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190000	https://math.stackexchange.com/questions/1423159/0-101001000100001000001-is-irrational	real analysis - $0.101001000100001000001$... is irrational. - 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 0.101001000100001000001 0.101001000100001000001... is irrational. [duplicate] Ask Question Asked 10 years ago Modified1 year, 1 month ago Viewed 3k times This question shows research effort; it is useful and clear 15 Save this question. Show activity on this post. This question already has answers here: Is 0.1010010001000010000010000001… transcendental? (1 answer) Prove that 2.101001000100001... is an irrational number. (5 answers) Closed 10 years ago. How do I show that 0.101001000100001000001... is irrational? How do I generalize this to decimal expansions of a similar type, i.e. what is a criterion for decimal expansions with 0's and 1's involving long gaps of zeros and which guarantees irrationality? real-analysis proof-writing rationality-testing Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Sep 14, 2015 at 4:19 Martin Sleziak 56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges asked Sep 5, 2015 at 15:46 user268209 user268209 6 5 There is no periodicity in this development. A rational number has a periodic development.Bernard –Bernard 2015-09-05 15:49:29 +00:00 Commented Sep 5, 2015 at 15:49 1 Can you show that a number is rational if and only if it has a repeating decimal expansion?Cameron L. Williams –Cameron L. Williams 2015-09-05 15:49:39 +00:00 Commented Sep 5, 2015 at 15:49 @Bernard Is there any closed form for this number?SMA.D –SMA.D 2015-09-05 15:51:41 +00:00 Commented Sep 5, 2015 at 15:51 An algebraic formula? I'm afraid not. At least, I know none.Bernard –Bernard 2015-09-05 15:56:30 +00:00 Commented Sep 5, 2015 at 15:56 5 @SMA.D Turns out it's transcendentalPeter Woolfitt –Peter Woolfitt 2015-09-05 16:00:17 +00:00 Commented Sep 5, 2015 at 16:00 \|Show 1 more comment 5 Answers 5 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. A well-known, general criterion to decide if a number is irrational or not from its decimal expression is that a number is rational if and only if its decimal expression is finite or periodic. The decimal expression of the number mentioned in your question has arbitrarily long gaps of zeroes. So it can not be periodic, or much less finite. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 5, 2015 at 15:51 ajotatxeajotatxe 67k 3 3 gold badges 60 60 silver badges 110 110 bronze badges 1 Can you give link with proof of this fact? "a number is irrational or not from its decimal expression is that a number is rational if and only if its decimal expression is finite or periodic"RFZ –RFZ 2015-09-05 15:59:12 +00:00 Commented Sep 5, 2015 at 15:59 Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. If 0<p<q, where p,q∈Z, the decimal expansion for p/q is eventually periodic. Indeed, the first k digits after the decimal point are given by ⌊10 k p/q⌋. Write m Δ n if 10 m and 10 n have the same remainder when we divide by q; the pigeonhole principle implies m Δ n,for some 1≤m<n≤q+1. If m Δ n, then q\|(10 m p−10 n p)⟹q\|(10(10 m p−10 n p))⟹q\|(10 m+1 p−10 n+1 p)⟹(m+1)Δ(n+1), and by induction, (m+k)Δ(n+k) for all k, Or in other form a Δ b if b−a=n−m which m≤a and n≤b(equivalent to 0≤k) . Next, note that if m Δ n, and r is the remainder when 10 m p and 10 n p are divided by q, then ⌊10 r/q⌋ is the (m+1)st digit of the decimal expansion of p/q, and the (n+1)st. Thus a Δ b similarly results that the (a+1)th digit and the (b+1)th digit of the expansion of p/q are equal, and because it occurs for any a and b that b−a=n−m which m≤a and n≤b, Thus p/q has eventually periodic expansion with period n−m. Now, the given decimal does not have period P for any P because for arbitrarily large N, there is a 1 in the N th decimal place followed by P zeros. More generally, if the decimal expansion of any x∈Q has arbitrarily long strings of zeros, and repeats with period K, then for any N, there exists K consecutive zeros after the N th spot, so that the K repeating digits are zeros. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 2, 2024 at 14:55 Mohammadali 3 3 3 bronze badges answered Sep 5, 2015 at 16:23 user268197 user268197 1 1 (1) I went ahead and placed your implications into an align environment, so hopefully you find it an improvement. (2) The m Δ n notation seems atypical to me; more standard is 10 m=10 n mod q.Semiclassical –Semiclassical 2015-09-05 22:36:36 +00:00 Commented Sep 5, 2015 at 22:36 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. If a=0.1010010001… is rational, then it either has a finite decimal representation, or it's decimal representation repeats with some period p. This decimal is obviously not finite, so if it is rational, the decimal repeats with period p. However, somewhere after the repeating starts, we get a string of 0's followed by a 1 of length greater than p. Hence the number is irrational. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 5, 2015 at 15:51 Peter WoolfittPeter Woolfitt 21.5k 6 6 gold badges 59 59 silver badges 91 91 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The given number x=∞∑n=1 10−n(n+1)/2 is irrational. Given any rational number r>0, we can rewrite it into the form r=1 10 a(b+p q) where a,b,p,q∈N, 0≤p<q and gcd(10,q)=1. If p=0 or q=1, everything is obvious. Consider the case 0<p<q and let φ(q) be the Euler's totient function. By Fermat-Euler theorem, we have 10 φ(q)≡1(mod q)⟹10 φ(q)−1=q s for some integer s This leads to a decimal expansion of 1 q, 1 q=s 10 φ(q)−1=∞∑k=1 s×10−k φ(q) which repeats every φ(q) digits. Multiply this expansion by p gives us a decimal expansion of p q which again repeat every φ(q) digits. Adding b and then divide by 10 a will introduce some garbage in the front and shift the whole expansion to the right for a digits. At the end, we continue to have an expansion which is ultimately periodic with a period which is a divisor of φ(q). Back to our number x, one find the sequence of '0' in its decimal expansion can be arbitrary long. Given any q, one can always a find a sequence of '0' longer than φ(q). This rules out the possibilities that x is a rational number with a corresponding q. As a result, x cannot be rational. In fact, the number x is known to be transcendental. See the refs in my answer to a related question. 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 12:21 CommunityBot 1 answered Sep 5, 2015 at 16:52 achille huiachille hui 126k 7 7 gold badges 189 189 silver badges 365 365 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. what is a criterion for decimal expansions with 0's and 1's involving long gaps of zeros and which guarantees irrationality? To give one answer to this more general part of your question, a sufficient criterion for irrationality is the existence of arbitrarily long sequences of 0 s and 1 s in the expansion. From this, you can show aperiodicity, and in turn irrationality. To show that such sequences must be aperiodic, consider such a sequence. Now, assume towards a contradiction that it eventually has a periodic repeating element p. First, you can easily show that p must contain at least one 1 and one 0, otherwise the sequence would end in all 0 s or all 1 s, contradicting the existence of arbitrary length 0 s and 1 s. But then, by similar reasoning to the other answers, we use the arbitrary-length 0 s condition to show that p must be contained within some sequence of all 0 s, contradicting the fact that it contains a 1. So such a sequence is aperiodic, and hence irrational. Note that this is a sufficient condition, not a necessary one. It can be strengthened. In fact, your sequence doesn't meet this condition (it only has arbitrary length 0 s, not 1 s). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 5, 2015 at 16:05 Colm BhandalColm Bhandal 4,847 16 16 silver badges 40 40 bronze badges Add a comment\| 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 52Is 0.1010010001000010000010000001… transcendental? 13Prove that 2.101001000100001... is an irrational number. 0show that 0,10100100010000100000···∉Q Related 3Using Ruler to Measure Irrational Number 0"Multivariable" version of this lemma about showing analytically that a number is irrational. 5If a is rational and b is irrational, is a b necessarily irrational? 0How To Show The Irrationality Of 0.0110001111... 1Proof that there exists a rational^irrational=irrational 1Show a function assumes all values in [0,1] on any open interval (a,b)⊂(0,1) Hot Network Questions Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? 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190001	https://sites.pitt.edu/~super1/Pirzado/CMAM%20Pakistan%20OPD%20Job-Aids.pdf	Compiled and Edited By Dr. Abdul Rehman Pirzado JOB AIDS Community Management of Acute Malnutrition (CMAM) Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 1 of 38 Bilateral Pitting Edema [All ages] Bilateral pitting oedema, or kwashiorkor, usually starts in the feet and ankles. Oedema in only one foot is not of nutritional origin. Hold the child’s feet and press thumbs on top of both feet. Count to 3 and then lift your thumbs. If no pit shows or if a pit only shows in one foot, the child does not have bilateral pitting oedema. If a pit shows in both feet, go to Step 2. Continue the same test on the lower legs, hands and lower arms. If no pitting appears in these areas, then the child is said to have Grade + or mild bilateral pitting oedema. Mild bilateral pitting oedema only shows in the feet. If pitting appears in these areas, go to Step 3. Look for swelling in the face, especially around the eyes. If no swelling appears in the face, then the child is said to have Grade ++, or moderate, bilateral pitting oedema. If swelling appears in the face, the child is said to have Grade +++ or severe bilateral pitting oedema. Have a second person repeat the test to confirm results. 1 2 3 4 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 MUAC (without aid) -[6-59 months] Have the person bend his/her left arm to 90-degree angle. Determine both the top of the shoulder and the tip of the elbow bone and put the start point of the MUAC tape to top of shoulder and mark the distance to the tip of elbow placing the right thumb on the tape (endpoint). Find the middle of the upper arm, by carefully folding the endpoint to the start point of the MUAC tape and place the left thumb on the mid-point of the folded ends of the tape With the person’s arm relaxed and falling alongside his/her body, wrap MUAC tape around the arm at the mid-point. There shouldn’t be any space between the skin and tape, but don’t wrap the tape too tight. Read MUAC in millimeters (mm), centimeters (cm) or in color (red, yellow, green) from middle window exactly where arrows point inward. MUAC is recorded with a precision of 1 mm (0.1 cm). 1 2 3 4 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 3 of 38 MUAC (with pen & string) [6-59 months] MUAC is always taken on the left arm. Measure the length of the person’s left upper arm, between the bone at the top of the shoulder and the tip of the elbow bone (the arm should be bent). Fold the end points of the string together on the shoulder top and mark the middle of the person’s upper arm with a pen. With the person’s arm relaxed and falling alongside his/her body, wrap the MUAC tape around the arm at the mid-point. There shouldn’t be any space between the skin and tape, but don’t wrap the tape too tight. For numbered tapes, feed the end of the tape down through the first opening and up through the third opening. For the simple three-color tape (red, yellow, green), slide the end through the first opening and then through the second opening. Read the measurement from the middle window where the two arrows meet. Depending on the tape used, the measurement will be in millimeters (mm), centimeters (cm) or in color (red, yellow, green). MUAC is recorded with a precision of 1 mm (0.1 cm). 1 2 3 4 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Measure Using Length Board [Under 2 years OR height below 87 cm] Place height board on the ground and remove child’s shoes. Place child on his/her back in middle of board, head facing straight up, arms at child’s sides and feet at right angles to board. While holding child’s ankles or knees, move sliding board up against bottom of child’s feet. Take measurement to nearest 0.1 cm and read out loud. Assistant, holding head in place, repeats the measurement for verification. If child is 2 years or older or 87 cm or above standing up, subtract 0.7 cm from measurement. 1 2 3 4 5 6 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 5 of 38 Measure Using Height Board [2 years or older OR height 87 cm or greater] Place child on height board, standing upright in middle of board with arms at his/her sides. Child’s feet, with shoes removed, should be close together with heels flat on floor. Child’s ankles and knees should be firmly pressed against board. Child’s heels, back of legs, buttocks, shoulders and head should be touching back of board. Measurer holds child’s head straight. A line between child’s eyes should be parallel to the floor. Measurer reads measurement out loud to nearest 0.1 cm. Assistant, holding child’s legs and feet, repeats the measurement for verification. Measurer records height to nearest 0.1 cm. 1 2 3 4 5 6 7 8 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Weigh Using Electronic Scale (Solar) [All ages] “Tared weighing” means that the scale can be re-set to zero (“tared”) with the person just weighed still on it. Stress that the mother must stay on the scale until her child has been weighed in her arms. If the child is 2 years or older, you will weigh the child alone if the child will stand still. Be sure that the scale is placed on a flat, hard, even surface. Since the scale is solar powered, there must be enough light to operate the scale. To turn on the scale, cover the solar panel for a second. When the number 0.0 appears, the scale is ready. Check to see that the mother has removed her shoes. You or someone else should hold the naked baby wrapped in a blanket. Ask the mother to stand in the middle of the scale, feet slightly apart (on the footprints, if marked), and remain still. The mother’s clothing must not cover the display or solar panel. Remind her to stay on the scale even after her weight appears, until the baby has been weighed in her arms. With the mother still on the scale and her weight displayed, tare the scale by covering the solar panel for a second. The scale is tared when it displays a figure of a mother and baby and the number 0.0. Gently hand the naked baby to the mother and ask her to remain still. The baby’s weight will appear on the display. Record the weight. Be careful to read the numbers in the correct order (as though you were viewing while standing on the scale rather than upside-down). 1 2 3 4 5 6 7 8 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 7 of 38 Weigh Using Hanging Scale (Pants) [Under 5] The scale should be checked daily against a known weight. To do this, set the scale to zero and weigh objects of known weight (e.g., 5.0 kg, 10.0 kg, 15.0 kg). If the measure does not match the weight to within 10 grams the springs must be changed or the scale should be replaced. The scale can be hooked to a rope on the ceiling or stand in a clinic, at eye level of the measurer. Before weighing the child, take all his/her clothes off. Make sure the weighing scale arrow is at 0 (zero the scale) with the weighing pants hooked on the scale. Place child in weighing pants and let child hang freely, touching nothing. Make sure the child is safely in the weighing pants with one arm in front and one arm behind the straps to help maintain balance. When arrow is steady, measurer reads child’s weight in kg at eye level and to the nearest 100 g (for example: 6.4 kg). The assistant repeats it for verification and records it. The child is then dressed. 1 2 3 4 5 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Weigh Using Hanging Scale (Bucket) [What age?] The scale can be hooked to a rope on the ceiling or stand in a clinic, at eye level of the measurer. Before weighing the child take all his/her clothes off. Make sure the weighing scale arrow is at 0 with the bucket hooked on the scale Place child in weighing bucket. When arrow is steady, measurer reads child’s weight in kg at eye level and assistant records it to nearest 100 g (for example: 5.2kg) 1 2 3 4 5 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 9 of 38 Weigh Using Hanging Scale (Hammock or Cloth) [Infants] The scale should be checked daily against a known weight. To do this, set the scale to zero and weigh objects of known weight (e.g., 5.0 kg, 10.0 kg, 15.0 kg). If the measure does not match the weight to within 10 grams the springs must be changed or the scale should be replaced. The scale can be hooked to a rope on the ceiling or stand in a clinic, at eye level of the measurer. Before weighing the child, take all his/her clothes off. Make sure the weighing scale arrow is at 0 (zero the scale) each time with the hammock or cloth that will be used hooked on the scale. Place child in hammock or cloth, hook it on the scale, and let child hang freely, touching nothing. Make sure the child is safely in the hammock or cloth. When arrow is steady, measurer reads child’s weight in kg at eye level and to the nearest 100 g (for example: 6.4 kg). The assistant repeats it for verification and records it. The child is then dressed. When arrow is steady, measurer reads child’s weight in kg at eye level and assistant records it to nearest 100 g (for example: 5.2kg) 1 2 3 4 5 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Admission and Discharge Criteria for CMAM in Children under Five Inpatient Care Outpatient Care ADMISSION CRITERIA CHILDREN 6-59 MONTHS Bilateral pitting oedema +++ CHILDREN 6-59 MONTHS Bilateral pitting oedema + and ++ OR Any grade of bilateral pitting oedema with severe wasting OR Severe wasting (MUAC < 115 mm or WFH < -3 z-score) AND  Appetite test passed  No medical complication  Child clinically well and alert OR SAM with any of the following medical complications:  Anorexia, poor appetite  Intractable vomiting  Convulsions  Lethargy, not alert  Unconsciousness  Hypoglycaemia  High fever  Hypothermia  Severe dehydration  Persistent diarrhoea  Lower respiratory tract infection  Severe anaemia  Eye signs of vitamin A deficiency  Skin lesion OR Referred from outpatient care according to action protocol INFANTS < 6 MONTHS Bilateral pitting oedema OR Visible wasting - Includes infants with SAM ≥ 6 months and < 4 kg REFERRAL/DISCHARGE CRITERIA CHILDREN 6-59 MONTHS Referred to outpatient care:  Appetite returned (passed appetite test)  Medical complication resolving  Severe bilateral pitting oedema decreasing  Child clinically well and alert (If admitted due to bilateral pitting oedema and severe wasting: criterion for referral is bilateral pitting oedema resolved) Discharged cured (special cases):  15 percent weight gain maintained for two consecutive days  Oedema free for two consecutive weeks  Child clinically well and alert INFANTS < 6 MONTHS Discharged cured: If successful re-lactation and appropriate weight gain maintained (minimum 20 g per day on breastfeeding alone for five days) and infant clinically well and alert (if infant has no access to breastfeeding, see other guidance for non-breastfed children on replacement feeding) CHILDREN 6-59 MONTHS Discharged cured:  15 percent weight gain maintained for two consecutive visits (of admission weight or weight free of oedema)  Oedema free for two consecutive visits  Child clinically well and alert Children are referred to receive supplementary feeding if available Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 11 of 38 Weight-for-Length Look-Up Table for Children 6-23 Months [WHO 2006 Child Growth Standards] Length is measured for children under 2 years or less than 87 cm height. For children 2 years or older or 87 cm height or greater, height is measured. Recumbent length is, on average, 0.7 cm greater than standing height; although the difference is of no importance to individual children, a correction may be made by subtracting 0.7 cm from all lengths above 86.9 cm if standing height cannot be measured. Boys' weight (kg) Length Girls' weight (kg) -3 SD -2 SD -1 SD Median (cm) Median -1 SD -2 SD -3 SD 1.9 2.0 2.2 2.4 45 2.5 2.3 2.1 1.9 2.0 2.2 2.4 2.6 46 2.6 2.4 2.2 2.0 2.1 2.3 2.5 2.8 47 2.8 2.6 2.4 2.2 2.3 2.5 2.7 2.9 48 3.0 2.7 2.5 2.3 2.4 2.6 2.9 3.1 49 3.2 2.9 2.6 2.4 2.6 2.8 3.0 3.3 50 3.4 3.1 2.8 2.6 2.7 3.0 3.2 3.5 51 3.6 3.3 3.0 2.8 2.9 3.2 3.5 3.8 52 3.8 3.5 3.2 2.9 3.1 3.4 3.7 4.0 53 4.0 3.7 3.4 3.1 3.3 3.6 3.9 4.3 54 4.3 3.9 3.6 3.3 3.6 3.8 4.2 4.5 55 4.5 4.2 3.8 3.5 3.8 4.1 4.4 4.8 56 4.8 4.4 4.0 3.7 4.0 4.3 4.7 5.1 57 5.1 4.6 4.3 3.9 4.3 4.6 5.0 5.4 58 5.4 4.9 4.5 4.1 4.5 4.8 5.3 5.7 59 5.6 5.1 4.7 4.3 4.7 5.1 5.5 6.0 60 5.9 5.4 4.9 4.5 4.9 5.3 5.8 6.3 61 6.1 5.6 5.1 4.7 5.1 5.6 6.0 6.5 62 6.4 5.8 5.3 4.9 5.3 5.8 6.2 6.8 63 6.6 6.0 5.5 5.1 5.5 6.0 6.5 7.0 64 6.9 6.3 5.7 5.3 5.7 6.2 6.7 7.3 65 7.1 6.5 5.9 5.5 5.9 6.4 6.9 7.5 66 7.3 6.7 6.1 5.6 6.1 6.6 7.1 7.7 67 7.5 6.9 6.3 5.8 6.3 6.8 7.3 8.0 68 7.7 7.1 6.5 6.0 6.5 7.0 7.6 8.2 69 8.0 7.3 6.7 6.1 6.6 7.2 7.8 8.4 70 8.2 7.5 6.9 6.3 6.8 7.4 8.0 8.6 71 8.4 7.7 7.0 6.5 7.0 7.6 8.2 8.9 72 8.6 7.8 7.2 6.6 7.2 7.7 8.4 9.1 73 8.8 8.0 7.4 6.8 7.3 7.9 8.6 9.3 74 9.0 8.2 7.5 6.9 7.5 8.1 8.8 9.5 75 9.1 8.4 7.7 7.1 7.6 8.3 8.9 9.7 76 9.3 8.5 7.8 7.2 7.8 8.4 9.1 9.9 77 9.5 8.7 8.0 7.4 7.9 8.6 9.3 10.1 78 9.7 8.9 8.2 7.5 8.1 8.7 9.5 10.3 79 9.9 9.1 8.3 7.7 8.2 8.9 9.6 10.4 80 10.1 9.2 8.5 7.8 8.4 9.1 9.8 10.6 81 10.3 9.4 8.7 8.0 8.5 9.2 10.0 10.8 82 10.5 9.6 8.8 8.1 8.7 9.4 10.2 11.0 83 10.7 9.8 9.0 8.3 8.9 9.6 10.4 11.3 84 11.0 10.1 9.2 8.5 9.1 9.8 10.6 11.5 85 11.2 10.3 9.4 8.7 9.3 10.0 10.8 11.7 86 11.5 10.5 9.7 8.9 9.5 10.2 11.1 12.0 87 11.7 10.7 9.9 9.1 9.7 10.5 11.3 12.2 88 12.0 11.0 10.1 9.3 9.9 10.7 11.5 12.5 89 12.2 11.2 10.3 9.5 10.1 10.9 11.8 12.7 90 12.5 11.4 10.5 9.7 10.3 11.1 12.0 13.0 91 12.7 11.7 10.7 9.9 10.5 11.3 12.2 13.2 92 13.0 11.9 10.9 10.1 10.7 11.5 12.4 13.4 93 13.2 12.1 11.1 10.2 10.8 11.7 12.6 13.7 94 13.5 12.3 11.3 10.4 11.0 11.9 12.8 13.9 95 13.7 12.6 11.5 10.6 11.2 12.1 13.1 14.1 96 14.0 12.8 11.7 10.8 11.4 12.3 13.3 14.4 97 14.2 13.0 12.0 11.0 11.6 12.5 13.5 14.6 98 14.5 13.3 12.2 11.2 11.8 12.7 13.7 14.9 99 14.8 13.5 12.4 11.4 12.0 12.9 14.0 15.2 100 15.0 13.7 12.6 11.6 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Weight-for-Height Look-Up Table, Children 24-59 Months [WHO 2006 Child Growth Standards] Length is measured for children under 2 years or less than 87 cm height. For children 2 years or older or 87 cm height or more, height is measured. Recumbent length is, on average, 0.7 cm greater than standing height; although the difference is of no importance to individual children, a correction may be made by subtracting 0.7 cm from all lengths greater than 86.9 cm if standing height cannot be measured. Boys' weight (kg) Length Girls' weight (kg) -3 SD -2 SD -1 SD Median (cm) Median -1 SD -2 SD -3 SD 5.9 6.3 6.9 7.4 65 7.2 6.6 6.1 5.6 6.1 6.5 7.1 7.7 66 7.5 6.8 6.3 5.8 6.2 6.7 7.3 7.9 67 7.7 7.0 6.4 5.9 6.4 6.9 7.5 8.1 68 7.9 7.2 6.6 6.1 6.6 7.1 7.7 8.4 69 8.1 7.4 6.8 6.3 6.8 7.3 7.9 8.6 70 8.3 7.6 7.0 6.4 6.9 7.5 8.1 8.8 71 8.5 7.8 7.1 6.6 7.1 7.7 8.3 9.0 72 8.7 8.0 7.3 6.7 7.3 7.9 8.5 9.2 73 8.9 8.1 7.5 6.9 7.4 8.0 8.7 9.4 74 9.1 8.3 7.6 7.0 7.6 8.2 8.9 9.6 75 9.3 8.5 7.8 7.2 7.7 8.4 9.1 9.8 76 9.5 8.7 8.0 7.3 7.9 8.5 9.2 10.0 77 9.6 8.8 8.1 7.5 8.0 8.7 9.4 10.2 78 9.8 9.0 8.3 7.6 8.2 8.8 9.6 10.4 79 10.0 9.2 8.4 7.8 8.3 9.0 9.7 10.6 80 10.2 9.4 8.6 7.9 8.5 9.2 9.9 10.8 81 10.4 9.6 8.8 8.1 8.7 9.3 10.1 11.0 82 10.7 9.8 9.0 8.3 8.8 9.5 10.3 11.2 83 10.9 10.0 9.2 8.5 9.0 9.7 10.5 11.4 84 11.1 10.2 9.4 8.6 9.2 10.0 10.8 11.7 85 11.4 10.4 9.6 8.8 9.4 10.2 11.0 11.9 86 11.6 10.7 9.8 9.0 9.6 10.4 11.2 12.2 87 11.9 10.9 10.0 9.2 9.8 10.6 11.5 12.4 88 12.1 11.1 10.2 9.4 10.0 10.8 11.7 12.6 89 12.4 11.4 10.4 9.6 10.2 11.0 11.9 12.9 90 12.6 11.6 10.6 9.8 10.4 11.2 12.1 13.1 91 12.9 11.8 10.9 10.0 10.6 11.4 12.3 13.4 92 13.1 12.0 11.1 10.2 10.8 11.6 12.6 13.6 93 13.4 12.3 11.3 10.4 11.0 11.8 12.8 13.8 94 13.6 12.5 11.5 10.6 11.1 12.0 13.0 14.1 95 13.9 12.7 11.7 10.8 11.3 12.2 13.2 14.3 96 14.1 12.9 11.9 10.9 11.5 12.4 13.4 14.6 97 14.4 13.2 12.1 11.1 11.7 12.6 13.7 14.8 98 14.7 13.4 12.3 11.3 11.9 12.9 13.9 15.1 99 14.9 13.7 12.5 11.5 12.1 13.1 14.2 15.4 100 15.2 13.9 12.8 11.7 12.3 13.3 14.4 15.6 101 15.5 14.2 13.0 12.0 12.5 13.6 14.7 15.9 102 15.8 14.5 13.3 12.2 12.8 13.8 14.9 16.2 103 16.1 14.7 13.5 12.4 13.0 14.0 15.2 16.5 104 16.4 15.0 13.8 12.6 13.2 14.3 15.5 16.8 105 16.8 15.3 14.0 12.9 13.4 14.5 15.8 17.2 106 17.1 15.6 14.3 13.1 13.7 14.8 16.1 17.5 107 17.5 15.9 14.6 13.4 13.9 15.1 16.4 17.8 108 17.8 16.3 14.9 13.7 14.1 15.3 16.7 18.2 109 18.2 16.6 15.2 13.9 14.4 15.6 17.0 18.5 110 18.6 17.0 15.5 14.2 14.6 15.9 17.3 18.9 111 19.0 17.3 15.8 14.5 14.9 16.2 17.6 19.2 112 19.4 17.7 16.2 14.8 15.2 16.5 18.0 19.6 113 19.8 18.0 16.5 15.1 15.4 16.8 18.3 20.0 114 20.2 18.4 16.8 15.4 15.7 17.1 18.6 20.4 115 20.7 18.8 17.2 15.7 16.0 17.4 19.0 20.8 116 21.1 19.2 17.5 16.0 16.2 17.7 19.3 21.2 117 21.5 19.6 17.8 16.3 16.5 18.0 19.7 21.6 118 22.0 19.9 18.2 16.6 16.8 18.3 20.0 22.0 119 22.4 20.3 18.5 16.9 17.1 18.6 20.4 22.4 120 22.8 20.7 18.9 17.3 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 13 of 38 Guidance Table to Identify Target Weight for Discharge for Children 6-59 Months Source: WHO child growth standards and the identification of severe acute malnutrition in infants and children 2009 Weight on admission or weight free of oedema. If weight on admission is pair, round the weight up with 0.1 kg. Example: weight on admission is 9.2 kg, use 9.3 kg as weight on admission. Weight on admission Target weight: 15% weight gain Weight on admission Target weight: 15% weight gain 4.1 4.7 11.1 12.8 4.3 4.9 11.3 13.0 4.5 5.2 11.5 13.2 4.7 5.4 11.7 13.5 4.9 5.6 11.9 13.7 5.1 5.9 12.1 13.9 5.3 6.1 12.3 14.1 5.5 6.3 12.5 14.4 5.7 6.6 12.7 14.6 5.9 6.8 12.9 14.8 6.1 7.0 13.1 15.1 6.3 7.2 13.3 15.3 6.5 7.5 13.5 15.5 6.7 7.7 13.7 15.8 6.9 7.9 13.9 16.0 7.1 8.2 14.1 16.2 7.3 8.4 14.3 16.4 7.5 8.6 14.5 16.7 7.7 8.9 14.7 16.9 7.9 9.1 14.9 17.1 8.1 9.3 15.1 17.4 8.3 9.5 15.3 17.6 8.5 9.8 15.5 17.8 8.7 10.0 15.7 18.1 8.9 10.2 15.9 18.3 9.1 10.5 16.1 18.5 9.3 10.7 16.3 18.7 9.5 10.9 16.5 19.0 9.7 11.2 16.7 19.2 9.9 11.4 16.9 19.4 10.1 11.6 17.1 19.7 10.3 11.8 10.5 12.1 10.7 12.3 10.9 12.5 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Routine Medicines Protocols Adapted from Valid International. 2006. Community-based Therapeutic Care (CTC): A Field Manual. Oxford, UK: Valid International. MEDICINE/SUPPLEMENT WHEN TO GIVE AGE / WEIGHT PRESCRIPTION DOSE ANTIBIOTIC On admission All beneficiaries Amoxicillin 50-100 mg/kg bodyweight/day 3 times a day for 5 days (8 hourly) ANTIMALARIAL (artemisinin-based combination therapy [ACT]) Test on admission; Repeat test later if initial test negative and malaria suspected. If no test, treat based on symptoms. All beneficiaries Artesunate (AS) 50 mg and Amodiaquine (AQ) 153 mg: ½ AS and ½ AQ 1 AS and 1 AQ Once a day for 3 days ANTIHELMINTHIC DRUG After 1 week If signs of re-infection appear, an anthelminthic drug can be given again after three months. < 12 months DO NOT GIVE NONE <10 kg ≥10 kg Albendazole 200 mg or Mebendazole 250 mg Albendazole 400 mg or Mebendazole 500 mg Single dose MEASLES VACCINATION Inpatient care: On admission and discharge Outpatient care: On week 4 (or upon discharge) In case of measles epidemic, a measles vaccination is given on admission and repeated on week four (or upon discharge). If the child receives a measles vaccination before the age of 12 months, a repeat vaccination is given at the age of 12 months. From 6 months Standard Two doses as per EPI guidelines VITAMIN A SUPPLEMENTATION On week 4 (or upon discharge) Children with bilateral pitting oedema should not be given vitamin A until discharged. In case there are signs of vitamin A deficiency, children are referred and treated in inpatient care (see National Guidelines). 6 months ≥12 months 100,000 IU 200,000 IU Single dose Iron and Folic Acid: - In outpatient care iron and folic acid should not be given. In case severe anaemia is identified according to the national Integrated Management of Childhood Illness (IMCI) guidelines, the child is referred and treated in inpatient care (see National Guidelines). - In inpatient care, if the child is on a therapeutic milk diet, he/she is given folic acid 5 mg on day one and 1 mg per day from day two onwards until discharge and iron 3 mg/kg bodyweight/day after two days on F100, when gaining weight. If the child is on a RUTF diet, neither folic acid nor iron is given. - Iron and folic acid should never be provided together with a malaria treatment. Malaria is treated first. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 15 of 38 Supplementary Medicines Protocols Refer to the Job Aid: Drug Doses. Adapted from Valid International. 2006. Community-based Therapeutic Care (CTC): A Field Manual. Oxford, UK: Valid International. MEDICINE/SUPPLEMENT PURPOSE PRESCRIPTION SPECIAL INSTRUCTIONS CHLORAMPHENICOL GENTAMYCIN Given as second-line antibiotic for children not responding to amoxicillin for continued fever that is not due to malaria Child with SAM who needs second-line antibiotic is referred to inpatient care See Job Aid: Drug Doses Continue for 7 days TETRACYCLINE EYE OINTMENT Treatment of eye infection Apply 3 times a day: morning, afternoon and at night before sleep Wash hands before and after use. Wash eyes before application. Continue for 2 days after infection is gone NYSTATIN Treatment of candida 100,000 units (1 ml) 4 times a day after food (use dropper; show caregiver how to use it) Continue for 7 days PARACETAMOL For children with fever over 38.5°C See Job Aid: Drug Doses Single dose only—DO NOT give to take home BENZYL BENZOATE Treatment of scabies Apply over whole body. Repeat without bathing on following day. Wash off 24 hours later. Avoid contact with eyes. Do not use on broken or skin with secondary infection. GENTIAN VIOLET Treatment of minor abrasions or fungal infections of the skin Apply on lesion. Can be repeated at next visit and continued until condition resolved QUININE Second-line antimalarial for children who have not responded to artemisinin-based combination therapy (ACT) See Job Aid: Drug Doses FERROUS SULPHATE/FOLATE Treatment of anaemia identified according to national Integrated Management of Childhood Illness (IMCI) guidelines See Job Aid: Drug Doses Given ONLY after 14 days in the programme METRONIDAZOLE For treatment of giardiasis and amoebiasis See Job Aid: Drug Doses Continue for 5 days Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Appetite Test [For children 6-59 months] The appetite is tested upon admission and at each follow-up visit to the health facility in outpatient care and in inpatient care in the transition phase. Conduct in a quiet separate area to give the caregiver and child time to get accustomed to the ready-to-use therapeutic food (RUTF). Explain the test’s procedure (steps #3 and #4) and purpose to the caregiver. The purpose of the appetite test is to determine if the child with SAM has appetite and can be treated in outpatient care, or the child with SAM has poor appetite and needs to be referred for treatment in inpatient care Advise the caregiver to wash hands before giving the RUTF. Advise the caregiver to sit with the child in his/her lap and gently offer the RUTF. The caregiver should encourage the child to eat the RUTF without force-feeding. The child should receive plenty of clean water to drink from a cup while eating the RUTF. Observe the child for 30 minutes and then answer the following question: 1 2 3 4 5 Did the child eat at least one third of a 92g packet of RUTF or three teaspoons from a pot? NO YES A child who fails the appetite test should be referred and treated in inpatient care. A child who passes the appetite test should be treated in outpatient care Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 17 of 38  Sugar Water Protocol [How to Make & Who to Give To] Sugar Water Should Be Given To:  All children who have travelled for long distances or have waited a long time for attention should be given sugar water as soon as they arrive. If possible, especially when it is very hot, give sugar water to all children awaiting treatment in outpatient care.  