cm-thick fiberglass insulation (k 0.035 W/m · °C), and (c) insulated with 2-cm-thick superinsulation that has an effective thermal conductivity of 0.00005 W/m · °C. N2 vapor Tair = 20°C 40 km/h 1 atm Liquid N2 –196°C Tsky = 100 K Tair = 10°C Concrete roof 60 km/h 20 m ε = 0.9 15 cm 15 m · Q Insulation FIGURE P7–101 Tin = 20°C FIGURE P7–99 7–100 Steam at 250°C flows in a stainless steel pipe (k 15 W/m · °C) whose inner and outer diameters are 4 cm and 4.6 cm, respectively. The pipe is covered with 3.5-cm-thick glass wool insulation (k 0.038 W/m · °C) whose outer surface has an emissivity of 0.3. Heat is lost to the surrounding air and surfaces at 3°C by convection and radiation. Taking the heat transfer coefficient inside the pipe to be 80 W/m2 · °C, determine the rate of heat loss from the steam per unit length of the pipe when air is flowing across the pipe at 4 m/s. 7–102 Repeat Problem 7–101 for liquid oxygen, which has a boiling temperature of 183°C, a heat of vaporization of 213 kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure. 7–103 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. The board is impregnated with copper fillings and has an effective thermal conductivity of 16 W/m · °C. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air at 30°C, which is forced to flow over the surface by a fan at a free-stream velocity of 400 m/min. Determine the temperatures on the two sides of the circuit board. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 418 418 HEAT TRANSFER 7–104E It is well known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the “chilling effect” of the wind. This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities. The equivalent windchill temperature in °F is given by (1993 ASHRAE Handbook of Fundamentals, Atlanta, GA, p. 8.15) Tequiv 91.4 (91.4 Tambient)(0.475 0.0203 0.304 ) where is the wind velocity in mph and Tambient is the ambient air temperature in °F in calm air, which is taken to be air with light winds at speeds up to 4 mph. The constant 91.4°F in the above equation is the mean skin temperature of a resting person in a comfortable environment. Windy air at a temperature Tambient and velocity will feel as cold as calm air at a temperature Tequiv. The equation above is valid for winds up to 43 mph. Winds at higher velocities produce little additional chilling effect. Determine the equivalent wind chill temperature of an environment at 10°F at wind speeds of 10, 20, 30, and 40 mph. Exposed flesh can freeze within one minute at a temperature below 25°F in calm weather. Does a person need to be concerned about this possibility in any of the cases above? Winds 40°F 35 mph It feels like 11°F FIGURE P7–104E 7–l05E Reconsider Problem 7–104E. Using EES (or other) software, plot the equivalent wind chill temperatures in ºF as a function of wind velocity in the range of 4 mph to 100 mph for ambient temperatures of 20ºF, 40ºF and 60ºF. Discuss the results. Design and Essay Problems 7–106 On average, superinsulated homes use just 15 percent of the fuel required to heat the same size conventional home built before the energy crisis in the 1970s. Write an essay on superinsulated homes, and identify the features that make them so energy efficient as well as the problems associated with them. Do you think superinsulated homes will be economically attractive in your area? 7–107 Conduct this experiment to determine the heat loss coefficient of your house or apartment in W/ºC or But/h ºF. First make sure that the conditions in the house are steady and the house is at the set temperature of the thermostat. Use an outdoor thermometer to monitor outdoor temperature. One evening, using a watch or timer, determine how long the heater was on during a 3-h period and the average outdoor temperature during that period. Then using the heat output rating of your heater, determine the amount of heat supplied. Also, estimate the amount of heat generation in the house during that period by noting the number of people, the total wattage of lights that were on, and the heat generated by the appliances and equipment. Using that information, calculate the average rate of heat loss from the house and the heat loss coefficient. 7–108 The decision of whether to invest in an energy-saving measure is made on the basis of the length of time for it to pay for itself in projected energy (and thus cost) savings. The easiest way to reach a decision is to calculate the simple payback period by simply dividing the installed cost of the measure by the annual cost savings and comparing it to the lifetime of the installation. This approach is adequate for short payback periods (less than 5 years) in stable economies with low interest rates (under 10 percent) since the error involved is no larger than the uncertainties. However, if the payback period is long, it may be necessary to consider the interest rate if the money is to be borrowed, or the rate of return if the money is invested elsewhere instead of the energy conservation measure. For example, a simple payback period of five years corresponds to 5.0, 6.12, 6.64, 7.27, 8.09, 9.919, 10.84, and 13.91 for an interest rate (or return on investment) of 0, 6, 8, 10, 12, 14, 16, and 18 percent, respectively. Finding out the proper relations from engineering economics books, determine the payback periods for the interest rates given above corresponding to simple payback periods of 1 through 10 years. 7–109 Obtain information on frostbite and the conditions under which it occurs. Using the relation in Problem 7–104E, prepare a table that shows how long people can stay in cold and windy weather for specified temperatures and wind speeds before the exposed flesh is in danger of experiencing frostbite. 7–110 Write an article on forced convection cooling with air, helium, water, and a dielectric liquid. Discuss the advantages and disadvantages of each fluid in heat transfer. Explain the circumstances under which a certain fluid will be most suitable for the cooling job. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 419 CHAPTER INTERNAL FORCED CONVECTION iquid or gas flow through pipes or ducts is commonly used in heating and cooling applications. The fluid in such applications is forced to flow by a fan or pump through a tube that is sufficiently long to accomplish the desired heat transfer. In this chapter we will pay particular attention to the determination of the friction factor and convection coefficient since they are directly related to the pressure drop and heat transfer rate, respectively. These quantities are then used to determine the pumping power requirement and the required tube length. There is a fundamental difference between external and internal flows. In external flow, considered in Chapter 7, the fluid has a free surface, and thus the boundary layer over the surface is free to grow indefinitely. In internal flow, however, the fluid is completely confined by the inner surfaces of the tube, and thus there is a limit on how much the boundary layer can grow. We start this chapter with a general physical description of internal flow, and the mean velocity and mean temperature. We continue with the discussion of the hydrodynamic and thermal entry lengths, developing flow, and fully developed flow. We then obtain the velocity and temperature profiles for fully developed laminar flow, and develop relations for the friction factor and Nusselt number. Finally we present empirical relations for developing and fully developed flows, and demonstrate their use. L 8 CONTENTS 8–1 Introduction 420 8–2 Mean Velocity and Mean Temperature 420 8–3 The Entrance Region 423 8–4 General Thermal Analysis 426 8–5 Laminar Flow in Tubes 431 8–6 Turbulent Flow in Tubes 441 419 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 420 420 HEAT TRANSFER 8–1 Circular pipe Water 50 atm Rectangular duct Air 1.2 atm FIGURE 8–1 Circular pipes can withstand large pressure differences between the inside and the outside without undergoing any distortion, but the noncircular pipes cannot. ■ INTRODUCTION You have probably noticed that most fluids, especially liquids, are transported in circular pipes. This is because pipes with a circular cross section can withstand large pressure differences between the inside and the outside without undergoing any distortion. Noncircular pipes are usually used in applications such as the heating and cooling systems of buildings where the pressure difference is relatively small and the manufacturing and installation costs are lower (Fig. 8–1). For a fixed surface area, the circular tube gives the most heat transfer for the least pressure drop, which explains the overwhelming popularity of circular tubes in heat transfer equipment. The terms pipe, duct, tube, and conduit are usually used interchangeably for flow sections. In general, flow sections of circular cross section are referred to as pipes (especially when the fluid is a liquid), and the flow sections of noncircular cross section as ducts (especially when the fluid is a gas). Small diameter pipes are usually referred to as tubes. Given this uncertainty, we will use more descriptive phrases (such as a circular pipe or a rectangular duct) whenever necessary to avoid any misunderstandings. Although the theory of fluid flow is reasonably well understood, theoretical solutions are obtained only for a few simple cases such as the fully developed laminar flow in a circular pipe. Therefore, we must rely on the experimental results and the empirical relations obtained for most fluid flow problems rather than closed form analytical solutions. Noting that the experimental results are obtained under carefully controlled laboratory conditions, and that no two systems are exactly alike, we must not be so naive as to view the results obtained as “exact.” An error of 10 percent (or more) in friction or convection coefficient calculated using the relations in this chapter is the “norm” rather than the “exception.” Perhaps we should mention that the friction between the fluid layers in a tube may cause a slight rise in fluid temperature as a result of mechanical energy being converted to thermal energy. But this frictional heating is too small to warrant any consideration in calculations, and thus is disregarded. For example, in the absence of any heat transfer, no noticeable difference will be detected between the inlet and exit temperatures of a fluid flowing in a tube. The primary consequence of friction in fluid flow is pressure drop. Thus, it is reasonable to assume that any temperature change in the fluid is due to heat transfer. But frictional heating must be considered for flows that involve highly viscous fluids with large velocity gradients. In most practical applications, the flow of a fluid through a pipe or duct can be approximated to be one-dimensional, and thus the properties can be assumed to vary in one direction only (the direction of flow). As a result, all properties are uniform at any cross section normal to the flow direction, and the properties are assumed to have bulk average values over the cross section. But the values of the properties at a cross section may change with time unless the flow is steady. 8–2 ■ MEAN VELOCITY AND MEAN TEMPERATURE In external flow, the free-stream velocity served as a convenient reference velocity for use in the evaluation of the Reynolds number and the friction cen58933_ch08.qxd 9/4/2002 11:29 AM Page 421 421 CHAPTER 8 coefficient. In internal flow, there is no free stream and thus we need an alternative. The fluid velocity in a tube changes from zero at the surface because of the no-slip condition, to a maximum at the tube center. Therefore, it is convenient to work with an average or mean velocity m, which remains constant for incompressible flow when the cross sectional area of the tube is constant. The mean velocity in actual heating and cooling applications may change somewhat because of the changes in density with temperature. But, in practice, we evaluate the fluid properties at some average temperature and treat them as constants. The convenience in working with constant properties usually more than justifies the slight loss in accuracy. The value of the mean velocity m in a tube is determined from the requirement that the conservation of mass principle be satisfied (Fig. 8–2). That is, m· m Ac (r, x)dA (8-1) c =0 max (a) Actual m (b) Idealized FIGURE 8–2 Actual and idealized velocity profiles for flow in a tube (the mass flow rate of the fluid is the same for both cases). Ac where m· is the mass flow rate, is the density, Ac is the cross sectional area, and (r, x) is the velocity profile. Then the mean velocity for incompressible flow in a circular tube of radius R can be expressed as m (r, x)dA R c Ac Ac 0 (r, x)2rdr R 2 2 R2 R 0 (r, x)rdr (8-2) Therefore, when we know the mass flow rate or the velocity profile, the mean velocity can be determined easily. When a fluid is heated or cooled as it flows through a tube, the temperature of the fluid at any cross section changes from Ts at the surface of the wall to some maximum (or minimum in the case of heating) at the tube center. In fluid flow it is convenient to work with an average or mean temperature Tm that remains uniform at a cross section. Unlike the mean velocity, the mean temperature Tm will change in the flow direction whenever the fluid is heated or cooled. The value of the mean temperature Tm is determined from the requirement that the conservation of energy principle be satisfied. That is, the energy transported by the fluid through a cross section in actual flow must be equal to the energy that would be transported through the same cross section if the fluid were at a constant temperature Tm. This can be expressed mathematically as (Fig. 8–3) · E fluid m· CpTm C T m· C T dA m· p p Ac c Tm m· p m· Cp R 0 CpT(2rdr) m(R )Cp 2 2 mR2 R 0 T(r, x) (r, x) rdr Tmin (a) Actual (8-3) where Cp is the specific heat of the fluid. Note that the product m·CpTm at any cross section along the tube represents the energy flow with the fluid at that cross section. Then the mean temperature of a fluid with constant density and specific heat flowing in a circular pipe of radius R can be expressed as C T m· Ts (8-4) Tm (b) Idealized FIGURE 8–3 Actual and idealized temperature profiles for flow in a tube (the rate at which energy is transported with the fluid is the same for both cases). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 422 422 HEAT TRANSFER Note that the mean temperature Tm of a fluid changes during heating or cooling. Also, the fluid properties in internal flow are usually evaluated at the bulk mean fluid temperature, which is the arithmetic average of the mean temperatures at the inlet and the exit. That is, Tb (Tm, i Tm, e)/2. Laminar and Turbulent Flow In Tubes Flow in a tube can be laminar or turbulent, depending on the flow conditions. Fluid flow is streamlined and thus laminar at low velocities, but turns turbulent as the velocity is increased beyond a critical value. Transition from laminar to turbulent flow does not occur suddenly; rather, it occurs over some range of velocity where the flow fluctuates between laminar and turbulent flows before it becomes fully turbulent. Most pipe flows encountered in practice are turbulent. Laminar flow is encountered when highly viscous fluids such as oils flow in small diameter tubes or narrow passages. For flow in a circular tube, the Reynolds number is defined as Re Circular tube: Dh = 4(πD2/4) =D πD Rectangular duct: Dh = 4Ac Dh p a a 4a2 Dh = =a 4a b 4ab 2ab = 2(a + b) a+b Circular tubes: FIGURE 8–4 The hydraulic diameter Dh 4Ac /p is defined such that it reduces to ordinary diameter for circular tubes. Die trace (8-6) where Ac is the cross sectional area of the tube and p is its perimeter. The hydraulic diameter is defined such that it reduces to ordinary diameter D for circular tubes since a Laminar (8-5) where m is the mean fluid velocity, D is the diameter of the tube, and / is the kinematic viscosity of the fluid. For flow through noncircular tubes, the Reynolds number as well as the Nusselt number and the friction factor are based on the hydraulic diameter Dh defined as (Fig. 8–4) D Square duct: mD m D Turbulent Pipe wall FIGURE 8–5 In the transitional flow region of 2300 Re 4000, the flow switches between laminar and turbulent randomly. 4Ac 4D2/4 Dh p D D It certainly is desirable to have precise values of Reynolds numbers for laminar, transitional, and turbulent flows, but this is not the case in practice. This is because the transition from laminar to turbulent flow also depends on the degree of disturbance of the flow by surface roughness, pipe vibrations, and the fluctuations in the flow. Under most practical conditions, the flow in a tube is laminar for Re 2300, turbulent for Re 10,000, and transitional in between. That is, Re 2300 2300 Re 10,000 Re 10,000 laminar flow transitional flow turbulent flow In transitional flow, the flow switches between laminar and turbulent randomly (Fig. 8–5). It should be kept in mind that laminar flow can be maintained at much higher Reynolds numbers in very smooth pipes by avoiding flow disturbances and tube vibrations. In such carefully controlled cen58933_ch08.qxd 9/4/2002 11:29 AM Page 423 423 CHAPTER 8 experiments, laminar flow has been maintained at Reynolds numbers of up to 100,000. 8–3 ■ THE ENTRANCE REGION Consider a fluid entering a circular tube at a uniform velocity. As in external flow, the fluid particles in the layer in contact with the surface of the tube will come to a complete stop. This layer will also cause the fluid particles in the adjacent layers to slow down gradually as a result of friction. To make up for this velocity reduction, the velocity of the fluid at the midsection of the tube will have to increase to keep the mass flow rate through the tube constant. As a result, a velocity boundary layer develops along the tube. The thickness of this boundary layer increases in the flow direction until the boundary layer reaches the tube center and thus fills the entire tube, as shown in Figure 8–6. The region from the tube inlet to the point at which the boundary layer merges at the centerline is called the hydrodynamic entrance region, and the length of this region is called the hydrodynamic entry length Lh. Flow in the entrance region is called hydrodynamically developing flow since this is the region where the velocity profile develops. The region beyond the entrance region in which the velocity profile is fully developed and remains unchanged is called the hydrodynamically fully developed region. The velocity profile in the fully developed region is parabolic in laminar flow and somewhat flatter in turbulent flow due to eddy motion in radial direction. Now consider a fluid at a uniform temperature entering a circular tube whose surface is maintained at a different temperature. This time, the fluid particles in the layer in contact with the surface of the tube will assume the surface temperature. This will initiate convection heat transfer in the tube and the development of a thermal boundary layer along the tube. The thickness of this boundary layer also increases in the flow direction until the boundary layer reaches the tube center and thus fills the entire tube, as shown in Figure 8–7. The region of flow over which the thermal boundary layer develops and reaches the tube center is called the thermal entrance region, and the length of this region is called the thermal entry length Lt. Flow in the thermal entrance region is called thermally developing flow since this is the region where the temperature profile develops. The region beyond the thermal entrance region in which the dimensionless temperature profile expressed as (Ts T)/ (Ts Tm) remains unchanged is called the thermally fully developed region. The region in which the flow is both hydrodynamically and thermally developed and thus both the velocity and dimensionless temperature profiles remain unchanged is called fully developed flow. That is, Velocity boundary layer Velocity profile r x Hydrodynamic entrance region Hydrodynamically fully developed region FIGURE 8–6 The development of the velocity boundary layer in a tube. (The developed mean velocity profile will be parabolic in laminar flow, as shown, but somewhat blunt in turbulent flow.) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 424 424 HEAT TRANSFER Thermal boundary layer Ti FIGURE 8–7 The development of the thermal boundary layer in a tube. (The fluid in the tube is being cooled.) Temperature profile Ts x Thermal entrance region Thermally fully developed region (r, x) 0 → (r) x Ts(x) T(r, x) 0 x Ts(x) Tm(x) Hydrodynamically fully developed: Thermally fully developed: (8-7) (8-8) The friction factor is related to the shear stress at the surface, which is related to the slope of the velocity profile at the surface. Noting that the velocity profile remains unchanged in the hydrodynamically fully developed region, the friction factor also remains constant in that region. A similar argument can be given for the heat transfer coefficient in the thermally fully developed region. In a thermally fully developed region, the derivative of (Ts T)/(Ts Tm) with respect to x is zero by definition, and thus (Ts T)/(Ts Tm) is independent of x. Then the derivative of (Ts T)/(Ts Tm) with respect r must also be independent of x. That is, Ts T r Ts Tm rR (T/r)rR f(x) Ts Tm (8-9) Surface heat flux can be expressed as T q·s hx(Ts Tm) k r → hx rR k(T/r)rR Ts Tm (8-10) which, from Eq. 8–9, is independent of x. Thus we conclude that in the thermally fully developed region of a tube, the local convection coefficient is constant (does not vary with x). Therefore, both the friction and convection coefficients remain constant in the fully developed region of a tube. Note that the temperature profile in the thermally fully developed region may vary with x in the flow direction. That is, unlike the velocity profile, the temperature profile can be different at different cross sections of the tube in the developed region, and it usually is. However, the dimensionless temperature profile defined above remains unchanged in the thermally developed region when the temperature or heat flux at the tube surface remains constant. During laminar flow in a tube, the magnitude of the dimensionless Prandtl number Pr is a measure of the relative growth of the velocity and thermal boundary layers. For fluids with Pr 1, such as gases, the two boundary layers essentially coincide with each other. For fluids with Pr 1, such as oils, the velocity boundary layer outgrows the thermal boundary layer. As a result, cen58933_ch08.qxd 9/4/2002 11:29 AM Page 425 425 CHAPTER 8 the hydrodynamic entry length is smaller than the thermal entry length. The opposite is true for fluids with Pr 1 such as liquid metals. Consider a fluid that is being heated (or cooled) in a tube as it flows through it. The friction factor and the heat transfer coefficient are highest at the tube inlet where the thickness of the boundary layers is zero, and decrease gradually to the fully developed values, as shown in Figure 8–8. Therefore, the pressure drop and heat flux are higher in the entrance regions of a tube, and the effect of the entrance region is always to enhance the average friction and heat transfer coefficients for the entire tube. This enhancement can be significant for short tubes but negligible for long ones. h or f hx fx Entrance Fully region developed region x Entry Lengths Lh The hydrodynamic entry length is usually taken to be the distance from the tube entrance where the friction coefficient reaches within about 2 percent of the fully developed value. In laminar flow, the hydrodynamic and thermal entry lengths are given approximately as [see Kays and Crawford (1993), Ref. 13, and Shah and Bhatti (1987), Ref. 25] Lh, laminar 0.05 Re D Lt, laminar 0.05 Re Pr D Pr Lh, laminar (8-11) (8-12) (8-13) The hydrodynamic entry length is much shorter in turbulent flow, as expected, and its dependence on the Reynolds number is weaker. It is 11D at Re 10,000, and increases to 43D at Re 105. In practice, it is generally agreed that the entrance effects are confined within a tube length of 10 diameters, and the hydrodynamic and thermal entry lengths are approximately taken to be Lh, turbulent Lt, turbulent 10D Fully developed flow Thermal boundary layer Velocity boundary layer For Re 20, the hydrodynamic entry length is about the size of the diameter, but increases linearly with the velocity. In the limiting case of Re 2300, the hydrodynamic entry length is 115D. In turbulent flow, the intense mixing during random fluctuations usually overshadows the effects of momentum and heat diffusion, and therefore the hydrodynamic and thermal entry lengths are of about the same size and independent of the Prandtl number. Also, the friction factor and the heat transfer coefficient remain constant in fully developed laminar or turbulent flow since the velocity and normalized temperature profiles do not vary in the flow direction. The hydrodynamic entry length for turbulent flow can be determined from [see Bhatti and Shah (1987), Ref. 1, and Zhi-qing (1982), Ref. 31] Lh, turbulent 1.359 Re1/4 Lt (8-14) The variation of local Nusselt number along a tube in turbulent flow for both uniform surface temperature and uniform surface heat flux is given in Figure 8–9 for the range of Reynolds numbers encountered in heat transfer equipment. We make these important observations from this figure: • The Nusselt numbers and thus the convection heat transfer coefficients are much higher in the entrance region. FIGURE 8–8 Variation of the friction factor and the convection heat transfer coefficient in the flow direction for flow in a tube (Pr 1). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 426 426 HEAT TRANSFER 800 700 Nux, T (Ts = constant) Nux, H (q· s = constant) Nux, T Nux, H 600 500 D 400 Re = 2 105 300 105 200 FIGURE 8–9 Variation of local Nusselt number along a tube in turbulent flow for both uniform surface temperature and uniform surface heat flux [Deissler (1953), Ref. 4]. 6 104 3 104 104 100 0 2 4 6 8 10 12 14 16 18 20 x/D • The Nusselt number reaches a constant value at a distance of less than 10 diameters, and thus the flow can be assumed to be fully developed for x 10D. • The Nusselt numbers for the uniform surface temperature and uniform surface heat flux conditions are identical in the fully developed regions, and nearly identical in the entrance regions. Therefore, Nusselt number is insensitive to the type of thermal boundary condition, and the turbulent flow correlations can be used for either type of boundary condition. Precise correlations for the friction and heat transfer coefficients for the entrance regions are available in the literature. However, the tubes used in practice in forced convection are usually several times the length of either entrance region, and thus the flow through the tubes is often assumed to be fully developed for the entire length of the tube. This simplistic approach gives reasonable results for long tubes and conservative results for short ones. 8–4 . GENERAL THERMAL ANALYSIS You will recall that in the absence of any work interactions (such as electric resistance heating), the conservation of energy equation for the steady flow of a fluid in a tube can be expressed as (Fig. 8–10) Q Ti · m Cp Ti ■ Te m· Cp Te Energy balance: · Q = m· Cp(Te – Ti ) FIGURE 8–10 The heat transfer to a fluid flowing in a tube is equal to the increase in the energy of the fluid. · Q m· Cp(Te Ti) (W) (8-15) where Ti and Te are the mean fluid temperatures at the inlet and exit of the · tube, respectively, and Q is the rate of heat transfer to or from the fluid. Note that the temperature of a fluid flowing in a tube remains constant in the absence of any energy interactions through the wall of the tube. The thermal conditions at the surface can usually be approximated with reasonable accuracy to be constant surface temperature (Ts constant) or constant surface heat flux (q·s constant). For example, the constant surface cen58933_ch08.qxd 9/4/2002 11:29 AM Page 427 427 CHAPTER 8 temperature condition is realized when a phase change process such as boiling or condensation occurs at the outer surface of a tube. The constant surface heat flux condition is realized when the tube is subjected to radiation or electric resistance heating uniformly from all directions. Surface heat flux is expressed as q·s hx (Ts Tm) (W/m2) (8-16) where hx is the local heat transfer coefficient and Ts and Tm are the surface and the mean fluid temperatures at that location. Note that the mean fluid temperature Tm of a fluid flowing in a tube must change during heating or cooling. Therefore, when hx h constant, the surface temperature Ts must change when q·s constant, and the surface heat flux q·s must change when Ts constant. Thus we may have either Ts constant or q·s constant at the surface of a tube, but not both. Next we consider convection heat transfer for these two common cases. Constant Surface Heat Flux (q·s constant) T In the case of q·s constant, the rate of heat transfer can also be expressed as · Q q·s As m· Cp(Te Ti) (W) Entrance region Fully developed region Ts (8-17) Te Then the mean fluid temperature at the tube exit becomes q·s As Te Ti · m Cp Ti Note that the mean fluid temperature increases linearly in the flow direction in the case of constant surface heat flux, since the surface area increases linearly in the flow direction (As is equal to the perimeter, which is constant, times the tube length). The surface temperature in the case of constant surface heat flux q·s can be determined from q·s h(Ts Tm) → q·s Ts Tm h q· s = constant L Ti x Te FIGURE 8–11 Variation of the tube surface and the mean fluid temperatures along the tube for the case of constant surface heat flux. · δ Q = h(Ts – Tm )dA Tm (8-20) where p is the perimeter of the tube. Noting that both q·s and h are constants, the differentiation of Eq. 8–19 with respect to x gives dTm dTs dx dx 0 (8-19) In the fully developed region, the surface temperature Ts will also increase linearly in the flow direction since h is constant and thus Ts Tm constant (Fig. 8–11). Of course this is true when the fluid properties remain constant during flow. The slope of the mean fluid temperature Tm on a T-x diagram can be determined by applying the steady-flow energy balance to a tube slice of thickness dx shown in Figure 8–12. It gives dTm q·s p m· Cp dTm q·s(pdx) → · constant m Cp dx Tm q· ∆T = Ts – Tm = ––s h (8-18) (8-21) Tm + dTm . m· Cp(Tm + d Tm) m CpTm Ts dx FIGURE 8–12 Energy interactions for a differential control volume in a tube. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 428 428 HEAT TRANSFER Also, the requirement that the dimensionless temperature profile remains unchanged in the fully developed region gives Ts T 0 x Ts Tm → Ts T 1 0 x Ts Tm x → T dTs x dx (8-22) since Ts Tm constant. Combining Eqs. 8–20, 8–21, and 8–22 gives T dTs dTm q·s p · constant x m Cp dx dx T (r) T (r) Ts2 Ts1 Then we conclude that in fully developed flow in a tube subjected to constant surface heat flux, the temperature gradient is independent of x and thus the shape of the temperature profile does not change along the tube (Fig. 8–13). For a circular tube, p 2R and m· m Ac m(R2), and Eq. 8–23 becomes 2q·s T dTs dTm constant x dx dx mCp R Circular tube: x1 x q· s x2 FIGURE 8–13 The shape of the temperature profile remains unchanged in the fully developed region of a tube subjected to constant surface heat flux. (8-23) (8-24) where m is the mean velocity of the fluid. Constant Surface Temperature (Ts constant) From Newton’s law of cooling, the rate of heat transfer to or from a fluid flowing in a tube can be expressed as · Q hAsTave hAs(Ts Tm)ave (W) (8-25) where h is the average convection heat transfer coefficient, As is the heat transfer surface area (it is equal to DL for a circular pipe of length L), and Tave is some appropriate average temperature difference between the fluid and the surface. Below we discuss two suitable ways of expressing Tave. In the constant surface temperature (Ts constant) case, Tave can be expressed approximately by the arithmetic mean temperature difference Tam as Ti Te (Ts Ti) (Ts Te) Ti Te Ts 2 2 2 Ts Tb Tave Tam (8-26) where Tb (Ti Te)/2 is the bulk mean fluid temperature, which is the arithmetic average of the mean fluid temperatures at the inlet and the exit of the tube. Note that the arithmetic mean temperature difference Tam is simply the average of the temperature differences between the surface and the fluid at the inlet and the exit of the tube. Inherent in this definition is the assumption that the mean fluid temperature varies linearly along the tube, which is hardly ever the case when Ts constant. This simple approximation often gives acceptable results, but not always. Therefore, we need a better way to evaluate Tave. Consider the heating of a fluid in a tube of constant cross section whose inner surface is maintained at a constant temperature of Ts. We know that the cen58933_ch08.qxd 9/4/2002 11:29 AM Page 429 429 CHAPTER 8 mean temperature of the fluid Tm will increase in the flow direction as a result of heat transfer. The energy balance on a differential control volume shown in Figure 8–12 gives m· Cp dTm h(Ts Tm)dAs (8-27) That is, the increase in the energy of the fluid (represented by an increase in its mean temperature by dTm) is equal to the heat transferred to the fluid from the tube surface by convection. Noting that the differential surface area is dAs pdx, where p is the perimeter of the tube, and that dTm d(Ts Tm), since Ts is constant, the relation above can be rearranged as d(Ts Tm) hp · dx Ts Tm mCp T (8-28) (8-29) (Tm approaches Ts asymptotically) 0 L x Te Ti Ts = constant (8-30) This relation can also be used to determine the mean fluid temperature Tm(x) at any x by replacing As pL by px. Note that the temperature difference between the fluid and the surface decays exponentially in the flow direction, and the rate of decay depends on the magnitude of the exponent hAx /m· Cp, as shown in Figure 8–14. This dimensionless parameter is called the number of transfer units, denoted by NTU, and is a measure of the effectiveness of the heat transfer systems. For NUT 5, the exit temperature of the fluid becomes almost equal to the surface temperature, Te Ts (Fig. 8–15). Noting that the fluid temperature can approach the surface temperature but cannot cross it, an NTU of about 5 indicates that the limit is reached for heat transfer, and the heat transfer will not increase no matter how much we extend the length of the tube. A small value of NTU, on the other hand, indicates more opportunities for heat transfer, and the heat transfer will continue increasing as the tube length is increased. A large NTU and thus a large heat transfer surface area (which means a large tube) may be desirable from a heat transfer point of view, but it may be unacceptable from an economic point of view. The selection of heat transfer equipment usually reflects a compromise between heat transfer performance and cost. Solving Eq. 8–29 for m· Cp gives hAs m· Cp ln[(Ts Te)/(Ts Ti)] ∆T = Ts – Tm Ti where As pL is the surface area of the tube and h is the constant average convection heat transfer coefficient. Taking the exponential of both sides and solving for Te gives the following relation which is very useful for the determination of the mean fluid temperature at the tube exit: Te Ts (Ts Ti) exp( hAs /m· Cp) ∆Te Tm ∆Ti Integrating from x 0 (tube inlet where Tm Ti) to x L (tube exit where Tm Te) gives Ts Te hAs ln · Ts Ti mCp Ts = constant Ts (8-31) FIGURE 8–14 The variation of the mean fluid temperature along the tube for the case of constant temperature. Ts = 100°C Ti = 20°C m· , Cp Te As, h NTU = hAs / m· Cp Te , °C 0.01 0.05 0.10 0.50 1.00 5.00 10.00 20.8 23.9 27.6 51.5 70.6 99.5 100.0 FIGURE 8–15 An NTU greater than 5 indicates that the fluid flowing in a tube will reach the surface temperature at the exit regardless of the inlet temperature. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 430 430 HEAT TRANSFER Substituting this into Eq. 8–17, we obtain · Q hAsTln (8-32) where Tln Te Ti Ti Te ln[(Ts Te)/(Ts Ti)] ln(Te /Ti) (8-33) is the logarithmic mean temperature difference. Note that Ti Ts Ti and Te Ts Te are the temperature differences between the surface and the fluid at the inlet and the exit of the tube, respectively. This Tln relation appears to be prone to misuse, but it is practically fail-safe, since using Ti in place of Te and vice versa in the numerator and/or the denominator will, at most, affect the sign, not the magnitude. Also, it can be used for both heating (Ts Ti and Te) and cooling (Ts Ti and Te) of a fluid in a tube. The logarithmic mean temperature difference Tln is obtained by tracing the actual temperature profile of the fluid along the tube, and is an exact representation of the average temperature difference between the fluid and the surface. It truly reflects the exponential decay of the local temperature difference. When Te differs from Ti by no more than 40 percent, the error in using the arithmetic mean temperature difference is less than 1 percent. But the error increases to undesirable levels when Te differs from Ti by greater amounts. Therefore, we should always use the logarithmic mean temperature difference when determining the convection heat transfer in a tube whose surface is maintained at a constant temperature Ts. EXAMPLE 8–1 Water enters a 2.5-cm-internal-diameter thin copper tube of a heat exchanger at 15°C at a rate of 0.3 kg/s, and is heated by steam condensing outside at 120°C. If the average heat transfer coefficient is 800 W/m2 C, determine the length of the tube required in order to heat the water to 115°C (Fig. 8–16). Steam Ts = 120°C 115°C Water 15°C 0.3 kg/s D = 2.5 cm FIGURE 8–16 Schematic for Example 8–1. Heating of Water in a Tube by Steam SOLUTION Water is heated by steam in a circular tube. The tube length required to heat the water to a specified temperature is to be determined. Assumptions 1 Steady operating conditions exist. 2 Fluid properties are constant. 3 The convection heat transfer coefficient is constant. 4 The conduction resistance of copper tube is negligible so that the inner surface temperature of the tube is equal to the condensation temperature of steam. Properties The specific heat of water at the bulk mean temperature of (15 115)/2 65°C is 4187 J/kg °C. The heat of condensation of steam at 120°C is 2203 kJ/kg (Table A-9). Analysis Knowing the inlet and exit temperatures of water, the rate of heat transfer is determined to be · Q m· Cp(Te Ti) (0.3 kg/s)(4.187 kJ/kg °C)(115°C 15°C) 125.6 kW The logarithmic mean temperature difference is cen58933_ch08.qxd 9/4/2002 11:29 AM Page 431 431 CHAPTER 8 Te Ts Te 120°C 115°C 5°C Ti Ts Ti 120°C 15°C 105°C Te Ti 5 105 Tln 32.85°C ln(Te /Ti) ln(5/105) The heat transfer surface area is · Q hAsTln → As Q· 125.6 kW 4.78 m2 hTln (0.8 kW/m2 · °C)(32.85°C) Then the required length of tube becomes As DL → L As 4.78 m2 61 m D (0.025 m) Discussion The bulk mean temperature of water during this heating process is 65°C, and thus the arithmetic mean temperature difference is Tam 120 65 55°C. Using Tam instead of Tln would give L 36 m, which is grossly in error. This shows the importance of using the logarithmic mean temperature in calculations. 8–5 ■ LAMINAR FLOW IN TUBES We mentioned earlier that flow in tubes is laminar for Re 2300, and that the flow is fully developed if the tube is sufficiently long (relative to the entry length) so that the entrance effects are negligible. In this section we consider the steady laminar flow of an incompressible fluid with constant properties in the fully developed region of a straight circular tube. We obtain the momentum and energy equations by applying momentum and energy balances to a differential volume element, and obtain the velocity and temperature profiles by solving them. Then we will use them to obtain relations for the friction factor and the Nusselt number. An important aspect of the analysis below is that it is one of the few available for viscous flow and forced convection. In fully developed laminar flow, each fluid particle moves at a constant axial velocity along a streamline and the velocity profile (r) remains unchanged in the flow direction. There is no motion in the radial direction, and thus the velocity component v in the direction normal to flow is everywhere zero. There is no acceleration since the flow is steady. Now consider a ring-shaped differential volume element of radius r, thickness dr, and length dx oriented coaxially with the tube, as shown in Figure 8–17. The pressure force acting on a submerged plane surface is the product of the pressure at the centroid of the surface and the surface area. The volume element involves only pressure and viscous effects, and thus the pressure and shear forces must balance each other. A force balance on the volume element in the flow direction gives (2rdrP)x (2rdrP)x dx (2rdx)r (2rdx)r dr 0 (8-34) τr dr Px Px dx τr (r) R dr dx r x max FIGURE 8–17 Free body diagram of a cylindrical fluid element of radius r, thickness dr, and length dx oriented coaxially with a horizontal tube in fully developed steady flow. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 432 432 HEAT TRANSFER which indicates that in fully developed flow in a tube, the viscous and pressure forces balance each other. Dividing by 2drdx and rearranging, r Pxdx Px (r )xdr (r )r 0 dx dr (8-35) Taking the limit as dr, dx → 0 gives dP d(r ) 0 dx dr r (8-36) Substituting (d /dr) and rearranging gives the desired equation, d d dP r dr r dr dx (8-37) The quantity d /dr is negative in tube flow, and the negative sign is included to obtain positive values for . (Or, d /dr d /dy since y R r.) The left side of this equation is a function of r and the right side is a function of x. The equality must hold for any value of r and x, and an equality of the form f(r) g(x) can happen only if both f(r) and g(x) are equal to constants. Thus we conclude that dP/dx constant. This can be verified by writing a force balance on a volume element of radius R and thickness dx (a slice of the tube), which gives dP/dx 2s/R. Here s is constant since the viscosity and the velocity profile are constants in the fully developed region. Therefore, dP/dx constant. Equation 8–37 can be solved by rearranging and integrating it twice to give (r) 1 dP C1lnr C2 4 dx (8-38) The velocity profile (r) is obtained by applying the boundary conditions /r 0 at r 0 (because of symmetry about the centerline) and 0 at r R (the no-slip condition at the tube surface). We get (r) R2 dP r2 1 2 4 dx R (8-39) Therefore, the velocity profile in fully developed laminar flow in a tube is parabolic with a maximum at the centerline and minimum at the tube surface. Also, the axial velocity is positive for any r, and thus the axial pressure gradient dP/dx must be negative (i.e., pressure must decrease in the flow direction because of viscous effects). The mean velocity is determined from its definition by substituting Eq. 8–39 into Eq. 8–2, and performing the integration. It gives m 2 R2 R 0 rdr 2 R2 4R dPdx1 Rr rdr 8R dPdx R 2 2 2 2 0 (8-40) Combining the last two equations, the velocity profile is obtained to be (r) 2m 1 r2 R2 (8-41) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 433 433 CHAPTER 8 This is a convenient form for the velocity profile since m can be determined easily from the flow rate information. The maximum velocity occurs at the centerline, and is determined from Eq. 8–39 by substituting r 0, max 2m (8-42) Therefore, the mean velocity is one-half of the maximum velocity. Pressure Drop A quantity of interest in the analysis of tube flow is the pressure drop P since it is directly related to the power requirements of the fan or pump to maintain flow. We note that dP/dx constant, and integrate it from x 0 where the pressure is P1 to x L where the pressure is P2. We get dP P2 P1 P L L dx (8-43) Note that in fluid mechanics, the pressure drop P is a positive quantity, and is defined as P P1 P2. Substituting Eq. 8–43 into the m expression in Eq. 8–40, the pressure drop can be expressed as Laminar flow: P 8L m 32L m R2 D2 (8-44) In practice, it is found convenient to express the pressure drop for all types of internal flows (laminar or turbulent flows, circular or noncircular tubes, smooth or rough surfaces) as (Fig. 8–18) L m D 2 2 P f (8-45) ∆P m D L where the dimensionless quantity f is the friction factor (also called the Darcy friction factor after French engineer Henry Darcy, 1803–1858, who first studied experimentally the effects of roughness on tube resistance). It should not be confused with the friction coefficient Cf (also called the Fanning friction factor), which is defined as Cf s(m2/2) f/4. Equation 8–45 gives the pressure drop for a flow section of length L provided that (1) the flow section is horizontal so that there are no hydrostatic or gravity effects, (2) the flow section does not involve any work devices such as a pump or a turbine since they change the fluid pressure, and (3) the cross sectional area of the flow section is constant and thus the mean flow velocity is constant. Setting Eqs. 8–44 and 8–45 equal to each other and solving for f gives the friction factor for the fully developed laminar flow in a circular tube to be Circular tube, laminar: f 64 64 Dm Re (8-46) This equation shows that in laminar flow, the friction factor is a function of the Reynolds number only and is independent of the roughness of the tube ρm Pressure drop: ∆P = f L D 2 2 FIGURE 8–18 The relation for pressure drop is one of the most general relations in fluid mechanics, and it is valid for laminar or turbulent flows, circular or noncircular pipes, and smooth or rough surfaces. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 434 434 HEAT TRANSFER surface. Once the pressure drop is available, the required pumping power is determined from · · Wpump VP (8-47) · where V is the volume flow rate of flow, which is expressed as · R4P D4P PR2 V ave Ac R2 8L 8L 128L (8-48) This equation is known as the Poiseuille’s Law, and this flow is called the Hagen–Poiseuille flow in honor of the works of G. Hagen (1797–1839) and J. Poiseuille (1799–1869) on the subject. Note from Eq. 8–48 that for a specified flow rate, the pressure drop and thus the required pumping power is proportional to the length of the tube and the viscosity of the fluid, but it is inversely proportional to the fourth power of the radius (or diameter) of the tube. Therefore, the pumping power requirement for a piping system can be reduced by a factor of 16 by doubling the tube diameter (Fig. 8–19). Of course the benefits of the reduction in the energy costs must be weighed against the increased cost of construction due to using a larger diameter tube. The pressure drop is caused by viscosity, and it is directly related to the wall shear stress. For the ideal inviscid flow, the pressure drop is zero since there are no viscous effects. Again, Eq. 8–47 is valid for both laminar and turbulent flows in circular and noncircular tubes. ⋅ Wpump = 16 hp D ⋅ Wpump = 1 hp 2D FIGURE 8–19 The pumping power requirement for a laminar flow piping system can be reduced by a factor of 16 by doubling the pipe diameter. dx mCpTx dx r dr Qr Temperature Profile and the Nusselt Number In the analysis above, we have obtained the velocity profile for fully developed flow in a circular tube from a momentum balance applied on a volume element, determined the friction factor and the pressure drop. Below we obtain the energy equation by applying the energy balance to a differential volume element, and solve it to obtain the temperature profile for the constant surface temperature and the constant surface heat flux cases. Reconsider steady laminar flow of a fluid in a circular tube of radius R. The fluid properties , k, and Cp are constant, and the work done by viscous stresses is negligible. The fluid flows along the x-axis with velocity . The flow is fully developed so that is independent of x and thus (r). Noting that energy is transferred by mass in the x-direction, and by conduction in the r-direction (heat conduction in the x-direction is assumed to be negligible), the steady-flow energy balance for a cylindrical shell element of thickness dr and length dx can be expressed as (Fig. 8–20) · · m· CpTx m· CpTx dx Q r Q r dr 0 mCpTx Qr dr FIGURE 8–20 The differential volume element used in the derivation of energy balance relation. (8-49) where m· Ac (2rdr). Substituting and dividing by 2rdrdx gives, after rearranging, Cp · · Txdx Tx 1 Q rdr Q r dx 2rdx dr (8-50) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 435 435 CHAPTER 8 or Q· T 1 x 2Cprdx r (8-51) where we used the definition of derivative. But Q· T T r k2rdx 2kdx r r r r r (8-52) Substituting and using k/Cp gives T α T r r x dr r (8-53) which states that the rate of net energy transfer to the control volume by mass flow is equal to the net rate of heat conduction in the radial direction. Constant Surface Heat Flux For fully developed flow in a circular pipe subjected to constant surface heat flux, we have, from Eq. 8–24, 2q·s T dTs dTm constant x dx dx mCpR (8-54) If heat conduction in the x-direction were considered in the derivation of Eq. 8–53, it would give an additional term 2T/x2, which would be equal to zero since T/x constant and thus T T(r). Therefore, the assumption that there is no axial heat conduction is satisfied exactly in this case. Substituting Eq. 8–54 and the relation for velocity profile (Eq. 8–41) into Eq. 8–53 gives 4q·s dT r2 1 d 1 2 r r kR dr dr R (8-55) which is a second-order ordinary differential equation. Its general solution is obtained by separating the variables and integrating twice to be T q·s 2 r2 r 2 C1r C2 kR 4R (8-56) The desired solution to the problem is obtained by applying the boundary conditions T/x 0 at r 0 (because of symmetry) and T Ts at r R. We get T Ts q·s R 3 r 2 r4 2 4 k 4 R 4R (8-57) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 436 436 HEAT TRANSFER The bulk mean temperature Tm is determined by substituting the velocity and temperature profile relations (Eqs. 8–41 and 8–57) into Eq. 8–4 and performing the integration. It gives Tm Ts 11 q·s R 24 k (8-58) Combining this relation with q·s h(Ts Tm) gives h k 24 k 48 k 4.36 D 11 R 11 D (8-59) or Circular tube, laminar (q·x constant): Nu hD 4.36 k (8-60) Therefore, for fully developed laminar flow in a circular tube subjected to constant surface heat flux, the Nusselt number is a constant. There is no dependence on the Reynolds or the Prandtl numbers. Constant Surface Temperature A similar analysis can be performed for fully developed laminar flow in a circular tube for the case of constant surface temperature Ts. The solution procedure in this case is more complex as it requires iterations, but the Nusselt number relation obtained is equally simple (Fig. 8–21): Ts = constant 64 f = ––– Re D Nu = 3.66 max m Fully developed laminar flow FIGURE 8–21 In laminar flow in a tube with constant surface temperature, both the friction factor and the heat transfer coefficient remain constant in the fully developed region. Circular tube, laminar (Ts constant): Nu hD 3.66 k (8-61) The thermal conductivity k for use in the Nu relations above should be evaluated at the bulk mean fluid temperature, which is the arithmetic average of the mean fluid temperatures at the inlet and the exit of the tube. For laminar flow, the effect of surface roughness on the friction factor and the heat transfer coefficient is negligible. Laminar Flow in Noncircular Tubes The friction factor f and the Nusselt number relations are given in Table 8–1 for fully developed laminar flow in tubes of various cross sections. The Reynolds and Nusselt numbers for flow in these tubes are based on the hydraulic diameter Dh 4Ac /p, where Ac is the cross sectional area of the tube and p is its perimeter. Once the Nusselt number is available, the convection heat transfer coefficient is determined from h kNu/Dh. Developing Laminar Flow in the Entrance Region For a circular tube of length L subjected to constant surface temperature, the average Nusselt number for the thermal entrance region can be determined from (Edwards et al., 1979) Entry region, laminar: Nu 3.66 0.065 (D/L) Re Pr 1 0.04[(D/L) Re Pr]2/3 (8-62) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 437 437 CHAPTER 8 TABLE 8–1 Nusselt number and friction factor for fully developed laminar flow in tubes of various cross sections (Dh 4Ac /p, Re m Dh /v, and Nu hDh /k) Tube Geometry Circle a/b or ° Nusselt Number Ts Const. q· s Const. Friction Factor f — 3.66 4.36 64.00/Re a/b 1 2 3 4 6 8 2.98 3.39 3.96 4.44 5.14 5.60 7.54 3.61 4.12 4.79 5.33 6.05 6.49 8.24 56.92/Re 62.20/Re 68.36/Re 72.92/Re 78.80/Re 82.32/Re 96.00/Re a/b 1 2 4 8 16 3.66 3.74 3.79 3.72 3.65 4.36 4.56 4.88 5.09 5.18 64.00/Re 67.28/Re 72.96/Re 76.60/Re 78.16/Re 10° 30° 60° 90° 120° 1.61 2.26 2.47 2.34 2.00 2.45 2.91 3.11 2.98 2.68 50.80/Re 52.28/Re 53.32/Re 52.60/Re 50.96/Re D Rectangle b a Ellipse b a Triangle θ Note that the average Nusselt number is larger at the entrance region, as expected, and it approaches asymptotically to the fully developed value of 3.66 as L → . This relation assumes that the flow is hydrodynamically developed when the fluid enters the heating section, but it can also be used approximately for flow developing hydrodynamically. When the difference between the surface and the fluid temperatures is large, it may be necessary to account for the variation of viscosity with temperature. The average Nusselt number for developing laminar flow in a circular tube in that case can be determined from [Sieder and Tate (1936), Ref. 26] Nu 1.86 1/3 b s 0.14 Re Pr D L (8-63) All properties are evaluated at the bulk mean fluid temperature, except for s, which is evaluated at the surface temperature. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 438 438 HEAT TRANSFER The average Nusselt number for the thermal entrance region of flow between isothermal parallel plates of length L is expressed as (Edwards et al., 1979) Entry region, laminar: Nu 7.54 0.03 (Dh /L) Re Pr 1 0.016[(Dh /L) Re Pr]2/3 (8-64) where Dh is the hydraulic diameter, which is twice the spacing of the plates. This relation can be used for Re 2800. EXAMPLE 8–2 3 ft/s 0.15 in. Pressure Drop in a Pipe Water at 40°F ( 62.42 lbm/ft3 and 3.74 lbm/ft h) is flowing in a 0.15in.-diameter 30-ft-long pipe steadily at an average velocity of 3 ft/s (Fig. 8–22). Determine the pressure drop and the pumping power requirement to overcome this pressure drop. 30 ft FIGURE 8–22 Schematic for Example 8–2. SOLUTION The average flow velocity in a pipe is given. The pressure drop and the required pumping power are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. Properties The density and dynamic viscosity of water are given to be 62.42 lbm/ft3 and 3.74 lbm/ft h 0.00104 lbm/ft s. Analysis First we need to determine the flow regime. The Reynolds number is Re m D (62.42 lbm/ft3)(3 ft/s)(0.12/12 ft) 3600 s 1803 3.74 lbm/ft · h 1h which is less than 2300. Therefore, the flow is laminar. Then the friction factor and the pressure drop become 64 64 0.0355 Re 1803 2 30 ft (62.42 lbm/ft3)(3 ft/s)2 1 lbf L m P f 0.0355 D 2 2 0.12/12 ft 32.174 lbm · ft/s2 930 lbf/ft2 6.46 psi f The volume flow rate and the pumping power requirements are · V m Ac m (D2/4) (3 ft/s)[(0.12/12 ft)2/4] 0.000236 ft3/s 0.7371lbfW · ft/s 0.30 W · · Wpump VP (0.000236 ft3/s)(930 lbf/ft2) Therefore, mechanical power input in the amount of 0.30 W is needed to overcome the frictional losses in the flow due to viscosity. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 439 439 CHAPTER 8 EXAMPLE 8–3 Flow of Oil in a Pipeline through a Lake Consider the flow of oil at 20°C in a 30-cm-diameter pipeline at an average velocity of 2 m/s (Fig. 8–23). A 200-m-long section of the pipeline passes through icy waters of a lake at 0°C. Measurements indicate that the surface temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance of the pipe material, determine (a) the temperature of the oil when the pipe leaves the lake, (b) the rate of heat transfer from the oil, and (c) the pumping power required to overcome the pressure losses and to maintain the flow of the oil in the pipe. Icy lake, 0°C 20°C Oil 2 m/s D = 0.3 m Te 0°C 200 m FIGURE 8–23 Schematic for Example 8–3. SOLUTION Oil flows in a pipeline that passes through icy waters of a lake at 0°C. The exit temperature of the oil, the rate of heat loss, and the pumping power needed to overcome pressure losses are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is very nearly 0°C. 3 The thermal resistance of the pipe is negligible. 4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically developed when the pipeline reaches the lake. Properties We do not know the exit temperature of the oil, and thus we cannot determine the bulk mean temperature, which is the temperature at which the properties of oil are to be evaluated. The mean temperature of the oil at the inlet is 20°C, and we expect this temperature to drop somewhat as a result of heat loss to the icy waters of the lake. We evaluate the properties of the oil at the inlet temperature, but we will repeat the calculations, if necessary, using properties at the evaluated bulk mean temperature. At 20°C we read (Table A-14) 888 kg/m3 k 0.145 W/m °C 901 10 6 m2/s Cp 1880 J/kg °C Pr 10,400 Analysis (a) The Reynolds number is Re m Dh (2 m/s)(0.3 m) 901 10 6 m2/s 666 which is less than the critical Reynolds number of 2300. Therefore, the flow is laminar, and the thermal entry length in this case is roughly Lt 0.05 Re Pr D 0.05 666 10,400 (0.3 m) 104,000 m which is much greater than the total length of the pipe. This is typical of fluids with high Prandtl numbers. Therefore, we assume thermally developing flow and determine the Nusselt number from Nu 0.065 (D/L) Re Pr hD 3.66 k 1 0.04 [(D/L) Re Pr]2/3 3.66 37.3 0.065(0.3/200) 666 10,400 1 0.04[(0.3/200) 666 10,400]2/3 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 440 440 HEAT TRANSFER Note that this Nusselt number is considerably higher than the fully developed value of 3.66. Then, h 0.145 W/m k Nu (37.3) 18.0 W/m2 °C D 0.3 m Also, As pL DL (0.3 m)(200 m) 188.5 m2 m· A (888 kg/m3)1(0.3 m)2 125.5 kg/s c m 4 Next we determine the exit temperature of oil from Te Ts (Ts Ti) exp ( hAs /m· Cp) (18.0 W/m2 · °C)(188.5 m2) 0°C [(0 20)°C] exp (125.5 kg/s)(1880 J/kg · °C) 19.71°C Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the lake. This makes the bulk mean oil temperature 19.86°C, which is practically identical to the inlet temperature of 20°C. Therefore, we do not need to reevaluate the properties. (b) The logarithmic mean temperature difference and the rate of heat loss from the oil are Ti Te 20 19.71 19.85°C Ts Te 0 19.71 ln ln 0 20 Ts Ti · Q hAs Tln (18.0 W/m2 °C)(188.5 m2)( 19.85°C) 6.74 104 Tln Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe in the icy waters of the lake. Note that Tln is identical to the arithmetic mean temperature in this case, since Ti Te. (c) The laminar flow of oil is hydrodynamically developed. Therefore, the friction factor can be determined from f 64 64 0.0961 Re 666 Then the pressure drop in the pipe and the required pumping power become 3 2 2 200 m (888 kg/m )(2 m/s) L m 0.0961 1.14 105 N/m2 D 2 0.3 m 2 5 2 · m· P (125.5 kg/s)(1.14 10 N/m ) Wpump 16.1 kW 3 888 kg/m P f Discussion We will need a 16.1-kW pump just to overcome the friction in the pipe as the oil flows in the 200-m-long pipe through the lake. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 441 441 CHAPTER 8 8–6 ■ TURBULENT FLOW IN TUBES We mentioned earlier that flow in smooth tubes is fully turbulent for Re 10,000. Turbulent flow is commonly utilized in practice because of the higher heat transfer coefficients associated with it. Most correlations for the friction and heat transfer coefficients in turbulent flow are based on experimental studies because of the difficulty in dealing with turbulent flow theoretically. For smooth tubes, the friction factor in turbulent flow can be determined from the explicit first Petukhov equation [Petukhov (1970), Ref. 21] given as f (0.790 ln Re 1.64) 2 Smooth tubes: 104 Re 106 (8-65) The Nusselt number in turbulent flow is related to the friction factor through the Chilton–Colburn analogy expressed as Nu 0.125 f RePr1/3 (8-66) Once the friction factor is available, this equation can be used conveniently to evaluate the Nusselt number for both smooth and rough tubes. For fully developed turbulent flow in smooth tubes, a simple relation for the Nusselt number can be obtained by substituting the simple power law relation f 0.184 Re 0.2 for the friction factor into Eq. 8–66. It gives Nu 0.023 Re0.8 Pr1/3 0.7 Pr 160 Re 10,000 (8-67) which is known as the Colburn equation. The accuracy of this equation can be improved by modifying it as Nu 0.023 Re0.8 Pr n (8-68) where n 0.4 for heating and 0.3 for cooling of the fluid flowing through the tube. This equation is known as the Dittus–Boelter equation [Dittus and Boelter (1930), Ref. 6] and it is preferred to the Colburn equation. The fluid properties are evaluated at the bulk mean fluid temperature Tb (Ti Te)/2. When the temperature difference between the fluid and the wall is very large, it may be necessary to use a correction factor to account for the different viscosities near the wall and at the tube center. The Nusselt number relations above are fairly simple, but they may give errors as large as 25 percent. This error can be reduced considerably to less than 10 percent by using more complex but accurate relations such as the second Petukhov equation expressed as Nu ( f/8) Re Pr 1.07 12.7( f/8)0.5 (Pr2/3 1) 100.5 PrRe 2000 5 10 4 6 (8-69) The accuracy of this relation at lower Reynolds numbers is improved by modifying it as [Gnielinski (1976), Ref. 8] Nu ( f/8)(Re 1000) Pr 1 12.7( f/8)0.5 (Pr2/3 1) 30.5 10Pr Re2000 5 10 3 6 (8-70) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 442 442 HEAT TRANSFER Relative Roughness, /L Friction Factor, f 0.0 0.00001 0.0001 0.0005 0.001 0.005 0.01 0.05 0.0119 0.0119 0.0134 0.0172 0.0199 0.0305 0.0380 0.0716 where the friction factor f can be determined from an appropriate relation such as the first Petukhov equation. Gnielinski’s equation should be preferred in calculations. Again properties should be evaluated at the bulk mean fluid temperature. The relations above are not very sensitive to the thermal conditions at the tube surfaces and can be used for both Ts constant and q·s constant cases. Despite their simplicity, the correlations already presented give sufficiently accurate results for most engineering purposes. They can also be used to obtain rough estimates of the friction factor and the heat transfer coefficients in the transition region 2300 Re 10,000, especially when the Reynolds number is closer to 10,000 than it is to 2300. The relations given so far do not apply to liquid metals because of their very low Prandtl numbers. For liquid metals (0.004 Pr 0.01), the following relations are recommended by Sleicher and Rouse (1975, Ref. 27) for 104 Re 106: Liquid metals, Ts constant: Liquid metals, q·s constant: Standard sizes for Schedule 40 steel pipes Nominal Size, in. Actual Inside Diameter, in. ⁄8 ⁄4 3 ⁄8 1 ⁄2 3 ⁄4 1 11⁄2 2 21⁄2 3 5 10 0.269 0.364 0.493 0.622 0.824 1.049 1.610 2.067 2.469 3.068 5.047 10.02 1 1 (8-72) Rough Surfaces Smooth surface. All values are for Re 10 , and are calculated from Eq. 8–73. TABLE 8–2 (8-71) where the subscript s indicates that the Prandtl number is to be evaluated at the surface temperature. 6 FIGURE 8–24 The friction factor is minimum for a smooth pipe and increases with roughness. Nu 4.8 0.0156 Re0.85 Pr0.93 s Nu 6.3 0.0167 Re0.85 Pr0.93 s Any irregularity or roughness on the surface disturbs the laminar sublayer, and affects the flow. Therefore, unlike laminar flow, the friction factor and the convection coefficient in turbulent flow are strong functions of surface roughness. The friction factor in fully developed turbulent flow depends on the Reynolds number and the relative roughness /D. In 1939, C. F. Colebrook (Ref. 3) combined all the friction factor data for transition and turbulent flow in smooth as well as rough pipes into the following implicit relation known as the Colebrook equation. /D 2.51 1 2.0 log 3.7 Re f f (turbulent flow) (8-73) In 1944, L. F. Moody (Ref. 17) plotted this formula into the famous Moody chart given in the Appendix. It presents the friction factors for pipe flow as a function of the Reynolds number and /D over a wide range. For smooth tubes, the agreement between the Petukhov and Colebrook equations is very good. The friction factor is minimum for a smooth pipe (but still not zero because of the no-slip condition), and increases with roughness (Fig. 8–24). Although the Moody chart is developed for circular pipes, it can also be used for noncircular pipes by replacing the diameter by the hydraulic diameter. At very large Reynolds numbers (to the right of the dashed line on the chart) the friction factor curves corresponding to specified relative roughness curves are nearly horizontal, and thus the friction factors are independent of the Reynolds number. In calculations, we should make sure that we use the internal diameter of the pipe, which may be different than the nominal diameter. For example, the internal diameter of a steel pipe whose nominal diameter is 1 in. is 1.049 in. (Table 8–2). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 443 443 CHAPTER 8 Commercially available pipes differ from those used in the experiments in that the roughness of pipes in the market is not uniform, and it is difficult to give a precise description of it. Equivalent roughness values for some commercial pipes are given in Table 8–3, as well as on the Moody chart. But it should be kept in mind that these values are for new pipes, and the relative roughness of pipes may increase with use as a result of corrosion, scale buildup, and precipitation. As a result, the friction factor may increase by a factor of 5 to 10. Actual operating conditions must be considered in the design of piping systems. Also, the Moody chart and its equivalent Colebrook equation involve several uncertainties (the roughness size, experimental error, curve fitting of data, etc.), and thus the results obtained should not be treated as “exact.” It is usually considered to be accurate to 15 percent over the entire range in the figure. The Colebrook equation is implicit in f, and thus the determination of the friction factor requires tedious iteration unless an equation solver is used. An approximate explicit relation for f is given by S. E. Haaland in 1983 (Ref. 9) as 6.9 /D 1 1.8 log Re 3.7 f 1.11 (8-74) The results obtained from this relation are within 2 percent of those obtained from Colebrook equation, and we recommend using this relation rather than the Moody chart to avoid reading errors. In turbulent flow, wall roughness increases the heat transfer coefficient h by a factor of 2 or more [Dipprey and Sabersky (1963), Ref. 5]. The convection heat transfer coefficient for rough tubes can be calculated approximately from the Nusselt number relations such as Eq. 8–70 by using the friction factor determined from the Moody chart or the Colebrook equation. However, this approach is not very accurate since there is no further increase in h with f for f 4fsmooth [Norris (1970), Ref. 20] and correlations developed specifically for rough tubes should be used when more accuracy is desired. TABLE 8–3 Equivalent roughness values for new commercial pipes Roughness, Material Glass, plastic Concrete Wood stave Rubber, smoothed Copper or brass tubing Cast iron Galvanized iron Wrought iron Stainless steel Commercial steel ft mm 0 (smooth) 0.003–0.03 0.9–9 0.0016 0.5 0.000033 0.01 0.000005 0.00085 0.0015 0.26 0.0005 0.00015 0.000007 0.15 0.046 0.002 0.00015 0.045 The uncertainty in these values can be as much as 60 percent. Developing Turbulent Flow in the Entrance Region The entry lengths for turbulent flow are typically short, often just 10 tube diameters long, and thus the Nusselt number determined for fully developed turbulent flow can be used approximately for the entire tube. This simple approach gives reasonable results for pressure drop and heat transfer for long tubes and conservative results for short ones. Correlations for the friction and heat transfer coefficients for the entrance regions are available in the literature for better accuracy. r 0 Turbulent Flow in Noncircular Tubes The velocity and temperature profiles in turbulent flow are nearly straight lines in the core region, and any significant velocity and temperature gradients occur in the viscous sublayer (Fig. 8–25). Despite the small thickness of laminar sublayer (usually much less than 1 percent of the pipe diameter), the characteristics of the flow in this layer are very important since they set the stage for flow in the rest of the pipe. Therefore, pressure drop and heat transfer characteristics of turbulent flow in tubes are dominated by the very thin (r) Turbulent layer Overlap layer Laminar sublayer FIGURE 8–25 In turbulent flow, the velocity profile is nearly a straight line in the core region, and any significant velocity gradients occur in the viscous sublayer. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 444 444 HEAT TRANSFER viscous sublayer next to the wall surface, and the shape of the core region is not of much significance. Consequently, the turbulent flow relations given above for circular tubes can also be used for noncircular tubes with reasonable accuracy by replacing the diameter D in the evaluation of the Reynolds number by the hydraulic diameter Dh 4Ac /p. Flow through Tube Annulus Tube Di Do Annulus Some simple heat transfer equipments consist of two concentric tubes, and are properly called double-tube heat exchangers (Fig. 8–26). In such devices, one fluid flows through the tube while the other flows through the annular space. The governing differential equations for both flows are identical. Therefore, steady laminar flow through an annulus can be studied analytically by using suitable boundary conditions. Consider a concentric annulus of inner diameter Di and outer diameter Do. The hydraulic diameter of annulus is FIGURE 8–26 A double-tube heat exchanger that consists of two concentric tubes. TABLE 8–4 Nusselt number for fully developed laminar flow in an annulus with one surface isothermal and the other adiabatic (Kays and Perkins, Ref. 14) Di /Do Nui Nuo 0 0.05 0.10 0.25 0.50 1.00 — 17.46 11.56 7.37 5.74 4.86 3.66 4.06 4.11 4.23 4.43 4.86 (a) Finned surface Fin 4Ac 4(D2o D2i )/4 Dh p Do Di (Do Di) Annular flow is associated with two Nusselt numbers—Nui on the inner tube surface and Nuo on the outer tube surface—since it may involve heat transfer on both surfaces. The Nusselt numbers for fully developed laminar flow with one surface isothermal and the other adiabatic are given in Table 8–4. When Nusselt numbers are known, the convection coefficients for the inner and the outer surfaces are determined from Nui hi Dh k FIGURE 8–27 Tube surfaces are often roughened, corrugated, or finned in order to enhance convection heat transfer. Nuo ho Dh k (8-76) 0.16 D F 0.86 D Fi 0.86 Di Do i Roughness and For fully developed turbulent flow, the inner and outer convection coefficients are approximately equal to each other, and the tube annulus can be treated as a noncircular duct with a hydraulic diameter of Dh Do Di. The Nusselt number in this case can be determined from a suitable turbulent flow relation such as the Gnielinski equation. To improve the accuracy of Nusselt numbers obtained from these relations for annular flow, Petukhov and Roizen (1964, Ref. 22) recommend multiplying them by the following correction factors when one of the tube walls is adiabatic and heat transfer is through the other wall: o (b) Roughened surface (8-75) (outer wall adiabatic) (8-77) (inner wall adiabatic) (8-78) 0.16 o Heat Transfer Enhancement Tubes with rough surfaces have much higher heat transfer coefficients than tubes with smooth surfaces. Therefore, tube surfaces are often intentionally roughened, corrugated, or finned in order to enhance the convection heat transfer coefficient and thus the convection heat transfer rate (Fig. 8–27). Heat transfer in turbulent flow in a tube has been increased by as much as cen58933_ch08.qxd 9/4/2002 11:29 AM Page 445 445 CHAPTER 8 400 percent by roughening the surface. Roughening the surface, of course, also increases the friction factor and thus the power requirement for the pump or the fan. The convection heat transfer coefficient can also be increased by inducing pulsating flow by pulse generators, by inducing swirl by inserting a twisted tape into the tube, or by inducing secondary flows by coiling the tube. EXAMPLE 8–4 Pressure Drop in a Water Pipe Water at 60°F ( 62.36 lbm/ft3 and 2.713 lbm/ft h) is flowing steadily in a 2-in.-diameter horizontal pipe made of stainless steel at a rate of 0.2 ft3/s (Fig. 8–28). Determine the pressure drop and the required pumping power input for flow through a 200-ft-long section of the pipe. 0.2 ft3/s water 2 in. 200 ft FIGURE 8–28 Schematic for Example 8–4. SOLUTION The flow rate through a specified water pipe is given. The pressure drop and the pumping power requirements are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves no work devices such as a pump or a turbine. Properties The density and dynamic viscosity of water are given by 62.36 lbm/ft3 and 2.713 lbm/ft h 0.0007536 lbm/ft s, respectively. Analysis First we calculate the mean velocity and the Reynolds number to determine the flow regime: · · V V 0.2 ft3/s 9.17 ft/s Ac D2/4 (2/12 ft)2/4 D (62.36 lbm/ft3)(9.17 ft/s)(2/12 ft) 3600 s Re 126,400 2.713 lbm/ft · h 1h which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is /D 0.000007 ft 0.000042 2/12 ft The friction factor corresponding to this relative roughness and the Reynolds number can simply be determined from the Moody chart. To avoid the reading error, we determine it from the Colebrook equation: /D 2.51 1 2.0 log 3.7 Re f f → 0.000042 2.51 1 2.0 log 3.7 126,400 f f Using an equation solver or an iterative scheme, the friction factor is determined to be f 0.0174. Then the pressure drop and the required power input become 200 ft (62.36 lbm/ft3)(9.17 ft/s)2 1 lbf L 0.0174 D 2 2 2/12 ft 32.2 lbm · ft/s2 1700 lbf/ft2 11.8 psi · · 1W Wpump VP (0.2 ft3/s)(1700 lbf/ft2) 461 W 0.737 lbf · ft/s 2 P f cen58933_ch08.qxd 9/4/2002 11:29 AM Page 446 446 HEAT TRANSFER Therefore, power input in the amount of 461 W is needed to overcome the frictional losses in the pipe. Discussion The friction factor also could be determined easily from the explicit Haaland relation. It would give f 0.0172, which is sufficiently close to 0.0174. Also, the friction factor corresponding to 0 in this case is 0.0171, which indicates that stainless steel pipes can be assumed to be smooth with negligible error. EXAMPLE 8–5 q· s = constant 15°C Water D = 3 cm 65°C Heating of Water by Resistance Heaters in a Tube Water is to be heated from 15°C to 65°C as it flows through a 3-cm-internaldiameter 5-m-long tube (Fig. 8–29). The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 10 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit. 5m FIGURE 8–29 Schematic for Example 8–5. SOLUTION Water is to be heated in a tube equipped with an electric resistance heater on its surface. The power rating of the heater and the inner surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. Properties The properties of water at the bulk mean temperature of Tb (Ti Te)/2 (15 65)/2 40°C are (Table A-9). 992.1 kg/m3 k 0.631 W/m °C / 0.658 10 6 m2/s Cp 4179 J/kg °C Pr 4.32 Analysis The cross sectional and heat transfer surface areas are Ac 14D2 14(0.03 m)2 7.069 10 4 m2 As pL DL (0.03 m)(5 m) 0.471 m2 · The volume flow rate of water is given as V 10 L/min 0.01 m3/min. Then the mass flow rate becomes · m· V (992.1 kg/m3)(0.01 m3/min) 9.921 kg/min 0.1654 kg/s To heat the water at this mass flow rate from 15°C to 65°C, heat must be supplied to the water at a rate of · Q m· Cp(Te Ti) (0.1654 kg/s)(4.179 kJ/kg °C)(65 15)°C 34.6 kJ/s 34.6 kW cen58933_ch08.qxd 9/4/2002 11:29 AM Page 447 447 CHAPTER 8 All of this energy must come from the resistance heater. Therefore, the power rating of the heater must be 34.6 kW. The surface temperature Ts of the tube at any location can be determined from q·s q·s h(Ts Tm) → Ts Tm h where h is the heat transfer coefficient and Tm is the mean temperature of the fluid at that location. The surface heat flux is constant in this case, and its value can be determined from Q· 34.6 kW q·s 73.46 kW/m2 As 0.471 m2 To determine the heat transfer coefficient, we first need to find the mean velocity of water and the Reynolds number: V· 0.010 m3/min 14.15 m/min 0.236 m/s Ac 7.069 10 4 m2 m D (0.236 m/s)(0.03 m) Re 10,760 0.658 10 6 m2/s m which is greater than 10,000. Therefore, the flow is turbulent and the entry length is roughly Lh Lt 10D 10 0.03 0.3 m which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe and determine the Nusselt number from Nu hD 0.023 Re0.8 Pr0.4 0.023(10,760)0.8 (4.34)0.4 69.5 k Then, h 0.631 W/m · °C k Nu (69.5) 1462 W/m2 °C D 0.03 m and the surface temperature of the pipe at the exit becomes Ts Tm q·s 73,460 W/m2 65°C 115°C h 1462 W/m2 · °C Discussion Note that the inner surface temperature of the pipe will be 50°C higher than the mean water temperature at the pipe exit. This temperature difference of 50°C between the water and the surface will remain constant throughout the fully developed flow region. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 448 448 HEAT TRANSFER EXAMPLE 8–6 Ts = 60°C 0.2 m Te Air 1 atm 80°C Heat Loss from the Ducts of a Heating System Hot air at atmospheric pressure and 80°C enters an 8–m-long uninsulated square duct of cross section 0.2 m 0.2 m that passes through the attic of a house at a rate of 0.15 m3/s (Fig. 8–30). The duct is observed to be nearly isothermal at 60°C. Determine the exit temperature of the air and the rate of heat loss from the duct to the attic space. 0.2 m 8m FIGURE 8–30 SOLUTION Heat loss from uninsulated square ducts of a heating system in the attic is considered. The exit temperature and the rate of heat loss are to be determined. Schematic for Example 8–6. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Air is an ideal gas. Properties We do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk mean temperature of air, which is the temperature at which the properties are to be determined. The temperature of air at the inlet is 80°C and we expect this temperature to drop somewhat as a result of heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we read (Table A-15) 0.9994 kg/m3 k 0.02953 W/m °C 2.097 10 5 m2/s Cp 1008 J/kg °C Pr 0.7154 Analysis The characteristic length (which is the hydraulic diameter), the mean velocity, and the Reynolds number in this case are 4Ac 4a2 Dh p a 0.2 m 4a V· 0.15 m3/s m 3.75 m/s Ac (0.2 m)2 m Dh (3.75 m/s)(0.2 m) Re 35,765 2.097 10 5 m2/s which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly Lh Lt 10D 10 0.2 m 2 m which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct and determine the Nusselt number from Nu hDh 0.023 Re0.8 Pr0.3 0.023(35,765)0.8 (0.7154)0.3 91.4 k cen58933_ch08.qxd 9/4/2002 11:29 AM Page 449 449 CHAPTER 8 Then, 0.02953 W/m · °C k Nu (91.4) 13.5 W/m2 °C Dh 0.2 m As pL 4aL 4 (0.2 m)(8 m) 6.4 m2 · m· V (1.009 kg/m3)(0.15 m3/s) 0.151 kg/s h Next, we determine the exit temperature of air from Te Ts (Ts Ti) exp ( hAs /m· Cp) 60°C [(60 80)°C] exp (13.5 W/m2 · °C)(6.4 m2) (0.151 kg/s)(1008 J/kg · °C) 71.3°C Then the logarithmic mean temperature difference and the rate of heat loss from the air become Ti Te 80 71.3 15.2°C Ts Te 60 71.3 ln ln 60 80 Ts Ti · Q hAs Tln (13.5 W/m2 °C)(6.4 m2)( 15.2°C) 1313 W Tln Therefore, air will lose heat at a rate of 1313 W as it flows through the duct in the attic. Discussion The average fluid temperature is (80 71.3)/2 75.7°C, which is sufficiently close to 80°C at which we evaluated the properties of air. Therefore, it is not necessary to re-evaluate the properties at this temperature and to repeat the calculations. SUMMARY Internal flow is characterized by the fluid being completely confined by the inner surfaces of the tube. The mean velocity and mean temperature for a circular tube of radius R are expressed as m 2 R2 R 0 (r, x)rdr and Tm 2 m R2 R 0 Trdr The Reynolds number for internal flow and the hydraulic diameter are defined as Re m D m D and 4Ac Dh p The flow in a tube is laminar for Re 2300, turbulent for Re 10,000, and transitional in between. The length of the region from the tube inlet to the point at which the boundary layer merges at the centerline is the hydro- dynamic entry length Lh. The region beyond the entrance region in which the velocity profile is fully developed is the hydrodynamically fully developed region. The length of the region of flow over which the thermal boundary layer develops and reaches the tube center is the thermal entry length Lt. The region in which the flow is both hydrodynamically and thermally developed is the fully developed flow region. The entry lengths are given by Lh, laminar 0.05 Re D Lt, laminar 0.05 Re Pr D Pr Lh, laminar Lh, turbulent Lt, turbulent 10D For q·s constant, the rate of heat transfer is expressed as · Q q·s As m· Cp(Te Ti) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 450 450 HEAT TRANSFER For Ts constant, we have · Q hAs Tln m· Cp(Te Ti) Te Ts (Ts Ti)exp( hAs /m· Cp) Ti Te Te Ti Tln ln[(Ts Te)/(Ts Ti)] ln(Te /Ti) The pressure drop and required pumping power for a volume · flow rate of V are L m D 2 2 P · · Wpump VP and For fully developed laminar flow in a circular pipe, we have: (r) 2m 1 r2 r2 max 1 2 2 R R 64 64 Dm Re · R4 P R4 P PR2 V ave Ac R2 8L 8L 128L f Circular tube, laminar (q·s constant): Circular tube, laminar (Ts constant): hD 4.36 k hD Nu 3.66 k Nu For developing laminar flow in the entrance region with constant surface temperature, we have Circular tube: Circular tube: 0.065(D/L) Re Pr 1 0.04[(D/L) Re Pr]2/3 Re Pr D 1/3 b 0.14 Nu 1.86 s L Nu 3.66 Parallel plates: Nu 7.54 For fully developed turbulent flow with smooth surfaces, we have f (0.790 ln Re 1.64) 2 Nu 0.125f Re Pr1/3 104 Re 106 0.7 Pr 160 Re 10,000 Nu 0.023 Re0.8 Pr1/3 Nu 0.023 Re0.8 Prn with n 0.4 for heating and 0.3 for cooling of fluid ( f/8)(Re 1000) Pr 0.5 Pr 2000 Nu 1 12.7( f/8)0.5 (Pr2/3 1) 3 103 Re 5 106 The fluid properties are evaluated at the bulk mean fluid temperature Tb (Ti Te)/2. For liquid metal flow in the range of 104 Re 106 we have: Ts constant: q·s constant: Nu 4.8 0.0156 Re0.85 Pr0.93 s Nu 6.3 0.0167 Re0.85 Pr0.93 s For fully developed turbulent flow with rough surfaces, the friction factor f is determined from the Moody chart or 6.9 /D /D 2.51 1 2.0 log 1.8 log Re 3.7 Re f 3.7 f 1.11 For a concentric annulus, the hydraulic diameter is Dh Do Di, and the Nusselt numbers are expressed as Nui hi Dh k and Nuo ho Dh k where the values for the Nusselt numbers are given in Table 8–4. 0.03(Dh /L) Re Pr 1 0.016[(Dh /L) Re Pr]2/3 REFERENCES AND SUGGESTED READING 1. M. S. Bhatti and R. K. Shah. “Turbulent and Transition Flow Convective Heat Transfer in Ducts.” In Handbook of Single-Phase Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung. New York: Wiley Interscience, 1987. A. P. Colburn. Transactions of the AIChE 26 (1933), p. 174. 3. C. F. Colebrook. “Turbulent flow in Pipes, with Particular Reference to the Transition between the Smooth and cen58933_ch08.qxd 9/4/2002 11:29 AM Page 451 451 CHAPTER 8 Rough Pipe Laws.” Journal of the Institute of Civil Engineers London. 11 (1939), pp. 133–156. Surfaces.” In Augmentation of Convective Heat Transfer, ed. A. E. Bergles and R. L. Webb. New York: ASME, 1970. R. G. Deissler. “Analysis of Turbulent Heat Transfer and Flow in the Entrance Regions of Smooth Passages.” 1953. Referred to in Handbook of Single-Phase Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung. New York: Wiley Interscience, 1987. B. S. Petukhov. “Heat Transfer and Friction in Turbulent Pipe Flow with Variable Physical Properties.” In Advances in Heat Transfer, ed. T. F. Irvine and J. P. Hartnett, Vol. 6. New York: Academic Press, 1970. D. F. Dipprey and D. H. Sabersky. “Heat and Momentum Transfer in Smooth and Rough Tubes at Various Prandtl Numbers.” International Journal of Heat Mass Transfer 6 (1963), pp. 329–353. B. S. Petukhov and L. I. Roizen. “Generalized Relationships for Heat Transfer in a Turbulent Flow of a Gas in Tubes of Annular Section.” High Temperature (USSR) 2 (1964), pp. 65–68. F. W. Dittus and L. M. K. Boelter. University of California Publications on Engineering 2 (1930), p. 433. O. Reynolds. “On the Experimental Investigation of the Circumstances Which Determine Whether the Motion of Water Shall Be Direct or Sinuous, and the Law of Resistance in Parallel Channels.” Philosophical Transactions of the Royal Society of London 174 (1883), pp. 935–982. D. K. Edwards, V. E. Denny, and A. F. Mills. Transfer Processes. 2nd ed. Washington, DC: Hemisphere, 1979. 8. V. Gnielinski. “New Equations for Heat and Mass Transfer in Turbulent Pipe and Channel Flow.” International Chemical Engineering 16 (1976), pp. 359–368. 9. S. E. Haaland. “Simple and Explicit Formulas for the Friction Factor in Turbulent Pipe Flow.” Journal of Fluids Engineering (March 1983), pp. 89–90. 10. J. P. Holman. Heat Transfer. 8th ed. New York: McGraw-Hill, 1997. 11. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 3rd ed. New York: John Wiley & Sons, 1996. H. Schlichting. Boundary Layer Theory. 7th ed. New York: McGraw-Hill, 1979. 25. R. K. Shah and M. S. Bhatti. “Laminar Convective Heat Transfer in Ducts.” In Handbook of Single-Phase Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung. New York: Wiley Interscience, 1987. 26. E. N. Sieder and G. E. Tate. “Heat Transfer and Pressure Drop of Liquids in Tubes.” Industrial Engineering Chemistry 28 (1936), pp. 1429–1435. S. Kakaç, R. K. Shah, and W. Aung, eds. Handbook of Single-Phase Convective Heat Transfer. New York: Wiley Interscience, 1987. C. A. Sleicher and M. W. Rouse. “A Convenient Correlation for Heat Transfer to Constant and Variable Property Fluids in Turbulent Pipe Flow.” International Journal of Heat Mass Transfer 18 (1975), pp. 1429–1435. W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West, 1995. W. M. Kays and H. C. Perkins. Chapter 7. In Handbook of Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett. New York: McGraw-Hill, 1972. F. M. White. Heat and Mass Transfer. Reading, MA: Addison-Wesley, 1988. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. 16. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed. Upper Saddle River, NJ: Prentice Hall, 1999. 17. L. F. Moody. “Friction Factors for Pipe Flows.” Transactions of the ASME 66 (1944), pp. 671–684. 18. M. Molki and E. M. Sparrow. “An Empirical Correlation for the Average Heat Transfer Coefficient in Circular Tubes.” Journal of Heat Transfer 108 (1986), pp. 482–484. 19. B. R. Munson, D. F. Young, and T. Okiishi. Fundamentals of Fluid Mechanics. 4th ed. New York: Wiley, 2002. 20. R. H. Norris. “Some Simple Approximate Heat Transfer Correlations for Turbulent Flow in Ducts with Rough S. Whitaker. “Forced Convection Heat Transfer Correlations for Flow in Pipes, Past Flat Plates, Single Cylinders, and for Flow in Packed Beds and Tube Bundles.” AIChE Journal 18 (1972), pp. 361–371. 31. W. Zhi-qing. “Study on Correction Coefficients of Laminar and Turbulent Entrance Region Effects in Round Pipes.” Applied Mathematical Mechanics 3 (1982), p. 433. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 452 452 HEAT TRANSFER PROBLEMS General Flow Analysis 8–1C Why are liquids usually transported in circular pipes? 8–2C Show that the Reynolds number for flow in a circular tube of diameter D can be expressed as Re 4m· /(D). 8–3C Which fluid at room temperature requires a larger pump to move at a specified velocity in a given tube: water or engine oil? Why? 8–4C What is the generally accepted value of the Reynolds number above which the flow in smooth pipes is turbulent? 8–5C What is hydraulic diameter? How is it defined? What is it equal to for a circular tube of diameter? 8–6C How is the hydrodynamic entry length defined for flow in a tube? Is the entry length longer in laminar or turbulent flow? 8–7C Consider laminar flow in a circular tube. Will the friction factor be higher near the inlet of the tube or near the exit? Why? What would your response be if the flow were turbulent? 8–8C How does surface roughness affect the pressure drop in a tube if the flow is turbulent? What would your response be if the flow were laminar? 8–9C How does the friction factor f vary along the flow direction in the fully developed region in (a) laminar flow and (b) turbulent flow? 8–10C What fluid property is responsible for the development of the velocity boundary layer? For what kinds of fluids will there be no velocity boundary layer in a pipe? 8–11C What is the physical significance of the number of transfer units NTU hA/m· Cp? What do small and large NTU values tell about a heat transfer system? 8–12C What does the logarithmic mean temperature difference represent for flow in a tube whose surface temperature is constant? Why do we use the logarithmic mean temperature instead of the arithmetic mean temperature? 8–13C How is the thermal entry length defined for flow in a tube? In what region is the flow in a tube fully developed? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 8–14C Consider laminar forced convection in a circular tube. Will the heat flux be higher near the inlet of the tube or near the exit? Why? 8–15C Consider turbulent forced convection in a circular tube. Will the heat flux be higher near the inlet of the tube or near the exit? Why? 8–16C In the fully developed region of flow in a circular tube, will the velocity profile change in the flow direction? How about the temperature profile? 8–17C Consider the flow of oil in a tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar? How would they compare if the flow were turbulent? 8–18C Consider the flow of mercury (a liquid metal) in a tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar? How would they compare if the flow were turbulent? 8–19C What do the mean velocity m and the mean temperature Tm represent in flow through circular tubes of constant diameter? 8–20C Consider fluid flow in a tube whose surface temperature remains constant. What is the appropriate temperature difference for use in Newton’s law of cooling with an average heat transfer coefficient? 8–21 Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a mean velocity of 7 m/s, and is cooled by the water outside. If the average heat transfer coefficient is 85 W/m2 °C and the tube temperature is nearly equal to the water temperature of 5°C, determine the exit temperature of air and the rate of heat transfer. 8–22 Cooling water available at 10°C is used to condense steam at 30°C in the condenser of a power plant at a rate of 0.15 kg/s by circulating the cooling water through a bank of 5-m-long 1.2-cm-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of 4 m/s, and leaves at a temperature of 24°C. The tubes are nearly isothermal at 30°C. Determine the average heat transfer coefficient between the water and the tubes, and the number of tubes needed to achieve the indicated heat transfer rate in the condenser. 8–23 Repeat Problem 8–22 for steam condensing at a rate of 0.60 kg/s. 8–24 Combustion gases passing through a 3-cm-internaldiameter circular tube are used to vaporize waste water at atmospheric pressure. Hot gases enter the tube at 115 kPa and 250°C at a mean velocity of 5 m/s, and leave at 150°C. If the average heat transfer coefficient is 120 W/m2 °C and the inner surface temperature of the tube is 110°C, determine (a) the tube length and (b) the rate of evaporation of water. 8–25 Repeat Problem 8–24 for a heat transfer coefficient of 60 W/m2 °C. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 453 453 CHAPTER 8 Laminar and Turbulent Flow in Tubes 8–26C How is the friction factor for flow in a tube related to the pressure drop? How is the pressure drop related to the pumping power requirement for a given mass flow rate? 8–27C Someone claims that the shear stress at the center of a circular pipe during fully developed laminar flow is zero. Do you agree with this claim? Explain. terline) is measured to be 6 m/s. Determine the velocity at the Answer: 8 m/s center of the pipe. 8–37 The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 2 cm, in m/s, is given by (r) 4(1 r2/R2). Determine the mean and maximum velocities in the pipe, and the volume flow rate. ( r2 (r) = 4 1 – –– R2 8–28C Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Do you agree with this claim? Explain. ) R = 2 cm 8–29C Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will (a) double, (b) more than double, (c) less than double, (d) reduce by half, or (e) remain constant. FIGURE P8–37 8–30C Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross sectional area, and dividing the result by 2. Do you agree? Explain. 8–39 Water at 10°C ( 999.7 kg/m3 and 1.307 10 3 kg/m s) is flowing in a 0.20-cm-diameter 15-m-long pipe steadily at an average velocity of 1.2 m/s. Determine (a) the pressure drop and (b) the pumping power requirement to overcome this pressure drop. 8–31C Someone claims that the average velocity in a circular pipe in fully developed laminar flow can be determined by simply measuring the velocity at R/2 (midway between the wall surface and the centerline). Do you agree? Explain. 8–32C Consider fully developed laminar flow in a circular pipe. If the diameter of the pipe is reduced by half while the flow rate and the pipe length are held constant, the pressure drop will (a) double, (b) triple, (c) quadruple, (d) increase by a factor of 8, or (e) increase by a factor of 16. 8–33C Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the flow rate is held constant, how will the pressure drop change? 8–34C How does surface roughness affect the heat transfer in a tube if the fluid flow is turbulent? What would your response be if the flow in the tube were laminar? 8–35 Water at 15°C ( 999.1 kg/m3 and 1.138 10 3 kg/m s) is flowing in a 4-cm-diameter and 30-m long horizontal pipe made of stainless steel steadily at a rate of 5 L/s. Determine (a) the pressure drop and (b) the pumping power requirement to overcome this pressure drop. 5 L/s 8–38 Answers: (a) 188 kPa,(b) 0.71 W 8–40 Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 7-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 8 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit. 8–41 Hot air at atmospheric pressure and 85°C enters a 10-m-long uninsulated square duct of cross section 0.15 m 0.15 m that passes through the attic of a house at a rate of 0.10 m3/s. The duct is observed to be nearly isothermal at 70°C. Determine the exit temperature of the air and the rate of heat loss from the duct to the air space in the attic. Answers: 75.7°C, 941 W Attic space Air 85°C 70°C 0.1 m3/s 4 cm 30 m FIGURE P8–35 8–36 In fully developed laminar flow in a circular pipe, the velocity at R/2 (midway between the wall surface and the cen- Repeat Problem 8–37 for a pipe of inner radius 5 cm. FIGURE P8–41 8–42 Reconsider Problem 8–41. Using EES (or other) software, investigate the effect of the volume flow rate of air on the exit temperature of air and the rate of heat loss. Let the flow rate vary from 0.05 m3/s to 0.15 m3/s. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 454 454 HEAT TRANSFER Plot the exit temperature and the rate of heat loss as a function of flow rate, and discuss the results. Parabolic solar collector 8–43 Consider an air solar collector that is 1 m wide and 5 m long and has a constant spacing of 3 cm between the glass cover and the collector plate. Air enters the collector at 30°C at a rate of 0.15 m3/s through the 1-m-wide edge and flows along the 5-m-long passage way. If the average temperatures of the glass cover and the collector plate are 20°C and 60°C, respectively, determine (a) the net rate of heat transfer to the air in the collector and (b) the temperature rise of air as it flows through the collector. Air 30°C 0.15 m3/s 5m Glass cover 20°C Insulation Collector plate, 60°C FIGURE P8–43 8–44 Consider the flow of oil at 10°C in a 40-cm-diameter pipeline at an average velocity of 0.5 m/s. A 300-m-long section of the pipeline passes through icy waters of a lake at 0°C. Measurements indicate that the surface temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance of the pipe material, determine (a) the temperature of the oil when the pipe leaves the lake, (b) the rate of heat transfer from the oil, and (c) the pumping power required to overcome the pressure losses and to maintain the flow oil in the pipe. 8–45 Consider laminar flow of a fluid through a square channel maintained at a constant temperature. Now the mean velocity of the fluid is doubled. Determine the change in the pressure drop and the change in the rate of heat transfer between the fluid and the walls of the channel. Assume the flow regime remains unchanged. 8–46 Repeat Problem 8–45 for turbulent flow. 8–47E The hot water needs of a household are to be met by heating water at 55°F to 200°F by a parabolic solar collector at a rate of 4 lbm/s. Water flows through a 1.25-in.-diameter thin aluminum tube whose outer surface is blackanodized in order to maximize its solar absorption ability. The centerline of the tube coincides with the focal line of the collector, and a glass Water 200°F 4 lbm/s Glass tube Water tube FIGURE P8–47E sleeve is placed outside the tube to minimize the heat losses. If solar energy is transferred to water at a net rate of 350 Btu/h per ft length of the tube, determine the required length of the parabolic collector to meet the hot water requirements of this house. Also, determine the surface temperature of the tube at the exit. 8–48 A 15-cm 20-cm printed circuit board whose components are not allowed to come into direct contact with air for reliability reasons is to be cooled by passing cool air through a 20-cm-long channel of rectangular cross section 0.2 cm 14 cm drilled into the board. The heat generated by the electronic components is conducted across the thin layer of the board to the channel, where it is removed by air that enters the channel at 15°C. The heat flux at the top surface of the channel can be considered to be uniform, and heat transfer through other surfaces is negligible. If the velocity of the air at the inlet of the channel is not to exceed 4 m/s and the surface temperature of the channel is to remain under 50°C, determine the maximum total power of the electronic components that can safely be mounted on this circuit board. Air 15°C Air channel 0.2 cm × 14 cm Electronic components FIGURE P8–48 8–49 Repeat Problem 8–48 by replacing air with helium, which has six times the thermal conductivity of air. 8–50 Reconsider Problem 8–48. Using EES (or other) software, investigate the effects of air velocity at the inlet of the channel and the maximum surface temperature on the maximum total power dissipation of electronic components. Let the air velocity vary from 1 m/s to 10 m/s and the surface temperature from 30°C to 90°C. Plot the power dissipation as functions of air velocity and surface temperature, and discuss the results. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 455 455 CHAPTER 8 8–51 Air enters a 7-m-long section of a rectangular duct of cross section 15 cm 20 cm at 50°C at an average velocity of 7 m/s. If the walls of the duct are maintained at 10°C, determine (a) the outlet temperature of the air, (b) the rate of heat transfer from the air, and (c) the fan power needed to overcome the pressure losses in this section of the duct. Answers: (a) 32.8°C, (b) 3674 W, (c) 4.2 W 8–52 Reconsider Problem 8–51. Using EES (or other) software, investigate the effect of air velocity on the exit temperature of air, the rate of heat transfer, and the fan power. Let the air velocity vary from 1 m/s to 10 m/s. Plot the exit temperature, the rate of heat transfer, and the fan power as a function of the air velocity, and discuss the results. 8–53 Hot air at 60°C leaving the furnace of a house enters a 12-m-long section of a sheet metal duct of rectangular cross section 20 cm 20 cm at an average velocity of 4 m/s. The thermal resistance of the duct is negligible, and the outer surface of the duct, whose emissivity is 0.3, is exposed to the cold air at 10°C in the basement, with a convection heat transfer coefficient of 10 W/m2 °C. Taking the walls of the basement to be at 10°C also, determine (a) the temperature at which the hot air will leave the basement and (b) the rate of heat loss from the hot air in the duct to the basement. 10°C ho = 10 W/ m2·°C 12 m Hot air 60°C 4 m /s Air duct 20 cm × 20 cm ε = 0.3 mine (a) the exit temperature of air and (b) the highest component surface temperature in the duct. 8–56 Repeat Problem 8–55 for a circular horizontal duct of 15-cm diameter. 8–57 Consider a hollow-core printed circuit board 12 cm high and 18 cm long, dissipating a total of 20 W. The width of the air gap in the middle of the PCB is 0.25 cm. The cooling air enters the 12-cm-wide core at 32°C at a rate of 0.8 L/s. Assuming the heat generated to be uniformly distributed over the two side surfaces of the PCB, determine (a) the temperature at which the air leaves the hollow core and (b) the highest temperature on the inner surface of the core. Answers: (a) 54.0°C, (b) 72.8°C 8–58 Repeat Problem 8–57 for a hollow-core PCB dissipating 35 W. 8–59E Water at 54°F is heated by passing it through 0.75-in.internal-diameter thin-walled copper tubes. Heat is supplied to the water by steam that condenses outside the copper tubes at 250°F. If water is to be heated to 140°F at a rate of 0.7 lbm/s, determine (a) the length of the copper tube that needs to be used and (b) the pumping power required to overcome pressure losses. Assume the entire copper tube to be at the steam temperature of 250°F. 8–60 A computer cooled by a fan contains eight PCBs, each dissipating 10 W of power. The height of the PCBs is 12 cm and the length is 18 cm. The clearance between the tips of the components on the PCB and the back surface of the adjacent PCB is 0.3 cm. The cooling air is supplied by a 10-W fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed 10°C, determine (a) the flow rate of the air that the fan needs to deliver, (b) the fraction of the temperature rise of air that is due to the heat generated by the fan and its motor, and (c) the highest allowable inlet air temperature if the surface temperature of the Air outlet FIGURE P8–53 0.3 cm 8–54 Reconsider Problem 8–53. Using EES (or other) software, investigate the effects of air velocity and the surface emissivity on the exit temperature of air and the rate of heat loss. Let the air velocity vary from 1 m/s to 10 m/s and the emissivity from 0.1 to 1.0. Plot the exit temperature and the rate of heat loss as functions of air velocity and emissivity, and discuss the results. 8–55 The components of an electronic system dissipating 90 W are located in a 1-m-long horizontal duct whose cross section is 16 cm 16 cm. The components in the duct are cooled by forced air, which enters at 32°C at a rate of 0.65 m3/min. Assuming 85 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 15 percent is lost through the outer surfaces of the duct, deter- 18 cm Air inlet FIGURE P8–60 PCB, 10 W cen58933_ch08.qxd 9/4/2002 11:29 AM Page 456 456 HEAT TRANSFER components is not to exceed 70°C anywhere in the system. Use air properties at 25°C. Air, 0.27 m3/s 10°C, 95 kPa Review Problems 8–61 A geothermal district heating system involves the transport of geothermal water at 110°C from a geothermal well to a city at about the same elevation for a distance of 12 km at a rate of 1.5 m3/s in 60-cm-diameter stainless steel pipes. The fluid pressures at the wellhead and the arrival point in the city are to be the same. The minor losses are negligible because of the large length-to-diameter ratio and the relatively small number of components that cause minor losses. (a) Assuming the pump-motor efficiency to be 65 percent, determine the electric power consumption of the system for pumping. (b) Determine the daily cost of power consumption of the system if the unit cost of electricity is $0.06/kWh. (c) The temperature of geothermal water is estimated to drop 0.5°C during this long flow. Determine if the frictional heating during flow can make up for this drop in temperature. 8–62 Repeat Problem 8–61 for cast iron pipes of the same diameter. 20 cm Air Compressor 150 hp FIGURE P8–65 water, determine the outlet temperature of air as it leaves the underwater portion of the duct. Also, for an overall fan efficiency of 55 percent, determine the fan power input needed to overcome the flow resistance in this section of the duct. 8–63 The velocity profile in fully developed laminar flow in a circular pipe, in m/s, is given by (r) 6(1 100r2) where r is the radial distance from the centerline of the pipe in m. Determine (a) the radius of the pipe, (b) the mean velocity through the pipe, and (c) the maximum velocity in the pipe. Air 25°C, 3 m/s 8–64E The velocity profile in fully developed laminar flow of water at 40°F in a 80-ft-long horizontal circular pipe, in ft/s, is given by (r) 0.8(1 625r2) where r is the radial distance from the centerline of the pipe in ft. Determine (a) the volume flow rate of water through the pipe, (b) the pressure drop across the pipe, and (c) the useful pumping power required to overcome this pressure drop. 8–65 The compressed air requirements of a manufacturing facility are met by a 150-hp compressor located in a room that is maintained at 20°C. In order to minimize the compressor work, the intake port of the compressor is connected to the outside through an 11-m-long, 20-cm-diameter duct made of thin aluminum sheet. The compressor takes in air at a rate of 0.27 m3/s at the outdoor conditions of 10°C and 95 kPa. Disregarding the thermal resistance of the duct and taking the heat transfer coefficient on the outer surface of the duct to be 10 W/m2 °C, determine (a) the power used by the compressor to overcome the pressure drop in this duct, (b) the rate of heat transfer to the incoming cooler air, and (c) the temperature rise of air as it flows through the duct. 8–66 A house built on a riverside is to be cooled in summer by utilizing the cool water of the river, which flows at an average temperature of 15°C. A 15-m-long section of a circular duct of 20-cm diameter passes through the water. Air enters the underwater section of the duct at 25°C at a velocity of 3 m/s. Assuming the surface of the duct to be at the temperature of the 11 m 15°C Air River, 15°C FIGURE P8–66 8–67 Repeat Problem 8–66 assuming that a 0.15-mm-thick layer of mineral deposit (k 3 W/m °C) formed on the inner surface of the pipe. 8–68E The exhaust gases of an automotive engine leave the combustion chamber and enter a 8-ft-long and 3.5-in.-diameter thin-walled steel exhaust pipe at 800°F and 15.5 psia at a rate of 0.2 lbm/s. The surrounding ambient air is at a temperature of 80°F, and the heat transfer coefficient on the outer surface of the exhaust pipe is 3 Btu/h ft2 °F. Assuming the exhaust gases to have the properties of air, determine (a) the velocity of the exhaust gases at the inlet of the exhaust pipe and (b) the temperature at which the exhaust gases will leave the pipe and enter the air. 8–69 Hot water at 90°C enters a 15-m section of a cast iron pipe (k 52 W/m °C) whose inner and outer diameters are 4 and 4.6 cm, respectively, at an average velocity of 0.8 m/s. The outer surface of the pipe, whose emissivity is 0.7, is exposed to the cold air at 10°C in a basement, with a convection heat cen58933_ch08.qxd 9/4/2002 11:29 AM Page 457 457 CHAPTER 8 8–72 Liquid-cooled systems have high heat transfer coefficients associated with them, but they have the inherent disadvantage that they present potential leakage problems. Therefore, air is proposed to be used as the microchannel coolant. Repeat Problem 8–71 using air as the cooling fluid instead of water, entering at a rate of 0.5 L/s. Tambient = 10°C ε = 0.7 Hot water 90°C 0.8 m/s 15 m FIGURE P8–69 transfer coefficient of l5 W/m2 °C. Taking the walls of the basement to be at 10°C also, determine (a) the rate of heat loss from the water and (b) the temperature at which the water leaves the basement. 8–70 Repeat Problem 8–69 for a pipe made of copper (k 386 W/m °C) instead of cast iron. 8–71 D. B. Tuckerman and R. F. Pease of Stanford University demonstrated in the early 1980s that integrated circuits can be cooled very effectively by fabricating a series of microscopic channels 0.3 mm high and 0.05 mm wide in the back of the substrate and covering them with a plate to confine the fluid flow within the channels. They were able to dissipate 790 W of power generated in a 1-cm2 silicon chip at a junctionto-ambient temperature difference of 71°C using water as the coolant flowing at a rate of 0.01 L/s through 100 such channels under a 1-cm 1-cm silicon chip. Heat is transferred primarily through the base area of the channel, and it was found that the increased surface area and thus the fin effect are of lesser importance. Disregarding the entrance effects and ignoring any heat transfer from the side and cover surfaces, determine (a) the temperature rise of water as it flows through the microchannels and (b) the average surface temperature of the base of the microchannels for a power dissipation of 50 W. Assume the water enters the channels at 20°C. Cover plate 1 cm 0.3 mm 0.05 mm Silicon substrate Electronic circuits on this side FIGURE P8–71 Microscopic channels 8–73 Hot exhaust gases leaving a stationary diesel engine at 450°C enter a l5-cm-diameter pipe at an average velocity of 3.6 m/s. The surface temperature of the pipe is 180°C. Determine the pipe length if the exhaust gases are to leave the pipe at 250°C after transferring heat to water in a heat recovery unit. Use properties of air for exhaust gases. 8–74 Geothermal steam at 165°C condenses in the shell side of a heat exchanger over the tubes through which water flows. Water enters the 4-cm-diameter, 14-m-long tubes at 20°C at a rate of 0.8 kg/s. Determine the exit temperature of water and the rate of condensation of geothermal steam. 8–75 Cold air at 5°C enters a l2-cm-diameter 20-m-long isothermal pipe at a velocity of 2.5 m/s and leaves at 19°C. Estimate the surface temperature of the pipe. 8–76 Oil at 10°C is to be heated by saturated steam at 1 atm in a double-pipe heat exchanger to a temperature of 30°C. The inner and outer diameters of the annular space are 3 cm and 5 cm, respectively, and oil enters at with a mean velocity of 0.8 m/s. The inner tube may be assumed to be isothermal at 100°C, and the outer tube is well insulated. Assuming fully developed flow for oil, determine the tube length required to heat the oil to the indicated temperature. In reality, will you need a shorter or longer tube? Explain. Design and Essay Problems 8–77 Electronic boxes such as computers are commonly cooled by a fan. Write an essay on forced air cooling of electronic boxes and on the selection of the fan for electronic devices. 8–78 Design a heat exchanger to pasteurize milk by steam in a dairy plant. Milk is to flow through a bank of 1.2-cm internal diameter tubes while steam condenses outside the tubes at 1 atm. Milk is to enter the tubes at 4°C, and it is to be heated to 72°C at a rate of 15 L/s. Making reasonable assumptions, you are to specify the tube length and the number of tubes, and the pump for the heat exchanger. 8–79 A desktop computer is to be cooled by a fan. The electronic components of the computer consume 80 W of power under full-load conditions. The computer is to operate in environments at temperatures up to 50°C and at elevations up to 3000 m where the atmospheric pressure is 70.12 kPa. The exit temperature of air is not to exceed 60°C to meet the reliability requirements. Also, the average velocity of air is not to exceed 120 m/min at the exit of the computer case, where the fan is installed to keep the noise level down. Specify the flow rate of the fan that needs to be installed and the diameter of the casing of the fan. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 458 cen58933_ch09.qxd 9/4/2002 12:25 PM Page 459 CHAPTER N AT U R A L C O N V E C T I O N n Chapters 7 and 8, we considered heat transfer by forced convection, where a fluid was forced to move over a surface or in a tube by external means such as a pump or a fan. In this chapter, we consider natural convection, where any fluid motion occurs by natural means such as buoyancy. The fluid motion in forced convection is quite noticeable, since a fan or a pump can transfer enough momentum to the fluid to move it in a certain direction. The fluid motion in natural convection, however, is often not noticeable because of the low velocities involved. The convection heat transfer coefficient is a strong function of velocity: the higher the velocity, the higher the convection heat transfer coefficient. The fluid velocities associated with natural convection are low, typically less than 1 m/s. Therefore, the heat transfer coefficients encountered in natural convection are usually much lower than those encountered in forced convection. Yet several types of heat transfer equipment are designed to operate under natural convection conditions instead of forced convection, because natural convection does not require the use of a fluid mover. We start this chapter with a discussion of the physical mechanism of natural convection and the Grashof number. We then present the correlations to evaluate heat transfer by natural convection for various geometries, including finned surfaces and enclosures. Finally, we discuss simultaneous forced and natural convection. I 9 CONTENTS 9–1 Physical Mechanism of Natural Convection 460 9–2 Equation of Motion and the Grashof Number 463 9–3 Natural Convection over Surfaces 466 9–4 Natural Convection from Finned Surfaces and PCBs 473 9–5 Natural Convection inside Enclosures 477 9–6 Combined Natural and Forced Convection 486 Topic of Special Interest: Heat Transfer Through Windows 489 459 cen58933_ch09.qxd 9/4/2002 12:25 PM Page 460 460 HEAT TRANSFER 9–1 Warm air Cool air Heat HOT transfer EGG FIGURE 9–1 The cooling of a boiled egg in a cooler environment by natural convection. Warm air Heat transfer COLD SODA Cool air FIGURE 9–2 The warming up of a cold drink in a warmer environment by natural convection. ■ PHYSICAL MECHANISM OF NATURAL CONVECTION Many familiar heat transfer applications involve natural convection as the primary mechanism of heat transfer. Some examples are cooling of electronic equipment such as power transistors, TVs, and VCRs; heat transfer from electric baseboard heaters or steam radiators; heat transfer from the refrigeration coils and power transmission lines; and heat transfer from the bodies of animals and human beings. Natural convection in gases is usually accompanied by radiation of comparable magnitude except for low-emissivity surfaces. We know that a hot boiled egg (or a hot baked potato) on a plate eventually cools to the surrounding air temperature (Fig. 9–1). The egg is cooled by transferring heat by convection to the air and by radiation to the surrounding surfaces. Disregarding heat transfer by radiation, the physical mechanism of cooling a hot egg (or any hot object) in a cooler environment can be explained as follows: As soon as the hot egg is exposed to cooler air, the temperature of the outer surface of the egg shell will drop somewhat, and the temperature of the air adjacent to the shell will rise as a result of heat conduction from the shell to the air. Consequently, the egg will soon be surrounded by a thin layer of warmer air, and heat will then be transferred from this warmer layer to the outer layers of air. The cooling process in this case would be rather slow since the egg would always be blanketed by warm air, and it would have no direct contact with the cooler air farther away. We may not notice any air motion in the vicinity of the egg, but careful measurements indicate otherwise. The temperature of the air adjacent to the egg is higher, and thus its density is lower, since at constant pressure the density of a gas is inversely proportional to its temperature. Thus, we have a situation in which some low-density or “light” gas is surrounded by a high-density or “heavy” gas, and the natural laws dictate that the light gas rise. This is no different than the oil in a vinegar-and-oil salad dressing rising to the top (since oil vinegar). This phenomenon is characterized incorrectly by the phrase “heat rises,” which is understood to mean heated air rises. The space vacated by the warmer air in the vicinity of the egg is replaced by the cooler air nearby, and the presence of cooler air in the vicinity of the egg speeds up the cooling process. The rise of warmer air and the flow of cooler air into its place continues until the egg is cooled to the temperature of the surrounding air. The motion that results from the continual replacement of the heated air in the vicinity of the egg by the cooler air nearby is called a natural convection current, and the heat transfer that is enhanced as a result of this natural convection current is called natural convection heat transfer. Note that in the absence of natural convection currents, heat transfer from the egg to the air surrounding it would be by conduction only, and the rate of heat transfer from the egg would be much lower. Natural convection is just as effective in the heating of cold surfaces in a warmer environment as it is in the cooling of hot surfaces in a cooler environment, as shown in Figure 9–2. Note that the direction of fluid motion is reversed in this case. In a gravitational field, there is a net force that pushes upward a light fluid placed in a heavier fluid. The upward force exerted by a fluid on a body cen58933_ch09.qxd 9/4/2002 12:25 PM Page 461 461 CHAPTER 9 completely or partially immersed in it is called the buoyancy force. The magnitude of the buoyancy force is equal to the weight of the fluid displaced by the body. That is, Fbuoyancy fluid gVbody (9-1) where fluid is the average density of the fluid (not the body), g is the gravitational acceleration, and Vbody is the volume of the portion of the body immersed in the fluid (for bodies completely immersed in the fluid, it is the total volume of the body). In the absence of other forces, the net vertical force acting on a body is the difference between the weight of the body and the buoyancy force. That is, Fnet W Fbuoyancy body gVbody fluid gVbody (body fluid) gVbody (9-2) Note that this force is proportional to the difference in the densities of the fluid and the body immersed in it. Thus, a body immersed in a fluid will experience a “weight loss” in an amount equal to the weight of the fluid it displaces. This is known as Archimedes’ principle. To have a better understanding of the buoyancy effect, consider an egg dropped into water. If the average density of the egg is greater than the density of water (a sign of freshness), the egg will settle at the bottom of the container. Otherwise, it will rise to the top. When the density of the egg equals the density of water, the egg will settle somewhere in the water while remaining completely immersed, acting like a “weightless object” in space. This occurs when the upward buoyancy force acting on the egg equals the weight of the egg, which acts downward. The buoyancy effect has far-reaching implications in life. For one thing, without buoyancy, heat transfer between a hot (or cold) surface and the fluid surrounding it would be by conduction instead of by natural convection. The natural convection currents encountered in the oceans, lakes, and the atmosphere owe their existence to buoyancy. Also, light boats as well as heavy warships made of steel float on water because of buoyancy (Fig. 9–3). Ships are designed on the basis of the principle that the entire weight of a ship and its contents is equal to the weight of the water that the submerged volume of the ship can contain. The “chimney effect” that induces the upward flow of hot combustion gases through a chimney is also due to the buoyancy effect, and the upward force acting on the gases in the chimney is proportional to the difference between the densities of the hot gases in the chimney and the cooler air outside. Note that there is no gravity in space, and thus there can be no natural convection heat transfer in a spacecraft, even if the spacecraft is filled with atmospheric air. In heat transfer studies, the primary variable is temperature, and it is desirable to express the net buoyancy force (Eq. 9-2) in terms of temperature differences. But this requires expressing the density difference in terms of a temperature difference, which requires a knowledge of a property that represents the variation of the density of a fluid with temperature at constant pressure. The property that provides that information is the volume expansion coefficient , defined as (Fig. 9–4) W Vsubmerged Fbuoyancy FIGURE 9–3 It is the buoyancy force that keeps the ships afloat in water (W Fbuoyancy for floating objects). ∂v ( ––– ∂T ) 20°C 100 kPa 1 kg P 21°C 100 kPa 1 kg (a) A substance with a large β ∂v ––– ∂T P ( ) 20°C 100 kPa 1 kg 21°C 100 kPa 1 kg (b) A substance with a small β FIGURE 9–4 The coefficient of volume expansion is a measure of the change in volume of a substance with temperature at constant pressure. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 462 462 HEAT TRANSFER 1 T P 1 T (1/K) (9-3) P In natural convection studies, the condition of the fluid sufficiently far from the hot or cold surface is indicated by the subscript “infinity” to serve as a reminder that this is the value at a distance where the presence of the surface is not felt. In such cases, the volume expansion coefficient can be expressed approximately by replacing differential quantities by differences as 1 1 T T T (at constant P) (9-4) or (T T) (at constant P) (9-5) where is the density and T is the temperature of the quiescent fluid away from the surface. We can show easily that the volume expansion coefficient of an ideal gas (P RT) at a temperature T is equivalent to the inverse of the temperature: ideal gas 1 T (1/K) (9-6) where T is the absolute temperature. Note that a large value of for a fluid means a large change in density with temperature, and that the product T represents the fraction of volume change of a fluid that corresponds to a temperature change T at constant pressure. Also note that the buoyancy force is proportional to the density difference, which is proportional to the temperature difference at constant pressure. Therefore, the larger the temperature difference between the fluid adjacent to a hot (or cold) surface and the fluid away from it, the larger the buoyancy force and the stronger the natural convection currents, and thus the higher the heat transfer rate. The magnitude of the natural convection heat transfer between a surface and a fluid is directly related to the flow rate of the fluid. The higher the flow rate, the higher the heat transfer rate. In fact, it is the very high flow rates that increase the heat transfer coefficient by orders of magnitude when forced convection is used. In natural convection, no blowers are used, and therefore the flow rate cannot be controlled externally. The flow rate in this case is established by the dynamic balance of buoyancy and friction. As we have discussed earlier, the buoyancy force is caused by the density difference between the heated (or cooled) fluid adjacent to the surface and the fluid surrounding it, and is proportional to this density difference and the volume occupied by the warmer fluid. It is also well known that whenever two bodies in contact (solid–solid, solid–fluid, or fluid–fluid) move relative to each other, a friction force develops at the contact surface in the direction opposite to that of the motion. This opposing force slows down the fluid and thus reduces the flow rate of the fluid. Under steady conditions, the air flow rate driven by buoyancy is established at the point where these two effects balance each other. The friction force increases as more and more solid surfaces are introduced, seriously disrupting the fluid flow and heat transfer. For that reason, heat sinks with closely spaced fins are not suitable for natural convection cooling. Most heat transfer correlations in natural convection are based on experimental measurements. The instrument often used in natural convection cen58933_ch09.qxd 9/4/2002 12:25 PM Page 463 463 CHAPTER 9 experiments is the Mach–Zehnder interferometer, which gives a plot of isotherms in the fluid in the vicinity of a surface. The operation principle of interferometers is based on the fact that at low pressure, the lines of constant temperature for a gas correspond to the lines of constant density, and that the index of refraction of a gas is a function of its density. Therefore, the degree of refraction of light at some point in a gas is a measure of the temperature gradient at that point. An interferometer produces a map of interference fringes, which can be interpreted as lines of constant temperature as shown in Figure 9–5. The smooth and parallel lines in (a) indicate that the flow is laminar, whereas the eddies and irregularities in (b) indicate that the flow is turbulent. Note that the lines are closest near the surface, indicating a higher temperature gradient. 9–2 ■ EQUATION OF MOTION AND THE GRASHOF NUMBER In this section we derive the equation of motion that governs the natural convection flow in laminar boundary layer. The conservation of mass and energy equations derived in Chapter 6 for forced convection are also applicable for natural convection, but the momentum equation needs to be modified to incorporate buoyancy. Consider a vertical hot flat plate immersed in a quiescent fluid body. We assume the natural convection flow to be steady, laminar, and two-dimensional, and the fluid to be Newtonian with constant properties, including density, with one exception: the density difference is to be considered since it is this density difference between the inside and the outside of the boundary layer that gives rise to buoyancy force and sustains flow. (This is known as the Boussinesq approximation.) We take the upward direction along the plate to be x, and the direction normal to surface to be y, as shown in Figure 9–6. Therefore, gravity acts in the x-direction. Noting that the flow is steady and two-dimensional, the x- and y-components of velocity within boundary layer are u u(x, y) and v v(x, y), respectively. The velocity and temperature profiles for natural convection over a vertical hot plate are also shown in Figure 9–6. Note that as in forced convection, the thickness of the boundary layer increases in the flow direction. Unlike forced convection, however, the fluid velocity is zero at the outer edge of the velocity boundary layer as well as at the surface of the plate. This is expected since the fluid beyond the boundary layer is motionless. Thus, the fluid velocity increases with distance from the surface, reaches a maximum, and gradually decreases to zero at a distance sufficiently far from the surface. At the surface, the fluid temperature is equal to the plate temperature, and gradually decreases to the temperature of the surrounding fluid at a distance sufficiently far from the surface, as shown in the figure. In the case of cold surfaces, the shape of the velocity and temperature profiles remains the same but their direction is reversed. Consider a differential volume element of height dx, length dy, and unit depth in the z-direction (normal to the paper) for analysis. The forces acting on this volume element are shown in Figure 9–7. Newton’s second law of motion for this control volume can be expressed as (a) Laminar flow (b) Turbulent flow FIGURE 9–5 Isotherms in natural convection over a hot plate in air. Ts Temperature profile T Velocity profile u= 0 u= 0 Boundary layer Stationary fluid at T Ts x y FIGURE 9–6 Typical velocity and temperature profiles for natural convection flow over a hot vertical plate at temperature Ts inserted in a fluid at temperature T. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 464 464 HEAT TRANSFER P ∂P ∂x m ax Fx dx dy τ τ dx W ∂τ ∂y dy (9-7) where m (dx dy 1) is the mass of the fluid element within the control volume. The acceleration in the x-direction is obtained by taking the total differential of u(x, y), which is du ( u/ x)dx ( u/ y)dy, and dividing it by dt. We get ax du u dx u dy u u v u x y dt x dt y dt (9-8) P FIGURE 9–7 Forces acting on a differential control volume in the natural convection boundary layer over a vertical flat plate. The forces acting on the differential volume element in the vertical direction are the pressure forces acting on the top and bottom surfaces, the shear stresses acting on the side surfaces (the normal stresses acting on the top and bottom surfaces are small and are disregarded), and the force of gravity acting on the entire volume element. Then the net surface force acting in the x-direction becomes dx (dy 1) g(dx dy 1) y dy(dx 1) P x u P g(dx dy 1) x y Fx 2 (9-9) 2 since ( u/ y). Substituting Eqs. 9-8 and 9-9 into Eq. 9-7 and dividing by dx dy 1 gives the conservation of momentum in the x-direction as u u u 2u P v 2 g x y x y (9-10) The x-momentum equation in the quiescent fluid outside the boundary layer can be obtained from the relation above as a special case by setting u 0. It gives P g x (9-11) which is simply the relation for the variation of hydrostatic pressure in a quiescent fluid with height, as expected. Also, noting that v u in the boundary layer and thus v/ x v/ y 0, and that there are no body forces (including gravity) in the y-direction, the force balance in that direction gives P/ y 0. That is, the variation of pressure in the direction normal to the surface is negligible, and for a given x the pressure in the boundary layer is equal to the pressure in the quiescent fluid. Therefore, P P(x) P(x) and P/ x P/ x g. Substituting into Eq. 9-10, u u u 2u v 2 ( )g x y y (9-12) The last term represents the net upward force per unit volume of the fluid (the difference between the buoyant force and the fluid weight). This is the force that initiates and sustains convection currents. From Eq. 9-5, we have (T T). Substituting it into the last equation and dividing both sides by gives the desired form of the x-momentum equation, cen58933_ch09.qxd 9/4/2002 12:25 PM Page 465 465 CHAPTER 9 u u 2u u v v 2 g(T T) x y y (9-13) This is the equation that governs the fluid motion in the boundary layer due to the effect of buoyancy. Note that the momentum equation involves the temperature, and thus the momentum and energy equations must be solved simultaneously. The set of three partial differential equations (the continuity, momentum, and the energy equations) that govern natural convection flow over vertical isothermal plates can be reduced to a set of two ordinary nonlinear differential equations by the introduction of a similarity variable. But the resulting equations must still be solved numerically [Ostrach (1953), Ref. 27]. Interested reader is referred to advanced books on the topic for detailed discussions [e.g., Kays and Crawford (1993), Ref. 23]. The Grashof Number The governing equations of natural convection and the boundary conditions can be nondimensionalized by dividing all dependent and independent variables by suitable constant quantities: all lengths by a characteristic length Lc , all velocities by an arbitrary reference velocity (which, from the definition of Reynolds number, is taken to be ReL /Lc), and temperature by a suitable temperature difference (which is taken to be Ts T) as x x Lc y y Lc u u v v and T T T Ts T where asterisks are used to denote nondimensional variables. Substituting them into the momentum equation and simplifying give u g(Ts T)L3c T u u 1 2u v 2 2 ReL ReL y 2 x y v (9-14) The dimensionless parameter in the brackets represents the natural convection effects, and is called the Grashof number GrL , GrL g(Ts T)L3c v2 (9-15) where g gravitational acceleration, m/s coefficient of volume expansion, 1/K ( 1/T for ideal gases) Ts temperature of the surface, ˚C T temperature of the fluid sufficiently far from the surface, ˚C Lc characteristic length of the geometry, m kinematic viscosity of the fluid, m2/s 2 We mentioned in the preceding chapters that the flow regime in forced convection is governed by the dimensionless Reynolds number, which represents the ratio of inertial forces to viscous forces acting on the fluid. The flow regime in natural convection is governed by the dimensionless Grashof number, which represents the ratio of the buoyancy force to the viscous force acting on the fluid (Fig. 9–8). Hot surface Friction force Cold fluid Warm fluid Buoyancy force FIGURE 9–8 The Grashof number Gr is a measure of the relative magnitudes of the buoyancy force and the opposing viscous force acting on the fluid. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 466 466 HEAT TRANSFER The role played by the Reynolds number in forced convection is played by the Grashof number in natural convection. As such, the Grashof number provides the main criterion in determining whether the fluid flow is laminar or turbulent in natural convection. For vertical plates, for example, the critical Grashof number is observed to be about 10 9. Therefore, the flow regime on a vertical plate becomes turbulent at Grashof numbers greater than 109. When a surface is subjected to external flow, the problem involves both natural and forced convection. The relative importance of each mode of heat transfer is determined by the value of the coefficient GrL /Re L2: Natural convection effects are negligible if GrL /Re L2 1, free convection dominates and the forced convection effects are negligible if GrL/ReL2 1, and both effects are significant and must be considered if GrL /Re L2 1. 9–3 ■ NATURAL CONVECTION OVER SURFACES Natural convection heat transfer on a surface depends on the geometry of the surface as well as its orientation. It also depends on the variation of temperature on the surface and the thermophysical properties of the fluid involved. Although we understand the mechanism of natural convection well, the complexities of fluid motion make it very difficult to obtain simple analytical relations for heat transfer by solving the governing equations of motion and energy. Some analytical solutions exist for natural convection, but such solutions lack generality since they are obtained for simple geometries under some simplifying assumptions. Therefore, with the exception of some simple cases, heat transfer relations in natural convection are based on experimental studies. Of the numerous such correlations of varying complexity and claimed accuracy available in the literature for any given geometry, we present here the ones that are best known and widely used. The simple empirical correlations for the average Nusselt number Nu in natural convection are of the form (Fig. 9–9) Constant coefficient Nu = C RanL Nusselt number Nu Constant exponent Rayleigh number FIGURE 9–9 Natural convection heat transfer correlations are usually expressed in terms of the Rayleigh number raised to a constant n multiplied by another constant C, both of which are determined experimentally. hLc C(GrL Pr)n C RanL k (9-16) where RaL is the Rayleigh number, which is the product of the Grashof and Prandtl numbers: Ra L GrL Pr g(Ts T)L3c Pr v2 (9-17) The values of the constants C and n depend on the geometry of the surface and the flow regime, which is characterized by the range of the Rayleigh number. The value of n is usually 14 for laminar flow and 13 for turbulent flow. The value of the constant C is normally less than 1. Simple relations for the average Nusselt number for various geometries are given in Table 9–1, together with sketches of the geometries. Also given in this table are the characteristic lengths of the geometries and the ranges of Rayleigh number in which the relation is applicable. All fluid properties are to be evaluated at the film temperature Tf 12(Ts T). When the average Nusselt number and thus the average convection coefficient is known, the rate of heat transfer by natural convection from a solid cen58933_ch09.qxd 9/4/2002 12:25 PM Page 467 467 CHAPTER 9 surface at a uniform temperature Ts to the surrounding fluid is expressed by Newton’s law of cooling as Q˙ conv hAs(Ts T) (W) (9-18) where As is the heat transfer surface area and h is the average heat transfer coefficient on the surface. Vertical Plates (Ts constant) For a vertical flat plate, the characteristic length is the plate height L. In Table 9–1 we give three relations for the average Nusselt number for an isothermal vertical plate. The first two relations are very simple. Despite its complexity, we suggest using the third one (Eq. 9-21) recommended by Churchill and Chu (1975, Ref. 13) since it is applicable over the entire range of Rayleigh number. This relation is most accurate in the range of 101 RaL 109. Vertical Plates (˙qs constant) In the case of constant surface heat flux, the rate of heat transfer is known (it · is simply Q q· s A s ), but the surface temperature Ts is not. In fact, Ts increases with height along the plate. It turns out that the Nusselt number relations for the constant surface temperature and constant surface heat flux cases are nearly identical [Churchill and Chu (1975), Ref. 13]. Therefore, the relations for isothermal plates can also be used for plates subjected to uniform heat flux, provided that the plate midpoint temperature TL / 2 is used for Ts in the evaluation of the film temperature, Rayleigh number, and the Nusselt number. Noting that h q· s / (TL / 2 T), the average Nusselt number in this case can be expressed as Nu q˙s L hL k k(TL / 2 T ) (9-27) The midpoint temperature TL / 2 is determined by iteration so that the Nusselt numbers determined from Eqs. 9-21 and 9-27 match. Vertical Cylinders An outer surface of a vertical cylinder can be treated as a vertical plate when the diameter of the cylinder is sufficiently large so that the curvature effects are negligible. This condition is satisfied if D 35L Gr1/4 L Hot plate (9-28) When this criteria is met, the relations for vertical plates can also be used for vertical cylinders. Nusselt number relations for slender cylinders that do not meet this criteria are available in the literature [e.g., Cebeci (1974), Ref. 8]. Boundary layer flow y θ x Fy F θ Fx g Inclined Plates Consider an inclined hot plate that makes an angle from the vertical, as shown in Figure 9–10, in a cooler environment. The net force F g( ) (the difference between the buoyancy and gravity) acting on a unit volume of the fluid in the boundary layer is always in the vertical direction. In the case FIGURE 9–10 Natural convection flows on the upper and lower surfaces of an inclined hot plate. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 468 468 HEAT TRANSFER TABLE 9–1 Empirical correlations for the average Nusselt number for natural convection over surfaces Characteristic length Lc Geometry Vertical plate Ts Range of Ra Nu 104–109 109–1013 Nu 0.59Ra1/4 L Nu 0.1Ra1/3 L Entire range Nu L L 0.825 6 0.387Ra1/ L [1 (0.492/Pr) (9-19) (9-20) 2 9/16 8/27 ] (9-21) (complex but more accurate) Use vertical plate equations for the upper surface of a cold plate and the lower surface of a hot plate Inclined plate L L θ Replace g by g cos Horiontal plate (Surface area A and perimeter p) (a) Upper surface of a hot plate (or lower surface of a cold plate) Ra 109 for 104–107 107–1011 Nu 0.54Ra1/4 L Nu 0.15Ra1/3 L (9-22) (9-23) 105–1011 Nu 0.27Ra1/4 L (9-24) Ts Hot surface A s /p (b) Lower surface of a hot plate (or upper surface of a cold plate) Ts Hot surface Vertical cylinder A vertical cylinder can be treated as a vertical plate when Ts L L D Horizontal cylinder Ts D 35L Gr1/4 L RaD 1012 Nu 0.6 RaD 1011 Nu 2 [1 (0.559/Pr) 2 0.387Ra 1/6 D 9/16 8/27 ] (9-25) D Sphere D D (Pr 0.7) 0.589RaD1/4 [1 (0.469/Pr)9/16]4/9 (9-26) cen58933_ch09.qxd 9/4/2002 12:25 PM Page 469 469 CHAPTER 9 of inclined plate, this force can be resolved into two components: Fy F cos parallel to the plate that drives the flow along the plate, and Fy F sin normal to the plate. Noting that the force that drives the motion is reduced, we expect the convection currents to be weaker, and the rate of heat transfer to be lower relative to the vertical plate case. The experiments confirm what we suspect for the lower surface of a hot plate, but the opposite is observed on the upper surface. The reason for this curious behavior for the upper surface is that the force component Fy initiates upward motion in addition to the parallel motion along the plate, and thus the boundary layer breaks up and forms plumes, as shown in the figure. As a result, the thickness of the boundary layer and thus the resistance to heat transfer decreases, and the rate of heat transfer increases relative to the vertical orientation. In the case of a cold plate in a warmer environment, the opposite occurs as expected: The boundary layer on the upper surface remains intact with weaker boundary layer flow and thus lower rate of heat transfer, and the boundary layer on the lower surface breaks apart (the colder fluid falls down) and thus enhances heat transfer. When the boundary layer remains intact (the lower surface of a hot plate or the upper surface of a cold plate), the Nusselt number can be determined from the vertical plate relations provided that g in the Rayleigh number relation is replaced by g cos for 60˚. Nusselt number relations for the other two surfaces (the upper surface of a hot plate or the lower surface of a cold plate) are available in the literature [e.g., Fujiii and Imura (1972), Ref. 18]. Horizontal Plates The rate of heat transfer to or from a horizontal surface depends on whether the surface is facing upward or downward. For a hot surface in a cooler environment, the net force acts upward, forcing the heated fluid to rise. If the hot surface is facing upward, the heated fluid rises freely, inducing strong natural convection currents and thus effective heat transfer, as shown in Figure 9–11. But if the hot surface is facing downward, the plate will block the heated fluid that tends to rise (except near the edges), impeding heat transfer. The opposite is true for a cold plate in a warmer environment since the net force (weight minus buoyancy force) in this case acts downward, and the cooled fluid near the plate tends to descend. The average Nusselt number for horizontal surfaces can be determined from the simple power-law relations given in Table 9–1. The characteristic length for horizontal surfaces is calculated from As Lc p (9-29) where As is the surface area and p is the perimeter. Note that Lc a/4 for a horizontal square surface of length a, and D/4 for a horizontal circular surface of diameter D. Horizontal Cylinders and Spheres The boundary layer over a hot horizontal cylinder start to develop at the bottom, increasing in thickness along the circumference, and forming a rising Natural convection currents Natural convection currents Hot plate FIGURE 9–11 Natural convection flows on the upper and lower surfaces of a horizontal hot plate. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 470 470 HEAT TRANSFER plume at the top, as shown in Figure 9–12. Therefore, the local Nusselt number is highest at the bottom, and lowest at the top of the cylinder when the boundary layer flow remains laminar. The opposite is true in the case of a cold horizontal cylinder in a warmer medium, and the boundary layer in this case starts to develop at the top of the cylinder and ending with a descending plume at the bottom. The average Nusselt number over the entire surface can be determined from Eq. 9-26 [Churchill and Chu (1975), Ref. 13] for an isothermal horizontal cylinder, and from Eq. 9-27 for an isothermal sphere [Churchill (1983}, Ref. 11] both given in Table 9–1. Boundary layer flow Hot cylinder FIGURE 9–12 Natural convection flow over a horizontal hot cylinder. T = 20°C 70°C D = 8 cm 6m FIGURE 9–13 Schematic for Example 9–1. EXAMPLE 9–1 Heat Loss from Hot Water Pipes A 6-m-long section of an 8-cm-diameter horizontal hot water pipe shown in Figure 9–13 passes through a large room whose temperature is 20˚C. If the outer surface temperature of the pipe is 70˚C, determine the rate of heat loss from the pipe by natural convection. SOLUTION A horizontal hot water pipe passes through a large room. The rate of heat loss from the pipe by natural convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. Properties The properties of air at the film temperature of Tf (Ts T)/2 (70 20)/2 45˚C and 1 atm are (Table A–15) k 0.02699 W/m ˚C 1.749 105 m2/s Pr 0.7241 1 1 Tf 318 K Analysis The characteristic length in this case is the outer diameter of the pipe, Lc D 0.08 m. Then the Rayleigh number becomes g(Ts T)D3 Pr v2 2 (9.81 m /s )1/(318 K)(0.08 m)3 (0.7241) 1.869 10 6 (1.749 105 m 2/s) 2 RaD The natural convection Nusselt number in this case can be determined from Eq. 9-25 to be Nu 0.6 0.387 Ra1/6 D [1 (0.559/Pr)9/16]8/27 17.40 2 0.6 0.387(1869 106)1/6 [1 (0.559/0.7241)9/16]8/27 Then, 0.02699 W/m ºC k Nu (17.40) 5.869 W/m ˚C D 0.08 m As DL (0.08 m)(6 m) 1.508 m2 h and · Q hAs(Ts T) (5.869 W/m2 ˚C)(1.508 m2)(70 20)˚C 443 W 2 cen58933_ch09.qxd 9/4/2002 12:25 PM Page 471 471 CHAPTER 9 Therefore, the pipe will lose heat to the air in the room at a rate of 443 W by natural convection. Discussion The pipe will lose heat to the surroundings by radiation as well as by natural convection. Assuming the outer surface of the pipe to be black (emissivity 1) and the inner surfaces of the walls of the room to be at room temperature, the radiation heat transfer is determined to be (Fig. 9–14) T = 20°C . · 4 Q rad As(T s4 T surr ) (1)(1.508 m2)(5.67 108 W/m2 K4)[(70 273 K)4 (20 273 K)4] 553 W Q nat conv = 443 W Ts = 70°C . which is larger than natural convection. The emissivity of a real surface is less than 1, and thus the radiation heat transfer for a real surface will be less. But radiation will still be significant for most systems cooled by natural convection. Therefore, a radiation analysis should normally accompany a natural convection analysis unless the emissivity of the surface is low. EXAMPLE 9–2 Q rad, max = 553 W FIGURE 9–14 Radiation heat transfer is usually comparable to natural convection in magnitude and should be considered in heat transfer analysis. Cooling of a Plate in Different Orientations Consider a 0.6-m 0.6-m thin square plate in a room at 30˚C. One side of the plate is maintained at a temperature of 90˚C, while the other side is insulated, as shown in Figure 9–15. Determine the rate of heat transfer from the plate by natural convection if the plate is (a) vertical, (b) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. SOLUTION A hot plate with an insulated back is considered. The rate of heat loss by natural convection is to be determined for different orientations. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. Properties The properties of air at the film temperature of Tf (Ts T)/2 (90 30)/2 60˚C and 1 atm are (Table A-15) k 0.02808 W/m ˚C 1.896 105 m2/s Pr 0.7202 1 1 Tf 333 K 90°C T = 30°C L = 0.6 m (a) Vertical (b) Hot surface facing up Analysis (a) Vertical. The characteristic length in this case is the height of the plate, which is L 0.6 m. The Rayleigh number is g(Ts T)L3 Pr v2 (9.81 m/s2)1/(333 K)(0.6 m)3 (0.722) 7.656 108 (1.896 105 m2/s)2 RaL Then the natural convection Nusselt number can be determined from Eq. 9-21 to be Nu 0.825 0.825 0.387 Ra1/6 L [1 (0.492 / Pr)9/16]8/27 2 0.387(7.656 108)1/6 1 (0.492/0.7202)9/16]8/27 2 113.4 (c) Hot surface facing down FIGURE 9–15 Schematic for Example 9–2. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 472 472 HEAT TRANSFER Note that the simpler relation Eq. 9-19 would give Nu 0.59 RaL1/4 98.14, which is 13 percent lower. Then, 0.02808 W/m ºC k Nu (113.4) 5.306 W/m2 ˚C L 0.6 m As L2 (0.6 m)2 0.36 m2 h and · Q hAs(Ts T) (5.306 W/m2 ˚C)(0.36 m2)(90 30)˚C 115 W (b) Horizontal with hot surface facing up. The characteristic length and the Rayleigh number in this case are As L2 L 0.6 m Lc p 0.15 m 4L 4 4 g(Ts T )L 3c Pr v2 (9.81 m/s2)1/(333 K)(0.15 m)3 (0.7202) 1.196 107 (1.896 105 m2/s)2 RaL The natural convection Nusselt number can be determined from Eq. 9-22 to be 7 1/4 Nu 0.54 Ra 1/4 31.76 L 0.54(1.196 10 ) Then, 0.0280 W/m ºC k Nu (31.76) 5.946 W/m2 ˚C Lc 0.15 m As L2 (0.6 m)2 0.36 m2 h and · Q hAs(Ts T) (5.946 W/m2 ˚C)(0.36 m2)(90 30)˚C 128 W (c) Horizontal with hot surface facing down. The characteristic length, the heat transfer surface area, and the Rayleigh number in this case are the same as those determined in (b). But the natural convection Nusselt number is to be determined from Eq. 9-24, 7 1/4 Nu 0.27 Ra 1/4 15.86 L 0.27(1.196 10 ) Then, h 0.02808 W/m ºC k Nu (15.86) 2.973 W/m2 ˚C Lc 0.15 m and · Q hAs(Ts T) (2.973 W/m2 ˚C)(0.36 m2)(90 30)˚C 64.2 W Note that the natural convection heat transfer is the lowest in the case of the hot surface facing down. This is not surprising, since the hot air is “trapped” under the plate in this case and cannot get away from the plate easily. As a result, the cooler air in the vicinity of the plate will have difficulty reaching the plate, which results in a reduced rate of heat transfer. Discussion The plate will lose heat to the surroundings by radiation as well as by natural convection. Assuming the surface of the plate to be black (emissivity cen58933_ch09.qxd 9/4/2002 12:25 PM Page 473 473 CHAPTER 9 1) and the inner surfaces of the walls of the room to be at room temperature, the radiation heat transfer in this case is determined to be · 4 Q rad As(Ts4 Tsurr ) 2 (1)(0.36 m )(5.67 108 W/m2 K4)[(90 273 K)4 (30 273 K)4] 182 W which is larger than that for natural convection heat transfer for each case. Therefore, radiation can be significant and needs to be considered in surfaces cooled by natural convection. 9–4 ■ NATURAL CONVECTION FROM FINNED SURFACES AND PCBs Natural convection flow through a channel formed by two parallel plates as shown in Figure 9–16 is commonly encountered in practice. When the plates are hot (Ts T), the ambient fluid at T enters the channel from the lower end, rises as it is heated under the effect of buoyancy, and the heated fluid leaves the channel from the upper end. The plates could be the fins of a finned heat sink, or the PCBs (printed circuit boards) of an electronic device. The plates can be approximated as being isothermal (Ts constant) in the first case, and isoflux (q·s constant) in the second case. Boundary layers start to develop at the lower ends of opposing surfaces, and eventually merge at the midplane if the plates are vertical and sufficiently long. In this case, we will have fully developed channel flow after the merger of the boundary layers, and the natural convection flow is analyzed as channel flow. But when the plates are short or the spacing is large, the boundary layers of opposing surfaces never reach each other, and the natural convection flow on a surface is not affected by the presence of the opposing surface. In that case, the problem should be analyzed as natural convection from two independent plates in a quiescent medium, using the relations given for surfaces, rather than natural convection flow through a channel. Natural Convection Cooling of Finned Surfaces (Ts constant) Finned surfaces of various shapes, called heat sinks, are frequently used in the cooling of electronic devices. Energy dissipated by these devices is transferred to the heat sinks by conduction and from the heat sinks to the ambient air by natural or forced convection, depending on the power dissipation requirements. Natural convection is the preferred mode of heat transfer since it involves no moving parts, like the electronic components themselves. However, in the natural convection mode, the components are more likely to run at a higher temperature and thus undermine reliability. A properly selected heat sink may considerably lower the operation temperature of the components and thus reduce the risk of failure. Natural convection from vertical finned surfaces of rectangular shape has been the subject of numerous studies, mostly experimental. Bar-Cohen and Fully developed flow Isothermal plate at Ts L Boundary layer Ambient fluid T S FIGURE 9–16 Natural convection flow through a channel between two isothermal vertical plates. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 474 474 HEAT TRANSFER Rohsenow (1984, Ref. 5) have compiled the available data under various boundary conditions, and developed correlations for the Nusselt number and optimum spacing. The characteristic length for vertical parallel plates used as fins is usually taken to be the spacing between adjacent fins S, although the fin height L could also be used. The Rayleigh number is expressed as RaS g(Ts T)S 3 Pr v2 RaL and g(Ts T)L 3 L3 Pr RaS 3 2 v S (9-30) The recommended relation for the average Nusselt number for vertical isothermal parallel plates is Ts constant: FIGURE 9–17 Heat sinks with (a) widely spaced and (b) closely packed fins (courtesy of Vemaline Products). W H Quiescent air, T Ts constant: S t FIGURE 9–18 Various dimensions of a finned surface oriented vertically. 0.5 (9-31) S L Ra Sopt 2.714 0.25 3 S 2.714 L Ra 0.25 L (9-32) It can be shown by combining the three equations above that when S Sopt, the Nusselt number is a constant and its value is 1.307, S Sopt: Ts hS 576 2.873 k (Ra S S/L)2 ( Ra S S/L)0.5 A question that often arises in the selection of a heat sink is whether to select one with closely packed fins or widely spaced fins for a given base area (Fig. 9–17). A heat sink with closely packed fins will have greater surface area for heat transfer but a smaller heat transfer coefficient because of the extra resistance the additional fins introduce to fluid flow through the interfin passages. A heat sink with widely spaced fins, on the other hand, will have a higher heat transfer coefficient but a smaller surface area. Therefore, there must be an optimum spacing that maximizes the natural convection heat transfer from the heat sink for a given base area WL, where W and L are the width and height of the base of the heat sink, respectively, as shown in Figure 9–18. When the fins are essentially isothermal and the fin thickness t is small relative to the fin spacing S, the optimum fin spacing for a vertical heat sink is determined by Bar-Cohen and Rohsenow to be L g Nu Nu h S opt 1.307 k (9-33) The rate of heat transfer by natural convection from the fins can be determined from · Q h(2nLH)(Ts T) (9-34) where n W/(S t) W/S is the number of fins on the heat sink and Ts is the surface temperature of the fins. All fluid properties are to be evaluated at the average temperature Tave (Ts T)/2. Natural Convection Cooling of Vertical PCBs (q˙ s constant) Arrays of printed circuit boards used in electronic systems can often be modeled as parallel plates subjected to uniform heat flux q·s (Fig. 9–19). The plate temperature in this case increases with height, reaching a maximum at the cen58933_ch09.qxd 9/4/2002 12:25 PM Page 475 475 CHAPTER 9 upper edge of the board. The modified Rayleigh number for uniform heat flux on both plates is expressed as Ra S g q˙s S 4 Pr kv 2 (9-35) The Nusselt number at the upper edge of the plate where maximum temperature occurs is determined from [Bar-Cohen and Rohsenow (1984), Ref. 5] hLS 48 2.51 NuL k Ra S S/L (Ra L S/L)0.4 0.5 (9-36) The optimum fin spacing for the case of uniform heat flux on both plates is given as q·s constant: Sopt 2.12 S L Ra 4 0.2 (9-37) S The total rate of heat transfer from the plates is · Q q·s As q·s (2nLH) (9-38) where n W/(S t) W/S is the number of plates. The critical surface temperature TL occurs at the upper edge of the plates, and it can be determined from q·s hL(TL T) W (9-39) All fluid properties are to be evaluated at the average temperature Tave (TL T)/2. Mass Flow Rate through the Space between Plates As we mentioned earlier, the magnitude of the natural convection heat transfer is directly related to the mass flow rate of the fluid, which is established by the dynamic balance of two opposing effects: buoyancy and friction. The fins of a heat sink introduce both effects: inducing extra buoyancy as a result of the elevated temperature of the fin surfaces and slowing down the fluid by acting as an added obstacle on the flow path. As a result, increasing the number of fins on a heat sink can either enhance or reduce natural convection, depending on which effect is dominant. The buoyancy-driven fluid flow rate is established at the point where these two effects balance each other. The friction force increases as more and more solid surfaces are introduced, seriously disrupting fluid flow and heat transfer. Under some conditions, the increase in friction may more than offset the increase in buoyancy. This in turn will tend to reduce the flow rate and thus the heat transfer. For that reason, heat sinks with closely spaced fills are not suitable for natural convection cooling. When the heat sink involves closely spaced fins, the narrow channels formed tend to block or “suffocate” the fluid, especially when the heat sink is long. As a result, the blocking action produced overwhelms the extra buoyancy and downgrades the heat transfer characteristics of the heat sink. Then, at a fixed power setting, the heat sink runs at a higher temperature relative to the no-shroud case. When the heat sink involves widely spaced fins, the L q·s H T S FIGURE 9–19 Arrays of vertical printed circuit boards (PCBs) cooled by natural convection. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 476 476 HEAT TRANSFER shroud does not introduce a significant increase in resistance to flow, and the buoyancy effects dominate. As a result, heat transfer by natural convection may improve, and at a fixed power level the heat sink may run at a lower temperature. When extended surfaces such as fins are used to enhance natural convection heat transfer between a solid and a fluid, the flow rate of the fluid in the vicinity of the solid adjusts itself to incorporate the changes in buoyancy and friction. It is obvious that this enhancement technique will work to advantage only when the increase in buoyancy is greater than the additional friction introduced. One does not need to be concerned with pressure drop or pumping power when studying natural convection since no pumps or blowers are used in this case. Therefore, an enhancement technique in natural convection is evaluated on heat transfer performance alone. The failure rate of an electronic component increases almost exponentially with operating temperature. The cooler the electronic device operates, the more reliable it is. A rule of thumb is that the semiconductor failure rate is halved for each 10˚C reduction in junction operating temperature. The desire to lower the operating temperature without having to resort to forced convection has motivated researchers to investigate enhancement techniques for natural convection. Sparrow and Prakash (Ref. 31) have demonstrated that, under certain conditions, the use of discrete plates in lieu of continuous plates of the same surface area increases heat transfer considerably. In other experimental work, using transistors as the heat source, Çengel and Zing (Ref. 9) have demonstrated that temperature recorded on the transistor case dropped by as much as 30˚C when a shroud was used, as opposed to the corresponding noshroud case. EXAMPLE 9–3 W = 0.12 m H = 2.4 cm L = 0.18 m Ts = 80°C t = 1 mm S FIGURE 9–20 Schematic for Example 9–3. Optimum Fin Spacing of a Heat Sink A 12-cm-wide and 18-cm-high vertical hot surface in 30˚C air is to be cooled by a heat sink with equally spaced fins of rectangular profile (Fig. 9–20). The fins are 0.1 cm thick and 18 cm long in the vertical direction and have a height of 2.4 cm from the base. Determine the optimum fin spacing and the rate of heat transfer by natural convection from the heat sink if the base temperature is 80˚C. SOLUTION A heat sink with equally spaced rectangular fins is to be used to cool a hot surface. The optimum fin spacing and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The atmospheric pressure at that location is 1 atm. 4 The thickness t of the fins is very small relative to the fin spacing S so that Eq. 9-32 for optimum fin spacing is applicable. 5 All fin surfaces are isothermal at base temperature. Properties The properties of air at the film temperature of Tf (Ts T)/2 (80 30)/2 55˚C and 1 atm pressure are (Table A-15) k 0.02772 W/m ˚C 1.846 105 m2/s Pr 0.7215 1/Tf 1/328 K Analysis We take the characteristic length to be the length of the fins in the vertical direction (since we do not know the fin spacing). Then the Rayleigh number becomes cen58933_ch09.qxd 9/4/2002 12:25 PM Page 477 477 CHAPTER 9 g(Ts T)L 3 Pr v2 (981 m/s2)1/(328 K)(0.18 m)3 (0.7215) 1.846 107 (1.846 105 m2/s) 2 RaL The optimum fin spacing is determined from Eq. 7-32 to be Sopt 2.714 0.8 m L 2.714 7.45 103 m 7.45 mm (1.846 10 7) 0.25 Ra 0.25 L which is about seven times the thickness of the fins. Therefore, the assumption of negligible fin thickness in this case is acceptable. The number of fins and the heat transfer coefficient for this optimum fin spacing case are n W 0.12 m 15 fins S t (0.00745 0.0001) m The convection coefficient for this optimum in spacing case is, from Eq. 9-33, h Nuopt 0.02772 W/m ºC k 1.307 0.2012 W/m2 ˚C Sopt 0.00745 m Cold surface Hot surface Then the rate of natural convection heat transfer becomes · Q hAs(Ts T) h(2nLH)(Ts T) (0.2012 W/m2 ˚C)2 15(0.18 m)(0.024 m)˚C 1.30 W . Velocity profile Q Therefore, this heat sink can dissipate heat by natural convection at a rate of 1.30 W. L 9–5 ■ NATURAL CONVECTION INSIDE ENCLOSURES A considerable portion of heat loss from a typical residence occurs through the windows. We certainly would insulate the windows, if we could, in order to conserve energy. The problem is finding an insulating material that is transparent. An examination of the thermal conductivities of the insulting materials reveals that air is a better insulator than most common insulating materials. Besides, it is transparent. Therefore, it makes sense to insulate the windows with a layer of air. Of course, we need to use another sheet of glass to trap the air. The result is an enclosure, which is known as a double-pane window in this case. Other examples of enclosures include wall cavities, solar collectors, and cryogenic chambers involving concentric cylinders or spheres. Enclosures are frequently encountered in practice, and heat transfer through them is of practical interest. Heat transfer in enclosed spaces is complicated by the fact that the fluid in the enclosure, in general, does not remain stationary. In a vertical enclosure, the fluid adjacent to the hotter surface rises and the fluid adjacent to the cooler one falls, setting off a rotationary motion within the enclosure that enhances heat transfer through the enclosure. Typical flow patterns in vertical and horizontal rectangular enclosures are shown in Figures 9–2l and 9–22. FIGURE 9–21 Convective currents in a vertical rectangular enclosure. Light fluid Hot (No fluid motion) Heavy fluid Cold (a) Hot plate at the top Heavy fluid Light fluid Cold Hot (b) Hot plate at the bottom FIGURE 9–22 Convective currents in a horizontal enclosure with (a) hot plate at the top and (b) hot plate at the bottom. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 478 478 HEAT TRANSFER The characteristics of heat transfer through a horizontal enclosure depend on whether the hotter plate is at the top or at the bottom, as shown in Figure 9–22. When the hotter plate is at the top, no convection currents will develop in the enclosure, since the lighter fluid will always be on top of the heavier fluid. Heat transfer in this case will be by pure conduction, and we will have Nu 1. When the hotter plate is at the bottom, the heavier fluid will be on top of the lighter fluid, and there will be a tendency for the lighter fluid to topple the heavier fluid and rise to the top, where it will come in contact with the cooler plate and cool down. Until that happens, however, the heat transfer is still by pure conduction and Nu 1. When Ra 1708, the buoyant force overcomes the fluid resistance and initiates natural convection currents, which are observed to be in the form of hexagonal cells called Bénard cells. For Ra 3 105, the cells break down and the fluid motion becomes turbulent. The Rayleigh number for an enclosure is determined from RaL g(T1 T2)L3c Pr v2 (9-40) where the characteristic length Lc is the distance between the hot and cold surfaces, and T1 and T2 are the temperatures of the hot and cold surfaces, respectively. All fluid properties are to be evaluated at the average fluid temperature Tave (T1 T2)/2. Effective Thermal Conductivity When the Nusselt number is known, the rate of heat transfer through the enclosure can be determined from T1 T2 · Q hAs(T1 T2) kNuAs Lc (9-41) since h kNu/L. The rate of steady heat conduction across a layer of thickness Lc , area As, and thermal conductivity k is expressed as Hot k Cold Nu = 3 Hot keff = 3k Cold . . Q = 10 W Q = 30 W (No motion) Pure conduction Natural convection FIGURE 9–23 A Nusselt number of 3 for an enclosure indicates that heat transfer through the enclosure by natural convection is three times that by pure conduction. T1 T2 · Q cond kAs Lc (9-42) where T1 and T2 are the temperatures on the two sides of the layer. A comparison of this relation with Eq. 9-41 reveals that the convection heat transfer in an enclosure is analogous to heat conduction across the fluid layer in the enclosure provided that the thermal conductivity k is replaced by kNu. That is, the fluid in an enclosure behaves like a fluid whose thermal conductivity is kNu as a result of convection currents. Therefore, the quantity kNu is called the effective thermal conductivity of the enclosure. That is, keff kNu (9-43) Note that for the special case of Nu 1, the effective thermal conductivity of the enclosure becomes equal to the conductivity of the fluid. This is expected since this case corresponds to pure conduction (Fig. 9–23). Natural convection heat transfer in enclosed spaces has been the subject of many experimental and numerical studies, and numerous correlations for the Nusselt number exist. Simple power-law type relations in the form of cen58933_ch09.qxd 9/4/2002 12:25 PM Page 479 479 CHAPTER 9 Nu CRaLn , where C and n are constants, are sufficiently accurate, but they are usually applicable to a narrow range of Prandtl and Rayleigh numbers and aspect ratios. The relations that are more comprehensive are naturally more complex. Next we present some widely used relations for various types of enclosures. Horizontal Rectangular Enclosures We need no Nusselt number relations for the case of the hotter plate being at the top, since there will be no convection currents in this case and heat transfer will be downward by conduction (Nu 1). When the hotter plate is at the bottom, however, significant convection currents set in for RaL 1708, and the rate of heat transfer increases (Fig. 9–24). For horizontal enclosures that contain air, Jakob (1949, Ref. 22) recommends the following simple correlations Nu 0.195RaL1/4 104 RaL 4 105 (9-44) Nu 0.068RaL1/3 4 105 RaL 107 (9-45) T2 L These relations can also be used for other gases with 0.5 Pr 2. Using water, silicone oil, and mercury in their experiments, Globe and Dropkin (1959) obtained this correlation for horizontal enclosures heated from below, Nu 0.069RaL1/3 Pr0.074 · Q T1 > T2 3 105 RaL 7 109 (9-46) Fluid T1 H FIGURE 9–24 A horizontal rectangular enclosure with isothermal surfaces. Based on experiments with air, Hollands et al (1976, Ref. 19) recommend this correlation for horizontal enclosures, Nu 1 1.44 1 1708 Ra L Ra18 1/3 L 1 RaL 108 (9-47) The notation [ ] indicates that if the quantity in the bracket is negative, it should be set equal to zero. This relation also correlates data well for liquids with moderate Prandtl numbers for RaL 105, and thus it can also be used for water. Inclined Rectangular Enclosures T2 Air spaces between two inclined parallel plates are commonly encountered in flat-plate solar collectors (between the glass cover and the absorber plate) and the double-pane skylights on inclined roofs. Heat transfer through an inclined enclosure depends on the aspect ratio H/L as well as the tilt angle from the horizontal (Fig. 9–25). For large aspect ratios (H/L 12), this equation [Hollands et al., 1976, Ref. 19] correlates experimental data extremely well for tilt angles up to 70˚, Nu 1 1.44 1 1708 RaL cos 1.8) (Ra 1 1708(sin Ra cos 1.6 L L cos )1/3 1 18 T1 H · Q L T1 > T2 (9-48) for RaL 105, 0 70˚, and H/L 12. Again any quantity in [ ] should be set equal to zero if it is negative. This is to ensure that Nu 1 for RaL cos 1708. Note that this relation reduces to Eq. 9-47 for horizontal enclosures for 0˚, as expected. θ FIGURE 9–25 An inclined rectangular enclosure with isothermal surfaces. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 480 480 HEAT TRANSFER For enclosures with smaller aspect ratios (H/L 12), the next correlation can be used provided that the tilt angle is less than the critical value cr listed in Table 9–2 [Catton (1978), Ref. 7] TABLE 9–2 Nu Nu 0º Critical angles for inclined rectangular enclosures Aspect ratio, H/L Critical angle, cr 1 3 6 12 12 25˚ 53˚ 60˚ 67˚ 70˚ /cr Nu90º Nu0º (sincr)/(4cr) 0º cr (9-49) For tilt angles greater than the critical value (cr 90˚), the Nusselt number can be obtained by multiplying the Nusselt number for a vertical enclosure by (sin )1/4 [Ayyaswamy and Catton (1973), Ref. 3], Nu Nu 90˚(sin )1/4 cr 90˚, any H/L (9-50) For enclosures tilted more than 90˚, the recommended relation is [Arnold et al., (1974), Ref. 2] Nu 1 (Nu 90˚ 1)sin 90˚ 180˚, any H/L (9-51) More recent but more complex correlations are also available in the literature [e.g., and ElSherbiny et al. (1982), Ref. 17]. Vertical Rectangular Enclosures T1 For vertical enclosures (Fig. 9–26), Catton (1978, Ref. 7) recommends these two correlations due to Berkovsky and Polevikov (1977, Ref. 6), T2 · Q H L T1 > T2 Nu 0.22 1 H/L 2 any Prandtl number RaL Pr/(0.2 Pr) 10 3 Pr Ra 0.2 Pr L Nu 0.18 Pr Ra 0.2 Pr L 0.29 0.28 H L 1/4 2 H/L 10 any Prandtl number RaL 10 10 (9-52) (9-53) For vertical enclosures with larger aspect ratios, the following correlations can be used [MacGregor and Emery (1969), Ref. 26] FIGURE 9–26 A vertical rectangular enclosure with isothermal surfaces. Pr Nu 0.46Ra1/3 L Outer cylinder at To Di Nu 0.42 Ra 1/4 L Do 0.012 H L 0.3 10 H/L 40 1 Pr 2 10 4 104 RaL 10 7 (9-54) 1 H/L 40 1 Pr 20 106 RaL 10 9 (9-55) Again all fluid properties are to be evaluated at the average temperature (T1 T2)/2. Concentric Cylinders Inner cylinder at Ti FIGURE 9–27 Two concentric horizontal isothermal cylinders. Consider two long concentric horizontal cylinders maintained at uniform but different temperatures of Ti and To, as shown in Figure 9–27. The diameters of the inner and outer cylinders are Di and Do, respectively, and the characteristic length is the spacing between the cylinders, Lc (Do Di)/2. The rate of heat transfer through the annular space between the natural convection unit is expressed as cen58933_ch09.qxd 9/4/2002 12:25 PM Page 481 481 CHAPTER 9 2k eff · Q ( T To ) ln(Do /Di ) i (W/m) (9-56) The recommended relation for effective thermal conductivity is [Raithby and Hollands (1975), Ref. 28] keff Pr 0.386 k 0.861 Pr 1/4 (Fcyl Ra L )1/4 (9-57) where the geometric factor for concentric cylinders Fcyl is Fcyl [ln(Do /Di )]4 L3c (D 3/5 D 3/5 )5 i o (9-58) The keff relation in Eq. 9-57 is applicable for 0.70 Pr 6000 and 102 FcylRaL 107. For FcylRaL 100, natural convection currents are negligible and thus keff k. Note that keff cannot be less than k, and thus we should set keff k if keff/k 1. The fluid properties are evaluated at the average temperature of (Ti To)/2. Concentric Spheres For concentric isothermal spheres, the rate of heat transfer through the gap between the spheres by natural convection is expressed as (Fig. 9–28) Di Do · (Ti To) Q k eff Lc (W) 1/4 (Fsph Ra L)1/4 Fsph (Di Do ) 4 (D7/5 D7/5 )5 i o (9-61) The keff relation in Eq. 9-60 is applicable for 0.70 Pr 4200 and 102 FsphRaL 104. If keff/k 1, we should set keff k. Combined Natural Convection and Radiation Gases are nearly transparent to radiation, and thus heat transfer through a gas layer is by simultaneous convection (or conduction, if the gas is quiescent) and radiation. Natural convection heat transfer coefficients are typically very low compared to those for forced convection. Therefore, radiation is usually disregarded in forced convection problems, but it must be considered in natural convection problems that involve a gas. This is especially the case for surfaces with high emissivities. For example, about half of the heat transfer through the air space of a double pane window is by radiation. The total rate of heat transfer is determined by adding the convection and radiation components, · · · Q total Q conv Q rad Lc (9-60) where the geometric factor for concentric spheres Fsph is Lc Di , Ti (9-59) where Lc (Do Di)/2 is the characteristic length. The recommended relation for effective thermal conductivity is [Raithby and Hollands (1975), Ref. 28] keff Pr 0.74 k 0.861 Pr D0 , T0 (9-62) FIGURE 9–28 Two concentric isothermal spheres. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 482 482 HEAT TRANSFER Radiation heat transfer from a surface at temperature Ts surrounded by surfaces at a temperature Tsurr (both in absolute temperature unit K) is determined from · 4 Q rad As(Ts4 Tsurr ) (W) (9-63) where is the emissivity of the surface, As is the surface area, and 5.67 108 W/m2 K4 is the Stefan–Boltzmann constant. When the end effects are negligible, radiation heat transfer between two large parallel plates at absolute temperatures T1 and T2 is expressed as (see Chapter 12 for details) As(T 41 T 42 ) · Q rad effective As(T 14 T 42 ) 1/1 1/ 2 1 (W) (9-64) where 1 and 2 are the emissivities of the plates, and effective is the effective emissivity defined as effective 1 1/1 1/2 1 (9-65) The emissivity of an ordinary glass surface, for example, is 0.84. Therefore, the effective emissivity of two parallel glass surfaces facing each other is 0.72. Radiation heat transfer between concentric cylinders and spheres is discussed in Chapter 12. Note that in some cases the temperature of the surrounding medium may be below the surface temperature (T Ts), while the temperature of the surrounding surfaces is above the surface temperature (Tsurr Ts). In such cases, convection and radiation heat transfers are subtracted from each other instead of being added since they are in opposite directions. Also, for a metal surface, the radiation effect can be reduced to negligible levels by polishing the surface and thus lowering the surface emissivity to a value near zero. EXAMPLE 9–4 Glass H = 0.8 m Air L = 2 cm Glass Heat Loss through a Double-Pane Window The vertical 0.8-m-high, 2-m-wide double-pane window shown in Fig. 9–29 consists of two sheets of glass separated by a 2-cm air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be 12˚C and 2˚C, determine the rate of heat transfer through the window. SOLUTION Two glasses of a double-pane window are maintained at specified temperatures. The rate of heat transfer through the window is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Radiation heat transfer is not considered. Properties The properties of air at the average temperature of Tave (T1 T2)/2 (12 2)/2 7˚C and 1 atm pressure are (Table A-15) k 0.02416 W/m ˚C 1.399 105 m2/s FIGURE 9–29 Schematic for Example 9–4. Pr 0.7344 1 1 Tave 280 K Analysis We have a rectangular enclosure filled with air. The characteristic length in this case is the distance between the two glasses, L 0.02 m. Then the Rayleigh number becomes cen58933_ch09.qxd 9/4/2002 12:25 PM Page 483 483 CHAPTER 9 g(T1 T2)L3 v2 (9.81 m/s2)1/(280 K)(0.02 m)3 (0.7344) 1.051 104 (1.399 105 m 2 /s)2 RaL The aspect ratio of the geometry is H/L 0.8/0.02 40. Then the Nusselt number in this case can be determined from Eq. 9-54 to be Nu 0.42Ra L1/4 Pr 0.012 HL 0.3 0.8 0.02 0.42(1.051 104)1/4(0.7344)0.012 0.3 1.401 Then, As H W (0.8 m)(2 m) 1.6 m2 and T1 T2 · Q hAs(T1 T2) kNuAs L (0.02416 W/m ˚C)(1.401)(1.6 m2) (12 2)ºC 27.1 W 0.02 m Therefore, heat will be lost through the window at a rate of 27.1 W. Discussion Recall that a Nusselt number of Nu 1 for an enclosure corresponds to pure conduction heat transfer through the enclosure. The air in the enclosure in this case remains still, and no natural convection currents occur in the enclosure. The Nusselt number in our case is 1.32, which indicates that heat transfer through the enclosure is 1.32 times that by pure conduction. The increase in heat transfer is due to the natural convection currents that develop in the enclosure. D0 = 30 cm T0 = 280 K EXAMPLE 9–5 Heat Transfer through a Spherical Enclosure The two concentric spheres of diameters Di 20 cm and Do 30 cm shown in Fig. 9–30 are separated by air at 1 atm pressure. The surface temperatures of the two spheres enclosing the air are Ti 320 K and To 280 K, respectively. Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection. Lc = 5 cm Di = 20 cm Ti = 320 K SOLUTION Two surfaces of a spherical enclosure are maintained at specified FIGURE 9–30 temperatures. The rate of heat transfer through the enclosure is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Radiation heat transfer is not considered. Properties The properties of air at the average temperature of Tave (Ti To)/2 (320 280)/2 300 K 27˚C and 1 atm pressure are (Table A-15) Schematic for Example 9–5. k 0.02566 W/m ˚C 1.580 105 m2/s Pr 0.7290 1 1 Tave 300 K cen58933_ch09.qxd 9/4/2002 12:25 PM Page 484 484 HEAT TRANSFER Analysis We have a spherical enclosure filled with air. The characteristic length in this case is the distance between the two spheres, Lc (Do Di)/2 (0.3 0.2)/2 0.05 m The Rayleigh number is g(Ti To)L 3 Pr v2 (9.81 m/s2)1/(300 K)(0.05 m)3 (0.729) 4.776 105 (1.58 105 m2/s)2 RaL The effective thermal conductivity is Lc (Di Do ) 4(D 7/5 D 7/5 )5 o i 0.05 m 0.005229 [(0.2 m)(0.3 m)]4[(0.2 m7/5 (0.3 m)7/5]5 Fsph 0.861Pr Pr 1/4 keff 0.74k (FsphRaL)1/4 (0.005229 4.776 10 ) 0.8610.729 0.729 0.74(0.02566 W/m ˚C) 5 1/4 0.1104 W/m ˚C Then the rate of heat transfer between the spheres becomes Di Do · Q keff (Ti To) Lc (0.1104 W/m ˚C) (0.2 m)(0.3 m) (320 280)K 16.7 W 0.05 m Therefore, heat will be lost from the inner sphere to the outer one at a rate of 16.7 W. Discussion Note that the air in the spherical enclosure will act like a stationary fluid whose thermal conductivity is keff/k 0.1104/0.02566 4.3 times that of air as a result of natural convection currents. Also, radiation heat transfer between spheres is usually very significant, and should be considered in a complete analysis. Solar energy Glass cover EXAMPLE 9–6 70°F 4 in. 2 in. Aluminum tube Water FIGURE 9–31 Schematic for Example 9–6. Heating Water in a Tube by Solar Energy A solar collector consists of a horizontal aluminum tube having an outer diameter of 2 in. enclosed in a concentric thin glass tube of 4-in.-diameter (Fig. 9–31). Water is heated as it flows through the tube, and the annular space between the aluminum and the glass tubes is filled with air at 1 atm pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of 30 Btu/h per foot length, and the temperature of the ambient air outside is 70˚F. Disregarding any heat loss by radiation, determine the temperature of the aluminum tube when steady operation is established (i.e., when the rate of heat loss from the tube equals the amount of solar energy gained by the tube). cen58933_ch09.qxd 9/4/2002 12:25 PM Page 485 485 CHAPTER 9 SOLUTION The circulating pump of a solar collector that consists of a horizontal tube and its glass cover fails. The equilibrium temperature of the tube is to be determined. Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are isothermal. 3 Air is an ideal gas. 4 Heat loss by radiation is negligible. Properties The properties of air should be evaluated at the average temperature. But we do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk fluid and glass cover temperatures at this point, and thus we cannot evaluate the average temperatures. Therefore, we will assume the glass temperature to be 110˚F, and use properties at an anticipated average temperature of (70 110)/2 90˚F (Table A-15E), k 0.01505 Btu/h ft ˚F Pr 0.7275 0.6310 ft2/h 1.753 104 ft2/s 1 1 Tave 550 K Analysis We have a horizontal cylindrical enclosure filled with air at 1 atm pressure. The problem involves heat transfer from the aluminum tube to the glass cover and from the outer surface of the glass cover to the surrounding ambient air. When steady operation is reached, these two heat transfer rates must equal the rate of heat gain. That is, · · · Q tube-glass Q glass-ambient Q so1ar gain 30 Btu/h (per foot of tube) The heat transfer surface area of the glass cover is Ao Aglass (Do L) (4/12 ft)(1 ft) 1.047 ft2 (per foot of tube) To determine the Rayleigh number, we need to know the surface temperature of the glass, which is not available. Therefore, it is clear that the solution will require a trial-and-error approach. Assuming the glass cover temperature to be 100˚F, the Rayleigh number, the Nusselt number, the convection heat transfer coefficient, and the rate of natural convection heat transfer from the glass cover to the ambient air are determined to be g(Ts T)Do3 Pr v2 (32.2 ft/s2)1/(550 R)(4/12 ft)3 (0.7275) 2.054 106 (1.753 104 ft2/s)2 RaDo Nu 0.6 0.387 Ra 1/6 D [1 (0.559/Pr)9/16 ] 8/27 17.89 ho 2 0.6 0.387(2.054 106 )1/6 [1 (0.559/0.7275)9/16]8/27 2 0.0150 Btu/h ft ºF k Nu (17.89) 0.8075 Btu/h ft2 ˚F D0 4/12 ft · Q o hoAo(To T) (0.8075 Btu/h ft2 ˚F)(1.047 ft2)(110 70)˚F 33.8 Btu/h which is more than 30 Btu/h. Therefore, the assumed temperature of 110˚F for the glass cover is high. Repeating the calculations with lower temperatures, the glass cover temperature corresponding to 30 Btu/h is determined to be 106˚F. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 486 486 HEAT TRANSFER The temperature of the aluminum tube is determined in a similar manner using the natural convection relations for two horizontal concentric cylinders. The characteristic length in this case is the distance between the two cylinders, which is Lc (Do Di)/2 (4 2)/2 1 in. 1/12 ft We start the calculations by assuming the tube temperature to be 200˚F, and thus an average temperature of (106 200)/2 154˚F 614 R. This gives g(Ti To)L3c Pr v2 2 (32.2 ft/s )1/614 R)(1/12 ft)3 (0.7184) 4.579 104 (2.117 104 ft2/s)2 RaL The effective thermal conductivity is [ln(Do /Di )]4 L3c (D3/5 D3/5 )5 o i [ln(4/2)]4 0.1466 (1/12 ft)3[(2/12 ft)3/5 (4/12 ft)3/5]5 Fcyl 0.861Pr Pr keff 0.386k 1/4 (FcylRaL)1/4 (0.1466 4.579 10 ) 0.8610.7184 0.7184 0.386(0.01653 Btu/h ft ˚F) 4 1/4 0.04743 Btu/h ft ˚F Then the rate of heat transfer between the cylinders becomes 2keff · Q (T To) ln(Do /Di ) i 2(0.04743 Btu/h ft ºF) (200 106)˚F 40.4 Btu/h ln(4/2) which is more than 30 Btu/h. Therefore, the assumed temperature of 200˚F for the tube is high. By trying other values, the tube temperature corresponding to 30 Btu/h is determined to be 180˚F. Therefore, the tube will reach an equilibrium temperature of 180˚F when the pump fails. Discussion Note that we have not considered heat loss by radiation in the calculations, and thus the tube temperature determined above is probably too high. This problem is considered again in Chapter 12 by accounting for the effect of radiation heat transfer. 9–6 ■ COMBINED NATURAL AND FORCED CONVECTION The presence of a temperature gradient in a fluid in a gravity field always gives rise to natural convection currents, and thus heat transfer by natural convection. Therefore, forced convection is always accompanied by natural convection. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 487 487 CHAPTER 9 10 1.0 Nux/Rex1/2 We mentioned earlier that the convection heat transfer coefficient, natural or forced, is a strong function of the fluid velocity. Heat transfer coefficients encountered in forced convection are typically much higher than those encountered in natural convection because of the higher fluid velocities associated with forced convection. As a result, we tend to ignore natural convection in heat transfer analyses that involve forced convection, although we recognize that natural convection always accompanies forced convection. The error involved in ignoring natural convection is negligible at high velocities but may be considerable at low velocities associated with forced convection. Therefore, it is desirable to have a criterion to assess the relative magnitude of natural convection in the presence of forced convection. For a given fluid, it is observed that the parameter Gr/Re2 represents the importance of natural convection relative to forced convection. This is not surprising since the convection heat transfer coefficient is a strong function of the Reynolds number Re in forced convection and the Grashof number Gr in natural convection. A plot of the nondimensionalized heat transfer coefficient for combined natural and forced convection on a vertical plate is given in Fig. 9–32 for different fluids. We note from this figure that natural convection is negligible when Gr/Re2 0.1, forced convection is negligible when Gr/Re2 10, and neither is negligible when 0.1 Gr/Re2 10. Therefore, both natural and forced convection must be considered in heat transfer calculations when the Gr and Re2 are of the same order of magnitude (one is within a factor of 10 times the other). Note that forced convection is small relative to natural convection only in the rare case of extremely low forced flow velocities. Natural convection may help or hurt forced convection heat transfer, depending on the relative directions of buoyancy-induced and the forced convection motions (Fig. 9–33): Experiment Approximate solution Pure forced convection Pure free convection 100 10 Pr = 0.72 (air) 0.1 0.03 0.01 Pr = 0.003 0.01 0.01 0.1 1.0 10 Grx /Rex2 FIGURE 9–32 Variation of the local Nusselt number NUx for combined natural and forced convection from a hot isothermal vertical plate (from Lloyd and Sparrow, Ref. 25). Hot plate Cold plate Buoyant flow Buoyant flow Forced flow (a) Assisting flow Buoyant flow Forced flow Forced flow (b) Opposing flow (c) Transverse flow FIGURE 9–33 Natural convection can enhance or inhibit heat transfer, depending on the relative directions of buoyancy-induced motion and the forced convection motion. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 488 488 HEAT TRANSFER In assisting flow, the buoyant motion is in the same direction as the forced motion. Therefore, natural convection assists forced convection and enhances heat transfer. An example is upward forced flow over a hot surface. 2. In opposing flow, the buoyant motion is in the opposite direction to the forced motion. Therefore, natural convection resists forced convection and decreases heat transfer. An example is upward forced flow over a cold surface. 3. In transverse flow, the buoyant motion is perpendicular to the forced motion. Transverse flow enhances fluid mixing and thus enhances heat transfer. An example is horizontal forced flow over a hot or cold cylinder or sphere. When determining heat transfer under combined natural and forced convection conditions, it is tempting to add the contributions of natural and forced convection in assisting flows and to subtract them in opposing flows. However, the evidence indicates differently. A review of experimental data suggests a correlation of the form Nucombined (Nunforced Nunnatural)1/n (9–41) where Nuforced and Nunatural are determined from the correlations for pure forced and pure natural convection, respectively. The plus sign is for assisting and transverse flows and the minus sign is for opposing flows. The value of the exponent n varies between 3 and 4, depending on the geometry involved. It is observed that n 3 correlates experimental data for vertical surfaces well. Larger values of n are better suited for horizontal surfaces. A question that frequently arises in the cooling of heat-generating equipment such as electronic components is whether to use a fan (or a pump if the cooling medium is a liquid)—that is, whether to utilize natural or forced convection in the cooling of the equipment. The answer depends on the maximum allowable operating temperature. Recall that the convection heat transfer rate from a surface at temperature Ts in a medium at T is given by · Q conv hAs(Ts T) where h is the convection heat transfer coefficient and As is the surface area. Note that for a fixed value of power dissipation and surface area, h and Ts are inversely proportional. Therefore, the device will operate at a higher temperature when h is low (typical of natural convection) and at a lower temperature when h is high (typical of forced convection). Natural convection is the preferred mode of heat transfer since no blowers or pumps are needed and thus all the problems associated with these, such as noise, vibration, power consumption, and malfunctioning, are avoided. Natural convection is adequate for cooling low-power-output devices, especially when they are attached to extended surfaces such as heat sinks. For highpower-output devices, however, we have no choice but to use a blower or a pump to keep the operating temperature below the maximum allowable level. For very-high-power-output devices, even forced convection may not be sufficient to keep the surface temperature at the desirable levels. In such cases, we may have to use boiling and condensation to take advantage of the very high heat transfer coefficients associated with phase change processes. cen58933_ch09.qxd 9/4/2002 12:25 PM Page 489 489 CHAPTER 9 TOPIC OF SCPECIAL INTEREST Heat Transfer Through Windows Windows are glazed apertures in the building envelope that typically consist of single or multiple glazing (glass or plastic), framing, and shading. In a building envelope, windows offer the least resistance to heat flow. In a typical house, about one-third of the total heat loss in winter occurs through the windows. Also, most air infiltration occurs at the edges of the windows. The solar heat gain through the windows is responsible for much of the cooling load in summer. The net effect of a window on the heat balance of a building depends on the characteristics and orientation of the window as well as the solar and weather data. Workmanship is very important in the construction and installation of windows to provide effective sealing around the edges while allowing them to be opened and closed easily. Despite being so undesirable from an energy conservation point of view, windows are an essential part of any building envelope since they enhance the appearance of the building, allow daylight and solar heat to come in, and allow people to view and observe outside without leaving their home. For low-rise buildings, windows also provide easy exit areas during emergencies such as fire. Important considerations in the selection of windows are thermal comfort and energy conservation. A window should have a good light transmittance while providing effective resistance to heat flow. The lighting requirements of a building can be minimized by maximizing the use of natural daylight. Heat loss in winter through the windows can be minimized by using airtight double- or triple-pane windows with spectrally selective films or coatings, and letting in as much solar radiation as possible. Heat gain and thus cooling load in summer can be minimized by using effective internal or external shading on the windows. Even in the absence of solar radiation and air infiltration, heat transfer through the windows is more complicated than it appears to be. This is because the structure and properties of the frame are quite different than the glazing. As a result, heat transfer through the frame and the edge section of the glazing adjacent to the frame is two-dimensional. Therefore, it is customary to consider the window in three regions when analyzing heat transfer through it: (1) the center-of-glass, (2) the edge-of-glass, and (3) the frame regions, as shown in Figure 9–34. Then the total rate of heat transfer through the window is determined by adding the heat transfer through each region as · · · · Q window Q center Q edge Q frame Uwindow Awindow (Tindoors Toutdoors) (9-67) where Uwindow (Ucenter Acenter Uedge Aedge Uframe Aframe)/Awindow Frame Glazing (glass or plastic) Edge of glass Center of glass (9-68) is the U-factor or the overall heat transfer coefficient of the window; Awindow is the window area; Acenter, Aedge, and Aframe are the areas of the This section can be skipped without a loss of continuity. FIGURE 9–34 The three regions of a window considered in heat transfer analysis. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 490 490 HEAT TRANSFER Glass center, edge, and frame sections of the window, respectively; and Ucenter, Uedge, and Uframe are the heat transfer coefficients for the center, edge, and frame sections of the window. Note that Awindow Acenter Aedge Aframe, and the overall U-factor of the window is determined from the areaweighed U-factors of each region of the window. Also, the inverse of the U-factor is the R-value, which is the unit thermal resistance of the window (thermal resistance for a unit area). Consider steady one-dimensional heat transfer through a single-pane glass of thickness L and thermal conductivity k. The thermal resistance network of this problem consists of surface resistances on the inner and outer surfaces and the conduction resistance of the glass in series, as shown in Figure 9–35, and the total resistance on a unit area basis can be expressed as L hi Ti k Rinside Rglass 1 — hi L — k Rtotal Rinside Rglass Routside ho Routside T o 1 — ho 1 Lglass 1 hi kglass ho (9-69) Using common values of 3 mm for the thickness and 0.92 W/m · °C for the thermal conductivity of the glass and the winter design values of 8.29 and 34.0 W/m2 · °C for the inner and outer surface heat transfer coefficients, the thermal resistance of the glass is determined to be 0.003 m 1 1 8.29 W/m2 · °C 0.92 W/m · °C 34.0 W/m2 · °C 0.121 0.003 0.029 0.153 m2 · °C/W Rtotal FIGURE 9–35 The thermal resistance network for heat transfer through a single glass. Glass Air space Ti Rinside Rspace 1 — hi 1 —— hspace Routside T o 1 — ho FIGURE 9–36 The thermal resistance network for heat transfer through the center section of a double-pane window (the resistances of the glasses are neglected). Note that the ratio of the glass resistance to the total resistance is Rglass 0.003 m2 · °C/W 2.0% Rtotal 0.153 m2 · °C/W That is, the glass layer itself contributes about 2 percent of the total thermal resistance of the window, which is negligible. The situation would not be much different if we used acrylic, whose thermal conductivity is 0.19 W/m · °C, instead of glass. Therefore, we cannot reduce the heat transfer through the window effectively by simply increasing the thickness of the glass. But we can reduce it by trapping still air between two layers of glass. The result is a double-pane window, which has become the norm in window construction. The thermal conductivity of air at room temperature is kair 0.025 W/m · °C, which is one-thirtieth that of glass. Therefore, the thermal resistance of 1-cm-thick still air is equivalent to the thermal resistance of a 30cm-thick glass layer. Disregarding the thermal resistances of glass layers, the thermal resistance and U-factor of a double-pane window can be expressed as (Fig. 9–36) 1 1 1 1 Udouble-pane (center region) hi hspace ho (9-70) where hspace hrad, space hconv, space is the combined radiation and convection heat transfer coefficient of the space trapped between the two glass layers. Roughly half of the heat transfer through the air space of a double-pane window is by radiation and the other half is by conduction (or convection, cen58933_ch09.qxd 9/4/2002 12:26 PM Page 491 491 CHAPTER 9 if there is any air motion). Therefore, there are two ways to minimize hspace and thus the rate of heat transfer through a double-pane window: 1. Minimize radiation heat transfer through the air space. This can be done by reducing the emissivity of glass surfaces by coating them with low-emissivity (or “low-e” for short) material. Recall that the effective emissivity of two parallel plates of emissivities 1 and 2 is given by effective 1 1/1 1/2 1 (9-71) The emissivity of an ordinary glass surface is 0.84. Therefore, the effective emissivity of two parallel glass surfaces facing each other is 0.72. But when the glass surfaces are coated with a film that has an emissivity of 0.1, the effective emissivity reduces to 0.05, which is one-fourteenth of 0.72. Then for the same surface temperatures, radiation heat transfer will also go down by a factor of 14. Even if only one of the surfaces is coated, the overall emissivity reduces to 0.1, which is the emissivity of the coating. Thus it is no surprise that about one-fourth of all windows sold for residences have a low-e coating. The heat transfer coefficient hspace for the air space trapped between the two vertical parallel glass layers is given in Table 9–3 for 13-mm- (21 -in.) and 6-mm- (14 -in.) thick air spaces for various effective emissivities and temperature differences. It can be shown that coating just one of the two parallel surfaces facing each other by a material of emissivity reduces the effective emissivity nearly to . Therefore, it is usually more economical to coat only one of the facing surfaces. Note from Figure 9–37 that coating one of the interior surfaces of a double-pane window with a material having an emissivity of 0.1 TABLE 9–3 The heat transfer coefficient hspace for the air space trapped between the two vertical parallel glass layers for 13-mm- and 6-mm-thick air spaces (from Building Materials and Structures, Report 151, U.S. Dept. of Commerce). (a) Air space thickness 13 mm (b) Air space thickness 6 mm 2 Tave, T, °C °C hspace, W/m · °C effective 0.72 0.4 0.2 0.1 0 0 0 5 15 30 5.3 5.3 5.5 3.8 3.8 4.0 2.9 2.9 3.1 2.4 2.4 2.6 10 10 10 5 15 30 5.7 5.7 6.0 4.1 4.1 4.3 3.0 3.1 3.3 2.5 2.5 2.7 30 30 30 5 15 30 5.7 5.7 6.0 4.6 4.7 4.9 3.4 3.4 3.6 2.7 2.8 3.0 Multiply by 0.176 to convert to Btu/h · ft2 · °F. Tave, T, °C °C hspace, W/m2 · °C effective 0.72 0.4 0.2 0.1 0 0 5 50 7.2 7.2 5.7 5.7 4.8 4.8 4.3 4.3 10 10 5 50 7.7 7.7 6.0 6.1 5.0 5.0 4.5 4.5 30 30 5 50 8.8 8.8 6.8 6.8 5.5 5.5 4.9 4.9 50 50 5 50 10.0 10.0 7.5 7.5 6.0 6.0 5.2 5.2 cen58933_ch09.qxd 9/4/2002 12:26 PM Page 492 492 HEAT TRANSFER 4.5 4.5 Gas fill in gap Air Argon Krypton 4 3.5 2 1 0.5 0 3 ε = 0.10 on surface 2 or 3 Inner glass 6 Inner glass Gas fill in gap Air Argon Krypton 2.5 ε = 0.84 2 1 4 Double-pane glazing 2 0 Triple-pane glazing 1.5 Outer glass 1 5 4 Outer glass 3 Center-of-glass U-factor, W/m2·K Center-of-glass U-factor, W/m2·K 2 2.5 1.5 3 3.5 ε = 0.84 3 1 4 5 (a) Double-pane window 10 15 Gap width, mm 0.5 20 25 0 ε = 0.10 on surfaces 2 or 3 and 4 or 5 0 5 10 15 Gap width, mm 20 25 (b) Triple-pane window FIGURE 9–37 The variation of the U-factor for the center section of double- and triple-pane windows with uniform spacing between the panes (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Fig. 1). reduces the rate of heat transfer through the center section of the window by half. 2. Minimize conduction heat transfer through air space. This can be done by increasing the distance d between the two glasses. However, this cannot be done indefinitely since increasing the spacing beyond a critical value initiates convection currents in the enclosed air space, which increases the heat transfer coefficient and thus defeats the purpose. Besides, increasing the spacing also increases the thickness of the necessary framing and the cost of the window. Experimental studies have shown that when the spacing d is less than about 13 mm, there is no convection, and heat transfer through the air is by conduction. But as the spacing is increased further, convection currents appear in the air space, and the increase in heat transfer coefficient offsets any benefit obtained by the thicker air layer. As a result, the heat transfer coefficient remains nearly constant, as shown in Figure 9–37. Therefore, it makes no sense to use an air space thicker than 13 mm in a double-pane window unless a thin polyester film is used to divide the air space into two halves to suppress convection currents. The film provides added insulation without adding much to the weight or cost of the double-pane window. The thermal resistance of the window can be increased further by using triple- or quadruple-pane windows whenever it is economical to do so. Note that using a triple-pane window instead of a double-pane reduces the rate of heat transfer through the center section of the window by about one-third. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 493 493 CHAPTER 9 Another way of reducing conduction heat transfer through a double-pane window is to use a less-conducting fluid such as argon or krypton to fill the gap between the glasses instead of air. The gap in this case needs to be well sealed to prevent the gas from leaking outside. Of course, another alternative is to evacuate the gap between the glasses completely, but it is not practical to do so. Edge-of-Glass U-Factor of a Window Frame U-Factor The framing of a window consists of the entire window except the glazing. Heat transfer through the framing is difficult to determine because of the different window configurations, different sizes, different constructions, and different combination of materials used in the frame construction. The type of glazing such as single pane, double pane, and triple pane affects the thickness of the framing and thus heat transfer through the frame. Most frames are made of wood, aluminum, vinyl, or fiberglass. However, using a combination of these materials (such as aluminum-clad wood and vinylclad aluminum) is also common to improve appearance and durability. Aluminum is a popular framing material because it is inexpensive, durable, and easy to manufacture, and does not rot or absorb water like wood. However, from a heat transfer point of view, it is the least desirable framing material because of its high thermal conductivity. It will come as no surprise that the U-factor of solid aluminum frames is the highest, and thus a window with aluminum framing will lose much more heat than a comparable window with wood or vinyl framing. Heat transfer through the aluminum framing members can be reduced by using plastic inserts between components to serve as thermal barriers. The thickness of these inserts greatly affects heat transfer through the frame. For aluminum frames without the plastic strips, the primary resistance to heat transfer is due to the interior surface heat transfer coefficient. The U-factors for various 5 Edge-of-glass U-factor, W/ m2·K The glasses in double- and triple-pane windows are kept apart from each other at a uniform distance by spacers made of metals or insulators like aluminum, fiberglass, wood, and butyl. Continuous spacer strips are placed around the glass perimeter to provide an edge seal as well as uniform spacing. However, the spacers also serve as undesirable “thermal bridges” between the glasses, which are at different temperatures, and this shortcircuiting may increase heat transfer through the window considerably. Heat transfer in the edge region of a window is two-dimensional, and lab measurements indicate that the edge effects are limited to a 6.5-cm-wide band around the perimeter of the glass. The U-factor for the edge region of a window is given in Figure 9–38 relative to the U-factor for the center region of the window. The curve would be a straight diagonal line if the two U-values were equal to each other. Note that this is almost the case for insulating spacers such as wood and fiberglass. But the U-factor for the edge region can be twice that of the center region for conducting spacers such as those made of aluminum. Values for steel spacers fall between the two curves for metallic and insulating spacers. The edge effect is not applicable to single-pane windows. Spacer type Metallic Insulating Ideal 4 3 2 1 0 0 1 2 3 4 Center-of-glass U-factor, W/m2·K 5 FIGURE 9–38 The edge-of-glass U-factor relative to the center-of-glass U-factor for windows with various spacers (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Fig. 2). cen58933_ch09.qxd 9/4/2002 12:26 PM Page 494 494 HEAT TRANSFER TABLE 9–4 Representative frame U-factors for fixed vertical windows (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 2) U-factor, W/m2 · °C Frame material Aluminum: Single glazing (3 mm) Double glazing (18 mm) Triple glazing (33 mm) 10.1 10.1 10.1 Wood or vinyl: Single glazing (3 mm) Double glazing (18 mm) Triple glazing (33 mm) 2.9 2.8 2.7 frames are listed in Table 9–4 as a function of spacer materials and the glazing unit thicknesses. Note that the U-factor of metal framing and thus the rate of heat transfer through a metal window frame is more than three times that of a wood or vinyl window frame. Interior and Exterior Surface Heat Transfer Coefficients Heat transfer through a window is also affected by the convection and radiation heat transfer coefficients between the glass surfaces and surroundings. The effects of convection and radiation on the inner and outer surfaces of glazings are usually combined into the combined convection and radiation heat transfer coefficients hi and ho, respectively. Under still air conditions, the combined heat transfer coefficient at the inner surface of a vertical window can be determined from hi hconv hrad 1.77(Tg Ti)0.25 g(Tg4 Ti4) Tg Ti Multiply by 0.176 to convert to Btu/h · ft2 · °F (W/m2 · °C) (9-72) where Tg glass temperature in K, Ti indoor air temperature in K, g emissivity of the inner surface of the glass exposed to the room (taken to be 0.84 for uncoated glass), and 5.67 108 W/m2 · K4 is the Stefan– Boltzmann constant. Here the temperature of the interior surfaces facing the window is assumed to be equal to the indoor air temperature. This assumption is reasonable when the window faces mostly interior walls, but it becomes questionable when the window is exposed to heated or cooled surfaces or to other windows. The commonly used value of hi for peak load calculation is hi 8.29 W/m2 · °C 1.46 Btu/h · ft2 · °F TABLE 9–5 Combined convection and radiation heat transfer coefficient hi at the inner surface of a vertical glass under still air conditions (in W/m2 · °C) Glass emissivity, g Ti, °C Tg, °C 0.05 0.20 0.84 20 20 20 20 20 20 20 17 15 10 5 0 5 10 2.6 2.9 3.4 3.7 4.0 4.2 4.4 3.5 3.8 4.2 4.5 4.8 5.0 5.1 7.1 7.3 7.7 7.9 8.1 8.2 8.3 Multiply by 0.176 to convert to Btu/h · ft2 · °F. (winter and summer) which corresponds to the winter design conditions of Ti 22°C and Tg 7°C for uncoated glass with g 0.84. But the same value of hi can also be used for summer design conditions as it corresponds to summer conditions of Ti 24°C and Tg 32°C. The values of hi for various temperatures and glass emissivities are given in Table 9–5. The commonly used values of ho for peak load calculations are the same as those used for outer wall surfaces (34.0 W/m2 · °C for winter and 22.7 W/m2 · °C for summer). Overall U-Factor of Windows The overall U-factors for various kinds of windows and skylights are evaluated using computer simulations and laboratory testing for winter design conditions; representative values are given in Table 9–6. Test data may provide more accurate information for specific products and should be preferred when available. However, the values listed in the table can be used to obtain satisfactory results under various conditions in the absence of product-specific data. The U-factor of a fenestration product that differs considerably from the ones in the table can be determined by (1) determining the fractions of the area that are frame, center-of-glass, and edge-ofglass (assuming a 65-mm-wide band around the perimeter of each glazing), cen58933_ch09.qxd 9/4/2002 12:26 PM Page 495 495 CHAPTER 9 TABLE 9–6 Overall U-factors (heat transfer coefficients) for various windows and skylights in W/m2 · °C (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 5) Aluminum frame (without thermal break) Glass section (glazing) only Type → Frame width → Spacer type → Centerof-glass Edge-ofglass (Not applicable) — Metal Insul. 6.30 5.28 5.79 6.30 5.28 5.79 Double Glazing (no coating) 6.4 mm air space 3.24 12.7 mm air space 2.78 6.4 mm argon space 2.95 12.7 mm argon space 2.61 3.71 3.40 3.52 3.28 Double Fixed door 32 mm 53 mm (141 in.) (2 in.) Sloped skylight 19 mm (34 in.) Wood or vinyl frame Fixed 41 mm (158 in.) Double door 88 mm 7 in.) (318 Sloped skylight 23 mm (87 in.) All All All Metal Insul. Metal Insul. Metal Insul. — — — 6.63 5.69 6.16 7.16 6.27 6.71 9.88 8.86 9.94 5.93 5.02 5.48 — — — 5.57 4.77 5.17 — — — 7.57 6.57 7.63 — — — 3.34 2.91 3.07 2.76 3.90 3.51 3.66 3.36 4.55 4.18 4.32 4.04 6.70 6.65 6.47 6.47 3.26 2.88 3.03 2.74 3.16 2.76 2.91 2.61 3.20 2.86 2.98 2.73 3.09 2.74 2.87 2.60 4.37 4.32 4.14 4.14 4.22 4.17 3.97 3.97 Glazing Type Single Glazing 3 mm (18 in.) glass 6.4 mm (41 in.) acrylic 3 mm (18 in.) acrylic Double Glazing [ 0.1, coating on one inside)] 6.4 mm air space 2.44 3.16 12.7 mm air space 1.82 2.71 6.4 mm argon space 1.99 2.83 12.7 mm argon space 1.53 2.49 Triple Glazing (no coating) 6.4 mm air space 2.16 12.7 mm air space 1.76 6.4 mm argon space 1.93 12.7 mm argon space 1.65 Triple Glazing [ 0.1, coating inside)] 6.4 mm air space 1.53 12.7 mm air space 0.97 6.4 mm argon space 1.19 12.7 mm argon space 0.80 2.96 2.67 2.79 2.58 of the surfaces of air space (surface 2 or 3, counting from the outside toward 2.60 2.06 2.21 1.83 3.21 2.67 2.82 2.42 3.89 3.37 3.52 3.14 6.04 6.04 5.62 5.71 2.59 2.06 2.21 1.82 2.46 1.92 2.07 1.67 2.60 2.13 2.26 1.91 2.47 1.99 2.12 1.78 3.73 3.73 3.32 3.41 3.53 3.53 3.09 3.19 2.35 2.02 2.16 1.92 2.97 2.62 2.77 2.52 3.66 3.33 3.47 3.23 5.81 5.67 5.57 5.53 2.34 2.01 2.15 1.91 2.18 1.84 1.99 1.74 2.36 2.07 2.19 1.98 2.21 1.91 2.04 1.82 3.48 3.34 3.25 3.20 3.24 3.09 3.00 2.95 on one of the surfaces of air spaces (surfaces 3 and 5, counting from the outside toward 2.49 2.05 2.23 1.92 1.83 1.38 1.56 1.25 2.42 1.92 2.12 1.77 3.14 2.66 2.85 2.51 5.24 5.10 4.90 4.86 1.81 1.33 1.52 1.18 1.64 1.15 1.35 1.01 1.89 1.46 1.64 1.33 1.73 1.30 1.47 1.17 2.92 2.78 2.59 2.55 2.66 2.52 2.33 2.28 Notes: (1) Multiply by 0.176 to obtain U-factors in Btu/h · ft2 · °F. (2) The U-factors in this table include the effects of surface heat transfer coefficients and are based on winter conditions of 18°C outdoor air and 21°C indoor air temperature, with 24 km/h (15 mph) winds outdoors and zero solar flux. Small changes in indoor and outdoor temperatures will not affect the overall U-factors much. Windows are assumed to be vertical, and the skylights are tilted 20° from the horizontal with upward heat flow. Insulation spacers are wood, fiberglass, or butyl. Edge-of-glass effects are assumed to extend the 65-mm band around perimeter of each glazing. The product sizes are 1.2 m 1.8 m for fixed windows, 1.8 m 2.0 m for double-door windows, and 1.2 m 0.6 m for the skylights, but the values given can also be used for products of similar sizes. All data are based on 3-mm (18 -in.) glass unless noted otherwise. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 496 496 HEAT TRANSFER (2) determining the U-factors for each section (the center-of-glass and edge-of-glass U-factors can be taken from the first two columns of Table 9–6 and the frame U-factor can be taken from Table 9–5 or other sources), and (3) multiplying the area fractions and the U-factors for each section and adding them up (or from Eq. 9-68 for Uwindow). Glazed wall systems can be treated as fixed windows. Also, the data for double-door windows can be used for single-glass doors. Several observations can be made from the data in the table: 1. Skylight U-factors are considerably greater than those of vertical windows. This is because the skylight area, including the curb, can be 13 to 240 percent greater than the rough opening area. The slope of the skylight also has some effect. 2. The U-factor of multiple-glazed units can be reduced considerably by filling cavities with argon gas instead of dry air. The performance of CO2-filled units is similar to those filled with argon. The U-factor can be reduced even further by filling the glazing cavities with krypton gas. 3. Coating the glazing surfaces with low-e (low-emissivity) films reduces the U-factor significantly. For multiple-glazed units, it is adequate to coat one of the two surfaces facing each other. 4. The thicker the air space in multiple-glazed units, the lower the U-factor, for a thickness of up to 13 mm (12 in.) of air space. For a specified number of glazings, the window with thicker air layers will have a lower U-factor. For a specified overall thickness of glazing, the higher the number of glazings, the lower the U-factor. Therefore, a triple-pane window with air spaces of 6.4 mm (two such air spaces) will have a lower U-value than a double-pane window with an air space of 12.7 mm. 5. Wood or vinyl frame windows have a considerably lower U-value than comparable metal-frame windows. Therefore, wood or vinyl frame windows are called for in energy-efficient designs. EXAMPLE 9–7 Glass ε = 0.84 Air space 1 — hi 1 —— hspace 1 — ho 6 mm FIGURE 9–39 Schematic of Example 9–7. U-Factor for Center-of-Glass Section of Windows Determine the U-factor for the center-of-glass section of a double-pane window with a 6-mm air space for winter design conditions (Fig. 9–39). The glazings are made of clear glass that has an emissivity of 0.84. Take the average air space temperature at design conditions to be 0°C. SOLUTION The U-factor for the center-of-glass section of a double-pane window is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 The thermal resistance of glass sheets is negligible. Properties The emissivity of clear glass is 0.84. Analysis Disregarding the thermal resistance of glass sheets, which are small, the U-factor for the center region of a double-pane window is determined from 1 1 1 1 Ucenter hi hspace ho cen58933_ch09.qxd 9/4/2002 12:26 PM Page 497 497 CHAPTER 9 where hi, hspace, and ho are the heat transfer coefficients at the inner surface of the window, the air space between the glass layers, and the outer surface of the window, respectively. The values of hi and ho for winter design conditions were given earlier to be hi 8.29 W/m2 · °C and ho 34.0 W/m2 · °C. The effective emissivity of the air space of the double-pane window is effective 1 1 0.72 1/1 1/2 1 1/0.84 1/0.84 1 For this value of emissivity and an average air space temperature of 0°C, we read hspace 7.2 W/m2 · °C from Table 9–3 for 6-mm-thick air space. Therefore, 1 1 1 1 → Ucenter 3.46 W/m2 · °C Ucenter 8.29 7.2 34.0 Discussion The center-of-glass U-factor value of 3.24 W/m2 · °C in Table 9–6 (fourth row and second column) is obtained by using a standard value of ho 29 W/m2 · °C (instead of 34.0 W/m2 · °C) and hspace 6.5 W/m2 · °C at an average air space temperature of 15°C. EXAMPLE 9-8 Heat Loss through Aluminum Framed Windows A fixed aluminum-framed window with glass glazing is being considered for an opening that is 4 ft high and 6 ft wide in the wall of a house that is maintained at 72°F (Fig. 9–40). Determine the rate of heat loss through the window and the inner surface temperature of the window glass facing the room when the outdoor air temperature is 15°F if the window is selected to be (a) 18-in. single glazing, (b) double glazing with an air space of 21 in., and (c) low-e-coated triple glazing with an air space of 12 in. SOLUTION The rate of heat loss through an aluminum framed window and the inner surface temperature are to be determined from the cases of single-pane, double-pane, and low-e triple-pane windows. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties of the windows and the heat transfer coefficients are constant. Properties The U-factors of the windows are given in Table 9–6. Analysis The rate of heat transfer through the window can be determined from · Q window Uoverall Awindow(Ti To) where Ti and To are the indoor and outdoor air temperatures, respectively; Uoverall is the U-factor (the overall heat transfer coefficient) of the window; and Awindow is the window area, which is determined to be Awindow Height Width (4 ft)(6 ft) 24 ft2 The U-factors for the three cases can be determined directly from Table 9–6 to be 6.63, 3.51, and 1.92 W/m2 · °C, respectively, to be multiplied by the factor 0.176 to convert them to Btu/h · ft2 · °F. Also, the inner surface temperature of the window glass can be determined from Newton’s law 6 ft 4 ft Aluminum frame Glazing (a) Single (b) Double (c) Low-e triple FIGURE 9–40 Schematic for Example 9–8. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 498 498 HEAT TRANSFER · Q window · Q window hi Awindow (Ti Tglass) → Tglass Ti hi A window where hi is the heat transfer coefficient on the inner surface of the window, which is determined from Table 9-5 to be hi 8.3 W/m2 · °C 1.46 Btu/h · ft2 · °F. Then the rate of heat loss and the interior glass temperature for each case are determined as follows: (a) Single glazing: · Q window (6.63 0.176 Btu/h · ft2 · °F)(24 ft2)(72 15)°F 1596 Btu/h Q· window 1596 Btu/h Tglass Ti 72°F 26.5°F hi Awindow (1.46 Btu/h · ft2 · °F)(24 ft2) (b) Double glazing (12 in. air space): · Q window (3.51 0.176 Btu/h · ft2 · °F)(24 ft2)(72 15)°F 845 Btu/h Tglass Ti · Q window 845 Btu/h 72°F 47.9°F hi Awindow (1.46 Btu/h · ft2 · °F)(24 ft2) (c) Triple glazing (12 in. air space, low-e coated): · Q window (1.92 0.176 Btu/h · ft2 · °F)(24 ft2)(72 15)°F 462 Btu/h Tglass Ti · Q window 462 Btu/h 72°F 58.8°F hi Awindow (1.46 Btu/h · ft2 · °F)(24 ft2) Therefore, heat loss through the window will be reduced by 47 percent in the case of double glazing and by 71 percent in the case of triple glazing relative to the single-glazing case. Also, in the case of single glazing, the low inner-glass surface temperature will cause considerable discomfort in the occupants because of the excessive heat loss from the body by radiation. It is raised from 26.5°F, which is below freezing, to 47.9°F in the case of double glazing and to 58.8°F in the case of triple glazing. Frame Edge of glass Center of glass EXAMPLE 9–9 U-Factor of a Double-Door Window Determine the overall U-factor for a double-door-type, wood-framed doublepane window with metal spacers, and compare your result to the value listed in Table 9–6. The overall dimensions of the window are 1.80 m 2.00 m, and the dimensions of each glazing are 1.72 m 0.94 m (Fig. 9–41). 1.8 m 1.72 m 0.94 m 0.94 m 2m FIGURE 9–41 Schematic for Example 9–9. SOLUTION The overall U-factor for a double-door type window is to be determined, and the result is to be compared to the tabulated value. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. Properties The U-factors for the various sections of windows are given in Tables 9–4 and 9–6. Analysis The areas of the window, the glazing, and the frame are Awindow Height Width (1.8 m)(2.0 m) 3.60 m2 cen58933_ch09.qxd 9/4/2002 12:26 PM Page 499 499 CHAPTER 9 Aglazing 2 (Height Width) 2(1.72 m)(0.94 m) 3.23 m2 Aframe Awindow Aglazing 3.60 3.23 0.37 m2 The edge-of-glass region consists of a 6.5-cm-wide band around the perimeter of the glazings, and the areas of the center and edge sections of the glazing are determined to be Acenter 2 (Height Width) 2(1.72 0.13 m)(0.94 0.13 m) 2.58 m2 Aedge Aglazing Acenter 3.23 2.58 0.65 m2 The U-factor for the frame section is determined from Table 9–4 to be Uframe 2.8 W/m2 · °C. The U-factors for the center and edge sections are determined from Table 9–6 (fifth row, second and third columns) to be Ucenter 3.24 W/m2 · °C and Uedge 3.71 W/m2 · °C. Then the overall U-factor of the entire window becomes Uwindow (Ucenter Acenter Uedge Aedge Uframe Aframe)/Awindow (3.24 2.58 3.71 0.65 2.8 0.37)/3.60 3.28 W/m2 · °C The overall U-factor listed in Table 9–6 for the specified type of window is 3.20 W/m2 · °C, which is sufficiently close to the value obtained above. SUMMARY In this chapter, we have considered natural convection heat transfer where any fluid motion occurs by natural means such as buoyancy. The volume expansion coefficient of a substance represents the variation of the density of that substance with temperature at constant pressure, and for an ideal gas, it is expressed as 1/T, where T is the absolute temperature in K or R. The flow regime in natural convection is governed by a dimensionless number called the Grashof number, which represents the ratio of the buoyancy force to the viscous force acting on the fluid and is expressed as GrL g(Ts T)L3c v2 where Lc is the characteristic length, which is the height L for a vertical plate and the diameter D for a horizontal cylinder. The correlations for the Nusselt number Nu hLc /k in natural convection are expressed in terms of the Rayleigh number defined as g(Ts T)L3c Pr RaL GrL Pr v2 Nusselt number relations for various surfaces are given in Table 9–1. All fluid properties are evaluated at the film tempera- ture of Tf 12(Ts T). The outer surface of a vertical cylinder can be treated as a vertical plate when the curvature effects are negligible. The characteristic length for a horizontal surface is Lc As/p, where As is the surface area and p is the perimeter. The average Nusselt number for vertical isothermal parallel plates of spacing S and height L is given as Nu hS 576 2.873 (Ra S S/L)2 (Ra S S /L)0.5 k 0.5 The optimum fin spacing for a vertical heat sink and the Nusselt number for optimally spaced fins is Sopt 2.714 S L Ra 3 0.25 S 2.714 hSopt L and Nu 1.307 0.25 k Ra L In a horizontal rectangular enclosure with the hotter plate at the top, heat transfer is by pure conduction and Nu 1. When the hotter plate is at the bottom, the Nusselt is Nu 1 1.44 1 1708 Ra L Ra 1/3 L 1 18 RaL 108 The notation [ ] indicates that if the quantity in the bracket is negative, it should be set equal to zero. For vertical horizontal enclosures, the Nusselt number can be determined from cen58933_ch09.qxd 9/4/2002 12:26 PM Page 500 500 HEAT TRANSFER Nu 0.18 1 H/L 2 any Prandtl number RaL Pr/(0.2 Pr) 10 3 Pr Ra 0.2 Pr L 0.29 Pr Nu 0.22 Ra 0.2 Pr L 0.28 H L 1/4 2 H/L 10 any Prandtl number RaL 1010 For aspect ratios greater than 10, Eqs. 9-54 and 9-55 should be used. For inclined enclosures, Eqs. 9-48 through 9-51 should be used. For concentric horizontal cylinders, the rate of heat transfer through the annular space between the cylinders by natural convection per unit length is 2k eff · Q (T To) ln(Do /Di ) i where keff Pr 0.386 k 0.861 Pr 1/4 (FcylRaL)1/4 and Fcyl [ln(Do /Di )]4 L3c (D 3/5 D3/5 )5 i o For a spherical enclosure, the rate of heat transfer through the space between the spheres by natural convection is expressed as Di Do · Q keff (Ti To) Lc where keff Pr 0.74 k 0.861 Pr 1/4 (FsphRaL)1/4 Lc (Do Di)/2 Lc Fsph (Di Do ) 4(D 7/5 D7/5 )5 i o The quantity kNu is called the effective thermal conductivity of the enclosure, since a fluid in an enclosure behaves like a quiescent fluid whose thermal conductivity is kNu as a result of convection currents. The fluid properties are evaluated at the average temperature of (Ti To)/2. For a given fluid, the parameter Gr/Re2 represents the importance of natural convection relative to forced convection. Natural convection is negligible when Gr/Re2 0.1, forced convection is negligible when Gr/Re2 10, and neither is negligible when 0.1 Gr/Re2 10. REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers. Handbook of Fundamentals. Atlanta: ASHRAE, 1993. 2. J. N Arnold, I. Catton, and D. K. Edwards. “Experimental Investigation of Natural Convection in Inclined Rectangular Region of Differing Aspects Ratios.” ASME Paper No. 75-HT-62, 1975. 3. P. S. Ayyaswamy and I. Catton. “The Boundary-Layer Regime for Natural Convection in a Differently Heated Tilted Rectangular Cavity.” Journal of Heat Transfer 95 (1973), p. 543. 4. A. Bar-Cohen. “Fin Thickness for an Optimized Natural Convection Array of Rectangular Fins.” Journal of Heat Transfer 101 (1979), pp. 564–566. 5. A. Bar-Cohen and W. M. Rohsenow. “Thermally Optimum Spacing of Vertical Natural Convection Cooled Parallel Plates.” Journal of Heat Transfer 106 (1984), p. 116. 6. B. M. Berkovsky and V. K. Polevikov. “Numerical Study of Problems on High-Intensive Free Convection.” In Heat Transfer and Turbulent Buoyant Convection, ed. D. B. Spalding and N. Afgan, pp. 443–445. Washington, DC: Hemisphere, 1977. 7. I. Catton. “Natural Convection in Enclosures.” Proceedings of Sixth International Heat Transfer Conference, Toronto, Canada, 1978, Vol. 6, pp. 13–31. 8. T. Cebeci. “Laminar Free Convection Heat Transfer from the Outer Surface of a Vertical Slender Circular Cylinder.” Proceedings Fifth International Heat Transfer Conference paper NCI.4, 1974 pp. 15–19. 9. Y. A. Çengel and P. T. L. Zing. “Enhancement of Natural Convection Heat Transfer from Heat Sinks by Shrouding.” Proceedings of ASME/JSME Thermal Engineering Conference, Honolulu, HA, 1987, Vol. 3, pp. 451–475. 10. S. W. Churchill. “A Comprehensive Correlating Equation for Laminar Assisting Forced and Free Convection.” AIChE Journal 23 (1977), pp. 10–16. 11. S. W. Churchill. “Free Convection Around Immersed Bodies.” In Heat Exchanger Design Handbook, ed. E. U. Schlünder, Section 2.5.7. New York: Hemisphere, 1983. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 501 501 CHAPTER 9 S. W. Churchill. “Combined Free and Forced Convection around Immersed Bodies.” In Heat Exchanger Design Handbook, Section 2.5.9. New York: Hemisphere Publishing, 1986. W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993. S. W. Churchill and H. H. S. Chu. “Correlating Equations for Laminar and Turbulent Free Convection from a Horizontal Cylinder.” International Journal of Heat Mass Transfer 18 (1975), p. 1049. J. R. Lloyd and E. M. Sparrow. “Combined Forced and Free Convection Flow on Vertical Surfaces.” International Journal of Heat Mass Transfer 13 (1970), p. 434. S. W. Churchill and H. H. S. Chu. “Correlating Equations for Laminar and Turbulent Free Convection from a Vertical Plate.” International Journal of Heat Mass Transfer 18 (1975), p. 1323. R. K. Macgregor and A. P. Emery. “Free Convection Through Vertical Plane Layers: Moderate and High Prandtl Number Fluids.” Journal of Heat Transfer 91 (1969), p. 391. E. R. G. Eckert and E. Soehngen. “Studies on Heat Transfer in Laminar Free Convection with Zehnder–Mach Interferometer.” USAF Technical Report 5747, December 1948. S. Ostrach. “An Analysis of Laminar Free Convection Flow and Heat Transfer About a Flat Plate Parallel to the Direction of the Generating Body Force.” National Advisory Committee for Aeronautics, Report 1111, 1953. E. R. G. Eckert and E. Soehngen. “Interferometric Studies on the Stability and Transition to Turbulence of a Free Convection Boundary Layer.” Proceedings of General Discussion, Heat Transfer ASME-IME, London, 1951. G. D. Raithby and K. G. T. Hollands. “A General Method of Obtaining Approximate Solutions to Laminar and Turbulent Free Convection Problems.” In Advances in Heat Transfer, ed. F. Irvine and J. P. Hartnett, Vol. II, pp. 265–315. New York: Academic Press, 1975. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. S. M. ElSherbiny, G. D. Raithby, and K. G. T. Hollands. “Heat Transfer by Natural Convection Across Vertical and Inclined Air Layers. Journal of Heat Transfer 104 (1982), pp. 96–102. E. M. Sparrow and J. L. Gregg. “Laminar Free Convection from a Vertical Flat Plate.” Transactions of the ASME 78 (1956), p. 438. T. Fujiii and H. Imura. “Natural Convection Heat Transfer from a Plate with Arbitrary Inclination.” International Journal of Heat Mass Transfer 15 (1972), p. 755. E. M. Sparrow and J. L. Gregg. “Laminar Free Convection Heat Transfer from the Outer Surface of a Vertical Circular Cylinder.” ASME 78 (1956), p. 1823. K. G. T. Hollands, T. E. Unny, G. D. Raithby, and L. Konicek. “Free Convective Heat Transfer Across Inclined Air Layers.” Journal of Heat Transfer 98 (1976), pp. 189–193. E. M. Sparrow and C. Prakash. “Enhancement of Natural Convection Heat Transfer by a Staggered Array of Vertical Plates.” Journal of Heat Transfer 102 (1980), pp. 215–220. J. P. Holman. Heat Transfer. 7th ed. New York: McGrawHill, 1990. E. M. Sparrow and S. B. Vemuri. “Natural Convection/Radiation Heat Transfer from Highly Populated Pin Fin Arrays.” Journal of Heat Transfer 107 (1985), pp. 190–197. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 3rd ed. New York: John Wiley & Sons, 1996. 22. M. Jakob. Heat Transfer. New York: Wiley, 1949. PROBLEMS Physical Mechanism of Natural Convection 9–1C What is natural convection? How does it differ from forced convection? What force causes natural convection currents? 9–2C In which mode of heat transfer is the convection heat transfer coefficient usually higher, natural convection or forced convection? Why? 9–3C Consider a hot boiled egg in a spacecraft that is filled with air at atmospheric pressure and temperature at all times. Will the egg cool faster or slower when the spacecraft is in space instead of on the ground? Explain. Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. cen58933_ch09.qxd 9/4/2002 12:26 PM Page 502 502 HEAT TRANSFER 9–4C What is buoyancy force? Compare the relative magnitudes of the buoyancy force acting on a body immersed in these mediums: (a) air, (b) water, (c) mercury, and (d) an evacuated chamber. regarding any heat transfer from the base surface, determine the surface temperature of the transistor. Use air properties at 100°C. Answer: 183°C 9–5C When will the hull of a ship sink in water deeper: when the ship is sailing in fresh water or in sea water? Why? 35°C 9–6C A person weighs himself on a waterproof spring scale placed at the bottom of a 1-m-deep swimming pool. Will the person weigh more or less in water? Why? Power transistor 0.18 W ε = 0.1 9–7C Consider two fluids, one with a large coefficient of volume expansion and the other with a small one. In what fluid will a hot surface initiate stronger natural convection currents? Why? Assume the viscosity of the fluids to be the same. 9–8C Consider a fluid whose volume does not change with temperature at constant pressure. What can you say about natural convection heat transfer in this medium? 9–9C What do the lines on an interferometer photograph represent? What do closely packed lines on the same photograph represent? 9–10C Physically, what does the Grashof number represent? How does the Grashof number differ from the Reynolds number? 9–11 Show that the volume expansion coefficient of an ideal gas is 1/T, where T is the absolute temperature. Natural Convection over Surfaces 9–12C How does the Rayleigh number differ from the Grashof number? 9–13C Under what conditions can the outer surface of a vertical cylinder be treated as a vertical plate in natural convection calculations? 9–14C Will a hot horizontal plate whose back side is insulated cool faster or slower when its hot surface is facing down instead of up? 9–15C Consider laminar natural convection from a vertical hot plate. Will the heat flux be higher at the top or at the bottom of the plate? Why? 9–16 A 10-m-long section of a 6-cm-diameter horizontal hot water pipe passes through a large room whose temperature is 22°C. If the temperature and the emissivity of the outer surface of the pipe are 65°C and 0.8, respectively, determine the rate of heat loss from the pipe by (a) natural convection and (b) radiation. 9–17 Consider a wall-mounted power transistor that dissipates 0.18 W of power in an environment at 35°C. The transistor is 0.45 cm long and has a diameter of 0.4 cm. The emissivity of the outer surface of the transistor is 0.1, and the average temperature of the surrounding surfaces is 25°C. Dis- 0.4 cm 0.45 cm FIGURE P9–17 9–18 Reconsider Problem 9–17. Using EES (or other) software, investigate the effect of ambient temperature on the surface temperature of the transistor. Let the environment temperature vary from 10˚C to 40˚C and assume that the surrounding surfaces are 10˚C colder than the environment temperature. Plot the surface temperature of the transistor versus the environment temperature, and discuss the results. 9–19E Consider a 2-ft 2-ft thin square plate in a room at 75°F. One side of the plate is maintained at a temperature of 130°F, while the other side is insulated. Determine the rate of heat transfer from the plate by natural convection if the plate is (a) vertical, (b) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. 9–20E Reconsider Problem 9–19E. Using EES (or other) software, plot the rate of natural convection heat transfer for different orientations of the plate as a function of the plate temperature as the temperature varies from 80˚F to 180˚F, and discuss the results. 9–21 A 400-W cylindrical resistance heater is 1 m long and 0.5 cm in diameter. The resistance wire is placed horizontally in a fluid at 20°C. Determine the outer surface temperature of the resistance wire in steady operation if the fluid is (a) air and (b) water. Ignore any heat transfer by radiation. Use properties at 500°C for air and 40°C for water. 9–22 Water is boiling in a 12-cm-deep pan with an outer diameter of 25 cm that is placed on top of a stove. The ambient air and the surrounding surfaces are at a temperature of 25°C, and the emissivity of the outer surface of the pan is 0.95. Assuming the entire pan to be at an average temperature of 98°C, determine the rate of heat loss from the cylindrical side surface of the pan to the surroundings by (a) natural convection and (b) radiation. (c) If water is boiling at a rate of 2 kg/h at 100°C, cen58933_ch09.qxd 9/4/2002 12:26 PM Page 503 503 CHAPTER 9 determine the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water. The heat of vaporization of water at 100°C is 2257 kJ/kg. Answers: 46.2 W, 56.1 W, 0.082 Vapor 2 kg/h 25°C Water 100°C 98°C ε = 0.95 FIGURE P9–22 radiation. An estimate is obtained from a local insulation contractor, who proposes to do the insulation job for $350, including materials and labor. Would you support this proposal? How long will it take for the insulation to pay for itself from the energy it saves? 9–26 Consider a 15-cm 20-cm printed circuit board (PCB) that has electronic components on one side. The board is placed in a room at 20°C. The heat loss from the back surface of the board is negligible. If the circuit board is dissipating 8 W of power in steady operation, determine the average temperature of the hot surface of the board, assuming the board is (a) vertical, (b) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. Take the emissivity of the surface of the board to be 0.8 and assume the surrounding surfaces to be at the same temperature as the air Answers: (a) 46.6°C, (b) 42.6°C, (c) 50.7°C in the room. 9–23 Repeat Problem 9–22 for a pan whose outer surface is polished and has an emissivity of 0.1. 9–24 In a plant that manufactures canned aerosol paints, the cans are temperature-tested in water baths at 55°C before they are shipped to ensure that they will withstand temperatures up to 55°C during transportation and shelving. The cans, moving on a conveyor, enter the open hot water bath, which is 0.5 m deep, 1 m wide, and 3.5 m long, and move slowly in the hot water toward the other end. Some of the cans fail the test and explode in the water bath. The water container is made of sheet metal, and the entire container is at about the same temperature as the hot water. The emissivity of the outer surface of the container is 0.7. If the temperature of the surrounding air and surfaces is 20°C, determine the rate of heat loss from the four side surfaces of the container (disregard the top surface, which is open). The water is heated electrically by resistance heaters, and the cost of electricity is $0.085/kWh. If the plant operates 24 h a day 365 days a year and thus 8760 h a year, determine the annual cost of the heat losses from the container for this facility. Aerosol can Water bath 55°C FIGURE P9–24 9–25 Reconsider Problem 9–24. In order to reduce the heating cost of the hot water, it is proposed to insulate the side and bottom surfaces of the container with 5-cm-thick fiberglass insulation (k 0.035 W/m · °C) and to wrap the insulation with aluminum foil ( 0.1) in order to minimize the heat loss by FIGURE P9–26 9–27 Reconsider Problem 9–26. Using EES (or other) software, investigate the effects of the room temperature and the emissivity of the board on the temperature of the hot surface of the board for different orientations of the board. Let the room temperature vary from 5˚C to 35˚C and the emissivity from 0.1 to 1.0. Plot the hot surface temperature for different orientations of the board as the functions of the room temperature and the emissivity, and discuss the results. 9–28 A manufacturer makes absorber plates that are 1.2 m 0.8 m in size for use in solar collectors. The back side of the plate is heavily insulated, while its front surface is coated with black chrome, which has an absorptivity of 0.87 for solar radiation and an emissivity of 0.09. Consider such a plate placed horizontally outdoors in calm air at 25°C. Solar radiation is incident on the plate at a rate of 700 W/m2. Taking the effective sky temperature to be 10°C, determine the equilibrium temperature of the absorber plate. What would your answer be if the absorber plate is made of ordinary aluminum plate that has a solar absorptivity of 0.28 and an emissivity of 0.07? cen58933_ch09.qxd 9/4/2002 12:26 PM Page 504 504 HEAT TRANSFER electricity associated with heating the pipe during a 10-h period under the above conditions if the price of electricity is Answers: 29.1 kW, $26.2 $0.09/kWh. Solar radiation 700 W/m2 Tsky = –30°C Absorber plate αs = 0.87 ε = 0.09 0°C 30 cm Ts = 25°C ε = 0.8 Insulation FIGURE P9–28 Asphalt 9–29 Repeat Problem 9–28 for an aluminum plate painted flat black (solar absorptivity 0.98 and emissivity 0.98) and also for a plate painted white (solar absorptivity 0.26 and emissivity 0.90). 9–30 The following experiment is conducted to determine the natural convection heat transfer coefficient for a horizontal cylinder that is 80 cm long and 2 cm in diameter. A 80-cm-long resistance heater is placed along the centerline of the cylinder, and the surfaces of the cylinder are polished to minimize the radiation effect. The two circular side surfaces of the cylinder are well insulated. The resistance heater is turned on, and the power dissipation is maintained constant at 40 W. If the average surface temperature of the cylinder is measured to be 120°C in the 20°C room air when steady operation is reached, determine the natural convection heat transfer coefficient. If the emissivity of the outer surface of the cylinder is 0.1 and a 5 percent error is acceptable, do you think we need to do any correction for the radiation effect? Assume the surrounding surfaces to be at 20°C also. 120°C 20°C Insulated Resistance heater 40 W Insulated Resistance heater FIGURE P9–31 9–32 Reconsider Problem 9–31. To reduce the heating cost of the pipe, it is proposed to insulate it with sufficiently thick fiberglass insulation (k 0.035 W/m · °C) wrapped with aluminum foil ( 0.1) to cut down the heat losses by 85 percent. Assuming the pipe temperature to remain constant at 25°C, determine the thickness of the insulation that needs to be used. How much money will the insulation save during this 10-h Answers: 1.3 cm, $22.3 period? 9–33E Consider an industrial furnace that resembles a 13-ftlong horizontal cylindrical enclosure 8 ft in diameter whose end surfaces are well insulated. The furnace burns natural gas at a rate of 48 therms/h (1 therm 100,000 Btu). The combustion efficiency of the furnace is 82 percent (i.e., 18 percent of the chemical energy of the fuel is lost through the flue gases as a result of incomplete combustion and the flue gases leaving the furnace at high temperature). If the heat loss from the outer surfaces of the furnace by natural convection and radiation is not to exceed 1 percent of the heat generated inside, determine the highest allowable surface temperature of the furnace. Assume the air and wall surface temperature of the room to be 75°F, and take the emissivity of the outer surface of the furnace to be 0.85. If the cost of natural gas is $0.65/therm and the furnace operates 2800 h per year, determine the annual cost of this heat loss to the plant. 75°F FIGURE P9–30 9–31 Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the fluids and thus to reduce the pumping costs. Consider the flow of such a fluid through a 100-m-long pipe of outer diameter 30 cm in calm ambient air at 0°C. The pipe is heated electrically, and a thermostat keeps the outer surface temperature of the pipe constant at 25°C. The emissivity of the outer surface of the pipe is 0.8, and the effective sky temperature is 30°C, Determine the power rating of the electric resistance heater, in kW, that needs to be used. Also, determine the cost of Furnace ε = 0.85 Ts = ? 8 ft 13 ft FIGURE P9–33 cen58933_ch09.qxd 9/4/2002 12:26 PM Page 505 505 CHAPTER 9 9–34 Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermal conductivity k 0.78 W/m · °C, and emissivity 0.9. The room and the walls that face the window are maintained at 25°C, and the average temperature of the inner surface of the window is measured to be 5°C. If the temperature of the outdoors is 5°C, determine (a) the convection heat transfer coefficient on the inner surface of the window, (b) the rate of total heat transfer through the window, and (c) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case? Wall Room 25°C Glass 1.2 m – 5°C 5°C ε = 0.9 FIGURE P9–34 9–35 A 3-mm-diameter and 12-m-long electric wire is tightly wrapped with a 1.5-mm-thick plastic cover whose thermal conductivity and emissivity are k 0.15 W/m · °C and 0.9. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to calm atmospheric air at T 30°C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Take the surrounding surfaces to be at about the same temperature as the air. 9–36 During a visit to a plastic sheeting plant, it was observed that a 60-m-long section of a 2-in. nominal (6.03-cm outer-diameter) steam pipe extended from one end of the plant to the other with no insulation on it. The temperature measurements at several locations revealed that the average temperature of the exposed surfaces of the steam pipe was 170°C, while the temperature of the surrounding air was 20°C. The outer surface of the pipe appeared to be oxidized, and its emissivity can be taken to be 0.7. Taking the temperature of the surrounding surfaces to be 20°C also, determine the rate of heat loss from the steam pipe. Steam is generated in a gas furnace that has an efficiency of 78 percent, and the plant pays $0.538 per therm (1 therm 105,500 kJ) of natural gas. The plant operates 24 h a day 365 days a year, and thus 8760 h a year. Determine the annual cost of the heat losses from the steam pipe for this facility. 20°C 170°C ε = 0.7 m 3c 6.0 60 m Steam FIGURE P9–36 9–37 Reconsider Problem 9–36. Using EES (or other) software, investigate the effect of the surface temperature of the steam pipe on the rate of heat loss from the pipe and the annual cost of this heat loss. Let the surface temperature vary from 100˚C to 200˚C. Plot the rate of heat loss and the annual cost as a function of the surface temperature, and discuss the results. 9–38 Reconsider Problem 9–36. In order to reduce heat losses, it is proposed to insulate the steam pipe with 5-cm-thick fiberglass insulation (k 0.038 W/m · °C) and to wrap it with aluminum foil ( 0.1) in order to minimize the radiation losses. Also, an estimate is obtained from a local insulation contractor, who proposed to do the insulation job for $750, including materials and labor. Would you support this proposal? How long will it take for the insulation to pay for itself from the energy it saves? Assume the temperature of the steam pipe to remain constant at 170°C. 9–39 A 30-cm 30-cm circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at 25°C. Each chip dissipates 0.05 W of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperAnswer: 33.4°C ature of the chips. 9–40 Repeat Prob. 9–35 assuming the circuit board to be positioned horizontally with (a) chips facing up and (b) chips facing down. 9–41 The side surfaces of a 2-m-high cubic industrial furnace burning natural gas are not insulated, and the temperature at the outer surface of this section is measured to be 110°C. The temperature of the furnace room, including its surfaces, is 30°C, and the emissivity of the outer surface of the furnace is 0.7. It is proposed that this section of the furnace wall be insulated with glass wool insulation (k 0.038 W/m · °C) wrapped by a reflective sheet ( 0.2) in order to reduce the heat loss by 90 percent. Assuming the outer surface temperature of the metal section still remains at about 110°C, determine the thickness of the insulation that needs to be used. The furnace operates continuously throughout the year and has an efficiency of 78 percent. The price of the natural gas is cen58933_ch09.qxd 9/4/2002 12:26 PM Page 506 506 HEAT TRANSFER $0.55/therm (1 therm 105,500 kJ of energy content). If the installation of the insulation will cost $550 for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves. Hot gases 30°C 2m Furnace 110°C ε = 0.7 9–44 An incandescent lightbulb is an inexpensive but highly inefficient device that converts electrical energy into light. It converts about 10 percent of the electrical energy it consumes into light while converting the remaining 90 percent into heat. The glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by convection and radiation. Consider an 8-cm-diameter 60-W light bulb in a room at 25°C. The emissivity of the glass is 0.9. Assuming that 10 percent of the energy passes through the glass bulb as light with negligible absorption and the rest of the energy is absorbed and dissipated by the bulb itself by natural convection and radiation, determine the equilibrium temperature of the glass bulb. Assume the interior surfaces of the room Answer: 169°C to be at room temperature. 25°C 2m 2m FIGURE P9–41 9–42 A 1.5-m-diameter, 5-m-long cylindrical propane tank is initially filled with liquid propane, whose density is 581 kg/m3. The tank is exposed to the ambient air at 25°C in calm weather. The outer surface of the tank is polished so that the radiation heat transfer is negligible. Now a crack develops at the top of the tank, and the pressure inside drops to 1 atm while the temperature drops to 42°C, which is the boiling temperature of propane at 1 atm. The heat of vaporization of propane at 1 atm is 425 kJ/kg. The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at 42°C through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the tank to empty if it is not insulated. Propane vapor 25°C 1.5 m Propane tank – 42°C 4m FIGURE P9–42 9–43E An average person generates heat at a rate of 287 Btu/h while resting in a room at 77°F. Assuming one-quarter of this heat is lost from the head and taking the emissivity of the skin to be 0.9, determine the average surface temperature of the head when it is not covered. The head can be approximated as a 12-in.-diameter sphere, and the interior surfaces of the room can be assumed to be at the room temperature. 60 W ε = 0.9 Light, 6 W FIGURE P9–44 9–45 A 40-cm-diameter, 110-cm-high cylindrical hot water tank is located in the bathroom of a house maintained at 20˚C. The surface temperature of the tank is measured to be 44˚C and its emissivity is 0.4. Taking the surrounding surface temperature to be also 20˚C, determine the rate of heat loss from all surfaces of the tank by natural convection and radiation. 9–46 A 28-cm-high, 18-cm-long, and 18-cm-wide rectangular container suspended in a room at 24˚C is initially filled with cold water at 2˚C. The surface temperature of the container is observed to be nearly the same as the water temperature inside. The emissivity of the container surface is 0.6, and the temperature of the surrounding surfaces is about the same as the air temperature. Determine the water temperature in the container after 3 h, and the average rate of heat transfer to the water. Assume the heat transfer coefficient on the top and bottom surfaces to be the same as that on the side surfaces. 9–47 Reconsider Problem 9–46. Using EES (or other) software, plot the water temperature in the container as a function of the heating time as the time varies from 30 min to 10 h, and discuss the results. 9–48 A room is to be heated by a coal-burning stove, which is a cylindrical cavity with an outer diameter of 32 cm and a height of 70 cm. The rate of heat loss from the room is estimated to be 1.2 kW when the air temperature in the room is maintained constant at 24˚C. The emissivity of the stove surface is 0.85 and the average temperature of the surrounding cen58933_ch09.qxd 9/4/2002 12:26 PM Page 507 507 CHAPTER 9 wall surfaces is 17˚C. Determine the surface temperature of the stove. Neglect the transfer from the bottom surface and take the heat transfer coefficient at the top surface to be the same as that on the side surface. The heating value of the coal is 30,000 kJ/kg, and the combustion efficiency is 65 percent. Determine the amount of coal burned a day if the stove operates 14 h a day. spaces are converted into rectangular channels. The base temperature of the heat sink in this case was measured to be 108°C. Noting that the shroud loses heat to the ambient air from both sides, determine the average natural convection heat transfer coefficient in this shrouded case. (For complete details, see Çengel and Zing, Ref. 9). Air flow 9–49 The water in a 40-L tank is to be heated from 15˚C to 45˚C by a 6-cm-diameter spherical heater whose surface temperature is maintained at 85˚C. Determine how long the heater should be kept on. Heat sink 7.62 cm Natural Convection from Finned Surfaces and PCBs 9–50C Why are finned surfaces frequently used in practice? Why are the finned surfaces referred to as heat sinks in the electronics industry? Shroud 9–51C Why are heat sinks with closely packed fins not suitable for natural convection heat transfer, although they increase the heat transfer surface area more? 9–52C Consider a heat sink with optimum fin spacing. Explain how heat transfer from this heat sink will be affected by (a) removing some of the fins on the heat sink and (b) doubling the number of fins on the heat sink by reducing the fin spacing. The base area of the heat sink remains unchanged at all times. 9–53 Aluminum heat sinks of rectangular profile are commonly used to cool electronic components. Consider a 7.62-cm-long and 9.68-cm-wide commercially available heat sink whose cross section and dimensions are as shown in Figure P9–53. The heat sink is oriented vertically and is used to cool a power transistor that can dissipate up to 125 W of power. The back surface of the heat sink is insulated. The surfaces of the heat sink are untreated, and thus they have a low emissivity (under 0.1). Therefore, radiation heat transfer from the heat sink can be neglected. During an experiment conducted in room air at 22°C, the base temperature of the heat sink was measured to be 120°C when the power dissipation of the transistor was 15 W. Assuming the entire heat sink to be at the base temperature, determine the average natural convection heat transfer coefficient for this case. Answer: 7.1 W/m2 °C Transistor 3.17 cm FIGURE P9–54 9–55E A 6-in.-wide and 8-in.-high vertical hot surface in 78°F air is to be cooled by a heat sink with equally spaced fins of rectangular profile. The fins are 0.08 in. thick and 8 in. long in the vertical direction and have a height of 1.2 in. from the base. Determine the optimum fin spacing and the rate of heat transfer by natural convection from the heat sink if the base temperature is 180°F. 9–56E Reconsider Problem 9–55E. Using EES (or other) software, investigate the effect of the length of the fins in the vertical direction on the optimum fin spacing and the rate of heat transfer by natural convection. Let the fin length vary from 2 in. to 10 in. Plot the optimum fin spacing and the rate of convection heat transfer as a function of the fin length, and discuss the results. 9–57 A 12.1-cm-wide and 18-cm-high vertical hot surface in 25°C air is to be cooled by a heat sink with equally spaced fins of rectangular profile. The fins are 0.1 cm thick and 18 cm long in the vertical direction. Determine the optimum fin height and the rate of heat transfer by natural convection from the heat sink if the base temperature is 65°C. 1.45 cm Natural Convection inside Enclosures 1.52 cm 0.48 cm Heat sink 9.68 cm FIGURE P9–53 9–54 Reconsider the heat sink in Problem 9–53. In order to enhance heat transfer, a shroud (a thin rectangular metal plate) whose surface area is equal to the base area of the heat sink is placed very close to the tips of the fins such that the interfin 9–58C The upper and lower compartments of a wellinsulated container are separated by two parallel sheets of glass with an air space between them. One of the compartments is to be filled with a hot fluid and the other with a cold fluid. If it is desired that heat transfer between the two compartments be minimal, would you recommend putting the hot fluid into the upper or the lower compartment of the container? Why? 9–59C Someone claims that the air space in a double-pane window enhances the heat transfer from a house because of the natural convection currents that occur in the air space and cen58933_ch09.qxd 9/4/2002 12:26 PM Page 508 508 HEAT TRANSFER recommends that the double-pane window be replaced by a single sheet of glass whose thickness is equal to the sum of the thicknesses of the two glasses of the double-pane window to save energy. Do you agree with this claim? 9–60C Consider a double-pane window consisting of two glass sheets separated by a 1-cm-wide air space. Someone suggests inserting a thin vinyl sheet in the middle of the two glasses to form two 0.5-cm-wide compartments in the window in order to reduce natural convection heat transfer through the window. From a heat transfer point of view, would you be in favor of this idea to reduce heat losses through the window? 9–61C What does the effective conductivity of an enclosure represent? How is the ratio of the effective conductivity to thermal conductivity related to the Nusselt number? 9–62 Show that the thermal resistance of a rectangular enclosure can be expressed as R /(Ak Nu), where k is the thermal conductivity of the fluid in the enclosure. 9–63E A vertical 4-ft-high and 6-ft-wide double-pane window consists of two sheets of glass separated by a 1-in. air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be 65°F and 40°F, determine the rate of heat transfer through the window by (a) natural convection and (b) radiation. Also, determine the R-value of insulation of this window such that multiplying the inverse of the R-value by the surface area and the temperature difference gives the total rate of heat transfer through the window. The effective emissivity for use in radiation calculations between two large parallel glass plates can be taken to be 0.82. 9–65 Two concentric spheres of diameters 15 cm and 25 cm are separated by air at 1 atm pressure. The surface temperatures of the two spheres enclosing the air are T1 350 K and T2 275 K, respectively. Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection. 9–66 Reconsider Problem 9–65. Using EES (or other) software, plot the rate of natural convection heat transfer as a function of the hot surface temperature of the sphere as the temperature varies from 300 K to 500 K, and discuss the results. 9–67 Flat-plate solar collectors are often tilted up toward the sun in order to intercept a greater amount of direct solar radiation. The tilt angle from the horizontal also affects the rate of heat loss from the collector. Consider a 2-m-high and 3-mwide solar collector that is tilted at an angle from the horizontal. The back side of the absorber is heavily insulated. The absorber plate and the glass cover, which are spaced 2.5 cm from each other, are maintained at temperatures of 80°C and 40°C, respectively. Determine the rate of heat loss from the absorber plate by natural convection for 0°, 20°, and 90°. Glass cover Solar radiation 40°C 80°C Absorber plate Air space 65°F Insulation θ 4 ft FIGURE P9–67 40°F 9–68 A simple solar collector is built by placing a 5-cmdiameter clear plastic tube around a garden hose whose outer diameter is 1.6 cm. The hose is painted black to maximize solar absorption, and some plastic rings are used to keep the spacing between the hose and the clear plastic cover constant. During a clear day, the temperature of the hose is measured to be 65°C, Glass 1 in. Frame Solar radiation FIGURE P9–63E Reconsider Problem 9–63E. Using EES (or other) software, investigate the effect of the air gap thickness on the rates of heat transfer by natural convection and radiation, and the R-value of insulation. Let the air gap thickness vary from 0.2 in. to 2.0 in. Plot the rates of heat transfer by natural convection and radiation, and the R-value of insulation as a function of the air gap thickness, and discuss the results. 26°C Clear plastic tube 9–64E Water Spacer Garden hose 65°C FIGURE P9–68 cen58933_ch09.qxd 9/4/2002 12:26 PM Page 509 509 CHAPTER 9 while the ambient air temperature is 26°C. Determine the rate of heat loss from the water in the hose per meter of its length by natural convection. Also, discuss how the performance of Answer: 8.2 W this solar collector can be improved. Reconsider Problem 9–68. Using EES (or other) software, plot the rate of heat loss from the water by natural convection as a function of the ambient air temperature as the temperature varies from 4˚C to 40˚C, and discuss the results. 30°C 2m 9–69 9–70 A vertical 1.3-m-high, 2.8-m-wide double-pane window consists of two layers of glass separated by a 2.2-cm air gap at atmospheric pressure. The room temperature is 26˚C while the inner glass temperature is 18˚C. Disregarding radiation heat transfer, determine the temperature of the outer glass layer and the rate of heat loss through the window by natural convection. 9–71 Consider two concentric horizontal cylinders of diameters 55 cm and 65 cm, and length 125 cm. The surfaces of the inner and outer cylinders are maintained at 46˚C and 74˚C, respectively. Determine the rate of heat transfer between the cylinders by natural convection if the annular space is filled with (a) water and (b) air. Combined Natural and Forced Convection 9–72C When is natural convection negligible and when is it not negligible in forced convection heat transfer? 9–73C Under what conditions does natural convection enhance forced convection, and under what conditions does it hurt forced convection? 9–74C When neither natural nor forced convection is negligible, is it correct to calculate each independently and add them to determine the total convection heat transfer? 9–75 Consider a 5-m-long vertical plate at 85°C in air at 30°C. Determine the forced motion velocity above which natural convection heat transfer from this plate is negligible. Answer: 9.04 m/s 9–76 Reconsider Problem 9–75. Using EES (or other) software, plot the forced motion velocity above which natural convection heat transfer is negligible as a function of the plate temperature as the temperature varies from 50˚C to 150˚C, and discuss the results. 9–77 Consider a 5-m-long vertical plate at 60°C in water at 25°C. Determine the forced motion velocity above which natural convection heat transfer from this plate is negligible. Take 0.0004 K1 for water. 9–78 In a production facility, thin square plates 2 m 2 m in size coming out of the oven at 270°C are cooled by blowing ambient air at 30°C horizontally parallel to their surfaces. Determine the air velocity above which the natural convection effects on heat transfer are less than 10 percent and thus are negligible. 2m Hot plates 270°C FIGURE P9–78 9–79 A 12-cm-high and 20-cm-wide circuit board houses 100 closely spaced logic chips on its surface, each dissipating 0.05 W. The board is cooled by a fan that blows air over the hot surface of the board at 35°C at a velocity of 0.5 m/s. The heat transfer from the back surface of the board is negligible. Determine the average temperature on the surface of the circuit board assuming the air flows vertically upwards along the 12cm-long side by (a) ignoring natural convection and (b) considering the contribution of natural convection. Disregard any heat transfer by radiation. Special Topic: Heat Transfer through Windows 9–80C Why are the windows considered in three regions when analyzing heat transfer through them? Name those regions and explain how the overall U-value of the window is determined when the heat transfer coefficients for all three regions are known. 9–81C Consider three similar double-pane windows with air gap widths of 5, 10, and 20 mm. For which case will the heat transfer through the window will be a minimum? 9–82C In an ordinary double-pane window, about half of the heat transfer is by radiation. Describe a practical way of reducing the radiation component of heat transfer. 9–83C Consider a double-pane window whose air space width is 20 mm. Now a thin polyester film is used to divide the air space into two 10-mm-wide layers. How will the film affect (a) convection and (b) radiation heat transfer through the window? 9–84C Consider a double-pane window whose air space is flashed and filled with argon gas. How will replacing the air in the gap by argon affect (a) convection and (b) radiation heat transfer through the window? 9–85C Is the heat transfer rate through the glazing of a double-pane window higher at the center or edge section of the glass area? Explain. 9–86C How do the relative magnitudes of U-factors of windows with aluminum, wood, and vinyl frames compare? Assume the windows are identical except for the frames. 9–87 Determine the U-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter cen58933_ch09.qxd 9/4/2002 12:26 PM Page 510 510 HEAT TRANSFER design conditions. The glazings are made of clear glass having an emissivity of 0.84. Take the average air space temperature at design conditions to be 10°C and the temperature difference across the air space to be 15°C. 9–88 A double-door wood-framed window with glass glazing and metal spacers is being considered for an opening that is 1.2 m high and 1.8 m wide in the wall of a house maintained at 20°C. Determine the rate of heat loss through the window and the inner surface temperature of the window glass facing the room when the outdoor air temperature is 8°C if the window is selected to be (a) 3-mm single glazing, (b) double glazing with an air space of 13 mm, and (c) low-e-coated triple glazing with an air space of 13 mm. Double-door window winds of 12 km/h outside. What will the U-factor be when the Answer: 2.88 W/m2 · °C wind velocity outside is doubled? 9–93 The owner of an older house in Wichita, Kansas, is considering replacing the existing double-door type wood-framed single-pane windows with vinyl-framed double-pane windows with an air space of 6.4 mm. The new windows are of doubledoor type with metal spacers. The house is maintained at 22°C at all times, but heating is needed only when the outdoor temperature drops below 18°C because of the internal heat gain from people, lights, appliances, and the sun. The average winter temperature of Wichita is 7.1°C, and the house is heated by electric resistance heaters. If the unit cost of electricity is $0.07/kWh and the total window area of the house is 12 m2, determine how much money the new windows will save the home owner per month in winter. Wood frame Single pane Double pane Glass Glass FIGURE P9–88 9–89 Determine the overall U-factor for a double-door-type wood-framed double-pane window with 13-mm air space and metal spacers, and compare your result to the value listed in Table 9–6. The overall dimensions of the window are 2.00 m 2.40 m, and the dimensions of each glazing are 1.92 m 1.14 m. 9–90 Consider a house in Atlanta, Georgia, that is maintained at 22°C and has a total of 20 m2 of window area. The windows are double-door-type with wood frames and metal spacers. The glazing consists of two layers of glass with 12.7 mm of air space with one of the inner surfaces coated with reflective film. The winter average temperature of Atlanta is 11.3°C. Determine the average rate of heat loss through the windows in winter. Answer: 456 W FIGURE P9–93 Review Problems 9–94E A 0.1-W small cylindrical resistor mounted on a lower part of a vertical circuit board is 0.3 in. long and has a diameter of 0.2 in. The view of the resistor is largely blocked by another circuit board facing it, and the heat transfer through the connecting wires is negligible. The air is free to flow through the large parallel flow passages between the boards as a result of natural convection currents. If the air temperature at the vicinity of the resistor is 120°F, determine the approximate surAnswer: 212°F face temperature of the resistor. 9–91E Consider an ordinary house with R-13 walls (walls that have an R-value of 13 h · ft2 · °F/Btu). Compare this to the R-value of the common double-door windows that are double pane with 14 in. of air space and have aluminum frames. If the windows occupy only 20 percent of the wall area, determine if more heat is lost through the windows or through the remaining 80 percent of the wall area. Disregard infiltration losses. 9–92 The overall U-factor of a fixed wood-framed window with double glazing is given by the manufacturer to be U 2.76 W/m2 · °C under the conditions of still air inside and FIGURE P9–94E cen58933_ch09.qxd 9/4/2002 12:26 PM Page 511 511 CHAPTER 9 9–95 An ice chest whose outer dimensions are 30 cm 40 cm 40 cm is made of 3-cm-thick styrofoam (k 0.033 W/m · °C). Initially, the chest is filled with 30 kg of ice at 0°C, and the inner surface temperature of the ice chest can be taken to be 0°C at all times. The heat of fusion of water at 0°C is 333.7 kJ/kg, and the surrounding ambient air is at 20°C. Disregarding any heat transfer from the 40 cm 40 cm base of the ice chest, determine how long it will take for the ice in the chest to melt completely if the ice chest is subjected to (a) calm air and (b) winds at 50 km/h. Assume the heat transfer coefficient on the front, back, and top surfaces to be the same as that on the side surfaces. 9–96 An electronic box that consumes 180 W of power is cooled by a fan blowing air into the box enclosure. The dimensions of the electronic box are 15 cm 50 cm 50 cm, and all surfaces of the box are exposed to the ambient except the base surface. Temperature measurements indicate that the box is at an average temperature of 32°C when the ambient temperature and the temperature of the surrounding walls are 25°C. If the emissivity of the outer surface of the box is 0.85, determine the fraction of the heat lost from the outer surfaces of the electronic box. 25°C 32°C ε = 0.85 15 cm 50 cm 50 cm FIGURE P9–96 9–97 A 6-m-internal-diameter spherical tank made of 1.5cm-thick stainless steel (k 15 W/m · °C) is used to store iced water at 0°C in a room at 20°C. The walls of the room are also at 20°C. The outer surface of the tank is black (emissivity 1), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. Assuming the entire steel tank to be at 0°C and thus the thermal resistance of the tank to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period. Answers: (a) 15.4 kW, (b) 3988 kg Wall 15 cm m 80 c Oil Ts = 45°C ε = 0.8 50 cm Electric heater Heating element FIGURE P9–99 15 cm wide filled with 45 kg of oil. The heater is to be placed against a wall, and thus heat transfer from its back surface is negligible for safety considerations. The surface temperature of the heater is not to exceed 45°C in a room at 25°C. Disregarding heat transfer from the bottom and top surfaces of the heater in anticipation that the top surface will be used as a shelf, determine the power rating of the heater in W. Take the emissivity of the outer surface of the heater to be 0.8 and the average temperature of the ceiling and wall surfaces to be the same as the room air temperature. Also, determine how long it will take for the heater to reach steady operation when it is first turned on (i.e., for the oil temperature to rise from 25°C to 45°C). State your assumptions in the calculations. 9–100 Skylights or “roof windows” are commonly used in homes and manufacturing facilities since they let natural light in during day time and thus reduce the lighting costs. However, they offer little resistance to heat transfer, and large amounts of energy are lost through them in winter unless they are equipped with a motorized insulating cover that can be used in cold weather and at nights to reduce heat losses. Consider a 1-m-wide and 2.5-m-long horizontal skylight on the roof of a house that is kept at 20°C. The glazing of the skylight is made of a single layer of 0.5-cm-thick 9–98 Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k 0.78 W/m · °C) separated by a 3-cm-wide air space. Determine the steady rate of heat transfer through this window and the temperature of its inner surface for a day during which the room is maintained at 20°C while the temperature of the outdoors is 0°C. Take the heat transfer coefficients on the inner and outer surfaces of the window to be h1 10 W/m2 · °C and h2 25 W/m2 · °C and disregard any heat transfer by radiation. Tin = 20°C 9–99 An electric resistance space heater is designed such that it resembles a rectangular box 50 cm high, 80 cm long, and FIGURE P9–100 Tsky = –30°C Tair = – 10°C 2.5 m Skylight ε = 0.9 1m cen58933_ch09.qxd 9/4/2002 12:26 PM Page 512 512 HEAT TRANSFER glass (k 0.78 W/m °C and 0.9). Determine the rate of heat loss through the skylight when the air temperature outside is 10°C and the effective sky temperature is 30°C. Compare your result with the rate of heat loss through an equivalent surface area of the roof that has a common R-5.34 construction in SI units (i.e., a thickness–to–effectivethermal-conductivity ratio of 5.34 m2 · °C/W). 9–101 A solar collector consists of a horizontal copper tube of outer diameter 5 cm enclosed in a concentric thin glass tube of 9 cm diameter. Water is heated as it flows through the tube, and the annular space between the copper and glass tube is filled with air at 1 atm pressure. During a clear day, the temperatures of the tube surface and the glass cover are measured to be 60°C and 32°C, respectively. Determine the rate of heat loss from the collector by natural convection per meter length of the tube. Answer: 17.4 W 9 cm m 5c Glass cover FIGURE P9–101 9–102 A solar collector consists of a horizontal aluminum tube of outer diameter 4 cm enclosed in a concentric thin glass tube of 7 cm diameter. Water is heated as it flows through the aluminum tube, and the annular space between the aluminum and glass tubes is filled with air at 1 atm pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of 20 W per meter length, and the temperature of the ambient air outside is 30°C. Approximating the surfaces of the tube and the glass cover as being black (emissivity 1) in radiation calculations and taking the effective sky temperature to be 20°C, determine the temperature of the aluminum tube when equilibrium is established (i.e., when the net heat loss from the tube by convection and radiation equals the amount of solar energy absorbed by the tube). 9–103E The components of an electronic system dissipating 180 W are located in a 4-ft-long horizontal duct whose crosssection is 6 in. 6 in. The components in the duct are cooled by forced air, which enters at 85°F at a rate of 22 cfm and leaves at 100°F. The surfaces of the sheet metal duct are not painted, and thus radiation heat transfer from the outer surfaces is negligible. If the ambient air temperature is 80°F, determine (a) the heat transfer from the outer surfaces of the duct to the ambient air by natural convection and (b) the average temperature of the duct. Natural convection 100°F 80°F 180 W 4 ft 85°F 22 cfm FIGURE P9–103E 9–104E Repeat Problem 9–103E for a circular horizontal duct of diameter 4 in. 9–105E Repeat Problem 9–103E assuming the fan fails and thus the entire heat generated inside the duct must be rejected to the ambient air by natural convection through the outer surfaces of the duct. 9–106 Consider a cold aluminum canned drink that is initially at a uniform temperature of 5°C. The can is 12.5 cm high and has a diameter of 6 cm. The emissivity of the outer surface of the can is 0.6. Disregarding any heat transfer from the bottom surface of the can, determine how long it will take for the average temperature of the drink to rise to 7°C if the surroundAnswer: 12.1 min ing air and surfaces are at 25°C. 9–107 Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and maintains the hot water at 60°C. The tank is located in a small room at 20°C whose walls and the ceiling are at about the same temperature. The tank is placed in a 46-cm-diameter sheet metal shell of negligible thickness, and the space between the tank and the shell is filled with foam insulation. The average temperature and emissivity of the outer surface of the shell are 40°C and 0.7, respectively. The price of 3 cm 40 cm 20°C 2m Tw = 60°C Foam insulation 40°C ε = 0.7 FIGURE P9–107 Water heater cen58933_ch09.qxd 9/4/2002 12:26 PM Page 513 513 CHAPTER 9 electricity is $0.08/kWh. Hot water tank insulation kits large enough to wrap the entire tank are available on the market for about $30. If such an insulation is installed on this water tank by the home owner himself, how long will it take for this additional insulation to pay for itself? Disregard any heat loss from the top and bottom surfaces, and assume the insulation to reduce the heat losses by 80 percent. 9–108 During a plant visit, it was observed that a 1.5-m-high and 1-m-wide section of the vertical front section of a natural gas furnace wall was too hot to touch. The temperature measurements on the surface revealed that the average temperature of the exposed hot surface was 110°C, while the temperature of the surrounding air was 25°C. The surface appeared to be oxidized, and its emissivity can be taken to be 0.7. Taking the temperature of the surrounding surfaces to be 25°C also, determine the rate of heat loss from this furnace. The furnace has an efficiency of 79 percent, and the plant pays $0.75 per therm of natural gas. If the plant operates 10 h a day, 310 days a year, and thus 3100 h a year, determine the annual cost of the heat loss from this vertical hot surface on the front section of the furnace wall. Furnace Ts = 110°C ε = 0.7 1.5m 1m Tair = 25°C FIGURE P9–108 9–109 A group of 25 power transistors, dissipating 1.5 W each, are to be cooled by attaching them to a black-anodized square aluminum plate and mounting the plate on the wall of a room at 30°C. The emissivity of the transistor and the plate surfaces is 0.9. Assuming the heat transfer from the back side of the plate to be negligible and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the size of the plate if the average surface temperature Answer: 43 cm 3 43 cm of the plate is not to exceed 50°C. Black-anodized aluminum plate Power transistor, 1.5 W FIGURE P9–109 is 0.5, and the walls of the basement are also at about 60°F. If the inlet temperature of the water is 150°F and the heat transfer coefficient on the inner surface of the pipe is 30 Btu/h · ft2 · °F, determine the temperature drop of water as it passes through the basement. 9–112 Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is 1.5 m wide and 6 m long, and the average temperature of the exposed surface of the collector is 42°C. Determine the rate of heat loss from the collector by natural convection during a calm day when the ambient air temperature is 15°C. Also, determine the heat loss by radiation by taking the emissivity of the collector surface to be 0.9 and the effective sky temperature to be Answers: 1295 W, 2921 W 30°C. 9–113 Solar radiation is incident on the glass cover of a solar collector at a rate of 650 W/m2. The glass transmits 88 percent of the incident radiation and has an emissivity of 0.90. The hot water needs of a family in summer can be met completely by a Glass cover Solar radiation 650 W/ m2 9–110 Repeat Problem 9–109 assuming the plate to be positioned horizontally with (a) transistors facing up and (b) transistors facing down. 9–111E Hot water is flowing at an average velocity of 4 ft/s through a cast iron pipe (k 30 Btu/h · ft · °F) whose inner and outer diameters are 1.0 in. and 1.2 in., respectively. The pipe passes through a 50-ft-long section of a basement whose temperature is 60°F. The emissivity of the outer surface of the pipe Absorber plate Insulation 40° FIGURE P9–113 cen58933_ch09.qxd 9/4/2002 12:26 PM Page 514 514 HEAT TRANSFER collector 1.5 m high and 2 m wide, and tilted 40° from the horizontal. The temperature of the glass cover is measured to be 40°C on a calm day when the surrounding air temperature is 20°C. The effective sky temperature for radiation exchange between the glass cover and the open sky is 40°C. Water enters the tubes attached to the absorber plate at a rate of 1 kg/min. Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss occurs through the glass cover, determine (a) the total rate of heat loss from the collector, (b) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and (c) the temperature rise of water as it flows through the collector. Design and Essay Problems 9–114 Write a computer program to evaluate the variation of temperature with time of thin square metal plates that are removed from an oven at a specified temperature and placed vertically in a large room. The thickness, the size, the initial temperature, the emissivity, and the thermophysical properties of the plate as well as the room temperature are to be specified by the user. The program should evaluate the temperature of the plate at specified intervals and tabulate the results against time. The computer should list the assumptions made during calculations before printing the results. For each step or time interval, assume the surface temperature to be constant and evaluate the heat loss during that time interval and the temperature drop of the plate as a result of this heat loss. This gives the temperature of the plate at the end of a time interval, which is to serve as the initial temperature of the plate for the beginning of the next time interval. Try your program for 0.2-cm-thick vertical copper plates of 40 cm 40 cm in size initially at 300°C cooled in a room at 25°C. Take the surface emissivity to be 0.9. Use a time interval of 1 s in calculations, but print the results at 10-s intervals for a total cooling period of 15 min. 9–115 Write a computer program to optimize the spacing between the two glasses of a double-pane window. Assume the spacing is filled with dry air at atmospheric pressure. The program should evaluate the recommended practical value of the spacing to minimize the heat losses and list it when the size of the window (the height and the width) and the temperatures of the two glasses are specified. 9–116 Contact a manufacturer of aluminum heat sinks and obtain their product catalog for cooling electronic components by natural convection and radiation. Write an essay on how to select a suitable heat sink for an electronic component when its maximum power dissipation and maximum allowable surface temperature are specified. 9–117 The top surfaces of practically all flat-plate solar collectors are covered with glass in order to reduce the heat losses from the absorber plate underneath. Although the glass cover reflects or absorbs about 15 percent of the incident solar radiation, it saves much more from the potential heat losses from the absorber plate, and thus it is considered to be an essential part of a well-designed solar collector. Inspired by the energy efficiency of double-pane windows, someone proposes to use double glazing on solar collectors instead of a single glass. Investigate if this is a good idea for the town in which you live. Use local weather data and base your conclusion on heat transfer analysis and economic considerations. cen58933_ch10.qxd 9/4/2002 12:37 PM Page 515 CHAPTER BOILING AND C O N D E N S AT I O N e know from thermodynamics that when the temperature of a liquid at a specified pressure is raised to the saturation temperature Tsat at that pressure, boiling occurs. Likewise, when the temperature of a vapor is lowered to Tsat, condensation occurs. In this chapter we study the rates of heat transfer during such liquid-to-vapor and vapor-to-liquid phase transformations. Although boiling and condensation exhibit some unique features, they are considered to be forms of convection heat transfer since they involve fluid motion (such as the rise of the bubbles to the top and the flow of condensate to the bottom). Boiling and condensation differ from other forms of convection in that they depend on the latent heat of vaporization hfg of the fluid and the surface tension at the liquid–vapor interface, in addition to the properties of the fluid in each phase. Noting that under equilibrium conditions the temperature remains constant during a phase-change process at a fixed pressure, large amounts of heat (due to the large latent heat of vaporization released or absorbed) can be transferred during boiling and condensation essentially at constant temperature. In practice, however, it is necessary to maintain some difference between the surface temperature Ts and Tsat for effective heat transfer. Heat transfer coefficients h associated with boiling and condensation are typically much higher than those encountered in other forms of convection processes that involve a single phase. We start this chapter with a discussion of the boiling curve and the modes of pool boiling such as free convection boiling, nucleate boiling, and film boiling. We then discuss boiling in the presence of forced convection. In the second part of this chapter, we describe the physical mechanism of film condensation and discuss condensation heat transfer in several geometrical arrangements and orientations. Finally, we introduce dropwise condensation and discuss ways of maintaining it. W 10 CONTENTS 10–1 Boiling Heat Transfer 516 10–2 Pool Boiling 518 10–3 Flow Boiling 530 10–4 Condensation Heat Transfer 532 10–5 Film Condensation 532 10–6 Film Condensation Inside Horizontal Tubes 545 10–7 Dropwise Condensation 545 Topic of Special Interest: Heat Pipes 546 515 cen58933_ch10.qxd 9/4/2002 12:37 PM Page 516 516 HEAT TRANSFER Evaporation Air 10–1 Water 20°C Boiling Water 100°C Heating FIGURE 10–1 A liquid-to-vapor phase change process is called evaporation if it occurs at a liquid–vapor interface and boiling if it occurs at a solid–liquid interface. P = 1 atm Water Tsat = 100°C Bubbles 110°C Heating element FIGURE 10–2 Boiling occurs when a liquid is brought into contact with a surface at a temperature above the saturation temperature of the liquid. ■ BOILING HEAT TRANSFER Many familiar engineering applications involve condensation and boiling heat transfer. In a household refrigerator, for example, the refrigerant absorbs heat from the refrigerated space by boiling in the evaporator section and rejects heat to the kitchen air by condensing in the condenser section (the long coils behind the refrigerator). Also, in steam power plants, heat is transferred to the steam in the boiler where water is vaporized, and the waste heat is rejected from the steam in the condenser where the steam is condensed. Some electronic components are cooled by boiling by immersing them in a fluid with an appropriate boiling temperature. Boiling is a liquid-to-vapor phase change process just like evaporation, but there are significant differences between the two. Evaporation occurs at the liquid–vapor interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature. Water in a lake at 20°C, for example, will evaporate to air at 20°C and 60 percent relative humidity since the saturation pressure of water at 20°C is 2.3 kPa and the vapor pressure of air at 20°C and 60 percent relative humidity is 1.4 kPa (evaporation rates are determined in Chapter 14). Other examples of evaporation are the drying of clothes, fruits, and vegetables; the evaporation of sweat to cool the human body; and the rejection of waste heat in wet cooling towers. Note that evaporation involves no bubble formation or bubble motion (Fig. 10–1). Boiling, on the other hand, occurs at the solid–liquid interface when a liquid is brought into contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid (Fig. 10–2). At 1 atm, for example, liquid water in contact with a solid surface at 110°C will boil since the saturation temperature of water at 1 atm is 100°C. The boiling process is characterized by the rapid formation of vapor bubbles at the solid–liquid interface that detach from the surface when they reach a certain size and attempt to rise to the free surface of the liquid. When cooking, we do not say water is boiling until we see the bubbles rising to the top. Boiling is a complicated phenomenon because of the large number of variables involved in the process and the complex fluid motion patterns caused by the bubble formation and growth. As a form of convection heat transfer, the boiling heat flux from a solid surface to the fluid is expressed from Newton’s law of cooling as q·boiling h(Ts Tsat) hTexcess (W/m2) (10-1) where Texcess Ts Tsat is called the excess temperature, which represents the excess of the surface above the saturation temperature of the fluid. In the preceding chapters we considered forced and free convection heat transfer involving a single phase of a fluid. The analysis of such convection processes involves the thermophysical properties , , k, and Cp of the fluid. The analysis of boiling heat transfer involves these properties of the liquid (indicated by the subscript l) or vapor (indicated by the subscript v) as well as the properties hfg (the latent heat of vaporization) and (the surface tension). The hfg represents the energy absorbed as a unit mass of liquid vaporizes at a specified temperature or pressure and is the primary quantity of energy cen58933_ch10.qxd 9/4/2002 12:38 PM Page 517 517 CHAPTER 10 transferred during boiling heat transfer. The hfg values of water at various temperatures are given in Table A-9. Bubbles owe their existence to the surface-tension at the liquid–vapor interface due to the attraction force on molecules at the interface toward the liquid phase. The surface tension decreases with increasing temperature and becomes zero at the critical temperature. This explains why no bubbles are formed during boiling at supercritical pressures and temperatures. Surface tension has the unit N/m. The boiling processes in practice do not occur under equilibrium conditions, and normally the bubbles are not in thermodynamic equilibrium with the surrounding liquid. That is, the temperature and pressure of the vapor in a bubble are usually different than those of the liquid. The pressure difference between the liquid and the vapor is balanced by the surface tension at the interface. The temperature difference between the vapor in a bubble and the surrounding liquid is the driving force for heat transfer between the two phases. When the liquid is at a lower temperature than the bubble, heat will be transferred from the bubble into the liquid, causing some of the vapor inside the bubble to condense and the bubble to collapse eventually. When the liquid is at a higher temperature than the bubble, heat will be transferred from the liquid to the bubble, causing the bubble to grow and rise to the top under the influence of buoyancy. Boiling is classified as pool boiling or flow boiling, depending on the presence of bulk fluid motion (Fig. 10–3). Boiling is called pool boiling in the absence of bulk fluid flow and flow boiling (or forced convection boiling) in the presence of it. In pool boiling, the fluid is stationary, and any motion of the fluid is due to natural convection currents and the motion of the bubbles under the influence of buoyancy. The boiling of water in a pan on top of a stove is an example of pool boiling. Pool boiling of a fluid can also be achieved by placing a heating coil in the fluid. In flow boiling, the fluid is forced to move in a heated pipe or over a surface by external means such as a pump. Therefore, flow boiling is always accompanied by other convection effects. Pool and flow boiling are further classified as subcooled boiling or saturated boiling, depending on the bulk liquid temperature (Fig. 10–4). Boiling is said to be subcooled (or local) when the temperature of the main body of the liquid is below the saturation temperature Tsat (i.e., the bulk of the liquid is subcooled) and saturated (or bulk) when the temperature of the liquid is equal to Tsat (i.e., the bulk of the liquid is saturated). At the early stages of boiling, the bubbles are confined to a narrow region near the hot surface. This is because the liquid adjacent to the hot surface vaporizes as a result of being heated above its saturation temperature. But these bubbles disappear soon after they move away from the hot surface as a result of heat transfer from the bubbles to the cooler liquid surrounding them. This happens when the bulk of the liquid is at a lower temperature than the saturation temperature. The bubbles serve as “energy movers” from the hot surface into the liquid body by absorbing heat from the hot surface and releasing it into the liquid as they condense and collapse. Boiling in this case is confined to a region in the locality of the hot surface and is appropriately called local or subcooled boiling. When the entire liquid body reaches the saturation temperature, the bubbles start rising to the top. We can see bubbles throughout the bulk of the liquid, Heating Heating (a) Pool boiling (b) Flow boiling FIGURE 10–3 Classification of boiling on the basis of the presence of bulk fluid motion. P = 1 atm P = 1 atm Subcooled 80°C water 107°C Saturated 100°C water 107°C Bubble Heating Heating (a) Subcooled boiling (b) Saturated boiling FIGURE 10–4 Classification of boiling on the basis of the presence of bulk liquid temperature. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 518 518 HEAT TRANSFER and boiling in this case is called the bulk or saturated boiling. Next, we consider different boiling regimes in detail. 10–2 ■ POOL BOILING So far we presented some general discussions on boiling. Now we turn our attention to the physical mechanisms involved in pool boiling, that is, the boiling of stationary fluids. In pool boiling, the fluid is not forced to flow by a mover such as a pump, and any motion of the fluid is due to natural convection currents and the motion of the bubbles under the influence of buoyancy. As a familiar example of pool boiling, consider the boiling of tap water in a pan on top of a stove. The water will initially be at about 15°C, far below the saturation temperature of 100°C at standard atmospheric pressure. At the early stages of boiling, you will not notice anything significant except some bubbles that stick to the surface of the pan. These bubbles are caused by the release of air molecules dissolved in liquid water and should not be confused with vapor bubbles. As the water temperature rises, you will notice chunks of liquid water rolling up and down as a result of natural convection currents, followed by the first vapor bubbles forming at the bottom surface of the pan. These bubbles get smaller as they detach from the surface and start rising, and eventually collapse in the cooler water above. This is subcooled boiling since the bulk of the liquid water has not reached saturation temperature yet. The intensity of bubble formation increases as the water temperature rises further, and you will notice waves of vapor bubbles coming from the bottom and rising to the top when the water temperature reaches the saturation temperature (100°C at standard atmospheric conditions). This full scale boiling is the saturated boiling. 100°C 100°C 103°C 110°C Heating (a) Natural convection boiling Heating (b) Nucleate boiling Vapor film Vapor pockets 100°C 100°C 180°C 400°C Heating (c) Transition boiling Heating (d) Film boiling FIGURE 10–5 Different boiling regimes in pool boiling. Boiling Regimes and the Boiling Curve Boiling is probably the most familiar form of heat transfer, yet it remains to be the least understood form. After hundreds of papers written on the subject, we still do not fully understand the process of bubble formation and we must still rely on empirical or semi-empirical relations to predict the rate of boiling heat transfer. The pioneering work on boiling was done in 1934 by S. Nukiyama, who used electrically heated nichrome and platinum wires immersed in liquids in his experiments. Nukiyama noticed that boiling takes different forms, depending on the value of the excess temperature Texcess. Four different boiling regimes are observed: natural convection boiling, nucleate boiling, transition boiling, and film boiling (Fig. 10–5). These regimes are illustrated on the boiling curve in Figure 10–6, which is a plot of boiling heat flux versus the excess temperature. Although the boiling curve given in this figure is for water, the general shape of the boiling curve remains the same for different fluids. The specific shape of the curve depends on the fluid–heating surface material combination and the fluid pressure, but it is practically independent of the geometry of the heating surface. We will describe each boiling regime in detail. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 519 519 CHAPTER 10 Natural convection boiling Bubbles collapse in the liquid 106 q· boiling, W/m2 Nucleate boiling Transition boiling C Film boiling Maximum (critical) heat flux, q· max E 105 B 104 103 Bubbles rise to the free surface A 1 ~5 10 D Leidenfrost point, q· min ~30 100 ~120 ∆Texcess = Ts – Tsat , °C 1000 Natural Convection Boiling (to Point A on the Boiling Curve) We learned in thermodynamics that a pure substance at a specified pressure starts boiling when it reaches the saturation temperature at that pressure. But in practice we do not see any bubbles forming on the heating surface until the liquid is heated a few degrees above the saturation temperature (about 2 to 6°C for water). Therefore, the liquid is slightly superheated in this case (a metastable condition) and evaporates when it rises to the free surface. The fluid motion in this mode of boiling is governed by natural convection currents, and heat transfer from the heating surface to the fluid is by natural convection. Nucleate Boiling (between Points A and C ) The first bubbles start forming at point A of the boiling curve at various preferential sites on the heating surface. The bubbles form at an increasing rate at an increasing number of nucleation sites as we move along the boiling curve toward point C. The nucleate boiling regime can be separated into two distinct regions. In region A–B, isolated bubbles are formed at various preferential nucleation sites on the heated surface. But these bubbles are dissipated in the liquid shortly after they separate from the surface. The space vacated by the rising bubbles is filled by the liquid in the vicinity of the heater surface, and the process is repeated. The stirring and agitation caused by the entrainment of the liquid to the heater surface is primarily responsible for the increased heat transfer coefficient and heat flux in this region of nucleate boiling. In region B–C, the heater temperature is further increased, and bubbles form at such great rates at such a large number of nucleation sites that they form numerous continuous columns of vapor in the liquid. These bubbles move all the way up to the free surface, where they break up and release their vapor content. The large heat fluxes obtainable in this region are caused by the combined effect of liquid entrainment and evaporation. FIGURE 10–6 Typical boiling curve for water at 1 atm pressure. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 520 520 HEAT TRANSFER At large values of Texcess, the rate of evaporation at the heater surface reaches such high values that a large fraction of the heater surface is covered by bubbles, making it difficult for the liquid to reach the heater surface and wet it. Consequently, the heat flux increases at a lower rate with increasing Texcess, and reaches a maximum at point C. The heat flux at this point is called the critical (or maximum) heat flux, q·max. For water, the critical heat flux exceeds 1 MW/m2. Nucleate boiling is the most desirable boiling regime in practice because high heat transfer rates can be achieved in this regime with relatively small values of Texcess, typically under 30°C for water. The photographs in Figure 10–7 show the nature of bubble formation and bubble motion associated with nucleate, transition, and film boiling. Transition Boiling (between Points C and D on the Boiling Curve) As the heater temperature and thus the Texcess is increased past point C, the heat flux decreases, as shown in Figure 10–6. This is because a large fraction of the heater surface is covered by a vapor film, which acts as an insulation due to the low thermal conductivity of the vapor relative to that of the liquid. In the transition boiling regime, both nucleate and film boiling partially occur. Nucleate boiling at point C is completely replaced by film boiling at point D. Operation in the transition boiling regime, which is also called the unstable film boiling regime, is avoided in practice. For water, transition boiling occurs over the excess temperature range from about 30°C to about 120°C. Film Boiling (beyond Point D ) q· W– — m2 q· max Sudden jump in temperature Bypassed part of the boiling curve 106 Sudden drop in temperature 1 q· min 1000 10 100 ∆Texcess = Ts – Tsat, °C FIGURE 10–8 The actual boiling curve obtained with heated platinum wire in water as the heat flux is increased and then decreased. In this region the heater surface is completely covered by a continuous stable vapor film. Point D, where the heat flux reaches a minimum, is called the Leidenfrost point, in honor of J. C. Leidenfrost, who observed in 1756 that liquid droplets on a very hot surface jump around and slowly boil away. The presence of a vapor film between the heater surface and the liquid is responsible for the low heat transfer rates in the film boiling region. The heat transfer rate increases with increasing excess temperature as a result of heat transfer from the heated surface to the liquid through the vapor film by radiation, which becomes significant at high temperatures. A typical boiling process will not follow the boiling curve beyond point C, as Nukiyama has observed during his experiments. Nukiyama noticed, with surprise, that when the power applied to the nichrome wire immersed in water exceeded q·max even slightly, the wire temperature increased suddenly to the melting point of the wire and burnout occurred beyond his control. When he repeated the experiments with platinum wire, which has a much higher melting point, he was able to avoid burnout and maintain heat fluxes higher than q·max. When he gradually reduced power, he obtained the cooling curve shown in Figure 10–8 with a sudden drop in excess temperature when q·min is reached. Note that the boiling process cannot follow the transition boiling part of the boiling curve past point C unless the power applied is reduced suddenly. The burnout phenomenon in boiling can be explained as follows: In order to move beyond point C where q·max occurs, we must increase the heater surface temperature Ts. To increase Ts, however, we must increase the heat flux. But cen58933_ch10.qxd 9/4/2002 12:38 PM Page 521 521 CHAPTER 10 (a) (b) (c) the fluid cannot receive this increased energy at an excess temperature just beyond point C. Therefore, the heater surface ends up absorbing the increased energy, causing the heater surface temperature Ts to rise. But the fluid can receive even less energy at this increased excess temperature, causing the heater surface temperature Ts to rise even further. This continues until the surface FIGURE 10–7 Various boiling regimes during boiling of methanol on a horizontal 1-cm-diameter steam-heated copper tube: (a) nucleate boiling, (b) transition boiling, and (c) film boiling (from J. W. Westwater and J. G. Santangelo, University of Illinois at Champaign-Urbana). cen58933_ch10.qxd 9/4/2002 12:38 PM Page 522 522 HEAT TRANSFER q· W– — m2 q· max q· max = constant C E Sudden jump in temperature Ts, °C Tmelting FIGURE 10–9 An attempt to increase the boiling heat flux beyond the critical value often causes the temperature of the heating element to jump suddenly to a value that is above the melting point, resulting in burnout. temperature reaches a point at which it no longer rises and the heat supplied can be transferred to the fluid steadily. This is point E on the boiling curve, which corresponds to very high surface temperatures. Therefore, any attempt to increase the heat flux beyond q·max will cause the operation point on the boiling curve to jump suddenly from point C to point E. However, surface temperature that corresponds to point E is beyond the melting point of most heater materials, and burnout occurs. Therefore, point C on the boiling curve is also called the burnout point, and the heat flux at this point the burnout heat flux (Fig. 10–9). Most boiling heat transfer equipment in practice operate slightly below q·max to avoid any disastrous burnout. However, in cryogenic applications involving fluids with very low boiling points such as oxygen and nitrogen, point E usually falls below the melting point of the heater materials, and steady film boiling can be used in those cases without any danger of burnout. Heat Transfer Correlations in Pool Boiling Boiling regimes discussed above differ considerably in their character, and thus different heat transfer relations need to be used for different boiling regimes. In the natural convection boiling regime, boiling is governed by natural convection currents, and heat transfer rates in this case can be determined accurately using natural convection relations presented in Chapter 9. Nucleate Boiling In the nucleate boiling regime, the rate of heat transfer strongly depends on the nature of nucleation (the number of active nucleation sites on the surface, the rate of bubble formation at each site, etc.), which is difficult to predict. The type and the condition of the heated surface also affect the heat transfer. These complications made it difficult to develop theoretical relations for heat transfer in the nucleate boiling regime, and people had to rely on relations based on experimental data. The most widely used correlation for the rate of heat transfer in the nucleate boiling regime was proposed in 1952 by Rohsenow, and expressed as g(l v) q·nucleate l hfg Cp(Ts Tsat) Csf hfg Prnl 1/2 3 where q·nucleate nucleate boiling heat flux, W/m2 l viscosity of the liquid, kg/m · s hfg enthalpy of vaporization, J/kg g gravitational acceleration, m/s2 l density of the liquid, kg/m3 v density of the vapor, kg/m3 surface tension of liquid–vapor interface, N/m Cpl specific heat of the liquid, J/kg · °C Ts surface temperature of the heater, °C Tsat saturation temperature of the fluid, °C Csf experimental constant that depends on surface–fluid combination Prl Prandtl number of the liquid n experimental constant that depends on the fluid (10-2) cen58933_ch10.qxd 9/4/2002 12:38 PM Page 523 523 CHAPTER 10 It can be shown easily that using property values in the specified units in the Rohsenow equation produces the desired unit W/m2 for the boiling heat flux, thus saving one from having to go through tedious unit manipulations (Fig. 10–10). The surface tension at the vapor–liquid interface is given in Table 10–1 for water, and Table 10–2 for some other fluids. Experimentally determined values of the constant Csf are given in Table 10–3 for various fluid–surface combinations. These values can be used for any geometry since it is found that the rate of heat transfer during nucleate boiling is essentially independent of the geometry and orientation of the heated surface. The fluid properties in Eq. 10–2 are to be evaluated at the saturation temperature Tsat. The condition of the heater surface greatly affects heat transfer, and the Rohsenow equation given above is applicable to clean and relatively smooth surfaces. The results obtained using the Rohsenow equation can be in error by 100% for the heat transfer rate for a given excess temperature and by 30% for the excess temperature for a given heat transfer rate. Therefore, care should be exercised in the interpretation of the results. Recall from thermodynamics that the enthalpy of vaporization hfg of a pure substance decreases with increasing pressure (or temperature) and reaches zero at the critical point. Noting that hfg appears in the denominator of the Rohsenow equation, we should see a significant rise in the rate of heat transfer at high pressures during nucleate boiling. Peak Heat Flux In the design of boiling heat transfer equipment, it is extremely important for the designer to have a knowledge of the maximum heat flux in order to avoid the danger of burnout. The maximum (or critical) heat flux in nucleate pool boiling was determined theoretically by S. S. Kutateladze in Russia in 1948 and N. Zuber in the United States in 1958 using quite different approaches, and is expressed as (Fig. 10–11) q·max Ccr hfg[g 2 (l )]1/4 (10-3) where Ccr is a constant whose value depends on the heater geometry. Exhaustive experimental studies by Lienhard and his coworkers indicated that the value of Ccr is about 0.15. Specific values of Ccr for different heater geometries are listed in Table 10–4. Note that the heaters are classified as being large or small based on the value of the parameter L. Equation 10–3 will give the maximum heat flux in W/m2 if the properties are used in the units specified earlier in their descriptions following Eq. 10–2. The maximum heat flux is independent of the fluid–heating surface combination, as well as the viscosity, thermal conductivity, and the specific heat of the liquid. Note that v increases but and hfg decrease with increasing pressure, and thus the change in q·max with pressure depends on which effect dominates. The experimental studies of Cichelli and Bonilla indicate that q·max increases with pressure up to about one-third of the critical pressure, and then starts to decrease and becomes zero at the critical pressure. Also note that q·max is proportional to hfg, and large maximum heat fluxes can be obtained using fluids with a large enthalpy of vaporization, such as water. kg J q· m · s kg m kg s2 m3 N m W 1 m m2 W/m2 1/23 J °C kg · °C J kg 3 1/2 (1)3 FIGURE 10–10 Equation 10–2 gives the boiling heat flux in W/m2 when the quantities are expressed in the units specified in their descriptions. TABLE 10–1 Surface tension of liquid–vapor interface for water T, °C , N/m 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 374 0.0757 0.0727 0.0696 0.0662 0.0627 0.0589 0.0550 0.0509 0.0466 0.0422 0,0377 0.0331 0.0284 0.0237 0.0190 0.0144 0.0099 0.0056 0.0019 0.0 Multiply by 0.06852 to convert to lbf/ft or by 2.2046 to convert to lbm/s2. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 524 524 HEAT TRANSFER TABLE 10–2 TABLE 10–3 Surface tension of some fluids (from Suryanarayana, Ref. 26; originally based on data from Jasper, Ref. 14) Values of the coefficient Csf and n for various fluid–surface combinations Substance and Temp. Range Water–copper (polished) Water–copper (scored) Water–stainless steel (mechanically polished) Water–stainless steel (ground and polished) Water–stainless steel (teflon pitted) Water–stainless steel (chemically etched) Water–brass Water–nickel Water–platinum n-Pentane–copper (polished) n-Pentane–chromium Benzene–chromium Ethyl alcohol–chromium Carbon tetrachloride–copper Isopropanol–copper Fluid-Heating Surface Combination Surface Tension, , N/m (T in °C) Ammonia, 75 to 40°C: 0.0264 0.000223T Benzene, 10 to 80°C: 0.0315 0.000129T Butane, 70 to 20°C: 0.0149 0.000121T Carbon dioxide, 30 to 20°C: 0.0043 0.000160T Ethyl alcohol, 10 to 70°C: 0.0241 0.000083T Mercury, 5 to 200°C: 0.4906 0.000205T Methyl alcohol, 10 to 60°C: 0.0240 0.000077T Pentane, 10 to 30°C: 0.0183 0.000110T Propane, 90 to 10°C: 0.0092 0.000087T Multiply by 0.06852 to convert to lbf/ft or by 2.2046 to convert to lbm/s2. Csf n 0.0130 0.0068 0.0130 0.0060 0.0058 0.0130 0.0060 0.0060 0.0130 0.0154 0.0150 0.1010 0.0027 0.0130 0.0025 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.7 1.7 1.7 1.7 1.7 1.7 Minimum Heat Flux Minimum heat flux, which occurs at the Leidenfrost point, is of practical interest since it represents the lower limit for the heat flux in the film boiling regime. Using the stability theory, Zuber derived the following expression for the minimum heat flux for a large horizontal plate, g(l ) q·min 0.09 hfg (l )2 1/4 (10-4) where the constant 0.09 was determined by Berenson in 1961. He replaced the theoretically determined value of 24 by 0.09 to match the experimental data better. Still, the relation above can be in error by 50 percent or more. TABLE 10–4 Values of the coefficient Ccr for use in Eq. 10–3 for maximum heat flux (dimensionless parameter L L[g(l )/]1/2) Heater Geometry Large horizontal flat heater Small horizontal flat heater1 Large horizontal cylinder Small horizontal cylinder Large sphere Small sphere K1 /[g(l v)Aheater] 1 Ccr Charac. Dimension of Heater, L Range of L 0.149 Width or diameter L 27 18.9K1 Width or diameter 9 L 20 0.12 Radius L 1.2 0.12L0.25 Radius 0.15 L 1.2 0.11 Radius L 4.26 0.227L0.5 Radius 0.15 L 4.26 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 525 525 CHAPTER 10 q· Film Boiling Using an analysis similar to Nusselt’s theory on filmwise condensation presented in the next section, Bromley developed a theory for the prediction of heat flux for stable film boiling on the outside of a horizontal cylinder. The heat flux for film boiling on a horizontal cylinder or sphere of diameter D is given by q·film Cfilm gk3 (l )[hfg 0.4Cp (Ts Tsat) D(Ts Tsat) (Ts Tsat) (10-5) 0.62 for horizontal cylinders 0.67 for spheres (10-6) where is the emissivity of the heating surface and 5.67 108 W/m2 · K4 is the Stefan–Boltzman constant. Note that the temperature in this case must be expressed in K, not °C, and that surface tension and the Stefan–Boltzman constant share the same symbol. You may be tempted to simply add the convection and radiation heat transfers to determine the total heat transfer during film boiling. However, these two mechanisms of heat transfer adversely affect each other, causing the total heat transfer to be less than their sum. For example, the radiation heat transfer from the surface to the liquid enhances the rate of evaporation, and thus the thickness of the vapor film, which impedes convection heat transfer. For q·rad q·film, Bromley determined that the relation 3 q·total q·film q·rad 4 Natural convection relations Minimum heat flux relation Ts – Tsat Other properties are as listed before in connection with Eq. 10–2. We used a modified latent heat of vaporization in Eq. 10–5 to account for the heat transfer associated with the superheating of the vapor. The vapor properties are to be evaluated at the film temperature, given as Tf (Ts Tsat)/2, which is the average temperature of the vapor film. The liquid properties and hfg are to be evaluated at the saturation temperature at the specified pressure. Again, this relation will give the film boiling heat flux in W/m2 if the properties are used in the units specified earlier in their descriptions following Eq. 10–2. At high surface temperatures (typically above 300°C), heat transfer across the vapor film by radiation becomes significant and needs to be considered (Fig. 10–12). Treating the vapor film as a transparent medium sandwiched between two large parallel plates and approximating the liquid as a blackbody, radiation heat transfer can be determined from 4 q·rad (Ts4 Tsat ) Nucleate boiling relations 1/4 where kv is the thermal conductivity of the vapor in W/m · °C and Cfilm Film boiling relations Critical heat flux relation (10-7) correlates experimental data well. Operation in the transition boiling regime is normally avoided in the design of heat transfer equipment, and thus no major attempt has been made to develop general correlations for boiling heat transfer in this regime. FIGURE 10–11 Different relations are used to determine the heat flux in different boiling regimes. P = 1 atm 100°C 400°C Vapor q· film boiling q· rad Heating FIGURE 10–12 At high heater surface temperatures, radiation heat transfer becomes significant during film boiling. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 526 526 HEAT TRANSFER Note that the gravitational acceleration g, whose value is approximately 9.81 m/s2 at sea level, appears in all of the relations above for boiling heat transfer. The effects of low and high gravity (as encountered in aerospace applications and turbomachinery) are studied experimentally. The studies confirm that the critical heat flux and heat flux in film boiling are proportional to g1/4. However, they indicate that heat flux in nucleate boiling is practically independent of gravity g, instead of being proportional to g1/2, as dictated by Eq. 10–2. Enhancement of Heat Transfer in Pool Boiling Liquid Vapor Nucleation sites for vapor FIGURE 10–13 The cavities on a rough surface act as nucleation sites and enhance boiling heat transfer. P = 1 atm 100°C The pool boiling heat transfer relations given above apply to smooth surfaces. Below we will discuss some methods to enhance heat transfer in pool boiling. We pointed out earlier that the rate of heat transfer in the nucleate boiling regime strongly depends on the number of active nucleation sites on the surface, and the rate of bubble formation at each site. Therefore, any modification that will enhance nucleation on the heating surface will also enhance heat transfer in nucleate boiling. It is observed that irregularities on the heating surface, including roughness and dirt, serve as additional nucleation sites during boiling, as shown in Figure 10–13. For example, the first bubbles in a pan filled with water are most likely to form at the scratches at the bottom surface. These scratches act like “nests” for the bubbles to form and thus increase the rate of bubble formation. Berensen has shown that heat flux in the nucleate boiling regime can be increased by a factor of 10 by roughening the heating surface. However, these high heat transfer rates cannot be sustained for long since the effect of surface roughness is observed to decay with time, and the heat flux to drop eventually to values encountered on smooth surfaces. The effect of surface roughness is negligible on the critical heat flux and the heat flux in film boiling. Surfaces that provide enhanced heat transfer in nucleate boiling permanently are being manufactured and are available in the market. Enhancement in nucleation and thus heat transfer in such special surfaces is achieved either by coating the surface with a thin layer (much less than 1 mm) of very porous material or by forming cavities on the surface mechanically to facilitate continuous vapor formation. Such surfaces are reported to enhance heat transfer in the nucleate boiling regime by a factor of up to 10, and the critical heat flux by a factor of 3. The enhancement provided by one such material prepared by machine roughening, the thermoexcel-E, is shown in Figure 10–14. The use of finned surfaces is also known to enhance nucleate boiling heat transfer and the critical heat flux. Boiling heat transfer can also be enhanced by other techniques such as mechanical agitation and surface vibration. These techniques are not practical, however, because of the complications involved. Water 108°C Heating FIGURE 10–15 Schematic for Example 10–1. EXAMPLE 10–1 Nucleate Boiling of Water in a Pan Water is to be boiled at atmospheric pressure in a mechanically polished stainless steel pan placed on top of a heating unit, as shown in Figure 10–15. The inner surface of the bottom of the pan is maintained at 108°C. If the diameter of the bottom of the pan is 30 cm, determine (a) the rate of heat transfer to the water and (b) the rate of evaporation of water. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 527 527 CHAPTER 10 Tsat = 0°C 105 0.5 be n tu Finn 104 Pore Tunnel Plai Th ed t ube erm qc′′ (kcal/(m 2 h) oe xc elE Vapor Liquid 1 2 5 (Ts – Tsat ) (°C) FIGURE 10–14 The enhancement of boiling heat transfer in Freon-12 by a mechanically roughened surface, thermoexcel-E. 10 SOLUTION Water is boiled at 1 atm pressure on a stainless steel surface. The rate of heat transfer to the water and the rate of evaporation of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the pan are negligible. Properties The properties of water at the saturation temperature of 100°C are 0.0589 N/m (Table 10–1) and, from Table A-9, l 957.9 kg/m3 v 0.6 kg/m3 Prl 1.75 hfg 2257.0 103 J/kg l 0.282 103 kg · m/s Cpl 4217 J/kg · °C Also, Csf 0.0130 and n 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10–3). Note that we expressed the properties in units specified under Eq. 10–2 in connection with their definitions in order to avoid unit manipulations. Analysis (a) The excess temperature in this case is T Ts Tsat 108 100 8°C which is relatively low (less than 30°C). Therefore, nucleate boiling will occur. The heat flux in this case can be determined from the Rohsenow relation to be Cpl (Ts Tsat) 3 Csf hfg Prnl 9.81 (957.9 0.6) (0.282 103)(2257 103) 0.0589 3 4217(108 100) 0.0130(2257 103)1.75 g(l ) q·nucleate l hfg 1/2 7.20 104 W/m2 The surface area of the bottom of the pan is A D2/4 (0.3 m)2/4 0.07069 m2 1/2 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 528 528 HEAT TRANSFER Then the rate of heat transfer during nucleate boiling becomes · Q boiling Aq·nucleate (0.07069 m2)(7.20 104 W/m2) 5093 W (b) The rate of evaporation of water is determined from m· evaporation Q· boiling 5093 J/s 2.26 103 kg/s hfg 2257 103 J/kg That is, water in the pan will boil at a rate of more than 2 grams per second. EXAMPLE 10–2 Water in a tank is to be boiled at sea level by a 1-cm-diameter nickel plated steel heating element equipped with electrical resistance wires inside, as shown in Figure 10–16. Determine the maximum heat flux that can be attained in the nucleate boiling regime and the surface temperature of the heater surface in that case. P = 1 atm Water, 100°C d Ts = ? Heating element FIGURE 10–16 Schematic for Example 10–2. Peak Heat Flux in Nucleate Boiling q· max SOLUTION Water is boiled at 1 atm pressure on a nickel plated steel surface. The maximum (critical) heat flux and the surface temperature are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. Properties The properties of water at the saturation temperature of 100°C are 0.0589 N/m (Table 10–1) and, from Table A-9, l 957.9 kg/m3 v 0.6 kg/m3 Prl 1.75 hfg 2257 103 J/kg l 0.282 103 kg · m/s Cpl 4217 J/kg · °C Also, Csf 0.0060 and n 1.0 for the boiling of water on a nickel plated surface (Table 10–3). Note that we expressed the properties in units specified under Eqs. 10–2 and 10–3 in connection with their definitions in order to avoid unit manipulations. Analysis The heating element in this case can be considered to be a short cylinder whose characteristic dimension is its radius. That is, L r 0.005 m. The dimensionless parameter L and the constant Ccr are determined from Table 10–4 to be g(l ) L L 1/2 (0.005) (9.81)(957.8 0.6) 0.0589 1/2 2.00 1.2 which corresponds to Ccr 0.12. Then the maximum or critical heat flux is determined from Eq. 10–3 to be q·max Ccr hfg [g 2 (l )]1/4 0.12(2257 103)[0.0589 9.8 (0.6)2(957.9 0.6)]1/4 1.02 106 W/m2 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 529 529 CHAPTER 10 The Rohsenow relation, which gives the nucleate boiling heat flux for a specified surface temperature, can also be used to determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into Eq. 10–2 together with other properties gives g(l ) q·nucleate l hfg Cpl (Ts Tsat) Csf hfg Prnl 1/2 3 0.6) 9.81(957.9 0.0589 1/2 1,017,200 (0.282 103)(2257 103) 4217(T 100) 0.0130(2257 10 ) 1.75 s 3 Ts 119°C Discussion Note that heat fluxes on the order of 1 MW/m2 can be obtained in nucleate boiling with a temperature difference of less than 20°C. P = 1 atm EXAMPLE 10–3 Film Boiling of Water on a Heating Element Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter D 5 mm and emissivity 0.05 immersed in water, as shown in Figure 10–17. If the surface temperature of the heating wire is 350°C, determine the rate of heat transfer from the wire to the water per unit length of the wire. SOLUTION Water is boiled at 1 atm by a horizontal polished copper heating element. The rate of heat transfer to the water per unit length of the heater is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. Properties The properties of water at the saturation temperature of 100°C are hfg 2257 103 J/kg and l 957.9 kg/m3 (Table A-9). The properties of vapor at the film temperature of Tf (Tsat Ts)/2 (100 350)/2 225°C 498 K (which is sufficiently close to 500 K) are, from Table A-16, 0.441 kg/m3 1.73 105 kg/m · s Cp 1977 J/kg · °C k 0.0357 W/m · °C Note that we expressed the properties in units that will cancel each other in boiling heat transfer relations. Also note that we used vapor properties at 1 atm pressure from Table A-16 instead of the properties of saturated vapor from Table A-9 at 250°C since the latter are at the saturation pressure of 4.0 MPa. Analysis The excess temperature in this case is T Ts Tsat 350 100 250°C, which is much larger than 30°C for water. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from Eq. 10–5 to be 100°C Heating element Vapor film FIGURE 10–17 Schematic for Example 10–3. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 530 530 HEAT TRANSFER q·film 0.62 gk3 (l )[hfg 0.4Cp (Ts Tsat)] D(Ts Tsat) 9.81(0.0357)3 (0.441)(957.9 0.441) [(2257 103 0.4 1977(250)] 0.62 (1.73 105)(5 103)(250) 5.93 104 W/m2 1/4 (Ts Tsat) 1/4 250 The radiation heat flux is determined from Eq. 10–6 to be 4 q·rad (Ts4 Tsat ) (0.05)(5.67 108 W/m2 · K4)[(250 273 K)4 (100 273 K)4] 157 W/m2 Note that heat transfer by radiation is negligible in this case because of the low emissivity of the surface and the relatively low surface temperature of the heating element. Then the total heat flux becomes (Eq. 10–7) 3 3 q·total q·film q·rad 5.93 104 157 5.94 104 W/m2 4 4 Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat transfer surface area, · Q total Aq·total ( DL)q·total ( 0.005 m 1 m)(5.94 104 W/m2) 933 W q· q· max High velocity Low velocity Discussion Note that the 5-mm-diameter copper heating element will consume about 1 kW of electric power per unit length in steady operation in the film boiling regime. This energy is transferred to the water through the vapor film that forms around the wire. 10–3 lo c h ve Hig it y Nucleate pool boiling regime io n oc vel w o L ity Fre e onv ec ct ∆Texcess FIGURE 10–18 The effect of forced convection on external flow boiling for different flow velocities. ■ FLOW BOILING The pool boiling we considered so far involves a pool of seemingly motionless liquid, with vapor bubbles rising to the top as a result of buoyancy effects. In flow boiling, the fluid is forced to move by an external source such as a pump as it undergoes a phase-change process. The boiling in this case exhibits the combined effects of convection and pool boiling. The flow boiling is also classified as either external and internal flow boiling depending on whether the fluid is forced to flow over a heated surface or inside a heated tube. External flow boiling over a plate or cylinder is similar to pool boiling, but the added motion increases both the nucleate boiling heat flux and the critical heat flux considerably, as shown in Figure 10–18. Note that the higher the velocity, the higher the nucleate boiling heat flux and the critical heat flux. In experiments with water, critical heat flux values as high as 35 MW/m2 have been obtained (compare this to the pool boiling value of 1.3 MW/m2 at 1 atm pressure) by increasing the fluid velocity. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 531 531 CHAPTER 10 Internal flow boiling is much more complicated in nature because there is no free surface for the vapor to escape, and thus both the liquid and the vapor are forced to flow together. The two-phase flow in a tube exhibits different flow boiling regimes, depending on the relative amounts of the liquid and the vapor phases. This complicates the analysis even further. The different stages encountered in flow boiling in a heated tube are illustrated in Figure 10–19 together with the variation of the heat transfer coefficient along the tube. Initially, the liquid is subcooled and heat transfer to the liquid is by forced convection. Then bubbles start forming on the inner surfaces of the tube, and the detached bubbles are drafted into the mainstream. This gives the fluid flow a bubbly appearance, and thus the name bubbly flow regime. As the fluid is heated further, the bubbles grow in size and eventually coalesce into slugs of vapor. Up to half of the volume in the tube in this slugflow regime is occupied by vapor. After a while the core of the flow consists of vapor only, and the liquid is confined only in the annular space between the vapor core and the tube walls. This is the annular-flow regime, and very high heat transfer coefficients are realized in this regime. As the heating continues, the annular liquid layer gets thinner and thinner, and eventually dry spots start to appear on the inner surfaces of the tube. The appearance of dry spots is accompanied by a sharp decrease in the heat transfer coefficient. This transition regime continues until the inner surface of the tube is completely dry. Any liquid at this moment is in the form of droplets suspended in the vapor core, which resembles a mist, and we have a mist-flow regime until all the liquid droplets are vaporized. At the end of the mist-flow regime we have saturated vapor, which becomes superheated with any further heat transfer. Note that the tube contains a liquid before the bubbly flow regime and a vapor after the mist-flow regime. Heat transfer in those two cases can be determined using the appropriate relations for single-phase convection heat transfer. Many correlations are proposed for the determination of heat transfer High x=1 Forced convection Liquid droplets Low Mist flow Vapor core Bubbles in liquid Liquid core Quality Transition flow Annular flow Slug flow Bubbly flow x=0 Forced convection Coefficient of heat transfer FIGURE 10–19 Different flow regimes encountered in flow boiling in a tube under forced convection. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 532 532 HEAT TRANSFER in the two-phase flow (bubbly flow, slug-flow, annular-flow, and mist-flow) cases, but they are beyond the scope of this introductory text. A crude estimate for heat flux in flow boiling can be obtained by simply adding the forced convection and pool boiling heat fluxes. 10–4 80°C 80°C Droplets Liquid film (a) Film condensation (b) Dropwise condensation FIGURE 10–20 When a vapor is exposed to a surface at a temperature below Tsat, condensation in the form of a liquid film or individual droplets occurs on the surface. Cold 0 plate y x g · m(x) Vapor, Liquid-vapor interface T(y) Ts Tsat Temperature profile Tv, Velocity (y) profile Liquid, l FIGURE 10–21 Film condensation on a vertical plate. ■ CONDENSATION HEAT TRANSFER Condensation occurs when the temperature of a vapor is reduced below its saturation temperature Tsat. This is usually done by bringing the vapor into contact with a solid surface whose temperature Ts is below the saturation temperature Tsat of the vapor. But condensation can also occur on the free surface of a liquid or even in a gas when the temperature of the liquid or the gas to which the vapor is exposed is below Tsat. In the latter case, the liquid droplets suspended in the gas form a fog. In this chapter, we will consider condensation on solid surfaces only. Two distinct forms of condensation are observed: film condensation and dropwise condensation. In film condensation, the condensate wets the surface and forms a liquid film on the surface that slides down under the influence of gravity. The thickness of the liquid film increases in the flow direction as more vapor condenses on the film. This is how condensation normally occurs in practice. In dropwise condensation, the condensed vapor forms droplets on the surface instead of a continuous film, and the surface is covered by countless droplets of varying diameters (Fig. 10–20). In film condensation, the surface is blanketed by a liquid film of increasing thickness, and this “liquid wall” between solid surface and the vapor serves as a resistance to heat transfer. The heat of vaporization hfg released as the vapor condenses must pass through this resistance before it can reach the solid surface and be transferred to the medium on the other side. In dropwise condensation, however, the droplets slide down when they reach a certain size, clearing the surface and exposing it to vapor. There is no liquid film in this case to resist heat transfer. As a result, heat transfer rates that are more than 10 times larger than those associated with film condensation can be achieved with dropwise condensation. Therefore, dropwise condensation is the preferred mode of condensation in heat transfer applications, and people have long tried to achieve sustained dropwise condensation by using various vapor additives and surface coatings. These attempts have not been very successful, however, since the dropwise condensation achieved did not last long and converted to film condensation after some time. Therefore, it is common practice to be conservative and assume film condensation in the design of heat transfer equipment. 10–5 ■ FILM CONDENSATION We now consider film condensation on a vertical plate, as shown in Figure 10–21. The liquid film starts forming at the top of the plate and flows downward under the influence of gravity. The thickness of the film increases in the flow direction x because of continued condensation at the liquid–vapor interface. Heat in the amount hfg (the latent heat of vaporization) is released during condensation and is transferred through the film to the plate surface at temperature Ts. Note that Ts must be below the saturation temperature Tsat of the vapor for condensation to occur. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 533 533 CHAPTER 10 Typical velocity and temperature profiles of the condensate are also given in Figure 10–21. Note that the velocity of the condensate at the wall is zero because of the “no-slip” condition and reaches a maximum at the liquid–vapor interface. The temperature of the condensate is Tsat at the interface and decreases gradually to Ts at the wall. As was the case in forced convection involving a single phase, heat transfer in condensation also depends on whether the condensate flow is laminar or turbulent. Again the criterion for the flow regime is provided by the Reynolds number, which is defined as Re Dh l l 4 Ac l l 4 l l 4m· p l pl l l (10-8) where Dh 4Ac /p 4 hydraulic diameter of the condensate flow, m p wetted perimeter of the condensate, m Ac p wetted perimeter film thickness, m2, cross-sectional area of the condensate flow at the lowest part of the flow l density of the liquid, kg/m3 l viscosity of the liquid, kg/m · s average velocity of the condensate at the lowest part of the flow, m/s m· l l Ac mass flow rate of the condensate at the lowest part, kg/s The evaluation of the hydraulic diameter Dh for some common geometries is illustrated in Figure 10–22. Note that the hydraulic diameter is again defined such that it reduces to the ordinary diameter for flow in a circular tube, as was done in Chapter 8 for internal flow, and it is equivalent to 4 times the thickness of the condensate film at the location where the hydraulic diameter is evaluated. That is, Dh 4. The latent heat of vaporization hfg is the heat released as a unit mass of vapor condenses, and it normally represents the heat transfer per unit mass of condensate formed during condensation. However, the condensate in an actual L D L D δ δ p=L Ac = L δ 4A Dh = ——c = 4δ p (a) Vertical plate p = πD Ac = π Dδ 4A Dh = ——c = 4δ p (b) Vertical cylinder δ p = 2L Ac = 2Lδ 4A Dh = ——c = 4δ p (c) Horizontal cylinder FIGURE 10–22 The wetted perimeter p, the condensate cross-sectional area Ac, and the hydraulic diameter Dh for some common geometries. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 534 534 HEAT TRANSFER condensation process is cooled further to some average temperature between Tsat and Ts, releasing more heat in the process. Therefore, the actual heat transfer will be larger. Rohsenow showed in 1956 that the cooling of the liquid below the saturation temperature can be accounted for by replacing hfg by the modified latent heat of vaporization hfg, defined as hfg hfg 0.68Cpl (Tsat Ts) (10-9a) where Cpl is the specific heat of the liquid at the average film temperature. We can have a similar argument for vapor that enters the condenser as superheated vapor at a temperature T instead of as saturated vapor. In this case the vapor must be cooled first to Tsat before it can condense, and this heat must be transferred to the wall as well. The amount of heat released as a unit mass of superheated vapor at a temperature T is cooled to Tsat is simply Cp (T Tsat), where Cp is the specific heat of the vapor at the average temperature of (T Tsat)/2. The modified latent heat of vaporization in this case becomes hfg hfg 0.68Cpl (Tsat Ts) Cp (T Tsat) (10-9b) With these considerations, the rate of heat transfer can be expressed as · Q conden hAs(Tsat Ts) mhfg (10-10) where As is the heat transfer area (the surface area on which condensation occurs). Solving for m· from the equation above and substituting it into Eq. 10–8 gives yet another relation for the Reynolds number, Re Re = 0 Laminar (wave-free) Re ≅ 30 4Q· conden 4As h(Tsat Ts) pl hfg pl hfg (10-11) This relation is convenient to use to determine the Reynolds number when the condensation heat transfer coefficient or the rate of heat transfer is known. The temperature of the liquid film varies from Tsat on the liquid–vapor interface to Ts at the wall surface. Therefore, the properties of the liquid should be evaluated at the film temperature Tf (Tsat Ts)/2, which is approximately the average temperature of the liquid. The hfg, however, should be evaluated at Tsat since it is not affected by the subcooling of the liquid. Flow Regimes Laminar (wavy) Re ≅ 1800 Turbulent FIGURE 10–23 Flow regimes during film condensation on a vertical plate. The Reynolds number for condensation on the outer surfaces of vertical tubes or plates increases in the flow direction due to the increase of the liquid film thickness . The flow of liquid film exhibits different regimes, depending on the value of the Reynolds number. It is observed that the outer surface of the liquid film remains smooth and wave-free for about Re 30, as shown in Figure 10–23, and thus the flow is clearly laminar. Ripples or waves appear on the free surface of the condensate flow as the Reynolds number increases, and the condensate flow becomes fully turbulent at about Re 1800. The condensate flow is called wavy-laminar in the range of 450 Re 1800 and turbulent for Re 1800. However, some disagreement exists about the value of Re at which the flow becomes wavy-laminar or turbulent. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 535 535 CHAPTER 10 δ–y Heat Transfer Correlations for Film Condensation Below we discuss relations for the average heat transfer coefficient h for the case of laminar film condensation for various geometries. dx Shear force du µl — (bdx) dy 1 Vertical Plates Consider a vertical plate of height L and width b maintained at a constant temperature Ts that is exposed to vapor at the saturation temperature Tsat. The downward direction is taken as the positive x-direction with the origin placed at the top of the plate where condensation initiates, as shown in Figure 10–24. The surface temperature is below the saturation temperature (Ts Tsat) and thus the vapor condenses on the surface. The liquid film flows downward under the influence of gravity. The film thickness and thus the mass flow rate of the condensate increases with x as a result of continued condensation on the existing film. Then heat transfer from the vapor to the plate must occur through the film, which offers resistance to heat transfer. Obviously the thicker the film, the larger its thermal resistance and thus the lower the rate of heat transfer. The analytical relation for the heat transfer coefficient in film condensation on a vertical plate described above was first developed by Nusselt in 1916 under the following simplifying assumptions: 1. Both the plate and the vapor are maintained at constant temperatures of Ts and Tsat, respectively, and the temperature across the liquid film varies linearly. 2. Heat transfer across the liquid film is by pure conduction (no convection currents in the liquid film). 3. The velocity of the vapor is low (or zero) so that it exerts no drag on the condensate (no viscous shear on the liquid–vapor interface). 4. The flow of the condensate is laminar and the properties of the liquid are constant. 5. The acceleration of the condensate layer is negligible. Then Newton’s second law of motion for the volume element shown in Figure 10–24 in the vertical x-direction can be written as F x max 0 since the acceleration of the fluid is zero. Noting that the only force acting downward is the weight of the liquid element, and the forces acting upward are the viscous shear (or fluid friction) force at the left and the buoyancy force, the force balance on the volume element becomes Fdownward ↓ Fupward ↑ Weight Viscous shear force Buoyancy force du (bdx) g( y)(bdx) l g( y)(bdx) l dy Canceling the plate width b and solving for du/dy gives du g(l )g( y) l dy Weight ρl g(δ – y) (bdx) Buoyancy force ρv g(δ – y) (bdx) y 0 x δ dx =0 at y = 0 y Idealized velocity profile No vapor drag Idealized temperature profile Ts g Liquid, l Tsat Linear FIGURE 10–24 The volume element of condensate on a vertical plate considered in Nusselt’s analysis. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 536 536 HEAT TRANSFER Integrating from y 0 where u 0 (because of the no-slip boundary condition) to y y where u u(y) gives u(y) g(l )g y2 y l 2 (10-12) The mass flow rate of the condensate at a location x, where the boundary layer thickness is , is determined from m· (x) u(y)dA l A y0 l u(y)bdy (10-13) Substituting the u(y) relation from Equation 10–12 into Eq. 10–13 gives gbl(l ) m· (x) 3l 3 (10-14) whose derivative with respect to x is gbl(l )2 d dm· l dx dx (10-15) which represents the rate of condensation of vapor over a vertical distance dx. The rate of heat transfer from the vapor to the plate through the liquid film is simply equal to the heat released as the vapor is condensed and is expressed as Tsat Ts kl b Tsat Ts · dm· dQ hfg dm· kl (bdx) → dx hfg (10-16) Equating Eqs. 10–15 and 10–16 for dm· /dx to each other and separating the variables give 3 d l kl (Tsat Ts) dx gl (l )hfg (10-17) Integrating from x 0 where 0 (the top of the plate) to x x where (x), the liquid film thickness at any location x is determined to be (x) 4g k((T )hT )x l l l sat l 1/4 s (10-18) fg The heat transfer rate from the vapor to the plate at a location x can be expressed as Tsat Ts kl q·x hx(Tsat Ts) kl → hx (x) (10-19) Substituting the (x) expression from Eq. 10–18, the local heat transfer coefficient hx is determined to be hx gl (l )hfg k3l l sat Ts)x 4 (T 1/4 (10-20) cen58933_ch10.qxd 9/4/2002 12:38 PM Page 537 537 CHAPTER 10 The average heat transfer coefficient over the entire plate is determined from its definition by substituting the hx relation and performing the integration. It gives h have 1 L L 0 gl (l )hfg k3l 4 hx dx hx L 0.943 3 l (Tsat Ts)L 1/4 (10-21) Equation 10–21, which is obtained with the simplifying assumptions stated earlier, provides good insight on the functional dependence of the condensation heat transfer coefficient. However, it is observed to underpredict heat transfer because it does not take into account the effects of the nonlinear temperature profile in the liquid film and the cooling of the liquid below the saturation temperature. Both of these effects can be accounted for by replacing hfg by hfg given by Eq. 10–9. With this modification, the average heat transfer coefficient for laminar film condensation over a vertical flat plate of height L is determined to be hvert 0.943 gl (l )hfg k3l l sat Ts)L (T 1/4 (W/m2 · °C), 0 Re 30 (10-22) where g gravitational acceleration, m/s2 l, densities of the liquid and vapor, respectively, kg/m3 l viscosity of the liquid, kg/m · s hfg hfg 0.68Cpl (Tsat Ts) modified latent heat of vaporization, J/kg kl thermal conductivity of the liquid, W/m · °C L height of the vertical plate, m Ts surface temperature of the plate, °C Tsat saturation temperature of the condensing fluid, °C At a given temperature, l and thus l l except near the critical point of the substance. Using this approximation and substituting Eqs. 10–14 kl and and 10–18 at x L into Eq. 10–8 by noting that x L hxL 4 hvert hx L (Eqs. 10–19 and 10–21) give 3 Re 4gl (l )3 4g2l kl 32l 32l hxL 3 kl 4g 2 3h 3 l vert/4 3 (10-23) Then the heat transfer coefficient hvert in terms of Re becomes hvert 1.47kl Re1/3 1/3 g 2 l , 0 Re 30 l (10-24) The results obtained from the theoretical relations above are in excellent agreement with the experimental results. It can be shown easily that using property values in Eqs. 10–22 and 10–24 in the specified units gives the condensation heat transfer coefficient in W/m2 · °C, thus saving one from having cen58933_ch10.qxd 9/4/2002 12:38 PM Page 538 538 HEAT TRANSFER 3 1/4 W m kg kg J s2 m3 m3 kg m · °C hvert kg m · s · °C · m J W3 m 1 s 6 3 m m · °C3 °C 1/4 W4 8 m · °C4 W/m2 · °C FIGURE 10–25 Equation 10–22 gives the condensation heat transfer coefficient in W/m2 · °C when the quantities are expressed in the units specified in their descriptions. to go through tedious unit manipulations each time (Fig. 10–25). This is also true for the equations below. All properties of the liquid are to be evaluated at the film temperature Tf (Tsat Ts)/2. The hfg and are to be evaluated at the saturation temperature Tsat. Wavy Laminar Flow on Vertical Plates At Reynolds numbers greater than about 30, it is observed that waves form at the liquid–vapor interface although the flow in liquid film remains laminar. The flow in this case is said to be wavy laminar. The waves at the liquid– vapor interface tend to increase heat transfer. But the waves also complicate the analysis and make it very difficult to obtain analytical solutions. Therefore, we have to rely on experimental studies. The increase in heat transfer due to the wave effect is, on average, about 20 percent, but it can exceed 50 percent. The exact amount of enhancement depends on the Reynolds number. Based on his experimental studies, Kutateladze (1963, Ref. 15) recommended the following relation for the average heat transfer coefficient in wavy laminar condensate flow for l and 30 Re 1800, hvert, wavy 1/3 Re kl g 1.08 Re1.22 5.2 2l , 30 Re 1800 l (10-25) A simpler alternative to the relation above proposed by Kutateladze (1963, Ref. 15) is hvert, wavy 0.8 Re0.11 hvert (smooth) (10-26) which relates the heat transfer coefficient in wavy laminar flow to that in wave-free laminar flow. McAdams (1954, Ref. 2) went even further and suggested accounting for the increase in heat transfer in the wavy region by simply increasing the heat transfer coefficient determined from Eq. 10–22 for the laminar case by 20 percent. Holman (1990) suggested using Eq. 10–22 for the wavy region also, with the understanding that this is a conservative approach that provides a safety margin in thermal design. In this book we will use Eq. 10–25. A relation for the Reynolds number in the wavy laminar region can be determined by substituting the h relation in Eq. 10–25 into the Re relation in Eq. 10–11 and simplifying. It yields Revert, wavy 4.81 3.70 Lkl (Tsat Ts) g l hfg 2l 1/3 0.820 , v l (10-27) Turbulent Flow on Vertical Plates At a Reynolds number of about 1800, the condensate flow becomes turbulent. Several empirical relations of varying degrees of complexity are proposed for the heat transfer coefficient for turbulent flow. Again assuming l for simplicity, Labuntsov (1957, Ref. 17) proposed the following relation for the turbulent flow of condensate on vertical plates: hvert, turbulent 1/3 Re kl g 8750 58 Pr0.5 (Re0.75 253) 2l , Re 1800 l (10-28) cen58933_ch10.qxd 9/4/2002 12:38 PM Page 539 539 CHAPTER 10 1.0 Pr = 10 Eq. 10-24 5 h (νl2/g)1/ 3 ———— — kl 3 Eq. 10-25 2 1 Eq. 10-28 Wave-free laminar 0.1 10 Wavy laminar 30 100 Turbulent 1000 1800 Re 10,000 FIGURE 10–26 Nondimensionalized heat transfer coefficients for the wave-free laminar, wavy laminar, and turbulent flow of condensate on vertical plates. The physical properties of the condensate are again to be evaluated at the film temperature Tf (Tsat Ts)/2. The Re relation in this case is obtained by substituting the h relation above into the Re relation in Eq. 10–11, which gives Revert, turbulent 0.0690 Lk Prh (T 0.5 l l fg sat Ts) g 2l 1/3 4/3 151 Pr0.5 253 (10-29) Nondimensionalized heat transfer coefficients for the wave-free laminar, wavy laminar, and turbulent flow of condensate on vertical plates are plotted in Figure 10–26. 2 Inclined Plates Equation 10–12 was developed for vertical plates, but it can also be used for laminar film condensation on the upper surfaces of plates that are inclined by an angle from the vertical, by replacing g in that equation by g cos (Fig. 10–27). This approximation gives satisfactory results especially for 60°. Note that the condensation heat transfer coefficients on vertical and inclined plates are related to each other by hinclined hvert (cos )1/4 (laminar) Vapor θ (10-30) Equation 10–30 is developed for laminar flow of condensate, but it can also be used for wavy laminar flows as an approximation. Inclined plate Condensate 3 Vertical Tubes Equation 10–22 for vertical plates can also be used to calculate the average heat transfer coefficient for laminar film condensation on the outer surfaces of vertical tubes provided that the tube diameter is large relative to the thickness of the liquid film. 4 Horizontal Tubes and Spheres Nusselt’s analysis of film condensation on vertical plates can also be extended to horizontal tubes and spheres. The average heat transfer coefficient for film condensation on the outer surfaces of a horizontal tube is determined to be FIGURE 10–27 Film condensation on an inclined plate. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 540 540 HEAT TRANSFER hhoriz 0.729 gl (l ) hfg k3l l(Tsat Ts)D 1/4 (W/m2 · °C) (10-31) where D is the diameter of the horizontal tube. Equation 10–31 can easily be modified for a sphere by replacing the constant 0.729 by 0.815. A comparison of the heat transfer coefficient relations for a vertical tube of height L and a horizontal tube of diameter D yields 1/4 hvert D 1.29 L hhoriz (10-32) Setting hvertical hhorizontal gives L 1.294 D 2.77D, which implies that for a tube whose length is 2.77 times its diameter, the average heat transfer coefficient for laminar film condensation will be the same whether the tube is positioned horizontally or vertically. For L 2.77D, the heat transfer coefficient will be higher in the horizontal position. Considering that the length of a tube in any practical application is several times its diameter, it is common practice to place the tubes in a condenser horizontally to maximize the condensation heat transfer coefficient on the outer surfaces of the tubes. 5 Horizontal Tube Banks Horizontal tubes stacked on top of each other as shown in Figure 10–28 are commonly used in condenser design. The average thickness of the liquid film at the lower tubes is much larger as a result of condensate falling on top of them from the tubes directly above. Therefore, the average heat transfer coefficient at the lower tubes in such arrangements is smaller. Assuming the condensate from the tubes above to the ones below drain smoothly, the average film condensation heat transfer coefficient for all tubes in a vertical tier can be expressed as hhoriz, N tubes 0.729 FIGURE 10–28 Film condensation on a vertical tier of horizontal tubes. gl (l ) hfg k3l l (Tsat Ts) ND 1/4 1 h N1/4 horiz, 1 tube (10-33) Note that Eq. 10–33 can be obtained from the heat transfer coefficient relation for a horizontal tube by replacing D in that relation by ND. This relation does not account for the increase in heat transfer due to the ripple formation and turbulence caused during drainage, and thus generally yields conservative results. Effect of Vapor Velocity In the analysis above we assumed the vapor velocity to be small and thus the vapor drag exerted on the liquid film to be negligible, which is usually the case. However, when the vapor velocity is high, the vapor will “pull” the liquid at the interface along since the vapor velocity at the interface must drop to the value of the liquid velocity. If the vapor flows downward (i.e., in the same direction as the liquid), this additional force will increase the average velocity of the liquid and thus decrease the film thickness. This, in turn, will decrease the thermal resistance of the liquid film and thus increase heat transfer. Upward vapor flow has the opposite effects: the vapor exerts a force on the cen58933_ch10.qxd 9/4/2002 12:38 PM Page 541 541 CHAPTER 10 liquid in the opposite direction to flow, thickens the liquid film, and thus decreases heat transfer. Condensation in the presence of high vapor flow is studied [e.g., Shekriladze and Gomelauri (1966), Ref. 23] and heat transfer relations are obtained, but a detailed analysis of this topic is beyond the scope of this introductory text. The Presence of Noncondensable Gases in Condensers Most condensers used in steam power plants operate at pressures well below the atmospheric pressure (usually under 0.1 atm) to maximize cycle thermal efficiency, and operation at such low pressures raises the possibility of air (a noncondensable gas) leaking into the condensers. Experimental studies show that the presence of noncondensable gases in the vapor has a detrimental effect on condensation heat transfer. Even small amounts of a noncondensable gas in the vapor cause significant drops in heat transfer coefficient during condensation. For example, the presence of less than 1 percent (by mass) of air in steam can reduce the condensation heat transfer coefficient by more than half. Therefore, it is common practice to periodically vent out the noncondensable gases that accumulate in the condensers to ensure proper operation. The drastic reduction in the condensation heat transfer coefficient in the presence of a noncondensable gas can be explained as follows: When the vapor mixed with a noncondensable gas condenses, only the noncondensable gas remains in the vicinity of the surface (Fig. 10–29). This gas layer acts as a barrier between the vapor and the surface, and makes it difficult for the vapor to reach the surface. The vapor now must diffuse through the noncondensable gas first before reaching the surface, and this reduces the effectiveness of the condensation process. Experimental studies show that heat transfer in the presence of a noncondensable gas strongly depends on the nature of the vapor flow and the flow velocity. As you would expect, a high flow velocity is more likely to remove the stagnant noncondensable gas from the vicinity of the surface, and thus improve heat transfer. EXAMPLE 10–4 Vapor + Noncondensable gas Cold surface Condensate Noncondensable gas Vapor FIGURE 10–29 The presence of a noncondensable gas in a vapor prevents the vapor molecules from reaching the cold surface easily, and thus impedes condensation heat transfer. Condensation of Steam on a Vertical Plate Saturated steam at atmospheric pressure condenses on a 2-m-high and 3-mwide vertical plate that is maintained at 80°C by circulating cooling water through the other side (Fig. 10–30). Determine (a) the rate of heat transfer by condensation to the plate and (b) the rate at which the condensate drips off the plate at the bottom. SOLUTION Saturated steam at 1 atm condenses on a vertical plate. The rates of heat transfer and condensation are to be determined. Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the entire plate (will be verified). 4 The density of vapor is much smaller than the density of liquid, l. Properties The properties of water at the saturation temperature of 100°C are hfg 2257 103 J/kg and 0.60 kg/m3. The properties of liquid water at the film temperature of Tf (Tsat Ts)/2 (100 80)/2 90°C are (Table A-9) 1 atm 3m Ts = 80°C 2m Condensate FIGURE 10–30 Schematic for Example 10–4. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 542 542 HEAT TRANSFER l 965.3 kg/m3 l 0.315 103 kg/m · s l l /l 0.326 106 m2/s Cpl 4206 J/kg · °C kl 0.675 W/m · °C Analysis (a) The modified latent heat of vaporization is hfg hfg 0.68Cpl (Tsat Ts) 2257 103 J/kg 0.68 (4206 J/kg · °C)(100 80)°C 2314 103 J/kg For wavy-laminar flow, the Reynolds number is determined from Eq. 10–27 to be 3.70 Lkl (Tsat Ts) g 1/3 0.820 l hfg 2l 3.70(3 m)(0.675 W/m · °C)(100 90)°C 4.81 (0.315 103 kg/m · s)(2314 103 J/kg) 1/3 0.82 9.81 m/s2 6 2 2 (0.326 10 m /s) 1287 Re Revertical, wavy 4.81 which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer coefficient is determined from Eq. 10–25 to be Re kl g 1/3 1.08 Re1.22 5.2 2l 1287 (0.675 W/m · °C) 9.81 m/s2 1.08(1287)1.22 5.2 (0.326 106 m2/s)2 h hvertical, wavy 1/3 5848 W/m2 · °C The heat transfer surface area of the plate is As W L (3 m)(2 m) 6 m2. Then the rate of heat transfer during this condensation process becomes · Q hAs(Tsat Ts) (5848 W/m2 · °C)(6 m2)(100 80)°C 7.02 105 W (b) The rate of condensation of steam is determined from Steam 100°C Q· 7.02 105 J/s m· condensation 0.303 kg/s hfg 2314 103 J/kg 30° That is, steam will condense on the surface at a rate of 303 grams per second. 80°C EXAMPLE 10–5 FIGURE 10–31 Schematic for Example 10–5. Condensation of Steam on a Tilted Plate What would your answer be to the preceding example problem if the plate were tilted 30° from the vertical, as shown in Figure 10–31? cen58933_ch10.qxd 9/4/2002 12:38 PM Page 543 543 CHAPTER 10 SOLUTION (a) The heat transfer coefficient in this case can be determined from the vertical plate relation by replacing g by g cos . But we will use Eq. 10–30 instead since we already know the value for the vertical plate from the preceding example: h hinclined hvert (cos )1/4 (5848 W/m2 · °C)(cos 30°)1/4 5641 W/m2 · °C The heat transfer surface area of the plate is still 6 m2. Then the rate of condensation heat transfer in the tilted plate case becomes · Q hAs(Tsat Ts) (5641 W/m2 · °C)(6 m2)(100 80)°C 6.77 105 W (b) The rate of condensation of steam is again determined from Q· 6.77 105 J/s m· condensation 0.293 kg/s hfg 2314 103 J/kg Discussion Note that the rate of condensation decreased by about 3.6 percent when the plate is tilted. EXAMPLE 10–6 Condensation of Steam on Horizontal Tubes The condenser of a steam power plant operates at a pressure of 7.38 kPa. Steam at this pressure condenses on the outer surfaces of horizontal pipes through which cooling water circulates. The outer diameter of the pipes is 3 cm, and the outer surfaces of the pipes are maintained at 30°C (Fig. 10–32). Determine (a) the rate of heat transfer to the cooling water circulating in the pipes and (b) the rate of condensation of steam per unit length of a horizontal pipe. SOLUTION Saturated steam at a pressure of 7.38 kPa (Table A-9) condenses on a horizontal tube at 30°C. The rates of heat transfer and condensation are to be determined. Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. Properties The properties of water at the saturation temperature of 40°C corresponding to 7.38 kPa are hfg 2407 103 J/kg and 0.05 kg/m3. The properties of liquid water at the film temperature of Tf (Tsat Ts)/2 (40 30)/2 35°C are (Table A-9) l 994 kg/m3 l 0.720 103 kg/m · s Cpl 4178 J/kg · °C kl 0.623 W/m · °C Analysis (a) The modified latent heat of vaporization is hfg hfg 0.68Cpl (Tsat Ts) 2407 103 J/kg 0.68 (4178 J/kg · °C)(40 30)°C 2435 103 J/kg Noting that l (since 0.05 994), the heat transfer coefficient for condensation on a single horizontal tube is determined from Eq. 10–31 to be Steam, 40°C 30°C Cooling water FIGURE 10–32 Schematic for Example 10–6. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 544 544 HEAT TRANSFER 1/4 gl (l ) hfg k3l 1/4 g2l hfg k3l 0.729 (Tsat Ts) D 1 (Tsat Ts) D (9.81 m/s2)(994 kg/m3)2 (2435 103 J/kg)(0.623 W/m · °C)3 0.729 (0.720 103 kg/m ·s)(40 30)°C(0.03 m) 2 9292 W/m · °C h hhorizontal 0.729 1/4 The heat transfer surface area of the pipe per unit of its length is As DL (0.03 m)(1 m) 0.09425 m2. Then the rate of heat transfer during this condensation process becomes · Q hAs(Tsat Ts) (9292 W/m2 · °C)(0.09425 m2)(40 30)°C 8758 W (b) The rate of condensation of steam is Q· 8578 J/s m· condensation 0.00360 kg/s hfg 2435 103 J/kg Therefore, steam will condense on the horizontal tube at a rate of 3.6 g/s or 12.9 kg/h per meter of its length. EXAMPLE 10–7 Condensation of Steam on Horizontal Tube Banks Repeat the proceeding example problem for the case of 12 horizontal tubes arranged in a rectangular array of 3 tubes high and 4 tubes wide, as shown in Figure 10–33. SOLUTION (a) Condensation heat transfer on a tube is not influenced by the presence of other tubes in its neighborhood unless the condensate from other tubes drips on it. In our case, the horizontal tubes are arranged in four vertical tiers, each tier consisting of 3 tubes. The average heat transfer coefficient for a vertical tier of N horizontal tubes is related to the one for a single horizontal tube by Eq. 10–33 and is determined to be Condensate flow FIGURE 10–33 Schematic for Example 10–7. hhoriz, N tubes 1 1 h (9292 W/m2 · °C) 7060 W/m2 · °C N1/4 horiz, 1 tube 31/4 Each vertical tier consists of 3 tubes, and thus the heat transfer coefficient determined above is valid for each of the four tiers. In other words, this value can be taken to be the average heat transfer coefficient for all 12 tubes. The surface area for all 12 tubes per unit length of the tubes is As Ntotal DL 12 (0.03 m)(1 m) 1.1310 m2 Then the rate of heat transfer during this condensation process becomes · Q hAs(Tsat Ts) (7060 W/m2 · °C)(1.131 m2)(40 30)°C 79,850 W (b) The rate of condensation of steam is again determined from cen58933_ch10.qxd 9/4/2002 12:38 PM Page 545 545 CHAPTER 10 Q· 79,850 J/s m· condensation 0.0328 kg/s hfg 2435 103 J/kg Therefore, steam will condense on the horizontal pipes at a rate of 32.8 g/s per meter length of the tubes. 10–6 ■ FILM CONDENSATION INSIDE HORIZONTAL TUBES So far we have discussed film condensation on the outer surfaces of tubes and other geometries, which is characterized by negligible vapor velocity and the unrestricted flow of the condensate. Most condensation processes encountered in refrigeration and air-conditioning applications, however, involve condensation on the inner surfaces of horizontal or vertical tubes. Heat transfer analysis of condensation inside tubes is complicated by the fact that it is strongly influenced by the vapor velocity and the rate of liquid accumulation on the walls of the tubes (Fig. 10–34). For low vapor velocities, Chato recommends this expression for condensation Liquid Vapor hinternal 0.555 g ((T T ) h l l l sat ) k3l fg s 3 Cpl(Tsat Ts) 8 1/4 (10-34) for Tube D Revapor inlet 35,000 (10-35) FIGURE 10–34 Condensate flow in a horizontal tube with large vapor velocities. where the Reynolds number of the vapor is to be evaluated at the tube inlet conditions using the internal tube diameter as the characteristic length. Heat transfer coefficient correlations for higher vapor velocities are given by Rohsenow. 10–7 ■ DROPWISE CONDENSATION Dropwise condensation, characterized by countless droplets of varying diameters on the condensing surface instead of a continuous liquid film, is one of the most effective mechanisms of heat transfer, and extremely large heat transfer coefficients can be achieved with this mechanism (Fig. 10–35). In dropwise condensation, the small droplets that form at the nucleation sites on the surface grow as a result of continued condensation, coalesce into large droplets, and slide down when they reach a certain size, clearing the surface and exposing it to vapor. There is no liquid film in this case to resist heat transfer. As a result, with dropwise condensation, heat transfer coefficients can be achieved that are more than 10 times larger than those associated with film condensation. Large heat transfer coefficients enable designers to achieve a specified heat transfer rate with a smaller surface area, and thus a smaller (and FIGURE 10–35 Dropwise condensation of steam on a vertical surface (from Hampson and Özi¸sik, Ref. 11). cen58933_ch10.qxd 9/4/2002 12:38 PM Page 546 546 HEAT TRANSFER less expensive) condenser. Therefore, dropwise condensation is the preferred mode of condensation in heat transfer applications. The challenge in dropwise condensation is not to achieve it, but rather, to sustain it for prolonged periods of time. Dropwise condensation is achieved by adding a promoting chemical into the vapor, treating the surface with a promoter chemical, or coating the surface with a polymer such as teflon or a noble metal such as gold, silver, rhodium, palladium, or platinum. The promoters used include various waxes and fatty acids such as oleic, stearic, and linoic acids. They lose their effectiveness after a while, however, because of fouling, oxidation, and the removal of the promoter from the surface. It is possible to sustain dropwise condensation for over a year by the combined effects of surface coating and periodic injection of the promoter into the vapor. However, any gain in heat transfer must be weighed against the cost associated with sustaining dropwise condensation. Dropwise condensation has been studied experimentally for a number of surface–fluid combinations. Of these, the studies on the condensation of steam on copper surfaces has attracted the most attention because of their widespread use in steam power plants. P. Griffith (1983) recommends these simple correlations for dropwise condensation of steam on copper surfaces: hdropwise 51,104 2044Tsat , 255,310 22°C Tsat 100°C Tsat 100°C (10-36) (10-37) where Tsat is in °C and the heat transfer coefficient hdropwise is in W/m2 · °C. The very high heat transfer coefficients achievable with dropwise condensation are of little significance if the material of the condensing surface is not a good conductor like copper or if the thermal resistance on the other side of the surface is too large. In steady operation, heat transfer from one medium to another depends on the sum of the thermal resistances on the path of heat flow, and a large thermal resistance may overshadow all others and dominate the heat transfer process. In such cases, improving the accuracy of a small resistance (such as one due to condensation or boiling) makes hardly any difference in overall heat transfer calculations. TOPIC OF SPECIAL INTEREST Heat Pipes A heat pipe is a simple device with no moving parts that can transfer large quantities of heat over fairly large distances essentially at a constant temperature without requiring any power input. A heat pipe is basically a sealed slender tube containing a wick structure lined on the inner surface and a small amount of fluid such as water at the saturated state, as shown in Figure 10–36. It is composed of three sections: the evaporator section at one end, where heat is absorbed and the fluid is vaporized; a condenser section at the other end, where the vapor is condensed and heat is rejected; and the adiabatic section in between, where the vapor and the liquid phases of the fluid flow in opposite directions through the core and the wick, This section can be skipped without a loss in continuity. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 547 547 CHAPTER 10 Wick Tube wall Liquid flow Copper tube Heat in Heat out Vapor core Vapor flow Wick (liquid flow passage) Cross-section of a heat pipe Evaporation section Adiabatic section Condenser section FIGURE 10–36 Schematic and operation of a heat pipe. respectively, to complete the cycle with no significant heat transfer between the fluid and the surrounding medium. The type of fluid and the operating pressure inside the heat pipe depend on the operating temperature of the heat pipe. For example, the criticaland triple-point temperatures of water are 0.01°C and 374.1°C, respectively. Therefore, water can undergo a liquid-to-vapor or vapor-to-liquid phase change process in this temperature range only, and thus it will not be a suitable fluid for applications involving temperatures beyond this range. Furthermore, water will undergo a phase-change process at a specified temperature only if its pressure equals the saturation pressure at that temperature. For example, if a heat pipe with water as the working fluid is designed to remove heat at 70°C, the pressure inside the heat pipe must be maintained at 31.2 kPa, which is the boiling pressure of water at this temperature. Note that this value is well below the atmospheric pressure of 101 kPa, and thus the heat pipe will operate in a vacuum environment in this case. If the pressure inside is maintained at the local atmospheric pressure instead, heat transfer would result in an increase in the temperature of the water instead of evaporation. Although water is a suitable fluid to use in the moderate temperature range encountered in electronic equipment, several other fluids can be used in the construction of heat pipes to enable them to be used in cryogenic as well as high-temperature applications. The suitable temperature ranges for some common heat pipe fluids are given in Table 10–5. Note that the overall temperature range extends from almost absolute zero for cryogenic fluids such as helium to over 1600°C for liquid metals such as lithium. The ultimate temperature limits for a fluid are the triple- and critical-point temperatures. However, a narrower temperature range is used in practice to avoid the extreme pressures and low heats of vaporization that occur near the critical point. Other desirable characteristics of the candidate fluids are having a high surface tension to enhance the capillary effect and being compatible with the wick material, as well as being readily available, chemically stable, nontoxic, and inexpensive. The concept of heat pipe was originally conceived by R. S. Gaugler of the General Motors Corporation, who filed a patent application for it in TABLE 10–5 Suitable temperature ranges for some fluids used in heat pipes Fluid Helium Hydrogen Neon Nitrogen Methane Ammonia Water Mercury Cesium Sodium Lithium Temperature Range, °C 271 259 248 210 182 78 5 200 400 500 850 to to to to to to to to to to to 268 240 230 150 82 130 230 500 1000 1200 1600 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 548 548 HEAT TRANSFER However, it did not receive much attention until 1962, when it was suggested for use in space applications. Since then, heat pipes have found a wide range of applications, including the cooling of electronic equipment. The Operation of a Heat Pipe The operation of a heat pipe is based on the following physical principles: • At a specified pressure, a liquid will vaporize or a vapor will condense at a certain temperature, called the saturation temperature. Thus, fixing the pressure inside a heat pipe fixes the temperature at which phase change will occur. • At a specified pressure or temperature, the amount of heat absorbed as a unit mass of liquid vaporizes is equal to the amount of heat rejected as that vapor condenses. • The capillary pressure developed in a wick will move a liquid in the wick even against the gravitational field as a result of the capillary effect. • A fluid in a channel flows in the direction of decreasing pressure. Initially, the wick of the heat pipe is saturated with liquid and the core section is filled with vapor. When the evaporator end of the heat pipe is brought into contact with a hot surface or is placed into a hot environment, heat will transfer into the heat pipe. Being at a saturated state, the liquid in the evaporator end of the heat pipe will vaporize as a result of this heat transfer, causing the vapor pressure there to rise. This resulting pressure difference drives the vapor through the core of the heat pipe from the evaporator toward the condenser section. The condenser end of the heat pipe is in a cooler environment, and thus its surface is slightly cooler. The vapor that comes into contact with this cooler surface condenses, releasing the heat a vaporization, which is rejected to the surrounding medium. The liquid then returns to the evaporator end of the heat pipe through the wick as a result of capillary action in the wick, completing the cycle. As a result, heat is absorbed at one end of the heat pipe and is rejected at the other end, with the fluid inside serving as a transport medium for heat. The boiling and condensation processes are associated with extremely high heat transfer coefficients, and thus it is natural to expect the heat pipe to be a very effective heat transfer device, since its operation is based on alternative boiling and condensation of the working fluid. Indeed, heat pipes have effective conductivities several hundred times that of copper or silver. That is, replacing a copper bar between two mediums at different temperatures by a heat pipe of equal size can increase the rate of heat transfer between those two mediums by several hundred times. A simple heat pipe with water as the working fluid has an effective thermal conductivity of the order of 100,000 W/m · °C compared with about 400 W/m · °C for copper. For a heat pipe, it is not unusual to have an effective conductivity of 400,000 W/m · °C, which is 1000 times that of copper. A 15-cm-long, 0.6-cm-diameter horizontal cylindrical heat pipe with water inside, for example, can transfer heat at a rate of 300 W. Therefore, heat pipes are preferred in some critical applications, despite their high initial cost. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 549 549 CHAPTER 10 There is a small pressure difference between the evaporator and condenser ends, and thus a small temperature difference between the two ends of the heat pipe. This temperature difference is usually between 1°C and 5°C. The Construction of a Heat Pipe The wick of a heat pipe provides the means for the return of the liquid to the evaporator. Therefore, the structure of the wick has a strong effect on the performance of a heat pipe, and the design and construction of the wick are the most critical aspects of the manufacturing process. The wicks are often made of porous ceramic or woven stainless wire mesh. They can also be made together with the tube by extruding axial grooves along its inner surface, but this approach presents manufacturing difficulties. The performance of a wick depends on its structure. The characteristics of a wick can be changed by changing the size and the number of the pores per unit volume and the continuity of the passageway. Liquid motion in the wick depends on the dynamic balance between two opposing effects: the capillary pressure, which creates the suction effect to draw the liquid, and the internal resistance to flow as a result of friction between the mesh surfaces and the liquid. A small pore size increases the capillary action, since the capillary pressure is inversely proportional to the effective capillary radius of the mesh. But decreasing the pore size and thus the capillary radius also increases the friction force opposing the motion. Therefore, the core size of the mesh should be reduced so long as the increase in capillary force is greater than the increase in the friction force. Note that the optimum pore size will be different for different fluids and different orientations of the heat pipe. An improperly designed wick will result in an inadequate liquid supply and eventual failure of the heat pipe. Capillary action permits the heat pipe to operate in any orientation in a gravity field. However, the performance of a heat pipe will be best when the capillary and gravity forces act in the same direction (evaporator end down) and will be worst when these two forces act in opposite directions (evaporator end up). Gravity does not affect the capillary force when the heat pipe is in the horizontal position. The heat removal capacity of a horizontal heat pipe can be doubled by installing it vertically with evaporator end down so that gravity helps the capillary action. In the opposite case, vertical orientation with evaporator end up, the performance declines considerably relative the horizontal case since the capillary force in this case must work against the gravity force. Most heat pipes are cylindrical in shape. However, they can be manufactured in a variety of shapes involving 90° bends, S-turns, or spirals. They can also be made as a flat layer with a thickness of about 0.3 cm. Flat heat pipes are very suitable for cooling high-power-output (say, 50 W or greater) PCBs. In this case, flat heat pipes are attached directly to the back surface of the PCB, and they absorb and transfer the heat to the edges. Cooling fins are usually attached to the condenser end of the heat pipe to improve its effectiveness and to eliminate a bottleneck in the path of heat flow from the components to the environment when the ultimate heat sink is the ambient air. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 550 550 HEAT TRANSFER Evaporator end FIGURE 10–37 Variation of the heat removal capacity of a heat pipe with tilt angle from the horizontal when the liquid flows in the wick against gravity (from Steinberg, Ref. 18). TABLE 10–6 Typical heat removal capacity of various heat pipes Outside Diameter, cm (in.) 0.64(14) 0.95(38) 1.27(12) Length, cm (in.) Heat Removal Rate, W 15.2(6) 30.5(12) 45.7(18) 15.2(6) 30.5(12) 45.7(18) 15.2(6) 30.5(12) 45.7(18) 300 175 150 500 375 350 700 575 550 ∆T = 3°C 180 W Heat pipe 0.6 cm L = 30 cm FIGURE 10–38 Schematic for Example 10–8. Power handling capability, % 100 Coarse wick Medium wick θ 80 Fine wick 60 Condenser end 40 20 0 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° Angle θ The decline in the performance of a 122–cm-long water heat pipe with the tilt angle from the horizontal is shown in Figure 10–37 for heat pipes with coarse, medium, and fine wicks. Note that for the horizontal case, the heat pipe with a coarse wick performs best, but the performance drops off sharply as the evaporator end is raised from the horizontal. The heat pipe with a fine wick does not perform as well in the horizontal position but maintains its level of performance greatly at tilted positions. It is clear from this figure that heat pipes that work against gravity must be equipped with fine wicks. The heat removal capacities of various heat pipes are given in Table 10–6. A major concern about the performance of a heat pipe is degradation with time. Some heat pipes have failed within just a few months after they are put into operation. The major cause of degradation appears to be contamination that occurs during the sealing of the ends of the heat pipe tube and affects the vapor pressure. This form of contamination has been minimized by electron beam welding in clean rooms. Contamination of the wick prior to installation in the tube is another cause of degradation. Cleanliness of the wick is essential for its reliable operation for a long time. Heat pipes usually undergo extensive testing and quality control process before they are put into actual use. An important consideration in the design of heat pipes is the compatibility of the materials used for the tube, wick, and fluid. Otherwise, reaction between the incompatible materials produces noncondensable gases, which degrade the performance of the heat pipe. For example, the reaction between stainless steel and water in some early heat pipes generated hydrogen gas, which destroyed the heat pipe. 180 W EXAMPLE 10–8 Replacing a Heat Pipe by a Copper Rod A 30-cm-long cylindrical heat pipe having a diameter of 0.6 cm is dissipating heat at a rate of 180 W, with a temperature difference of 3°C across the heat pipe, as shown in Figure 10–38. If we were to use a 30-cm-long copper rod in- cen58933_ch10.qxd 9/4/2002 12:38 PM Page 551 551 CHAPTER 10 stead to remove heat at the same rate, determine the diameter and the mass of the copper rod that needs to be installed. SOLUTION A cylindrical heat pipe dissipates heat at a specified rate. The diameter and mass of a copper rod that can conduct heat at the same rate are to be determined. Assumptions Steady operating conditions exist. Properties The properties of copper at room temperature are 8950 kg/m3 and k 386 W/m · °C. · Analysis The rate of heat transfer Q through the copper rod can be expressed as · T Q kA L where k is the thermal conductivity, L is the length, and T is the temperature difference across the copper bar. Solving for the cross-sectional area A and substituting the specified values gives A 0.3 m L · Q (180 W) 0.04663 m2 466.3 cm2 (386 W/m · °C)(3°C) kT Then the diameter and the mass of the copper rod become 1 A D 2 4 → D 4 A/ 4(466.3 cm2)/ 24.4 cm m V AL (8590 kg/m3)(0.04663 m2)(0.3 m) 125.2 kg Therefore, the diameter of the copper rod needs to be almost 25 times that of the heat pipe to transfer heat at the same rate. Also, the rod would have a mass of 125.2 kg, which is impossible for an average person to lift. SUMMARY Boiling occurs when a liquid is in contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid. Boiling is classified as pool boiling or flow boiling depending on the presence of bulk fluid motion. Boiling is called pool boiling in the absence of bulk fluid flow and flow boiling (or forced convection boiling) in its presence. Pool and flow boiling are further classified as subcooled boiling and saturated boiling depending on the bulk liquid temperature. Boiling is said to be subcooled (or local) when the temperature of the main body of the liquid is below the saturation temperature Tsat and saturated (or bulk) when the temperature of the liquid is equal to Tsat. Boiling exhibits different regimes depending on the value of the excess temperature Texcess. Four different boiling regimes are observed: natural convection boiling, nucleate boiling, transition boiling, and film boiling. These regimes are illustrated on the boiling curve. The rate of evaporation and the rate of heat transfer in nucleate boiling increase with increasing Texcess and reach a maximum at some point. The heat flux at this point is called the critical (or maximum) heat flux, q·max. The rate of heat transfer in nucleate pool boiling is determined from g(l ) q·nucleate l hfg Cpl (Ts Tsat) Csf hfg Prln 1/2 3 The maximum (or critical) heat flux in nucleate pool boiling is determined from q·max Ccr hfg[g2 (l )]1/4 where the value of the constant Ccr is about 0.15. The minimum heat flux is given by cen58933_ch10.qxd 9/4/2002 12:38 PM Page 552 552 HEAT TRANSFER g(l ) q·min 0.09 hfg 2 (l ) and fully turbulent for Re 1800. Heat transfer coefficients in the wavy-laminar and turbulent flow regions are determined from 1/4 The heat flux for stable film boiling on the outside of a horizontal cylinder or sphere of diameter D is given by (l )[hfg 0.4Cp (Ts Tsat) D(Ts Tsat) (Ts Tsat) q·film Cfilm gk3 1/4 where the constant Cfilm 0.62 for horizontal cylinders and 0.67 for spheres, and k is the thermal conductivity of the vapor. The vapor properties are to be evaluated at the film temperature Tf (Tsat Ts)/2, which is the average temperature of the vapor film. The liquid properties and hfg are to be evaluated at the saturation temperature at the specified pressure. Two distinct forms of condensation are observed in nature: film condensation and dropwise condensation. In film condensation, the condensate wets the surface and forms a liquid film on the surface that slides down under the influence of gravity. In dropwise condensation, the condensed vapor forms countless droplets of varying diameters on the surface instead of a continuous film. The Reynolds number for the condensate flow is defined as Re Dh l l 4 Al l 4m· l pl pl Re 4Q· conden 4 As h(Tsat Ts) pl hfg pl hfg and where hfg is the modified latent heat of vaporization, defined as hfg hfg 0.68Cpl (Tsat Ts) and represents heat transfer during condensation per unit mass of condensate. Here Cpl is the specific heat of the liquid in J/kg · °C. Using some simplifying assumptions, the average heat transfer coefficient for film condensation on a vertical plate of height L is determined to be hvert 0.943 gl (l ) hfg k3l l (Ts Tsat)L 1/4 (W/m2 · °C) All properties of the liquid are to be evaluated at the film temperature Tf (Tsat Ts)/2. The hfg and are to be evaluated at Tsat. Condensate flow is smooth and wave-free laminar for about Re 30, wavy-laminar in the range of 30 Re 1800, Re kl g 1/3 30 Re 1800 , l 1.08 Re1.22 5.2 2l Re kl g 1/3 hvert, turbulent , 8750 58 Pr0.5 (Re0.75 253) 2l Re 1800 l hvert, wavy Equations for vertical plates can also be used for laminar film condensation on the upper surfaces of the plates that are inclined by an angle from the vertical, by replacing g in that equation by g cos . Vertical plate equations can also be used to calculate the average heat transfer coefficient for laminar film condensation on the outer surfaces of vertical tubes provided that the tube diameter is large relative to the thickness of the liquid film. The average heat transfer coefficient for film condensation on the outer surfaces of a horizontal tube is determined to be hhoriz 0.729 gl (l ) hfg k3l l s sat)D (T T 1/4 (W/m2 · °C) where D is the diameter of the horizontal tube. This relation can easily be modified for a sphere by replacing the constant 0.729 by 0.815. It can also be used for N horizontal tubes stacked on top of each other by replacing D in the denominator by ND. For low vapor velocities, film condensation heat transfer inside horizontal tubes can be determined from hinternal 0.555 g ((T T) k) h l l 3 l l sat fg s 3 Cpl (Tsat Ts) 8 1/4 and Revapor D inlet 35,000 where the Reynolds number of the vapor is to be evaluated at the tube inlet conditions using the internal tube diameter as the characteristic length. Finally, the heat transfer coefficient for dropwise condensation of steam on copper surfaces is given by hdropwise 51,104 2044Tsat , 255,310 22°C Tsat 100°C Tsat 100°C where Tsat is in °C and the heat transfer coefficient hdropwise is in W/m2 · °C. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 553 553 CHAPTER 10 REFERENCES AND SUGGESTED READING 1. N. Arai, T. Fukushima, A. Arai, T. Nakajima, K. Fujie, and Y. Nakayama. “Heat Transfer Tubes Enhancing Boiling and Condensation in Heat Exchangers of a Refrigeration Machine.” ASHRAE Journal 83 (1977), p. 58. 2. P. J. Berensen. “Film Boiling Heat Transfer for a Horizontal Surface.” Journal of Heat Transfer 83 (1961), pp. 351–358. 3. P. J. Berensen. “Experiments in Pool Boiling Heat Transfer.” International Journal of Heat Mass Transfer 5 (1962), pp. 985–999. 4. L. A. Bromley. “Heat Transfer in Stable Film Boiling.” Chemical Engineering Prog. 46 (1950), pp. 221–227. 5. J. C. Chato. “Laminar Condensation inside Horizontal and Inclined Tubes.” ASHRAE Journal 4 (1962), p. 52. 6. S. W. Chi. Heat Theory and Practice. Washington, D.C.: Hemisphere, 1976. 7. M. T. Cichelli and C. F. Bonilla. “Heat Transfer to Liquids Boiling under Pressure.” Transactions of AIChE 41 (1945), pp. 755–787. 8. R. A. Colclaser, D. A. Neaman, and C. F. Hawkins. Electronic Circuit Analysis. New York: John Wiley & Sons, 1984. 9. J. W. Dally. Packaging of Electronic Systems. New York: McGraw-Hill, 1960. 10. P. Griffith. “Dropwise Condensation.” In Heat Exchanger Design Handbook, ed. E. U. Schlunder, Vol 2, Ch. 2.6.5. New York: Hemisphere, 1983. 11. H. Hampson and N. Özis¸ik. “An Investigation into the Condensation of Steam.” Proceedings of the Institute of Mechanical Engineers, London 1B (1952), pp. 282–294. 12. J. P. Holman. Heat Transfer. 8th ed. New York: McGraw-Hill, 1997. 13. F. P. Incropera and D. P. DeWitt. lntroduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. 14. J. J. Jasper. “The Surface Tension of Pure Liquid Compounds.” Journal of Physical and Chemical Reference Data 1, No. 4 (1972), pp. 841–1009. R. Kemp. “The Heat Pipe—A New Tune on an Old Pipe.” Electronics and Power (August 9, 1973), p. 326. 16. S. S. Kutateladze. Fundamentals of Heat Transfer. New York: Academic Press, 1963. 17. S. S. Kutateladze. “On the Transition to Film Boiling under Natural Convection.” Kotloturbostroenie 3 (1948), p. 48. 18. D. A. Labuntsov. “Heat Transfer in Film Condensation of Pure Steam on Vertical Surfaces and Horizontal Tubes.” Teploenergetika 4 (l957), pp. 72–80. 19. J. H. Lienhard and V. K. Dhir. “Extended Hydrodynamic Theory of the Peak and Minimum Pool Boiling Heat Fluxes.” NASA Report, NASA-CR-2270, July 1973. 20. J. H. Lienhard and V. K. Dhir. “Hydrodynamic Prediction of Peak Pool Boiling Heat Fluxes from Finite Bodies.” Journal of Heat Transfer 95 (1973), pp. 152–158. 21. W. H. McAdams. Heat Transmission. 3rd ed. New York: McGraw-Hill, 1954. 22. W. M. Rohsenow. “A Method of Correlating Heat Transfer Data for Surface Boiling of Liquids.” ASME Transactions 74 (1952), pp. 969–975. 23. D. S. Steinberg. Cooling Techniques for Electronic Equipment. New York: John Wiley & Sons, 1980. 24. W. M. Rohsenow. “Film Condensation.” In Handbook of Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett, Ch. 12A. New York: McGraw-Hill, 1973. 25. I. G. Shekriladze, I. G. Gomelauri, and V. I. Gomelauri. “Theoretical Study of Laminar Film Condensation of Flowing Vapor.” International Journal of Heat Mass Transfer 9 (1966), pp. 591–592. 26. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West Publishing, 1995. 27. J. W. Westwater and J. G. Santangelo. Industrial Engineering Chemistry 47 (1955), p. 1605. 28. N. Zuber. “On the Stability of Boiling Heat Transfer.” ASME Transactions 80 (1958), pp. 711–720. PROBLEMS Boiling Heat Transfer 10–1C What is boiling? What mechanisms are responsible for the very high heat transfer coefficients in nucleate boiling? 10–2C Does the amount of heat absorbed as 1 kg of saturated liquid water boils at 100°C have to be equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100°C? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 554 554 HEAT TRANSFER 10–3C What is the difference between evaporation and boiling? 10–4C What is the difference between pool boiling and flow boiling? 10–5C What is the difference between subcooled and saturated boiling? 10–6C Draw the boiling curve and identify the different boiling regimes. Also, explain the characteristics of each regime. 10–7C How does film boiling differ from nucleate boiling? Is the boiling heat flux necessarily higher in the stable film boiling regime than it is in the nucleate boiling regime? 10–8C Draw the boiling curve and identify the burnout point on the curve. Explain how burnout is caused. Why is the burnout point avoided in the design of boilers? 10–9C Discuss some methods of enhancing pool boiling heat transfer permanently. 10–10C Name the different boiling regimes in the order they occur in a vertical tube during flow boiling. from 70 kPa and 101.3 kPa. Plot the maximum heat flux and the temperature difference as a function of the atmospheric pressure, and discuss the results. 10–14E Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter D 0.5 in. and emissivity 0.08 immersed in water. If the surface temperature of the heating element is 788°F, determine the rate of heat transfer to the water per unit length of the heating element. Answer: 2465 Btu/h 10–15E Repeat Problem 10–14E for a heating element temperature of 988°F. 10–16 Water is to be boiled at sea level in a 30-cm-diameter mechanically polished AISI 304 stainless steel pan placed on top of a 3-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the temperature of the inner surface of the bottom of the pan. Also, determine the temperature difference between the inner and outer surfaces of the bottom of the pan if it is 6 mm thick. 10–11 Water is to be boiled at atmospheric pressure in a mechanically polished steel pan placed on top of a heating unit. The inner surface of the bottom of the pan is maintained at 110°C. If the diameter of the bottom of the pan is 25 cm, determine (a) the rate of heat transfer to the water and (b) the rate of evaporation. 1 atm 3 kW 1 atm FIGURE P10–16 110°C 10–17 Repeat Problem 10–16 for a location at an elevation of 1500 m where the atmospheric pressure is 84.5 kPa and thus the boiling temperature of water is 95°C. Answers: 100.9°C, 10.3°C 10–18 Water is boiled at sea level in a coffee maker equipped with a 20-cm long 0.4-cm-diameter immersion-type electric 1 atm FIGURE P10–11 10–12 Water is to be boiled at atmospheric pressure on a 3-cm-diameter mechanically polished steel heater. Determine the maximum heat flux that can be attained in the nucleate boiling regime and the surface temperature of the heater surface in that case. Reconsider Problem 10–12. Using EES (or other) software, investigate the effect of local atmospheric pressure on the maximum heat flux and the temperature difference Ts Tsat. Let the atmospheric pressure vary Coffee maker 1L 10–13 FIGURE P10–18 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 555 555 CHAPTER 10 heating element made of mechanically polished stainless steel. The coffee maker initially contains 1 L of water at 18°C. Once boiling starts, it is observed that half of the water in the coffee maker evaporates in 25 min. Determine the power rating of the electric heating element immersed in water and the surface temperature of the heating element. Also determine how long it will take for this heater to raise the temperature of 1 L of cold water from 18°C to the boiling temperature. 10–19 Vent Boiler 150°C Repeat Problem 10–18 for a copper heating element. 10–20 A 65-cm-long, 2-cm-diameter brass heating element is to be used to boil water at 120°C. If the surface temperature of the heating element is not to exceed 125°C, determine the highest rate of steam production in the boiler, in kg/h. Water 165°C Answer: 19.4 kg/h 10–21 To understand the burnout phenomenon, boiling experiments are conducted in water at atmospheric pressure using an electrically heated 30-cm-long, 3-mm-diameter nickelplated horizontal wire. Determine (a) the critical heat flux and (b) the increase in the temperature of the wire as the operating point jumps from the nucleate boiling to the film boiling regime at the critical heat flux. Take the emissivity of the wire to be 0.5. 10–22 Reconsider Problem 10–21. Using EES (or other) software, investigate the effects of the local atmospheric pressure and the emissivity of the wire on the critical heat flux and the temperature rise of wire. Let the atmospheric pressure vary from 70 kPa and 101.3 kPa and the emissivity from 0.1 to 1.0. Plot the critical heat flux and the temperature rise as functions of the atmospheric pressure and the emissivity, and discuss the results. 10–23 Water is boiled at 1 atm pressure in a 20-cm-internaldiameter teflon-pitted stainless steel pan on an electric range. If it is observed that the water level in the pan drops by 10 cm in 30 min, determine the inner surface temperature of the pan. Answer: 111.5°C 10–24 Repeat Problem 10–23 for a polished copper pan. 10–25 In a gas-fired boiler, water is boiled at 150°C by hot gases flowing through 50-m-long, 5-cm-outer-diameter mechanically polished stainless steel pipes submerged in water. If the outer surface temperature of the pipes is 165°C, determine (a) the rate of heat transfer from the hot gases to water, (b) the rate of evaporation, (c) the ratio of the critical heat flux to the present heat flux, and (d) the surface temperature of the pipe at which critical heat flux occurs. Answers: (a) 10,865 kW, (b) 5.139 kg/s, (c) 1.34, (d) 166.5°C 10–26 Repeat Problem 10–25 for a boiling temperature of 160°C. 10–27E Water is boiled at 250°F by a 2-ft-long and 0.5-in.diameter nickel-plated electric heating element maintained at 280°F. Determine (a) the boiling heat transfer coefficient, Hot gases FIGURE P10–25 (b) the electric power consumed by the heating element, and (c) the rate of evaporation of water. 10–28E Repeat Problem 10–27E for a platinum-plated heating element. 10–29E Reconsider Problem 10–27E. Using EES (or other) software, investigate the effect of surface temperature of the heating element on the boiling heat transfer coefficient, the electric power, and the rate of evaporation of water. Let the surface temperature vary from 260°F to 300°F. Plot the boiling heat transfer coefficient, the electric power consumption, and the rate of evaporation of water as a function of the surface temperature, and discuss the results. 10–30 Cold water enters a steam generator at 15°C and leaves as saturated steam at 100°C. Determine the fraction of heat used to preheat the liquid water from 15°C to the saturation temperature of 100°C in the steam generator. Answer: 13.6 percent 10–31 Cold water enters a steam generator at 20°C and leaves as saturated steam at the boiler pressure. At what pressure will the amount of heat needed to preheat the water to saturation temperature be equal to the heat needed to vaporize the liquid at the boiler pressure? 10–32 Reconsider Problem 10–31. Using EES (or other) software, plot the boiler pressure as a function of the cold water temperature as the temperature varies from 0°C to 30°C, and discuss the results. 10–33 A 50-cm-long, 2-mm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The wire temperature is measured to be 130°C when a wattmeter cen58933_ch10.qxd 9/4/2002 12:38 PM Page 556 556 HEAT TRANSFER will the average heat transfer coefficient for the entire stack of tubes be equal to half of what it is for a single horizontal tube? 1 atm 3.8 kW 130°C FIGURE P10–33 indicates the electric power consumed to be 3.8 kW. Using Newton’s law of cooling, determine the boiling heat transfer coefficient. Answer: 16 10–44 Saturated steam at 1 atm condenses on a 3-m-high and 5-m-wide vertical plate that is maintained at 90°C by circulating cooling water through the other side. Determine (a) the rate of heat transfer by condensation to the plate, and (b) the rate at which the condensate drips off the plate at the bottom. Answers: (a) 942 kW, (b) 0.412 kg/s 1 atm Steam 5m Condensation Heat Transfer 10–34C What is condensation? How does it occur? 10–35C What is the difference between film and dropwise condensation? Which is a more effective mechanism of heat transfer? 90°C 3m 10–36C In condensate flow, how is the wetted perimeter defined? How does wetted perimeter differ from ordinary perimeter? 10–37C What is the modified latent heat of vaporization? For what is it used? How does it differ from the ordinary latent heat of vaporization? m· FIGURE P10–44 10–38C Consider film condensation on a vertical plate. Will the heat flux be higher at the top or at the bottom of the plate? Why? 10–45 Repeat Problem 10–44 for the case of the plate being tilted 60° from the vertical. 10–39C Consider film condensation on the outer surfaces of a tube whose length is 10 times its diameter. For which orientation of the tube will the heat transfer rate be the highest: horizontal or vertical? Explain. Disregard the base and top surfaces of the tube. 10–46 Saturated steam at 30°C condenses on the outside of a 4-cm-outer-diameter, 2-m-long vertical tube. The temperature of the tube is maintained at 20°C by the cooling water. Determine (a) the rate of heat transfer from the steam to the cooling water, (b) the rate of condensation of steam, and (c) the approximate thickness of the liquid film at the bottom of the tube. 10–40C Consider film condensation on the outer surfaces of four long tubes. For which orientation of the tubes will the condensation heat transfer coefficient be the highest: (a) vertical, (b) horizontal side by side, (c) horizontal but in a vertical tier (directly on top of each other), or (d) a horizontal stack of two tubes high and two tubes wide? 10–41C How does the presence of a noncondensable gas in a vapor influence the condensation heat transfer? 10–42 The Reynolds number for condensate flow is defined as Re 4m· /pl, where p is the wetted perimeter. Obtain simplified relations for the Reynolds number by expressing p and m· by their equivalence for the following geometries: (a) a vertical plate of height L and width w, (b) a tilted plate of height L and width w inclined at an angle from the vertical, (c) a vertical cylinder of length L and diameter D, (d) a horizontal cylinder of length L and diameter D, and (e) a sphere of diameter D. 10–43 Consider film condensation on the outer surfaces of N horizontal tubes arranged in a vertical tier. For what value of N 4 cm Steam 30°C Condensate L=2m 20°C FIGURE P10–46 10–47E Saturated steam at 95°F is condensed on the outer surfaces of an array of horizontal pipes through which cooling water circulates. The outer diameter of the pipes is 1 in. and the outer surfaces of the pipes are maintained at 85°F. Determine (a) the rate of heat transfer to the cooling water circulating in the pipes and (b) the rate of condensation of steam per unit length of a single horizontal pipe. cen58933_ch10.qxd 9/4/2002 12:38 PM Page 557 557 CHAPTER 10 10–48E Repeat Problem 10–47E for the case of 32 horizontal pipes arranged in a rectangular array of 4 pipes high and 8 pipes wide. 10–49 Saturated steam at 55°C is to be condensed at a rate of 10 kg/h on the outside of a 3-cm-outer-diameter vertical tube whose surface is maintained at 45°C by the cooling water. Determine the tube length required. 10–50 Repeat Problem 10–49 for a horizontal tube. Answer: 0.70 m 10–51 Saturated steam at 100°C condenses on a 2-m 2-m plate that is tilted 40° from the vertical. The plate is maintained at 80°C by cooling it from the other side. Determine (a) the average heat transfer coefficient over the entire plate and (b) the rate at which the condensate drips off the plate at the bottom. 10–52 Reconsider Problem 10–51. Using EES (or other) software, investigate the effects of plate temperature and the angle of the plate from the vertical on the average heat transfer coefficient and the rate at which the condensate drips off. Let the plate temperature vary from 40°C to 90°C and the plate angle from 0° to 60°. Plot the heat transfer coefficient and the rate at which the condensate drips off as the functions of the plate temperature and the tilt angle, and discuss the results. 10–53 Saturated ammonia vapor at 10°C condenses on the outside of a 2-cm-outer-diameter, 8-m-long horizontal tube whose outer surface is maintained at 10°C. Determine (a) the rate of heat transfer from the ammonia and (b) the rate of condensation of ammonia. 10–54 The condenser of a steam power plant operates at a pressure of 4.25 kPa. The condenser consists of 100 horizontal tubes arranged in a 10 10 square array. The tubes are 8 m Cooling water 20°C L=8m Reconsider Problem 10–54. Using EES (or other) software, investigate the effect of the condenser pressure on the rate of heat transfer and the rate of condensation of the steam. Let the condenser pressure vary from 3 kPa to 15 kPa. Plot the rate of heat transfer and the rate of condensation of the steam as a function of the condenser pressure, and discuss the results. 10–56 A large heat exchanger has several columns of tubes, with 20 tubes in each column. The outer diameter of the tubes is 1.5 cm. Saturated steam at 50°C condenses on the outer surfaces of the tubes, which are maintained at 20°C. Determine (a) the average heat transfer coefficient and (b) the rate of condensation of steam per m length of a column. 10–57 Saturated refrigerant-134a vapor at 30°C is to be condensed in a 5-m-long, 1-cm-diameter horizontal tube that is maintained at a temperature of 20°C. If the refrigerant enters the tube at a rate of 2.5 kg/min, determine the fraction of the refrigerant that will have condensed at the end of the tube. 10–58 Repeat Problem 10–57 for a tube length of 8 m. Answer: 17.2 percent 10–59 Reconsider Problem 10–57. Using EES (or other) software, plot the fraction of the refrigerant condensed at the end of the tube as a function of the temperature of the saturated R-134a vapor as the temperature varies from 25°C to 50°C, and discuss the results. Special Topic: Heat Pipes 10–61C A heat pipe with water as the working fluid is said to have an effective thermal conductivity of 100,000 W/m · °C, which is more than 100,000 times the conductivity of water. How can this happen? P = 4.25 kPa n = 100 tubes 10–55 10–60C What is a heat pipe? How does it operate? Does it have any moving parts? Saturated steam N = 10 long and have an outer diameter of 3 cm. If the tube surfaces are at 20°C, determine (a) the rate of heat transfer from the steam to the cooling water and (b) the rate of condensation of steam in the condenser. Answers: (a) 3678 kW, (b) 1.496 kg/s 10–62C What is the effect of a small amount of noncondensable gas such as air on the performance of a heat pipe? 10–63C Why do water-based heat pipes used in the cooling of electronic equipment operate below atmospheric pressure? 10–64C What happens when the wick of a heat pipe is too coarse or too fine? 10–65C Does the orientation of a heat pipe affect its performance? Does it matter if the evaporator end of the heat pipe is up or down? Explain. FIGURE P10–54 10–66C How can the liquid in a heat pipe move up against gravity without a pump? For heat pipes that work against gravity, is it better to have coarse or fine wicks? Why? cen58933_ch10.qxd 9/4/2002 12:38 PM Page 558 558 HEAT TRANSFER 10–67C What are the important considerations in the design and manufacture of heat pipes? coefficient, (b) the rate of heat transfer, and (c) the rate of condensation of ammonia. 10–68C What is the major cause for the premature degradation of the performance of some heat pipes? 10–74 Saturated isobutane vapor in a binary geothermal power plant is to be condensed outside an array of eight horizontal tubes. Determine the ratio of the condensation rate for the cases of the tubes being arranged in a horizontal tier versus Answer: 1.68 in a vertical tier of horizontal tubes. 10–69 A 40-cm-long cylindrical heat pipe having a diameter of 0.5 cm is dissipating heat at a rate of 150 W, with a temperature difference of 4°C across the heat pipe. If we were to use a 40-cm-long copper rod (k 386 W/m · °C and 8950 kg/m3) instead to remove heat at the same rate, determine the diameter and the mass of the copper rod that needs to be installed. 10–70 Repeat Problem 10–69 for an aluminum rod instead of copper. 10–71E A plate that supports 10 power transistors, each dissipating 35 W, is to be cooled with 1-ft-long heat pipes having a diameter of 14 in. Using Table 10–6, determine how many pipes need to be attached to this plate. Answer: 2 Heat sink Heat pipe 10–75E The condenser of a steam power plant operates at a pressure of 0.95 psia. The condenser consists of 144 horizontal tubes arranged in a 12 12 square array. The tubes are 15 ft long and have an outer diameter of 1.2 in. If the outer surfaces of the tubes are maintained at 80°F, determine (a) the rate of heat transfer from the steam to the cooling water and (b) the rate of condensation of steam in the condenser. 10–76E Repeat Problem 10–75E for a tube diameter of 2 in. 10–77 Water is boiled at 100°C electrically by a 80-cm-long, 2-mm-diameter horizontal resistance wire made of chemically etched stainless steel. Determine (a) the rate of heat transfer to the water and the rate of evaporation of water if the temperature of the wire is 115°C and (b) the maximum rate of evaporation in the nucleate boiling regime. Answers: (a) 2387 W, 3.81 kg/h, (b) 1280 kW/m2 Transistor Steam Water 100°C FIGURE P10–71E 115°C Review Problems 10–72 Steam at 40°C condenses on the outside of a 3-cm diameter thin horizontal copper tube by cooling water that enters the tube at 25°C at an average velocity of 2 m/s and leaves at 35°C. Determine the rate of condensation of steam, the average overall heat transfer coefficient between the steam and the cooling water, and the tube length. Steam 40°C Cooling water 35°C 25°C FIGURE P10–72 10–73 Saturated ammonia vapor at 25°C condenses on the outside of a 2-m-long, 3.2-cm-outer-diameter vertical tube maintained at 15°C. Determine (a) the average heat transfer FIGURE P10–77 10–78E Saturated steam at 100°F is condensed on a 6-ft-high vertical plate that is maintained at 80°F. Determine the rate of heat transfer from the steam to the plate and the rate of condensation per foot width of the plate. 10–79 Saturated refrigerant-134a vapor at 35°C is to be condensed on the outer surface of a 7-m-long, 1.5-cm-diameter horizontal tube that is maintained at a temperature of 25°C. Determine the rate at which the refrigerant will condense, in kg/min. 10–80 Repeat Problem 10–79 for a tube diameter of 3 cm. 10–81 Saturated steam at 270.1 kPa condenses inside a horizontal, 6-m-long, 3-cm-internal-diameter pipe whose surface is maintained at 110°C. Assuming low vapor velocity, determine cen58933_ch10.qxd 9/4/2002 12:38 PM Page 559 559 CHAPTER 10 the average heat transfer coefficient and the rate of condensation of the steam inside the pipe. Answers: 3345 W/m2 · °C, 0.0174 kg/s 10–82 A 1.5-cm-diameter silver sphere initially at 30°C is suspended in a room filled with saturated steam at 100°C. Using the lumped system analysis, determine how long it will take for the temperature of the ball to rise to 50°C. Also, determine the amount of steam that condenses during this process and verify that the lumped system analysis is applicable. 10–83 Repeat Problem 10–82 for a 3-cm-diameter copper ball. 10–84 You have probably noticed that water vapor that condenses on a canned drink slides down, clearing the surface for further condensation. Therefore, condensation in this case can be considered to be dropwise. Determine the condensation heat transfer coefficient on a cold canned drink at 5°C that is placed in a large container filled with saturated steam at 95°C. Steam 95°C 5°C FIGURE P10–84 behind the refrigerator. Heat transfer from the outer surface of the coil to the surroundings is by natural convection and radiation. Obtaining information about the operating conditions of the refrigerator, including the pressures and temperatures of the refrigerant at the inlet and the exit of the coil, show that the coil is selected properly, and determine the safety margin in the selection. 10–88 Water-cooled steam condensers are commonly used in steam power plants. Obtain information about water-cooled steam condensers by doing a literature search on the topic and also by contacting some condenser manufacturers. In a report, describe the various types, the way they are designed, the limitation on each type, and the selection criteria. 10–89 Steam boilers have long been used to provide process heat as well as to generate power. Write an essay on the history of steam boilers and the evolution of modern supercritical steam power plants. What was the role of the American Society of Mechanical Engineers in this development? 10–90 The technology for power generation using geothermal energy is well established, and numerous geothermal power plants throughout the world are currently generating electricity economically. Binary geothermal plants utilize a volatile secondary fluid such as isobutane, n-pentane, and R-114 in a closed loop. Consider a binary geothermal plant with R-114 as the working fluid that is flowing at a rate of 600 kg/s. The R-114 is vaporized in a boiler at 115°C by the geothermal fluid that enters at 165°C and is condensed at 30°C outside the tubes by cooling water that enters the tubes at 18°C. Design the condenser of this binary plant. Specify (a) the length, diameter, and number of tubes and their arrangement in the condenser, (b) the mass flow rate of cooling water, and (c) the flow rate of make-up water needed if a cooling tower is used to reject the waste heat from the cooling water. The liquid velocity is to remain under 6 m/s and the length of the tubes is limited to 8 m. 10–85 A resistance heater made of 2-mm-diameter nickel wire is used to heat water at 1 atm pressure. Determine the highest temperature at which this heater can operate safely without the danger of burning out. Answer: 109.6°C 10–91 A manufacturing facility requires saturated steam at 120°C at a rate of 1.2 kg/min. Design an electric steam boiler for this purpose under these constraints: Computer, Design, and Essay Problems • The boiler will be in cylindrical shape with a height-todiameter ratio of 1.5. The boiler can be horizontal or vertical. 10–86 Design the condenser of a steam power plant that has a thermal efficiency of 40 percent and generates 10 MW of net electric power. Steam enters the condenser as saturated vapor at 10 kPa, and it is to be condensed outside horizontal tubes through which cooling water from a nearby river flows. The temperature rise of the cooling water is limited to 8°C, and the velocity of the cooling water in the pipes is limited to 6 m/s to keep the pressure drop at an acceptable level. Specify the pipe diameter, total pipe length, and the arrangement of the pipes to minimize the condenser volume. 10–87 The refrigerant in a household refrigerator is condensed as it flows through the coil that is typically placed Boiler FIGURE P10–91 cen58933_ch10.qxd 9/4/2002 12:38 PM Page 560 560 HEAT TRANSFER • The boiler will operate in the nucleate boiling regime, and the design heat flux will not exceed 60 percent of the critical heat flux to provide an adequate safety margin. • A commercially available plug-in type electrical heating element made of mechanically polished stainless steel will be used. The diameter of the heater cannot be between 0.5 cm and 3 cm. • Half of the volume of the boiler should be occupied by steam, and the boiler should be large enough to hold enough water for 2 h supply of steam. Also, the boiler will be well insulated. late the surface area of the heater. First, boil water in a pan using the heating element and measure the temperature of the boiling water away from the heating element. Based on your reading, estimate the elevation of your location, and compare it to the actual value. Then glue the tip of the thermocouple wire of the thermometer to the midsection of the heater surface. The temperature reading in this case will give the surface temperature of the heater. Assuming the rated power of the heater to be the actual power consumption during heating (you can check this by measuring the electric current and voltage), calculate the heat transfer coefficients from Newton’s law of cooling. You are to specify the following: (1) The height and inner diameter of the tank, (2) the length, diameter, power rating, and surface temperature of the electric heating element, (3) the maximum rate of steam production during short periods of overload conditions, and how it can be accomplished. 10–92 Repeat Problem 10–91 for a boiler that produces steam at 150°C at a rate of 2.5 kg/min. 10–93 Conduct this experiment to determine the boiling heat transfer coefficient. You will need a portable immersion-type electric heating element, an indoor-outdoor thermometer, and metal glue (all can be purchased for about $15 in a hardware store). You will also need a piece of string and a ruler to calcu- FIGURE P10–93 cen58933_ch11.qxd 9/9/2002 9:38 AM Page 561 CHAPTER F U N D A M E N TA L S O F T H E R M A L R A D I AT I O N o far, we have considered the conduction and convection modes of heat transfer, which are related to the nature of the materials involved and the presence of fluid motion, among other things. We now turn our attention to the third mechanism of heat transfer: radiation, which is characteristically different from the other two. We start this chapter with a discussion of electromagnetic waves and the electromagnetic spectrum, with particular emphasis on thermal radiation. Then we introduce the idealized blackbody, blackbody radiation, and blackbody radiation function, together with the Stefan–Boltzmann law, Planck’s law, and Wien’s displacement law. Radiation is emitted by every point on a plane surface in all directions into the hemisphere above the surface. The quantity that describes the magnitude of radiation emitted or incident in a specified direction in space is the radiation intensity. Various radiation fluxes such as emissive power, irradiation, and radiosity are expressed in terms of intensity. This is followed by a discussion of radiative properties of materials such as emissivity, absorptivity, reflectivity, and transmissivity and their dependence on wavelength, direction, and temperature. The greenhouse effect is presented as an example of the consequences of the wavelength dependence of radiation properties. The last section is devoted to the discussions of atmospheric and solar radiation because of their importance. S 11 CONTENTS 11-1 Introduction 562 11-2 Thermal Radiation 563 11-3 Blackbody Radiation 565 11-4 Radiation Intensity 571 11-5 Radiative Properties 577 11-6 Atmospheric and Solar Radiation 586 Topic of Special Interest: Solar Heat Gain Through Windows 590 561 cen58933_ch11.qxd 9/9/2002 9:38 AM Page 562 562 HEAT TRANSFER 11–1 Vacuum chamber Hot object Radiation FIGURE 11–1 A hot object in a vacuum chamber loses heat by radiation only. Person 30°C Fire 900°C Air 5°C Radiation FIGURE 11–2 Unlike conduction and convection, heat transfer by radiation can occur between two bodies, even when they are separated by a medium colder than both of them. ■ INTRODUCTION Consider a hot object that is suspended in an evacuated chamber whose walls are at room temperature (Fig. 11–1). The hot object will eventually cool down and reach thermal equilibrium with its surroundings. That is, it will lose heat until its temperature reaches the temperature of the walls of the chamber. Heat transfer between the object and the chamber could not have taken place by conduction or convection, because these two mechanisms cannot occur in a vacuum. Therefore, heat transfer must have occurred through another mechanism that involves the emission of the internal energy of the object. This mechanism is radiation. Radiation differs from the other two heat transfer mechanisms in that it does not require the presence of a material medium to take place. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. Also, radiation transfer occurs in solids as well as liquids and gases. In most practical applications, all three modes of heat transfer occur concurrently at varying degrees. But heat transfer through an evacuated space can occur only by radiation. For example, the energy of the sun reaches the earth by radiation. You will recall that heat transfer by conduction or convection takes place in the direction of decreasing temperature; that is, from a high-temperature medium to a lower-temperature one. It is interesting that radiation heat transfer can occur between two bodies separated by a medium colder than both bodies (Fig. 11–2). For example, solar radiation reaches the surface of the earth after passing through cold air layers at high altitudes. Also, the radiationabsorbing surfaces inside a greenhouse reach high temperatures even when its plastic or glass cover remains relatively cool. The theoretical foundation of radiation was established in 1864 by physicist James Clerk Maxwell, who postulated that accelerated charges or changing electric currents give rise to electric and magnetic fields. These rapidly moving fields are called electromagnetic waves or electromagnetic radiation, and they represent the energy emitted by matter as a result of the changes in the electronic configurations of the atoms or molecules. In 1887, Heinrich Hertz experimentally demonstrated the existence of such waves. Electromagnetic waves transport energy just like other waves, and all electromagnetic waves travel at the speed of light in a vacuum, which is C0 2.9979 108 m/s. Electromagnetic waves are characterized by their frequency or wavelength . These two properties in a medium are related by c (11-1) where c is the speed of propagation of a wave in that medium. The speed of propagation in a medium is related to the speed of light in a vacuum by c c0 /n, where n is the index of refraction of that medium. The refractive index is essentially unity for air and most gases, about 1.5 for glass, and about 1.33 for water. The commonly used unit of wavelength is the micrometer (m) or micron, where 1 m 106 m. Unlike the wavelength and the speed of propagation, the frequency of an electromagnetic wave depends only on the source and is independent of the medium through which the wave travels. The frequency (the number of oscillations per second) of an electromagnetic wave can range cen58933_ch11.qxd 9/9/2002 9:38 AM Page 563 563 CHAPTER 11 from less than a million Hz to a septillion Hz or higher, depending on the source. Note from Eq. 11-1 that the wavelength and the frequency of electromagnetic radiation are inversely proportional. It has proven useful to view electromagnetic radiation as the propagation of a collection of discrete packets of energy called photons or quanta, as proposed by Max Planck in 1900 in conjunction with his quantum theory. In this view, each photon of frequency is considered to have an energy of e h hc λ (11-2) where h 6.6256 1034 J · s is Planck’s constant. Note from the second part of Eq. 11-2 that the energy of a photon is inversely proportional to its wavelength. Therefore, shorter-wavelength radiation possesses larger photon energies. It is no wonder that we try to avoid very-short-wavelength radiation such as gamma rays and X-rays since they are highly destructive. λ , µm Electrical power waves 1010 109 108 107 Radio and TV waves 106 105 11–2 ■ 104 THERMAL RADIATION Although all electromagnetic waves have the same general features, waves of different wavelength differ significantly in their behavior. The electromagnetic radiation encountered in practice covers a wide range of wavelengths, varying from less than 1010 m for cosmic rays to more than 1010 m for electrical power waves. The electromagnetic spectrum also includes gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, thermal radiation, microwaves, and radio waves, as shown in Figure 11–3. Different types of electromagnetic radiation are produced through various mechanisms. For example, gamma rays are produced by nuclear reactions, X-rays by the bombardment of metals with high-energy electrons, microwaves by special types of electron tubes such as klystrons and magnetrons, and radio waves by the excitation of some crystals or by the flow of alternating current through electric conductors. The short-wavelength gamma rays and X-rays are primarily of concern to nuclear engineers, while the long-wavelength microwaves and radio waves are of concern to electrical engineers. The type of electromagnetic radiation that is pertinent to heat transfer is the thermal radiation emitted as a result of energy transitions of molecules, atoms, and electrons of a substance. Temperature is a measure of the strength of these activities at the microscopic level, and the rate of thermal radiation emission increases with increasing temperature. Thermal radiation is continuously emitted by all matter whose temperature is above absolute zero. That is, everything around us such as walls, furniture, and our friends constantly emits (and absorbs) radiation (Fig. 11–4). Thermal radiation is also defined as the portion of the electromagnetic spectrum that extends from about 0.1 to 100 m, since the radiation emitted by bodies due to their temperature falls almost entirely into this wavelength range. Thus, thermal radiation includes the entire visible and infrared (IR) radiation as well as a portion of the ultraviolet (UV) radiation. What we call light is simply the visible portion of the electromagnetic spectrum that lies between 0.40 and 0.76 m. Light is characteristically no different than other electromagnetic radiation, except that it happens to trigger the 103 Microwaves 102 10 1 10–1 Thermal Infrared radiation Visible Ultraviolet 10–2 10–3 X-rays 10–4 10–5 10–6 γ-rays 10–7 10–8 10–9 Cosmic rays FIGURE 11–3 The electromagnetic wave spectrum. Plants Walls People Furniture FIGURE 11–4 Everything around us constantly emits thermal radiation. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 564 564 HEAT TRANSFER TABLE 11–1 The wavelength ranges of different colors Color Violet Blue Green Yellow Orange Red Wavelength band 0.40–0.44 0.44–0.49 0.49–0.54 0.54–0.60 0.60–0.67 0.63–0.76 m m m m m m FIGURE 11–5 Food is heated or cooked in a microwave oven by absorbing the electromagnetic radiation energy generated by the magnetron of the oven. sensation of seeing in the human eye. Light, or the visible spectrum, consists of narrow bands of color from violet (0.40–0.44 m) to red (0.63–0.76 m), as shown in Table 11–1. A body that emits some radiation in the visible range is called a light source. The sun is obviously our primary light source. The electromagnetic radiation emitted by the sun is known as solar radiation, and nearly all of it falls into the wavelength band 0.3–3 m. Almost half of solar radiation is light (i.e., it falls into the visible range), with the remaining being ultraviolet and infrared. The radiation emitted by bodies at room temperature falls into the infrared region of the spectrum, which extends from 0.76 to 100 m. Bodies start emitting noticeable visible radiation at temperatures above 800 K. The tungsten filament of a lightbulb must be heated to temperatures above 2000 K before it can emit any significant amount of radiation in the visible range. The ultraviolet radiation includes the low-wavelength end of the thermal radiation spectrum and lies between the wavelengths 0.01 and 0.40 m. Ultraviolet rays are to be avoided since they can kill microorganisms and cause serious damage to humans and other living organisms. About 12 percent of solar radiation is in the ultraviolet range, and it would be devastating if it were to reach the surface of the earth. Fortunately, the ozone (O3) layer in the atmosphere acts as a protective blanket and absorbs most of this ultraviolet radiation. The ultraviolet rays that remain in sunlight are still sufficient to cause serious sunburns to sun worshippers, and prolonged exposure to direct sunlight is the leading cause of skin cancer, which can be lethal. Recent discoveries of “holes” in the ozone layer have prompted the international community to ban the use of ozone-destroying chemicals such as the refrigerant Freon-12 in order to save the earth. Ultraviolet radiation is also produced artificially in fluorescent lamps for use in medicine as a bacteria killer and in tanning parlors as an artificial tanner. The connection between skin cancer and ultraviolet rays has caused dermatologists to issue strong warnings against its use for tanning. Microwave ovens utilize electromagnetic radiation in the microwave region of the spectrum generated by microwave tubes called magnetrons. Microwaves in the range of 102–105 m are very suitable for use in cooking since they are reflected by metals, transmitted by glass and plastics, and absorbed by food (especially water) molecules. Thus, the electric energy converted to radiation in a microwave oven eventually becomes part of the internal energy of the food. The fast and efficient cooking of microwave ovens has made them as one of the essential appliances in modern kitchens (Fig. 11–5). Radars and cordless telephones also use electromagnetic radiation in the microwave region. The wavelength of the electromagnetic waves used in radio and TV broadcasting usually ranges between 1 and 1000 m in the radio wave region of the spectrum. In heat transfer studies, we are interested in the energy emitted by bodies because of their temperature only. Therefore, we will limit our consideration to thermal radiation, which we will simply call radiation. The relations developed below are restricted to thermal radiation only and may not be applicable to other forms of electromagnetic radiation. The electrons, atoms, and molecules of all solids, liquids, and gases above absolute zero temperature are constantly in motion, and thus radiation is constantly emitted, as well as being absorbed or transmitted throughout the entire volume of matter. That is, radiation is a volumetric phenomenon. However, cen58933_ch11.qxd 9/9/2002 9:38 AM Page 565 565 CHAPTER 11 for opaque (nontransparent) solids such as metals, wood, and rocks, radiation is considered to be a surface phenomenon, since the radiation emitted by the interior regions can never reach the surface, and the radiation incident on such bodies is usually absorbed within a few microns from the surface (Fig. 11–6). Note that the radiation characteristics of surfaces can be changed completely by applying thin layers of coatings on them. Radiation emitted Gas or vacuum Solid 11–3 ■ BLACKBODY RADIATION A body at a temperature above absolute zero emits radiation in all directions over a wide range of wavelengths. The amount of radiation energy emitted from a surface at a given wavelength depends on the material of the body and the condition of its surface as well as the surface temperature. Therefore, different bodies may emit different amounts of radiation per unit surface area, even when they are at the same temperature. Thus, it is natural to be curious about the maximum amount of radiation that can be emitted by a surface at a given temperature. Satisfying this curiosity requires the definition of an idealized body, called a blackbody, to serve as a standard against which the radiative properties of real surfaces may be compared. A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation, regardless of wavelength and direction. Also, a blackbody emits radiation energy uniformly in all directions per unit area normal to direction of emission. (Fig. 11–7). That is, a blackbody is a diffuse emitter. The term diffuse means “independent of direction.” The radiation energy emitted by a blackbody per unit time and per unit surface area was determined experimentally by Joseph Stefan in 1879 and expressed as Eb(T ) T 4 (W/m2) FIGURE 11–6 Radiation in opaque solids is considered a surface phenomenon since the radiation emitted only by the molecules at the surface can escape the solid. Uniform Nonuniform Blackbody Real body (11-3) where 5.67 108 W/m2 · K4 is the Stefan–Boltzmann constant and T is the absolute temperature of the surface in K. This relation was theoretically verified in 1884 by Ludwig Boltzmann. Equation 11-3 is known as the Stefan–Boltzmann law and Eb is called the blackbody emissive power. Note that the emission of thermal radiation is proportional to the fourth power of the absolute temperature. Although a blackbody would appear black to the eye, a distinction should be made between the idealized blackbody and an ordinary black surface. Any surface that absorbs light (the visible portion of radiation) would appear black to the eye, and a surface that reflects it completely would appear white. Considering that visible radiation occupies a very narrow band of the spectrum from 0.4 to 0.76 m, we cannot make any judgments about the blackness of a surface on the basis of visual observations. For example, snow and white paint reflect light and thus appear white. But they are essentially black for infrared radiation since they strongly absorb long-wavelength radiation. Surfaces coated with lampblack paint approach idealized blackbody behavior. Another type of body that closely resembles a blackbody is a large cavity with a small opening, as shown in Figure 11–8. Radiation coming in through the opening of area A will undergo multiple reflections, and thus it will have several chances to be absorbed by the interior surfaces of the cavity before any FIGURE 11–7 A blackbody is said to be a diffuse emitter since it emits radiation energy uniformly in all directions. Small opening of area A Large cavity T FIGURE 11–8 A large isothermal cavity at temperature T with a small opening of area A closely resembles a blackbody of surface area A at the same temperature. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 566 566 HEAT TRANSFER part of it can possibly escape. Also, if the surface of the cavity is isothermal at temperature T, the radiation emitted by the interior surfaces will stream through the opening after undergoing multiple reflections, and thus it will have a diffuse nature. Therefore, the cavity will act as a perfect absorber and perfect emitter, and the opening will resemble a blackbody of surface area A at temperature T, regardless of the actual radiative properties of the cavity. The Stefan–Boltzmann law in Eq. 11-3 gives the total blackbody emissive power Eb, which is the sum of the radiation emitted over all wavelengths. Sometimes we need to know the spectral blackbody emissive power, which is the amount of radiation energy emitted by a blackbody at an absolute temperature T per unit time, per unit surface area, and per unit wavelength about the wavelength . For example, we are more interested in the amount of radiation an incandescent light bulb emits in the visible wavelength spectrum than we are in the total amount emitted. The relation for the spectral blackbody emissive power Eb was developed by Max Planck in 1901 in conjunction with his famous quantum theory. This relation is known as Planck’s law and is expressed as Eb(, T ) C1 [exp (C2 /T ) 1] 5 (W/m2 · m) (11-4) where C1 2 hc 02 3.742 108 W · m4/m2 C2 hc0 /k 1.439 104 m · K Also, T is the absolute temperature of the surface, is the wavelength of the radiation emitted, and k 1.38065 1023 J/K is Boltzmann’s constant. This relation is valid for a surface in a vacuum or a gas. For other mediums, it needs to be modified by replacing C1 by C1/n2, where n is the index of refraction of the medium. Note that the term spectral indicates dependence on wavelength. The variation of the spectral blackbody emissive power with wavelength is plotted in Figure 11–9 for selected temperatures. Several observations can be made from this figure: 1. The emitted radiation is a continuous function of wavelength. At any specified temperature, it increases with wavelength, reaches a peak, and then decreases with increasing wavelength. 2. At any wavelength, the amount of emitted radiation increases with increasing temperature. 3. As temperature increases, the curves shift to the left to the shorterwavelength region. Consequently, a larger fraction of the radiation is emitted at shorter wavelengths at higher temperatures. 4. The radiation emitted by the sun, which is considered to be a blackbody at 5780 K (or roughly at 5800 K), reaches its peak in the visible region of the spectrum. Therefore, the sun is in tune with our eyes. On the other hand, surfaces at T 800 K emit almost entirely in the infrared region and thus are not visible to the eye unless they reflect light coming from other sources. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 567 Violet Red 567 CHAPTER 11 1014 Visible light region 5800 K (Solar) 4000 K 1012 Locus of maximum power: λT = 2898 µm·K 2000 K 1010 Ebλ , W/ m2 ·µm 1000 K 500 K 108 300 K 106 100 K 104 102 1 0.01 0.1 1 10 100 1000 Wavelength λ, µm As the temperature increases, the peak of the curve in Figure 11–9 shifts toward shorter wavelengths. The wavelength at which the peak occurs for a specified temperature is given by Wien’s displacement law as (T )max power 2897.8 m · K (11-5) This relation was originally developed by Willy Wien in 1894 using classical thermodynamics, but it can also be obtained by differentiating Eq. 11-4 with respect to while holding T constant and setting the result equal to zero. A plot of Wien’s displacement law, which is the locus of the peaks of the radiation emission curves, is also given in Figure 11–9. The peak of the solar radiation, for example, occurs at 2897.8/ 5780 0.50 m, which is near the middle of the visible range. The peak of the radiation emitted by a surface at room temperature (T 298 K) occurs at 9.72 m, which is well into the infrared region of the spectrum. An electrical resistance heater starts radiating heat soon after it is plugged in, and we can feel the emitted radiation energy by holding our hands facing the heater. But this radiation is entirely in the infrared region and thus cannot FIGURE 11–9 The variation of the blackbody emissive power with wavelength for several temperatures. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 568 568 HEAT TRANSFER Incident light Reflected Re d d w Re ello en Y re e G lu B w llo n Ye ree e G lu B Absorbed FIGURE 11–10 A surface that reflects red while absorbing the remaining parts of the incident light appears red to the eye. be sensed by our eyes. The heater would appear dull red when its temperature reaches about 1000 K, since it will start emitting a detectable amount (about 1 W/m2 · m) of visible red radiation at that temperature. As the temperature rises even more, the heater appears bright red and is said to be red hot. When the temperature reaches about 1500 K, the heater emits enough radiation in the entire visible range of the spectrum to appear almost white to the eye, and it is called white hot. Although it cannot be sensed directly by the human eye, infrared radiation can be detected by infrared cameras, which transmit the information to microprocessors to display visual images of objects at night. Rattlesnakes can sense the infrared radiation or the “body heat” coming off warm-blooded animals, and thus they can see at night without using any instruments. Similarly, honeybees are sensitive to ultraviolet radiation. A surface that reflects all of the light appears white, while a surface that absorbs all of the light incident on it appears black. (Then how do we see a black surface?) It should be clear from this discussion that the color of an object is not due to emission, which is primarily in the infrared region, unless the surface temperature of the object exceeds about 1000 K. Instead, the color of a surface depends on the absorption and reflection characteristics of the surface and is due to selective absorption and reflection of the incident visible radiation coming from a light source such as the sun or an incandescent lightbulb. A piece of clothing containing a pigment that reflects red while absorbing the remaining parts of the incident light appears “red” to the eye (Fig. 11–10). Leaves appear “green” because their cells contain the pigment chlorophyll, which strongly reflects green while absorbing other colors. It is left as an exercise to show that integration of the spectral blackbody emissive power Eb over the entire wavelength spectrum gives the total blackbody emissive power Eb: Eb(T) 0 Eb(, T) d T 4 (W/m2) (11-6) Thus, we obtained the Stefan–Boltzmann law (Eq. 11-3) by integrating Planck’s law (Eq. 11-4) over all wavelengths. Note that on an Eb– chart, Eb corresponds to any value on the curve, whereas Eb corresponds to the area under the entire curve for a specified temperature (Fig. 11–11). Also, the term total means “integrated over all wavelengths.” Ebλ Ebλ(λ, T ) Eb(T ) EXAMPLE 11–1 λ FIGURE 11–11 On an Eb– chart, the area under a curve for a given temperature represents the total radiation energy emitted by a blackbody at that temperature. Radiation Emission from a Black Ball Consider a 20-cm-diameter spherical ball at 800 K suspended in air as shown in Figure 11–12. Assuming the ball closely approximates a blackbody, determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 m. SOLUTION An isothermal sphere is suspended in air. The total blackbody emissive power, the total radiation emitted in 5 minutes, and the spectral blackbody emissive power at 3 mm are to be determined. Assumptions The ball behaves as a blackbody. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 569 569 CHAPTER 11 Analysis (a) The total blackbody emissive power is determined from the Stefan–Boltzmann law to be Eb T 4 (5.67 108 W/m2 · K4)(800 K)4 23.2 103 W/m2 23.2 kW/m2 800 K That is, the ball emits 23.2 kJ of energy in the form of electromagnetic radiation per second per m2 of the surface area of the ball. 20 cm (b) The total amount of radiation energy emitted from the entire ball in 5 min is determined by multiplying the blackbody emissive power obtained above by the total surface area of the ball and the given time interval: Ball As D2 (0.2 m)2 0.1257 m2 FIGURE 11–12 The spherical ball considered in Example 11–1. 60 s t (5 min) 300 s 1 min 10001 kJW · s Qrad EbAs t (23.2 kW/m2)(0.1257 m2)(300 s) 876 kJ That is, the ball loses 876 kJ of its internal energy in the form of electromagnetic waves to the surroundings in 5 min, which is enough energy to raise the temperature of 1 kg of water by 50°C. Note that the surface temperature of the ball cannot remain constant at 800 K unless there is an equal amount of energy flow to the surface from the surroundings or from the interior regions of the ball through some mechanisms such as chemical or nuclear reactions. (c) The spectral blackbody emissive power at a wavelength of 3 m is determined from Planck’s distribution law to be Eb C1 T 1 5 exp C2 3.743 108 W · m4/m2 1.4387 104 m · K 1 (3 m)(800 K) (3 m)5 exp 3848 W/m2 · m The Stefan–Boltzmann law Eb(T ) T 4 gives the total radiation emitted by a blackbody at all wavelengths from 0 to . But we are often interested in the amount of radiation emitted over some wavelength band. For example, an incandescent lightbulb is judged on the basis of the radiation it emits in the visible range rather than the radiation it emits at all wavelengths. The radiation energy emitted by a blackbody per unit area over a wavelength band from 0 to is determined from (Fig. 11–13) Eb, 0–(T ) E (, T ) d 0 b (W/m2) Ebλ λ1 Eb, 0 – λ (T ) = Ebλ(λ, T )d λ 1 0 Ebλ(λ, T ) (11-7) It looks like we can determine Eb, 0– by substituting the Eb relation from Eq. 11-4 and performing this integration. But it turns out that this integration does not have a simple closed-form solution, and performing a numerical integration each time we need a value of Eb, 0– is not practical. Therefore, we define a dimensionless quantity f called the blackbody radiation function as λ1 λ FIGURE 11–13 On an Eb– chart, the area under the curve to the left of the 1 line represents the radiation energy emitted by a blackbody in the wavelength range 0–1 for the given temperature. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 570 570 HEAT TRANSFER TABLE 11–2 Blackbody radiation functions f T, m · K f T, m · K 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000 5200 5400 5600 5800 6000 0.000000 0.000000 0.000000 0.000016 0.000321 0.002134 0.007790 0.019718 0.039341 0.066728 0.100888 0.140256 0.183120 0.227897 0.273232 0.318102 0.361735 0.403607 0.443382 0.480877 0.516014 0.548796 0.579280 0.607559 0.633747 0.658970 0.680360 0.701046 0.720158 0.737818 6200 6400 6600 6800 7000 7200 7400 7600 7800 8000 8500 9000 9500 10,000 10,500 11,000 11,500 12,000 13,000 14,000 15,000 16,000 18,000 20,000 25,000 30,000 40,000 50,000 75,000 100,000 f 0.754140 0.769234 0.783199 0.796129 0.808109 0.819217 0.829527 0.839102 0.848005 0.856288 0.874608 0.890029 0.903085 0.914199 0.923710 0.931890 0.939959 0.945098 0.955139 0.962898 0.969981 0.973814 0.980860 0.985602 0.992215 0.995340 0.997967 0.998953 0.999713 0.999905 Ebλ —— Eb fλ 1 – λ2 = f0 – λ – f0 – λ 2 1 Ebλ(λ, T ) ———–— Eb(T ) λ1 λ2 FIGURE 11–14 Graphical representation of the fraction of radiation emitted in the wavelength band from 1 to 2. λ f(T) 0 Eb( , T ) d T 4 (11-8) The function f represents the fraction of radiation emitted from a blackbody at temperature T in the wavelength band from 0 to . The values of f are listed in Table 11–2 as a function of T, where is in m and T is in K. The fraction of radiation energy emitted by a blackbody at temperature T over a finite wavelength band from 1 to 2 is determined from (Fig. 11–14) f1–2(T ) f2(T ) f1(T ) (11-9) where f1(T ) and f2(T ) are blackbody radiation functions corresponding to 1T and 2T, respectively. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 571 571 CHAPTER 11 EXAMPLE 11–2 Emission of Radiation from a Lightbulb The temperature of the filament of an incandescent lightbulb is 2500 K. Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks. SOLUTION The temperature of the filament of an incandescent lightbulb is given. The fraction of visible radiation emitted by the filament and the wavelength at which the emission peaks are to be determined. Assumptions The filament behaves as a blackbody. Analysis The visible range of the electromagnetic spectrum extends from 1 0.4 m to 2 0.76 m. Noting that T 2500 K, the blackbody radiation functions corresponding to 1T and 2T are determined from Table 11–2 to be 1T (0.40 m)(2500 K) 1000 m · K 2T (0.76 m)(2500 K) 1900 m · K → f1 0.000321 → f2 0.053035 That is, 0.03 percent of the radiation is emitted at wavelengths less than 0.4 m and 5.3 percent at wavelengths less than 0.76 m. Then the fraction of radiation emitted between these two wavelengths is (Fig. 11–15) f1–2 f2 f1 0.053035 0.000321 0.0527135 Ebλ —— Eb f0.4 – 0.76 = f0 – 0.76 – f0 – 0.4 Therefore, only about 5 percent of the radiation emitted by the filament of the lightbulb falls in the visible range. The remaining 95 percent of the radiation appears in the infrared region in the form of radiant heat or “invisible light,” as it used to be called. This is certainly not a very efficient way of converting electrical energy to light and explains why fluorescent tubes are a wiser choice for lighting. The wavelength at which the emission of radiation from the filament peaks is easily determined from Wien’s displacement law to be (T )max power 2897.8 m · K → max power 2897.8 m · K 1.16 m 2500 K Discussion Note that the radiation emitted from the filament peaks in the infrared region. 11–4 ■ RADIATION INTENSITY Radiation is emitted by all parts of a plane surface in all directions into the hemisphere above the surface, and the directional distribution of emitted (or incident) radiation is usually not uniform. Therefore, we need a quantity that describes the magnitude of radiation emitted (or incident) in a specified direction in space. This quantity is radiation intensity, denoted by I. Before we can describe a directional quantity, we need to specify direction in space. The direction of radiation passing through a point is best described in spherical coordinates in terms of the zenith angle and the azimuth angle , as shown in 0.4 0.76 1.16 λ, µm FIGURE 11–15 Graphical representation of the fraction of radiation emitted in the visible range in Example 11–2. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 572 572 HEAT TRANSFER Figure 11–16. Radiation intensity is used to describe how the emitted radiation varies with the zenith and azimuth angles. If all surfaces emitted radiation uniformly in all directions, the emissive power would be sufficient to quantify radiation, and we would not need to deal with intensity. The radiation emitted by a blackbody per unit normal area is the same in all directions, and thus there is no directional dependence. But this is not the case for real surfaces. Before we define intensity, we need to quantify the size of an opening in space. z Emitted radiation I(θ , φ) θ dA y φ x FIGURE 11–16 Radiation intensity is used to describe the variation of radiation energy with direction. Arc length Plain angle, A slice of pizza of plain angle Solid Angle Let us try to quantify the size of a slice of pizza. One way of doing that is to specify the arc length of the outer edge of the slice, and to form the slice by connecting the endpoints of the arc to the center. A more general approach is to specify the angle of the slice at the center, as shown in Figure 11–17. An angle of 90˚ (or /2 radians), for example, always represents a quarter pizza, no matter what the radius is. For a circle of unit radius, the length of an arc is equivalent in magnitude to the plane angle it subtends (both are 2 for a complete circle of radius r 1). Now consider a watermelon, and let us attempt to quantify the size of a slice. Again we can do it by specifying the outer surface area of the slice (the green part), or by working with angles for generality. Connecting all points at the edges of the slice to the center in this case will form a three-dimensional body (like a cone whose tip is at the center), and thus the angle at the center in this case is properly called the solid angle. The solid angle is denoted by , and its unit is the steradian (sr). In analogy to plane angle, we can say that the area of a surface on a sphere of unit radius is equivalent in magnitude to the solid angle it subtends (both are 4 for a sphere of radius r 1). This can be shown easily by considering a differential surface area on a sphere dS r2 sin d d, as shown in Figure 11–18, and integrating it from 0 to , and from 0 to 2 . We get S dS Solid angle, sphere Surface area, S A slice of watermelon of solid angle FIGURE 11–17 Describing the size of a slice of pizza by a plain angle, and the size of a watermelon slice by a solid angle. 2 r 2 sin d 2 r 2 0 0 sin d 4 r 2 0 (11-10) which is the formula for the area of a sphere. For r 1 it reduces to S 4 , and thus the solid angle associated with a sphere is 4 sr. For a hemisphere, which is more relevant to radiation emitted or received by a surface, it is 2 sr. The differential solid angle d subtended by a differential area dS on a sphere of radius r can be expressed as d dS sin d d r2 (11-11) Note that the area dS is normal to the direction of viewing since dS is viewed from the center of the sphere. In general, the differential solid angle d subtended by a differential surface area dA when viewed from a point at a distance r from dA is expressed as d dAn d A cos r2 r2 (11-12) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 573 573 CHAPTER 11 Radiation emitted into direction (θ ,φ) 0 θ π/2 0 φ 2π dS (r sin θ dφ )(r dθ ) r 2 sin θ dθ dφ Ie (θ ,φ) r dA Solid angle: d dS/r 2 sin θ d θ dφ θ φ r dθ r sin θ dS dφ r sin θ dφ θ P dθ dφ Solid angle for a hemisphere: π/2 2π d 0θ 0φsin θ r dφ d θ d φ 2π Hemisphere where is the angle between the normal of the surface and the direction of viewing, and thus dAn dA cos is the normal (or projected) area to the direction of viewing. Small surfaces viewed from relatively large distances can approximately be treated as differential areas in solid angle calculations. For example, the solid angle subtended by a 5 cm2 plane surface when viewed from a point O at a distance of 80 cm along the normal of the surface is An 5 cm 2 7.81 104 sr 2 (80 cm) 2 r If the surface is tilted so that the normal of the surface makes an angle of 60˚ with the line connecting point O to the center of the surface, the projected area would be dAn dA cos (5 cm2)cos 60˚ 2.5 cm2, and the solid angle in this case would be half of the value just determined. Intensity of Emitted Radiation Consider the emission of radiation by a differential area element dA of a surface, as shown in Figure 11–18. Radiation is emitted in all directions into the hemispherical space, and the radiation streaming though the surface area dS is proportional to the solid angle d subtended by dS. It is also proportional to the radiating area dA as seen by an observer on dS, which varies from a maximum of dA when dS is at the top directly above dA ( 0˚) to a minimum of zero when dS is at the bottom ( 90˚). Therefore, the effective area of dA for emission in the direction of is the projection of dA on a plane normal to , which is dA cos . Radiation intensity in a given direction is based on a unit area normal to that direction to provide a common basis for the comparison of radiation emitted in different directions. The radiation intensity for emitted radiation Ie( , ) is defined as the rate · at which radiation energy dQ e is emitted in the ( , ) direction per unit area normal to this direction and per unit solid angle about this direction. That is, Ie( , ) dQ˙ e dQ˙ e d A cos d d A cos sin d d (W/m2 sr) (11-13) FIGURE 11–18 The emission of radiation from a differential surface element into the surrounding hemispherical space through a differential solid angle. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 574 574 HEAT TRANSFER The radiation flux for emitted radiation is the emissive power E (the rate at which radiation energy is emitted per unit area of the emitting surface), which can be expressed in differential form as dE dQ˙ e Ie( , ) cos sin d d dA (11-14) Noting that the hemisphere above the surface will intercept all the radiation rays emitted by the surface, the emissive power from the surface into the hemisphere surrounding it can be determined by integration as E dE hemisphere 2 /2 0 0 Ie( , ) cos sin d d (W/m 2) (11-15) The intensity of radiation emitted by a surface, in general, varies with direction (especially with the zenith angle ). But many surfaces in practice can be approximated as being diffuse. For a diffusely emitting surface, the intensity of the emitted radiation is independent of direction and thus Ie constant. 2 Noting that /2 0 0 cos sin d d , the emissive power relation in Eq. 11-15 reduces in this case to A Projected area An A cos θ n θ Solid angle, FIGURE 11–19 Radiation intensity is based on projected area, and thus the calculation of radiation emission from a surface involves the projection of the surface. (W/m2) (11-16) Note that the factor in Eq. 11-16 is . You might have expected it to be 2 since intensity is radiation energy per unit solid angle, and the solid angle associated with a hemisphere is 2 . The reason for the factor being is that the emissive power is based on the actual surface area whereas the intensity is based on the projected area (and thus the factor cos that accompanies it), as shown in Figure 11–19. For a blackbody, which is a diffuse emitter, Eq. 11-16 can be expressed as Eb Ib Blackbody: (11-17) where Eb T4 is the blackbody emissive power. Therefore, the intensity of the radiation emitted by a blackbody at absolute temperature T is Ib(T) Blackbody: Incident radiation E Ie Diffusely emitting surface: Eb(T ) T 4 (W/m2 sr) (11-18) I i (θ , φ ) θ dA φ FIGURE 11–20 Radiation incident on a surface in the direction ( , ). Incident Radiation All surfaces emit radiation, but they also receive radiation emitted or reflected by other surfaces. The intensity of incident radiation Ii( , ) is defined as the rate at which radiation energy dG is incident from the ( , ) direction per unit area of the receiving surface normal to this direction and per unit solid angle about this direction (Fig. 11–20). Here is the angle between the direction of incident radiation and the normal of the surface. The radiation flux incident on a surface from all directions is called irradiation G, and is expressed as G dG hemisphere 2 /2 0 0 Ii ( , ) cos sin d d (W/m2) (11-19) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 575 575 CHAPTER 11 Therefore irradiation represents the rate at which radiation energy is incident on a surface per unit area of the surface. When the incident radiation is diffuse and thus Ii constant, Eq. 11-19 reduces to G Ii Diffusely incident radiation: (W/m2) (11-20) Again note that irradiation is based on the actual surface area (and thus the factor cos ), whereas the intensity of incident radiation is based on the projected area. Radiosity Surfaces emit radiation as well as reflecting it, and thus the radiation leaving a surface consists of emitted and reflected components, as shown in Figure 11–21. The calculation of radiation heat transfer between surfaces involves the total radiation energy streaming away from a surface, with no regard for its origin. Thus, we need to define a quantity that represents the rate at which radiation energy leaves a unit area of a surface in all directions. This quantity is called the radiosity J, and is expressed as J 2 /2 Ier( , ) cos sin d d 0 0 (W/m2) (11-21) where Ier is the sum of the emitted and reflected intensities. For a surface that is both a diffuse emitter and a diffuse reflector, Ier constant, and the radiosity relation reduces to J Ier Diffuse emitter and reflector: (W/m2) (11-22) For a blackbody, radiosity J is equivalent to the emissive power Eb since a blackbody absorbs the entire radiation incident on it and there is no reflected component in radiosity. Spectral Quantities So far we considered total radiation quantities (quantities integrated over all wavelengths), and made no reference to wavelength dependence. This lumped approach is adequate for many radiation problems encountered in practice. But sometimes it is necessary to consider the variation of radiation with wavelength as well as direction, and to express quantities at a certain wavelength or per unit wavelength interval about . Such quantities are referred to as spectral quantities to draw attention to wavelength dependence. The modifier “spectral” is used to indicate “at a given wavelength.” The spectral radiation intensity I(, , ), for example, is simply the total radiation intensity I( , ) per unit wavelength interval about . The spectral intensity for emitted radiation I, e(, , ) can be defined as the rate at which · radiation energy dQ e is emitted at the wavelength in the ( , ) direction per unit area normal to this direction, per unit solid angle about this direction, and it can be expressed as I, e(, , ) dQ˙ e d A cos d d λ (W/m2 sr m) (11-23) Radiosity, J (Reflected irradiation) Irradiation, G Emissive power, E FIGURE 11–21 The three kinds of radiation flux (in W/m2): emissive power, irradiation, and radiosity. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 576 576 HEAT TRANSFER Then the spectral emissive power becomes 2 E 0 0 Iλ, e Iλ, e (, , )cos sin d d (W/m2) (11-24) Similar relations can be obtained for spectral irradiation G, and spectral radiosity J by replacing I, e in this equation by I, i and I, er, respectively. When the variation of spectral radiation intensity I with wavelength is known, the total radiation intensity I for emitted, incident, and emitted reflected radiation can be determined by integration over the entire wavelength spectrum as (Fig. 11–22) Area Iλ, e dλ Ie 0 Iλ, e λ dλ /2 FIGURE 11–22 Integration of a “spectral” quantity for all wavelengths gives the “total” quantity. Ie I 0 λ, e d λ, Ii I 0 λ, i d λ , Ier and I 0 λ, er d λ (11-25) These intensities can then be used in Eqs. 11-15, 11-19, and 11-21 to determine the emissive power E, irradiation G, and radiosity J, respectively. Similarly, when the variations of spectral radiation fluxes E, G, and J with wavelength are known, the total radiation fluxes can be determined by integration over the entire wavelength spectrum as E E d λ, 0 λ G G d λ, 0 J and λ J dλ 0 λ (11-26) When the surfaces and the incident radiation are diffuse, the spectral radiation fluxes are related to spectral intensities as E λ Iλ, e , Gλ Iλ, i , Jλ I λ, er and (11-27) Note that the relations for spectral and total radiation quantities are of the same form. The spectral intensity of radiation emitted by a blackbody at an absolute temperature T at a wavelength has been determined by Max Planck, and is expressed as Ibλ( λ, T ) 2hc 20 λ5[exp(hc 0 / λkT ) 1] (W/m2 sr m) (11-28) where h 6.6256 1034 J s is the Planck constant, k 1.38065 1023 J/K is the Boltzmann constant, and c0 2.9979 108 m/s is the speed of light in a vacuum. Then the spectral blackbody emissive power is, from Eq. 11-27, Eb (, T ) Ib(, T ) A2 5 cm2 (11-29) A simplified relation for Eb is given by Eq. 11-4. θ 2 40° θ 1 55° r 75 cm A1 3 cm2 T1 600 K FIGURE 11–23 Schematic for Example 11–3. EXAMPLE 11–3 Radiation Incident on a Small Surface A small surface of area A1 3 cm2 emits radiation as a blackbody at T1 600 K. Part of the radiation emitted by A1 strikes another small surface of area A2 5 cm2 oriented as shown in Figure 11–23. Determine the solid angle subtended by A2 when viewed from A1, and the rate at which radiation emitted by A1 that strikes A2. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 577 577 CHAPTER 11 SOLUTION A surface is subjected to radiation emitted by another surface. The solid angle subtended and the rate at which emitted radiation is received are to be determined. Assumptions 1 Surface A1 emits diffusely as a blackbody. 2 Both A1 and A2 can be approximated as differential surfaces since both are very small compared to the square of the distance between them. Analysis Approximating both A1 and A2 as differential surfaces, the solid angle subtended by A2 when viewed from A1 can be determined from Eq. 11-12 to be 2–1 An, 2 r 2 A2 cos 2 (5 cm2) cos 40º 6.81 104 sr (75 cm) 2 r2 since the normal of A2 makes 40˚ with the direction of viewing. Note that solid angle subtended by A2 would be maximum if A2 were positioned normal to the direction of viewing. Also, the point of viewing on A1 is taken to be a point in the middle, but it can be any point since A1 is assumed to be very small. The radiation emitted by A1 that strikes A2 is equivalent to the radiation emitted by A1 through the solid angle 2–1. The intensity of the radiation emitted by A1 is I1 Eh(T1) T 41 (5.67 108 W/m2 K4)(600 K)4 2339 W/m 2 sr This value of intensity is the same in all directions since a blackbody is a diffuse emitter. Intensity represents the rate of radiation emission per unit area normal to the direction of emission per unit solid angle. Therefore, the rate of radiation energy emitted by A1 in the direction of 1 through the solid angle 2–1 is determined by multiplying I1 by the area of A1 normal to 1 and the solid angle 2–1. That is, · Q 1–2 I1(A1 cos 1)2–1 (2339 W/m2 sr)(3 104 cos 55˚ m2)(6.81 104 sr) 2.74 104 W Therefore, the radiation emitted from surface A1 will strike surface A2 at a rate of 2.74 104 W. · Discussion The total rate of radiation emission from surface A1 is Q e A1T14 2.204 W. Therefore, the fraction of emitted radiation that strikes A2 is 2.74 104/2.204 0.00012 (or 0.012 percent). Noting that the solid angle associated with a hemisphere is 2 , the fraction of the solid angle subtended by A2 is 6.81 104/(2 ) 0.000108 (or 0.0108 percent), which is 0.9 times the fraction of emitted radiation. Therefore, the fraction of the solid angle a surface occupies does not represent the fraction of radiation energy the surface will receive even when the intensity of emitted radiation is constant. This is because radiation energy emitted by a surface in a given direction is proportional to the projected area of the surface in that direction, and reduces from a maximum at 0˚ (the direction normal to surface) to zero at 90˚ (the direction parallel to surface). 11–5 ■ RADIATIVE PROPERTIES Most materials encountered in practice, such as metals, wood, and bricks, are opaque to thermal radiation, and radiation is considered to be a surface cen58933_ch11.qxd 9/9/2002 9:38 AM Page 578 578 HEAT TRANSFER phenomenon for such materials. That is, thermal radiation is emitted or absorbed within the first few microns of the surface, and thus we speak of radiative properties of surfaces for opaque materials. Some other materials, such as glass and water, allow visible radiation to penetrate to considerable depths before any significant absorption takes place. Radiation through such semitransparent materials obviously cannot be considered to be a surface phenomenon since the entire volume of the material interacts with radiation. On the other hand, both glass and water are practically opaque to infrared radiation. Therefore, materials can exhibit different behavior at different wavelengths, and the dependence on wavelength is an important consideration in the study of radiative properties such as emissivity, absorptivity, reflectivity, and transmissivity of materials. In the preceding section, we defined a blackbody as a perfect emitter and absorber of radiation and said that no body can emit more radiation than a blackbody at the same temperature. Therefore, a blackbody can serve as a convenient reference in describing the emission and absorption characteristics of real surfaces. Emissivity The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature. The emissivity of a surface is denoted by , and it varies between zero and one, 0 1. Emissivity is a measure of how closely a surface approximates a blackbody, for which 1. The emissivity of a real surface is not a constant. Rather, it varies with the temperature of the surface as well as the wavelength and the direction of the emitted radiation. Therefore, different emissivities can be defined for a surface, depending on the effects considered. The most elemental emissivity of a surface at a given temperature is the spectral directional emissivity, which is defined as the ratio of the intensity of radiation emitted by the surface at a specified wavelength in a specified direction to the intensity of radiation emitted by a blackbody at the same temperature at the same wavelength. That is, , (, , , T) Iλ, e (λ, , , T ) Ibλ(λ , T ) (11-30) where the subscripts and are used to designate spectral and directional quantities, respectively. Note that blackbody radiation intensity is independent of direction, and thus it has no functional dependence on and . The total directional emissivity is defined in a like manner by using total intensities (intensities integrated over all wavelengths) as ( , , T) Ie( , , T ) Ib(T ) (11-31) In practice, it is usually more convenient to work with radiation properties averaged over all directions, called hemispherical properties. Noting that the integral of the rate of radiation energy emitted at a specified wavelength per unit surface area over the entire hemisphere is spectral emissive power, the spectral hemispherical emissivity can be expressed as cen58933_ch11.qxd 9/9/2002 9:38 AM Page 579 579 CHAPTER 11 Eλ(λ, T ) E bλ(λ, T ) (, T) (11-32) Note that the emissivity of a surface at a given wavelength can be different at different temperatures since the spectral distribution of emitted radiation (and thus the amount of radiation emitted at a given wavelength) changes with temperature. Finally, the total hemispherical emissivity is defined in terms of the radiation energy emitted over all wavelengths in all directions as E(T ) Eb(T ) (T) (11-33) Therefore, the total hemispherical emissivity (or simply the “average emissivity”) of a surface at a given temperature represents the ratio of the total radiation energy emitted by the surface to the radiation emitted by a blackbody of the same surface area at the same temperature. Noting from Eqs. 11-26 and 11-32 that E E d and E (, T ) 0 (, T )Eb(, T ), and the total hemispherical emissivity can also be expressed as ( λ, T )E E(T ) (T ) E b(T ) bλ( λ, λ 0 T )d λ (11-34) T 4 since Eb(T ) T 4. To perform this integration, we need to know the variation of spectral emissivity with wavelength at the specified temperature. The integrand is usually a complicated function, and the integration has to be performed numerically. However, the integration can be performed quite easily by dividing the spectrum into a sufficient number of wavelength bands and assuming the emissivity to remain constant over each band; that is, by expressing the function (, T) as a step function. This simplification offers great convenience for little sacrifice of accuracy, since it allows us to transform the integration into a summation in terms of blackbody emission functions. As an example, consider the emissivity function plotted in Figure 11–24. It seems like this function can be approximated reasonably well by a step function of the form 1 constant, • 2 constant, 3 constant, 0 1 1 2 2 ελ (11-35) ε2 Then the average emissivity can be determined from Eq. 11-34 by breaking the integral into three parts and utilizing the definition of the blackbody radiation function as ε1 1 (T) λ1 0 Eb 2 Ebλdλ λ2 λ1 E 3 Ebλdλ λ2 bλdλ Eb Eb 1 f01(T) 2 f12(T) 3 f2 (T) ε2 ε1 ε3 ε3 λ1 (11-36) Radiation is a complex phenomenon as it is, and the consideration of the wavelength and direction dependence of properties, assuming sufficient data Actual variation λ2 FIGURE 11–24 Approximating the actual variation of emissivity with wavelength by a step function. λ cen58933_ch11.qxd 9/9/2002 9:38 AM Page 580 580 HEAT TRANSFER Real surface: εθ ≠ constant ελ ≠ constant Diffuse surface: εθ = constant Gray surface: ελ = constant Diffuse, gray surface: ε = ελ = εθ = constant FIGURE 11–25 The effect of diffuse and gray approximations on the emissivity of a surface. 1 Nonconductor εθ 0.5 Conductor 0 0° 15° 30° 45° θ 60° 75° 90° FIGURE 11–26 Typical variations of emissivity with direction for electrical conductors and nonconductors. exist, makes it even more complicated. Therefore, the gray and diffuse approximations are often utilized in radiation calculations. A surface is said to be diffuse if its properties are independent of direction, and gray if its properties are independent of wavelength. Therefore, the emissivity of a gray, diffuse surface is simply the total hemispherical emissivity of that surface because of independence of direction and wavelength (Fig. 11–25). A few comments about the validity of the diffuse approximation are in order. Although real surfaces do not emit radiation in a perfectly diffuse manner as a blackbody does, they often come close. The variation of emissivity with direction for both electrical conductors and nonconductors is given in Figure 11–26. Here is the angle measured from the normal of the surface, and thus 0 for radiation emitted in a direction normal to the surface. Note that remains nearly constant for about 40˚ for conductors such as metals and for 70˚ for nonconductors such as plastics. Therefore, the directional emissivity of a surface in the normal direction is representative of the hemispherical emissivity of the surface. In radiation analysis, it is common practice to assume the surfaces to be diffuse emitters with an emissivity equal to the value in the normal ( 0) direction. The effect of the gray approximation on emissivity and emissive power of a real surface is illustrated in Figure 11–27. Note that the radiation emission from a real surface, in general, differs from the Planck distribution, and the emission curve may have several peaks and valleys. A gray surface should emit as much radiation as the real surface it represents at the same temperature. Therefore, the areas under the emission curves of the real and gray surfaces must be equal. The emissivities of common materials are listed in Table A–18 in the appendix, and the variation of emissivity with wavelength and temperature is illustrated in Figure 11–28. Typical ranges of emissivity of various materials are given in Figure 11–29. Note that metals generally have low emissivities, as low as 0.02 for polished surfaces, and nonmetals such as ceramics and organic materials have high ones. The emissivity of metals increases with temperature. Also, oxidation causes significant increases in the emissivity of metals. Heavily oxidized metals can have emissivities comparable to those of nonmetals. ελ Eλ Blackbody, ε = 1 1 Blackbody, Ebλ Gray surface, ε = const. Gray surface, Eλ = εEbλ ε Real surface, ελ FIGURE 11–27 Comparison of the emissivity (a) and emissive power (b) of a real surface with those of a gray surface and a blackbody at the same temperature. T = const. 