All children refusing ready-to-use therapeutic food (RUTF) or being referred to inpatient care.  All children that are at risk of hypothermia or septic shock should be given sugar water whether or not they have a low blood glucose level.  A child who has taken the diet during the day will not develop hypoglycaemia overnight and does not need to be woken for night-time feeding. If the diet has not been taken during the day, the mother should give at least one feed during the night. To Make Sugar Water 10 Percent Dilution:  Take clean drinking water (slightly warm if possible to help dilution)  Add the required amount of sugar to clean drinking water and shake or stir vigorously. Water Sugar 100 ml 10 g 2 heaped teaspoons 200 ml (average cup) 20 g 4 heaped teaspoons 500 ml (small bottle) 50 g 10 heaped teaspoons 1 litre 100 g 20 heaped teaspoons Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado Action Protocol in Outpatient Care REFERRAL TO HOME VISIT SIGN or STATUS REFERRAL TO INPATIENT CARE GENERAL CONDITION Deteriorating Child is not losing oedema BILATERAL PITTING OEDEMA Grade +++ Any grade of bilateral pitting oedema with severe wasting (marasmic kwashiorkor) Increase in bilateral pitting oedema Bilateral pitting oedema not reducing by week 3 ANOREXIA Poor appetite or unable to eat – Failed appetite test VOMITING Intractable vomiting CONVULSIONS Ask mother if the child had convulsions during the since the previous visit LETHARGY, NOT ALERT Child is difficult to awake UNCONSCIOUSNESS Child does not respond to painful stimuli HYPOGLYCAEMIA A clinical sign in a child with SAM is eye-lid retraction: child sleeps with eyes slightly open. Low level of blood glucose < 3 mmol/l, < 54 mg/dl DEHYDRATION Severe dehydration based primarily on recent history of diarrhoea, vomiting, fever or sweating and on recent appearance of clinical signs of dehydration as reported by the mother/caregiver HIGH FEVER Axillary temperature ≥ 38.5° C, rectal temperature ≥ 39° C taking into consideration the ambient temperature HYPOTHERMIA Axillary temperature < 35° C, rectal temperature < 35.5° C taking into consideration the ambient temperature RESPIRATION RATE ≥ 60 breaths/minute for children under 2 months ≥ 50 breaths/minute from 2-12 months ≥ 40 breath /minute from 1-5 years ≥ 30 breaths /minute for children over 5 years Any chest in-drawing ANAEMIA Palmer pallor or unusual paleness of skin SKIN LESION Broken skin, fissures, flaking of skin SUPERFICIAL INFECTION Any infection requiring intramuscular antibiotic treatment Child is not gaining weight or losing weight on follow-up visit WEIGHT CHANGES Below admission weight on week 3 Weight loss for three consecutive visits Static weight for three consecutive visits Child has returned from inpatient care or refuses referral to inpatient care REQUEST Mother/caregiver requests treatment of child in inpatient care for social reasons (decided by supervisor) Child is absent or defaulting NOT RESPONDING Child that is not responding to treatment is referred to inpatient care or hospital for further medical investigation. Integrated Management of Childhood Illness (IMCI) danger sign Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 19 of 38 Key Messages Upon Admission to Outpatient Care Adapted from Valid International. 2006. Community-based Therapeutic Care (CTC): A Field Manual. Oxford, UK: Valid International. 1. RUTF is a food and medicine for very thin and swollen children only. Do not share it. 2. Sick children often don’t like to eat. Give small regular meals of RUTF and encourage the child to eat often (if possible, 8 meals a day). Your child should have _ packets a day. 3. For young children, continue to breastfeed. Offer breast milk first before every RUTF feed. 4. RUTF is the only food sick and thin/swollen children need to recover during their time in Outpatient Care. Always give RUTF before other foods, such as porridge (use local name). 5. Always offer the child plenty of clean water to drink while eating RUTF. Children will need more water than normal. 6. Wash the child’s hands and face with soap before feeding. Keep food clean and covered. 7. Sick children get cold quickly. Always keep the child covered and warm. 8. For children with diarrhoea, continue feeding. Give them extra food and water. 9. Return to the health facility whenever the child’s condition deteriorates or if the child is not eating sufficiently. Note: Ask the caregiver to repeat the messages to be sure they have been correctly understood. Upon the next visits to the health facility, the health and nutrition messages are expanded (see Job Aid. Health and Nutrition Education Messages) Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Look-Up Table for Use of RUTF in Outpatient Care Amounts of RUTF to Give to a Child per Day or Week Based on a Dose of 200 kcal/kg Bodyweight/Day Using 92 g Packets Containing 500 kcal Child’s Weight (kg) Packets per Day Packets per Week 4.0 – 4.9 2 14 5.0 – 6.9 2.5 18 7.0 – 8.4 3 21 8.5 – 9.4 3.5 25 9.5 – 10.4 4 28 10.5 – 11.9 4.5 32  12 5 35 Infants < 4 kg are referred to inpatient care Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado Danger Signs for Inpatient Care Alert a physician if danger sign occur: Normal Ranges: Danger sign: Suggests: Pulse and Respirations 2 to 12 months:  Pulse 80 up to 160 in child  Respirations 20 up to 60 12 to 60 months (5 years):  Pulse 80 up to 120  Respirations 20 up to 40 Confirmed increase in pulse rate of 25 or more beats/ minute along with Confirmed increase in respiratory rate of 5 or more breaths/ minute  Infection or  Heart failure (possibly from overhydration due to feeding or rehydrating too fast) Respirations Only 2 to 12 months:  Respirations 20 up to 60 12 to 60 months (5 years):  Respirations 20 up to 40 2 months up to 12 months:  Fast breathing is considered 50 breaths/minute or more 12 months up to 60 months (5 years):  Fast breathing is considered 40 breaths/ minute or more  Pneumonia Temperature  Any sudden increase or decrease  Rectal temperature below 35.5C (95.9F)  Infection  Hypothermia (possibly due to infection, a missed feed, or child being uncovered) In addition to watching for increasing pulse or respirations and changes in temperature, watch for other danger signs such as:  Anorexia (loss of appetite)  Change in mental state (e.g., becomes lethargic)  Jaundice (yellowish skin or eyes)  Cyanosis (tongue/lips turning blue from lack of oxygen)  Difficult breathing  Difficulty feeding or waking (drowsy)  Abdominal distention  New edema  Large weight changes  Increased vomiting  Petechiae (bruising) Some children 2-12 months of age will normally breathe fast (i.e., 50-60 breaths per minute) without having pneumonia. However, unless the child’s normal respiratory rate is known to be high, he/she should be assumed to have either overhydration or pneumonia. Careful evaluation, taking into account prior fluid administration, will help differentiate the two conditions and plan appropriate treatment. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com RUTF Packets Per Day to Give to a Child with SAM in Inpatient Care TRANSITION PHASE REHABILITATION PHASE 150 kcal/kg bodyweight/day 200 kcal/kg bodyweight/day Child’s weight (kg) Packets per Day (92 g Packets Containing 500 kcal) Packets per Day (92 g Packets Containing 500 kcal) 3.5 – 3.9 1.2 1.5 4.0 – 4.9 1.5 2 5.0 – 6.9 2.1 2.5 7.0 – 8.4 2.5 3 8.5 – 9.4 2.8 3.5 9.5 – 10.4 3.1 4 10.5 – 11.9 3.6 4.5  12 4 5 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 23 of 38 F75 for use with Severe Wasting in Stabilization Phase Weight of child (kg) Volume of F-75 per feed (ml)a Daily total (130 ml/kg) 80% of daily total a (minimum) Every 2 hours (12 feeds) Every 3 hoursc (8 feeds) Every 4 hours (6 feeds) 2.0 20 30 45 260 210 2.2 25 35 50 286 230 2.4 25 40 55 312 250 2.6 30 45 55 338 265 2.8 30 45 60 364 290 3.0 35 50 65 390 310 3.2 35 55 70 416 335 3.4 35 55 75 442 355 3.6 40 60 80 468 375 3.8 40 60 85 494 395 4.0 45 65 90 520 415 4.2 45 70 90 546 435 4.4 50 70 95 572 460 4.6 50 75 100 598 480 4.8 55 80 105 624 500 5.0 55 80 110 650 520 5.2 55 85 115 676 540 5.4 60 90 120 702 560 5.6 60 90 125 728 580 5.8 65 95 130 754 605 6.0 65 100 130 780 625 6.2 70 100 135 806 645 6.4 70 105 140 832 665 6.6 75 110 145 858 685 6.8 75 110 150 884 705 7.0 75 115 155 910 730 7.2 80 120 160 936 750 7.4 80 120 160 962 770 7.6 85 125 165 988 790 7.8 85 130 170 1014 810 8.0 90 130 175 1040 830 8.2 90 135 180 1066 855 8.4 90 140 185 1092 875 8.6 95 140 190 1118 895 8.8 95 145 195 1144 915 9.0 100 145 200 1170 935 9.2 100 150 200 1196 960 9.4 105 155 205 1222 980 9.6 105 155 210 1248 1000 9.8 110 160 215 1274 1020 10.0 110 160 220 1300 1040 a Volumes in these columns are rounded to the nearest 5 ml. b Feed two-hourly for at least the first day. Then, when the child has little or no vomiting, modest diarrhea (< 5 watery stools per day), and is finishing most feeds, change to three-hourly feeds. c After a day on three-hourly feeds: If no vomiting, less diarrhea, and finishing most feeds, change to four-hourly feeds. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com F75 for use with Severe (+++) Bilateral Pitting Edema in Stabilization Phase Weight with +++ edema (kg) Volume of F-75 per feed (ml)a Daily total (100 ml/kg) 80% of daily totala (minimum) Every 2 hours (12 feeds) Every 3 hoursc (8 feeds) Every 4 hours (6 feeds) 3.0 25 40 50 300 240 3.2 25 40 55 320 255 3.4 30 45 60 340 270 3.6 30 45 60 360 290 3.8 30 50 65 380 305 4.0 35 50 65 400 320 4.2 35 55 70 420 335 4.4 35 55 75 440 350 4.6 40 60 75 460 370 4.8 40 60 80 480 385 5.0 40 65 85 500 400 5.2 45 65 85 520 415 5.4 45 70 90 540 430 5.6 45 70 95 560 450 5.8 50 75 95 580 465 6.0 50 75 100 600 480 6.2 50 80 105 620 495 6.4 55 80 105 640 510 6.6 55 85 110 660 530 6.8 55 85 115 680 545 7.0 60 90 115 700 560 7.2 60 90 120 720 575 7.4 60 95 125 740 590 7.6 65 95 125 760 610 7.8 65 100 130 780 625 8.0 65 100 135 800 640 8.2 70 105 135 820 655 8.4 70 105 140 840 670 8.6 70 110 145 860 690 8.8 75 110 145 880 705 9.0 75 115 150 900 720 9.2 75 115 155 920 735 9.4 80 120 155 940 750 9.6 80 120 160 960 770 9.8 80 125 165 980 785 10.0 85 125 165 1000 800 10.2 85 130 170 1020 815 10.4 85 130 175 1040 830 10.6 90 135 175 1060 850 10.8 90 135 180 1080 865 11.0 90 140 185 1100 880 11.2 95 140 185 1120 895 11.4 95 145 190 1140 910 11.6 95 145 195 1160 930 11.8 100 150 195 1180 945 12.0 100 150 200 1200 960 a Volumes in these columns are rounded to the nearest 5 ml. b Feed two-hourly for at least the first day. Then, when the child has little or no vomiting, modest diarrhea (< 5 watery stools per day), and is finishing most feeds, change to three-hourly feeds. c After a day on three-hourly feeds: If no vomiting, less diarrhea, and finishing most feeds, change to four-hourly feeds. Infants Under 6 months in Inpatient Care With Breastfeeding Caregiver: F100-Diluted or F75 Look-Up Table for Maintenance Amounts of F100-Diluted (Severe Wasting) or F75 (Bilateral Pitting Oedema Until the Oedema is Resolved) for Breastfed Infants Child’s Weight (kg) F100-Diluted or F75 in case of oedema (ml per feed if 12 feeds per day) F100-Diluted or F75 in case of oedema (ml per feed if 8 feeds per day) ≥ 1.2 20 25 1.3 – 1.5 25 30 1.6 – 1.7 30 35 1.8 – 2.1 30 40 2.2 – 2.4 35 45 2.5 – 2.7 40 50 2.8 – 2.9 40 55 3.0 – 3.4 45 60 3.5 – 3.9 50 65 4.0 – 4.4 50 70 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Infants under 6 months in Inpatient Care With No Prospect of Breastfeeding: F100-Diluted or F75 Stabilisation Phase Look-Up Table for Volume of F100-Diluted (Severe Wasting) or F75 (Bilateral Pitting Oedema) for Non-Breastfed Infants Under 6 Months Child’s Weight (kg) F100-Diluted or F75 in case of oedema (ml per feed if 12 feeds per day) F100-Diluted or F75 in case of oedema (ml per feed if 8 feeds per day) ≤ 1.5 25 30 1.6 – 1.8 30 35 1.9 – 2.1 30 40 2.2 – 2.4 35 45 2.5 – 2.7 40 50 2.8 – 2.9 40 55 3.0 – 3.4 45 60 3.5 – 3.9 50 65 4.0 – 4.4 50 70 Transition Phase Look-Up Table for Volume of F100-Diluted for Non-Breastfed Infants Child’s Weight (kg) F100-Diluted (ml per feed if 8 feeds per day) ≤ 1.5 45 1.6 – 1.8 53 1.9 – 2.1 60 2.2 – 2.4 68 2.5 – 2.7 75 2.8 – 2.9 83 3.0 – 3.4 90 3.5 – 3.9 96 4.0 – 4.4 105 Rehabilitation Phase Look-Up Table for Volume of F100-Diluted for Non-Breastfed Infants Child’s Weight (kg) F100-Diluted (ml per feed if 6-8 feeds per day) ≤ 1.5 60 1.6 – 1.8 70 1.9 – 2.1 80 2.2 – 2.4 90 2.5 – 2.7 100 2.8 – 2.9 110 3.0 – 3.4 120 3.5 – 3.9 130 4.0 – 4.4 140 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 27 of 38 F100 for Use in Transition Phase Transition Phase Look-Up Table for Volume of F100 per Feed if No RUTF is Taken, Based on 150 kcal/kg Bodyweight/Day Weight (kg) F100 (ml per feed if 8 feeds per day) F100 (ml per feed if 6 feeds per day) F100 (ml per feed if 5 feeds per day) 3.0 - 3.4 60 75 85 3.5 – 3.9 65 80 95 4.0 – 4.4 70 85 110 4.5 – 4.9 80 95 120 5.0 – 5.4 90 110 130 5.5 – 5.9 100 120 150 6.0 – 6.9 110 140 175 7.0 – 7.9 125 160 200 8.0 – 8.9 140 180 225 9.0 – 9.9 155 190 250 10 – 10.9 170 200 275 11 – 11.9 190 230 275 12 - 12.9 205 250 300 13 - 13.9 230 275 350 14 - 14.9 250 290 375 15 - 19.9 260 300 400 20 - 24.9 290 320 450 25 - 29.9 300 350 450 30 - 39.9 320 370 500 40 - 60 350 400 500 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com  F100 for use in Rehabilitation Phase (Free Feeding) Weight of Child (kg) Range of volumes per four-hourly feed of F-100 (6 feeds daily) Range of daily volumes of F-100 Minimum (ml) Maximum (ml) a Minimum (150 ml/kg/day) Maximum (220 ml/kg/day) 2.0 50 75 300 440 2.2 55 80 330 484 2.4 60 90 360 528 2.6 65 95 390 572 2.8 70 105 420 616 3.0 75 110 450 660 3.2 80 115 480 704 3.4 85 125 510 748 3.6 90 130 540 792 3.8 95 140 570 836 4.0 100 145 600 880 4.2 105 155 630 924 4.4 110 160 660 968 4.6 115 170 690 1012 4.8 120 175 720 1056 5.0 125 185 750 1100 5.2 130 190 780 1144 5.4 135 200 810 1188 5.6 140 205 840 1232 5.8 145 215 870 1276 6.0 150 220 900 1320 6.2 155 230 930 1364 6.4 160 235 960 1408 6.6 165 240 990 1452 6.8 170 250 1020 1496 7.0 175 255 1050 1540 7.2 180 265 1080 1588 7.4 185 270 1110 1628 7.6 190 280 1140 1672 7.8 195 285 1170 1716 8.0 200 295 1200 1760 8.2 205 300 1230 1804 8.4 210 310 1260 1848 8.6 215 315 1290 1892 8.8 220 325 1320 1936 9.0 225 330 1350 1980 9.2 230 335 1380 2024 9.4 235 345 1410 2068 9.6 240 350 1440 2112 9.8 245 360 1470 2156 10.0 250 365 1500 2200 a Volumes per feed are rounded to the nearest 5 ml. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 29 of 38  Infants Under 6 months in Inpatient Care With Breastfeeding Caregiver: F100-Diluted or F75 Look-Up Table for Maintenance Amounts of F100-Diluted (Severe Wasting) or F75 (Bilateral Pitting Oedema Until the Oedema is Resolved) for Breastfed Infants Child’s Weight (kg) F100-Diluted or F75 in case of oedema (ml per feed if 12 feeds per day) F100-Diluted or F75 in case of oedema (ml per feed if 8 feeds per day) ≥ 1.2 20 25 1.3 – 1.5 25 30 1.6 – 1.7 30 35 1.8 – 2.1 30 40 2.2 – 2.4 35 45 2.5 – 2.7 40 50 2.8 – 2.9 40 55 3.0 – 3.4 45 60 3.5 – 3.9 50 65 4.0 – 4.4 50 70 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com  Infants under 6 months in Inpatient Care With No Prospect of Breastfeeding: F100-Diluted or F75 Stabilisation Phase Look-Up Table for Volume of F100-Diluted (Severe Wasting) or F75 (Bilateral Pitting Oedema) for Non-Breastfed Infants Under 6 Months Child’s Weight (kg) F100-Diluted or F75 in case of oedema (ml per feed if 12 feeds per day) F100-Diluted or F75 in case of oedema (ml per feed if 8 feeds per day) ≤ 1.5 25 30 1.6 – 1.8 30 35 1.9 – 2.1 30 40 2.2 – 2.4 35 45 2.5 – 2.7 40 50 2.8 – 2.9 40 55 3.0 – 3.4 45 60 3.5 – 3.9 50 65 4.0 – 4.4 50 70 Transition Phase Look-Up Table for Volume of F100-Diluted for Non-Breastfed Infants Child’s Weight (kg) F100-Diluted (ml per feed if 8 feeds per day) ≤ 1.5 45 1.6 – 1.8 53 1.9 – 2.1 60 2.2 – 2.4 68 2.5 – 2.7 75 2.8 – 2.9 83 3.0 – 3.4 90 3.5 – 3.9 96 4.0 – 4.4 105 Rehabilitation Phase Look-Up Table for Volume of F100-Diluted for Non-Breastfed Infants Child’s Weight (kg) F100-Diluted (ml per feed if 6-8 feeds per day) ≤ 1.5 60 1.6 – 1.8 70 1.9 – 2.1 80 2.2 – 2.4 90 2.5 – 2.7 100 2.8 – 2.9 110 3.0 – 3.4 120 3.5 – 3.9 130 4.0 – 4.4 140 Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 31 of 38 Entry and Exit Categories for Individual and Service Monitoring of CMAM for Children 6-59 Months Inpatient Care Outpatient Care ENTRY CATEGORIES 1. New admission: New case of child 6-59 months who meets the admission criteria -- including relapse after cure 2. Other new admissions: New case: infants, child, adolescent, adult (< 6 months or ≥ 5 years) who needs treatment of SAM in inpatient care 3. Referral from outpatient care: Condition of child deteriorated in outpatient care (according to action protocol) and child needs inpatient care Or Returned after defaulting (or Moved from other inpatient care site) 1. New admission: New case of child 6-59 months who meets the admission criteria -- including relapse after cure 2. Other new admissions: New case who does not meet pre-set admission criteria but needs treatment of SAM in outpatient care (special case, based on decision of supervisor) 3. Referral from inpatient care: Case of child 6-59 months referred from inpatient care after stabilisation and continues treatment in outpatient care Or Returned after defaulting (or Moved from other outpatient care site) EXIT CATEGORIES 1. Discharged cured: Child 6-59 months who meets discharge criteria, i.e., special cases that were not referred to outpatient care earlier 2. Discharged died: Child 6-59 months who dies while in inpatient care 3. Discharged defaulted: Child 6-59 months who is absent for two days 4. Discharged non-recovered: Child 6-59 months who remained in inpatient care does not reach discharge criteria after two months in treatment 5. Referred to outpatient care: Condition of child has stabilised, child’s appetite has returned, the medical complication is resolving and the child is referred to outpatient care to continue treatment 1. Discharged cured: Child 6-59 months who meets discharge criteria 2. Discharged died: Child 6-59 months who dies while in outpatient care 3. Discharged defaulted: Child 6-59 months who is absent for three consecutive visits 4. Discharged non-recovered: Child 6-59 months who does not reach discharge criteria after four months in treatment 5. Referred to inpatient care: Condition of child has deteriorated or child is not responding to treatment (per the action protocol), and child is referred to inpatient care Movement between sites is possible in mobile populations or during emergencies. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Health and Nutrition Education Messages KEY BEHAVIORS TO PROMOTE1 Essential Nutrition Actions  Optimal breastfeeding during the first 6 months of life  Optimal complementary feeding starting at 6 months with continued breastfeeding to 2 years of age and beyond  Continued feeding when the child is ill  Optimal nutrition care of malnourished children  Prevention of vitamin A deficiency for women and children  Adequate iron and folic acid intake, and the prevention and control of anaemia for women and children  Adequate iodine intake by all members of the household  Optimal nutrition for women Household Hygiene Actions  Treatment and safe storage of drinking water  Handwashing with soap or ash at critical times: after defecation, after handling children’s faeces, before preparing food, before feeding children, before eating  Safe disposal of faeces  Proper storage and handling of food to prevent contamination Other Care Practices  Antenatal care attendance, including: at least four visits, tetanus toxoid vaccine, iron/folic acid supplementation  Full course of immunisations for all children before their first birthday  Children and women sleeping under insecticide-treated bed nets  Recognition when a sick child needs treatment outside of the home and seeking care from appropriate providers  Recognition of pregnancy danger signs 2. IMPORTANCE OF BREASTFEEDING Importance of Breastfeeding for the Infant/Young Child Breast Milk:  Saves infants’ lives  Is a whole food for the infant, and contains balanced proportions and sufficient quantity of all the needed nutrients for the first 6 months  Promotes adequate growth and development, thus preventing stunting  Is always clean  Contains antibodies that protect against diseases, especially against diarrhoea and respiratory infections  Is always ready and at the right temperature  Is easy to digest; nutrients are well absorbed  Protects against allergies; breast milk antibodies protect the baby’s gut preventing harmful substances to pass into the blood  Contains enough water for the baby’s needs (87% of water and minerals)  Helps jaw and teeth development; suckling develops facial muscles  Provides frequent skin-to-skin contact between mother and infant, which leads to better psychomotor, affective and social development of the infant  Provides the infant with benefits from the colostrum, which protects him/her from diseases; the amount is perfect for newborn stomach size  Promotes brain development; increased Intelligence Quotient (IQ) scores Importance of Breastfeeding for the Mother  Putting the baby to the breast immediately after birth facilitates the expulsion of placenta because the baby’s suckling stimulates uterine contractions  Reduces risks of bleeding after delivery  When the baby is immediately breastfed after birth, breast milk production is stimulated 1 Adapted from Ghana Sustainable Change Project. 2008. Priority Nutrition Messages. Accra, Ghana: AED. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 33 of 38  Breastfeeding is more than 98% effective as a contraceptive method during the first 6 months provided that breastfeeding is exclusive and amenorrhea persists  Immediate and frequent suckling prevents engorgement  Reduces the mother’s workload (no time is involved in boiling water, gathering fuel, preparing milk)  Breastmilk is available at anytime and anywhere, is always clean, nutritious and at the right temperature  It is economical  Stimulates bond between mother and baby  Reduces risks of breast and ovarian cancer Importance of Breastfeeding for the Family  The child receives the best possible quality of food, no matter what the family’s economic situation  No expenses in buying formula, firewood or other fuel to boil water, milk or utensils; the money saved can be used to meet the family’s other needs  No medical expenses due to sickness that formula could cause; the mothers and their children are healthier  As illness episodes are reduced in number, the family encounters few emotional problems associated with the baby’s illness  Births are spaced thanks to the contraceptive effect  Time is saved  Feeding the baby reduces work because the milk is always available and ready Importance of Breastfeeding for the Community  Not importing formula and utensils necessary for its preparation saves hard currencies that could be used for something else  Healthy babies make a healthy nation  Savings are made in the health area; a decrease in the number of child illnesses leads to decreased national medical expenses  Improves child survival; reduces child morbidity and mortality  Protects the environment (trees are not used for firewood to boil water, milk and utensils, thus protecting the environment); breast milk is a natural renewable resource Recommended Breastfeeding Practices and Possible Points of Discussion for Counselling Recommended Breastfeeding Practice Possible Points of Discussion for Counselling (Choose most relevant to mother’s situation) Put infant skin-to-skin with mother immediately after birth.  Skin-to-skin with mother keeps newborn warm.  Skin-to-skin with mother helps stimulate brain development. Initiate breastfeeding within the first hour of birth.  This first milk ’local word’ is called colostrum. It is yellow and full of antibodies which help protect your baby.  Colostrum provides the first immunization against many diseases.  Breastfeeding from birth helps the milk ‘come in’ and ensures plenty of breast milk. Exclusively breastfeed (no other food or drink) for 6 months.  Breast milk is all the infant needs for the first 6 months.  Do not give anything else to the infant before 6 months, not even water  Giving water will fill the infant and cause less suckling; less breast milk will be produced. Breastfeed frequently, day and night.  Breastfeed the baby often, at least 8-12 times for a newborn and 8 or more times after breastfeeding is well-established, day and night, to produce lots of breast milk.  More suckling (with good attachment) makes more breast milk. Breastfeed on demand (or cue) every time the baby asks to breastfeed.  Crying is a late sign of hunger.  Early signs that baby wants to breastfeed:  Restlessness  Opening mouth and turning head from side to side  Putting tongue in and out  Sucking on fingers or fists Let infant finish one breast and come off by him/herself before switching to the other breast.  Switching back and forth from one breast to the other prevents the infant from getting the nutritious ‘hind milk’.  The ‘fore milk’ has more water content and quenches infant’s thirst; the ‘hind milk’ has more fat content and satisfies the infant’s hunger. Continue breastfeeding until 2 years of age or longer.  Breast milk contributes a significant proportion of energy and nutrients during the complementary feeding period and helps protect babies from illness.  In the first year, breastfeed before giving foods to maintain breast milk supply. Mother needs to eat and drink to satisfy hunger and thirst.  No one special food or diet is required to provide adequate quantity or quality of breast milk.  Breast milk production is not affected by maternal diet.  No foods are forbidden.  Mothers should be encouraged to eat supplemental foods where they are accessible. Avoid feeding bottles.  Foods or liquids should be given by a spoon or cup to reduce nipple confusion and the possible introduction of contaminants. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Recommended Complementary Feeding Practices Age Frequency (per day) Amount of food an average child will usually eat at each serving (in addition to breast milk) Texture (thickness/ consistency) Variety 6-8 months 2-3 times food 2-3 tablespoons ‘Tastes’ up to ½ cup (250 ml) Thick porridge/pap Mashed/ pureed family foods Breastfeeding + staples (porridge, other local examples) Legumes (local examples) Vegetables/fruits (local examples) Animal foods (local examples) 9-11 months 4 times foods and snacks ½ cup/bowl (250 ml) Finely chopped family foods Finger foods Sliced foods 12-23 months 5 times foods and snacks ¾ -1 cup/bowl (250 ml) Family foods Sliced foods Note: If baby is not breastfed Add 1-2 extra times food and snacks Add 1-2 cups of milk per day Responsive/active feeding Be patient and actively encourage your baby to eat. Hygiene  Feed your baby using a clean cup and spoon, never a bottle, as this is difficult to clean and may cause your baby to get diarrhoea.  Wash your hands with soap and water before preparing food, eating and feeding young children. Adapt the chart to use a suitable cup/bowl to show the amount. The amounts assume an energy density of 0.8 – 1 kcal/g. Use iodized salt in preparing family foods. Recommended Complementary Feeding Practices and Possible Points of Discussion for Counselling Recommended Complementary Feeding Practice Possible Points of Discussion for Counselling (choose most relevant to mother’s situation) At 6 months of age, add complementary foods (e.g., thick porridge 2-3 times a day) to breastfeeds.  Give local examples of first types of complementary foods. As baby grows older, increase feeding frequency, amount, texture and variety.  FATV: Gradually increase the frequency (F), amount (A), texture (T) (thickness/consistency) and variety (V) of foods. From 6-8 months of age, breastfeed plus give 2-3 servings of foods.  Start with 2-3 tablespoonfuls of cooked porridge or mashed foods (give examples of cereals and family foods).  At 6 months, these foods are more like ‘tastes’ than actual servings.  Increase gradually to ½ cup (250 ml cup). Show amount in cup brought by mother. From 9-11 months of age, breastfeed plus give 4 servings of food or snacks per day.  Give finely chopped, mashed foods and finger foods.  Increase gradually to ½ cup (250 ml cup). Show amount in cup brought by mother. From 12-23 months of age, give 5 servings of food or snacks per day, plus breastfeed.  Give family foods.  Give ¾ to one cup (250 ml cup/bowl). Show amount in cup brought by mother.  Other solid foods (snacks) can be given as many times as possible each day and can include [give examples].  Foods given to the child must be stored in hygienic conditions to avoid diarrhoea and illness. Give baby 2-3 different family foods: staple, legumes, vegetables/fruits, and animal foods at each serving.  Try to feed different foods at each serving. Continue breastfeeding until 2 years of age or longer.  During the first and second years, breast milk is an important source of nutrients for your baby.  During the first year, breastfeed first to maintain breast milk supply. Be patient and actively encourage baby to eat all his/her food.  At first, baby may need time to get used to eating foods other than breast milk.  Use a separate plate to feed the child to make sure he/she eats all the food given. Wash hands with soap and water before preparing food, eating and feeding young children.  Foods given to the child must be stored in hygienic conditions to avoid diarrhoea and illness. Feed baby using a clean cup and spoon.  Cups are easy to keep clean. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com Page 35 of 38 Emotional & Physical Stimulation for Children Adapted from WHO Management of Severe Malnutrition: A Manual for Physicians and Other Senior Health Workers 1999 At each play session and for all ages:  Teach local songs and finger and toe games.  Have the child vocalise/repeat what he/she hears.  Describe all activites.  Teach action words with activities (e.g., say “bang bang” while beating a drum).  Encourage next motor milestone while playing: bouncing, crawling, standing, walking. For Children Who Cannot Easily Move:  Passive limb movements  Splashing in a warm bath AGE TOY TO MAKE 6+ months Ring on a string String together cotton reels and other small objects with holes. Tie the string in a ring, leaving a long piece of string hanging. 9+ months In-and-out toy Use plastic or cardboard container and small objects that child can’t swallow. 9+ months Blocks Use small wood blocks and sanded smooth. Decorate with carving or non-toxic paint if possible. 9+ months Nesting toys Cut off bottom of two containers of identical shape but different size. 12+ months Rattle Cut long strips of plastic from coloured plastic bottles. Place in small clear plastic bottle. If available, use non-toxic glue to secure the top on firmly . Do not use this toy if the child is at risk of choking on the rattle’s contents. Job Aids for Community-Based Management of Acute Malnutrition (CMAM), March 2018 Edited by Dr. Abdul Rehman Pirzado pirzado@gmail.com AGE TOY TO MAKE 12+ months Drum Use a tin with a tight lid. 18+ months Posting Bottle Use a large clear plastic bottle with a small neck and place objects like spools that child can’t swallow but can fit through the neck. Don’t seal lid. 18+ months Pull-along toy Make a hole in the centre of top and bottom of a round tin. Thread string (about 60 cm long) through the holes and tie the ends inside the tin. To make noise while pulling, put metal bottle tops inside of the tin and close lid. 18+ months Doll Cut out two doll shapes from a piece of cloth and sew the edges together, leaving a small opening. Turn doll inside out and stuff with scraps of materials. Stitch opening closed and draw or sew a face on doll. 18+ months Stacking bottle tops Cut at least 3 same-sized round plastic bottles in half and stack tops. 18+ months Mirror A tin lid with no sharp edges. If you cannot smooth the edges of the tin lid, do not use this toy as it could harm the child. 18+ months Puzzle Draw a figure (e.g., a doll) in color on a square or rectangular piece of cardboard. Cut the figure into a few larger pieces. 18+ months Book Cut out 3 equal-sized rectangular pieces of cardboard. Glue or draw simple and familiar pictures on both sides of each piece. Make 2 holes on one side of each piece and thread string through.
190002	https://electronbunker.ca/DLpublic/GMD.pdf	Geometric Mean Distance — Its Derivation and Application in Inductance Calculations Robert Weaver, Saskatoon, Canada 2016-03-05 (Revised 2016-03-06 ) 1 Abstract This article begins with an informal introduction to the use of Geometric Mean Distance (GMD) in inductance calculations, and why it is important. This is followed by brief discussion of the logarithm function. Because it pervades the subject of GMD calculation, it is important to have a good working knowledge of various logarithmic identities and integrals of formulae involving logarithms. The issue of integrating across logarithmic anomalies (i.e., log(0) ) is also treated. The self GMD is derived in complete detail for some very simple shapes, in order to demonstrate the general analytical method of calculation. This is followed by a discussion of the use of numerical methods—The Monte Carlo Method, in particular—for the GMD calculation of more difﬁcult shapes. Detailed examples are given, which result in empirical formulae for the self GMD of a triangular area, the mutual GMD of elliptical areas and the self GMD of elliptical loci. The use of GMD in inductance calculations is often accompanied by the proviso, that the "diameter of the conductor should be considerably smaller than the diameter of the winding." However, there is almost nothing available in the literature giving quantitive information about the limits of accuracy of the GMD method. This is treated for the cases of self inductance of a circular loop, and the mutual inductance of two parallel circular loops. Empirical functions are developed, which give the estimated error due to the approximation of the GMD method. These error formulae may be used to calculate correction factors. An example of this is given for a multi-turn solenoid coil. The ﬁnal part of the article deals with the calculation of the self GMD of groups of objects. In particular, closed form formulae are developed for the self GMD of linear arrays of circular conductors and linear arrays of thin strips of conductors. These are further developed into a new closed form formula for Rosa's round wire inductance corrections, and formulae for the inductance of short coils which account for conductor shape. For revision history, see the last page of this document. The latest version of this document may be downloaded, free 1 of charge, from the author's website: – ! – 1 Table of Contents Abstract 1 Part 1 - Basic Concepts 3 1.1 Introduction 3 ......................................................................................................................................... 1.2 Conventions 3 ......................................................................................................................................... 1.3 Inductance Calculations using GMD 4 ............................................................................................... 1.4 Why Geometric Mean and not Arithmetic Mean? 5 ......................................................................... 1.5 The Logarithm Function - A Primer 7 ................................................................................................. Part 2 - Analytical Calculation of Simple Shapes 12 2.1 GMD of two co-linear lines of equal length 12 .................................................................................. 2.2 GMD of a Line from Itself (Self GMD of a line) 18 ............................................................................ Part 3 - Calculation of GMD using numerical methods 23 3.1 The Monte Carlo Method 23 ................................................................................................................. 3.2 The GMD of a Triangle 24 ..................................................................................................................... 3.3 The GMD of an Elliptical Area 27 ........................................................................................................ 3.4 The Self GMD of an Elliptical Line 37 ................................................................................................. Part 4 - Limits of Accuracy of the GMD method 44 4.1 Self Inductance 44 .................................................................................................................................. 4.2 Mutual Inductance 49 ............................................................................................................................ 4.3 Example Calculation 60 ......................................................................................................................... 4.4 Summary 61 ............................................................................................................................................ Part 5 - Combined GMD of Multiple Objects 62 5.1 Basic Principles 62 .................................................................................................................................. 5.2 Formula for the GMD of a Linear Array of Circular Conductors (LACC) 63 ............................... 5.3 Formula for the GMD of a Linear Array of Co-linear Straight Line Segments (LALS) 71 .......... 5.4 Inductance Calculation Based on Aggregate GMD 75 ..................................................................... 5.5 Rosa's Round Wire Inductance Corrections 77 .................................................................................. 5.6 Flat Conductor Corrections 87 ............................................................................................................. Afterword 89 References 90 Appendices 92 A – Program Listing for calculation of the GMD of an Elliptical Locus 92 ......................................... B – A Method for Finding and Eliminating Errors in Mathematical Derivations 96 ......................... Revision History 99 – ! – 2 Part 1 - Basic Concepts 1.1 Introduction It is impossible to study the techniques of inductance calculation to any great extent without encountering the term geometric mean distance. Often a simple but inadequate explanation is given, resulting in more questions than answers. For that reason, this article has been prepared to address some of the basics and some of the common questions that arise. This is largely a non-rigourous treatment. However, in different sections, the degree of rigour will vary, according to what information could be readily located or calculated with reasonable effort. For some sections, the reader will require some knowledge of basic calculus. 1.2 Conventions In this article, SI units will be used. Where non-SI formulae are quoted from various references they will be adjusted to SI. Hence inductance will be expressed in henries, and length, diameter and radius in metres. Logarithms will be encountered frequently, and in all cases will be natural logarithms. The logarithm function will be indicated as log(). In other literature, Ln() is often used, but it means the same thing. The following symbols will be used: ϵ error µ0 Permeability of free space, or vacuum, equal to 4π⨉10-7 henries/metre µR Relative permeability of the medium under consideration µ Absolute permeability of the medium under consideration (µ = µ0 ⨉ µR) 𝛾 self GMD shape characteristic factor a, b, c... principal dimensions (length, width etc.) of a shape under study D Diameter of coil or loop of wire d Diameter of conductor g geometric mean distance (GMD) of one object from another gS geometric mean distance of an object from itself (self GMD)2 gM mutual GMD, geometric mean distance between objects k Miscellaneous numeric constant, often distinguished by a numeric subscript ℓ Length (of conductor or coil depending on subscript) ℓW Conductor (wire) length In the technical literature, the term geometric mean radius or GMR is often used for the geometric mean distance of an 2 object from itself. It is an inaccurate term, and hence, self GMD will be used in this discussion. – ! – 3 ℓC Coil length n Number of turns in coil, or number of conductors in array R Radius of coil or loop of wire r Radius of conductor (the subscript is w for wire, since the subscript c has already been used for coil) s length of line or line segment u Shape factor, aspect ratio, or miscellaneous dimensionless ratio uc Coil shape factor ℓC/R x distance separating two conductors or ﬁlaments xij distance separating conductor or ﬁlament i from conductor or ﬁlament j A word of warning should be given about the use of symbols in this document. Because many of the derivations will be developed in great detail, a consequence of this is that many numeric constants will appear during the derivation, and then later disappear as formulae become simpliﬁed. For example, a formula that employs constants k0, k1, and k2 may end up having k0 and k1 drop out, leaving an orphan k2. Leaving such a constant named k2 in the ﬁnal result is guaranteed to invite confusion, especially if the formula is later presented out of context. For that reason, once a formula reaches its ﬁnal form, some symbol names may change. Every attempt will be made to clearly point this out whenever it happens. In any event, the reader is hereby forewarned. Similarly, the symbol u is used for ratios or shape factors such as length/diameter, length/width. Depending in which section it is used it may have different meanings. However, its deﬁnition will be given when it is introduced in a discussion. 1.3 Inductance Calculations using GMD A formula for the mutual inductance between a pair parallel of identical inﬁnitesimally thin ﬁlaments can be turned into a formula for self inductance of a ﬁnite conductor by one of two methods: 1. Difﬁcult: Integrate the ﬁlamentary mutual inductance formula across the cross sectional area of the conductor (a double integral involving area, which devolves into a quadruple integral involving length); 2. Moderately Easy: Determine the self-geometric mean distance (self GMD) of the cross section of the conductor, and then use that value to do a single mutual inductance calculation. In the ﬁrst case we are treating the conductor as an inﬁnite number of parallel inﬁnitesimal ﬁlaments, and then calculating the mutual inductance of each ﬁlament paired with each other ﬁlament, multiplying it by a weighting factor dA which is the inﬁnitesimal cross sectional area of the ﬁlament. These mutual inductance calculations are summed and then divided by the total cross sectional area A of the conductor to yield the self inductance. – ! – 4 In the second case, we are doing essentially the same thing, with one important difference: In each mutual inductance calculation, rather than using the actual distance, xij, (between the ﬁlaments i and j), which varies with each ﬁlament pair, instead we use a mean of the distances separating the ﬁlament pairs. Using this single value allows for a massive simpliﬁcation of the calculation—the integration becomes trivial—with virtually no loss of accuracy for most practical applications. The GMD principle is strictly correct for an inﬁnitely long straight conductor, but is approximately correct for a straight conductor of ﬁnite length, as long as the end effects are relatively small (i.e., the conductor diameter is signiﬁcantly smaller than the conductor length). The GMD principle can also be applied to conductor geometries other than straight lines as long as the conductor diameter is small relative to the overall conductor length, and the radius of curvature is large compared to the cross sectional diameter of the conductor. This is because the GMD is really a two dimensional attribute related to the cross sectional geometry of the conductor (in, say, the X–Y plane), and assumes that what goes on in the third dimension (Z–axis) is constant (ideally), or changes very gradually and smoothly relative to what happens in the other dimensions. In this way the effect of what happens in the third dimension has minimal effect on the ﬁeld lines in the other dimensions. For the remainder of this discussion we will assume that conductor size is small relative to these other parameters. An exception will be where these conditions are deliberately violated in order to ﬁnd the limits of accuracy of the GMD method. 1.4 Why Geometric Mean and not Arithmetic Mean? Mutual inductance of two ﬁlaments is determined from the Neumann integral or one of several other equivalent formulae. The Neumann integral is: ! (1.4.1) where x is the distance separating inﬁnitesimal segments of ﬁlaments ds1 and ds2. When the integral is evaluated, the principle term will be log(x), because the integral ∫ 1⁄x dx is equal to log(x). Depending on the geometric details of the ﬁlaments, there will also be other terms, but generally having a smaller contribution. To take this result and then calculate the self inductance of a conductor of ﬁnite cross section, it is necessary to integrate this log term across the cross sectional area of the conductor. Since integration is a linear operation, it is permissible to simplify the calculation by taking the arithmetic mean of the integrand and multiplying by the limits of the integral. But since the arithmetic mean of log(x) is the same as the antilog of the geometric mean of x, we are essentially taking the geometric mean of x. (There are some cases where one or two of the secondary non logarithmic terms are signiﬁcant enough that an arithmetic mean or mean square should be applied to those terms rather than a geometric mean.) For a straight conductor, we can take the formula for the mutual inductance between two straight ﬁlaments and using the self GMD of the conductor cross section, combine the two into a formula M = µ0 4⇡ Z ds1 ⇧ds2 x – ! – 5 for the self inductance of a straight conductor. The formula for the mutual inductance of two parallel straight ﬁlaments (page 31), adjusted to SI units, is: ! (1.4.2) where ℓ is the length of the ﬁlaments, and x is their separation distance. Assuming the conductor has a round cross section, the self GMD is given by Maxwell (Art. 691) to be simply: e-¼⨉r or 0.77880⨉r (1.4.3) where r is the radius of the conductor. We replace the ﬁnite round conductor with two ﬁctitious inﬁnitesimal ﬁlaments separated by the conductor’s self GMD, and use that self GMD value, as the separation distance, in the mutual inductance formula giving: ! (1.4.4) In the case of a circular loop of round wire, we can use the mutual inductance formula for two circular ﬁlaments, in combination with the GMD of round wire to quickly come up with a formula for the self inductance of the loop. However, there’s a bit of a catch. So far, nothing has been said about the direction of offset of the two ﬁctitious ﬁlaments. In the case of the straight conductor it makes no difference. If the conductor cross section is in the X-Y plane, and the length of the conductor is in the Z dimension, then the two ﬁlaments can be separated in either the X or Y dimension or a combination of both, because the conductor is radially symmetric. However, in the case of a circular loop we have the choice of using two ﬁlaments which are offset axially or radially (see diagram below). The axial offset would seem to make the most sense, as there is no question about what the radius of the ﬁlaments should be (i.e., the same as the radius to the centre of the conductor). Indeed, that is generally the approach that is taken. In the loop cross section shown in the leftmost ﬁgure, the two black dots indicate the spacial relationship of the proxy ﬁlaments in true scale relative to the conductor cross section for which they are substituted. It turns out that we can also use radially offset ﬁlaments (rightmost ﬁgure) as long as the mean radius of the two ﬁlaments is the same as radius of the centre of the conductor. But what kind of mean? Arithmetic or Geometric? Not surprisingly, in the context of inductance calculations, it turns out that it is the geometric mean. Two radially offset circular ﬁlaments whose M = µ02⇡ 2 4log 0 @ x + s 1 + ✓x ◆2 1 A − r 1 + ⇣x ⌘2 + x 3 5 L = µ0 2⇡ 2 4log 0 @ 0.7788r + s 1 + ✓ 0.7788r ◆2 1 A − s 1 + ✓0.7788r ◆2 + 0.7788r 3 5 – ! – 6 geometric mean radius is the same as the radius of the centre of the conductor will give exactly the same inductance value as two axially offset ﬁlaments. (We are assuming conductors of round cross section.) If the radius of the loop of conductor is R (measured from axis to centre of conductor), and the radii of the two proxy ﬁlaments are RA (inner) and RB (outer), then we are saying: R = √(RA RB) (1.4.5) And since the ﬁlaments are separated by a distance equal to gS, then: gS = RB − RA (1.4.6) Combining the two formulae, leads to a quadratic equation: RA 2 + gS RA − R2 = 0 (1.4.7) With the solution: RA = ( −gS+√(gS2 + 4 R2))/2 (1.4.8) RB = RA + gS (1.4.9) In the above right diagram, the difference between arithmetic mean positioning and geometric mean positioning is too small to be seen, and in most cases the numerical difference is also nearly insigniﬁcant, but for the most accurate calculations, the geometric mean is the correct one. The above illustrations are scale drawings of loops with a radius of 8 (any units) and conductor radius of 1. For a geometric mean positioning, RA and RB are 7.62007 and 8.39887 respectively, while for an arithmetic mean positioning, RA and RB would be 7.61060 and 8.38940 respectively. The difference would be more signiﬁcant if the ratio of conductor radius to loop radius were larger. However, with these larger ratios, the accuracy of the overall GMD approximation would deteriorate. As an important consequence from the above, it can be seen that there is an equivalence between axially offset circular ﬁlaments and radially offset circular ﬁlaments (or for that matter ﬁlaments that are offset both radially and axially). This means that it is possible to convert a pair of radially offset ﬁlaments into an equivalent pair of axially offset ﬁlaments having the same mutual inductance. This will prove useful in developing certain inductance formulae. 1.5 The Logarithm Function - A Primer Since the logarithm function ﬁgures prominently in GMD calculations, it is worthwhile to do a quick review of some of its properties. In calculus, the logarithm function, log(x), is deﬁned as a deﬁnite integral with a variable upper limit: ! (1.5.1) log x = x Z 1 dt t ⇥ x > 0 ⇤ . – ! – 7 Alert readers will note that when calculating geometric mean distances, the argument of the logarithm function is a distance, and not surprisingly, has the units of distance. It’s often stated that the argument of a logarithm function must be dimensionless, and in many cases it is the ratio of two values which have the same dimensions so that they do cancel out and make for a dimensionless argument. In the case of GMD calculations we clearly see arguments which have dimensions of length. This leads to the question: What are the units of the result of the logarithm function? From the above integral formula, it can be seen that if the variable x has dimensional units such as length, then both t and dt must take on the same units. Since dt is in the numerator and t is in the denominator, the units cancel out, and the integrand is therefore dimensionless. Because integration is a linear operation, the units of the integrand, if any, become the units of the overall integral. Therefore, the argument of the logarithm function may have dimensional units, but from a dimensional analysis point of view, the resulting value of the logarithm function is dimensionless. Of course, this leads to the question of what happens when we take the anti-log of a dimensionless value. This seems to imply that a random dimension can magically appear as a result. However, common sense tells us that if we take the logarithm of an argument having some particular dimension, and then subsequently take the anti-log, as part of the same derivation, the result should have the same dimension as the original argument. Dimensional analysis does have its limits. The integral deﬁnition of the log function has no analytical solution, which is why we simply refer to it as log(x) and leave the dirty work to a calculator or a spreadsheet program. However, it may be evaluated numerically using one of several series expansions such as: ! (1.5.2) Several different series may be found in the literature. Different expansions will provide different rates of convergence, depending on the value of argument. Some fundamental logarithm function identities follow: ! (1.5.3) ! (1.5.4) ! (1.5.5) The calculation of GMD involves integrating the logarithm function over various ranges. Hence, it is important to remember that log(r) is undeﬁned for r ≤ 0. Since the calculation is only concerned log x = x Z 1 dt t = (x−1)−(x −1)2 2 + (x −1)3 3 −(x −1)4 4 +· · · . log (ab) = log(a) + log(b) . log (a/b) = log(a) −log(b) . log (an) = n log(a) . – ! – 8 with the magnitude of distances, we can write log(\|r\|). Often, the absolute value is obtained automatically such as in the case where r is calculated from cartesian coordinates as r = √(x2+y2) In cases of integration, it may be necessary to split the integral into two parts: ∫log(r) dr + ∫log(-r) dr in order to cover the range of r < 0 separately from the range of r > 0. While this resolves how to deal with negative values, it still leaves the situation of calculating a self GMD which will always involve integration across an area where r will, at some point, have a value of zero. How is this singularity resolved? To understand what happens in this situation, it is useful to look at a simpler but equivalent situation, and that is the following integral: ! In graphical terms (see diagram below), we are taking the area bounded by the x and y axes, the function log(x), and the vertical line x = a. ! Remembering that an integral can be deﬁned as a sum of areas f(x)∆x as ∆x approaches zero, it is equivalent to: ! Z a 0 log(x)dx . lim ∆x!0 a P x=o log (x)∆x . – ! – 9 where we are taking the area from x = 0 to x = a, and the areas are long skinny slices with height equal to the value of the function and width equal to ∆x. The value ∆x is made inﬁnitely small. The red rectangle denotes the area log(x+½∆x) ⨉ ∆x and is summed as x varies from 0 to a−∆x in increments of ∆x. And of course ∆x is made inﬁnitesimal. As x approaches zero, it can be seen that log(x) tends towards minus inﬁnity. The area we are concerned with is where x = 0, and thus the value of log(½∆x)⨉∆x as ∆x becomes inﬁnitesimal. ! As ∆x approaches zero the incremental area approaches 0 ⨉ -∞, leaving us to wonder which factor wins out: the zero or the minus inﬁnity. To resolve this, we can apply L’Hôpital’s rule which states that in the case of a limit involving a quotient of two functions that approaches 0/0 or ∞/∞, then the limit is equal to the quotient of the derivatives of the two functions. In this case it is possible to rewrite log(½∆x)⨉∆x as log(½∆x)/(1/∆x) which yields the ∞/∞ form, and then: ! (1.5.6) And so, it is clear that, as x goes to zero, −x⁄2 also goes to zero. Therefore, there is no singularity in the integral due to log(0). This is perfectly logical in the context of GMD calculations, because it is lim x!0+ x log ✓x 2 ◆ = d dx log ✓x 2 ◆ d dxx−1 = 1 2x −x−2 = −x2 2x = −x 2 . – ! – 10 saying, essentially, that when two points have zero distance separating them, their mean distance is zero, not minus inﬁnity. In fact using the same analysis, it can be shown that ! (1.5.7) for any k > 0 and n > 0, including fractional powers. The zero limit can be illustrated by plotting the formula, x log(x): ! In this graph, the smallest value of x that was calculated and plotted is 10−8, and it can readily be seen that as x gets closer to zero, the value of x log(x) clearly tends to zero. Having resolved this issue, we can safely proceed to calculate GMD integrals with no further worries about log (0) singularities. Following are the solutions to several integrals involving the logarithm function which will prove useful in the upcoming derivations: ! (1.5.8) ! (1.5.9) ! (1.5.10) The next few sections will give several examples of the calculation of GMD for some simple shapes. lim x!0+ xn log(kx) = 0 . Z x log (x)dx = x2 2 log (x) −x2 4 . Z log (a + bx)dx = a + bx b log (a + bx) −x . Z x log (a + bx)dx = b2x2 −a2 2b2 log (a + bx) + ax 2b −x2 4 . – ! – 11 Part 2 - Analytical Calculation of Simple Shapes 2.1 GMD of two co-linear lines of equal length The following diagram shows two co-linear lines of length s with the ﬁrst line’s leftmost point at the origin (for convenience), and the second line offset by a distance r. ! If Py is a point on the ﬁrst line located at a distance y from the origin, and Px is a point on the second line located at a distance x from the origin, then the distance between these points is (x − y). To calculate the GMD, we take the sum of the logarithms of the distance between every combination of points and divide by the length of the ﬁrst line and then divide again by the length of the second line. The result is the logarithm of the GMD. Hence: log(g) = 1/s x 1/s x ∑∑log(xi-yj) Because there is an inﬁnite number of points on each line, the formula for the GMD between the two lines becomes a double integral: ! (2.1.1) where x is a distance over the right line segment and y is a distance over the left line segment. log (g) = 1 s2 r+s Z r s Z 0 log (x −y)dy dx . – ! – 12 Evaluating the ﬁrst integral with respect to y, using the integral formulae given in the previous section, we get: ! (2.1.2) Before doing the second integration, it’s a good idea to split the terms into separate integrals rather than trying to deal with the entire integral, which would quickly become very messy. Therefore we now have: ! (2.1.3) where: ! Additionally, we will split the terms of A into sub-terms A1 and A2: ! log (g) = 1 s2 r+s Z r  (y −x) log (x −y) −y #s 0 dx = 1 s2 r+s Z r  (s −x) log (x −s) −s # −  −x log (x) # dx = 1 s2 r+s Z r (s −x) log (x −s) −s + x log (x) dx . log (g) = 1 s2 ✓r+s R r A dx − r+s R r B dx + r+s R r C dx ◆ . A = (s −x) log (x −s) = s log (x −s) −x log (x −s) B = s C = x log (x) . A1 = s log (x −s) A2 = x log (x −s) . – ! – 13 Evaluating these integrals one by one: ! ! r+s Z r A1 dx = r+s Z r s log (x −s) dx =  s(x −s) log (x −s) −sx #r+s r =  s(r + s −s) log (r + s −s) −s(r + s) − # −  s(r −s) log (r −s) −sr # = sr log (r) −sr −s2 −(sr −s2) log (r −s) + sr = sr log (r) −(sr −s2) log (r −s) −s2 . r+s Z r A2 dx = r+s Z r x log (x −s) dx = x2 −s2 2 log (x −s) −sx 2 −x2 4 #r+s r = (r + s)2 −s2 2 log (r + s −s) −s(r + s) 2 −(r + s)2 4 # − r2 −s2 2 log (r −s) −sr 2 −r2 4 # = ✓r2 2 + sr ◆ log (r) −sr −3s2 4 −r2 4 + (s2 −r2) 2 log (r −s) + sr 2 + r2 4 = ✓r2 2 + sr ◆ log (r) −sr 2 −3s2 4 + (s2 −r2) 2 log (r −s) . – ! – 14 Combining the above A1 and A2 integrals: ! The B integral is mercifully simple: ! And ﬁnally, the C integral: ! r+s Z r A dx = r+s Z r A1 −A2 dx = sr log (r) −(sr −s2) log (r −s) −s2 − ✓r2 2 + sr ◆ log (r) −sr 2 −3s2 4 + (s2 −r2) 2 log (r −s) % = sr log (r) −(sr −s2) log (r −s) −s2 − ✓r2 2 + sr ◆ log (r) + sr 2 + 3s2 4 −(s2 −r2) 2 log (r −s) = −r2 2 log (r) − ✓ sr −s2 + r2 2 ◆ log (r −s) −s2 4 + sr 2 . r+s Z r B dx = r+s Z r s dx =  sx #r+s r = ⇥ s(r + s) ⇤ − ⇥ sr ⇤ = sr + s2 −sr = s2 . r+s Z r C dx = r+s Z r x log (x) dx = x2 2 log (x) −x2 4 #r+s r = (r + s)2 2 log (r + s) −(r + s)2 4 # − r2 2 log (r) −r2 4 # = ✓ sr + r2 + s2 2 ◆ log (r + s) −r2 2 log (r) −s2 4 −sr 2 . – ! – 15 At last, combining all the terms: ! And so, we have derived the GMD formula for two equal co-linear lines: ! …(2.1.4) This is equivalent to formula (130) given by Rosa and Grover 4. We might be tempted to set r = 0 in order to superimpose the two lines, and thus get the self GMD of a line. However, this fails, because the argument of the log function in the second term goes negative, resulting in an undeﬁned value. There is also a singularity in the case of s = r, i.e., the point where the end of the ﬁrst line coincides with the start of the second line. In that case, the argument of the log function in the second term is equal to zero, and the log term goes to −∞. However, the factor outside log function goes to zero, and it forces the term to zero (as with the discussion of limits in the previous section). Hence, the second term may be eliminated in the case where s = r, and for two equal co-linear lines just touching, we get: ! (2.1.5) log (g) = 1 s2 r+s Z r A dx − r+s Z r B dx + r+s Z r C dx # = 1 s2 −r2 2 log (r) − ✓ sr −s2 + r2 2 ◆ log (r −s) −s2 4 + sr 2 −s2 + ✓ sr + r2 + s2 2 ◆ log (r + s) −r2 2 log (r) −s2 4 −sr 2 # = 1 s2 ✓ sr+ r2 + s2 2 ◆ log (r + s)− ✓ sr−r2 + s2 2 ◆ log (r −s)−r2 log (r)−3s2 2 # . log (g) = 1 s2 ✓ sr+ r2 + s2 2 ◆ log (r + s)− ✓ sr−r2 + s2 2 ◆ log (r −s)−r2 log (r)−3s2 2 $ . log (g) = 1 s2 ✓ sr + r2 + s2 2 ◆ log (r + s) −r2 log (r) −3s2 2 $ = log(4s) −3 2 ) g = 0.8925s – ! – 16 One can also use the result of the ﬁrst integration, above, to ﬁnd the GMD from a line to a co-linear point. ! It’s simply a matter of substituting into the the result of the ﬁrst integral, the distance a between the point and the line (and adjusting signs), which then gives: ! (2.1.6) In the case of a point located exactly at the end of a line, i.e., a = 0, then the last term disappears (as previously explained), and the right hand side of the formula simpliﬁes to: ! (2.1.7) Hence, the GMD of a point at the end of a line to the line itself is: ! (2.1.8) Before moving on to the next example, something should be said about the practical difﬁculty in working through the above derivation. Even though this example has dealt with very simple shapes, and even though the integrals themselves have straightforward solutions, nevertheless, the number of terms that are generated during the derivation, make it a rather daunting task. The probability of committing errors during the evaluation and subsequent simpliﬁcation, is enormous. (Software packages which perform symbolic mathematical operations are available. However, they can be costly, and therefore are unavailable to many people. There are also free symbolic math packages such as Maxima, but still, they can be rather unwieldy to use for various reasons.) The method which was used here, is described in Appendix B. – ! – 17 2.2 GMD of a Line from Itself (Self GMD of a line) The two lines of length s from the previous example are now superimposed as shown in the following diagram. The line’s leftmost point is again at the origin. ! Px and Py are two points both located on the line segment at distances x and y, respectively, from the origin. The distance between points Px and Py is \|x−y\|. We will begin by calculating the GMD of point Px to the entire line. Hence: ! (2.2.1) The variable x is held constant, and the variable of integration is y. The requirement for the absolute value of the log function argument poses a minor problem. It is necessary to split the integral into two separate integrals: the ﬁrst to handle the range 0 ≤ y ≤ x, and the second to handle the range x ≤ y ≤ s: ! (2.2.2) log (g) = 1 s s Z 0 log \|y −x\| dy . log (g) = 1 s x Z 0 log (x −y) dy + 1 s s Z x log (y −x) dy . – ! – 18 This ensures that the argument of the log function will never be negative. These integrals can be solved using the integral formulae given previously. ! (2.2.3) which is the formula for the GMD of a point on a line to the line itself. The symmetry of the formula is immediately apparent: a point located at a distance x from either end of the line will have exactly same GMD, just as we would expect. Now, integrating a second time from 0 to s with respect to x and multiplying by 1/s, we will get the self GMD of the line. ! (2.2.4) Unfortunately, the second integration gets rather messy. So, as was done in the case of the GMD of two co-linear lines, we will split the formula into terms A, B, C, etc., integrate them separately, and then add them back together. Thus: ! And term B will be split into B1 and B2: ! log (G) = 1 s  (y −x) log (x −y) −y "x 0 + 1 s  (y −x) log (y −x) −y "s x = 1 s # (x −x) log(x −x) −x $ − # −x log(x) $ + # (s −x) log(s −x) −s $ − # (x −x) log(x −x) −x $" = 1 s ⇥ (x log(x) −x) + ((s −x) log(s −x) −s + x) ⇤ = 1 s ⇥ x log(x) + (s −x) log(s −x) −s ⇤ log (g) = 1 s ⇥1 s s Z 0 x log(x) + (s −x) log(s −x) −s dx . A = x log(x) B = (s −x) log(s −x) C = s . B1 = s log(s −x) B2 = x log(s −x) . – ! – 19 Integrating the individual terms: ! ! ! s Z 0 A dx = s Z 0 x log(x) dx = x2 2 log (x) −x2 4 #s 0 = s2 2 log (s) −s2 4 . s Z 0 B1 dx = s Z 0 s log(s −x) dx = s s Z 0 log(s −x) dx = s  (x −s) log (s −x) −x #s 0 = s $ (s −s) log (s −s) −s % − $ (0 −s) log (s −0) −0 %# = s ⇥ s log (s) −s ⇤ = s2 log (s) −s2 . s Z 0 B2 dx = s Z 0 x log(s −x) dx = x2 −s2 2 log (s −x) −sx 2 −x2 4 #s 0 = $s2 −s2 2 log (s −s) −s2 2 −s2 4 % − $−s2 2 log (s) −0 2 −0 4 %# = $ −s2 2 −s2 4 % − $−s2 2 log (s) %# = s2 2 log (s) −3 4s2 . – ! – 20 ! ! Now recombining the terms and simplifying: ! (2.2.5) Taking the anti-log of both sides we get: ! (2.2.6) which is the well known formula 4 for the self GMD of a straight line of length s. s Z 0 B dx = s Z 0 B1 dx − s Z 0 B2 dx = ✓ s2 log (s) −s2 ◆ − ✓s2 2 log (s) −3 4s2 ◆ = s2 2 log (s) −1 4s2 . s Z 0 C dx = s Z 0 s dx =  sx #s 0 = s2 −0 = s2 . log (g) = 1 s2 s Z 0 A dx + s Z 0 B dx − s Z 0 C dx # = 1 s2 ✓s2 2 log (s) −s2 4 ◆ + ✓s2 2 log (s) −s2 4 ◆ − ✓ s2 ◆# = 1 s2 ✓ s2 log (s) −3 2s2 ◆ = log (s) −3 2 . g = elog (s)−3 2 = elog (s) ⇥e−3 2 = se−3 2 = 0.22313s . – ! – 21 Further GMD formulae for straight lines are given by Gray , including the case of parallel lines of equal length. By means of additional integrations, he then expands this to ﬁnd the GMD of rectangles (of which squares are a special case). Gray’s derivation is presented in good detail and is recommended reading. Derivations of the GMD for circles and circular areas are given by Gray and in more detail by Seneff – ! – 22 Part 3 - Calculation of GMD using numerical methods 3.1 The Monte Carlo Method From the previous section, it can be seen that the derivation of formulae for GMD for even the simplest shapes can get quite involved. For more complex shapes, it may be more practical to use numerical methods to determine GMD. The Monte Carlo method is suitable for this, and can ﬁnd the GMD of complex shapes to useable accuracies with relatively little computing power. These methods can also be used to check analytical GMD derivations. The Monte Carlo method involves the generation of random data points in the object under consideration which are then used to calculate the desired result. This is repeated many times with more random data points, and the results are averaged to produce a good approximation of the true value. For the case of the self GMD of a straight line, the method is almost trivial. Simply generate two random numbers in the range from 0 to 1 (representing random points on a line of length = 1), take the logarithm of the absolute value of their difference and then repeat this many times, keeping a running total of the result of each iteration. Finally, divide the total by the number of trials and take the exponential. The program code is even shorter than this brief explanation: N = 1000000000 Sum = 0 For i = 1 to N Sum = Sum+log(abs(Rnd()-Rnd())) next GMD = exp(Sum/N) Print GMD One quick run of the program immediately veriﬁes the value of 0.22313, as was derived in the previous section. The accuracy of the Monte Carlo method is approximately proportional to the square root of the number of random samples. Therefore, to get an improvement in accuracy of one additional signiﬁcant digit, the number of random samples must be increased by a factor of 100. So, although the method is usually very simple to implement, it may require a lot of computing power if more than ﬁve signiﬁcant ﬁgures are required in the result. – ! – 23 3.2 The GMD of a Triangle For a slightly more complicated example, suppose we wish to ﬁnd the self GMD of the area enclosed by an equilateral triangular. (Doing a search of the technical literature, it appears that no one has investigated the GMD of a triangular area, making this exercise somewhat worthwhile.) In the physical world, an analogous methodology might be as follows: 1. Draw an equilateral triangle on the wall. 2. Throw darts at the wall randomly, until two of the darts fall within the bounds of the triangle. 3. Measure the distance between the points where the two darts landed. Take the logarithm of this distance, and record it. 4. Repeat steps 2 and 3 many many many times. 5. Calculate the total of the recorded log(distance) measurements, and then divide this total by the count of measurements. Take the anti-log of this value. The result will be, approximately, the self GMD. There are two primary requirements for this method to produce acceptable results: 1. The darts must have an equal probability of hitting any spot on the triangular area. That is, the x and y coordinates of the dart positions must have a uniform random distribution. (This requires very special dart playing skills, which are, unfortunately, disadvantageous for tournament play.) 2. Sufﬁcient trials must be conducted in order to average out the random noise. To get reasonable results, we likely will want to throw several million darts. Therefore, it may be more appealing to give this job to a computer. Most computer languages provide a random number generating function, typically generating numbers in the range 0 ≤ NR < 1, where NR is the random number. Assuming that the random number function produces uniformly distributed random numbers , we must make sure that we 3 don’t subsequently manipulate them in such a way that we render them non-uniform, while ﬁtting them into the test shape that we are analyzing. A reliable method is to place the shape inside a bounding rectangle. Then, generate a random x coordinate scaled to the width of the bounding rectangle, and a random y coordinate scaled to the height of the bounding rectangle. The random point will then fall within the bounding rectangle. Then, it must be determined if the resulting random point also falls inside the bounds of the test shape. This is accomplished by applying the appropriate geometric formulae which deﬁne the test shape. If the point does not fall within the bounds of the test shape, then it is discarded and the process is repeated. Random number generators that are built into spreadsheets and various programming languages tend to be deﬁcient, 3 and begin to display non uniform behaviour for very large sample sizes. For this reason it is a good idea to implement an independent and well proven pseudo-random number generator routine. Eg., the Wichman and Hill algorithm . – ! – 24 Continuing with the example of an equilateral triangle, refer to the diagram below. Let’s assume a side length of one, and one of the sides lying on the x axis, leftmost vertex at the origin, and the rightmost vertex at x = 1, y = 0, and then it follows that the third vertex is at x = 0.5, y = 0.866 = sin(60°). The triangle is shown in red. The bounding rectangle, shown in green, will have a width of 1 and a height of 0.866. ! The three lines forming the triangle are deﬁned as: y1 = 0 (3.2.1) y2 = 2x sin(60°) = 1.732x (3.2.2) y3 = (2 − 2x) sin(60°) = 1.732 − 1.732x (3.2.3) To generate the random points, the x coordinate random number, nx, is scaled to the range of 0..1, and the y coordinate random number ny, is scaled to the range of 0..0.866. To determine whether the random point is within the bounds of the triangle, the following conditions must be true: ny ≤ 1.732 nx ny ≤ 1.732 − 1.732 nx – ! – 25 There is no need to test that the point lies on or above the line y = 0 because the y coordinate random number will always be greater than or equal to zero. Following, is the listing for a BASIC program which performs the calculation. 'calculate self GMD of equilateral triangle with side length equal to 1 a = sin(60/180pi) 'Height (altitude) of triangle b = 1 'Base of triangle N = 1000000000 'Number of random pairs of points inside triangle Sum = 0 'Sum of all log(distance) calculations Npt = 0 'Count of all random points generated For i = 1 to N 'Generate test point T do 'Generate point tx,ty that falls inside the circumscribed rectangle. tx = rndb ty = rnda tya = 2atx 'eqn of first bounding line tyb = 2a-2atx 'eqn of second bounding line Npt = Npt+1 'Test whether the point falls within the 'bounds of the triangle, and repeat until true loop until (ty< = tya and ty< = tyb) 'Now generate the second test point S do sx = rndb sy = rnda sya = 2asx syb = 2a-2asx Npt = Npt+1 Loop Until (sy< = sya and sy< = syb) 'Calculate the distance between points T and S, ' then take the log and add to the sum Sum = Sum+log(Sqrt((sx-tx)^2+(sy-ty)^2)) next 'Calculate GMD from the sum of Log(distance) GMD = exp(Sum/N) 'Accuracy check based on comparison of Monte Carlo ' area calculation compared to analytical area. 'Monte Carlo area is equal to inside/outside points ' ratio (2N/Npt) times bounding rectangle area. 'The bounding rectangle area is ab Aarea = ab2N/Npt 'True area calculation Tarea = 0.5ab Referring to the last few lines of the program, to get an indication of the accuracy of the Monte Carlo calculation, we can keep count of the total number of points that fall inside the triangle compared to the total number of points generated. The ratio will approximately equal the ratio of the area of the triangle to the area of the rectangle—the approximation improving as the number of test points increases. The true ratio in the present example will be exactly ½. By checking the number of signiﬁcant digits that are in agreement with this value, we can get a reasonable estimate of the number of signiﬁcant digits of the GMD. – ! – 26 The following ﬁgure shows a plot of points from a program run of 108 random points. The points falling inside the triangular area are green, and the remainder are red. (Only a fraction of the total points have been plotted.) ! In a subsequent program run, the number of points falling inside the triangle was 20000000000 out of a total of 39999997361 generated points, suggesting an accuracy of about seven signiﬁcant ﬁgures. The self GMD value produced in the same run, was 0.30838221. It should then be safe to assume a self GMD of perhaps six signiﬁcant ﬁgures, or 0.308382. Hence, we can conclude that the self GMD of an equilateral triangular area with a side length s, will be 0.308382s. g = 0.308382 s (3.2.4) Naturally, having arrived at the self GMD value of 0.308382, we must immediately be curious whether this happens to be the anti-logarithm of a simple fraction, as the factor would be in the case of a circular area (−1⁄4) or a straight line (−3⁄2). Alas, the natural logarithm of 0.308382 is −1.17642, which doesn’t suggest any obvious simple fraction. The nearest fraction with a reasonably small denominator is −20⁄17 = −1.17647. It’s likely unwise to draw any conclusions about a simple fraction in this case. 3.3 The GMD of an Elliptical Area As a another example, we can consider the case of the self GMD of a circle or ellipse. The program code is virtually the same as the case of the triangle. The only difference being the formulae of the shape’s bounding lines. For an ellipse with a horizontal axis of 2b and vertical axis of 2a, the formula for the upper half of the curve is: y = a √(1 − (x/b)2) (3.3.1) And the formula for the lower half of the curve is: y = −a √(1 − (x/b)2) (3.3.2) – ! – 27 We will depart from the usual convention of referring to the ellipse axes as “major“ and “minor” and use “horizontal” and “vertical” instead. The horizontal axis may be either the major or minor axis, and we will deﬁne the horizontal axis to be the line in which the ellipses will be offset. If a = b, then the ﬁgure will be a circle. If a≠b, then the ﬁgure will be an ellipse. The program code follows: ' Monte Carlo method for determining the Self or Mutual GMD of ellipses a = 3 'vertical semi-axis b = 1 'horizontal semi-axis x0 = 0 'horizontal offset of the conductors (zero for self GMD) y0 = 0 'vertical offset of the conductors (zero for self GMD) N = 1000000000 'Number of random pairs of points inside ellipse to be evaluated Npt = 0 'Count of all random points generated Sum = 0 'Sum of all log(distance) calculations For i = 1 to N 'Generate test point T do 'Generate point tx,ty that falls inside a circumscribed rectangle. tx = rnd2b-b ty = rnd2a-a tya = asqrt(1-(tx/b)^2) 'eqn of upper bound of ellipse tyb = -tya 'lower bound of ellipse Npt = Npt+1 'Test whether the point falls within the 'bounds of the ellipse; repeat until true loop until (ty< = tye and ty> = tye1) 'Generate test point S do sx = rnd2b-b sy = rnd2a-a sya = asqrt(1-(sx/b)^2) syb = -sya Npt = Npt+1 Loop Until (sy< = sye and sy> = sye1) 'Calculate the distance between points T and S, ' then take the log and add to the sum Sum = Sum+log(Sqrt((sx-tx+x0)^2+(sy-ty+y0)^2)) next 'Calculate GMD from the sum of Log(distance) GMD = exp(Sum1/N) 'Accuracy check based on comparison of Monte Carlo ‘ area calculation compared to analytical area 'Monte Carlo area is equal to inside/outside points ‘ ratio (2N/Npt) times bounding rectangle area. 