0 T = const. Real surface, Eλ = ελEbλ λ 0 (a) λ (b) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 581 581 CHAPTER 11 1.0 0.8 Silicon carbide, 1000 K 0.6 Tungsten 1600 K Aluminum oxide, 1400 K 0.4 Stainless steel, 1200 K heavily oxidized 0.2 2800 K 0 0.1 0.2 0.4 0.6 1 2 4 6 10 Total normal emissivity, ε n Spectral, normal emissivity, ε λ, n 1.0 Stainless steel, 800 K lightly oxidized 20 Heavily oxidized stainless steel 0.8 0.6 Aluminum oxide 0.4 Lightly oxidized stainless steel 0.2 Tungsten 0 40 60 100 0 500 0 1000 Wavelength, λ, µm 1500 2000 2500 3000 3500 Temperature, K (a) (b) FIGURE 11–28 The variation of normal emissivity with (a) wavelength and (b) temperature for various materials. Vegetation, water, skin Care should be exercised in the use and interpretation of radiation property data reported in the literature, since the properties strongly depend on the surface conditions such as oxidation, roughness, type of finish, and cleanliness. Consequently, there is considerable discrepancy and uncertainty in the reported values. This uncertainty is largely due to the difficulty in characterizing and describing the surface conditions precisely. EXAMPLE 11–4 Carbon Ceramics Oxidized metals Metals, unpolished Polished metals Emissivity of a Surface and Emissive Power The spectral emissivity function of an opaque surface at 800 K is approximated as (Fig. 11–30) 1 0.3, 2 0.8, 3 0.1, 0 3 m 3 m 7 m 7 m Determine the average emissivity of the surface and its emissive power. SOLUTION The variation of emissivity of a surface at a specified temperature with wavelength is given. The average emissivity of the surface and its emissive power are to be determined. Analysis The variation of the emissivity of the surface with wavelength is given as a step function. Therefore, the average emissivity of the surface can be determined from Eq. 11-34 by breaking the integral into three parts, (T ) Building materials, paints Rocks, soil Glasses, minerals 1 1 0 2 Eb d T 4 2 1 Eb d T 4 3 2 1 f0–1(T ) 2 f1–2(T ) 3 f2– (T ) 1 f1 2(f2 f1) 3(1 f2) Eb d T 4 0 0.2 0.4 0.6 0.8 1.0 FIGURE 11–29 Typical ranges of emissivity for various materials. ελ 1.0 0.8 0.3 0.1 0 0 3 7 λ, µm FIGURE 11–30 The spectral emissivity of the surface considered in Example 11–4. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 582 582 HEAT TRANSFER where f1 and f2 are blackbody radiation functions corresponding to 1T and 2T. These functions are determined from Table 11–2 to be 1T (3 m)(800 K) 2400 m · K 2T (7 m)(800 K) 5600 m · K → f1 0.140256 → f2 0.701046 Note that f0–1 f1 f0 f1, since f0 0, and f2– f f2 1 f2, since f 1. Substituting, 0.3 0.140256 0.8(0.701046 0.140256) 0.1(1 0.701046) 0.521 That is, the surface will emit as much radiation energy at 800 K as a gray surface having a constant emissivity of 0.521. The emissive power of the surface is E T 4 0.521(5.67 108 W/m2 · K4)(800 K)4 12,100 W/m2 Discussion Note that the surface emits 12.1 kJ of radiation energy per second per m2 area of the surface. Absorptivity, Reflectivity, and Transmissivity Incident radiation G, W/m 2 Reflected ρG Everything around us constantly emits radiation, and the emissivity represents the emission characteristics of those bodies. This means that every body, including our own, is constantly bombarded by radiation coming from all directions over a range of wavelengths. Recall that radiation flux incident on a surface is called irradiation and is denoted by G. When radiation strikes a surface, part of it is absorbed, part of it is reflected, and the remaining part, if any, is transmitted, as illustrated in Figure 11–31. The fraction of irradiation absorbed by the surface is called the absorptivity , the fraction reflected by the surface is called the reflectivity , and the fraction transmitted is called the transmissivity . That is, Absorptivity: Semitransparent material Absorbed αG Reflectivity: Transmissivity: Transmitted τG FIGURE 11–31 The absorption, reflection, and transmission of incident radiation by a semitransparent material. Absorbed radiation Gabs , G Incident radiation G Reflected radiation ref , G Incident radiation Transmitted radiation Gtr , G Incident radiation 0 1 (11-37) 0 1 (11-38) 0 1 (11-39) where G is the radiation energy incident on the surface, and Gabs, Gref, and Gtr are the absorbed, reflected, and transmitted portions of it, respectively. The first law of thermodynamics requires that the sum of the absorbed, reflected, and transmitted radiation energy be equal to the incident radiation. That is, Gabs Gref Gtr G (11-40) Dividing each term of this relation by G yields 1 For opaque surfaces, 0, and thus (11-41) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 583 583 CHAPTER 11 1 (11-42) This is an important property relation since it enables us to determine both the absorptivity and reflectivity of an opaque surface by measuring either of these properties. These definitions are for total hemispherical properties, since G represents the radiation flux incident on the surface from all directions over the hemispherical space and over all wavelengths. Thus, , , and are the average properties of a medium for all directions and all wavelengths. However, like emissivity, these properties can also be defined for a specific wavelength and/or direction. For example, the spectral directional absorptivity and spectral directional reflectivity of a surface are defined, respectively, as the absorbed and reflected fractions of the intensity of radiation incident at a specified wavelength in a specified direction as λ, (λ, , ) Iλ, abs (λ, , ) Iλ, i (λ, , ) and λ, (λ, , ) Iλ, ref (λ, , ) Iλ, i (λ, , ) (11-43) Likewise, the spectral hemispherical absorptivity and spectral hemispherical reflectivity of a surface are defined as G λ, ref (λ) Gλ( λ) (11-44) where G is the spectral irradiation (in W/m2 m) incident on the surface, and G, abs and G, ref are the reflected and absorbed portions of it, respectively. Similar quantities can be defined for the transmissivity of semitransparent materials. For example, the spectral hemispherical transmissivity of a medium can be expressed as () G λ, tr (λ) G d , G d 0 0 G d , G d 0 0 G d G d 0 Normal Incident ray Re fle cte dr (b) ay s 0 Normal Reflected rays (a) (11-45) G λ(λ) The average absorptivity, reflectivity, and transmissivity of a surface can also be defined in terms of their spectral counterparts as Incident ray (11-46) The reflectivity differs somewhat from the other properties in that it is bidirectional in nature. That is, the value of the reflectivity of a surface depends not only on the direction of the incident radiation but also the direction of reflection. Therefore, the reflected rays of a radiation beam incident on a real surface in a specified direction will form an irregular shape, as shown in Figure 11–32. Such detailed reflectivity data do not exist for most surfaces, and even if they did, they would be of little value in radiation calculations since this would usually add more complication to the analysis than it is worth. In practice, for simplicity, surfaces are assumed to reflect in a perfectly specular or diffuse manner. In specular (or mirrorlike) reflection, the angle of reflection equals the angle of incidence of the radiation beam. In diffuse reflection, radiation is reflected equally in all directions, as shown in Figure Incident ray (c) Normal θ ra y () θ ted and Gλ( λ) lec G λ, abs( λ) Re f () FIGURE 11–32 Different types of reflection from a surface: (a) actual or irregular, (b) diffuse, and (c) specular or mirrorlike. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 584 584 HEAT TRANSFER 1.0 7 9 Absorptivity, α 0.8 6 0.6 0.4 0.2 White fireclay 2. Asbestos 3. Cork 4. Wood 5. Porcelain 6. Concrete 7. Roof shingles 8. Aluminum 9. Graphite 5 2 3 4 8 1 0 300400 600 1000 2000 4000 6000 Source temperature, K FIGURE 11–33 Variation of absorptivity with the temperature of the source of irradiation for various common materials at room temperature. 11–32. Reflection from smooth and polished surfaces approximates specular reflection, whereas reflection from rough surfaces approximates diffuse reflection. In radiation analysis, smoothness is defined relative to wavelength. A surface is said to be smooth if the height of the surface roughness is much smaller than the wavelength of the incident radiation. Unlike emissivity, the absorptivity of a material is practically independent of surface temperature. However, the absorptivity depends strongly on the temperature of the source at which the incident radiation is originating. This is also evident from Figure 11–33, which shows the absorptivities of various materials at room temperature as functions of the temperature of the radiation source. For example, the absorptivity of the concrete roof of a house is about 0.6 for solar radiation (source temperature: 5780 K) and 0.9 for radiation originating from the surrounding trees and buildings (source temperature: 300 K), as illustrated in Figure 11–34. Notice that the absorptivity of aluminum increases with the source temperature, a characteristic for metals, and the absorptivity of electric nonconductors, in general, decreases with temperature. This decrease is most pronounced for surfaces that appear white to the eye. For example, the absorptivity of a white painted surface is low for solar radiation, but it is rather high for infrared radiation. Kirchhoff’s Law Sun α = 0.9 α = 0.6 Consider a small body of surface area As , emissivity , and absorptivity at temperature T contained in a large isothermal enclosure at the same temperature, as shown in Figure 11–35. Recall that a large isothermal enclosure forms a blackbody cavity regardless of the radiative properties of the enclosure surface, and the body in the enclosure is too small to interfere with the blackbody nature of the cavity. Therefore, the radiation incident on any part of the surface of the small body is equal to the radiation emitted by a blackbody at temperature T. That is, G Eb(T ) T 4, and the radiation absorbed by the small body per unit of its surface area is Gabs G T 4 The radiation emitted by the small body is FIGURE 11–34 The absorptivity of a material may be quite different for radiation originating from sources at different temperatures. Eemit T 4 Considering that the small body is in thermal equilibrium with the enclosure, the net rate of heat transfer to the body must be zero. Therefore, the radiation emitted by the body must be equal to the radiation absorbed by it: As T 4 As T 4 Thus, we conclude that (T ) (T ) (11-47) That is, the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature. This relation, which greatly simplifies the radiation analysis, was first developed by Gustav Kirchhoff in 1860 and is now called Kirchhoff’s law. Note that this relation is derived under the condition cen58933_ch11.qxd 9/9/2002 9:38 AM Page 585 585 CHAPTER 11 that the surface temperature is equal to the temperature of the source of irradiation, and the reader is cautioned against using it when considerable difference (more than a few hundred degrees) exists between the surface temperature and the temperature of the source of irradiation. The derivation above can also be repeated for radiation at a specified wavelength to obtain the spectral form of Kirchhoff’s law: (T ) (T ) T T As , ε, α (11-48) This relation is valid when the irradiation or the emitted radiation is independent of direction. The form of Kirchhoff’s law that involves no restrictions is the spectral directional form expressed as , (T ) , (T ). That is, the emissivity of a surface at a specified wavelength, direction, and temperature is always equal to its absorptivity at the same wavelength, direction, and temperature. It is very tempting to use Kirchhoff’s law in radiation analysis since the relation together with 1 enables us to determine all three properties of an opaque surface from a knowledge of only one property. Although Eq. 11-47 gives acceptable results in most cases, in practice, care should be exercised when there is considerable difference between the surface temperature and the temperature of the source of incident radiation. G Eemit FIGURE 11–35 The small body contained in a large isothermal enclosure used in the development of Kirchhoff’s law. Visible 1.0 0.8 The Greenhouse Effect You have probably noticed that when you leave your car under direct sunlight on a sunny day, the interior of the car gets much warmer than the air outside, and you may have wondered why the car acts like a heat trap. The answer lies in the spectral transmissivity curve of the glass, which resembles an inverted U, as shown in Figure 11–36. We observe from this figure that glass at thicknesses encountered in practice transmits over 90 percent of radiation in the visible range and is practically opaque (nontransparent) to radiation in the longer-wavelength infrared regions of the electromagnetic spectrum (roughly 3 m). Therefore, glass has a transparent window in the wavelength range 0.3 m 3 m in which over 90 percent of solar radiation is emitted. On the other hand, the entire radiation emitted by surfaces at room temperature falls in the infrared region. Consequently, glass allows the solar radiation to enter but does not allow the infrared radiation from the interior surfaces to escape. This causes a rise in the interior temperature as a result of the energy build-up in the car. This heating effect, which is due to the nongray characteristic of glass (or clear plastics), is known as the greenhouse effect, since it is utilized primarily in greenhouses (Fig. 11–37). The greenhouse effect is also experienced on a larger scale on earth. The surface of the earth, which warms up during the day as a result of the absorption of solar energy, cools down at night by radiating its energy into deep space as infrared radiation. The combustion gases such as CO2 and water vapor in the atmosphere transmit the bulk of the solar radiation but absorb the infrared radiation emitted by the surface of the earth. Thus, there is concern that the energy trapped on earth will eventually cause global warming and thus drastic changes in weather patterns. In humid places such as coastal areas, there is not a large change between the daytime and nighttime temperatures, because the humidity acts as a barrier τλ 0.6 Thickness 0.038 cm 0.318 cm 0.635 cm 0.4 0.2 0 0.25 0.4 0.6 1.5 3.1 4.7 6.3 7.9 0.7 Wavelength λ, µm FIGURE 11–36 The spectral transmissivity of low-iron glass at room temperature for different thicknesses. Solar radiation Greenhouse Infrared radiation FIGURE 11–37 A greenhouse traps energy by allowing the solar radiation to come in but not allowing the infrared radiation to go out. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 586 586 HEAT TRANSFER on the path of the infrared radiation coming from the earth, and thus slows down the cooling process at night. In areas with clear skies such as deserts, there is a large swing between the daytime and nighttime temperatures because of the absence of such barriers for infrared radiation. 11–6 n^ G0 = Gs cos θ Earth’s surface ys a ’s r Sun θ Gs , W/m2 ’s e rth er Eaosph atm FIGURE 11–38 Solar radiation reaching the earth’s atmosphere and the total solar irradiance. ■ ATMOSPHERIC AND SOLAR RADIATION The sun is our primary source of energy. The energy coming off the sun, called solar energy, reaches us in the form of electromagnetic waves after experiencing considerable interactions with the atmosphere. The radiation energy emitted or reflected by the constituents of the atmosphere form the atmospheric radiation. Below we give an overview of the solar and atmospheric radiation because of their importance and relevance to daily life. Also, our familiarity with solar energy makes it an effective tool in developing a better understanding for some of the new concepts introduced earlier. Detailed treatment of this exciting subject can be found in numerous books devoted to this topic. The sun is a nearly spherical body that has a diameter of D 1.39 109 m and a mass of m 2 1030 kg and is located at a mean distance of L 1.50 1011 m from the earth. It emits radiation energy continuously at a rate of Esun 3.8 1026 W. Less than a billionth of this energy (about 1.7 1017 W) strikes the earth, which is sufficient to keep the earth warm and to maintain life through the photosynthesis process. The energy of the sun is due to the continuous fusion reaction during which two hydrogen atoms fuse to form one atom of helium. Therefore, the sun is essentially a nuclear reactor, with temperatures as high as 40,000,000 K in its core region. The temperature drops to about 5800 K in the outer region of the sun, called the convective zone, as a result of the dissipation of this energy by radiation. The solar energy reaching the earth’s atmosphere is called the total solar irradiance Gs , whose value is Gs 1373 W/m2 (4πL2)Gs (4πr 2)Eb Sun r 1m2 Eb = σT 4sun Gs (11-49) The total solar irradiance (also called the solar constant) represents the rate at which solar energy is incident on a surface normal to the sun’s rays at the outer edge of the atmosphere when the earth is at its mean distance from the sun (Fig. 11–38). The value of the total solar irradiance can be used to estimate the effective surface temperature of the sun from the requirement that 4 (4 L2 )Gs (4 r 2 ) Tsun (11-50) L Earth 1m2 FIGURE 11–39 The total solar energy passing through concentric spheres remains constant, but the energy falling per unit area decreases with increasing radius. where L is the mean distance between the sun’s center and the earth and r is the radius of the sun. The left-hand side of this equation represents the total solar energy passing through a spherical surface whose radius is the mean earth–sun distance, and the right-hand side represents the total energy that leaves the sun’s outer surface. The conservation of energy principle requires that these two quantities be equal to each other, since the solar energy experiences no attenuation (or enhancement) on its way through the vacuum (Fig. 11–39). The effective surface temperature of the sun is determined from Eq. 11-50 to be Tsun 5780 K. That is, the sun can be treated as a cen58933_ch11.qxd 9/9/2002 9:38 AM Page 587 587 CHAPTER 11 2500 5780 K blackbody Solar irradiation Spectral irradiation, W/ m2·µm blackbody at a temperature of 5780 K. This is also confirmed by the measurements of the spectral distribution of the solar radiation just outside the atmosphere plotted in Figure 11–40, which shows only small deviations from the idealized blackbody behavior. The spectral distribution of solar radiation on the ground plotted in Figure 11–40 shows that the solar radiation undergoes considerable attenuation as it passes through the atmosphere as a result of absorption and scattering. About 99 percent of the atmosphere is contained within a distance of 30 km from the earth’s surface. The several dips on the spectral distribution of radiation on the earth’s surface are due to absorption by the gases O2, O3 (ozone), H2O, and CO2. Absorption by oxygen occurs in a narrow band about 0.76 m. The ozone absorbs ultraviolet radiation at wavelengths below 0.3 m almost completely, and radiation in the range 0.3–0.4 m considerably. Thus, the ozone layer in the upper regions of the atmosphere protects biological systems on earth from harmful ultraviolet radiation. In turn, we must protect the ozone layer from the destructive chemicals commonly used as refrigerants, cleaning agents, and propellants in aerosol cans. The use of these chemicals is now banned in many countries. The ozone gas also absorbs some radiation in the visible range. Absorption in the infrared region is dominated by water vapor and carbon dioxide. The dust particles and other pollutants in the atmosphere also absorb radiation at various wavelengths. As a result of these absorptions, the solar energy reaching the earth’s surface is weakened considerably, to about 950 W/m2 on a clear day and much less on cloudy or smoggy days. Also, practically all of the solar radiation reaching the earth’s surface falls in the wavelength band from 0.3 to 2.5 m. Another mechanism that attenuates solar radiation as it passes through the atmosphere is scattering or reflection by air molecules and the many other kinds of particles such as dust, smog, and water droplets suspended in the atmosphere. Scattering is mainly governed by the size of the particle relative to the wavelength of radiation. The oxygen and nitrogen molecules primarily scatter radiation at very short wavelengths, comparable to the size of the molecules themselves. Therefore, radiation at wavelengths corresponding to violet and blue colors is scattered the most. This molecular scattering in all directions is what gives the sky its bluish color. The same phenomenon is responsible for red sunrises and sunsets. Early in the morning and late in the afternoon, the sun’s rays pass through a greater thickness of the atmosphere than they do at midday, when the sun is at the top. Therefore, the violet and blue colors of the light encounter a greater number of molecules by the time they reach the earth’s surface, and thus a greater fraction of them are scattered (Fig. 11–41). Consequently, the light that reaches the earth’s surface consists primarily of colors corresponding to longer wavelengths such as red, orange, and yellow. The clouds appear in reddish-orange color during sunrise and sunset because the light they reflect is reddish-orange at those times. For the same reason, a red traffic light is visible from a longer distance than is a green light under the same circumstances. The solar energy incident on a surface on earth is considered to consist of direct and diffuse parts. The part of solar radiation that reaches the earth’s surface without being scattered or absorbed by the atmosphere is called direct solar radiation GD. The scattered radiation is assumed to reach the earth’s surface uniformly from all directions and is called diffuse solar radiation Gd. 2000 Extraterrestrial 1500 O3 O2 1000 Earth’s surface H2O 500 O3 H2O H2O 0 0 0.5 H2O CO2 1.0 1.5 2.0 Wavelength, µm 2.5 3.0 FIGURE 11–40 Spectral distribution of solar radiation just outside the atmosphere, at the surface of the earth on a typical day, and comparison with blackbody radiation at 5780 K. Mostly red White Sun Red Orange Yellow Blue Violet Air molecules Atmosphere Earth FIGURE 11–41 Air molecules scatter blue light much more than they do red light. At sunset, the light travels through a thicker layer of atmosphere, which removes much of the blue from the natural light, allowing the red to dominate. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 588 588 HEAT TRANSFER Then the total solar energy incident on the unit area of a horizontal surface on the ground is (Fig. 11–42) Diffuse solar radiation D s ire ra olar ct di at io n n^ θ Gd , W/m2 GD , W/m2 FIGURE 11–42 The direct and diffuse radiation incident on a horizontal surface at the earth’s surface. Gsolar GD cos Gd (W/m2) (11-51) where is the angle of incidence of direct solar radiation (the angle that the sun’s rays make with the normal of the surface). The diffuse radiation varies from about 10 percent of the total radiation on a clear day to nearly 100 percent on a totally cloudy day. The gas molecules and the suspended particles in the atmosphere emit radiation as well as absorbing it. The atmospheric emission is primarily due to the CO2 and H2O molecules and is concentrated in the regions from 5 to 8 m and above 13 m. Although this emission is far from resembling the distribution of radiation from a blackbody, it is found convenient in radiation calculations to treat the atmosphere as a blackbody at some lower fictitious temperature that emits an equivalent amount of radiation energy. This fictitious temperature is called the effective sky temperature Tsky. Then the radiation emission from the atmosphere to the earth’s surface is expressed as 4 Gsky Tsky (W/m2 ) (11-52) The value of Tsky depends on the atmospheric conditions. It ranges from about 230 K for cold, clear-sky conditions to about 285 K for warm, cloudy-sky conditions. Note that the effective sky temperature does not deviate much from the room temperature. Thus, in the light of Kirchhoff’s law, we can take the absorptivity of a surface to be equal to its emissivity at room temperature, . Then the sky radiation absorbed by a surface can be expressed as 4 4 Esky, absorbed Gsky Tsky Tsky (W/m2) (11-53) The net rate of radiation heat transfer to a surface exposed to solar and atmospheric radiation is determined from an energy balance (Fig. 11–43): Sun Atmosphere Gsolar Gsky Eemitted αs Gsolar εGsky Eabsorbed FIGURE 11–43 Radiation interactions of a surface exposed to solar and atmospheric radiation. q· net, rad Eabsorbed Eemitted Esolar, absorbed Esky, absorbed Eemitted 4 s Gsolar Tsky Ts4 4 s Gsolar (Tsky Ts4 ) (W/m2) (11-54) where Ts is the temperature of the surface in K and is its emissivity at room temperature. A positive result for q· net, rad indicates a radiation heat gain by the surface and a negative result indicates a heat loss. The absorption and emission of radiation by the elementary gases such as H2, O2, and N2 at moderate temperatures are negligible, and a medium filled with these gases can be treated as a vacuum in radiation analysis. The absorption and emission of gases with larger molecules such as H2O and CO2, however, can be significant and may need to be considered when considerable amounts of such gases are present in a medium. For example, a 1-m-thick layer of water vapor at 1 atm pressure and 100°C emits more than 50 percent of the energy that a blackbody would emit at the same temperature. In solar energy applications, the spectral distribution of incident solar radiation is very different than the spectral distribution of emitted radiation by cen58933_ch11.qxd 9/9/2002 9:39 AM Page 589 589 CHAPTER 11 the surfaces, since the former is concentrated in the short-wavelength region and the latter in the infrared region. Therefore, the radiation properties of surfaces will be quite different for the incident and emitted radiation, and the surfaces cannot be assumed to be gray. Instead, the surfaces are assumed to have two sets of properties: one for solar radiation and another for infrared radiation at room temperature. Table 11–3 lists the emissivity and the solar absorptivity s of the surfaces of some common materials. Surfaces that are intended to collect solar energy, such as the absorber surfaces of solar collectors, are desired to have high s but low values to maximize the absorption of solar radiation and to minimize the emission of radiation. Surfaces that are intended to remain cool under the sun, such as the outer surfaces of fuel tanks and refrigerator trucks, are desired to have just the opposite properties. Surfaces are often given the desired properties by coating them with thin layers of selective materials. A surface can be kept cool, for example, by simply painting it white. We close this section by pointing out that what we call renewable energy is usually nothing more than the manifestation of solar energy in different forms. Such energy sources include wind energy, hydroelectric power, ocean thermal energy, ocean wave energy, and wood. For example, no hydroelectric power plant can generate electricity year after year unless the water evaporates by absorbing solar energy and comes back as a rainfall to replenish the water source (Fig. 11–44). Although solar energy is sufficient to meet the entire energy needs of the world, currently it is not economical to do so because of the low concentration of solar energy on earth and the high capital cost of harnessing it. EXAMPLE 11–5 Selective Absorber and Reflective Surfaces TABLE 11–3 Comparison of the solar absorptivity s of some surfaces with their emissivity at room temperature Surface Aluminum Polished Anodized Foil Copper Polished Tarnished Stainless steel Polished Dull Plated metals Black nickel oxide Black chrome Concrete White marble Red brick Asphalt Black paint White paint Snow Human skin (caucasian) Consider a surface exposed to solar radiation. At a given time, the direct and diffuse components of solar radiation are GD 400 and Gd 300 W/m2, and the direct radiation makes a 20° angle with the normal of the surface. The surface temperature is observed to be 320 K at that time. Assuming an effective sky temperature of 260 K, determine the net rate of radiation heat transfer for these cases (Fig. 11–45): (a) s 0.9 and 0.9 (gray absorber surface) (b) s 0.1 and 0.1 (gray reflector surface) (c) s 0.9 and 0.1 (selective absorber surface) (d) s 0.1 and 0.9 (selective reflector surface) s 0.09 0.14 0.15 0.03 0.84 0.05 0.18 0.65 0.03 0.75 0.37 0.50 0.60 0.21 0.92 0.87 0.60 0.46 0.63 0.90 0.97 0.14 0.28 0.08 0.09 0.88 0.95 0.93 0.90 0.97 0.93 0.97 0.62 0.97 Winds Clouds Rain Reservoir Power lines Evaporation SOLUTION A surface is exposed to solar and sky radiation. The net rate of radiation heat transfer is to be determined for four different combinations of emissivities and solar absorptivities. Analysis The total solar energy incident on the surface is Gsolar GD cos Gd (400 W/m2) cos 20° (300 W/m2) 676 W/m2 Then the net rate of radiation heat transfer for each of the four cases is determined from: 4 q· net, rad s Gsolar (Tsky Ts4) HPP Solar energy FIGURE 11–44 The cycle that water undergoes in a hydroelectric power plant. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 590 590 HEAT TRANSFER ε (a) s 0.9 and 0.9 (gray absorber surface): 0.9 q· net, rad 0.9(676 W/m2) 0.9(5.67 108 W/m2 · K4)[(260 K)4 (320 K)4] 307 W/m2 λ (a) ε (b) s 0.1 and 0.1 (gray reflector surface): q· net, rad 0.1(676 W/m2) 0.1(5.67 108 W/m2 · K4)[(260 K)4 (320 K)4] 34 W/m2 (c) s 0.9 and 0.1 (selective absorber surface): q· net, rad 0.9(676 W/m2) 0.1(5.67 108 W/m2 · K4)[(260 K)4 (320 K)4] 575 W/m2 0.1 λ (b) ε q· net, rad 0.1(676 W/m2) 0.9(5.67 108 W/m2 · K4)[(260 K)4 (320 K)4] 234 W/m2 0.9 0.1 (c) ε 3 µm λ 3 µm λ 0.9 0.1 (d) (d) s 0.1 and 0.9 (selective reflector surface): FIGURE 11–45 Graphical representation of the spectral emissivities of the four surfaces considered in Example 11–5. Discussion Note that the surface of an ordinary gray material of high absorptivity gains heat at a rate of 307 W/m2. The amount of heat gain increases to 575 W/m2 when the surface is coated with a selective material that has the same absorptivity for solar radiation but a low emissivity for infrared radiation. Also note that the surface of an ordinary gray material of high reflectivity still gains heat at a rate of 34 W/m2. When the surface is coated with a selective material that has the same reflectivity for solar radiation but a high emissivity for infrared radiation, the surface loses 234 W/m2 instead. Therefore, the temperature of the surface will decrease when a selective reflector surface is used. TOPIC OF SPECIAL INTEREST Solar Heat Gain Through Windows The sun is the primary heat source of the earth, and the solar irradiance on a surface normal to the sun’s rays beyond the earth’s atmosphere at the mean earth–sun distance of 149.5 million km is called the total solar irradiance or solar constant. The accepted value of the solar constant is 1373 W/m2 (435.4 Btu/h · ft2), but its value changes by 3.5 percent from a maximum of 1418 W/m2 on January 3 when the earth is closest to the sun, to a minimum of 1325 W/m2 on July 4 when the earth is farthest away from the sun. The spectral distribution of solar radiation beyond the earth’s atmosphere resembles the energy emitted by a blackbody at 5780°C, with about 9 percent of the energy contained in the ultraviolet region (at wavelengths between 0.29 to 0.4 m), 39 percent in the visible region (0.4 to 0.7 m), and the remaining 52 percent in the near-infrared region (0.7 to 3.5 m). The peak radiation occurs at a wavelength of about 0.48 m, which corresponds to the green color portion of the visible spectrum. Obviously a This section can be skipped without a loss in continuity. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 591 glazing material that transmits the visible part of the spectrum while absorbing the infrared portion is ideally suited for an application that calls for maximum daylight and minimum solar heat gain. Surprisingly, the ordinary window glass approximates this behavior remarkably well (Fig. 11–46). Part of the solar radiation entering the earth’s atmosphere is scattered and absorbed by air and water vapor molecules, dust particles, and water droplets in the clouds, and thus the solar radiation incident on earth’s surface is less than the solar constant. The extent of the attenuation of solar radiation depends on the length of the path of the rays through the atmosphere as well as the composition of the atmosphere (the clouds, dust, humidity, and smog) along the path. Most ultraviolet radiation is absorbed by the ozone in the upper atmosphere, and the scattering of short wavelength radiation in the blue range by the air molecules is responsible for the blue color of the clear skies. At a solar altitude of 41.8°, the total energy of direct solar radiation incident at sea level on a clear day consists of about 3 percent ultraviolet, 38 percent visible, and 59 percent infrared radiation. The part of solar radiation that reaches the earth’s surface without being scattered or absorbed is called direct radiation. Solar radiation that is scattered or reemitted by the constituents of the atmosphere is called diffuse radiation. Direct radiation comes directly from the sun following a straight path, whereas diffuse radiation comes from all directions in the sky. The entire radiation reaching the ground on an overcast day is diffuse radiation. The radiation reaching a surface, in general, consists of three components: direct radiation, diffuse radiation, and radiation reflected onto the surface from surrounding surfaces (Fig. 11–47). Common surfaces such as grass, trees, rocks, and concrete reflect about 20 percent of the radiation while absorbing the rest. Snow-covered surfaces, however, reflect 70 percent of the incident radiation. Radiation incident on a surface that does not have a direct view of the sun consists of diffuse and reflected radiation. Therefore, at solar noon, solar radiations incident on the east, west, and north surfaces of a south-facing house are identical since they all consist of diffuse and reflected components. The difference between the radiations incident on the south and north walls in this case gives the magnitude of direct radiation incident on the south wall. When solar radiation strikes a glass surface, part of it (about 8 percent for uncoated clear glass) is reflected back to outdoors, part of it (5 to 50 percent, depending on composition and thickness) is absorbed within the glass, and the remainder is transmitted indoors, as shown in Figure 11–48. The conservation of energy principle requires that the sum of the transmitted, reflected, and absorbed solar radiations be equal to the incident solar radiation. That is, s s s 1 where s is the transmissivity, s is the reflectivity, and s is the absorptivity of the glass for solar energy, which are the fractions of incident solar radiation transmitted, reflected, and absorbed, respectively. The standard 3-mm- (18 -in.) thick single-pane double-strength clear window glass transmits 86 percent, reflects 8 percent, and absorbs 6 percent of the solar energy incident on it. The radiation properties of materials are usually given for normal incidence, but can also be used for radiation incident at other Spectral transmittance 591 CHAPTER 11 1.00 1 0.80 2 0.60 3 0.40 0.20 0 0.2 2 0.4 0.6 1 Wave length, µm 3 4 5 3 mm regular sheet 2. 6 mm gray heat-absorbing plate/float 3. 6 mm green heat-absorbing plate/float FIGURE 11–46 The variation of the transmittance of typical architectural glass with wavelength (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Fig. 11). Sun Direct radiation Window Diffuse radiation Reflected radiation FIGURE 11–47 Direct, diffuse, and reflected components of solar radiation incident on a window. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 592 592 HEAT TRANSFER Sun Incident solar radiation 100% 6-mm thick clear glass Transmitted 80% Reflected 8% Absorbed 12% Outward transfer of absorbed radiation 8% Inward transfer of absorbed radiation 4% FIGURE 11–48 Distribution of solar radiation incident on a clear glass. angles since the transmissivity, reflectivity, and absorptivity of the glazing materials remain essentially constant for incidence angles up to about 60° from the normal. The hourly variation of solar radiation incident on the walls and windows of a house is given in Table 11–4. Solar radiation that is transmitted indoors is partially absorbed and partially reflected each time it strikes a surface, but all of it is eventually absorbed as sensible heat by the furniture, walls, people, and so forth. Therefore, the solar energy transmitted inside a building represents a heat gain for the building. Also, the solar radiation absorbed by the glass is subsequently transferred to the indoors and outdoors by convection and radiation. The sum of the transmitted solar radiation and the portion of the absorbed radiation that flows indoors constitutes the solar heat gain of the building. The fraction of incident solar radiation that enters through the glazing is called the solar heat gain coefficient (SHGC) and is expressed as Solar heat gain through the window Solar radiation incident on the window q· solar, gain · s fis q solar, incident SHGC (11–55) where s is the solar absorptivity of the glass and fi is the inward flowing fraction of the solar radiation absorbed by the glass. Therefore, the dimensionless quantity SHGC is the sum of the fractions of the directly transmitted (s) and the absorbed and reemitted ( fis) portions of solar radiation incident on the window. The value of SHGC ranges from 0 to 1, with 1 corresponding to an opening in the wall (or the ceiling) with no glazing. When the SHGC of a window is known, the total solar heat gain through that window is determined from · Q solar, gain SHGC Aglazing q· solar, incident (W) (11–56) where Aglazing is the glazing area of the window and q· solar, incident is the solar heat flux incident on the outer surface of the window, in W/m2. Another way of characterizing the solar transmission characteristics of different kinds of glazing and shading devices is to compare them to a wellknown glazing material that can serve as a base case. This is done by taking the standard 3-mm-(18 -in.)-thick double-strength clear window glass sheet whose SHGC is 0.87 as the reference glazing and defining a shading coefficient SC as Solar heat gain of product Solar heat gain of reference glazing SHGC SHGC 1.15 SHGC SHGCref 0.87 SC (11–57) Therefore, the shading coefficient of a single-pane clear glass window is SC 1.0. The shading coefficients of other commonly used fenestration products are given in Table 11–5 for summer design conditions. The values for winter design conditions may be slightly lower because of the higher heat transfer coefficients on the outer surface due to high winds and thus cen58933_ch11.qxd 9/9/2002 9:39 AM Page 593 TABLE 11–4 Hourly variation of solar radiation incident on various surfaces and the daily totals throughout the year at 40° latitude (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 15) Solar Radiation Incident on the Surface, W/m2 Solar Time Direction of Surface 5 Jan. N NE E SE S SW W NW Horizontal Direct 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Apr. N NE E SE S SW W NW Horizontal Direct 0 0 0 0 0 0 0 0 0 0 July N NE E SE S SW W NW Horizontal Direct Oct. N NE E SE S SW W NW Horizontal Direct Date 8 9 10 11 12 noon 13 14 15 16 0 0 0 0 0 0 0 0 0 0 20 63 402 483 271 20 20 20 51 446 43 47 557 811 579 48 43 43 198 753 66 66 448 875 771 185 59 59 348 865 68 68 222 803 884 428 68 68 448 912 71 71 76 647 922 647 76 71 482 926 68 68 68 428 884 803 222 68 448 912 66 59 59 185 771 875 448 66 348 865 43 43 43 48 579 811 557 47 198 753 20 20 20 20 271 483 402 63 51 446 0 0 0 0 0 0 0 0 0 0 41 262 321 189 18 17 17 17 39 282 57 508 728 518 59 52 52 52 222 651 79 462 810 682 149 77 77 77 447 794 97 291 732 736 333 97 97 97 640 864 110 134 552 699 437 116 110 110 786 901 120 123 293 582 528 187 120 120 880 919 122 122 131 392 559 392 392 122 911 925 120 120 120 187 528 582 293 123 880 919 110 110 110 116 437 699 552 134 786 901 97 97 97 97 333 736 732 291 640 864 79 77 77 77 149 682 810 462 447 794 3 8 7 2 0 0 0 0 1 7 133 454 498 248 39 39 39 39 115 434 109 590 739 460 76 71 71 71 320 656 103 540 782 580 108 95 95 95 528 762 117 383 701 617 190 114 114 114 702 818 126 203 531 576 292 131 126 126 838 850 134 144 294 460 369 155 134 134 922 866 138 138 149 291 395 291 149 138 949 871 134 134 134 155 369 460 294 144 922 866 126 126 126 131 292 576 531 203 838 850 117 114 114 114 190 617 701 383 702 818 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 74 163 152 44 7 7 7 14 152 40 178 626 680 321 40 40 40 156 643 62 84 652 853 547 66 62 62 351 811 77 80 505 864 711 137 87 87 509 884 87 87 256 770 813 364 87 87 608 917 90 90 97 599 847 599 97 90 640 927 87 87 87 364 813 770 256 87 608 917 77 87 87 137 711 864 505 80 509 884 62 62 62 66 547 853 652 84 351 811 6 7 19 Daily Total 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 446 489 1863 4266 5897 4266 1863 489 2568 — 57 52 52 52 59 518 728 508 222 651 41 17 17 17 18 189 321 262 39 282 0 0 0 0 0 0 0 0 0 0 1117 2347 4006 4323 3536 4323 4006 2347 6938 — 103 95 95 95 108 580 782 540 528 762 109 71 71 71 76 460 739 590 320 656 133 39 39 39 39 248 498 454 115 434 3 0 0 0 0 2 7 8 1 7 1621 3068 4313 3849 2552 3849 4313 3068 3902 — 40 40 40 40 321 680 626 178 156 643 7 7 7 7 44 152 163 74 14 152 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 453 869 2578 4543 5731 4543 2578 869 3917 — 17 18 Multiply by 0.3171 to convert to Btu/h · ft2. Values given are for the 21st of the month for average days with no clouds. The values can be up to 15 percent higher at high elevations under very clear skies and up to 30 percent lower at very humid locations with very dusty industrial atmospheres. Daily totals are obtained using Simpson’s rule for integration with 10-min time intervals. Solar reflectance of the ground is assumed to be 0.2, which is valid for old concrete, crushed rock, and bright green grass. For a specified location, use solar radiation data obtained for that location. The direction of a surface indicates the direction a vertical surface is facing. For example, W represents the solar radiation incident on a west-facing wall per unit area of the wall. Solar time may deviate from the local time. Solar noon at a location is the time when the sun is at the highest location (and thus when the shadows are shortest). Solar radiation data are symmetric about the solar noon: the value on a west wall before the solar noon is equal to the value on an east wall two hours after the solar noon. 593 cen58933_ch11.qxd 9/9/2002 9:39 AM Page 594 594 HEAT TRANSFER TABLE 11–5 Shading coefficient SC and solar transmissivity solar for some common glass types for summer design conditions (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 11). Type of Glazing Nominal Thickness mm in. solar SC (a) Single Glazing Clear 3 6 10 13 Heat absorbing 3 6 10 13 (b) Double Glazing Clear in, 3a clear out 6 Clear in, heat absorbing outc 6 1 8 1 4 3 8 1 2 1 8 1 4 3 8 1 2 0.86 0.78 0.72 0.67 0.64 0.46 0.33 0.24 1 8 1 4 0.71b 0.88 0.61 0.82 1 4 0.36 0.58 1.0 0.95 0.92 0.88 0.85 0.73 0.64 0.58 Multiply by 0.87 to obtain SHGC. a The thickness of each pane of glass. Combined transmittance for assembled unit. c Refers to gray-, bronze-, and green-tinted heat-absorbing float glass. b higher rate of outward flow of solar heat absorbed by the glazing, but the difference is small. Note that the larger the shading coefficient, the smaller the shading effect, and thus the larger the amount of solar heat gain. A glazing material with a large shading coefficient will allow a large fraction of solar radiation to come in. Shading devices are classified as internal shading and external shading, depending on whether the shading device is placed inside or outside. External shading devices are more effective in reducing the solar heat gain since they intercept the sun’s rays before they reach the glazing. The solar heat gain through a window can be reduced by as much as 80 percent by exterior shading. Roof overhangs have long been used for exterior shading of windows. The sun is high in the horizon in summer and low in winter. A properly sized roof overhang or a horizontal projection blocks off the sun’s rays completely in summer while letting in most of them in winter, as shown in Figure 11–49. Such shading structures can reduce the solar heat gain on the south, southeast, and southwest windows in the northern hemisphere considerably. A window can also be shaded from outside by vertical or horizontal architectural projections, insect or shading screens, and sun screens. To be effective, air must be able to move freely around the exterior device to carry away the heat absorbed by the shading and the glazing materials. Some type of internal shading is used in most windows to provide privacy and aesthetic effects as well as some control over solar heat gain. Internal shading devices reduce solar heat gain by reflecting transmitted solar radiation back through the glazing before it can be absorbed and converted into heat in the building. Draperies reduce the annual heating and cooling loads of a building by 5 to 20 percent, depending on the type and the user habits. In summer, they reduce heat gain primarily by reflecting back direct solar radiation (Fig. 11–50). The semiclosed air space formed by the draperies serves as an additional barrier against heat transfer, resulting in a lower U-factor for the window and thus a lower rate of heat transfer in summer and winter. The solar optical properties of draperies can be measured accurately, or they can be obtained directly from the manufacturers. The shading coefficient of draperies depends on the openness factor, which is the ratio of the open area between the fibers that permits the sun’s rays to pass freely, to the total area of the fabric. Tightly woven fabrics allow little direct radiation to pass through, and thus they have a small openness factor. The reflectance of the surface of the drapery facing the glazing has a major effect on the amount of solar heat gain. Light-colored draperies made of closed or tightly woven fabrics maximize the back reflection and thus minimize the solar gain. Dark-colored draperies made of open or semi-open woven fabrics, on the other hand, minimize the back reflection and thus maximize the solar gain. The shading coefficients of drapes also depend on the way they are hung. Usually, the width of drapery used is twice the width of the draped area to allow folding of the drapes and to give them their characteristic “full” or “wavy” appearance. A flat drape behaves like an ordinary window shade. A flat drape has a higher reflectance and thus a lower shading coefficient than a full drape. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 595 595 CHAPTER 11 External shading devices such as overhangs and tinted glazings do not require operation, and provide reliable service over a long time without significant degradation during their service life. Their operation does not depend on a person or an automated system, and these passive shading devices are considered fully effective when determining the peak cooling load and the annual energy use. The effectiveness of manually operated shading devices, on the other hand, varies greatly depending on the user habits, and this variation should be considered when evaluating performance. The primary function of an indoor shading device is to provide thermal comfort for the occupants. An unshaded window glass allows most of the incident solar radiation in, and also dissipates part of the solar energy it absorbs by emitting infrared radiation to the room. The emitted radiation and the transmitted direct sunlight may bother the occupants near the window. In winter, the temperature of the glass is lower than the room air temperature, causing excessive heat loss by radiation from the occupants. A shading device allows the control of direct solar and infrared radiation while providing various degrees of privacy and outward vision. The shading device is also at a higher temperature than the glass in winter, and thus reduces radiation loss from occupants. Glare from draperies can be minimized by using off-white colors. Indoor shading devices, especially draperies made of a closed-weave fabric, are effective in reducing sounds that originate in the room, but they are not as effective against the sounds coming from outside. The type of climate in an area usually dictates the type of windows to be used in buildings. In cold climates where the heating load is much larger than the cooling load, the windows should have the highest transmissivity for the entire solar spectrum, and a high reflectivity (or low emissivity) for the far infrared radiation emitted by the walls and furnishings of the room. Low-e windows are well suited for such heating-dominated buildings. Properly designed and operated windows allow more heat into the building over a heating season than it loses, making them energy contributors rather then energy losers. In warm climates where the cooling load is much larger than the heating load, the windows should allow the visible solar radiation (light) in, but should block off the infrared solar radiation. Such windows can reduce the solar heat gain by 60 percent with no appreciable loss in daylighting. This behavior is approximated by window glazings that are coated with a heat-absorbing film outside and a low-e film inside (Fig. 11–51). Properly selected windows can reduce the cooling load by 15 to 30 percent compared to windows with clear glass. Note that radiation heat transfer between a room and its windows is proportional to the emissivity of the glass surface facing the room, glass, and can be expressed as · 4 4 Q rad, room-window glass Aglass (Troom Tglass ) (11-58) Therefore, a low-e interior glass will reduce the heat loss by radiation in winter (Tglass Troom) and heat gain by radiation in summer (Tglass Troom). Tinted glass and glass coated with reflective films reduce solar heat gain in summer and heat loss in winter. The conductive heat gains or losses can be minimized by using multiple-pane windows. Double-pane windows are Summer Sun Winter Overhang Sun Window FIGURE 11–49 A properly sized overhang blocks off the sun’s rays completely in summer while letting them in in winter. Sun Drape Reflected by glass Reflected by drapes Window FIGURE 11–50 Draperies reduce heat gain in summer by reflecting back solar radiation, and reduce heat loss in winter by forming an air space before the window. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 596 596 HEAT TRANSFER Glass (colder than room) Sun · Qrad ~ ε No reflective film Low-e film (high infrared reflectivity) (a) Cold climates Glass (warmer than room) Sun · Qrad ~ ε Infrared Reflective film Visible Low-e film (b) Warm climates FIGURE 11–51 Radiation heat transfer between a room and its windows is proportional to the emissivity of the glass surface, and low-e coatings on the inner surface of the windows reduce heat loss in winter and heat gain in summer. usually called for in climates where the winter design temperature is less than 7°C (45°F). Double-pane windows with tinted or reflective films are commonly used in buildings with large window areas. Clear glass is preferred for showrooms since it affords maximum visibility from outside, but bronze-, gray-, and green-colored glass are preferred in office buildings since they provide considerable privacy while reducing glare. EXAMPLE 11–6 Installing Reflective Films on Windows A manufacturing facility located at 40° N latitude has a glazing area of 40 m2 that consists of double-pane windows made of clear glass (SHGC 0.766). To reduce the solar heat gain in summer, a reflective film that will reduce the SHGC to 0.261 is considered. The cooling season consists of June, July, August, and September, and the heating season October through April. The average daily solar heat fluxes incident on the west side at this latitude are 1.86, 2.66, 3.43, 4.00, 4.36, 5.13, 4.31, 3.93, 3.28, 2.80, 1.84, and 1.54 kWh/day · m2 for January through December, respectively. Also, the unit cost of the electricity and natural gas are $0.08/kWh and $0.50/therm, respectively. If the coefficient of performance of the cooling system is 2.5 and efficiency of the furnace is 0.8, determine the net annual cost savings due to installing reflective coating on the windows. Also, determine the simple payback period if the installation cost of reflective film is $20/m2 (Fig. 11–52). SOLUTION The net annual cost savings due to installing reflective film on the west windows of a building and the simple payback period are to be determined. Assumptions 1 The calculations given below are for an average year. 2 The unit costs of electricity and natural gas remain constant. Analysis Using the daily averages for each month and noting the number of days of each month, the total solar heat flux incident on the glazing during summer and winter months are determined to be Qsolar, summer 5.13 30 4.31 31 3.93 31 3.28 30 508 kWh/year Qsolar, winter 2.80 31 1.84 30 1.54 31 1.86 31 2.66 28 3.43 31 4.00 30 548 kWh/year Then the decrease in the annual cooling load and the increase in the annual heating load due to the reflective film become Cooling load decrease Qsolar, summer Aglazing (SHGCwithout film SHGCwith film) (508 kWh/year)(40 m2)(0.766 0.261) 10,262 kWh/year Heating load increase Qsolar, winter Aglazing (SHGCwithout film SHGCwith film) (548 kWh/year)(40 m2)(0.766 0.261) 11,070 kWh/year 377.7 therms/year since 1 therm 29.31 kWh. The corresponding decrease in cooling costs and the increase in heating costs are cen58933_ch11.qxd 9/9/2002 9:39 AM Page 597 597 CHAPTER 11 Glass Decrease in cooling costs (Cooling load decrease)(Unit cost of electricity)/COP (10,262 kWh/year)($0.08/kWh)/2.5 $328/year Sun Air space Increase in heating costs (Heating load increase)(Unit cost of fuel)/Efficiency (377.7 therms/year)($0.50/therm)/0.80 $236/year Then the net annual cost savings due to the reflective film become Cost savings Decrease in cooling costs Increase in heating costs $328 $236 $92/year Reflected The implementation cost of installing films is Reflective film Implementation cost ($20/m )(40 m ) $800 2 2 FIGURE 11–52 Schematic for Example 11–6. This gives a simple payback period of Simple payback period Transmitted Implementation cost $800 8.7 years Annual cost savings $92/year Discussion The reflective film will pay for itself in this case in about nine years. This may be unacceptable to most manufacturers since they are not usually interested in any energy conservation measure that does not pay for itself within three years. But the enhancement in thermal comfort and thus the resulting increase in productivity often makes it worthwhile to install reflective film. SUMMARY Radiation propagates in the form of electromagnetic waves. The frequency and wavelength of electromagnetic waves in a medium are related by c/, where c is the speed of propagation in that medium. All matter whose temperature is above absolute zero continuously emits thermal radiation as a result of vibrational and rotational motions of molecules, atoms, and electrons of a substance. Temperature is a measure of the strength of these activities at the microscopic level. A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation, regardless of wavelength and direction. The radiation energy emitted by a blackbody per unit time and per unit surface area is called the blackbody emissive power Eb and is expressed by the Stefan–Boltzmann law as Eb(T ) T 4 where 5.670 108 W/m2 K4 is the Stefan–Boltzmann constant and T is the absolute temperature of the surface in K. At any specified temperature, the spectral blackbody emissive power Eb increases with wavelength, reaches a peak, and then decreases with increasing wavelength. The wavelength at which the peak occurs for a specified temperature is given by Wien’s displacement law as (T)max power 2897.8 m K The blackbody radiation function f represents the fraction of radiation emitted by a blackbody at temperature T in the wavelength band from 0 to . The fraction of radiation energy emitted by a blackbody at temperature T over a finite wavelength band from 1 to 2 is determined from fλ 1λ 2(T ) fλ 2(T ) f λ 1(T ) where fλ 1(T) and fλ 2(T) are the blackbody radiation functions corresponding to 1T and 2T, respectively. The magnitude of a viewing angle in space is described by solid angle expressed as d dAn/r2. The radiation intensity for emitted radiation Ie( , ) is defined as the rate at which radiation energy is emitted in the ( , ) direction per unit area normal to this direction and per unit solid angle about this direction. The radiation flux for emitted radiation is the emissive power E, and is expressed as E dE hemisphere 2 /2 0 0 Ie( , ) cos sin d d For a diffusely emitting surface, intensity is independent of direction and thus cen58933_ch11.qxd 9/9/2002 9:39 AM Page 598 598 HEAT TRANSFER E Ie For a blackbody, we have Eb Ib Ib(T) and E b(T ) T 4 The radiation flux incident on a surface from all directions is irradiation G, and for diffusely incident radiation of intensity Ii it is expressed as G Ii The rate at which radiation energy leaves a unit area of a surface in all directions is radiosity J, and for a surface that is both a diffuse emitter and a diffuse reflector it is expressed as where Ier is the sum of the emitted and reflected intensities. The spectral emitted quantities are related to total quantities as I 0 λ, e d λ and E E dλ 0 λ The last relation reduces for a diffusely emitting surface and for a blackbody to E I, e Eb(, T) Ib(, T) and , (, , ) () J Ier Ie The total hemispherical emissivity of a surface is the average emissivity over all directions and wavelengths. Radiation energy incident on a surface per unit surface area per unit time is called irradiation G. When radiation strikes a surface, part of it is absorbed, part of it is reflected, and the remaining part, if any, is transmitted. The fraction of incident radiation (intensity Ii or irradiation G) absorbed by the surface is called the absorptivity, the fraction reflected by the surface is called the reflectivity, and the fraction transmitted is called the transmissivity. Various absorptivities, reflectivities, and transmissivities for a medium are expressed as The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature. Different emissivities are defined as Iλ, abs (λ, , ) Iλ, i (λ, , ) Gλ, abs(λ) Gλ(λ) , () Gabs , G Ie ( , , T ) Ib (T ) Spectral hemispherical emissivity: E λ(λ, T ) (, T) E bλ(λ, T ) 0 Eb(T ) λ bλ( λ, Gref , and G Gλ, tr(λ) Gλ(λ) Gtr G For opaque surfaces, 0, and thus 1 , (T) , (T), Total hemispherical emissivity: , and () Surfaces are usually assumed to reflect in a perfectly specular or diffuse manner for simplicity. In specular (or mirrorlike) reflection, the angle of reflection equals the angle of incidence of the radiation beam. In diffuse reflection, radiation is reflected equally in all directions. Reflection from smooth and polished surfaces approximates specular reflection, whereas reflection from rough surfaces approximates diffuse reflection. Kirchhoff’s law of radiation is expressed as Ibλ( λ, T ) ( λ, T )E E(T ) (T) Gλ(λ) Iλ, i ( λ, , ) 1 Iλ, e (λ, , , T ) Total directional emissivity: ( , , T) Gλ, ref (λ) Iλ, ref ( λ, , ) The consideration of wavelength and direction dependence of properties makes radiation calculations very complicated. Therefore, the gray and diffuse approximations are commonly utilized in radiation calculations. A surface is said to be diffuse if its properties are independent of direction and gray if its properties are independent of wavelength. The sum of the absorbed, reflected, and transmitted fractions of radiation energy must be equal to unity, Spectral directional emissivity: , (, , , T) and , (, , ) T )d λ T 4 Emissivity can also be expressed as a step function by dividing the spectrum into a sufficient number of wavelength bands of constant emissivity as (T ) 1 f0λ1(T ) 2 fλ1λ 2(T ) 3 f λ 2 (T ) (T) (T), and (T) (T) That is, the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature. Gas molecules and the suspended particles in the atmosphere emit radiation as well as absorbing it. The atmosphere can be treated as a blackbody at some lower fictitious temperature, called the effective sky temperature Tsky that emits an equivalent amount of radiation energy. Then the radiation emitted by the atmosphere is expressed as cen58933_ch11.qxd 9/9/2002 9:39 AM Page 599 599 CHAPTER 11 4 Gsky T sky The net rate of radiation heat transfer to a surface exposed to solar and atmospheric radiation is determined from an energy balance expressed as 4 q·net, rad sGsolar (T sky T s4 ) where Ts is the surface temperature in K, and is the surface emissivity at room temperature. REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers, Handbook of Fundamentals, Atlanta, ASHRAE, 1993. M. F. Modest, Radiative Heat Transfer. New York: McGraw-Hill, 1993. 9. M. Planck. The Theory of Heat Radiation. New York: Dover, 1959. A. G. H. Dietz. “Diathermanous Materials and Properties of Surfaces.” In Space Heating with Solar Energy, ed. R. W. Hamilton. Cambridge, MA: MIT Press, 1954. W. Sieber. Zeitschrift für Technische Physics 22 (1941), pp. 130–135. J. A. Duffy and W. A. Beckman. Solar Energy Thermal Process. New York: John Wiley & Sons, 1974. R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. 3rd ed. Washington, D.C.: Hemisphere, 1992. J. P. Holman. Heat Transfer. 9th ed. New York: McGrawHill, 2002. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West, 1995. H. C. Hottel. “Radiant Heat Transmission,” In Heat Transmission, 3rd ed., ed. W. H. McAdams. New York: McGraw-Hill, 1954. Y. S. Touloukain and D. P. DeWitt. “Nonmetallic Solids.” In Thermal Radiative Properties. Vol. 8. New York: IFI/Plenum, 1970. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. Y. S. Touloukian and D. P. DeWitt. “Metallic Elements and Alloys.” In Thermal Radiative Properties, Vol. 7. New York: IFI/Plenum, 1970. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. PROBLEMS 11–1C What is an electromagnetic wave? How does it differ from a sound wave? 11–6C What is the cause of color? Why do some objects appear blue to the eye while others appear red? Is the color of a surface at room temperature related to the radiation it emits? 11–2C By what properties is an electromagnetic wave characterized? How are these properties related to each other? 11–7C Why is radiation usually treated as a surface phenomenon? 11–3C What is visible light? How does it differ from the other forms of electromagnetic radiation? 11–8C Why do skiers get sunburned so easily? Electromagnetic and Thermal Radiation 11–4C How do ultraviolet and infrared radiation differ? Do you think your body emits any radiation in the ultraviolet range? Explain. 11–5C What is thermal radiation? How does it differ from the other forms of electromagnetic radiation? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 11–9C How does microwave cooking differ from conventional cooking? 11–10 Electricity is generated and transmitted in power lines at a frequency of 60 Hz (1 Hz 1 cycle per second). Determine the wavelength of the electromagnetic waves generated by the passage of electricity in power lines. 11–11 A microwave oven is designed to operate at a frequency of 2.8 109 Hz. Determine the wavelength of these microwaves and the energy of each microwave. 11–12 A radio station is broadcasting radio waves at a wavelength of 200 m. Determine the frequency of these waves. Answer: 1.5 106 Hz 11–13 A cordless telephone is designed to operate at a frequency of 8.5 108 Hz. Determine the wavelength of these telephone waves. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 600 600 HEAT TRANSFER Blackbody Radiation 11–14C exist? What is a blackbody? Does a blackbody actually 11–15C Define the total and spectral blackbody emissive powers. How are they related to each other? How do they differ? 11–16C Why did we define the blackbody radiation function? What does it represent? For what is it used? 11–17C Consider two identical bodies, one at 1000 K and the other at 1500 K. Which body emits more radiation in the shorter-wavelength region? Which body emits more radiation at a wavelength of 20 m? 11–18 Consider a 20-cm 20-cm 20-cm cubical body at 1000 K suspended in the air. Assuming the body closely approximates a blackbody, determine (a) the rate at which the cube emits radiation energy, in W, and (b) the spectral blackbody emissive power at a wavelength of 4 m. 11–19E The sun can be treated as a blackbody at an effective surface temperature of 10,400 R. Determine the rate at which infrared radiation energy ( 0.76–100 m) is emitted by the sun, in Btu/h · ft2. 11–20 The sun can be treated as a blackbody at 5780 K. Using EES (or other) software, calculate and plot the spectral blackbody emissive power Eb of the sun versus wavelength in the range of 0.01 m to 1000 m. Discuss the results. 11–21 The temperature of the filament of an incandescent lightbulb is 3200 K. Treating the filament as a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks. 11–22 Reconsider Problem 11–21. Using EES (or other) software, investigate the effect of temperature on the fraction of radiation emitted in the visible range. Let the surface temperature vary from 1000 K to 4000 K, and plot fraction of radiation emitted in the visible range versus the surface temperature. 11–23 An incandescent lightbulb is desired to emit at least 15 percent of its energy at wavelengths shorter than 1 m. Determine the minimum temperature to which the filament of the lightbulb must be heated. 11–24 It is desired that the radiation energy emitted by a light source reach a maximum in the blue range ( 0.47 m). Determine the temperature of this light source and the fraction of radiation it emits in the visible range ( 0.40–0.76 m). 11–25 A 3-mm-thick glass window transmits 90 percent of the radiation between 0.3 and 3.0 m and is essentially opaque for radiation at other wavelengths. Determine the rate of radiation transmitted through a 2-m 2-m glass window from blackbody sources at (a) 5800 K and (b) 1000 K. Answers: (a) 218,400 kW, (b) 55.8 kW Radiation Intensity 11–26C What does a solid angle represent, and how does it differ from a plane angle? What is the value of a solid angle associated with a sphere? 11–27C How is the intensity of emitted radiation defined? For a diffusely emitting surface, how is the emissive power related to the intensity of emitted radiation? 11–28C For a surface, how is irradiation defined? For diffusely incident radiation, how is irradiation on a surface related to the intensity of incident radiation? 11–29C For a surface, how is radiosity defined? For diffusely emitting and reflecting surfaces, how is radiosity related to the intensities of emitted and reflected radiation? 11–30C When the variation of spectral radiation quantity with wavelength is known, how is the corresponding total quantity determined? 11–31 A small surface of area A1 4 cm2 emits radiation as a blackbody at T1 800 K. Part of the radiation emitted by A1 strikes another small surface of area A2 4 cm2 oriented as shown in the figure. Determine the solid angle subtended by A2 when viewed from A1, and the rate at which radiation emitted by A1 that strikes A2 directly. What would your answer be if A2 were directly above A1 at a distance of 80 cm? A2 4 cm2 θ 2 60° θ 1 45° r 80 cm A1 4 cm2 T1 800 K FIGURE P11–31 11–32 A small circular surface of area A1 2 cm2 located at the center of a 2-m-diameter sphere emits radiation as a blackbody at T1 1000 K. Determine the rate at which radiation energy is streaming through a D2 1-cm-diameter hole located (a) on top of the sphere directly above A1 and (b) on the side of sphere such that the line that connects the centers of A1 and A2 makes 45˚ with surface A1. 11–33 Repeat Problem 11–32 for a 4-m-diameter sphere. 11–34 A small surface of area A 1 cm2 emits radiation as a blackbody at 1500 K. Determine the rate at which radiation energy is emitted through a band defined by 0 2 and 45 cen58933_ch11.qxd 9/9/2002 9:39 AM Page 601 601 CHAPTER 11 60˚ where is the angle a radiation beam makes with the normal of the surface and is the azimuth angle. determine the absorptivity and reflectivity of the filament at both temperatures. 11–35 A small surface of area A 1 cm2 is subjected to incident radiation of constant intensity Ii 2.2 104 W/m2 sr over the entire hemisphere. Determine the rate at which radiation energy is incident on the surface through (a) 0 45˚ and (b) 45 90˚, where is the angle a radiation beam makes with the normal of the surface. 11–45 The variations of the spectral emissivity of two surfaces are as given in Figure P11–45. Determine the average emissivity of each surface at T 3000 K. Also, determine the average absorptivity and reflectivity of each surface for radiation coming from a source at 3000 K. Which surface is more suitable to serve as a solar absorber? Radiation Properties ελ 11–36C Define the properties emissivity and absorptivity. When are these two properties equal to each other? 11–37C Define the properties reflectivity and transmissivity and discuss the different forms of reflection. 11–38C What is a graybody? How does it differ from a blackbody? What is a diffuse gray surface? 11–39C What is the greenhouse effect? Why is it a matter of great concern among atmospheric scientists? 11–40C We can see the inside of a microwave oven during operation through its glass door, which indicates that visible radiation is escaping the oven. Do you think that the harmful microwave radiation might also be escaping? 11–41 The spectral emissivity function of an opaque surface at 1000 K is approximated as 1 0.4, 2 0.7, 3 0.3, 0 2 m 2 m 6 m 6 m Determine the average emissivity of the surface and the rate of radiation emission from the surface, in W/m2. 1.0 0.9 0.8 (1) 0.5 0.2 0 0 0.1 (2) λ, µm 3 FIGURE P11–45 11–46 The emissivity of a surface coated with aluminum oxide can be approximated to be 0.2 for radiation at wavelengths less than 5 m and 0.9 for radiation at wavelengths greater than 5 m. Determine the average emissivity of this surface at (a) 5800 K and (b) 300 K. What can you say about the absorptivity of this surface for radiation coming from sources at 5800 K and 300 K? Answers: (a) 0.203, (b) 0.89 11–47 The variation of the spectral absorptivity of a surface is as given in Figure P11–47. Determine the average absorptivity and reflectivity of the surface for radiation that originates from a source at T 2500 K. Also, determine the average emissivity of this surface at 3000 K. Answers: 0.575, 32.6 kW/m2 11–42 The reflectivity of aluminum coated with lead sulfate is 0.35 for radiation at wavelengths less than 3 m and 0.95 for radiation greater than 3 m. Determine the average reflectivity of this surface for solar radiation (T 5800 K) and radiation coming from surfaces at room temperature (T 300 K). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors? 11–43 A furnace that has a 25-cm 25-cm glass window can be considered to be a blackbody at 1200 K. If the transmissivity of the glass is 0.7 for radiation at wavelengths less than 3 m and zero for radiation at wavelengths greater than 3 m, determine the fraction and the rate of radiation coming from the furnace and transmitted through the window. 11–44 The emissivity of a tungsten filament can be approximated to be 0.5 for radiation at wavelengths less than 1 m and 0.15 for radiation at greater than 1 m. Determine the average emissivity of the filament at (a) 2000 K and (b) 3000 K. Also, αλ 0.7 0.2 2 λ, µm FIGURE P11–47 11–48E A 5-in.-diameter spherical ball is known to emit radiation at a rate of 120 Btu/h when its surface temperature is 950 R. Determine the average emissivity of the ball at this temperature. 11–49 The variation of the spectral transmissivity of a 0.6-cm-thick glass window is as given in Figure P11–49. Determine the average transmissivity of this window for solar radiation (T 5800 K) and radiation coming from surfaces at room temperature (T 300 K). Also, determine the amount of cen58933_ch11.qxd 9/9/2002 9:39 AM Page 602 602 HEAT TRANSFER solar radiation transmitted through the window for incident solar radiation of 650 W/m2. Answers: 0.848, 0.00015, 551.1 W/m2 τλ 0 Answer: 413.3 R 11–60 The air temperature on a clear night is observed to remain at about 4°C. Yet water is reported to have frozen that night due to radiation effect. Taking the convection heat transfer coefficient to be 18 W/m2 · °C, determine the value of the maximum effective sky temperature that night. 0.92 0 0.3 into space at 0 R. If there is no net heat transfer into the spaceship, determine the equilibrium temperature of the surface. 3 λ, µm FIGURE P11–49 Atmospheric and Solar Radiation 11–50C What is the solar constant? How is it used to determine the effective surface temperature of the sun? How would the value of the solar constant change if the distance between the earth and the sun doubled? 11–61 The absorber surface of a solar collector is made of aluminum coated with black chrome (s 0.87 and 0.09). Solar radiation is incident on the surface at a rate of 600 W/m2. The air and the effective sky temperatures are 25°C and 15°C, respectively, and the convection heat transfer coefficient is 10 W/m2 · °C. For an absorber surface temperature of 70°C, determine the net rate of solar energy delivered by the absorber plate to the water circulating behind it. 11–51C What changes would you notice if the sun emitted radiation at an effective temperature of 2000 K instead of 5762 K? Sun 11–52C Explain why the sky is blue and the sunset is yelloworange. Gsolar = 600 W/m2 11–53C When the earth is closest to the sun, we have winter in the northern hemisphere. Explain why. Also explain why we have summer in the northern hemisphere when the earth is farthest away from the sun. 11–54C What is the effective sky temperature? 11–55C You have probably noticed warning signs on the highways stating that bridges may be icy even when the roads are not. Explain how this can happen. T = 25°C Tsky = 15°C Ts = 70°C 11–56C Unless you live in a warm southern state, you have probably had to scrape ice from the windshield and windows of your car many mornings. You may have noticed, with frustration, that the thickest layer of ice always forms on the windshield instead of the side windows. Explain why this is the case. Absorber plate Water tubes Insulation FIGURE P11–61 11–57C Explain why surfaces usually have quite different absorptivities for solar radiation and for radiation originating from the surrounding bodies. 11–62 A surface has an absorptivity of s 0.85 for solar radiation and an emissivity of 0.5 at room temperature. The surface temperature is observed to be 350 K when the direct and the diffuse components of solar radiation are GD 350 and Gd 400 W/m2, respectively, and the direct radiation makes a 30° angle with the normal of the surface. Taking the effective sky temperature to be 280 K, determine the net rate of radiation heat transfer to the surface at that time. 11–63 Determine the equilibrium temperature of the absorber surface in Problem 11–61 if the back side of the absorber is insulated. 11–58 11–59E Solar radiation is incident on the outer surface of a spaceship at a rate of 400 Btu/h · ft2. The surface has an absorptivity of s 0.10 for solar radiation and an emissivity of 0.8 at room temperature. The outer surface radiates heat Reconsider Problem 11–61. Using EES (or other) software, plot the net rate of solar energy transferred to water as a function of the absorptivity of the absorber plate. Let the absorptivity vary from 0.5 to 1.0, and discuss the results. Special Topic: Solar Heat Gain through Windows 11–64C What fraction of the solar energy is in the visible range (a) outside the earth’s atmosphere and (b) at sea level on earth? Answer the same question for infrared radiation. 11–65C Describe the solar radiation properties of a window that is ideally suited for minimizing the air-conditioning load. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 603 603 CHAPTER 11 11–66C Define the SHGC (solar heat gain coefficient), and explain how it differs from the SC (shading coefficient). What are the values of the SHGC and SC of a single-pane clear-glass window? 11–67C What does the SC (shading coefficient) of a device represent? How do the SCs of clear glass and heat-absorbing glass compare? Venetian blinds Double-pane window Light colored 11–68C What is a shading device? Is an internal or external shading device more effective in reducing the solar heat gain through a window? How does the color of the surface of a shading device facing outside affect the solar heat gain? 11–69C What is the effect of a low-e coating on the inner surface of a window glass on the (a) heat loss in winter and (b) heat gain in summer through the window? 11–70C What is the effect of a reflective coating on the outer surface of a window glass on the (a) heat loss in winter and (b) heat gain in summer through the window? 11–71 A manufacturing facility located at 32° N latitude has a glazing area of 60 m2 facing west that consists of doublepane windows made of clear glass (SHGC 0.766). To reduce the solar heat gain in summer, a reflective film that will reduce the SHGC to 0.35 is considered. The cooling season consists of June, July, August, and September, and the heating season, October through April. The average daily solar heat fluxes incident on the west side at this latitude are 2.35, 3.03, 3.62, 4.00, 4.20, 4.24, 4.16, 3.93, 3.48, 2.94, 2.33, and 2.07 kWh/day · m2 for January through December, respectively. Also, the unit costs of electricity and natural gas are $0.09/kWh and $0.45/therm, respectively. If the coefficient of performance of the cooling system is 3.2 and the efficiency of the furnace is 0.90, determine the net annual cost savings due to installing reflective coating on the windows. Also, determine the simple payback period if the installation cost of reAnswers: $53, 23 years flective film is $20/m2. 11–72 A house located in Boulder, Colorado (40° N latitude), has ordinary double-pane windows with 6-mm-thick glasses and the total window areas are 8, 6, 6, and 4 m2 on the south, west, east, and north walls, respectively. Determine the total solar heat gain of the house at 9:00, 12:00, and 15:00 solar time in July. Also, determine the total amount of solar heat gain per day for an average day in January. 11–73 Repeat Problem 11–72 for double-pane windows that are gray-tinted. 11–74 Consider a building in New York (40° N latitude) that has 200 m2 of window area on its south wall. The windows are double-pane heat-absorbing type, and are equipped with lightcolored venetian blinds with a shading coefficient of SC 0.30. Determine the total solar heat gain of the building through the south windows at solar noon in April. What would your answer be if there were no blinds at the windows? Heat-absorbing glass FIGURE P11–74 11–75 A typical winter day in Reno, Nevada (39° N latitude), is cold but sunny, and thus the solar heat gain through the windows can be more than the heat loss through them during daytime. Consider a house with double-door-type windows that are double paned with 3-mm-thick glasses and 6.4 mm of air space and have aluminum frames and spacers. The house is maintained at 22°C at all times. Determine if the house is losing more or less heat than it is gaining from the sun through an east window on a typical day in January for a 24-h period if the Answer: less average outdoor temperature is 10°C. Double-pane window Sun Solar heat gain 10°C 22°C Heat loss FIGURE P11–75 11–76 Repeat Problem 11–75 for a south window. 11–77E Determine the rate of net heat gain (or loss) through a 9-ft-high, 15-ft-wide, fixed 18 -in. single-glass window with aluminum frames on the west wall at 3 PM solar time during a typical day in January at a location near 40° N latitude when the indoor and outdoor temperatures are 70°F and 45°F, Answer: 16,840 Btu/h gain respectively. 11–78 Consider a building located near 40° N latitude that has equal window areas on all four sides. The building owner is considering coating the south-facing windows with reflective film to reduce the solar heat gain and thus the cooling load. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 604 604 HEAT TRANSFER But someone suggests that the owner will reduce the cooling load even more if she coats the west-facing windows instead. What do you think? 11–83 The surface in Problem 11–82 receives solar radiation at a rate of 820 W/m2. Determine the solar absorptivity of the surface and the rate of absorption of solar radiation. Review Problems 11–84 The spectral transmissivity of a glass cover used in a solar collector is given as 11–79 The spectral emissivity of an opaque surface at 1200 K is approximated as for 2 m 1 0 2 0.85 for 2 6 m for 6 m 3 0 Determine the total emissivity and the emissive flux of the surface. 11–80 The spectral transmissivity of a 3-mm-thick regular glass can be expressed as 1 0 2 0.85 3 0 for 0.35 m for 0.35 2.5 m for 2.5 m Determine the transmissivity of this glass for solar radiation. What is the transmissivity of this glass for light? 11–81 A 1-m-diameter spherical cavity is maintained at a uniform temperature of 600 K. Now a 5-mm-diameter hole is drilled. Determine the maximum rate of radiation energy streaming through the hole. What would your answer be if the diameter of the cavity were 3 m? 11–82 The spectral absorptivity of an opaque surface is as shown on the graph. Determine the absorptivity of the surface for radiation emitted by a source at (a) 1000 K and (b) 3000 K. αλ 0.8 0.1 0 0.3 FIGURE P11–82 1.2 λ, µm 1 0 2 0.9 3 0 for 0.3 m for 0.3 3 m for 3 m Solar radiation is incident at a rate of 950 W/m2, and the absorber plate, which can be considered to be black, is maintained at 340 K by the cooling water. Determine (a) the solar flux incident on the absorber plate, (b) the transmissivity of the glass cover for radiation emitted by the absorber plate, and (c) the rate of heat transfer to the cooling water if the glass cover temperature is also 340 K. 11–85 Consider a small black surface of area A 2 cm2 maintained at 600 K. Determine the rate at which radiation energy is emitted by the surface through a ring-shaped opening defined by 0 2 and 40 50˚ where is the azimuth angle and is the angle a radiation beam makes with the normal of the surface. Design and Essay Problems 11–86 Write an essay on the radiation properties of selective surfaces used on the absorber plates of solar collectors. Find out about the various kinds of such surfaces, and discuss the performance and cost of each type. Recommend a selective surface that optimizes cost and performance. 11–87 According to an Atomic Energy Commission report, a hydrogen bomb can be approximated as a large fireball at a temperature of 7200 K. You are to assess the impact of such a bomb exploded 5 km above a city. Assume the diameter of the fireball to be 1 km, and the blast to last 15 s. Investigate the level of radiation energy people, plants, and houses will be exposed to, and how adversely they will be affected by the blast. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 605 CHAPTER R A D I AT I O N H E AT T R A N S F E R n Chapter 11, we considered the fundamental aspects of radiation and the radiation properties of surfaces. We are now in a position to consider radiation exchange between two or more surfaces, which is the primary quantity of interest in most radiation problems. We start this chapter with a discussion of view factors and the rules associated with them. View factor expressions and charts for some common configurations are given, and the crossed-strings method is presented. We then discuss radiation heat transfer, first between black surfaces and then between nonblack surfaces using the radiation network approach. We continue with radiation shields and discuss the radiation effect on temperature measurements and comfort. Finally, we consider gas radiation, and discuss the effective emissivities and absorptivities of gas bodies of various shapes. We also discuss radiation exchange between the walls of combustion chambers and the high-temperature emitting and absorbing combustion gases inside. I 12 CONTENTS 12–1 The View Factor 606 12–2 View Factor Relations 609 12–3 Radiation Heat Transfer: Black Surfaces 620 12–4 Radiation Heat Transfer: Diffuse, Gray Surfaces 623 12–5 Radiation Shields and the Radiation Effect 635 12–6 Radiation Exchange with Emitting and Absorbing Gases 639 Topic of Special Interest: Heat Transfer from the Human Body 649 605 cen58933_ch12.qxd 9/9/2002 9:48 AM Page 606 606 HEAT TRANSFER 12–1 Surface 2 Surface 1 Surface 3 Point source FIGURE 12–1 Radiation heat exchange between surfaces depends on the orientation of the surfaces relative to each other, and this dependence on orientation is accounted for by the view factor. ■ THE VIEW FACTOR Radiation heat transfer between surfaces depends on the orientation of the surfaces relative to each other as well as their radiation properties and temperatures, as illustrated in Figure 12–1. For example, a camper will make the most use of a campfire on a cold night by standing as close to the fire as possible and by blocking as much of the radiation coming from the fire by turning her front to the fire instead of her side. Likewise, a person will maximize the amount of solar radiation incident on him and take a sunbath by lying down on his back instead of standing up on his feet. To account for the effects of orientation on radiation heat transfer between two surfaces, we define a new parameter called the view factor, which is a purely geometric quantity and is independent of the surface properties and temperature. It is also called the shape factor, configuration factor, and angle factor. The view factor based on the assumption that the surfaces are diffuse emitters and diffuse reflectors is called the diffuse view factor, and the view factor based on the assumption that the surfaces are diffuse emitters but specular reflectors is called the specular view factor. In this book, we will consider radiation exchange between diffuse surfaces only, and thus the term view factor will simply mean diffuse view factor. The view factor from a surface i to a surface j is denoted by Fi → j or just Fij, and is defined as Fij the fraction of the radiation leaving surface i that strikes surface j directly n2 θ2 n1 r θ1 dA2 A2 d21 dA1 A1 FIGURE 12–2 Geometry for the determination of the view factor between two surfaces. The notation Fi → j is instructive for beginners, since it emphasizes that the view factor is for radiation that travels from surface i to surface j. However, this notation becomes rather awkward when it has to be used many times in a problem. In such cases, it is convenient to replace it by its shorthand version Fij. Therefore, the view factor F12 represents the fraction of radiation leaving surface 1 that strikes surface 2 directly, and F21 represents the fraction of the radiation leaving surface 2 that strikes surface 1 directly. Note that the radiation that strikes a surface does not need to be absorbed by that surface. Also, radiation that strikes a surface after being reflected by other surfaces is not considered in the evaluation of view factors. To develop a general expression for the view factor, consider two differential surfaces dA1 and dA2 on two arbitrarily oriented surfaces A1 and A2, respectively, as shown in Figure 12–2. The distance between dA1 and dA2 is r, and the angles between the normals of the surfaces and the line that connects dA1 and dA2 are 1 and 2, respectively. Surface 1 emits and reflects radiation diffusely in all directions with a constant intensity of I1, and the solid angle subtended by dA2 when viewed by dA1 is d21. The rate at which radiation leaves dA1 in the direction of 1 is I1 cos 1dA1. Noting that d21 dA2 cos 2 /r 2, the portion of this radiation that strikes dA2 is dA2 cos 2 · Q d A1 → d A2 I1 cos 1dA1d21 I1 cos 1dA1 r2 (12-1) cen58933_ch12.qxd 9/9/2002 9:48 AM Page 607 607 CHAPTER 12 The total rate at which radiation leaves dA1 (via emission and reflection) in all directions is the radiosity (which is J1 I1) times the surface area, · Q d A1 J1dA1 I1dA1 (12-2) Then the differential view factor dFd A1 → d A2, which is the fraction of radiation leaving dA1 that strikes dA2 directly, becomes dFd A1 → d A2 Q· dA1 → dA2 cos 1 cos 2 dA2 r 2 Q· (12-3) dA1 The differential view factor dFd A2 → d A1 can be determined from Eq. 12–3 by interchanging the subscripts 1 and 2. The view factor from a differential area dA1 to a finite area A2 can be determined from the fact that the fraction of radiation leaving dA1 that strikes A2 is the sum of the fractions of radiation striking the differential areas dA2. Therefore, the view factor Fd A1 → A2 is determined by integrating dFd A1 → d A2 over A2, Fd A1 → A2 cos rcos dA 1 2 2 A2 (12-4) 2 The total rate at which radiation leaves the entire A1 (via emission and reflection) in all directions is · Q A1 J1A1 I1A1 (12-5) The portion of this radiation that strikes dA2 is determined by considering the radiation that leaves dA1 and strikes dA2 (given by Eq. 12–1), and integrating it over A1, · Q A1 → d A2 Q· A1 d A1 → d A2 I cos rcos dA dA 1 1 2 2 2 A1 (12-6) 1 Integration of this relation over A2 gives the radiation that strikes the entire A2, · Q A1 → A2 Q· A2 A1 → d A2 I cos r cos dA dA 1 A2 A1 1 2 2 1 2 (12-7) Dividing this by the total radiation leaving A1 (from Eq. 12–5) gives the fraction of radiation leaving A1 that strikes A2, which is the view factor FA1 → A2 (or F12 for short), Q· A1 → A2 1 F12 F A1 → A2 · A1 Q A1 cos rcos dA dA 1 2 1 2 A2 A1 2 (12-8) The view factor F A2 → A1 is readily determined from Eq. 12–8 by interchanging the subscripts 1 and 2, F21 F A2 → A1 Q· A2 → A1 1 A2 Q· A2 cos rcos dA dA 1 A2 A1 2 2 1 2 (12-9) cen58933_ch12.qxd 9/9/2002 9:48 AM Page 608 608 HEAT TRANSFER Note that I1 is constant but r, 1, and 2 are variables. Also, integrations can be performed in any order since the integration limits are constants. These relations confirm that the view factor between two surfaces depends on their relative orientation and the distance between them. Combining Eqs. 12–8 and 12–9 after multiplying the former by A1 and the latter by A2 gives 1 F1 → 1 = 0 (a) Plane surface A1F12 A2F21 2 F2 → 2 = 0 (12-10) which is known as the reciprocity relation for view factors. It allows the calculation of a view factor from a knowledge of the other. The view factor relations developed above are applicable to any two surfaces i and j provided that the surfaces are diffuse emitters and diffuse reflectors (so that the assumption of constant intensity is valid). For the special case of j i, we have (b) Convex surface 3 F3 → 3 ≠ 0 Fi → i the fraction of radiation leaving surface i that strikes itself directly (c) Concave surface FIGURE 12–3 The view factor from a surface to itself is zero for plane or convex surfaces and nonzero for concave surfaces. Outer sphere 2 Inner sphere 1 F1 → 2 = 1 FIGURE 12–4 In a geometry that consists of two concentric spheres, the view factor F1 → 2 1 since the entire radiation leaving the surface of the smaller sphere will be intercepted by the larger sphere. Noting that in the absence of strong electromagnetic fields radiation beams travel in straight paths, the view factor from a surface to itself will be zero unless the surface “sees” itself. Therefore, Fi → i 0 for plane or convex surfaces and Fi → i 0 for concave surfaces, as illustrated in Figure 12–3. The value of the view factor ranges between zero and one. The limiting case Fi → j 0 indicates that the two surfaces do not have a direct view of each other, and thus radiation leaving surface i cannot strike surface j directly. The other limiting case Fi → j 1 indicates that surface j completely surrounds surface i, so that the entire radiation leaving surface i is intercepted by surface j. For example, in a geometry consisting of two concentric spheres, the entire radiation leaving the surface of the smaller sphere (surface 1) will strike the larger sphere (surface 2), and thus F1 → 2 1, as illustrated in Figure 12–4. The view factor has proven to be very useful in radiation analysis because it allows us to express the fraction of radiation leaving a surface that strikes another surface in terms of the orientation of these two surfaces relative to each other. The underlying assumption in this process is that the radiation a surface receives from a source is directly proportional to the angle the surface subtends when viewed from the source. This would be the case only if the radiation coming off the source is uniform in all directions throughout its surface and the medium between the surfaces does not absorb, emit, or scatter radiation. That is, it will be the case when the surfaces are isothermal and diffuse emitters and reflectors and the surfaces are separated by a nonparticipating medium such as a vacuum or air. The view factor F1 → 2 between two surfaces A1 and A2 can be determined in a systematic manner first by expressing the view factor between two differential areas dA1 and dA2 in terms of the spatial variables and then by performing the necessary integrations. However, this approach is not practical, since, even for simple geometries, the resulting integrations are usually very complex and difficult to perform. View factors for hundreds of common geometries are evaluated and the results are given in analytical, graphical, and tabular form in several publications. View factors for selected geometries are given in Tables 12–1 and 12–2 in analytical form and in Figures 12–5 to 12–8 in graphical form. The view cen58933_ch12.qxd 9/9/2002 9:48 AM Page 609 609 CHAPTER 12 TABLE 12–1 View factor expressions for some common geometries of finite size (3D) Geometry Aligned parallel rectangles –– –– 2 1/2 2 (1 + X 2)(1 + Y— ) Fi → j = —— –– –– ln —————— – 2 –– 2 πXY 1+X +Y –– –– –– X + X(1 + Y 2)1/ 2 tan–1 ———–— –– (1 + Y 2)1/2 – – –– –– Y + Y(1 + X 2)1/ 2 tan–1 ———–— –– 2 1/2 (1 + X ) j L i Y Relation –– –– X = X/L and Y = Y/L X –– –– –– –– – X tan–1 X – Y tan–1 Y Coaxial parallel disks rj j ri L i Perpendicular rectangles with a common edge Ri = ri /L and Rj = rj /L 1 + R 2j S = 1 + ——– R 2i rj 1 Fi → j = – S – S 2 – 4 — 2 ri () 2 1/ 2 H = Z /X and W = Y/X ( 1 1 1 Fi → j = —– W tan–1 — + H tan–1 — πW W H 1 – (H 2 + W 2)1/ 2 tan–1 ———–—— (H 2 + W 2)1/2 j Z i Y X (1 + W 2)(1 + H 2) + 1– ln ——————— 4 1 + W2 + H2 W 2(1 + W 2 + H 2) × ———————— (1 + W 2)(W 2 + H 2) W2 H 2(1 + H 2 + W 2) × ———————— (1 + H 2)(H 2 + W 2) H2 ) factors in Table 12–1 are for three-dimensional geometries. The view factors in Table 12–2, on the other hand, are for geometries that are infinitely long in the direction perpendicular to the plane of the paper and are therefore two-dimensional. 12–2 ■ VIEW FACTOR RELATIONS Radiation analysis on an enclosure consisting of N surfaces requires the evaluation of N2 view factors, and this evaluation process is probably the most time-consuming part of a radiation analysis. However, it is neither practical nor necessary to evaluate all of the view factors directly. Once a sufficient number of view factors are available, the rest of them can be determined by utilizing some fundamental relations for view factors, as discussed next. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 610 610 HEAT TRANSFER TABLE 12–2 View factor expressions for some infinitely long (2D) geometries Geometry Relation Parallel plates with midlines connected by perpendicular line wi Wi = wi /L and Wj = wj /L i [(Wi + Wj)2 + 4]1/ 2 – (Wj – Wi)2 + 4]1/ 2 Fi → j = ——————————————— 2Wi L j wj Inclined plates of equal width and with a common edge j 1 Fi → j = 1 – sin – α 2 w α i w Perpendicular plates with a common edge j wj wj 1 Fi → j = – 1 + — – 1 + — 2 wi wi () wj 2 1/2 i wi Three-sided enclosure wk wj k wi + wj – wk Fi → j = ————— 2wi j i wi Infinite plane and row of cylinders s D j () ( ) D Fi → j = 1 – 1 – — s 2 1/ 2 s2 – D2 D + — tan–1 ——— s D2 i 1/ 2 1 The Reciprocity Relation The view factors Fi → j and Fj → i are not equal to each other unless the areas of the two surfaces are. That is, Fj → i Fi → j Fj → i Fi → j when when Ai Aj Ai Aj cen58933_ch12.qxd 9/9/2002 9:48 AM Page 611 611 CHAPTER 12 1.0 0.9 0.8 0.7 0.6 0.5 L2 L1 10 5 43 A2 D 0.4 R A1 o a ti /D L1 2 1.5 1 0.9 0.8 0.7 0.6 0.5 0.3 0.2 10 0.4 0.3 0.25 0.2 0.18 0.16 0.14 0.12 0.1 F1 → 2 0.1 0.09 0.08 0.07 0.06 0.05 0.5 0.4 0.04 0.3 0.03 0.2 0.02 0.1 0.2 A2 0.3 L2 W A1 0.4 0.5 0.6 0.8 1 2 Ratio L 2 / D 3 4 5 6 8 10 L1 0.1 0.15 0.2 0.4 20 Asymptote 0.01 0.1 FIGURE 12–5 View factor between two aligned parallel rectangles of equal size. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.5 3 L 1 /W 0.3 Ra tio F1 → 2 0.2 0.5 0.1 4 5 6 8 10 20 1 2 0 0.1 5 10 0.2 0.3 0.4 0.5 0.6 0.8 1 2 Ratio L2/W 3 4 5 6 8 10 20 FIGURE 12–6 View factor between two perpendicular rectangles with a common edge. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 612 612 HEAT TRANSFER 1.0 0.9 r2 r2 /L = 8 6 2 5 4 0.8 L r1 3 0.7 1 2 1.5 0.6 1.25 F1 → 2 0.5 1.0 0.4 0.3 0.8 0.2 0.6 0.5 0.4 0.1 FIGURE 12–7 View factor between two coaxial parallel disks. 0 0.1 0.2 0.3 0.4 0.6 1.0 L /r1 r2 /L = 0.3 2 3 4 5 6 8 10 1.0 A2 L r1 0.8 r2 A1 1.0 0.9 0.6 = 0.8 L/ r2 F2 → 1 2 0.7 L/ r2 0.6 1 4 0. 5 0.4 0. 25 0.5 F2 → 2 1 2 0.4 1 0.3 0. 0.2 = 0.5 0.2 0.25 0.1 0 0 0.2 0.4 0.6 r1/r2 0.8 1.0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 r1/r2 FIGURE 12–8 View factors for two concentric cylinders of finite length: (a) outer cylinder to inner cylinder; (b) outer cylinder to itself. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 613 613 CHAPTER 12 We have shown earlier the pair of view factors Fi → j and Fj → i are related to each other by AiFi → j AjFj → i (12-11) This relation is referred to as the reciprocity relation or the reciprocity rule, and it enables us to determine the counterpart of a view factor from a knowledge of the view factor itself and the areas of the two surfaces. When determining the pair of view factors Fi → j and Fj → i, it makes sense to evaluate first the easier one directly and then the more difficult one by applying the reciprocity relation. 2 The Summation Rule The radiation analysis of a surface normally requires the consideration of the radiation coming in or going out in all directions. Therefore, most radiation problems encountered in practice involve enclosed spaces. When formulating a radiation problem, we usually form an enclosure consisting of the surfaces interacting radiatively. Even openings are treated as imaginary surfaces with radiation properties equivalent to those of the opening. The conservation of energy principle requires that the entire radiation leaving any surface i of an enclosure be intercepted by the surfaces of the enclosure. Therefore, the sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including to itself, must equal unity. This is known as the summation rule for an enclosure and is expressed as (Fig. 12–9) Surface i N F i→j 1 (12-12) j1 where N is the number of surfaces of the enclosure. For example, applying the summation rule to surface 1 of a three-surface enclosure yields 3 F 1→j F1 → 1 F1 → 2 F1 → 3 1 j1 The summation rule can be applied to each surface of an enclosure by varying i from 1 to N. Therefore, the summation rule applied to each of the N surfaces of an enclosure gives N relations for the determination of the view factors. Also, the reciprocity rule gives 12 N(N 1) additional relations. Then the total number of view factors that need to be evaluated directly for an N-surface enclosure becomes N2 [N 21 N(N 1)] 12 N(N 1) For example, for a six-surface enclosure, we need to determine only 1 2 2 6(6 1) 15 of the 6 36 view factors directly. The remaining 21 view factors can be determined from the 21 equations that are obtained by applying the reciprocity and the summation rules. FIGURE 12–9 Radiation leaving any surface i of an enclosure must be intercepted completely by the surfaces of the enclosure. Therefore, the sum of the view factors from surface i to each one of the surfaces of the enclosure must be unity. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 614 614 HEAT TRANSFER EXAMPLE 12–1 View Factors Associated with Two Concentric Spheres Determine the view factors associated with an enclosure formed by two spheres, shown in Figure 12–10. SOLUTION The view factors associated with two concentric spheres are to be r1 r2 1 2 FIGURE 12–10 The geometry considered in Example 12–1. determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The outer surface of the smaller sphere (surface 1) and inner surface of the larger sphere (surface 2) form a two-surface enclosure. Therefore, N 2 and this enclosure involves N 2 22 4 view factors, which are F11, F12, F21, and F22. In this two-surface enclosure, we need to determine only 1 N(N 2 1) 12 2(2 1) 1 view factor directly. The remaining three view factors can be determined by the application of the summation and reciprocity rules. But it turns out that we can determine not only one but two view factors directly in this case by a simple inspection: F11 0, F12 1, since no radiation leaving surface 1 strikes itself since all radiation leaving surface 1 strikes surface 2 Actually it would be sufficient to determine only one of these view factors by inspection, since we could always determine the other one from the summation rule applied to surface 1 as F11 F12 1. The view factor F21 is determined by applying the reciprocity relation to surfaces 1 and 2: A1F12 A2F21 which yields F21 A1 4r12 r1 F12 1 r 2 A2 4r22 2 Finally, the view factor F22 is determined by applying the summation rule to surface 2: F21 F22 1 and thus 2 r1 F22 1 F21 1 r 2 Discussion Note that when the outer sphere is much larger than the inner sphere (r2 r1), F22 approaches one. This is expected, since the fraction of radiation leaving the outer sphere that is intercepted by the inner sphere will be negligible in that case. Also note that the two spheres considered above do not need to be concentric. However, the radiation analysis will be most accurate for the case of concentric spheres, since the radiation is most likely to be uniform on the surfaces in that case. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 615 615 CHAPTER 12 3 The Superposition Rule Sometimes the view factor associated with a given geometry is not available in standard tables and charts. In such cases, it is desirable to express the given geometry as the sum or difference of some geometries with known view factors, and then to apply the superposition rule, which can be expressed as the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j. Note that the reverse of this is not true. That is, the view factor from a surface j to a surface i is not equal to the sum of the view factors from the parts of surface j to surface i. Consider the geometry in Figure 12–11, which is infinitely long in the direction perpendicular to the plane of the paper. The radiation that leaves surface 1 and strikes the combined surfaces 2 and 3 is equal to the sum of the radiation that strikes surfaces 2 and 3. Therefore, the view factor from surface 1 to the combined surfaces of 2 and 3 is F1 → (2, 3) F1 → 2 F1 → 3 (12-13) Suppose we need to find the view factor F1 → 3. A quick check of the view factor expressions and charts in this section will reveal that such a view factor cannot be evaluated directly. However, the view factor F1 → 3 can be determined from Eq. 12–13 after determining both F1 → 2 and F1 → (2, 3) from the chart in Figure 12–12. Therefore, it may be possible to determine some difficult view factors with relative ease by expressing one or both of the areas as the sum or differences of areas and then applying the superposition rule. To obtain a relation for the view factor F(2, 3) → 1, we multiply Eq. 12–13 by A1, 3 3 = 2 2 1 1 1 F1 → (2, 3) =F1 → 2 + F1 → 3 FIGURE 12–11 The view factor from a surface to a composite surface is equal to the sum of the view factors from the surface to the parts of the composite surface. 3 r2 = 5 cm r3 = 8 cm 2 A1 F1 → (2, 3) A1 F1 → 2 A1 F1 → 3 and apply the reciprocity relation to each term to get (A2 A3)F(2, 3) → 1 A2 F2 → 1 A3 F3 → 1 or F(2, 3) → 1 A2 F2 → 1 A3 F3 → 1 A2 A3 (12-14) Areas that are expressed as the sum of more than two parts can be handled in a similar manner. EXAMPLE 12–2 Fraction of Radiation Leaving through an Opening Determine the fraction of the radiation leaving the base of the cylindrical enclosure shown in Figure 12–12 that escapes through a coaxial ring opening at its top surface. The radius and the length of the enclosure are r1 10 cm and L 10 cm, while the inner and outer radii of the ring are r2 5 cm and r3 8 cm, respectively. 1 r1 = 10 cm FIGURE 12–12 The cylindrical enclosure considered in Example 12–2. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 616 616 HEAT TRANSFER SOLUTION The fraction of radiation leaving the base of a cylindrical enclosure through a coaxial ring opening at its top surface is to be determined. Assumptions The base surface is a diffuse emitter and reflector. Analysis We are asked to determine the fraction of the radiation leaving the base of the enclosure that escapes through an opening at the top surface. Actually, what we are asked to determine is simply the view factor F1 → ring from the base of the enclosure to the ring-shaped surface at the top. We do not have an analytical expression or chart for view factors between a circular area and a coaxial ring, and so we cannot determine F1 → ring directly. However, we do have a chart for view factors between two coaxial parallel disks, and we can always express a ring in terms of disks. Let the base surface of radius r1 10 cm be surface 1, the circular area of r2 5 cm at the top be surface 2, and the circular area of r3 8 cm be surface 3. Using the superposition rule, the view factor from surface 1 to surface 3 can be expressed as F1 → 3 F1 → 2 F1 → ring since surface 3 is the sum of surface 2 and the ring area. The view factors F1 → 2 and F1 → 3 are determined from the chart in Figure 12–7. L 10 cm r1 10 cm 1 and L 10 cm r1 10 cm 1 and r2 (Fig. 12–7) 5 cm 0.5 → F1 → 2 0.11 L 10 cm r3 (Fig. 12–7) 8 cm 0.8 → F1 → 3 0.28 L 10 cm Therefore, F1 → ring F1 → 3 F1 → 2 0.28 0.11 0.17 which is the desired result. Note that F1 → 2 and F1 → 3 represent the fractions of radiation leaving the base that strike the circular surfaces 2 and 3, respectively, and their difference gives the fraction that strikes the ring area. 4 The Symmetry Rule 3 2 1 F1 → 2 = F1 → 3 (Also, F2 → 1 = F3 → 1 ) FIGURE 12–13 Two surfaces that are symmetric about a third surface will have the same view factor from the third surface. The determination of the view factors in a problem can be simplified further if the geometry involved possesses some sort of symmetry. Therefore, it is good practice to check for the presence of any symmetry in a problem before attempting to determine the view factors directly. The presence of symmetry can be determined by inspection, keeping the definition of the view factor in mind. Identical surfaces that are oriented in an identical manner with respect to another surface will intercept identical amounts of radiation leaving that surface. Therefore, the symmetry rule can be expressed as two (or more) surfaces that possess symmetry about a third surface will have identical view factors from that surface (Fig. 12–13). The symmetry rule can also be expressed as if the surfaces j and k are symmetric about the surface i then Fi → j Fi → k. Using the reciprocity rule, we can show that the relation Fj → i Fk → i is also true in this case. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 617 617 CHAPTER 12 EXAMPLE 12–3 View Factors Associated with a Tetragon Determine the view factors from the base of the pyramid shown in Figure 12–14 to each of its four side surfaces. The base of the pyramid is a square, and its side surfaces are isosceles triangles. 3 4 2 5 SOLUTION The view factors from the base of a pyramid to each of its four side surfaces for the case of a square base are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The base of the pyramid (surface 1) and its four side surfaces (surfaces 2, 3, 4, and 5) form a five-surface enclosure. The first thing we notice about this enclosure is its symmetry. The four side surfaces are symmetric about the base surface. Then, from the symmetry rule, we have 1 FIGURE 12–14 The pyramid considered in Example 12–3. F12 F13 F14 F15 Also, the summation rule applied to surface 1 yields 5 F 1j F11 F12 F13 F14 F15 1 j 1 However, F11 0, since the base is a flat surface. Then the two relations above yield F12 F13 F14 F15 0.25 Discussion Note that each of the four side surfaces of the pyramid receive one-fourth of the entire radiation leaving the base surface, as expected. Also note that the presence of symmetry greatly simplified the determination of the view factors. EXAMPLE 12–4 View Factors Associated with a Triangular Duct Determine the view factor from any one side to any other side of the infinitely long triangular duct whose cross section is given in Figure 12–15. SOLUTION The view factors associated with an infinitely long triangular duct are to be determined. Assumptions The surfaces are diffuse emitters and reflectors. Analysis The widths of the sides of the triangular cross section of the duct are L1, L2, and L3, and the surface areas corresponding to them are A1, A2, and A3, respectively. Since the duct is infinitely long, the fraction of radiation leaving any surface that escapes through the ends of the duct is negligible. Therefore, the infinitely long duct can be considered to be a three-surface enclosure, N 3. This enclosure involves N 2 32 9 view factors, and we need to determine 1 N(N 2 1) 12 3(3 1) 3 3 2 L3 L2 1 L1 FIGURE 12–15 The infinitely long triangular duct considered in Example 12–4. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 618 618 HEAT TRANSFER of these view factors directly. Fortunately, we can determine all three of them by inspection to be F11 F22 F33 0 since all three surfaces are flat. The remaining six view factors can be determined by the application of the summation and reciprocity rules. Applying the summation rule to each of the three surfaces gives F11 F12 F13 1 F21 F22 F23 1 F31 F32 F33 1 Noting that F11 F22 F33 0 and multiplying the first equation by A1, the second by A2, and the third by A3 gives A1F12 A1F13 A1 A2F21 A2F23 A2 A3F31 A3F32 A3 Finally, applying the three reciprocity relations A1F12 A2F21, A1F13 A3F31, and A2F23 A3F32 gives A1F12 A1F13 A1 A1F12 A2F23 A2 A1F13 A2F23 A3 This is a set of three algebraic equations with three unknowns, which can be solved to obtain A1 A2 A3 L1 L2 L3 2A1 2L1 A1 A3 A2 L1 L3 L2 F13 2A1 2L1 A2 A3 A1 L2 L3 L1 F23 2A2 2L2 F12 (12-15) Discussion Note that we have replaced the areas of the side surfaces by their corresponding widths for simplicity, since A Ls and the length s can be factored out and canceled. We can generalize this result as the view factor from a surface of a very long triangular duct to another surface is equal to the sum of the widths of these two surfaces minus the width of the third surface, divided by twice the width of the first surface. View Factors between Infinitely Long Surfaces: The Crossed-Strings Method Many problems encountered in practice involve geometries of constant cross section such as channels and ducts that are very long in one direction relative cen58933_ch12.qxd 9/9/2002 9:48 AM Page 619 619 CHAPTER 12 to the other directions. Such geometries can conveniently be considered to be two-dimensional, since any radiation interaction through their end surfaces will be negligible. These geometries can subsequently be modeled as being infinitely long, and the view factor between their surfaces can be determined by the amazingly simple crossed-strings method developed by H. C. Hottel in the 1950s. The surfaces of the geometry do not need to be flat; they can be convex, concave, or any irregular shape. To demonstrate this method, consider the geometry shown in Figure 12–16, and let us try to find the view factor F1 → 2 between surfaces 1 and 2. The first thing we do is identify the endpoints of the surfaces (the points A, B, C, and D) and connect them to each other with tightly stretched strings, which are indicated by dashed lines. Hottel has shown that the view factor F1 → 2 can be expressed in terms of the lengths of these stretched strings, which are straight lines, as (L5 L6) (L3 L4) F1 → 2 2L1 (12-16) 2 L2 C D L5 L6 L4 L3 L1 B A 1 FIGURE 12–16 Determination of the view factor F1 → 2 by the application of the crossed-strings method. Note that L5 L6 is the sum of the lengths of the crossed strings, and L3 L4 is the sum of the lengths of the uncrossed strings attached to the endpoints. Therefore, Hottel’s crossed-strings method can be expressed verbally as Fi → j (Crossed strings) (Uncrossed strings) 2 (String on surface i) (12-17) The crossed-strings method is applicable even when the two surfaces considered share a common edge, as in a triangle. In such cases, the common edge can be treated as an imaginary string of zero length. The method can also be applied to surfaces that are partially blocked by other surfaces by allowing the strings to bend around the blocking surfaces. EXAMPLE 12–5 The Crossed-Strings Method for View Factors Two infinitely long parallel plates of widths a 12 cm and b 5 cm are located a distance c 6 cm apart, as shown in Figure 12–17. (a) Determine the view factor F1 → 2 from surface 1 to surface 2 by using the crossed-strings method. (b) Derive the crossed-strings formula by forming triangles on the given geometry and using Eq. 12–15 for view factors between the sides of triangles. SOLUTION The view factors between two infinitely long parallel plates are to be determined using the crossed-strings method, and the formula for the view factor is to be derived. Assumptions The surfaces are diffuse emitters and reflectors. Analysis (a) First we label the endpoints of both surfaces and draw straight dashed lines between the endpoints, as shown in Figure 12–17. Then we identify the crossed and uncrossed strings and apply the crossed-strings method (Eq. 12–17) to determine the view factor F1 → 2: F1 → 2 (Crossed strings) (Uncrossed strings) (L5 L6) (L3 L4) 2L1 2 (String on surface 1) b = L 2 = 5 cm C 2 D L4 L3 c = 6 cm A L5 L6 1 a = L 1 = 12 cm B FIGURE 12–17 The two infinitely long parallel plates considered in Example 12–5. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 620 620 HEAT TRANSFER where L1 a 12 cm L2 b 5 cm L3 c 6 cm L4 72 62 9.22 cm L5 52 62 7.81 cm L6 122 62 13.42 cm Substituting, F1 → 2 [(7.81 13.42) (6 9.22)] cm 0.250 2 12 cm (b) The geometry is infinitely long in the direction perpendicular to the plane of the paper, and thus the two plates (surfaces 1 and 2) and the two openings (imaginary surfaces 3 and 4) form a four-surface enclosure. Then applying the summation rule to surface 1 yields F11 F12 F13 F14 1 But F11 0 since it is a flat surface. Therefore, F12 1 F13 F14 where the view factors F13 and F14 can be determined by considering the triangles ABC and ABD, respectively, and applying Eq. 12–15 for view factors between the sides of triangles. We obtain F13 L1 L 3 L 6 , 2L1 F14 L1 L4 L5 2L1 Substituting, F12 1 L1 L3 L6 L1 L4 L5 2L1 2L1 (L5 L6) (L3 L4) 2L1 which is the desired result. This is also a miniproof of the crossed-strings method for the case of two infinitely long plain parallel surfaces. 12–3 ■ RADIATION HEAT TRANSFER: BLACK SURFACES So far, we have considered the nature of radiation, the radiation properties of materials, and the view factors, and we are now in a position to consider the rate of heat transfer between surfaces by radiation. The analysis of radiation exchange between surfaces, in general, is complicated because of reflection: a radiation beam leaving a surface may be reflected several times, with partial reflection occurring at each surface, before it is completely absorbed. The analysis is simplified greatly when the surfaces involved can be approximated cen58933_ch12.qxd 9/9/2002 9:48 AM Page 621 621 CHAPTER 12 as blackbodies because of the absence of reflection. In this section, we consider radiation exchange between black surfaces only; we will extend the analysis to reflecting surfaces in the next section. Consider two black surfaces of arbitrary shape maintained at uniform temperatures T1 and T2, as shown in Figure 12–18. Recognizing that radiation leaves a black surface at a rate of Eb T 4 per unit surface area and that the view factor F1 → 2 represents the fraction of radiation leaving surface 1 that strikes surface 2, the net rate of radiation heat transfer from surface 1 to surface 2 can be expressed as T1 A1 2 1 Radiation leaving Radiation leaving the entire surface 1 the entire surface 2 that strikes surface 2 that strikes surface 1 A1 Eb1 F1 → 2 A2 Eb2 F2 → 1 (W) · Q1 → 2 T2 A2 · Q12 (12-18) FIGURE 12–18 Two general black surfaces maintained at uniform temperatures T1 and T2. Applying the reciprocity relation A1F1 → 2 A2F2 → 1 yields · Q 1 → 2 A1 F1 → 2 (T14 T24) (W) (12-19) · which is the desired relation. A negative value for Q 1 → 2 indicates that net radiation heat transfer is from surface 2 to surface 1. Now consider an enclosure consisting of N black surfaces maintained at specified temperatures. The net radiation heat transfer from any surface i of this enclosure is determined by adding up the net radiation heat transfers from surface i to each of the surfaces of the enclosure: · Qi N Q · j1 N i→j AF i j1 (Ti4 Tj4) i → j (W) (12-20) · Again a negative value for Q indicates that net radiation heat transfer is to surface i (i.e., surface i gains radiation energy instead of losing). Also, the net heat transfer from a surface to itself is zero, regardless of the shape of the surface. EXAMPLE 12–6 Radiation Heat Transfer in a Black Furnace Consider the 5-m 5-m 5-m cubical furnace shown in Figure 12–19, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Determine (a) the net rate of radiation heat transfer between the base and the side surfaces, (b) the net rate of radiation heat transfer between the base and the top surface, and (c) the net radiation heat transfer from the base surface. SOLUTION The surfaces of a cubical furnace are black and are maintained at uniform temperatures. The net rate of radiation heat transfer between the base and side surfaces, between the base and the top surface, from the base surface are to be determined. Assumptions The surfaces are black and isothermal. 2 T2 = 1500 K 3 T3 = 500 K 1 T1 = 800 K FIGURE 12–19 The cubical furnace of black surfaces considered in Example 12–6. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 622 622 HEAT TRANSFER Analysis (a) Considering that the geometry involves six surfaces, we may be tempted at first to treat the furnace as a six-surface enclosure. However, the four side surfaces possess the same properties, and thus we can treat them as a single side surface in radiation analysis. We consider the base surface to be surface 1, the top surface to be surface 2, and the side surfaces to be surface · · · 3. Then the problem reduces to determining Q 1 → 3, Q 1 → 2, and Q 1. · The net rate of radiation heat transfer Q 1 → 3 from surface 1 to surface 3 can be determined from Eq. 12–19, since both surfaces involved are black, by replacing the subscript 2 by 3: · Q 1 → 3 A1F1 → 3 (T14 T34) But first we need to evaluate the view factor F1 → 3. After checking the view factor charts and tables, we realize that we cannot determine this view factor directly. However, we can determine the view factor F1 → 2 directly from Figure 12–5 to be F1 → 2 0.2, and we know that F1 → 1 0 since surface 1 is a plane. Then applying the summation rule to surface 1 yields F1 → 1 F1 → 2 F1 → 3 1 or F1 → 3 1 F1 → 1 F1 → 2 1 0 0.2 0.8 Substituting, · Q 1 → 3 (25 m2)(0.8)(5.67 10 8 W/m2 · K4)[(800 K)4 (500 K)4] 394 103 W 394 kW · (b) The net rate of radiation heat transfer Q 1 → 2 from surface 1 to surface 2 is determined in a similar manner from Eq. 12–19 to be · Q 1 → 2 A1F1 → 2 (T14 T24) (25 m2)(0.2)(5.67 10 8 W/m2 · K4)[(800 K)4 (1500 K)4] 1319 103 W 1319 kW The negative sign indicates that net radiation heat transfer is from surface 2 to surface 1. · (c) The net radiation heat transfer from the base surface Q 1 is determined from Eq. 12–20 by replacing the subscript i by 1 and taking N 3: · Q1 3 Q · 1→j · · · Q1 → 1 Q1 → 2 Q1 → 3 j 1 0 ( 1319 kW) (394 kW) 925 kW Again the negative sign indicates that net radiation heat transfer is to surface 1. That is, the base of the furnace is gaining net radiation at a rate of about 925 kW. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 623 623 CHAPTER 12 12–4 ■ RADIATION HEAT TRANSFER: DIFFUSE, GRAY SURFACES The analysis of radiation transfer in enclosures consisting of black surfaces is relatively easy, as we have seen above, but most enclosures encountered in practice involve nonblack surfaces, which allow multiple reflections to occur. Radiation analysis of such enclosures becomes very complicated unless some simplifying assumptions are made. To make a simple radiation analysis possible, it is common to assume the surfaces of an enclosure to be opaque, diffuse, and gray. That is, the surfaces are nontransparent, they are diffuse emitters and diffuse reflectors, and their radiation properties are independent of wavelength. Also, each surface of the enclosure is isothermal, and both the incoming and outgoing radiation are uniform over each surface. But first we review the concept of radiosity discussed in Chap. 11. Radiosity Surfaces emit radiation as well as reflect it, and thus the radiation leaving a surface consists of emitted and reflected parts. The calculation of radiation heat transfer between surfaces involves the total radiation energy streaming away from a surface, with no regard for its origin. The total radiation energy leaving a surface per unit time and per unit area is the radiosity and is denoted by J (Fig. 12–20). For a surface i that is gray and opaque ( i i and i i 1), the radiosity can be expressed as (blackbody) Reflected Emitted radiation radiation ε Eb G (12-21) where Ebi Ti4 is the blackbody emissive power of surface i and Gi is irradiation (i.e., the radiation energy incident on surface i per unit time per unit area). For a surface that can be approximated as a blackbody ( i 1), the radiosity relation reduces to Ji Ebi Ti4 Incident radiation ρG Radiation emitted Radiation reflected by surface i by surface i iEbi iGi iEbi (1 i)Gi (W/m2) Ji Radiosity, J (12-22) That is, the radiosity of a blackbody is equal to its emissive power. This is expected, since a blackbody does not reflect any radiation, and thus radiation coming from a blackbody is due to emission only. Net Radiation Heat Transfer to or from a Surface During a radiation interaction, a surface loses energy by emitting radiation and gains energy by absorbing radiation emitted by other surfaces. A surface experiences a net gain or a net loss of energy, depending on which quantity is larger. The net rate of radiation heat transfer from a surface i of surface area Ai · is denoted by Q i and is expressed as Surface FIGURE 12–20 Radiosity represents the sum of the radiation energy emitted and reflected by a surface. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 624 624 HEAT TRANSFER · Radiation leaving Radiation incident Qi entire surface i on entire surface i Ai(Ji Gi) (W) (12-23) Solving for Gi from Eq. 12–21 and substituting into Eq. 12–23 yields Ji i Ebi Ai i · Q i Ai Ji (E Ji) 1 i 1 i bi (W) (12-24) In an electrical analogy to Ohm’s law, this equation can be rearranged as Ebi Ji · Qi Ri (W) (12-25) where Ri . Qi Ebi Surface i 1 – εi Ri = ——– Ai εi Ji FIGURE 12–21 Electrical analogy of surface resistance to radiation. 1 i Ai i (12-26) is the surface resistance to radiation. The quantity Ebi Ji corresponds to a potential difference and the net rate of radiation heat transfer corresponds to current in the electrical analogy, as illustrated in Figure 12–21. The direction of the net radiation heat transfer depends on the relative magnitudes of Ji (the radiosity) and Ebi (the emissive power of a blackbody at the temperature of the surface). It will be from the surface if Ebi Ji and to the · surface if Ji Ebi. A negative value for Q i indicates that heat transfer is to the surface. All of this radiation energy gained must be removed from the other side of the surface through some mechanism if the surface temperature is to remain constant. The surface resistance to radiation for a blackbody is zero since i 1 and Ji Ebi. The net rate of radiation heat transfer in this case is determined directly from Eq. 12–23. Some surfaces encountered in numerous practical heat transfer applications are modeled as being adiabatic since their back sides are well insulated and the net heat transfer through them is zero. When the convection effects on the front (heat transfer) side of such a surface is negligible and steady-state conditions are reached, the surface must lose as much radiation energy as it gains, · and thus Q i 0. In such cases, the surface is said to reradiate all the radiation energy it receives, and such a surface is called a reradiating surface. Setting · Q i 0 in Eq. 12–25 yields Ji Ebi Ti4 (W/m2) (12-27) Therefore, the temperature of a reradiating surface under steady conditions can easily be determined from the equation above once its radiosity is known. Note that the temperature of a reradiating surface is independent of its emissivity. In radiation analysis, the surface resistance of a reradiating surface is disregarded since there is no net heat transfer through it. (This is like the fact that there is no need to consider a resistance in an electrical network if no current is flowing through it.) cen58933_ch12.qxd 9/9/2002 9:48 AM Page 625 625 CHAPTER 12 Net Radiation Heat Transfer between Any Two Surfaces Ebj Consider two diffuse, gray, and opaque surfaces of arbitrary shape maintained at uniform temperatures, as shown in Figure 12–22. Recognizing that the radiosity J represents the rate of radiation leaving a surface per unit surface area and that the view factor Fi → j represents the fraction of radiation leaving surface i that strikes surface j, the net rate of radiation heat transfer from surface i to surface j can be expressed as · Qi → j Radiation leaving Radiation leaving the entire surface i the entire surface j that strikes surface j that strikes surface i Ai Ji Fi → j Aj Jj Fj → i Rj Jj . Qij 1 Rij = —— Ai Fij Ji Ri (12-28) Ebi (W) Surface i FIGURE 12–22 Electrical analogy of space resistance to radiation. Applying the reciprocity relation Ai Fi → j Aj Fj → i yields · Q i → j Ai Fi → j (Ji Jj) Surface j (W) (12-29) Again in analogy to Ohm’s law, this equation can be rearranged as Ji Jj · Qi → j Ri → j (W) (12-30) where Ri → j 1 Ai Fi → j (12-31) Ji Jj · · Qi Qi → j Ai Fi → j (Ji Jj) Ri → j j1 j1 j1 N N (W) (12-32) (W) (12-33) Ji J2 R i2 R i(N – R iN Ebi Ri The network representation of net radiation heat transfer from surface i to the remaining surfaces of an N-surface enclosure is given in Figure 12–23. Note · that Q i → i (the net rate of heat transfer from a surface to itself) is zero regardless of the shape of the surface. Combining Eqs. 12–25 and 12–32 gives N J J Ebi Ji i j Ri R i→j j1 Qi R . N i1 J 1 is the space resistance to radiation. Again the quantity Ji Jj corresponds to a potential difference, and the net rate of heat transfer between two surfaces corresponds to current in the electrical analogy, as illustrated in Figure 12–22. The direction of the net radiation heat transfer between two surfaces de· pends on the relative magnitudes of Ji and Jj. A positive value for Q i → j indicates that net heat transfer is from surface i to surface j. A negative value indicates the opposite. In an N-surface enclosure, the conservation of energy principle requires that the net heat transfer from surface i be equal to the sum of the net heat transfers from surface i to each of the N surfaces of the enclosure. That is, 1) JN Surface i –1 JN FIGURE 12–23 Network representation of net radiation heat transfer from surface i to the remaining surfaces of an N-surface enclosure. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 626 626 HEAT TRANSFER which has the electrical analogy interpretation that the net radiation flow from a surface through its surface resistance is equal to the sum of the radiation flows from that surface to all other surfaces through the corresponding space resistances. Methods of Solving Radiation Problems In the radiation analysis of an enclosure, either the temperature or the net rate of heat transfer must be given for each of the surfaces to obtain a unique solution for the unknown surface temperatures and heat transfer rates. There are two methods commonly used to solve radiation problems. In the first method, Eqs. 12–32 (for surfaces with specified heat transfer rates) and 12–33 (for surfaces with specified temperatures) are simplified and rearranged as N F Surfaces with specified · net heat transfer rate Q i · Q i Ai Surfaces with specified temperature Ti Ti4 Ji j1 i → j(Ji 1 i i Jj) (12-34) N F j1 i → j(Ji Jj) (12-35) · Note that Q i 0 for insulated (or reradiating) surfaces, and Ti4 Ji for black surfaces since i 1 in that case. Also, the term corresponding to j i will drop out from either relation since Ji Jj Ji Ji 0 in that case. The equations above give N linear algebraic equations for the determination of the N unknown radiosities for an N-surface enclosure. Once the radiosities J1, J2, . . . , JN are available, the unknown heat transfer rates can be determined from Eq. 12–34 while the unknown surface temperatures can be determined from Eq. 12–35. The temperatures of insulated or reradiating surfaces can be · determined from Ti4 Ji. A positive value for Q i indicates net radiation heat transfer from surface i to other surfaces in the enclosure while a negative value indicates net radiation heat transfer to the surface. The systematic approach described above for solving radiation heat transfer problems is very suitable for use with today’s popular equation solvers such as EES, Mathcad, and Matlab, especially when there are a large number of surfaces, and is known as the direct method (formerly, the matrix method, since it resulted in matrices and the solution required a knowledge of linear algebra). The second method described below, called the network method, is based on the electrical network analogy. The network method was first introduced by A. K. Oppenheim in the 1950s and found widespread acceptance because of its simplicity and emphasis on the physics of the problem. The application of the method is straightforward: draw a surface resistance associated with each surface of an enclosure and connect them with space resistances. Then solve the radiation problem by treating it as an electrical network problem where the radiation heat transfer replaces the current and radiosity replaces the potential. The network method is not practical for enclosures with more than three or four surfaces, however, because of the increased complexity of the network. Next we apply the method to solve radiation problems in two- and threesurface enclosures. cen58933_ch12.qxd 9/9/2002 9:48 AM Page 627 627 CHAPTER 12 Radiation Heat Transfer in Two-Surface Enclosures Consider an enclosure consisting of two opaque surfaces at specified temperatures T1 and T2, as shown in Fig. 12–24, and try to determine the net rate of radiation heat transfer between the two surfaces with the network method. Surfaces 1 and 2 have emissivities 1 and 2 and surface areas A1 and A2 and are maintained at uniform temperatures T1 and T2, respectively. There are only two surfaces in the enclosure, and thus we can write · · · Q 12 Q 1 Q 2 ε2 A2 T2 2 1 . Eb1 That is, the net rate of radiation heat transfer from surface 1 to surface 2 must equal the net rate of radiation heat transfer from surface 1 and the net rate of radiation heat transfer to surface 2. The radiation network of this two-surface enclosure consists of two surface resistances and one space resistance, as shown in Figure 12–24. In an electrical network, the electric current flowing through these resistances connected in series would be determined by dividing the potential difference between points A and B by the total resistance between the same two points. The net rate of radiation transfer is determined in the same manner and is expressed as · Q 12 . Q12 ε1 A1 T1 . Q1 J1 1–ε R1 = ——–1 A1ε1 Q12 1 R12 = ——– A1F12 . J2 Q2 Eb2 1–ε R2 = ——–2 A2ε2 FIGURE 12–24 Schematic of a two-surface enclosure and the radiation network associated with it. Eb1 Eb2 · · Q 1 Q 2 R1 R12 R2 or · Q 12 (T14 T24) 1 1 1 2 1 A1 1 A1 F12 A2 2 (W) (12-36) This important result is applicable to any two gray, diffuse, opaque surfaces that form an enclosure. The view factor F12 depends on the geometry and must be determined first. Simplified forms of Eq. 12–36 for some familiar arrangements that form a two-surface enclosure are given in Table 12–3. Note that F12 1 for all of these special cases. EXAMPLE 12–7 Radiation Heat Transfer between Parallel Plates Two very large parallel plates are maintained at uniform temperatures T1 800 K and T2 500 K and have emissivities 1 0.2 and 2 0.7, respectively, as shown in Figure 12–25. Determine the net rate of radiation heat transfer between the two surfaces per unit surface area of the plates. SOLUTION Two large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between the plates is to be determined. Assumptions Both surfaces are opaque, diffuse, and gray. Analysis The net rate of radiation heat transfer between the two plates per unit area is readily determined from Eq. 12–38 to be 1 ε1 = 0.2 . T1 = 800 K Q12 2 ε 2 = 0.7 T2 = 500 K FIGURE 12–25 The two parallel plates considered in Example 12–7. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 628 628 HEAT TRANSFER TABLE 12–3 Small object in a large cavity A1 —– ≈0 A2 A1, T1, ε1 . Q12 = A1σε1(T 41 – T 24) (12-37) F12 = 1 A2, T2, ε2 Infinitely large parallel plates A1, T1, ε1 A2, T2, ε 2 A1 = A2 = A F12 = 1 . Aσ(T 41 – T 24) Q12 = ——————— 1 1 — ε +— ε –1 (12-38) . A1σ(T 41 – T 24) Q12 = ————————— r1 1 1 – ε2 —– — ε1 + ––— ε2 r2 (12-39) . A1σ(T 41 – T 24) Q12 = ————————— r1 2 1 1 – ε2 —– — ε1 + ––— ε2 r2 (12-40) 1 2 Infinitely long concentric cylinders r1 r2 A1 r1 —– = —– A2 r2 F12 = 1 Concentric spheres r1 r2 A1 r1 2 —– = —– A2 r2 F12 = 1 · Q 12 (T14 T24) (5.67 10 8 W/m2 · K4)[(800 K)4 (500 K)4] q· 12 A 1 1 1 1 1 1 2 1 0.2 0.7 3625 W/m2 Discussion Note that heat at a net rate of 3625 W is transferred from plate 1 to plate 2 by radiation per unit surface area of either plate. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 629 629 CHAPTER 12 Radiation Heat Transfer in Three-Surface Enclosures We now consider an enclosure consisting of three opaque, diffuse, gray surfaces, as shown in Figure 12–26. Surfaces 1, 2, and 3 have surface areas A1, A2, and A3; emissivities 1, 2, and 3; and uniform temperatures T1, T2, and T3, respectively. The radiation network of this geometry is constructed by following the standard procedure: draw a surface resistance associated with each of the three surfaces and connect these surface resistances with space resistances, as shown in the figure. Relations for the surface and space resistances are given by Eqs. 12–26 and 12–31. The three endpoint potentials Eb1, Eb2, and Eb3 are considered known, since the surface temperatures are specified. Then all we need to find are the radiosities J1, J2, and J3. The three equations for the determination of these three unknowns are obtained from the requirement that the algebraic sum of the currents (net radiation heat transfer) at each node must equal zero. That is, Eb1 J1 J2 J1 J3 J1 0 R1 R12 R13 J1 J2 Eb2 J2 J3 J2 0 R12 R2 R23 J1 J3 J2 J3 Eb3 J3 0 R13 R23 R3 (12-41) Once the radiosities J1, J2, and J3 are available, the net rate of radiation heat transfers at each surface can be determined from Eq. 12–32. The set of equations above simplify further if one or more surfaces are “special” in some way. For example, Ji Ebi Ti4 for a black or reradiating sur· face. Also, Q i 0 for a reradiating surface. Finally, when the net rate of · radiation heat transfer Q i is specified at surface i instead of the temperature, · the term (Ebi Ji)/Ri should be replaced by the specified Q i. ε1, A1, T1 ε 2, A2, T2 1 2 . Q1 Eb1 J1 1–ε R1 = ——–1 A1ε1 1 R12 = ——– A1F12 . Q12 1 R13 = ——– A1F13 3 . Q23 . Q13 Eb2 J2 1–ε R2 = ——–2 A2ε2 . Q2 1 R23 = ——– A2 F23 J3 ε 3 , A3, T3 1–ε R3 = ——–3 A 3ε 3 Eb3 . Q3 FIGURE 12–26 Schematic of a three-surface enclosure and the radiation network associated with it. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 630 630 HEAT TRANSFER T1 = 700 K ε1 = 0.8 1 ro 3 Black T3 = 400 K H EXAMPLE 12–8 Radiation Heat Transfer in a Cylindrical Furnace Consider a cylindrical furnace with r0 H 1 m, as shown in Figure 12–27. The top (surface 1) and the base (surface 2) of the furnace has emissivities 1 0.8 and 2 0.4, respectively, and are maintained at uniform temperatures T1 700 K and T2 500 K. The side surface closely approximates a blackbody and is maintained at a temperature of T3 400 K. Determine the net rate of radiation heat transfer at each surface during steady operation and explain how these surfaces can be maintained at specified temperatures. SOLUTION The surfaces of a cylindrical furnace are maintained at uniform temperatures. The net rate of radiation heat transfer at each surface during steady operation is to be determined. 2 T2 = 500 K ε2 = 0.4 FIGURE 12–27 The cylindrical furnace considered in Example 12–8. Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. Analysis We will solve this problem systematically using the direct method to demonstrate its use. The cylindrical furnace can be considered to be a threesurface enclosure with surface areas of A1 A2 ro2 (1 m)2 3.14 m2 A3 2roH 2(1 m)(1 m) 6.28 m2 The view factor from the base to the top surface is, from Figure 12–7, F12 0.38. Then the view factor from the base to the side surface is determined by applying the summation rule to be F11 F12 F13 1 → F13 1 F11 F12 1 0 0.38 0.62 since the base surface is flat and thus F11 0. Noting that the top and bottom surfaces are symmetric about the side surface, F21 F12 0.38 and F23 F13 0.62. The view factor F31 is determined from the reciprocity relation, A1F13 A3F31 → F31 F13(A1/A3) (0.62)(0.314/0.628) 0.31 Also, F32 F31 0.31 because of symmetry. Now that all the view factors are available, we apply Eq. 12–35 to each surface to determine the radiosities: 1 1 1 [F1 → 2 (J1 J2) F1 → 3 (J1 J3)] 1 2 Bottom surface (i 2): T24 J2 [F2 → 1 (J2 J1) F2 → 3 (J2 J3)] 2 Side surface (i 3): T34 J3 0 (since surface 3 is black and thus 3 1) Top surface (i 1): T14 J1 Substituting the known quantities, (5.67 10 8 W/m2 · K4)(700 K)4 J1 1 0.8 [0.38(J1 J2) 0.68(J1 J3)] 0.8 (5.67 10 8 W/m2 · K4)(500 K)4 J2 1 0.4 [0.28(J2 J1) 0.68(J2 J3)] 0.4 (5.67 10 8 W/m2 · K4)(400 K)4 J3 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 631 631 CHAPTER 12 Solving the equations above for J1, J2, and J3 gives J1 11,418 W/m2, J2 4562 W/m2, and J3 1452 W/m2 Then the net rates of radiation heat transfer at the three surfaces are determined from Eq. 12–34 to be · Q 1 A1[F1 → 2 (J1 J2) F1 → 3 (J1 J3)] (3.14 m2)[0.38(11,418 4562) 0.62(11,418 1452)] W/m2 27.6 103 W 27.6 kW · Q 2 A2[F2 → 1 (J2 J1) F2 → 3 (J2 J3)] (3.12 m2)[0.38(4562 11,418) 0.62(4562 1452)] W/m2 2.13 103 W 2.13 kW · Q 3 A3[F3 → 1 (J3 J1) F3 → 2 (J3 J2)] (6.28 m2)[0.31(1452 11,418) 0.31(1452 4562)] W/m2 25.5 103 W 25.5 kW Note that the direction of net radiation heat transfer is from the top surface to the base and side surfaces, and the algebraic sum of these three quantities must be equal to zero. That is, · · · Q 1 Q 2 Q 3 27.6 ( 2.13) ( 25.5) 0 Discussion To maintain the surfaces at the specified temperatures, we must supply heat to the top surface continuously at a rate of 27.6 kW while removing 2.13 kW from the base and 25.5 kW from the side surfaces. The direct method presented here is straightforward, and it does not require the evaluation of radiation resistances. Also, it can be applied to enclosures with any number of surfaces in the same manner. EXAMPLE 12–9 Radiation Heat Transfer in a Triangular Furnace A furnace is shaped like a long equilateral triangular duct, as shown in Figure 12–28. The width of each side is 1 m. The base surface has an emissivity of 0.7 and is maintained at a uniform temperature of 600 K. The heated left-side surface closely approximates a blackbody at 1000 K. The right-side surface is well insulated. Determine the rate at which heat must be supplied to the heated side externally per unit length of the duct in order to maintain these operating conditions. . Q1 Eb1 2 T2 = 1000 K Black J1 R12 J2 = Eb2 R1 3 . . Q2 = – Q1 Insulated R13 1 ε1 = 0.7 T1 = 600 K R23 (J3 = Eb3 ) . Q3 = 0 FIGURE 12–28 The triangular furnace considered in Example 12–9. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 632 632 HEAT TRANSFER SOLUTION Two of the surfaces of a long equilateral triangular furnace are maintained at uniform temperatures while the third surface is insulated. The external rate of heat transfer to the heated side per unit length of the duct during steady operation is to be determined. Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. Analysis The furnace can be considered to be a three-surface enclosure with a radiation network as shown in the figure, since the duct is very long and thus the end effects are negligible. We observe that the view factor from any surface to any other surface in the enclosure is 0.5 because of symmetry. Surface 3 is a reradiating surface since the net rate of heat transfer at that surface is zero. · · Then we must have Q 1 Q 2, since the entire heat lost by surface 1 must be gained by surface 2. The radiation network in this case is a simple series– · parallel connection, and we can determine Q 1 directly from Eb1 Eb2 · Q1 R1 1 1 R12 R13 R23 1 Eb1 Eb2 1 1 1 A1 F12 A1 1 1/A1 F13 1/A2 F23 1 where A1 A2 A3 wL 1 m 1 m 1 m2 (per unit length of the duct) (symmetry) F12 F13 F23 0.5 Eb1 T14 (5.67 10 8 W/m2 · K4)(600 K)4 7348 W/m2 Eb2 T24 (5.67 10 8 W/m2 · K4)(1000 K)4 56,700 W/m2 Substituting, · Q1 (56,700 7348) W/m2 1 0.7 1 (0.5 1 m2) 0.7 1 m2 1/(0.5 1 m2) 1/(0.5 1 m2) 28.0 103 28.0 kW 1 Therefore, heat at a rate of 28 kW must be supplied to the heated surface per unit length of the duct to maintain steady operation in the furnace. Solar energy Glass cover ε = 0.9 EXAMPLE 12–10 Heat Transfer through a Tubular Solar Collector 70°F 4 in. 2 in. Aluminum tube ε = 0.95 Water FIGURE 12–29 Schematic for Example 12–10. A solar collector consists of a horizontal aluminum tube having an outer diameter of 2 in. enclosed in a concentric thin glass tube of 4-in. diameter, as shown in Figure 12–29. Water is heated as it flows through the tube, and the space between the aluminum and the glass tubes is filled with air at 1 atm pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of 30 Btu/h per foot length, and the temperature of the ambient air outside is 70°F. The emissivities of the tube and the glass cover are 0.95 and 0.9, respectively. Taking the effective sky temperature to be 50°F, determine the cen58933_ch12.qxd 9/9/2002 9:49 AM Page 633 633 CHAPTER 12 temperature of the aluminum tube when steady operating conditions are established (i.e., when the rate of heat loss from the tube equals the amount of solar energy gained by the tube). SOLUTION The circulating pump of a solar collector that consists of a horizontal tube and its glass cover fails. The equilibrium temperature of the tube is to be determined. Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are isothermal. 3 Air is an ideal gas. 4 The surfaces are opaque, diffuse, and gray for infrared radiation. 5 The glass cover is transparent to solar radiation. Properties The properties of air should be evaluated at the average temperature. But we do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk fluid and glass cover temperatures at this point, and thus we cannot evaluate the average temperatures. Therefore, we will assume the glass temperature to be 110°F, and use properties at an anticipated average temperature of (70 110)/2 90°F (Table A-15E), k 0.01505 Btu/h ft °F 0.6310 ft2/h 1.753 10 4 ft2/s Pr 0.7275 1 1 Tave 550 R Analysis This problem was solved in Chapter 9 by disregarding radiation heat transfer. Now we will repeat the solution by considering natural convection and radiation occurring simultaneously. We have a horizontal cylindrical enclosure filled with air at 1 atm pressure. The problem involves heat transfer from the aluminum tube to the glass cover and from the outer surface of the glass cover to the surrounding ambient air. When steady operation is reached, these two heat transfer rates must equal the rate of heat gain. That is, · · · Q tube-glass Q glass-ambient Q solar gain 30 Btu/h (per foot of tube) The heat transfer surface area of the glass cover is Ao Aglass (Do L) (4/12 ft)(1 ft) 1.047 ft2 (per foot of tube) To determine the Rayleigh number, we need to know the surface temperature of the glass, which is not available. Therefore, it is clear that the solution will require a trial-and-error approach. Assuming the glass cover temperature to be 110°F, the Rayleigh number, the Nusselt number, the convection heat transfer coefficient, and the rate of natural convection heat transfer from the glass cover to the ambient air are determined to be g(To T) D3o Pr 2 (32.2 ft/s2)1/(550 R)(4/12 ft)3 (0.7275) 2.054 106 (1.753 10 4 ft2/s)2 2 0.387 Ra1/6 0.387(2.054 106)1/6 Do Nu 0.6 0.6 9/16 8/27 [1 (0.559/Pr) ] [1 (0.559/0.7275)9/16]8/27 17.89 RaDo 2 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 634 634 HEAT TRANSFER 0.01505 Btu/h · ft · °F k Nu (17.89) 0.8075 Btu/h ft2 °F Do 4/12 ft · Q o, conv ho Ao(To T) (0.8075 Btu/h ft2 °F)(1.047 ft2)(110 70)°F 33.8 Btu/h ho Also, · 4 Q o, rad o Ao(To4 Tsky ) (0.9)(0.1714 10 8 Btu/h ft2 R4)(1.047 ft2)[(570 R)4 (510 R)4] 61.2 Btu/h Then the total rate of heat loss from the glass cover becomes · · · Q o, total Q o, conv Q o, rad 33.8 61.2 95.0 Btu/h which is much larger than 30 Btu/h. Therefore, the assumed temperature of 110°F for the glass cover is high. Repeating the calculations with lower temperatures (including the evaluation of properties), the glass cover temperature corresponding to 30 Btu/h is determined to be 78°F (it would be 106°F if radiation were ignored). The temperature of the aluminum tube is determined in a similar manner using the natural convection and radiation relations for two horizontal concentric cylinders. The characteristic length in this case is the distance between the two cylinders, which is Lc (Do Di)/2 (4 2)/2 1 in. 1/12 ft Also, Ai Atube (Di L) (2 /12 ft)(1 ft) 0.5236 ft2 (per foot of tube) We start the calculations by assuming the tube temperature to be 122°F, and thus an average temperature of (78 122)/2 100°F 640 R. Using properties at 100°F, g(Ti To)L3c Pr 2 (32.2 ft/s2)1/(640 R)(1/12 ft)3 (0.726) 3.249 104 (1.809 10 4 ft2/s)2 RaL The effective thermal conductivity is [ln(Do /Di)]4 (Di3/5 Do3/5)5 [ln(4/2)]4 0.1466 3 (1/12 ft) [(2/12 ft) 3/5 (4/12 ft) 3/5]5 1/4 Pr (FcycRaL)1/4 keff 0.386k 0.861 Pr 0.726 0.386(0.01529 Btu/h ft °F) (0.1466 3.249 104)1/4 0.861 0.726 0.04032 Btu/h ft °F Fcyc L3c cen58933_ch12.qxd 9/9/2002 9:49 AM Page 635 635 CHAPTER 12 Then the rate of heat transfer between the cylinders by convection becomes 2keff · Q i, conv (T To) ln(Do /Di) i 2(0.04032 Btu/h · ft °F) (122 78)°F 16.1 Btu/h ln(4/2) Also, Ai (Ti4 To4) · Q i, rad 1 1 o Di i o Do (0.1714 10 8 Btu/h · ft2 · R4)(0.5236 ft2)[(582 R)4 (538 R)4] 1 0.9 2 in. 1 0.9 4 in. 0.95 25.1 Btu/h Then the total rate of heat loss from the glass cover becomes · · · Q i, total Q i, conv Q i, rad 16.1 25.1 41.1 Btu/h which is larger than 30 Btu/h. Therefore, the assumed temperature of 122°F for the tube is high. By trying other values, the tube temperature corresponding to 30 Btu/h is determined to be 112°F (it would be 180°F if radiation were ignored). Therefore, the tube will reach an equilibrium temperature of 112°F when the pump fails. Discussion It is clear from the results obtained that radiation should always be considered in systems that are heated or cooled by natural convection, unless the surfaces involved are polished and thus have very low emissivities. 12–5 ■ RADIATION SHIELDS AND THE RADIATION EFFECT Radiation heat transfer between two surfaces can be reduced greatly by inserting a thin, high-reflectivity (low-emissivity) sheet of material between the two surfaces. Such highly reflective thin plates or shells are called radiation shields. Multilayer radiation shields constructed of about 20 sheets per cm thickness separated by evacuated space are commonly used in cryogenic and space applications. Radiation shields are also used in temperature measurements of fluids to reduce the error caused by the radiation effect when the temperature sensor is exposed to surfaces that are much hotter or colder than the fluid itself. The role of the radiation shield is to reduce the rate of radiation heat transfer by placing additional resistances in the path of radiation heat flow. The lower the emissivity of the shield, the higher the resistance. Radiation heat transfer between two large parallel plates of emissivities 1 and 2 maintained at uniform temperatures T1 and T2 is given by Eq. 12–38: cen58933_ch12.qxd 9/9/2002 9:49 AM Page 636 636 HEAT TRANSFER (1) Shield T1 ε1 T3 ε3, 1 . 1–ε ——–1 ε1 A1 T3 ε3, 2 T2 ε2 . Q12 FIGURE 12–30 The radiation shield placed between two parallel plates and the radiation network associated with it. (2) . Q12 1 – ε3,1 –——– ε3,1 A3 1 ——– A1 F13 1 – ε3, 2 –——– ε3, 2 A3 Q12 1 – ε2 –——– ε2 A2 1 ——– A3 F32 Eb1 Eb2 A (T14 T24) · Q 12, no shield 1 1 1 2 1 Now consider a radiation shield placed between these two plates, as shown in Figure 12–30. Let the emissivities of the shield facing plates 1 and 2 be 3, 1 and 3, 2, respectively. Note that the emissivity of different surfaces of the shield may be different. The radiation network of this geometry is constructed, as usual, by drawing a surface resistance associated with each surface and connecting these surface resistances with space resistances, as shown in the figure. The resistances are connected in series, and thus the rate of radiation heat transfer is · Q 12, one shield Eb1 Eb2 1 1 3, 2 1 1 1 2 3, 1 1 1 A1 1 A1 F12 A3 3, 1 A3 3, 2 A3 F32 A2 2 (12-42) Noting that F13 F23 1 and A1 A2 A3 A for infinite parallel plates, Eq. 12–42 simplifies to · Q 12, one shield A (T14 T24) 1 1 1 1 1 2 1 3, 1 3, 2 1 (12-43) where the terms in the second set of parentheses in the denominator represent the additional resistance to radiation introduced by the shield. The appearance of the equation above suggests that parallel plates involving multiple radiation shields can be handled by adding a group of terms like those in the second set of parentheses to the denominator for each radiation shield. Then the radiation heat transfer through large parallel plates separated by N radiation shields becomes · Q 12, N shields A (T14 T24) 1 1 1 1 1 2 3, 1 1 1 1 1 · · · 1 3, 2 N, 1 N, 2 (12-44) cen58933_ch12.qxd 9/9/2002 9:49 AM Page 637 637 CHAPTER 12 If the emissivities of all surfaces are equal, Eq. 12–44 reduces to · Q 12, N shields A (T14 T24) 1 1 (N 1) 1 · 1 Q N 1 12, no shield (12-45) Therefore, when all emissivities are equal, 1 shield reduces the rate of radiation heat transfer to one-half, 9 shields reduce it to one-tenth, and 19 shields reduce it to one-twentieth (or 5 percent) of what it was when there were no shields. The equilibrium temperature of the radiation shield T3 in Figure 12–30 can · · be determined by expressing Eq. 12–43 for Q 13 or Q 23 (which involves T3) · · · · after evaluating Q 12 from Eq. 12–43 and noting that Q 12 Q 13 Q 23 when steady conditions are reached. Radiation shields used to reduce the rate of radiation heat transfer between concentric cylinders and spheres can be handled in a similar manner. In case of one shield, Eq. 12–42 can be used by taking F13 F23 1 for both cases and by replacing the A’s by the proper area relations. Radiation Effect on Temperature Measurements A temperature measuring device indicates the temperature of its sensor, which is supposed to be, but is not necessarily, the temperature of the medium that the sensor is in. When a thermometer (or any other temperature measuring device such as a thermocouple) is placed in a medium, heat transfer takes place between the sensor of the thermometer and the medium by convection until the sensor reaches the temperature of the medium. But when the sensor is surrounded by surfaces that are at a different temperature than the fluid, radiation exchange will take place between the sensor and the surrounding surfaces. When the heat transfers by convection and radiation balance each other, the sensor will indicate a temperature that falls between the fluid and surface temperatures. Below we develop a procedure to account for the radiation effect and to determine the actual fluid temperature. Consider a thermometer that is used to measure the temperature of a fluid flowing through a large channel whose walls are at a lower temperature than the fluid (Fig. 12–31). Equilibrium will be established and the reading of the thermometer will stabilize when heat gain by convection, as measured by the sensor, equals heat loss by radiation (or vice versa). That is, on a unitarea basis, Tw . qconv (K) Tw FIGURE 12–31 A thermometer used to measure the temperature of a fluid in a channel. or th (Tth4 Tw4) h . qrad Tf q· conv, to sensor q· rad, from sensor h(Tf Tth) th (Tth4 Tw4) Tf Tth Tth (12-46) cen58933_ch12.qxd 9/9/2002 9:49 AM Page 638 638 HEAT TRANSFER where Tf actual temperature of the fluid, K Tth temperature value measured by the thermometer, K Tw temperature of the surrounding surfaces, K h convection heat transfer coefficient, W/m2 · K emissivity of the sensor of the thermometer The last term in Eq. 12–46 is due to the radiation effect and represents the radiation correction. Note that the radiation correction term is most significant when the convection heat transfer coefficient is small and the emissivity of the surface of the sensor is large. Therefore, the sensor should be coated with a material of high reflectivity (low emissivity) to reduce the radiation effect. Placing the sensor in a radiation shield without interfering with the fluid flow also reduces the radiation effect. The sensors of temperature measurement devices used outdoors must be protected from direct sunlight since the radiation effect in that case is sure to reach unacceptable levels. The radiation effect is also a significant factor in human comfort in heating and air-conditioning applications. A person who feels fine in a room at a specified temperature may feel chilly in another room at the same temperature as a result of the radiation effect if the walls of the second room are at a considerably lower temperature. For example, most people will feel comfortable in a room at 22°C if the walls of the room are also roughly at that temperature. When the wall temperature drops to 5°C for some reason, the interior temperature of the room must be raised to at least 27°C to maintain the same level of comfort. Therefore, well-insulated buildings conserve energy not only by reducing the heat loss or heat gain, but also by allowing the thermostats to be set at a lower temperature in winter and at a higher temperature in summer without compromising the comfort level. EXAMPLE 12–11 1 3 2 ε1 = 0.2 T1 = 800 K ε3 = 0.1 ε2 = 0.7 T2 = 500 K . q12 A thin aluminum sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures T1 800 K and T2 500 K and have emissivities 1 0.2 and 2 0.7, respectively, as shown in Fig. 12–32. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result to that without the shield. SOLUTION A thin aluminum sheet is placed between two large parallel plates maintained at uniform temperatures. The net rates of radiation heat transfer between the two plates with and without the radiation shield are to be determined. Assumptions FIGURE 12–32 Schematic for Example 12–11. Radiation Shields The surfaces are opaque, diffuse, and gray. Analysis The net rate of radiation heat transfer between these two plates without the shield was determined in Example 12–7 to be 3625 W/m2. Heat transfer in the presence of one shield is determined from Eq. 12–43 to be cen58933_ch12.qxd 9/9/2002 9:49 AM Page 639 639 CHAPTER 12 · Q 12, one shield (T14 T24) ·q 12, one shield A 1 1 1 1 1 2 1 3, 1 3, 2 1 (5.67 10 8 W/m2 · K4)[(800 K)4 (500 K)4] 0.21 0.71 1 0.11 0.11 1 806 W/m2 Discussion Note that the rate of radiation heat transfer reduces to about onefourth of what it was as a result of placing a radiation shield between the two parallel plates. EXAMPLE 12–12 Radiation Effect on Temperature Measurements A thermocouple used to measure the temperature of hot air flowing in a duct whose walls are maintained at Tw 400 K shows a temperature reading of Tth 650 K (Fig. 12–33). Assuming the emissivity of the thermocouple junction to be 0.6 and the convection heat transfer coefficient to be h 80 W/m2 · °C, determine the actual temperature of the air. SOLUTION The temperature of air in a duct is measured. The radiation effect on the temperature measurement is to be quantified, and the actual air temperature is to be determined. Assumptions The surfaces are opaque, diffuse, and gray. Analysis The walls of the duct are at a considerably lower temperature than the air in it, and thus we expect the thermocouple to show a reading lower than the actual air temperature as a result of the radiation effect. The actual air temperature is determined from Eq. 12–46 to be Tf Tth th (Tth4 Tw4) h (650 K) 0.6 (5.67 10 8 W/m2 · K4)[(650 K)4 (400 K)4] 80 W/m2 · °C 715 K Note that the radiation effect causes a difference of 65°C (or 65 K since °C K for temperature differences) in temperature reading in this case. 12–6 ■ RADIATION EXCHANGE WITH EMITTING AND ABSORBING GASES So far we considered radiation heat transfer between surfaces separated by a medium that does not emit, absorb, or scatter radiation—a nonparticipating medium that is completely transparent to thermal radiation. A vacuum satisfies this condition perfectly, and air at ordinary temperatures and pressures Tth = 650 K Tf ε = 0.6 Tw = 400 K FIGURE 12–33 Schematic for Example 12–12. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 640 640 HEAT TRANSFER comes very close. Gases that consist of monatomic molecules such as Ar and He and symmetric diatomic molecules such as N2 and O2 are essentially transparent to radiation, except at extremely high temperatures at which ionization occurs. Therefore, atmospheric air can be considered to be a nonparticipating medium in radiation calculations. Gases with asymmetric molecules such as H2O, CO2, CO, SO2, and hydrocarbons HnCm may participate in the radiation process by absorption at moderate temperatures, and by absorption and emission at high temperatures such as those encountered in combustion chambers. Therefore, air or any other medium that contains such gases with asymmetric molecules at sufficient concentrations must be treated as a participating medium in radiation calculations. Combustion gases in a furnace or a combustion chamber, for example, contain sufficient amounts of H2O and CO2, and thus the emission and absorption of gases in furnaces must be taken into consideration. The presence of a participating medium complicates the radiation analysis considerably for several reasons: • A participating medium emits and absorbs radiation throughout its entire volume. That is, gaseous radiation is a volumetric phenomena, and thus it depends on the size and shape of the body. This is the case even if the temperature is uniform throughout the medium. • Gases emit and absorb radiation at a number of narrow wavelength bands. This is in contrast to solids, which emit and absorb radiation over the entire spectrum. Therefore, the gray assumption may not always be appropriate for a gas even when the surrounding surfaces are gray. • The emission and absorption characteristics of the constituents of a gas mixture also depends on the temperature, pressure, and composition of the gas mixture. Therefore, the presence of other participating gases affects the radiation characteristics of a particular gas. The propagation of radiation through a medium can be complicated further by presence of aerosols such as dust, ice particles, liquid droplets, and soot (unburned carbon) particles that scatter radiation. Scattering refers to the change of direction of radiation due to reflection, refraction, and diffraction. Scattering caused by gas molecules themselves is known as the Rayleigh scattering, and it has negligible effect on heat transfer. Radiation transfer in scattering media is considered in advanced books such as the ones by Modest (1993, Ref. 12) and Siegel and Howell (1992, Ref. 14). The participating medium can also be semitransparent liquids or solids such as water, glass, and plastics. To keep complexities to a manageable level, we will limit our consideration to gases that emit and absorb radiation. In particular, we will consider the emission and absorption of radiation by H2O and CO2 only since they are the participating gases most commonly encountered in practice (combustion products in furnaces and combustion chambers burning hydrocarbon fuels contain both gases at high concentrations), and they are sufficient to demonstrate the basic principles involved. Radiation Properties of a Participating Medium Consider a participating medium of thickness L. A spectral radiation beam of intensity I, 0 is incident on the medium, which is attenuated as it propagates cen58933_ch12.qxd 9/9/2002 9:49 AM Page 641 641 CHAPTER 12 due to absorption. The decrease in the intensity of radiation as it passes through a layer of thickness dx is proportional to the intensity itself and the thickness dx. This is known as Beer’s law, and is expressed as (Fig. 12–34) dI(x) I(x)dx Iλ,0 where the constant of proportionality is the spectral absorption coefficient of the medium whose unit is m 1 (from the requirement of dimensional homogeneity). This is just like the amount of interest earned by a bank account during a time interval being proportional to the amount of money in the account and the time interval, with the interest rate being the constant of proportionality. Separating the variables and integrating from x 0 to x L gives I, L e L I, 0 (12-48) where we have assumed the absorptivity of the medium to be independent of x. Note that radiation intensity decays exponentially in accordance with Beer’s law. The spectral transmissivity of a medium can be defined as the ratio of the intensity of radiation leaving the medium to that entering the medium. That is, I, L e L I, 0 (12-49) Note that 1 when no radiation is absorbed and thus radiation intensity remains constant. Also, the spectral transmissivity of a medium represents the fraction of radiation transmitted by the medium at a given wavelength. Radiation passing through a nonscattering (and thus nonreflecting) medium is either absorbed or transmitted. Therefore 1, and the spectral absorptivity of a medium of thickness L is 1 1 e L (12-50) From Kirchoff’s law, the spectral emissivity of the medium is 1 e L IIλλ(x) (x) Iλ,L (12-47) (12-51) Note that the spectral absorptivity, transmissivity, and emissivity of a medium are dimensionless quantities, with values less than or equal to 1. The spectral absorption coefficient of a medium (and thus , , and ), in general, vary with wavelength, temperature, pressure, and composition. For an optically thick medium (a medium with a large value of L), Eq. 12–51 gives 1. For L 5, for example, 0.993. Therefore, an optically thick medium emits like a blackbody at the given wavelength. As a result, an optically thick absorbing-emitting medium with no significant scattering at a given temperature Tg can be viewed as a “black surface” at Tg since it will absorb essentially all the radiation passing through it, and it will emit the maximum possible radiation that can be emitted by a surface at Tg, which is Eb(Tg). 0 x L dx FIGURE 12–34 The attenuation of a radiation beam while passing through an absorbing medium of thickness L. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 642 642 HEAT TRANSFER Emissivity and Absorptivity of Gases and Gas Mixtures The spectral absorptivity of CO2 is given in Figure 12–35 as a function of wavelength. The various peaks and dips in the figure together with discontinuities show clearly the band nature of absorption and the strong nongray characteristics. The shape and the width of these absorption bands vary with temperature and pressure, but the magnitude of absorptivity also varies with the thickness of the gas layer. Therefore, absorptivity values without specified thickness and pressure are meaningless. The nongray nature of properties should be considered in radiation calculations for high accuracy. This can be done using a band model, and thus performing calculations for each absorption band. However, satisfactory results can be obtained by assuming the gas to be gray, and using an effective total absorptivity and emissivity determined by some averaging process. Charts for the total emissivities of gases are first presented by Hottel (Ref. 6), and they have been widely used in radiation calculations with reasonable accuracy. Alternative emissivity charts and calculation procedures have been developed more recently by Edwards and Matavosian (Ref. 2). Here we present the Hottel approach because of its simplicity. Even with gray assumption, the total emissivity and absorptivity of a gas depends on the geometry of the gas body as well as the temperature, pressure, and composition. Gases that participate in radiation exchange such as CO2 and H2O typically coexist with nonparticipating gases such as N2 and O2, and thus radiation properties of an absorbing and emitting gas are usually reported for a mixture of the gas with nonparticipating gases rather than the pure gas. The emissivity and absorptivity of a gas component in a mixture depends primarily on its density, which is a function of temperature and partial pressure of the gas. The emissivity of H2O vapor in a mixture of nonparticipating gases is plotted in Figure 12–36a for a total pressure of P 1 atm as a function of gas temperature Tg for a range of values for Pw L, where Pw is the partial pressure of water vapor and L is the mean distance traveled by the radiation beam. Band designation λ, µm 15 4.3 1.0 2.7 10.4 Absorptivity α λ 0.8 FIGURE 12–35 Spectral absorptivity of CO2 at 830 K and 10 atm for a path length of 38.8 cm (from Siegel and Howell, 1992). 9.4 0.6 4.8 2.0 0.4 0.2 0 20 10 8 6 5 4 3 Wavelength λ, µm 2.5 2 1.67 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 643 643 CHAPTER 12 0.8 P 1 atm 0.6 Pw L 0.4 20 ft • 0.2 0.06 0.02 Emissivity, εc Emissivity, εw 0.04 0.03 0.2 0.06 0.01 0.02 0.1 0.04 0.06 0.04 0.03 0.006 0.004 0.003 0.02 0.015 0.01 0.007 0.01 0.008 0.006 0.005 0.004 0.003 0.01 0.008 0.02 0.006 300 2.0 1.0 0.8 0.4 0.2 0.1 0.06 0.04 0.1 0.08 1 0.6 0.4 0.1 0.08 Pc L 4.0 ft • atm 0.2 atm 10 5 3 2 0.3 0.3 0.002 0.001 0.002 0.005 600 900 1200 1500 1800 2100 0.001 300 600 900 Gas temperature, Tg (K) 1200 1500 1800 2100 Gas temperature, Tg (K) (a) H20 (b) C02 FIGURE 12–36 Emissivities of H2O and CO2 gases in a mixture of nonparticipating gases at a total pressure of 1 atm for a mean beam length of L (1 m atm 3.28 ft atm) (from Hottel, 1954, Ref. 6). Cw 1.8 5 ft 1.6 PwL 1.4 0 0 0.0 1.2 • atm 0.25 0.50 1.00 2.50 5.00 10.0 Cc 2.0 1.0 1.5 0.8 1.0 0.8 0.6 0.6 0.5 0.4 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 .5 ft • atm PcL 2 1.0 0.5 0.25 0.12 0.05 0–0.02 0.3 0.05 0.08 0.1 (Pw P)/2 (atm) (a) H20 0.2 0.3 0.5 0.8 1.0 2.0 3.0 50 Total pressure, P (atm) (b) C02 FIGURE 12–37 Correction factors for the emissivities of H2O and CO2 gases at pressures other than 1 atm for use in the relations w Cw w, 1 atm and c Cc c, 1 atm (1 m atm 3.28 ft atm) (from Hottel, 1954, Ref. 6). Emissivity at a total pressure P other than P 1 atm is determined by multiplying the emissivity value at 1 atm by a pressure correction factor Cw obtained from Figure 12–37a for water vapor. That is, cen58933_ch12.qxd 9/9/2002 9:49 AM Page 644 644 HEAT TRANSFER w Cw w, 1 atm (12-52) Note that Cw 1 for P 1 atm and thus (Pw P)/2 0.5 (a very low concentration of water vapor is used in the preparation of the emissivity chart in Fig. 12–36a and thus Pw is very low). Emissivity values are presented in a similar manner for a mixture of CO2 and nonparticipating gases in Fig. 12–36b and 12–37b. Now the question that comes to mind is what will happen if the CO2 and H2O gases exist together in a mixture with nonparticipating gases. The emissivity of each participating gas can still be determined as explained above using its partial pressure, but the effective emissivity of the mixture cannot be determined by simply adding the emissivities of individual gases (although this would be the case if different gases emitted at different wavelengths). Instead, it should be determined from g c w Cc c, 1 atm Cw w, 1 atm (12-53) where is the emissivity correction factor, which accounts for the overlap of emission bands. For a gas mixture that contains both CO2 and H2O gases, is plotted in Figure 12–38. The emissivity of a gas also depends on the mean length an emitted radiation beam travels in the gas before reaching a bounding surface, and thus the shape and the size of the gas body involved. During their experiments in the 1930s, Hottel and his coworkers considered the emission of radiation from a hemispherical gas body to a small surface element located at the center of the base of the hemisphere. Therefore, the given charts represent emissivity data for the emission of radiation from a hemispherical gas body of radius L toward the center of the base of the hemisphere. It is certainly desirable to extend the reported emissivity data to gas bodies of other geometries, and this 0.07 0.07 T 400 K T 800 K T 1200 K and above 0.06 Pc L Pw L 5 ft • atm 0.05 0.04 Pc L Pw L 5 ft • atm 0.06 3.0 2.0 1.5 1.0 0.05 Pc L Pw L 5 ft • atm ∆ε 3.0 0.03 3.0 2.0 0.02 1.5 1.0 0.75 1.5 0.01 0 0 0.2 0.4 1.0 0.75 0.5 0.3 0.2 0.6 0.8 Pw Pc Pw 1.0 0 0.2 0.5 0.3 0.2 0.4 0.6 Pw Pc Pw ∆ε 0.03 0.75 2.0 0.04 0.02 0.5 0.01 0.8 1.0 0 0.2 0.3 0.2 0.4 0.6 0.8 0 1.0 Pw Pc Pw FIGURE 12–38 Emissivity correction for use in g w c when both CO2 and H2O vapor are present in a gas mixture (1 m atm 328 ft atm) (from Hottel, 1954, Ref. 6). cen58933_ch12.qxd 9/9/2002 9:49 AM Page 645 645 CHAPTER 12 is done by introducing the concept of mean beam length L, which represents the radius of an equivalent hemisphere. The mean beam lengths for various gas geometries are listed in Table 12–4. More extensive lists are available in the literature [such as Hottel (1954, Ref. 6), and Siegel and Howell, (1992, Ref. 14)]. The emissivities associated with these geometries can be determined from Figures 12–36 through 12–38 by using the appropriate mean beam length. Following a procedure recommended by Hottel, the absorptivity of a gas that contains CO2 and H2O gases for radiation emitted by a source at temperature Ts can be determined similarly from g c w (12-54) where and is determined from Figure 12–38 at the source temperature Ts. The absorptivities of CO2 and H2O can be determined from the emissivity charts (Figs. 12–36 and 12–37) as CO2: c Cc (Tg / Ts)0.65 c(Ts, Pc LTs / Tg) (12-55) w Cw (Tg / Ts)0.45 w(Ts, Pw LTs / Tg) (12-56) and H2O: The notation indicates that the emissivities should be evaluated using Ts instead of Tg (both in K or R), Pc LTs / Tg instead of Pc L, and Pw LTs / Tg instead of Pw L. Note that the absorptivity of the gas depends on the source temperature Ts as well as the gas temperature Tg. Also, when Ts Tg, as expected. The pressure correction factors Cc and Cw are evaluated using Pc L and Pw L, as in emissivity calculations. When the total emissivity of a gas g at temperature Tg is known, the emissive power of the gas (radiation emitted by the gas per unit surface area) can TABLE 12 –4 Mean beam length L for various gas volume shapes Gas Volume Geometry Hemisphere of radius R radiating to the center of its base Sphere of diameter D radiating to its surface Infinite circular cylinder of diameter D radiating to curved surface Semi-infinite circular cylinder of diameter D radiating to its base Semi-infinite circular cylinder of diameter D radiating to center of its base Infinite semicircular cylinder of radius R radiating to center of its base Circular cylinder of height equal to diameter D radiating to entire surface Circular cylinder of height equal to diameter D radiating to center of its base Infinite slab of thickness D radiating to either bounding plane Cube of side length L radiating to any face Arbitrary shape of volume V and surface area As radiating to surface L R 0.65D 0.95D 0.65D 0.90D 1.26R 0.60D 0.71D 1.80D 0.66L 3.6V /As cen58933_ch12.qxd 9/9/2002 9:49 AM Page 646 646 HEAT TRANSFER be expressed as Eg g Tg4. Then the rate of radiation energy emitted by a gas to a bounding surface of area As becomes · Q g, e g As Tg4 (12-57) If the bounding surface is black at temperature Ts, the surface will emit radiation to the gas at a rate of As Ts4 without reflecting any, and the gas will absorb this radiation at a rate of g As Ts4, where g is the absorptivity of the gas. Then the net rate of radiation heat transfer between the gas and a black surface surrounding it becomes Black enclosure: · Q net As ( gTg4 gTs4) (12-58) If the surface is not black, the analysis becomes more complicated because of the radiation reflected by the surface. But for surfaces that are nearly black with an emissivity s 0.7, Hottel (1954, Ref. 6), recommends this modification, s 1 s 1 · · Q net, gray Q net, black As ( gTg4 gTs4) 2 2 (12-59) The emissivity of wall surfaces of furnaces and combustion chambers are typically greater than 0.7, and thus the relation above provides great convenience for preliminary radiation heat transfer calculations. EXAMPLE 12–13 H5m D5m Tg 1200 K FIGURE 12–39 Schematic for Example 12–13. Effective Emissivity of Combustion Gases A cylindrical furnace whose height and diameter are 5 m contains combustion gases at 1200 K and a total pressure of 2 atm. The composition of the combustion gases is determined by volumetric analysis to be 80 percent N2, 8 percent H2O, 7 percent O2, and 5 percent CO2. Determine the effective emissivity of the combustion gases (Fig. 12–39). SOLUTION The temperature, pressure, and composition of a gas mixture is given. The emissivity of the mixture is to be determined. Assumptions 1 All the gases in the mixture are ideal gases. 2 The emissivity determined is the mean emissivity for radiation emitted to all surfaces of the cylindrical enclosure. Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O are Pc yCO2 P 0.05(2 atm) 0.10 atm Pw yH2O P 0.08(2 atm) 0.16 atm The mean beam length for a cylinder of equal diameter and height for radiation emitted to all surfaces is, from Table 12–4, L 0.60D 0.60(5 m) 3 m cen58933_ch12.qxd 9/9/2002 9:49 AM Page 647 647 CHAPTER 12 Then, Pc L (0.10 atm)(3 m) 0.30 m atm 0.98 ft atm Pw L (0.16 atm)(3 m) 0.48 m atm 1.57 ft atm The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg 1200 K and 1 atm are, from Figure 12–36, c, 1 atm 0.16 and w, 1 atm 0.23 These are the base emissivity values at 1 atm, and they need to be corrected for the 2 atm total pressure. Noting that (Pw P)/2 (0.16 2)/2 1.08 atm, the pressure correction factors are, from Figure 12–37, Cc 1.1 and Cw 1.4 Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity correction factor at T Tg 1200 K is, from Figure 12–38, Pc L Pw L 0.98 1.57 2.55 Pw 0.16 0.615 Pw Pc 0.16 0.10 0.048 Then the effective emissivity of the combustion gases becomes g Cc c, 1 atm Cw w, 1 atm 1.1 0.16 1.4 0.23 0.048 0.45 Discussion This is the average emissivity for radiation emitted to all surfaces of the cylindrical enclosure. For radiation emitted towards the center of the base, the mean beam length is 0.71D instead of 0.60D, and the emissivity value would be different. EXAMPLE 12–14 Radiation Heat Transfer in a Cylindrical Furnace Reconsider the cylindrical furnace discussed in Example 12–13. For a wall temperature of 600 K, determine the absorptivity of the combustion gases and the rate of radiation heat transfer from the combustion gases to the furnace walls (Fig. 12–40). SOLUTION The temperatures for the wall surfaces and the combustion gases are given for a cylindrical furnace. The absorptivity of the gas mixture and the rate of radiation heat transfer are to be determined. Assumptions 1 All the gases in the mixture are ideal gases. 2 All interior surfaces of furnace walls are black. 3 Scattering by soot and other particles is negligible. Analysis The average emissivity of the combustion gases at the gas temperature of Tg 1200 K was determined in the preceding example to be g 0.45. H5m Tg 1200 K D5m Ts 600 K · Qnet FIGURE 12–40 Schematic for Example 12–14. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 648 648 HEAT TRANSFER For a source temperature of Ts 600 K, the absorptivity of the gas is again determined using the emissivity charts as Pc L Ts 600 K (0.10 atm)(3 m) 0.15 m atm 0.49 ft atm Tg 1200 K Pw L Ts 600 K (0.16 atm)(3 m) 0.24 m atm 0.79 ft atm Tg 1200 K The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts 600 K and 1 atm are, from Figure 12–36, c, 1 atm 0.11 w, 1 atm 0.25 and The pressure correction factors were determined in the preceding example to be Cc 1.1 and Cw 1.4, and they do not change with surface temperature. Then the absorptivities of CO2 and H2O become 0.65 Tg 0.65 1200 K c, 1 atm (1.1) (0.11) 0.19 Ts 600 K 0.45 Tg 0.45 1200 K w Cw w, 1 atm (1.4) (0.25) 0.48 Ts 600 K c Cc Also , but the emissivity correction factor is to be evaluated from Figure 12–38 at T Ts 600 K instead of Tg 1200 K. There is no chart for 600 K in the figure, but we can read values at 400 K and 800 K, and take their average. At Pw /(Pw Pc) 0.615 and Pc L Pw L 2.55 we read 0.027. Then the absorptivity of the combustion gases becomes g c w 0.19 0.48 0.027 0.64 The surface area of the cylindrical surface is As DH 2 (5 m)2 D2 (5 m)(5 m) 2 118 m2 4 4 Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes · Q net As ( gTg4 gTs4) (118 m2)(5.67 10 8 W/m2 K4)[0.45(1200 K)4 0.64(600 K)4] 2.79 104 W Discussion The heat transfer rate determined above is for the case of black wall surfaces. If the surfaces are not black but the surface emissivity s is greater than 0.7, the heat transfer rate can be determined by multiplying the rate of heat transfer already determined by ( s 1)/2. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 649 649 CHAPTER 12 TOPIC OF SPECIAL INTEREST Heat Transfer from the Human Body The metabolic heat generated in the body is dissipated to the environment through the skin and the lungs by convection and radiation as sensible heat and by evaporation as latent heat (Fig. 12–41). Latent heat represents the heat of vaporization of water as it evaporates in the lungs and on the skin by absorbing body heat, and latent heat is released as the moisture condenses on cold surfaces. The warming of the inhaled air represents sensible heat transfer in the lungs and is proportional to the temperature rise of inhaled air. The total rate of heat loss from the body can be expressed as Convection 27% Radiation 40% Air motion · · · Q body, total Q skin Q lungs · · · · (Q sensible Q latent)skin (Q sensible Q latent)lungs · · · · · (Q convection Q radiation Q latent)skin (Q convection Q latent)lungs (12-60) Therefore, the determination of heat transfer from the body by analysis alone is difficult. Clothing further complicates the heat transfer from the body, and thus we must rely on experimental data. Under steady conditions, the total rate of heat transfer from the body is equal to the rate of metabolic heat generation in the body, which varies from about 100 W for light office work to roughly 1000 W during heavy physical work. Sensible heat loss from the skin depends on the temperatures of the skin, the environment, and the surrounding surfaces as well as the air motion. The latent heat loss, on the other hand, depends on the skin wettedness and the relative humidity of the environment as well. Clothing serves as insulation and reduces both the sensible and latent forms of heat loss. The heat transfer from the lungs through respiration obviously depends on the frequency of breathing and the volume of the lungs as well as the environmental factors that affect heat transfer from the skin. Sensible heat from the clothed skin is first transferred to the clothing and then from the clothing to the environment. The convection and radiation heat losses from the outer surface of a clothed body can be expressed as · Q conv hconv Aclothing(Tclothing Tambient) · Q rad hrad Aclothing(Tclothing Tsurr) (W) (12-61) (12-62) where hconv convection heat transfer coefficient, as given in Table 12–5 hrad radiation heat transfer coefficient, 4.7 W/m2 · °C for typical indoor conditions; the emissivity is assumed to be 0.95, which is typical Aclothing outer surface area of a clothed person Tclothing average temperature of exposed skin and clothing Tambient ambient air temperature Tsurr average temperature of the surrounding surfaces This section can be skipped without a loss in continuity. Evaporation 30% Conduction 3% Floor FIGURE 12–41 Mechanisms of heat loss from the human body and relative magnitudes for a resting person. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 650 TABLE 12 –5 Convection heat transfer coefficients for a clothed body at 1 atm ( is in m/s) (compiled from various sources) hconv, W/m2 · °C Activity Seated in air moving at 0 0.2 m/s 3.1 0.2 4 m/s 8.3 0.6 Walking in still air at 0.5 2 m/s 8.6 0.53 Walking on treadmill in still air at 0.5 2 m/s 6.5 0.39 Standing in moving air at 0 0.15 m/s 4.0 0.15 1.5 m/s 14.8 0.69 At pressures other than 1 atm, multiply by P 0.55, where P is in atm. The convection heat transfer coefficients at 1 atm pressure are given in Table 12–5. Convection coefficients at pressures P other than 1 atm are obtained by multiplying the values at atmospheric pressure by P 0.55 where P is in atm. Also, it is recognized that the temperatures of different surfaces surrounding a person are probably different, and Tsurr represents the mean radiation temperature, which is the temperature of an imaginary isothermal enclosure in which radiation heat exchange with the human body equals the radiation heat exchange with the actual enclosure. Noting that most clothing and building materials are essentially black, the mean radiation temperature of an enclosure that consists of N surfaces at different temperatures can be determined from Tsurr Fperson-1 T1 Fperson-2 T2 · · · · Fperson-N TN (12-63) where Ti is the temperature of the surface i and Fperson-i is the view factor between the person and surface i. Total sensible heat loss can also be expressed conveniently by combining the convection and radiation heat losses as · Q convrad hcombined Aclothing (Tclothing Toperative) (hconv hrad)Aclothing (Tclothing Toperative) (W) (12-64) (12-65) Tsurr · Qrad · Qconv Tambient (a) Convection and radiation, separate Toperative · Qconv + rad (b) Convection and radiation, combined FIGURE 12–42 Heat loss by convection and radiation from the body can be combined into a single term by defining an equivalent operative temperature. 650 where the operative temperature Toperative is the average of the mean radiant and ambient temperatures weighed by their respective convection and radiation heat transfer coefficients and is expressed as (Fig. 12–42) Toperative hconv Tambient hrad Tsurr Tambient Tsurr 2 hconv hrad (12-66) Note that the operative temperature will be the arithmetic average of the ambient and surrounding surface temperatures when the convection and radiation heat transfer coefficients are equal to each other. Another environmental index used in thermal comfort analysis is the effective temperature, which combines the effects of temperature and humidity. Two environments with the same effective temperature will evoke the same thermal response in people even though they are at different temperatures and humidities. Heat transfer through the clothing can be expressed as Aclothing (Tskin Tclothing) · Q conv rad Rclothing (12-67) where Rclothing is the unit thermal resistance of clothing in m2 · °C/W, which involves the combined effects of conduction, convection, and radiation between the skin and the outer surface of clothing. The thermal resistance of clothing is usually expressed in the unit clo where 1 clo 0.155 m2 · °C/W 0.880 ft2 · °F · h/Btu. The thermal resistance of trousers, long-sleeve shirt, long-sleeve sweater, and T-shirt is 1.0 clo, or 0.155 m2 · °C/W. Summer clothing such as light slacks and short-sleeved shirt has an insulation value of 0.5 clo, whereas winter clothing such as heavy slacks, long-sleeve shirt, and a sweater or jacket has an insulation value of 0.9 clo. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 651 651 CHAPTER 12 Then the total sensible heat loss can be expressed in terms of the skin temperature instead of the inconvenient clothing temperature as (Fig. 12–43) Aclothing (Tskin Toperative) · Q conv rad Rclothing 1 Tcloth Toperative (12-68) Rcombined hcombined Rcloth At a state of thermal comfort, the average skin temperature of the body is observed to be 33°C (91.5°F). No discomfort is experienced as the skin temperature fluctuates by 1.5°C (2.5°F). This is the case whether the body is clothed or unclothed. Evaporative or latent heat loss from the skin is proportional to the difference between the water vapor pressure at the skin and the ambient air, and the skin wettedness, which is a measure of the amount of moisture on the skin. It is due to the combined effects of the evaporation of sweat and the diffusion of water through the skin, and can be expressed as · Q latent m· vapor hfg Skin Tskin FIGURE 12–43 Simplified thermal resistance network for heat transfer from a clothed person. (12-69) where m· vapor the rate of evaporation from the body, kg/s hfg the enthalpy of vaporization of water 2430 kJ/kg at 30°C Heat loss by evaporation is maximum when the skin is completely wetted. Also, clothing offers resistance to evaporation, and the rate of evaporation in clothed bodies depends on the moisture permeability of the clothes. The maximum evaporation rate for an average man is about 1 L/h (0.3 g/s), which represents an upper limit of 730 W for the evaporative cooling rate. A person can lose as much as 2 kg of water per hour during a workout on a hot day, but any excess sweat slides off the skin surface without evaporating (Fig. 12–44). During respiration, the inhaled air enters at ambient conditions and exhaled air leaves nearly saturated at a temperature close to the deep body temperature (Fig. 12–45). Therefore, the body loses both sensible heat by convection and latent heat by evaporation from the lungs, and these can be expressed as · Q conv, lungs m· air, lungs Cp, air(Texhale Tambient) · Q latent, lungs m· vapor, lungs hfg m· air, lungs (wexhale wambient)hfg where m· air, lungs rate of air intake to the lungs, kg/s Cp, air specific heat of air 1.0 kJ/kg · °C Texhale temperature of exhaled air w humidity ratio (the mass of moisture per unit mass of dry air) Water vapor m· vapor, max = 0.3 g/s (12-70) (12-71) · Qlatent, max = m· latent, max hfg @ 30°C = (0.3 g/s)(2430 kJ/kg) = 730 W FIGURE 12–44 An average person can lose heat at a rate of up to 730 W by evaporation. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 652 652 HEAT TRANSFER Cool ambient air 20°C Warm and moist exhaled air 35°C 37°C The rate of air intake to the lungs is directly proportional to the metabolic · rate Q met. The rate of total heat loss from the lungs through respiration can be expressed approximately as · · · Q conv latent, lungs 0.0014Q met (34 Tambient) 0.0173Q met (5.87 P, ambient) (12-72) where P, ambient is the vapor pressure of ambient air in kPa. The fraction of sensible heat varies from about 40 percent in the case of heavy work to about 70 percent during light work. The rest of the energy is rejected from the body by perspiration in the form of latent heat. Lungs Heat and moisture EXAMPLE 12–15 FIGURE 12–45 Part of the metabolic heat generated in the body is rejected to the air from the lungs during respiration. Effect of Clothing on Thermal Comfort It is well established that a clothed or unclothed person feels comfortable when the skin temperature is about 33°C. Consider an average man wearing summer clothes whose thermal resistance is 0.6 clo. The man feels very comfortable while standing in a room maintained at 22°C. The air motion in the room is negligible, and the interior surface temperature of the room is about the same as the air temperature. If this man were to stand in that room unclothed, determine the temperature at which the room must be maintained for him to feel thermally comfortable. SOLUTION A man wearing summer clothes feels comfortable in a room at 22°C. The room temperature at which this man would feel thermally comfortable when unclothed is to be determined. Assumptions 1 Steady conditions exist. 2 The latent heat loss from the person remains the same. 3 The heat transfer coefficients remain the same. Analysis The body loses heat in sensible and latent forms, and the sensible heat consists of convection and radiation heat transfer. At low air velocities, the convection heat transfer coefficient for a standing man is given in Table 12–5 to be 4.0 W/m2 · °C. The radiation heat transfer coefficient at typical indoor conditions is 4.7 W/m2 · °C. Therefore, the surface heat transfer coefficient for a standing person for combined convection and radiation is hcombined hconv hrad 4.0 4.7 8.7 W/m2 · °C 22°C 22°C The thermal resistance of the clothing is given to be Rclothing 0.6 clo 0.6 0.155 m2 · °C/W 0.093 m2 · °C/W 33°C Noting that the surface area of an average man is 1.8 m2, the sensible heat loss from this person when clothed is determined to be (Fig. 12–46) As(Tskin Tambient) (1.8 m2)(33 22)°C · Q sensible, clothed 1 1 Rclothing 0.093 m2 · °C/ W hcombined 8.7 W/m2 · °C 95.2 W FIGURE 12–46 Schematic for Example 12–15. From a heat transfer point of view, taking the clothes off is equivalent to removing the clothing insulation or setting Rcloth 0. The heat transfer in this case can be expressed as cen58933_ch12.qxd 9/9/2002 9:49 AM Page 653 653 CHAPTER 12 As(Tskin Tambient) (1.8 m2)(33 Tambient)°C · Q sensible, unclothed 1 1 hcombined 8.7 W/m2 · °C Troom = 22°C Troom = 27°C Tskin = 33°C Tskin = 33°C To maintain thermal comfort after taking the clothes off, the skin temperature of the person and the rate of heat transfer from him must remain the same. Then setting the equation above equal to 95.2 W gives Tambient 26.9°C Clothed person Therefore, the air temperature needs to be raised from 22 to 26.9°C to ensure that the person will feel comfortable in the room after he takes his clothes off (Fig. 12–47). Note that the effect of clothing on latent heat is assumed to be negligible in the solution above. We also assumed the surface area of the clothed and unclothed person to be the same for simplicity, and these two effects should counteract each other. Unclothed person FIGURE 12–47 Clothing serves as insulation, and the room temperature needs to be raised when a person is unclothed to maintain the same comfort level. SUMMARY Radiaton heat transfer between surfaces depends on the orientation of the surfaces relative to each other, the effects of orientation are accounted for by the geometric parameter view factor. The view factor from a surface i to a surface j is denoted by Fi → j or Fij, and is defined as the fraction of the radiation leaving surface i that strikes surface j directly. The view factors between differential and finite surfaces are expressed as differential view factor dFdA1 → dA2 is expressed as Q· dA1 → dA2 cos 1 cos 2 dFdA1 → dA2 dA2 r 2 Q· The net radiation heat transfer from any surface i of a black enclosure is determined by adding up the net radiation heat transfers from surface i to each of the surfaces of the enclosure: dA1 cos rcos dA Q· cos cos 1 · dA dA A r Q FdA1 → A2 F12 F A1 → A2 1 2 2 2 A2 A1 → A2 A1 1 1 A2 A1 2 2 1 unity. This is known as the summation rule for an enclosure. The superposition rule is expressed as the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j. The symmetry rule is expressed as if the surfaces j and k are symmetric about the surface i then Fi → j Fi → k. The rate of net radiation heat transfer between two black surfaces is determined from · Q 1 → 2 A1 F1 → 2 (T14 T24) (W) 2 where r is the distance between dA1 and dA2, and 1 and 2 are the angles between the normals of the surfaces and the line that connects dA1 and dA2. The view factor Fi → i represents the fraction of the radiation leaving surface i that strikes itself directly; Fi → i 0 for plane or convex surfaces and Fi → i 0 for concave surfaces. For view factors, the reciprocity rule is expressed as Ai Fi → j Aj Fj → i The sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including to itself, must equal · Qi N j1 · Qi → j N AF i j1 i→j (Ti4 Tj4) (W) The total radiation energy leaving a surface per unit time and per unit area is called the radiosity and is denoted by J. The net rate of radiation heat transfer from a surface i of surface area Ai is expressed as Ebi Ji · Qi Ri where Ri 1 i Ai i (W) cen58933_ch12.qxd 9/9/2002 9:49 AM Page 654 654 HEAT TRANSFER is the surface resistance to radiation. The net rate of radiation heat transfer from surface i to surface j can be expressed as Ji Jj · Qi → j Tf Tth (W) Ri → j where Ri → j The radiation effect in temperature measurements can be properly accounted for by the relation 1 Ai Fi → j is the space resistance to radiation. The network method is applied to radiation enclosure problems by drawing a surface resistance associated with each surface of an enclosure and connecting them with space resistances. Then the problem is solved by treating it as an electrical network problem where the radiation heat transfer replaces the current and the radiosity replaces the potential. The direct method is based on the following two equations: N Surfaces with specified · Fi → j(Ji Jj) · Q i Ai net heat transfer rate Q i j1 1 i Surfaces with specified Ti4 Ji Fi → j(Ji Jj) i temperature Ti j1 th (Tth4 Tw4) h (K) where Tf is the actual temperature of the fluid, Tth is the temperature value measured by the thermometer, and Tw is the temperature of the surrounding walls, all in K. Gases with asymmetric molecules such as H2O, CO2 CO, SO2, and hydrocarbons HnCm participate in the radiation process by absorption and emission. The spectral transmissivity, absorptivity, and emissivity of a medium are expressed as 1 1 e L, e L, 1 e L and where is the spectral absorption coefficient of the medium. The emissivities of H2O and CO2 gases are given in Figure 12–36 for a total pressure of P 1 atm. Emissivities at other pressures are determined from w Cw w, 1 atm and c Cc c, 1 atm N The first group (for surfaces with specified heat transfer rates) and the second group (for surfaces with specified temperatures) of equations give N linear algebraic equations for the determination of the N unknown radiosities for an N-surface enclosure. Once the radiosities J1, J2, . . . , JN are available, the unknown surface temperatures and heat transfer rates can be determined from the equations just shown. The net rate of radiation transfer between any two gray, diffuse, opaque surfaces that form an enclosure is given by · Q 12 (T14 T24) 1 1 1 2 1 A1 1 A1 F12 A2 2 (W) A (T14 T24) 1 1 1 1 1 1 1 1 · · · 1 1 2 3, 1 N, 1 N, 2 g c w Cc c, 1 atm Cw w, 1 atm where is the emissivity correction factor, which accounts for the overlap of emission bands. The gas absorptivities for radiation emitted by a source at temperature Ts are determined similarly from g c w where at the source temperature Ts and CO2: H2O: Radiation heat transfer between two surfaces can be reduced greatly by inserting between the two surfaces thin, highreflectivity (low-emissivity) sheets of material called radiation shields. Radiation heat transfer between two large parallel plates separated by N radiation shields is · Q 12, N shields where Cw and Cc are the pressure correction factors. For gas mixtures that contain both of H2O and CO2, the emissivity is determined from 3, 2 c Cc (Tg / Ts)0.65 c(Ts, Pc LTs / Tg) w Cw (Tg / Ts)0.45 w(Ts, Pw LTs / Tg) The rate of radiation heat transfer between a gas and a surrounding surface is Black enclosure: Gray enclosure, with s 0.7: · Q net As ( gTg4 gTs4) s 1 · Q net, gray As ( gTg4 gTs4) 2 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 655 655 CHAPTER 12 REFERENCES AND SUGGESTED READING 1. D. K. Edwards. Radiation Heat Transfer Notes. Washington, D.C.: Hemisphere, 1981. 2. D. K. Edwards and R. Matavosian. “Scaling Rules for Total Absorptivity and Emissivity of Gases.” Journal of Heat Transfer 106 (1984), pp. 684–689. H. C. Hottel and R. B. Egbert. “Radiant Heat Transmission from Water Vapor.” Transactions of the AIChE 38 (1942), pp. 531–565. 9. J. R. Howell. A Catalog of Radiation Configuration Factors. New York: McGraw-Hill, 1982. D. K. Edwards and R. Matavosian. “Emissivity Data for Gases.” Section 5.5.5, in Hemisphere Handbook of Heat Exchanger Design, ed. G. F. Hewitt. New York: Hemisphere, 1990. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. D. C. Hamilton and W. R. Morgan. “Radiation Interchange Configuration Factors.” National Advisory Committee for Aeronautics, Technical Note 2836, 1952. M. F. Modest. Radiative Heat Transfer. New York: McGraw-Hill, 1993. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. J. P. Holman. Heat Transfer. 9th ed. New York: McGraw-Hill, 2002. A. K. Oppenheim. “Radiation Analysis by the Network Method.” Transactions of the ASME 78 (1956), pp. 725–735. H. C. Hottel. “Radiant Heat Transmission.” In Heat Transmission, ed. W. H. McAdams. 3rd ed. New York: McGraw-Hill, 1954. R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. 3rd ed. Washington, D.C.: Hemisphere, 1992. H. C. Hottel. “Heat Transmission by Radiation from Non-luminous Gases,” Transaction of the AIChE (1927), pp. 173–205. N. V. Suryanaraya. Engineering Heat Transfer. St. Paul, Minn.: West, 1995. PROBLEMS The View Factor 12–1C What does the view factor represent? When is the view factor from a surface to itself not zero? 12–2C How can you determine the view factor F12 when the view factor F21 and the surface areas are available? 12–6 Consider an enclosure consisting of five surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of the reciprocity and summation rules? 12–3C What are the summation rule and the superposition rule for view factors? 12–7 Consider an enclosure consisting of 12 surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of the reciprocity and the summation rules? Answers: 144, 78 12–4C What is the crossed-strings method? For what kind of geometries is the crossed-strings method applicable? 12–8 Determine the view factors F13 and F23 between the rectangular surfaces shown in Figure P12–8. 12–5 Consider an enclosure consisting of six surfaces. How many view factors does this geometry involve? How many of these view factors can be determined by the application of the reciprocity and the summation rules? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 2m A2 1m A1 1m A3 1m FIGURE P12–8 12–9 Consider a cylindrical enclosure whose height is twice the diameter of its base. Determine the view factor from the side surface of this cylindrical enclosure to its base surface. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 656 656 HEAT TRANSFER b 2 3 FIGURE P12–11 (a) Semicylindrical duct. (b) Triangular duct. (c) Rectangular duct. 2 2 1 1 a a D (a) b 50° 1 50° (b) (c) a D b b b a (a) (b) (c) FIGURE P12–12 (a) Semicylindrical groove. (b) Triangular groove. (c) Rectangular groove. 12–10 Consider a hemispherical furnace with a flat circular base of diameter D. Determine the view factor from the dome Answer: 0.5 of this furnace to its base. 12–15 Determine the four view factors associated with an enclosure formed by two very long concentric cylinders of radii r1 and r2. Neglect the end effects. 12–11 Determine the view factors F12 and F21 for the very long ducts shown in Figure P12–11 without using any view factor tables or charts. Neglect end effects. 12–16 Determine the view factor F12 between the rectangular surfaces shown in Figure P12–16. 12–12 Determine the view factors from the very long grooves shown in Figure P12–12 to the surroundings without using any view factor tables or charts. Neglect end effects. 12–13 Determine the view factors from the base of a cube to each of the other five surfaces. 12–14 Consider a conical enclosure of height h and base diameter D. Determine the view factor from the conical side surface to a hole of diameter d located at the center of the base. 12–17 Two infinitely long parallel cylinders of diameter D are located a distance s apart from each other. Determine the view factor F12 between these two cylinders. 12–18 Three infinitely long parallel cylinders of diameter D are located a distance s apart from each other. Determine the view factor between the cylinder in the middle and the surroundings. h 1 s d D FIGURE P12–14 s D D D FIGURE P12–18 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 657 657 CHAPTER 12 1 1m 1m 1 1m 1 1m 2m 1m 1m 1m 1m 2 1m 1m 3m (a) 2m 1m 2 1m 3m (b) 2 2m FIGURE P12–16 (c) Radiation Heat Transfer between Surfaces 12–19C Why is the radiation analysis of enclosures that consist of black surfaces relatively easy? How is the rate of radiation heat transfer between two surfaces expressed in this case? 12–20C How does radiosity for a surface differ from the emitted energy? For what kind of surfaces are these two quantities identical? ture vary from 500 K to 1000 K and the emissivity from 0.1 to 0.9. Plot the net rate of radiation heat transfer as functions of temperature and emissivity, and discuss the results. 12–28 A furnace is of cylindrical shape with R H 2 m. The base, top, and side surfaces of the furnace are all black and are maintained at uniform temperatures of 500, 700, and 1200 K, respectively. Determine the net rate of radiation heat transfer to or from the top surface during steady operation. 12–21C What are the radiation surface and space resistances? How are they expressed? For what kind of surfaces is the radiation surface resistance zero? 12–22C What are the two methods used in radiation analysis? How do these two methods differ? 1 3 12–23C What is a reradiating surface? What simplifications does a reradiating surface offer in the radiation analysis? 12–24E Consider a 10-ft 10-ft 10-ft cubical furnace whose top and side surfaces closely approximate black surfaces and whose base surface has an emissivity 0.7. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 R, 1600 R, and 2400 R, respectively. Determine the net rate of radiation heat transfer between (a) the base and the side surfaces and (b) the base and the top surfaces. Also, determine the net rate of radiation heat transfer to the base surface. Reconsider Problem 12–24E. Using EES (or other) software, investigate the effect of base surface emissivity on the net rates of radiation heat transfer between the base and the side surfaces, between the base and top surfaces, and to the base surface. Let the emissivity vary from 0.1 to 0.9. Plot the rates of heat transfer as a function of emissivity, and discuss the results. H=2n R=2m 2 FIGURE P12–28 12–29 Consider a hemispherical furnace of diameter D 5 m with a flat base. The dome of the furnace is black, and the base has an emissivity of 0.7. The base and the dome of the furnace are maintained at uniform temperatures of 400 and 1000 K, respectively. Determine the net rate of radiation heat transfer from the dome to the base surface during steady operation. Answer: 759 kW 12–25E 12–26 Two very large parallel plates are maintained at uniform temperatures of T1 600 K and T2 400 K and have emissivities 1 0.5 and 2 0.9, respectively. Determine the net rate of radiation heat transfer between the two surfaces per unit area of the plates. 12–27 Reconsider Problem 12–26. Using EES (or other) software, investigate the effects of the temperature and the emissivity of the hot plate on the net rate of radiation heat transfer between the plates. Let the tempera- Black 2 1 ε = 0.7 5m FIGURE P12–29 12–30 Two very long concentric cylinders of diameters D1 0.2 m and D2 0.5 m are maintained at uniform temperatures of T1 950 K and T2 500 K and have emissivities 1 1 and 2 0.7, respectively. Determine the net rate of radiation heat transfer between the two cylinders per unit length of the cylinders. 12–31 This experiment is conducted to determine the emissivity of a certain material. A long cylindrical rod of diameter cen58933_ch12.qxd 9/9/2002 9:49 AM Page 658 658 HEAT TRANSFER D1 0.01 m is coated with this new material and is placed in an evacuated long cylindrical enclosure of diameter D2 0.1 m and emissivity 2 0.95, which is cooled externally and maintained at a temperature of 200 K at all times. The rod is heated by passing electric current through it. When steady operating conditions are reached, it is observed that the rod is dissipating electric power at a rate of 8 W per unit of its length and its surface temperature is 500 K. Based on these measurements, determine the emissivity of the coating on the rod. 12–32E A furnace is shaped like a long semicylindrical duct of diameter D 15 ft. The base and the dome of the furnace have emissivities of 0.5 and 0.9 and are maintained at uniform temperatures of 550 and 1800 R, respectively. Determine the net rate of radiation heat transfer from the dome to the base surface per unit length during steady operation. 1 the net rate of radiation heat transfer between the floor and the ceiling of the furnace. 12–37 Two concentric spheres of diameters D1 0.3 m and D2 0.8 m are maintained at uniform temperatures T1 700 K and T2 400 K and have emissivities 1 0.5 and 2 0.7, respectively. Determine the net rate of radiation heat transfer between the two spheres. Also, determine the convection heat transfer coefficient at the outer surface if both the surrounding medium and the surrounding surfaces are at 30°C. Assume the emissivity of the outer surface is 0.35. 12–38 A spherical tank of diameter D 2 m that is filled with liquid nitrogen at 100 K is kept in an evacuated cubic enclosure whose sides are 3 m long. The emissivities of the spherical tank and the enclosure are 1 0.1 and 2 0.8, respectively. If the temperature of the cubic enclosure is measured to be 240 K, determine the net rate of radiation heat Answer: 228 W transfer to the liquid nitrogen. 2 D = 15 ft Liquid N2 FIGURE P12–32E 12–33 Two parallel disks of diameter D 0.6 m separated by L 0.4 m are located directly on top of each other. Both disks are black and are maintained at a temperature of 700 K. The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at T 300 K. Determine the net rate of radiation heat transfer Answer: 5505 W from the disks to the environment. 12–34 A furnace is shaped like a long equilateral-triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is 1 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure? 12–35 Reconsider Problem 12–34. Using EES (or other) software, investigate the effects of the rate of the heat transfer at the base surface and the temperature of the side surfaces on the temperature of the base surface. Let the rate of heat transfer vary from 500 W/m2 to 1000 W/m2 and the temperature from 300 K to 700 K. Plot the temperature of the base surface as functions of the rate of heat transfer and the temperature of the side surfaces, and discuss the results. 12–36 Consider a 4-m 4-m 4-m cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of 550 K and 1100 K, respectively. Determine FIGURE P12–38 12–39 Repeat Problem 12–38 by replacing the cubic enclosure by a spherical enclosure whose diameter is 3 m. 12–40 Reconsider Problem 12–38. Using EES (or other) software, investigate the effects of the side length and the emissivity of the cubic enclosure, and the emissivity of the spherical tank on the net rate of radiation heat transfer. Let the side length vary from 2.5 m to 5.0 m and both emissivities from 0.1 to 0.9. Plot the net rate of radiation heat transfer as functions of side length and emissivities, and discuss the results. 12–41 Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 1100 K, while the wire mesh on top of the grill is covered with steaks initially at 5°C. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface. Answers: 1674 W, 3757 W cen58933_ch12.qxd 9/9/2002 9:49 AM Page 659 659 CHAPTER 12 Steaks T2 = 275 K ε2 = 1 3 T3 = 300 K ε3 = 1 D 0.20 m Coal bricks D 2 s 1 T1 = 425 K ε1 = 1 FIGURE P12–44 FIGURE P12–41 12–42E A 19–ft-high room with a base area of 12 ft 12 ft is to be heated by electric resistance heaters placed on the ceiling, which is maintained at a uniform temperature of 90°F at all times. The floor of the room is at 65°F and has an emissivity of 0.8. The side surfaces are well insulated. Treating the ceiling as a blackbody, determine the rate of heat loss from the room through the floor. 12–43 Consider two rectangular surfaces perpendicular to each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is black and is maintained at 550 K. The back sides of the surfaces are insulated. The surrounding surfaces are at 290 K, and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. T2 = 550 K ε2 = 1 W = 1.6 m L2 = 1.2 m L1 = 0.8 m A2 2 A1 1 12–45 Consider a long semicylindrical duct of diameter 1.0 m. Heat is supplied from the base surface, which is black, at a rate of 1200 W/m2, while the side surface with an emissivity of 0.4 are is maintained at 650 K. Neglecting the end effects, determine the temperature of the base surface. 12–46 Consider a 20-cm-diameter hemispherical enclosure. The dome is maintained at 600 K and heat is supplied from the dome at a rate of 50 W while the base surface with an emissivity is 0.55 is maintained at 400 K. Determine the emissivity of the dome. Radiation Shields and the Radiation Effect 12–47C What is a radiation shield? Why is it used? 12–48C What is the radiation effect? How does it influence the temperature measurements? 12–49C Give examples of radiation effects that affect human comfort. 12–50 Consider a person whose exposed surface area is 1.7 m2, emissivity is 0.85, and surface temperature is 30°C. Determine the rate of heat loss from that person by radiation in a large room whose walls are at a temperature of (a) 300 K and (b) 280 K. 3 T3 = 290 K ε3 = 0.85 T1 = 400 K ε1 = 0.75 FIGURE P12–43 12–44 Two long parallel 16-cm-diameter cylinders are located 50 cm apart from each other. Both cylinders are black, and are maintained at temperatures 425 K and 275 K. The surroundings can be treated as a blackbody at 300 K. For a 1-m-long section of the cylinders, determine the rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings. 12–51 A thin aluminum sheet with an emissivity of 0.15 on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures T1 900 K and T2 650 K and have emissivities 1 0.5 and 2 0.8, respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield. 1 T1 = 900 K ε1 = 0.5 3 ε3 = 0.15 2 T2 = 650 K ε2 = 0.8 FIGURE P12–51 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 660 660 HEAT TRANSFER 12–52 Reconsider Problem 12–51. Using EES (or other) software, plot the net rate of radiation heat transfer between the two plates as a function of the emissivity of the aluminum sheet as the emissivity varies from 0.05 to 0.25, and discuss the results. 12–53 Two very large parallel plates are maintained at uniform temperatures of T1 1000 K and T2 800 K and have emissivities of 1 2 0.2, respectively. It is desired to reduce the net rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum sheets with an emissivity of 0.15 on both sides between the plates. Determine the number of sheets that need to be inserted. 12–54 Five identical thin aluminum sheets with emissivities of 0.1 on both sides are placed between two very large parallel plates, which are maintained at uniform temperatures of T1 800 K and T2 450 K and have emissivities of 1 2 0.1, respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result to that without the shield. 12–55 Reconsider Problem 12–54. Using EES (or other) software, investigate the effects of the number of the aluminum sheets and the emissivities of the plates on the net rate of radiation heat transfer between the two plates. Let the number of sheets vary from 1 to 10 and the emissivities of the plates from 0.1 to 0.9. Plot the rate of radiation heat transfer as functions of the number of sheets and the emissivities of the plates, and discuss the results. 12–56E Two parallel disks of diameter D 3 ft separated by L 2 ft are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is 0.15. Both disks are black and are maintained at temperatures of 1200 R and 700 R, respectively. The environment that the disks are in can be considered to be a blackbody at 540 R. Determine the net rate of radiation heat transfer through the shield under Answer: 866 Btu/h steady conditions. 12–57 A radiation shield that has the same emissivity 3 on both sides is placed between two large parallel plates, which are maintained at uniform temperatures of T1 650 K and T2 400 K and have emissivities of 1 0.6 and 2 0.9, respectively. Determine the emissivity of the radiation shield if the radiation heat transfer between the plates is to be reduced to 15 percent of that without the radiation shield. 12–58 Reconsider Problem 12–57. Using EES (or other) software, investigate the effect of the percent reduction in the net rate of radiation heat transfer between the plates on the emissivity of the radiation shields. Let the percent reduction vary from 40 to 95 percent. Plot the emissivity versus the percent reduction in heat transfer, and discuss the results. 12–59 Two coaxial cylinders of diameters D1 0.10 m and D2 0.30 m and emissivities 1 0.7 and 2 0.4 are maintained at uniform temperatures of T1 750 K and T2 500 K, respectively. Now a coaxial radiation shield of diameter D3 0.20 m and emissivity 3 0.2 is placed between the two cylinders. Determine the net rate of radiation heat transfer between the two cylinders per unit length of the cylinders and compare the result with that without the shield. 12–60 Reconsider Problem 12–59. Using EES (or other) software, investigate the effects of the diameter of the outer cylinder and the emissivity of the radiation shield on the net rate of radiation heat transfer between the two cylinders. Let the diameter vary from 0.25 m to 0.50 m and the emissivity from 0.05 to 0.35. Plot the rate of radiation heat transfer as functions of the diameter and the emissivity, and discuss the results. Radiation Exchange with Absorbing and Emitting Gases 12–61C How does radiation transfer through a participating medium differ from that through a nonparticipating medium? 12–62C Define spectral transmissivity of a medium of thickness L in terms of (a) spectral intensities and (b) the spectral absorption coefficient. 12–63C Define spectral emissivity of a medium of thickness L in terms of the spectral absorption coefficient. 1 T1 = 1200 R 1 ft 3 ε3 = 0.15 1 ft 2 T2 = 700 R FIGURE P12–56E 12–64C How does the wavelength distribution of radiation emitted by a gas differ from that of a surface at the same temperature? 12–65 Consider an equimolar mixture of CO2 and O2 gases at 500 K and a total pressure of 0.5 atm. For a path length of 1.2 m, determine the emissivity of the gas. 12–66 A cubic furnace whose side length is 6 m contains combustion gases at 1000 K and a total pressure of 1 atm. The composition of the combustion gases is 75 percent N2, 9 percent H2O, 6 percent O2, and 10 percent CO2. Determine the effective emissivity of the combustion gases. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 661 661 CHAPTER 12 12–67 A cylindrical container whose height and diameter are 8 m is filled with a mixture of CO2 and N2 gases at 600 K and 1 atm. The partial pressure of CO2 in the mixture is 0.15 atm. If the walls are black at a temperature of 450 K, determine the rate of radiation heat transfer between the gas and the container walls. 12–68 Repeat Problem 12–67 by replacing CO2 by the H2O gas. 12–69 A 2-m-diameter spherical furnace contains a mixture of CO2 and N2 gases at 1200 K and 1 atm. The mole fraction of CO2 in the mixture is 0.15. If the furnace wall is black and its temperature is to be maintained at 600 K, determine the net rate of radiation heat transfer between the gas mixture and the furnace walls. 12–70 A flow-through combustion chamber consists of 15-cm diameter long tubes immersed in water. Compressed air is routed to the tube, and fuel is sprayed into the compressed air. The combustion gases consist of 70 percent N2, 9 percent H2O, 15 percent O2, and 6 percent CO2, and are maintained at 1 atm and 1500 K. The tube surfaces are near black, with an emissivity of 0.9. If the tubes are to be maintained at a temperature of 600 K, determine the rate of heat transfer from combustion gases to tube wall by radiation per m length of tube. 12–71 Repeat Problem 12–70 for a total pressure of 3 atm. 12–72 In a cogeneration plant, combustion gases at 1 atm and 800 K are used to preheat water by passing them through 6-m-long 10-cm-diameter tubes. The inner surface of the tube is black, and the partial pressures of CO2 and H2O in combustion gases are 0.12 atm and 0.18 atm, respectively. If the tube temperature is 500 K, determine the rate of radiation heat transfer from the gases to the tube. 12–73 A gas at 1200 K and 1 atm consists of 10 percent CO2, 10 percent H2O, 10 percent N2, and 70 percent N2 by volume. The gas flows between two large parallel black plates maintained at 600 K. If the plates are 20 cm apart, determine the rate of heat transfer from the gas to each plate per unit surface area. Special Topic: Heat Transfer from the Human Body 12–74C Consider a person who is resting or doing light work. Is it fair to say that roughly one-third of the metabolic heat generated in the body is dissipated to the environment by convection, one-third by evaporation, and the remaining onethird by radiation? 12–75C What is sensible heat? How is the sensible heat loss from a human body affected by (a) skin temperature, (b) environment temperature, and (c) air motion? 12–76C What is latent heat? How is the latent heat loss from the human body affected by (a) skin wettedness and (b) relative humidity of the environment? How is the rate of evaporation from the body related to the rate of latent heat loss? 12–77C How is the insulating effect of clothing expressed? How does clothing affect heat loss from the body by convection, radiation, and evaporation? How does clothing affect heat gain from the sun? 12–78C Explain all the different mechanisms of heat transfer from the human body (a) through the skin and (b) through the lungs. 12–79C What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature? 12–80 The convection heat transfer coefficient for a clothed person while walking in still air at a velocity of 0.5 to 2 m/s is given by h 8.60.53, where is in m/s and h is in W/m2 · °C. Plot the convection coefficient against the walking velocity, and compare the convection coefficients in that range to the average radiation coefficient of about 5 W/m2 · °C. 12–81 A clothed or unclothed person feels comfortable when the skin temperature is about 33°C. Consider an average man wearing summer clothes whose thermal resistance is 0.7 clo. The man feels very comfortable while standing in a room maintained at 20°C. If this man were to stand in that room unclothed, determine the temperature at which the room must be maintained for him to feel thermally comfortable. Assume the latent heat loss from the person to remain the same. Answer: 26.4°C 12–82E An average person produces 0.50 lbm of moisture while taking a shower and 0.12 lbm while bathing in a tub. Consider a family of four who shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be 1050 Btu/lbm, determine the contribution of showers to the latent heat load of the air conditioner in summer per day. 12–83 An average (1.82 kg or 4.0 lbm) chicken has a basal metabolic rate of 5.47 W and an average metabolic rate of 10.2 W (3.78 W sensible and 6.42 W latent) during normal activity. If there are 100 chickens in a breeding room, determine the rate of total heat generation and the rate of moisture production in the room. Take the heat of vaporization of water to be 2430 kJ/kg. 12–84 Consider a large classroom with 150 students on a hot summer day. All the lights with 4.0 kW of rated power are kept on. The room has no external walls, and thus heat gain through the walls and the roof is negligible. Chilled air is available at 15°C, and the temperature of the return air is not to exceed 25°C. Determine the required flow rate of air, in kg/s, that Answer: 1.45 kg/s needs to be supplied to the room. 12–85 A person feels very comfortable in his house in light clothing when the thermostat is set at 22°C and the mean radiation temperature (the average temperature of the surrounding surfaces) is also 22°C. During a cold day, the average mean radiation temperature drops to 18°C. To what level must the cen58933_ch12.qxd 9/9/2002 9:49 AM Page 662 662 HEAT TRANSFER Thermocouple Tth = 850 K 22°C ε = 0.6 Air 22°C Tw = 500 K FIGURE P12–88 0.6 and the convection heat transfer coefficient to be h 60 W/m2 · °C, determine the actual temperature of air. Answer: 1111 K FIGURE P12–85 indoor air temperature be raised to maintain the same level of comfort in the same clothing? 12–86 Repeat Problem 12–85 for a mean radiation temperature of 12°C. 12–87 A car mechanic is working in a shop whose interior space is not heated. Comfort for the mechanic is provided by two radiant heaters that radiate heat at a total rate of 10 kJ/s. About 5 percent of this heat strikes the mechanic directly. The shop and its surfaces can be assumed to be at the ambient temperature, and the emissivity and absorptivity of the mechanic can be taken to be 0.95 and the surface area to be 1.8 m2. The mechanic is generating heat at a rate of 350 W, half of which is latent, and is wearing medium clothing with a thermal resistance of 0.7 clo. Determine the lowest ambient temperature in which the mechanic can work comfortably. Radiant heater 12–89 A thermocouple shielded by aluminum foil of emissivity 0.15 is used to measure the temperature of hot gases flowing in a duct whose walls are maintained at Tw 380 K. The thermometer shows a temperature reading of Tth 530 K. Assuming the emissivity of the thermocouple junction to be 0.7 and the convection heat transfer coefficient to be h 120 W/m2 · °C, determine the actual temperature of the gas. What would the thermometer reading be if no radiation shield was used? 12–90E Consider a sealed 8-in.-high electronic box whose base dimensions are 12 in. 12 in. placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 130°F, determine the highest temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surAnswer: 43°F face of the box to the stand to be negligible. 12 in. 12 in. 100 W ε = 0.95 Ts = 130°F Electronic box 8 in. Stand FIGURE P12–87 Review Problems 12–88 A thermocouple used to measure the temperature of hot air flowing in a duct whose walls are maintained at Tw 500 K shows a temperature reading of Tth 850 K. Assuming the emissivity of the thermocouple junction to be FIGURE P12–90E 12–91 A 2-m-internal-diameter double-walled spherical tank is used to store iced water at 0°C. Each wall is 0.5 cm thick, and the 1.5-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of 0.15. The temperature of the outer cen58933_ch12.qxd 9/9/2002 9:49 AM Page 663 663 CHAPTER 12 0°C 20°C ε = 0.15 Iced water Vacuum 2m 1.5 cm 0.5 cm 0.5 cm FIGURE P12–91 wall of the tank is measured to be 20°C. Assuming the inner wall of the steel tank to be at 0°C, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-h period. 12–92 Two concentric spheres of diameters D1 15 cm and D2 25 cm are separated by air at 1 atm pressure. The surface temperatures of the two spheres enclosing the air are T1 350 K and T2 275 K, respectively, and their emissivities are 0.5. Determine the rate of heat transfer from the inner sphere to the outer sphere by (a) natural convection and (b) radiation. Water is heated as it flows through the tube, and the annular space between the aluminum and the glass tube is filled with air at 0.5 atm pressure. The pump circulating the water fails during a clear day, and the water temperature in the tube starts rising. The aluminum tube absorbs solar radiation at a rate of 30 Btu/h per foot length, and the temperature of the ambient air outside is 75°F. The emissivities of the tube and the glass cover are 0.9. Taking the effective sky temperature to be 60°F, determine the temperature of the aluminum tube when thermal equilibrium is established (i.e., when the rate of heat loss from the tube equals the amount of solar energy gained by the tube). 12–95 A vertical 2-m-high and 3-m-wide double-pane window consists of two sheets of glass separated by a 5-cm-thick air gap. In order to reduce heat transfer through the window, the air space between the two glasses is partially evacuated to 0.3 atm pressure. The emissivities of the glass surfaces are 0.9. Taking the glass surface temperatures across the air gap to be 15°C and 5°C, determine the rate of heat transfer through the window by natural convection and radiation. 2m 15°C 5°C 12–93 Consider a 1.5-m-high and 3-m-wide solar collector that is tilted at an angle 20° from the horizontal. The distance between the glass cover and the absorber plate is 3 cm, and the back side of the absorber is heavily insulated. The absorber plate and the glass cover are maintained at temperatures of 80°C and 32°C, respectively. The emissivity of the glass surface is 0.9 and that of the absorber plate is 0.8. Determine the rate of heat loss from the absorber plate by natural convection Answers: 750 W, 1289 W and radiation. Glass 1.5 cm Frame FIGURE P12–95 12–96 Glass cover Solar radiation 32°C A simple solar collector is built by placing a 6-cm-diameter clear plastic tube around a garden hose whose outer diameter is 2 cm. The hose is painted black to maximize solar absorption, and some plastic rings are used to keep the spacing between the hose and the clear plastic cover constant. The emissivities of the hose surface and the glass cover are 0.9, and the effective sky temperature is 80°C Absorber plate Solar radiation Air space θ = 20° Tsky = 15°C 25°C Insulation Clear plastic tube 40°C Water FIGURE P12–93 12–94E A solar collector consists of a horizontal aluminum tube having an outer diameter of 2.5 in. enclosed in a concentric thin glass tube of diameter 5 in. Spacer Garden hose FIGURE P12–96 cen58933_ch12.qxd 9/9/2002 9:49 AM Page 664 664 HEAT TRANSFER estimated to be 15°C. The temperature of the plastic tube is measured to be 40°C, while the ambient air temperature is 25°C. Determine the rate of heat loss from the water in the hose by natural convection and radiation per meter of its length under steady conditions. Answers: 5.2 W, 26.2 W 12–97 A solar collector consists of a horizontal copper tube of outer diameter 5 cm enclosed in a concentric thin glass tube of diameter 9 cm. Water is heated as it flows through the tube, and the annular space between the copper and the glass tubes is filled with air at 1 atm pressure. The emissivities of the tube surface and the glass cover are 0.85 and 0.9, respectively. During a clear day, the temperatures of the tube surface and the glass cover are measured to be 60°C and 40°C, respectively. Determine the rate of heat loss from the collector by natural convection and radiation per meter length of the tube. 12–98 A furnace is of cylindrical shape with a diameter of 1.2 m and a length of 1.2 m. The top surface has an emissivity of 0.70 and is maintained at 500 K. The bottom surface has an emissivity of 0.50 and is maintained at 650 K. The side surface has an emissivity of 0.40. Heat is supplied from the base surface at a net rate of 1400 W. Determine the temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces. 12–100 A thin aluminum sheet with an emissivity of 0.12 on both sides is placed between two very large parallel plates maintained at uniform temperatures of T1 750 K and T2 550 K. The emissivities of the plates are 1 0.8 and 2 0.9. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates, and the temperature of the radiation shield in steady operation. 12–101 Two thin radiation shields with emissivities of 3 0.10 and 4 0.15 on both sides are placed between two very large parallel plates, which are maintained at uniform temperatures T1 600 K and T2 300 K and have emissivities 1 0.6 and 2 0.7, respectively. Determine the net rates of radiation heat transfer between the two plates with and without the shields per unit surface area of the plates, and the temperatures of the radiation shields in steady operation. T1 = 600 K ε1 = 0.6 ε3 = 0.10 T2 = 300 K ε2 = 0.7 ε4 = 0.15 FIGURE P12–101 T1 = 500 K ε1 = 0.70 r1 = 0.6 m h = 1.2 m T3 ε3 = 0.40 T2 = 650 K ε2 = 0.50 r2 = 0.6 m FIGURE P12–98 12–99 Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has an emissivity of 0.90 and is maintained at 950 K. The side surface is black and is maintained at 450 K. Heat is supplied from the base surface at a rate of 340 kW. Determine the emissivity of the top surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces. 12–102 In a natural-gas fired boiler, combustion gases pass through 6-m-long 15-cm-diameter tubes immersed in water at 1 atm pressure. The tube temperature is measured to be 105°C, and the emissivity of the inner surfaces of the tubes is estimated to be 0.9. Combustion gases enter the tube at 1 atm and 1200 K at a mean velocity of 3 m/s. The mole fractions of CO2 and H2O in combustion gases are 8 percent and 16 percent, respectively. Assuming fully developed flow and using properties of air for combustion gases, determine (a) the rates of heat transfer by convection and by radiation from the combustion gases to the tube wall and (b) the rate of evaporation of water. 12–103 Repeat Problem 12–102 for a total pressure of 3 atm for the combustion gases. Computer, Design, and Essay Problems 12–104 Consider an enclosure consisting of N diffuse and gray surfaces. The emissivity and temperature of each surface as well as all the view factors between the surfaces are specified. Write a program to determine the net rate of radiation heat transfer for each surface. 12–105 Radiation shields are commonly used in the design of superinsulations for use in space and cryogenic applications. Write an essay on superinsulations and how they are used in different applications. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 665 665 CHAPTER 12 12–106 Thermal comfort in a house is strongly affected by the so-called radiation effect, which is due to radiation heat transfer between the person and surrounding surfaces. A person feels much colder in the morning, for example, because of the lower surface temperature of the walls at that time, although the thermostat setting of the house is fixed. Write an essay on the radiation effect, how it affects human comfort, and how it is accounted for in heating and air-conditioning applications. cen58933_ch12.qxd 9/9/2002 9:49 AM Page 666 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 667 CHAPTER H E AT E X C H A N G E R S eat exchangers are devices that facilitate the exchange of heat between two fluids that are at different temperatures while keeping them from mixing with each other. Heat exchangers are commonly used in practice in a wide range of applications, from heating and air-conditioning systems in a household, to chemical processing and power production in large plants. Heat exchangers differ from mixing chambers in that they do not allow the two fluids involved to mix. In a car radiator, for example, heat is transferred from the hot water flowing through the radiator tubes to the air flowing through the closely spaced thin plates outside attached to the tubes. Heat transfer in a heat exchanger usually involves convection in each fluid and conduction through the wall separating the two fluids. In the analysis of heat exchangers, it is convenient to work with an overall heat transfer coefficient U that accounts for the contribution of all these effects on heat transfer. The rate of heat transfer between the two fluids at a location in a heat exchanger depends on the magnitude of the temperature difference at that location, which varies along the heat exchanger. In the analysis of heat exchangers, it is usually convenient to work with the logarithmic mean temperature difference LMTD, which is an equivalent mean temperature difference between the two fluids for the entire heat exchanger. Heat exchangers are manufactured in a variety of types, and thus we start this chapter with the classification of heat exchangers. We then discuss the determination of the overall heat transfer coefficient in heat exchangers, and the LMTD for some configurations. We then introduce the correction factor F to account for the deviation of the mean temperature difference from the LMTD in complex configurations. Next we discuss the effectiveness–NTU method, which enables us to analyze heat exchangers when the outlet temperatures of the fluids are not known. Finally, we discuss the selection of heat exchangers. H 13 CONTENTS 13–1 Types of Heat Exchangers 668 13–2 The Overall Heat Transfer Coefficient 671 13–3 Analysis of Heat Exchangers 678 13–4 The Log Mean Temperature Difference Method 680 13–5 The Effectiveness–NTU Method 690 13–6 Selection of Heat Exchangers 700 667 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 668 668 HEAT TRANSFER 13–1 ■ TYPES OF HEAT EXCHANGERS Different heat transfer applications require different types of hardware and different configurations of heat transfer equipment. The attempt to match the heat transfer hardware to the heat transfer requirements within the specified constraints has resulted in numerous types of innovative heat exchanger designs. The simplest type of heat exchanger consists of two concentric pipes of different diameters, as shown in Figure 13–1, called the double-pipe heat exchanger. One fluid in a double-pipe heat exchanger flows through the smaller pipe while the other fluid flows through the annular space between the two pipes. Two types of flow arrangement are possible in a double-pipe heat exchanger: in parallel flow, both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction. In counter flow, on the other hand, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions. Another type of heat exchanger, which is specifically designed to realize a large heat transfer surface area per unit volume, is the compact heat exchanger. The ratio of the heat transfer surface area of a heat exchanger to its volume is called the area density . A heat exchanger with 700 m2/m3 (or 200 ft2/ft3) is classified as being compact. Examples of compact heat exchangers are car radiators ( 1000 m2/m3), glass ceramic gas turbine heat exchangers ( 6000 m2/m3), the regenerator of a Stirling engine ( 15,000 m2/m3), and the human lung ( 20,000 m2/m3). Compact heat exchangers enable us to achieve high heat transfer rates between two fluids in T T Ho t fl Hot flui Cold d Co ld f FIGURE 13–1 Different flow regimes and associated temperature profiles in a double-pipe heat exchanger. Cold in Hot out Cold in (a) Parallel flow luid fluid Cold out Hot in uid Hot out Hot in Cold out (b) Counter flow cen58933_ch13.qxd 9/9/2002 9:57 AM Page 669 669 CHAPTER 13 a small volume, and they are commonly used in applications with strict limitations on the weight and volume of heat exchangers (Fig. 13–2). The large surface area in compact heat exchangers is obtained by attaching closely spaced thin plate or corrugated fins to the walls separating the two fluids. Compact heat exchangers are commonly used in gas-to-gas and gas-toliquid (or liquid-to-gas) heat exchangers to counteract the low heat transfer coefficient associated with gas flow with increased surface area. In a car radiator, which is a water-to-air compact heat exchanger, for example, it is no surprise that fins are attached to the air side of the tube surface. In compact heat exchangers, the two fluids usually move perpendicular to each other, and such flow configuration is called cross-flow. The cross-flow is further classified as unmixed and mixed flow, depending on the flow configuration, as shown in Figure 13–3. In (a) the cross-flow is said to be unmixed since the plate fins force the fluid to flow through a particular interfin spacing and prevent it from moving in the transverse direction (i.e., parallel to the tubes). The cross-flow in (b) is said to be mixed since the fluid now is free to move in the transverse direction. Both fluids are unmixed in a car radiator. The presence of mixing in the fluid can have a significant effect on the heat transfer characteristics of the heat exchanger. FIGURE 13–2 A gas-to-liquid compact heat exchanger for a residential airconditioning system. Cross-flow (unmixed) Cross-flow (mixed) Tube flow (unmixed) (a) Both fluids unmixed Tube flow (unmixed) (b) One fluid mixed, one fluid unmixed FIGURE 13–3 Different flow configurations in cross-flow heat exchangers. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 670 670 HEAT TRANSFER Tube outlet FIGURE 13–4 The schematic of a shell-and-tube heat exchanger (one-shell pass and one-tube pass). Shell inlet Baffles Front-end header Rear-end header Tubes Shell Shell outlet Shell-side fluid In Tube-side fluid Out In Out (a) One-shell pass and two-tube passes Shell-side fluid In Out Tubeside fluid In Out (b) Two-shell passes and four-tube passes FIGURE 13–5 Multipass flow arrangements in shelland-tube heat exchangers. Tube inlet Perhaps the most common type of heat exchanger in industrial applications is the shell-and-tube heat exchanger, shown in Figure 13–4. Shell-and-tube heat exchangers contain a large number of tubes (sometimes several hundred) packed in a shell with their axes parallel to that of the shell. Heat transfer takes place as one fluid flows inside the tubes while the other fluid flows outside the tubes through the shell. Baffles are commonly placed in the shell to force the shell-side fluid to flow across the shell to enhance heat transfer and to maintain uniform spacing between the tubes. Despite their widespread use, shelland-tube heat exchangers are not suitable for use in automotive and aircraft applications because of their relatively large size and weight. Note that the tubes in a shell-and-tube heat exchanger open to some large flow areas called headers at both ends of the shell, where the tube-side fluid accumulates before entering the tubes and after leaving them. Shell-and-tube heat exchangers are further classified according to the number of shell and tube passes involved. Heat exchangers in which all the tubes make one U-turn in the shell, for example, are called one-shell-pass and twotube-passes heat exchangers. Likewise, a heat exchanger that involves two passes in the shell and four passes in the tubes is called a two-shell-passes and four-tube-passes heat exchanger (Fig. 13–5). An innovative type of heat exchanger that has found widespread use is the plate and frame (or just plate) heat exchanger, which consists of a series of plates with corrugated flat flow passages (Fig. 13–6). The hot and cold fluids flow in alternate passages, and thus each cold fluid stream is surrounded by two hot fluid streams, resulting in very effective heat transfer. Also, plate heat exchangers can grow with increasing demand for heat transfer by simply mounting more plates. They are well suited for liquid-to-liquid heat exchange applications, provided that the hot and cold fluid streams are at about the same pressure. Another type of heat exchanger that involves the alternate passage of the hot and cold fluid streams through the same flow area is the regenerative heat exchanger. The static-type regenerative heat exchanger is basically a porous mass that has a large heat storage capacity, such as a ceramic wire mesh. Hot and cold fluids flow through this porous mass alternatively. Heat is transferred from the hot fluid to the matrix of the regenerator during the flow of the hot fluid, and from the matrix to the cold fluid during the flow of the cold fluid. Thus, the matrix serves as a temporary heat storage medium. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 671 671 CHAPTER 13 FIGURE 13–6 A plate-and-frame liquid-to-liquid heat exchanger (courtesy of Trante Corp.). The dynamic-type regenerator involves a rotating drum and continuous flow of the hot and cold fluid through different portions of the drum so that any portion of the drum passes periodically through the hot stream, storing heat, and then through the cold stream, rejecting this stored heat. Again the drum serves as the medium to transport the heat from the hot to the cold fluid stream. Heat exchangers are often given specific names to reflect the specific application for which they are used. For example, a condenser is a heat exchanger in which one of the fluids is cooled and condenses as it flows through the heat exchanger. A boiler is another heat exchanger in which one of the fluids absorbs heat and vaporizes. A space radiator is a heat exchanger that transfers heat from the hot fluid to the surrounding space by radiation. 13–2 ■ Cold fluid Hot fluid Heat transfer Ti Hot fluid Ai hi THE OVERALL HEAT TRANSFER COEFFICIENT A heat exchanger typically involves two flowing fluids separated by a solid wall. Heat is first transferred from the hot fluid to the wall by convection, through the wall by conduction, and from the wall to the cold fluid again by convection. Any radiation effects are usually included in the convection heat transfer coefficients. The thermal resistance network associated with this heat transfer process involves two convection and one conduction resistances, as shown in Figure 13–7. Here the subscripts i and o represent the inner and outer surfaces of the Cold fluid Wall Ao ho To Ti 1 Ri = ––– hi A i Rwall 1 Ro = ––– ho Ao FIGURE 13–7 Thermal resistance network associated with heat transfer in a double-pipe heat exchanger. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 672 672 HEAT TRANSFER inner tube. For a double-pipe heat exchanger, we have Ai DiL and Ao DoL, and the thermal resistance of the tube wall in this case is Rwall R Rtotal Ri Rwall Ro Do Di Outer tube Outer fluid Inner fluid (13-1) where k is the thermal conductivity of the wall material and L is the length of the tube. Then the total thermal resistance becomes L Heat transfer ln (Do /Di ) 2kL Inner tube Ao = πDo L Ai = πDi L FIGURE 13–8 The two heat transfer surface areas associated with a double-pipe heat exchanger (for thin tubes, Di Do and thus Ai Ao). ln (Do /Di ) 1 1 hi Ai 2kL ho Ao (13-2) The Ai is the area of the inner surface of the wall that separates the two fluids, and Ao is the area of the outer surface of the wall. In other words, Ai and Ao are surface areas of the separating wall wetted by the inner and the outer fluids, respectively. When one fluid flows inside a circular tube and the other outside of it, we have Ai DiL and Ao DoL (Fig. 13–8). In the analysis of heat exchangers, it is convenient to combine all the thermal resistances in the path of heat flow from the hot fluid to the cold one into a single resistance R, and to express the rate of heat transfer between the two fluids as · T Q UA T Ui Ai T Uo Ao T R (13-3) where U is the overall heat transfer coefficient, whose unit is W/m2 · °C, which is identical to the unit of the ordinary convection coefficient h. Canceling T, Eq. 13-3 reduces to 1 1 1 1 1 R Rwall UAs Ui Ai Uo Ao hi Ai ho Ao (13-4) Perhaps you are wondering why we have two overall heat transfer coefficients Ui and Uo for a heat exchanger. The reason is that every heat exchanger has two heat transfer surface areas Ai and Ao, which, in general, are not equal to each other. Note that Ui Ai Uo Ao, but Ui Uo unless Ai Ao. Therefore, the overall heat transfer coefficient U of a heat exchanger is meaningless unless the area on which it is based is specified. This is especially the case when one side of the tube wall is finned and the other side is not, since the surface area of the finned side is several times that of the unfinned side. When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, as is usually the case, the thermal resistance of the tube is negligible (Rwall 0) and the inner and outer surfaces of the tube are almost identical (Ai Ao As). Then Eq. 13-4 for the overall heat transfer coefficient simplifies to 1 1 1 U hi ho (13-5) where U Ui Uo. The individual convection heat transfer coefficients inside and outside the tube, hi and ho, are determined using the convection relations discussed in earlier chapters. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 673 673 CHAPTER 13 The overall heat transfer coefficient U in Eq. 13-5 is dominated by the smaller convection coefficient, since the inverse of a large number is small. When one of the convection coefficients is much smaller than the other (say, hi ho), we have 1/hi 1/ho, and thus U hi. Therefore, the smaller heat transfer coefficient creates a bottleneck on the path of heat flow and seriously impedes heat transfer. This situation arises frequently when one of the fluids is a gas and the other is a liquid. In such cases, fins are commonly used on the gas side to enhance the product UAs and thus the heat transfer on that side. Representative values of the overall heat transfer coefficient U are given in Table 13–1. Note that the overall heat transfer coefficient ranges from about 10 W/m2 · °C for gas-to-gas heat exchangers to about 10,000 W/m2 · °C for heat exchangers that involve phase changes. This is not surprising, since gases have very low thermal conductivities, and phase-change processes involve very high heat transfer coefficients. When the tube is finned on one side to enhance heat transfer, the total heat transfer surface area on the finned side becomes As Atotal Afin Aunfinned (13-6) where Afin is the surface area of the fins and Aunfinned is the area of the unfinned portion of the tube surface. For short fins of high thermal conductivity, we can use this total area in the convection resistance relation Rconv 1/hAs since the fins in this case will be very nearly isothermal. Otherwise, we should determine the effective surface area A from As Aunfinned fin Afin (13-7) TABLE 13–1 Representative values of the overall heat transfer coefficients in heat exchangers Type of heat exchanger Water-to-water Water-to-oil Water-to-gasoline or kerosene Feedwater heaters Steam-to-light fuel oil Steam-to-heavy fuel oil Steam condenser Freon condenser (water cooled) Ammonia condenser (water cooled) Alcohol condensers (water cooled) Gas-to-gas Water-to-air in finned tubes (water in tubes) Steam-to-air in finned tubes (steam in tubes) Multiply the listed values by 0.176 to convert them to Btu/h · ft2 · °F. † Based on air-side surface area. ‡ Based on water- or steam-side surface area. U, W/m2 · °C 850–1700 100–350 300–1000 1000–8500 200–400 50–200 1000–6000 300–1000 800–1400 250–700 10–40 30–60† 400–850† 30–300† 400–4000‡ cen58933_ch13.qxd 9/9/2002 9:57 AM Page 674 674 HEAT TRANSFER where fin is the fin efficiency. This way, the temperature drop along the fins is accounted for. Note that fin 1 for isothermal fins, and thus Eq. 13-7 reduces to Eq. 13-6 in that case. Fouling Factor The performance of heat exchangers usually deteriorates with time as a result of accumulation of deposits on heat transfer surfaces. The layer of deposits represents additional resistance to heat transfer and causes the rate of heat transfer in a heat exchanger to decrease. The net effect of these accumulations on heat transfer is represented by a fouling factor Rf , which is a measure of the thermal resistance introduced by fouling. The most common type of fouling is the precipitation of solid deposits in a fluid on the heat transfer surfaces. You can observe this type of fouling even in your house. If you check the inner surfaces of your teapot after prolonged use, you will probably notice a layer of calcium-based deposits on the surfaces at which boiling occurs. This is especially the case in areas where the water is hard. The scales of such deposits come off by scratching, and the surfaces can be cleaned of such deposits by chemical treatment. Now imagine those mineral deposits forming on the inner surfaces of fine tubes in a heat exchanger (Fig. 13–9) and the detrimental effect it may have on the flow passage area and the heat transfer. To avoid this potential problem, water in power and process plants is extensively treated and its solid contents are removed before it is allowed to circulate through the system. The solid ash particles in the flue gases accumulating on the surfaces of air preheaters create similar problems. Another form of fouling, which is common in the chemical process industry, is corrosion and other chemical fouling. In this case, the surfaces are fouled by the accumulation of the products of chemical reactions on the surfaces. This form of fouling can be avoided by coating metal pipes with glass or using plastic pipes instead of metal ones. Heat exchangers may also be fouled by the growth of algae in warm fluids. This type of fouling is called biological fouling and can be prevented by chemical treatment. In applications where it is likely to occur, fouling should be considered in the design and selection of heat exchangers. In such applications, it may be FIGURE 13–9 Precipitation fouling of ash particles on superheater tubes (from Steam, Its Generation, and Use, Babcock and Wilcox Co., 1978). cen58933_ch13.qxd 9/9/2002 9:57 AM Page 675 675 CHAPTER 13 necessary to select a larger and thus more expensive heat exchanger to ensure that it meets the design heat transfer requirements even after fouling occurs. The periodic cleaning of heat exchangers and the resulting down time are additional penalties associated with fouling. The fouling factor is obviously zero for a new heat exchanger and increases with time as the solid deposits build up on the heat exchanger surface. The fouling factor depends on the operating temperature and the velocity of the fluids, as well as the length of service. Fouling increases with increasing temperature and decreasing velocity. The overall heat transfer coefficient relation given above is valid for clean surfaces and needs to be modified to account for the effects of fouling on both the inner and the outer surfaces of the tube. For an unfinned shell-and-tube heat exchanger, it can be expressed as Rf, i ln (Do /Di ) Rf, o 1 1 1 1 1 R UAs Ui Ai Uo Ao hi Ai Ai 2kL Ao ho Ao (13-8) where Ai Di L and Ao Do L are the areas of inner and outer surfaces, and Rf, i and Rf, o are the fouling factors at those surfaces. Representative values of fouling factors are given in Table 13–2. More comprehensive tables of fouling factors are available in handbooks. As you would expect, considerable uncertainty exists in these values, and they should be used as a guide in the selection and evaluation of heat exchangers to account for the effects of anticipated fouling on heat transfer. Note that most fouling factors in the table are of the order of 10 4 m2 · °C/W, which is equivalent to the thermal resistance of a 0.2-mm-thick limestone layer (k 2.9 W/m · °C) per unit surface area. Therefore, in the absence of specific data, we can assume the surfaces to be coated with 0.2 mm of limestone as a starting point to account for the effects of fouling. EXAMPLE 13–1 Overall Heat Transfer Coefficient of a Heat Exchanger Hot oil is to be cooled in a double-tube counter-flow heat exchanger. The copper inner tubes have a diameter of 2 cm and negligible thickness. The inner diameter of the outer tube (the shell) is 3 cm. Water flows through the tube at a rate of 0.5 kg/s, and the oil through the shell at a rate of 0.8 kg/s. Taking the average temperatures of the water and the oil to be 45°C and 80°C, respectively, determine the overall heat transfer coefficient of this heat exchanger. SOLUTION Hot oil is cooled by water in a double-tube counter-flow heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1 The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its thickness is negligible. 2 Both the oil and water flow are fully developed. 3 Properties of the oil and water are constant. Properties The properties of water at 45°C are (Table A–9) 990 kg/m3 k 0.637 W/m · °C Pr 3.91 / 0.602 10 6 m2/s TABLE 13–2 Representative fouling factors (thermal resistance due to fouling for a unit surface area) (Source: Tubular Exchange Manufacturers Association.) Fluid Distilled water, sea water, river water, boiler feedwater: Below 50°C Above 50°C Fuel oil Steam (oil-free) Refrigerants (liquid) Refrigerants (vapor) Alcohol vapors Air Rf , m2 · °C/W 0.0001 0.0002 0.0009 0.0001 0.0002 0.0004 0.0001 0.0004 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 676 676 HEAT TRANSFER The properties of oil at 80°C are (Table A–16). 852 kg/m3 k 0.138 W/m · °C Hot oil 0.8 kg/s Analysis The schematic of the heat exchanger is given in Figure 13–10. The overall heat transfer coefficient U can be determined from Eq. 13-5: Cold water 2 cm Pr 490 37.5 10 6 m2/s 1 1 1 U hi ho 3 cm 0.5 kg/s FIGURE 13–10 Schematic for Example 13–1. where hi and ho are the convection heat transfer coefficients inside and outside the tube, respectively, which are to be determined using the forced convection relations. The hydraulic diameter for a circular tube is the diameter of the tube itself, Dh D 0.02 m. The mean velocity of water in the tube and the Reynolds number are m m· m· 0.5 kg/s 1.61 m/s 1 Ac ( 4 D 2 ) (990 kg/m3)[14 (0.02 m)2] and Re m Dh (1.61 m/s)(0.02 m) 53,490 0.602 10 6 m2/s which is greater than 4000. Therefore, the flow of water is turbulent. Assuming the flow to be fully developed, the Nusselt number can be determined from Nu hDh 0.023 Re0.8 Pr 0.4 0.023(53,490)0.8(3.91)0.4 240.6 k Then, h k 0.637 W/m · °C Nu (240.6) 7663 W/m2 · °C Dh 0.02 m Now we repeat the analysis above for oil. The properties of oil at 80°C are 852 kg/m3 k 0.138 W/m · °C 37.5 10 6 m2/s Pr 490 The hydraulic diameter for the annular space is Dh Do Di 0.03 0.02 0.01 m TABLE 13–3 The mean velocity and the Reynolds number in this case are Nusselt number for fully developed laminar flow in a circular annulus with one surface insulated and the other isothermal (Kays and Perkins, Ref. 8.) Di /Do Nui Nuo 0.00 0.05 0.10 0.25 0.50 1.00 — 17.46 11.56 7.37 5.74 4.86 3.66 4.06 4.11 4.23 4.43 4.86 m · 0.8 kg/s m· m 1 2.39 m/s 1 2 2 3 Ac [ (Do D i )] (852 kg/m )[ (0.032 0.022)] m2 4 4 and Re m Dh (2.39 m/s)(0.01 m) 37.5 10 6 m2/s 637 which is less than 4000. Therefore, the flow of oil is laminar. Assuming fully developed flow, the Nusselt number on the tube side of the annular space Nui corresponding to Di /Do 0.02/0.03 0.667 can be determined from Table 13–3 by interpolation to be Nu 5.45 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 677 677 CHAPTER 13 and ho 0.138 W/m · °C k Nu (5.45) 75.2 W/m2 · °C Dh 0.01 m Then the overall heat transfer coefficient for this heat exchanger becomes U 1 1 74.5 W/m2 · °C 1 1 1 1 hi ho 7663 W/m2 · °C 75.2 W/m2 · °C Discussion Note that U ho in this case, since hi ho. This confirms our earlier statement that the overall heat transfer coefficient in a heat exchanger is dominated by the smaller heat transfer coefficient when the difference between the two values is large. To improve the overall heat transfer coefficient and thus the heat transfer in this heat exchanger, we must use some enhancement techniques on the oil side, such as a finned surface. EXAMPLE 13–2 Effect of Fouling on the Overall Heat Transfer Coefficient A double-pipe (shell-and-tube) heat exchanger is constructed of a stainless steel (k 15.1 W/m · °C) inner tube of inner diameter Di 1.5 cm and outer diameter Do 1.9 cm and an outer shell of inner diameter 3.2 cm. The convection heat transfer coefficient is given to be hi 800 W/m2 · °C on the inner surface of the tube and ho 1200 W/m2 · °C on the outer surface. For a fouling factor of Rf, i 0.0004 m2 · °C/ W on the tube side and Rf, o 0.0001 m2 · °C/ W on the shell side, determine (a) the thermal resistance of the heat exchanger per unit length and (b) the overall heat transfer coefficients, Ui and Uo based on the inner and outer surface areas of the tube, respectively. SOLUTION The heat transfer coefficients and the fouling factors on the tube Cold fluid and shell sides of a heat exchanger are given. The thermal resistance and the overall heat transfer coefficients based on the inner and outer areas are to be determined. Assumptions The heat transfer coefficients and the fouling factors are constant and uniform. Analysis (a) The schematic of the heat exchanger is given in Figure 13–11. The thermal resistance for an unfinned shell-and-tube heat exchanger with fouling on both heat transfer surfaces is given by Eq. 13-8 as Outer layer of fouling R Rf, i ln (Do /Di ) Rf, o 1 1 1 1 1 UAs Ui Ai Uo Ao hi Ai Ai Ao 2kL ho Ao where Ai Di L (0.015 m)(1 m) 0.0471 m2 Ao Do L (0.019 m)(1 m) 0.0597 m2 Substituting, the total thermal resistance is determined to be Tube wall Hot fluid Inner layer of fouling Cold fluid Hot fluid Di = 1.5 cm hi = 800 W/m2·°C Rf, i = 0.0004 m2·°C/ W Do = 1.9 cm ho = 1200 W/ m2·°C Rf, o = 0.0001 m2·°C/ W FIGURE 13–11 Schematic for Example 13–2. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 678 678 HEAT TRANSFER R 0.0004 m2 · °C/ W 1 (800 W/m2 · °C)(0.0471 m2) 0.0471 m2 ln (0.019/0.015) 2(15.1 W/m · °C)(1 m) 0.0001 m2 · °C/ W 1 0.0597 m2 (1200 W/m2 · °C)(0.0597 m2) (0.02654 0.00849 0.0025 0.00168 0.01396)°C/ W 0.0532°C/ W Note that about 19 percent of the total thermal resistance in this case is due to fouling and about 5 percent of it is due to the steel tube separating the two fluids. The rest (76 percent) is due to the convection resistances on the two sides of the inner tube. (b) Knowing the total thermal resistance and the heat transfer surface areas, the overall heat transfer coefficient based on the inner and outer surfaces of the tube are determined again from Eq. 13-8 to be Ui 1 1 399 W/m2 · °C RAi (0.0532 °C/ W)(0.0471 m2) Uo 1 1 315 W/m2 · °C RAo (0.0532 °C/ W)(0.0597 m2) and Discussion Note that the two overall heat transfer coefficients differ significantly (by 27 percent) in this case because of the considerable difference between the heat transfer surface areas on the inner and the outer sides of the tube. For tubes of negligible thickness, the difference between the two overall heat transfer coefficients would be negligible. 13–3 ■ ANALYSIS OF HEAT EXCHANGERS Heat exchangers are commonly used in practice, and an engineer often finds himself or herself in a position to select a heat exchanger that will achieve a specified temperature change in a fluid stream of known mass flow rate, or to predict the outlet temperatures of the hot and cold fluid streams in a specified heat exchanger. In upcoming sections, we will discuss the two methods used in the analysis of heat exchangers. Of these, the log mean temperature difference (or LMTD) method is best suited for the first task and the effectiveness–NTU method for the second task as just stated. But first we present some general considerations. Heat exchangers usually operate for long periods of time with no change in their operating conditions. Therefore, they can be modeled as steady-flow devices. As such, the mass flow rate of each fluid remains constant, and the fluid properties such as temperature and velocity at any inlet or outlet remain the same. Also, the fluid streams experience little or no change in their velocities and elevations, and thus the kinetic and potential energy changes are negligible. The specific heat of a fluid, in general, changes with temperature. But, in cen58933_ch13.qxd 9/9/2002 9:57 AM Page 679 679 CHAPTER 13 a specified temperature range, it can be treated as a constant at some average value with little loss in accuracy. Axial heat conduction along the tube is usually insignificant and can be considered negligible. Finally, the outer surface of the heat exchanger is assumed to be perfectly insulated, so that there is no heat loss to the surrounding medium, and any heat transfer occurs between the two fluids only. The idealizations stated above are closely approximated in practice, and they greatly simplify the analysis of a heat exchanger with little sacrifice of accuracy. Therefore, they are commonly used. Under these assumptions, the first law of thermodynamics requires that the rate of heat transfer from the hot fluid be equal to the rate of heat transfer to the cold one. That is, · Q m· cCpc(Tc, out Tc, in) (13-9) · Q m· hCph(Th, in Th, out) (13-10) and where the subscripts c and h stand for cold and hot fluids, respectively, and m· c, m· h mass flow rates Cpc, Cph specific heats Tc, out, Th, out outlet temperatures Tc, in, Th, in inlet temperatures · Note that the heat transfer rate Q is taken to be a positive quantity, and its direction is understood to be from the hot fluid to the cold one in accordance with the second law of thermodynamics. In heat exchanger analysis, it is often convenient to combine the product of the mass flow rate and the specific heat of a fluid into a single quantity. This quantity is called the heat capacity rate and is defined for the hot and cold fluid streams as Ch m· hCph and Cc m· cCpc (13-11) The heat capacity rate of a fluid stream represents the rate of heat transfer needed to change the temperature of the fluid stream by 1°C as it flows through a heat exchanger. Note that in a heat exchanger, the fluid with a large heat capacity rate will experience a small temperature change, and the fluid with a small heat capacity rate will experience a large temperature change. Therefore, doubling the mass flow rate of a fluid while leaving everything else unchanged will halve the temperature change of that fluid. With the definition of the heat capacity rate above, Eqs. 13-9 and 13-10 can also be expressed as · Q Cc(Tc, out Tc, in) T Hot fluid ∆T1 Ch ∆T ∆T2 Cold fluid Cc = Ch ∆T = ∆T1 = ∆T2 = constant (13-12) and · Q Ch(Th, in Th, out) x (13-13) That is, the heat transfer rate in a heat exchanger is equal to the heat capacity rate of either fluid multiplied by the temperature change of that fluid. Note that the only time the temperature rise of a cold fluid is equal to the temperature drop of the hot fluid is when the heat capacity rates of the two fluids are equal to each other (Fig. 13–12). Inlet Outlet FIGURE 13–12 Two fluids that have the same mass flow rate and the same specific heat experience the same temperature change in a well-insulated heat exchanger. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 680 680 HEAT TRANSFER Two special types of heat exchangers commonly used in practice are condensers and boilers. One of the fluids in a condenser or a boiler undergoes a phase-change process, and the rate of heat transfer is expressed as · Q m· hfg T where m· is the rate of evaporation or condensation of the fluid and hfg is the enthalpy of vaporization of the fluid at the specified temperature or pressure. An ordinary fluid absorbs or releases a large amount of heat essentially at constant temperature during a phase-change process, as shown in Figure 13–13. The heat capacity rate of a fluid during a phase-change process must approach infinity since the temperature change is practically zero. That is, · C m· Cp → when T → 0, so that the heat transfer rate Q m· Cp T is a finite quantity. Therefore, in heat exchanger analysis, a condensing or boiling fluid is conveniently modeled as a fluid whose heat capacity rate is infinity. The rate of heat transfer in a heat exchanger can also be expressed in an analogous manner to Newton’s law of cooling as Condensing fluid . Q Cold fluid · Q UAs Tm Inlet (13-14) Outlet (a) Condenser (Ch → ) T Hot fluid . Q (13-15) where U is the overall heat transfer coefficient, As is the heat transfer area, and Tm is an appropriate average temperature difference between the two fluids. Here the surface area As can be determined precisely using the dimensions of the heat exchanger. However, the overall heat transfer coefficient U and the temperature difference T between the hot and cold fluids, in general, are not constant and vary along the heat exchanger. The average value of the overall heat transfer coefficient can be determined as described in the preceding section by using the average convection coefficients for each fluid. It turns out that the appropriate form of the mean temperature difference between the two fluids is logarithmic in nature, and its determination is presented in Section 13–4. Boiling fluid Inlet Outlet (b) Boiler (Cc → ) FIGURE 13–13 Variation of fluid temperatures in a heat exchanger when one of the fluids condenses or boils. 13–4 ■ THE LOG MEAN TEMPERATURE DIFFERENCE METHOD Earlier, we mentioned that the temperature difference between the hot and cold fluids varies along the heat exchanger, and it is convenient to have a · mean temperature difference Tm for use in the relation Q UAs Tm. In order to develop a relation for the equivalent average temperature difference between the two fluids, consider the parallel-flow double-pipe heat exchanger shown in Figure 13–14. Note that the temperature difference T between the hot and cold fluids is large at the inlet of the heat exchanger but decreases exponentially toward the outlet. As you would expect, the temperature of the hot fluid decreases and the temperature of the cold fluid increases along the heat exchanger, but the temperature of the cold fluid can never exceed that of the hot fluid no matter how long the heat exchanger is. Assuming the outer surface of the heat exchanger to be well insulated so that any heat transfer occurs between the two fluids, and disregarding any cen58933_ch13.qxd 9/9/2002 9:57 AM Page 681 681 CHAPTER 13 changes in kinetic and potential energy, an energy balance on each fluid in a differential section of the heat exchanger can be expressed as · Q m· h Cph dTh T Th, in . δ Q = U(Th – Tc ) d As Th (13-16) ∆T ∆T1 and · Q m· c Cpc dTc Q˙ dTh ˙ hCph m (13-18) ∆T2 Th,out Tc,out dTc (13-17) That is, the rate of heat loss from the hot fluid at any section of a heat exchanger is equal to the rate of heat gain by the cold fluid in that section. The temperature change of the hot fluid is a negative quantity, and so a negative · sign is added to Eq. 13-16 to make the heat transfer rate Q a positive quantity. Solving the equations above for dTh and dTc gives dTh . δQ Tc Tc, in 1 Hot fluid ∆T1 = Th, in – Tc,in ∆T2 = Th, out – Tc, out 2 dAs Tc, out dAs As Th, out Th, in and Q˙ dTc ˙ c Cpc m (13-19) Taking their difference, we get · 1 1 dTh dTc d(Th Tc) Q · m h Cph m· c Cpc (13-20) The rate of heat transfer in the differential section of the heat exchanger can also be expressed as · Q U(Th Tc) dAs (13-21) Substituting this equation into Eq. 13-20 and rearranging gives d(Th Tc) 1 1 U dAs · Th Tc m h Cph m· c Cpc (13-22) Integrating from the inlet of the heat exchanger to its outlet, we obtain ln Th, out Tc, out 1 1 UAs · Th, in Tc, in m h Cph m· c Cpc (13-23) Finally, solving Eqs. 13-9 and 13-10 for m· cCpc and m· hCph and substituting into Eq. 13-23 gives, after some rearrangement, · Q UAs Tlm (13-24) where Tlm T1 T2 ln (T1/T2) (13-25) is the log mean temperature difference, which is the suitable form of the average temperature difference for use in the analysis of heat exchangers. Here T1 and T2 represent the temperature difference between the two fluids Cold fluid Tc, in FIGURE 13–14 Variation of the fluid temperatures in a parallel-flow double-pipe heat exchanger. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 682 682 HEAT TRANSFER Tc, out ∆T2 Hot fluid Th,in ∆T1 Th,out Cold fluid Tc, in ∆T1 = Th,in – Tc, in ∆T2 = Th,out – Tc, out (a) Parallel-flow heat exchangers Cold fluid Tc, in ∆T2 Hot fluid Th,in Th,out ∆T1 Tc, out at the two ends (inlet and outlet) of the heat exchanger. It makes no difference which end of the heat exchanger is designated as the inlet or the outlet (Fig. 13–15). The temperature difference between the two fluids decreases from T1 at the inlet to T2 at the outlet. Thus, it is tempting to use the arithmetic mean temperature Tam 12 (T1 T2) as the average temperature difference. The logarithmic mean temperature difference Tlm is obtained by tracing the actual temperature profile of the fluids along the heat exchanger and is an exact representation of the average temperature difference between the hot and cold fluids. It truly reflects the exponential decay of the local temperature difference. Note that Tlm is always less than Tam. Therefore, using Tam in calculations instead of Tlm will overestimate the rate of heat transfer in a heat exchanger between the two fluids. When T1 differs from T2 by no more than 40 percent, the error in using the arithmetic mean temperature difference is less than 1 percent. But the error increases to undesirable levels when T1 differs from T2 by greater amounts. Therefore, we should always use the logarithmic mean temperature difference when determining the rate of heat transfer in a heat exchanger. ∆T1 = Th,in – Tc, out ∆T2 = Th,out – Tc, in (b) Counter-flow heat exchangers FIGURE 13–15 The T1 and T2 expressions in parallel-flow and counter-flow heat exchangers. Counter-Flow Heat Exchangers The variation of temperatures of hot and cold fluids in a counter-flow heat exchanger is given in Figure 13–16. Note that the hot and cold fluids enter the heat exchanger from opposite ends, and the outlet temperature of the cold fluid in this case may exceed the outlet temperature of the hot fluid. In the limiting case, the cold fluid will be heated to the inlet temperature of the hot fluid. However, the outlet temperature of the cold fluid can never exceed the inlet temperature of the hot fluid, since this would be a violation of the second law of thermodynamics. The relation above for the log mean temperature difference is developed using a parallel-flow heat exchanger, but we can show by repeating the analysis above for a counter-flow heat exchanger that is also applicable to counterflow heat exchangers. But this time, T1 and T2 are expressed as shown in Figure 13–15. For specified inlet and outlet temperatures, the log mean temperature difference for a counter-flow heat exchanger is always greater than that for a parallel-flow heat exchanger. That is, Tlm, CF Tlm, PF, and thus a smaller surface area (and thus a smaller heat exchanger) is needed to achieve a specified heat transfer rate in a counter-flow heat exchanger. Therefore, it is common practice to use counter-flow arrangements in heat exchangers. In a counter-flow heat exchanger, the temperature difference between the hot and the cold fluids will remain constant along the heat exchanger when the heat capacity rates of the two fluids are equal (that is, T constant when Ch Cc or m· hCph m· cCpc). Then we have T1 T2, and the last log mean temperature difference relation gives Tlm 00 , which is indeterminate. It can be shown by the application of l’Hôpital’s rule that in this case we have Tlm T1 T2, as expected. A condenser or a boiler can be considered to be either a parallel- or counterflow heat exchanger since both approaches give the same result. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 683 683 CHAPTER 13 Multipass and Cross-Flow Heat Exchangers: Use of a Correction Factor T Th, in The log mean temperature difference Tlm relation developed earlier is limited to parallel-flow and counter-flow heat exchangers only. Similar relations are also developed for cross-flow and multipass shell-and-tube heat exchangers, but the resulting expressions are too complicated because of the complex flow conditions. In such cases, it is convenient to relate the equivalent temperature difference to the log mean temperature difference relation for the counter-flow case as Tlm F Tlm, CF t2 t1 T1 t1 Th ∆T Cold fluid Tc Tc, in Tc, in Th, in Th, out Tc, out FIGURE 13–16 The variation of the fluid temperatures in a counter-flow double-pipe heat exchanger. Cold Tc,in fluid Hot fluid and Th,in (13-28) Cold fluid Hot fluid (13-27) T1 T2 (m· Cp)tube side R t t · (m C ) 2 1 Th, out (13-26) where F is the correction factor, which depends on the geometry of the heat exchanger and the inlet and outlet temperatures of the hot and cold fluid streams. The Tlm, CF is the log mean temperature difference for the case of a counter-flow heat exchanger with the same inlet and outlet temperatures and is determined from Eq. 13-25 by taking Tl Th, in Tc, out and T2 Th, out Tc, in (Fig. 13–17). The correction factor is less than unity for a cross-flow and multipass shelland-tube heat exchanger. That is, F 1. The limiting value of F 1 corresponds to the counter-flow heat exchanger. Thus, the correction factor F for a heat exchanger is a measure of deviation of the Tlm from the corresponding values for the counter-flow case. The correction factor F for common cross-flow and shell-and-tube heat exchanger configurations is given in Figure 13–18 versus two temperature ratios P and R defined as P Hot fluid Tc,out ∆T1 Cross-flow or multipass shell-and-tube heat exchanger ∆T2 Th,out Tc, out p shell side where the subscripts 1 and 2 represent the inlet and outlet, respectively. Note that for a shell-and-tube heat exchanger, T and t represent the shell- and tube-side temperatures, respectively, as shown in the correction factor charts. It makes no difference whether the hot or the cold fluid flows through the shell or the tube. The determination of the correction factor F requires the availability of the inlet and the outlet temperatures for both the cold and hot fluids. Note that the value of P ranges from 0 to 1. The value of R, on the other hand, ranges from 0 to infinity, with R 0 corresponding to the phase-change (condensation or boiling) on the shell-side and R → to phase-change on the tube side. The correction factor is F 1 for both of these limiting cases. Therefore, the correction factor for a condenser or boiler is F 1, regardless of the configuration of the heat exchanger. Heat transfer rate: . Q = UAsF ∆Tlm, CF where ∆Tlm, CF = ∆T1 – ∆T2 ln(∆T1/∆T2) ∆T1 = Th,in – Tc,out ∆T2 = Th,out – Tc,in and F = … (Fig. 13–18) FIGURE 13–17 The determination of the heat transfer rate for cross-flow and multipass shell-and-tube heat exchangers using the correction factor. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 684 684 HEAT TRANSFER Correction factor F 1.0 T1 t2 t1 0.9 0.8 R = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 T2 0.2 0.7 T1 – T2 R = ——– t2 – t1 0.6 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (a) One-shell pass and 2, 4, 6, etc. (any multiple of 2), tube passes t2 – t1 P = ——– T1 – t1 Correction factor F 1.0 T1 0.9 t2 0.8 R = 4.0 3.0 2.0 1.5 t1 1.0 0.8 0.6 0.4 0.2 0.7 T2 0.6 0.5 T1 – T2 R = ——– t2 – t1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t2 – t1 P = ——– T1 – t1 (b) Two-shell passes and 4, 8, 12, etc. (any multiple of 4), tube passes Correction factor F 1.0 T1 0.9 0.8 R = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 0.2 t1 0.7 T1 – T2 R = ——– t2 – t1 0.6 0.5 0 0.1 0.2 T2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (c) Single-pass cross-flow with both fluids unmixed Correction factor F 1.0 FIGURE 13–18 Correction factor F charts for common shell-and-tube and cross-flow heat exchangers (from Bowman, Mueller, and Nagle, Ref. 2). t2 t2 – t1 P = ——– T1 – t1 T1 0.9 0.8 R = 4.0 3.0 2.0 1.5 1.0 0.8 0.6 0.4 t1 0.2 t2 0.7 0.6 0.5 0 T1 – T2 R = ——– t2 – t1 0.1 0.2 T2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (d) Single-pass cross-flow with one fluid mixed and the other unmixed t2 – t1 P = ——– T1 – t1 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 685 685 CHAPTER 13 EXAMPLE 13–3 The Condensation of Steam in a Condenser Steam in the condenser of a power plant is to be condensed at a temperature of 30°C with cooling water from a nearby lake, which enters the tubes of the condenser at 14°C and leaves at 22°C. The surface area of the tubes is 45 m2, and the overall heat transfer coefficient is 2100 W/m2 · °C. Determine the mass flow rate of the cooling water needed and the rate of condensation of the steam in the condenser. SOLUTION Steam is condensed by cooling water in the condenser of a power plant. The mass flow rate of the cooling water and the rate of condensation are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties The heat of vaporization of water at 30°C is hfg 2431 kJ/kg and the specific heat of cold water at the average temperature of 18°C is Cp 4184 J/kg · °C (Table A–9). Analysis The schematic of the condenser is given in Figure 13–19. The condenser can be treated as a counter-flow heat exchanger since the temperature of one of the fluids (the steam) remains constant. The temperature difference between the steam and the cooling water at the two ends of the condenser is Steam 30°C Cooling water T1 Th, in Tc, out (30 22)°C 8°C T2 Th, out Tc, in (30 14)°C 16°C 14°C That is, the temperature difference between the two fluids varies from 8°C at one end to 16°C at the other. The proper average temperature difference between the two fluids is the logarithmic mean temperature difference (not the arithmetic), which is determined from Tlm T1 T2 8 16 11.5°C ln (T1/T2) ln (8/16) This is a little less than the arithmetic mean temperature difference of 1 (8 16) 12°C. Then the heat transfer rate in the condenser is determined 2 from · Q UAs Tlm (2100 W/m2 · °C)(45 m2)(11.5°C) 1.087 106 W 1087 kW Therefore, the steam will lose heat at a rate of 1,087 kW as it flows through the condenser, and the cooling water will gain practically all of it, since the condenser is well insulated. The mass flow rate of the cooling water and the rate of the condensation of the · steam are determined from Q [m· Cp (Tout Tin)]cooling water (m· hfg)steam to be · m cooling water · Q Cp (Tout Tin) 1,087 kJ/s 32.5 kg/s (4.184 kJ/kg · °C)(22 14)°C 22°C 30°C FIGURE 13–19 Schematic for Example 13–3. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 686 686 HEAT TRANSFER and · Q 1,087 kJ/s · m 0.45 kg/s steam hfg 2431 kJ/kg Therefore, we need to circulate about 72 kg of cooling water for each 1 kg of steam condensing to remove the heat released during the condensation process. EXAMPLE 13–4 Heating Water in a Counter-Flow Heat Exchanger A counter-flow double-pipe heat exchanger is to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be accomplished by geothermal water available at 160°C at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. If the overall heat transfer coefficient of the heat exchanger is 640 W/m2 · °C, determine the length of the heat exchanger required to achieve the desired heating. Hot geothermal 160°C water 2 kg/s Cold water 20°C 1.2 kg/s 80°C D = 1.5 cm SOLUTION Water is heated in a counter-flow double-pipe heat exchanger by geothermal water. The required length of the heat exchanger is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 There is no fouling. 5 Fluid properties are constant. Properties We take the specific heats of water and geothermal fluid to be 4.18 and 4.31 kJ/kg · °C, respectively. Analysis The schematic of the heat exchanger is given in Figure 13–20. The rate of heat transfer in the heat exchanger can be determined from · · C (T T )] Q [m p out in water (1.2 kg/s)(4.18 kJ/kg · °C)(80 20)°C 301 kW Noting that all of this heat is supplied by the geothermal water, the outlet temperature of the geothermal water is determined to be · Q · · C (T T )] Q [m → Tout Tin · p in out geothermal m Cp 160°C FIGURE 13–20 Schematic for Example 13–4. 301 kW (2 kg/s)(4.31 kJ/kg ·°C) 125°C Knowing the inlet and outlet temperatures of both fluids, the logarithmic mean temperature difference for this counter-flow heat exchanger becomes T1 Th, in Tc, out (160 80)°C 80°C T2 Th, out Tc, in (125 20)°C 105°C and Tlm T1 T2 80 105 92.0°C ln (T1/T2) ln (80/105) Then the surface area of the heat exchanger is determined to be · Q UAs Tlm → As · Q 301,000 W 5.11 m2 U Tlm (640 W/m2 · °C)(92.0°C) cen58933_ch13.qxd 9/9/2002 9:57 AM Page 687 687 CHAPTER 13 To provide this much heat transfer surface area, the length of the tube must be As DL → L As 5.11 m2 108 m D (0.015 m) Discussion The inner tube of this counter-flow heat exchanger (and thus the heat exchanger itself) needs to be over 100 m long to achieve the desired heat transfer, which is impractical. In cases like this, we need to use a plate heat exchanger or a multipass shell-and-tube heat exchanger with multiple passes of tube bundles. EXAMPLE 13–5 Heating of Glycerin in a Multipass Heat Exchanger A 2-shell passes and 4-tube passes heat exchanger is used to heat glycerin from 20°C to 50°C by hot water, which enters the thin-walled 2-cm-diameter tubes at 80°C and leaves at 40°C (Fig. 13–21). The total length of the tubes in the heat exchanger is 60 m. The convection heat transfer coefficient is 25 W/m2 · °C on the glycerin (shell) side and 160 W/m2 · °C on the water (tube) side. Determine the rate of heat transfer in the heat exchanger (a) before any fouling occurs and (b) after fouling with a fouling factor of 0.0006 m2 · °C/ W occurs on the outer surfaces of the tubes. SOLUTION Glycerin is heated in a 2-shell passes and 4-tube passes heat exchanger by hot water. The rate of heat transfer for the cases of fouling and no fouling are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Heat transfer coefficients and fouling factors are constant and uniform. 5 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Analysis The tubes are said to be thin-walled, and thus it is reasonable to assume the inner and outer surface areas of the tubes to be equal. Then the heat transfer surface area becomes As DL (0.02 m)(60 m) 3.77 m2 The rate of heat transfer in this heat exchanger can be determined from · Q UAs F Tlm, CF where F is the correction factor and Tlm, CF is the log mean temperature difference for the counter-flow arrangement. These two quantities are determined from T1 Th, in Tc, out (80 50)°C 30°C T2 Th, out Tc, in (40 20)°C 20°C T1 T2 30 20 24.7°C Tlm, CF ln (T1/T2) ln (30/20) Cold glycerin 20°C 40°C Hot water 80°C 50°C FIGURE 13–21 Schematic for Example 13–5. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 688 688 HEAT TRANSFER and P t2 t1 40 80 0.67 T1 t1 20 80 T1 T2 20 50 0.75 R t t 40 80 2 1 u F 0.91 (Fig. 13–18b) (a) In the case of no fouling, the overall heat transfer coefficient U is determined from U 1 1 21.6 W/m2 · °C 1 1 1 1 hi ho 160 W/m2 · °C 25 W/m2 · °C Then the rate of heat transfer becomes · Q UAs F Tlm, CF (21.6 W/m2 · °C)(3.77m2)(0.91)(24.7°C) 1830 W (b) When there is fouling on one of the surfaces, the overall heat transfer coefficient U is U 1 1 1 1 1 1 0.0006 m2 · °C/ W Rf hi ho 160 W/m2 · °C 25 W/m2 · °C 21.3 W/m2 · °C The rate of heat transfer in this case becomes · Q UAs F Tlm, CF (21.3 W/m2 · °C)(3.77 m2)(0.91)(24.7°C) 1805 W Discussion Note that the rate of heat transfer decreases as a result of fouling, as expected. The decrease is not dramatic, however, because of the relatively low convection heat transfer coefficients involved. EXAMPLE 13–6 Cooling of an Automotive Radiator A test is conducted to determine the overall heat transfer coefficient in an automotive radiator that is a compact cross-flow water-to-air heat exchanger with both fluids (air and water) unmixed (Fig. 13–22). The radiator has 40 tubes of internal diameter 0.5 cm and length 65 cm in a closely spaced plate-finned matrix. Hot water enters the tubes at 90°C at a rate of 0.6 kg/s and leaves at 65°C. Air flows across the radiator through the interfin spaces and is heated from 20°C to 40°C. Determine the overall heat transfer coefficient Ui of this radiator based on the inner surface area of the tubes. SOLUTION During an experiment involving an automotive radiator, the inlet and exit temperatures of water and air and the mass flow rate of water are measured. The overall heat transfer coefficient based on the inner surface area is to be determined. Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are negligible. 3 Fluid properties are constant. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 689 689 CHAPTER 13 90°C Air flow (unmixed) 20°C 40°C 65°C Water flow (unmixed) Properties The specific heat of water at the average temperature of (90 65)/ 2 77.5°C is 4.195 kJ/kg · °C. Analysis The rate of heat transfer in this radiator from the hot water to the air is determined from an energy balance on water flow, · · C (T T )] Q [m p in out water (0.6 kg/s)(4.195 kJ/kg · °C)(90 65)°C 62.93 kW The tube-side heat transfer area is the total surface area of the tubes, and is determined from Ai nDi L (40)(0.005 m)(0.65 m) 0.408 m2 Knowing the rate of heat transfer and the surface area, the overall heat transfer coefficient can be determined from · Q Ui Ai F Tlm, CF → Ui · Q Ai F Tlm, CF where F is the correction factor and Tlm, CF is the log mean temperature difference for the counter-flow arrangement. These two quantities are found to be T1 Th, in Tc, out (90 40)°C 50°C T2 Th, out Tc, in (65 20)°C 45°C T1 T2 50 45 47.6°C Tlm, CF ln (T1/T2) ln (50/45) and P t2 t1 65 90 0.36 T1 t1 20 90 u F 0.97 T1 T2 20 40 R t t 0.80 65 90 2 1 (Fig. 13–18c) Substituting, the overall heat transfer coefficient Ui is determined to be · Q 62,930 W Ui 3341 W/m2 · °C Ai F Tlm, CF (0.408 m2)(0.97)(47.6°C) Note that the overall heat transfer coefficient on the air side will be much lower because of the large surface area involved on that side. FIGURE 13–22 Schematic for Example 13–6. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 690 690 HEAT TRANSFER 13–5 ■ THE EFFECTIVENESS–NTU METHOD The log mean temperature difference (LMTD) method discussed in Section 13–4 is easy to use in heat exchanger analysis when the inlet and the outlet temperatures of the hot and cold fluids are known or can be determined from an energy balance. Once Tlm, the mass flow rates, and the overall heat transfer coefficient are available, the heat transfer surface area of the heat exchanger can be determined from · Q UAs Tlm Therefore, the LMTD method is very suitable for determining the size of a heat exchanger to realize prescribed outlet temperatures when the mass flow rates and the inlet and outlet temperatures of the hot and cold fluids are specified. With the LMTD method, the task is to select a heat exchanger that will meet the prescribed heat transfer requirements. The procedure to be followed by the selection process is: 1. Select the type of heat exchanger suitable for the application. 2. Determine any unknown inlet or outlet temperature and the heat transfer rate using an energy balance. 3. Calculate the log mean temperature difference Tlm and the correction factor F, if necessary. 4. Obtain (select or calculate) the value of the overall heat transfer coefficient U. 5. Calculate the heat transfer surface area As . The task is completed by selecting a heat exchanger that has a heat transfer surface area equal to or larger than As . A second kind of problem encountered in heat exchanger analysis is the determination of the heat transfer rate and the outlet temperatures of the hot and cold fluids for prescribed fluid mass flow rates and inlet temperatures when the type and size of the heat exchanger are specified. The heat transfer surface area A of the heat exchanger in this case is known, but the outlet temperatures are not. Here the task is to determine the heat transfer performance of a specified heat exchanger or to determine if a heat exchanger available in storage will do the job. The LMTD method could still be used for this alternative problem, but the procedure would require tedious iterations, and thus it is not practical. In an attempt to eliminate the iterations from the solution of such problems, Kays and London came up with a method in 1955 called the effectiveness–NTU method, which greatly simplified heat exchanger analysis. This method is based on a dimensionless parameter called the heat transfer effectiveness , defined as Q· Actual heat transfer rate Qmax Maximum possible heat transfer rate (13-29) The actual heat transfer rate in a heat exchanger can be determined from an energy balance on the hot or cold fluids and can be expressed as · Q Cc(Tc, out Tc, in) Ch(Th, in Th, out) (13-30) cen58933_ch13.qxd 9/9/2002 9:57 AM Page 691 691 CHAPTER 13 where Cc m· cCpc and Ch m· cCph are the heat capacity rates of the cold and the hot fluids, respectively. To determine the maximum possible heat transfer rate in a heat exchanger, we first recognize that the maximum temperature difference in a heat exchanger is the difference between the inlet temperatures of the hot and cold fluids. That is, Tmax Th, in Tc, in (13-31) The heat transfer in a heat exchanger will reach its maximum value when (1) the cold fluid is heated to the inlet temperature of the hot fluid or (2) the hot fluid is cooled to the inlet temperature of the cold fluid. These two limiting conditions will not be reached simultaneously unless the heat capacity rates of the hot and cold fluids are identical (i.e., Cc Ch). When Cc Ch, which is usually the case, the fluid with the smaller heat capacity rate will experience a larger temperature change, and thus it will be the first to experience the maximum temperature, at which point the heat transfer will come to a halt. Therefore, the maximum possible heat transfer rate in a heat exchanger is (Fig. 13–23) · Q max Cmin(Th, in Tc, in) Hot oil 130°C 40 kg/s . Upper Limit for Heat Transfer in a Heat Exchanger Cold water enters a counter-flow heat exchanger at 10°C at a rate of 8 kg/s, where it is heated by a hot water stream that enters the heat exchanger at 70°C at a rate of 2 kg/s. Assuming the specific heat of water to remain constant at Cp 4.18 kJ/kg · °C, determine the maximum heat transfer rate and the outlet temperatures of the cold and the hot water streams for this limiting case. SOLUTION Cold and hot water streams enter a heat exchanger at specified temperatures and flow rates. The maximum rate of heat transfer in the heat exchanger is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Heat transfer coefficients and fouling factors are constant and uniform. 5 The thermal resistance of the inner tube is negligible since the tube is thin-walled and highly conductive. Properties The specific heat of water is given to be Cp 4.18 kJ/kg · °C. Analysis A schematic of the heat exchanger is given in Figure 13–24. The heat capacity rates of the hot and cold fluids are determined from · C (2 kg/s)(4.18 kJ/kg · °C) 8.36 kW/°C Ch m h ph Cold water (13-32) where Cmin is the smaller of Ch m· hCph and Cc m· cCpc. This is further clarified by the following example. EXAMPLE 13–7 20°C 25 kg/s Cc = mcCpc = 104.5 kW/°C . Ch = mcCph = 92 kW/°C Cmin = 92 kW/°C ∆Tmax = Th,in – Tc,in = 110°C . Qmax = Cmin ∆Tmax = 10,120 kW FIGURE 13–23 The determination of the maximum rate of heat transfer in a heat exchanger. 10°C 8 kg/s Cold water Hot water 70°C 2 kg/s and · C (8 kg/s)(4.18 kJ/kg · °C) 33.4 kW/°C Cc m c pc FIGURE 13–24 Schematic for Example 13–7. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 692 692 HEAT TRANSFER Therefore Cmin Ch 8.36 kW/°C which is the smaller of the two heat capacity rates. Then the maximum heat transfer rate is determined from Eq. 13-32 to be · Q max Cmin(Th, in Tc, in) (8.36 kW/°C)(70 10)°C 502 kW That is, the maximum possible heat transfer rate in this heat exchanger is 502 kW. This value would be approached in a counter-flow heat exchanger with a very large heat transfer surface area. The maximum temperature difference in this heat exchanger is Tmax Th, in Tc, in (70 10)°C 60°C. Therefore, the hot water cannot be cooled by more than 60°C (to 10°C) in this heat exchanger, and the cold water cannot be heated by more than 60°C (to 70°C), no matter what we do. The outlet temperatures of the cold and the hot streams in this limiting case are determined to be · Q · 502 kW Q Cc(Tc, out Tc, in) → Tc, out Tc, in 10°C 25°C Cc 33.4 kW/°C · Q · 502 kW Q Ch(Th, in Th, out) → Th, out Th, in 70°C 10°C Ch 8.38 kW/°C . mc ,Cpc Cold fluid Hot fluid . mh ,Cph . . Q = mh Cph ∆Th . = mc Cpc ∆Tc If . . mc Cpc = mh Cph Discussion Note that the hot water is cooled to the limit of 10°C (the inlet temperature of the cold water stream), but the cold water is heated to 25°C only when maximum heat transfer occurs in the heat exchanger. This is not surprising, since the mass flow rate of the hot water is only one-fourth that of the cold water, and, as a result, the temperature of the cold water increases by 0.25°C for each 1°C drop in the temperature of the hot water. You may be tempted to think that the cold water should be heated to 70°C in the limiting case of maximum heat transfer. But this will require the temperature of the hot water to drop to 170°C (below 10°C), which is impossible. Therefore, heat transfer in a heat exchanger reaches its maximum value when the fluid with the smaller heat capacity rate (or the smaller mass flow rate when both fluids have the same specific heat value) experiences the maximum temperature change. This example explains why we use Cmin in the evaluation of · Q max instead of Cmax. We can show that the hot water will leave at the inlet temperature of the cold water and vice versa in the limiting case of maximum heat transfer when the mass flow rates of the hot and cold water streams are identical (Fig. 13–25). We can also show that the outlet temperature of the cold water will reach the 70°C limit when the mass flow rate of the hot water is greater than that of the cold water. then ∆Th = ∆Tc FIGURE 13–25 The temperature rise of the cold fluid in a heat exchanger will be equal to the temperature drop of the hot fluid when the mass flow rates and the specific heats of the hot and cold fluids are identical. · The determination of Q max requires the availability of the inlet temperature of the hot and cold fluids and their mass flow rates, which are usually specified. Then, once the effectiveness of the heat exchanger is known, the actual · heat transfer rate Q can be determined from · · Q Q max Cmin(Th, in Tc, in) (13-33) cen58933_ch13.qxd 9/9/2002 9:57 AM Page 693 693 CHAPTER 13 Therefore, the effectiveness of a heat exchanger enables us to determine the heat transfer rate without knowing the outlet temperatures of the fluids. The effectiveness of a heat exchanger depends on the geometry of the heat exchanger as well as the flow arrangement. Therefore, different types of heat exchangers have different effectiveness relations. Below we illustrate the development of the effectiveness relation for the double-pipe parallel-flow heat exchanger. Equation 13-23 developed in Section 13–4 for a parallel-flow heat exchanger can be rearranged as ln Th, out Tc, out UAs Cc 1 Th, in Tc, in Cc Ch (13-34) Also, solving Eq. 13-30 for Th, out gives Th, out Th, in Cc (T Tc, in) Ch c, out (13-35) Substituting this relation into Eq. 13-34 after adding and subtracting Tc, in gives Th, in Tc, in Tc, in Tc, out ln Cc (T Tc, in) Ch c, out Th, in Tc, in UAs Cc 1 Cc Ch which simplifies to ln 1 1 UAs Cc Tc, out Tc, in Cc 1 Ch Th, in Tc, in Cc Ch (13-36) We now manipulate the definition of effectiveness to obtain Cc(Tc, out Tc, in) Q· · C Q max min(Th, in Tc, in) Tc, out Tc, in Cmin Th, in Tc, in Cc → Substituting this result into Eq. 13-36 and solving for gives the following relation for the effectiveness of a parallel-flow heat exchanger: 1 exp parallel flow 1 UAs Cc 1 Cc Ch Cc Cmin Ch Cc (13-37) Taking either Cc or Ch to be Cmin (both approaches give the same result), the relation above can be expressed more conveniently as 1 exp parallel flow UAs Cmin 1 Cmin Cmax 1 Cmin Cmax (13-38) Again Cmin is the smaller heat capacity ratio and Cmax is the larger one, and it makes no difference whether Cmin belongs to the hot or cold fluid. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 694 694 HEAT TRANSFER Effectiveness relations of the heat exchangers typically involve the dimensionless group UAs /Cmin. This quantity is called the number of transfer units NTU and is expressed as NTU UAs UAs Cmin (m· Cp)min (13-39) where U is the overall heat transfer coefficient and As is the heat transfer surface area of the heat exchanger. Note that NTU is proportional to As . Therefore, for specified values of U and Cmin, the value of NTU is a measure of the heat transfer surface area As . Thus, the larger the NTU, the larger the heat exchanger. In heat exchanger analysis, it is also convenient to define another dimensionless quantity called the capacity ratio c as c Cmin Cmax (13-40) It can be shown that the effectiveness of a heat exchanger is a function of the number of transfer units NTU and the capacity ratio c. That is, function (UAs /Cmin, Cmin /Cmax) function (NTU, c) Effectiveness relations have been developed for a large number of heat exchangers, and the results are given in Table 13–4. The effectivenesses of some common types of heat exchangers are also plotted in Figure 13–26. More TABLE 13–4 Effectiveness relations for heat exchangers: NTU UAs /Cmin and c Cmin/Cmax (m· Cp)min/(m· Cp)max (Kays and London, Ref. 5.) Heat exchanger type 1 Double pipe: Parallel-flow Counter-flow 2 Shell and tube: One-shell pass 2, 4, . . . tube passes Effectiveness relation 1 exp [ NTU(1 c )] 1c 1 exp [ NTU(1 c )] 1 c exp [ NTU(1 c )] 2 1 c 1 c 2 1 exp [ NTU 1 c 2] 1 exp [ NTU 1 c 2] 3 Cross-flow (single-pass) Both fluids unmixed 1 exp Cmax mixed, Cmin unmixed 1 c (1 exp {1 c[1 exp ( NTU)]}) Cmin mixed, Cmax unmixed 1 1 exp c [1 exp ( c NTU)] 4 All heat exchangers with c 0 NTU0.22 [exp ( c NTU0.78) 1] c 1 exp( NTU) 1 9:57 AM Page 695 695 CHAPTER 13 100 100 5 0.71.00 / mi n 55 0.2 C Tube fluid 60 =0 0.5 0 Effectiveness ε, % mi n ax Cm 80 0.25 0.50 0.75 1.00 60 40 =0 / ax Cm C Effectiveness ε, % 80 Shell fluid 40 Tube fluid 20 20 Shell fluid 0 3 4 5 1 2 Number of transfer units NTU = AsU/Cmin (a) Parallel-flow (b) Counter-flow 100 40 Shell fluid C 60 0.50 0.75 1.00 80 m n Effectiveness ε, % /C 100 =0 0.25 x ma m i 80 3 4 5 1 2 Number of transfer units NTU = AsU/Cmin C 0 Effectiveness ε, % =0 ax m C / 0.250 in 0.5 0.75 1.00 60 Shell fluid 40 20 20 Tube fluid Tube fluid 1 2 3 4 0 5 Number of transfer units NTU = AsU/Cmin (c) One-shell pass and 2, 4, 6, … tube passes 60 =0 5 0.2 0 0.5 5 0.7 00 1. 80 5 d xe mi un , =0 0.25 4 0.5 2 0.75 1.33 1 Hot fluid C mi xe Cold fluid 40 4 (d ) Two-shell passes and 4, 8, 12, … tube passes Effectiveness ε, % mi n / ax Cm 3 2 Number of transfer units NTU = AsU/Cmin 100 100 80 1 d /C 0 C 9/9/2002 Effectiveness ε, % cen58933_ch13.qxd 60 Mixed fluid 40 20 20 0 0 Unmixed fluid 1 2 3 4 5 Number of transfer units NTU = AsU/Cmin (e) Cross-flow with both fluids unmixed 1 2 3 4 5 Number of transfer units NTU = AsU/Cmin ( f ) Cross-flow with one fluid mixed and the other unmixed FIGURE 13–26 Effectiveness for heat exchangers (from Kays and London, Ref. 5). cen58933_ch13.qxd 9/9/2002 9:57 AM Page 696 696 HEAT TRANSFER extensive effectiveness charts and relations are available in the literature. The dashed lines in Figure 13–26f are for the case of Cmin unmixed and Cmax mixed and the solid lines are for the opposite case. The analytic relations for the effectiveness give more accurate results than the charts, since reading errors in charts are unavoidable, and the relations are very suitable for computerized analysis of heat exchangers. We make these following observations from the effectiveness relations and charts already given: 1. The value of the effectiveness ranges from 0 to 1. It increases rapidly 1 Counter-flow ε Cross-flow with both fluids unmixed 0.5 Parallel-flow (for c = 1) 0 0 1 2 3 4 NTU = UAs /Cmin 5 FIGURE 13–27 For a specified NTU and capacity ratio c, the counter-flow heat exchanger has the highest effectiveness and the parallel-flow the lowest. ε 0 ε = 1 – e– NTU (All heat exchangers with c = 0) 0 1 2 3 4 NTU = UAs /Cmin FIGURE 13–28 The effectiveness relation reduces to max 1 exp( NTU) for all heat exchangers when the capacity ratio c 0. max 1 exp( NTU) (13-41) regardless of the type of heat exchanger (Fig. 13–28). Note that the temperature of the condensing or boiling fluid remains constant in this case. The effectiveness is the lowest in the other limiting case of c Cmin/Cmax 1, which is realized when the heat capacity rates of the two fluids are equal. 1 0.5 with NTU for small values (up to about NTU 1.5) but rather slowly for larger values. Therefore, the use of a heat exchanger with a large NTU (usually larger than 3) and thus a large size cannot be justified economically, since a large increase in NTU in this case corresponds to a small increase in effectiveness. Thus, a heat exchanger with a very high effectiveness may be highly desirable from a heat transfer point of view but rather undesirable from an economical point of view. 2. For a given NTU and capacity ratio c Cmin /Cmax, the counter-flow heat exchanger has the highest effectiveness, followed closely by the cross-flow heat exchangers with both fluids unmixed. As you might expect, the lowest effectiveness values are encountered in parallel-flow heat exchangers (Fig. 13–27). 3. The effectiveness of a heat exchanger is independent of the capacity ratio c for NTU values of less than about 0.3. 4. The value of the capacity ratio c ranges between 0 and 1. For a given NTU, the effectiveness becomes a maximum for c 0 and a minimum for c 1. The case c Cmin /Cmax → 0 corresponds to Cmax → , which is realized during a phase-change process in a condenser or boiler. All effectiveness relations in this case reduce to 5 Once the quantities c Cmin /Cmax and NTU UAs /Cmin have been evaluated, the effectiveness can be determined from either the charts or (preferably) the effectiveness relation for the specified type of heat exchanger. Then · the rate of heat transfer Q and the outlet temperatures Th, out and Tc, out can be determined from Eqs. 13-33 and 13-30, respectively. Note that the analysis of heat exchangers with unknown outlet temperatures is a straightforward matter with the effectiveness–NTU method but requires rather tedious iterations with the LMTD method. We mentioned earlier that when all the inlet and outlet temperatures are specified, the size of the heat exchanger can easily be determined using the LMTD method. Alternatively, it can also be determined from the effectiveness–NTU method by first evaluating the effectiveness from its definition (Eq. 13-29) and then the NTU from the appropriate NTU relation in Table 13–5. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 697 697 CHAPTER 13 TABLE 13–5 NTU relations for heat exchangers NTU UAs /Cmin and c Cmin /Cmax · C ) /(m · C ) (Kays and London, Ref. 5.) (m p min p max Heat exchanger type NTU relation 1 Double-pipe: Parallel-flow NTU NTU Counter-flow ln [1 (1 c )] 1c 1 1 ln c 1 c 1 2 Shell and tube: One-shell pass 2, 4, . . . tube passes NTU 3 Cross-flow (single-pass) Cmax mixed, Cmin unmixed NTU ln 1 Cmin mixed, Cmax unmixed 4 All heat exchangers with c 0 ln [c ln (1 ) 1] NTU c NTU ln(1 ) 2/ 1 c 1 c 2 1 ln 2 2/ 1 c 1 c 2 1 c ln (1 c ) c Note that the relations in Table 13–5 are equivalent to those in Table 13–4. Both sets of relations are given for convenience. The relations in Table 13–4 give the effectiveness directly when NTU is known, and the relations in Table 13–5 give the NTU directly when the effectiveness is known. EXAMPLE 13–8 Using the Effectiveness–NTU Method Repeat Example 13–4, which was solved with the LMTD method, using the effectiveness–NTU method. SOLUTION The schematic of the heat exchanger is redrawn in Figure 13–29, and the same assumptions are utilized. Analysis In the effectiveness–NTU method, we first determine the heat capacity rates of the hot and cold fluids and identify the smaller one: Ch m· hCph (2 kg/s)(4.31 kJ/kg · °C) 8.62 kW/°C Cc m· cCpc (1.2 kg/s)(4.18 kJ/kg · °C) 5.02 kW/°C Cold water 20°C 1.2 kg/s Hot geothermal 160°C brine 2 kg/s 80°C D = 1.5 cm Therefore, Cmin Cc 5.02 kW/°C and c Cmin /Cmax 5.02/8.62 0.583 Then the maximum heat transfer rate is determined from Eq. 13-32 to be · Q max Cmin(Th, in Tc, in) (5.02 kW/°C)(160 20)°C 702.8 kW FIGURE 13–29 Schematic for Example 13–8. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 698 698 HEAT TRANSFER That is, the maximum possible heat transfer rate in this heat exchanger is 702.8 kW. The actual rate of heat transfer in the heat exchanger is · Q [m· Cp(Tout Tin)]water (1.2 kg/s)(4.18 kJ/kg · °C)(80 20)°C 301.0 kW Thus, the effectiveness of the heat exchanger is · Q 301.0 kW · 0.428 Q max 702.8 kW Knowing the effectiveness, the NTU of this counter-flow heat exchanger can be determined from Figure 13–26b or the appropriate relation from Table 13–5. We choose the latter approach for greater accuracy: NTU 0.428 1 1 1 1 ln 0.651 ln c 1 c 1 0.583 1 0.428 0.583 1 Then the heat transfer surface area becomes NTU UAs Cmin → As NTU Cmin (0.651)(5020 W/°C) 5.11 m2 U 640 W/m2 · °C To provide this much heat transfer surface area, the length of the tube must be As DL → L As 5.11 m2 108 m D (0.015 m) Discussion Note that we obtained the same result with the effectiveness–NTU method in a systematic and straightforward manner. EXAMPLE 13–9 150°C Hot oil is to be cooled by water in a 1-shell-pass and 8-tube-passes heat exchanger. The tubes are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 5 m, and the overall heat transfer coefficient is 310 W/m2 · °C. Water flows through the tubes at a rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/s. The water and the oil enter at temperatures of 20°C and 150°C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil. Oil 0.3 kg/s 20°C Water 0.2 kg/s FIGURE 13–30 Schematic for Example 13–9. Cooling Hot Oil by Water in a Multipass Heat Exchanger SOLUTION Hot oil is to be cooled by water in a heat exchanger. The mass flow rates and the inlet temperatures are given. The rate of heat transfer and the outlet temperatures are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 The thickness of the tube is negligible since it is thin-walled. 4 Changes in the kinetic and potential energies of fluid streams are negligible. 5 The overall heat transfer coefficient is constant and uniform. Analysis The schematic of the heat exchanger is given in Figure 13–30. The outlet temperatures are not specified, and they cannot be determined from an cen58933_ch13.qxd 9/9/2002 9:57 AM Page 699 699 CHAPTER 13 energy balance. The use of the LMTD method in this case will involve tedious iterations, and thus the –NTU method is indicated. The first step in the –NTU method is to determine the heat capacity rates of the hot and cold fluids and identify the smaller one: Ch m· hCph (0.3 kg/s)(2.13 kJ/kg · °C) 0.639 kW/°C Cc m· cCpc (0.2 kg/s)(4.18 kJ/kg · °C) 0.836 kW/°C Therefore, Cmin Ch 0.639 kW/°C and c Cmin 0.639 0.764 Cmax 0.836 Then the maximum heat transfer rate is determined from Eq. 13-32 to be · Q max Cmin(Th, in Tc, in) (0.639 kW/°C)(150 20)°C 83.1 kW That is, the maximum possible heat transfer rate in this heat exchanger is 83.1 kW. The heat transfer surface area is As n(DL) 8(0.014 m)(5 m) 1.76 m2 Then the NTU of this heat exchanger becomes NTU UAs (310 W/m2 · °C)(1.76 m2) 0.853 Cmin 639 W/°C The effectiveness of this heat exchanger corresponding to c 0.764 and NTU 0.853 is determined from Figure 13–26c to be 0.47 We could also determine the effectiveness from the third relation in Table 13–4 more accurately but with more labor. Then the actual rate of heat transfer becomes · · Q Q max (0.47)(83.1 kW) 39.1 kW Finally, the outlet temperatures of the cold and the hot fluid streams are determined to be · Q · Q Cc(Tc, out Tc, in) → Tc, out Tc, in Cc 39.1 kW 20°C 66.8°C 0.836 kW/°C · Q · Q Ch(Th, in Th, out) → Th, out Th, in Ch 150°C 39.1 kW 88.8°C 0.639 kW/°C Therefore, the temperature of the cooling water will rise from 20°C to 66.8°C as it cools the hot oil from 150°C to 88.8°C in this heat exchanger. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 700 700 HEAT TRANSFER 13–6 ■ SELECTION OF HEAT EXCHANGERS Heat exchangers are complicated devices, and the results obtained with the simplified approaches presented above should be used with care. For example, we assumed that the overall heat transfer coefficient U is constant throughout the heat exchanger and that the convection heat transfer coefficients can be predicted using the convection correlations. However, it should be kept in mind that the uncertainty in the predicted value of U can even exceed 30 percent. Thus, it is natural to tend to overdesign the heat exchangers in order to avoid unpleasant surprises. Heat transfer enhancement in heat exchangers is usually accompanied by increased pressure drop, and thus higher pumping power. Therefore, any gain from the enhancement in heat transfer should be weighed against the cost of the accompanying pressure drop. Also, some thought should be given to which fluid should pass through the tube side and which through the shell side. Usually, the more viscous fluid is more suitable for the shell side (larger passage area and thus lower pressure drop) and the fluid with the higher pressure for the tube side. Engineers in industry often find themselves in a position to select heat exchangers to accomplish certain heat transfer tasks. Usually, the goal is to heat or cool a certain fluid at a known mass flow rate and temperature to a desired temperature. Thus, the rate of heat transfer in the prospective heat exchanger is · Q max m· Cp(Tin Tout) which gives the heat transfer requirement of the heat exchanger before having any idea about the heat exchanger itself. An engineer going through catalogs of heat exchanger manufacturers will be overwhelmed by the type and number of readily available off-the-shelf heat exchangers. The proper selection depends on several factors. Heat Transfer Rate This is the most important quantity in the selection of a heat exchanger. A heat exchanger should be capable of transferring heat at the specified rate in order to achieve the desired temperature change of the fluid at the specified mass flow rate. Cost Budgetary limitations usually play an important role in the selection of heat exchangers, except for some specialized cases where “money is no object.” An off-the-shelf heat exchanger has a definite cost advantage over those made to order. However, in some cases, none of the existing heat exchangers will do, and it may be necessary to undertake the expensive and time-consuming task of designing and manufacturing a heat exchanger from scratch to suit the needs. This is often the case when the heat exchanger is an integral part of the overall device to be manufactured. The operation and maintenance costs of the heat exchanger are also important considerations in assessing the overall cost. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 701 701 CHAPTER 13 Pumping Power In a heat exchanger, both fluids are usually forced to flow by pumps or fans that consume electrical power. The annual cost of electricity associated with the operation of the pumps and fans can be determined from Operating cost (Pumping power, kW) (Hours of operation, h) (Price of electricity, $/kWh) where the pumping power is the total electrical power consumed by the motors of the pumps and fans. For example, a heat exchanger that involves a 1-hp pump and a 13 -hp fan (1 hp 0.746 kW) operating 8 h a day and 5 days a week will consume 2017 kWh of electricity per year, which will cost $161.4 at an electricity cost of 8 cents/kWh. Minimizing the pressure drop and the mass flow rate of the fluids will minimize the operating cost of the heat exchanger, but it will maximize the size of the heat exchanger and thus the initial cost. As a rule of thumb, doubling the mass flow rate will reduce the initial cost by half but will increase the pumping power requirements by a factor of roughly eight. Typically, fluid velocities encountered in heat exchangers range between 0.7 and 7 m/s for liquids and between 3 and 30 m/s for gases. Low velocities are helpful in avoiding erosion, tube vibrations, and noise as well as pressure drop. Size and Weight Normally, the smaller and the lighter the heat exchanger, the better it is. This is especially the case in the automotive and aerospace industries, where size and weight requirements are most stringent. Also, a larger heat exchanger normally carries a higher price tag. The space available for the heat exchanger in some cases limits the length of the tubes that can be used. Type The type of heat exchanger to be selected depends primarily on the type of fluids involved, the size and weight limitations, and the presence of any phasechange processes. For example, a heat exchanger is suitable to cool a liquid by a gas if the surface area on the gas side is many times that on the liquid side. On the other hand, a plate or shell-and-tube heat exchanger is very suitable for cooling a liquid by another liquid. Materials The materials used in the construction of the heat exchanger may be an important consideration in the selection of heat exchangers. For example, the thermal and structural stress effects need not be considered at pressures below 15 atm or temperatures below 150°C. But these effects are major considerations above 70 atm or 550°C and seriously limit the acceptable materials of the heat exchanger. A temperature difference of 50°C or more between the tubes and the shell will probably pose differential thermal expansion problems and needs to be considered. In the case of corrosive fluids, we may have to select expensive cen58933_ch13.qxd 9/9/2002 9:57 AM Page 702 702 HEAT TRANSFER corrosion-resistant materials such as stainless steel or even titanium if we are not willing to replace low-cost heat exchangers frequently. Other Considerations There are other considerations in the selection of heat exchangers that may or may not be important, depending on the application. For example, being leak-tight is an important consideration when toxic or expensive fluids are involved. Ease of servicing, low maintenance cost, and safety and reliability are some other important considerations in the selection process. Quietness is one of the primary considerations in the selection of liquid-to-air heat exchangers used in heating and air-conditioning applications. EXAMPLE 13–10 Installing a Heat Exchanger to Save Energy and Money In a dairy plant, milk is pasteurized by hot water supplied by a natural gas furnace. The hot water is then discharged to an open floor drain at 80°C at a rate of 15 kg/min. The plant operates 24 h a day and 365 days a year. The furnace has an efficiency of 80 percent, and the cost of the natural gas is $0.40 per therm (1 therm 105,500 kJ). The average temperature of the cold water entering the furnace throughout the year is 15°C. The drained hot water cannot be returned to the furnace and recirculated, because it is contaminated during the process. In order to save energy, installation of a water-to-water heat exchanger to preheat the incoming cold water by the drained hot water is proposed. Assuming that the heat exchanger will recover 75 percent of the available heat in the hot water, determine the heat transfer rating of the heat exchanger that needs to be purchased and suggest a suitable type. Also, determine the amount of money this heat exchanger will save the company per year from natural gas savings. SOLUTION A water-to-water heat exchanger is to be installed to transfer energy Hot 80°C water Cold water 15°C from drained hot water to the incoming cold water to preheat it. The rate of heat transfer in the heat exchanger and the amount of energy and money saved per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The effectiveness of the heat exchanger remains constant. Properties We use the specific heat of water at room temperature, Cp 4.18 kJ/ kg · °C (Table A–9), and treat it as a constant. Analysis A schematic of the prospective heat exchanger is given in Figure 13–31. The heat recovery from the hot water will be a maximum when it leaves the heat exchanger at the inlet temperature of the cold water. Therefore, · · C (T T ) Q max m h p h, in c, in FIGURE 13–31 Schematic for Example 13–10. 60 kg/s(4.18 kJ/kg · °C)(80 15)°C 15 67.9 kJ/s That is, the existing hot water stream has the potential to supply heat at a rate of 67.9 kJ/s to the incoming cold water. This value would be approached in a counter-flow heat exchanger with a very large heat transfer surface area. A heat exchanger of reasonable size and cost can capture 75 percent of this heat cen58933_ch13.qxd 9/9/2002 9:57 AM Page 703 703 CHAPTER 13 transfer potential. Thus, the heat transfer rating of the prospective heat exchanger must be · · Q Q max (0.75)(67.9 kJ/s) 50.9 kJ/s That is, the heat exchanger should be able to deliver heat at a rate of 50.9 kJ/s from the hot to the cold water. An ordinary plate or shell-and-tube heat exchanger should be adequate for this purpose, since both sides of the heat exchanger involve the same fluid at comparable flow rates and thus comparable heat transfer coefficients. (Note that if we were heating air with hot water, we would have to specify a heat exchanger that has a large surface area on the air side.) The heat exchanger will operate 24 h a day and 365 days a year. Therefore, the annual operating hours are Operating hours (24 h/day)(365 days/year) 8760 h/year Noting that this heat exchanger saves 50.9 kJ of energy per second, the energy saved during an entire year will be Energy saved (Heat transfer rate)(Operation time) (50.9 kJ/s)(8760 h/year)(3600 s/h) 1.605 109 kJ/year The furnace is said to be 80 percent efficient. That is, for each 80 units of heat supplied by the furnace, natural gas with an energy content of 100 units must be supplied to the furnace. Therefore, the energy savings determined above result in fuel savings in the amount of Energy saved 1.605 109 kJ/year 1 therm 0.80 Furnace efficiency 105,500 kJ 19,020 therms/year Fuel saved Noting that the price of natural gas is $0.40 per therm, the amount of money saved becomes Money saved (Fuel saved) (Price of fuel) (19,020 therms/year)($0.40/therm) $7607/ year Therefore, the installation of the proposed heat exchanger will save the company $7607 a year, and the installation cost of the heat exchanger will probably be paid from the fuel savings in a short time. SUMMARY Heat exchangers are devices that allow the exchange of heat between two fluids without allowing them to mix with each other. Heat exchangers are manufactured in a variety of types, the simplest being the double-pipe heat exchanger. In a parallel-flow type, both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction, whereas in a counter-flow type, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions. In compact heat exchangers, the two fluids move perpendicular to each other, and such a flow configuration is called cross-flow. Other common types of heat exchangers in industrial applications are the plate and the shell-and-tube heat exchangers. Heat transfer in a heat exchanger usually involves convection in each fluid and conduction through the wall separating the two fluids. In the analysis of heat exchangers, it is convenient to cen58933_ch13.qxd 9/9/2002 9:57 AM Page 704 704 HEAT TRANSFER work with an overall heat transfer coefficient U or a total thermal resistance R, expressed as 1 1 1 1 1 R Rwall UAs Ui Ai Uo Ao hi Ai ho Ao where the subscripts i and o stand for the inner and outer surfaces of the wall that separates the two fluids, respectively. When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, the last relation simplifies to 1 1 1 U hi ho where U Ui Uo. The effects of fouling on both the inner and the outer surfaces of the tubes of a heat exchanger can be accounted for by 1 1 1 R UAs Ui Ai Uo Ao Rf, i ln (Do /Di ) Rf, o 1 1 Ai Ao hi Ai 2kL ho Ao where Ai Di L and Ao Do L are the areas of the inner and outer surfaces and Rf, i and Rf, o are the fouling factors at those surfaces. In a well-insulated heat exchanger, the rate of heat transfer from the hot fluid is equal to the rate of heat transfer to the cold one. That is, · Q m· cCpc(Tc, out Tc, in) Cc(Tc, out Tc, in) and · Q m· hCph(Th, in Th, out) Ch(Th, in Th, out) where the subscripts c and h stand for the cold and hot fluids, respectively, and the product of the mass flow rate and the specific heat of a fluid m· Cp is called the heat capacity rate. Of the two methods used in the analysis of heat exchangers, the log mean temperature difference (or LMTD) method is best suited for determining the size of a heat exchanger when all the inlet and the outlet temperatures are known. The effectiveness–NTU method is best suited to predict the outlet temperatures of the hot and cold fluid streams in a specified heat exchanger. In the LMTD method, the rate of heat transfer is determined from · Q UAs Tlm where Tlm T1 T2 ln (T1/T2) is the log mean temperature difference, which is the suitable form of the average temperature difference for use in the analysis of heat exchangers. Here T1 and T2 represent the temperature differences between the two fluids at the two ends (inlet and outlet) of the heat exchanger. For cross-flow and multipass shell-and-tube heat exchangers, the logarithmic mean temperature difference is related to the counter-flow one Tlm, CF as Tlm F Tlm, CF where F is the correction factor, which depends on the geometry of the heat exchanger and the inlet and outlet temperatures of the hot and cold fluid streams. The effectiveness of a heat exchanger is defined as Q· Actual heat transfer rate Qmax Maximum possible heat transfer rate where · Q max Cmin(Th, in Tc, in) and Cmin is the smaller of Ch m· hCph and Cc m· cCpc. The effectiveness of heat exchangers can be determined from effectiveness relations or charts. The selection or design of a heat exchanger depends on several factors such as the heat transfer rate, cost, pressure drop, size, weight, construction type, materials, and operating environment. REFERENCES AND SUGGESTED READING 1. N. Afgan and E. U. Schlunder. Heat Exchanger: Design and Theory Sourcebook. Washington D.C.: McGrawHill/Scripta, 1974. K. A. Gardner. “Variable Heat Transfer Rate Correction in Multipass Exchangers, Shell Side Film Controlling.” Transactions of the ASME 67 (1945), pp. 31–38. R. A. Bowman, A. C. Mueller, and W. M. Nagle. “Mean Temperature Difference in Design.” Transactions of the ASME 62 (1940), p. 283. W. M. Kays and A. L. London. Compact Heat Exchangers. 3rd ed. New York: McGraw-Hill, 1984. A. P. Fraas. Heat Exchanger Design. 2d ed. New York: John Wiley & Sons, 1989. W. M. Kays and H. C. Perkins. In Handbook of Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett. New York: McGraw-Hill, 1972, Chap. 7. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 705 705 CHAPTER 13 A. C. Mueller. “Heat Exchangers.” In Handbook of Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett. New York: McGraw-Hill, 1972, Chap. 18. 8. M. N. Özis,ik. Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. 9. E. U. Schlunder. Heat Exchanger Design Handbook. Washington, D.C.: Hemisphere, 1982. 10. Standards of Tubular Exchanger Manufacturers Association. New York: Tubular Exchanger Manufacturers Association, latest ed. R. A. Stevens, J. Fernandes, and J. R. Woolf. “Mean Temperature Difference in One, Two, and Three Pass Crossflow Heat Exchangers.” Transactions of the ASME 79 (1957), pp. 287–297. 12. J. Taborek, G. F. Hewitt, and N. Afgan. Heat Exchangers: Theory and Practice. New York: Hemisphere, 1983. 13. G. Walker. Industrial Heat Exchangers. Washington, D.C.: Hemisphere, 1982. PROBLEMS Types of Heat Exchangers 13–1C Classify heat exchangers according to flow type and explain the characteristics of each type. 13–2C Classify heat exchangers according to construction type and explain the characteristics of each type. 13–3C When is a heat exchanger classified as being compact? Do you think a double-pipe heat exchanger can be classified as a compact heat exchanger? 13–4C How does a cross-flow heat exchanger differ from a counter-flow one? What is the difference between mixed and unmixed fluids in cross-flow? 13–10C Under what conditions is the thermal resistance of the tube in a heat exchanger negligible? 13–11C Consider a double-pipe parallel-flow heat exchanger of length L. The inner and outer diameters of the inner tube are D1 and D2, respectively, and the inner diameter of the outer tube is D3. Explain how you would determine the two heat transfer surface areas Ai and Ao. When is it reasonable to assume Ai Ao As? 13–12C Is the approximation hi ho h for the convection heat transfer coefficient in a heat exchanger a reasonable one when the thickness of the tube wall is negligible? 13–5C What is the role of the baffles in a shell-and-tube heat exchanger? How does the presence of baffles affect the heat transfer and the pumping power requirements? Explain. 13–13C Under what conditions can the overall heat transfer coefficient of a heat exchanger be determined from U (1/hi 1/ho) 1? 13–6C Draw a 1-shell-pass and 6-tube-passes shell-and-tube heat exchanger. What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter? 13–14C What are the restrictions on the relation UAs Ui Ai Uo Ao for a heat exchanger? Here As is the heat transfer surface area and U is the overall heat transfer coefficient. 13–7C Draw a 2-shell-passes and 8-tube-passes shell-andtube heat exchanger. What is the primary reason for using so many tube passes? 13–8C What is a regenerative heat exchanger? How does a static type of regenerative heat exchanger differ from a dynamic type? The Overall Heat Transfer Coefficient 13–9C What are the heat transfer mechanisms involved during heat transfer from the hot to the cold fluid? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 13–15C In a thin-walled double-pipe heat exchanger, when is the approximation U hi a reasonable one? Here U is the overall heat transfer coefficient and hi is the convection heat transfer coefficient inside the tube. 13–16C What are the common causes of fouling in a heat exchanger? How does fouling affect heat transfer and pressure drop? 13–17C How is the thermal resistance due to fouling in a heat exchanger accounted for? How do the fluid velocity and temperature affect fouling? 13–18 A double-pipe heat exchanger is constructed of a copper (k 380 W/m · °C) inner tube of internal diameter Di 1.2 cm and external diameter Do 1.6 cm and an outer tube of diameter 3.0 cm. The convection heat transfer coefficient is reported to be hi 700 W/m2 · °C on the inner surface of the tube and ho 1400 W/m2 · °C on its outer surface. For a fouling factor Rf, i 0.0005 m2 · °C/W on the tube side and Rf, o 0.0002 m2 · °C/W on the shell side, determine (a) the thermal resistance of the heat exchanger per unit length and (b) the cen58933_ch13.qxd 9/9/2002 9:57 AM Page 706 706 HEAT TRANSFER overall heat transfer coefficients Ui and Uo based on the inner and outer surface areas of the tube, respectively. 13–19 Reconsider Problem 13–18. Using EES (or other) software, investigate the effects of pipe conductivity and heat transfer coefficients on the thermal resistance of the heat exchanger. Let the thermal conductivity vary from 10 W/m · ºC to 400 W/m · ºC, the convection heat transfer coefficient from 500 W/m2 · ºC to 1500 W/m2 · ºC on the inner surface, and from 1000 W/m2 · ºC to 2000 W/m2 · ºC on the outer surface. Plot the thermal resistance of the heat exchanger as functions of thermal conductivity and heat transfer coefficients, and discuss the results. 13–20 Water at an average temperature of 107°C and an average velocity of 3.5 m/s flows through a 5-m-long stainless steel tube (k 14.2 W/m · °C) in a boiler. The inner and outer diameters of the tube are Di 1.0 cm and Do 1.4 cm, respectively. If the convection heat transfer coefficient at the outer surface of the tube where boiling is taking place is ho 8400 W/m2 · °C, determine the overall heat transfer coefficient Ui of this boiler based on the inner surface area of the tube. 13–21 Repeat Problem 13–20, assuming a fouling factor Rf, i 0.0005 m2 · °C/W on the inner surface of the tube. 13–22 Reconsider Problem 13–21. Using EES (or other) software, plot the overall heat transfer coefficient based on the inner surface as a function of fouling factor Fi as it varies from 0.0001 m2 · ºC/W to 0.0008 m2 · ºC/W, and discuss the results. 13–23 A long thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used to condense refrigerant 134a by water at 20°C. The refrigerant flows through the tube, with a convection heat transfer coefficient of hi 5000 W/m2 · °C. Water flows through the shell at a rate of 0.3 kg/s. Determine the overall heat transfer Answer: 2020 W/m2 · °C coefficient of this heat exchanger. 13–24 Repeat Problem 13–23 by assuming a 2-mm-thick layer of limestone (k 1.3 W/m · °C) forms on the outer surface of the inner tube. 13–25 Reconsider Problem 13–24. Using EES (or other) software, plot the overall heat transfer coefficient as a function of the limestone thickness as it varies from 1 mm to 3 mm, and discuss the results. 13–26E Water at an average temperature of 140°F and an average velocity of 8 ft/s flows through a thin-walled 34 -in.diameter tube. The water is cooled by air that flows across the tube with a velocity of 12 ft/s at an average temperature of 80°F. Determine the overall heat transfer coefficient. Analysis of Heat Exchangers 13–27C What are the common approximations made in the analysis of heat exchangers? 13–28C Under what conditions is the heat transfer relation · Q m· cCpc(Tc, out Tc, in) m· hCph (Th, in Th, out) valid for a heat exchanger? 13–29C What is the heat capacity rate? What can you say about the temperature changes of the hot and cold fluids in a heat exchanger if both fluids have the same capacity rate? What does a heat capacity of infinity for a fluid in a heat exchanger mean? 13–30C Consider a condenser in which steam at a specified temperature is condensed by rejecting heat to the cooling water. If the heat transfer rate in the condenser and the temperature rise of the cooling water is known, explain how the rate of condensation of the steam and the mass flow rate of the cooling water can be determined. Also, explain how the total thermal resistance R of this condenser can be evaluated in this case. 13–31C Under what conditions will the temperature rise of the cold fluid in a heat exchanger be equal to the temperature drop of the hot fluid? The Log Mean Temperature Difference Method · 13–32C In the heat transfer relation Q UAs Tlm for a heat exchanger, what is Tlm called? How is it calculated for a parallel-flow and counter-flow heat exchanger? 13–33C How does the log mean temperature difference for a heat exchanger differ from the arithmetic mean temperature difference (AMTD)? For specified inlet and outlet temperatures, which one of these two quantities is larger? 13–34C The temperature difference between the hot and cold fluids in a heat exchanger is given to be T1 at one end and T2 at the other end. Can the logarithmic temperature difference Tlm of this heat exchanger be greater than both T1 and T2? Explain. 13–35C Can the logarithmic mean temperature difference Tlm of a heat exchanger be a negative quantity? Explain. 13–36C Can the outlet temperature of the cold fluid in a heat exchanger be higher than the outlet temperature of the hot fluid in a parallel-flow heat exchanger? How about in a counter-flow heat exchanger? Explain. 13–37C For specified inlet and outlet temperatures, for what kind of heat exchanger will the Tlm be greatest: double-pipe parallel-flow, double-pipe counter-flow, cross-flow, or multipass shell-and-tube heat exchanger? · 13–38C In the heat transfer relation Q UAs F Tlm for a heat exchanger, what is the quantity F called? What does it represent? Can F be greater than one? 13–39C When the outlet temperatures of the fluids in a heat exchanger are not known, is it still practical to use the LMTD method? Explain. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 707 707 CHAPTER 13 13–40C Explain how the LMTD method can be used to determine the heat transfer surface area of a multipass shell-andtube heat exchanger when all the necessary information, including the outlet temperatures, is given. 13–41 Steam in the condenser of a steam power plant is to be condensed at a temperature of 50°C (hfg 2305 kJ/kg) with cooling water (Cp 4180 J/kg · °C) from a nearby lake, which enters the tubes of the condenser at 18°C and leaves at 27°C. The surface area of the tubes is 58 m2, and the overall heat transfer coefficient is 2400 W/m2 · °C. Determine the mass flow rate of the cooling water needed and the rate of condensation of Answers: 101 kg/s, 1.65 kg/s the steam in the condenser. Steam 50°C 27°C 13–45 A test is conducted to determine the overall heat transfer coefficient in a shell-and-tube oil-to-water heat exchanger that has 24 tubes of internal diameter 1.2 cm and length 2 m in a single shell. Cold water (Cp 4180 J/kg · °C) enters the tubes at 20°C at a rate of 5 kg/s and leaves at 55°C. Oil (Cp 2150 J/kg · °C) flows through the shell and is cooled from 120°C to 45°C. Determine the overall heat transfer coefficient Ui of this heat exchanger based on the inner surface area Answer: 13.9 kW/m2 · °C of the tubes. 13–46 A double-pipe counter-flow heat exchanger is to cool ethylene glycol (Cp 2560 J/kg · °C) flowing at a rate of 3.5 kg/s from 80°C to 40°C by water (Cp 4180 J/kg · °C) that enters at 20°C and leaves at 55°C. The overall heat transfer coefficient based on the inner surface area of the tube is 250 W/m2 · °C. Determine (a) the rate of heat transfer, (b) the mass flow rate of water, and (c) the heat transfer surface area on the inner side of the tube. Cold water 20°C Hot glycol 18°C Water 80°C 3.5 kg/s 40°C 50°C FIGURE P13–41 FIGURE P13–46 13–42 A double-pipe parallel-flow heat exchanger is to heat water (Cp 4180 J/kg · °C) from 25°C to 60°C at a rate of 0.2 kg/s. The heating is to be accomplished by geothermal water (Cp 4310 J/kg · °C) available at 140°C at a mass flow rate of 0.3 kg/s. The inner tube is thin-walled and has a diameter of 0.8 cm. If the overall heat transfer coefficient of the heat exchanger is 550 W/m2 · °C, determine the length of the heat exchanger required to achieve the desired heating. 13–43 Reconsider Problem 13–42. Using EES (or other) software, investigate the effects of temperature and mass flow rate of geothermal water on the length of the heat exchanger. Let the temperature vary from 100ºC to 200ºC, and the mass flow rate from 0.1 kg/s to 0.5 kg/s. Plot the length of the heat exchanger as functions of temperature and mass flow rate, and discuss the results. 13–44E A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin (Cp 0.60 Btu/lbm · °F) from 65°F to 140°F by hot water (Cp 1.0 Btu/lbm · °F) that enters the thinwalled 0.5-in.-diameter tubes at 175°F and leaves at 120°F. The total length of the tubes in the heat exchanger is 500 ft. The convection heat transfer coefficient is 4 Btu/h · ft2 · °F on the glycerin (shell) side and 50 Btu/h · ft2 · °F on the water (tube) side. Determine the rate of heat transfer in the heat exchanger (a) before any fouling occurs and (b) after fouling with a fouling factor of 0.002 h · ft2 · °F/Btu occurs on the outer surfaces of the tubes. 13–47 Water (Cp 4180 J/kg · °C) enters the 2.5-cminternal-diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rate of 3 kg/s. It is heated by steam condensing at 120°C (hfg 2203 kJ/kg) in the shell. If the overall heat transfer coefficient of the heat exchanger is 1500 W/m2 · °C, determine the length of the tube required in order to heat the water to 80°C. 13–48 A thin-walled double-pipe counter-flow heat exchanger is to be used to cool oil (Cp 2200 J/kg · °C) from 150°C to 40°C at a rate of 2 kg/s by water (Cp 4180 J/kg · °C) that enters at 22°C at a rate of 1.5 kg/s. The diameter of the tube is 2.5 cm, and its length is 6 m. Determine the overall heat transfer coefficient of this heat exchanger. 13–49 Reconsider Problem 13–48. Using EES (or other) software, investigate the effects of oil exit temperature and water inlet temperature on the overall heat transfer coefficient of the heat exchanger. Let the oil exit temperature vary from 30ºC to 70ºC and the water inlet temperature from 5ºC to 25ºC. Plot the overall heat transfer coefficient as functions of the two temperatures, and discuss the results. 13–50 Consider a water-to-water double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold water enters at 20°C and leaves at 50°C, while the hot water enters at 80°C and leaves at cen58933_ch13.qxd 9/9/2002 9:57 AM Page 708 708 HEAT TRANSFER 45°C. Do you think this is a parallel-flow or counter-flow heat exchanger? Explain. 13–51 Cold water (Cp 4180 J/kg · °C) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (Cp 4190 J/kg · °C) that enters at 100°C at a rate of 3 kg/s. If the overall heat transfer coefficient is 1210 W/m2 · °C, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger. 13–52 Engine oil (Cp 2100 J/kg · °C) is to be heated from 20°C to 60°C at a rate of 0.3 kg/s in a 2-cm-diameter thinwalled copper tube by condensing steam outside at a temperature of 130°C (hfg 2174 kJ/kg). For an overall heat transfer coefficient of 650 W/m2 · °C, determine the rate of heat transfer and the length of the tube required to achieve it. Answers: 25.2 kW, 7.0 m Steam 130°C Oil 20°C 0.3 kg/s 60°C 55°C FIGURE P13–52 13–53E Geothermal water (Cp 1.03 Btu/lbm · °F) is to be used as the heat source to supply heat to the hydronic heating system of a house at a rate of 30 Btu/s in a double-pipe counter-flow heat exchanger. Water (Cp 1.0 Btu/lbm · °F) is heated from 140°F to 200°F in the heat exchanger as the geothermal water is cooled from 310°F to 180°F. Determine the mass flow rate of each fluid and the total thermal resistance of this heat exchanger. 13–54 Glycerin (Cp 2400 J/kg · °C) at 20°C and 0.3 kg/s is to be heated by ethylene glycol (Cp 2500 J/kg · °C) at 60°C in a thin-walled double-pipe parallel-flow heat exchanger. The temperature difference between the two fluids is 15°C at the outlet of the heat exchanger. If the overall heat transfer coefficient is 240 W/m2 · °C and the heat transfer surface area is 3.2 m2, determine (a) the rate of heat transfer, (b) the outlet temperature of the glycerin, and (c) the mass flow rate of the ethylene glycol. 13–55 Air (Cp 1005 J/kg · °C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. Air enters the heat exchanger at 95 kPa and 20°C at a rate of 0.8 m3/s. The combustion gases (Cp 1100 J/kg · °C) enter at 180°C at a rate of 1.1 kg/s and leave at 95°C. The product of the overall heat transfer coefficient and the heat transfer surface area is AU 1200 W/°C. Assuming both fluids to be unmixed, determine the rate of heat transfer and the outlet temperature of the air. Air 95 kPa 20°C 0.8 m3/s Exhaust gases 1.1 kg/s 95°C FIGURE P13–55 13–56 A shell-and-tube heat exchanger with 2-shell passes and 12-tube passes is used to heat water (Cp 4180 J/kg · °C) in the tubes from 20°C to 70°C at a rate of 4.5 kg/s. Heat is supplied by hot oil (Cp 2300 J/kg · °C) that enters the shell side at 170°C at a rate of 10 kg/s. For a tube-side overall heat transfer coefficient of 600 W/m2 · °C, determine the heat transAnswer: 15 m2 fer surface area on the tube side. 13–57 Repeat Problem 13–56 for a mass flow rate of 2 kg/s for water. 13–58 A shell-and-tube heat exchanger with 2-shell passes and 8-tube passes is used to heat ethyl alcohol (Cp 2670 J/kg · °C) in the tubes from 25°C to 70°C at a rate of 2.1 kg/s. The heating is to be done by water (Cp 4190 J/kg · °C) that enters the shell side at 95°C and leaves at 45°C. If the overall heat transfer coefficient is 950 W/m2 · °C, determine the heat transfer surface area of the heat exchanger. Water 95°C 70°C Ethyl alcohol 25°C 2.1 kg/s (8-tube passes) 45°C FIGURE P13–58 13–59 A shell-and-tube heat exchanger with 2-shell passes and 12-tube passes is used to heat water (Cp 4180 J/kg · °C) with ethylene glycol (Cp 2680 J/kg · °C). Water enters the tubes at 22°C at a rate of 0.8 kg/s and leaves at 70°C. Ethylene glycol enters the shell at 110°C and leaves at 60°C. If the overall heat transfer coefficient based on the tube side is 280 W/m2 · °C, determine the rate of heat transfer and the heat transfer surface area on the tube side. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 709 709 CHAPTER 13 13–60 Reconsider Problem 13–59. Using EES (or other) software, investigate the effect of the mass flow rate of water on the rate of heat transfer and the tube-side surface area. Let the mass flow rate vary from 0.4 kg/s to 2.2 kg/s. Plot the rate of heat transfer and the surface area as a function of the mass flow rate, and discuss the results. 13–61E Steam is to be condensed on the shell side of a 1-shell-pass and 8-tube-passes condenser, with 50 tubes in each pass at 90°F (hfg 1043 Btu/lbm). Cooling water (Cp 1.0 Btu/lbm · °F) enters the tubes at 60°F and leaves at 73°F. The tubes are thin-walled and have a diameter of 3/4 in. and length of 5 ft per pass. If the overall heat transfer coefficient is 600 Btu/h · ft2 · °F, determine (a) the rate of heat transfer, (b) the rate of condensation of steam, and (c) the mass flow rate of cold water. Steam 90°F 20 lbm/s 73°F 60°F Water 90°F FIGURE P13–61E 13–62E Reconsider Problem 13–61E. Using EES (or other) software, investigate the effect of the condensing steam temperature on the rate of heat transfer, the rate of condensation of steam, and the mass flow rate of cold water. Let the steam temperature vary from 80ºF to 120ºF. Plot the rate of heat transfer, the condensation rate of steam, and the mass flow rate of cold water as a function of steam temperature, and discuss the results. 13–63 A shell-and-tube heat exchanger with 1-shell pass and 20–tube passes is used to heat glycerin (Cp 2480 J/kg · °C) in the shell, with hot water in the tubes. The tubes are thinwalled and have a diameter of 1.5 cm and length of 2 m per pass. The water enters the tubes at 100°C at a rate of 5 kg/s and leaves at 55°C. The glycerin enters the shell at 15°C and leaves at 55°C. Determine the mass flow rate of the glycerin and the overall heat transfer coefficient of the heat exchanger. 13–64 In a binary geothermal power plant, the working fluid isobutane is to be condensed by air in a condenser at 75°C (hfg 255.7 kJ/kg) at a rate of 2.7 kg/s. Air enters the condenser at 21ºC and leaves at 28ºC. The heat transfer surface Air 28°C Isobutane 75°C 2.7 kg/s Air 21°C FIGURE P13–64 area based on the isobutane side is 24 m2. Determine the mass flow rate of air and the overall heat transfer coefficient. 13–65 Hot exhaust gases of a stationary diesel engine are to be used to generate steam in an evaporator. Exhaust gases (Cp 1051 J/kg · ºC) enter the heat exchanger at 550ºC at a rate of 0.25 kg/s while water enters as saturated liquid and evaporates at 200ºC (hfg 1941 kJ/kg). The heat transfer surface area of the heat exchanger based on water side is 0.5 m2 and overall heat transfer coefficient is 1780 W/m2 · ºC. Determine the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water. 13–66 Reconsider Problem 13–65. Using EES (or other) software, investigate the effect of the exhaust gas inlet temperature on the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water. Let the temperature of exhaust gases vary from 300ºC to 600ºC. Plot the rate of heat transfer, the exit temperature of exhaust gases, and the rate of evaporation of water as a function of the temperature of the exhaust gases, and discuss the results. 13–67 In a textile manufacturing plant, the waste dyeing water (Cp 4295 J/g · ºC) at 75°C is to be used to preheat fresh water (Cp 4180 J/kg · ºC) at 15ºC at the same flow rate in a double-pipe counter-flow heat exchanger. The heat transfer surface area of the heat exchanger is 1.65 m2 and the overall heat transfer coefficient is 625 W/m2 · ºC. If the rate of heat transfer in the heat exchanger is 35 kW, determine the outlet temperature and the mass flow rate of each fluid stream. Fresh water 15°C Dyeing water 75°C Th, out Tc, out FIGURE P13–67 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 710 710 HEAT TRANSFER The Effectiveness–NTU Method 13–68C Under what conditions is the effectiveness–NTU method definitely preferred over the LMTD method in heat exchanger analysis? 13–69C What does the effectiveness of a heat exchanger represent? Can effectiveness be greater than one? On what factors does the effectiveness of a heat exchanger depend? 13–70C For a specified fluid pair, inlet temperatures, and mass flow rates, what kind of heat exchanger will have the highest effectiveness: double-pipe parallel-flow, double-pipe counter-flow, cross-flow, or multipass shell-and-tube heat exchanger? 13–71C Explain how you can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined. 13–72C Can the temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? Explain. 13–73C Can the temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? Explain. 13–74C Consider a heat exchanger in which both fluids have the same specific heats but different mass flow rates. Which fluid will experience a larger temperature change: the one with the lower or higher mass flow rate? 13–75C Explain how the maximum possible heat transfer · rate Q max in a heat exchanger can be determined when the mass flow rates, specific heats, and the inlet temperatures of the two · fluids are specified. Does the value of Q max depend on the type of the heat exchanger? 13–76C Consider two double-pipe counter-flow heat exchangers that are identical except that one is twice as long as the other one. Which heat exchanger is more likely to have a higher effectiveness? 13–77C Consider a double-pipe counter-flow heat exchanger. In order to enhance heat transfer, the length of the heat exchanger is now doubled. Do you think its effectiveness will also double? 13–78C Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hot and cold water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same. 13–79C Under what conditions can a counter-flow heat exchanger have an effectiveness of one? What would your answer be for a parallel-flow heat exchanger? 13–80C How is the NTU of a heat exchanger defined? What does it represent? Is a heat exchanger with a very large NTU (say, 10) necessarily a good one to buy? 13–81C Consider a heat exchanger that has an NTU of 4. Someone proposes to double the size of the heat exchanger and thus double the NTU to 8 in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal? 13–82C Consider a heat exchanger that has an NTU of 0.1. Someone proposes to triple the size of the heat exchanger and thus triple the NTU to 0.3 in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal? 13–83 Air (Cp 1005 J/kg · °C) enters a cross-flow heat exchanger at 10°C at a rate of 3 kg/s, where it is heated by a hot water stream (Cp 4190 J/kg · °C) that enters the heat exchanger at 95°C at a rate of 1 kg/s. Determine the maximum heat transfer rate and the outlet temperatures of the cold and the hot water streams for that case. 13–84 Hot oil (Cp 2200 J/kg · °C) is to be cooled by water (Cp 4180 J/kg · °C) in a 2-shell-pass and 12-tube-pass heat exchanger. The tubes are thin-walled and are made of copper with a diameter of 1.8 cm. The length of each tube pass in the heat exchanger is 3 m, and the overall heat transfer coefficient is 340 W/m2 · °C. Water flows through the tubes at a total rate of 0.1 kg/s, and the oil through the shell at a rate of 0.2 kg/s. The water and the oil enter at temperatures 18°C and 160°C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil. Answers: 36.2 kW, 104.6°C, 77.7°C Oil 160°C 0.2 kg/s Water 18°C 0.1 kg/s (12-tube passes) FIGURE P13–84 13–85 Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C. Do you think this is a parallel-flow or counter-flow heat exchanger? Why? Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger. 13–86E Hot water enters a double-pipe counter-flow waterto-oil heat exchanger at 220°F and leaves at 100°F. Oil enters at 70°F and leaves at 150°F. Determine which fluid has the smaller heat capacity rate and calculate the effectiveness of this heat exchanger. 13–87 A thin-walled double-pipe parallel-flow heat exchanger is used to heat a chemical whose specific heat is 1800 cen58933_ch13.qxd 9/9/2002 9:57 AM Page 711 711 CHAPTER 13 J/kg · °C with hot water (Cp 4180 J/kg · °C). The chemical enters at 20°C at a rate of 3 kg/s, while the water enters at 110°C at a rate of 2 kg/s. The heat transfer surface area of the heat exchanger is 7 m2 and the overall heat transfer coefficient is 1200 W/m2 · °C. Determine the outlet temperatures of the chemical and the water. Chemical 20°C 3 kg/s Hot water 110°C 2 kg/s by hot oil (Cp 2200 J/kg · °C) that enters at 120°C. If the heat transfer surface area and the overall heat transfer coefficients are 6.2 m2 and 320 W/m2 · °C, respectively, determine the outlet temperature and the mass flow rate of oil using (a) the LMTD method and (b) the –NTU method. 13–92 Water (Cp 4180 J/kg · °C) is to be heated by solarheated hot air (Cp 1010 J/kg · °C) in a double-pipe counterflow heat exchanger. Air enters the heat exchanger at 90°C at a rate of 0.3 kg/s, while water enters at 22°C at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given to be 80 W/m2 · °C. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm. Determine the outlet temperatures of the water and the air. 13–93 FIGURE P13–87 13–88 Reconsider Problem 13–87. Using EES (or other) software, investigate the effects of the inlet temperatures of the chemical and the water on their outlet temperatures. Let the inlet temperature vary from 10ºC to 50ºC for the chemical and from 80ºC to 150ºC for water. Plot the outlet temperature of each fluid as a function of the inlet temperature of that fluid, and discuss the results. 13–89 A cross-flow air-to-water heat exchanger with an effectiveness of 0.65 is used to heat water (Cp 4180 J/kg · °C) with hot air (Cp 1010 J/kg · °C). Water enters the heat exchanger at 20°C at a rate of 4 kg/s, while air enters at 100°C at a rate of 9 kg/s. If the overall heat transfer coefficient based on the water side is 260 W/m2 · °C, determine the heat transfer surface area of the heat exchanger on the water side. Assume Answer: 52.4 m2 both fluids are unmixed. 13–90 Water (Cp 4180 J/kg · °C) enters the 2.5-cminternal-diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rate of 3 kg/s. Water is heated by steam condensing at 120°C (hfg 2203 kJ/kg) in the shell. If the overall heat transfer coefficient of the heat exchanger is 900 W/m2 · °C, determine the length of the tube required in order to heat the water to 80°C using (a) the LMTD method and (b) the –NTU method. Reconsider Problem 13–92. Using EES (or other) software, investigate the effects of the mass flow rate of water and the tube length on the outlet temperatures of water and air. Let the mass flow rate vary from 0.05 kg/s to 1.0 kg/s and the tube length from 5 m to 25 m. Plot the outlet temperatures of the water and the air as the functions of the mass flow rate and the tube length, and discuss the results. 13–94E A thin-walled double-pipe heat exchanger is to be used to cool oil (Cp 0.525 Btu/lbm · °F) from 300°F to 105°F at a rate of 5 lbm/s by water (Cp 1.0 Btu/lbm · °F) that enters at 70°F at a rate of 3 lbm/s. The diameter of the tube is 1 in. and its length is 20 ft. Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and (b) the –NTU method. 13–95 Cold water (Cp 4180 J/kg · °C) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (Cp 4190 J/kg · °C) that enters at 100°C at a rate of 3 kg/s. If the overall heat transfer coefficient is 950 W/m2 · °C, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the –NTU Answers: 31.35 kW, 0.482 m2 method. Cold water 15°C 0.25 kg/s Hot water 13–91 Ethanol is vaporized at 78°C (hfg 846 kJ/kg) in a double-pipe parallel-flow heat exchanger at a rate of 0.03 kg/s 100°C 3 kg/s Oil 120°C 45°C FIGURE P13–95 Ethanol 78°C 0.03 kg/s FIGURE P13–91 13–96 Reconsider Problem 13–95. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and surface area. Let the inlet temperature vary from 60ºC to 120ºC and the overall heat cen58933_ch13.qxd 9/9/2002 9:57 AM Page 712 712 HEAT TRANSFER transfer coefficient from 750 W/m2 · °C to 1250 W/m2 · °C. Plot the rate of heat transfer and surface area as functions of inlet temperature and the heat transfer coefficient, and discuss the results. Steam 30°C 13–97 Glycerin (Cp 2400 J/kg · °C) at 20°C and 0.3 kg/s is to be heated by ethylene glycol (Cp 2500 J/kg · °C) at 60°C and the same mass flow rate in a thin-walled doublepipe parallel-flow heat exchanger. If the overall heat transfer coefficient is 380 W/m2 · °C and the heat transfer surface area is 5.3 m2, determine (a) the rate of heat transfer and (b) the outlet temperatures of the glycerin and the glycol. 13–98 A cross-flow heat exchanger consists of 40 thinwalled tubes of 1-cm diameter located in a duct of 1 m 1 m cross-section. There are no fins attached to the tubes. Cold water (Cp 4180 J/kg · °C) enters the tubes at 18°C with an average velocity of 3 m/s, while hot air (Cp 1010 J/kg · °C) enters the channel at 130°C and 105 kPa at an average velocity of 12 m/s. If the overall heat transfer coefficient is 130 W/m2 · °C, determine the outlet temperatures of both fluids and the rate of heat transfer. 1m Hot air 130°C 105 kPa 12 m/s 1m Water 18°C 3 m/s FIGURE P13–98 13–99 A shell-and-tube heat exchanger with 2-shell passes and 8-tube passes is used to heat ethyl alcohol (Cp 2670 J/kg · °C) in the tubes from 25°C to 70°C at a rate of 2.1 kg/s. The heating is to be done by water (Cp 4190 J/kg · °C) that enters the shell at 95°C and leaves at 60°C. If the overall heat transfer coefficient is 800 W/m2 · °C, determine the heat transfer surface area of the heat exchanger using (a) the LMTD method and (b) the –NTU method. Answer (a): 11.4 m2 13–100 Steam is to be condensed on the shell side of a 1-shell-pass and 8-tube-passes condenser, with 50 tubes in each pass, at 30°C (hfg 2430 kJ/kg). Cooling water (Cp 4180 J/kg · °C) enters the tubes at 15°C at a rate of 1800 kg/h. The tubes are thin-walled, and have a diameter of 1.5 cm and length of 2 m per pass. If the overall heat transfer coefficient is 3000 W/m2 · °C, determine (a) the rate of heat transfer and (b) the rate of condensation of steam. 15°C Water 1800 kg/h 30°C FIGURE P13–100 13–101 Reconsider Problem 13–100. Using EES (or other) software, investigate the effects of the condensing steam temperature and the tube diameters on the rate of heat transfer and the rate of condensation of steam. Let the steam temperature vary from 20ºC to 70ºC and the tube diameter from 1.0 cm to 2.0 cm. Plot the rate of heat transfer and the rate of condensation as functions of steam temperature and tube diameter, and discuss the results. 13–102 Cold water (Cp 4180 J/kg · °C) enters the tubes of a heat exchanger with 2-shell-passes and 13–tube-passes at 20°C at a rate of 3 kg/s, while hot oil (Cp 2200 J/kg · °C) enters the shell at 130°C at the same mass flow rate. The overall heat transfer coefficient based on the outer surface of the tube is 300 W/m2 · °C and the heat transfer surface area on that side is 20 m2. Determine the rate of heat transfer using (a) the LMTD method and (b) the –NTU method. Selection of Heat Exchangers 13–103C A heat exchanger is to be selected to cool a hot liquid chemical at a specified rate to a specified temperature. Explain the steps involved in the selection process. 13–104C There are two heat exchangers that can meet the heat transfer requirements of a facility. One is smaller and cheaper but requires a larger pump, while the other is larger and more expensive but has a smaller pressure drop and thus requires a smaller pump. Both heat exchangers have the same life expectancy and meet all other requirements. Explain which heat exchanger you would choose under what conditions. 13–105C There are two heat exchangers that can meet the heat transfer requirements of a facility. Both have the same pumping power requirements, the same useful life, and the same price tag. But one is heavier and larger in size. Under what conditions would you choose the smaller one? 13–106 A heat exchanger is to cool oil (Cp 2200 J/kg · °C) at a rate of 13 kg/s from 120°C to 50°C by air. Determine the heat transfer rating of the heat exchanger and propose a suitable type. cen58933_ch13.qxd 9/9/2002 9:57 AM Page 713 713 CHAPTER 13 13–107 A shell-and-tube process heater is to be selected to heat water (Cp 4190 J/kg · °C) from 20°C to 90°C by steam flowing on the shell side. The heat transfer load of the heater is 600 kW. If the inner diameter of the tubes is 1 cm and the velocity of water is not to exceed 3 m/s, determine how many tubes need to be used in the heat exchanger. 60°C. If the overall heat transfer coefficient based on the outer surface of the tube is 300 W/m2 · °C, determine (a) the rate of heat transfer and (b) the heat transfer surface area on the outer Answers: (a) 462 kW, (b) 29.2 m2 side of the tube. Hot oil 130°C 3 kg/s Steam 90°C Cold water 20°C 3 kg/s (20-tube passes) 60°C 20°C Water FIGURE P13–107 13–108 Reconsider Problem 13–107. Using EES (or other) software, plot the number of tube passes as a function of water velocity as it varies from 1 m/s to 8 m/s, and discuss the results. 13–109 The condenser of a large power plant is to remove 500 MW of heat from steam condensing at 30°C (hfg 2430 kJ/kg). The cooling is to be accomplished by cooling water (Cp 4180 J/kg · °C) from a nearby river, which enters the tubes at 18°C and leaves at 26°C. The tubes of the heat exchanger have an internal diameter of 2 cm, and the overall heat transfer coefficient is 3500 W/m2 · °C. Determine the total length of the tubes required in the condenser. What type of heat Answer: 312.3 km exchanger is suitable for this task? 13–110 Repeat Problem 13–109 for a heat transfer load of 300 MW. FIGURE P13–113 13–114E Water (Cp 1.0 Btu/lbm · °F) is to be heated by solar-heated hot air (Cp 0.24 Btu/lbm · °F) in a double-pipe counter-flow heat exchanger. Air enters the heat exchanger at 190°F at a rate of 0.7 lbm/s and leaves at 135°F. Water enters at 70°F at a rate of 0.35 lbm/s. The overall heat transfer coefficient based on the inner side of the tube is given to be 20 Btu/h · ft2 · °F. Determine the length of the tube required for a tube internal diameter of 0.5 in. 13–115 By taking the limit as T2 → T1, show that when T1 T2 for a heat exchanger, the Tlm relation reduces to Tlm T1 T2. 13–116 The condenser of a room air conditioner is designed to reject heat at a rate of 15,000 kJ/h from Refrigerant-134a as the refrigerant is condensed at a temperature of 40°C. Air (Cp 1005 J/kg · °C) flows across the finned condenser coils, entering at 25°C and leaving at 35°C. If the overall heat transfer coefficient based on the refrigerant side is 150 W/m2 · °C, determine the heat transfer area on the refrigerant side. Answer: 3.05 m2 R-134a 40°C Review Problems 13–111 Hot oil is to be cooled in a multipass shell-and-tube heat exchanger by water. The oil flows through the shell, with a heat transfer coefficient of ho 35 W/m2 · °C, and the water flows through the tube with an average velocity of 3 m/s. The tube is made of brass (k 110 W/m · °C) with internal and external diameters of 1.3 cm and 1.5 cm, respectively. Using water properties at 25°C, determine the overall heat transfer coefficient of this heat exchanger based on the inner surface. 35°C Air 25°C 13–112 Repeat Problem 13–111 by assuming a fouling factor Rf, o 0.0004 m2 · °C/W on the outer surface of the tube. 13–113 Cold water (Cp 4180 J/kg · °C) enters the tubes of a heat exchanger with 2-shell passes and 20–tube passes at 20°C at a rate of 3 kg/s, while hot oil (Cp 2200 J/kg · °C) enters the shell at 130°C at the same mass flow rate and leaves at 40°C FIGURE P13–116 13–117 Air (Cp 1005 J/kg · °C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters cen58933_ch13.qxd 9/9/2002 9:57 AM Page 714 714 HEAT TRANSFER the furnace. Air enters the heat exchanger at 95 kPa and 20°C at a rate of 0.8 m3/s. The combustion gases (Cp 1100 J/kg · °C) enter at 180°C at a rate of 1.1 kg/s and leave at 95°C. The product of the overall heat transfer coefficient and the heat transfer surface area is UAs 1620 W/°C. Assuming both fluids to be unmixed, determine the rate of heat transfer. 13–118 In a chemical plant, a certain chemical is heated by hot water supplied by a natural gas furnace. The hot water (Cp 4180 J/kg · °C) is then discharged at 60°C at a rate of 8 kg/min. The plant operates 8 h a day, 5 days a week, 52 weeks a year. The furnace has an efficiency of 78 percent, and the cost of the natural gas is $0.54 per therm (1 therm 100,000 Btu 105,500 kJ). The average temperature of the cold water entering the furnace throughout the year is 14°C. In order to save energy, it is proposed to install a water-to-water heat exchanger to preheat the incoming cold water by the drained hot water. Assuming that the heat exchanger will recover 72 percent of the available heat in the hot water, determine the heat transfer rating of the heat exchanger that needs to be purchased and suggest a suitable type. Also, determine the amount of money this heat exchanger will save the company per year from natural gas savings. 13–119 A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at 120ºC (hfg 2203 kJ/kg) on the shell side. The tubes are thin-walled and have a diameter of 2.4 cm and length of 3.2 m per pass. Water (Cp 4180 J/kg · ºC) enters the tubes at 22ºC at a rate of 3.9 kg/s. If the temperature difference between the two fluids at the exit is 46ºC, determine (a) the rate of heat transfer, (b) the rate of condensation of steam, and (c) the overall heat transfer coefficient. Steam 120°C mass flow rate of geothermal water and the outlet temperatures of both fluids. 13–121 Air at 18ºC (Cp 1006 J/kg · ºC) is to be heated to 70ºC by hot oil at 80ºC (Cp 2150 J/kg · ºC) in a cross-flow heat exchanger with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is 750 W/m2 · ºC and the mass flow rate of air is twice that of oil. Determine (a) the effectiveness of the heat exchanger, (b) the mass flow rate of air, and (c) the rate of heat transfer. 13–122 Consider a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at 95ºC while cold water enters at 20ºC. The exit temperature of hot water is 15ºC greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is 1400 W/m2 · ºC. Taking the specific heat of both cold and hot water to be Cp 4180 J/kg · ºC, determine (a) the outlet temperature of the cold water, (b) the effectiveness of the heat exchanger, (c) the mass flow rate of the cold water, and (d) the heat transfer rate. Cold water 20°C Hot water 95°C FIGURE P13–122 Computer, Design, and Essay Problems 22°C Water 3.9 kg/s 14 tubes 120°C FIGURE P13–119 13–120 Geothermal water (Cp 4250 J/kg · ºC) at 95ºC is to be used to heat fresh water (Cp 4180 J/kg · ºC) at 12ºC at a rate of 1.2 kg/s in a double-pipe counter-flow heat exchanger. The heat transfer surface area is 25 m2, the overall heat transfer coefficient is 480 W/m2 · ºC, and the mass flow rate of geothermal water is larger than that of fresh water. If the effectiveness of the heat exchanger is desired to be 0.823, determine the 13–123 Write an interactive computer program that will give the effectiveness of a heat exchanger and the outlet temperatures of both the hot and cold fluids when the type of fluids, the inlet temperatures, the mass flow rates, the heat transfer surface area, the overall heat transfer coefficient, and the type of heat exchanger are specified. The program should allow the user to select from the fluids water, engine oil, glycerin, ethyl alcohol, and ammonia. Assume constant specific heats at about room temperature. 13–124 Water flows through a shower head steadily at a rate of 8 kg/min. The water is heated in an electric water heater from 15°C to 45°C. In an attempt to conserve energy, it is proposed to pass the drained warm water at a temperature of 38°C through a heat exchanger to preheat the incoming cold water. Design a heat exchanger that is suitable for this task, and discuss the potential savings in energy and money for your area. 13–125 Open the engine compartment of your car and search for heat exchangers. How many do you have? What type are they? Why do you think those specific types are selected? If cen58933_ch13.qxd 9/9/2002 9:58 AM Page 715 715 CHAPTER 13 you were redesigning the car, would you use different kinds? Explain. 13–126 Write an essay on the static and dynamic types of regenerative heat exchangers and compile information about the manufacturers of such heat exchangers. Choose a few models by different manufacturers and compare their costs and performance. 13–127 Design a hydrocooling unit that can cool fruits and vegetables from 30°C to 5°C at a rate of 20,000 kg/h under the following conditions: The unit will be of flood type that will cool the products as they are conveyed into the channel filled with water. The products will be dropped into the channel filled with water at one end and picked up at the other end. The channel can be as wide as 3 m and as high as 90 cm. The water is to be circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the coils is to be –2°C, and the water temperature is not to drop below 1°C and not to exceed 6°C. Assuming reasonable values for the average product density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for the quantities related to the thermal aspects of the hydrocooler, including (a) how long the fruits and vegetables need to remain in the channel, (b) the length of the channel, (c) the water velocity through the channel, (d) the velocity of the conveyor and thus the fruits and vegetables through the channel, (e) the refrigeration capacity of the refrigeration system, and (f) the type of heat exchanger for the evaporator and the surface area on the water side. 13–128 Design a scalding unit for slaughtered chicken to loosen their feathers before they are routed to feather-picking machines with a capacity of 1200 chickens per hour under the following conditions: The unit will be of immersion type filled with hot water at an average temperature of 53°C at all times. Chickens with an average mass of 2.2 kg and an average temperature of 36°C will be dipped into the tank, held in the water for 1.5 min, and taken out by a slow-moving conveyor. Each chicken is expected to leave the tank 15 percent heavier as a result of the water that sticks to its surface. The center-to-center distance between chickens in any direction will be at least 30 cm. The tank can be as wide as 3 m and as high as 60 cm. The water is to be circulated through and heated by a natural gas furnace, but the temperature rise of water will not exceed 5°C as it passes through the furnace. The water loss is to be made up by the city water at an average temperature of 16°C. The ambient air temperature can be taken to be 20°C. The walls and the floor of the tank are to be insulated with a 2.5-cm-thick urethane layer. The unit operates 24 h a day and 6 days a week. Assuming reasonable values for the average properties, recommend reasonable values for the quantities related to the thermal aspects of the scalding tank, including (a) the mass flow rate of the make-up water that must be supplied to the tank; (b) the length of the tank; (c) the rate of heat transfer from the water to the chicken, in kW; (d) the velocity of the conveyor and thus the chickens through the tank; (e) the rate of heat loss from the exposed surfaces of the tank and if it is significant; ( f ) the size of the heating system in kJ/h; (g) the type of heat exchanger for heating the water with flue gases of the furnace and the surface area on the water side; and (h) the operating cost of the scalding unit per month for a unit cost of $0.56 therm of natural gas (1 therm 105,000 kJ). 13–129 A company owns a refrigeration system whose refrigeration capacity is 200 tons (1 ton of refrigeration 211 kJ/min), and you are to design a forced-air cooling system for fruits whose diameters do not exceed 7 cm under the following conditions: The fruits are to be cooled from 28°C to an average temperature of 8°C. The air temperature is to remain above –2°C and below 10°C at all times, and the velocity of air approaching the fruits must remain under 2 m/s. The cooling section can be as wide as 3.5 m and as high as 2 m. Assuming reasonable values for the average fruit density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for the quantities related to the thermal aspects of the forced-air cooling, including (a) how long the fruits need to remain in the cooling section; (b) the length of the cooling section; (c) the air velocity approaching the cooling section; (d) the product cooling capacity of the system, in kg · fruit/h; (e) the volume flow rate of air; and ( f ) the type of heat exchanger for the evaporator and the surface area on the air side. cen58933_ch13.qxd 9/9/2002 9:58 AM Page 716 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 717 CHAPTER MASS TRANSFER o this point we have restricted our attention to heat transfer problems that did not involve any mass transfer. However, many significant heat transfer problems encountered in practice involve mass transfer. For example, about one-third of the heat loss from a resting person is due to evaporation. It turns out that mass transfer is analogous to heat transfer in many respects, and there is close resemblance between heat and mass transfer relations. In this chapter we discuss the mass transfer mechanisms and develop relations for the mass transfer rate for some situations commonly encountered in practice. Distinction should be made between mass transfer and the bulk fluid motion (or fluid flow) that occurs on a macroscopic level as a fluid is transported from one location to another. Mass transfer requires the presence of two regions at different chemical compositions, and mass transfer refers to the movement of a chemical species from a high concentration region toward a lower concentration one relative to the other chemical species present in the medium. The primary driving force for fluid flow is the pressure difference, whereas for mass transfer it is the concentration difference. Therefore, we do not speak of mass transfer in a homogeneous medium. We begin this chapter by pointing out numerous analogies between heat and mass transfer and draw several parallels between them. We then discuss boundary conditions associated with mass transfer and one-dimensional steady and transient mass diffusion. Following is a discussion of mass transfer in a moving medium. Finally, we consider convection mass transfer and simultaneous heat and mass transfer. T 14 CONTENTS 14–1 Introduction 718 14–2 Analogy between Heat and Mass Transfer 719 14–3 Mass Diffusion 721 14–4 Boundary Conditions 727 14–5 Steady Mass Diffusion through a Wall 732 14–6 Water Vapor Migration in Buildings 736 14–7 Transient Mass Diffusion 740 14–8 Diffusion in a Moving Medium 743 14–9 Mass Convection 754 14–10 Simultaneous Heat and Mass Transfer 763 717 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 718 718 HEAT TRANSFER 14–1 Water Salty water Salt (a) Before (b) After FIGURE 14–1 Whenever there is concentration difference of a physical quantity in a medium, nature tends to equalize things by forcing a flow from the high to the low concentration region. ■ INTRODUCTION It is a common observation that whenever there is an imbalance of a commodity in a medium, nature tends to redistribute it until a “balance” or “equality” is established. This tendency is often referred to as the driving force, which is the mechanism behind many naturally occurring transport phenomena. If we define the amount of a commodity per unit volume as the concentration of that commodity, we can say that the flow of a commodity is always in the direction of decreasing concentration; that is, from the region of high concentration to the region of low concentration (Fig. 14–1). The commodity simply creeps away during redistribution, and thus the flow is a diffusion process. The rate of flow of the commodity is proportional to the concentration gradient dC/dx, which is the change in the concentration C per unit length in the flow direction x, and the area A normal to flow direction and is expressed as Flow rate (Normal area)(Concentration gradient) or · dC Q kdiff A dx 1 Initial N2 concentration 0.79 Initial O2 concentration 0.21 0 x N2 N2 Air O2 FIGURE 14–2 A tank that contains N2 and air in its two compartments, and the diffusion of N2 into the air when the partition is removed. (14-1) Here the proportionality constant kdiff is the diffusion coefficient of the medium, which is a measure of how fast a commodity diffuses in the medium, and the negative sign is to make the flow in the positive direction a positive quantity (note that dC/dx is a negative quantity since concentration decreases in the flow direction). You may recall that Fourier’s law of heat conduction, Ohm’s law of electrical conduction, and Newton’s law of viscosity are all in the form of Equation 14–1. To understand the diffusion process better, consider a tank that is divided into two equal parts by a partition. Initially, the left half of the tank contains nitrogen N2 gas while the right half contains air (about 21 percent O2 and 79 percent N2) at the same temperature and pressure. The O2 and N2 molecules are indicated by dark and light circles, respectively. When the partition is removed, we know that the N2 molecules will start diffusing into the air while the O2 molecules diffuse into the N2, as shown in Figure 14–2. If we wait long enough, we will have a homogeneous mixture of N2 and O2 in the tank. This mass diffusion process can be explained by considering an imaginary plane indicated by the dashed line in the figure as: Gas molecules move randomly, and thus the probability of a molecule moving to the right or to the left is the same. Consequently, half of the molecules on one side of the dashed line at any given moment will move to the other side. Since the concentration of N2 is greater on the left side than it is on the right side, more N2 molecules will move toward the right than toward the left, resulting in a net flow of N2 toward the right. As a result, N2 is said to be transferred to the right. A similar argument can be given for O2 being transferred to the left. The process continues until uniform concentrations of N2 and O2 are established throughout the tank so that the number of N2 (or O2) molecules moving to the right equals cen58933_ch14.qxd 9/9/2002 10:05 AM Page 719 719 CHAPTER 14 the number moving to the left, resulting in zero net transfer of N2 or O2 across an imaginary plane. The molecules in a gas mixture continually collide with each other, and the diffusion process is strongly influenced by this collision process. The collision of like molecules is of little consequence since both molecules are identical and it makes no difference which molecule crosses a certain plane. The collisions of unlike molecules, however, influence the rate of diffusion since unlike molecules may have different masses and thus different momentums, and thus the diffusion process will be dominated by the heavier molecules. The diffusion coefficients and thus diffusion rates of gases depend strongly on temperature since the temperature is a measure of the average velocity of gas molecules. Therefore, the diffusion rates will be higher at higher temperatures. Mass transfer can also occur in liquids and solids as well as in gases. For example, a cup of water left in a room will eventually evaporate as a result of water molecules diffusing into the air (liquid-to-gas mass transfer). A piece of solid CO2 (dry ice) will also get smaller and smaller in time as the CO2 molecules diffuse into the air (solid-to-gas mass transfer). A spoon of sugar in a cup of coffee will eventually move up and sweeten the coffee although the sugar molecules are much heavier than the water molecules, and the molecules of a colored pencil inserted into a glass of water will diffuse into the water as evidenced by the gradual spread of color in the water (solid-to-liquid mass transfer). Of course, mass transfer can also occur from a gas to a liquid or solid if the concentration of the species is higher in the gas phase. For example, a small fraction of O2 in the air diffuses into the water and meets the oxygen needs of marine animals. The diffusion of carbon into iron during case-hardening, doping of semiconductors for transistors, and the migration of doped molecules in semiconductors at high temperature are examples of solidto-solid diffusion processes (Fig. 14–3). Another factor that influences the diffusion process is the molecular spacing. The larger the spacing, in general, the higher the diffusion rate. Therefore, the diffusion rates are typically much higher in gases than they are in liquids and much higher in liquids than in solids. Diffusion coefficients in gas mixtures are a few orders of magnitude larger than these of liquid or solid solutions. 14–2 ■ ANALOGY BETWEEN HEAT AND MASS TRANSFER We have spent a considerable amount of time studying heat transfer, and we could spend just as much time (perhaps more) studying mass transfer. However, the mechanisms of heat and mass transfer are analogous to each other, and thus we can develop an understanding of mass transfer in a short time with little effort by simply drawing parallels between heat and mass transfer. Establishing those “bridges” between the two seemingly unrelated areas will make it possible to use our heat transfer knowledge to solve mass transfer problems. Alternately, gaining a working knowledge of mass transfer will help us to better understand the heat transfer processes by thinking of heat as a massless substance as they did in the nineteenth century. The short-lived caloric theory of heat is the origin of most heat transfer terminology used Air Air Water vapor CO2 Liquid water Dry ice (a) Liquid to gas (b) Solid to gas Coffee Iron Sugar Carbon (c) Solid to liquid (d) Solid to solid FIGURE 14–3 Some examples of mass transfer that involve a liquid and/or a solid. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 720 720 HEAT TRANSFER 70°C Heat 10°C Heat concentration today and served its purpose well until it was replaced by the kinetic theory. Mass is, in essence, energy since mass and energy can be converted to each other according to Einstein’s formula E mc2, where c is the speed of light. Therefore, we can look at mass and heat as two different forms of energy and exploit this to advantage without going overboard. Temperature Mass concentration 70% CO2 Mass 10% CO2 FIGURE 14–4 Analogy between heat and mass transfer. The driving force for heat transfer is the temperature difference. In contrast, the driving force for mass transfer is the concentration difference. We can view temperature as a measure of “heat concentration,” and thus a high temperature region as one that has a high heat concentration (Fig. 14–4). Therefore, both heat and mass are transferred from the more concentrated regions to the less concentrated ones. If there is no temperature difference between two regions, then there is no heat transfer. Likewise, if there is no difference between the concentrations of a species at different parts of a medium, there will be no mass transfer. Conduction Thermal radiation No mass radiation Hot body Mass You will recall that heat is transferred by conduction, convection, and radiation. Mass, however, is transferred by conduction (called diffusion) and convection only, and there is no such thing as “mass radiation” (unless there is something Scotty knows that we don’t when he “beams” people to anywhere in space at the speed of light) (Fig. 14–5). The rate of heat conduction in a direction x is proportional to the temperature gradient dT/dx in that direction and is expressed by Fourier’s law of heat conduction as FIGURE 14–5 Unlike heat radiation, there is no such thing as mass radiation. Temperature profile A · dT Q cond kA dx where k is the thermal conductivity of the medium and A is the area normal to the direction of heat transfer. Likewise, the rate of mass diffusion m· diff of a chemical species A in a stationary medium in the direction x is proportional to the concentration gradient dC/dx in that direction and is expressed by Fick’s law of diffusion by (Fig. 14–6) dCA m· diff DAB A dx · dT Qcond = – kA — dx x A Concentration profile of species A dC m· diff = – DAB A ——A dx FIGURE 14–6 Analogy between heat conduction and mass diffusion. (14-2) (14-3) where DAB is the diffusion coefficient (or mass diffusivity) of the species in the mixture and CA is the concentration of the species in the mixture at that location. It can be shown that the differential equations for both heat conduction and mass diffusion are of the same form. Therefore, the solutions of mass diffusion equations can be obtained from the solutions of corresponding heat conduction equations for the same type of boundary conditions by simply switching the corresponding coefficients and variables. Heat Generation Heat generation refers to the conversion of some form of energy such as electrical, chemical, or nuclear energy into sensible heat energy in the medium. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 721 721 CHAPTER 14 Heat generation occurs throughout the medium and exhibits itself as a rise in temperature. Similarly, some mass transfer problems involve chemical reactions that occur within the medium and result in the generation of a species throughout. Therefore, species generation is a volumetric phenomenon, and the rate of generation may vary from point to point in the medium. Such reactions that occur within the medium are called homogeneous reactions and are analogous to internal heat generation. In contrast, some chemical reactions result in the generation of a species at the surface as a result of chemical reactions occurring at the surface due to contact between the medium and the surroundings. This is a surface phenomenon, and as such it needs to be treated as a boundary condition. In mass transfer studies, such reactions are called heterogeneous reactions and are analogous to specified surface heat flux. Convection You will recall that heat convection is the heat transfer mechanism that involves both heat conduction (molecular diffusion) and bulk fluid motion. Fluid motion enhances heat transfer considerably by removing the heated fluid near the surface and replacing it by the cooler fluid further away. In the limiting case of no bulk fluid motion, convection reduces to conduction. Likewise, mass convection (or convective mass transfer) is the mass transfer mechanism between a surface and a moving fluid that involves both mass diffusion and bulk fluid motion. Fluid motion also enhances mass transfer considerably by removing the high concentration fluid near the surface and replacing it by the lower concentration fluid further away. In mass convection, we define a concentration boundary layer in an analogous manner to the thermal boundary layer and define new dimensionless numbers that are counterparts of the Nusselt and Prandtl numbers. The rate of heat convection for external flow was expressed conveniently by Newton’s law of cooling as (14-4) where hconv is the heat transfer coefficient, As is the surface area, and Ts T is the temperature difference across the thermal boundary layer. Likewise, the rate of mass convection can be expressed as (Fig. 14–7) (14-5) MASS DIFFUSION Fick’s law of diffusion, proposed in 1855, states that the rate of diffusion of a chemical species at a location in a gas mixture (or liquid or solid solution) is proportional to the concentration gradient of that species at that location. →  14–3 ■ →   where hmass is the mass transfer coefficient, As is the surface area, and Cs C is a suitable concentration difference across the concentration boundary layer. Various aspects of the analogy between heat and mass convection are explored in Section 14–9. The analogy is valid for low mass transfer rate cases in which the flow rate of species undergoing mass flow is low (under 10 percent) relative to the total flow rate of the liquid or gas mixture. Concentration difference Mass convection: 678 m· conv hmass As(Cs C) Heat convection: · Q conv hconv As(Ts T) 123  → m· conv hmass As(Cs C) Mass transfer coefficient Heat transfer coefficient  → · Q conv hconv As(Ts T) Temperature difference FIGURE 14–7 Analogy between convection heat transfer and convection mass transfer. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 722 722 HEAT TRANSFER A + B mixture B A V = VA = VB m = mA + mB ρ = ρA + ρB C = CA + CB Mass basis: ρ m m ρA = —–A , ρ = — , wA = —A– V V ρ Mole basis: N C N CA = —A– , C = — , yA = —A– V V C Relation between them: ρ M CA = —A– , wA = yA —–A MA M FIGURE 14–8 Different ways of expressing the concentration of species A of a binary mixture A and B. Although a higher concentration for a species means more molecules of that species per unit volume, the concentration of a species can be expressed in several ways. Next we describe two common ways. 1 Mass Basis On a mass basis, concentration is expressed in terms of density (or mass concentration), which is mass per unit volume. Considering a small volume V at a location within the mixture, the densities of a species (subscript i) and of the mixture (no subscript) at that location are given by (Fig. 14–8) Partial density of species i: Total density of mixture: i mi /V m/V m /V i (kg/m3) i Therefore, the density of a mixture at a location is equal to the sum of the densities of its constituents at that location. Mass concentration can also be expressed in dimensionless form in terms of mass fraction w as Mass fraction of species i: mi mi /V i wi m m /V (14-6) Note that the mass fraction of a species ranges between 0 and 1, and the conservation of mass requires that the sum of the mass fractions of the constituents of a mixture be equal to 1. That is, wi 1. Also note that the density and mass fraction of a constituent in a mixture, in general, vary with location unless the concentration gradients are zero. 2 Mole Basis On a mole basis, concentration is expressed in terms of molar concentration (or molar density), which is the amount of matter in kmol per unit volume. Again considering a small volume V at a location within the mixture, the molar concentrations of a species (subscript i) and of the mixture (no subscript) at that location are given by Partial molar concentration of species i: Ci Ni /V Total molar concentration of mixture: C N/V N /V C i (kmol/m3) i Therefore, the molar concentration of a mixture at a location is equal to the sum of the molar concentrations of its constituents at that location. Molar concentration can also be expressed in dimensionless form in terms of mole fraction y as Mole fraction of species i: yi Ni Ni /V Ci N C N/V (14-7) Again the mole fraction of a species ranges between 0 and 1, and the sum of the mole fractions of the constituents of a mixture is unity, yi 1. The mass m and mole number N of a substance are related to each other by m NM (or, for a unit volume, CM) where M is the molar mass (also called the molecular weight) of the substance. This is expected since the mass of 1 kmol of the substance is M kg, and thus the mass of N kmol is NM kg. Therefore, the mass and molar concentrations are related to each other by cen58933_ch14.qxd 9/9/2002 10:05 AM Page 723 723 CHAPTER 14 Ci i (for species i) and Mi C M (for the mixture) (14-8) where M is the molar mass of the mixture which can be determined from M m N N M i N i Ni N M y M i i i (14-9) The mass and mole fractions of species i of a mixture are related to each other by i Ci Mi Mi wi yi M CM (14-10) Two different approaches are presented above for the description of concentration at a location, and you may be wondering which approach is better to use. Well, the answer depends on the situation on hand. Both approaches are equivalent, and the better approach for a given problem is the one that yields the desired solution more easily. Special Case: Ideal Gas Mixtures At low pressures, a gas or gas mixture can conveniently be approximated as an ideal gas with negligible error. For example, a mixture of dry air and water vapor at atmospheric conditions can be treated as an ideal gas with an error much less than 1 percent. The total pressure of a gas mixture P is equal to the sum of the partial pressures Pi of the individual gases in the mixture and is expressed as P Pi. Here Pi is called the partial pressure of species i, which is the pressure species i would exert if it existed alone at the mixture temperature and volume. This is known as Dalton’s law of additive pressures. Then using the ideal gas relation PV NRuT where Ru is the universal gas constant for both the species i and the mixture, the pressure fraction of species i can be expressed as (Fig. 14–9) Pi Ni RuT/V Ni yi P N NRuT/V (14-11) Therefore, the pressure fraction of species i of an ideal gas mixture is equivalent to the mole fraction of that species and can be used in place of it in mass transfer analysis. Fick’s Law of Diffusion: Stationary Medium Consisting of Two Species We mentioned earlier that the rate of mass diffusion of a chemical species in a stagnant medium in a specified direction is proportional to the local concentration gradient in that direction. This linear relationship between the rate of diffusion and the concentration gradient proposed by Fick in 1855 is known as Fick’s law of diffusion and can be expressed as Mass flux Constant of proportionality Concentration gradient 2 mol A 6 mol B P 120 kPa A mixture of two ideal gases A and B NA 2 0.25 N 26 PA yAP 0.25 120 30 kPa yA FIGURE 14–9 For ideal gas mixtures, pressure fraction of a gas is equal to its mole fraction. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 724 724 HEAT TRANSFER Higher concentration of species A Lower concentration of species A dC slope = ——A dx Area A But the concentration of a species in a gas mixture or liquid or solid solution can be defined in several ways such as density, mass fraction, molar concentration, and mole fraction, as already discussed, and thus Fick’s law can be expressed mathematically in many ways. It turns out that it is best to express the concentration gradient in terms of the mass or mole fraction, and the most appropriate formulation of Fick’s law for the diffusion of a species A in a stationary binary mixture of species A and B in a specified direction x is given by (Fig. 14–10) m· diff, A d(A /) dwA DAB DAB A dx dx N· diff, A dyA d(CA/C) Mole basis: ¯jdiff, A CDAB CDAB A dx dx Mass basis: jdiff, A CA(x) Concentration profile of species A (kg/s · m2) (mol/s · m2) (14-12) x Mass basis: dwA m· diff = – ρADAB —— dx d(ρA/ρ) = – ρADAB ——— dx dρA = –ADAB —— (if ρ = constant) dx Here jdiff, A is the (diffusive) mass flux of species A (mass transfer by diffusion per unit time and per unit area normal to the direction of mass transfer, in kg/s · m2) and ¯jdiff, A is the (diffusive) molar flux (in kmol/s · m2). The mass flux of a species at a location is proportional to the density of the mixture at that location. Note that A B is the density and C CA CB is the molar concentration of the binary mixture, and in general, they may vary throughout the mixture. Therefore, d(A /) dA or Cd(CA /C) dCA. But in the special case of constant mixture density or constant molar concentration C, the relations above simplify to Mole basis: d(CA/C) – = – CADAB ——— dx FIGURE 14–10 Various expressions of Fick’s law of diffusion for a binary mixture. Mass diffusivity Concentration gradient  → dA → m· A DAB A dx · dT Q kA → dx   → Heat conduction: jdiff, A DAB Mole basis (C constant): dCA = –ADAB —— (if C = constant) dx Mass diffusion: dA dx dCA ¯j diff, A DAB dx Mass basis ( constant): dyA · Ndiff, A = – CADAB —— dx  Thermal Temperature conductivity gradient FIGURE 14–11 Analogy between Fourier’s law of heat conduction and Fick’s law of mass diffusion. (kg/s · m2) (kmol/s · m2) (14-13) The constant density or constant molar concentration assumption is usually appropriate for solid and dilute liquid solutions, but often this is not the case for gas mixtures or concentrated liquid solutions. Therefore, Eq. 14–12 should be used in the latter case. In this introductory treatment we will limit our consideration to one-dimensional mass diffusion. For two- or three-dimensional cases, Fick’s law can conveniently be expressed in vector form by simply replacing the derivatives in the above relations by the corresponding gradients (such as jA DAB wA). Remember that the constant of proportionality in Fourier’s law was defined as the transport property thermal conductivity. Similarly, the constant of proportionality in Fick’s law is defined as another transport property called the binary diffusion coefficient or mass diffusivity, DAB. The unit of mass diffusivity is m2/s, which is the same as the units of thermal diffusivity or momentum diffusivity (also called kinematic viscosity) (Fig. 14–11). Because of the complex nature of mass diffusion, the diffusion coefficients are usually determined experimentally. The kinetic theory of gases indicates that the diffusion coefficient for dilute gases at ordinary pressures is essentially independent of mixture composition and tends to increase with temperature while decreasing with pressure as DAB T 3/2 P or DAB, 1 P2 T1 DAB, 2 P1 T2 3/2 (14-14) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 725 725 CHAPTER 14 This relation is useful in determining the diffusion coefficient for gases at different temperatures and pressures from a knowledge of the diffusion coefficient at a specified temperature and pressure. More general but complicated relations that account for the effects of molecular collisions are also available. The diffusion coefficients of some gases in air at 1 atm pressure are given in Table 14–1 at various temperatures. The diffusion coefficients of solids and liquids also tend to increase with temperature while exhibiting a strong dependence on the composition. The diffusion process in solids and liquids is a great deal more complicated than that in gases, and the diffusion coefficients in this case are almost exclusively determined experimentally. The binary diffusion coefficient for several binary gas mixtures and solid and liquid solutions are given in Tables 14–2 and 14–3. We make these two observations from these tables: 1. The diffusion coefficients, in general, are highest in gases and lowest in solids. The diffusion coefficients of gases are several orders of magnitude greater than those of liquids. 2. Diffusion coefficients increase with temperature. The diffusion coefficient (and thus the mass diffusion rate) of carbon through iron during a hardening process, for example, increases by 6000 times as the temperature is raised from 500°C to 1000°C. Due to its practical importance, the diffusion of water vapor in air has been the topic of several studies, and some empirical formulas have been developed for the diffusion coefficient DH2O–air. Marrero and Mason proposed this popular formula (Table 14–4): TABLE 14–1 Binary diffusion coefficients of some gases in air at 1 atm pressure (from Mills, Ref. 13, Table A.17a, p. 869) Binary Diffusion Coefficient, m2/s 105 T, K O2 200 300 400 500 600 700 800 900 1000 1200 1400 1600 1800 2000 0.95 1.88 5.25 4.75 6.46 8.38 10.5 12.6 15.2 20.6 26.6 33.2 40.3 48.0 CO2 H2 0.74 3.75 1.57 7.77 2.63 12.5 3.85 17.1 5.37 24.4 6.84 31.7 8.57 39.3 10.5 47.7 12.4 56.9 16.9 77.7 21.7 99.0 27.5 125 32.8 152 39.4 180 NO 0.88 1.80 3.03 4.43 6.03 7.82 9.78 11.8 14.1 19.2 24.5 30.4 37.0 44.8 Multiply by 10.76 to convert to ft2/s. TABLE 14–2 Binary diffusion coefficients of dilute gas mixtures at 1 atm (from Barrer, Ref. 2; Geankoplis, Ref. 5; Perry, Ref. 14; and Reid et al., Ref. 15) Substance A Substance B Air Air Air Air Air Air Air Air Air Air Air Air Air Air Air Acetone Ammonia, NH3 Benzene Carbon dioxide Chlorine Ethyl alcohol Ethyl ether Helium, He Hydrogen, H2 Iodine, I2 Methanol Mercury Napthalene Oxygen, O2 Water vapor T, K DAB or DBA, m2/s 273 298 298 298 273 298 298 298 298 298 298 614 300 298 298 1.1 105 2.6 105 0.88 105 1.6 105 1.2 105 1.2 105 0.93 105 7.2 105 7.2 105 0.83 105 1.6 105 4.7 105 0.62 105 2.1 105 2.5 105 Substance B Substance A Argon, Ar Carbon dioxide, Carbon dioxide, Carbon dioxide, Carbon dioxide, Carbon dioxide, Hydrogen, H2 Hydrogen, H2 Oxygen, O2 Oxygen, O2 Oxygen, O2 Oxygen, O2 Water vapor Water vapor Water vapor CO2 CO2 CO2 CO2 CO2 Nitrogen, N2 Benzene Hydrogen, H2 Nitrogen, N2 Oxygen, O2 Water vapor Nitrogen, N2 Oxygen, O2 Ammonia Benzene Nitrogen, N2 Water vapor Argon, Ar Helium, He Nitrogen, N2 T, K DAB or DBA, m2/s 293 318 273 293 273 298 273 273 293 296 273 298 298 298 298 1.9 105 0.72 105 5.5 105 1.6 105 1.4 105 1.6 105 6.8 105 7.0 105 2.5 105 0.39 105 1.8 105 2.5 105 2.4 105 9.2 105 2.5 105 Note: The effect of pressure and temperature on DAB can be accounted for through DAB ~ T3/2/P. Also, multiply DAB values by 10.76 to convert them to ft2/s. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 726 726 HEAT TRANSFER TABLE 14–3 Binary diffusion coefficients of dilute liquid solutions and solid solutions at 1 atm (from Barrer, Ref. 2; Reid et al., Ref. 15; Thomas, Ref. 19; and van Black, Ref. 20) (a) Diffusion through Liquids Substance A (Solute) Substance B (Solvent) Ammonia Benzene Carbon dioxide Chlorine Ethanol Ethanol Ethanol Glucose Hydrogen Methane Methane Methane Methanol Nitrogen Oxygen Water Water Water Chloroform Water Water Water Water Water Water Water Water Water Water Water Water Water Water Water Ethanol Ethylene glycol Methanol Methanol (b) Diffusion through Solids T, K DAB, m2/s 285 293 298 285 283 288 298 298 298 275 293 333 288 298 298 298 298 298 288 1.6 109 1.0 109 2.0 109 1.4 109 0.84 109 1.0 109 1.2 109 0.69 109 6.3 109 0.85 109 1.5 109 3.6 109 1.3 109 2.6 109 2.4 109 1.2 109 0.18 109 1.8 109 2.1 109 Substance A (Solute) Substance B (Solvent) Carbon dioxide Nitrogen Oxygen Helium Helium Helium Hydrogen Hydrogen Hydrogen Cadmium Zinc Zinc Antimony Bismuth Mercury Copper Copper Carbon Carbon Natural rubber Natural rubber Natural rubber Pyrex Pyrex Silicon dioxide Iron Nickel Nickel Copper Copper Copper Silver Lead Lead Aluminum Aluminum Iron (fcc) Iron (fcc) DH2O–Air 1.87 1010 TABLE 14–4 In a binary ideal gas mixture of species A and B, the diffusion coefficient of A in B is equal to the diffusion coefficient of B in A, and both increase with temperature T, °C DH2O–Air or DAir–H2O at 1 atm, in m2/s (from Eq. 14–15) 0 5 10 15 20 25 30 35 40 50 100 150 2.09 105 2.17 105 2.25 105 2.33 105 2.42 105 2.50 105 2.59 105 2.68 105 2.77 105 2.96 105 3.99 105 5.18 105 T 2.072 P T, K DAB, m2/s 298 298 298 773 293 298 298 358 438 293 773 1273 293 293 293 773 1273 773 1273 1.1 1010 1.5 1010 2.1 1010 2.0 1012 4.5 1015 4.0 1014 2.6 1013 1.2 1012 1.0 1011 2.7 1019 4.0 1018 5.0 1013 3.5 1025 1.1 1020 2.5 1019 4.0 1014 1.0 1010 5.0 1015 3.0 1011 (m2/s), 280 K T 450 K (14-15) where P is total pressure in atm and T is the temperature in K. The primary driving mechanism of mass diffusion is the concentration gradient, and mass diffusion due to a concentration gradient is known as the ordinary diffusion. However, diffusion may also be caused by other effects. Temperature gradients in a medium can cause thermal diffusion (also called the soret effect), and pressure gradients may result in pressure diffusion. Both of these effects are usually negligible, however, unless the gradients are very large. In centrifuges, the pressure gradient generated by the centrifugal effect is used to separate liquid solutions and gaseous isotopes. An external force field such as an electric or magnetic field applied on a mixture or solution can be used successfully to separate electrically charged or magnetized molecules (as in an electrolyte or ionized gas) from the mixture. This is called forced diffusion. Also, when the pores of a porous solid such as silica-gel are smaller than the mean free path of the gas molecules, the molecular collisions may be negligible and a free molecule flow may be initiated. This is known as Knudsen diffusion. When the size of the gas molecules is comparable to the pore size, adsorbed molecules move along the pore walls. This is known as surface diffusion. Finally, particles whose diameter is under 0.1 m such as mist and soot particles act like large molecules, and the diffusion process of such particles due to the concentration gradient is called Brownian motion. Large particles (those whose diameter is greater than 1 m) are not affected cen58933_ch14.qxd 9/9/2002 10:05 AM Page 727 727 CHAPTER 14 by diffusion as the motion of such particles is governed by Newton’s laws. In our elementary treatment of mass diffusion, we will assume these additional effects to be nonexistent or negligible, as is usually the case, and refer the interested reader to advanced books on these topics. EXAMPLE 14–1 Determining Mass Fractions from Mole Fractions The composition of dry standard atmosphere is given on a molar basis to be 78.1 percent N2, 20.9 percent O2, and 1.0 percent Ar and other constituents (Fig. 14–12). Treating other constituents as Ar, determine the mass fractions of the constituents of air. AIR 78.1% N2 20.9% O2 1.0% Ar SOLUTION The molar fractions of the constituents of air are given. The mass fractions are to be determined. Assumptions The small amounts of other gases in air are treated as argon. Properties The molar masses of N2, O2, and Ar are 28.0, 32.0, and 39.9 kg/kmol, respectively (Table A–1). Analysis The molar mass of air is determined to be FIGURE 14–12 Schematic for Example 14–1. M yi Mi 0.781 28.0 0.209 32.0 0.01 39.9 29.0 kg/kmol Then the mass fractions of constituent gases are determined from Eq. 14–10 to be MN2 28.0 (0.781) 0.754 M 29.0 MO2 32.0 wO2 yO2 (0.209) 0.231 M 29.0 MAr 39.9 (0.01) 0.014 wAr yAr M 29.0 wN2 yN2 N2: O2: Ar: Therefore, the mass fractions of N2, O2, and Ar in dry standard atmosphere are 75.4 percent, 23.1 percent, and 1.4 percent, respectively. 14–4 ■ x Air BOUNDARY CONDITIONS We mentioned earlier that the mass diffusion equation is analogous to the heat diffusion (conduction) equation, and thus we need comparable boundary conditions to determine the species concentration distribution in a medium. Two common types of boundary conditions are the (1) specified species concentration, which corresponds to specified temperature, and (2) specified species flux, which corresponds to specified heat flux. Despite their apparent similarity, an important difference exists between temperature and concentration: temperature is necessarily a continuous function, but concentration, in general, is not. The wall and air temperatures at a wall surface, for example, are always the same. The concentrations of air on the two sides of a water–air interface, however, are obviously very different (in fact, the concentration of air in water is close to zero). Likewise, the concentrations of water on the two sides of a water–air interface are also different even when air is saturated (Fig. 14–13). Therefore, when specifying a yH 2O, 0 Jump in concentration gas side (0) yH 2O, liquid side = 1.0 Water Concentration profile FIGURE 14–13 Unlike temperature, the concentration of species on the two sides of a liquid–gas (or solid–gas or solid–liquid) interface are usually not the same. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 728 728 HEAT TRANSFER boundary condition, specifying the location is not enough. We also need to specify the side of the boundary. To do this, we consider two imaginary surfaces on the two sides of the interface that are infinitesimally close to the interface. Whenever there is a doubt, we indicate the desired side of the interface by specifying its phase as a subscript. For example, the water (liquid or vapor) concentration at the liquid and gas sides of a water–air interface at x 0 can be expressed on a molar basis is yH2O, liquid side (0) y1 and yH2O, gas side (0) y2 (14-16) Using Fick’s law, the constant species flux boundary condition for a diffusing species A at a boundary at x 0 is expressed, in the absence of any blowing or suction, as T(x) Insulated surface dT(0) ——– = 0 dx · Q(0) = 0 x Impermeable surface CA(x) dCA(0) ——– —=0 dx · mA(0) = 0 FIGURE 14–14 An impermeable surface in mass transfer is analogous to an insulated surface in heat transfer. Air 92 kPa, 15°C Saturated air yH O, air side = 0.0185 2 yH Lake 15°C 2O, liquid side ≅ 1.0 CDAB dyA dx ¯jA, 0 or DAB x0 dwA dx jA, 0 (14-17) x0 where ¯jA, 0 and jA, 0 are the specified mole and mass fluxes of species A at the boundary, respectively. The special case of zero mass flux (¯jA, 0 jA, 0 0) corresponds to an impermeable surface for which dyA(0)/dx dwA (0)/ dx 0 (Fig. 14–14). To apply the specified concentration boundary condition, we must know the concentration of a species at the boundary. This information is usually obtained from the requirement that thermodynamic equilibrium must exist at the interface of two phases of a species. In the case of air–water interface, the concentration values of water vapor in the air are easily determined from saturation data, as shown in Example 14–2. EXAMPLE 14–2 Mole Fraction of Water Vapor at the Surface of a Lake Determine the mole fraction of the water vapor at the surface of a lake whose temperature is 15°C and compare it to the mole fraction of water in the lake (Fig. 14–15). Take the atmospheric pressure at lake level to be 92 kPa. SOLUTION The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared. Assumptions 1 Both the air and water vapor are ideal gases. 2 The mole fraction of dissolved air in water is negligible. Properties The saturation pressure of water at 15°C is 1.705 kPa (Table A–9). Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 15°C, FIGURE 14–15 Schematic for Example 14–2. Pvapor Psat @ 15°C 1.705 kPa Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air at the surface of the lake is determined from Eq. 14–11 to be yvapor Pvapor 1.705 kPa 0.0185 P 92 kPa (or 1.85 percent) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 729 729 CHAPTER 14 Water contains some dissolved air, but the amount is negligible. Therefore, we can assume the entire lake to be liquid water. Then its mole fraction becomes ywater, liquid side 1.0 (or 100 percent) Discussion Note that the concentration of water on a molar basis is 100 percent just beneath the air–water interface and 1.85 percent just above it, even though the air is assumed to be saturated (so this is the highest value at 15°C). Therefore, huge discontinuities can occur in the concentrations of a species across phase boundaries. The situation is similar at solid–liquid interfaces. Again, at a given temperature, only a certain amount of solid can be dissolved in a liquid, and the solubility of the solid in the liquid is determined from the requirement that thermodynamic equilibrium exists between the solid and the solution at the interface. The solubility represents the maximum amount of solid that can be dissolved in a liquid at a specified temperature and is widely available in chemistry handbooks. In Table 14–5 we present sample solubility data for sodium chloride (NaCl) and calcium bicarbonate [Ca(HCO3)2] at various temperatures. For example, the solubility of salt (NaCl) in water at 310 K is 36.5 kg per 100 kg of water. Therefore, the mass fraction of salt in the brine at the interface is simply 36.5 kg msalt wsalt, liquid side m 0.267 (100 36.5) kg Pi, gas side (at interface) H Solubility of two inorganic compounds in water at various temperatures, in kg, in 100 kg of water [from Handbook of Chemistry (New York: McGraw-Hill, 1961)] (or 26.7 percent) Solute whereas the mass fraction of salt in the pure solid salt is w 1.0. Note that water becomes saturated with salt when 36.5 kg of salt are dissolved in 100 kg of water at 310 K. Many processes involve the absorption of a gas into a liquid. Most gases are weakly soluble in liquids (such as air in water), and for such dilute solutions the mole fractions of a species i in the gas and liquid phases at the interface are observed to be proportional to each other. That is, yi, gas side yi, liquid side or Pi, gas side P yi, liquid side since yi, gas side Pi, gas side /P for ideal gas mixtures. This is known as Henry’s law and is expressed as yi, liquid side TABLE 14–5 (14-18) where H is Henry’s constant, which is the product of the total pressure of the gas mixture and the proportionality constant. For a given species, it is a function of temperature only and is practically independent of pressure for pressures under about 5 atm. Values of Henry’s constant for a number of aqueous solutions are given in Table 14–6 for various temperatures. From this table and the equation above we make the following observations: 1. The concentration of a gas dissolved in a liquid is inversely proportional to Henry’s constant. Therefore, the larger Henry’s constant, the smaller the concentration of dissolved gases in the liquid. Temperature, K Salt, NaCl Calcium Bicarbonate, Ca(HCO3)2 273.15 280 290 300 310 320 330 340 350 360 370 373.15 35.7 35.8 35.9 36.2 36.5 36.9 37.2 37.6 38.2 38.8 39.5 39.8 16.15 16.30 16.53 16.75 16.98 17.20 17.43 17.65 17.88 18.10 18.33 18.40 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 730 730 HEAT TRANSFER TABLE 14–6 Henry’s constant H (in bars) for selected gases in water at low to moderate pressures (for gas i, H Pi, gas side /yi, water side) (from Mills, Ref. 13, Table A.21, p. 874) Solute 290 K 300 K 310 K 320 K 330 K 340 K H2S CO2 O2 H2 CO Air N2 440 1280 38,000 67,000 51,000 62,000 76,000 560 1710 45,000 72,000 60,000 74,000 89,000 700 2170 52,000 75,000 67,000 84,000 101,000 830 2720 57,000 76,000 74,000 92,000 110,000 980 3220 61,000 77,000 80,000 99,000 118,000 1140 — 65,000 76,000 84,000 104,000 124,000 Gas A yA, gas side yA, liquid side Gas: A Liquid: B yA, gas side yA, liquid side or PA, gas side ———— yA, liquid side P or PA, gas side = HyA, liquid side FIGURE 14–16 Dissolved gases in a liquid can be driven off by heating the liquid. Air Saturated air Lake 17°C Pdry air, gas side ydry air, liquid side FIGURE 14–17 Schematic for Example 14–3. Henry’s constant increases (and thus the fraction of a dissolved gas in the liquid decreases) with increasing temperature. Therefore, the dissolved gases in a liquid can be driven off by heating the liquid (Fig. 14–16). 3. The concentration of a gas dissolved in a liquid is proportional to the partial pressure of the gas. Therefore, the amount of gas dissolved in a liquid can be increased by increasing the pressure of the gas. This can be used to advantage in the carbonation of soft drinks with CO2 gas. Strictly speaking, the result obtained from Eq. 14–18 for the mole fraction of dissolved gas is valid for the liquid layer just beneath the interface and not necessarily the entire liquid. The latter will be the case only when thermodynamic phase equilibrium is established throughout the entire liquid body. EXAMPLE 14–3 Mole Fraction of Dissolved Air in Water Determine the mole fraction of air dissolved in water at the surface of a lake whose temperature is 17°C (Fig. 14–17). Take the atmospheric pressure at lake level to be 92 kPa. SOLUTION The mole fraction of air dissolved in water at the surface of a lake is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water so that Henry’s law is applicable. Properties The saturation pressure of water at 17°C is 1.92 kPa (Table A–9). Henry’s constant for air dissolved in water at 290 K is H 62,000 bar (Table 14–6). Analysis This example is similar to the previous example. Again the air at the water surface will be saturated, and thus the partial pressure of water vapor in the air at the lake surface will be the saturation pressure of water at 17°C, Pvapor Psat @ 17°C 1.92 kPa Assuming both the air and vapor to be ideal gases, the partial pressure of dry air is determined to be Pdry air P Pvapor 92 1.92 90.08 kPa 0.9008 bar cen58933_ch14.qxd 9/9/2002 10:05 AM Page 731 731 CHAPTER 14 Note that with little loss in accuracy (an error of about 2 percent), we could have ignored the vapor pressure since the amount of vapor in air is so small. Then the mole fraction of air in the water becomes ydry air, liquid state Pdry air, gas side 0.9008 bar 1.45 105 H 62,000 bar which is very small, as expected. Therefore, the concentration of air in water just below the air–water interface is 1.45 moles per 100,0000 moles. But obviously this is enough oxygen for fish and other creatures in the lake. Note that the amount of air dissolved in water will decrease with increasing depth. We mentioned earlier that the use of Henry’s law is limited to dilute gas– liquid solutions; that is, a liquid with a small amount of gas dissolved in it. Then the question that arises naturally is, what do we do when the gas is highly soluble in the liquid (or solid), such as ammonia in water? In this case the linear relationship of Henry’s law does not apply, and the mole fraction of a gas dissolved in the liquid (or solid) is usually expressed as a function of the partial pressure of the gas in the gas phase and the temperature. An approximate relation in this case for the mole fractions of a species on the liquid and gas sides of the interface is given by Raoult’s law as Pi, gas side yi, gas side P yi, liquid side Pi, sat(T) (14-19) where Pi, sat(T) is the saturation pressure of the species i at the interface temperature and P is the total pressure on the gas phase side. Tabular data are available in chemical handbooks for common solutions such as the ammonia– water solution that is widely used in absorption-refrigeration systems. Gases may also dissolve in solids, but the diffusion process in this case can be very complicated. The dissolution of a gas may be independent of the structure of the solid, or it may depend strongly on its porosity. Some dissolution processes (such as the dissolution of hydrogen in titanium, similar to the dissolution of CO2 in water) are reversible, and thus maintaining the gas content in the solid requires constant contact of the solid with a reservoir of that gas. Some other dissolution processes are irreversible. For example, oxygen gas dissolving in titanium forms TiO2 on the surface, and the process does not reverse itself. The concentration of the gas species i in the solid at the interface Ci, solid side is proportional to the partial pressure of the species i in the gas Pi, gas side on the gas side of the interface and is expressed as Ci, solid side Pi, gas side (kmol/m3) (14-20) where is the solubility. Expressing the pressure in bars and noting that the unit of molar concentration is kmol of species i per m3, the unit of solubility is kmol/m3 · bar. Solubility data for selected gas–solid combinations are given in Table 14–7. The product of the solubility of a gas and the diffusion coefficient of the gas in a solid is referred to as the permeability , which is a measure of the ability of the gas to penetrate a solid. That is, DAB where DAB is TABLE 14–7 Solubility of selected gases and solids (for gas i, Ci, solid side /Pi, gas side) (from Barrer, Ref. 2) Gas Solid T, K kmol/m3 · bar O2 Rubber 298 N2 Rubber 298 CO2 Rubber 298 He SiO2 293 H2 Ni 358 0.00312 0.00156 0.04015 0.00045 0.00901 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 732 732 HEAT TRANSFER the diffusivity of the gas in the solid. Permeability is inversely proportional to thickness and has the unit kmol/s · bar. Finally, if a process involves the sublimation of a pure solid (such as ice or solid CO2) or the evaporation of a pure liquid (such as water) in a different medium such as air, the mole (or mass) fraction of the substance in the liquid or solid phase is simply taken to be 1.0, and the partial pressure and thus the mole fraction of the substance in the gas phase can readily be determined from the saturation data of the substance at the specified temperature. Also, the assumption of thermodynamic equilibrium at the interface is very reasonable for pure solids, pure liquids, and solutions, except when chemical reactions are occurring at the interface. EXAMPLE 14–4 Consider a nickel plate that is in contact with hydrogen gas at 358 K and 300 kPa. Determine the molar and mass density of hydrogen in the nickel at the interface (Fig. 14–18). Nickel plate SOLUTION A nickel plate is exposed to hydrogen. The molar and mass density of hydrogen in the nickel at the interface is to be determined. Assumptions Nickel and hydrogen are in thermodynamic equilibrium at the interface. Properties The molar mass of hydrogen is M 2 kg/kmol (Table A–1). The solubility of hydrogen in nickel at 358 K is 0.00901 kmol/m3 · bar (Table 14–7). Analysis Noting that 300 kPa 3 bar, the molar density of hydrogen in the nickel at the interface is determined from Eq. 14–20 to be Air H2 358 K 300 kPa 0 Diffusion of Hydrogen Gas into a Nickel Plate CH2, solid side PH2, gas side (0.00901 kmol/m3 · bar)(3 bar) 0.027 kmol/m3 L x FIGURE 14–18 Schematic for Example 14–4. It corresponds to a mass density of H2, solid side CH2, solid side MH2 (0.027 kmol/m3)(2) 0.054 kg/m3 That is, there will be 0.027 kmol (or 0.054 kg) of H2 gas in each m3 volume of nickel adjacent to the interface. TABLE 14–8 Analogy between heat conduction and mass diffusion in a stationary medium Mass Diffusion Heat Conduction Mass Basis Molar Basis T k q· L wi DAB ji DAB L yi CDAB j¯i DAB L 14–5 ■ STEADY MASS DIFFUSION THROUGH A WALL Many practical mass transfer problems involve the diffusion of a species through a plane-parallel medium that does not involve any homogeneous chemical reactions under one-dimensional steady conditions. Such mass transfer problems are analogous to the steady one-dimensional heat conduction problems in a plane wall with no heat generation and can be analyzed similarly. In fact, many of the relations developed in Chapter 3 can be used for mass transfer by replacing temperature by mass (or molar) fraction, thermal conductivity by DAB (or CDAB), and heat flux by mass (or molar) flux (Table 14–8). cen58933_ch14.qxd 9/9/2002 10:05 AM Page 733 733 CHAPTER 14 Consider a solid plane wall (medium B) of area A, thickness L, and density . The wall is subjected on both sides to different concentrations of a species A to which it is permeable. The boundary surfaces at x 0 and x L are located within the solid adjacent to the interfaces, and the mass fractions of A at those surfaces are maintained at wA, 1 and wA, 2, respectively, at all times (Fig. 14–19). The mass fraction of species A in the wall will vary in the x-direction only and can be expressed as wA(x). Therefore, mass transfer through the wall in this case can be modeled as steady and one-dimensional. Here we determine the rate of mass diffusion of species A through the wall using a similar approach to that used in Chapter 3 for heat conduction. The concentration of species A at any point will not change with time since operation is steady, and there will be no production or destruction of species A since no chemical reactions are occurring in the medium. Then the conservation of mass principle for species A can be expressed as the mass flow rate of species A through the wall at any cross section is the same. That is m· diff, A jAA constant L CA(x) dx wA, 2 wA, 1 DAB dwA wA, 2 A, 1 A, 2 wA, 1 wA, 2 DAB A L L FIGURE 14–19 Schematic for steady one-dimensional mass diffusion of species A through a plane wall. (14-21) (kg/s) (14-22) · T1 – T2 Q = ——— R T1 T2 R (a) Heat flow This relation can be rearranged as m· diff, A, wall x L where the mass transfer rate m· diff, A and the wall area A are taken out of the integral sign since both are constants. If the density and the mass diffusion coefficient DAB vary little along the wall, they can be assumed to be constant. The integration can be performed in that case to yield m· diff, A, wall DAB A m· diff, A dCA 0 Separating the variables in this equation and integrating across the wall from x 0, where w(0) wA, 1, to x L, where w(L) wA, 2, we get 0 wA, 1 ρ ≅ constant dx m· diff, A dwA jA DAB constant A dx A (kg/s) Then Fick’s law of diffusion becomes m· diff, A A Medium B wA, 1 wA, 2 wA, 1 wA, 2 Rdiff, wall L/DAB A 1 – 2 I = ——— Re (14-23) 1 where 2 Re Rdiff, wall L DAB A is the diffusion resistance of the wall, in s/kg, which is analogous to the electrical or conduction resistance of a plane wall of thickness L and area A (Fig. 14–20). Thus, we conclude that the rate of mass diffusion through a plane wall is proportional to the average density, the wall area, and the concentration difference across the wall, but is inversely proportional to the wall thickness. Also, once the rate of mass diffusion is determined, the mass fraction wA(x) at any location x can be determined by replacing wA, 2 in Eq. 14–22 by wA(x) and L by x. (b) Current flow wA, 1 – wA, 2 — m· diff, A = ———— Rmass wA, 2 wA, 1 Rmass (c) Mass flow FIGURE 14–20 Analogy between thermal, electrical, and mass diffusion resistance concepts. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 734 734 HEAT TRANSFER The preceding analysis can be repeated on a molar basis with this result, yA, 1 yA, 2 CA, 1 CA, 2 yA, 1 yA, 2 · Ndiff, A, wall CDAB A DAB A L L Rdiff, wall m· diff, A B r2 r1 wA, 1 wA, 2 FIGURE 14–21 One-dimensional mass diffusion through a cylindrical or spherical shell. (14-24) where Rdiff, wall L/CDAB A is the molar diffusion resistance of the wall in s/kmol. Note that mole fractions are accompanied by molar concentrations and mass fractions are accompanied by density. Either relation can be used to determine the diffusion rate of species A across the wall, depending on whether the mass or molar fractions of species A are known at the boundaries. Also, the concentration gradients on both sides of an interface are different, and thus diffusion resistance networks cannot be constructed in an analogous manner to thermal resistance networks. In developing these relations, we assumed the density and the diffusion coefficient of the wall to be nearly constant. This assumption is reasonable when a small amount of species A diffuses through the wall and thus the concentration of A is small. The species A can be a gas, a liquid, or a solid. Also, the wall can be a plane layer of a liquid or gas provided that it is stationary. The analogy between heat and mass transfer also applies to cylindrical and spherical geometries. Repeating the approach outlined in Chapter 3 for heat conduction, we obtain the following analogous relations for steady onedimensional mass transfer through nonreacting cylindrical and spherical layers (Fig. 14–21) m· diff, A, cyl 2LDAB wA, 1 wA, 2 A, 1 A, 2 2LDAB ln(r2/r1) ln(r2/r1) m· diff, A, sph 4r1r2DAB A, 1 A, 2 wA, 1 wA, 2 r2 r1 4r1r2DAB r2 r1 (14-25) (14-26) or, on a molar basis, CA, 1 CA, 2 yA, 1 yA, 2 · Ndiff, A, cyl 2LCDAB 2LDAB ln(r2/r1) ln(r2/r1) yA, 1 yA, 2 CA, 1 CA, 2 · Ndiff, A, sph 4r1r2CDAb r r 4r1r2DAB r r 2 1 2 1 PA, 1 Solid wall PA, 2 Gas Gas A 0 PA, 1 – PA, 2 AB A ————– L · Ndiff, A L x FIGURE 14–22 The diffusion rate of a gas species through a solid can be determined from a knowledge of the partial pressures of the gas on both sides and the permeability of the solid to that gas. (14-27) (14-28) Here, L is the length of the cylinder, r1 is the inner radius, and r2 is the outer radius for the cylinder or the sphere. Again, the boundary surfaces at r r1 and r r2 are located within the solid adjacent to the interfaces, and the mass fractions of A at those surfaces are maintained at wA, 1 and wA, 2, respectively, at all times. (We could make similar statements for the density, molar concentration, and mole fraction of species A at the boundaries.) We mentioned earlier that the concentration of the gas species in a solid at the interface is proportional to the partial pressure of the adjacent gas and was expressed as CA, solid side AB PA, gas side where AB is the solubility (in kmol/m3 bar) of the gas A in the solid B. We also mentioned that the product of solubility and the diffusion coefficient is called the permeability, Ab AB DAB (in kmol/m · s · bar). Then the molar flow rate of a gas through a solid under steady one-dimensional conditions can be expressed in terms of the partial pressures of the adjacent gas on the two sides of the solid by replacing CA in these relations by AB PA or AB PA /DAB. In the case of a plane wall, for example, it gives (Fig. 14–22) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 735 735 CHAPTER 14 PA, 1 PA, 2 PA, 1 PA, 2 · Ndiff, A, wall DABAB A AB A L L (kmol/s) (14-29) where PA, 1 and PA, 2 are the partial pressures of gas A on the two sides of the wall. Similar relations can be obtained for cylindrical and spherical walls by following the same procedure. Also, if the permeability is given on a mass basis (in kg/m · s · bar), then Eq. 14–29 will give the diffusion mass flow rate. Noting that 1 kmol of an ideal gas at the standard conditions of 0°C and 1 atm occupies a volume of 22.414 m3, the volume flow rate of the gas through the wall by diffusion can be determined from · · Vdiff, A 22.414Ndiff, A (standard m3/s, at 0°C and 1 atm) The volume flow rate at other conditions can be determined from the ideal gas · · relation PAV NA RuT. EXAMPLE 14–5 Diffusion of Hydrogen through a Spherical Container Pressurized hydrogen gas is stored at 358 K in a 4.8-m-outer-diameter spherical container made of nickel (Fig. 14–23). The shell of the container is 6 cm thick. The molar concentration of hydrogen in the nickel at the inner surface is determined to be 0.087 kmol/m3. The concentration of hydrogen in the nickel at the outer surface is negligible. Determine the mass flow rate of hydrogen by diffusion through the nickel container. SOLUTION Pressurized hydrogen gas is stored in a spherical container. The kmol– CA, 1 = 0.087 —— m3 CA, 2 = 0 Pressurized H2 gas 358 K m· diff diffusion rate of hydrogen through the container is to be determined. Assumptions 1 Mass diffusion is steady and one-dimensional since the hydrogen concentration in the tank and thus at the inner surface of the container is practically constant, and the hydrogen concentration in the atmosphere and thus at the outer surface is practically zero. Also, there is thermal symmetry about the center. 2 There are no chemical reactions in the nickel shell that result in the generation or depletion of hydrogen. Properties The binary diffusion coefficient for hydrogen in the nickel at the specified temperature is 1.2 1012 m2/s (Table 14–3b). Analysis We can consider the total molar concentration to be constant (C CA CB CB constant), and the container to be a stationary medium since · there is no diffusion of nickel molecules (NB 0) and the concentration of the hydrogen in the container is extremely low (CA 1).Then the molar flow rate of hydrogen through this spherical shell by diffusion can readily be determined from Eq. 14–28 to be CA, 1 CA, 2 · Ndiff 4r1r2DAB r r 2 1 4(2.34 m)(2.40 m)(1.2 1012 m2/s) 1.228 1010 kmol/s (0.087 0) kmol/m3 2.40 2.34 Nickel container FIGURE 14–23 Schematic for Example 14–5. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 736 736 HEAT TRANSFER The mass flow rate is determined by multiplying the molar flow rate by the molar mass of hydrogen, which is M 2 kg/kmol, · m· diff MNdiff (2 kg/kmol)(1.228 1010 kmol/s) 2.46 1010 kg/s Therefore, hydrogen will leak out through the shell of the container by diffusion at a rate of 2.46 1010 kg/s or 7.8 g/year. Note that the concentration of hydrogen in the nickel at the inner surface depends on the temperature and pressure of the hydrogen in the tank and can be determined as explained in Example 14–4. Also, the assumption of zero hydrogen concentration in nickel at the outer surface is reasonable since there is only a trace amount of hydrogen in the atmosphere (0.5 part per million by mole numbers). 14–6 Dry insulation Wet insulation · Q · 1.25 Q 0% moisture 5% moisture FIGURE 14–24 A 5 percent moisture content can increase heat transfer through wall insulation by 25 percent. ■ WATER VAPOR MIGRATION IN BUILDINGS Moisture greatly influences the performance and durability of building materials, and thus moisture transmission is an important consideration in the construction and maintenance of buildings. The dimensions of wood and other hygroscopic substances change with moisture content. For example, a variation of 4.5 percent in moisture content causes the volume of white oak wood to change by 2.5 percent. Such cyclic changes of dimensions weaken the joints and can jeopardize the structural integrity of building components, causing “squeaking” at the minimum. Excess moisture can also cause changes in the appearance and physical properties of materials: corrosion and rusting in metals, rotting in woods, and peeling of paint on the interior and exterior wall surfaces. Soaked wood with a water content of 24 to 31 percent is observed to decay rapidly at temperatures 10 to 38°C. Also, molds grow on wood surfaces at relative humidities above 85 percent. The expansion of water during freezing may damage the cell structure of porous materials. Moisture content also affects the effective conductivity of porous mediums such as soils, building materials, and insulations, and thus heat transfer through them. Several studies have indicated that heat transfer increases almost linearly with moisture content, at a rate of 3 to 5 percent for each percent increase in moisture content by volume. Insulation with 5 percent moisture content by volume, for example, increases heat transfer by 15 to 25 percent relative to dry insulation (ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 20) (Fig. 14–24). Moisture migration may also serve as a transfer mechanism for latent heat by alternate evaporation and condensation. During a hot and humid day, for example, water vapor may migrate through a wall and condense on the inner side, releasing the heat of vaporization, with the process reversing during a cool night. Moisture content also affects the specific heat and thus the heat storage characteristics of building materials. Moisture migration in the walls, floors, or ceilings of buildings and in other applications is controlled by either vapor barriers or vapor retarders. Vapor barriers are materials that are impermeable to moisture, such as sheet metals, cen58933_ch14.qxd 9/9/2002 10:05 AM Page 737 737 CHAPTER 14 heavy metal foils, and thick plastic layers, and they effectively bar the vapor from migrating. Vapor retarders, on the other hand, retard or slow down the flow of moisture through the structures but do not totally eliminate it. Vapor retarders are available as solid, flexible, or coating materials, but they usually consist of a thin sheet or coating. Common forms of vapor retarders are reinforced plastics or metals, thin foils, plastic films, treated papers, coated felts, and polymeric or asphaltic paint coatings. In applications such as the building of walls where vapor penetration is unavoidable because of numerous openings such as electrical boxes, telephone lines, and plumbing passages, vapor retarders are used instead of vapor barriers to allow the vapor that somehow leaks in to exit to the outside instead of trapping it in. Vapor retarders with a permeance of 57.4 109 kg/s · m2 are commonly used in residential buildings. The insulation on chilled water lines and other impermeable surfaces that are always cold must be wrapped with a vapor barrier jacket, or such cold surfaces must be insulated with a material that is impermeable to moisture. This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside. The accumulation of moisture in such cases may render the insulation useless, resulting in excessive energy consumption. Atmospheric air can be viewed as a mixture of dry air and water vapor, and the atmospheric pressure is the sum of the pressure of dry air and the pressure of water vapor, which is called the vapor pressure P. Air can hold a certain amount of moisture only, and the ratio of the actual amount of moisture in the air at a given temperature to the maximum amount air can hold at that temperature is called the relative humidity . The relative humidity ranges from 0 for dry air to 100 percent for saturated air (air that cannot hold any more moisture). The partial pressure of water vapor in saturated air is called the saturation pressure Psat. Table 14–9 lists the saturation pressure at various temperatures. The amount of moisture in the air is completely specified by the temperature and the relative humidity, and the vapor pressure is related to relative humidity by P Psat (14-30) where Psat is the saturation (or boiling) pressure of water at the specified temperature. Then the mass flow rate of moisture through a plain layer of thickness L and normal area A can be expressed as m· A 1 Psat, 1 2 Psat, 2 P, 1 P, 2 A L L (kg/s) (14-31) where is the vapor permeability of the material, which is usually expressed on a mass basis in the unit ng/s · m · Pa, where ng 1012 kg and 1 Pa 105 bar. Note that vapor migrates or diffuses from a region of higher vapor pressure toward a region of lower vapor pressure. TABLE 14–9 Saturation pressure of water at various temperatures Temperature, °C Saturation Pressure, Pa 40 36 32 28 24 20 16 12 8 4 0 5 10 15 20 25 30 35 40 50 100 200 300 13 20 31 47 70 104 151 218 310 438 611 872 1,228 1,705 2,339 3,169 4,246 5,628 7,384 12,349 101,330 1.55 106 8.58 106 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 738 738 HEAT TRANSFER The permeability of most construction materials is usually expressed for a given thickness instead of per unit thickness. It is called the permeance , which is the ratio of the permeability of the material to its thickness. That is, TABLE 14–10 Permeance Typical vapor permeance of common building materials (from ASHRAE, Ref. 1, Chap. 22, Table 9) Materials and Its Thickness Permeance ng/s · m2 · Pa Concrete (1:2:4 mix, 1 m) 4.7 Brick, masonry, 100 mm 46 Plaster on metal lath, 19 mm 860 Plaster on wood lath, 19mm 630 Gypsum wall board, 9.5 mm 2860 Plywood, 6.4 mm 40–109 Still air, 1 m 174 Mineral wool insulation (unprotected), 1 m 245 Expanded polyurethane insulation board, 1 m 0.58–2.3 Aluminum foil, 0.025 mm 0.0 Aluminum foil, 0.009 mm 2.9 Polyethylene, 0.051 mm 9.1 Polyethylene, 0.2 mm 2.3 Polyester, 0.19 mm 4.6 Vapor retarder latex paint, 0.070 mm 26 Exterior acrylic house and trim paint, 0.040 mm 313 Building paper, unit mass of 0.16–0.68 kg/m2 0.1–2400 Data vary greatly. Consult manufacturer for more accurate data. Multiply by 1.41 106 to convert to lbm/s · ft2 · psi. Also, 1 ng 1012 kg. Permeability Thickness L (kg/s · m2 · Pa) (14-32) The reciprocal of permeance is called (unit) vapor resistance and is expressed as Vapor resistance R 1 Permeance 1 L (s · m2 · Pa/kg) (14-33) Note that vapor resistance represents the resistance of a material to water vapor transmission. It should be pointed out that the amount of moisture that enters or leaves a building by diffusion is usually negligible compared to the amount that enters with infiltrating air or leaves with exfiltrating air. The primary cause of interest in the moisture diffusion is its impact on the performance and longevity of building materials. The overall vapor resistance of a composite building structure that consists of several layers in series is the sum of the resistances of the individual layers and is expressed as R, total R, 1 R, 2 · · · R, n R , i (14-34) Then the rate of vapor transmission through a composite structure can be determined in an analogous manner to heat transfer from P m· A R, total (kg/s) (14-35) Vapor permeance of common building materials is given in Table 14–10. EXAMPLE 14–6 Condensation and Freezing of Moisture in Walls The condensation and even freezing of moisture in walls without effective vapor retarders is a real concern in cold climates, and it undermines the effectiveness of insulations. Consider a wood frame wall that is built around 38 mm 90 mm (2 4 nominal) wood studs. The 90-mm-wide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13-mm gypsum wallboard and the outside with 13-mm wood fiberboard and 13-mm 200-mm wood bevel lapped siding. Using manufacturer’s data, the thermal and vapor resistances of various components for a unit wall area are determined to be as: cen58933_ch14.qxd 9/9/2002 10:05 AM Page 739 739 CHAPTER 14 Construction 1. 2. 3. 4. 5. 6. Outside surface, 24 km/h wind Painted wood bevel lapped siding Wood fiberboard sheeting, 13 mm Glass fiber insulation, 90 mm Painted gypsum wallboard, 13 mm Inside surface, still air TOTAL R-Value, m2 · °C/W R-Value, s · m2 · Pa/ng 0.030 0.14 0.23 2.45 0.079 0.12 — 0.019 0.0138 0.0004 0.012 — 3.05 0.0452 The indoor conditions are 20°C and 60 percent relative humidity while the outside conditions are 16°C and 70 percent relative humidity. Determine if condensation or freezing of moisture will occur in the insulation. SOLUTION The thermal and vapor resistances of different layers of a wall are given. The possibility of condensation or freezing of moisture in the wall is to be investigated. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal and vapor resistances of different layers of the wall and the heat transfer coefficients are constant. Properties The thermal and vapor resistances are as given in the problem statement. The saturation pressures of water at 20°C and 16°C are 2339 Pa and 151 Pa, respectively (Table 14–9). Analysis The schematic of the wall as well as the different elements used in its construction are shown in Figure 14–25. Condensation is most likely to occur at the coldest part of insulation, which is the part adjacent to the exterior sheathing. Noting that the total thermal resistance of the wall is 3.05 m2 · °C/W, the rate of heat transfer through a unit area A 1 m2 of the wall is Ti To [20 (16)°C] · Q wall A (1 m2) 11.8 W Rtotal 3.05 m2 · °C/W The thermal resistance of the exterior part of the wall beyond the insulation is 0.03 0.14 0.23 0.40 m2 · °C/W. Then the temperature of the insulation–outer sheathing interface is 6 · TI To Q wall Rext 16°C (11.8 W)(0.40°C/W) 11.3°C The saturation pressure of water at 11.3°C is 234 Pa, as shown in Table 14–9, and if there is condensation or freezing, the vapor pressure at the insulation–outer sheathing interface will have to be this value. The vapor pressure at the indoors and the outdoors is P, 1 1Psat, 1 0.60 (2340 Pa) 1404 Pa P, 2 2Psat, 2 0.70 (151 Pa) 106 Pa Then the rate of moisture flow through the interior and exterior parts of the wall becomes 3 4 5 2 1 FIGURE 14–25 Schematic for Example 14–6. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 740 740 HEAT TRANSFER P, 1 P, l R, interior interior (1404 234) Pa (1 m2) 94,355 ng/s 94.4 g/s (0.012 0.0004) Pa · m2 · s/ng P, 1 P, l P m· , exterior A A R exterior R, exterior (234 106) Pa (1 m2) 3902 ng/s 3.9 g/s (0.019 0.0138) Pa · m2 · s/ng P m· , interior A R A That is, moisture is flowing toward the interface at a rate of 94.4 g/s but flowing from the interface to the outdoors at a rate of only 3.9 g/s. Noting that the interface pressure cannot exceed 234 Pa, these results indicate that moisture is freezing in the insulation at a rate of m· , freezing m· , interior m· , exterior 94.4 3.9 90.5 g/s This corresponds to 7.82 g during a 24-h period, which can be absorbed by the insulation or sheathing, and then flows out when the conditions improve. However, excessive condensation (or frosting at temperatures below 0°C) of moisture in the walls during long cold spells can cause serious problems. This problem can be avoided or minimized by installing vapor barriers on the interior side of the wall, which will limit the moisture flow rate to 3.9 g/s. Note that if there were no condensation or freezing, the flow rate of moisture through a 1 m2 section of the wall would be 28.7 g/s (can you verify this?). 14–7 Carbonaceous material Hardened surface Steel shaft Soft core Carbon FIGURE 14–26 The surface hardening of a mild steel component by the diffusion of carbon molecules is a transient mass diffusion process. ■ TRANSIENT MASS DIFFUSION The steady analysis discussed earlier is useful when determining the leakage rate of a species through a stationary layer. But sometimes we are interested in the diffusion of a species into a body during a limited time before steady operating conditions are established. Such problems are studied using transient analysis. For example, the surface of a mild steel component is commonly hardened by packing the component in a carbonaceous material in a furnace at high temperature. During the short time period in the furnace, the carbon molecules diffuse through the surface of the steel component, but they penetrate to a depth of only a few millimeters. The carbon concentration decreases exponentially from the surface to the inner parts, and the result is a steel component with a very hard surface and a relatively soft core region (Fig. 14–26). The same process is used in the gem industry to color clear stones. For example, a clear sapphire is given a brilliant blue color by packing it in titanium and iron oxide powders and baking it in an oven at about 2000°C for about a month. The titanium and iron molecules penetrate less than 0.5 mm in the sapphire during this process. Diffusion in solids is usually done at high temperatures to take advantage of the high diffusion coefficients at high temperatures and thus to keep the diffusion time at a reasonable level. Such diffusion or “doping” is also commonly practiced in the production of n- or p-type semiconductor materials used in the manufacture of electronic components. Drying processes such as the drying of coal, timber, food, and textiles constitute another major application area of transient mass diffusion. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 741 741 CHAPTER 14 Transient mass diffusion in a stationary medium is analogous to transient heat transfer provided that the solution is dilute and thus the density of the medium is constant. In Chapter 4 we presented analytical and graphical solutions for one-dimensional transient heat conduction problems in solids with constant properties, no heat generation, and uniform initial temperature. The analogous one-dimensional transient mass diffusion problems satisfy these requirements: 1. The diffusion coefficient is constant. This is valid for an isothermal medium since DAB varies with temperature (corresponds to constant thermal diffusivity). 2. There are no homogeneous reactions in the medium that generate or deplete the diffusing species A (corresponds to no heat generation). 3. Initially (t 0) the concentration of species A is constant throughout the medium (corresponds to uniform initial temperature). Then the solution of a mass diffusion problem can be obtained directly from the analytical or graphical solution of the corresponding heat conduction problem given in Chapter 4. The analogous quantities between heat and mass transfer are summarized in Table 14–11 for easy reference. For the case of a semi-infinite medium with constant surface concentration, for example, the solution can be expressed in an analogous manner to Eq. 4-24 as CA(x, t) CA, i x erfc CA, s CA, i 2 DABt TABLE 14–11 Analogy between the quantities that appear in the formulation and solution of transient heat conduction and transient mass diffusion in a stationary medium Heat Conduction Mass Diffusion T C, y, or w DAB T (x, t ) T , mass Ti T wA(x, t ) wA, w w A, i A, T (x, t ) Ts Ti Ts wA(x, t ) wA wA, i wA (14-36) x 2 t mass where CA, i is the initial concentration of species A at time t 0 and CA, s is the concentration at the inner side of the exposed surface of the medium. By using the definitions of molar fraction, mass fraction, and density, it can be shown that for dilute solutions, Bi hconv L k Bimass CA(x, t) CA, i A(x, t) A, i wA(x, t) wA, i yA(x, t) yA, i w w y y A, s A, i A, s A, i A, s A, i CA, s CA, i CA, s CA, i CA, s CA, i D t (dCA/dx)x0 (CA, s CA, i)/ DABt ABz t L2 hmass L DAB DABt L2 (14-37) since the total density or total molar concentration of dilute solutions is usually constant ( constant or C constant). Therefore, other measures of concentration can be used in Eq. 14–36. A quantity of interest in mass diffusion processes is the depth of diffusion at a given time. This is usually characterized by the penetration depth defined as the location x where the tangent to the concentration profile at the surface (x 0) intercepts the CA CA, i line, as shown in Figure 14–27. Obtaining the concentration gradient at x 0 by differentiating Eq. 14–36, the penetration depth is determined to be diff x 2 DABt (14-38) Therefore, the penetration depth is proportional to the square root of both the diffusion coefficient and time. The diffusion coefficient of zinc in copper at 1000°C, for example, is 5.0 1012 m2/s. Then the penetration depth of zinc in copper in 10 h is CA, s CA(x, t) CA, i 0 δdiff x Tangent line to concentration gradient at x = 0 Semi-infinite medium Slope of tangent line dCA —– dx x=0 CA, s – CA, i = – —–——— δdiff FIGURE 14–27 The concentration profile of species A in a semi-infinite medium during transient mass diffusion and the penetration depth. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 742 742 HEAT TRANSFER diff DABt (5.0 1012 m2/s)(10 3600 s) 0.00038 m 0.38 mm That is, zinc will penetrate to a depth of about 0.38 mm in an appreciable amount in 10 h, and there will hardly be any zinc in the copper block beyond a depth of 0.38 mm. The diffusion coefficients in solids are typically very low (on the order of 109 to 1015 m2/s), and thus the diffusion process usually affects a thin layer at the surface. A solid can conveniently be treated as a semi-infinite medium during transient mass diffusion regardless of its size and shape when the penetration depth is small relative to the thickness of the solid. When this is not the case, solutions for one-dimensional transient mass diffusion through a plane wall, cylinder, and sphere can be obtained from the solutions of analogous heat conduction problems using the Heisler charts or one-term solutions presented in Chapter 4. EXAMPLE 14–7 The surface of a mild steel component is commonly hardened by packing the component in a carbonaceous material in a furnace at a high temperature for a predetermined time. Consider such a component with a uniform initial carbon concentration of 0.15 percent by mass. The component is now packed in a carbonaceous material and is placed in a high-temperature furnace. The diffusion coefficient of carbon in steel at the furnace temperature is given to be 4.8 1010 m2/s, and the equilibrium concentration of carbon in the iron at the interface is determined from equilibrium data to be 1.2 percent by mass. Determine how long the component should be kept in the furnace for the mass concentration of carbon 0.5 mm below the surface to reach 1 percent (Fig. 14–28). Furnace wA, s = 1.2% Carbonaceous material wA(x, t) = 1% 0 0.5 mm Hardening of Steel by the Diffusion of Carbon wA, i = 0.15% Steel component Carbon FIGURE 14–28 Schematic for Example 14–7. x SOLUTION A steel component is to be surface hardened by packing it in a carbonaceous material in a furnace. The length of time the component should be kept in the furnace is to be determined. Assumptions Carbon penetrates into a very thin layer beneath the surface of the component, and thus the component can be modeled as a semi-infinite medium regardless of its thickness or shape. Properties The relevant properties are given in the problem statement. Analysis This problem is analogous to the one-dimensional transient heat conduction problem in a semi-infinite medium with specified surface temperature, and thus can be solved accordingly. Using mass fraction for concentration since the data are given in that form, the solution can be expressed as wA(x, t) wA, i x wA, s wA, i erfc 2 D t AB Substituting the specified quantities gives 0.01 0.0015 x 0.81 erfc 0.012 0.0015 2 DABt cen58933_ch14.qxd 9/9/2002 10:05 AM Page 743 743 CHAPTER 14 The argument whose complementary error function is 0.81 is determined from Table 4–3 to be 0.17. That is, x 0.17 2 DABt Then solving for the time t gives t (0.0005 m)2 x2 4505 s 1 h 15 min 2 4DAB(0.17) 4 (4.8 1010 m2/s)(0.17)2 Discussion The steel component in this case must be held in the furnace for 1 h and 15 min to achieve the desired level of hardening. The diffusion coefficient of carbon in steel increases exponentially with temperature, and thus this process is commonly done at high temperatures to keep the diffusion time at a reasonable level. 14–8 ■ DIFFUSION IN A MOVING MEDIUM To this point we have limited our consideration to mass diffusion in a stationary medium, and thus the only motion involved was the creeping motion of molecules in the direction of decreasing concentration, and there was no motion of the mixture as a whole. Many practical problems, such as the evaporation of water from a lake under the influence of the wind or the mixing of two fluids as they flow in a pipe, involve diffusion in a moving medium where the bulk motion is caused by an external force. Mass diffusion in such cases is complicated by the fact that chemical species are transported both by diffusion and by the bulk motion of the medium (i.e., convection). The velocities and mass flow rates of species in a moving medium consist of two components: one due to molecular diffusion and one due to convection (Fig. 14–29). Diffusion in a moving medium, in general, is difficult to analyze since various species can move at different velocities in different directions. Turbulence will complicate the things even more. To gain a firm understanding of the physical mechanism while keeping the mathematical complexities to a minimum, we limit our consideration to systems that involve only two components (species A and B) in one-dimensional flow (velocity and other properties change in one direction only, say the x-direction). We also assume the total density (or molar concentration) of the medium remains constant. That is, A B constant (or C CA CB constant) but the densities of species A and B may vary in the x-direction. Several possibilities are summarized in Figure 14–30. In the trivial case (case a) of a stationary homogeneous mixture, there will be no mass transfer by molecular diffusion or convection since there is no concentration gradient or bulk motion. The next case (case b) corresponds to the flow of a well-mixed fluid mixture through a pipe. Note that there is no concentration gradients and thus molecular diffusion in this case, and all species move at the bulk flow velocity of . The mixture in the third case (case c) is stationary ( 0) and thus it corresponds to ordinary molecular diffusion in stationary mediums, which we discussed before. Note that the velocity of a species at a location in Air Convection Diffusion Lake FIGURE 14–29 In a moving medium, mass transfer is due to both diffusion and convection. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 744 744 HEAT TRANSFER A B (a) Homogeneous mixture without bulk motion (no concentration gradients and thus no diffusion) =0 (b) Homogeneous mixture with bulk motion (no concentration gradients and thus no diffusion) Species Density Velocity Species A ρA = constant A = 0 Mass flow rate m· = 0 Species B ρB = constant B = 0 Mixture of A and B ρ = ρA + ρB =0 m· = 0 Species A ρA = constant A = m· A = ρAA A Species B ρB = constant B = m· B = ρBB A Mixture of A and B ρ = ρA + ρB = · = ρA m = m· + m· = constant = constant (c) Nonhomogeneous mixture without bulk motion (stationary medium with concentration gradients) =0 diff, A diff, B (d) Nonhomogeneous mixture with bulk motion (moving medium with concentration gradients) diff, A diff, B A m· B = 0 A B Species A ρA ≠ constant A = diff, A m· A = ρAdiff, A A Species B ρB ≠ constant B = diff, B m· B = ρBdiff, B A Mixture of A and B ρ = ρA + ρB =0 · = ρA = 0 m · = –m · ) (thus m A B Species A ρA ≠ constant A = + diff, A m· A = ρAdiff, A A Species B ρB ≠ constant B = + diff, B m· B = ρBdiff, B A Mixture of A and B ρ = ρA + ρB = constant = = constant · = ρA m = m· + m· A B FIGURE 14–30 Various quantities associated with a mixture of two species A and B at a location x under one-dimensional flow or no-flow conditions. (The density of the mixture A B is assumed to remain constant.) this case is simply the diffusion velocity, which is the average velocity of a group of molecules at that location moving under the influence of concentration gradient. Finally, the last case (case d) involves both molecular diffusion and convection, and the velocity of a species in this case is equal to the sum of the bulk flow velocity and the diffusion velocity. Note that the flow and the diffusion velocities can be in the same or opposite directions, depending on the direction of the concentration gradient. The diffusion velocity of a species is negative when the bulk flow is in the positive x-direction and the concentration gradient is positive (i.e., the concentration of the species increases in the x-direction). Noting that the mass flow rate at any flow section is expressed as m· A where is the density, is the velocity, and A is the cross sectional area, the conservation of mass relation for the flow of a mixture that involves two species A and B can be expressed as m· m· A m· B or A AAA BB A Canceling A and solving for gives cen58933_ch14.qxd 9/9/2002 10:05 AM Page 745 745 CHAPTER 14 AA BB A B A B wAA wBB (14-39) where is called the mass-average velocity of the flow, which is the velocity that would be measured by a velocity sensor such as a pitot tube, a turbine device, or a hot wire anemometer inserted into the flow. The special case 0 corresponds to a stationary medium, which can now be defined more precisely as a medium whose mass-average velocity is zero. Therefore, mass transport in a stationary medium is by diffusion only, and zero mass-average velocity indicates that there is no bulk fluid motion. When there is no concentration gradient (and thus no molecular mass diffusion) in the fluid, the velocity of all species will be equal to the massaverage velocity of the flow. That is, A B. But when there is a concentration gradient, there will also be a simultaneous flow of species in the direction of decreasing concentration at a diffusion velocity of diff. Then the average velocity of the species A and B can be determined by superimposing the average flow velocity and the diffusion velocity as (Fig. 14–31) A diff, A B diff, B diff, A = 0 A = (14-40) A Similarly, we apply the superposition principle to the species mass flow rates to get m· A AA A A( diff, A)A AA Adiff, A A m· conv, A m· diff, A m· B BB A B( diff, B)A BA Bdiff, B A m· conv, B m· diff, B (14-41) Flow velocity (a) No concentration gradient Using Fick’s law of diffusion, the total mass fluxes j m· /A can be expressed as diff, A ≠ 0 A = + diff, A A dwA dwA jA A Adiff, A DAB wA(jA jB) DAB dx dx B dwB dwB jB B Bdiff, B DBA wB(jA jB) DBA dx dx (14-42) Note that the diffusion velocity of a species is negative when the molecular diffusion occurs in the negative x-direction (opposite to flow direction). The mass diffusion rates of the species A and B at a specified location x can be expressed as m· diff, A Adiff, A A A(A )A A ( )A m· diff, B B diff, B B B (14-43) By substituting the relation from Eq. 14–39 into Eq. 11–43, it can be shown that at any cross section m· diff, A m· diff, B 0 → m· diff, A m· diff, B → ADAB dwA dwB ADBA dx dx (14-44) which indicates that the rates of diffusion of species A and B must be equal in magnitude but opposite in sign. This is a consequence of the assumption diff, A A Flow velocity (b) Mass concentration gradient and thus mass diffusion FIGURE 14–31 The velocity of a species at a point is equal to the sum of the bulk flow velocity and the diffusion velocity of that species at that point. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 746 746 HEAT TRANSFER A B constant, and it indicates that anytime the species A diffuses in one direction, an equal amount of species B must diffuse in the opposite direction to maintain the density (or the molar concentration) constant. This behavior is closely approximated by dilute gas mixtures and dilute liquid or solid solutions. For example, when a small amount of gas diffuses into a liquid, it is reasonable to assume the density of the liquid to remain constant. Note that for a binary mixture, wA wB 1 at any location x. Taking the derivative with respect to x gives dwA dwB dx dx Thus we conclude from Eq. 14–44 that (Fig. 14–32) w = wA + wB = 1 1 wA (14-45) DAB DBA wB w (14-46) That is, in the case of constant total concentration, the diffusion coefficient of species A into B is equal to the diffusion coefficient of species B into A. We now repeat the analysis presented above with molar concentration C and · the molar flow rate N. The conservation of matter in this case is expressed as 0 x · · · N NA NB m· diff, A m· diff, B or A AA A BB A B A wA = – wB dwB dwA —— = – —— dx dx m· = – m· diff, A (14-47) Canceling A and solving for gives CB CAA CBB CA yAA yBB C C A C B (14-48) diff, B DAB = DBA FIGURE 14–32 In a binary mixture of species A and B with A B constant, the rates of mass diffusion of species A and B are equal magnitude and opposite in direction. where is called the molar-average velocity of the flow. Note that unless the mass and molar fractions are the same. The molar flow rates of species are determined similarly to be · · · NA CAA A CA( diff, A)A CAA CAdiff, A A Nconv, A Ndiff, A · · · NB CBB A CB( diff, B)A CBA CBdiff, B A Nconv, B Ndiff, B (14-49) · Using Fick’s law of diffusion, the total molar fluxes ¯j N /A and diffusion · molar flow rates Ndiff can be expressed as dyA dyA CA ¯j C C C CDAB yA( ¯jA ¯jB) CDAB A A A diff, A C dx dx dy dy CB B B ¯j C C C CDBA yB( ¯jA ¯jB) CDBA B B B diff, B C dx dx (14-50) and · Ndiff, A CAdiff, A A CA(A )A · Ndiff, B CBdiff, B A CB(B )A (14-51) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 747 747 CHAPTER 14 By substituting the relation from Eq. 14–48 into these two equations, it can be shown that · · Ndiff, A Ndiff, B 0 → · · Ndiff, A Ndiff, B (14-52) which again indicates that the rates of diffusion of species A and B must be equal in magnitude but opposite in sign. It is important to note that when working with molar units, a medium is said to be stationary when the molar-average velocity is zero. The average velocity of the molecules will be zero in this case, but the apparent velocity of the mixture as measured by a velocimeter placed in the flow will not necessarily be zero because of the different masses of different molecules. In a massbased stationary medium, for each unit mass of species A moving in one direction, a unit mass of species B moves in the opposite direction. In a molebased stationary medium, however, for each mole of species A moving in one direction, one mole of species B moves in the opposite direction. But this may result in a net mass flow rate in one direction that can be measured by a velocimeter since the masses of different molecules are different. You may be wondering whether to use the mass analysis or molar analysis in a problem. The two approaches are equivalent, and either approach can be used in mass transfer analysis. But sometimes it may be easier to use one of the approaches, depending on what is given. When mass-average velocity is known or can easily be obtained, obviously it is more convenient to use the mass-based formulation. When the total pressure and temperature of a mixture are constant, however, it is more convenient to use the molar formulation, as explained next. Special Case: Gas Mixtures at Constant Pressure and Temperature Consider a gas mixture whose total pressure and temperature are constant throughout. When the mixture is homogeneous, the mass density , the molar density (or concentration) C, the gas constant R, and the molar mass M of the mixture are the same throughout the mixture. But when the concentration of one or more gases in the mixture is not constant, setting the stage for mass diffusion, then the mole fractions yi of the species will vary throughout the mixture. As a result, the gas constant R, the molar mass M, and the mass density of the mixture will also vary since, assuming ideal gas behavior, M y M, i i R Ru , M and P RT where Ru 8.314 kJ/kmol · K is the universal gas constant. Therefore, the assumption of constant mixture density ( constant) in such cases will not be accurate unless the gas or gases with variable concentrations constitute a very small fraction of the mixture. However, the molar density C of a mixture remains constant when the mixture pressure P and temperature T are constant since P RT Ru T CRuT M (14-53) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 748 748 HEAT TRANSFER The condition C constant offers considerable simplification in mass transfer analysis, and thus it is more convenient to use the molar formulation when dealing with gas mixtures at constant total pressure and temperature (Fig. 14–33). Gas mixture T constant P constant Diffusion of Vapor through a Stationary Gas: Stefan Flow Independent of composition of mixture P P RT (Ru /M)T → →   P C RuT Depends on composition of mixture FIGURE 14–33 When the total pressure P and temperature T of a binary mixture of ideal gases is held constant, then the molar concentration C of the mixture remains constant. Liquid A L Diffusion of B Bulk flow Diffusion of A Gas mixture A+B Many engineering applications such as heat pipes, cooling ponds, and the familiar perspiration involve condensation, evaporation, and transpiration in the presence of a noncondensable gas, and thus the diffusion of a vapor through a stationary (or stagnant) gas. To understand and analyze such processes, consider a liquid layer of species A in a tank surrounded by a gas of species B, such as a layer of liquid water in a tank open to the atmospheric air (Fig. 14–34), at constant pressure P and temperature T. Equilibrium exists between the liquid and vapor phases at the interface (x 0), and thus the vapor pressure at the interface must equal the saturation pressure of species A at the specified temperature. We assume the gas to be insoluble in the liquid, and both the gas and the vapor to behave as ideal gases. If the surrounding gas at the top of the tank (x L) is not saturated, the vapor pressure at the interface will be greater than the vapor pressure at the top of the tank (PA, 0 PA, L and thus yA, 0 yA, L since yA PA/P), and this pressure (or concentration) difference will drive the vapor upward from the air–water interface into the stagnant gas. The upward flow of vapor will be sustained by the evaporation of water at the interface. Under steady conditions, the molar (or mass) flow rate of vapor throughout the stagnant gas column remains constant. That is, ¯j N· /A constant A A 0 FIGURE 14–34 Diffusion of a vapor A through a stagnant gas B. (or jA m· A/A constant) The pressure and temperature of the gas–vapor mixture are said to be constant, and thus the molar density of the mixture must be constant throughout the mixture, as shown earlier. That is, C CA CB constant, and it is more convenient to work with mole fractions or molar concentrations in this case instead of mass fractions or densities since constant. Noting that yA yB 1 and that yA, 0 yA, L, we must have yB, 0 yB, L. That is, the mole fraction of the gas must be decreasing downward by the same amount that the mole fraction of the vapor is increasing. Therefore, gas must be diffusing from the top of the column toward the liquid interface. However, the gas is said to be insoluble in the liquid, and thus there can be no net mass flow of the gas downward. Then under steady conditions, there must be an upward bulk fluid motion with an average velocity that is just large enough to balance the diffusion of air downward so that the net molar (or mass) flow rate of the gas at any point is zero. In other words, the upward bulk motion offsets the downward diffusion, and for each air molecule that moves downward, there is another air molecule that moves upward. As a result, the air appears to be stagnant (it does not move). That is, ¯j N· /A 0 B B (or jB m· B /A 0) The diffusion medium is no longer stationary because of the bulk motion. The implication of the bulk motion of the gas is that it transports vapor as well as cen58933_ch14.qxd 9/9/2002 10:05 AM Page 749 749 CHAPTER 14 the gas upward with a velocity of , which results in additional mass flow of vapor upward. Therefore, the molar flux of the vapor can be expressed as dyA ¯j N· /A ¯j ¯ ¯ ¯ A A A, conv jA, diff yA( jA jB) CDAB dx (14-54) Noting that ¯jB 0, it simplifies to ¯j y ¯j CD dyA A A A AB dx (14-55) ¯ ¯j CDAB dyA → 1 dyA jA constant A CDAB 1 yA dx 1 yA dx (14-56) Solving for ¯jA gives since ¯jA constant, C constant, and DAB constant. Separating the variables and integrating from x 0, where yA(0) yA, 0, to x L, where yA(L) yA, L gives yA, L A, 0 dyA 1 yA L 0 j¯A dx CDAB (14-57) Performing the integrations, ln 1 yA, L j¯A L CD 1 yA, 0 AB (14-58) Then the molar flux of vapor A, which is the evaporation rate of species A per unit interface area, becomes 1 yA, L ¯j N· /A CDAB ln A A L 1 yA, 0 (kmol/s · m2) (14-59) This relation is known as Stefan’s law, and the induced convective flow described that enhances mass diffusion is called the Stefan flow. An expression for the variation of the mole fraction of A with x can be determined by performing the integration in Eq. 14–57 to the upper limit of x where yA(x) yA (instead of to L where yA(L) yA, L). It yields ln j¯A 1 yA x 1 yA, 0 CDAB (14-60) Substituting the ¯jA expression from Eq. 14–59 into this relation and rearranging gives 1 yA, L 1 yA 1 yA, 0 1 yA, 0 x/L and yB, L yB yB, 0 yB, 0 x/L (14-61) The second relation for the variation of the mole fraction of the stationary gas B is obtained from the first one by substituting 1 yA yB since yA yB 1. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 750 750 HEAT TRANSFER To maintain isothermal conditions in the tank during evaporation, heat must be supplied to the tank at a rate of · Q m· A hfg, A jA As hfg, A ( ¯jAMA)As hfg, A (kJ/s) (14-62) where As is the surface area of the liquid–vapor interface, hfg, A is the latent heat of vaporization, and MA is the molar mass of species A. Equimolar Counterdiffusion Gas mixture A+B yA > yB A B · NA 0 x T, P · NB Gas mixture A+B yA < yB L T, P FIGURE 14–35 Equimolar isothermal counterdiffusion of two gases A and B. Consider two large reservoirs connected by a channel of length L, as shown in Figure 14–35. The entire system contains a binary mixture of gases A and B at a uniform temperature T and pressure P throughout. The concentrations of species are maintained constant in each of the reservoirs such that yA, 0 yA, L and yB, 0 yB, L. The resulting concentration gradients will cause the species A to diffuse in the positive x-direction and the species B in the opposite direction. Assuming the gases to behave as ideal gases and thus P CRuT, the total molar concentration of the mixture C will remain constant throughout the mixture since P and T are constant. That is, C CA CB constant (kmol/m3) This requires that for each molecule of A that moves to the right, a molecule of B moves to the left, and thus the molar flow rates of species A and B must be equal in magnitude and opposite in sign. That is, · · NA NB · · NA NB 0 or (kmol/s) This process is called equimolar counterdiffusion for obvious reasons. The net molar flow rate of the mixture for such a process, and thus the molaraverage velocity, is zero since · · · N NA NB 0 → CA 0 → 0 Therefore, the mixture is stationary on a molar basis and thus mass transfer is by diffusion only (there is no mass transfer by convection) so that ¯j N· /A CD dyA A A AB dx and ¯j N· /A CD dyB B B BA dx (14-63) Under steady conditions, the molar flow rates of species A and B can be determined directly from Eq. 14–24 developed earlier for one-dimensional steady diffusion in a stationary medium, noting that P CRuT and thus C P/RuT for each constituent gas and the mixture. For one-dimensional flow through a channel of uniform cross sectional area A with no homogeneous chemical reactions, they are expressed as yA, 1 yA, 2 CA, 1 CA, 2 DAB PA, 0 PA, L · Ndiff, A CDAB A DAB A A L L RuT L yB, 1 yB, 2 CB, 1 CB, 2 DBA PB, 0 PB, L · Ndiff, B CDBA A DBA A A L L RuT L (14-64) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 751 751 CHAPTER 14 These relations imply that the mole fraction, molar concentration, and the partial pressure of either gas vary linearly during equimolar counterdiffusion. It is interesting to note that the mixture is stationary on a molar basis, but it is not stationary on a mass basis unless the molar masses of A and B are equal. Although the net molar flow rate through the channel is zero, the net mass flow rate of the mixture through the channel is not zero and can be determined from · · · m· m· A m· B NAMA NB MB NA(MA MB) (14-65) · · since N B N A. Note that the direction of net mass flow rate is the flow direction of the gas with the larger molar mass. A velocity measurement device such as an anemometer placed in the channel will indicate a velocity of m· /A where is the total density of the mixture at the site of measurement. EXAMPLE 14–8 Venting of Helium into the Atmosphere by Diffusion The pressure in a pipeline that transports helium gas at a rate of 2 kg/s is maintained at 1 atm by venting helium to the atmosphere through a 5-mm-internaldiameter tube that extends 15 m into the air, as shown in Figure 14–36. Assuming both the helium and the atmospheric air to be at 25°C, determine (a) the mass flow rate of helium lost to the atmosphere through an individual tube, (b) the mass flow rate of air that infiltrates into the pipeline, and (c) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation. Air 1 atm 25°C Air x He L = 15 m SOLUTION The pressure in a helium pipeline is maintained constant by venting to the atmosphere through a long tube. The mass flow rates of helium and air through the tube and the net flow velocity at the bottom are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium and atmospheric air are ideal gases. 3 No chemical reactions occur in the tube. 4 Air concentration in the pipeline and helium concentration in the atmosphere are negligible so that the mole fraction of the helium is 1 in the pipeline and 0 in the atmosphere (we will check this assumption later). Properties The diffusion coefficient of helium in air (or air in helium) at normal atmospheric conditions is DAB 7.20 105 m2/s (Table 14–2). The molar masses of air and helium are 29 and 4 kg/kmol, respectively (Table A–1). Analysis This is a typical equimolar counterdiffusion process since the problem involves two large reservoirs of ideal gas mixtures connected to each other by a channel, and the concentrations of species in each reservoir (the pipeline and the atmosphere) remain constant. (a) The flow area, which is the cross sectional area of the tube, is A D2/4 (0.005 m)2/4 1.963 105 m2 Noting that the pressure of helium is 1 atm at the bottom of the tube (x 0) and 0 at the top (x L), its molar flow rate is determined from Eq. 14–64 to be 5 mm He 0 Helium (A) Air 2 kg/s 1 atm 25°C FIGURE 14–36 Schematic for Example 14–8 cen58933_ch14.qxd 9/9/2002 10:05 AM Page 752 752 HEAT TRANSFER DAB A PA, 0 PA, L · · Nhelium Ndiff, A RuT L 5 2 (7.20 10 m /s)(1.963 105 m2) 1 atm 0 101.3 kPa 1 atm 15 m (8.314 kPa · m3/kmol · K)(298 K) 12 3.85 10 kmol/s Therefore, · m· helium (NM)helium (3.85 1012 kmol/s)(4 kg/kmol) 1.54 1011 kg/s which corresponds to about 0.5 g per year. · · (b) Noting that N B N A during an equimolar counterdiffusion process, the molar flow rate of air into the helium pipeline is equal to the molar flow rate of helium. The mass flow rate of air into the pipeline is · m· air (NM)air (–3.85 1012 kmol/s)(29 kg/kmol) 112 1012 kg/s The mass fraction of air in the helium pipeline is m· air 112 1012 kg/s wair · 5.6 1011 0 m total (2 112 1012 1.54 1011) kg/s which validates our original assumption of negligible air in the pipeline. (c) The net mass flow rate through the tube is m· net m· helium m· air 1.54 1011 112 1012 9.66 1011 kg/s The mass fraction of air at the bottom of the tube is very small, as shown above, and thus the density of the mixture at x 0 can simply be taken to be the density of helium, which is helium 101.325 kPa P 0.1637 kg/m3 RT (2.0769 kPa · m3/kg · K)(298 K) Then the average flow velocity at the bottom part of the tube becomes 9.66 1011 kg/s m· 3.02 105 m/s A (0.01637 kg/m3)(1.963 105 m2) which is difficult to measure by even the most sensitive velocity measurement devices. The negative sign indicates flow in the negative x-direction (toward the pipeline). EXAMPLE 14–9 Measuring Diffusion Coefficient by the Stefan Tube A 3-cm-diameter Stefan tube is used to measure the binary diffusion coefficient of water vapor in air at 20°C at an elevation of 1600 m where the atmospheric cen58933_ch14.qxd 9/9/2002 10:05 AM Page 753 753 CHAPTER 14 SOLUTION The amount of water that evaporates from a Stefan tube at a specified temperature and pressure over a specified time period is measured. The diffusion coefficient of water vapor in air is to be determined. Assumptions 1 Water vapor and atmospheric air are ideal gases. 2 The amount of air dissolved in liquid water is negligible. 3 Heat is transferred to the water from the surroundings to make up for the latent heat of vaporization so that the temperature of water remains constant at 20°C. Properties The saturation pressure of water at 20°C is 2.34 kPa (Table A–9). Analysis The vapor pressure at the air–water interface is the saturation pressure of water at 20°C, and the mole fraction of water vapor (species A) at the interface is determined from yvapor, 0 yA, 0 Pvapor, 0 2.34 kPa 0.0280 P 83.5 kPa Dry air is blown on top of the tube and, thus, yvapor, L yA, L 0. Also, the total molar density throughout the tube remains constant because of the constant temperature and pressure conditions and is determined to be C 83.5 kPa P 0.0343 kmol/m3 RuT (8.314 kPa · m3/kmol · K)(293 K) The cross-sectional area of the tube is A D2/4 (0.03 m)2/4 7.069 104 m2 The evaporation rate is given to be 1.23 g per 15 days. Then the molar flow rate of vapor is determined to be m· vapor 1.23 103 kg · · NA Nvapor 5.27 1011 kmol/s Mvapor (15 24 3600 s)(18 kg/kmol) Finally, substituting the information above into Eq. 14–59 we get 3 5.27 1011 kmol/s (0.0343 kmol/m )DAB 10 ln 0.4 m 1 0.028 7.069 104 m2 which gives DAB 3.06 105 m2/s for the binary diffusion coefficient of water vapor in air at 20°C and 83.5 kPa. Air, B yA, L Water, A Diffusion of air L Bulk flow of air and vapor Diffusion of vapor pressure is 83.5 kPa. The tube is partially filled with water, and the distance from the water surface to the open end of the tube is 40 cm (Fig. 14–37). Dry air is blown over the open end of the tube so that water vapor rising to the top is removed immediately and the concentration of vapor at the top of the tube is zero. In 15 days of continuous operation at constant pressure and temperature, the amount of water that has evaporated is measured to be 1.23 g. Determine the diffusion coefficient of water vapor in air at 20°C and 83.5 kPa. yA 0 0 yB yA, 0 1 FIGURE 14–37 Schematic for Example 14–9. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 754 754 HEAT TRANSFER 14–9 ρA, 0 Concentration boundary layer y Concentration profile ρA, ρA, s x Species A FIGURE 14–38 The development of a concentration boundary layer for species A during external flow on a flat surface. Concentration entry length Fully developed region Species A Concentration boundary layer Thermal boundary layer Velocity boundary layer FIGURE 14–39 The development of the velocity, thermal, and concentration boundary layers in internal flow. ■ MASS CONVECTION So far we have considered mass diffusion, which is the transfer of mass due to a concentration gradient. Now we consider mass convection (or convective mass transfer), which is the transfer of mass between a surface and a moving fluid due to both mass diffusion and bulk fluid motion. We mentioned earlier that fluid motion enhances heat transfer considerably by removing the heated fluid near the surface and replacing it by the cooler fluid further away. Likewise, fluid motion enhances mass transfer considerably by removing the high-concentration fluid near the surface and replacing it by the lowerconcentration fluid further away. In the limiting case of no bulk fluid motion, mass convection reduces to mass diffusion, just as convection reduces to conduction. The analogy between heat and mass convection holds for both forced and natural convection, laminar and turbulent flow, and internal and external flow. Like heat convection, mass convection is also complicated because of the complications associated with fluid flow such as the surface geometry, flow regime, flow velocity, and the variation of the fluid properties and composition. Therefore, we will have to rely on experimental relations to determine mass transfer. Also, mass convection is usually analyzed on a mass basis rather than on a molar basis. Therefore, we will present formulations in terms of mass concentration (density or mass fraction w) instead of molar concentration (molar density C or mole fraction y). But the formulations on a molar basis can be obtained using the relation C /M where M is the molar mass. Also, for simplicity, we will restrict our attention to convection in fluids that are (or can be treated as) binary mixtures. Consider the flow of air over the free surface of a water body such as a lake under isothermal conditions. If the air is not saturated, the concentration of water vapor will vary from a maximum at the water surface where the air is always saturated to the free steam value far from the surface. In heat convection, we defined the region in which temperature gradients exist as the thermal boundary layer. Similarly, in mass convection, we define the region of the fluid in which concentration gradients exist as the concentration boundary layer, as shown in Figure 14–38. In external flow, the thickness of the concentration boundary layer c for a species A at a specified location on the surface is defined as the normal distance y from the surface at which A, s A A, s A, 0.99 where A, s and A, are the densities of species A at the surface (on the fluid side) and the free stream, respectively. In internal flow, we have a concentration entrance region where the concentration profile develops, in addition to the hydrodynamic and thermal entry regions (Fig. 14–39). The concentration boundary layer continues to develop in the flow direction until its thickness reaches the tube center and the boundary layers merge. The distance from the tube inlet to the location where this merging occurs is called the concentration entry length Lc, and the region beyond that point is called the fully developed region, which is characterized by cen58933_ch14.qxd 9/9/2002 10:05 AM Page 755 755 CHAPTER 14 A, s A 0 x A, s A, b (14-66) where A, b is the bulk mean density of species A defined as A, b 1 Acave dA Ac A (14-67) c Therefore, the nondimensionalized concentration difference profile as well as the mass transfer coefficient remain constant in the fully developed region. This is analogous to the friction and heat transfer coefficients remaining constant in the fully developed region. In heat convection, the relative magnitudes of momentum and heat diffusion in the velocity and thermal boundary layers are expressed by the dimensionless Prandtl number, defined as (Fig. 14–40) Prandtl number: Momentum diffusivity Pr Thermal diffusivity (14-68) The corresponding quantity in mass convection is the dimensionless Schmidt number, defined as Schmidt number: Sc Momentum diffusivity DAB Mass diffusivity (14-69) Thermal diffusivity Sc Pr DAB Mass diffusivity (14-70) thermal Prn, velocity concentration Scn, and thermal concentration Sc Le Len (14-71) where n 13 for most applications in all three relations. These relations, in general, are not applicable to turbulent boundary layers since turbulent mixing in this case may dominate the diffusion processes. DAB Thermal diffusivity Sc Pr DAB → The relative thicknesses of velocity, thermal, and concentration boundary layers in laminar flow are expressed as velocity Mass transfer: →  Le Pr FIGURE 14–40 In mass transfer, the Schmidt number plays the role of the Prandtl number in heat transfer. which represents the relative magnitudes of molecular momentum and mass diffusion in the velocity and concentration boundary layers, respectively. The relative growth of the velocity and thermal boundary layers in laminar flow is governed by the Prandtl number, whereas the relative growth of the velocity and concentration boundary layers is governed by the Schmidt number. A Prandtl number of near unity (Pr 1) indicates that momentum and heat transfer by diffusion are comparable, and velocity and thermal boundary layers almost coincide with each other. A Schmidt number of near unity (Sc 1) indicates that momentum and mass transfer by diffusion are comparable, and velocity and concentration boundary layers almost coincide with each other. It seems like we need one more dimensionless number to represent the relative magnitudes of heat and mass diffusion in the thermal and concentration boundary layers. That is the Lewis number, defined as (Fig. 14–41) Lewis number: Heat transfer: Mass diffusivity FIGURE 14–41 Lewis number is a measure of heat diffusion relative to mass diffusion. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 756 756 HEAT TRANSFER Note that species transfer at the surface (y 0) is by diffusion only because of the no-slip boundary condition, and mass flux of species A at the surface can be expressed by Fick’s law as (Fig. 14–42) Concentration profile wA, wA, dwA —– dy y y=0 wA jA m· A/A DAB y Mass diffusion wA, s Species A 0 ∂C – DAB —–A ∂y y=0 = hmass(wA, s – wA, ) FIGURE 14–42 Mass transfer at a surface occurs by diffusion because of the no-slip boundary condition, just like heat transfer occurring by conduction. (kg/s · m2) (14-72) y0 This is analogous to heat transfer at the surface being by conduction only and expressing it by Fourier’s law. The rate of heat convection for external flow was expressed conveniently by Newton’s law of cooling as · Q conv hconv A(Ts T) where hconv is the average heat transfer coefficient, A is the surface area, and Ts T is the temperature difference across the thermal boundary layer. Likewise, the rate of mass convection can be expressed as m· conv hmass A(A, s A, ) hmassA(wA, s wA, ) (kg/s) (14-73) where hmass is the average mass transfer coefficient, in m/s; A is the surface area; A, s A, is the mass concentration difference of species A across the concentration boundary layer; and is the average density of the fluid in the boundary layer. The product hmass, whose unit is kg/m2 · s, is called the mass transfer conductance. If the local mass transfer coefficient varies in the flow direction, the average mass transfer coefficient can be determined from hmass, ave 1 A h massdA A (14-74) In heat convection analysis, it is often convenient to express the heat transfer coefficient in a nondimensionalized form in terms of the dimensionless Nusselt number, defined as Nusselt number: Heat transfer: Nu hconv L k Mass transfer: Sh hmass L DAB FIGURE 14–43 In mass transfer, the Sherwood number plays the role the Nusselt number plays in heat transfer. Nu hconv L k (14-75) where L is the characteristic length of k is the thermal conductivity of the fluid. The corresponding quantity in mass convection is the dimensionless Sherwood number, defined as (Fig. 14–43) Sherwood number: Sh hmass L DAB (14-76) where hmass is the mass transfer coefficient and DAB is the mass diffusivity. The Nusselt and Sherwood numbers represent the effectiveness of heat and mass convection at the surface, respectively. Sometimes it is more convenient to express the heat and mass transfer coefficients in terms of the dimensionless Stanton number as Heat transfer Stanton number: St hconv 1 Nu Re Pr Cp (14-77) cen58933_ch14.qxd 9/9/2002 10:05 AM Page 757 757 CHAPTER 14 and Mass transfer Stanton number: Stmass hmass 1 Sh Re Sc (14-78) where is the free steam velocity in external flow and the bulk mean fluid velocity in internal flow. For a given geometry, the average Nusselt number in forced convection depends on the Reynolds and Prandtl numbers, whereas the average Sherwood number depends on the Reynolds and Schmidt numbers. That is, Nu f(Re, Pr) Sh f(Re, Sc) Nusselt number: Sherwood number: TABLE 14–12 where the functional form of f is the same for both the Nusselt and Sherwood numbers in a given geometry, provided that the thermal and concentration boundary conditions are of the same type. Therefore, the Sherwood number can be obtained from the Nusselt number expression by simply replacing the Prandtl number by the Schmidt number. This shows what a powerful tool analogy can be in the study of natural phenomena (Table 14–12). In natural convection mass transfer, the analogy between the Nusselt and Sherwood numbers still holds, and thus Sh f(Gr, Sc). But the Grashof number in this case should be determined directly from Gr g( s) L3c g(/) L3c 2 2 (14-79) which is applicable to both temperature- and/or concentration-driven natural convection flows. Note that in homogeneous fluids (i.e., fluids with no concentration gradients), density differences are due to temperature differences only, and thus we can replace / by T for convenience, as we did in natural convection heat transfer. However, in nonhomogeneous fluids, density differences are due to the combined effects of temperature and concentration differences, and / cannot be replaced by T in such cases even when all we care about is heat transfer and we have no interest in mass transfer. For example, hot water at the bottom of a pond rises to the top. But when salt is placed at the bottom, as it is done in solar ponds, the salty water (brine) at the bottom will not rise because it is now heavier than the fresh water at the top (Fig. 14–44). Concentration-driven natural convection flows are based on the densities of different species in a mixture being different. Therefore, at isothermal conditions, there will be no natural convection in a gas mixture that is composed of gases with identical molar masses. Also, the case of a hot surface facing up corresponds to diffusing fluid having a lower density than the mixture (and thus rising under the influence of buoyancy), and the case of a hot surface facing down corresponds to the diffusing fluid having a higher density. For example, the evaporation of water into air corresponds to a hot surface facing up since water vapor is lighter than the air and it will tend to rise. But this will not be the case for gasoline unless the temperature of the gasoline–air mixture at the gasoline surface is so high that thermal expansion overwhelms the density differential due to higher gasoline concentration near the surface. Analogy between the quantities that appear in the formulation and solution of heat convection and mass convection Heat Convection Mass Convection T C, y, , or w hmass hconv thermal Lc Re Gr g ( s) L3c g (Ts T) L3c , Gr 2 2 Pr St Nu concentration Lc Re h conv C p hconv Lc k Sc Stmass Sh D AB h mass hmass Lc DAB Nu f (Re, Pr) Sh f (Re, Sc) Nu f (Gr, Pr) Sh f (Gr, Sc) 20°C Fresh water No convection currents 70°C SOLAR POND ρbrine > ρwater Brine Salt FIGURE 14–44 A hot fluid at the bottom will rise and initiate natural convection currents only if its density is lower. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 758 758 HEAT TRANSFER CA, T Velocity, temperature, or concentration profile y Tangent line at y = 0 Analogy between Friction, Heat Transfer, and Mass Transfer Coefficients Consider the flow of a fluid over a flat plate of length L with free steam conditions of T, , and wA, (Fig. 14–45). Noting that convection at the surface (y 0) is equal to diffusion because of the no-slip condition, the friction, heat transfer, and mass transfer conditions at the surface can be expressed as 0 FIGURE 14–45 The friction, heat, and mass transfer coefficients for flow over a surface are proportional to the slope of the tangent line of the velocity, temperature, and concentration profiles, respectively, at the surface. Heat transfer: Mass transfer: T q· k y w D y s Wall friction: y s jA, s y0 y0 A AB y0 f 2 2 (14-80) hheat(Ts T) (14-81) hmass(wA, s wA, ) (14-82) These relations can be rewritten for internal flow by using bulk mean properties instead of free stream properties. After some simple mathematical manipulations, the three relations above can be rearranged as d(/ ) d(y/L) Wall friction: s Heat transfer: Mass transfer: d[(T T )/(T T )] d(y/L) w )/(w w )] d(y/L) d[(wA A, s A, y0 s y0 A, s y0 f L f Re 2 2 (14-83) hheat L Nu k (14-84) hmass L Sh DAB (14-85) The left sides of these three relations are the slopes of the normalized velocity, temperature, and concentration profiles at the surface, and the right sides are the dimensionless numbers discussed earlier. wA, T Normalized velocity, temperature, or concentration profile Special Case: Pr Sc 1 (Reynolds Analogy) Velocity, temperature, or concentration boundary layer Now consider the hypothetical case in which the molecular diffusivities of momentum, heat, and mass are identical. That is, DAB and thus Pr Sc Le 1. In this case the normalized velocity, temperature, and concentration profiles will coincide, and thus the slope of these three curves at the surface (the left sides of Eqs. 14–83 through 14–85) will be identical (Fig. 14–46). Then we can set the right sides of those three equations equal to each other and obtain f Re Nu Sh 2 Reynolds analogy ν = α = DAB (or Pr = Sc = Le) FIGURE 14–46 When the molecular diffusivities of momentum, heat, and mass are equal to each other, the velocity, temperature, and concentration boundary layers coincide. or f L hheat L hmass L DAB 2 k (14-86) Noting that Pr Sc 1, we can also write this equation as f Sh Nu 2 Re Pr Re Sc or f St Stmass 2 (14-87) This relation is known as the Reynolds analogy, and it enables us to determine the seemingly unrelated friction, heat transfer, and mass transfer coefficients when only one of them is known or measured. (Actually the original cen58933_ch14.qxd 9/9/2002 10:05 AM Page 759 759 CHAPTER 14 Reynolds analogy proposed by O. Reynolds in 1874 is St f /2, which is then extended to include mass transfer.) However, it should always be remembered that the analogy is restricted to situations for which Pr Sc 1. Of course the first part of the analogy between friction and heat transfer coefficients can always be used for gases since their Prandtl number is close to unity. General Case: Pr Sc 1 (Chilton–Colburn Analogy) The Reynolds analogy is a very useful relation, and it is certainly desirable to extend it to a wider range of Pr and Sc numbers. Several attempts have been made in this regard, but the simplest and the best known is the one suggested by Chilton and Colburn in 1934 as f St Pr2/3 StmassSc2/3 2 (14-88) for 0.6 Pr 60 and 0.6 Sc 3000. This equation is known as the Chilton–Colburn analogy. Using the definition of heat and mass Stanton numbers, the analogy between heat and mass transfer can be expressed more conveniently as (Fig. 14–47) Chilton–Colburn Analogy General: 2/3 or St Sc Pr Stmass hheat Sc Cp Pr hmass hmass 2/3 2/3 Cp D AB Cp Le2/3 (14-89) For air–water vapor mixtures at 298 K, the mass and thermal diffusivities are DAB 2.5 105 m2/s and 2.18 105 m2/s and thus the Lewis number is Le /DAB 0.872. (We simply use the value of dry air instead of the moist air since the fraction of vapor in the air at atmospheric conditions is low.) Then ( /DAB)2/3 0.8722/3 0.913, which is close to unity. Also, the Lewis number is relatively insensitive to variations in temperature. Therefore, for air–water vapor mixtures, the relation between heat and mass transfer coefficients can be expressed with a good accuracy as hheat Cp hmass (air–water vapor mixtures) (14-90) where and Cp are the density and specific heat of air at mean conditions (or Cp is the specific heat of air per unit volume). Equation 14–90 is known as the Lewis relation and is commonly used in air-conditioning applications. Another important consequence of Le 1 is that the wet-bulb and adiabatic saturation temperatures of moist air are nearly identical. In turbulent flow, the Lewis relation can be used even when the Lewis number is not 1 since eddy mixing in turbulent flow overwhelms any molecular diffusion, and heat and mass are transported at the same rate. The Chilton–Colburn analogy has been observed to hold quite well in laminar or turbulent flow over plane surfaces. But this is not always the case for internal flow and flow over irregular geometries, and in such cases specific relations developed should be used. When dealing with flow over blunt bodies, it is important to note that f in these relations is the skin friction coefficient, not the total drag coefficient, which also includes the pressure drag. 2/3 hheat DAB Cp 2/3 DAB 1 f 2 Special case: DAB hmass hheat 1 f Cp 2 FIGURE 14–47 When the friction or heat transfer coefficient is known, the mass transfer coefficient can be determined directly from the Chilton–Colburn analogy. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 760 760 HEAT TRANSFER Limitation on the Heat–Mass Convection Analogy Air Saturated air Evaporation Lake FIGURE 14–48 Evaporation from the free surface of water into air. Caution should be exercised when using the analogy in Eq. 14–88 since there are a few factors that put some shadow on the accuracy of that relation. For one thing, the Nusselt numbers are usually evaluated for smooth surfaces, but many mass transfer problems involve wavy or roughened surfaces. Also, many Nusselt relations are obtained for constant surface temperature situations, but the concentration may not be constant over the entire surface because of the possible surface dryout. The blowing or suction at the surface during mass transfer may also cause some deviation, especially during high speed blowing or suction. Finally, the heat–mass convection analogy is valid for low mass flux cases in which the flow rate of species undergoing mass flow is low relative to the total flow rate of the liquid or gas mixture so that the mass transfer between the fluid and the surface does not affect the flow velocity. (Note that convection relations are based on zero fluid velocity at the surface, which is true only when there is no net mass transfer at the surface.) Therefore, the heat–mass convection analogy is not applicable when the rate of mass transfer of a species is high relative to the flow rate of that species. Consider, for example, the evaporation and transfer of water vapor into air in an air washer, an evaporative cooler, a wet cooling tower, or just at the free surface of a lake or river (Fig. 14–48). Even at a temperature of 40°C, the vapor pressure at the water surface is the saturation pressure of 7.4 kPa, which corresponds to a mole fraction of 0.074 or a mass fraction of wA, s 0.047 for the vapor. Then the mass fraction difference across the boundary layer will be, at most, w wA, s wA, 0.047 0 0.047. For the evaporation of water into air, the error involved in the low mass flux approximation is roughly w/2, which is 2.5 percent in the worst case considered above. Therefore, in processes that involve the evaporation of water into air, we can use the heat–mass convection analogy with confidence. However, the mass fraction of vapor approaches 1 as the water temperature approaches the saturation temperature, and thus the low mass flux approximation is not applicable to mass transfer in boilers, condensers, and the evaporation of fuel droplets in combustion chambers. In this chapter we limit our consideration to low mass flux applications. Mass Convection Relations Under low mass flux conditions, the mass convection coefficients can be determined by either (1) determining the friction or heat transfer coefficient and then using the Chilton–Colburn analogy or (2) picking the appropriate Nusselt number relation for the given geometry and analogous boundary conditions, replacing the Nusselt number by the Sherwood number and the Prandtl number by the Schmidt number, as shown in Table 14–13 for some representative cases. The first approach is obviously more convenient when the friction or heat transfer coefficient is already known. Otherwise, the second approach should be preferred since it is generally more accurate, and the Chilton– Colburn analogy offers no significant advantage in this case. Relations for convection mass transfer in other geometries can be written similarly using the corresponding heat transfer relation in Chapters 6 and 7. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 761 761 CHAPTER 14 TABLE 14–13 Sherwood number relations in mass convection for specified concentration at the surface corresponding to the Nusselt number relations in heat convection for specified surface temperature Convective Heat Transfer Convective Mass Transfer Forced Convection over a Flat Plate (a) Laminar flow (Re 5 105) 1/3 Nu 0.664 Re0.5 Pr 0.6 L Pr , 1/3 Sh 0.664 Re0.5 L Sc , Sc 0.5 1/3 Sh 0.037 Re0.8 L Sc , Sc 0.5 (b) Turbulent flow (5 105 Re 107) 1/3 Nu 0.037 Re0.8 Pr 0.6 L Pr , 2. Fully Developed Flow in Smooth Circular Pipes (a) Laminar flow (Re 2300) Nu 3.66 (b) Turbulent flow (Re 10,000) Nu 0.023 Re0.8 Pr0.4, 0.7 Pr 160 3. Natural Convection over Surfaces (a) Vertical plate Nu 0.59(Gr Pr)1/4, 105 Gr Pr 109 1/3 Nu 0.1(Gr Pr) , 109 Gr Pr 1013 Sh 3.66 Sh 0.023 Re0.8 Sc0.4, Sh 0.59(Gr Sc)1/4, Sh 0.1(Gr Sc)1/3, 0.7 Sc 160 105 Gr Sc 109 109 Gr Sc 1013 (b) Upper surface of a horizontal plate Surface is hot (Ts T) Nu 0.54(Gr Pr)1/4, 104 Gr Pr 107 Nu 0.15(Gr Pr)1/3, 107 Gr Pr 1011 Fluid near the surface is light (s ) Sh 0.54(Gr Sc)1/4, 104 Gr Sc 107 Sh 0.15(Gr Sc)1/3, 107 Gr Sc 1011 (c) Lower surface of a horizontal plate Surface is hot (Ts T) Nu 0.27(Gr Pr)1/4, 105 Gr Pr 1011 Fluid near the surface is light (s ) Sh 0.27(Gr Sc)1/4, 105 Gr Sc 1011 EXAMPLE 14–10 Mass Convection inside a Circular Pipe Consider a circular pipe of inner diameter D 0.015 m whose inner surface is covered with a layer of liquid water as a result of condensation (Fig. 14–49). In order to dry the pipe, air at 300 K and 1 atm is forced to flow through it with an average velocity of 1.2 m/s. Using the analogy between heat and mass transfer, determine the mass transfer coefficient inside the pipe for fully developed flow. SOLUTION The liquid layer on the inner surface of a circular pipe is dried by blowing air through it. The mass transfer coefficient is to be determined. Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 The flow is fully developed. Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 300 K and 1 atm, for which 1.58 105 m2/s (Table A–15). The mass diffusivity of water vapor in the air at 300 K is determined from Eq. 14–15 to be DAB DH2O-air 1.87 1010 3002.072 T 2.072 1.87 1010 2.54 105 m2/s P 1 Wet pipe Air 300 K 1 atm = 1.2 m/s FIGURE 14–49 Schematic for Example 14–10. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 762 762 HEAT TRANSFER Analysis The Reynolds number for this internal flow is D (1.2 m/s)(0.015 m) Re 1139 1.58 105 m2/s which is less than 2300 and thus the flow is laminar. Therefore, based on the analogy between heat and mass transfer, the Nusselt and the Sherwood numbers in this case are Nu Sh 3.66. Using the definition of Sherwood number, the mass transfer coefficient is determined to be hmass ShDAB (3.66)(2.54 105 m2/s) 0.00620 m/s D 0.015 m The mass transfer rate (or the evaporation rate) in this case can be determined by defining the logarithmic mean concentration difference in an analogous manner to the logarithmic mean temperature difference. EXAMPLE 14–11 Naphthalene vapor Air 1 atm T = 25°C = 2 m/s Body covered with a layer of naphthalene FIGURE 14–50 Schematic for Example 14–11. 0.3 m2 Analogy between Heat and Mass Transfer Heat transfer coefficients in complex geometries with complicated boundary conditions can be determined by mass transfer measurements on similar geometries under similar flow conditions using volatile solids such as naphthalene and dichlorobenzene and utilizing the Chilton–Colburn analogy between heat and mass transfer at low mass flux conditions. The amount of mass transfer during a specified time period is determined by weighing the model or measuring the surface recession. During a certain experiment involving the flow of dry air at 25°C and 1 atm at a free stream velocity of 2 m/s over a body covered with a layer of naphthalene, it is observed that 12 g of naphthalene has sublimated in 15 min (Fig. 14–50). The surface area of the body is 0.3 m2. Both the body and the air were kept at 25°C during the study. The vapor pressure of naphthalene at 25°C is 11 Pa and the mass diffusivity of naphthalene in air at 25°C is DAB 0.61 105 m2/s. Determine the heat transfer coefficient under the same flow conditions over the same geometry. SOLUTION Air is blown over a body covered with a layer of naphthalene, and the rate of sublimation is measured. The heat transfer coefficient under the same flow conditions over the same geometry is to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton–Colburn analogy between heat and mass transfer is applicable (will be verified). 2 Both air and naphthalene vapor are ideal gases. Properties The molar mass of naphthalene is 128.2 kg/kmol. Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 25°C and 1 atm, at which 1.184 kg/m3, CP 1007 J/kg · K, and 2.141 105 m2/s (Table A–15). Analysis The incoming air is free of naphthalene, and thus the mass fraction of naphthalene at free stream conditions is zero, wA, 0. Noting that the vapor pressure of naphthalene at the surface is 11 Pa, its mass fraction at the surface is determined to be cen58933_ch14.qxd 9/9/2002 10:05 AM Page 763 763 CHAPTER 14 wA, s PA, s MA 128.2 kg/kmol 11 Pa 4.8 104 P Mair 101,325 Pa 29 kg/kmol which confirms that the low mass flux approximation is valid. The rate of evaporation of naphthalene in this case is 0.012 kg m m· evap 1.33 105 kg/s t (15 60 s) Then the mass convection coefficient becomes hmass 1.33 105 kg/s m· 0.0780 m/s As(wA, s wA, ) (1.184 kg/m3)(0.3 m2)(4.8 104 0) Using the analogy between heat and mass transfer, the average heat transfer coefficient is determined from Eq. 14–89 to be 2/3 hheat Cp hmass D AB (1.184 kg/m3)(1007 J/kg · °C)(0.0776 m/s) 10 2.141 0.61 10 5 5 2/3 m2/s m2/s 215 W/m2 · °C Discussion Because of the convenience it offers, naphthalene has been used in numerous heat transfer studies to determine convection heat transfer coefficients. 14–10 ■ SIMULTANEOUS HEAT AND MASS TRANSFER Many mass transfer processes encountered in practice occur isothermally, and thus they do not involve any heat transfer. But some engineering applications involve the vaporization of a liquid and the diffusion of this vapor into the surrounding gas. Such processes require the transfer of the latent heat of vaporization hfg to the liquid in order to vaporize it, and thus such problems involve simultaneous heat and mass transfer. To generalize, any mass transfer problem involving phase change (evaporation, sublimation, condensation, melting, etc.) must also involve heat transfer, and the solution of such problems needs to be analyzed by considering simultaneous heat and mass transfer. Some examples of simultaneous heat and mass problems are drying, evaporative cooling, transpiration (or sweat) cooling, cooling by dry ice, combustion of fuel droplets, and ablation cooling of space vehicles during reentry, and even ordinary events like rain, snow, and hail. In warmer locations, for example, the snow melts and the rain evaporates before reaching the ground (Fig. 14–51). To understand the mechanism of simultaneous heat and mass transfer, consider the evaporation of water from a swimming pool into air. Let us assume that the water and the air are initially at the same temperature. If the air is saturated (a relative humidity of 100 percent), there will be no heat or mass transfer as long as the isothermal conditions remain. But if the air is not Plastic or glass Heat Evaporation Heat Space vehicle during reentry (a) Ablation Evaporation (b) Evaporation of rain droplet Heat rejection Condensation Vapor Liquid Evaporation Heat (c) Drying of clothes Heat absorption (d) Heat pipes FIGURE 14–51 Many problems encountered in practice involve simultaneous heat and mass transfer. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 764 764 HEAT TRANSFER Surroundings 20°C Air 20°C · Qrad Evaporation · (Qlatent ) Water 20°C · Qconv 18°C · Qcond saturated ( 100 percent), there will be a difference between the concentration of water vapor at the water–air interface (which is always saturated) and some distance above the interface (the concentration boundary layer). Concentration difference is the driving force for mass transfer, and thus this concentration difference will drive the water into the air. But the water must vaporize first, and it must absorb the latent heat of vaporization in order to vaporize. Initially, the entire heat of vaporization will come from the water near the interface since there is no temperature difference between the water and the surroundings and thus there cannot be any heat transfer. The temperature of water near the surface must drop as a result of the sensible heat loss, which also drops the saturation pressure and thus vapor concentration at the interface. This temperature drop creates temperature differences within the water at the top as well as between the water and the surrounding air. These temperature differences drive heat transfer toward the water surface from both the air and the deeper parts of the water, as shown in Figure 14–52. If the evaporation rate is high and thus the demand for the heat of vaporization is higher than the amount of heat that can be supplied from the lower parts of the water body and the surroundings, the deficit will be made up from the sensible heat of the water at the surface, and thus the temperature of water at the surface will drop further. The process will continue until the latent heat of vaporization is equal to the rate of heat transfer to the water at the surface. Once the steady operation conditions are reached and the interface temperature stabilizes, the energy balance on a thin layer of liquid at the surface can be expressed as · · Q sensible, transferred Q latent, absorbed Lake FIGURE 14–52 Various mechanisms of heat transfer involved during the evaporation of water from the surface of a lake. or · Q m· hfg (14-91) where m· is the rate of evaporation and hfg is the latent heat of vaporization of water at the surface temperature. Various expressions for m· under various approximations are given in Table 14–14. The mixture properties such as the specific heat Cp and molar mass M should normally be evaluated at the mean film composition and mean film temperature. However, when dealing with air–water vapor mixtures at atmospheric conditions or other low mass flux situations, we can simply use the properties of the gas with reasonable accuracy. TABLE 14–14 Various expressions for evaporation rate of a liquid into a gas through an interface area As under various approximations (subscript stands for vapor, s for liquid–gas interface, and away from surface) Assumption General Assuming vapor to be an ideal gas, P RT Using Chilton–Colburn analogy, hheat CphmassLe2/3 Ts T 1 1 1 , where T Ts T T 2 and P RT (Ru /M)T Using Evaporation Rate m· hmass As(, s , ) hmass As P, s P, m· Ts T R hmass As P, s P, m· T CpLe2/3R Ts hmass As M P, s P, m· P CpLe2/3 M cen58933_ch14.qxd 9/9/2002 10:05 AM Page 765 765 CHAPTER 14 · The Q in Eq. 14–91 represents all forms of heat from all sources transferred to the surface, including convection and radiation from the surroundings and conduction from the deeper parts of the water due to the sensible energy of the water itself or due to heating the water body by a resistance heater, heating coil, or even chemical reactions in the water. If heat transfer from the water body to the surface as well as radiation from the surroundings is negligible, which is often the case, then the heat loss by evaporation must equal heat gain by convection. That is, · Q conv m· hfg or hconv As(T Ts) hconv As hfg M P, s P, P Cp Le2/3 M Canceling hconv As from both sides of the second equation gives Ts T M P, s P, P Cp Le2/3 M hfg (14-92) which is a relation for the temperature difference between the liquid and the surrounding gas under steady conditions. EXAMPLE 14–12 Evaporative Cooling of a Canned Drink During a hot summer day, a canned drink is to be cooled by wrapping it in a cloth that is kept wet continually, and blowing air to it with a fan (Fig. 14–53). If the environment conditions are 1 atm, 30°C, and 40 percent relative humidity, determine the temperature of the drink when steady conditions are reached. 1 atm 30°C 40% RH Wet cloth SOLUTION Air is blown over a canned drink wrapped in a wet cloth to cool it by simultaneous heat and mass transfer. The temperature of the drink when steady conditions are reached is to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton–Colburn analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 Radiation effects are negligible. Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the average temperature of (T Ts)/2 which cannot be determined at this point because of the unknown surface temperature Ts. We know that Ts T and, for the purpose of property evaluation, we take Ts to be 20°C. Then the properties of water at 20°C and the properties of dry air at the average temperature of 25°C and 1 atm are (Tables A–9 and A–15) Water: hfg 2454 kJ/kg, P 2.34 kPa; also, P 4.25 kPa at 30°C Dry air: CP 1.007 kJ/kg · °C, 2.141 105 m2/s The molar masses of water and air are 18 and 29 kg/kmol, respectively (Table A–1). Also, the mass diffusivity of water vapor in air at 25°C is DH2O-air 2.50 105 m2/s (Table 14–4). Analysis Utilizing the Chilton–Colburn analogy, the surface temperature of the drink can be determined from Eq. 14–92, FIGURE 14–53 Schematic for Example 14–12. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 766 766 HEAT TRANSFER Ts T hfg 2/3 Cp Le M P, s P, M P where the Lewis number is Le 2.141 105 m2/s 0.856 DAB 2.5 105 m2/s Note that we could take the Lewis number to be 1 for simplicity, but we chose to incorporate it for better accuracy. The air at the surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (2.34 kPa). The vapor pressure of air far from the surface is determined from P, Psat @ T (0.40)Psat @ 30°C (0.40)(4.25 kPa) 1.70 kPa Noting that the atmospheric pressure is 1 atm 101.3 kPa, substituting gives Ts 30°C 2454 kJ/kg 18 kg/kmol (2.34 1.70) kPa 2/3 29 kg/kmol 101.3 kPa (1.007 kJ/kg · °C)(0.872) 19.4°C Therefore, the temperature of the drink can be lowered to 19.4°C by this process. EXAMPLE 14–13 Surrounding surfaces 20°C · Qrad · Qevap · Qconv Air 25°C 92 kPa 52% RH Aerosol can Water bath 50°C Heat supplied Resistance heater FIGURE 14–54 Schematic for Example 14–13. Heat Loss from Uncovered Hot Water Baths Hot water baths with open tops are commonly used in manufacturing facilities for various reasons. In a plant that manufactures spray paints, the pressurized paint cans are temperature tested by submerging them in hot water at 50°C in a 40-cm-deep rectangular bath and keeping them there until the cans are heated to 50°C to ensure that the cans can withstand temperatures up to 50°C during transportation and storage (Fig. 14–54). The water bath is 1 m wide and 3.5 m long, and its top surface is open to ambient air to facilitate easy observation for the workers. If the average conditions in the plant are 92 kPa, 25°C, and 52 percent relative humidity, determine the rate of heat loss from the top surface of the water bath by (a) radiation, (b) natural convection, and (c) evaporation. Assume the water is well agitated and maintained at a uniform temperature of 50°C at all times by a heater, and take the average temperature of the surrounding surfaces to be 20°C. SOLUTION Spray paint cans are temperature tested by submerging them in an uncovered hot water bath. The rates of heat loss from the top surface of the bath by radiation, natural convection, and evaporation are to be determined. Assumptions 1 The low mass flux conditions exist so that the Chilton–Colburn analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water vapor at specified conditions are ideal gases (the error involved in this assumption is less than 1 percent). 3 Water is maintained at a uniform temperature of 50°C. cen58933_ch14.qxd 9/9/2002 10:05 AM Page 767 767 CHAPTER 14 Properties Relevant properties for each mode of heat transfer are determined below in respective sections. Analysis (a) The emissivity of liquid water is given in Table A–18 to be 0.95. Then the radiation heat loss from the water to the surrounding surfaces becomes · 4 Q rad As(Ts4 Tsurr ) 2 (0.95)(3.5 m )(5.67 108 W/m2 K4)[(323 K)4 (293K)4] 663 W (b) The air–water vapor mixture is dilute and thus we can use dry air properties for the mixture at the average temperature of (T Ts)/2 (25 50)/2 37.5°C. Noting that the total atmospheric pressure is 92/101.3 0.9080 atm, the properties of dry air at 37.5°C and 0.908 atm are k 0.02644 W/m · °C, Pr 0.7261 (independent of pressure) (2.311 105 m2/s)/0.9080 2.545 105 m2/s (1.678 105 m2/s)/0.9080 1.848 105 m2/s The properties of water at 50°C are hfg 2383 kJ/kg and P 12.35 kPa The air at the surface is saturated, and thus the vapor pressure at the surface is simply the saturation pressure of water at the surface temperature (12.35 kPa). The vapor pressure of air far from the water surface is determined from P, Psat @ T (0.52)Psat @ 25°C (0.52)(3.17 kPa) 1.65 kPa Treating the water vapor and the air as ideal gases and noting that the total atmospheric pressure is the sum of the vapor and dry air pressures, the densities of the water vapor, dry air, and their mixture at the water–air interface and far from the surface are determined to be At the surface: P, s 12.35 kPa 0.0828 kg/m3 RTs (0.4615 kPa · m3/kg · K)(323 K) Pa, s (92 12.35) kPa a, s 0.8592 kg/m3 RaTs (0.287 kPa · m3/kg · K)(323 K) , s S , s a, s 0.0828 0.8592 0.9420 kg/m3 and P, Away from 1.65 kPa 0.0120 kg/m3 , RT (0.4615 kPa · m3/kg · K)(298 K) the surface: Pa, (92 1.65) kPa a, 1.0564 kg/m3 RaT (0.287 kPa · m3/kg · K)(298 K) , a, 0.0120 1.0564 1.0684 kg/m3 The area of the top surface of the water bath is As (3.5 m)(1 m) 3.5 m2 and its perimeter is p 2(3.5 1) 9 m. Therefore, the characteristic length is cen58933_ch14.qxd 9/9/2002 10:05 AM Page 768 768 HEAT TRANSFER As 3.5 m2 Lc p 0.3889 m 9m Then using densities (instead of temperatures) since the mixture is not homogeneous, the Grashof number is determined to be g( S)L3c 2 (9.81 m/s2)(1.0684 0.9420 kg/m3)(0.3889 m)3 (0.9420 1.0684)/2 kg/m32 2.125 108 Gr Recognizing that this is a natural convection problem with hot horizontal surface facing up, the Nusselt number and the convection heat transfer coefficients are determined to be Nu 0.15(Gr Pr)1/3 0.15(2.125 108 0.7261)1/3 80.45 and hconv Nuk (80.45)(0.02644 W/m ·°C) 5.47 W/m2 · °C L 0.3889 m Then the natural convection heat transfer rate becomes · Q conv hconv As(Ts T) (5.47 W/m2 · °C)(3.5 m2)(50 25)°C 479 W Note that the magnitude of natural convection heat transfer is comparable to that of radiation, as expected. (c) Utilizing the analogy between heat and mass convection, the mass transfer coefficient is determined the same way by replacing Pr by Sc. The mass diffusivity of water vapor in air at the average temperature of 310.5 K is determined from Eq. 14–15 to be DAB DH2O–air 1.87 1010 310.52.072 T 2.072 1.87 1010 P 0.908 3.00 105 m2/s The Schmidt number is determined from its definition to be Sc 1.848 105 m2/s 0.616 DAB 3.00 105 m2/s The Sherwood number and the mass transfer coefficients are determined to be Sh 0.15(Gr Sc)1/3 0.15(2.125 108 0.616)1/3 76.2 and hmass ShDAB (76.2)(3.00 105 m2/s) 0.00588 m/s Lc 0.3889 m cen58933_ch14.qxd 9/9/2002 10:05 AM Page 769 769 CHAPTER 14 Then the evaporation rate and the rate of heat transfer by evaporation become m· hmass As(, s , ) (0.00588 m/s)(3.5 m2)(0.0828 0.0120)kg/m3 0.00146 kg/s 5.25 kg/h and · Q evap m· hfg (0.00146 kg/s)(2383 kJ/kg) 3.479 kW 3479 W which is more than seven times the rate of heat transfer by natural convection. Finally, noting that the direction of heat transfer is always from high to low temperature, all forms of heat transfer determined above are in the same direction, and the total rate of heat loss from the water to the surrounding air and surfaces is · · · · Q total Q rad Q conv Q evap 663 479 3479 4621 W Discussion Note that if the water bath is heated electrically, a 4.6 kW resistance heater will be needed just to make up for the heat loss from the top surface. The total heater size will have to be larger to account for the heat losses from the side and bottom surfaces of the bath as well as the heat absorbed by the spray paint cans as they are heated to 50°C. Also note that water needs to be supplied to the bath at a rate of 5.25 kg/h to make up for the water loss by evaporation. Also, in reality, the surface temperature will probably be a little lower than the bulk water temperature, and thus the heat transfer rates will be somewhat lower than indicated here. SUMMARY Mass transfer is the movement of a chemical species from a high concentration region toward a lower concentration one relative to the other chemical species present in the medium. Heat and mass transfer are analogous to each other, and several parallels can be drawn between them. The driving forces are the temperature difference in heat transfer and the concentration difference in mass transfer. Fick’s law of mass diffusion is of the same form as Fourier’s law of heat conduction. The species generation in a medium due to homogeneous reactions is analogous to heat generation. Also, mass convection due to bulk fluid motion is analogous to heat convection. Constant surface temperature corresponds to constant concentration at the surface, and an adiabatic wall corresponds to an impermeable wall. However, concentration is usually not a continuous function at a phase interface. The concentration of a species A can be expressed in terms of density A or molar concentration CA. It can also be expressed in dimensionless form in terms of mass or molar fraction as Mass fraction of species A: mA mA /V A wA m m /V Mole fraction of species A: yA NA NA /V CA N C N/V In the case of an ideal gas mixture, the mole fraction of a gas is equal to its pressure fraction. Fick’s law for the diffusion of a species A in a stationary binary mixture of species A and B in a specified direction x is expressed as Mass basis: Mole basis: m· diff, A d(A/) DAB A dx dwA DAB dx N· diff, A d(CA/c) ¯j CDAB diff, A A dx dyA CDAB dx jdiff, A cen58933_ch14.qxd 9/9/2002 10:06 AM Page 770 770 HEAT TRANSFER where DAB is the diffusion coefficient (or mass diffusivity) of the species in the mixture, jdiff, A is the diffusive mass flux of species A, and ¯jdiff, A is the molar flux. The mole fractions of a species i in the gas and liquid phases at the interface of a dilute mixture are proportional to each other and are expressed by Henry’s law as where PA, 1 and PA, 2 are the partial pressures of gas A on the two sides of the wall. During mass transfer in a moving medium, chemical species are transported both by molecular diffusion and by the bulk fluid motion, and the velocities of the species are expressed as A diff, A B diff, B Pi, gas side yi, liquid side H where H is Henry’s constant. When the mixture is not dilute, an approximate relation for the mole fractions of a species on the liquid and gas sides of the interface are expressed approximately by Raoult’s law as Pi, gas side yi, gas side P yi, liquid side Pi, sat(T) where Pi, sat(T) is the saturation pressure of the species i at the interface temperature and P is the total pressure on the gas phase side. The concentration of the gas species i in the solid at the interface Ci, solid side is proportional to the partial pressure of the species i in the gas Pi, gas side on the gas side of the interface and is expressed as Ci, solid side Pi, gas side where is the solubility. The product of the solubility of a gas and the diffusion coefficient of the gas in a solid is referred to as the permeability , which is a measure of the ability of the gas to penetrate a solid. In the absence of any chemical reactions, the mass transfer rates m· diff, A through a plane wall of area A and thickness L and cylindrical and spherical shells of inner and outer radii r1 and r2 under one-dimensional steady conditions are expressed as wA, 1 wA, 2 A, 1 A, 2 DAB A L L A, 1 A, 2 wA, 1 wA, 2 2LDAB m· diff, A, cyl 2LDAB ln(r2/r1) ln(r2/r1) where is the mass-average velocity of the flow. It is the velocity that would be measured by a velocity sensor and is expressed as wAA wBB The special case 0 corresponds to a stationary medium. Using Fick’s law of diffusion, the total mass fluxes j m· /A in a moving medium are expressed as dwA dx dwB jB B Bdiff, B wB( jA jB) DBA dx jA A Adiff, A wA( jA jB) DAB The rate of mass convection of species A in a binary mixture is expressed in an analogous manner to Newton’s law of cooling as m· conv hmass As(A, s A, ) hmass As(wA, s wA, ) where hmass is the average mass transfer coefficient, in m/s. The counterparts of the Prandtl and Nusselt numbers in mass convection are the Schmidt number Sc and the Sherwood number Sh, defined as Momentum diffusivity DAB Mass diffusivity hmass L DAB m· diff, A, wall DAB A Sc m· diff, A, sph 4r1r2DAB The relative magnitudes of heat and mass diffusion in the thermal and concentration boundary layers are represented by the Lewis number, defined as 4r1r2DAB wA, 1 wA, 2 r2 r1 A, 1 A, 2 r2 r1 The mass flow rate of a gas through a solid plane wall under steady one-dimensional conditions can also be expressed in terms of the partial pressures of the adjacent gas on the two sides of the solid as PA, 1 PA, 2 L PA, 1 PA, 2 AB A L m· diff, A, wall DAB AB A Le and Sh Thermal diffusivity Sc Pr DAB Mass diffusivity Heat and mass transfer coefficients are sometimes expressed in terms of the dimensionless Stanton number, defined as Stheat hconv 1 Nu Re Pr Cp and Stmass hconv 1 Sh Re Sc where is the free-stream velocity in external flow and the bulk mean fluid velocity in internal flow. For a given geometry and boundary conditions, the Sherwood number in natural or cen58933_ch14.qxd 9/9/2002 10:06 AM Page 771 771 CHAPTER 14 forced convection can be determined from the corresponding Nusselt number expression by simply replacing the Prandtl number by the Schmidt number. But in natural convection, the Grashof number should be expressed in terms of density difference instead of temperature difference. When the molecular diffusivities of momentum, heat, and mass are identical, we have DAB, and thus Pr Sc Le 1. The similarity between momentum, heat, and mass transfer in this case is given by the Reynolds analogy, expressed as f Re Nu Sh or 2 f L hheat L hmass L DAB 2 k or f St Stmass 2 which is known as the Chilton–Colburn analogy. The analogy between heat and mass transfer is expressed more conveniently as hheat CpLe2/3 hmass Cp( /DAB)2/3hmass For air–water vapor mixtures, Le 1, and thus this relation simplifies further. The heat–mass convection analogy is limited to low mass flux cases in which the flow rate of species undergoing mass flow is low relative to the total flow rate of the liquid or gas mixture. The mass transfer problems that involve phase change (evaporation, sublimation, condensation, melting, etc.) also involve heat transfer, and such problems are analyzed by considering heat and mass transfer simultaneously. For the general case of Pr Sc 1, it is modified as f St Pr2/3 StmassSc2/3 2 REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers. Handbook of Fundamentals. Atlanta: ASHRAE, 1993. W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. 2nd ed. New York: McGraw-Hill, 1980. R. M. Barrer. Diffusion in and through Solids. New York: Macmillan, 1941. T. R. Marrero and E. A. Mason. “Gaseous Diffusion Coefficients.” Journal of Phys. Chem. Ref. Data 1 (1972), pp. 3–118. R. B. Bird. “Theory of Diffusion.” Advances in Chemical Engineering 1 (1956), p. 170. A. F. Mills. Basic Heat and Mass Transfer. Burr Ridge, IL: Richard D. Irwin, 1995. R. B. Bird, W. E. Stewart, and E. N. Lightfoot. Transport Phenomena. New York: John Wiley & Sons, 1960. J. H. Perry, ed. Chemical Engineer’s Handbook. 4th ed. New York: McGraw-Hill, 1963. C. J. Geankoplis. Mass Transport Phenomena. New York: Holt, Rinehart, and Winston, 1972. R. D. Reid, J. M. Prausnitz, and T. K. Sherwood. The Properties of Gases and Liquids. 3rd ed. New York: McGraw-Hill, 1977. Handbook of Chemistry and Physics 56th ed. Cleveland, OH: Chemical Rubber Publishing Co., 1976. 7. J. O. Hirshfelder, F. Curtis, and R. B. Bird. Molecular Theory of Gases and Liquids. New York: John Wiley & Sons, 1954. A. H. P. Skelland. Diffusional Mass Transfer. New York: John Wiley & Sons, 1974. 17. D. B. Spalding. Convective Mass Transfer. New York: McGraw-Hill, 1963. J. P. Holman. Heat Transfer. 7th ed. New York: McGraw-Hill, 1990. W. F. Stoecker and J. W. Jones. Refrigeration and Air Conditioning. New York: McGraw-Hill, 1982. F. P. Incropera and D. P. De Witt. Fundamentals of Heat and Mass Transfer. 2nd ed. New York: John Wiley & Sons, 1985. L. C. Thomas. Mass Transfer Supplement—Heat Transfer. Englewood Cliffs, NJ: Prentice Hall, 1991. International Critical Tables. Vol. 3. New York: McGraw-Hill, 1928. L. Van Black. Elements of Material Science and Engineering. Reading, MA: Addison-Wesley, 1980. cen58933_ch14.qxd 9/9/2002 10:06 AM Page 772 772 HEAT TRANSFER PROBLEMS Analogy between Heat and Mass Transfer 14–1C How does mass transfer differ from bulk fluid flow? Can mass transfer occur in a homogeneous medium? 14–2C How is the concentration of a commodity defined? How is the concentration gradient defined? How is the diffusion rate of a commodity related to the concentration gradient? 14–3C Give examples for (a) liquid-to-gas, (b) solid-toliquid, (c) solid-to-gas, and (d) gas-to-liquid mass transfer. 14–4C Someone suggests that thermal (or heat) radiation can also be viewed as mass radiation since, according to Einstein’s formula, an energy transfer in the amount of E corresponds to a mass transfer in the amount of m E/c2. What do you think? 14–5C What is the driving force for (a) heat transfer, (b) electric current flow, (c) fluid flow, and (d) mass transfer? 14–6C What do (a) homogeneous reactions and (b) heterogeneous reactions represent in mass transfer? To what do they correspond in heat transfer? Mass Diffusion 14–7C Both Fourier’s law of heat conduction and Fick’s law · of mass diffusion can be expressed as Q kA(dT/dx). What · do the quantities Q , k, A, and T represent in (a) heat conduction and (b) mass diffusion? 14–8C Mark these statements as being True or False for a binary mixture of substances A and B. (a) The density of a mixture is always equal to the sum of the densities of its constituents. (b) The ratio of the density of component A to the density of component B is equal to the mass fraction of component A. (c) If the mass fraction of component A is greater than 0.5, then at least half of the moles of the mixture are component A. (d) If the molar masses of A and B are equal to each other, then the mass fraction of A will be equal to the mole fraction of A. (e) If the mass fractions of A and B are both 0.5, then the molar mass of the mixture is simply the arithmetic average of the molar masses of A and B. 14–9C Mark these statements as being True or False for a binary mixture of substances A and B. (a) The molar concentration of a mixture is always equal to the sum of the molar concentrations of its constituents. (b) The ratio of the molar concentration of A to the molar concentration of B is equal to the mole fraction of component A. (c) If the mole fraction of component A is greater than 0.5, then at least half of the mass of the mixture is component A. (d) If both A and B are ideal gases, then the pressure fraction of A is equal to its mole fraction. (e) If the mole fractions of A and B are both 0.5, then the molar mass of the mixture is simply the arithmetic average of the molar masses of A and B. 14–10C Fick’s law of diffusion is expressed on the mass · and mole basis as m· diff, A ADAB(dwA/dx) and Ndiff, A CADAB(dyA/dx), respectively. Are the diffusion coefficients DAB in the two relations the same or different? 14–11C How does the mass diffusivity of a gas mixture change with (a) temperature and (b) pressure? 14–12C At a given temperature and pressure, do you think the mass diffusivity of air in water vapor will be equal to the mass diffusivity of water vapor in air? Explain. 14–13C At a given temperature and pressure, do you think the mass diffusivity of copper in aluminum will be equal to the mass diffusivity of aluminum in copper? Explain. 14–14C In a mass production facility, steel components are to be hardened by carbon diffusion. Would you carry out the hardening process at room temperature or in a furnace at a high temperature, say 900°C? Why? 14–15C Someone claims that the mass and the mole fractions for a mixture of CO2 and N2O gases are identical. Do you agree? Explain. 14–16 The composition of moist air is given on a molar basis to be 78 percent N2, 20 percent O2, and 2 percent water vapor. Determine the mass fractions of the constituents of air. Answers: 76.4 percent N2, 22.4 percent O2, 1.2 percent H2O Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 14–17E A gas mixture consists of 5 lbm of O2, 8 lbm of N2, and 10 lbm of CO2. Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass of the mixture. 14–18 A gas mixture consists of 8 kmol of H2 and 2 kmol of N2. Determine the mass of each gas and the apparent gas constant of the mixture. cen58933_ch14.qxd 9/9/2002 10:06 AM Page 773 773 CHAPTER 14 14–19 The molar analysis of a gas mixture at 290 K and 250 kPa is 65 percent N2, 20 percent O2, and 15 percent CO2. Determine the mass fraction and partial pressure of each gas. 14–20 Determine the binary diffusion coefficient of CO2 in air at (a) 200 K and 1 atm, (b) 400 K and 0.8 atm, and (c) 600 K and 3 atm. 14–21 Repeat Problem 14–20 for O2 in N2. 14–22E The relative humidity of air at 80°F and 14.7 psia is increased from 30 percent to 90 percent during a humidification process at constant temperature and pressure. Determine the percent error involved in assuming the density of air to have remained constant. Answer: 2.1 percent 80°F 14.7 psia 30% RH Humidifier FIGURE P14–22E 14–23 The diffusion coefficient of hydrogen in steel is given as a function of temperature as DAB 1.65 106 exp(–4630/T) (m2/s) where T is in K. Determine the diffusion coefficients from 200 K to 1200 K in 200 K increments and plot the results. 14–24 Reconsider Problem 14–23. Using EES (or other) software, plot the diffusion coefficient as a function of the temperature in the range of 200 K to 1200 K. the same as the mole fraction of water in the lake (which is nearly 1)? 14–28C When prescribing a boundary condition for mass transfer at a solid–gas interface, why do we need to specify the side of the surface (whether the solid or the gas side)? Why did we not do it in heat transfer? 14–29C Using properties of saturated water, explain how you would determine the mole fraction of water vapor at the surface of a lake when the temperature of the lake surface and the atmospheric pressure are specified. 14–30C Using solubility data of a solid in a specified liquid, explain how you would determine the mass fraction of the solid in the liquid at the interface at a specified temperature. 14–31C Using solubility data of a gas in a solid, explain how you would determine the molar concentration of the gas in the solid at the solid–gas interface at a specified temperature. 14–32C Using Henry’s constant data for a gas dissolved in a liquid, explain how you would determine the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature. 14–33C What is permeability? How is the permeability of a gas in a solid related to the solubility of the gas in that solid? 14–34E Determine the mole fraction of the water vapor at the surface of a lake whose temperature at the surface is 60°F, and compare it to the mole fraction of water in the lake. Take the atmospheric pressure at lake level to be 13.8 psia. 14–35 Determine the mole fraction of dry air at the surface of a lake whose temperature is 15°C. Take the atmospheric presAnswer: 98.3 percent sure at lake level to be 100 kPa. 14–36 Reconsider Problem 14–35. Using EES (or other) software, plot the mole fraction of dry air at the surface of the lake as a function of the lake temperature as the temperatue varies from 5°C to 25°C, and discuss the results. 14–37 Consider a rubber plate that is in contact with nitrogen gas at 298 K and 250 kPa. Determine the molar and mass densities of nitrogen in the rubber at the interface. Answers: 0.0039 kmol/m3, 0.1092 kg/m3 Rubber plate Boundary Conditions 14–25C Write three boundary conditions for mass transfer (on a mass basis) for species A at x 0 that correspond to specified temperature, specified heat flux, and convection boundary conditions in heat transfer. 14–26C What is an impermeable surface in mass transfer? How is it expressed mathematically (on a mass basis)? To what does it correspond in heat transfer? 14–27C Consider the free surface of a lake exposed to the atmosphere. If the air at the lake surface is saturated, will the mole fraction of water vapor in air at the lake surface be N2 298 K 250 kPa ρN2 = ? FIGURE P14–37 cen58933_ch14.qxd 9/9/2002 10:06 AM Page 774 774 HEAT TRANSFER 14–38 A wall made of natural rubber separates O2 and N2 gases at 25°C and 500 kPa. Determine the molar concentrations of O2 and N2 in the wall. (d) Other things being equal, doubling the mass fraction of the diffusing species at the high concentration side will double the rate of mass transfer. 14–39 Consider a glass of water in a room at 20°C and 97 kPa. If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium, determine (a) the mole fraction of the water vapor in the air and (b) the mole fraction of air in the water. 14–44C Consider one-dimensional mass diffusion of species A through a plane wall of thickness L. Under what conditions will the concentration profile of species A in the wall be a straight line? 14–40E Water is sprayed into air at 80°F and 14.3 psia, and the falling water droplets are collected in a container on the floor. Determine the mass and mole fractions of air dissolved in the water. 14–41 Consider a carbonated drink in a bottle at 27°C and 130 kPa. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as water, determine (a) the mole fraction of the water vapor in the CO2 gas and (b) the mass of dissolved CO2 in a Answers: (a) 2.77 percent, (b) 0.36 g 200-ml drink. 14–45C Consider one-dimensional mass diffusion of species A through a plane wall. Does the species A content of the wall change during steady mass diffusion? How about during transient mass diffusion? 14–46 Helium gas is stored at 293 K in a 3-m-outer-diameter spherical container made of 5-cm-thick Pyrex. The molar concentration of helium in the Pyrex is 0.00073 kmol/m3 at the inner surface and negligible at the outer surface. Determine the mass flow rate of helium by diffusion through the Pyrex Answer: 7.2 1015 kg/s container. 5 cm CO2 H2O Pyrex He gas 293 K Air He diffusion FIGURE P14–46 27°C 130 kPa FIGURE P14–41 Steady Mass Diffusion through a Wall 14–42C Write down the relations for steady one-dimensional heat conduction and mass diffusion through a plane wall, and identify the quantities in the two equations that correspond to each other. 14–43C Consider steady one-dimensional mass diffusion through a wall. Mark these statements as being True or False. (a) Other things being equal, the higher the density of the wall, the higher the rate of mass transfer. (b) Other things being equal, doubling the thickness of the wall will double the rate of mass transfer. (c) Other things being equal, the higher the temperature, the higher the rate of mass transfer. 14–47 A thin plastic membrane separates hydrogen from air. The molar concentrations of hydrogen in the membrane at the inner and outer surfaces are determined to be 0.065 and 0.003 kmol/m3, respectively. The binary diffusion coefficient of hydrogen in plastic at the operation temperature is 5.3 1010 m2/s. Determine the mass flow rate of hydrogen by diffusion through the membrane under steady conditions if the thickness of the membrane is (a) 2 mm and (b) 0.5 mm. 14–48 The solubility of hydrogen gas in steel in terms of its mass fraction is given as wH2 2.09 104 exp(–3950/T)P0,5 H2 where PH2 is the partial pressure of hydrogen in bars and T is the temperature in K. If natural gas is transported in a 1-cmthick, 3-m-internal-diameter steel pipe at 500 kPa pressure and the mole fraction of hydrogen in the natural gas is 8 percent, determine the highest rate of hydrogen loss through a 100-mlong section of the pipe at steady conditions at a temperature of 293 K if the pipe is exposed to air. Take the diffusivity of hydrogen in steel to be 2.9 1013 m2/s. Answer: 3.98 1014 kg/s cen58933_ch14.qxd 9/9/2002 10:06 AM Page 775 775 CHAPTER 14 14–49 Reconsider Problem 14–48. Using EES (or other) software, plot the highest rate of hydrogen loss as a function of the mole fraction of hydrogen in natural gas as the mole fraction varies from 5 to 15 percent, and discuss the results. 14–50 Helium gas is stored at 293 K and 500 kPa in a 1-cmthick, 2-m-inner-diameter spherical tank made of fused silica (SiO2). The area where the container is located is well ventilated. Determine (a) the mass flow rate of helium by diffusion through the tank and (b) the pressure drop in the tank in one week as a result of the loss of helium gas. 14–51 You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of 0.1-mm-thick soft rubber and has a diameter of 15 cm when inflated. The pressure and temperature inside the balloon are initially 110 kPa and 25°C. The permeability of rubber to helium, oxygen, and nitrogen at 25°C are 9.4 1013, 7.05 1013, and 2.6 1013 kmol/m · s · bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first 5 h assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be 100 kPa and 25°C. 110 kPa 25°C He Air determine how long it will take for the pressure inside the balloon to drop to 100 kPa. 14–53 Pure N2 gas at 1 atm and 25°C is flowing through a 10-m-long, 3-cm-inner diameter pipe made of 1-mm-thick rubber. Determine the rate at which N2 leaks out of the pipe if the medium surrounding the pipe is (a) a vacuum and (b) atmospheric air at 1 atm and 25°C with 21 percent O2 and 79 percent N2. Answers: (a) 4.48 1010 kmol/s, (b) 9.4 1011 kmol/s Vacuum N2 N2 gas 1 atm 25°C Rubber pipe FIGURE P14–53 Water Vapor Migration in Buildings 14–54C Consider a tank that contains moist air at 3 atm and whose walls are permeable to water vapor. The surrounding air at 1 atm pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain. 14–55C Express the mass flow rate of water vapor through a wall of thickness L in terms of the partial pressure of water vapor on both sides of the wall and the permeability of the wall to the water vapor. 14–56C How does the condensation or freezing of water vapor in the wall affect the effectiveness of the insulation in the wall? How does the moisture content affect the effective thermal conductivity of soil? 14–57C Moisture migration in the walls, floors, and ceilings of buildings is controlled by vapor barriers or vapor retarders. Explain the difference between the two, and discuss which is more suitable for use in the walls of residential buildings. 14–58C What are the adverse effects of excess moisture on the wood and metal components of a house and the paint on the walls? 14–59C Why are the insulations on the chilled water lines always wrapped with vapor barrier jackets? FIGURE P14–51 14–52 Reconsider the balloon discussed in Problem 14–51. Assuming the volume to remain constant and disregarding the diffusion of air into the balloon, obtain a relation for the variation of pressure in the balloon with time. Using the results obtained and the numerical values given in the problem, 14–60C Explain how vapor pressure of the ambient air is determined when the temperature, total pressure, and relative humidity of the air are given. 14–61 The diffusion of water vapor through plaster boards and its condensation in the wall insulation in cold weather are of concern since they reduce the effectiveness of insulation. Consider a house that is maintained at 20°C and 60 percent cen58933_ch14.qxd 9/9/2002 10:06 AM Page 776 776 HEAT TRANSFER 25°C 88 kPa 50% RH Plaster board Moisture migration 9.5 mm Room Outdoors 20°C 97 kPa 60% RH Aluminum foil Vapor diffusion Milk 25°C FIGURE P14–61 FIGURE P14–66 relative humidity at a location where the atmospheric pressure is 97 kPa. The inside of the walls is finished with 9.5-mm-thick gypsum wallboard. Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a 3-m 8-m section of a wall during a 24-h period. The permeance of the 9.5-mm-thick gypsum wallboard to water vapor is 2.86 109 kg/s · m2 · Pa. 14–62 Reconsider Problem 14–61. In order to reduce the migration of water vapor through the wall, it is proposed to use a 0.2-mm-thick polyethylene film with a permeance of 2.3 1012 kg/s · m2 · Pa. Determine the amount of water vapor that will diffuse through the wall in this case during a 24-h period. Answer: 6.7 g 14–63 The roof of a house is 15 m 8 m and is made of a 20-cm-thick concrete layer. The interior of the house is maintained at 25°C and 50 percent relative humidity and the local atmospheric pressure is 100 kPa. Determine the amount of water vapor that will migrate through the roof in 24 h if the average outside conditions during that period are 3°C and 30 percent relative humidity. The permeability of concrete to water vapor is 24.7 1012 kg/s · m · Pa. 14–64 Reconsider Problem 14–63. Using EES (or other) software, investigate the effects of temperature and relative humidity of air inside the house on the amount of water vapor that will migrate through the roof. Let the temperature vary from 15°C to 30°C and the relative humidity from 30 to 70 percent. Plot the amount of water vapor that will migrate as functions of the temperature and the relative humidity, and discuss the results. permeance is 2.9 1012 kg/s · m2 · Pa. The inner diameter of the glass is 12 cm. Assuming the air in the glass to be saturated at all times, determine how much the level of the milk in the Answer: 0.00079 mm glass will recede in 12 h. Transient Mass Diffusion 14–67C In transient mass diffusion analysis, can we treat the diffusion of a solid into another solid of finite thickness (such as the diffusion of carbon into an ordinary steel component) as a diffusion process in a semi-infinite medium? Explain. 14–68C Define the penetration depth for mass transfer, and explain how it can be determined at a specified time when the diffusion coefficient is known. 14–69C When the density of a species A in a semi-infinite medium is known at the beginning and at the surface, explain how you would determine the concentration of the species A at a specified location and time. 14–70 A steel part whose initial carbon content is 0.12 percent by mass is to be case-hardened in a furnace at 1150 K by exposing it to a carburizing gas. The diffusion coefficient of carbon in steel is strongly temperature dependent, and at the furnace temperature it is given to be DAB 7.2 1012 m2/s. 1150 K Carbon 14–65 Reconsider Problem 14–63. In order to reduce the migration of water vapor, the inner surface of the wall is painted with vapor retarder latex paint whose permeance is 26 1012 kg/s · m2 · Pa. Determine the amount of water vapor that will diffuse through the roof in this case during a 24-h period. 14–66 A glass of milk left on top of a counter in the kitchen at 25°C, 88 kPa, and 50 percent relative humidity is tightly sealed by a sheet of 0.009-mm-thick aluminum foil whose Steel part FIGURE P14–70 cen58933_ch14.qxd 9/9/2002 10:06 AM Page 777 777 CHAPTER 14 Also, the mass fraction of carbon at the exposed surface of the steel part is maintained at 0.011 by the carbon-rich environment in the furnace. If the hardening process is to continue until the mass fraction of carbon at a depth of 0.7 mm is raised to 0.32 percent, determine how long the part should be held in Answer: 9 h the furnace. 14–71 Repeat Problem 14–70 for a furnace temperature of 500 K at which the diffusion coefficient of carbon in steel is DAB 2.1 1020 m2/s. 14–72 A pond with an initial oxygen content of zero is to be oxygenated by forming a tent over the water surface and filling the tent with oxygen gas at 25°C and 130 kPa. Determine the mole fraction of oxygen at a depth of 2 cm from the surface after 12 h. Tent O2 gas 25°C 130 kPa O2 diffusion Pond (a) The rates of mass diffusion of species A and B are equal in magnitude and opposite in direction. (b) DAB DBA. (c) During equimolar counterdiffusion through a tube, equal numbers of moles of A and B move in opposite directions, and thus a velocity measurement device placed in the tube will read zero. (d) The lid of a tank containing propane gas (which is heavier than air) is left open. If the surrounding air and the propane in the tank are at the same temperature and pressure, no propane will escape the tank and no air will enter. 14–78C What is Stefan flow? Write the expression for Stefan’s law and indicate what each variable represents. 14–79E The pressure in a pipeline that transports helium gas at a rate of 5 lbm/s is maintained at 14.5 psia by venting helium to the atmosphere through a 14 -in. internal diameter tube that extends 30 ft into the air. Assuming both the helium and the atmospheric air to be at 80°F, determine (a) the mass flow rate of helium lost to the atmosphere through an individual tube, (b) the mass flow rate of air that infiltrates into the pipeline, and (c) the flow velocity at the bottom of the tube where it is attached to the pipeline that will be measured by an anemometer in steady operation. FIGURE P14–72 Air 80°F 14–73 A long nickel bar with a diameter of 5 cm has been stored in a hydrogen-rich environment at 358 K and 300 kPa for a long time, and thus it contains hydrogen gas throughout uniformly. Now the bar is taken into a well-ventilated area so that the hydrogen concentration at the outer surface remains at almost zero at all times. Determine how long it will take for the hydrogen concentration at the center of the bar to drop by half. The diffusion coefficient of hydrogen in the nickel bar at the room temperature of 298 K can be taken to be DAB 1.2 Answer: 3.3 years 1012 m2/s. Diffusion in a Moving Medium 14–74C Define the following terms: mass-average velocity, diffusion velocity, stationary medium, and moving medium. 14–75C What is diffusion velocity? How does it affect the mass-average velocity? Can the velocity of a species in a moving medium relative to a fixed reference point be zero in a moving medium? Explain. 14–76C What is the difference between mass-average velocity and mole-average velocity during mass transfer in a moving medium? If one of these velocities is zero, will the other also necessarily be zero? Under what conditions will these two velocities be the same for a binary mixture? 14–77C Consider one-dimensional mass transfer in a moving medium that consists of species A and B with A B constant. Mark these statements as being True or False. He Air 0.25 in. 30 ft He Helium Air 14.5 psia 80°F 5 lbm/s FIGURE P14–79E 14–80E Repeat Problem 14–79E for a pipeline that transports carbon dioxide instead of helium. 14–81 A tank with a 2-cm thick shell contains hydrogen gas at the atmospheric conditions of 25°C and 90 kPa. The charging valve of the tank has an internal diameter of 3 cm and extends 8 cm above the tank. If the lid of the tank is left open so that hydrogen and air can undergo equimolar counterdiffusion through the 10-cm-long passageway, determine the mass flow rate of hydrogen lost to the atmosphere through the valve at the Answer: 4.20 108 kg/s initial stages of the process. cen58933_ch14.qxd 9/9/2002 10:06 AM Page 778 778 HEAT TRANSFER 14–82 Reconsider Problem 14–81. Using EES (or other) software, plot the mass flow rate of hydrogen lost as a function of the diameter of the charging valve as the diameter varies from 1 cm to 5 cm, and discuss the results. 14–83E A 1-in.-diameter Stefan tube is used to measure the binary diffusion coefficient of water vapor in air at 70°F and 13.8 psia. The tube is partially filled with water with a distance from the water surface to the open end of the tube of 10 in. Dry air is blown over the open end of the tube so that water vapor rising to the top is removed immediately and the concentration of vapor at the top of the tube is zero. During 10 days of continuous operation at constant pressure and temperature, the amount of water that has evaporated is measured to be 0.0015 lbm. Determine the diffusion coefficient of water vapor in air at 70°F and 13.8 psia. 14–84 An 8-cm-internal-diameter, 30-cm-high pitcher half filled with water is left in a dry room at 15°C and 87 kPa with its top open. If the water is maintained at 15°C at all times also, determine how long it will take for the water to evaporate Answer: 1125 days completely. Room 15°C 87 kPa Water vapor 14–87C What is a concentration boundary layer? How is it defined for flow over a plate? 14–88C What is the physical significance of the Schmidt number? How is it defined? To what dimensionless number does it correspond in heat transfer? What does a Schmidt number of 1 indicate? 14–89C What is the physical significance of the Sherwood number? How is it defined? To what dimensionless number does it correspond in heat transfer? What does a Sherwood number of 1 indicate for a plain fluid layer? 14–90C What is the physical significance of the Lewis number? How is it defined? What does a Lewis number of 1 indicate? 14–91C In natural convection mass transfer, the Grashof number is evaluated using density difference instead of temperature difference. Can the Grashof number evaluated this way be used in heat transfer calculations also? 14–92C Using the analogy between heat and mass transfer, explain how the mass transfer coefficient can be determined from the relations for the heat transfer coefficient. 14–93C It is well known that warm air in a cooler environment rises. Now consider a warm mixture of air and gasoline (C8H18) on top of an open gasoline can. Do you think this gas mixture will rise in a cooler environment? 14–94C Consider two identical cups of coffee, one with no sugar and the other with plenty of sugar at the bottom. Initially, both cups are at the same temperature. If left unattended, which cup of coffee will cool faster? 14–95C Under what conditions will the normalized velocity, thermal, and concentration boundary layers coincide during flow over a flat plate? Water 15°C FIGURE P14–84 14–85 A large tank containing ammonia at 1 atm and 25°C is vented to the atmosphere through a 3-m-long tube whose internal diameter is 1 cm. Determine the rate of loss of ammonia and the rate of infiltration of air into the tank. Mass Convection 14–86C Heat convection is expressed by Newton’s law of · cooling as Q hA(Ts T). Express mass convection in an analogous manner on a mass basis, and identify all the quantities in the expression and state their units. 14–96C What is the relation ( f/2) Re Nu Sh known as? Under what conditions is it valid? What is the practical importance of it? 14–97C What is the name of the relation f/2 St Pr2/3 StmassSc2/3 and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering? 14–98C What is the relation hheat Cp hmass known as? For what kind of mixtures is it valid? What is the practical importance of it? 14–99C What is the low mass flux approximation in mass transfer analysis? Can the evaporation of water from a lake be treated as a low mass flux process? 14–100E Consider a circular pipe of inner diameter D 0.5 in. whose inner surface is covered with a thin layer of liquid water as a result of condensation. In order to dry the pipe, air at 540 R and 1 atm is forced to flow through it with an average velocity of 4 ft/s. Using the analogy between heat and cen58933_ch14.qxd 9/9/2002 10:06 AM Page 779 779 CHAPTER 14 mass transfer, determine the mass transfer coefficient inside the Answer: 0.024 ft/s pipe for fully developed flow. 14–101 The average heat transfer coefficient for air flow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton–Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at 1 atm at a free stream velocity of 2 m/s over a body covered with a layer of naphthalene. The surface area of the body is 0.75 m2, and it is observed that 100 g of naphthalene has sublimated in 45 min. During the experiment, both the body and the air were kept at 25°C, at which the vapor pressure and mass diffusivity of naphthalene are 11 Pa and DAB 0.61 105 m2/s, respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry. 0.75 m2 Air 1 atm 2 m /s 25°C Body 25°C Naphthalene vapor 14–105 Consider a 5-m 5-m wet concrete patio with an average water film thickness of 0.3 mm. Now wind at 50 km/h is blowing over the surface. If the air is at 1 atm, 15°C, and 35 percent relative humidity, determine how long it will take Answer: 18.6 min for the patio to dry completely. 14–106E A 2-in.-diameter spherical naphthalene ball is suspended in a room at 1 atm and 80°F. Determine the average mass transfer coefficient between the naphthalene and the air if air is forced to flow over naphthalene with a free stream velocity of 15 ft/s. The Schmidt number of naphthalene in air at Answer: 0.0525 ft/s room temperature is 2.35. 14–107 Consider a 3-mm-diameter raindrop that is falling freely in atmospheric air at 25°C. Taking the temperature of the raindrop to be 9°C, determine the terminal velocity of the raindrop at which the drag force equals the weight of the drop and the average mass transfer coefficient at that time. 14–108 In a manufacturing facility, wet brass plates coming out of a water bath are to be dried by passing them through a section where dry air at 1 atm and 25°C is blown parallel to their surfaces. If the plates are at 20°C and there are no dry spots, determine the rate of evaporation from both sides of a plate. FIGURE P14–101 14–102 Consider a 15-cm-internal-diameter, 10-m-long circular duct whose interior surface is wet. The duct is to be dried by forcing dry air at 1 atm and 15°C through it at an average velocity of 3 m/s. The duct passes through a chilled room, and it remains at an average temperature of 15°C at all times. Determine the mass transfer coefficient in the duct. 14–103 Reconsider Problem 14–102. Using EES (or other) software, plot the mass transfer coefficient as a function of the air velocity as the velocity varies from 1 m/s to 8 m/s, and discuss the results. 14–104 Dry air at 15°C and 92 kPa flows over a 2-m-long wet surface with a free stream velocity of 4 m/s. Determine the average mass transfer coefficient. Answer: 0.00514 m/s Dry air 15°C, 92 kPa 4 m /s Air 25°C 4 m/s 40 cm 40 cm Brass plate 20°C FIGURE P14–108 14–109E Air at 80°F, 1 atm, and 30 percent relative humidity is blown over the surface of a 15-in. 15-in. square pan filled with water at a free stream velocity of 10 ft/s. If the water is maintained at a uniform temperature of 80°F, determine the rate of evaporation of water and the amount of heat that needs to be supplied to the water to maintain its temperature constant. 14–110E Repeat Problem 14–109E for temperature of 60°F for both the air and water. Evaporation Wet FIGURE P14–104 Simultaneous Heat and Mass Transfer 14–111C Does a mass transfer process have to involve heat transfer? Describe a process that involves both heat and mass transfer. 14–112C Consider a shallow body of water. Is it possible for this water to freeze during a cold and dry night even when the ambient air and surrounding surface temperatures never drop to 0°C? Explain. cen58933_ch14.qxd 9/9/2002 10:06 AM Page 780 780 HEAT TRANSFER 14–113C During evaporation from a water body to air, under what conditions will the latent heat of vaporization be equal to convection heat transfer from the air? 14–114 Jugs made of porous clay were commonly used to cool water in the past. A small amount of water that leaks out keeps the outer surface of the jug wet at all times, and hot and relatively dry air flowing over the jug causes this water to evaporate. Part of the latent heat of evaporation comes from the water in the jug, and the water is cooled as a result. If the environment conditions are 1 atm, 30°C, and 35 percent relative humidity, determine the temperature of the water when steady conditions are reached. the emissivities of sheet metal and water to be 0.61 and 0.95, respectively. Answers: (a) 61,337 W, (b) 1480 W, (c) 3773 W, (d ) 79,960 W, 44.9 kg/h 14–118 Repeat Problem 14–117 for a water bath temperature of 50°C. 14–119 One way of increasing heat transfer from the head on a hot summer day is to wet it. This is especially effective in windy weather, as you may have noticed. Approximating the head as a 30-cm-diameter sphere at 30°C with an emissivity of 0.95, determine the total rate of heat loss from the head at ambient air conditions of 1 atm, 25°C, 40 percent relative humidity, and 25 km/h winds if the head is (a) dry and (b) wet. Take the surrounding temperature to be 25°C. Answers: (a) 40.6 W, (b) 352 W Water that leaks out Evaporation Hot, dry air 30°C 35% RH Wet 30°C 1 atm 25°C 40% RH 25 km/h FIGURE P14–114 14–115 Reconsider Problem 14–114. Using EES (or other) software, plot the water temperature as a function of the relative humidity of air as the relative humidity varies from 10 to 100 percent, and discuss the results. 14–116E During a hot summer day, a 2-L bottle drink is to be cooled by wrapping it in a cloth kept wet continually and blowing air to it with a fan. If the environment conditions are 1 atm, 80°F, and 30 percent relative humidity, determine the temperature of the drink when steady conditions are reached. 14–117 A glass bottle washing facility uses a wellagitated hot water bath at 55°C with an open top that is placed on the ground. The bathtub is 1 m high, 2 m wide, and 4 m long and is made of sheet metal so that the outer side surfaces are also at about 55°C. The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of 150 g and removes 0.6 g of water as it leaves the bath wet. Makeup water is supplied at 15°C. If the average conditions in the plant are 1 atm, 25°C, and 50 percent relative humidity, and the average temperature of the surrounding surfaces is 15°C, determine (a) the amount of heat and water removed by the bottles themselves per second; (b) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation; (c) the rate of heat loss from the side surfaces by natural convection and radiation; and (d) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take FIGURE P14–119 14–120 A 2-m-deep 20-m 20-m heated swimming pool is maintained at a constant temperature of 30°C at a location where the atmospheric pressure is 1 atm. If the ambient air is at 20°C and 60 percent relative humidity and the effective sky temperature is 0°C, determine the rate of heat loss from the top surface of the pool by (a) radiation, (b) natural convection, and (c) evaporation. (d) Assuming the heat losses to the ground to be negligible, determine the size of the heater. 14–121 Repeat Problem 14–120 for a pool temperature of 25°C. Review Problems 14–122C Mark these statements as being True or False. (a) The units of mass diffusivity, heat diffusivity, and momentum diffusivity are all the same. (b) If the molar concentration (or molar density) C of a mixture is constant, then its density must also be constant. (c) If the mass-average velocity of a binary mixture is zero, then the mole-average velocity of the mixture must also be zero. (d) If the mole fractions of A and B of a mixture are both 0.5, then the molar mass of the mixture is simply the arithmetic average of the molar masses of A and B. 14–123 Using Henry’s law, show that the dissolved gases in a liquid can be driven off by heating the liquid. cen58933_ch14.qxd 9/9/2002 10:06 AM Page 781 781 CHAPTER 14 14–124 Show that for an ideal gas mixture maintained at a constant temperature and pressure, the molar concentration C of the mixture remains constant but this is not necessarily the case for the density of the mixture. CO2 Water 14–125E A gas mixture in a tank at 600 R and 20 psia consists of 1 lbm of CO2 and 3 lbm of CH4. Determine the volume of the tank and the partial pressure of each gas. 14–126 Dry air whose molar analysis is 78.1 percent N2, 20.9 percent O2, and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at 1 atm and 25°C during the process, determine (a) the molar analysis of the saturated air and (b) the density of air before and after the process. What do you conclude from your results? 14–127 Consider a glass of water in a room at 25°C and 100 kPa. If the relative humidity in the room is 70 percent and the water and the air are at the same temperature, determine (a) the mole fraction of the water vapor in the room air, (b) the mole fraction of the water vapor in the air adjacent to the water surface, and (c) the mole fraction of air in the water near the surface. Answers: (a) 2.22 percent, (b) 3.17 percent, (c) 1.34 105 percent 25°C 100 kPa 70% RH Air-water interface FIGURE P14–129 14–130 Consider a brick house that is maintained at 20°C and 60 percent relative humidity at a location where the atmospheric pressure is 85 kPa. The walls of the house are made of 20-cm thick brick whose permeance is 23 109 kg/s · m2 · Pa. Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a 4-m 7-m section of a wall during a 24-h period. 14–131E Consider a masonry cavity wall that is built around 6-in.-thick concrete blocks. The outside is finished with 4-in. face brick with 21 -in. cement mortar between the bricks and concrete blocks. The inside finish consists of 12 -in. gypsum wallboard separated from the concrete block by 34 -in.-thick air space. The thermal and vapor resistances of various components for a unit wall area are as follows: Water 25°C FIGURE P14–127 14–128 The diffusion coefficient of carbon in steel is given as DAB 2.67 105 exp(–17,400/T) m2/s where T is in K. Determine the diffusion coefficient from 300 K to 1500 K in 100 K increments and plot the results. 14–129 A carbonated drink is fully charged with CO2 gas at 17°C and 600 kPa such that the entire bulk of the drink is in thermodynamic equilibrium with the CO2–water vapor mixture. Now consider a 2-L soda bottle. If the CO2 gas in that bottle were to be released and stored in a container at 25°C and 100 kPa, determine the volume of the container. Answer: 12.7 L 4 2 1 FIGURE P14–131E 3 5 6 7 cen58933_ch14.qxd 9/9/2002 10:06 AM Page 782 782 HEAT TRANSFER R-Value, h · ft2 · °F/Btu Construction 1. Outside surface, 15 mph wind 2. Face brick, 4 in. 3. Cement mortar, 0.5 in. 4. Concrete block, 6-in. 5. Air space, 3 -in. 4 6. Gypsum wallboard, 0.5 in. 7. Inside surface, still air R-Value, s · ft2 · psi/lbm 0.17 0.43 — 15,000 0.10 1930 4.20 23,000 1.02 77.6 0.45 332 0.68 — The indoor conditions are 70°F and 65 percent relative humidity while the outdoor conditions are 32°F and 40 percent relative humidity. Determine the rates of heat and water vapor transfer through a 9-ft 25-ft section of the wall. Answers: 1436 Btu/h, 4.03 lbm/h 14–132 The oxygen needs of fish in aquariums are usually met by forcing air to the bottom of the aquarium by a compressor. The air bubbles provide a large contact area between the water and the air, and as the bubbles rise, oxygen and nitrogen gases in the air dissolve in water while some water evaporates into the bubbles. Consider an aquarium that is maintained at room temperature of 25°C at all times. The air bubbles are observed to rise to the free surface of water in 2 s. If the air entering the aquarium is completely dry and the diameter of the air bubbles is 4 mm, determine the mole fraction of water vapor at the center of the bubble when it leaves the aquarium. Assume no fluid motion in the bubble so that water vapor propagates in the bubble by diffusion only. Answer: 3.13 percent 1 atm 25°C Air bubbles Aquarium 25°C 14–134 Consider a 30-cm-diameter pan filled with water at 15°C in a room at 20°C, 1 atm, and 30 percent relative humidity. Determine (a) the rate of heat transfer by convection, (b) the rate of evaporation of water, and (c) the rate of heat transfer to the water needed to maintain its temperature at 15°C. Disregard any radiation effects. 14–135 Repeat Problem 14–134 assuming a fan blows air over the water surface at a velocity of 3 m/s. Take the radius of the pan to be the characteristic length. 14–136 Naphthalene is commonly used as a repellent against moths to protect clothing during storage. Consider a 1-cmdiameter spherical naphthalene ball hanging in a closet at 25°C and 1 atm. Considering the variation of diameter with time, determine how long it will take for the naphthalene to sublimate completely. The density and vapor pressure of naphthalene at 25°C are 0.11 Pa and 1100 kg/m3 and 11 Pa, respectively, and the mass diffusivity of naphthalene in air at 25°C is DAB 0.61 105 m2/s. Answer: 45.7 days Closet 25°C 1 atm Sublimation Naphthalene 25°C FIGURE P14–136 14–137E A swimmer extends his wet arms into the windy air outside at 1 atm, 40°F, 50 percent relative humidity, and 20 mph. If the average skin temperature is 80°F, determine the rate at which water evaporates from both arms and the corresponding rate of heat transfer by evaporation. The arm can be modeled as a 2-ft-long and 3-in.-diameter cylinder with adiabatic ends. 14–138 A thick part made of nickel is put into a room filled with hydrogen at 3 atm and 85°C. Determine the hydrogen concentration at a depth of 2-mm from the surface after 24 h. Answer: 4.1 107 kmol/m3 FIGURE P14–132 14–133 Oxygen gas is forced into an aquarium at 1 atm and 25°C, and the oxygen bubbles are observed to rise to the free surface in 2 s. Determine the penetration depth of oxygen into water from a bubble during this time period. 14–139 A membrane made of 0.1-mm-thick soft rubber separates pure O2 at 1 atm and 25°C from air at 1.2 atm pressure. Determine the mass flow rate of O2 through the membrane per unit area and the direction of flow. 14–140E The top section of an 8-ft-deep 100-ft 100-ft heated solar pond is maintained at a constant temperature of 80°F at a location where the atmospheric pressure is 1 atm. If the ambient air is at 70°F and 100 percent relative humidity cen58933_ch14.qxd 9/9/2002 10:06 AM Page 783 783 CHAPTER 14 and wind is blowing at an average velocity of 40 mph, determine the rate of heat loss from the top surface of the pond by (a) forced convection, (b) radiation, and (c) evaporation. Take the average temperature of the surrounding surfaces to be 60°F. 14–141E Repeat Problem 14–140E for a solar pond surface temperature of 90°F. Answers: (a) 299,400 Btu/h, (b) 1,057,000 Btu/h, (c) 3,396,000 Btu/h Computer, Design, and Essay Problems 14–142 Write an essay on diffusion caused by effects other than the concentration gradient such as thermal diffusion, pressure diffusion, forced diffusion, knodsen diffusion, and surface diffusion. 14–143 Write a computer program that will convert the mole fractions of a gas mixture to mass fractions when the molar masses of the components of the mixture are specified. 14–144 One way of generating electricity from solar energy involves the collection and storage of solar energy in large artificial lakes of a few meters deep, called solar ponds. Solar energy is stored at the bottom part of the pond at temperatures close to boiling, and the rise of hot water to the top is prevented by planting salt to the bottom of the pond. Write an essay on the operation of solar pond power plants, and find out how much salt is used per year per m2. If the cost is not a factor, can sugar be used instead of salt to maintain the concentration gradient? Explain. 14–145 The condensation and even freezing of moisture in building walls without effective vapor retarders is a real con- cern in cold climates as it undermines the effectiveness of the insulation. Investigate how the builders in your area are coping with this problem, whether they are using vapor retarders or vapor barriers in the walls, and where they are located in the walls. Prepare a report on your findings and explain the reasoning for the current practice. 14–146 You are asked to design a heating system for a swimming pool that is 2 m deep, 25 m long, and 25 m wide. Your client desires that the heating system be large enough to raise the water temperature from 20°C to 30°C in 3 h. The heater must also be able to maintain the pool at 30°C at the outdoor design conditions of 15°C, 1 atm, 35 percent relative humidity, 40 mph winds, and effective sky temperature of 10°C. Heat losses to the ground are expected to be small and can be disregarded. The heater considered is a natural gas furnace whose efficiency is 80 percent. What heater size (in Btu/h input) would you recommend that your client buy? Evaporation 15°C 1 atm 35% RH 30°C Heating fluid FIGURE P14–146 Pool Heat loss cen58933_ch14.qxd 9/9/2002 10:06 AM Page 784 cen58933_ch15.qxd 9/9/2002 10:20 AM Page 785 CHAPTER COOLING OF ELECTRONIC EQUIPMENT lectronic equipment has made its way into practically every aspect of modern life, from toys and appliances to high-power computers. The reliability of the electronics of a system is a major factor in the overall reliability of the system. Electronic components depend on the passage of electric current to perform their duties, and they become potential sites for excessive heating, since the current flow through a resistance is accompanied by heat generation. Continued miniaturization of electronic systems has resulted in a dramatic increase in the amount of heat generated per unit volume, comparable in magnitude to those encountered at nuclear reactors and the surface of the sun. Unless properly designed and controlled, high rates of heat generation result in high operating temperatures for electronic equipment, which jeopardizes its safety and reliability. The failure rate of electronic equipment increases exponentially with temperature. Also, the high thermal stresses in the solder joints of electronic components mounted on circuit boards resulting from temperature variations are major causes of failure. Therefore, thermal control has become increasingly important in the design and operation of electronic equipment. In this chapter, we discuss several cooling techniques commonly used in electronic equipment such as conduction cooling, natural convection and radiation cooling, forced-air cooling, liquid cooling, and immersion cooling. This chapter is intended to familiarize the reader with these techniques and put them into perspective. The reader interested in an in-depth coverage of any of these topics can consult numerous other sources available, such as those listed in the references. E 15 CONTENTS 15–1 Introduction and History 786 15–2 Manufacturing of Electronic Equipment 787 15–3 Cooling Load of Electronic Equipment 793 15–4 Thermal Environment 794 15–5 Electronics Cooling in Different Applications 795 15–6 Conduction Cooling 797 15–7 Air Cooling: Natural Convection and Radiation 812 15–8 Air Cooling: Forced Convection 820 15–9 Liquid Cooling 833 15–10 Immersion Cooling 836 785 cen58933_ch15.qxd 9/9/2002 10:20 AM Page 786 786 HEAT TRANSFER 15–1 1010 GSI 109 108 ULSI Components per chip 107 106 VLSI 105 104 VLSI 103 102 MSI 101 100 SSI 1960 1970 1980 Year 1990 FIGURE 15–1 The increase in the number of components packed on a chip over the years. 2000 ■ INTRODUCTION AND HISTORY The field of electronics deals with the construction and utilization of devices that involve current flow through a vacuum, a gas, or a semiconductor. This exciting field of science and engineering dates back to 1883, when Thomas Edison invented the vacuum diode. The vacuum tube served as the foundation of the electronics industry until the 1950s, and played a central role in the development of radio, TV, radar, and the digital computer. Of the several computers developed in this era, the largest and best known is the ENIAC (Electronic Numerical Integrator and Computer), which was built at the University of Pennsylvania in 1946. It had over 18,000 vacuum tubes and occupied a room 7 m 14 m in size. It consumed a large amount of power, and its reliability was poor because of the high failure rate of the vacuum tubes. The invention of the bipolar transistor in 1948 marked the beginning of a new era in the electronics industry and the obsolescence of vacuum tube technology. Transistor circuits performed the functions of the vacuum tubes with greater reliability, while occupying negligible space and consuming negligible power compared with vacuum tubes. The first transistors were made from germanium, which could not function properly at temperatures above 100°C. Soon they were replaced by silicon transistors, which could operate at much higher temperatures. The next turning point in electronics occurred in 1959 with the introduction of the integrated circuits (IC), where several components such as diodes, transistors, resistors, and capacitors are placed in a single chip. The number of components packed in a single chip has been increasing steadily since then at an amazing rate, as shown in Figure 15–1. The continued miniaturization of electronic components has resulted in medium-scale integration (MSI) in the 1960s with 50–1000 components per chip, large-scale integration (LSI) in the 1970s with 1000–100,000 components per chip, and very large-scale integration (VLSI) in the 1980s with 100,000–10,000,000 components per chip. Today it is not unusual to have a chip 3 cm 3 cm in size with several million components on it. The development of the microprocessor in the early 1970s by the Intel Corporation marked yet another beginning in the electronics industry. The accompanying rapid development of large-capacity memory chips in this decade made it possible to introduce capable personal computers for use at work or at home at an affordable price. Electronics has made its way into practically everything from watches to household appliances to automobiles. Today it is difficult to imagine a new product that does not involve any electronic parts. The current flow through a resistance is always accompanied by heat generation in the amount of I 2R, where I is the electric current and R is the resistance. When the transistor was first introduced, it was touted in the newspapers as a device that “produces no heat.” This certainly was a fair statement, considering the huge amount of heat generated by vacuum tubes. Obviously, the little heat generated in the transistor was no match to that generated in its predecessor. But when thousands or even millions of such components are packed in a small volume, the heat generated increases to such high levels that its removal becomes a formidable task and a major concern for the safety and reliability of the electronic devices. The heat fluxes encountered in electronic devices range from less than 1 W/cm2 to more than 100 W/cm2. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 787 787 CHAPTER 15 15–2 ■ MANUFACTURING OF ELECTRONIC EQUIPMENT The narrow band where two different regions of a semiconductor (such as the p-type and n-type regions) come in contact is called a junction. A transistor, for example, involves two such junctions, and a diode, which is the simplest semiconductor device, is based on a single p-n junction. In heat transfer analysis, the circuitry of an electronic component through which electrons flow and thus heat is generated is also referred to as the junction. That is, junctions are the sites of heat generation and thus the hottest spots in a component. In silicon-based semiconductor devices, the junction temperature is limited to 125°C for safe operation. However, lower junction temperatures are desirable for extended life and lower maintenance costs. In a typical application, numerous electronic components, some smaller than 1 m in size, are formed from a silicon wafer into a chip. 10 failure rate at T fT = —–—————— failure rate at 75°C 9 Failure factor fT Heat is generated in a resistive element for as long as current continues to flow through it. This creates a heat build-up and a subsequent temperature rise at and around the component. The temperature of the component will continue rising until the component is destroyed unless heat is transferred away from it. The temperature of the component will remain constant when the rate of heat removal from it equals the rate of heat generation. Individual electronic components have no moving parts, and thus nothing to wear out with time. Therefore, they are inherently reliable, and it seems as if they can operate safely for many years. Indeed, this would be the case if components operated at room temperature. But electronic components are observed to fail under prolonged use at high temperatures. Possible causes of failure are diffusion in semiconductor materials, chemical reactions, and creep in the bonding materials, among other things. The failure rate of electronic devices increases almost exponentially with the operating temperature, as shown in Figure 15–2. The cooler the electronic device operates, the more reliable it is. A rule of thumb is that the failure rate of electronic components is halved for each 10°C reduction in their junction temperature. 8 7 6 5 4 3 2 1 0 20 40 60 80 100 120 140 75 Temperature, °C FIGURE 15–2 The increase in the failure rate of bipolar digital devices with temperature (from Ref. 15). The Chip Carrier The chip is housed in a chip carrier or substrate made of ceramic, plastic, or glass in order to protect its delicate circuitry from the detrimental effects of the environment. The chip carrier provides a rugged housing for the safe handling of the chip during the manufacturing process, as well as the connectors between the chip and the circuit board. The various components of the chip carrier are shown in Figure 15–3. The chip is secured in the carrier by bonding it to the bottom surface. The thermal expansion coefficient of the plastic is about 20 times that of silicon. Therefore, bonding the silicon chip directly to the plastic case would result in such large thermal stresses that the reliability would be seriously jeopardized. To avoid this problem, a lead frame made of a copper alloy with a thermal expansion coefficient close to that of silicon is used as the bonding surface. The design of the chip carrier is the first level in the thermal control of electronic devices, since the transfer of heat from the chip to the chip carrier is the Lid Bond wires Case Leads Chip Lead frame Bond Pins FIGURE 15–3 The components of a chip carrier. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 788 788 HEAT TRANSFER FIGURE 15–4 The chip carrier for a high-power transistor attached to a flange for enhanced heat transfer. first step in the dissipation of the heat generated on the chip. The heat generated on the chip is transferred to the case of the chip carrier by a combination of conduction, convection, and radiation. However, it is obvious from the figure that the common chip carrier is designed with the electrical aspects in mind, and little consideration is given to the thermal aspects. First of all, the cavity of the chip carrier is filled with a gas, which is a poor heat conductor, and the case is often made of materials that are also poor conductors of heat. This results in a relatively large thermal resistance between the chip and the case, called the junction-to-case resistance, and thus a large temperature difference. As a result, the temperature of the chip will be much higher than that of the case for a specified heat dissipation rate. The junction-to-case thermal resistance depends on the geometry and the size of the chip and the chip carrier as well as the material properties of the bonding and the case. It varies considerably from one device to another and ranges from about 10°C/ W to more than 100°C/W. Moisture in the cavity of the chip carrier is highly undesirable, since it causes corrosion on the wiring. Therefore, chip carriers are made of materials that prevent the entry of moisture by diffusion and are hermetically sealed in order to prevent the direct entry of moisture through cracks. Materials that outgas are also not permitted in the chip cavity, because such gases can also cause corrosion. In products with strict hermeticity requirements, the more expensive ceramic cases are used instead of the plastic ones. A common type of chip carrier for high-power transistors is shown in Figure 15–4. The transistor is formed on a small silicon chip housed in the diskshaped cavity, and the I/O pins come out from the bottom. The case of the transistor carrier is usually attached directly to a flange, which provides a large surface area for heat dissipation and reduces the junction-to-case thermal resistance. It is often desirable to house more than one chip in a single chip carrier. The result is a hybrid or multichip package. Hybrid packages house several chips, individual electronic components, and ordinary circuit elements connected to each other in a single chip carrier. The result is improved performance due to the shortening of the wiring lengths, and enhanced reliability. Lower cost would be an added benefit of multichip packages if they are produced in sufficiently large quantity. EXAMPLE 15–1 Rjunction-case Junction . Q FIGURE 15–5 Schematic for Example 15–1. Case Predicting the Junction Temperature of a Transistor The temperature of the case of a power transistor that is dissipating 3 W is measured to be 50°C. If the junction-to-case resistance of this transistor is specified by the manufacturer to be 15°C/ W, determine the temperature at the junction of the transistor. SOLUTION The case temperature of a power transistor and the junction-tocase resistance are given. The junction temperature is to be determined. Assumptions Steady operating conditions exist. Analysis The schematic of the transistor is given in Figure 15–5. The rate of heat transfer between the junction and the case in steady operation can be expressed as cen58933_ch15.qxd 9/9/2002 10:20 AM Page 789 789 CHAPTER 15 · T Q R junction-case Tjunction Tcase Rjunction-case Then the junction temperature becomes · Tjunction Tcase Q Rjunction-case 50°C (3 W)(15°C/ W) 95°C Therefore, the temperature of the transistor junction will be 95°C when its case is at 50°C. EXAMPLE 15–2 Determining the Junction-to-Case Thermal Resistance This experiment is conducted to determine the junction-to-case thermal resistance of an electronic component. Power is supplied to the component from a 15-V source, and the variations in the electric current and in the junction and the case temperatures with time are observed. When things are stabilized, the current is observed to be 0.1 A and the temperatures to be 80°C and 55°C at the junction and the case, respectively. Calculate the junction-to-case resistance of this component. SOLUTION The power dissipated by an electronic component as well as the junction and case temperatures are measured. The junction-to-case resistance is to be determined. Assumptions Steady operating conditions exist. Analysis The schematic of the component is given in Figure 15–6. The electric power consumed by this electronic component is · We VI (15 V)(0.1 A) 1.5 W Rjunction-case . Q 80°C 55°C In steady operation, this is equivalent to the heat dissipated by the component. That is, · T Q R junction-case Tjunction Tcase 1.5 W Rjunction-case Then the junction-to-case resistance is determined to be Rjunction-case Tjunction Tcase (80 55)°C 16.7°C/ W · 1.5 W Q Discussion Note that a temperature difference of 16.7°C will occur between the electronic circuitry and the case of the chip carrier for each W of power consumed by the component. Printed Circuit Boards A printed circuit board (PCB) is a properly wired plane board made of polymers and glass–epoxy materials on which various electronic components such FIGURE 15–6 Schematic for Example 15–2. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 790 790 HEAT TRANSFER as the ICs, diodes, transistors, resistors, and capacitors are mounted to perform a certain task, as shown in Figure 15–7. The PCBs are commonly called cards, and they can be replaced easily during a repair. The PCBs are plane boards, usually 10 cm wide and 15 cm long and only a few millimeters thick, and they are not suitable for heavy components such as transformers. Usually a copper cladding is added on one or both sides of the board. The cladding on one side is subjected to an etching process to form wiring strips and attachment pads for the components. The power dissipated by a PCB usually ranges from 5 W to about 30 W. A typical electronic system involves several layers of PCBs. The PCBs are usually cooled by direct contact with a fluid such as air flowing between the boards. But when the boards are placed in a hermetically sealed enclosure, they must be cooled by a cold plate (a heat exchanger) in contact with the edge of the boards. The device-to-board edge thermal resistance of a PCB is usually high (about 20 to 60°C/W) because of the small thickness of the board and the low thermal conductivity of the board material. In such cases, even a thin layer of copper cladding on one side of the board can decrease the deviceto-board edge thermal resistance in the plane of the board and enhance heat transfer in that direction drastically. In the thermal design of a PCB, it is important to pay particular attention to the components that are not tolerant of high temperatures, such as certain high-performance capacitors, and to ensure their safe operation. Often when one component on a PCB fails, the whole board fails and must be replaced. Printed circuit boards come in three types: single-sided, double-sided, and multilayer boards. Each type has its own strengths and weaknesses. Singlesided PCBs have circuitry lines on one side of the board only and are suitable for low-density electronic devices (10 to 20 components). Double-sided PCBs FIGURE 15–7 A printed circuit board (PCB) with a variety of components on it (courtesy of Litton Systems, Inc.). cen58933_ch15.qxd 9/9/2002 10:20 AM Page 791 791 CHAPTER 15 have circuits on both sides and are best suited for intermediate-density devices. Multilayer PCBs contain several layers of circuitry and are suitable for high-density devices. They are equivalent to several PCBs sandwiched together. The single-sided PCB has the lowest cost, as expected, and it is easy to maintain, but it occupies a lot of space. The multilayer PCB, on the other hand, allows the placement of a large number of components in a threedimensional configuration, but it has the highest initial cost and is difficult to repair. Also, temperatures are most likely to be the highest in multilayer PCBs. In critical applications, the electronic components are placed on boards attached to a conductive metal, called the heat frame, that serves as a conduction path to the edge of the circuit board and thus to the cold plate for the heat generated in the components. Such boards are said to be conduction-cooled. The temperature of the components in this case will depend on the location of the components on the boards: it will be highest for the components in the middle and lowest for those near the edge, as shown in Figure 15–8. Materials used in the fabrication of circuit boards should be (1) effective electrical insulators to prevent electrical breakdown and (2) good heat conductors to conduct away the heat generated. They should also have (3) high material strength to withstand forces and to maintain dimensional stability; (4) thermal expansion coefficients that closely match that of copper, to prevent cracking in the copper cladding during thermal cycling; (5) resistance to moisture absorption, since moisture can affect both mechanical and electrical properties and degrade performance; (6) stability in properties at temperature levels encountered in electronic applications; (7) ready availability and manufacturability; and, of course, (8) low cost. As you might have already guessed, no existing material has all of these desirable characteristics. Glass–epoxy laminates made of an epoxy or polymide matrix reinforced by several layers of woven glass cloth are commonly used in the production of circuit boards. Polymide matrices are more expensive than epoxy but can withstand much higher temperatures. Polymer or polymide films are also used without reinforcement for flexible circuits. Maximum temperature T Cold plate temperature Electronic component Circuit board Heat frame Cold plate FIGURE 15–8 The path of heat flow in a conductioncooled PCB and the temperature distribution. The Enclosure An electronic system is not complete without a rugged enclosure (a case or a cabinet) that will house the circuit boards and the necessary peripheral equipment and connectors, protect them from the detrimental effects of the environment, and provide a cooling mechanism (Fig. 15–9). In a small electronic system such as a personal computer, the enclosure can simply be an inexpensive box made of sheet metal with proper connectors and a small fan. But for a large system with several hundred PCBs, the design and construction of the enclosure are challenges for both electronic and thermal designers. An enclosure must provide easy access for service personnel so that they can identify and replace any defective parts easily and quickly in order to minimize down time, which can be very costly. But, at the same time, the enclosure must prevent any easy access by unauthorized people in order to protect the sensitive electronics from them as well as the people from possible electrical hazards. Electronic circuits are powered by low voltages (usually under 15 V), but the currents involved may be very high (sometimes a few hundred amperes). FIGURE 15–9 A cabinet-style enclosure. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 792 792 HEAT TRANSFER Edge connector Wafer Substrate Chip Back panel Printed circuit board Chip carrier Chassis Cabinet FIGURE 15–10 Different stages involved in the production of an electronic system (from Dally, Ref. 3). Plug-in-type circuit boards make it very easy to replace a defective board and are commonly used in low-power electronic equipment. High-power circuit boards in large systems, however, are tightly attached to the racks of the cabinet with special brackets. A well-designed enclosure also includes switches, indicator lights, a screen to display messages and present information about the operation, and a key pad for user interface. The printed circuit boards in a large system are plugged into a back panel through their edge connectors. The back panel supplies power to the PCBs and interconnects them to facilitate the passage of current from one board to another. The PCBs are assembled in an orderly manner in card racks or chassis. One or more such assemblies are housed in a cabinet, as shown in Figure 15–10. Electronic enclosures come in a wide variety of sizes and shapes. Sheet metals such as thin-gauge aluminum or steel sheets are commonly used in the production of enclosures. The thickness of the enclosure walls depends on the shock and vibration requirements. Enclosures made of thick metal sheets or by casting can meet these requirements, but at the expense of increased weight and cost. Electronic boxes are sometimes sealed to prevent the fluid inside (usually air) from leaking out and the water vapor outside from leaking in. Sealing against moisture migration is very difficult because of the small size of the water molecule and the large vapor pressure outside the box relative to that within the box. Sealing adds to the size, weight, and cost of an electronic box, especially in space or high-altitude operation, since the box in this case must withstand the larger forces due to the higher pressure differential between inside and outside the box. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 793 793 CHAPTER 15 15–3 ■ COOLING LOAD OF ELECTRONIC EQUIPMENT The first step in the selection and design of a cooling system is the determination of the heat dissipation, which constitutes the cooling load. The easiest way to determine the power dissipation of electronic equipment is to measure the voltage applied V and the electric current I at the entrance of the electronic device under full-load conditions and to substitute them into the relation · We VI I 2R (W) (15-1) · where We is the electric power consumption of the electronic device, which constitutes the energy input to the device. The first law of thermodynamics requires that in steady operation the energy input into a system be equal to the energy output from the system. Considering that the only form of energy leaving the electronic device is heat generated as the current flows through resistive elements, we conclude that the heat dissipation or cooling load of an electronic device is equal to its · · power consumption. That is, Q We, as shown in Figure 15–11. The exception to this rule is equipment that outputs other forms of energy as well, such as the emitter tubes of a radar, radio, or TV installation emitting radiofrequency (RF) electromagnetic radiation. In such cases, the cooling load will be equal to the difference between the power consumption and the RF power emission. An equivalent but cumbersome way of determining the cooling load of an electronic device is to determine the heat dissipated by each component in the device and then to add them up. The discovery of superconductor materials that can operate at room temperature will cause drastic changes in the design of electronic devices and cooling techniques, since such devices will generate hardly any heat. As a result, more components can be packed into a smaller volume, resulting in enhanced speed and reliability without having to resort to exotic cooling techniques. Once the cooling load has been determined, it is common practice to inflate this number to leave some safety margin, or a “cushion,” and to make some allowance for future growth. It is not uncommon to add another card to an existing system (such as adding a fax/modem card to a PC) to perform an additional task. But we should not go overboard in being conservative, since an oversized cooling system will cost more, occupy more space, be heavier, and consume more power. For example, there is no need to install a large and noisy fan in an electronic system just to be “safe” when a smaller one will do. For the same reason, there is no need to use an expensive and failure-prone liquid cooling system when air cooling is adequate. We should always keep in mind that the most desirable form of cooling is natural convection cooling, since it does not require any moving parts, and thus it is inherently reliable, quiet, and, best of all, free. The cooling system of an electronic device must be designed considering the actual field operating conditions. In critical applications such as those in the military, the electronic device must undergo extensive testing to satisfy stringent requirements for safety and reliability. Several such codes exist to specify the minimum standards to be met in some applications. The duty cycle is another important consideration in the design and selection of a cooling technique. The actual power dissipated by a device can be 5W electrical energy in 5W Heat out FIGURE 15–11 In the absence of other energy interactions, the heat output of an electronic device in steady operation is equal to the power input to the device. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 794 794 HEAT TRANSFER T Steady operating temperature T(t) Transient operation Steady operation Environment temperature 0 Time, t FIGURE 15–12 The temperature change of an electronic component with time as it reaches steady operating temperature after it is turned on. much less than the rated power, depending on its duty cycle (the fraction of time it is on). A 5-W power transistor, for example, will dissipate an average of 2 W of power if it is active only 40 percent of the time. If the chip of this transistor is 1.5 mm wide, 1.5 mm high, and 0.1 mm thick, then the heat flux on the chip will be (2 W)/(0.15 cm)2 89 W/cm2. An electronic device that is not running is in thermal equilibrium with its surroundings, and thus is at the temperature of the surrounding medium. When the device is turned on, the temperature of the components and thus the device starts rising as a result of absorbing the heat generated. The temperature of the device stabilizes at some point when the heat generated equals the heat removed by the cooling mechanism. At this point, the device is said to have reached steady operating conditions. The warming-up period during which the component temperature rises is called the transient operation stage (Fig. 15–12). Another thermal factor that undermines the reliability of electronic devices is the thermal stresses caused by temperature cycling. In an experimental study (see Hilbert and Kube, Ref. 10), the failure rate of electronic devices subjected to deliberate temperature cycling of more than 20°C is observed to increase eightfold. Shock and vibration are other common causes of failure for electronic devices and should be considered in the design and manufacturing process for increased reliability. Most electronic devices operate for long periods of time, and thus their cooling mechanism is designed for steady operation. But electronic devices in some applications never run long enough to reach steady operation. In such cases, it may be sufficient to use a limited cooling technique, such as thermal storage for a short period, or not to use one at all. Transient operation can also be caused by large swings in the environmental conditions. A common cooling technique for transient operation is to use a double-wall construction for the enclosure of the electronic equipment, with the space between the walls filled with a wax with a suitable melting temperature. As the wax melts, it absorbs a large amount of heat and thus delays overheating of the electronic components considerably. During off periods, the wax solidifies by rejecting heat to the environment. 15–4 Warm air out Ventilation slits Cool air in FIGURE 15–13 Strategically located ventilation holes are adequate to cool low-power electronics such as a TV or VCR. ■ THERMAL ENVIRONMENT An important consideration in the selection of a cooling technique is the environment in which the electronic equipment is to operate. Simple ventilation holes on the case may be all we need for the cooling of low-power-density electronics such as a TV or a VCR in a room, and a fan may be adequate for the safe operation of a home computer (Fig. 15–13). But the thermal control of the electronics of an aircraft will challenge thermal designers, since the environmental conditions in this case will swing from one extreme to another in a matter of minutes. The expected duration of operation in a hostile environment is also an important consideration in the design process. The thermal design of the electronics for an aircraft that cruises for hours each time it takes off will be quite different than that of a missile that has an operation time of a few minutes. The thermal environment in marine applications is relatively stable, since the ultimate heat sink in this case is water with a temperature range of 0°C to 30°C. For ground applications, however, the ultimate heat sink is the atmospheric air, whose temperature varies from 50°C at polar regions to 50°C cen58933_ch15.qxd 9/9/2002 10:20 AM Page 795 795 CHAPTER 15 in desert climates, and whose pressure ranges from about 70 kPa (0.7 atm) at 3000 m elevation to 107 kPa (1.08 atm) at 500 m below sea level. The combined convection and radiation heat transfer coefficient can range from 10 W/m2 · °C in calm weather to 80 W/m2 · °C in 100 km/h (62 mph) winds. Also, the surfaces of the devices facing the sun directly can be subjected to solar radiation heat flux of 1000 W/m2 on a clear day. In airborne applications, the thermal environment can change from 1 atm and 35°C on the ground to 19 kPa (0.2 atm) and 60°C at a typical cruising altitude of 12,000 m in minutes (Fig. 15–14). At supersonic velocities, the surface temperature of some part of the aircraft may rise 200°C above the environment temperature. Electronic devices are rarely exposed to uncontrolled environmental conditions directly because of the wide variations in the environmental variables. Instead, a conditioned fluid such as air, water, or a dielectric fluid is used to serve as a local heat sink and as an intermediary between the electronic equipment and the environment, just like the air-conditioned air in a building providing thermal comfort to the human body. Conditioned air is the preferred cooling medium, since it is benign, readily available, and not prone to leakage. But its use is limited to equipment with low power densities, because of the low thermal conductivity of air. The thermal design of electronic equipment in military applications must comply with strict military standards in order to satisfy the utmost reliability requirements. 15–5 ■ ELECTRONICS COOLING IN DIFFERENT APPLICATIONS The cooling techniques used in the cooling of electronic equipment vary widely, depending on the particular application. Electronic equipment designed for airborne applications such as airplanes, satellites, space vehicles, and missiles offers challenges to designers because it must fit into odd-shaped spaces because of the curved shape of the bodies, yet be able to provide adequate paths for the flow of fluid and heat. Most such electronic equipment are cooled by forced convection using pressurized air bled off a compressor. This compressed air is usually at a high temperature, and thus it is cooled first by expanding it through a turbine. The moisture in the air is also removed before the air is routed to the electronic boxes. But the removal process may not be adequate under rainy conditions. Therefore, electronics in some cases are placed in sealed finned boxes that are externally cooled to eliminate any direct contact with electronic components. The electronics of short-range missiles do not need any cooling because of their short cruising times (Fig. 15–15). The missiles reach their destinations before the electronics reach unsafe temperatures. Long-range missiles such as cruise missiles, however, may have a flight time of several hours. Therefore, they must utilize some form of cooling mechanism. The first thing that comes to mind is to use forced convection with the air that rams the missile by utilizing its large dynamic pressure. However, the dynamic temperature of air, which is the rise in the temperature of the air as a result of the ramming effect, may be more than 50°C at speeds close to the speed of sound (Fig. 15–16). For example, at a speed of 320 m/s, the dynamic temperature of air is –273°C 0 atm 25°C 1 atm Ground FIGURE 15–14 The thermal environment of a spacecraft changes drastically in a short time, and this complicates the thermal control of the electronics. Missile FIGURE 15–15 The electronics of short-range missiles may not need any cooling because of the short flight time involved. Temperature rise during stagnation =0 Air 100 m/s 305 K 300 K Stationary object FIGURE 15–16 The temperature of a gas having a specific heat Cp flowing at a velocity of rises by 2/2Cp when it is brought to a complete stop. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 796 796 HEAT TRANSFER Tdynamic 1 J/kg (320 m/s)2 2 51°C 2Cp 2(1005 J/kg · °C) 1 m2/s2 (15-2) Therefore, the temperature of air at a velocity of 320 m/s and a temperature of 30°C will rise to 81°C as a result of the conversion of kinetic energy to internal energy. Air at such high temperatures is not suitable for use as a cooling medium. Instead, cruise missiles are often cooled by taking advantage of the cooling capacity of the large quantities of liquid fuel they carry. The electronics in this case are cooled by passing the fuel through the cold plate of the electronic enclosure as it flows toward the combustion chamber. Electronic equipment in space vehicles is usually cooled by a liquid circulated through the components, where heat is picked up, and then through a space radiator, where the waste heat is radiated into deep space at about 0 K. Note that radiation is the only heat transfer mechanism for rejecting heat to the vacuum environment of space, and radiation exchange depends strongly on surface properties. Desirable radiation properties on surfaces can be obtained by special coatings and surface treatments. When electronics in sealed boxes are cooled by a liquid flowing through the outer surface of the electronics box, it is important to run a fan in the box to circulate the air, since there are no natural convection currents in space because of the absence of a gravity field. Electronic equipment in ships and submarines is usually housed in rugged cabinets to protect it from vibrations and shock during stormy weather. Because of easy access to water, water-cooled heat exchangers are commonly used to cool shipboard electronics. This is usually done by cooling air in a closed- or open-loop air-to-water heat exchanger and forcing the cool air to the electronic cabinet by a fan. When forced-air cooling is used, it is important to establish a flow path for air such that no trapped hot-air pockets will be formed in the cabinets. Communication systems located at remote locations offer challenges to thermal designers because of the extreme conditions under which they operate. These electronic systems operate for long periods of time under adverse conditions such as rain, snow, high winds, solar radiation, high altitude, high humidity, and extremely high or low temperatures. Large communication systems are housed in specially built shelters. Sometimes it is necessary to aircondition these shelters to safely dissipate the large quantities of heat dissipated by the electronics of communication systems. Electronic components used in high-power microwave equipment such as radars generate enormous amounts of heat because of the low conversion efficiency of electrical energy to microwave energy. Klystron tubes of highpower radar systems where radio-frequency energy is generated can yield local heat fluxes as high as 2000 W/cm2, which is close to one-third of the heat flux on the sun’s surface. The safe and reliable dissipation of these high heat fluxes usually requires the immersion of such equipment in a suitable dielectric fluid that can remove large quantities of heat by boiling. The manufacturers of electronic devices usually specify the rate of heat dissipation and the maximum allowable component temperature for reliable operation. These two numbers help us determine the cooling techniques that are suitable for the device under consideration. The heat fluxes attainable at specified temperature differences are plotted in Figure 15–17 for some common heat transfer mechanisms. When the power cen58933_ch15.qxd 9/9/2002 10:20 AM Page 797 797 CHAPTER 15 DI AT IO N 103 8 6 ED RC I CT A EC T IR EC E V D IR N 4 N 0.02 0.04 R, W AT E IM 1 0.01 L IO M 2 RA FO S ER U AT ,N G LIN BOI NS ION BO ERS CAR IMM UORO FL RC 10 8 6 CO ED 2 N O IR T ,F A O IR 4 S N O B R CA RO O U FL CO N V EC TI ON CO ,N AT U RA L 102 8 6 D Temperature difference, °C 2 CO N V EC TI O N N V EC TI O N + RA 4 0.1 0.2 0.4 1 Surface heat flux, W/cm2 2 3 4 6 8 10 20 rating of a device or component is given, the heat flux is determined by dividing the power rating by the exposed surface area of the device or component. Then suitable heat transfer mechanisms can be determined from Figure 15–17 from the requirement that the temperature difference between the surface of the device and the surrounding medium not exceed the allowable maximum value. For example, a heat flux of 0.5 W/cm2 for an electronic component would result in a temperature difference of about 500°C between the component surface and the surrounding air if natural convection in air is used. Considering that the maximum allowable temperature difference is typically under 80°C, the natural convection cooling of this component in air is out of the question. But forced convection with air is a viable option if using a fan is acceptable. Note that at heat fluxes greater than 1 W/cm2, even forced convection with air will be inadequate, and we must use a sufficiently large heat sink or switch to a different cooling fluid such as water. Forced convection with water can be used effectively for cooling electronic components with high heat fluxes. Also note that dielectric liquids such as fluorochemicals can remove high heat fluxes by immersing the component directly in them. 15–6 ■ CONDUCTION COOLING Heat is generated in electronic components whenever electric current flows through them. The generated heat causes the temperature of the components to rise, and the resulting temperature difference drives the heat away from the components through a path of least thermal resistance. The temperature of the components stabilizes when the heat dissipated equals the heat generated. In order to minimize the temperature rise of the components, effective heat transfer FIGURE 15–17 Heat fluxes that can be attained at specified temperature differences with various heat transfer mechanisms and fluids (from Kraus and Bar-Cohen, Ref. 14, p. 22; reproduced with permission). cen58933_ch15.qxd 9/9/2002 10:20 AM Page 798 798 HEAT TRANSFER A . Q T1 T1 T2 paths must be established between the components and the ultimate heat sink, which is usually the atmospheric air. The selection of a cooling mechanism for electronic equipment depends on the magnitude of the heat generated, reliability requirements, environmental conditions, and cost. For low-cost electronic equipment, inexpensive cooling mechanisms such as natural or forced convection with air as the cooling medium are commonly used. For high-cost, high-performance electronic equipment, however, it is often necessary to resort to expensive and complicated cooling techniques. Conduction cooling is based on the diffusion of heat through a solid, liquid, or gas as a result of molecular interactions in the absence of any bulk fluid motion. Steady one-dimensional heat conduction through a plane medium of thickness L, heat transfer surface area A, and thermal conductivity k is given by (Fig. 15–18). · T T Q kA L R x 0 Q. L T2 L R = —– kA FIGURE 15–18 The thermal resistance of a medium is proportional to its length in the direction of heat transfer, and inversely proportional to its heat transfer surface area and thermal conductivity. (W) (15-3) where R Length L kA Thermal conductivity Heat transfer area (15-4) is the thermal resistance of the medium and T is the temperature difference across the medium. Note that this is analogous to the electric current being equal to the potential difference divided by the electrical resistance. The thermal resistance concept enables us to solve heat transfer problems in an analogous manner to electric circuit problems using the thermal resistance · network, as discussed in Chapter 3. When the rate of heat conduction Q is known, the temperature drop along a medium whose thermal resistance is R is simply determined from · T Q R (°C) (15-5) Therefore, the greatest temperature drops along the path of heat conduction will occur across portions of the heat flow path with the largest thermal resistances. Conduction in Chip Carriers The conduction analysis of an electronic device starts with the circuitry or junction of a chip, which is the site of heat generation. In order to understand the heat transfer mechanisms at the chip level, consider the DIP (dual in-line package) type chip carrier shown in Figure 15–19. The heat generated at the junction spreads throughout the chip and is conducted across the thickness of the chip. The spread of heat from the junction to the body of the chip is three-dimensional in nature, but can be approximated as one-dimensional by adding a constriction thermal resistance to the thermal resistance network. For a small heat generation area of diameter d on a considerably larger body, the constriction resistance is given by Rconstriction 1 dk (°C/W) where k is the thermal conductivity of the larger body. (15-6) cen58933_ch15.qxd 9/9/2002 10:20 AM Page 799 799 CHAPTER 15 Junction Air gap Bond wires Lid Case Chip Lead frame Leads Bond The chip is attached to the lead frame with a highly conductive bonding material that provides a low-resistance path for heat flow from the chip to the lead frame. There is no metal connection between the lead frame and the leads, since this would short-circuit the entire chip. Therefore, heat flow from the lead frame to the leads is through the dielectric case material such as plastic or ceramic. Heat is then transported outside the electronic device through the leads. When solving a heat transfer problem, it is often necessary to make some simplifying assumptions regarding the primary heat flow path and the magnitudes of heat transfer in other directions (Fig. 15–20). In the chip carrier discussed above, for example, heat transfer through the top is disregarded since it is very small because of the large thermal resistance of the stagnant air space between the chip and the lid. Heat transfer from the base of the electronic device is also considered to be negligible because of the low thermal conductivity of the case material and the lack of effective convection on the base surface. EXAMPLE 15–3 Analysis of Heat Conduction in a Chip A chip is dissipating 0.6 W of power in a DIP with 12 pin leads. The materials and the dimensions of various sections of this electronic device are as given in the table below. If the temperature of the leads is 40°C, estimate the temperature at the junction of the chip. Section and Material Junction constriction Silicon chip Eutectic bond Copper lead frame Plastic separator Copper leads Thermal Conductivity, W/m · °C Thickness, mm Heat Transfer Surface Area — 120† 296 386 1 386 — 0.4 0.03 0.25 0.2 5 diameter 0.4 mm 3 mm 3 mm 3 mm 3 mm 3 mm 3 mm 12 1 mm 0.25 mm 12 1 mm 0.25 mm † The thermal conductivity of silicon varies greatly with temperature from 153.5 W/m · °C at 27°C to 113.7 W/m · °C at 100°C, and the value 120 W/m · °C reflects the anticipation that the temperature of the silicon chip will be close to 100°C. FIGURE 15–19 The schematic for the internal geometry and the cross-sectional view of a DIP (dual in-line package) type electronic device with 14 leads. Junction Air gap Bond wires Lid Case Chip Lead frame Leads Bond FIGURE 15–20 Heat generated at the junction of an electronic device flows through the path of least resistance. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 800 800 HEAT TRANSFER SOLUTION The dimensions and power dissipation of a chip are given. The junction temperature of the chip is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through various components is one-dimensional. 3 Heat transfer through the air gap and the lid on top of the chip is negligible because of the very large thermal resistance involved along this path. Analysis The geometry of the device is as shown in Figure 15–20. We take the primary heat flow path to be the chip, the eutectic bond, the lead frame, the plastic insulator, and the 12 leads. When the constriction resistance between the junction and the chip is considered, the thermal resistance network for this problem becomes as shown in Figure 15–21. The various thermal resistances on the path of primary heat flow are determined as follows. Junction Rconstriction 1 1 5.88°C/ W 2 dk (0.4 10 3 m)(120 W/m· °C) L 0.4 10 3 m 0.37°C/ W Rchip kA chip (120 W/m · °C)(9 10 6 m2) Rconstriction Rchip Rbond Rbond KA L bond Rlead kA L kA Rlead frame frame 0.03 10 3 m 0.01°C/ W (296 W/m · °C)(9 10 6 m2) L lead frame Rplastic Rplastic plastic Rleads FIGURE 15–21 Thermal resistance network for the electronic device considered in Example 15–3. L Rleads kA leads 0.25 10 3 m 0.07°C/ W (386 W/m · °C)(9 10 6 m2) 0.2 10 3 m 66.67°C/ W (1 W/m · °C)(12 0.25 10 6 m2) 5 10 3 m 4.32°C/ W (386 W/m · °C)(12 0.25 10 6 m2) Note that for heat transfer purposes, all 12 leads can be considered as a single lead whose cross-sectional area is 12 times as large. The alternative is to find the resistance of a single lead and to calculate the equivalent resistance for 12 such resistances connected in parallel. Both approaches give the same result. All the resistances determined here are in series. Thus the total thermal resistance between the junction and the leads is determined by simply adding them up: Rtotal Rjunction-lead Rconstriction Rchip Rbond Rlead frame Rplastic Rleads (5.88 0.37 0.01 0.07 66.67 4.32)°C/ W 77.32°C/ W Heat transfer through the chip can be expressed as T · Q R junction-leads Tjunction Tleads Rjunction-leads Solving for Tjunction and substituting the given values, the junction temperature is determined to be · Tjunction Tleads Q Rjunction-leads 40°C (0.6 W)(77.32°C/ W) 86.4°C Note that the plastic layer between the lead frame and the leads accounts for 66.67/77.32 86 percent of the total thermal resistance and thus the 86 cen58933_ch15.qxd 9/9/2002 10:20 AM Page 801 801 CHAPTER 15 percent of the temperature drop (0.6 66.67 40°C) between the junction and the leads. In other words, the temperature of the junction would be just 86.5 40 46.5°C if the thermal resistance of the plastic was eliminated. Discussion The simplified analysis given here points out that any attempt to reduce the thermal resistance in the chip carrier and thus improve the heat flow path should start with the plastic layer. We also notice from the magnitudes of individual resistances that some sections, such as the eutectic bond and the lead frame, have negligible thermal resistances, and any attempt to improve them further will have practically no effect on the junction temperature of the chip. The analytical determination of the junction-to-case thermal resistance of an electronic device can be rather complicated and can involve considerable uncertainty, as already shown. Therefore, the manufacturers of electronic devices usually determine this value experimentally and list it as part of their product description. When the thermal resistance is known, the temperature difference between the junction and the outer surface of the device can be determined from · Tjunction-case Tjunction Tcase Q Rjunction-case (°C) (15-7) · where Q is the power consumed by the device. The determination of the actual junction temperature depends on the ambient temperature Tambient as well as the thermal resistance Rcase-ambient between the case and the ambient (Fig. 15–22). The magnitude of this resistance depends on the type of ambient (such as air or water) and the fluid velocity. The two thermal resistances discussed above are in series, and the total resistance between the junction and the ambient is determined by simply adding them up: Rtotal Rjunction-ambient Rjunction-case Rcase-ambient (°C/W) (°C) Junction Rcase-ambient . Q Case Rjunction-case (15-8) Many manufacturers of electronic devices go the extra step and list the total resistance between the junction and the ambient for various chip configurations and ambient conditions likely to be encountered. Once the total thermal resistance is available, the junction temperature corresponding to the specified · power consumption (or heat dissipation rate) of Q is determined from · Tjunction Tambient Q Rjunction-ambient Ambient (15-9) A typical chart for the total junction-to-ambient thermal resistance for a single DIP-type electronic device mounted on a circuit board is given in Figure 15–23 for various air velocities and lead numbers. The values at the intersections of the curves and the vertical axis represent the thermal resistances corresponding to natural convection conditions (zero air velocity). Note that the thermal resistance and thus the junction temperature decrease with increasing air velocity and the number of leads extending from the electronic device, as expected. Rtotal = Rjunction-case + Rcase-ambient . Tjunction = Tambient + QRtotal FIGURE 15–22 The junction temperature of a chip depends on the external case-toambient thermal resistance as well as the internal junction-to-case resistance. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 802 802 HEAT TRANSFER 200 FIGURE 15–23 Total thermal resistance between the junction of a plastic DIP device mounted on a circuit board and the ambient air as a function of the air velocity and the number of leads (courtesy of Motorola Semiconductor Products, Inc.). Average thermal resistance, °C/ W 180 z–axis (transverse) 160 140 Airflow direction 120 100 8–lead 80 14–lead 60 16–lead 24–lead 40 20 0 50 EXAMPLE 15–4 Air 30°C 1.2 W FIGURE 15–24 Schematic for Example 15–4. 100 150 200 Airflow velocity, m/min 250 300 Predicting the Junction Temperature of a Device A fan blows air at 30°C and a velocity of 200 m/min over a 1.2-W plastic DIP with 16 leads mounted on a PCB, as shown in Figure 15–24. Using data from Figure 15–23, determine the junction temperature of the electronic device. What would the junction temperature be if the fan were to fail? SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined. Assumptions Steady operating conditions exist. Analysis The junction-to-ambient thermal resistance of the device with 16 leads corresponding to an air velocity of 200 m/min is determined from Figure 15–23 to be Rjunction-ambient 55°C/ W Then the junction temperature can be determined from Eq. 15-9 to be · Tjunction Tambient Q Rjunction-ambient 30°C (1.2 W)(55°C/ W) 96°C When the fan fails, the airflow velocity over the device will be zero. The total thermal resistance in this case is determined from the same chart by reading the value at the intersection of the curve and the vertical axis to be Rjunction-ambient 70°C/ W which gives · Tjunction Tambient Q Rjunction-ambient 30°C (1.2 W)(70°C/ W) 114°C Discussion Note that the temperature of the junction will rise by 18°C when the fan fails. Of course, this analysis assumes the temperature of the surrounding air still to be 30°C, which may no longer be the case. Any increase in the ambient temperature as a result of inadequate airflow will reflect on the junction temperature, which will seriously jeopardize the safety of the electronic device. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 803 803 CHAPTER 15 Conduction in Printed Circuit Boards Heat-generating electronic devices are commonly mounted on thin rectangular boards, usually 10 cm 15 cm in size, made of electrically insulating materials such as glass–epoxy laminates, which are also poor conductors of heat. The resulting printed circuit boards are usually cooled by blowing air or passing a dielectric liquid through them. In such cases, the components on the PCBs are cooled directly, and we are not concerned about heat conduction along the PCBs. But in some critical applications such as those encountered in the military, the PCBs are contained in sealed enclosures, and the boards provide the only effective heat path between the components and the heat sink attached to the sealed enclosure. In such cases, heat transfer from the side faces of the PCBs is negligible, and the heat generated in the components must be conducted along the PCB toward its edges, which are clamped to cold plates for removing the heat externally. Heat transfer along a PCB is complicated in nature because of the multidimensional effects and nonuniform heat generation on the surfaces. We can still obtain sufficiently accurate results by using the thermal resistance network in one or more dimensions. Copper or aluminum cladding, heat frames, or cores are commonly used to enhance heat conduction along the PCBs. The thickness of the copper cladding on the PCB is usually expressed in terms of ounces of copper, which is the thickness of 1-ft2 copper sheet made of one ounce of copper. An ounce of copper is equivalent to 0.03556-mm (1.4-mil) thickness of a copper layer. When analyzing heat conduction along a PCB with copper (or aluminum) cladding on one or both sides, often the question arises whether heat transfer along the epoxy laminate can be ignored relative to that along the copper layer, since the thermal conductivity of copper is about 1500 times that of epoxy. The answer depends on the relative cross-sectional areas of each layer, since heat conduction is proportional to the cross-sectional area as well as the thermal conductivity. Consider a copper-cladded PCB of width w and length L, across which the temperature difference is T, as shown in Figure 15–25. Assuming heat conduction is along the length L only and heat conduction in other dimensions is negligible, the rate of heat conduction along this PCB is the sum of the heat conduction along the epoxy board and the copper layer and is expressed as · · · T Q PCB Q epoxy Q copper k A L [(kA)epoxy (kA)copper] [(kt)epoxy (kt)copper] T L epoxy kA T L w Epoxy L PCB . copper Copper tepoxy tcopper Q (15-10) wT L where t denotes the thickness. Therefore, the relative magnitudes of heat conduction along the two layers depend on the relative magnitudes of the thermal conductivity–thickness product kt of the layer. Therefore, if the kt product of the copper is 100 times that of epoxy, then neglecting heat conduction along the epoxy board will involve an error of just 1 percent, which is negligible. We can also define an effective thermal conductivity for metal-cladded PCBs as tPCB = tepoxy + tcopper FIGURE 15–25 Schematic of a copper-cladded epoxy board and heat conduction along it. cen58933_ch15.qxd 9/9/2002 10:20 AM Page 804 804 HEAT TRANSFER keff (kt)epoxy (kt)copper tepoxy tcopper (W/m · °C) (15-11) so that the rate of heat conduction along the PCB can be expressed as wT · T Q PCB keff(tepoxy tcopper) keff APCB L L (W) (15-12) where APCB w(tepoxy tcopper) is the area normal to the direction of heat transfer. When there are holes or discontinuities along the copper cladding, the above analysis needs to be modified to account for their effect. EXAMPLE 15–5 Heat is to be conducted along a PCB with copper cladding on one side. The PCB is 10 cm long and 10 cm wide, and the thickness of the copper and epoxy layers are 0.04 mm and 0.16 mm, respectively, as shown in Figure 15–26. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the copper (k 386 W/m · °C) and epoxy (k 0.26 W/m · °C) layers. Also, determine the effective thermal conductivity of the PCB. 10 cm 10 cm PCB . Q Heat Conduction along a PCB with Copper Cladding Epoxy Copper 0.16 mm 0.04 mm FIGURE 15–26 Schematic for Example 15–5. SOLUTION A PCB with copper cladding is given. The percentages of heat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the PCB are to be determined. Assumptions 1 Heat conduction along the PCB is one-dimensional since heat transfer from side surfaces is negligible. 2 The thermal properties of epoxy and copper layers are constant. Analysis The length and width of both layers are the same, and so is the temperature difference across each layer. Heat conduction along a layer is proportional to the thermal conductivity–thickness product kt, which is determined for each layer and the entire PCB to be (kt )copper (386 W/m · °C)(0.04 103 m) 15.44 103 W/°C (kt )epoxy (0.26 W/m · °C)(0.16 103 m) 0.04 103 W/°C (kt )PCB (kt )copper (kt )epoxy (15.44 0.04) 103 m 15.48 103 W/°C Therefore, heat conduction along the epoxy board will constitute f (kt )epoxy 0.04 10 3 W/°C 0.0026 (kt )PCB 15.48 10 3 W/°C or 0.26 percent of the thermal conduction along the PCB, which is negligible. Therefore, heat conduction along the epoxy layer in this case can be disregarded without any reservations. The effective thermal conductivity of the board is determined from Eq. 15-11 to be keff (kt )epoxy (kt )copper (15.44 0.04) 10 3 W/°C 77.4 W/m · °C tepoxy tcopper (0.16 0.04) 10 3 m That is, the entire PCB can be treated as a 0.20-mm-thick single homogeneous layer whose thermal conductivity is 77.4 W/m · °C for heat transfer along its length. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 805 805 CHAPTER 15 Temperature distribution Discussion Note that a very thin layer of copper cladding on a PCB improves heat conduction along the PCB drastically, and thus it is commonly used in conduction-cooled electronic devices. Electronic components Heat Frames In applications where direct cooling of circuit boards by passing air or a dielectric liquid over the electronic components is not allowed, and the junction temperatures are to be maintained relatively low to meet strict safety requirements, a thick heat frame is used instead of a thin layer of copper cladding. This is especially the case for multilayer PCBs that are packed with highpower output chips. The schematic of a PCB that is conduction-cooled via a heat frame is shown in Figure 15–27. Heat generated in the chips is conducted through the circuit board, through the epoxy adhesive, to the center of the heat frame, along the heat frame, and to a heat sink or cold plate, where heat is externally removed. The heat frame provides a low-resistance path for the flow of heat from the circuit board to the heat sink. The thicker the heat frame, the lower the thermal resistance, and thus the smaller the temperature difference between the center and the ends of the heat frame. When the heat load is evenly distributed on the PCB, there will be thermal symmetry about the centerline, and the temperature distribution along the heat frame and the PCB will be parabolic in nature, with the chips in the middle of the PCB (farthest away from the edges) operating at the highest temperatures and the chips near the edges operating at the lowest temperatures. Also, when the PCB is cooled from two edges, heat generated in the left half of the PCB will flow toward the left edge and heat generated in the right half will flow toward the right edge of the heat frame. But when the PCB is cooled from all four edges, the heat transfer along the heat frame as well as the resistance network will be twodimensional. When a heat frame is used, heat conduction in the epoxy layer of the PCB is through its thickness instead of along its length. The epoxy layer in this case offers a much smaller resistance to heat flow because of the short distance involved. This resistance can be made even smaller by drilling holes in the epoxy and filling them with copper, as shown in Figure 15–28. These copper fillings are usually 1 mm in diameter and their centers are a few millimeters apart. Such highly conductive fillings provide easy passageways for heat from one side of the PCB to the other and result in considerable reduction in the thermal resistance of the board along its thickness, as shown in Examples 15–6, 15-7, and 15–8. Cold plate FIGURE 15–27 Conduction cooling of a printed circuit board with a heat frame, and the typical temperature distribution along the frame. Epoxy board Copper fillings FIGURE 15–28 Planting the epoxy board with copper fillings decreases the thermal resistance across its thickness considerably. (b) 10 cm Epoxy board Thermal Resistance of an Epoxy Glass Board Consider a 10-cm 15-cm glass–epoxy laminate (k 0.26 W/m · °C) whose thickness is 0.8 mm, as shown in Figure 15–29. Determine the thermal resistance of this epoxy layer for heat flow (a) along the 15-cm-long side and (b) across its thickness. Heat frame Epoxy adhesive 0.8 mm EXAMPLE 15–6 PCB 15 cm (a) FIGURE 15–29 Schematic for Example 15–6. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 806 806 HEAT TRANSFER SOLUTION The dimensions of an epoxy–glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined. Assumptions 1 Heat conduction in the laminate is one-dimensional in either case. 2 Thermal properties of the laminate are constant. Analysis The thermal resistance of a plane parallel medium in the direction of heat conduction is given by R L kA where L is the length in the direction of heat flow, k is the thermal conductivity, and A is the area normal to the direction of heat conduction. Substituting the given values, the thermal resistances of the board for both cases are determined to be (a) Ralong length kA L along length (b) Racross thickness 0.15 m 7212°C/ W (0.26 W/m · °C)(0.1 m)(0.8 10 3 m) kA L across thickness 0.8 10 3 m 0.21°C/ W (0.26 W/m · °C)(0.1 m)(0.15 m) Discussion Note that heat conduction at a rate of 1 W along this PCB would cause a temperature difference of 7212°C across a length of 15 cm. But the same rate of heat conduction would cause a temperature difference of only 0.21°C across the thickness of the epoxy board. 1 mm 2.5 mm EXAMPLE 15–7 Planting Cylindrical Copper Fillings in an Epoxy Board Reconsider the 10-cm 15-cm glass–epoxy laminate (k 0.26 W/m · °C) of thickness 0.8 mm discussed in Example 15–6. In order to reduce the thermal resistance across its thickness from the current value of 0.21°C/ W, cylindrical copper fillings (k 386 W/m · °C) of 1-mm diameter are to be planted throughout the board with a center-to-center distance of 2.5 mm, as shown in Figure 15–30. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification. Copper filling Epoxy board FIGURE 15–30 Schematic for Example 15–7. SOLUTION Cylindrical copper fillings are planted throughout an epoxy glass board. The thermal resistance of the board across its thickness is to be determined. Assumptions 1 Heat conduction along the board is one-dimensional. 2 Thermal properties of the board are constant. Analysis Heat flow through the thickness of the board in this case will take place partly through the copper fillings and partly through the epoxy in parallel paths. The thickness of both materials is the same and is given to be 0.8 mm. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 807 807 CHAPTER 15 But we also need to know the surface area of each material before we can determine the thermal resistances. It is stated that the distance between the centers of the copper fillings is 2.5 mm. That is, there is only one 1-mm-diameter copper filling in every 2.5-mm 2.5-mm square section of the board. The number of such squares and thus the number of copper fillings on the board are n (100 mm)(150 mm) Area of the board 2400 Area of one square (2.5 mm)(2.5 mm) Then the surface areas of the copper fillings and the remaining epoxy layer become (1 10 3 m)2 D 2 (2400) 0.001885 m2 4 4 Atotal (Length)(Width) (0.1 m)(0.15 m) 0.015 m2 Aepoxy Atotal Acopper (0.015 0.001885) m2 0.013115 m2 Acopper n The thermal resistance of each material is Repoxy 0.8 10 3 m 0.0011°C/ W (386 W/m · °C)(0.001885 m2) copper 0.8 10 3 m 0.2346°C/ W (0.26 W/m · °C)(0.013115 m2) epoxy kA L kA Rcopper L Noting that these two resistances are in parallel, the equivalent thermal resistance of the entire board is determined from 1 1 1 1 1 Rboard Rcopper Repoxy 0.0011°C/ W 0.2346°C/ W which gives Rboard 0.00109°C/ W Discussion Note that the thermal resistance of the epoxy board has dropped from 0.21°C/ W by a factor of almost 200 to just 0.00109°C/ W as a result of implanting 1-mm-diameter copper fillings into it. Therefore, implanting copper pins into the epoxy laminate has virtually eliminated the thermal resistance of the epoxy across its thickness. EXAMPLE 15–8 Conduction Cooling of PCBs by a Heat Frame A 10-cm 12-cm circuit board dissipating 24 W of heat is to be conductioncooled by a 1.2-mm-thick copper heat frame (k 386 W/m · °C) 10 cm 14 cm in size. The epoxy laminate (k 0.26 W/m · °C) has a thickness of 0.8 mm and is attached to the heat frame with conductive epoxy adhesive (k 1.8 W/m · °C) of 0.13-mm thickness, as shown in Figure 15–31. The PCB is attached to a heat sink by clamping a 5-mm-wide portion of the edge to the heat sink from both ends. The temperature of the heat frame at this point is 20°C. Heat is uniformly generated on the PCB at a rate of 2 W per 1-cm 10-cm strip. Considering only one-half of the PCB board because of symmetry, determine the cen58933_ch15.qxd 9/9/2002 10:21 AM Page 808 1 cm 1 cm 1 cm 1 cm 1 cm Symmetry plane 2W 2W 2W 2W 2W 2W cm T7 Repoxy Epoxy adhesive 10 Epoxy board 1 cm 0 0.8 5 .13 mm m m m m 5 m m 1. 2 m m 808 HEAT TRANSFER R Rcopper I Heat frame 4W T5 T6 6W T4 8W T3 10 W T2 Clamp area 12 W T1 T0 = 20°C Rcopper II FIGURE 15–31 The schematic and thermal resistance network for Example 15–8. maximum temperature on the PCB and the temperature distribution along the heat frame. SOLUTION A circuit board with uniform heat generation is to be conductioncooled by a copper heat frame. Temperature distribution along the heat frame and the maximum temperature in the PCB are to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties are constant. 3 There is no direct heat dissipation from the surface of the PCB, and thus all the heat generated is conducted by the heat frame to the heat sink. Analysis The PCB under consideration possesses thermal symmetry about the centerline. Therefore, the heat generated on the left half of the PCB is conducted to the left heat sink, and the heat generated on the right half is conducted to the right heat sink. Thus we need to consider only half of the board in the analysis. The maximum temperature will occur at a location furthest away from the heat sinks, which is the symmetry line. Therefore, the temperature of the electronic components located at the center of the PCB will be the highest, and their reliability will be the lowest. Heat generated in the components on each strip is conducted through the epoxy layer underneath. Heat is then conducted across the epoxy adhesive and to the middle of the copper heat frame. Finally, heat is conducted along the heat frame to the heat sink. The thermal resistance network associated with heat flow in the right half of the PCB is also shown in Figure 15–31. Note that all vertical resistances are identical and are equal to the sum of the three resistances in series. Also note that heat conduction toward the heat sink is assumed to be predominantly along the heat frame, and conduction along the epoxy adhesive is considered to be negligible. This assumption is quite reasonable, since the conductivity– thickness product of the heat frame is much larger than those of the other two layers. The properties and dimensions of various sections of the PCB are summarized in this table. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 809 809 CHAPTER 15 Section and Material Epoxy board Epoxy adhesive Copper heat frame, (normal to frame) Copper heat frame, (along the frame) Thermal Conductivity, W/m · °C Thickness, mm 0.26 1.8 0.8 0.13 10 mm 100 mm 10 mm 100 mm 0.6 10 mm 100 mm 386 386 10 Heat Transfer Surface area 1.2 mm 100 mm Using the values in the table, the various thermal resistances are determined to be Radhesive Rcopper, 0.8 10 3 m 3.077°C/ W (0.26 W/m · °C)(0.01 m 0.1 m) epoxy 3 0.13 10 m 0.072°C/ W (1.8 W/m · °C)(0.01 m 0.1 m) adhesive 0.6 10 3 m 0.002°C/ W copper, _ (386 W/m · °C)(0.01 m 0.1 m) L kA L kA Repoxy L kA Rframe Rcopper, kA L copper, 0.01 m (386 W/m · °C)(0.0012 m 0.1 m) 0.216°C/ W The combined resistance between the electronic components on each strip and the heat frame can be determined, by adding the three resistances in series, to be Rvertical Repoxy Radhesive Rcopper, (3.077 0.072 0.002)°C/ W 3.151°C/ W The various temperatures along the heat frame can be determined from the relation · T Thigh Tlow Q R · where R is the thermal resistance between two specified points, Q is the heat transfer rate through that resistance, and T is the temperature difference across that resistance. The temperature at the location where the heat frame is clamped to the heat sink is given as T0 20°C. Noting that the entire 12 W of heat generated on the right half of the PCB must pass through the last thermal resistance adjacent to the heat sink, the temperature T1 can be determined from · T1 T0 Q 1–0 R1–0 20°C (12 W)(0.216°C/ W) 22.59°C Following the same line of reasoning, the temperatures at specified locations along the heat frame are determined to be · T2 T1 Q 2–1 R2–1 22.59°C (10 W)(0.216°C/ W) 24.75°C · T3 T2 Q 3–2 R3–2 24.75°C (8 W)(0.216°C/ W) 26.48°C · T4 T3 Q 4–3 R4–3 26.48°C (6 W)(0.216°C/ W) 27.78°C · T5 T4 Q 5–4 R5–4 27.78°C (4 W)(0.216°C/ W) 28.64°C · T6 T5 Q 6–5 R6–5 28.64°C (2 W)(0.216°C/ W) 29.07°C cen58933_ch15.qxd 9/9/2002 10:21 AM Page 810 810 HEAT TRANSFER Finally, T7, which is the maximum temperature on the PCB, is determined from · T7 T6 Q vertical Rvertical 29.07°C (2 W)(3.151°C/ W) 35.37°C Discussion The maximum temperature difference between the PCB and the heat sink is only 15.37°C, which is very impressive considering that the PCB has no direct contact with the cooling medium. The junction temperatures in this case can be determined by calculating the temperature difference between the junction and the leads of the chip carrier at the point of contact to the PCB and adding 35.37°C to it. The maximum temperature rise of 15.37°C can be reduced, if necessary, by using a thicker heat frame. Metal core Epoxy lamina Electronic components Heat sink FIGURE 15–32 A two-sided printed circuit board with a metal core for conduction cooling. Conduction cooling can also be used when electronic components are mounted on both sides of the PCB by using a copper or aluminum core plate in the middle of the PCB, as shown in Figure 15–32. The heat load in this case will be twice that of a PCB that has components on one side only. Again, heat generated in the components will be conducted through the thickness of the epoxy layer to the metal core, which serves as a channel for effective heat removal. The thickness of the core is selected such that the maximum component temperatures remain below specified values to meet a prescribed reliability criterion. The thermal expansion coefficients of aluminum and copper are about twice as large as that of the glass–epoxy. This large difference in thermal expansion coefficients can cause warping on the PCBs if the epoxy and the metal are not bonded properly. One way of avoiding warping is to use PCBs with components on both sides, as discussed. Extreme care should be exercised during the bonding and curing process when components are mounted on only one side of the PCB. The Thermal Conduction Module (TCM) The heat flux for logic chips has been increasing steadily as a result of the increasing circuit density in the chips. For example, the peak flux at the chip level has increased from 2 W/cm2 on IBM System 370 to 20 W/cm2 on IBM System 3081, which was introduced in the early 1980s. The conventional forced-air cooling technique used in earlier machines was inadequate for removing such high heat fluxes, and it was necessary to develop a new and more effective cooling technique. The result was the thermal conduction module, shown in Figure 15–33. The TCM was different from previous chip packaging designs in that it incorporated both electrical and thermal considerations in early stages of chip design. Previously, a chip would be designed primarily by electrical designers, and the thermal designer would be told to come up with a cooling scheme for the chip. That approach resulted in unnecessarily high junction temperatures, and reduced reliability, since the thermal designer had no direct access to the chip. The TCM reflects a new philosophy in electronic packaging in that the thermal and electrical aspects are given equal treatment in the design process, and a successful thermal design starts at the chip level. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 811 811 CHAPTER 15 Cold plate Cold plate Cooling medium Rext Interposer Hat Rw–h Piston Spring Helium (or gas mixture) Rp–h Helium reservoir Piston C-ring seal Chip Ceramic substrate Rp–w Rt Spherical radius Chip Rgp Rc Rint Rc–p Substrate FIGURE 15–33 Cutaway view of the thermal conduction module (TCM), and the thermal resistance network between a single chip and the cooling fluid (courtesy of IBM Corporation). In the TCM, one side of the chip is reserved for electrical connections and the other side for heat rejection. The chip is cooled by direct contact to the cooling system to minimize the junction-to-case thermal resistance. The TCM houses 100 to 118 logic chips, which are bonded to a multilayer ceramic substrate 90 mm 90 mm in size with solder balls, which also provide the electrical connections between the chips and the substrate. Each chip dissipates about 4 W of power. The heat flow path from the chip to the metal casing is provided by a piston, which is pressed against the back surface of the chip by a spring. The tip of the piston is slightly curved to ensure good thermal contact even when the chip is tilted or misaligned. Heat conduction between the chip and the piston occurs primarily through the gas space between the chip and the piston because of the limited contact area between them. To maximize heat conduction through the gas, the air in the TCM cavity is evacuated and is replaced by helium gas, whose thermal conductivity is about six times that of air. Heat is then conducted through the piston, across the surrounding helium gas layer, through the module housing, and finally to the cooling water circulating through the cold plate attached to the top surface of the TCM. The total internal thermal resistance Rint of the TCM is about 8°C/W, which is rather impressive. This means that the temperature difference between the chip surface and the outer surface of the housing of the module will be only 24°C for a 3-W chip. The external thermal resistance Rext between the housing of the module and the cooling fluid is usually comparable in magnitude to Rint. Also, the thermal resistance between the junction and the surface of the chip can be taken to be 1°C/W. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 812 812 HEAT TRANSFER The compact design of the TCM significantly reduces the distance between the chips, and thus the signal transmission time between the chips. This, in turn, increases the operating speed of the electronic device. EXAMPLE 15–9 Cooling of Chips by the Thermal Conduction Module Consider a thermal conduction module with 100 chips, each dissipating 3 W of power. The module is cooled by water at 25°C flowing through the cold plate on top of the module. The thermal resistances in the path of heat flow are Rchip 1°C/ W between the junction and the surface of the chip, Rint 8°C/ W between the surface of the chip and the outer surface of the thermal conduction module, and Rext 6°C/ W between the outer surface of the module and the cooling water. Determine the junction temperature of the chip. Cold plate H2O Rext Rtotal Rjunction-water Rchip Rint Rext (1 8 6)°C/ W 15°C/ W Spring Module housing SOLUTION A thermal conduction module TCM with 100 chips is cooled by water. The junction temperature of the chip is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through various components is one-dimensional. Analysis Because of symmetry, we will consider only one of the chips in our analysis. The thermal resistance network for heat flow is given in Figure 15–34. Noting that all resistances are in series, the total thermal resistance between the junction and the cooling water is Rint Piston Noting that the total power dissipated by the chip is 3 W and the water temperature is 25°C, the junction temperature of the chip in steady operation can be determined from T · Q R Gas Chip Substrate Tjunction Twater Rjunction-water Solving for Tjunction and substituting the specified values gives · Tjunction Twater Q Rjunction-water 25°C (3 W)(15°C/ W) 70°C Rchip FIGURE 15–34 Thermal resistance network for Example 15–9. junction-water Therefore, the circuits of the chip will operate at about 70°C, which is considered to be a safe operating temperature for silicon chips. Cold plates are usually made of metal plates with fluid channels running through them, or copper tubes attached to them by brazing. Heat transferred to the cold plate is conducted to the tubes, and from the tubes to the fluid flowing through them. The heat carried away by the fluid is finally dissipated to the ambient in a heat exchanger. 15–7 ■ AIR COOLING: NATURAL CONVECTION AND RADIATION Low-power electronic systems are conveniently cooled by natural convection and radiation. Natural convection cooling is very desirable, since it does not involve any fans that may break down. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 813 813 CHAPTER 15 Natural convection is based on the fluid motion caused by the density differences in a fluid due to a temperature difference. A fluid expands when heated and becomes less dense. In a gravitational field, this lighter fluid rises and initiates a motion in the fluid called natural convection currents (Fig. 15–35). Natural convection cooling is most effective when the path of the fluid is relatively free of obstacles, which tend to slow down the fluid, and is least effective when the fluid has to pass through narrow flow passages and over many obstacles. The magnitude of the natural convection heat transfer between a surface and a fluid is directly related to the flow rate of the fluid. The higher the flow rate, the higher the heat transfer rate. In natural convection, no blowers are used and therefore the flow rate cannot be controlled externally. The flow rate in this case is established by the dynamic balance of buoyancy and friction. The larger the temperature difference between the fluid adjacent to a hot surface and the fluid away from it, the larger the buoyancy force, and the stronger the natural convection currents, and thus the higher the heat transfer rate. Also, whenever two bodies in contact move relative to each other, a friction force develops at the contact surface in the direction opposite to that of the motion. This opposing force slows down the fluid, and thus reduces the flow rate of the fluid. Under steady conditions, the airflow rate driven by buoyancy is established at the point where these two effects balance each other. The friction force increases as more and more solid surfaces are introduced, seriously disrupting the fluid flow and heat transfer. Electronic components or PCBs placed in enclosures such as a TV or VCR are cooled by natural convection by providing a sufficient number of vents on the case to enable the cool air to enter and the heated air to leave the case freely, as shown in Figure 15–36. From the heat transfer point of view, the vents should be as large as possible to minimize the flow resistance and should be located at the bottom of the case for air entering and at the top for air leaving. But equipment and human safety requirements dictate that the vents should be quite narrow to discourage unintended entry into the box. Also, concern about human habits such as putting a cup of coffee on the closest flat surface make it very risky to place vents on the top surface. The narrow clearance allowed under the case also offers resistance to airflow. Therefore, vents on the enclosures of natural convection–cooled electronic equipment are usually placed at the lower section of the side or back surfaces for air inlet and at the upper section of those surfaces for air exit. The heat transfer from a surface at temperature Ts to a fluid at temperature Tfluid by convection is expressed as · Q conv hconv AsT hconv As(Ts Tfluid) (W) (15-13) where hconv is the convection heat transfer coefficient and As is the heat transfer surface area. The value of hconv depends on the geometry of the surface and the type of fluid flow, among other things. Natural convection currents start out as laminar (smooth and orderly) and turn turbulent when the dimension of the body and the temperature difference between the hot surface and the fluid are large. For air, the flow remains laminar when the temperature differences involved are less than 100°C and the characteristic length of the body is less than 0.5 m, which is almost always the Warm air rising AIR MOTION Heated air (light) Velocity profile Hot object Friction at the surface Unheated air (dense) FIGURE 15–35 Natural convection currents around a hot object in air. Warm air Cool air FIGURE 15–36 Natural convection cooling of electronic components in an enclosure with air vents. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 814 814 HEAT TRANSFER case in electronic equipment. Therefore, the airflow in the analysis of electronic equipment can be assumed to be laminar. The natural convection heat transfer coefficient for laminar flow of air at atmospheric pressure is given by a simplified relation of the form T L hconv K Natural convection currents Cool surface tion al radia Therm 0.25 FIGURE 15–37 Simultaneous natural convection heat transfer to air and radiation heat transfer to the surrounding surfaces from a hot electronic component mounted on the wall of an enclosure. Warm air out Cool air in FIGURE 15–38 A chassis with an array of vertically oriented PCBs cooled by natural convection. (W/m2 · °C) (15-15) When hot surfaces are surrounded by cooler surfaces such as the walls and ceilings of a room or just the sky, the surfaces are also cooled by radiation, as shown in Figure 15–37. The magnitude of radiation heat transfer, in general, is comparable to the magnitude of natural convection heat transfer. This is especially the case for surfaces whose emissivity is close to unity, such as plastics and painted surfaces (regardless of color). Radiation heat transfer is negligible for polished metals because of their very low emissivity and for bodies surrounded by surfaces at about the same temperature. Radiation heat transfer between a surface at temperature Ts completely surrounded by a much larger surface at temperature Tsurr can be expressed as · 4 Q rad As (T s4 T surr ) Plug-in PCB (15-14) where T Ts Tfluid is the temperature difference between the surface and the fluid, L is the characteristic length (the length of the body along the heat flow path), and K is a constant whose value depends on the geometry and orientation of the body. The heat transfer coefficient relations are given in Table 15–1 for some common geometries encountered in electronic equipment in both SI and English unit systems. Once hconv has been determined from one of these relations, the rate of heat transfer can be determined from Eq. 15-13. The relations in Table 15–1 can also be used at pressures other than 1 atm by multiplying them by P, where P is the air pressure in atm (1 atm 101.325 kPa 14.696 psia). That is, hconv, P atm hconv, 1 atm P Hot component (W/m2 · °C) (W) (15-16) where is the emissivity of the surface, As is the heat transfer surface area, and is the Stefan–Boltzmann constant, whose value is 5.67 108 W/m2 · K4 0.1714 108 Btu/h · ft2 · R4. Here, both temperatures must be expressed in K or R. Also, if the hot surface analyzed has only a partial view of the surrounding cooler surface at Tsurr, the result obtained from Eq. 15-16 must be multiplied by a view factor, which is the fraction of the view of the hot surface blocked by the cooler surface. The value of the view factor ranges from 0 (the hot surface has no direct view of the cooler surface) to 1 (the hot surface is completely surrounded by the cooler surface). In preliminary analysis, the surface is usually assumed to be completely surrounded by a single hypothetical surface whose temperature is the equivalent average temperature of the surrounding surfaces. Arrays of low-power PCBs are often cooled by natural convection by mounting them within a chassis with adequate openings at the top and at the bottom to facilitate airflow, as shown in Figure 15–38. The air between the PCBs rises when heated by the electronic components and is replaced by the cooler air entering from below. This initiates the natural convection flow cen58933_ch15.qxd 9/9/2002 10:21 AM Page 815 815 CHAPTER 15 TABLE 15–1 Simplified relations for natural convection heat transfer coefficients for various geometries in air at atmospheric pressure for laminar flow conditions (From Refs. 4 and 5.) Natural convection heat transfer coefficient W/m2 ºC (T in ºC, L or D in m) Geometry Btu/h ft2 · ºF (T in ºF, L or D in ft) Vertical plate or cylinder L T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 hconv 1.42 L hconv 1.32 D hconv 1.32 L hconv 0.59 L hconv 2.44 L hconv 3.53 L hconv 1.92 D T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 T 0.25 hconv 0.29 L hconv 0.27 D hconv 0.27 L hconv 0.12 L hconv 0.50 L hconv 0.72 L hconv 0.39 D Horizontal cylinder D Horizontal plate (L 4A/p, where A is surface area and p is perimeter Hot surface A (a) Hot surface facing up Hot surface (b) Hot surface facing down Components on a circuit board Small components or wires in free air L L Sphere D cen58933_ch15.qxd 9/9/2002 10:21 AM Page 816 816 HEAT TRANSFER Warm air out PCB Electronic components Cool air in FIGURE 15–39 The PCBs in a chassis must be oriented vertically and spaced adequately to maximize heat transfer by natural convection. through the parallel flow passages formed by the PCBs. The PCBs must be placed vertically to take advantage of natural convection currents and to minimize trapped air pockets (Fig. 15–39). Placing the PCBs too far from each other wastes valuable cabinet space, and placing them too close tends to “choke” the flow because of the increased resistance. Therefore, there should be an optimum spacing between the PCBs. It turns out that a distance of about 2 cm between the PCBs provides adequate air flow for effective natural convection cooling. In the heat transfer analysis of PCBs, radiation heat transfer is disregarded, since the view of the components is largely blocked by other heat-generating components. As a result, hot components face other hot surfaces instead of a cooler surface. The exceptions are the two PCBs at the ends of the chassis that view the cooler side surfaces. Therefore, it is wise to mount any high-power components on the PCBs facing the walls of the chassis to take advantage of the additional cooling provided by radiation. Circuit boards that dissipate up to about 5 W of power (or that have a power density of about 0.02 W/cm2) can be cooled effectively by natural convection. Heat transfer from PCBs can be analyzed by treating them as rectangular plates with uniformly distributed heat sources on one side, and insulated on the other side, since heat transfer from the back surfaces of the PCBs is usually small. For PCBs with electronic components mounted on both sides, the rate of heat transfer and the heat transfer surface area will be twice as large. It should be remembered that natural convection currents occur only in gravitational fields. Therefore, there can be no heat transfer in space by natural convection. This will also be the case when the air passageways are blocked and hot air cannot rise. In such cases, there will be no air motion, and heat transfer through the air will be by convection. The heat transfer from hot surfaces by natural convection and radiation can be enhanced by attaching fins to the surfaces. The heat transfer in this case can best be determined by using the data supplied by the manufacturers, as discussed in Chapter 3, especially for complex geometries. EXAMPLE 15–10 Electronic box 35°C 30 cm 40 cm 75 W ε = 0.85 Ts ≤ 65°C 15 cm Stand FIGURE 15–40 Schematic for Example 15–10. Cooling of a Sealed Electronic Box Consider a sealed electronic box whose dimensions are 15 cm 30 cm 40 cm placed on top of a stand in a room at 35°C, as shown in Figure 15–40. The box is painted, and the emissivity of its outer surface is 0.85. If the electronic components in the box dissipate 75 W of power and the outer surface temperature of the box is not to exceed 65°C, determine if this box can be cooled by natural convection and radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible. SOLUTION The surface temperature of a sealed electronic box placed on top of a stand is not to exceed 65°C. It is to be determined if this box can be cooled by natural convection and radiation alone. Assumptions 1 The box is located at sea level so that the local atmospheric pressure is 1 atm. 2 The temperature of the surrounding surfaces is the same as the air temperature in the room. Analysis The sealed electronic box will lose heat from the top and the side surfaces by natural convection and radiation. All four side surfaces of the box cen58933_ch15.qxd 9/9/2002 10:21 AM Page 817 817 CHAPTER 15 can be treated as 0.15-m-high vertical surfaces. Then the natural convection heat transfer from these surfaces is determined to be L 0.15 m Aside (2 0.4 m 2 0.3 m)(0.15 m) 0.21 m2 hconv, side T L 0.25 65 35 0.15 1.42 0.25 5.34 W/m2 · °C · Q conv, side hconv, side Aside(Ts Tfluid) (5.34 W/m2 · °C)(0.21 m2)(65 35)°C 33.6 W Similarly, heat transfer from the horizontal top surface by natural convection is determined to be 4Atop 4(0.3 m)(0.4 m) L p 0.34 m 2(0.3 0.4) m Atop (0.3 m)(0.4 m) 0.12 m2 T 0.25 L hconv, top 1.32 65 35 0.34 1.32 0.25 4.05 W/m2 · °C · Q conv, top hconv, top Atop(Ts Tfluid) (4.05 W/m2 · °C)(0.12 m2)(65 35)°C 14.6 W Therefore, the natural convection heat transfer from the entire box is · · · Q conv Q conv, side Q conv, top 33.6 14.6 48.2 W The box is completely surrounded by the surfaces of the room, and it is stated that the temperature of the surfaces facing the box is equal to the air temperature in the room. Then the rate of heat transfer from the box by radiation can be determined from · 4 Q rad As (Ts4 Tsurr ) 0.85(0.21 0.12) m2 [(65 273)4 (35 273)4]K4 64.5 W Note that we must use absolute temperatures in radiation calculations. Then the total heat transfer from the box is simply · · · Q total Q conv Q rad 48.2 64.5 112.7 W which is greater than 75 W. Therefore, this box can be cooled by combined natural convection and radiation, and there is no need to install any fans. There is even some safety margin left for occasions when the air temperature rises above 35°C. PCB 0.2 W Resistor EXAMPLE 15–11 Cooling of a Component by Natural Convection A 0.2-W small cylindrical resistor mounted on a PCB is 1 cm long and has a diameter of 0.3 cm, as shown in Figure 15–41. The view of the resistor is FIGURE 15–41 Schematic for Example 15–11. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 818 818 HEAT TRANSFER largely blocked by the PCB facing it, and the heat transfer from the connecting wires is negligible. The air is free to flow through the parallel flow passages between the PCBs. If the air temperature at the vicinity of the resistor is 50°C, determine the surface temperature of the resistor. SOLUTION A small cylindrical resistor mounted on a PCB is being cooled by natural convection and radiation. The surface temperature of the resistor is to be determined. Assumptions 1 Steady operating conditions exist. 2 The device is located at sea level so that the local atmospheric pressure is 1 atm. 3 Radiation is negligible in this case since the resistor is surrounded by surfaces that are at about the same temperature, and the net radiation heat transfer between two surfaces at the same temperature is zero. This leaves natural convection as the only mechanism of heat transfer from the resistor. Analysis Using the relation for components on a circuit board from Table 15–1, the natural convection heat transfer coefficient for this cylindrical component can be determined from Ts Tfluid D hconv 2.44 0.25 where the diameter D 0.003 m, which is the length in the heat flow path, is the characteristic length. We cannot determine hconv yet since we do not know the surface temperature of the component and thus T. But we can substitute this relation into the heat transfer relation to get Ts Tfluid · Q conv hconv As(Ts Tfluid) 2.44 D 2.44As (Ts Tfluid)1.25 D 0.25 0.25 A(Ts Tfluid) The heat transfer surface area of the component is As 2 41 D 2 DL 2 14 (0.3 cm)2 (0.3 cm)(1 cm) 1.084 cm2 · Substituting this and other known quantities in proper units (W for Q , °C for T, 2 m for A, and m for D) into this equation and solving for Ts yields 0.2 2.44(1.084 104) 20 cm 15 cm FIGURE 15–42 Schematic for Example 15–12. → Ts 113°C Therefore, the surface temperature of the resistor on the PCB will be 113°C, which is considered to be a safe operating temperature for the resistors. Note that blowing air to the circuit board will lower this temperature considerably as a result of increasing the convection heat transfer coefficient and decreasing the air temperature at the vicinity of the components due to the larger flow rate of air. EXAMPLE 15–12 7W PCB (Ts 50)1.25 0.0030.25 Cooling of a PCB in a Box by Natural Convection A 15-cm 20-cm PCB has electronic components on one side, dissipating a total of 7 W, as shown in Figure 15–42. The PCB is mounted in a rack vertically together with other PCBs. If the surface temperature of the components is not to exceed 100°C, determine the maximum temperature of the environment in cen58933_ch15.qxd 9/9/2002 10:21 AM Page 819 819 CHAPTER 15 which this PCB can operate safely at sea level. What would your answer be if this rack is located at a location at 4000 m altitude where the atmospheric pressure is 61.66 kPa? SOLUTION The surface temperature of a PCB is not to exceed 100°C. The maximum environment temperatures for safe operaton at sea level and at 4000 m altitude are to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer is negligible since the PCB is surrounded by other PCBs at about the same temperature. 3 Heat transfer from the back surface of the PCB will be very small and thus negligible. Analysis The entire heat load of the PCB is dissipated to the ambient air by natural convection from its front surface, which can be treated as a vertical flat plate. Using the simplified relation for a vertical surface from Table 15–1, the natural convection heat transfer coefficient for this PCB can be determined from hconv 1.42 T L 0.25 1.42 Ts Tfluid L 0.25 The characteristic length in this case is the height (L 0.15 m) of the PCB, which is the length in the path of heat flow. We cannot determine hconv yet, since we do not know the ambient temperature and thus T. But we can substitute this relation into the heat transfer relation to get Ts Tfluid · Q conv hconv As(Ts Tfluid) 1.42 L 1.42As 0.25 A(Ts Tfluid) (Ts Tfluid)1.25 L 0.25 The heat transfer surface area of the PCB is As (Width)(Height) (0.2 m)(0.15 m) 0.03 m2 · Substituting this and other known quantities in proper units (W for Q , °C for T, m2 for As, and m for L) into this equation and solving for Tfluid yields 7 1.42(0.03) (100 Tfluid)1.25 → Tfluid 59.5°C 0.150.25 Therefore, the PCB will operate safely in environments with temperatures up to 59.4°C by relying solely on natural convection. At an altitude of 4000 m, the atmospheric pressure is 61.66 kPa, which is equivalent to P (61.66 kPa) 1 atm 0.609 atm 101.325 kPa The heat transfer coefficient in this case is obtained by multiplying the value at sea level by P , where P is in atm. Substituting 7 1.42(0.03) (100 Tfluid)1.25 0.609 0.150.25 → Tfluid 50.6°C which is about 10°C lower than the value obtained at 1 atm pressure. Therefore, the effect of altitude on convection should be considered in high-altitude applications. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 820 820 HEAT TRANSFER 15–8 Electronic components Air Tout exit . Qloss ≈0 ■ AIR COOLING: FORCED CONVECTION We mentioned earlier that convection heat transfer between a solid surface and a fluid is proportional to the velocity of the fluid. The higher the velocity, the larger the flow rate and the higher the heat transfer rate. The fluid velocities associated with natural convection currents are naturally low, and thus natural convection cooling is limited to low-power electronic systems. When natural convection cooling is not adequate, we simply add a fan and blow air through the enclosure that houses the electronic components. In other words, we resort to forced convection in order to enhance the velocity and thus the flow rate of the fluid as well as the heat transfer. By doing so, we can increase the heat transfer coefficient by a factor of up to about 10, depending on the size of the fan. This means we can remove heat at much higher rates for a specified temperature difference between the components and the air, or we can reduce the surface temperature of the components considerably for a specified power dissipation. The radiation heat transfer in forced-convection-cooled electronic systems is usually disregarded for two reasons. First, forced convection heat transfer is usually much larger than that due to radiation, and the consideration of radiation causes no significant change in the results. Second, the electronic components and circuit boards in convection-cooled systems are mounted so close to each other that a component is almost entirely surrounded by other components at about the same high temperature. That is, the components have hardly any direct view of a cooler surface. This results in little or no radiation heat transfer from the components. The components near the edges of circuit boards with a large view of a cooler surface may benefit somewhat from the additional cooling by radiation, and it is a good design practice to reserve those spots for high-power components to have a thermally balanced system. When heat transfer from the outer surface of the enclosure of the electronic equipment is negligible, the amount of heat absorbed by the air becomes equal to the amount of heat rejected (or power dissipated) by the electronic components in the enclosure, and can be expressed as (Fig. 15–43) · Q m· Cp(Tout Tin) Fan Tin Air inlet Power consumed: . . . Welectric = Qabsorbed = mCp(Tout – Tin) FIGURE 15–43 In steady operation, the heat absorbed by air per unit time as it flows through an electronic box is equal to the power consumed by the electronic components in the box. (W) (15-17) · where Q is the rate of heat transfer to the air; Cp is the specific heat of air; Tin and Tout are the average temperatures of air at the inlet and exit of the enclosure, respectively; and m· is the mass flow rate of air. Note that for a specified mass flow rate and power dissipation, the temperature rise of air, Tout Tin, remains constant as it flows through the enclosure. Therefore, the higher the inlet temperature of the air, the higher the exit temperature, and thus the higher the surface temperature of the components. It is considered a good design practice to limit the temperature rise of air to 10°C and the maximum exit temperature of air to 70°C. In a properly designed forced-air-cooled system, this results in a maximum component surface temperature of under 100°C. The mass flow rate of air required for cooling an electronic box depends on the temperature of air available for cooling. In cool environments, such as an air-conditioned room, a smaller flow rate will be adequate. However, in hot environments, we may need to use a larger flow rate to avoid overheating the components and the potential problems associated with it. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 821 821 CHAPTER 15 Forced convection is covered in detail in a separate chapter. For those who skipped that chapter because of time limitations, here we present a brief review of basic concepts and relations. The fluid flow over a body such as a transistor is called external flow, and flow through a confined space such as inside a tube or through the parallel passage area between two circuit boards in an enclosure is called internal flow (Fig. 15–44). Both types of flow are encountered in a typical electronic system. Fluid flow is also categorized as being laminar (smooth and streamlined) or turbulent (intense eddy currents and random motion of chunks of fluid). Turbulent flow is desirable in heat transfer applications since it results in a much larger heat transfer coefficient. But it also comes with a much larger friction coefficient, which requires a much larger fan (or pump for liquids). Numerous experimental studies have shown that turbulence tends to occur at larger velocities, during flow over larger bodies or flow through larger channels, and with fluids having smaller viscosities. These effects are combined into the dimensionless Reynolds number, defined as D Re (15-18) where velocity of the fluid ( free-stream velocity for external flow and average velocity for internal flow), m/s D characteristic length of the geometry (the length the fluid flows over in external flow, and the equivalent diameter in internal flow), m / kinematic viscosity of the fluid, m2/s. The Reynolds number at which the flow changes from laminar to turbulent is called the critical Reynolds number, whose value is 2300 for internal flow, 500,000 for flow over a flat plate, and 200,000 for flow over a cylinder or sphere. The equivalent (or hydraulic) diameter for internal flow is defined as 4Ac Dh p (m) (15-19) where Ac is the cross-sectional area of the flow passage and p is the perimeter. Note that for a circular pipe, the hydraulic diameter is equivalent to the ordinary diameter. The convection heat transfer is expressed by Newton’s law of cooling as · Q conv hAs(Ts Tfluid) (W) (15-20) where h average convection heat transfer coefficient, W/m2 · °C As heat transfer surface area, m2 Ts temperature of the surface, °C Tfluid temperature of the fluid sufficiently far from the surface for external flow, and average temperature of the fluid at a specified location in internal flow, °C When the heat load is distributed uniformly on the surfaces with a constant · heat flux q· , the total rate of heat transfer can also be expressed as Q q· As. External flow Internal flow FIGURE 15–44 Internal flow through a circular tube and external flow over it. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 822 822 HEAT TRANSFER In fully developed flow through a pipe or duct (i.e., when the entrance effects are negligible) subjected to constant heat flux on the surfaces, the convection heat transfer coefficient h remains constant. In this case, both the surface temperature Ts and the fluid temperature Tfluid increase linearly, as shown in Figure 15–45, but the difference between them, Ts Tfluid, remains constant. Then the temperature rise of the surface above the fluid temperature can be determined from Eq. 15-20 to be Ts T Entry region Fully developed region Tout Tfluid Tin 0 Tin . q ∆T = Ts – Tfluid = —s h . qs = constant L x Tout Trise, surface Ts Tfluid (°C) (15-21) Note that the temperature rise of the surface is inversely proportional to the convection heat transfer coefficient. Therefore, the greater the convection coefficient, the lower the surface temperature of the electronic components. When the exit temperature of the fluid, Tout, is known, the highest surface temperature that will occur at the end of the flow channel can be determined from Eq. 15-21 to be Ts, max Tfluid, max FIGURE 15–45 Under constant heat flux conditions, the surface and fluid temperatures increase linearly, but their difference remains constant in the fully developed region. Q· conv hAs Q· Q· Tout hAs hAs (°C) (15-22) If this temperature is within the safe range, then we don’t need to worry about temperatures at other locations. But if it is not, it may be necessary to use a larger fan to increase the flow rate of the fluid. In convection analysis, the convection heat transfer coefficient h is usually expressed in terms of the dimensionless Nusselt number Nu as h k Nu D (W/m2 · °C) (15-23) where k is the thermal conductivity of the fluid and D is the characteristic length of the geometry. Relations for the average Nusselt number based on experimental data are given in Table 15–2 for external flow and in Table 15–3 for laminar (Re 2300) internal flow under a uniform heat flux condition, which is closely approximated by electronic equipment. For turbulent flow (Re 2300) through smooth tubes and channels, the Nusselt number can be determined from the Dittus–Boelter correlation, Nu 0.023 Re0.8 Pr 0.4 (15-24) for any geometry. Here Pr is the dimensionless Prandtl number, and its value is about 0.7 for air at room temperature. The fluid properties in the above relations are to be evaluated at the bulk mean fluid temperature Tave 12(Tin Tout) for internal flow, which is the arithmetic average of the mean fluid temperatures at the inlet and the exit of the tube, and at the film temperature Tfilm 12(Ts Tfluid) for external flow, which is the arithmetic average of the surface temperature and free-stream temperature of the fluid. The relations in Table 15–3 for internal flow assume fully developed flow over the entire flow section, and disregard the heat transfer enhancement effects of the development region at the entrance. Therefore, the results obtained from these relations are on the conservative side. We don’t mind this much, however, since it is common practice in engineering design to have some safety margin to fall back to “just in case,” as long as it does not result cen58933_ch15.qxd 9/9/2002 10:21 AM Page 823 823 CHAPTER 15 TABLE 15–2 Empirical correlations for the average Nusselt number for forced convection over a flat plate and circular and noncircular cylinders in cross flow (From Jakob, Ref. 12, and Zukauskas, Ref. 17.) Cross-section of the cylinder Circle Fluid Range of Re Gas or liquid 0.4–4 4–40 40–4000 4000–40,000 40,000–400,000 Nu Nu Nu Nu Nu 5000–100,000 Nu 0.102 Re0.675Pr1/3 D Square Gas Nusselt number 0.989 0.911 0.683 0.193 0.027 Re0.330Pr1/3 Re0.385Pr1/3 Re0.466Pr1/3 Re0.618Pr1/3 Re0.805Pr1/3 TABLE 15–3 Nusselt number of fully developed laminar flow in circular tubes and rectangular channels Cross-section of the tube Aspect Nusselt ratio number a Circle a Square (tilted 45º) Gas 5000–100,000 — 4.36 Square — 3.61 Rectangle a /b — 1 2 3 4 6 8 3.61 4.12 4.79 5.33 6.05 6.49 8.24 Nu 0.246 Re0.588Pr1/3 D Flat plate Gas or liquid 0–5 105 Nu 0.664 Re1/2Pr1/3 5 105–107 Nu (0.037 Re4/5 871)Pr1/3 4000–15,000 Nu 0.228 Re0.731Pr1/3 L Vertical plate Gas b a L in a grossly overdesigned system. Also, it may sometimes be necessary to do some local analysis for critical components with small surface areas to assure reliability and to incorporate solutions to local problems such as attaching heat sinks to high power components. Fan Selection Air can be supplied to electronic equipment by one or several fans. Although the air is free and abundant, the fans are not. Therefore, a few words about the fan selection are in order. A fan at a fixed speed (or fixed rpm) will deliver a fixed volume of air regardless of the altitude and pressure. But the mass flow rate of air will be less at high altitude as a result of the lower density of air. For example, the atmospheric cen58933_ch15.qxd 9/9/2002 10:21 AM Page 824 824 HEAT TRANSFER Air outlet Exhaust fan Electronic box Crack Air inlet FIGURE 15–46 A fan placed at the exit of an electronic box draws in air as well as contaminants in the air through cracks. pressure of air drops by more than 50 percent at an altitude of 6000 m from its value at sea level. This means that the fan will deliver half as much air mass at this altitude at the same rpm and temperature, and thus the temperature rise of air cooling will double. This may create serious reliability problems and catastrophic failures of electronic equipment if proper precautions are not taken. Variable-speed fans that automatically increase speed when the air density decreases are available to avoid such problems. Expensive electronic systems are usually equipped with thermal cutoff switches to prevent overheating due to inadequate airflow rate or the failure of the cooling fan. Fans draw in not only cooling air but also all kinds of contaminants that are present in the air, such as lint, dust, moisture, and even oil. If unattended, these contaminants can pile up on the components and plug up narrow passageways, causing overheating. It should be remembered that the dust that settles on the electronic components acts as an insulation layer that makes it very difficult for the heat generated in the component to escape. To minimize the contamination problem, air filters are commonly used. It is good practice to use the largest air filter practical to minimize the pressure drop of air and to maximize the dust capacity. Often the question arises about whether to place the fan at the inlet or the exit of an electronic box. The generally preferred location is the inlet. A fan placed at the inlet draws air in and pressurizes the electronic box and prevents air infiltration into the box from cracks or other openings. Having only one location for air inlet makes it practical to install a filter at the inlet to clean the air from all the dust and dirt before they enter the box. This allows the electronic system to operate in a clean environment. Also, a fan placed at the inlet handles cooler and thus denser air, which results in a higher mass flow rate for the same volume flow rate or rpm. Since the fan is always subjected to cool air, this has the added benefit that it increased the reliability and extends the life of the fan. The major disadvantage associated with having a fan mounted at the inlet is that the heat generated by the fan and its motor is picked up by air on its way into the box, which adds to the heat load of the system. When the fan is placed at the exit, the heat generated by the fan and its motor is immediately discarded to the atmosphere without getting blown first into the electronic box. However, a fan at the exit creates a vacuum inside the box, which draws air into the box through inlet vents as well as any cracks and openings (Fig. 15–46). Therefore, the air is difficult to filter, and the dirt and dust that collect on the components undermine the reliability of the system. There are several types of fans available on the market for cooling electronic equipment, and the right choice depends on the situation on hand. There are two primary considerations in the selection of the fan: the static pressure head of the system, which is the total resistance an electronic system offers to air as it passes through, and the volume flow rate of the air. Axial fans are simple, small, light, and inexpensive, and they can deliver a large flow rate. However, they are suitable for systems with relatively small pressure heads. Also, axial fans usually run at very high speeds, and thus they are noisy. The radial or centrifugal fans, on the other hand, can deliver moderate flow rates to systems with high-static pressure heads at relatively low speeds. But they are larger, heavier, more complex, and more expensive than axial fans. The performance of a fan is represented by a set of curves called the characteristic curves, which are provided by fan manufacturers to help engineers cen58933_ch15.qxd 9/9/2002 10:21 AM Page 825 825 CHAPTER 15 Before deciding on forced-air cooling, check to see if natural convection cooling is adequate. If it is, which may be the case for low-power systems, incorporate it and avoid all the problems associated with fans such as cost, power consumption, noise, complexity, maintenance, and possible failure. 2. Select a fan that is neither too small nor too large. An undersized fan may cause the electronic system to overheat and fail. An oversized fan will definitely provide adequate cooling, but it will needlessly be larger and more expensive and will consume more power. 3. If the temperature rise of air due to the power consumed by the motor of the fan is acceptable, mount the fan at the inlet of the box to pressurize the box and filter the air to keep dirt and dust out (Fig. 15–48). 4. Position and size the air exit vents so that there is adequate airflow throughout the entire box. More air can be directed to a certain area by enlarging the size of the vent at that area. The total exit areas should be at least as large as the inlet flow area to avoid the choking of the airflow, which may result in a reduced airflow rate. 5. Place the most critical electronic components near the entrance, where the air is coolest. Place the non-critical components that consume a lot of power near the exit (Fig. 15–49). 6. Arrange the circuit boards and the electronic components in the box such that the resistance of the box to airflow is minimized and thus the flow rate of air through the box is maximized for the same fan speed. Make sure that no hot air pockets are formed during operation. Fan static pressure curve System resistance curve (head loss) Static pressure with the selection of fans. A typical static pressure head curve for a fan is given in Figure 15–47 together with a typical system flow resistance curve plotted against the flow rate of air. Note that a fan creates the highest pressure head at zero flow rate. This corresponds to the limiting case of blocked exit vents of the enclosure. The flow rate increases with decreasing static head and reaches its maximum value when the fan meets no flow resistance. Any electronic enclosure will offer some resistance to flow. The system resistance curve is parabolic in shape, and the pressure or head loss due to this resistance is nearly proportional to the square of the flow rate. The fan must overcome this resistance to maintain flow through the enclosure. The design of a forced convection cooling system requires the determination of the total system resistance characteristic curve. This curve can be generated accurately by measuring the static pressure drop at different flow rates. It can also be determined approximately by evaluating the pressure drops. A fan will operate at the point where the fan static head curve and the system resistance curve intersect. Note that a fan will deliver a higher flow rate to a system with a low flow resistance. The required airflow rate for a system can be determined from heat transfer requirements alone, using the design heat load of the system and the allowable temperature rise of air. Then the flow resistance of the system at this flow rate can be determined analytically or experimentally. Knowing the flow rate and the needed pressure head, it is easy to select a fan from manufacturers’ catalogs that will meet both of these requirements. Below we present some general guidelines associated with the forced-air cooling of electronic systems. Operating point of the fan 0 Airflow rate FIGURE 15–47 The airflow rate a fan delivers into an electronic enclosure depends on the flow resistance of that system as well as the variation of the static head of the fan with flow rate. Electronic box Air outlet Electronic components Filter Air inlet Fan Heat Motor FIGURE 15–48 Installing the fan at the inlet keeps the dirt and dust out, but the heat generated by the fan motor in. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 826 826 HEAT TRANSFER Air outlet High-power component Critical component Air inlet FIGURE 15–49 Sensitive components should be located near the inlet and high-power components near the exit. Consider the effect of altitude in high-altitude applications. 8. Try to avoid any flow sections that increase the flow resistance of the systems, such as unnecessary corners, sharp turns, sudden expansions and contractions, and very high velocities (greater than 7 m/s), since the flow resistance is nearly proportional to the flow rate. Also, avoid very low velocities since these result in a poor heat transfer performance and allow the dirt and the dust in the air to settle on the components. 9. Arrange the system such that natural convection helps forced convection instead of hurting it. For example, mount the PCBs vertically, and blow the air from the bottom toward the top instead of the other way around. 10. When the design calls for the use of two or more fans, a decision needs to be made about mounting the fans in parallel or in series. Fans mounted in series will boost the pressure head available and are best suited for systems with a high flow resistance. Fans connected in parallel will increase the flow rate of air and are best suited for systems with small flow resistance. Cooling Personal Computers The introduction of the 4004 chip, the first general-purpose microprocessor, by the Intel Corporation in the early 1970s marked the beginning of the electronics era in consumer goods, from calculators and washing machines to personal computers. The microprocessor, which is the “brain” of the personal computer, is basically a DIP-type LSE package that incorporates a central processing unit (CPU), memory, and some input/output capabilities. A typical desktop personal computer consists of a few circuit boards plugged into a mother board, which houses the microprocessor and the memory chips, as well as the network of interconnections enclosed in a formed sheet metal chassis, which also houses the disk and CD-ROM drives. Connected to this “magic” box are the monitor, a keyboard, a printer, and other auxiliary equipment (Fig. 15–50). The PCBs are normally mounted vertically on a mother board, since this facilitates better cooling. A small and quiet fan is usually mounted to the rear or side of the chassis to cool the electronic components. There are also louvers and openings on the side surfaces to facilitate air circulation. Such openings are not placed on the top surface, since many users would block them by putting books or other things there, which will jeopardize safety, and a coffee or soda spill can cause major damage to the system. FIGURE 15–50 A desktop personal computer with monitor and keyboard. EXAMPLE 15–13 Forced-Air Cooling of a Hollow-Core PCB Some strict specifications of electronic equipment require that the cooling air not come into direct contact with the electronic components, in order to protect them from exposure to the contaminants in the air. In such cases, heat generated in the components on a PCB must be conducted a long way to the walls of the enclosure through a metal core strip or a heat frame attached to the PCB. An alternative solution is the hollow-core PCB, which is basically a narrow duct cen58933_ch15.qxd 9/9/2002 10:21 AM Page 827 827 CHAPTER 15 of rectangular cross section made of thin glass–epoxy board with electronic components mounted on both sides, as shown in Figure 15–51. Heat generated in the components is conducted to the hollow core through a thin layer of epoxy board and is then removed by the cooling air flowing through the core. Effective sealing is provided to prevent air leakage into the component chamber. Consider a hollow-core PCB 12 cm high and 18 cm long, dissipating a total of 40 W. The width of the air gap between the two sides of the PCB is 0.3 cm. The cooling air enters the core at 20°C at a rate of 0.72 L/s. Assuming the heat generated to be uniformly distributed over the two side surfaces of the PCB, determine (a) the temperature at which the air leaves the hollow core and (b) the highest temperature on the inner surface of the core. SOLUTION A hollow-core PCB is cooled by forced air. The outlet temperature of air and the highest surface temperature are to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Air is an ideal gas. 4 Operation is at sea level and thus the atmospheric pressure is 1 atm. 5 The entire heat generated in electronic components is removed by the air flowing through the hollow core. Properties The temperature of air varies as it flows through the core, and so do its properties. We will perform the calculations using property values at 25°C from Table A–15 since the air enters at 20°C and its temperature will increase. 1.184 kg/m3 Cp 1007 J/kg · °C Pr 0.7296 k 0.02551 W/m · °C 1.562 105 m2/s After we calculate the exit temperature of air, we can repeat the calculations, if necessary, using properties at the average temperature. Analysis The cross-sectional area of the channel and its hydraulic diameter are Ac (Height)(Width) (0.12 m)(0.003 m) 3.6 104 m2 4Ac 4 (3.6 10 4 m3) Dh p 0.005854 m 2 (0.12 0.003) m The average velocity and the mass flow rate of air are · V 0.72 10 3 m2/s 2.0 m/s Ac 3.6 10 4 m2 · m· V (1.184 kg/m3)(0.72 103 m3/s) 0.8525 103 kg/s (a ) The temperature of air at the exit of the hollow core can be determined from · Q m· Cp(Tout Tin) Solving for Tout and substituting the given values, we obtain · Q 40 J/s Tout Tin · 20°C 66.6°C (0.8525 10 3 kg/s)(1007 J/kg · °C) m Cp (b ) The surface temperature of the channel at any location can be determined from · Q conv hAs(Ts Tfluid) where the heat transfer surface area is Cooling air Gasket Electronic component PCB Hollow core FIGURE 15–51 The hollow-core PCB discussed in Example 15–13. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 828 828 HEAT TRANSFER As 2Aside 2(Height)(Length) 2(0.12 m)(0.18 m) 0.0432 m2 To determine the convection heat transfer coefficient, we first need to calculate the Reynolds number: Dh (2 m/s)(0.005854 m) Re 750 2300 1.562 10 5 m2/s Therefore, the flow is laminar, and, assuming fully developed flow, the Nusselt number for the airflow in this rectangular cross section corresponding to the aspect ratio a/b (12 cm)/(0.3 cm) 40 is determined from Table 15–3 to be Nu 8.24 and thus h 0.02551 W/m · °C k (8.24) 35.9 W/m2 · °C Nu Dh 0.005854 m Then the surface temperature of the hollow core near the exit becomes Ts, max Tout · Q 40 W 66.6°C 92.4°C hAs (35.9 W/m2 · °C)(0.0432 m2) Discussion Note that the temperature difference between the surface and the air at the exit of the hollow core is 25.8°C. This temperature difference between the air and the surface remains at that value throughout the core, since the heat generated on the side surfaces is uniform and the convection heat transfer coefficient is constant. Therefore, the surface temperature of the core at the inlet will be 20°C 25.8°C 45.8°C. In reality, however, this temperature will be somewhat lower because of the entrance effects, which affect heat transfer favorably. The fully developed flow assumption gives somewhat conservative results but is commonly used in practice because it provides considerable simplification in calculations. EXAMPLE 15–14 PCB TO 71 transistor 0.44 cm Forced-Air Cooling of a Transistor Mounted on a PCB A TO 71 transistor with a height of 0.53 cm and a diameter of 0.44 cm is mounted on a circuit board, as shown in Figure 15–52. The transistor is cooled by air flowing over it at a velocity of 90 m/min. If the air temperature is 65°C and the transistor case temperature is not to exceed 95°C, determine the amount of power this transistor can dissipate safely. 0.53 cm SOLUTION A transistor mounted on a circuit board is cooled by air flowing Air flow = 90 m/min 65°C over it. The power dissipated when its case temperature is 90°C is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Operation is at sea level and thus the atmospheric pressure is 1 atm. Properties The properties of air at 1 atm pressure and the film temperature of (Ts Tfluid)/2 (95 65)/2 80°C are Tf FIGURE 15–52 Schematic for Example 15–14. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 829 829 CHAPTER 15 0.9994 kg/m3 Cp 1008 J/kg · °C Pr 0.7154 k 0.02953 W/m · °C 2.097 105 m2/s Analysis The transistor is cooled by forced convection through its cylindrical surface as well as its flat top and bottom surfaces. The characteristic length for flow over a cylinder is the diameter D 0.0044 m. Then the Reynolds number becomes 90 D ( 60 m /s)(0.0044 m) Re 315 2.097 10 5 m2/s which falls into the range 40 to 4000. Using the corresponding relation from Table 15–2 for the Nusselt number, we obtain Nu 0.683 Re0.466 Pr1/3 0.683(315)0.466(0.7154)1/ 3 8.91 and h 0.02953 W/m · °C k (8.91) 59.8 W/m2 · °C Nu D 0.0044 m Also, Acyl DL (0.0044 m)(0.0053 m) 0.7326 104 m2 Then the rate of heat transfer from the cylindrical surface becomes · Q cyl hAcyl(Ts Tfluid) (59.8 W/m2 · °C)(0.7326 104 m2)(95 65)°C 0.131 W We now repeat the calculations for the top and bottom surfaces of the transistor, which can be treated as flat plates of length L 0.0044 m in the flow direction (which is the diameter), and, using the proper relation from Table 15–2, 90 L ( 60 m/s)(0.0044 m) Re 315 2.097 10 5 m2/s Nu 0.664 Re1/2 Pr1/3 0.664(315)1/2(0.7154)1/3 10.5 and h 0.02953 W/m · °C k (10.5) 70.7 W/m2 · °C Nu D 0.0044 m Also, Aflat Atop Abottom 2 14 D 2 2 14 (0.0044 m)2 0.3041 104 m2 · Q flat hAflat(Ts Tfluid) (70.7 W/m2 · °C)(0.3041 104 m2)(95 65)°C 0.065 W Therefore, the total rate of heat that can be dissipated from all surfaces of the transistor is · · · Q total Q cyl Q flat (0.131 0.065) W 0.196 W which seems to be low. This value can be increased considerably by attaching a heat sink to the transistor to enhance the heat transfer surface area and thus heat transfer, or by increasing the air velocity, which will increase the heat transfer coefficient. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 830 830 HEAT TRANSFER EXAMPLE 15–15 Air outlet Tout Exhaust fan 75 W Tin Air inlet FIGURE 15–53 Schematic for Example 15–15. Choosing a Fan to Cool a Computer The desktop computer shown in Figure 15–53 is to be cooled by a fan. The electronics of the computer consume 75 W of power under full-load conditions. The computer is to operate in environments at temperatures up to 40°C and at elevations up to 2000 m, where the atmospheric pressure is 79.50 kPa. The exit temperature of air is not to exceed 70°C to meet reliability requirements. Also, the average velocity of air is not to exceed 75 m/min at the exit of the computer case, where the fan is installed, to keep the noise level down. Determine the flow rate of the fan that needs to be installed and the diameter of the casing of the fan. SOLUTION A desktop computer is to be cooled by a fan safely in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined. Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas. Properties The specific heat of air at the average temperature of (40 70)/2 55°C is 1007 J/kg · °C (Table A–15). Analysis We need to determine the flow rate of air for the worst-case scenario. Therefore, we assume the inlet temperature of air to be 40°C and the atmospheric pressure to be 79.50 kPa and disregard any heat transfer from the outer surfaces of the computer case. Note that any direct heat loss from the computer case will provide a safety margin in the design. Noting that all the heat dissipated by the electronic components is absorbed by air, the required mass flow rate of air to absorb heat at a rate of 75 W can be determined from · Q m· Cp(Tout Tin) Solving for m· and substituting the given values, we obtain · Q 75 J/s Cp(Tout Tin) (1007 J/kg · °C)(70 40)°C 0.00249 kg/s 0.149 kg/min m· In the worst case, the exhaust fan will handle air at 70°C. Then the density of air entering the fan and the volume flow rate become 79.50 kPa P 0.8076 kg/m3 RT (0.287 kPa · m3/kg · K)(70 275) K m· 0.149 kg/min · V 0.184 m3/min 0.8076 kg/m3 Therefore, the fan must be able to provide a flow rate of 0.184 m3/min or 6.5 cfm (cubic feet per minute). Note that if the fan were installed at the inlet instead of the exit, then we would need to determine the flow rate using the density of air at the inlet temperature of 40°C, and we would need to add the power consumed by the motor of the fan to the heat load of 75 W. The result may be a slightly smaller or larger fan, depending on which effect dominates. For an average velocity of 75 m/min, the diameter of the duct in which the fan is installed can be determined from 1 · V Ac D2 4 cen58933_ch15.qxd 9/9/2002 10:21 AM Page 831 831 CHAPTER 15 Solving for D and substituting the known values, we obtain · 4 0.184 m3/min) D 4V 0.056 m 5.6 cm (75 m/min) Therefore, a fan with a casing diameter of 5.6 cm and a flow rate of 0.184 m3/min will meet the design requirements. EXAMPLE 15–16 Cooling of a Computer by a Fan A computer cooled by a fan contains six PCBs, each dissipating 15 W of power, as shown in Figure 15–54. The height of the PCBs is 15 cm and the length is 20 cm. The clearance between the tips of the components on the PCB and the back surface of the adjacent PCB is 0.4 cm. The cooling air is supplied by a 20-W fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed 10°C, determine (a) the flow rate of the air the fan needs to deliver, (b) the fraction of the temperature rise of air due to the heat generated by the fan and its motor, and (c) the highest allowable inlet air temperature if the surface temperature of the components is not to exceed 90°C anywhere in the system. SOLUTION A computer cooled by a fan, and the temperature rise of air is limited to 10°C. The flow rate of the air, the fraction of the temperature rise of air caused by the fan and its motor, and maximum allowable air inlet temperature are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The computer is located at sea level so that the local atmospheric pressure is 1 atm. 4 The entire heat generated by the electronic components is removed by air flowing through the opening between the PCBs. 5 The entire power consumed by the fan motor is transferred as heat to the cooling air. This is a conservative approach since the fan and its motor are usually mounted to the chassis of the electronic system, and some of the heat generated in the motor may be conducted to the chassis through the mounting brackets. Properties We use properties of air at 30°C since the air enters at room temperature, and the temperature rise of air is limited to just 10°C: 1.164 kg/m3 Cp 1005 J/kg · °C Pr 0.7282 k 0.02588 W/m · °C 1.608 105 m2/s Analysis Because of symmetry, we consider the flow area between the two adjacent PCBs only. We assume the flow rate of air through all six channels to be identical, and to be equal to one-sixth of the total flow rate. (a) Noting that the temperature rise of air is limited to 10°C and that the power consumed by the fan is also absorbed by the air, the total mass flow rate of air through the computer can be determined from · Q m· Cp(Tout Tin) Solving for m· and substituting the given values, we obtain m· · Q (6 15 20) J/s 0.01092 kg/s Cp(Tout Tin) (1007 J/kg · °C)(10°C) 0.4 cm Air outlet 20 cm Fan, 20 W PCB, 15 W Air inlet FIGURE 15–54 Schematic of the computer discussed in Example 15–16. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 832 832 HEAT TRANSFER Then the volume flow rate of air and the air velocity become m· 0.01092 kg/s · V 0.009381 m3/s 0.563 m3/min 1.164 kg/m3 · 0.009381 m3/s V 2.606 m/s Ac 6 (6 10 4 m2) Therefore, the fan needs to supply air at a rate of 0.563 m3/min, or about 20 cfm. (b) The temperature rise of air due to the power consumed by the fan motor can be determined by assuming the entire 20 W of power consumed by the motor to be transferred to air as heat: · Q fan 20 J/s Tair, rise · 1.8°C m Cp (0.01092 kg/s)(1007 J/kg · °C) Therefore, 18 percent of the temperature rise of air is due to the heat generated by the fan motor. Note that the fraction of the power consumed by the fan is also 18 percent of the total, as expected. (c) The surface temperature of the channel at any location can be determined from · q· conv Q conv /As h(Ts Tfluid) where the heat transfer surface area is As Aside (Height)(Length) (0.15 m)(0.20 m) 0.03 m2 To determine the convection heat transfer coefficient, we first need to calculate the Reynolds number. The cross-sectional area of the channel and its hydraulic diameter are Ac (Height)(Width) (0.15 m)(0.004 m) 6 104 m2 4 Ac 4 (6 10 4 m2) Dh p 0.007792 m 2 (0.15 0.004) m Then the Reynolds number becomes Dh (2.606 m/s)(0.007792 m) Re 1263 2300 1.606 10 5 m2/s Therefore, the flow is laminar, and, assuming fully developed flow, the Nusselt number for the airflow in this rectangular cross-section corresponding to the aspect ratio a/b (15 cm)/(0.4 cm) 37.5 is determined from Table 15–3 to be Nu 8.24 and thus h 0.02588 W/m · °C k (8.24) 27.4 W/m2 · °C Nu Dh 0.007792 m Disregarding the entrance effects, the temperature difference between the surface of the PCB and the air anywhere along the channel is determined to be cen58933_ch15.qxd 9/9/2002 10:21 AM Page 833 833 CHAPTER 15 Ts Tfluid q· h PCB (15 W)/(0.03 m2) 18.3°C 27.4 W/m2 · °C That is, the surface temperature of the components on the PCB will be 18.3°C higher than the temperature of air passing by. The highest air and component temperatures will occur at the exit. Therefore, in the limiting case, the component surface temperature at the exit will be 90°C. The air temperature at the exit in this case will be Tout, max Ts, max Trise 90°C 18.3°C 71.7°C Noting that the air experiences a temperature rise of 10°C between the inlet and the exit, the inlet temperature of air is Tin, max Tout, max 10°C (71.7 10)°C 61.7°C This is the highest allowable air inlet temperature if the surface temperature of the components is not to exceed 90°C anywhere in the system. It should be noted that the analysis presented above is approximate since we have made some simplifying assumptions. However, the accuracy of the results obtained is usually adequate for engineering purposes. 15–9 ■ LIQUID COOLING Liquids normally have much higher thermal conductivities than gases, and thus much higher heat transfer coefficients associated with them. Therefore, liquid cooling is far more effective than gas cooling. However, liquid cooling comes with its own risks and potential problems, such as leakage, corrosion, extra weight, and condensation. Therefore, liquid cooling is reserved for applications involving power densities that are too high for safe dissipation by air cooling. Liquid cooling systems can be classified as direct cooling and indirect cooling systems. In direct cooling systems, the electronic components are in direct contact with the liquid, and thus the heat generated in the components is transferred directly to the liquid. In indirect cooling systems, however, there is no direct contact with the components. The heat generated in this case is first transferred to a medium such as a cold plate before it is carried away by the liquid. Liquid cooling systems are also classified as closed-loop and open-loop systems, depending on whether the liquid is discarded or recirculated after it is heated. In open-loop systems, tap water flows through the cooling system and is discarded into a drain after it is heated. The heated liquid in closed-loop systems is cooled in a heat exchanger and recirculated through the system. Closedloop systems facilitate better temperature control while conserving water. The electronic components in direct cooling systems are usually completely immersed in the liquid. The heat transfer from the components to the liquid can be by natural or forced convection or boiling, depending on the temperature levels involved and the properties of the fluids. Immersion cooling of electronic devices usually involves boiling and thus very high heat transfer coefficients, as discussed in the next section. Note that only dielectric fluids can be used in immersion or direct liquid cooling. This limitation immediately excludes water from consideration as a prospective fluid in immersion cooling. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 834 834 HEAT TRANSFER Expansion tank Air flow Heat exchanger Pump Electronic devices Liquid-cooled plate FIGURE 15–55 Schematic of an indirect liquid cooling system. Fluorocarbon fluids such as FC75 are well suited for direct cooling and are commonly used in such applications. Indirect liquid cooling systems of electronic devices operate just like the cooling system of a car engine, where the water (actually a mixture of water and ethylene glycol) circulates through the passages around the cylinders of the engine block, absorbing heat generated in the cylinders by combustion. The heated water is then routed by the water pump to the radiator, where it is cooled by air blown through the radiator coils by the cooling fan. The cooled water is then rerouted to the engine to transfer more heat. To appreciate the effectiveness of the cooling system of a car engine, it will suffice to say that the temperatures encountered in the engine cylinders are typically much higher than the melting temperatures of the engine blocks. In an electronic system, the heat is generated in the components instead of the combustion chambers. The components in this case are mounted on a metal plate made of a highly conducting material such as copper or aluminum. The metal plate is cooled by circulating a cooling fluid through tubes attached to it, as shown in Figure 15–55. The heated liquid is then cooled in a heat exchanger, usually by air (or sea water in marine applications), and is recirculated by a pump through the tubes. The expansion and storage tank accommodates any expansions and contractions of the cooling liquid due to temperature variations while acting as a liquid reservoir. The liquids used in the cooling of electronic equipment must meet several requirements, depending on the specific application. Desirable characteristics of cooling liquids include high thermal conductivity (yields high heat transfer coefficients), high specific heat (requires smaller mass flow rate), low viscosity (causes a smaller pressure drop, and thus requires a smaller pump), high surface tension (less likely to cause leakage problems), high dielectric strength (a must in direct liquid cooling), chemical inertness (does not react with surfaces with which it comes into contact), chemical stability (does not decompose under prolonged use), nontoxic (safe for personnel to handle), low freezing and high boiling points (extends the useful temperature range), and low cost. Different fluids may be selected in different applications because of the different priorities set in the selection process. The heat sinks or cold plates of an electronic enclosure are usually cooled by water by passing it through channels made for this purpose or through tubes attached to the cold plate. High heat removal rates can be achieved by circulating water through these channels or tubes. In high-performance systems, a refrigerant can be used in place of water to keep the temperature of the heat sink at subzero temperatures and thus reduce the junction temperatures of the electronic components proportionately. The heat transfer and pressure drop calculations in liquid cooling systems can be performed using appropriate relations. Liquid cooling can be used effectively to cool clusters of electronic devices attached to a tubed metal plate (or heat sink), as shown in Figure 15–56. Here 12 TO-3 cases, each dissipating up to 150 W of power, are mounted on a heat sink equipped with tubes on the back side through which a liquid flows. The thermal resistance between the case of the devices and the liquid is minimized in this case, since the electronic devices are mounted directly over the cooling lines. The case-to-liquid thermal resistance depends on the spacing between the devices, the quality of the thermal contact between the devices and the plate, the thickness of the plate, and the flow rate of the liquid, among other cen58933_ch15.qxd 9/9/2002 10:21 AM Page 835 835 CHAPTER 15 FIGURE 15–56 Liquid cooling of TO-3 packages placed on top of the coolant line (courtesy of Wakefield Engineering). things. The tubed metal plate shown is 15.2 cm 18 cm 2.5 cm in size and is capable of dissipating up to 2 kW of power. The thermal resistance network of a liquid cooling system is shown in Figure 15–57. The junction temperatures of silicon-based electronic devices are usually limited to 125°C. The junction-to-case thermal resistance of a device is provided by the manufacturer. The case-to-liquid thermal resistance can be determined experimentally by measuring the temperatures of the case and the liquid, and dividing the difference by the total power dissipated. The liquidto-air thermal resistance at the heat exchanger can be determined in a similar manner. That is, Rcase-liquid Tcase Tliquid, device , Q· Tliquid, hx Tair, in Rliquid-air Q· Junction Rjunction-case Case Rcase-liquid Liquid Rliquid-ambient (°C/W) (15-25) where Tliquid, device and Tliquid, hx are the inlet temperatures of the liquid to the electronic device and the heat exchanger, respectively. The required mass flow rate of the liquid corresponding to a specified temperature rise of the liquid as it flows through the electronic systems can be determined from Eq. 15-17. Electronic components mounted on liquid-cooled metal plates should be provided with good thermal contact in order to minimize the thermal resistance between the components and the plate. The thermal resistance can be minimized by applying silicone grease or beryllium oxide to the contact surfaces and fastening the components tightly to the metal plate. The liquid cooling of a cold plate with a large number of high-power components attached to it is illustrated in Example 15–17. Ambient FIGURE 15–57 Thermal resistance network for a liquid-cooled electronic device. Cold plate EXAMPLE 15–17 Cooling of Power Transistors on a Cold Plate by Water A cold plate that supports 20 power transistors, each dissipating 40 W, is to be cooled with water, as shown in Figure 15–58. Half of the transistors are attached to the back side of the cold plate. It is specified that the temperature rise of the water is not to exceed 3°C and the velocity of water is to remain under 1 m/s. Assuming that 20 percent of the heat generated is dissipated from Power transistor Water inlet Water exit FIGURE 15–58 Schematic for Example 15–17. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 836 836 HEAT TRANSFER the components to the surroundings by convection and radiation, and the remaining 80 percent is removed by the cooling water, determine the mass flow rate of water needed and the diameter of the pipe to satisfy the restriction imposed on the flow velocity. Also, determine the case temperature of the devices if the total case-to-liquid thermal resistance is 0.030°C/ W and the water enters the cold plate at 35°C. SOLUTION A cold plate is to be cooled by water. The mass flow rate of water, the diameter of the pipe, and the case temperature of the transistors are to be determined. Assumptions 1 Steady operating conditions exist. 2 About 20 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation. Properties The properties of water at room temperature are 1000 kg/m3 and Cp 4180 J/kg · °C. Analysis Noting that each of the 20 transistors dissipates 40 W of power and 80 percent of this power must be removed by the water, the amount of heat that must be removed by the water is · Q (20 transistors)(40 W/transistor)(0.80) 640 W In order to limit the temperature rise of the water to 3°C, the mass flow rate of water must be no less than m· · Q 640 J/s 0.051 kg/s 3.06 kg/min Cp Trise (4180 J/kg · °C)(3°C) The mass flow rate of a fluid through a circular pipe can be expressed as D 2 m· Ac 4 Then the diameter of the pipe to maintain the velocity of water under 1 m/s is determined to be D 4m· 4(0.051 kg/s) 0.0081 m 0.81 cm (1000 kg/m3)(1 m/s) Noting that the total case-to-liquid thermal resistance is 0.030°C/ W and the water enters the cold plate at 35°C, the case temperature of the devices is determined from Eq. 15-25 to be · Tcase Tliquid, device Q Rcase-liquid 35°C (640 W)(0.03°C/ W) 54.2°C The junction temperature of the device can be determined similarly by using the junction-to-case thermal resistance of the device supplied by the manufacturer. 15–10 ■ IMMERSION COOLING High-power electronic components can be cooled effectively by immersing them in a dielectric liquid and taking advantage of the very high heat transfer coefficients associated with boiling. Immersion cooling has been used since the 1940s in the cooling of electronic equipment, but for many years its cen58933_ch15.qxd 9/9/2002 10:21 AM Page 837 837 CHAPTER 15 use was largely limited to the electronics of high-power radar systems. The miniaturization of electronic equipment and the resulting high heat fluxes brought about renewed interest in immersion cooling, which had been largely viewed as an exotic cooling technique. You will recall from thermodynamics that, at a specified pressure, a fluid boils isothermally at the corresponding saturation temperature. A large amount of heat is absorbed during the boiling process, essentially in an isothermal manner. Therefore, immersion cooling also provides a constant-temperature bath for the electronic components and eliminates hot spots effectively. The simplest type of immersion cooling system involves an external reservoir that supplies liquid continually to the electronic enclosure. The vapor generated inside is simply allowed to escape to the atmosphere, as shown in Figure 15–59. A pressure relief valve on the vapor vent line keeps the pressure and thus the temperature inside at the preset value, just like the petcock of a pressure cooker. Note that without a pressure relief valve, the pressure inside the enclosure would be atmospheric pressure and the temperature would have to be the boiling temperature of the fluid at the atmospheric pressure. The open-loop-type immersion cooling system described here is simple, but there are several impracticalities associated with it. First of all, it is heavy and bulky because of the presence of an external liquid reservoir, and the fluid lost through evaporation needs to be replenished continually, which adds to the cost. Further, the release of the vapor into the atmosphere greatly limits the fluids that can be used in such a system. Therefore, the use of open-loop immersion systems is limited to applications that involve occasional use and thus have a light duty cycle. More sophisticated immersion cooling systems operate in a closed loop in that the vapor is condensed and returned to the electronic enclosure instead of being purged to the atmosphere. Schematics of two such systems are given in Figure 15–60. The first system involves a condenser external to the electronics Safety valve Liquid reservoir Pressure relief valve Safety valve Vapor Dielectric liquid FIGURE 15–59 A simple open-loop immersion cooling system. Air Air-cooled condenser Safety valve Fan Fins Coolant exit Vapor Vapor Coolant inlet Electronic components Dielectric liquid (a) System with external condenser Electronic components Dielectric liquid (b) System with internal condenser FIGURE 15–60 The schematics of two closed-loop immersion cooling systems. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 838 838 HEAT TRANSFER Expansion chamber Expansion chamber Pressure relief valve Coolant exit Pressure relief valve Fins Air outlet Coolant inlet Air inlet Dielectric liquid FIGURE 15–61 The schematics of two all-liquid immersion cooling systems. Electronic components (a) System with internal cooling Electronic Dielectric components liquid Fan (b) System with external cooling enclosure, and the vapor leaving the enclosure is cooled by a cooling fluid such as air or water outside the enclosure. The condensate is returned to the enclosure for reuse. The condenser in the second system is actually submerged in the electronic enclosure and is part of the electronic system. The cooling fluid in this case circulates through the condenser tube, removing heat from the vapor. The vapor that condenses drips on top of the liquid in the enclosure and continues to recirculate. The performance of closed-loop immersion cooling systems is most susceptible to the presence of noncondensable gases such as air in the vapor space. An increase of 0.5 percent of air by mass in the vapor can cause the condensation heat transfer coefficient to drop by a factor of up to 5. Therefore, the fluid used in immersion cooling systems should be degassed as much as practical, and care should be taken during the filling process to avoid introducing any air into the system. The problems associated with the condensation process and noncondensable gases can be avoided by submerging the condenser (actually, heat exchanger tubes in this case) in the liquid instead of the vapor in the electronic enclosure, as shown in Figure 15–61a. The cooling fluid, such as water, circulating through the tubes absorbs heat from the dielectric liquid at the top portion of the enclosure and subcools it. The liquid in contact with the electronic components is heated and may even be vaporized as a result of absorbing heat from the components. But these vapor bubbles collapse as they move up, as a result of transferring heat to the cooler liquid with which they come in contact. This system can still remove heat at high rates from the surfaces of electronic components in an isothermal manner by utilizing the boiling process, but its overall capacity is limited by the rate of heat that can be removed by the external cooling fluid in a liquid-to-liquid heat exchanger. Noting that the heat transfer coefficients associated with forced convection are far less than those associated with condensation, this all-liquid immersion cooling system is not suitable for electronic boxes with very high power dissipation rates per unit volume. A step further in the all-liquid immersion cooling systems is to remove the heat from the dielectric liquid directly from the outer surface of the electronics enclosure, as shown in Figure 15–61b. In this case, the dielectric liquid cen58933_ch15.qxd 9/9/2002 10:21 AM Page 839 839 CHAPTER 15 Air 1–3 atm Fluorochemical vapor Natural convection Silicone oil Transformer oil Fluorochemical liquids Air 1–3 atm Fluorochemical vapor Forced convection Transformer oil Fluorochemical liquids Boiling fluorochemical liquids 1 10 100 W/m 2 . °C 1000 10,000 inside the sealed enclosure is heated as a result of absorbing the heat dissipated by the electronic components. The heat is then transferred to the walls of the enclosure, where it is removed by external means. This immersion cooling technique is the most reliable of all since it does not involve any penetration into the electronics enclosure and the components reside in a completely sealed liquid environment. However, the use of this system is limited to applications that involve moderate power dissipation rates. The heat dissipation is limited by the ability of the system to reject the heat from the outer surface of the enclosure. To enhance this ability, the outer surfaces of the enclosures are often finned, especially when the enclosure is cooled by air. Typical ranges of heat transfer coefficients for various dielectric fluids suitable for use in the cooling of electronic equipment are given in Figure 15–62 for natural convection, forced convection, and boiling. Note that extremely high heat transfer coefficients (from about 1500 to 6000 W/m2 · °C) can be attained with the boiling of fluorocarbon fluids such as FC78 and FC86 manufactured by the 3M company. Fluorocarbon fluids, not to be confused with the ozone-destroying fluorochloro fluids, are found to be very suitable for immersion cooling of electronic equipment. They have boiling points ranging from 30°C to 174°C and freezing points below 50°C. They are nonflammable, chemically inert, and highly compatible with materials used in electronic equipment. Experimental results for the power dissipation of a chip having a heat transfer area of 0.457 cm2 and its substrate during immersion cooling in an FC86 bath are given in Figure 15–63. The FC86 liquid is maintained at a bulk temperature of 5°C during the experiments by the use of a heat exchanger. Heat transfer from the chip is by natural convection in regime A–B, and bubble formation and thus boiling begins in regime B–C. Note that the chip surface temperature suddenly drops with the onset of boiling because of the high heat transfer coefficients associated with boiling. Heat transfer is by nucleate boiling in regime C–D, and very high heat transfer rates can be achieved in this regime with relatively small temperature differences. FIGURE 15–62 Typical heat transfer coefficients for various dielectric fluids (from Hwang and Moran, Ref. 11). Chip power, W 20 Heat flux, W/m2 Tliquid = 5°C 43.8 D 21.9 10 9 8 7 6 5 4 17.5 C B 13.1 11.0 8.8 3 6.6 2 4.4 A 1 10 20 30 40 50 70 2.2 100 Tchip – Tliquid , °C FIGURE 15–63 Heat transfer from a chip immersed in the fluorocarbon fluid FC86 (from Hwang and Moran, Ref. 11). cen58933_ch15.qxd 9/9/2002 10:21 AM Page 840 840 HEAT TRANSFER EXAMPLE 15–18 4W Logic chip 80°C Dielectric fluid 20°C FIGURE 15–64 Schematic for Example 15–18. Immersion Cooling of a Logic Chip A logic chip used in an IBM 3081 computer dissipates 4 W of power and has a heat transfer surface area of 0.279 cm2, as shown in Figure 15–64. If the surface of the chip is to be maintained at 80°C while being cooled by immersion in a dielectric fluid at 20°C, determine the necessary heat transfer coefficient and the type of cooling mechanism that needs to be used to achieve that heat transfer coefficient. SOLUTION A logic chip is to be cooled by immersion in a dielectric fluid. The minimum heat transfer coefficient and the type of cooling mechanism are to be determined. Assumptions Steady operating conditions exist. Analysis The average heat transfer coefficient over the surface of the chip can be determined from Newton’s law of cooling, · Q hAs(Tchip Tfluid) Solving for h and substituting the given values, the convection heat transfer coefficient is determined to be h · Q 4W 2390 W/m2 · °C As(Tchip Tfluid) (0.279 10 4 m2)(80 20)°C which is rather high. Examination of Figure 15–62 reveals that we can obtain such high heat transfer coefficients with the boiling of fluorocarbon fluids. Therefore, a suitable cooling technique in this case is immersion cooling in such a fluid. A viable alternative to immersion cooling is the thermal conduction module discussed earlier. EXAMPLE 15–19 Cooling of a Chip by Boiling An 8-W chip having a surface area of 0.6 cm2 is cooled by immersing it into FC86 liquid that is maintained at a temperature of 15°C, as shown in Figure 15–65. Using the boiling curve in Figure 15–63, estimate the temperature of the chip surface. 8W Chip Dielectric liquid, 15°C FIGURE 15–65 Schematic for Example 15–19. SOLUTION A chip is cooled by boiling in a dielectric fluid. The surface temperature of the chip is to be determined. Assumptions The boiling curve in Figure 15–63 is prepared for a chip having a surface area of 0.457 cm2 being cooled in FC86 maintained at 5°C. The chart can be used for similar cases with reasonable accuracy. Analysis The heat flux is · Q 8W q· 13.3 W/cm2 As 0.6 cm2 Corresponding to this value on the chart is Tchip Tfluid 60°C. Therefore, Tchip Tfluid 60 15 60 75°C That is, the surface of this 8-W chip will be at 75°C as it is cooled by boiling in the dielectric fluid FC86. cen58933_ch15.qxd 9/9/2002 10:21 AM Page 841 841 CHAPTER 15 A liquid-based cooling system brings with it the possibility of leakage and associated reliability concerns. Therefore, the consideration of immersion cooling should be limited to applications that require precise temperature control and those that involve heat dissipation rates that are too high for effective removal by conduction or air cooling. SUMMARY Electric current flow through a resistance is always accompanied by heat generation, and the essence of thermal design is the safe removal of this internally generated heat by providing an effective path for heat flow from electronic components to the surrounding medium. In this chapter, we have discussed several cooling techniques commonly used in electronic equipment, such as conduction cooling, natural convection and radiation cooling, forced-air convection cooling, liquid cooling, immersion cooling, and heat pipes. In a chip carrier, heat generated at the junction is conducted along the thickness of the chip, the bonding material, the lead frame, the case material, and the leads. The junction-to-case thermal resistance Rjunction-case represents the total resistance to heat transfer between the junction of a component and its case. This resistance should be as low as possible to minimize the temperature rise of the junction above the case temperature. The epoxy board used in PCBs is a poor heat conductor, and so it is necessary to use copper cladding or to attach the PCB to a heat frame in conduction-cooled systems. Low-power electronic systems can be cooled effectively with natural convection and radiation. The heat transfer from a surface at temperature Ts to a fluid at temperature Tfluid by convection is expressed as · Q conv hAs(Ts Tfluid) (W) where h is the convection heat transfer coefficient and As is the heat transfer surface area. The natural convection heat transfer coefficient for laminar flow of air at atmospheric pressure is given by a simplified relation of the form T L 0.25 hK (W/m2 · °C) where T Ts Tfluid is the temperature difference between the surface and the fluid, L is the characteristic length (the length of the body along the heat flow path), and K is a constant, whose value is given in Table 15–1. The relations in Table 15–1 can also be used at pressures other than 1 atm by multiplying them by P, where P is the air pressure in atm. Radiation heat transfer between a surface at temperature Ts completely surrounded by a much larger surface at temperature Tsurr can be expressed as · 4 Q rad As (Ts4 Tsurr ) (W) where is the emissivity of the surface, As is the heat transfer surface area, and is the Stefan–Boltzmann constant, whose value is 5.67 108 W/m2 · K4 0.1714 108 Btu/h · ft2 · R4. Fluid flow over a body such as a transistor is called external flow, and flow through a confined space such as inside a tube or through the parallel passage area between two circuit boards in an enclosure is called internal flow. Fluid flow is also categorized as being laminar or turbulent, depending on the value of the Reynolds number. In convection analysis, the convection heat transfer coefficient is usually expressed in terms of the dimensionless Nusselt number Nu as h k Nu D (W/m2 · °C) where k is thermal conductivity of the fluid and D is the characteristic length of the geometry. Relations for the average Nusselt number based on experimental data are given in Table 15–2 for external flow and in Table 15–3 for laminar internal flow under the uniform heat flux condition, which is closely approximated by electronic equipment. In forced-air-cooled systems, the heat transfer can also be expressed as · Q m· Cp(Tout Tin) · (W) where Q is the rate of heat transfer to the air; Cp is the specific heat of air; Tin and Tout are the average temperatures of air at the inlet and exit of the enclosure, respectively; and m· is the mass flow rate of air. The heat transfer coefficients associated with liquids are usually an order of magnitude higher than those associated with gases. Liquid cooling systems can be classified as direct cooling and indirect cooling systems. In direct cooling systems, the electronic components are in direct contact with the liquid, and thus the heat generated in the components is transferred directly to the liquid. In indirect cooling systems, however, there is no direct contact with the components. Liquid cooling systems are also classified as closed-loop and open-loop systems, depending on whether the liquid is discharged or recirculated after it is heated. Only dielectric fluids can be used in immersion or direct liquid cooling. High-power electronic components can be cooled effectively by immersing them in a dielectric liquid and taking advantage of the very high heat transfer coefficients associated with cen58933_ch15.qxd 9/9/2002 10:21 AM Page 842 842 HEAT TRANSFER boiling. The simplest type of immersion cooling system involves an external reservoir that supplies liquid continually to the electronic enclosure. This open-loop-type immersion cooling system is simple but often impractical. Immersion cooling systems usually operate in a closed loop, in that the vapor is condensed and returned to the electronic enclosure instead of being purged to the atmosphere. REFERENCES AND SUGGESTED READING 1. E. P. Black and E. M. Daley. “Thermal Design Considerations for Electronic Components.” ASME Paper No. 70-DE-17, 1970. 2. R. A. Colclaser, D. A. Neaman, and C. F. Hawkins. Electronic Circuit Analysis. New York: John Wiley & Sons, 1984. 3. J. W. Dally. Packaging of Electronic Systems. New York: McGraw-Hill, 1990. 4. Design Manual of Cooling Methods for Electronic Equipment. NAVSHIPS 900-190. Department of the Navy, Bureau of Ships, March 1955. 5. Design Manual of Natural Methods of Cooling Electronic Equipment. NAVSHIPS 900-192. Department of the Navy, Bureau of Ships, November 1956. 6. G. N. Ellison. Thermal Computations for Electronic Equipment. New York: Van Nostrand Reinhold, 1984. 7. J. A. Gardner. “Liquid Cooling Safeguards High-Power Semiconductors.” Electronics, February 24, 1974, p. 103. 8. R. S. Gaugler. “Heat Transfer Devices.” U.S. Patent 2350348, 1944. 9. G. M. Grover, T. P. Cotter, and G. F. Erickson. “Structures of Very High Thermal Conductivity.” Journal of Applied Physics 35 (1964), pp. 1190–1191. W. F. Hilbert and F. H. Kube. “Effects on Electronic Equipment Reliability of Temperature Cycling in Equipment.” Report No. EC-69-400. Final Report, Grumman Aircraft Engineering Corporation, Bethpage, NY, February 1969. 11. U. P. Hwang and K. P. Moran. “Boiling Heat Transfer of Silicon Integrated Circuits Chip Mounted on a Substrate.” ASME HTD 20 (1981), pp. 53–59. 12. M. Jakob. Heat Transfer. Vol. 1. New York: John Wiley & Sons, 1949. 13. R. D. Johnson. “Enclosures—State of the Art.” New Electronics, July 29, 1983, p. 29. 14. A. D. Kraus and A. Bar-Cohen. Thermal Analysis and Control of Electronic Equipment. New York: McGraw-Hill/Hemisphere, 1983. 15. Reliability Prediction of Electronic Equipment. U.S. Department of Defense, MIL-HDBK-2178B, NTIS, Springfield, VA, 1974. 16. D. S. Steinberg. Cooling Techniques for Electronic Equipment. New York: John Wiley & Sons, 1980. 17. A. Zukauskas. “Heat Transfer from Tubes in Cross Flow.” In Advances in Heat Transfer, vol. 8, ed. J. P. Hartnett and T. F. Irvine, Jr. New York: Academic Press, 1972. PROBLEMS Introduction and History 15–1C What invention started the electronic age? Why did the invention of the transistor mark the beginning of a revolution in that age? 15–2C What is an integrated circuit? What is its significance in the electronics era? What do the initials MSI, LSI, and VLSI stand for? Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 15–3C When electric current I passes through an electrical element having a resistance R, why is heat generated in the element? How is the amount of heat generation determined? 15–4C Consider a TV that is wrapped in the blankets from all sides except its screen. Explain what will happen when the TV is turned on and kept on for a long time, and why. What will happen if the TV is kept on for a few minutes only? 15–5C Consider an incandescent light bulb that is completely wrapped in a towel. Explain what will happen when the light is turned on and kept on. (P.S. Do not try this at home!) 15–6C A businessman ties a large cloth advertisement banner in front of his car such that it completely blocks the airflow to the radiator. What do you think will happen when he starts the car and goes on a trip? cen58933_ch15.qxd 9/9/2002 10:21 AM Page 843 843 CHAPTER 15 15–7C Which is more likely to break: a car or a TV? Why? 15–8C Why do electronic components fail under prolonged use at high temperatures? resistor is 7.5 V and its surface temperature is not to exceed 150°C, determine the power at which it can operate safely in an Answer: 0.4 W ambient at 30°C. 15–9 The temperature of the case of a power transistor that is dissipating 12 W is measured to be 60°C. If the junction-tocase thermal resistance of this transistor is specified by the manufacturer to be 5°C/W, determine the junction temperature of the transistor. Answer: 120°C Reconsider Problem 15–14. Using EES (or other) software, plot the power at which the resistor can operate safely as a function of the ambient temperature as the temperature varies from 20ºC to 40ºC, and discuss the results. 15–10 Power is supplied to an electronic component from a 12-V source, and the variation in the electric current, the junction temperature, and the case temperatures with time are observed. When everything is stabilized, the current is observed to be 0.15 A and the temperatures to be 80°C and 55°C at the junction and the case, respectively. Calculate the junction-tocase thermal resistance of this component. Manufacturing of Electronic Equipment 15–11 A logic chip used in a computer dissipates 6 W of power in an environment at 55°C and has a heat transfer surface area of 0.32 cm2. Assuming the heat transfer from the surface to be uniform, determine (a) the amount of heat this chip dissipates during an 8-h work day, in kWh, and (b) the heat flux on the surface of the chip, in W/cm2. 15–12 A 15–cm 20-cm circuit board houses 90 closely spaced logic chips, each dissipating 0.1 W, on its surface. If the heat transfer from the back surface of the board is negligible, determine (a) the amount of heat this circuit board dissipates during a 10-h period, in kWh, and (b) the heat flux on the surface of the circuit board, in W/cm2. Chips 15 cm 20 cm FIGURE P15–12 15–13E A resistor on a circuit board has a total thermal resistance of 130°F/W. If the temperature of the resistor is not to exceed 360°F, determine the power at which it can operate safely in an ambient at 120°F. 15–14 Consider a 0.1-k resistor whose surface-to-ambient thermal resistance is 300°C/ W. If the voltage drop across the 15–15 15–16C Why is a chip in a chip carrier bonded to a lead frame instead of the plastic case of the chip carrier? 15–17C Draw a schematic of a chip carrier, and explain how heat is transferred from the chip to the medium outside of the chip carrier. 15–18C What does the junction-to-case thermal resistance represent? On what does it depend for a chip carrier? 15–19C What is a hybrid chip carrier? What are the advantages of hybrid electronic packages? 15–20C What is a PCB? Of what is the board of a PCB made? What does the “device-to-PCB edge” thermal resistance in conduction-cooled systems represent? Why is this resistance relatively high? 15–21C What are the three types of printed circuit boards? What are the advantages and disadvantages of each type? 15–22C What are the desirable characteristics of the materials used in the fabrication of the circuit boards? 15–23C What is an electronic enclosure? What is the primary function of the enclosure of an electronic system? Of what materials are the enclosures made? Cooling Load of Electronic Equipment and Thermal Environment 15–24C Consider an electronics box that consumes 120 W of power when plugged in. How is the heating load of this box determined? 15–25C Why is the development of superconducting materials generating so much excitement among designers of electronic equipment? 15–26C How is the duty cycle of an electronic system defined? How does the duty cycle affect the design and selection of a cooling technique for a system? 15–27C What is temperature cycling? How does it affect the reliability of electronic equipment? 15–28C What is the ultimate heat sink for (a) a TV, (b) an airplane, and (c) a ship? For each case, what is the range of temperature variation of the ambient? 15–29C What is the ultimate heat sink for (a) a VCR, (b) a spacecraft, and (c) a communication system on top of a mountain? For each case, what is the range of temperature variation of the ambient? cen58933_ch15.qxd 9/9/2002 10:21 AM Page 844 844 HEAT TRANSFER Electronics Cooling in Different Applications 15–30C How are the electronics of short-range and longrange missiles cooled? 15–31C What is dynamic temperature? What causes it? How is it determined? At what velocities is it significant? 15–32C How are the electronics of a ship or submarine cooled? 15–33C How are the electronics of the communication systems at remote areas cooled? 15–34C How are the electronics of high-power microwave equipment such as radars cooled? 15–35C How are the electronics of a space vehicle cooled? 15–36 Consider an airplane cruising in the air at a temperature of 25°C at a velocity of 850 km/h. Determine the temperature rise of air at the nose of the airplane as a result of the ramming effect of the air. 15–43C What is thermal resistance? To what is it analogous in electrical circuits? Can thermal resistance networks be analyzed like electrical circuits? Explain. 15–44C If the rate of heat conduction through a medium and the thermal resistance of the medium are known, how can the temperature difference between the two sides of the medium be determined? 15–45C Consider a wire of electrical resistance R, length L, and cross-sectional area A through which electric current I is flowing. How is the voltage drop across the wire determined? What happens to the voltage drop when L is doubled while I is held constant? · Now consider heat conduction at a rate of Q through the same wire having a thermal resistance of R. How is the temperature drop across the wire determined? What happens to the · temperature drop when L is doubled while Q is held constant? Wire A 850 km/h L FIGURE P15–45C FIGURE P15–36 15–37 The temperature of air in high winds is measured by a thermometer to be 12°C. Determine the true temperature of air Answer: 11.7°C if the wind velocity is 90 km/h. 15–38 Reconsider Problem 15–37. Using EES (or other) software, plot the true temperature of air as a function of the wind velocity as the velocity varies from 20 km/h to 120 km/h, and discuss the results. 15–39 Air at 25°C is flowing in a channel at a velocity of (a) 1, (b) 10, (c) 100, and (d) 1000 m/s. Determine the temperature that a stationary probe inserted into the channel will read for each case. 15–40 An electronic device dissipates 2 W of power and has a surface area of 5 cm2. If the surface temperature of the device is not to exceed the ambient temperature by more than 50°C, determine a suitable cooling technique for this device. Use Figure 15–17. 15–41E A stand-alone circuit board, 6 in. 8 in. in size, dissipates 20 W of power. The surface temperature of the board is not to exceed 165°F in an 85°F environment. Using Figure 15–17 as a guide, select a suitable cooling mechanism. 15–46C What is a heat frame? How does it enhance heat transfer along a PCB? Which components on a PCB attached to a heat frame operate at the highest temperatures: those at the middle of the PCB or those near the edge? 15–47C What is constriction resistance in heat flow? To what is it analogous in fluid flow through tubes and electric current flow in wires? 15–48C What does the junction-to-case thermal resistance of an electronic component represent? In practice, how is this value determined? How can the junction temperature of a component be determined when the case temperature, the power dissipation of the component, and the junction-to-case thermal resistance are known? 15–49C What does the case-to-ambient thermal resistance of an electronic component represent? In practice, how is this value determined? How can the case temperature of a component be determined when the ambient temperature, the power dissipation of the component, and the case-to-ambient thermal resistance are known? Conduction Cooling 15–50C Consider an electronic component whose junction-tocase thermal resistance Rjunction-case is provided by the manufacturer and whose case-to-ambient thermal resistance Rcase-ambient is determined by the thermal designer. When the power dissipation of the component and the ambient temperature are known, explain how the junction temperature can be determined. When Rjunction-case is greater than Rcase-ambient, will the case temperature be closer to the junction or ambient temperature? 15–42C What are the major considerations in the selection of a cooling technique for electronic equipment? 15–51C Why is the rate of heat conduction along a PCB very low? How can heat conduction from the mid-parts of a PCB to cen58933_ch15.qxd 9/9/2002 10:21 AM Page 845 845 CHAPTER 15 its outer edges be improved? How can heat conduction across the thickness of the PCB be improved? 12 cm 12 cm 15–52C Why is the warping of epoxy boards that are coppercladded on one side a major concern? What is the cause of this warping? How can the warping of PCBs be avoided? PCB 15–53C Why did the thermal conduction module receive so much attention from thermal designers of electronic equipment? How does the design of TCM differ from traditional chip carrier design? Why is the cavity in the TCM filled with helium instead of air? 15–54 Consider a chip dissipating 0.8 W of power in a DIP with 18 pin leads. The materials and the dimensions of various sections of this electronic device are given in the table. If the temperature of the leads is 50°C, estimate the temperature at the junction of the chip. Thermal Conductivity, Thickness, Heat Transfer W/m · °C mm Surface Area Section and Material Junction constriction — — Diameter 0.5 mm Silicon chip 120 0.5 4 mm 4 mm Eutectic bond 296 0.05 4 mm 4 mm Copper lead frame 386 0.25 4 mm 4 mm Plastic separator 1 Copper leads 386 Air gap 0.3 18 1 mm 0.25 mm 6 18 1 mm 0.25 mm Junction Bond wires Lid Case Chip Lead frame Leads Bond FIGURE P15–54 15–55 A fan blows air at 25°C over a 2-W plastic DIP with 16 leads mounted on a PCB at a velocity of 300 m/min. Using data from Figure 15–23, determine the junction temperature of the electronic device. What would the junction temperature be if the fan were to fail? 15–56 Heat is to be conducted along a PCB with copper cladding on one side. The PCB is 12 cm long and 12 cm wide, . Q Epoxy Copper 0.5 mm 0.06 mm FIGURE P15–56 and the thicknesses of the copper and epoxy layers are 0.06 mm and 0.5 mm, respectively. Disregarding heat transfer from the side surfaces, determine the percentages of heat conduction along the copper (k 386 W/m · °C) and epoxy (k 0.26 W/m · °C) layers. Also, determine the effective thermal conductivity of the PCB. Answers: 0.6 percent, 99.4 percent, 41.6 W/m · °C 15–57 Reconsider Problem 15–56. Using EES (or other) software, investigate the effect of the thickness of the copper layer on the percentage of heat conducted along the copper layer and the effective thermal conductivity of the PCB. Let the thickness vary from 0.02 mm to 0.10 mm. Plot the percentage of heat conducted along the copper layer and the effective thermal conductivity as a function of the thickness of the copper layer, and discuss the results. 15–58 The heat generated in the circuitry on the surface of a silicon chip (k 130 W/m · °C) is conducted to the ceramic substrate to which it is attached. The chip is 6 mm 6 mm in size and 0.5-mm thick and dissipates 3 W of power. Determine the temperature difference between the front and back surfaces of the chip in steady operation. 15–59E Consider a 6-in. 7-in. glass–epoxy laminate (k 0.15 Btu/h · ft · °F) whose thickness is 0.05 in. Determine the thermal resistance of this epoxy layer for heat flow (a) along the 7-in.-long side and (b) across its thickness. 15–60 Consider a 15–cm 18-cm glass–epoxy laminate (k 0.26 W/m · °C) whose thickness is 1.4 mm. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k 386 W/m · °C) of diameter 1 mm are to be planted throughout the board with a center-to-center distance of 3 mm. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification. 15–61 Reconsider Problem 15–60. Using EES (or other) software, investigate the effects of the thermal conductivity and the diameter of the filling material on the thermal resistance of the epoxy board. Let the thermal conductivity vary from 10 W/m · 0C to 400 W/m · 0C and the diameter from 0.5 mm to 2.0 mm. Plot the thermal resistance as cen58933_ch15.qxd 9/9/2002 10:21 AM Page 846 846 HEAT TRANSFER functions of the thermal conductivity and the diameter, and discuss the results. 15–62 A 12-cm 15–cm circuit board dissipating 45 W of heat is to be conduction-cooled by a 1.5-mm-thick copper heat frame (k 386 W/m · °C) 12 cm 17 cm in size. The epoxy laminate (k 0.26 W/m · °C) has a thickness of 2 mm and is attached to the heat frame with conductive epoxy adhesive (k 1.8 W/m · °C) of thickness 0.12 mm. The PCB is attached to a heat sink by clamping a 5-mm-wide portion of the edge to the heat sink from both ends. The temperature of the heat frame at this point is 30°C. Heat is uniformly generated on the PCB at a rate of 3 W per 1-cm 12-cm strip. Considering only onehalf of the PCB board because of symmetry, determine the maximum surface temperature on the PCB and the temperature distribution along the heat frame. Electronic components PCB Heat frame Cold plate Epoxy adhesive FIGURE P15–62 15–63 Consider a 15–cm 20-cm double-sided circuit board dissipating a total of 30 W of heat. The board consists of a 3-mm-thick epoxy layer (k 0.26 W/m · °C) with 1-mm-diameter aluminum wires (k 237 W/m · °C) inserted along the 20-cm-long direction, as shown in Figure P15–63. The distance between the centers of the aluminum wires is 2 mm. The circuit board is attached to a heat sink from both ends, and the temperature of the board at both ends is 30°C. Heat is considered to be uniformly generated on both sides of the epoxy layer Double-sided PCB Electronic components of the PCB. Considering only a portion of the PCB because of symmetry, determine the magnitude and location of the maxiAnswer: 88.7°C mum temperature that occurs in the PCB. 15–64 Repeat Problem 15–63, replacing the aluminum wires by copper wires (k 386 W/m · °C). 15–65 Repeat Problem 15–63 for a center-to-center distance of 4 mm instead of 2 mm between the wires. 15–66 Consider a thermal conduction module with 80 chips, each dissipating 4 W of power. The module is cooled by water at 18°C flowing through the cold plate on top of the module. The thermal resistances in the path of heat flow are Rchip 12°C/ W between the junction and the surface of the chip, Rint 9°C/ W between the surface of the chip and the outer surface of the thermal conduction module, and Rext 7°C/ W between the outer surface of the module and the cooling water. Determine the junction temperature of the chip. 15–67 Consider a 0.3-mm-thick epoxy board (k 0.26 W/m · °C) that is 15 cm 20 cm in size. Now a 0.1-mm-thick layer of copper (k 386 W/m · °C) is attached to the back surface of the PCB. Determine the effective thermal conductivity of the PCB along its 20-cm-long side. What fraction of the heat conducted along that side is conducted through copper? 15–68 A 0.5-mm-thick copper plate (k 386 W/m · °C) is sandwiched between two 3-mm-thick epoxy boards (k 0.26 W/m · °C) that are 12 cm 18 cm in size. Determine the effective thermal conductivity of the PCB along its 18-cm-long side. What fraction of the heat conducted along that side is conducted through copper? 18 cm Epoxy boards Copper plate 2 mm 12 cm 0.5 mm FIGURE P15–68 3 mm Aluminum wire FIGURE P15–63 15–69E A 6-in. 8-in. 0.06-in. copper heat frame is used to conduct 20 W of heat generated in a PCB along the cen58933_ch15.qxd 9/9/2002 10:21 AM Page 847 847 CHAPTER 15 8-in.-long side toward the ends. Determine the temperature difference between the midsection and either end of the heat Answer: 12.8°F frame. 15–70 A 12-W power transistor is cooled by mounting it on an aluminum bracket (k 237 W/m · °C) that is attached to a liquid-cooled plate by 0.2-mm-thick epoxy adhesive (k 1.8 W/m · °C), as shown in Figure P15–70. The thermal resistance of the plastic washer is given as 2.5°C/W. Preliminary calculations show that about 20 percent of the heat is dissipated by convection and radiation, and the rest is conducted to the liquid-cooled plate. If the temperature of the cold plate is 50°C, determine the temperature of the transistor case. Liquid channels Transistor 1 cm Aluminum bracket 15–78C For most effective natural convection cooling of a PCB array, should the PCBs be placed horizontally or vertically? Should they be placed close to each other or far from each other? 15–79C Why is radiation heat transfer from the components on the PCBs in an enclosure negligible? 15–80 Consider a sealed 20-cm-high electronic box whose base dimensions are 35 cm 50 cm that is placed on top of a stand in a room at 30°C. The emissivity of the outer surface of the box is 0.85. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 65°C, determine if this box can be cooled by natural convection and radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room. Electronic box 0.3 cm 35 cm 50 cm 2 cm 2 cm Plastic washer 100 W ε = 0.85 Ts ≤ 65°C 20 cm FIGURE P15–70 Air Cooling: Natural Convection and Radiation 15–71C A student puts his books on top of a VCR, completely blocking the air vents on the top surface. What do you think will happen as the student watches a rented movie played by that VCR? 15–72C Can a low-power electronic system in space be cooled by natural convection? Can it be cooled by radiation? Explain. 15–73C Why are there several openings on the various surfaces of a TV, VCR, and other electronic enclosures? What happens if a TV or VCR is enclosed in a cabinet with no free air space around? 15–74C Why should radiation heat transfer always be considered in the analysis of natural convection–cooled electronic equipment? 15–75C How does atmospheric pressure affect natural convection heat transfer? What are the best and worst orientations for heat transfer from a square surface? 15–76C What is view factor? How does it affect radiation heat transfer between two surfaces? 15–77C What is emissivity? How does it affect radiation heat transfer between two surfaces? Stand FIGURE P15–80 15–81 Repeat Problem 15–80, assuming the box is mounted on a wall instead of a stand such that it is 0.5 m high. Again, assume heat transfer from the bottom surface to the wall to be negligible. 15–82E A 0.15-W small cylindrical resistor mounted on a circuit board is 0.5 in. long and has a diameter of 0.15 in. The view of the resistor is largely blocked by the circuit board facing it, and the heat transfer from the connecting wires is negligible. The air is free to flow through the parallel flow passages between the PCBs as a result of natural convection currents. If the air temperature in the vicinity of the resistor is 130°F, determine the surface temperature of the resistor. Answer: 194°F 15–83 A 14-cm 20-cm PCB has electronic components on one side, dissipating a total of 7 W. The PCB is mounted in a rack vertically (height 14 cm) together with other PCBs. If the surface temperature of the components is not to exceed 90°C, determine the maximum temperature of the environment in which this PCB can operate safely at sea level. What would your answer be if this rack is located at a location at 3000 m altitude where the atmospheric pressure is 70.12 kPa? cen58933_ch15.qxd 9/9/2002 10:21 AM Page 848 848 HEAT TRANSFER surfaces is negligible. If the ambient air temperature is 25°C, determine (a) the heat transfer from the outer surfaces of the duct to the ambient air by natural convection and (b) the averAnswers: (a) 31.7 W, (b) 40°C age temperature of the duct. Warm air out Natural convection PCB 45°C 25°C Forced air 30°C 150 W 1m FIGURE P15–88 Cool air in FIGURE P15–83 15–89 Repeat Problem 15–88 for a circular horizontal duct of diameter 10 cm. 15–90 15–84 A cylindrical electronic component whose diameter is 2 cm and length is 4 cm is mounted on a board with its axis in the vertical direction and is dissipating 3 W of power. The emissivity of the surface of the component is 0.8, and the temperature of the ambient air is 30°C. Assuming the temperature of the surrounding surfaces to be 20°C, determine the average surface temperature of the component under combined natural convection and radiation cooling. 15–85 Repeat Problem 15–84, assuming the component is oriented horizontally. 15–86 Reconsider Problem 15–84. Using EES (or other) software, investigate the effects of surface emissivity and ambient temperature on the average surface temperature of the component. Let the emissivity vary from 0.1 to 1.0 and the ambient temperature from 15ºC to 35ºC. Take the temperature of the surrounding surfaces to be 10ºC smaller than the ambient air temperature. Plot the average surface temperature as functions of the emissivity and the ambient air temperature, and discuss the results. Reconsider Problem 15–88. Using EES (or other) software, investigate the effects of the volume flow rate of air and the side-length of the duct on heat transfer by natural convection and the average temperature of the duct. Let the flow rate vary from 0.1 m3/min to 0.5 m3/min, and the side-length from 10 cm to 20 cm. Plot the heat transfer rate by natural convection and the average duct temperature as functions of flow rate and side-length, and discuss the results. 15–91 Repeat Problem 15–88, assuming that the fan fails and thus the entire heat generated inside the duct must be rejected to the ambient air by natural convection from the outer surfaces of the duct. 15–92 A 20-cm 20-cm circuit board containing 81 square chips on one side is to be cooled by combined natural convection Chips 20 cm 15–87 Consider a power transistor that dissipates 0.1 W of power in an environment at 30°C. The transistor is 0.4 cm long and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the heat flux on the surface of the transistor, in W/cm2, and (b) the surface temperature of the transistor for a combined convection and radiation heat transfer coefficient of 18 W/m2 · °C. 15–88 The components of an electronic system dissipating 150 W are located in a 1-m-long horizontal duct whose cross section is 15 cm 15 cm. The components in the duct are cooled by forced air, which enters at 30°C at a rate of 0.4 m3/min and leaves at 45°C. The surfaces of the sheet metal duct are not painted, and thus radiation heat transfer from the outer 20 cm FIGURE P15–92 cen58933_ch15.qxd 9/9/2002 10:21 AM Page 849 849 CHAPTER 15 and radiation by mounting it on a vertical surface in a room at 25°C. Each chip dissipates 0.08 W of power, and the emissivity of the chip surfaces is 0.65. Assuming the heat transfer from the back side of the circuit board to be negligible and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. PCB, determine (a) the temperature at which the air leaves the hollow core and (b) the highest temperature on the inner surAnswers: (a) 56.4°C, (b) 67.6°C face of the core. Hollow core for air flow Electronic components 15–93 Repeat Problem 15–92, assuming the circuit board to be positioned horizontally with (a) chips facing up and (b) chips facing down. 15 cm Air Cooling: Forced Convection 15–94C Why is radiation heat transfer in forced-air-cooled systems disregarded? 15–95C If an electronic system can be cooled adequately by either natural convection or forced-air convection, which would you prefer? Why? FIGURE P15-104 15–96C Why is forced convection cooling much more effective than natural convection cooling? 15–105 Repeat Problem 15–104 for a hollow-core PCB dissipating 45 W. 15–97C Consider a forced-air-cooled electronic system dissipating a fixed amount of power. How will increasing the flow rate of air affect the surface temperature of the components? Explain. How will it affect the exit temperature of the air? 15–106 15–98C To what do internal and external flow refer in forced convection cooling? Give an example of a forced-air-cooled electronic system that involves both types of flow. 15–99C For a specified power dissipation and air inlet temperature, how does the convection heat transfer coefficient affect the surface temperature of the electronic components? Explain. 15–100C How does high altitude affect forced convection heat transfer? How would you modify your forced-air cooling system to operate at high altitudes safely? 15–101C What are the advantages and disadvantages of placing the cooling fan at the inlet or at the exit of an electronic box? 15–102C How is the volume flow rate of air in a forced-aircooled electronic system that has a constant-speed fan established? If a few more PCBs are added to the box while keeping the fan speed constant, will the flow rate of air through the system change? Explain. 15–103C What happens if we attempt to cool an electronic system with an undersized fan? What about if we do that with an oversized fan? 15–104 Consider a hollow-core PCB that is 15 cm high and 20 cm long, dissipating a total of 30 W. The width of the air gap in the middle of the PCB is 0.25 cm. The cooling air enters the core at 30°C at a rate of 1 L/s. Assuming the heat generated to be uniformly distributed over the two side surfaces of the Reconsider Problem 15–104. Using EES (or other) software, investigate the effects of the power rating of the PCB and the volume flow rate of the air on the exit temperature of the air and the maximum temperature on the inner surface of the core. Let the power vary from 20 W to 60 W and the flow rate from 0.5 L/s to 2.5 L/s. Plot the air exit temperature and the maximum surface temperature of the core as functions of power and flow rate, and discuss the results. 15–107E A transistor with a height of 0.25 in. and a diameter of 0.2 in. is mounted on a circuit board. The transistor is cooled by air flowing over it at a velocity of 400 ft/min. If the air temperature is 140°F and the transistor case temperature is not to exceed 175°F, determine the amount of power this transistor can dissipate safely. Answer: 0.15 W 15–108 A desktop computer is to be cooled by a fan. The electronic components of the computer consume 75 W of power under full-load conditions. The computer is to operate in environments at temperatures up to 45°C and at elevations up to 3400 m where the atmospheric pressure is 66.63 kPa. The exit temperature of air is not to exceed 60°C to meet reliability requirements. Also, the average velocity of air is not to exceed 110 m/min at the exit of the computer case, where the fan is installed to keep the noise level down. Determine the flow rate of the fan that needs to be installed and the diameter of the casing of the fan. 15–109 Repeat Problem 15–108 for a computer that consumes 100 W of power. 15–110 A computer cooled by a fan contains eight PCBs, each dissipating 12 W of power. The height of the PCBs is 12 cm and the length is 18 cm. The clearance between the tips of the components on the PCB and the back surface of the cen58933_ch15.qxd 9/9/2002 10:21 AM Page 850 850 HEAT TRANSFER adjacent PCB is 0.3 cm. The cooling air is supplied by a 15-W fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed 15°C, determine (a) the flow rate of the air that the fan needs to deliver, (b) the fraction of the temperature rise of air due to the heat generated by the fan and its motor, and (c) the highest allowable inlet air temperature if the surface temperature of the components is not to exceed 90°C anywhere in the system. 0.3 cm Air outlet 18 cm PCB, 12 W Air inlet 15–114 An enclosure contains an array of circuit boards, 15 cm high and 20 cm long. The clearance between the tips of the components on the PCB and the back surface of the adjacent PCB is 0.3 cm. Each circuit board contains 75 square chips on one side, each dissipating 0.15 W of power. Air enters the space between the boards through the 0.3-cm 15–cm cross section at 40°C with a velocity of 300 m/min. Assuming the heat transfer from the back side of the circuit board to be negligible, determine the exit temperature of the air and the highest surface temperature of the chips. 15–115 The components of an electronic system dissipating 120 W are located in a 1-m-long horizontal duct whose cross section is 20 cm 20 cm. The components in the duct are cooled by forced air, which enters at 30°C at a rate of 0.5 m3/min. Assuming 80 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 20 percent is lost through the outer surfaces of the duct, determine (a) the exit temperature of air and (b) the highest component surface temperature in the duct. 15–116 Repeat Problem 15–115 for a circular horizontal duct of diameter 10 cm. FIGURE P15–110 15–111 An array of power transistors, each dissipating 2 W of power, is to be cooled by mounting them on a 20-cm 20-cm square aluminum plate and blowing air over the plate with a fan at 30°C with a velocity of 3 m/s. The average temperature of the plate is not to exceed 60°C. Assuming the heat transfer from the back side of the plate to be negligible, determine the number of transistors that can be placed on this Answer: 9 plate. Aluminum plate number of transistors as functions of air velocity and maximum plate temperature, and discuss the results. Power transistors Air Liquid Cooling 15–117C If an electronic system can be cooled adequately by either forced-air cooling or liquid cooling, which one would you prefer? Why? 15–118C Explain how direct and indirect liquid cooling systems differ from each other. 15–119C Explain how closed-loop and open-loop liquid cooling systems operate. 15–120C What are the properties of a liquid ideally suited for cooling electronic equipment? 15–121 A cold plate that supports 10 power transistors, each dissipating 40 W, is to be cooled with water. It is specified that the temperature rise of the water not exceed 4°C and the velocity of water remain under 0.5 m/s. Assuming 25 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation, and the remaining 75 percent is removed by the cooling water, determine the mass FIGURE P15–111 Cold plate 15–112 Repeat Problem 15–111 for a location at an elevation of 1610 m where the atmospheric pressure is 83.4 kPa. Reconsider Problem 15–111. Using EES (or other) software, investigate the effects of air velocity and the maximum plate temperature on the number of transistors. Let the air velocity vary from 1 rn/s to 8 m/s and the maximum plate temperature from 40ºC to 80ºC. Plot the Power transistor Water inlet 15–113 Water exit FIGURE P15–121 cen58933_ch15.qxd 9/9/2002 10:21 AM Page 851 851 CHAPTER 15 flow rate of water needed and the diameter of the pipe to satisfy the restriction imposed on the flow velocity. Also, determine the case temperature of the devices if the total case-to-liquid thermal resistance is 0.04°C/ W and the water enters the cold plate at 25°C. 15–122 Reconsider Problem 15–121. Using EES (or other) software, investigate the effect of the maximum temperature rise of the water on the mass flow rate of water, the diameter of the pipe, and the case temperature. Let the maximum temperature rise vary from 1ºC to 10ºC. Plot the mass flow rate, the diameter, and the case temperature as a function of the temperature rise, and discuss the results. temperature of 25°C. Using the boiling curve in Figure 15–63, Answer: 82°C estimate the temperature of the chip surface. 15–132 A logic chip cooled by immersing it in a dielectric liquid dissipates 3.5 W of power in an environment at 50°C and has a heat transfer surface area of 0.8 cm2. The surface temperature of the chip is measured to be 95°C. Assuming the heat transfer from the surface to be uniform, determine (a) the heat flux on the surface of the chip, in W/cm2; (b) the heat transfer coefficient on the surface of the chip, in W/m2 · °C; and (c) the thermal resistance between the surface of the chip and the cooling medium, in °C/W. 15–123E Water enters the tubes of a cold plate at 95°F with an average velocity of 60 ft/min and leaves at 105°F. The diameter of the tubes is 0.25 in. Assuming 15 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation, and the remaining 85 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. 3.5 W Logic chip 95°C Answer: 263 W 15–124 A sealed electronic box is to be cooled by tap water flowing through channels on two of its sides. It is specified that the temperature rise of the water not exceed 3°C. The power dissipation of the box is 2 kW, which is removed entirely by water. If the box operates 24 h a day, 365 days a year, determine the mass flow rate of water flowing through the box and the amount of cooling water used per year. 15–125 3 kW. Repeat Problem 15–124 for a power dissipation of Immersion Cooling 15–126C What are the desirable characteristics of a liquid used in immersion cooling of electronic devices? 15–127C How does an open-loop immersion cooling system operate? How does it differ from closed-loop cooling systems? 15–128C How do immersion cooling systems with internal and external cooling differ? Why are externally cooled systems limited to relatively low-power applications? 15–129C Why is boiling heat transfer used in the cooling of very high-power electronic devices instead of forced air or liquid cooling? 15–130 A logic chip used in a computer dissipates 4 W of power and has a heat transfer surface area of 0.3 cm2. If the surface of the chip is to be maintained at 70°C while being cooled by immersion in a dielectric fluid at 20°C, determine the necessary heat transfer coefficient and the type of cooling mechanism that needs to be used to achieve that heat transfer coefficient. 15–131 A 6-W chip having a surface area of 0.5 cm2 is cooled by immersing it into FC86 liquid that is maintained at a Dielectric liquid, 50°C FIGURE P15–132 15–133 Reconsider Problem 15–132. Using EES (or other) software, investigate the effect of chip power on the heat flux, the heat transfer coefficient, and the convection resistance on chip surface. Let the power vary from 2 W to 10 W. Plot the heat flux, the heat transfer coefficient, and the thermal resistance as a function of power dissipated, and discuss the results. 15–134 A computer chip dissipates 5 W of power and has a heat transfer surface area of 0.4 cm2. If the surface of the chip is to be maintained at 55°C while being cooled by immersion in a dielectric fluid at 10°C, determine the necessary heat transfer coefficient and the type of cooling mechanism that needs to be used to achieve that heat transfer coefficient. 15–135 A 3-W chip having a surface area of 0.2 cm2 is cooled by immersing it into FC86 liquid that is maintained at a temperature of 45°C. Using the boiling curve in Figure 15–63, Answer: 108°C estimate the temperature of the chip surface. 15–136 A logic chip having a surface area of 0.3 cm2 is to be cooled by immersing it into FC86 liquid that is maintained at a temperature of 35°C. The surface temperature of the chip is not to exceed 60°C. Using the boiling curve in Figure 15–63, estimate the maximum power that the chip can dissipate safely. 15–137 A 2-kW electronic device that has a surface area of 120 cm2 is to be cooled by immersing it in a dielectric fluid with a boiling temperature of 60°C contained in a 1-m 1-m cen58933_ch15.qxd 9/9/2002 10:22 AM Page 852 852 HEAT TRANSFER 1-m cubic enclosure. Noting that the combined natural convection and the radiation heat transfer coefficients in air are typically about 10 W/m2 · °C, determine if the heat generated inside can be dissipated to the ambient air at 20°C by natural convection and radiation. If it cannot, explain what modification you could make to allow natural convection cooling. Also, determine the heat transfer coefficients at the surface of the electronic device for a surface temperature of 80°C. Assume the liquid temperature remains constant at 60°C throughout the enclosure. the maximum temperature that occurs in the PCB. Assume heat transfer from the top and bottom faces of the PCB to be negligiAnswer: 55.5°C ble. Copper Glass-epoxy Dielectric liquid Air 20°C 1m 1m 15 cm 15 cm FIGURE P15–140 60°C 80°C 1m 2 kW Electronic device FIGURE P15–137 Review Problems 15–138C Several power transistors are to be cooled by mounting them on a water-cooled metal plate. The total power dissipation, the mass flow rate of water through the tube, and the water inlet temperature are fixed. Explain what you would do for the most effective cooling of the transistors. 15–139C Consider heat conduction along a vertical copper bar whose sides are insulated. One person claims that the bar should be oriented such that the hot end is at the bottom and the cold end is at the top for better heat transfer, since heat rises. Another person claims that it makes no differences to heat conduction whether heat is conducted downward or upward, and thus the orientation of the bar is irrelevant. With which person do you agree? 15–140 Consider a 15–cm 15–cm multilayer circuit board dissipating 22.5 W of heat. The board consists of four layers of 0.1-mm-thick copper (k 386 W/m · °C) and three layers of 0.5-mm-thick glass–epoxy (k 0.26 W/m · °C) sandwiched together, as shown in Figure P15–140. The circuit board is attached to a heat sink from both ends, and the temperature of the board at those ends is 35°C. Heat is considered to be uniformly generated in the epoxy layers of the PCB at a rate of 0.5 W per 1-cm 15–cm epoxy laminate strip (or 1.5 W per 1-cm 15–cm strip of the board). Considering only a portion of the PCB because of symmetry, determine the magnitude and location of 15–141 Repeat Problem 15–140, assuming that the board consists of a single 1.5-mm-thick layer of glass–epoxy, with no copper layers. 15–142 The components of an electronic system that is dissipating 150 W are located in a 1-m-long horizontal duct whose cross section is 10 cm 10 cm. The components in the duct are cooled by forced air, which enters at 30°C and 50 m/min and leaves at 45°C. The surfaces of the sheet metal duct are not painted, and so radiation heat transfer from the outer surfaces is negligible. If the ambient air temperature is 30°C, determine (a) the heat transfer from the outer surfaces of the duct to the ambient air by natural convection, (b) the average temperature of the duct, and (c) the highest component surface temperature in the duct. 15–143 Two 10-W power transistors are cooled by mounting them on the two sides of an aluminum bracket (k 237 W/m · °C) that is attached to a liquid-cooled plate by 0.2-mm-thick epoxy adhesive (k 1.8 W/m · °C), as shown in Figure P15–143. The thermal resistance of each plastic washer is Liquid-cooled plate 40°C Transistor 3 cm Plastic washer 2 cm 1 cm FIGURE P15–143 0.3 cm cen58933_ch15.qxd 9/9/2002 10:22 AM Page 853 853 CHAPTER 15 given as 2°C/W, and the temperature of the cold plate is 40°C. The surface of the aluminum plate is untreated, and thus radiation heat transfer from it is negligible because of the low emissivity of aluminum surfaces. Disregarding heat transfer from the 0.3-cm-wide edges of the aluminum plate, determine the surface temperature of the transistor case. Also, determine the fraction of heat dissipation to the ambient air by natural convection and to the cold plate by conduction. Take the ambient temperature to be 25°C. 15–144E A fan blows air at 70°F and a velocity of 500 ft/min over a 1.5-W plastic DIP with 24 leads mounted on a PCB. Using data from Figure 15–23, determine the junction temperature of the electronic device. What would the junction temperature be if the fan were to fail? A 15–cm 18-cm double-sided circuit board dissipating a total of 18 W of heat is to be conduction-cooled by a 1.2-mm-thick aluminum core plate (k 237 W/m · °C) sandwiched between two epoxy laminates (k 0.26 W/m · °C). Each epoxy layer has a thickness of 0.5 mm and is attached to the aluminum core plate with conductive epoxy adhesive (k 1.8 W/m · °C) of thickness 0.1 mm. Heat is uniformly generated on each side of the PCB at a rate of 0.5 W per 1-cm 15–cm epoxy laminate strip. All of the heat is conducted along the 18-cm side since the PCB is cooled along the two 15–cm-long edges. Considering only part of the PCB board because of symmetry, determine the maximum temperature rise across the 9-cm distance between the Answer: 10.1°C center and the sides of the PCB. 15–145 15–146 Ten power transistors, each dissipating 2 W, are attached to a 7-cm 7-cm 0.2-cm aluminum plate with a square cutout in the middle in a symmetrical arrangement, as shown in Figure P15–146. The aluminum plate is cooled from two sides by liquid at 40°C. If 70 percent of the heat generated by the transistors is estimated to be conducted through the aluTransistors 40°C Aluminum plate minum plate, determine the temperature rise across the 1-cmwide section of the aluminum plate between the transistors and the heat sink. 15–147 The components of an electronic system are located in a 1.2-m-long horizontal duct whose cross section is 10 cm 20 cm. The components in the duct are not allowed to come into direct contact with cooling air, and so are cooled by air flowing over the duct at 30°C with a velocity of 250 m/min. The duct is oriented such that air strikes the 10-cm-high side of the duct normally. If the surface temperature of the duct is not to exceed 60°C, determine the total power rating of the electronic devices that can be mounted in the duct. What would your answer be if the duct is oriented such that air strikes the Answers: 481 W, 384 W 20-cm-high side normally? 15–148 Repeat Problem 15–147 for a location at an altitude of 5000 m, where the atmospheric pressure is 54.05 kPa. 15–149E A computer that consumes 65 W of power is cooled by a fan blowing air into the computer enclosure. The dimensions of the computer case are 6 in. 20 in. 24 in., and all surfaces of the case are exposed to the ambient, except for the base surface. Temperature measurements indicate that the case is at an average temperature of 95°F when the ambient temperature and the temperature of the surrounding walls are 80°F. If the emissivity of the outer surface of the case is 0.85, determine the fraction of heat lost from the outer surfaces of the computer case. Air 80°F 95°F Heat transfer 65 W 40°C Computer case FIGURE P15–149E Computer, Design, and Essay Problems 1 cm 5 cm FIGURE P15–146 7 cm Cut out section 4 cm × 4 cm 15–150 Bring an electronic device that is cooled by heat sinking to class and discuss how the heat sink enhances heat transfer. 15–151 Obtain a catalog from a heat sink manufacturer and select a heat sink model that can cool a 10-W power transistor safely under (a) natural convection and radiation and (b) forced convection conditions. 1 cm 15–152 Take the cover off a PC or another electronic box and identify the individual sections and components. Also, identify the cooling mechanisms involved and discuss how the current design facilitates effective cooling. cen58933_ch15.qxd 9/9/2002 10:22 AM Page 854 × Report "Heat Transfer - A Practical Approach - Yunus a. Cengel" Your name Email Reason Description Close Submit Contact information Ronald F. 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