'The bounding rectangle area is 2a2b Aarea = 8abN/Npt 'True area of ellipse Tarea = piab End The analytical formula for the self GMD of an ellipse is given by Grover (page 21) as g = 0.5(a+b) e−¼ (3.3.3) – ! – 28 A quick program run veriﬁes agreement with this formula. Grover is a bit ambiguous about whether this formula is supposed to apply to an elliptical area, or an elliptical line, or both. It is, in fact, the formula for an elliptical area as the above program run has demonstrated, and not for the GMD of an elliptical line. Since a circle is a degenerate form of an ellipse, we see that the formula will give the correct value for a circular area when a = b. But since the self GMD of a circular line is simply equal to its radius, Grover’s formula is obviously not applicable in the case of the elliptical line. Alas, developing a numerical method to ﬁnd the GMD of an elliptical line requires a more sophisticated Monte Carlo simulation than what has been discussed so far, but will be addressed in a later section. In addition to the calculation of self GMD, the Monte Carlo method can of course be applied, with almost no additional effort, to the calculation of GMD between two shapes external to each other. In fact, it is as simple as setting non-zero values to the variables x0 and y0 in the above program code. These variables are the horizontal and vertical offsets of the shapes being analyzed. Now, whereas the case of self GMD is dependent only upon the geometry of the shape, and it scales linearly with its size, the GMD between two objects is often a non linear function of the distance separating them, and also varies depending upon the relative orientation of the objects. The case of two circles is a notable exception, because the GMD is simply the distance between their centres, and a circle is identical with any orientation. But for non-circular shapes, the calculation must performed for every different spacing and orientation of the objects. It may be possible to analyze the data generated by these calculations, and then derive an empirical formula that predicts the GMD, but there is no guarantee that this will be a simple formula. If we are using this method to determine the GMD of only a small group of conductors, then it doesn’t matter too much, because there are only a few GMD values that need to be calculated, and we can calculate each of them directly using the Monte Carlo method. When dealing with a large number of evenly spaced conductors, the most practical solution may be to calculate the cases for adjacent conductors, and then for 1 ⨉, 2 ⨉, 3 ⨉ conductor pitch separations, and so on up to, perhaps, 5 ⨉ conductor pitch. These can be stored in a lookup table, and used as required in an inductance calculation. For any spacing greater than what is in the table, the centre to centre conductor spacing can be used, since the true GMD will quickly converge to this value as the spacing becomes signiﬁcantly larger than the conductor mean diameter. For example, the table to the right shows the results of calculating the GMD of ellipses with a horizontal semi-axis of 0.5 and a vertical semi-axis of 1.0, for offsets ranging from 0 to 20. The GMD for a zero offset is the self GMD of a single ellipse. – ! – 29 a b offset GMD 0.5 1 0 0.5840898753 0.5 1 2 1.8946815473 0.5 1 4 3.9520103933 0.5 1 6 5.9684639214 0.5 1 8 7.9762391919 0.5 1 10 9.9810447569 0.5 1 12 11.9845009509 0.5 1 14 13.9863041625 0.5 1 16 15.9881743067 0.5 1 18 17.9895128001 0.5 1 20 19.9905701963 For each calculation, 200 million points were generated, and it is probably safe to rely on ﬁve signiﬁcant ﬁgures of the GMD results. It can be seen that as the offset increases in value, the GMD quickly converges towards the offset value. If we are to come up with an empirical function, based on the offset, it seems reasonable to use the actual offset value as the principal term, and then determine an additional correction term to adjust the ﬁnal value. We will use the function f(x,u) for the correction term, where x is the offset (which is measured from centre to centre of the shapes), and u is a shape factor (see diagram to the right, and description below). Hence GMD = x + f(x,u). We must be careful, because we must ﬁnd the correction term by taking the difference between the Monte Carlo GMD value and the offset. This difference becomes very small very quickly as the offset value increases, which can result in noisy data for large offsets. Fortunately, this is the part of the range where the correction term will tend towards zero and has little signiﬁcance. The above table lists the GMD only for ellipses with an a⁄b shape factor of 0.5. An empirical formula must be general enough to accommodate different shape factors. We will deﬁne the variable u to be the shape factor and it will be the ratio of the vertical semi-axis to the horizontal semi-axis. Hence u = a⁄b. – ! – 30 The Monte Carlo program was run for shape factors varying from 0.125 to 8.0, and the GMD correction factor is plotted in the following graph: ! The leftmost point of each curve corresponds to two ellipses, of the speciﬁed shape factor, just touching each other. It can be seen that the correction term is largest at this point and then decays to zero as the spacing increases. Developing a ﬁtting function of more than a single argument is usually a major undertaking. It’s worthwhile to look at developing a function of one variable while the other is held ﬁxed. We can ﬁx either the shape factor u or the offset x. Let us look at developing a function with a ﬁxed offset x and variable shape factor u. The offset will be ﬁxed at 2b (double the horizontal semi-axis) so that the two ellipses will be just touching each other in every case. The rationale for proceeding in this order is that the most critical GMD values are the ones for the closest spacings, and if we develop a formula for closely spaced ellipses ﬁrst, and then later develop a secondary decay function to correct for spacing, then it is much easier to ensure that the overall function is most accurate at the minimum offset value. – ! – 31 Therefore, let us re-plot the above data to show the required correction for different values of u for a ﬁxed offset of 2b. We will include a larger range of shape factors varying from 10−13 to 2000. This larger range allows us to get a better idea of any possible asymptotic behaviour. ! The graph shows what may be a horizontal asymptote near zero, and the curve appears to asymptotically approach an inclined straight line at large values of u. The inclined asymptote is much more obvious when the graph is redrawn with a linear horizontal axis: ! – ! – 32 Similarly the horizontal asymptote is more obvious on a linear plot restricted to small values of u: ! Further analysis of the data conﬁrms the asymptotes. The horizontal asymptote is, in fact, at a value of about −0.149261, and the inclined asymptote has a slope of 0.38940. The existence of and the values of these asymptotes can be easily explained. In the case of very large shape factor values (see ﬁgure at left), the horizontal component of the distance between points on the ellipses becomes insigniﬁcant compared to the vertical coordinate. The two ellipses begin to approximate vertical lines which are nearly superimposed. As such, the GMD begins to approximate the self GMD of an ellipse of the same shape factor. That is: g = 0.5(a+b) e-¼ But because u = a/b, (hence, a = bu): g = 0.5(ub+b) e-¼ or g = 0.5b(u+1) e-¼ Note when u is very large, u+1≃u and asymptotic nature is apparent. At large values of u, we have an asymptote with a slope of 0.5e−¼ or 0.38940, just as the data show. – ! – 33 The horizontal asymptote for small values of u can be explained in a similar way. Very small values of u, indicate two extremely ﬂat looking ellipses located end to end (see diagram below). ! Because they are so ﬂat, these appear much like two co-linear lines of unit length just touching. We might think therefore, that they would approach the value calculated for touching co-linear lines previously derived as 2⨉4e–3/2 = 1.7850. (The additional factor of 2 accounts for the fact that the closest spacing is 2⨉b.) Instead they approach a limiting GMD value of 1.8507. The reason for this is that even though the ellipses appear to become straight horizontal lines, the distribution of points is not linear, as they would be in the case of lines. Instead, the points are more dense in the centres of the ellipses, and so this results in the slightly larger limiting value of GMD. However, the horizontal distribution of points remains the same regardless of the shape factor, and so a limiting value is reached at small values of u, when the vertical coordinate of the points becomes insigniﬁcant compared to the horizontal coordinate. Subtracting the ﬁxed offset of 2 from the limiting GMD value of 1.8507, gives the asymptote value of −0.1493. It is encouraging to ﬁnd that we are dealing with a doubly asymptotic function. We can come up with a simple ﬁtting function that gives these same asymptotes, and then we only have to worry about minimizing error in the intermediate region. A function which accomplishes this is: fv(u) = k0u+k1+k2/(1+k3u) k0 is the slope of the inclined asymptote and has a value of 0.38940 as previously determined. The sum of k1+k2 is equal to the horizontal asymptote of −0.1493. The values of k2 and k3 determine the shape of the curve in the transition region. It is also possible, if necessary, to add additional terms to the denominator of the last term to help minimize error. It turns out that additional terms are unnecessary, because the existing terms can be optimized to give a peak error of no more than 0.6%. The optimized values are: k0 = 0.38940 k1 = −0.71245 k2 = 0.56397 k3 = 0.73139 Thus, the correction term is: fv(u) = 0.38940u − 0.71245 + 0.56397/(1+ 0.73139u) And the complete GMD formula for touching ellipses is: g = 2 + [ 0.3894u − 0.71245 + 0.56397/(1+ 0.73139u)] (3.3.2) To get the complete correction term f(x,u), we must combine fv(u) with another function that is dependent on the offset x. We will now introduce a decay function, fx(x), to account for variable – ! – 34 offset values. This function must have a value of 1 at x = 2 (the minimum offset) and then decrease asymptotically towards zero as x becomes large. A simple function meeting the requirements is: fx(x) = k4/(k5+k6x) which, when combined with fv(u), gives a complete formula for the correction term: f(x,u) = [ k0v + k1 + k2/(1+k3v)] ⨉ [k4/(k5+k6x)] And ultimately, the complete normalized GMD formula: g = x + [ k0v + k1 + k2/(1+k3v)] ⨉ [k4/(k5+k6x)] The function fx(x) is known as a rational function because it is the ratio of two polynomials., although this is a fairly trivial example. The polynomial in the numerator is simply the constant k4, and the polynomial in the denominator is k5+k6x. Neither are overly complicated. However, rational functions can have a polynomial of any degree in both the numerator and denominator, and because they can be tailored to ﬁt a wide variety of curves they are popular as ﬁtting functions. (They will be employed several more times as ﬁtting functions in later sections.) It turns out that the simple form given here will ﬁt the data just as well as higher degree polynomials. The optimum values for the constants are: k4 = 3.55610 k5 = 2.54114 k6 = 0.49887 Adding more terms of higher power gives no improvement beyond what this basic function gives. These constants give a peak error of 1.78% in the overall GMD value, which is fair, considering that fv(u) and fx(x) are independent of each other and the functions are very simple. Unfortunately, the worst case error occurs close to the minimum spacing where it does the most harm. Therefore, it’s worth putting a bit more effort into the development of a decay function with better characteristics. Looking at the original set of curves, it can be seen that there is a point of inﬂection for each curve which occurs at a different position. The varying location of this inﬂection point makes it difﬁcult to apply a simple decay function based only on the offset distance. It appears that the inﬂection point occurs approximately at x = u. It may be beneﬁcial therefore to shift the argument of the decay function by a factor of u. Hence fx(x) now becomes fx(x+k7u). This does give some further improvement. In addition, once the constants k4 through k7, have been optimized, we can then do a ﬁnal re-optimization all of the constants k1 through k7, and during the optimization, weight the optimizer’s error criteria so that errors at the lowest offsets have the most inﬂuence on the ﬁnal result. During this optimization, it is also useful to rearrange the fv(u) function as: fv(u) = k0u+(k1+k2u)/(1+k3u) With suitable adjustment of the values k1 and k2, this is completely equivalent to the previous form of fv(u), but allows for more convenient constant values. The constant k1 is now equal to the horizontal asymptote. However, rather than restricting its value to the known asymptote, we will – ! – 35 adjust it in order to improve the accuracy of the function over the more useful range of the function, thus sacriﬁcing the correct asymptotic behaviour at very small values of u. However, this is somewhat compensated for by the simultaneous optimization of the decay function constants. Doing the ﬁnal optimization, we end up with a maximum peak error of about 0.5% and this peak error occurs at at more distant offsets than in the earlier optimization. The optimized function and the constants are: g = x + [k0u + (k1 + k2u)/(1+k3u)] ⨉ [k4/(k5+k6(k7u + x))] (3.3.3) k0 = 0.38940 k1 = −0.0025158 k2 = −0.39042 k3 = 0.0091452 k4 = 32.53253 k5 = −0.05161 k6 = 0.29995 k7 = 0.86873 This is a large number of constants, but the formula itself is fairly simple. Overlaying the values produced by the empirical function onto the Monte Carlo calculated data, we get the following graph: ! – ! – 36 It can be seen that the approximation is very good at the left side of the graph where the offset is less than ﬁve times the ellipses’ horizontal semi-axis, but then fails to decay to zero as quickly as the true data. Because this is a graph of the correction term only, the effect of the error appears to be worse than it is in practice. The overall error becomes a very small percentage of the net GMD value once the correction factor is added to the offset (0.5% peak error). Also, the way that GMD is used in multi-turn coil inductance calculations will further reduce the total error due to distant spacings. Additional work could be done to improve the decay function, but it is unlikely to provide any signiﬁcant overall improvement in the ﬁnal inductance calculations. It should be remembered that this function has been normalized such that the offset value x is expressed as a multiple of the horizontal semi-axis b. Therefore, x must be calculated by dividing the actual offset distance by the horizontal semi-axis of the ellipse, b, and the result of the formula must then be multiplied by b to get the true GMD value. It should also be remembered that this function cannot be used to ﬁnd the self GMD (by setting x = 0), as this is well beyond the range of the data that were used for ﬁtting the function. In addition to ellipses, the method developed here should be readily adaptable to the creation of a GMD function for other oblong shapes. For example rectangles would be logical candidates, were it not for the fact that an analytical rectangle GMD formula already exists. 3.4 The Self GMD of an Elliptical Line To distinguish between the area enclosed by an ellipse, and ellipse itself (excluding the internal area), we will refer to the hollow ellipse as an elliptical line or elliptical locus. In the previous section we discussed the GMD of the area enclosed by an ellipse. It was also brieﬂy mentioned that a more sophisticated Monte Carlo method would be required for the case of an elliptical line. The problem arises when we try to generate uniformly distributed random points on the locus of the ellipse (or any curve for that matter). We cannot use the rejection sampling method that we used for the triangular and elliptical areas, because the locus has no area and therefore there is zero probability that a random point generated in the enclosing area will fall on the locus. We need to ﬁnd another way to drop points randomly on the locus. For the degenerate case of a circle, the solution is quite simple. We simply use the trigonometric parametric equations for a circle: x = b cos(θ) (3.4.1) y = a sin(θ) (3.4.2) where a = b, and θ ranges from 0 to 2π radians. If we assign to θ, uniformly distributed random numbers in the range of 0 to 2π, then we achieve the uniformly distributed points on the circular locus. However, in the more general case where a≠b, then we have an ellipse, and this method fails. The diagram below shows what happens when we plot points at equal increments of θ on a circle (left), and then on an ellipse (centre) with b = 0.3a. As the ellipse becomes squashed, the points become concentrated towards the ends. – ! – 37 Hence, if we were to assign uniformly distributed random values to θ, the distribution of points on the ellipse would not be uniform. ! The ﬁgure on the far right, however, shows the desired uniform spacing of points. The fact that such a ﬁgure exists, with its evenly distributed points, means that there must be a way to achieve this distribution. To do this, we need to ﬁnd a function that takes the distance along the locus as an argument, and returns the corresponding angle θ. By feeding this function a set of uniformly distributed random numbers, we can then apply the resulting values of θ as the parameters of the ellipse, to get the uniform point distribution. We will refer to this distance to angle function as the mapping function. The inverse of the formula we are looking for, would take the angle θ as the argument and return the distance along the locus. This function exists; it is the Incomplete Elliptic Integral of the Second Kind E(ϕ,k). We can use this function to generate a dataset of angle and length pairs, then create an empirical ﬁtting function for these data that uses the arc length as argument and returns the angle, essentially the inverse of E(ϕ,k). The ﬁrst step then, is to generate the dataset. Unfortunately, an implementation of the function E(ϕ,k) is not readily available—certainly not in Open Ofﬁce Calc. One of the goals of this work was to avoid the use of special math software, but rather, to use simple software such as is available in a spreadsheet program, or if necessary, a simple programming language such as BASIC. Of course the function E(ϕ,k) could be coded into BASIC, but this is not really necessary in order to generate the data. Instead, we simply use a spreadsheet, and the parametric equations for the ellipse to generate a list of x and y coordinates for values of θ ranging from 0 to π⁄2, with θ changing in very small increments. We can then calculate the length of the arc segments from one increment to the next using the formula: ! (3.4.3) arc = p (xi −xi−1)2 + (yi −yi−1)2 . – ! – 38 Then, by taking the running total of these ℓarc values, we get the total length from the start of the locus to the speciﬁed point on the curve. This method has the advantage that it can also be used for shapes other than ellipses. We need only generate the data for one quadrant of the ellipse, and then ﬁnd a ﬁtting function for that single quadrant. Because of its symmetry, the data for the other three quadrants can be easily determined from the single quadrant data set. Dealing with a single quadrant, makes it easier to ﬁnd a ﬁtting function. A graph of the data for the ellipse with an aspect ratio of u = b⁄a = 0.3 is shown below: ! Incidentally, if we had used an aspect ratio of 1 (a circle), the above graph would be a straight line. As the aspect ratio gets smaller, the line becomes more curved. The arc length is measured from a starting point which we will deﬁne to be the point where the ellipse crosses the positive x axis. For an aspect ratio of 1 (a circle) where semi-major and semi-minor axes a = b = 1, the arc length will be equal to the angle in radians. So, for one quadrant the arc length is π⁄2. As the aspect ratio decreases, the quadrant arc length decreases until, at an aspect ratio of 0, the arc length is equal to the semi-major axis a. For this study we will normalize the quadrant arc length to a range of 0 to 1 for all aspect ratios. Hence the horizontal axis on the above graph ranges from 0 to 1. Additionally, the arc will be measured counterclockwise from the starting point. We also deﬁne the geometry such that, as the aspect ratio becomes smaller, the ellipse is ﬂattened in the x dimension. That is, the ellipse becomes tall and skinny. When the aspect ratio is one, the ﬁgure is a circle. When the aspect ratio is zero, the ﬁgure becomes a vertical line. This covers the entire range of shapes that the ellipse can assume. – ! – 39 To ﬁt these data, a rational expression of the following form was chosen for the mapping function: θ = (k1s +k2s2+k3s3+k4s4)/(1+k5s+k6s2+k7s3) (3.4.4) where s is the length of the arc measured from the starting point and θ is the parametric angle as discussed above. This function very likely has more terms than are necessary, but since it serves only a temporary purpose, maximum efﬁciency was not the goal. Optimizing the coefﬁcients was done using a solver macro in an Open Ofﬁce spreadsheet. For an aspect ratio of u = 0.3, the optimum coefﬁcients are: k1 = 1.09677644 k2 = 0.31383321 k3 = −2.92801438 k4 = 1.52766397 k5 = 0.29919089 k6 = −2.96132963 k7 = 1.66867001 Of course this gives us a function only for the case where the aspect ratio of the ellipse is equal to 0.3. Developing a more general empirical function that takes the additional argument of the aspect ratio would be a far bigger job than is warranted. Instead, different sets of coefﬁcients k1..k7 were obtained for aspect ratios of 0.125, 0.25, 0.3, 0.35, 0.4, 0.5, and 0.75. After looking at the results—in particular, the ﬁtting error—additional sets of coefﬁcients were obtained for aspect ratios of 0.02, 0.03, 0.063, 0.315, and 0.325. The peak errors for the various sets of coefﬁcients of the ﬁtting function are shown in the graph below. ! – ! – 40 It is not surprising to see the higher errors for smaller aspect ratios, because the mapping function becomes more curved for smaller ratios, and conceivably harder to ﬁt. However, it is surprising to see the second peak at an aspect ratio of 0.25. This may simply be due the solver having difﬁculty ﬁnding the best ﬁt in this case. The errors are still quite small however, and are not expected to have a signiﬁcant impact on the ﬁnal result. Having created the mapping function, we can now take a set of random numbers, pass them through this function and then use the resulting values of θ to create a set of uniformly distributed random points on an elliptical locus. All this effort to produce a uniform distribution of random points on the locus of the ellipse has overshadowed an otherwise simple Monte Carlo algorithm. However, this mapping function is crucial, because without a means to generate uniformly distributed random points, the Monte Carlo method would be unusable. We can now return to the primary issue. As with the previous calculations, the program code will generate random points on the elliptical locus, then measure the distance between pairs of them, take the logarithm, sum them, and hence, ﬁnd the geometric mean. The program code is given in Appendix A. With 109 pairs of random points for each run, we get the results shown in the table to the right. These are normalized to a semi-major axis value of 1.0. In addition, there are two limiting values that can be included in this list. We know that for an aspect ratio of 1, the ellipse becomes a circle, and therefore, its GMD is equal to 1. Also, but not quite as obvious, when the aspect ratio goes to zero, the ellipse becomes a line. But unlike the case of an elliptical area where the point density never becomes linear, in the case of an elliptical locus, it does indeed become linear, and we can use the self GMD of a line which is e −3/2 = 0.22313 times its length. With a semi-major axis of 1, the ellipse has a length of 2, and the GMD in this limiting case becomes 2 ⨉ 0.22313 = 0.44626. – ! – 41 Aspect Ratio GMD 0.020 0.4577052 0.030 0.4636561 0.063 0.4828448 0.125 0.5205759 0.185 0.5571103 0.250 0.5963800 0.275 0.6111260 0.300 0.6258739 0.315 0.6346541 0.325 0.6404787 0.350 0.6549459 0.400 0.6835489 0.500 0.7393943 0.750 0.8727630 Plotting the data from the table along with the two limiting values, we get the following graph: ! Despite the ordeal of obtaining the uniformly distributed random points, we see that the ﬁnal result is a nearly straight line. It turns out that, while not a perfectly straight line, these points can be ﬁt very well with a parabolic curve. Doing a least squares ﬁt, the optimized formula for self GMD of an elliptical line is: ! (3.4.5) where u = b/a (the aspect ratio). The ﬁrst term is analytical, as previously discussed, while the last two are empirical. The similarity between the two empirical coefﬁcients (and considering that 0.615 ≈ 8⁄13) suggests a simpler and easier to remember formula: % (3.4.6) gs = [2e−3 2 + 0.61578u −0.061591u2]a . gs = h 2e−3 2 + 8 13 ⇣ u −u2 10 ⌘i a . – ! – 42 This simpliﬁcation gives an insigniﬁcant increase, from 0.0000448 to 0.0000453, in the total squared error, but notably, a small decrease in peak error, from 0.407% down to 0.402%. Hence, this simpler formula is recommended. There may not be much demand for the GMD of an elliptical line . If there was, then either Grover, 4 or one of his contemporaries likely would have come up with an expression for it six decades ago. However, the main point is that if we do need to ﬁnd the GMD of some shape that has not been previously analyzed, then we have the means to do it. An elliptical line is a good example of one of the more complicated cases. Yet, we are able to analyze it, and come up with a practical expression for it. Appendix A gives the BASIC program listing for the Monte Carlo calculation of GMD of the elliptical lines discussed here. It consists of a small main program which implements the main part of the Monte Carlo simulation, and a subprogram which implements the mapping function. The sets of mapping function coefﬁcients for different aspect ratios are readily apparent in the program code. It should be noted that this program runs much slower than the program for ﬁnding the GMD of elliptical areas. This is due to the mapping function which adds a signiﬁcant number of mathematical operations inside the main calculation loop. Even though the earlier program used sample rejection to obtain the uniform distribution and therefore rejected 21% of the random points generated, the remaining operations were very few and very efﬁcient. As previously mentioned, the determination of the constants for the mapping function was done using a spreadsheet and a solver macro. The ﬁnal ﬁtting of the GMD data produced by the BASIC Monte Carlo program was also done using a spreadsheet and solver macro. In fact, when calculating the inductance of a coil constructed of round thin wall tubing, and having large pitch, the 4 cross section of the conductor, taken in a plane that passes through the coil axis, will be an elliptical line. The GMD of an elliptical line then becomes useful in the analysis of the coil. – ! – 43 Part 4 - Limits of Accuracy of the GMD method 4.1 Self Inductance It was stated earlier that the GMD principle can be applied to conductor geometries other than straight lines as long as the conductor diameter is small relative to the overall conductor length, and the radius of curvature is large compared to the cross sectional diameter of the conductor. That is a somewhat vague statement, and deserves further discussion. Unfortunately, any research done on the limits of accuracy of the application of GMD in inductance calculations, appears to be non-existent. In this section, inductance values calculated using the GMD method will be compared with those calculated by other methods, in order to determine the point at which the accuracy deteriorates. The simplest example for examining the effect of conductor curvature is the inductance of a circular loop of a conductor of ﬁnite diameter. We will look at what happens as the ratio of conductor diameter to loop diameter is changed. The GMD method is straightforward: we calculate the self GMD of the conductor, and then use that as the axial separation of two ﬁlamental loops of the same diameter, and plug these numbers into Maxwell’s loop mutual inductance formula [1b]. This must then be compared with a method which does not make the GMD simpliﬁcation. One such formula has been given by Max Wien , where he integrated the mutual inductance formula twice across the cross section of a round conductor to get a supposedly exact formula. However, his formula was later found to have errors in one of the terms. In addition, he discarded high order terms, which renders the formula unsuitable for this study. Other formulae followed Wien's. However, these too contained simpliﬁcations where high order terms were discarded, again making them unsuitable for use here. The exact method is to integrate Maxwell’s formula across the cross section of the conductor. Because the integrand is an elliptic integral, this must be carried out using numerical integration techniques. This poses a bit of a problem, because with numerical integration, we normally evaluate and sum a series of rectangular areas or, in the case of a circular area, circular segments. Both of these methods are rather cumbersome. We still need to calculate the mutual inductance of every possible combination of pairs of these oddly oriented and/or oddly shaped areas, and end up having to use the GMD between these areas for lack of any better method. Further complicating matters, there is no formula for the GMD of the oddly shaped pieces in all their myriad orientations, necessitating more approximations. We would expect that as the cross sectional area is divided into smaller and smaller parts, that the accuracy would improve. Yet, there is no guarantee that some systematic error will not manifest itself in the mutual inductance between very closely spaced pairs of areas, or especially the self inductance of an area. Nevertheless, this method was initially tried, and seemed to give reasonable results despite being computationally inefﬁcient. On the other hand, by adapting the Monte Carlo method, we replace the potential systematic error for random error. But at least the random error should have no particular bias as long as the random numbers themselves have no bias. The other advantage is that it is much simpler to code, – ! – 44 and is computationally much more efﬁcient. Consequently, the Monte Carlo method is used in the following discussion. The Monte Carlo method assumes that the conductor is composed of an inﬁnite number of inﬁnitesimal ﬁlaments. A representative sample of random pairs of these is taken, their mutual inductance is calculated, and the mean is taken. By taking a larger and larger sample we can reduce the random error to an acceptable value. Following is the BASIC program code which implements the Monte Carlo procedure. Function McL(Rc As Double, dw As Double, yOffset As Double, N As Int64) As double ' Calculate Self/Mutual Inductance of Loop Conductor by Monte Carlo method ' Rc = Loop Radius, dw = Conductor diameter ' yOffset = axial loop offset, N = Number of random point pairs dim i As Int64 dim tx,ty,tye,tye1,sx,sy,sye,sye1,r1,r2,rw,sum,axial As Double rw = dw/2 sum = 0 For i = 1 to N 'Generate two random test points S and T do tx = (rnd-0.5)dw ty = (rnd-0.5)dw tye = sqrt(rw^2-tx^2) tye1 = -tye loop until (ty< = tye and ty> = tye1) do sx = (rnd-0.5)dw sy = (rnd-0.5)dw sye = sqrt(rw^2-sx^2) sye1 = -sye Loop Until (sy< = sye and sy> = sye1) 'Add offsets to random points to get radius and axial displacement axial = abs(yOffset+sx-tx) r1 = Rc+ty r2 = Rc+sy sum = sum+Mut1(r1,r2,axial) next return sum/N End Function Function Mut(ByVal r1 as double,r2 as double,x as Double) As Double ' Maxwell's elliptic integral formula to calculate mutual inductance ' between two coaxial circular loops. ' Uses AGM method to calculate elliptic integrals ' r1,r2 are radii of the respective loops ' x is the axial distance separating them muo = pi()4e-7 a = sqr((r1+r2)^2+x^2) b = sqr((r1-r2)^2+x^2) m = 4(r1r2)/((r1+r2)^2+x^2) c = a-b ci = 1 cs = cc do – ! – 45 ao = (a+b)/2 b = sqr(ab) a = ao co = c c = a-b ci = 2ci cs = cs+cicc loop while c<co Mut = muo/8pi()cs/a end Function The results of the program are plotted in the following graph which shows the relative error of the GMD inductance value, compared to the Monte Carlo inductance value, as the ratio of conductor diameter to loop radius is varied. Note that the largest possible ratio d/D is 1. We will deﬁne the variable u as the ratio d/D in the following discussion. We will use the variable ϵ as the relative error, and deﬁne it as: ϵ = LGMD/LMC − 1 (4.1.1) where LGMD is the inductance value calculated by the GMD method, and LMC is the inductance calculated by the Monte Carlo method. By this deﬁnition, a positive value for ϵ means that LGMD is higher than the the "true" inductance value. ! Surprisingly, the error is only −0.1% (−0.001 per unit, or −1000 ppM) when the wire diameter is as much as 0.15 times the loop diameter, which indicates that the GMD approach to inductance calculation is an excellent approximation for practical values of u. Even in the worst case value of – ! – 46 u = 1.0, the error is −13% (−0.13 per unit), which is surprisingly good when one considers the puffy bagel-like torus in such a situation. Also note that the error is always negative, indicating that the GMD method will always yield a self inductance value that is slightly lower than the true value. If we do a log-log plot of the data we get a much straighter line, but still slightly curved. So, there 5 is no simple power relationship: ! The inherent randomness in the Monte Carlo method makes the line noisy for values of u less than 0.02 (i.e., log(u) < −4), which is noticeable at the left side of the graph, but this is of little consequence because the error at this point is only a few parts/million, and is heading towards zero. It turns out that these data can be ﬁt very well with a third degree polynomial. Solving for the best ﬁt coefﬁcients for the cubic regression line, the equation for the error is: ϵ = −2.9945⨉10−5 u − 0.027273 u2 − 0.10003 u3 (4.1.2) Rather than plot the data on a logarithmically scaled graph as is traditionally done, we simply plot the natural 5 logarithms of the data on a linearly scaled graph. The resulting line has the same curvature in either case. – ! – 47 Below are both linear and log graphs showing the ﬁtted function (in red) along with the original data points. ! ! Note that there is no zero order (constant) term. It must be zero to ensure that the error value properly goes to zero as u goes to zero. – ! – 48 This error function could be used as a correction factor applied to GMD calculated self-inductance values, if extreme accuracy is needed in theoretical calculations. However, in real world coil production, the normal variations in coil construction would result in a much larger inductance variation than the magnitude of this error value. 4.2 Mutual Inductance Having looked at the limits of accuracy for loop self inductance, we will now look at what happens to the mutual inductance of a co-axial pair of wire loops as u is varied, and as the axial separation is varied. The same program is used, but the yOffset variable is now set to values other than zero. In this investigation, we will normalize the the axial offset as multiples of the loop diameter, so that the results may be more easily applied to other conﬁgurations. Hence, the variable x will be deﬁned as the ratio of axial spacing to loop diameter. The following graph shows the set of curves generated by the program, each curve representing a different value of axial offset ratio x, and with u varying along the horizontal axis. Each curve is labeled with the corresponding value of x. ! Note that for each curve, the maximum possible value for u is the smaller of x and 1. – ! – 49 Re-plotting as a log-log graph shows (below), a series of straight lines. ! The log-log plot emphasizes the random scatter in the Monte Carlo calculation at very small values of u. This is a result of subtracting pairs of nearly equal values, differing only in the last couple of signiﬁcant digits, causing the Monte Carlo random error to be accentuated. But again, these correspond to only a few parts/million of actual GMD error and are of minor consequence. The curves lie very close together, indicating that the error is much more sensitive to variations of u than of x. It is also interesting to note that for very small values of x, the curves start to bunch together, and the same thing happens, for large values of x. This implies that values of x outside the range examined here, will produce error values that do not lie signiﬁcantly outside the range of the curves shown. – ! – 50 Because the lines are straight (ignoring the noise) on the log-log plot, they can be ﬁt by the log of the straight line formula: log(ϵ) = a + m log(u) (4.2.1) or taking the exponential function of both sides: ϵ = ea vm (4.2.2) Since ea is a constant, we will deﬁne a new constant b = ea, and hence: ϵ = b vm (4.2.3) By performing a regression analysis on each line, we get a set of y-intercept (b) and slope (m) values as shown in the table at the right. The slope is very close to 2 in every case. It drops very slightly where the extreme cases u = 1 can be included, suggesting that the value is relatively constant at 2, but then starts to change in the region where u approaches the value 1. In fact, this can be seen at the far right side of the log-log graph. Where u approaches 1, the lines begin to curve slightly. This evidence seems to be sufﬁciently compelling to conclude that the principal term in the error function is second order (i.e., proportional to u2). Hence, the GMD inductance approximation is correct to the ﬁrst order, but does not fully account for second order effects, and that higher order error effects are almost negligible, except when u approaches 1. Consequently, we will be presume the slope to be exactly 2, and we will then recalculate the intercept b values for best ﬁt. This recalculation is shown in the next table to the right. – ! – 51 x b m 0.02 -1.65984 2.01249 0.05 -1.59745 2.03071 0.1 -1.59625 2.00430 0.2 -1.36547 2.02736 0.5 -1.17197 2.00648 1.0 -1.09616 1.94528 2.0 -0.99331 1.92341 5.0 -0.94899 1.91409 10.0 -0.93911 1.91810 50.0 -0.93638 1.91813 x b m 0.02 -1.79005 2.00000 0.05 -1.73758 2.00000 0.1 -1.62895 2.00000 0.2 -1.47300 2.00000 0.5 -1.18791 2.00000 1 -1.05116 2.00000 2 -0.93378 2.00000 5 -0.88396 2.00000 10 -0.87750 2.00000 50 -0.87378 2.00000 If we plot these b values against x, we see an extremely nonlinear relationship: ! Given the hyperbola-like shape of the curve, it seems plausible that changing the ordinate of the graph from x to 1/x might straighten it out. However, it turns out that it makes virtually no improvement. On the other hand, a log-log plot is more enlightening: ! – ! – 52 This is a sigmoid (S-shaped) curve with an inﬂection point near log(x) = 0. The inﬂection point deserves some discussion. It occurs where the distance separating the loops is approximately equal to the loop diameter—a square aspect ratio. It signiﬁes a point where one factor becomes less inﬂuential to the overall behaviour, and another factor becomes more inﬂuential. In this case, when the loops are close together, the largest inﬂuence on the mutual inductance on a point on one loop is due to the parallel current ﬂowing in the nearest point of the other loop. This is much more signiﬁcant than any other factor. As the spacing reaches the inﬂection point, the inﬂuence of the anti-parallel currents from diametrically opposed points on the loops become just as inﬂuential as the parallel currents, and then remain that way as the separation increases further. Considering that the mutual inductance error curves are a function of two variables, with one of them being very nonlinear, it was questionable whether it would be feasible to try to ﬁt an empirical error function. Nevertheless, it seemed to be worthwhile to expend some effort to ﬁnd a ﬁtting function. Given that each curve is primarily a simple square law function, the main effort then would be to ﬁnd a ﬁtting function for a sigmoid curve that predicts the b coefﬁcients based on the value of x. There are numerous functions which have this shape, including the arctangent function and the hyperbolic tangent function, as well as some algebraic functions and some functions involving exponentials. Testing several of them, it was found that the hyperbolic tangent function ﬁts these data very well. Since tangent functions are related to aspect ratios, perhaps this should not be too surprising. Overlaying the ﬁtted hyperbolic tangent function (red), we get the following graph: ! – ! – 53 As can be seen, it ﬁts the data quite well. However, some caution is required. This is a ﬁtted function of the log of the data. We must then take the antilog of this function to ﬁt the original data, and doing so can exacerbate any errors in the ﬁtting function. The function and optimized coefﬁcients are as follows: log(-b) = k0+k1 tanh(k2+k3 log(x) ) (4.2.4) k0 = −1.34177 k1 = 0.47936 k2 = 0.74321 k3 = 0.62668 Applying the exponential function to both sides of the equation, and negating, we get: ! (4.2.5) Having developed a relationship for the b values as a function of x, we now have the basis of a function that will predict the GMD error for co-axial loop mutual inductance. Substituting the b function into the ϵ function we get: ! (4.2.6) b = −exp  k0 + k1 tanh " k2 + k3 log (x) #$ . ✏= −exp  k0 + k1 tanh " k2 + k3 log (x) #$ u2 . – ! – 54 Below is the log-log error graph with the predicted error function overlaid: ! With the lines grouped so closely together it is not easy to see much beyond the fact that they do seem to line up. It can be seen that as log(u) approaches 0 (i.e., u approaches 1), the error curves start to bend away from a straight line, as was mentioned earlier. – ! – 55 Re-plotting on a linear graph, we get the following: ! The correlation for 0 < u ≤ 0.5 is reasonable. However, the region of 0.5 < u ≤ 1 leaves much to be desired. While this range of geometry is an extreme that is unlikely to be encountered in practice, it's worth considering the inclusion of a correction term that comes into effect for u > 0.5. The correction may take either of two forms, an additive correction term, or a multiplying factor. After some experimentation, the additive term appeared to be the most promising, and that is the approach that was taken. We therefore want a term which is asymptotic to zero when u < 0.5, and then smoothly bends away from zero when u > 0.5. As was used in the development of the mutual GMD formula for elliptical areas, a rational function again appears to be suitable. A simple function that accomplishes this is: ! The exponents n and m determine the function's behaviour when u is large. if we set n = m+1, then for large u, the function will asymptotically approach a straight line with a slope equal to k4. The constant k5 determines the crossover point between the zero asymptote and the constant slope asymptote. Though it is proposed to set n to be one greater than m, there is nothing to suggest that k4 un um + k5 . – ! – 56 the difference cannot be greater, especially since the function is limited to a ﬁnite range of values for u, eliminating the risk of it blowing up for large values of u. It's always best to start with the simplest function, and if it's found to be unsuitable, then it can be further adjusted. With some experimentation, it was found that the optimum values for m and n are 3 and 4 respectively. Hence, the correction term is: ! The optimum values of k4 and k5 will be given later. While this correction term was a good starting point and gave reasonable improvement in the ﬁt, it was found not to be sufﬁcient to adequately correct the empirical function. It became clear that the correction term needs to be a function of both u and x. While u is conﬁned to the range of values from 0 to 1, x can take on any value from zero to inﬁnity, which adds some complication to ﬁnding a function that will be well behaved for any x. As previously mentioned, for values of x greater than about 5, the error curves begin to bunch together. In fact there is no discernible difference between the curve for x = 10 and for x = 50. The correction term involving x must not disturb this behaviour. Therefore, this correction factor must approach a ﬁxed value asymptotically as x becomes large. Again a rational function is chosen. In this case: ! This function has a value equal to k6 when x = 0, and then asymptotically approaches 1 as x becomes large. The product of the ﬁrst and second correction factors then becomes the complete correction term: ! And, the complete GMD error function is then ! (4.2.7) k4 u4 u3 + k5 . x2 + k6 x2 + 1 . k4 ✓ u4 u3 + k5 ◆✓x2 + k6 x2 + 1 ◆ . ✏= −exp  k0 + k1 tanh " k2 + k3 log (x) #$ u2 + k4 ✓ u4 u3 + k5 ◆✓x2 + k6 x2 + 1 ◆ . – ! – 57 With the candidate function having been set, all constants (including k0−k3) are then re-optimized, yielding the following values: k0 = −1.22339 k1 = 0.53125 k2 = 0.66240 k3 = 0.77879 k4 = 0.76927 k5 = 4.58965 k6 = 0.25572 The optimization process required certain compromises to be made. Primarily, it was a question of whether to do a least squares ﬁt based on absolute error or on relative error. Relative error produces a better ﬁt for small values of u, while absolute error produces a better ﬁt for large values of u. Normally, relative error would be the best choice, especially since the parameters are normalized so that they can be scaled for any application, and would then yield a consistent percentage error for any value. However, the large scatter of the points at small u resulted in an unsatisfactory ﬁt. Ultimately, it was decided to use relative error as the ﬁtting criterion, but with the data weighted by the square root of u. No statistical checks were done to verify the validity of this approach. However, this gave the most satisfactory ﬁt based on a visible inspection of the graphs of the ﬁtting function, when viewed at microscopic and macroscopic levels. The following graphs show the results of the ﬁtting process on both linear and log scales. The red curves are the ﬁtting function, while the blue points are the data (Monte Carlo calculations) that were ﬁtted. ! – ! – 58 ! While the ﬁnal formula may seem slightly unwieldy, the fact that it is reasonably accurate over several decades of u and x values, the complexity is not excessive. – ! – 59 4.3 Example Calculation At this point it is worthwhile to provide a worked example. We will show the total error in the GMD based inductance calculation of a solenoid coil of practical dimensions. For this example, the coil will have the following parameters: N = 30 turns d = 0.5 mm D = 25 mm Winding Pitch = 1 mm Below is a clipping from a spreadsheet showing the inductance calculation using the summation method. ! The self inductance of each turn is calculated on row 8 by setting the pair offset value to the self GMD of the wire cross section. The mutual inductance of every combination of pairs of turns is calculated on rows 11 through 39. All self and mutual inductance calculations are done using the – ! – 60 Mut() function previously described and with the results shown in column D in units of microhenries. The mutual inductance of each pair of turns from column D is multiplied by the number of occurrences of each pair. This total value is shown in column E. The total of column E from rows 8 through 39 gives the inductance of the complete coil and is shown in cell E40 as 13.36784 µH. Columns F and G show the x and u values which are used to calculate the predicted error ϵ (per unit) in column H. The absolute error in microhenries shown in column I is calculated by multiplying the value of self or mutual inductance from column E by ϵ/(ϵ+1). The total of column I is the amount, in microhenries, that the inductance is predicted to be in error. The total error value shown in cell I40 is −0.00106 µH. That is, the calculated inductance value is predicted to be low by −0.00106 µH. This amounts to only about 0.008% of the calculated inductance value, and is essentially insigniﬁcant. However, if we wish to adjust the calculated value to account for the predicted error, then we can subtract this error value from the calculated value to get an “adjusted” value which is shown in cell I42. What this example tells us is that the approximation involved in calculation by the GMD method is a very good one for coils of typical geometry. 4.4 Summary Having derived the error functions for both the self and mutual loop inductance GMD based calculations, these values can be used as part of an error analysis procedure when calculating the inductance of an entire coil using the summation method. If one were so inclined, and suitably ambitious, this study could be expanded to include loop conductors having non circular cross sections. – ! – 61 Part 5 - Combined GMD of Multiple Objects 5.1 Basic Principles Maxwell [1a], in addition to providing formulae for the GMD of simple geometric shapes using integration as discussed above, also showed how to combine the GMD of these simple shapes to determine the GMD of more complex shapes. If we can calculate the GMD between objects using any of the previously discussed methods, then we can also calculate the GMD between groups of objects simply by taking the individual GMDs and ﬁnding their overall geometric mean weighted by the areas of the objects. For example, if we have three areas A1, A2 and A3, and if the GMD between A1 and A2 is g12, the GMD between A1 and A3 is g13, then the GMD between A1 and the group A2 and A3 g1-23 is: log(g1-23) = (A1A2log(g12)+A1A3log(g13))/(A1A2+A1A3) (5.1.1) Area A1 is included in the formula in order to show how it applies to the weighted mean, but in this case it cancels out and disappears from the formula, leaving: log(g1-23) = (A2log(g12)+A3log(g13))/(A2+A3) (5.1.2) In many other cases, the areas of all objects will remain in the formula. In cases where all of the ﬁgures have the same area, then all of the areas disappear from the formula, and only the number of objects remains, thus becoming a simple unweighted mean. The most general case is for the GMD between two different groups of areas. If we designate the ﬁrst group a consisting of m areas Aai where i = 1..m, and the second group b of n areas Abj where j = 1..n, then the GMD between the two groups is: ! (5.1.3) Now, in the special case where group a is the same group of objects as group b, then the above formula will produce the self GMD of that group of objects. Thus, for the self GMD of a group of objects, the formula becomes: ! (5.1.4) log(g) = m P i=1 n P j=1 AaiAbj log (gij) m P i=1 n P j=1 AaiAbj log(gS) = n P i=1 n P j=1 AiAj log (gij) n P i=1 n P j=1 AiAj – ! – 62 These formulae are sufﬁciently general to cover just about any case. From here, we can move on to some special cases which result in simpler formulae. Perhaps the most common of these is where all areas are identical. This eliminates the need for weighting by area, and so the two previous formulae 5.1.3 and 5.1.4 become: ! (5.1.5) and ! (5.1.6) Since the self GMD of each component is calculated differently than the GMD between different components, it makes practical sense to separate the self GMD terms from the other GMD terms, and rewrite the expression as follows: ! (5.1.7) The leftmost summation is for the self GMD of each component, and the double summation is for the GMD between different components. We will adopt the term mutual GMD for the GMD between different objects and designate it as gM. The formula may also be written as: ! (5.1.8) because the ﬁrst summation includes n terms and the second double summation includes n2−n terms. 5.2 Formula for the GMD of a Linear Array of Circular Conductors (LACC) A linear array of conductors frequently occurs in electrical and electronic systems. For example, high current power distribution systems often employ multiple bus bars per phase because multiple small bars will have more surface area than single large bars, allowing for better cooling, and thus higher current carrying capacity than a large bar with the same total amount of copper. As another example, a linear array of circular conductors also appears as the cross section of a single layer solenoid coil. In this section, a closed form formula will be developed for the self GMD of a linear array of circular conductors. It will be assumed that the pitch (centre-centre spacing between conductors) is constant. log(g) = 1 mn m P i=1 n P j=1 log (gij) log(gS) = 1 n2 n P i=1 n P j=1 log (gij) log (gS) = 1 n2 " n P i=1 log(gii) + n−1 P i=1 n P j=i+1 log(gij) # log (gS) = 1 n n P i=1 log(gii) + 1 n2 −n n−1 P i=1 n P j=i+1 log(gij) – ! – 63 Formula (5.1.7) from the previous section will be the starting point. The leftmost summation in the formula is for the self GMD component. Formula (1.4.3) for the self GMD of a circular area has been given as: gS = r e −¼ or log(gS) = log(r) − ¼ where r is the radius of the circular area, (i.e., the conductor). By multiplying the self GMD by the number of occurrences n, we have the ﬁrst summation of the formula: n (log(r) − ¼) For the mutual GMD terms, the GMD is simply the centre to centre distance between the various pairs of conductors. The general formula indicates that we must perform n2−n calculations. However, this number can be reduced signiﬁcantly. Many of the calculations are identical due to the fact that there are only n−1 unique spacings between conductors in the array. Consequently, it is necessary only to do n−1 calculations, and then multiply these results by the number of times they occur. If we designate the pitch as p, then the distance between any two conductors will be i⨉p, where i is an integer between 1 and n−1. The number of occurrences of pairs of conductors with the various spacings is equal to n−i. However, we must double this number in order to include both gij and gji cases. Taking all of this into account, the mutual GMD part of the formula becomes: ! And the complete formula for the self GMD of the linear array becomes: ! (5.2.1) This summation formula can be coded as a Basic program as follows: Function NetGMD(ByVal rw as double, n as double, p as double) ' Return aggregate GMD of linear array of circular areas ' rw is radius of circle; n is number of circles in the array ' p is centre to centre spacing (pitch) of circles s = (Log(rw)-0.25)n for i = 1 to n-1 s = s+2(n-i)Log(pi) next NetGMD = exp(s/(nn)) End Function n−1 P i=1 2(n −i) log (p ⇥i) log (gS) = 1 n2  n(log (r) −0.25) + n−1 P i=1 2(n −i) log (p ⇥i) # – ! – 64 While this is useful in a hard coded computer application (and will be useful later for testing), the goal in this section is to develop the summation formula into a closed form expression with all of the summations eliminated, thus eliminating the need for For-Next Loops. The ﬁrst term: ! is already a closed form expression and requires no further treatment. For the second term, it can be seen that since log(p⨉i) = log(p)+log(i), and the summation can be split into two parts: ! (5.2.2) It is straightforward to show that because p is a constant for a given conﬁguration, the leftmost summation can be simpliﬁed: ! (5.2.3) This leaves a single summation which is a function of only a single variable n: ! (5.2.4) To simplify the remaining derivation, the terms inside the square brackets will be multiplied by the 1/n2 factor: ! (5.2.5) We will designate the remaining summation term (including the 1/n2 factor) as gM’ which refers to the mutual GMD component gM excluding the pitch effect. n(log (r) −0.25) log (gS) = 1 n2  n(log (r) −0.25) + n−1 P i=1 2(n −i) log (p) + n−1 P i=1 2(n −i) log (i) # n−1 P i=1 2(n −i) log (p) = (n2 −n) log(p) log (gS) = 1 n2  n(log (r) −0.25) + (n2 −n) log(p) + n−1 P i=1 2(n −i) log (i) # log (gS) = 1 n(log (r) −0.25) + ✓ 1 −1 n ◆ log(p) + 1 n2 n−1 P i=1 2(n −i) log (i) – ! – 65 The following graph shows values of gM’ for n up to 13000. ! A semi-log plot of the same function shows asymptotic behaviour: ! – ! – 66 For values of n > 10 the function approximates a straight line on the semi-log graph. This indicates that for large n, gM’ can be approximated by: exp(gM’) = k0 n + k1 (5.2.5) where k0 is the slope of the line and k1 is the intercept of the asymptote on the exp(gM’) axis. Analysis of the data shows the slope to be approximately 0.22313, which is readily explained. As n grows large, this term of the self GMD formula becomes dominant. At the same time, our linear array of circular areas begins to approximate a line of evenly spaced dots. Hence, the self GMD begins to approach that of a straight line which is e−3/2 times its length. Thus, the asymptote has a slope exactly equal to e−3/2 (0.223130160148…). Interestingly, the asymptote does not pass through the origin. The exp(gM’) axis intercept can be determined from the data to be 0.410086, which gives a function that is approximately correct for large n: exp(gM’) = e−3/2 n + 0.410086 (5.2.6) The correct minimum value of the function is 0, occurring for both n = 1 and n = 2. To account for the nonlinear behaviour when n < 10, it appears that a correction term of the following form can be included: k2/(1+k3nm) (5.2.7) This works acceptably well, but if we consider that n is never less than one, it will turn out to be more convenient (and coincidentally, a better ﬁt) to use this form instead: k2/(1+k3(n−1)m) (5.2.8) The complete expression for gM’ is then: exp(gM’) = e−3/2 n + k1 + k2/(1+k3(n−1)m) (5.2.9) In the limiting case when n = 1, the denominator of the correction term is 1 and the value of the correction term is equal to k2. It then becomes a straightforward calculation to determine the value k2 so that the overall expression will produce the correct value of exactly 1 when n = 1. Hence: k2 = 1 − e−3/2 − k1 Restricting the value of k2 in this way, and then optimizing only k3, will result in no signiﬁcant difference in overall error as compared to optimizing both k2 and k3. – ! – 67 Choosing different values of the exponent m and adjusting the k3 value for the best ﬁt, yields a best exponent value of m = 1, and optimum k3 = 1.583635, resulting in this complete formula for gM’: exp(gM’) = e−3/2 n+ k1 + (1 − e−3/2 − k1)/(1+ k3 (n−1) ) (5.2.10) or gM’ = log( e−3/2 n+ k1 + (1 − e−3/2 − k1)/(1+ k3 (n−1) ) ) (5.2.11) where k1 = 0.410086 k3 = 1.583635 This has a worst case error of 0.17% occurring at n = 2 and n = 5, as shown in the following graph: ! The contribution of this term to the overall GMD, at these values of n, results in about the same overall percentage error, i.e., 0.17%, but its contribution drops off rapidly as n increases, so that the overall error in GMD drops off much more rapidly than what is shown in the gM’ error graph. We have now replaced all of the summation expressions in the GMD formula with simple closed form expressions. Combining all of the terms, we have: ! (5.2.12) Finally, taking the exponential function of both sides, we have the formula for the GMD of a linear array of circular conductors: ! (5.2.13) log(gS) = log e−3 2 n + k1 + 1 −e−3 2 −k1 1 + k3(n −1) ! + ✓ 1 −1 n ◆ log(p) + 1 n % log(r) −0.25 & . gS = exp " log e−3 2 n + k1 + 1 −e−3 2 −k1 1 + k3(n −1) ! + ✓ 1 −1 n ◆ log(p) + 1 n & log(r) −0.25 ' # . – ! – 68 By applying various exponential and logarithmic identities, all of the log functions and the enclosing exponential function can be eliminated from the formula: ! (5.2.14) When n = 1, the empirical factor is also 1. Thus, in this form, it is easy to see how the formula transitions from r e−1/4 (the self GMD of a circular area) when n = 1, to n p e−3/2 (the self GMD of a line) when n is large. The constant term, k1 = 0.410086, inside the ﬁrst factor, is signiﬁcant and deserves some discussion. Because of its presence, the GMD of the linear array will always be slightly greater than the GMD of a straight line of the same overall length. The principal difference between the two geometries, is that a straight line comprises a continuous row of points of inﬁnitesimal spacing. The linear array of circular areas has its points concentrated at intervals of p. It can be concluded that this difference accounts for the small difference in GMD between the two, which is equal to k1 p when n is large. If the pitch is made very small, tending towards zero, corresponding to inﬁnitesimal spacing between points, then this factor disappears, and the formula gives the same value as the formula for the GMD of a straight line. This coefﬁcient is of sufﬁcient importance that it should have a better name than k1. We will refer to it as the density uniformity coefﬁcient, and give it the symbol kD. This will come up again very shortly. It should also be stated that even though we don't yet have a simple analytical expression for the value of kD, this will be addressed later. It is important to note that in formula (5.2.14), variables r and p must have the same units of length. The resulting value of gS is in the same units as r and p. Since the results are scaleable, it is possible to remove either p or r from the formula, and replace the remaining variable with the ratio of the two, which would then give a dimensionless result. Let u be the ratio of conductor radius to pitch, r/p. Then: ! Looking at only the rightmost factors: ! and substituting u⨉p for r: ! Expanding the factors: ! gS = e−3 2 n + k1 + 1 −e−3 2 −k1 1 + k3(n −1) ! e( −1 4n )p(1−1 n)r( 1 n) . r = u ⇥p p(1−1 n)r( 1 n) p(1−1 n)(u ⇥p)( 1 n) p(1−1 n)p( 1 n)u( 1 n) . – ! – 69 Combining the p factors: ! The exponent of p simpliﬁes to one, and the complete GMD formula becomes: ! (5.2.15) The last remaining empirical constant k3 has been renamed kF (for ﬁtting constant). The values of these constants are as previously given: kD = 0.410086 kF = 1.583635 Pitch can now be removed from the formula, if desired, leaving a dimensionless formula, where gS is now expressed in terms (i.e., multiples) of pitch p. We are not quite ﬁnished with this formula. It is accurate as shown, but as we continue through the next sections, it will become apparent that certain terms can be combined and simpliﬁed. Additionally, the exact value of kD will be derived. It will be shown that kD is exactly equal to log(2π) e−3/2. However, it will also be shown that e−3/2 is a common factor throughout the formula. It will be separated out, and with the revised formula, kD will be redeﬁned simply as log(2π). The formula will also be modiﬁed to allow for hollow tubular conductors as well as solid conductors. This is governed by the rightmost exponential function. By changing the constant in the exponent from ¼ to 0, it changes from a solid conductor formula to a hollow conductor formula. This ﬁxed constant will be replaced by the variable 𝛾. Thus, the formula becomes: ! (5.2.16) As the ﬁnal step here, the factor e−3/2 will be factored out, and taken outside of the parentheses. ! (5.2.17) kF = 1.583635 𝛾 = ¼ for solid conductors 𝛾 = 0 for thin tubular conductors p(1−1 n + 1 n)u( 1 n) gS = e−3 2 n + kD + 1 −e−3 2 −kD 1 + kF (n −1) ! e( −1 4n )u( 1 n)p gS = e−3 2 n + log (2⇡)e−3 2 + 1 −e−3 2 −log (2⇡)e−3 2 1 + kF (n −1) ! e(−γ n)u( 1 n)p gS = n + log (2⇡) + e 3 2 −log (2⇡) −1 1 + kF (n −1) ! e−3 2 e(−γ n)u( 1 n)p – ! – 70 5.3 Formula for the GMD of a Linear Array of Co-linear Straight Line Segments (LALS) Having worked through the derivation of the GMD for the Linear Array of Circular Conductors (LACC), and recognizing the simple correspondence of each term in the formula to either the GMD of a circular conductor, or the GMD of a straight line, it quickly becomes apparent that it is nearly trivial to arrive at the equivalent GMD formula for a linear array of any geometric shape. A linear array of co-linear straight line segments (henceforth LALS) is a worthwhile example, because it can represent a current sheet which has been very well studied, and it gives us the opportunity to make comparisons between the two. First of all, from formula (2.2.6), the self GMD factor 𝛾 will change from ¼ to 3⁄2. The mutual GMD factor will change from the simple distance between circle centres to the more complicated formula (2.1.4) for the GMD for co-linear line segments. Fortunately, we already know from the previous derivation, that the mutual GMD factor must quickly tend to a straight line with a slope of e−3/2, and therefore the inﬂuence of (2.1.4) can be distilled down to nothing more than the single constant, kD, the density uniformity constant. Starting with (5.2.15) and substituting 𝛾 for the ﬁxed self inductance constant, our GMD function looks like this: ! (5.3.1) The variable u is now the ratio of line segment length to pitch. u = s/p (5.3.2) The density uniformity constant kD is not expected to be the same as for the LACC (round conductor) case, but it can be readily determined. By calculating the summations using a discrete summation formula where segment length is equal to the pitch (u=1), the value of kD is determined to be approximately 0.334707. However, if the ratio of segment length to pitch approaches zero, then it is apparent that the segments have an effect on the mutual component the same as inﬁnitesimal points, and the value of kD must become the same as for the round conductor case, 0.410086. This has been veriﬁed by further computer calculations. Therefore, kD is not a ﬁxed constant, but is now a function of the ratio of segment length to pitch, u. These two values for kD are the limiting values, and kD varies between them in a smooth curve. The function kD(u) will be discussed shortly. Remaining to be determined is the empirical coefﬁcient kF which, as before, is used to correct for the transition region between the linear large n region, and the small n region where the value must diverge from linearity to the predeﬁned constant values. Because kD is no longer a ﬁxed constant, the optimum value of kF may also vary. gS = e−3 2 n + kD + 1 −e−3 2 −kD 1 + kF (n −1) ! e(−γ n)u( 1 n)p – ! – 71 For the special case where segment length/pitch ratio u is one, we can remove the u factor: ! (5.3.3) Because the segment length to pitch ratio u is equal to one, this is little more than an elaborate formula for the GMD of a line of length n p. Therefore, it should give the same value as the standard formula (2.2.6), i.e.: n p e−3/2. Allowing for the approximation error in (5.3.3) when n is small, these two formulae are indeed equivalent. The real value of (5.3.3) is that we can equate (5.3.3) to (2.2.6), taking the limit as n approaches inﬁnity, to solve for kD exactly. Doing so yields the following result: kD(1) = 3⁄2 e−3/2 = 0.33469524… (5.3.4) While this may seem obvious in retrospect, this value converged quite slowly when doing the numerical summations, and did not provide enough precision to use any of the available inverse symbolic calculation tools. Having now derived the exact value of kD for when u = 1, it is hoped that this may give some insight into the exact value for kD when u = 0. This value does not converge very quickly, but by using a convergence acceleration software tool, the value of kD(0) is estimated to be 0.4100858059 with an estimated error of ±1.34x10−9. A reasonable possibility is: ! (5.3.5) which is in agreement to 8 signiﬁcant ﬁgures, and the simplicity of the factors is quite compelling. The fundamental constant, π, has made a sudden surprise appearance. This is not entirely unexpected, because this exercise has been leading up to a comparison between round wires and current sheet segments, and the value, log(2π) − 3⁄2, is the limiting value, at n = ∞, of Rosa's round wire mutual inductance correction. For that reason, we should expect this term to show up at some point. Indeed, Rosa's formula for round wire corrections will be used later to demonstrate that kD(0) is exactly equal to log(2π) e−3/2. gS = e−3 2 n + kD + 1 −e−3 2 −kD 1 + kF (n −1) ! e( −3 2n )p kD(0) ? = log(2⇡)e−3 2 = 0.4100858042 . . . – ! – 72 Considering that the limiting values of kD are log(2π) e−3/2, and 3⁄2 e−3/2, it appears that the common factor of e−3/2 can be removed, and we can focus on the unique values, log(2π) and 3⁄2, and then try to fathom how kD transitions in a smooth curve between the two, as a function of u. It is not a straight line function of u. It changes rapidly in the region of u = 1, but ﬂattens out as it approaches its lower limiting value at u = 0, as shown in the graph below. ! From the graph, it can be seen that at u = 0, the slope is apparently zero, or very nearly so. There are many functions that can be adopted, having a derivative of zero at u = 0. However, given the frequency with which it has been encountered during this investigation, the exponential function seems to be a likely candidate. There is further evidence to support this hypothesis: a least squares ﬁt using polynomials, ranging from degree 2 all the way up to degree 7, produces coefﬁcients of roughly equal magnitude, hinting that there is an underlying power series. Unfortunately, the exponential function never has a zero derivative, no matter what value its argument takes. However, it can be very simply modiﬁed, thus: ! (5.3.6) where m ≥ 2. When u = 0, the derivative will be zero. We will start with the lowest order option, m = 2, and including a single adjustable ﬁtting constant kF2: ! (5.3.7) umeum kD = log(2⇡) −kF 2u2ekF 2u2 – ! – 73 In this form, kF2 has no effect on kD(0). Therefore, we will calculate kF2 to ﬁt the other limiting value of kD(1) = log(2π) − 3⁄2 Hence: ! (5.3.8) Solving for kF2 we get kF2 = 0.26041306 (5.3.9) And, plotting this prospective ﬁtting function (red line) with superimposed data points (blue), we get: ! The ﬁt is remarkably good considering that it has been ﬁt only to the end points. Considering that the mid region data is less accurate due to its slow convergence, it's probably best to leave things as they are, and consider our search for a ﬁtting function to be complete. Having found a function for kD, we still need to adjust empirical transitional coefﬁcient kF. As previously mentioned, a single value for kF will quite possibly not be optimum for all values of kD. In testing, it was found that the optimum value of kF varied from 1.584 for u=0 to 1.709 for u=1. However the peak error was typically around 0.3%, and it was found be a fairly broad peak. Choosing a ﬁxed value for kF = 1.6334 gave a peak error of no more than 0.44% for any value of u, and so there appears to be no reason to complicate matters by making kF a function of u. Therefore, kF will be kept ﬁxed. We now have: ! (5.3.10) kF 2ekF 2 = log(2⇡) −3 2 gS = n + log(2⇡) −kF2u2ekF2u2 + e 3 2 −log(2⇡) −kF2u2ekF2u2 −1 1 + kF (n −1) ! e−3 2 e(−γ n)u( 1 n)p – ! – 74 kF = 1.6334 kF2 = 0.26041306 𝛾 = 3⁄2 Formula (5.3.10) is a more general form than the round conductor formula (5.2.17), because it accounts for a varying density uniformity term, which is required for any geometric shape other than a circle. It is no doubt possible to adapt (5.3.10) to other oblong ﬁgures such as rectangles or ellipses, though it is not known at this point whether it is simply a matter of calculating new values for kF and kF2. The value of 𝛾 will certainly become a function of the ﬁgure's shape factor. 5.4 Inductance Calculation Based on Aggregate GMD Formulae (5.2.17) for the circular array LACC GMD, and (5.3.10) for the linear array (LASL), developed previously, can be directly applied in the standard formula for the mutual inductance between two parallel straight ﬁlaments (1.4.2) to determine the self-inductance of an array of parallel round or ﬂat conductors, such as for busbars. Below, the spacing x between ﬁlaments in formula (1.4.2) is replaced with the value gS calculated from formula (5.2.15). ! (5.4.1) What may be even more interesting, is to see how the LACC and LASL GMD formulae can be applied to calculate the self-inductance of a coil. In Part 4, it was shown that when the GMD inductance calculation method is applied to loops rather than straight conductors, the result will be accurate as long as the largest dimension of the conductor cross section is signiﬁcantly smaller than the diameter of the loop. This would suggest that for a short coil, i.e., one that is signiﬁcantly shorter than its diameter, the LACC and LASL GMD formulae could be used to determine the GMD of the complete cross section of the coil, and then applied just once in a parallel loop mutual inductance calculation, to determine the self-inductance of the coil. Instead of performing n mutual inductance calculations for an n turn coil, only one mutual inductance calculation would be required. The result would then be multiplied by n2 to account for effect of the multiple turns. It seems plausible that this procedure could produce useable inductance values for short coils. A ﬁlamentary loop mutual inductance formula is given by Maxwell [1c]. Adjusted for SI units, and using this document's symbol conventions, it is: ! (5.4.2) L = µ0 2⇡ 2 4log 0 @ gS + s 1 + ✓ gS ◆2 1 A − r 1 + ⇣gS ⌘2 + gS 3 5 M = µ0rc ✓ 1 + 3 16 x2 rc2 ◆ log 8rc x − ✓ 2 + x2 16rc2 ◆$ . – ! – 75 Replacing the axial separation, x, with the self GMD of the coil, gS, and including the adjustment for the number of turns n, we get: ! (5.4.3) Because gS appears at three different places in the formula, it's not practical to expand it out into a one line formulae. Instead, we will leave this as a two part formula using (5.4.3) with either (5.2.17) or (5.3.10). Nevertheless, it is a closed form formula, requiring only direct calculation of readily available math functions, with no need for iteration. Hence, they can be placed directly into a spreadsheet for calculation. What may be of interest is that we have only discussed the arrangement of the cross section as a linear array of circles (or line segments), without specifying its orientation with respect to the coil axis. In other words, we have not yet considered whether the coil is a helix or a spiral, and the results, so far, would indicate that in the case of short or shallow windings of large diameter, there should be no signiﬁcant difference in the inductance value between the two. This is indeed what has been found in practice, and will be demonstrated here. Having available an existing set of spreadsheet macros which use the mutual inductance summation method to calculate self inductance of a solenoid , it was a simple matter to compare the two calculation methods. In this discussion, the inductance value as calculated by the accurate summation macros will be represented by LT (subscript T for true inductance), and the value calculated by the experimental LACC method as LX. The following graph shows the ratio of LX/LT, for different ratios of coil length to coil diameter. We will also deﬁne the coil shape factor (i.e., coil length to diameter ratio): uc = ℓC/D where the coil length is: ℓC = n⨉p L = µ0rcn2 ✓ 1 + 3 16 gs2 rc2 ◆ log 8rc gs − ✓ 2 + gs2 16rc2 ◆$ . – ! – 76 Plotting LX/LT vs. the coil shape factor uc gives the following graph: ! The results of the comparison show that the experimental formula does produce good results (within 1%) for coils where the ratio of winding length to coil diameter is ⅓ or less. The graph appears to be doubly asymptotic, approaching 1 for small values of u, and approaching a straight line (on the log-log graph) with a constant negative slope for large values of u. Analysis of the data shows that the slope of this asymptote is equal to −2, indicating that the LX/LT is a function of 1/uc2 as uc gets larger. This would seem to indicate that the formula's accuracy could be extended for larger values of uc by including a simple correction term. Indeed, if the pitch and conductor diameter are held constant, with the only variable being the coil length to diameter ratio uc, then this is exactly the case. However, it must be pointed out that as uc becomes larger, the error in the inductance value calculated by the LACC formula becomes very large, and hence, the required correction factor becomes extremely large: an electromagnetic instance of the tail wagging the dog. This, by itself, would not be so bad, except that when we now include the effect of pitch and wire diameter, even these small inﬂuences are ampliﬁed when uc is large. As a result, a simple correction factor based only uc, does not appear to be practical. The better option is to accept these as short coil formulae, with the beneﬁt that they properly account for conductor shape. If calculations for longer coils are desired, then it is more practical to employ a method given in the following section. 5.5 Rosa's Round Wire Inductance Corrections In the ﬁrst decade of the 20th century, the science of inductance calculation was evolving rapidly. At that time, there were only two basic methods for calculating the inductance of a single layer solenoid coil: the current sheet method, and the summation method. (Now, more than a hundred years later, there are two basic methods for calculating the inductance of a single layer solenoid – ! – 77 coil: the current sheet method, and the summation method.) The current sheet method was best suited for long coils wound with many turns of ﬁne wire and of small pitch. The summation method was more accurate, but involved the calculation of the self inductance of one turn of the coil, as well as the mutual inductance between a turn and every other turn in the coil. For a coil of n turns, it was necessary to do (n−1) mutual inductance calculations, a very tedious process in the days before computers. So, this method was impractical for coils of more than a few turns. Though, Strasser derived such a formula and provided a table to aid in determining the mutual inductance component for coils of up to 30 turns. The current sheet formula approximates the coil with a very thin cylindrical sheet of the same length and diameter as the coil under consideration. The inductance of the current sheet can be solved exactly, and is approximately correct for the actual coil. In 1906, Edward Rosa, a physicist working for the U.S. National Bureau of Standards, investigated these two calculation methods . He found that in certain cases, there is a small but signiﬁcant difference in the results produced by these two methods. He reasoned that, by comparing the difference between the self inductance of one turn of a round wire and the self inductance of one turn of current sheet, and comparing the difference between the mutual inductance between turns of round wire and the mutual inductance between turns of current sheet that it would be possible to come up with a set of round wire correction factors that could be applied to the current sheet formula, making it as accurate as the summation method, but far easier to calculate. Rosa derived two correction factors. The ﬁrst, kS, corrects for the difference in self inductance, and the second, kM, corrects for the difference in mutual inductance. kS is a function of a single argument: the ratio of pitch to wire radius p/r. Similarly, kM is a function of a single argument: the number of turns n. Rosa's complete correction, adjusted to SI units, is: ∆L = µ R n (kS + kM) (5.5.1) where R is the coil radius and n is the number of turns. This is then applied to the current sheet inductance LS to get the corrected inductance value: L = LS − ∆L (5.5.2) Note that ∆L can be either positive or negative. Rosa based his corrections on Maxwell's formula for the mutual inductance of two circular loops: ! (5.5.3) R is the radius of the loops, and x is their axial separation. Rosa found that the second order terms amounted to only about 1 part/million, and would not contribute signiﬁcantly to the ultimate value. He eliminated them, leaving only the log term and the numeric constant inside the brackets: L = 4⇡R ✓ 1 + 3 16 x2 R2 ◆ log ✓ 8R x ◆ −2 −1 16 x2 R2 $ . – ! – 78 ! (5.5.4) To calculate the self inductance of one turn of current sheet, he replaced x with the self GMD of a straight line, which is its length multiplied by e−3/2. Rosa factored out the number of turns and coil length, so that the length of a current sheet segment is now simply equal to the pitch p. To calculate the self inductance of one turn of round wire, he replaced x with the self GMD of the conductor cross section, r e−1/4. Taking the difference between these two inductance values: ! (5.5.5) Since the difference between two logarithms is the same as the logarithm of the quotient of their arguments, this then becomes: ! (5.5.6) The exponentials can be combined and taken outside of the log function giving a constant term of 5⁄4. In addition, 4πR appears as a common factor in all of the correction terms. And so, Rosa removed it from the individual correction factors, and applied it in the summary calculation (see formula (5.5.1) ). When using SI units, the 4π term will be replaced by the permeability constant µ. Rosa also used the convention that the correction factor would be subtracted from the base inductance value, thus making the sign the opposite of what has been used here. Thus, Rosa's self inductance correction factor kS is: ! (5.5.7) Rosa's derivation of the mutual inductance correction term kM is more involved, requiring a summation of n − 1 mutual inductances for an n turn coil. For many years the most accessible form of Rosa's kM corrections was the tabulated data, precise to about four decimal places, in Grover's book 2. For this correction term, Grover used the symbol H rather than kM. In 2006, David Knight created an empirical ﬁtting function to reproduce the values in this table. In 2008, R. Weaver set about creating some computer code to calculate Rosa's mutual L = 4⇡R  log ✓ 8R x ◆ −2 $ . ∆LS = 4⇡R ✓ log ✓ 8 R re−1 4 ◆ −2 $ −  log ✓ 8 R pe−3 2 ◆ −2 $◆ = 4⇡R ✓ log ✓ 8 R re−1 4 ◆ −log ✓ 8 R pe−3 2 ◆◆ . ∆LS = 4⇡R log pe−3 2 re−1 4 ! . kS = 5 4 −log ⇣p r ⌘ . – ! – 79 inductance corrections using the original summation formula. This resulted in an iterative computer function that could calculate Rosa's kM correction to much higher precision. Because it was based on Rosa's summation method, it required one iteration for every turn of the coil, making it somewhat cumbersome, and slow for coils with many turns. However, it's value has been in its ability to provide a set of accurate data which could be used as a benchmark. With a bit more research, a series formula by Grover was located 13: ! (5.5.8) This formula is accurate for large n but is not suitable for n < 5, and notably, it gives the wrong value for n = 1. David Knight adapted this formula, adding a 7th order empirical term to correct the accuracy for n < 5 and a 9th order closing term to force it to the correct value of zero for n = 1. Knight's formula is: ! (5.5.9) This formula has a worst case absolute error of ±1.3⨉10−8, making it essentially exact for any practical purpose. The derivation of Rosa's summation formula for kM will not be given here, but the reader can refer to the references for details. The preceding synopsis should be sufﬁcient background for what is to follow. From the foregoing, it would seem apparent that by taking the difference of the logarithms of the LACC formula (5.2.15) and the co-linear line GMD formula (2.2.6) for of a line of equal length, then it should result in the same value as Rosa's kS and kM factors combined. Thus, we start with (5.2.15) and (2.2.6) combined: ! (5.5.10) Note that the factor n is included outside of the log function. This is due to a primary difference between the LACC formula and Rosa's formula. The LACC formula is a GMD calculation for a complete winding, and in order to turn it into an inductance value, it must be multiplied by n2. Rosa's formulae were developed as per-turn corrections and are only multiplied by n in his overall ∆L correction function. Therefore, an additional factor of n needs to be included in the LACC formula to yield the same results. km = log (2⇡) −3 2 −1 6n log(n) −0.330842 n − 1 120n3 + 1 504n5 . km = log (2⇡) −3 2 −1 6n log(n) −0.33084236 n − 1 120n3 + 1 504n5 −0.0011935 n7 + 0.000507 n9 . ks + km = n log " e−3 2 n + kD + 1 −e−3 2 −kD 1 + kF (n −1) ! e( −1 4n )u( 1 n)p # −n log ⇣ npe−3 2 ⌘ . – ! – 80 Converting the difference of logarithms to the logarithm of a quotient: ! (5.5.11) The pitch p appears in both numerator and denominator, and cancels out, resulting in a function of only two arguments, u and n. The value of kD, has previously been given as log(2π) e−3/2 (though not yet proven), and is now substituted for kD: ! (5.5.12) Inside the ﬁrst parentheses, the factor e−3/2 occurs in nearly every term, and will be taken outside: ! (5.5.13) The factor e−3/2 outside the parentheses now cancels with the factor e−3/2 in the denominator of the rightmost fraction. We will replace the ﬁxed coefﬁcient for circular areas, 1⁄4, in the self GMD exponent, with the variable 𝛾, to allow the formula to be used for both solid round wires and round tubular conductors. ! (5.5.15) There is no reason why the constant kF cannot be re-optimized to give the minimum error in this application. Doing so gives a marginally different value than before: kF = 1.59516 In testing, the formula has a worst case error of no more than about 1% in most cases. The error here is higher than the peak error of 0.17% of formula (5.2.15) by itself, because the correction formula involves the subtraction of terms of similar value, thus increasing the % error. In certain situations when Rosa's kS and kM corrections are roughly equal in magnitude and opposite in sign, while at the same time, n is less than 10 (the least accurate region of the empirical factor), then the % error can be considerably higher, though at the same time, the absolute error is extremely small. Therefore, it is of no great consequence, because in that case, the actual correction is negligible. However, this can be disconcerting when one looks only at relative error rather than absolute error. ks + km = n log " e−3 2 n + kD + 1 −e−3 2 −kD 1 + kF (n −1) ! e( −1 4n )u( 1 n)p npe−3 2 # . ks + km = n log " e−3 2 n + log(2⇡)e−3 2 + 1 −e−3 2 −log(2⇡)e−3 2 1 + kF (n −1) ! e( −1 4n )u( 1 n) ne−3 2 # . ks + km = n log " n + log(2⇡) + e 3 2 −1 −log(2⇡) 1 + kF (n −1) ! e−3 2 e( −1 4n )u( 1 n) ne−3 2 # . ks + km = n log " n + log (2⇡) + e 3 2 −log (2⇡) −1 kF (n −1) + 1 ! e( −γ n )u( 1 n) n # . – ! – 81 To verify that this error is due only to the approximation in the empirical factor, and not a fundamental error in the derivation, the formula was also tested with the empirical factor replaced with the exact numerical summation. In this case the agreement with Rosa's original calculations was found to be exact, thus validating this formula. — Now that we have veriﬁed that this formula is equivalent to Rosa's formula, we are ﬁnally at a point where we can demonstrate that the value of the density uniformity constant, kD, is in fact exactly log(2π). Grover has demonstrated that the limiting value at n = ∞ for Rosa's kM is log(2π)−3/2 13. Also, refer to formula (5.5.8). Therefore, we must show that the limit as n goes to ∞ of formula (5.5.15) also gives this same value. We will replace kM on the left side of the formula with the limiting value. The empirical fractional term on the right side inside the parentheses will go to zero due to n being in the denominator. So, this term can immediately be eliminated. Also, as we have not yet proven that kD=log(2π), we will put kD back into the formula. So, we begin with: ! (5.5.16) The self inductance correction kS will be dealt with presently. Next we take the factor n inside the log term making it an exponent: ! (5.5.17) Rearranging: ! (5.5.18) Making use of this exponential relationship: ! the n exponent can be combined with the exponents of e and u; n and 1⁄n cancel each other out, leaving these terms no longer a function of n: ! (5.5.19) log(2⇡) −3 2 + ks = lim n!1 " n log (n + kD) e( −γ n )u( 1 n) n !# . log(2⇡) −3 2 + ks = lim n!1 " log (n + kD) e( −γ n )u( 1 n) n !n# . log(2⇡) −3 2 + ks = lim n!1  log ✓✓n + kD n ◆n ⇣ e( −γ n )u( 1 n)⌘n◆& . ! ab"c = abc . log(2⇡) −3 2 + ks = lim n!1  log ✓✓n + kD n ◆n e−γu1 ◆$ . – ! – 82 At this point we will move the limit inside the log function. We can do this because the function is continuous. ! (5.5.20) This remaining limit is a well known exponential identity. For example, see Abramowitz and Stegun formula 4.2.21 (page 70): ! (5.5.21) Therefore the remaining limit can be replaced with a simple exponential ekD, giving: ! (5.5.22) We must now consider that (5.5.15) includes both self and mutual inductance corrections. Therefore, we will choose a value for u that gives a self inductance correction kS = 0. Then, the resulting value of the formula will be due to the mutual inductance correction factor alone. From Rosa: ! (5.5.23) Setting kS to zero we get: ! (5.5.24) Hence: ! (5.5.25) Substituting this into the main formula: ! (5.5.26) Combining the log and exponential terms on the righthand side of the formula: ! (5.5.27) For solid round wires, 𝛾 = ¼. So: ! (5.5.28) log(2⇡) −3 2 + ks = log ✓ lim n!1 ✓n + kD n ◆n$ e−γu ◆ . lim a!1 a + x a "a = ex . log(2⇡) −3 2 + ks = log ! ekDe−γu " . ks = 5 4 −log ✓p r ◆ = 5 4 −log ✓1 u ◆ . log ✓1 u ◆ = 5 4 . u = e−5 4 . log(2⇡) −3 2 = log ⇣ ekDe−γe−5 4 ⌘ . log(2⇡) −3 2 = kD −γ −5 4 . log(2⇡) −3 2 = kD −1 4 −5 4 . – ! – 83 The numeric constants cancel out, leaving: ! (5.5.29) Thus, we have derived the analytical value for kD. Now, returning to the topic of the accuracy of formula (5.5.15), it can be stated, that because the self inductance correction component is not affected by any approximations in the derivation of the formula, it is exact. So, all error in the formula is due strictly to the empirical portion of the mutual component. Therefore, to better judge accuracy, we can again set u equal to e−5/4, resulting in a zero value for kS. Then we will compare formula (5.5.15) with the formula for Rosa's kM correction. Doing this gives the following: ! The peak error is ±4% occurring in two places at n < 10. Bear in mind that a mutual inductance correction with an error of 4% when combined with an exact self inductance correction and then applied to an inductance calculation will result in a much smaller overall percentage error. Nevertheless, it does seem unfortunate that after the effort of deriving this function that we can't have a smaller peak error. Therefore, a second order term was added to the empirical part of the function, and re-optimized, now giving a peak error of ±1% as shown in the following graph. ! kD = log(2⇡) . – ! – 84 The resulting formula is: ! (5.5.30) And the empirical constants are: kF0 = 1.530574 kF1 = 0.0326627 In addition, the constants in the numerator of the empirical term, which have been kept in their symbolic form throughout the derivation, can be evaluated and combined, giving a value of e3/2 − log(2π) − 1 = 1.643812004. A peak error of ±1% sounds more impressive, but makes little difference in the overall scheme of things. Of course, for utmost accuracy, Rosa's kS formula (5.5.7) combined with Knight's kM formula (5.5.9) will give essentially exact results. The most useful test is in comparing actual coil inductance values calculated by: i) the current sheet inductance, corrected with formula (5.5.30); ii) the current sheet inductance, corrected with formulae (5.5.7) and (5.5.9); iii) the iterative summation method; The peak error in formulae (5.5.15) and (5.5.30) always occurs at n = 2, and so we will limit our comparison to coils of two turns. We will presume method iii) to give the most accurate inductance value against which the others will be compared. This is a fair assumption when the pitch and conductor diameter are small compared to the coil diameter. ks + km = n log " n + log (2⇡) + e 3 2 −log (2⇡) −1 kF0(n −1) + kF1(n −1)2 + 1 ! e( −γ n )u( 1 n) n # . – ! – 85 The following table compares the inductance calculated by the summation method with the inductance calculated by the current sheet method which is then corrected with formula (5.5.15), (5.5.30) and the original rosa corrections. ! Units are millimetres and microhenries, but the results are scaleable to any unit, and the error values will remain the same. The ratio of conductor to coil radius r/R = 0.01 is very modest. However, a larger ratio would introduce factors that would overshadow the GMD error. The ﬁrst column shows the ratio of pitch to conductor diameter (p/d). The second column is the ratio of conductor radius to pitch (u=r/p). The next two columns show the inductance values calculated by the summation method and current sheet methods. The inductance corrections (by three methods) are shown in the next three columns, and ﬁnally the error due to the correction methods are shown in the last three columns. Formula (5.5.15) shows an error of about 500 ppM (parts/million) for worst case 2 turn coils. Formula (5.5.30), with the additional ﬁtting term, shows an error of about 120 ppM, while the original Rosa corrections show an error of no more than −9 ppM. Again, these numbers are for the worst case 2 turn coil. For coils of more turns, the error in formulae (5.5.15) and (5.5.30) will drop signiﬁcantly, closer to the values of the original Rosa values. Now, compare these values with the error of the current sheet, having no correction at all, as shown in the far right column. These errors are up to 65,000 ppM, or 6.5%. This gives some perspective of just how well these round wire corrections do work. The above comparison is based on the assumption that the summation inductance calculation is completely accurate. However, the summation calculation assumes that the turns of the coil are perfect parallel circular rings spaced apart at distance equal to the pitch, ignoring the true helicity – ! – 86 of the coil. This is not usually a bad assumption, but if we are comparing the amount of error introduced by other approximations, then this one needs to be addressed as well. In the following table, the summation calculation is compared to a true helical inductance calculation for the same range of 2 turn coil parameters: ! It can be seen that by failing to take into account the helicity of the coil, the summation method is in error by as much as 400 ppM. This is approximately the same amount of error as the simpler correction formula (5.5.15), and nearly four times the error of the more accurate correction formula (5.5.30). 5.6 Flat Conductor Corrections Formulae in the general form of (5.5.15) or (5.5.30) can, of course, be simply adapted to handle corrections for other conductor shapes. This may be the most useful aspect of this form of correction expression. For example, formula (5.3.10) can be developed into a correction factor for straight co-linear conductors which have gaps between the conductors. The derivation is exactly the same as for formula (5.5.30), and so only the ﬁnal formula is given here: ! (5.6.1) kS + kM = n log " n + log(2⇡) −kF2u2ekF2u2 + e 3 2 −log(2⇡) −kF2u2ekF2u2 −1 1 + kF (n −1) ! e(−γ n)u( 1 n) n # – ! – 87 kF = 1.6334 kF2 = 0.26041306 𝛾 = 3⁄2 This correction formula would be applicable to ﬂat spiral inductors—etched on printed circuit boards, or fabricated on silicon—when combined with a circular disk current sheet inductance formula, such as that of Rayleigh and Niven 4: ! (5.6.2) where x is the radial depth of winding, and R is the mean radius of the winding. At the time of writing, this correction formula has been only brieﬂy tested, but it appears to give at least a tenfold improvement in accuracy over an uncorrected disk current sheet inductance calculation, as compared to a loop summation calculation. What is more in question at this time is the range of accuracy of available current sheet formulae. However, this will be the subject of a future article. LS = µ⇡Rn2  log 8R x −1 2 + x2 96R2 ✓ log 8R x + 43 12 ◆$ . – ! – 88 Afterword This article began as a result of my work developing a method for calculating the inductance of a helical coil that fully accounts for the effect of its helicity for any pitch: 0 to ∞. That formula made only one approximation: its use of the GMD of the cross section of the conductor, rather than integrating across the conductor cross section. That raised the question of whether the resulting formula gained accuracy due to its accounting for helicity, only to lose it again due to the use of GMD. Unfortunately, there appeared to be no information, readily available, that quantiﬁed the error caused by the use of GMD. That lack of information was the spark that launched the investigation into the accuracy of the GMD method, and which eventually became Part 4 of this document. The result of that work has shown that the error introduced by the use of GMD is almost always considerably smaller than the error introduced by the failure to account for the helicity of the coil. Neither the current sheet method nor the summation method account for helicity. The GMD approximation error becomes signiﬁcant only when the conductor diameter becomes a very signiﬁcant fraction of the coil diameter. However, a consequence of this is that the coil pitch must then also become large, and therefore, the non-helicity error also becomes prominent. For coils having a pitch angle of more than about 6° (equivalent to p/D = 0.33), the non-helicity error is appreciable. For ψ < 6°, neither helicity nor GMD error should be signiﬁcant, except for work requiring the utmost accuracy, and in that case the method of correction given in Section 4.3 may be used. While the accuracy question has been resolved, that research spawned other lines of investigation into related GMD topics, such as numerical GMD calculation methods, which either had not been treated elsewhere, or else their sources were very obscure. As a result, it seemed worthwhile to combine these other topics to make a more general article on the subject of GMD. There continue to be related areas of interest which will be explored in due course, but if we waited for that list to be exhausted, this document would never be ﬁnished. Finally, I would like to express my appreciation to Dr. David Knight, whose work is cited here, and who reviewed several drafts of this document, and provided valuable feedback. – ! – 89 References 1. Maxwell, James Clerk; A Treatise on Electricity and Magnetism, Vol. 2, Third Edition, Dover 1954; (a) Art. 691, On the Geometrical Mean Distance of Two Figures on a Plane, pp. 324-326. (b) Art. 701, To ﬁnd M by Elliptic Integrals, pp. 338-340. (c) Art. 705, Further approximation of M, pp. 343-345 2. Grover, Frederick W.; Inductance Calculations–Working Formulas and Tables, D. Van Nostrand, 1946; Reprint: Dover, 2004. 3. Dwight, Herbert B.; Tables of Integrals and other Mathematical Data; Fourth Edition, MacMillan, 1961. 4. Rosa, E. B. & Grover, F. W.; Formulas and Tables for the Calculation of Mutual and Self-Inductance. Scientiﬁc Paper 169, National Bureau of Standards, Washington DC, 1916. 5. Gray, Andrew; Absolute Measurements in Electricity and Magnetism; 2nd Edition; MacMillan and Co.; 1921; (Downloaded on 2014-07-25) 6. Seneff, E. Lockwood Jr.; Study of the Method of Geometric Mean Distances Used in Inductance Calculations; MScEE Thesis, University of Missouri School of Mines and Metallurgy; 1947; (Downloaded on 2014-01-29) 7. Weaver, Robert S.; Webpage: Maxwell and Coaxial Circular Conductors (Downloaded on 2014-09-02) 8. Higgins, Thomas James; Formulas for the Geometric Mean Distances of Rectangular Areas and of Line Segments; Journal of Applied Physics, Vol. 14, April 1943, pp. 188-195. 9. Wichman, B. A. and Hill, I. D.; Algorithm AS 183: An Efﬁcient and Portable Pseudo-Random Number Generator; Applied Statistics, 31, pp. 188-190, 1982. 10. Wien, Max; Ueber die Berechnung und Messung kleiner Selbstpotentiale [On the Calculation and Measurement of Small Self-potentials]; Annalen der Physik, Vol. 289, No. 13, 1894, pp. 928-947. jportal_derivate_00147928/18942891312_ftp.pdf (Downloaded on 2015-08-13) 11. Strasser, B.; Über die Bestimmung des Selbstinduktionskoefﬁzieten von Solenoiden [On the Determination of Self Induction coefﬁcients of Solenoids]; Annalen der Physik, 1905 pp. 763–771. jportal_derivate_00150036/19053220912_ftp.pdf (Downloaded on 2016-03-06) – ! – 90 12. Grover, Frederick W.; A Comparison of the Formulas for the Calculation of the inductance of coils and spirals wound with wire of large cross section; National Bureau of Standards Journal of Research vol. 3, RP90. 13. Rosa, Edward, B.; Calculation of the Self-Inductance of Single-Layer Coils; National Bureau of Standards, Bulletin 2; 1906. 14. Knight, David; An Introduction to the Art of Solenoid Inductance Calculation With Emphasis on Radio-Frequency Applications; Version 0.20; 2016. (Downloaded on 2016-02-25) 15. Weaver, Robert S.; Investigation of E.B. Rosa’s Round Wire Mutual Inductance Correction Formula; 2008. (Downloaded on 2016-02-25) 16. Abramowitz, Milton and Stegun, Irene A.; Handbook of Mathematical Functions, With Formulas, Graphs, and Mathematical Tables; National Bureau of Standards, 1964. (Downloaded on 2016-02-25) 17. Weaver, Robert S.; The Inductance of a Helix of Any Pitch; 2011. (Downloaded on 2016-02-28) 18. Grover, Frederick W.; Additions to the Formulas for the Calculation of Mutual and Self Inductance; National Bureau of Standards, Bulletin Volume 14, No. 4, July 12, 1919. – ! – 91 Appendices A – Program Listing for calculation of the GMD of an Elliptical Locus Program GMDelliptLine (n As int64) 'Calculate self gmd of elliptical lines of multiple aspect ratios. 'N is number of random pairs of points to be generated. dim a,x1,y1,x2,y2,theta1,theta2,gmd As Double dim i,j As Int64 dim aspect,aspectAry() As Integer aspect = ef_pitch.Text.Val 'Load the 'aspectAry()' array with the aspect ratios to be used. aspectAry = array(20,30,63,125,250,315,325,350,400,500,750) Print "Aspect, gmd" for j = 0 to Ubound(aspectAry) aspect = aspectAry(j) a = aspect/1000 gmd = 0 for i = 1 to n theta1 = ArcToAngle(rnd,aspect) theta2 = ArcToAngle(rnd,aspect) x1 = acos(theta1) y1 = sin(theta1) x2 = acos(theta2) y2 = sin(theta2) gmd = gmd+log(sqrt((x1-x2)^2+(y1-y2)^2)) next gmd = exp(gmd/n) Print aspect, gmd next end function ArcToAngle (arcLength As Double, aspect as Integer)As Double 'Given the partial arc length of an ellipse ' and aspect ratio, return the parametric angle. 'Note that the argument 'aspect' is an integer and is equal to the aspect ratio ' multiplied by 1000 to simplify its use as the parameter of the case statements. dim k1,k2,k3,k4,k5,k6,k7,a,b,c,d,f As Double 'Determine quadrant transformation parameters according to arc length. if arcLength<0.25 then 'Quadrant 1 a = arcLength b = 0 'Angle offset c = 1 'sign of function d = 0 'function adjustment ElseIf arcLength<0.5 then 'Quadrant 2 a = 0.5-arcLength b = pi/2 'Angle offset c = -1 'sign of function d = pi/2 'function adjustment ElseIf arcLength<0.75 then 'Quadrant 3 a = arcLength-.5 b = pi 'Angle offset – ! – 92 c = 1 'sign of function d = 0 'function adjustment Else 'Quadrant 4 a = 1-arcLength b = 1.5pi 'Angle offset c = -1 'sign of function d = pi/2 'function adjustment end if a = a4 'scale up the argument from the range 0< = a<0.25 to 0< = a<1 'Choose the mapping function parameters according to the aspect ratio. select case aspect Case 20 'Aspect = 0.0200 k1 = 0.9954197523 k2 = -1.5960238734 k3 = 0.5221336302 k4 = 0.0804990919 k5 = -1.6485276059 k6 = 0.5239512123 k7 = 0.1258679037 Case 30 'Aspect = 0.0300 k1 = 0.9984223767 k2 = -1.5947471825 k3 = 0.5303252429 k4 = 0.0687710369 k5 = -1.6373539169 k6 = 0.5175107258 k7 = 0.1216076254 Case 63 ' Aspect = 0.063 k1 = 1.0053774155 k2 = -1.0812115077 k3 = -0.3381538919 k4 = 0.4219678890 k5 = -1.1015803243 k6 = -0.4182729136 k7 = 0.5249334748 Case 125 ' Aspect = 0.125 k1 = 1.0226488149 k2 = -0.9357991585 k3 = -0.3526079227 k4 = 0.3054989960 k5 = -0.9158611270 k6 = -0.5115714778 k7 = 0.4527321137 Case 185 ' Aspect = 0.185 k1 = 1.0440215988 k2 = -1.7853069002 k3 = 1.4637691010 k4 = -0.6049651753 k5 = -1.7200884720 k6 = 1.3112327787 – ! – 93 k7 = -0.5163293003 Case 250 ' Aspect = 0.250 k1 = 1.0702064611 k2 = -1.5638577536 k3 = 6.7291168067 k4 = -4.6588962658 k5 = -1.4784593304 k6 = 6.2841564621 k7 = -4.8020393235 Case 275 ' Aspect = 0.275 k1 = 1.08297603 k2 = 0.32065304 k3 = -2.92460709 k4 = 1.54112028 k5 = 0.29906132 k6 = -2.96287518 k7 = 1.67663668 Case 300 ' Aspect = 0.300 k1 = 1.09677644 k2 = 0.31383321 k3 = -2.92801438 k4 = 1.52766397 k5 = 0.29919089 k6 = -2.96132963 k7 = 1.66867001 Case 315 ' Aspect = 0.315 k1 = 1.10471140 k2 = 0.34291534 k3 = -2.94324420 k4 = 1.51517585 k5 = 0.32691385 k6 = -2.96787128 k7 = 1.65340874 Case 325 ' Aspect = 0.325 k1 = 1.11002732 k2 = 0.34464876 k3 = -2.92094714 k4 = 1.49373693 k5 = 0.32828082 k6 = -2.93894663 k7 = 1.62815121 Case 350 ' Aspect = 0.350 k1 = 1.12318127 k2 = 0.00286285 k3 = -2.28947823 k4 = 1.20606362 k5 = 0.01527216 k6 = -2.31312279 k7 = 1.32498955 Case 400 – ! – 94 ' Aspect = 0.400 k1 = 1.15094948 k2 = -0.33417431 k3 = -1.63203136 k4 = 0.89775819 k5 = -0.28281778 k6 = -1.65893124 k7 = 0.99427163 Case 500 ' Aspect = 0.500 k1 = 1.21132835 k2 = -0.25885181 k3 = -1.52341812 k4 = 0.81547090 k5 = -0.20722133 k6 = -1.48854536 k7 = 0.85143913 Case 750 ' Aspect = 0.750 k1 = 1.3815773746 k2 = 0.26848230 k3 = -1.39594071 k4 = 0.91476754 k5 = 0.19716802 k6 = -1.17412808 k7 = 0.72109609 else 'For other cases return linear value. 'This is also the correct case for an aspect ratio of 1. return arcLength2pi end Select 'Calculate the mapping function value. f = (k1a+k2a^2+k3a^3+k4a^4)/(1+k5a+k6a^2+k7a^3) 'Apply the quadrant adjustments to the mapping function value and then return. return d+cf+b end – ! – 95 B – A Method for Finding and Eliminating Errors in Mathematical Derivations Following is a somewhat crude but effective technique to help eliminate blunders during the derivation of formulae having many terms, using nothing more than BASIC or a similar programming language. The idea is to take the immediate result of each step of the derivation before any simpliﬁcation, and code it as a BASIC function. For example, from the section, GMD of two co-linear lines of equal length, the result of evaluating the A1 integral is: ! This is coded into a program function as follows: function _A1(ByVal s as double, r as double)as double x = r+s b = s(x-s)log(x-s)-sx x = r a = s(x-s)log(x-s)-sx _A1 = b-a end function This is quick to do, and once this simple function template is set up, it can then be repeated for every term in the solution with just a bit of copying and pasting. Hence the complete, but messy formula for the GMD of two co-linear lines, coded into BASIC, is: function _A1(ByVal s as double, r as double)as double x = r+s b = s(x-s)log(x-s)-sx x = r a = s(x-s)log(x-s)-sx _A1 = b-a end function function _A2(ByVal s as double, r as double)as double x = r+s b = 0.5(x^2-s^2)log(x-s)-0.5sx-0.25x^2 x = r a = 0.5(x^2-s^2)log(x-s)-0.5sx-0.25x^2 _A2 = b-a end function function _B(ByVal s as double, r as double)as double x = r+s b = sx x = r a = sx _B = b-a end function r+s R r A1 dx =  s(x −s) log (x −s) −sx #r+s r . – ! – 96 function _C(ByVal s as double, r as double)as double x = r+s b = 0.5x^2log(x)-0.25x^2 x = r a = 0.5x^2log(x)-0.25x^2 _C = b-a end function function _GMD(ByVal s as double, r as double)as double A1 = _A1(s,r) A2 = _A2(s,r) A = A1-A2 B = _B(s,r) C = _C(s,r) D = A-B+C _GMD = exp(D/s^2) end function The very last function, _GMD, calls each individual function and sums theirs results to calculate the complete GMD value. Since no simpliﬁcation of terms has been done at this stage, the chance of errors is greatly diminished. The functions can be veriﬁed by solving the GMD for speciﬁc values of s and r. For this example, the ﬁnal formula was already known, because it has previously been published . So, it is easy to verify. In the case of a new derivation, it’s not difﬁcult to come up with a ballpark estimate of the correct GMD value, because it will fall somewhere between the distance between the closest points and the distance between the furthest points on the two objects. If a better estimate of the true GMD value is needed, then the Monte Carlo method can be used to calculate a couple of examples. However, errors in the formulae tend to affect the results dramatically. So, it’s not difﬁcult to tell when an error occurs. Once it is veriﬁed that the functions are producing correct results, each function can then be simpliﬁed, with veriﬁcation that the program still produces the same correct results at each stage. If an error occurs, then it is easy to pinpoint its location before continuing. The _A1 function is ﬁrst simpliﬁed by replacing the variable x with the limits of integration (r+s) and (r): function _A1(ByVal s as double, r as double)as double b = s((r+s)-s)log((r+s)-s)-s(r+s) a = s((r)-s)log((r)-s)-s(r) _A1 = b-a end function Next, the a and b assignment statements are combined, paying close attention to the sign of the terms: function _A1(ByVal s as double, r as double)as double b = s((r+s)-s)log((r+s)-s)-s(r+s) - s((r)-s)log((r)-s)+s(r) _A1 = b end function – ! – 97 With further simpliﬁcation, eventually we end up with this: function _A1(ByVal s as double, r as double)as double b = srlog(r)-s^2 - s(r-s)log(r-s) _A1 = b end function This simpliﬁcation is repeated with each function until the complete result is in its simplest form. – ! – 98 Revision History 2016-03-05: Original Issue 2016-03-06: -Corrected several typographical errors -Minor revisions to the text for clarity – ! – 99
190003	https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_11?srsltid=AfmBOopABEVnqcfbxrm8yDyPKkCK-6QLRNZxwQq8l7f6kFUPl4Qa_Qir	Art of Problem Solving 2022 AIME I Problems/Problem 11 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2022 AIME I Problems/Problem 11 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2022 AIME I Problems/Problem 11 Contents 1 Problem 2 Solution 1 (No trig) 3 Solution 2 4 Solution 3 5 Solution 4 6 Solution 5 7 Solution 6 (Short and Sweet) 8 Solution 7 (Intuitive, no trig, no weird auxiliary lines) 9 Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem) 10 Solution 9 11 Video Solution 12 Video Solution 2 (Mathematical Dexterity) 13 Video Solution 3 by OmegaLearn 14 See Also Problem Let be a parallelogram with A circle tangent to sides and intersects diagonal at points and with as shown. Suppose that and Then the area of can be expressed in the form where and are positive integers, and is not divisible by the square of any prime. Find Solution 1 (No trig) Let's redraw the diagram, but extend some helpful lines. We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of . Then . Similarly, the power of and . We let and label the diagram accordingly. Notice that because . Let be the center of the circle. Since and intersect and , respectively, at right angles, we have is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from to and to , and both are equal to . Since , . Since and . We can now use Pythagorean theorem on ; we have and . We know that because is a parallelogram. Using Pythagorean theorem on , . Therefore, base . Thus the area of the parallelogram is the base times the height, which is and the answer is ~KingRavi Solution 2 Let the circle tangent to at separately, denote that Using POP, it is very clear that , let , using LOC in ,, similarly, use LOC in , getting that . We use the second equation to minus the first equation, getting that , we can get . Now applying LOC in , getting , solving this equation to get , then , , the area is leads to ~bluesoul,HarveyZhang Solution 3 Denote by the center of the circle. Denote by the radius of the circle. Denote by , , the points that the circle meets , , at, respectively. Because the circle is tangent to , , , , , , . Because , , , are collinear. Following from the power of a point, . Hence, . Following from the power of a point, . Hence, . Denote . Because and are tangents to the circle, . Because is a right trapezoid, . Hence, . This can be simplified as In , by applying the law of cosines, we have Because , we get . Plugging this into Equation (1), we get . Therefore, Therefore, the answer is . ~Steven Chen (www.professorchenedu.com) Solution 4 Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. Note that PoP on and with respect to yields and . We can compute the area of in two ways: By the half-base-height formula, . We can drop altitudes from the center of to , , and , which have lengths , , and . Thus, . Equating the two expressions for and solving for yields . Let . By the Parallelogram Law, . Solving for yields . Thus, , for a final answer of . ~ Leo.Euler Solution 5 Let be the circle, let be the radius of , and let the points at which is tangent to , , and be , , and , respectively. PoP on and with respect to yields Let In Area is vladimir.shelomovskii@gmail.com, vvsss Solution 6 (Short and Sweet) Let be the center of the circle. Let points and be the tangent points of lines and respectively to the circle. By Power of a Point, . Similarly, . Notice that since quadrilateral is symmetrical. Let intersect at . Then, is similar to . Therefore, . Let the length of , then . Solving we get . Doing the Pythagorean theorem on triangles and for sides and respectively, we obtain the equation where denotes the radius of the circle. Solving, we get . Additionally, quadrilateral is symmetrical so . Let and extend a perpendicular foot from to and call it . Then, is right with , , and . Taking the difference of squares, we get . The area of is . Therefore, the answer is ~Magnetoninja Solution 7 (Intuitive, no trig, no weird auxiliary lines) Say that is tangent to the circle at and tangent at . Also, is the intersection of (diameter) and (diagonal). Then by power of a point with given info on and we get that and . Note that , and since we note that . Since , we get that and . This is the length information within the circle. The same triangle similarity also means that , so if the radius of the circle is then we have and . By power of a point on H, we can figure out : and we get that . Thus, we have that the height of the parallelogram is and we want to find . If is tangent to the circle at , then set . Using pythagorean theorem, and we can plug in diagram values: Solving, we get Finally, we have ~ Brocolimanx Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem) Let , , be the circle's point of tangency with sides , , and , respectively. Let be the center of the inscribed circle. By Power of a Point, , so . Similarly, , so . Construct , and let be the point of intersection of and . and . By AA, , and we have . We also know , so and . Using Pythagorean Theorem on and , we find that and . Thus, , and the radius of the circle is . Construct , . , so is cyclic. Similarly, is cyclic. Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on , we find that . By Ptolemy's Theorem, , from which we have and . From Thales' Circle, is a right triangle, and , so . Set . , so by Ptolemy's Theorem on , we have Solving yields . We know that , so the area of . The requested answer is . ~ adam_zheng Solution 9 Let points , , be the points where lines , , and are tangent to the circle respectively. Then extend line to point such that is perpendicular to . According to Power of a Point, . Similarly, . (Note that because they are intersecting tangents from the same circle. Furthermore, due to a simple upwards translation.) Therefore, , and we already know that , meaning that we can use the Pythagorean Theorem on to obtain: . Let . Then . Since they are intersecting tangents from the same circle, . Therefore, . We have all the side lengths of , so applying the Pythagorean Theorem: . Thus, . , so the answer is . ~sid2012 Video Solution ~Steven Chen (www.professorchenedu.com) Video Solution 2 (Mathematical Dexterity) Video Solution 3 by OmegaLearn ~ pi_is_3.14 See Also 2022 AIME I (Problems • Answer Key • Resources) Preceded by Problem 10Followed by Problem 12 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Intermediate Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190004	https://www.ncbi.nlm.nih.gov/books/NBK482303/	Biochemistry, Glycolysis - StatPearls - NCBI Bookshelf 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Biochemistry, Glycolysis Raheel Chaudhry; Matthew A. Varacallo. Author Information and Affiliations Authors Raheel Chaudhry 1; Matthew A. Varacallo 2. Affiliations 1 Ross University School Of Medicine 2 Penn Highlands Healthcare System Last Update: August 8, 2023. Go to: Introduction Glycolysis is a metabolic pathway and an anaerobic energy source that has evolved in nearly all types of organisms. Another name for the process is the Embden-Meyerhof pathway, in honor of the major contributors towards its discovery and understanding. Although it doesn't require oxygen, hence its purpose in anaerobic respiration, it is also the first step in cellular respiration. The process entails the oxidation of glucose molecules, the single most crucial organic fuel in plants, microbes, and animals. Most cells prefer glucose (although there are exceptions, such as acetic acid bacteria that prefer ethanol). In glycolysis, 2 ATP molecules are consumed, producing 4 ATP, 2 NADH, and 2 pyruvates per glucose molecule. The pyruvate can be used in the citric acid cycle or serve as a precursor for other reactions. Go to: Fundamentals Glycolysis ultimately splits glucose into two pyruvate molecules. One can think of glycolysis as having two phases that occur in the cytosol of cells. The first phase is the"investment" phase due to its usage of two ATP molecules, and the second is the "payoff" phase. These reactions are all catalyzed by their own enzyme, with phosphofructokinase being the most essential for regulation as it controls the speed of glycolysis. Glycolysis occurs in both aerobic and anaerobic states. In aerobic conditions, pyruvate enters the citric acid cycle and undergoes oxidative phosphorylation leading to the net production of 32 ATP molecules. In anaerobic conditions, pyruvate converts to lactate through anaerobic glycolysis. Anaerobic respiration results in the production of 2 ATP molecules.Glucose is a hexose sugar, meaning it is a monosaccharide with six carbon atoms and six oxygen atoms. The first carbon has an attached aldehyde group, and the other five carbons have one hydroxyl group each. During glycolysis, glucose ultimately breaks down into pyruvate and energy; a total of 2 ATP is derived in the process (Glucose + 2 NAD+ + 2 ADP + 2 Pi --> 2 Pyruvate + 2 NADH + 2 H+ + 2 ATP + 2 H2O). The hydroxyl groups allow for phosphorylation. The specific form of glucose used in glycolysis is glucose 6-phosphate. Go to: Cellular Level Glycolysis occurs in the cytosol of cells. Under aerobic conditions, pyruvate derived from glucose will enter the mitochondria to undergo oxidative phosphorylation. Anaerobic conditions result in pyruvate staying in the cytoplasm and being converted to lactate by the enzyme lactate dehydrogenase. Go to: Molecular Level Glucose first converts to glucose-6-phosphate by hexokinase or glucokinase, using ATP and a phosphate group. Glucokinase is a subtype of hexokinase found in humans.Glucokinase has a reduced affinity for glucose and is found only in the pancreas and liver, whereas hexokinase is present in all cells. Glucose 6-phosphate is then converted to fructose-6-phosphate, an isomer, by phosphoglucose isomerase. Phosphofructose-kinase then produces fructose-1,6-bisphosphate, using another ATP molecule. Dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate are then created from fructose-1,6-bisphosphate by fructose bisphosphate aldolase. DHAP will be converted to glyceraldehyde-3-phosphate by triosephosphate isomerase, where now the two glyceraldehyde-3-phosphate molecules will continue down the same pathway. Glyceraldehyde-3-phosphate will become oxidized in an exergonic reaction into 1,3-bisphosphoglycerate, reducing an NAD+molecule to NADH and H+. 1,3-bisphosphoglycerate will then turn into 3-phosphoglycerate with the help of phosphoglycerate kinase, along with the production of the first ATP molecule from glycolysis. 3-phosphoglycerate will then convert, with the help of phosphoglycerate mutase, into 2-phosphoglycerate. With the release of one molecule of H2O, Enolase will make phosphoenolpyruvate (PEP) from 2-phosphoglycerate. Due to the unstable state of PEP, pyruvate kinase will facilitate its loss of a phosphate group to create the second ATP in glycolysis. Thus, PEP will then undergo conversion to pyruvate. Go to: Function Glycolysis occurs in the cytosol of the cell. It is a metabolic pathway that creates ATP without the use of oxygen but can occur in the presence of oxygen. In cells that use aerobic respiration as the primary energy source, the pyruvate formed from the pathway can be used in the citric acid cycle and go through oxidative phosphorylation to undergo oxidation into carbon dioxide and water. Even if cells primarily use oxidative phosphorylation, glycolysis can serve as an emergency backup for energy or as the preparation step before oxidative phosphorylation. In highly oxidative tissue, such as the heart, pyruvate production is essential for acetyl-CoA synthesis and L-malate synthesis. It serves as a precursor to many molecules, such as lactate, alanine, and oxaloacetate. Glycolysis precedes lactic acid fermentation; the pyruvate made in the former process serves as the prerequisite for the lactate made in the latter process. Lactic acid fermentation is the primary source of ATP in animal tissues with low metabolic requirements and little to no mitochondria. In erythrocytes, lactic acid fermentation is the sole source of ATP, as they lack mitochondria and mature red blood cells have little demand for ATP. Another part of the body that relies entirely or almost entirely on anaerobic glycolysis is the eye's lens, which is devoid of mitochondria, as their presence would lead to light scattering. Though skeletal muscles prefer to catalyze glucose into carbon dioxide and water during heavy exercise where oxygen is inadequate, the muscles simultaneously undergo anaerobic glycolysis and oxidative phosphorylation. Regulation Glucose The amount of glucose available for the process regulates glycolysis, which becomes available primarily in two ways: regulation of glucose reuptake or regulation of the breakdown of glycogen. Glucose transporters (GLUT) transport glucose from the outside of the cell to the inside. Cells containing GLUT can increase the number of GLUT in the cell's plasma membrane from the intracellular matrix, therefore increasing the uptake of glucose and the supply of glucose available for glycolysis. There are five types of GLUTs. GLUT1 is present in RBCs, the blood-brain barrier,and the blood-placental barrier. GLUT2 is in the liver, beta-cells of the pancreas, kidney, and gastrointestinal (GI) tract. GLUT3 is present in neurons. GLUT4 is in adipocytes, heart, and skeletal muscle. GLUT5 specifically transports fructose into cells. Another form of regulation is the breakdown of glycogen. Cells can store extra glucose as glycogen when glucose levels are high in the cell plasma. Conversely, when levels are low, glycogen can be converted back into glucose. Two enzymes control the breakdown of glycogen: glycogen phosphorylase and glycogen synthase. The enzymes can be regulated through feedback loops of glucose or glucose 1-phosphate, or via allosteric regulation by metabolites, or from phosphorylation/dephosphorylation control. Allosteric Regulators and Oxygen As described before, many enzymes are involved in the glycolytic pathway by converting one intermediate to another. Control of these enzymes, such as hexokinase, phosphofructokinase, glyceraldehyde-3-phosphate dehydrogenase, and pyruvate kinase, can regulate glycolysis. The amount of oxygen available can also regulate glycolysis. The “Pasteur effect” describes how the availability of oxygen diminishes the effect of glycolysis, and decreased availability leads to an acceleration of glycolysis, at least initially. The mechanisms responsible for this effect include allosteric regulators of glycolysis (enzymes such as hexokinase). The “Pasteur effect” appears to mostly occur in tissue with high mitochondrial capacities, such as myocytes or hepatocytes. Still, this effect is not universal in oxidative tissue, such as pancreatic cells. Enzyme Induction Another mechanism for controlling glycolytic rates is transcriptional control of glycolytic enzymes. Altering the concentration of key enzymes allows the cell to change and adapt to alterations in hormonal status. For example, increasing glucose and insulin levels can increase hexokinase and pyruvate kinase activity, therefore increasing the production of pyruvate. PFK-1 Fructose 2,6-bisphosphate is an allosteric regulator of PFK-1. High levels of fructose 2,6-bisphosphate increase the activity of PFK-1. Its production occurs through the action of phosphofructokinase-2 (PFK-2). PFK-2 has both kinase and phosphorylase activity and can transform fructose 6 phosphates to fructose 2,6-bisphosphate and vice versa. Insulin dephosphorylates PFK-2,activating its kinase activity, which increases fructose 2,6-bisphosphate and subsequently activates PFK-1. Glucagon can also phosphorylate PFK-2, which activates phosphatase, transforming fructose 2,6-bisphosphate back to fructose 6-phosphate. This reaction decreases fructose 2,6-bisphosphate levels and decreases PFK-1 activity. Go to: Mechanism Glycolysis Phases Glycolysis has two phases: the investment phase and the payoff phase. The investment phase is where there is energy, as ATP, is put in, and the payoff phase is where the net creation of ATP and NADH molecules occurs. A total of 2 ATP goes into the investment phase, with the production of 4 ATP resulting in the payoff phase; thus, there is a net total of 2 ATP. The steps by which new ATP is created have the name of substrate-level phosphorylation. Investment Phase In this phase, there are two phosphates added to glucose. Glycolysis begins with hexokinase phosphorylating glucose into glucose-6 phosphate (G6P). This step is the first transfer of a phosphate group and where the consumption of the first ATP takes place. Also, this is an irreversible step. This phosphorylation traps the glucose molecule in the cell because it cannot readily pass the cell membrane. From there, phosphoglucose isomerase isomerizes G6P into fructose 6-phosphate (F6P). Then, phosphofructokinase (PFK-1) adds the second phosphate. PFK-1 uses the second ATP and phosphorylates the F6P into fructose 1,6-bisphosphate. This step is also irreversible and is the rate-limiting step. In the following step, fructose 1,6-bisphosphate undergoes lysis into two molecules, which are substrates for fructose-bisphosphate aldolase to convert it into dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (G3P). DHAP is turned into G3P by triosephosphate isomerase. DHAP and G3p are in equilibrium with each other, meaning they transform back and forth. Payoff Phase It is critical to remember that there are a total of two 3-carbon sugars for every one glucose at the beginning of this phase. The enzyme glyceraldehyde-3-phosphate dehydrogenase metabolizes the G3P into 1,3-diphosphoglycerate by reducing NAD+ into NADH. Next, the 1,3-diphosphoglycerate loses a phosphate group through phosphoglycerate kinase to make 3-phosphoglycerate and creates an ATP through substrate-level phosphorylation. At this point, there are 2 ATP produced, one from each 3-carbon molecule. The 3-phosphoglycerate turns into 2-phosphoglycerate by phosphoglycerate mutase, and then enolase turns the 2-phosphoglycerate into phosphoenolpyruvate (PEP). In the final step, pyruvate kinase turns PEP into pyruvate and phosphorylates ADP into ATP through substrate-level phosphorylation, thus creating two more ATP. This step is also irreversible. Overall, the input for 1 glucose molecule is 2 ATP, and the output is 4 ATP, 2 NADH, and 2 pyruvate molecules. In cells, NADH must recycle back to NAD+ to permit glycolysis to keep running. Absent NAD+, the payoff phase will come to a halt, resulting in a backup in glycolysis. In aerobic cells, NADH recycles back into NAD+ by way of oxidative phosphorylation. In anaerobic cells, it occurs through fermentation.There are two types of fermentation: lactic acid and alcohol fermentation. Go to: Clinical Significance Pyruvate kinase deficiency is an autosomal recessive mutation that causes hemolytic anemia. There is an inability to form ATP and causes cell damage. Cells become swollen and are taken up by the spleen, causing splenomegaly. Signs and symptoms include jaundice, icterus, elevated bilirubin, and splenomegaly. Arsenic poisoning also prevents ATP synthesis because arsenic takes the place of phosphate in the steps of glycolysis. Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Anaerobic glycolysis Image courtesy O.Chaigasame Go to: References 1. Akram M. Mini-review on glycolysis and cancer. J Cancer Educ. 2013 Sep;28(3):454-7. [PubMed: 23728993] 2. Matschinsky FM, Wilson DF. The Central Role of Glucokinase in Glucose Homeostasis: A Perspective 50 Years After Demonstrating the Presence of the Enzyme in Islets of Langerhans. Front Physiol. 2019;10:148. [PMC free article: PMC6435959] [PubMed: 30949058] 3. Jin L, Zhou Y. Crucial role of the pentose phosphate pathway in malignant tumors. Oncol Lett. 2019 May;17(5):4213-4221. [PMC free article: PMC6444344] [PubMed: 30944616] 4. Liu Y, Zhang Z, Wang J, Chen C, Tang X, Zhu J, Liu J. Metabolic reprogramming results in abnormal glycolysis in gastric cancer: a review. Onco Targets Ther. 2019;12:1195-1204. [PMC free article: PMC6389007] [PubMed: 30863087] 5. Granchi C, Bertini S, Macchia M, Minutolo F. Inhibitors of lactate dehydrogenase isoforms and their therapeutic potentials. Curr Med Chem. 2010;17(7):672-97. [PubMed: 20088761] 6. Xiong Y, Lei QY, Zhao S, Guan KL. Regulation of glycolysis and gluconeogenesis by acetylation of PKM and PEPCK. Cold Spring Harb Symp Quant Biol. 2011;76:285-9. [PMC free article: PMC4876600] [PubMed: 22096030] 7. Malaisse WJ, Zhang Y, Sener A. Enzyme-to-enzyme channeling in the early steps of glycolysis in rat pancreatic islets. Endocrine. 2004 Jul;24(2):105-9. [PubMed: 15347834] 8. Lenzen S. A fresh view of glycolysis and glucokinase regulation: history and current status. J Biol Chem. 2014 May 02;289(18):12189-94. [PMC free article: PMC4007419] [PubMed: 24637025] 9. Ren Y, Shen HM. Critical role of AMPK in redox regulation under glucose starvation. Redox Biol. 2019 Jul;25:101154. [PMC free article: PMC6859544] [PubMed: 30853530] 10. Fan T, Sun G, Sun X, Zhao L, Zhong R, Peng Y. Tumor Energy Metabolism and Potential of 3-Bromopyruvate as an Inhibitor of Aerobic Glycolysis: Implications in Tumor Treatment. Cancers (Basel). 2019 Mar 06;11(3) [PMC free article: PMC6468516] [PubMed: 30845728] 11. Orang AV, Petersen J, McKinnon RA, Michael MZ. Micromanaging aerobic respiration and glycolysis in cancer cells. Mol Metab. 2019 May;23:98-126. [PMC free article: PMC6479761] [PubMed: 30837197] 12. DeMaster EG, Mitchell A. A comparison of arsenate and vanadate as inhibitors or uncouplers of mitochondrial and glycolytic energy metabolism. Biochemistry. 1973 Sep 11;12(19):3616-21. [PubMed: 4788305] Disclosure:Raheel Chaudhry declares no relevant financial relationships with ineligible companies. Disclosure:Matthew Varacallo declares no relevant financial relationships with ineligible companies. Introduction Fundamentals Cellular Level Molecular Level Function Mechanism Clinical Significance Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK482303 PMID: 29493928 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Introduction Fundamentals Cellular Level Molecular Level Function Mechanism Clinical Significance Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Biochemistry, Anaerobic Glycolysis.[StatPearls. 2025]Biochemistry, Anaerobic Glycolysis.Melkonian EA, Schury MP. StatPearls. 2025 Jan Coexistence of the Entner-Doudoroff and Embden-Meyerhof-Parnas pathways enhances glucose consumption of ethanol-producing Corynebacterium glutamicum.[Biotechnol Biofuels. 2021]Coexistence of the Entner-Doudoroff and Embden-Meyerhof-Parnas pathways enhances glucose consumption of ethanol-producing Corynebacterium glutamicum.Jojima T, Igari T, Noburyu R, Watanabe A, Suda M, Inui M. Biotechnol Biofuels. 2021 Feb 16; 14(1):45. Epub 2021 Feb 16. Changes in energy metabolism during the early development of Xenopus laevis.[Exp Cell Biol. 1982]Changes in energy metabolism during the early development of Xenopus laevis.Løvtrup-Rein H, Nelson L. Exp Cell Biol. 1982; 50(3):162-8. Review Control and regulation of the pyrophosphate-dependent glucose metabolism in Entamoeba histolytica.[Mol Biochem Parasitol. 2019]Review Control and regulation of the pyrophosphate-dependent glucose metabolism in Entamoeba histolytica.Saavedra E, Encalada R, Vázquez C, Olivos-García A, Michels PAM, Moreno-Sánchez R. Mol Biochem Parasitol. 2019 Apr; 229:75-87. Epub 2019 Feb 14. Review The biochemistry of sugars.[Int Z Vitam Ernahrungsforsch B...]Review The biochemistry of sugars.Horecker BL. Int Z Vitam Ernahrungsforsch Beih. 1976; 15:1-21. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Biochemistry, Glycolysis - StatPearlsBiochemistry, Glycolysis - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... 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190005	https://www.bbc.co.uk/bitesize/guides/zdykw6f/revision/2	Bitesize GCSE WJEC Group 1 alkali metalsReactions of alkali metals with water The Group 1 elements, also known as the alkali metals, all react vigorously with water to produce an alkaline solution. Part of Chemistry (Single Science)Atomic structure and the periodic table Reactions of alkali metals with water All the alkali metals react vigorously with cold water. In each reaction, hydrogen gas is given off and the metal hydroxide is produced. The speed and violence of the reaction increases as you go down the group. This shows that the reactivity of the alkali metals increases as you go down Group 1. Note that you are expected to know some of the key safety points for these reactions. These include: use of a plastic safety screen using small pieces of the metal using a large volume of water using tweezers to hold the metal wearing thick rubber gloves Lithium When lithium is added to water, lithium floats. It fizzes steadily and becomes smaller, until it eventually disappears. lithium + water → lithium hydroxide + hydrogen 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) Sodium When sodium is added to water, the sodium melts to form a ball that moves around on the surface. It fizzes rapidly before it disappears. sodium + water → sodium hydroxide + hydrogen 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Potassium When potassium is added to water, the metal melts and floats. It moves around very quickly on the surface of the water. The metal self-ignites, which also ignites the hydrogen gas. This results in sparks and a lilac flame. There is sometimes a small explosion at the end of the reaction. potassium + water → potassium hydroxide + hydrogen 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) Test for hydrogen gas To identify the gas given off by these reactions as hydrogen, it is necessary to test the gas: Collect the gas in a test tube. Put the flame from a burning splint into the top of the test tube. The gas will burn quickly and give off a squeaky ‘pop’. Strong alkalis The hydroxides formed in all of these reactions dissolve in water to form alkalineHaving a pH greater than 7. solutions. These solutions turn universal indicator paperPaper stained with universal indicator, a chemical solution that produces many different colour changes corresponding to different pH levels. purple, showing they are strongly alkaline. Strong alkalis are corrosiveAble to damage metal, stonework, clothes and skin. Strong acids and alkalis are corrosive.. Care must be taken when they are used – goggles and gloves should be worn. Next page Reactions of alkali metals with oxygen Previous page Group 1 – the alkali metals More guides on this topic The atom The periodic table Group 7 halogens Group 0 and testing ions Related links Chemistry: Exam practice Chemistry revision resources Bitesize revision podcasts Personalise your Bitesize! Jobs that use Chemistry Save My Exams Subscription Quizlet Royal Society of Chemistry Revisio Subscription Science Museum
190006	https://www.emilyzabor.com/condsurv/articles/estimate_cs.html	Estimate conditional survival • condsurv Toggle navigationcondsurv1.0.0 Function Documentation Tutorials Estimate conditional survival Plot conditional survival curves Estimate conditional survival Emily C. Zabor Last updated: 2022-10-20 estimate_cs.Rmd ``` library(condsurv) library(dplyr) > Warning: package 'dplyr' was built under R version 4.2.1 library(survival) > Warning: package 'survival' was built under R version 4.2.1 ``` If S(t)S(t) represents the survival function at time t t, then conditional survival is defined as S(y\|x)=S(x+y)S(x)S(y\|x)=S(x+y)S(x) where y y is the number of additional survival years of interest and x x is the number of years a subject has already survived. Generating conditional survival estimates The conditional_surv_est function will generate this estimate along with 95% confidence intervals. The lung dataset from the survival package will be used to illustrate. ``` Scale the time variable to be in years rather than days lung2 <- mutate( lung, os_yrs = time / 365.25 ) ``` First generate a single conditional survival estimate. This is the conditional survival of surviving to 1 year conditioned on already having survived 6 months (0.5 0.5 year). This returns a list, where cs_est is the conditional survival estimate, cs_lci is the lower bound of the 95% confidence interval and cs_uci is the upper bound of the 95% confidence interval. ``` myfit <- survfit(Surv(os_yrs, status) ~ 1, data = lung2) conditional_surv_est( basekm = myfit, t1 = 0.5, t2 = 1 ) > $cs_est > 0.58 > > $cs_lci > 0.49 > > $cs_uci > 0.66 ``` You can easily use purrr::map_df to get a table of estimates for multiple timepoints. For example we could get the conditional survival estimate of surviving to a variety of different time points given that the subject has already survived for 6 months (0.5 years). ``` prob_times <- seq(1, 2.5, 0.5) purrr::map_df( prob_times, ~conditional_surv_est( basekm = myfit, t1 = 0.5, t2 = .x) ) %>% dplyr::mutate(years = prob_times) %>% dplyr::select(years, everything()) %>% knitr::kable() ``` \| years \| cs_est \| cs_lci \| cs_uci \| ---: ---: \| \| 1.0 \| 0.58 \| 0.49 \| 0.66 \| \| 1.5 \| 0.36 \| 0.27 \| 0.45 \| \| 2.0 \| 0.16 \| 0.10 \| 0.25 \| \| 2.5 \| 0.07 \| 0.02 \| 0.15 \| A note on confidence interval estimation The confidence intervals are based on a variation of the log-log transformation, also known as the “exponential” Greenwood formula, where the conditional survival estimate is substituted in for the traditional survival estimate in constructing the confidence interval. If S^(y\|x)S^(y\|x) is the estimated conditional survival to y y given having already survived to x x, then S^(y\|x)e x p(±1.96 L^(y\|x)√)S^(y\|x)e x p(±1.96 L^(y\|x)) where L^(y\|x)=1 log(S^(y\|x))2∑j:x≤τ j≤y d j(r j−d j)r j L^(y\|x)=1 log⁡(S^(y\|x))2∑j:x≤τ j≤y d j(r j−d j)r j and τ j τ j = distinct death time j j d j d j = number of failures at death time j j r j r j = number at risk at death time j j Contents Generating conditional survival estimates A note on confidence interval estimation Developed by Emily C. Zabor, Mithat Gonen. Site built with pkgdown 2.0.6.
190007	https://math.stackexchange.com/questions/2977217/a-integral-inequality-and-generalization	real analysis - A integral inequality and generalization - 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 A integral inequality and generalization Ask Question Asked 6 years, 11 months ago Modified6 years, 11 months ago Viewed 474 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. If f(x)f(x) is decreasing then ∫b a−1 f(x)d x≥∑b k=a f(k)≥∫b+1 a f(x)d x∫a−1 b f(x)d x≥∑k=a b f(k)≥∫a b+1 f(x)d x. And reverse for increasing functions. I try to generalize it Generalization: Let f(x)f(x) be a monotone decreasing function on (0,+∞)(0,+∞). Prove that: ∫b+1 a f(x)d x≤∑k=a b f(k)≤∫b a−1 f(x)d x≤∑k=a−1 b−1 f(k)a,b∈N∫a b+1 f(x)d x≤∑k=a b f(k)≤∫a−1 b f(x)d x≤∑k=a−1 b−1 f(k)a,b∈N When does equality hold? Note that 0∉N 0∉N I have 3 questions as follow : 1) Could you please prove generalization 2) Is this true ? Generalization (inversion) Let f(x)f(x) be a monotone increasing function on (0,+∞)(0,+∞). Prove that: ∫b+1 a f(x)d x≥∑k=a b f(k)≥∫b a−1 f(x)d x≥∑k=a−1 b−1 f(k)a,b∈N∫a b+1 f(x)d x≥∑k=a b f(k)≥∫a−1 b f(x)d x≥∑k=a−1 b−1 f(k)a,b∈N When does equality hold? 3) Can we have a stronger generalization ? Thank you in advance! real-analysis inequality definite-integrals Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 30, 2018 at 11:37 namnam asked Oct 30, 2018 at 6:02 namnam 357 1 1 silver badge 3 3 bronze badges 2 There might be some typos in your question: in line 1 the upper index of the sum should be b b i guess. In the other two inequalities the bounds of the second integral cannot be correct. Please edit!weee –weee 2018-10-30 07:01:01 +00:00 Commented Oct 30, 2018 at 7:01 I edited my post !nam –nam 2018-10-30 11:37:59 +00:00 Commented Oct 30, 2018 at 11:37 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. 1) Once you have the fact ∫b a−1 f(x)d x≥∑k=a b f(k)≥∫b+1 a f(x)d x,∀a,b∈N,∫a−1 b f(x)d x≥∑k=a b f(k)≥∫a b+1 f(x)d x,∀a,b∈N, there is not much to prove, for at the expression ∫b+1 a f(x)d x≤∑k=a b f(k)≤∫b a−1 f(x)d x≤∑k=a−1 b−1 f(k),∀a,b∈N,∫a b+1 f(x)d x≤∑k=a b f(k)≤∫a−1 b f(x)d x≤∑k=a−1 b−1 f(k),∀a,b∈N, the first two inequalities follow directly from the fact and the third one follows from the fact applied to a−1 a−1 and b−1 b−1 in the place of a a and b b, respectively. 2) The proof for increasing functions is analog to that for decreasing functions. Using the fact for increasing functions ∫b a−1 f(x)d x≤∑k=a b f(k)≤∫b+1 a f(x)d x,∀a,b∈N,∫a−1 b f(x)d x≤∑k=a b f(k)≤∫a b+1 f(x)d x,∀a,b∈N, at the expression ∫b+1 a f(x)d x≥∑k=a b f(k)≥∫b a−1 f(x)d x≥∑k=a−1 b−1 f(k),∀a,b∈N,∫a b+1 f(x)d x≥∑k=a b f(k)≥∫a−1 b f(x)d x≥∑k=a−1 b−1 f(k),∀a,b∈N, the first two inequalities follow directly from the fact and the third one follows from the fact applied to a−1 a−1 and b−1 b−1 in the place of a a and b b, respectively. 3) I don't know and I can't think of any suggestions. About the question of when the equality holds for (1), I have an idea. Note that ∫b a−1 f(x)d x=∫b+1 a f(x)d x∫a−1 b f(x)d x=∫a b+1 f(x)d x if, and only if, ∫a a−1 f(x)d x+∫b a f(x)d x=∫b a f(x)d x+∫b+1 b f(x)d x⇔∫a a−1 f(x)d x=∫b+1 b f(x)d x.∫a−1 a f(x)d x+∫a b f(x)d x=∫a b f(x)d x+∫b b+1 f(x)d x⇔∫a−1 a f(x)d x=∫b b+1 f(x)d x. So our problem is equivalent to analyze which conditions imply that ∫a a−1 f(x)d x=∫b+1 b f(x)d x.∫a−1 a f(x)d x=∫b b+1 f(x)d x. The simplest condition that guarantees the equality above is: Condition 1.a−1=b a−1=b; Other two conditions for which the equality immediately holds are: Condition 2.a−1<b a−1<b and f f is a constant at the interval (a−1,b+1)(a−1,b+1); Condition 3.a−1>b a−1>b and f f is a constant at the interval (b,a)(b,a); I guess these are the onliest cases for that the equality holds, the proof is not so difficult. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 30, 2018 at 19:36 André PortoAndré Porto 1,945 10 10 silver badges 19 19 bronze badges 2 To make it all correct you have to impose that a≤b a≤b otherwise the sums are possibly empty, whereas the integrals are possibly negative. Moreover for the second sum we need a>1 a>1 because f(0)f(0) is not defined.weee –weee 2018-10-30 20:36:36 +00:00 Commented Oct 30, 2018 at 20:36 Can't we use the algebraic rule ∫a b f(x)d x=−∫b a f(x)d x∫b a f(x)d x=−∫a b f(x)d x?André Porto –André Porto 2018-10-30 21:04:39 +00:00 Commented Oct 30, 2018 at 21:04 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. For the second inequality note that ∫b a−1 f(x)d x=∑b−1 k=a−1∫k+1 k f(x)d x∫a−1 b f(x)d x=∑k=a−1 b−1∫k k+1 f(x)d x. Now by monotonicity we get that ∫k+1 k f(x)d x≥f(k+1)∫k k+1 f(x)d x≥f(k+1) and hence the second inequality. For the third inequality we can use the same reasoning as above except that we use ∫k+1 k f(x)d x≤f(k)∫k k+1 f(x)d x≤f(k). As stated above the equality in the first inequality holds if we have a piecewise constant function which is constant on [k,k+1)[k,k+1) for k=a,…,b k=a,…,b. Now assume that we have a function g g that satisfies the equality. Then g g is piecewise constant on [k,k+1)[k,k+1) for k=a,…,b k=a,…,b. If it was not then there would exist a k∈{a,…,b}k∈{a,…,b} and some ξ∈[k,k+1)ξ∈[k,k+1) for which we have g(k)>g(ξ)g(k)>g(ξ) since g g is decreasing. This would however imply that ∫k+1 k g(x)d x<g(k)∫k k+1 g(x)d x<g(k) and hence we do not have equality. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 4, 2018 at 8:19 namnam 357 1 1 silver badge 3 3 bronze badges 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 definite-integrals See similar questions with these tags. 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190008	https://math.stackexchange.com/questions/458898/3-digit-numbers-that-the-sum-of-digits-are-even	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 3-digit numbers that the sum of digits are even. Ask Question Asked Modified 6 years, 9 months ago Viewed 21k times 0 $\begingroup$ How many three digit numbers are there such that the sum of the digits is even? So I guess we're taking the total number of three digit numbers, then eliminate the ones that doesn't satisfy the properties. But, can someone give me a hint on how to count the number of 3-digit numbers that has a sum of digits that's even? Ty! combinatorics Share asked Aug 3, 2013 at 16:53 Commander ShepardCommander Shepard 1,87244 gold badges2121 silver badges4040 bronze badges $\endgroup$ 3 $\begingroup$ If I give you any two digits X and Y, what digit Z can you give me back so that X+Y+Z is even? $\endgroup$ RghtHndSd – RghtHndSd 2013-08-03 16:55:56 +00:00 Commented Aug 3, 2013 at 16:55 $\begingroup$ Why would it be easier to count the ones with odd sums? $\endgroup$ Thomas Andrews – Thomas Andrews 2013-08-03 16:59:11 +00:00 Commented Aug 3, 2013 at 16:59 1 $\begingroup$ Basically, given the first two digits, exactly have of ${0,\dots,9}$ can be used for the last digit. $\endgroup$ Thomas Andrews – Thomas Andrews 2013-08-03 16:59:59 +00:00 Commented Aug 3, 2013 at 16:59 Add a comment \| 7 Answers 7 Reset to default 5 $\begingroup$ HINT: Suppose that the digits are $abc$. How many choices are there for $ab$? How many choices are there for $c$ if $a+b$ is even? How many choices are there for $c$ if $a+b$ is odd? Does it make any difference whether $a+b$ is odd or even? Share answered Aug 3, 2013 at 16:55 Brian M. ScottBrian M. Scott 633k5757 gold badges824824 silver badges1.4k1.4k bronze badges $\endgroup$ Add a comment \| 3 $\begingroup$ A simple way of seeing that it is exactly half, is to realize that we have a bijection between 3 digit numbers by sending $N$ to $1099- N$. Note that since the sum of these numbers is odd, hence the pair of numbers have different parity. Put explicitly, we are pairing up ${100, 999}, {101, 998} \ldots, {544, 545 } $. Share answered Aug 3, 2013 at 18:12 Calvin LinCalvin Lin 77.6k55 gold badges8686 silver badges170170 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ Ways to get an even sum: $2$ odd $1$ even. $3$ even. First case: First see that there are three posiblities for the place the even number takes: (the first number cant be 0 because then it would not be a 3 digit number. Therefore there are $(455)+2(555)$ ways second case: all even. there are $455$ ways to do it therefore the total is $100+250+100=450$ numbers Share answered Aug 3, 2013 at 16:59 AsinomásAsinomás 108k2525 gold badges139139 silver badges290290 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ Exactly half of them. If you allow leading zeroes, there are $1000$ three-digit numbers, if you are strict, there are $900$ three-digit numbers, so the final answer is $500$ or $450$, depending on your definition. Note that among any two consecutive three-digit numbers tha tdiffer only in their last digit, one has even and one has odd digit sum. Share answered Aug 3, 2013 at 16:55 Hagen von EitzenHagen von Eitzen 384k3333 gold badges379379 silver badges686686 bronze badges $\endgroup$ 4 $\begingroup$ So, if we're counting for 4-digit numbers, or 5-digit, so-on, it's still half of them that have a digit sum that's even/odd? $\endgroup$ Commander Shepard – Commander Shepard 2013-08-03 16:57:13 +00:00 Commented Aug 3, 2013 at 16:57 $\begingroup$ This is not a hint. $\endgroup$ RghtHndSd – RghtHndSd 2013-08-03 16:57:27 +00:00 Commented Aug 3, 2013 at 16:57 $\begingroup$ I'm not sure for your result: notice that we begin by 100 and end by 999 so I think that the number is the half-1=449. Isn't it? $\endgroup$ user63181 – user63181 2013-08-03 17:10:38 +00:00 Commented Aug 3, 2013 at 17:10 $\begingroup$ no. substract 99 to all terms to get {1,2,3...900} which clearly has 900 elements $\endgroup$ Asinomás – Asinomás 2013-08-03 17:13:04 +00:00 Commented Aug 3, 2013 at 17:13 Add a comment \| 1 $\begingroup$ HINT If the final digit of $n$ (units digit) is not $9$ then $n$ and $n+1$ have one odd digit sum and one even sum. Can you see how to pair numbers up so that one is odd and one is even, and every three-digit number is included in the pairing? Share answered Aug 3, 2013 at 16:59 Mark BennetMark Bennet 102k1414 gold badges119119 silver badges232232 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ HINT: So, the number digits with odd value is even If we take no digits with odd values, we shall have $4\cdot 5\cdot5=100$ combinations , If we take two digits with odd values, we have following combinations $(O,O,E),(O,E,O),(E,O,O)$ Now, for the first digit in even case, it can assume $4$ values namely, ${2,4,6,8}$ and in odd case, it can assume, $5$ values namely, ${1,3,5,7,9}$ Share edited Aug 3, 2013 at 17:05 answered Aug 3, 2013 at 17:00 lab bhattacharjeelab bhattacharjee 279k2020 gold badges213213 silver badges337337 bronze badges $\endgroup$ 4 $\begingroup$ {0,2,4,6,8} has 5 elements $\endgroup$ Asinomás – Asinomás 2013-08-03 17:02:36 +00:00 Commented Aug 3, 2013 at 17:02 1 $\begingroup$ @Omnitic, the first digit can not be $0,$ for a valid three digit number, right? $\endgroup$ lab bhattacharjee – lab bhattacharjee 2013-08-03 17:03:24 +00:00 Commented Aug 3, 2013 at 17:03 $\begingroup$ should it not be 455 then? $\endgroup$ Asinomás – Asinomás 2013-08-03 17:04:05 +00:00 Commented Aug 3, 2013 at 17:04 $\begingroup$ @Omnitic, sorry, Rectified. I wrongly included one extra condition: without repetition $\endgroup$ lab bhattacharjee – lab bhattacharjee 2013-08-03 17:06:47 +00:00 Commented Aug 3, 2013 at 17:06 Add a comment \| 0 $\begingroup$ Since the even number = sum 3 even or sum 2 even, 1 odd so we have 4 case: The number is $\, \overline{abc}\, $ where $a,b,c\in { 0,2,4,6,8}$. The number $a$ has $4$ choices, the number $b$ has $4$ choices, the number $c$ has $3$ choices. Therefore, we have $\, 443=48,$ numbers. The number is $\, \overline{abc}\, $ where $a\in { 2,4,6,8}$ and $b,c \in {1,3,5,7,9 }$. The number $a$ has $4$ choices, the number $b$ has $5$ choices, the number $c$ has $4$ choices. Therefore, we have $\, 454=80\,$ numbers. The number is $\, \overline{abc}\, $ where $a,c\in { 0,2,4,6,8}$ and $b\in {1,3,5,7,9}$. The number $a$ has $4$ choices, the number $c$ has $4$ choices, the number $b$ has $5$ choices. Therefore, we have $\, 445=80\,$ numbers. The number is $\, \overline{abc}\, $ where $c\in { 0,2,4,6,8}$ and $a, b\in {1,3,5,7,9}$. The number $a$ has $5$ choices, the number $b$ has $4$ choices, the number $c$ has $5$ choices. Therefore, we have $\, 545=100\,$ numbers. The conclution, we have $\, 48+80+80+100=308$ numbers. Share answered Dec 31, 2018 at 15:35 Huong Nguyen thiHuong Nguyen thi 6611 bronze badge $\endgroup$ 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 combinatorics See similar questions with these tags. 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190009	https://old.maa.org/press/maa-reviews/the-higher-arithmetic-an-introduction-to-the-theory-of-numbers-0	The Higher Arithmetic: An Introduction to the Theory of Numbers \| Mathematical Association of America Skip to main content Home Math Careers Contact Us Login Search form Search Login Join Give Events About MAA MAA History MAA Centennial MathDL Spotlight: Archives of American Mathematics MAA Officers MAA to the Power of New Governance Council and Committees Governance Documents Bylaws Policies and Procedures MAA Code of Conduct Policy on Conflict of Interest Statement about Conflict of Interest Recording or Broadcasting of MAA Events Policy for Establishing Endowments and Funds Avoiding Implicit Bias Copyright Agreement Principal Investigator's Manual Advocacy Our Partners Advertise with MAA Employment Opportunities Staff Directory Contact Us 2022 Impact Report In Memoriam Membership Membership Categories Membership Renewal Member Discount Programs MERCER Insurance MAA Member Directories New Member Benefits MAA Publications Periodicals The American Mathematical Monthly Mathematics Magazine The College Mathematics Journal Loci/JOMA Browse How to Cite Communications in Visual Mathematics Convergence About Convergence What's in Convergence? 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E. Shaw Group AMC 8 Awards & Certificates Maryam Mirzakhani AMC 10 A Awards & Certificates Two Sigma AMC 10 B Awards & Certificates Jane Street AMC 12 A Awards & Certificates Akamai AMC 12 B Awards & Certificates High School Teachers News Our Blog MAA Social Media RSS You are here Home » MAA Publications » MAA Reviews » The Higher Arithmetic: An Introduction to the Theory of Numbers The Higher Arithmetic: An Introduction to the Theory of Numbers H. Davenport Publisher: Cambridge University Press Publication Date: 2008 Number of Pages: 239 Format: Paperback Edition: 8 Price: 71.99 ISBN: 9780521722360 Category: Textbook BLL Rating: BLL The Basic Library List Committee recommends this book for acquisition by undergraduate mathematics libraries. MAA Review Table of Contents [Reviewed by Allen Stenger , on 01/23/2018 ] Harold Davenport (1907–1969) was a British mathematician who worked in many areas of number theory. He wrote this book in 1952 as an introduction to number theory for a general audience. It has expanded over the years, and recent editions have been edited and added to by his son, James H. Davenport, a computer scientist at the University of Bath. The prerequisites have been kept very low: just high-school algebra and no calculus (but you do need a strong grasp of the algebra). This is not a theorem-proof book. It’s more like a good expository talk, where you hear about the problems, see some examples, and then look at some statements of facts (results) along with arguments or explanations why they are true. There are in fact a lot of theorems and proofs here, but they are not packaged as such. The chapter on continued fractions is especially good. It creeps up on the subject gradually, starting with fractions and the Euclidean algorithm, and lots of examples, but developing the algorithms in more generality than this. Then it makes the leap to irrational numbers (infinite continued fractions), concentrating on the case of quadratic irrationals (this case is especially nice because the fractions are periodic). It also goes into a good bit of detail about the approximation properties of the convergents. I believe this is the only part of the book that uses limits or infinite processes, although the way it is developed you don’t really have to understand those to follow the narrative. There’s a lengthly and mostly up-to-date chapter on computers and number theory, that concentrates on several factorization methods and has a very clear section on cryptography with a good introduction (not just jumping directly to the Diffie-Hellman and RSA algorithms). This eighth edition, published in 2008, is very polished. It has a support website with four pages of notes on new developments since publication, although it hasn’t been updated since 2013. The market for this book is a little uncertain today. Originally it was aimed at the intelligent general reader who had a good knowledge of high-school mathematics, but I think there aren’t very many general readers today with a strong enough knowledge of algebra to follow the narrative. Most introductory number theory courses today are upper-division undergraduate courses and they are proofs courses, so this would not work well as a text. I do think it would work well as enrichment material; bright high-school math students would certainly be able to follow the book and many would be fascinated by it. It could also work for pre-service teachers. There are a large number of short introductory number theory books at about the same level, most of which are more traditional textbooks than this one. Good examples of these include Vinogradov’s Elements of Number Theory, Dudley’s two books Elementary Number Theory and A Guide to Elementary Number Theory, LeVeque’s Fundamentals of Number Theory, and Andrews’s Number Theory. Some other books at the same level and with the same prerequisites as the present book, that also take an informal and exploratory approach, include Burn’s A Pathway into Number Theory (very IBL-oriented), Ore’s Number Theory and its History (lots of examples, almost no prerequisites), and Ogilvy & Anderson’sExcursions in Number Theory (a 1966 book that was also intended for a general audience but is probably too advanced today). Allen Stenger is a math hobbyist and retired software developer. He is an editor of the Missouri Journal of Mathematical Sciences. His personal web page is allenstenger.com. His mathematical interests are number theory and classical analysis. (See also the more detailedtable of contents in pdf format.) Introduction Factorization and the primes Congruences Quadratic residues Continued fractions Sums of squares Quadratic forms Some Diophantine equations Computers and number theory Exercises Hints Answers Bibliography Index Additional notes Tags: Elementary Number Theory Number Theory Log in to post comments Dummy View - NOT TO BE DELETED Get Ready: Our Brand New Website is Coming Soon! 2024 MAA Awards & Prize Winners Announced! 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190010	https://socratic.garden/subjects/Science:%20Motion%20-%20Measuring%20time%20and%20distance%20using%20standard%20units	Instruct Science: Motion - Measuring time and distance using standard units faster with AI We use essential cookies to make our site work. With your consent, we may also use non-essential cookies to improve user experience and analyze website traffic. By clicking “Accept,” you agree to our website's cookie use as described in our Cookie Policy. You can change your cookie settings at any time by clicking “Preferences.” Preferences Decline Accept American EnglishPricingLog In Sign UpCompose American English Instructing Science: Motion - Measuring Time and Distance Using Standard Units In the study of motion in science, understanding how to measure time and distance is fundamental. Motion refers to the change in the position of an object over time, and accurate measurements are crucial for analyzing this change. Standard units such as meters (for distance) and seconds (for time) provide a consistent framework for measuring and communicating results. This subject covers methodologies for measuring distance—using tools like rulers or measuring tapes—and time, typically through stopwatches or timers. By mastering these concepts, students can effectively analyze motion, calculate speed, and understand the relationships between time, distance, and velocity in various physical contexts. Table of Contents Useful Resources for Instructing Science: Motion - Measuring time and distance using standard units Strategies for Engaging Learners on Science: Motion - Measuring time and distance using standard units Tips on Making Science: Motion - Measuring time and distance using standard units Feel Relevant Resources for Teaching Science: Motion - Measuring Time and Distance Using Standard Units 1. Textbooks and Reference Books Primary Textbook: A foundational science textbook that covers motion, distance, and time. Supplementary Books: Additional references focusing on physics concepts, experiments, and real-world applications. 2. Digital Resources Educational Websites: Websites like Khan Academy, National Geographic, and PBS LearningMedia that provide videos, articles, and exercises. Online Simulations: Platforms such as PhET offer interactive simulations for understanding motion and measurement. YouTube Channels: Channels dedicated to physics education, providing visual demonstrations of motion concepts. 3. Multimedia Tools Presentation Software: Tools such as PowerPoint or Google Slides to create engaging lessons. Video Clips: Short clips demonstrating motion experiments or real-life applications of motion, such as car racing or sports. Podcasts: Science-related podcasts that explain concepts in an engaging way. 4. Laboratory Equipment Measuring Instruments: Stopwatches, rulers, tape measures, and motion detectors for hands-on experiments. Motion Toys: Simple devices like toy cars, pendulums, or marbles for observational experiments. Scale Models: Miniatures of cars, planets, or other objects to illustrate motion concepts. 5. Classroom Supplies Whiteboards/Markers: For illustrating concepts and equations during lessons. Posters and Charts: Visual aids showing units of measurement, speed, velocity, and graphs related to motion. Worksheets: Printable materials for student practice and assessment. 6. Assessment Materials Quizzes and Tests: Pre-prepared evaluations to assess students' understanding of motion and measurement. Rubrics: Assessment criteria for projects and practical activities related to measuring time and distance. Feedback Forms: Tools for providing constructive feedback to students on their work. 7. Collaborative Tools Group Project Materials: Supplies for collaborative experiments, like graph paper, calculators, and drawing tools. Online Collaboration Platforms: Tools like Google Classroom or Padlet for sharing resources, assignments, and discussions. 8. Professional Development Resources Workshops and Seminars: Opportunities for training on the latest teaching strategies and technologies in science education. Online Courses: MOOCs (Massive Open Online Courses) on physics education topics for deepening content knowledge. Networking Opportunities: Join science educator organizations for resource sharing and support. 9. Community and Extracurricular Resources Local Science Museums: Arranging field trips to enhance learning experiences. Guest Speakers: Inviting local scientists or educators to share their expertise and engage students. Science Fairs: Participation in or organization of fairs to encourage student exploration of motion concepts. 10. Technology Tools Data Analysis Software: Programs like Logger Pro for analyzing motion data collected during experiments. Apps: Mobile applications focused on physics and motion measurement for interactive learning experiences. 11. Safety Equipment Safety Goggles: Ensure a safe laboratory environment. Safety Protocols: Guidelines and training on safe handling of equipment during experiments. 12. Visual Aids Diagrams and Charts: Flowcharts and infographics to illustrate motion concepts and measurement units. 3D Models: Physical models that depict motion principles, such as inertia and friction. 13. Student-Centered Resources Learning Stations: Setup for different activities on measuring time and distance to encourage exploration and inquiry. Interactive Notebooks: Encourage students to document their learning process and reflections on motion topics. These resources collectively support an engaging and effective learning environment for students studying the concepts of motion, time, and distance measurement in science. Prepare lessons on Science: Motion - Measuring time and distance using standard units using AI. The Socratic Garden saves instructors time & money with AI tools that help create, organize, and disseminate custom learning material for their audience. Try for free Engaging Learners in Science: Motion - Measuring Time and Distance Using Standard Units Use Real-Life Examples Incorporate relatable scenarios like the speed of a car, the time it takes to walk to school, or the distance covered during a sports event. Discussing practical applications helps students see the relevance of the concepts. Incorporate Hands-On Activities Engage learners with experiments where they measure time and distance. For instance, use a stopwatch to time how long it takes to roll a toy car down a ramp and calculate the distance it travels. This active participation can solidify understanding. Integrate Technology Utilize apps and simulations that allow students to visualize motion and experiment with variables. Graphing tools can also help them analyze data effectively and see the relationship between time and distance visually. Create Cooperative Learning Opportunities Encourage group projects where students can collaborate to measure distances in their community or school using standard units. Presenting their findings to the class promotes interaction and allows for diverse methods of understanding. Use Engaging Visuals Incorporate videos, animations, or dynamic presentations that illustrate motion concepts. Visual aids make complex ideas more accessible and capture students' attention. Gamify the Learning Process Design games that involve measuring distances and time, such as relay races where students calculate their speeds or virtual scavenger hunts that require them to estimate and measure distances effectively. Relate to Student Interests Find connections between the topic and students’ interests, like how timing affects athletes in sports or how vehicles are designed based on speed and distance. Tailoring examples to individual or group interests increases engagement. Encourage Critical Thinking Pose intriguing questions that prompt students to hypothesize about motion, like "What would happen if we doubled the distance?" or challenge them with problems that require them to apply their knowledge in new contexts. Foster a Competitive Spirit Set up challenges, such as timed measuring races or accuracy tests, where students can compete in small groups. This competitive energy can motivate students to engage more deeply with the subject matter. Use Storytelling Create narratives or scenarios involving characters making decisions based on measurements of time and distance. Storytelling makes the content more memorable and relatable, fostering a deeper connection with the material. You make Science: Motion - Measuring time and distance using standard units engaging. But the tools on the Socratic Garden can at least help you save time preparing courses and lessons. Try for free Tips for Making Science: Motion Relevant in Teaching Measuring Time and Distance 1. Connect Concepts to Real-World Examples Use Everyday Scenarios: Relate measuring time and distance to daily activities, like jogging, commuting, or sports. Discuss Current Events: Mention scientific projects, such as space exploration, where precise measurements of distance and time are crucial. 2. Incorporate Technology Utilize Apps and Devices: Introduce tools like GPS, apps for measuring distance, and stopwatches, so students can see the practical applications of measuring motion. Interactive Simulations: Use online simulations where students can experiment with different speeds and distances, observing real-time results. 3. Implement Hands-On Activities Perform Experiments: Conduct experiments like timing a rolling ball or measuring a distance on a track to gather hands-on experience. Group Challenges: Have students measure the time and distance of various activities, like a relay race or a “throw and catch” game. 4. Foster Critical Thinking Pose Real-World Problems: Challenge students to solve problems involving time and distance that they might encounter in careers, like engineering or transportation. Analyze Data: Have students collect data from experiments and make sense of it through analysis, encouraging them to draw conclusions about motion. 5. Use Storytelling Share Historical Anecdotes: Discuss how scientists like Galileo or Einstein changed our understanding of motion and measurement. Create Challenges with a Narrative: Present measurement-based problems within a story context, making it more engaging and relatable. 6. Integrate Interdisciplinary Learning Tie in Mathematics: Show the relationship between geometry and measuring distance, or use formulas that require math skills. Combine with Physical Education: Organize physical activities where students actively measure and report distances or times, merging fun with learning. 7. Showcase Career Relevance Introduce Various Professions: Discuss how fields such as astrophysics, sports science, and civil engineering utilize measurement of time and distance. Guest Speakers or Field Trips: Invite professionals to speak about their work or organize visits to places like science centers or sports facilities where measurement is key. 8. Encourage Student Engagement Create Interactive Discussions: Foster an environment where students feel comfortable discussing their ideas and experiences related to motion. Use Questions Effectively: Implement open-ended questions that prompt students to think critically about their measurements and what they mean. 9. Relate to Environmental Awareness Discuss Sustainability: Talk about how measuring distance and time can relate to issues like climate change, pollution tracking, or urban planning. Explore Physics in Nature: Analyze motion in nature by studying the movement of animals or plants, such as understanding growth patterns over time. 10. Leverage Visual Aids Utilize Charts and Graphs: Teach students how to represent their findings graphically, making it easier to comprehend relationships between distance and time. Use Movies and Videos: Show clips that depict motion in nature plus real-world examples of time and distance being measured, enhancing visual learning. By implementing these strategies, instructors can create an engaging and relevant learning environment that impacts students' understanding of motion, time, and distance. Make your lessons on Science: Motion - Measuring time and distance using standard units relevant with ease. 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190011	https://journals.lww.com/md-journal/fulltext/2021/04160/serious_bronchospasm_induced_by_cisatracurium.66.aspx	Medicine Log in or Register Get new issue alerts Get alerts;;) Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation RegisterLogin Browsing History Articles & Issues Current Release Previous Release Browse by Specialty Channels Breast Cancer Gastroenterology and Hepatology HIV/AIDS Infectious Diseases Oncology Radiology Surgery Cardiovascular For Authors Information for Authors Criteria For Acceptance Pre-Submission Checklist Author Services Medicine FAQ Author Advice Gallery Choosing Medicine For Reviewers Guidelines and Information CME For Reviewers Peer Review Training Journal Info About the Journal Editorial Board - Medicine Rights and Permissions Submit Collections High-impact research Articles Advanced Search April 16, 2021 - Volume 100 - Issue 15 Previous Article Next Article Outline 1 Introduction 2 Case presentation 3 Discussion 4 Conclusion Author contributions References Download PDF EPUB Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection Research Article: Clinical Case Report Serious bronchospasm induced by cisatracurium besylate A case report Wang, Ning MMed; Zhang, Yaozhong MMed; Hu, Yu MMed; Yang, Qiyao MMed; Su, Zhenbo PhD∗ Editor(s): Saranathan., Maya Author Information Department of Anesthesiology, China-Japan Union Hospital of Jilin University, Changchun, China. ∗Correspondence: Zhenbo Su, Department of Anesthesiology, China-Japan Union Hospital of Jilin University, No. 126, Xiantai Rd., Changchun 130000, China (e-mail: suzb@jlu.edu.cn). Abbreviations: ABP = arterial blood pressure, ASA = American Society of Anesthesiologists, BMI = body mass index, ETCO 2 = end tidal CO 2, FiO 2 = fraction inspired O 2, HR = heart rate, PaCO 2 = arterial partial pressure of carbon dioxide, PaO 2 = arterial partial pressure of oxygen, Paw = airway pressure, SpO 2 = peripheral oxygen saturation, V t = tidal volume. How to cite this article: Wang N, Zhang Y, Hu Y, Yang Q, Su Z. Serious bronchospasm induced by cisatracurium besylate: a case report. Medicine. 2021;100:15(e25516). Disclosure: Written informed consent was obtained from the patient for publication of this case report. No additional ethics approval was required, since the study did not comprise a clinical trial, nor use any unapproved medications. The manuscript is approved by all authors for publication. The authors have no conflicts of interest to disclose. Data sharing not applicable to this article as no datasets were generated or analyzed during the current study. This is an open access article distributed under the Creative Commons Attribution License 4.0 (CCBY), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Medicine 100(15):p e25516, April 16, 2021. \| DOI: 10.1097/MD.0000000000025516 Open Abstract Rationale: Cis-atracurium as an intermediate-acting non-depolarizing neuromuscular blocker is widely used clinically with less causing cyclic fluctuations and less histamine release. As the use rate increases, allergic reactions and anaphylactoid reactions caused by cis-atracurium increase. Patient concerns: A 23-year-old woman underwent laparoscopic bariatric surgery. Airway spasm occurred after anesthesia induction and the operation was suspended. After adjustment, the anesthesia was performed with the same anesthetic scheme again. After induction, skin flushing and airway resistance increased, then the symptoms were relieved. When the cis-atracurium was given again, the symptoms of airway spasm reappeared immediately, and after communicating with the family, the operation was successfully completed with rocuronium. Diagnoses: Serious bronchospasm induced by cisatracurium besylate. Interventions: The patient was undergone assisted ventilation with continuous positive airway pressure (CPAP) and aminophylline 250 mg, methylprednisolone 80 mg were given intravenously. Outcomes: There was no any obvious discomfort in the patient's self-report during the next day's visit. The patient was discharged 7 days later. No abnormalities were observed during following 4 weeks. Lessons: Although the anaphylactoid reactions caused by cis-atracurium are rare, the bronchospasm and anaphylactic shock caused by it greatly increase the risk of anesthesia, which should be taken seriously by clinicians. Increased vigilance in diagnosis, and treatment are essential to prevent aggravation and further complication. 1 Introduction Allergic reactions caused by anesthetic drugs are very rare. Allergens that are more common during perioperative period are allergic to artificial colloids, antibiotics, and implants. A 2-year study in France (January 1, 1997–December 31, 1998) showed that the most common allergen causing anaphylaxis during anesthesia was neuromuscular blockers (69.2%) and latex (12.1%). The 6th National Audit Project (NAP6) on perioperative anaphylaxis in Britain showed that antibiotics were the main allergens causing perioperative anaphylaxis, accounting for 48.9%, followed by neuromuscular blockers, accounting for 33.8%. Succinylcholine (60.6%) was the most frequently incriminated NMBA, whereas the low frequency of reactions involving cis-atracurium was confirmed (5.9%) when market shares of each NMBA were taken into account. We report a case of an anaphylactoid reaction following induction of anesthesia with cis-atracurium. We introduced the mechanism of anaphylactic reaction and anaphylactoid reaction in detail, with how to diagnose them. 2 Case presentation The patient, woman, 23 years old, height 170 cm, weight 137 kg, body mass index (BMI) 47.4 kg/m 2, entered our hospital on November 7, 2018 to perform laparoscopic bariatric surgery under general anesthesia. She was evaluated class III risk using the American Society of Anesthesiologists (ASA) criteria and did not have respiratory, circulatory diseases, liver, kidney, or other organ dysfunction. She had a history of penicillin allergy, appendectomy under local anesthesia, and cesarean section under spinal anesthesia. She has smoked for 7 years, 7 cigarettes/d, and has stopped smoking for 4 days before the operation. With preoperative prohibition of drinking for 6 hours, fasting for 12 hours, she has been monitored routinely and has been measured radial artery pressure when entered the operating room at 8:30 am on November 12, 2018. We provided the patient with 8 L/min high-flow mask oxygen. The arterial blood pressure (ABP) was 158/76 mm Hg. The heart rate (HR) was 78 beats/min. The peripheral oxygen saturation (SpO 2) measured using pulse oximetry was 99%. The results of arterial blood gas analysis were as follows: arterial partial pressure of carbon dioxide (PaCO 2), 44.3 mm Hg, arterial partial pressure of oxygen (PaO 2), 260 mm Hg; and arterial oxygen saturation (SaO 2) 100%. After 10 minutes of oxygen inhalation, anesthesia was induced with intravenous (IV) penehyclidine 1 mg (Chengdu Lisite Pharmaceutical Co., Ltd., Jinjiang Industrial Development Zone, Chengdu, H200220606, 180703), prednisolone 20 mg (Jiangxi Sinopharm Co., Ltd., No.888, Guoyao Avenue, Xiaolan Industrial Park, Nanchang City, Jiangxi Province, H36022365, 19031022), midazolam 10 mg (Jiangsu Enhua Pharmaceutical Co., Ltd., No.18, Yangshan Road, Xuzhou Economic Development Zone, H10980025, 20181203), propofol 150 mg (Sichuan Guorui Pharmaceutical Co., Ltd., North section of Fenghuang Road, Yujin Town, Qianwei County, Leshan City, H200330115, 1905161), cis-atracurium 40 mg (Jiangsu Hengrui Pharmaceutical Co., Ltd., No. 38 the Yellow River Road, Lianyungang economic and Technological Development Zone, H20060869 , 18072721), and Sufentanil 20 μg (Yichang Renfu Pharmaceutical Co., Ltd., No. 19 Dalian Road, Yichang Development Zone, Hubei, H20054171, 91A03051A1). After tracheal intubation, breathing sounds were not heard in both lungs, and end tidal CO 2 (ETCO 2) can be seen continuously waveform. Auscultating again, we could hear the weak breath of both lungs, and a lot of dry and wet rales, then we used negative pressure to suck sputum with only a small amount of clear secretions. A lot of dry and wet rales could still been heard at this time. With hand-controlled ventilation, airway pressure (Paw) was 40 cmH 2 O, and tidal volume (V t) was 350 mL. At this point PaCO 2 was 50.6 mmHg, PaO 2 81 mm Hg, SaO 2 was 95%, SpO 2 was 95%, and fraction inspired O 2(FiO 2) was 90%. We considered bronchospasm, followed by intravenous injection of aminophylline 250 mg, methylprednisolone 80 mg. After 15 minutes, the results of arterial blood gas analysis were as follows: PaCO 2 was 46.3 mm Hg, PaO 2 was 80 mm Hg, SaO 2 was 95%, SpO 2 was 95%, and FiO 2 was 90%. Hemodynamics showed no significant fluctuation and respiratory compliance improved. After 10 minutes, PaCO 2 was 40.6 mm Hg, PaO 2 was 103 mm Hg, SaO 2 was 98%, SpO 2 was 96%, and FiO 2 was 90%. Pressure controlled ventilation was used after intubation, Paw was 38 cmH 2 O, V t was 350 mL, respiratory rate (RR) was 12 breath/min, positive end-expiratory pressure (PEEP) was 5 cmH 2 O, and slope was 1.0 seconds. Please consult the Department of Respiratory Medicine. The consultation opinion indicates that the patient has a long history of smoking, and the airway sensitivity is likely to cause airway spasm after endotracheal intubation. We informed the patient's family related risks fully and obtained the consent of the patient's family to suspend the operation. After 1 hour, Paw was 35 cmH 2 O, V t was 450 mL, RR was 12 breath/min, PEEP was 5 cmH 2 O, slope was 1.0 seconds, and respiratory compliance was significantly improved. Return to intensive care unit safely under moderate sedation. Respiratory function recovered completely after 2 hours. We removed the tracheal tube. At this time, PaCO 2 was 40.1 mmHg, PaO 2 was 96 mm Hg, SaO 2 was 98%, SpO 2 was 96%, FiO 2 was 40%. The next day, the patient reported tracheal burning sensation, which was considered by the pressure injury of the airway. After preparation by atomization and oxygen therapy, the operation was performed again on November 21, 2018. Intramuscular injection of atropine 1 mg, intravenous drip of aminophylline 250 mg, and methylprednisolone 80 mg to prevent airway spasm. At this time, PaCO 2 was 44.2 mmHg, PaO 2 was 243 mmHg, SaO 2 was 100%, SpO 2 was 99%, FiO 2 was 50%. The drug and dose for anesthesia induction were the same as the first time, and the Hand-controlled ventilation was given after the spontaneous breathing disappears. During ventilation, airway resistance was high, skin of anterior chest and neck was flushed, and there was no obvious hemodynamic fluctuation. After tracheal intubation, anesthesia was maintained with sevoflurane in 50% oxygen in air. A large number of dry and wet rales could been heard in both lungs during auscultation. The airway resistance increased significantly. After 5 minutes, the symptoms of skin flushing gradually disappeared. At this moment, PaCO 2 was 41.6 mmHg, PaO 2 was 98 mmHg, SaO 2 was 98%, SPO 2 was 98%, FiO 2 was 90%. Then, the pressure ventilation mode was given, P plat was 28 cmH 2 O, RR was 12 breath/min, PEEP was 7 cmH 2 O, slope was 1.0 seconds, V t was 522 mL. After 1 hour, the rales gradually disappeared and the breath sound was clear. Re-examination of arterial blood gas analysis were as follows: PaCO 2 was 49.5 mm Hg, PaO 2 was 135 mm Hg, SaO 2 was 99%, SpO 2 was 99%, FiO 2 was 90%. At this time, the patient's breathing gradually recovered, and 5 mg of cis-atracurium was given again. The airway pressure increased immediately. Auscultation of the lungs again showed a large amount of dry rales, considering anaphylactoid reaction caused by cis-atracurium which induced the bronchial spasm. Respiratory consultation pointed out that under general anesthesia tracheal intubation, the left lung breath sounds weak, the right lower lung was covered with biphasic high-pitched dry sounds. Symptoms such as skin flushing and bronchospasm appeared immediately after induction of this patient, and cis-atracurium was considered to cause anaphylactoid reaction. When the symptoms were gradually relieved, we gave cisatracurium again with bronchospasm symptoms immediately developed, while bronchospasm did not occur again when rocuronium was given. It was equivalent to carry out drug provocation test without knowing it. Drug provocation test is the golden standard for finding allergens, so it was determined that the bronchospasm was caused by cis-atracurium. Considering airway sensitivity, airway spasm (not excluding drug allergy) under high-sensitivity state was at high risk of operation. After actively communicating with the patient's family and explaining the associated risks, the patient's family indicated that the risk was approved for surgery. Switch to the muscle relaxant rocuronium bromide (Eccorson) 20 mg (N.V.ORGANON, Molenstraat 110 P.O.BOX 20 OSS 5340 BH Netherlands, H20140847, R006942), no other allergic reactions such as skin flushing. After that, surgery was started. During the operation, rocuronium 20 to 30 mg/h and sufentanil 5 to 10 μg/h were added on time, and a total of 2000 mL Ringer lactated solution was infused. No allergic reaction occurred, no obvious hemodynamic fluctuation was observed, and airway resistance gradually reduced. After 4 hours, the operation was completed. At this time, PaCO 2 was 56.2 mmHg, PaO 2 was 138 mmHg, SaO 2 was 99%, SpO 2 was 99%, FiO 2 was 90%. Methylprednisolone 80 mg and sufentanil 50 μg were given again for the transition of respiratory function recovery. After the operation, the patient returned to the intensive care unit. After 2 hours, the tube was removed and the results of arterial blood gas analysis were as follows: PaCO 2 was 52.1 mm Hg, PaO 2 was 103 mm Hg, SaO 2 was 97%, SpO 2 was 98%, FiO 2 was 30%. There was no any obvious discomfort in the patient's self-report during the next day's visit, rales completely disappeared. The patient was discharged 7 days later. No abnormalities were observed during following 4 weeks. 3 Discussion The clinical manifestations of perioperative allergic reactions and anaphylactoid reactions are related to the amount of histamine releasing from mast cells, usually involving multiple systems and multiple organs, which can be life-threatening in severe cases. Hypersensitivity reactions occurring during anesthesia remain a major cause of concern for anesthesiologists. Neuromuscular blockers are the major allergens that cause allergic reactions and anaphylactoid reactions during anesthesia. Cis-atracurium can cause an allergic reaction as an antigen, and can also cause an anaphylactoid reaction by histamine release. The patient was not exposed to latex and antibiotics before the allergy occurred, so latex and antibiotic allergies were not considered. Since it was an allergic reaction that occurrence after induction, it was necessary to mainly consider the allergy of the inducing drug. The patient had undergone painless gastroscopy which used propofol and sufentanil before admission. During the operation, propofol and sufentanil were added regularly. After the operation, 50 μg of sufentanil was given again for transitional extubation. No allergic reactions occurred, and it was unlikely that allergic reactions to propofol and sufentanil are considered. Midazolam is a safe drug because it does not produce active metabolites that are thought to cause cross-allergic reactions with other benzodiazepines. A total of 2 cases of allergic reactions caused by intravenous injection have been reported worldwide,[5,6] an anaphylactoid reaction caused by intravenous injection, 1 case of allergic reaction caused by intramuscular injection. There are no reports of allergies in penehyclidine and prednisolone. Perioperative allergic reactions are type I allergic reactions. Allergens first enter the body to produce high levels of immunoglobulin E through T cells and B cells, and then bind to mast cells and basophils surface receptors to make the body sensitized. When the body contacts the same allergen again, it will produce degranulation of mast cells and basophils, and rapid release of a large amount of histamine, bradykinin, prostaglandin, and other active media causes a series of clinical symptoms. Clinically applied neuromuscular blockers have quaternary ammonium groups, and some foods, drugs, and cosmetics also contain similar groups. When entering the human body, it can presensitize mast cells and basophils. When cis-atracurium enters the body, allergic reactions can occur and there is a risk of cross-allergy. According to reports, incidence of allergic reactions are as follows: Rocuronium 8.0/100,000, vecuronium 2.8 /100,000, atracurium 4.01 /100,000; and in patients with Rocuronium allergy, the incidence of cross-allergy to other muscle relaxants is: succinylcholine (44%) and vecuronium (40%), while pancuronium 19% and atracurium 20%, cisatracurium 5%. Therefore, when a neuromuscular blocker is allergic, be careful to use other types of neuromuscular blockers or alternative anesthesia without nerve blockers. Goikoetxea et al have also reported that anaphylactic shock had occurrence immediately after induction with cis-atracurium in patients underwent radical resection of rectal cancer under general anesthesia. One week later, skin tests and basophilic activated granulocyte experiments confirmed that the allergic reaction was caused by cis-atracurium. Recent studies have shown that hairdressers who are frequently exposed to quaternary ammonium ion compounds have a significant increase in allergic reactions to neuromuscular blockers, so repeated exposure to quaternary ammonium-containing compounds is also one of the sensitizing risk factors for neuromuscular blockers. Most of the perioperative anaphylactoid reactions induced by neuromuscular blockers are caused by non-immune mechanisms that directly release the medium from mast cells and basophils, or directly activate complements, leading to systemic histamine release. Clinical symptoms are similar to anaphylactic reactions. Recent studies by Che et al have shown that cis-atracurium triggers LAD2 cell degranulation and release of inflammatory precursors through G protein-coupled receptor MRGPRX2 to produce anaphylactoid reactions. It is known that most neuromuscular blockers cause non-specific histamine release from mast cells, and benzylisoquinolines are more likely to cause histamine release than aminoguanidines. If anaphylactoid reactions occurs, we can use a neuromuscular blocker with a low rate of histamine release to replace the neuromuscular blocker with a higher rate of histamine release. In this case, no anaphylactoid reactions occurred after replacing the benzylisoquinoline cis-atracurium with aminoguanidine rocuronium. It was confirmed from the side that anaphylactoid reactions occurred in this case rather than allergic reactions. Treatment and diagnosis of allergic reaction: when suspected allergic reactions in patients, all suspected sensitizing drugs should be stopped immediately, anti-allergy treatment should be carried out. Meanwhile, blood samples should be taken for serum tryptase detection, because serum tryptase increases with the activation of mast cells, which is the only useful blood test during an acute allergic reaction. Although it can confirm the occurrence of anaphylaxis, it is not helpful for the detection of allergens. In order to determine whether an immune-mediated allergic reaction occurs, a prick test can be performed 4 to 6 weeks after the occurrence of allergic reaction. Because a large amount of inflammatory factors such as histamine will be consumed in the event of an allergic reaction, a false negative result may occur in an immediate prick test. However, some scholars believe that it is meaningful to do the prick experiment immediately. The intradermal test was performed when the prick test was negative. All suspected drugs should be tested during prick test and intradermal test, and all drugs of the same type should be tested to prevent cross-allergic reactions between drugs, and non-allergic drugs can be selected for emergencies. Drug provocation test is the golden standard for finding allergens, but drug provocation test is risky and generally not recommended. No matter what kind of test is carried out, it should be carried out in the professional allergy and immunity department, and adrenaline and other rescue drugs should be prepared to avoid unforeseen events. Diagnosis of anaphylactoid reactions: The diagnosis of anaphylactoid reactions is difficult, and there is no good laboratory diagnostic method. At present, there is no recognized standard model of anaphylactoid detection at home and abroad. Anaphylactoid reactions are generally an exclusive diagnosis. Histamine, β-hexosaminidase, and tryptase are common biomarkers for anaphylactoid reactions and anaphylactic reactions. When anaphylaxis occurs, the specific immunoglobulin E increases correspondingly except for the above biomarkers, while the level of immunoglobulin E does not increase in anaphylactoid reaction. Inquiry about medical history: There is a sensitization process in the history of anaphylactic reactions, whereas anaphylactoid reactions have no sensitization process. It is difficult to distinguish allergic reactions from anaphylactoid reactions in clinic. Under uncertain circumstances, other types of neuromuscular blockers cannot be easily substituted for each other. In this case, we inferred that anaphylactoid reaction occurred in the patient by inquiring about the medical history, and no anaphylactoid reactions occurred in this patient who used aminosterol rocuronium instead of benzylisoquinoline cis-atracurium during the operation, which confirmed our diagnosis. However, blood samples were not preserved during the operation, and patients refused further allergen diagnostic tests after surgery. We still lack laboratory evidence. 4 Conclusion In conclusion, although the anaphylactoid reactions caused by cis-atracurium are rare, the bronchospasm and anaphylactic shock caused by it greatly increase the risk of anesthesia, which should be taken seriously by clinicians. Once an allergic reaction or anaphylactoid reaction is suspected, the anesthesiologist should record the real-time relevant information in detail and immediately save the serum samples of the allergic person for subsequent testing and diagnosis. Increased vigilance in diagnosis, and treatment are essential to prevent aggravation and further complication. Author contributions Conceptualization: Yaozhong Zhang. Data curation: Qiyao Yang. Formal analysis: Yu Hu. Investigation: Yu Hu. Methodology: Ning Wang. Project administration: Yaozhong Zhang. Supervision: Qiyao Yang. Validation: Ning Wang. Writing – original draft: Ning Wang. Writing – review & editing: Zhenbo Su. References . Laxenaire MC, Mertes PM. Groupe D’etudes Des Reactions Anaphylactoides P. Anaphylaxis during anaesthesia. Results of a two-year survey in France. Br J Anaesth 2001;87:549–58. Cited Here \| Google Scholar . Harper NJN, Cook TM, Garcez T, et al. Anaesthesia, surgery, and life-threatening allergic reactions: epidemiology and clinical features of perioperative anaphylaxis in the 6th National Audit Project (NAP6). Br J Anaesth 2018;121:159–71. Cited Here \| Google Scholar . Dong SW, Mertes PM, Petitpain N, et al. Hypersensitivity reactions during anesthesia. Results from the ninth French survey (2005–2007). Minerva Anestesiol 2012;78:868–78. Cited Here \| Google Scholar . Lagopoulos V, Gigi E. Anaphylactic and anaphylactoid reactions during the perioperative period. Hippokratia 2011;15:138–40. Cited Here \| Google Scholar . Fujita Y, Ishikawa H, Yokota K. Anaphylactoid reaction to midazolam. Anesth Analg 1994;79:811–2. Cited Here \| Google Scholar . Shin JG, Hwang JH, Lee BS, et al. A case of midazolam anaphylaxis. Clin Endosc 2014;47:262–5. Cited Here \| Google Scholar . Ayuse T, Kurata S, Ayuse T. Anaphylactoid-like reaction to midazolam during oral and maxillofacial surgery. Anesth Prog 2015;62:64–5. Cited Here \| Google Scholar . Kim KN, Kim DW, Sin YH, et al. Anaphylactic shock caused by intramuscular injection of midazolam during the perioperative period: a case report. Korean J Anesthesiol 2016;69:510–3. Cited Here \| Google Scholar . Mertes PM, Aimone-Gastin I, Gueant-Rodriguez RM, et al. Hypersensitivity reactions to neuromuscular blocking agents. Curr Pharm Des 2008;14:2809–25. Cited Here \| Google Scholar . Sadleir PH, Clarke RC, Bunning DL, et al. Anaphylaxis to neuromuscular blocking drugs: incidence and cross-reactivity in Western Australia from 2002 to 2011. Br J Anaesth 2013;110:981–7. Cited Here \| Google Scholar . Takazawa T, Mitsuhata H, Mertes PM. Sugammadex and rocuronium-induced anaphylaxis. J Anesth 2016;30:290–7. Cited Here \| Google Scholar . Goikoetxea MJ, Lafuente A, De La Borbolla JM, et al. Early diagnosis of an allergic reaction to cisatracurium. J Investig Allergol Clin Immunol 2013;23:69–71. Cited Here \| Google Scholar . Dong S, Acouetey DS, Gueant-Rodriguez RM, et al. Prevalence of IgE against neuromuscular blocking agents in hairdressers and bakers. Clin Exp Allergy 2013;43:1256–62. Cited Here \| Google Scholar . Che D, Rui L, Cao J, et al. Cisatracurium induces mast cell activation and pseudo-allergic reactions via MRGPRX2. Int Immunopharmacol 2018;62:244–50. Cited Here \| Google Scholar . Claudius C, Garvey LH, Viby-Mogensen J. The undesirable effects of neuromuscular blocking drugs. Anaesthesia 2009;64: (suppl): 10–21. Cited Here \| Google Scholar . Garvey LH, Ebo DG, Kroigaard M, et al. The use of drug provocation testing in the investigation of suspected immediate perioperative allergic reactions: current status. Br J Anaesth 2019;123:e126–34. Cited Here \| Google Scholar View full references list Keywords: anaphylactoid reaction; anesthesia; bronchospasm; cis-atracurium Copyright © 2021 the Author(s). Published by Wolters Kluwer Health, Inc. View full article text Source Serious bronchospasm induced by cisatracurium besylate: A case report Medicine100(15):e25516, April 16, 2021. Full-Size Email Favorites Export View in Gallery Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. 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190012	https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-linear-functions-modeling/e/linear-equations-word-problems	Khan Academy \| 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 Search DonateLog inSign up Search for courses, skills, and videos Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! Site Navigation About News Impact Our team Our interns Our content specialists Our leadership Our supporters Our contributors Our finances Careers Internships Cookie Preferences Contact Help center Support community Share your story Press Download our apps Courses Language en CountryU.S.IndiaMexicoBrazil © 2025 Khan Academy Terms of use Privacy Policy Cookie Notice Accessibility Statement
190013	https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/11%3A_Solutions/11.3%3A_The_Dissolution_Process	Skip to main content 11.3: The Dissolution Process Last updated : May 19, 2019 Save as PDF 11.2: Concentration 11.4: Properties of Solutions Page ID : 15488 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To describe the dissolution process at the molecular level The Dissolution Process What occurs at the molecular level to cause a solute to dissolve in a solvent? The answer depends in part on the solute, but there are some similarities common to all solutes. Recall the rule that like dissolves like. This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called solvation and is illustrated in Figure 11.3.1. When the solvent is water, the word hydration, rather than solvation, is used. Ionic Compounds and Covalent Compounds as Solutes In the case of molecular solutes like glucose, the solute particles are individual molecules. However, if the solute is ionic, the individual ions separate from each other and become surrounded by solvent particles. That is, the cations and anions of an ionic solute separate when the solute dissolves. This process is referred to as dissociation (Figure 11.3.1). The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called strong electrolytes. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called weak electrolytes. Acetic acid (CH3COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Table sugar (C12H22O11) is an example of a nonelectrolyte. The term electrolyte is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na+, K+, Ca2+, Mg2+, and Cl−. Example 11.3.1 The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. potassium chloride (KCl) fructose (C6H12O6) isopropyl alcohol [CH3CH(OH)CH3] magnesium hydroxide [Mg(OH)2] Solution Each substance can be classified as an ionic solute or a nonionic solute. Ionic solutes are electrolytes, and nonionic solutes are nonelectrolytes. Potassium chloride is an ionic compound; therefore, when it dissolves, its ions separate, making it an electrolyte. Fructose is a sugar similar to glucose. (In fact, it has the same molecular formula as glucose.) Because it is a molecular compound, we expect it to be a nonelectrolyte. Isopropyl alcohol is an organic molecule containing the alcohol functional group. The bonding in the compound is all covalent, so when isopropyl alcohol dissolves, it separates into individual molecules but not ions. Thus, it is a nonelectrolyte Magnesium hydroxide is an ionic compound, so when it dissolves it dissociates. Thus, magnesium hydroxide is an electrolyte. Exercise 11.3.1 The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. acetone (CH3COCH3) iron(III) nitrate [Fe(NO3)3] elemental bromine (Br2) sodium hydroxide (NaOH) Answer : a. nonelectrolyte b. electrolyte c. nonelectrolyte d. electrolyte Electrolytes in Body Fluids Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the river of life, because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life. The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called macrominerals. Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities. When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is given when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, MD, assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry. Sports drinks are designed to rehydrate the body after excessive fluid depletion. Electrolytes in particular promote normal rehydration to prevent fatigue during physical exertion. Are they a good choice for achieving the recommended fluid intake? Are they performance and endurance enhancers like they claim? Who should drink them? Typically, eight ounces of a sports drink provides between fifty and eighty calories and 14 to 17 grams of carbohydrate, mostly in the form of simple sugars. Sodium and potassium are the most commonly included electrolytes in sports drinks, with the levels of these in sports drinks being highly variable. The American College of Sports Medicine says a sports drink should contain 125 milligrams of sodium per 8 ounces as it is helpful in replenishing some of the sodium lost in sweat and promotes fluid uptake in the small intestine, improving hydration. Gatorade In the summer of 1965, the assistant football coach of the University of Florida Gators requested scientists affiliated with the university study why the withering heat of Florida caused so many heat-related illnesses in football players and provide a solution to increase athletic performance and recovery post-training or game. The discovery was that inadequate replenishment of fluids, carbohydrates, and electrolytes was the reason for the “wilting” of their football players. Based on their research, the scientists concocted a drink for the football players containing water, carbohydrates, and electrolytes and called it “Gatorade.” In the next football season the Gators were nine and two and won the Orange Bowl. The Gators’ success launched the sports-drink industry, which is now a multibillion-dollar industry that is still dominated by Gatorade. Key Takeaway When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. 11.2: Concentration 11.4: Properties of Solutions
190014	https://www.facingourrisk.org/info/risk-management-and-treatment/cancer-treatment/by-cancer-type/endometrial/inherited-mutations	Get notified of page updates Risk Management & Treatment / Cancer Treatment / By Cancer Type / Endometrial / Inherited Mutations Endometrial Cancer: Genetic Testing for Inherited Mutations Learn about genetic testing and the genes associated with hereditary endometrial cancer. SIGN UP NOW Explore More Pages via Quick Links Overview Endometrial Cancer: Stages and Subtypes Endometrial Cancer: Standard Therapy Endometrial Cancer: Biomarker Testing Endometrial Cancer: Targeted and Immunotherapies Endometrial Cancer: Genetic Testing for Inherited Mutations Glossary on off Genetic Testing for People Diagnosed with Endometrial Cancer This section covers the following information about genetic testing in people diagnosed with endometrial cancer: Guidelines for genetic testing after tumor testing Additional guidelines for people with endometrial cancer Genes linked to hereditary endometrial cancer People with endometrial cancer should ask about genetic testing Endometrial cancer survivors and those in treatment should speak with a genetics expert to decide if genetic testing is right for them. About 10% of people diagnosed with endometrial cancer have an that caused their cancer. Genetic test results may affect your cancer treatment and prognosis. Some genetic test results can help guide treatment decisions. People with endometrial cancer who test positive for an may benefit from targeted or to treat their cancer. Genetic test results may help you and your relatives understand the risk for cancer and take action to stay healthy. If you test positive for an , you may have an increased risk for other cancers. Expert guidelines outline cancer screening and prevention options for people at high risk for cancer due to an . Guidelines for tumor testing followed by genetic testing Experts recommend tumor testing all endometrial cancers to look for biomarkers known as "" () or microsatellite instability-High () at the time of diagnosis. These two related abnormalities are commonly found in the cancers of people with . People with dMMR/MSI-High cancers are recommended to have genetic counseling and testing for . dMMR/MSI-High cancers can occur in people who do not have . Not all people with cancers will test positive for a mutation. People with inherited mutations who develop endometrial may have cancers that do not have these biomarkers. People with cancers that are not may still benefit from genetic counseling and testing. Cancers with these biomarkers tend to respond well to treatment with a certain type of treatment known as . Additional guidelines for genetic testing in people diagnosed with endometrial cancer Genetic counseling and testing for an is also recommended for people diagnosed with endometrial cancer who have any of the following: a blood relative with a known in a cancer gene. diagnosed before age 50. diagnosed with another type of cancer seen in people with , including colorectal, ovarian, pancreatic, stomach or other Lynch syndrome-related cancer. have a close relative with a Lynch syndrome-related cancer diagnosed under age 50. have two or more relatives with a Lynch syndrome-related cancer diagnosed at any age. Other endometrial cancer survivors who may benefit from genetic counseling and testing People with endometrial cancer may benefit from additional genetic counseling and expanded genetic testing if they had genetic testing in the past, tested negative, and: their situation matches any of the other guidelines above, and they had a test that only looked for one or a few genes, or they had genetic testing before 2014. Genetic testing has improved, and laboratories can now find gene mutations that may have previously been missed. All endometrial cancer survivors and those in treatment should speak with a genetics expert to decide if genetic testing is right for them. Relatives of people with endometrial cancer may also benefit from testing. Genetic counseling and testing is also recommended for anyone with the following family history of endometrial cancer: one or more 1st-degree relatives who has been diagnosed with endometrial cancer before age 50. one or more 1st-degree relatives wo have been diagnosed with endometrial cancer and a second cancer related to (e.g., colon cancer). two or more 1st- or 2nd- degree relatives who have been diagnosed with a cancer related to with at least one of the cancers diagnosed before age 50. three or more 1st- or 2nd- degree relatives who have been diagnosed with a cancer related to at any age. Genes linked to endometrial cancer Mutations in the following genes have been linked to increased risk for endometrial cancer. \| \| \| \| --- \| Syndrome (if applicable) \| \| \| \| \| , , , , \| is the most common inherited syndrome linked to endometrial cancer. People with who develop endometrial cancer are most likely to have tumors that are MSI-H or . People with have an increased risk for other types of cancers, too. \| \| Hamartoma Tumor syndrome \| \| People with Hamartoma Tumor syndrome have an increased risk for other types of cancers. \| \| \| \| Peutz-Jegher syndrome have an increased risk for other types of cancer. \| \| Hereditary breast and ovarian cancer syndrome () \| , \| and have been linked to a rare but aggressive form of endometrial cancer. The increase in risk for these cancers is very small. Researchers are continuing to study this link. \| More Resources Get Support or Navigation Find Information on a Gene Find an Expert or Service Find Financial Resources Last updated July 18, 2024 Knowledge Can Save Lives Receive FORCE Newsletters Stay up to date on hereditary cancer information. Knowledge Can Save LivesStay up to date on hereditary cancer information.
190015	https://topblogtenz.com/combined-gas-law-equation-p1-v1-t1-p2-v2-t2/	Skip to content Published By Ammara Waheed \| Last updated: June 12, 2025 Home » Chemistry » P1V1/T1=P2V2/T2, Equation, Examples, Units, Where to Use Chemistry / By Combined gas law equation; what is it, where to use P1V1/T1 = P2V2/T2 Page Contents Which three laws form the basis of the combined gas law? What does P1V1/T1 = P2V2/T2 represent? Where and how to use P1V1/T1 = P2V2/T2? – Examples More examples on P1V1/T1 = P2V2/T2 FAQ P1V1/T1 = P2V2/T2 represents the combined gas law, a mathematical relationship used to study the behavior of gases under changing temperature, pressure, and volume conditions. It is referred to as the combined gas law because it is derived by the combination of three different gas laws into a single equation. In this article, we will teach you what P1V1/T1 = P2V2/T2 represents and how to use it to solve numerous physics, chemistry, and mathematical problems. But first, let us discuss some background related to the combined gas law. Which three laws form the basis of the combined gas law? The combined gas law is a combination of: Charles law: It says that the volume of a gas is directly proportional to its temperature if the pressure is kept constant. V α T V = kT…..Equation (i) Where V = volume, T= temperature, k= proportionality constant Boyle’s law: It states that the pressure exerted by a gas is inversely proportional to its volume. P α 1/V P = k/V….Equation (ii) Where P = pressure, V=volume, k= constant Gay-Lussac’s law: It says that the pressure exerted by a gas is directly proportional to the temperature provided to it. P = kT……Equation (iii) Where P = pressure, T = temperature, k= constant Combining equations (i), (ii), and (iii) gives us PV/T = k, which implies that pressure and volume are inversely related to each other while both are in direct relationship with the temperature of the gas. Increasing the temperature of a gas in a container increases its volume, which in turn decreases the pressure so that the value of k stays unchanged. Therefore, at a time, the effect of only two variables can be investigated, keeping the third variable constant. What does P1V1/T1 = P2V2/T2 represent? For a gaseous system undergoing some change, P1V1/T1 = P2V2/T2 represents: P1 = Initial pressure of a gaseous system (Units = Pa, atm or mm Hg) V1 = Initial volume of gas (Units = L or m3) T1 = Initial temperature (Units = K) P2 = Final pressure of the gaseous system (Units = Pa, atm or mm Hg) V2 = Final volume of the gas (Units = L or m3) T2 = Final temperature (Units = K) Now let us practice some examples using P1V1/T1 = P1V1/T2. Where and how to use P1V1/T1 = P2V2/T2? – Examples You must note that to use this formula, five variables should ideally be known so that we can find the unknown variable, whether be it pressure, temperature, or volume. Hence, the formula P1V1/T1 = P2V2/T2 can be rearranged in the following six ways, depending upon which variable is unknown. Another important point is that consistency in units is very important. The pressure, volume, and temperature must be in the same units on either side of P1V1/T1 = P2V2/T2. For example, A gas sample in a container exerts an initial pressure of 2.0 atm; the volume of the gas in the container is 3.0 L at 300 K. What is the final volume of the gas if the pressure is increased to 3.5 atm on increasing the temperature to 350 K? Solution As per the question statement; P1 = 2.0 atm V1 = 3.0 L T1 = 300 K P2 =3.5 atm V2 = ? T2 = 350 K We need to insert the above data in the combined gas equation and solve for V2 by cross multiplication, as shown below: 6(350) = 1050 V2 2100 = 1050 V2 ∴ V2 = 2100/1050 = 2 Result: The final volume of gas in the container is 2.0 L. Another example is- A gas at 110 kPa and 30°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised by 50°C, the volume decreases to 0.58 L. Calculate the final pressure in the gas container. Solution As per the question statement; P1 = 110 kPa V1 = 2.0 L T1 = 30°C P2 = ? V2 = 0.58 L T2 = 30 + 50 = 80 °C Remember to convert both T1 and T2 from °C to Kelvin (K). 1 K = 1°C + 273 T1 = 30 + 273 = 303 K T2 = 80 + 273 = 353 K Now let’s plug in all the above data into the equation and find P2: Cross multiply and make P2 the subject of the formula: 220(353) = P2 (175.74) 77660 = 175.74 P2 ∴ P2 = 77660/175.74 = 441.9 kPa Note: As P1 is given in kilopascals (kPa), therefore P2 is also obtained in kPa, respectively. Result: The final pressure in the gas container is 441.9 kPa. More examples on P1V1/T1 = P2V2/T2 \| \| \| If you collect a gas at 620 mmHg and 13°C. At the time of collection, it takes up a volume of 1.3 L. What will be the temperature of the gas in Kelvin if the pressure is increased to 1.2 atm and the volume is decreased to 0.85 L? \| \| As per the question statement; P1 = 620 mm Hg V1 = 1.3 L T1 = 13° C (Converting the temperature into Kelvin gives us 13 + 273) = 286 K P2 = 1.2 atm V2 = 0.85 L T2 =? To bring consistency in units, P2 is also converted from atm into mm Hg as follows: 1 atm = 760 mm Hg 1.2 atm = 1.2 x 760 = 912 mm Hg Now substitute all the above data into the combined gas law equation to find the unknown variable, i.e., T2. Via cross multiplication: 806 (T2) = 286(912 x 0.85) 806 (T2) = 221707.2 ∴ T2 = 221707.2/806 = 275.1 K Result: The final temperature of the gas is 275.1 K \| \| \| \| A gas initially occupies a volume of 2.5 L when its temperature is raised twice from 25°C while the pressure changes from 1.75 atm to 3 atm. Find the initial volume of the gas by applying the formula P1V1/T1 = P2V2/T2. \| \| As per the question statement; P1 = 1.75 atm V1 = ? T1 = 25° C (Converting the temperature into Kelvin gives us 25 + 273) = 298 K P2 = 3.0 atm V2 = 2.5 L T2 = 323 K (as the temperature is raised twice, so 2(25) = 50°C = 50 + 273 = 323 K). Now substitute all the above data into the combined gas law to find the unknown variable, i.e., V1. Via cross multiplication: 565.25 (V1) = 298(3.0)(2.5) 565.25 (V1) = 2235 ∴ V1 = 2235/565.25= 3.95 L Result: The initial volume occupied by the gas is 3.95 L. \| FAQ \| \| \| How is the combined gas law P1V1/T1 = P2V2/T2 derived? \| \| The combined gas law is derived from the combination of three different gas laws, i.e., Charles’ law, Boyles’s law, and Gay-Lussac’s law. \| \| \| \| What do the components of P1V1/T1 = P2V2/T2 represent? \| \| In P1V1/T1 = P2V2/T2, P1 = Initial pressure (in atm, Pa, or mm Hg) V1 = Initial volume (in L or m3) T1 = Initial temperature (in K) P2 = final pressure (in atm, Pa, or mm Hg) V2 = final volume (in L or m3) T2 = final temperature (in K) \| \| \| \| In the combined gas law P1V1/T1 = P2V2/T2, whose law is: a) V1/T1 = V2/T2 b) P1V1 = P2V2 c) P1/T1 = P2/T2 \| \| a) V1/T1 = V2/T2 represents Charles gas law which says volume is directly proportional to temperature if the pressure is kept constant. b) P1V1 = P2V2 denotes Boyle’s law which says that the pressure and volume of a gas are inversely related to each other at a constant temperature. c) P1/T1 = P2/T2 represents Gay-Lussac’s law, according to which pressure and temperature are inversely proportional to each other at constant volume. \| \| \| \| Given the equation P1V1/T1 = P2V2/T2, which of the following is the formula for the final temperature? A) T2 = P2V2T1/P1V1 B) T2 = P1V2T1/P2V1 C) T2 = P2V1T1/P1V2 D) T2 = P1V1/P2V2T1 \| \| Option A is the correct answer. The equation can be rearranged to make T2 the subject of the formula, as shown below: ⇒ P1V1/T1 = P2V2/T2 Via cross multiplication: P1V1T2 = P2V2T1 ∴ T2 = P2V2T1/P1V1 \| Share it…0 Shares Related Posts V1/n1=V2/n2, Equation, Examples, Units, Where to Use Q=nRT ln(V2/V1), Derivation, Examples, Units, Equation, Uses S=klnW, Derivation, Examples, Units, Equation, How to use E=hc/lambda, Equation, How to solve, Units, Where to Use pKa + pKb = pKw, Equation, How to solve, Units, Where to Use pH = -log[H+], Equation, How to solve, Units, Where to Use 8.314, gas constant (R), Units, Where to use, Conversion 0.08206, Ideal gas constant (R), Units, Where to use C3H8O + O2 = CO2 + H2O, Balance the equation [Step by Step] C3H8 + O2 = CO2 + H2O, Balance the equation [Step by Step] C1V1 = C2V2, Equation, Examples, Where to use k = Ae^(-Ea/RT), Units, Equation, Problems, How to use? pH = pKa + log ([A-]/[HA]), Units, Equation, Problems? Kp = Kc(RT)^delta n, Units, Formula, Equation, How to use? Delta Hf, Equation, Problems, How to calculate in chemistry? q = n delta H, Units, Formula, Equation, Examples, How to use? Q = mc delta T, Units, Formula, Heat transfer equation Delta G = Delta H – T Delta S, Equation, Units, Formula Delta Tf = iKfm, Equation, Freezing point depression formula Delta Tb = imKb, Equation, Boiling point elevation formula Ammara Waheed Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - Leave a Comment Cancel Reply Scroll to Top
190016	https://kids.britannica.com/students/article/vegetarianism/277560	vegetarianism - Students \| Britannica Kids \| Homework Help Articles AnimalsArts and LiteratureEarth and GeographyHistoryLife ProcessesLiving Things (Other)Philosophy and ReligionPlantsScience and MathematicsSocietySports and RecreationTechnology Images & Video AnimalsArts and LiteratureEarth and GeographyHistoryLife ProcessesLiving Things (Other)Philosophy and ReligionPlantsScience and MathematicsSocietySports and RecreationTechnology BiographiesDictionaryCompare CountriesWorld AtlasPodcast Switch Level KidsStudentsScholars About Us students Kids Students Scholars Fundamentals NEW 7-day free trial Search Login ArticlesFeatured Article music All Categories Animals Arts and Literature Earth and Geography History Life Processes Living Things (Other) Philosophy and Religion Plants Science and Mathematics Society Sports and Recreation Technology Images & VideosFeatured Media wrestling All Categories Animals Arts and Literature Earth and Geography History Life Processes Living Things (Other) Philosophy and Religion Plants Science and Mathematics Society Sports and Recreation Technology Biographies Dictionary Compare Countries World Atlas Podcast × vegetarianism View article for: Kids Students Scholars Article Images & Videos Related Subscriber features Print (Subscriber Feature) Email (Subscriber Feature) Cite (Subscriber Feature) Translate (Subscriber Feature) Listen (Subscriber Feature) × Related resources for this article Articles Primary Sources & E-Books Websites View search results for: Search © Mercy for Animals The practice of vegetarianism involves eating vegetable products and eliminating meat, fish, and, in many instances, eggs and dairy products from the diet for ethical, religious, or nutritional reasons. Some vegetarians trace the term to the Latin vegetus, meaning “active, vigorous.” Vegetarianism traditionally has been associated with the philosophy of living a more peaceful life in harmony with natural laws and principles. Some aspects of the vegetarian diet currently are regarded as healthy alternatives to the modern diet consumed in many Western countries, which is high in animal fat and low in vegetable fiber. Encyclopædia Britannica, Inc. In general, a vegetarian diet consists of vegetables, fruits, grains, nuts, and seeds. Strict vegetarians, known as vegans, avoid all foods from the animal kingdom, including eggs and dairy products. Less strict vegetarians who consume milk and milk products are known as lactovegetarians, while those who include eggs in their diet are called ovovegetarians. Some consider themselves to be vegetarians if they eliminate only red meat from their diets. Vegetarians tend to prefer food in its most natural state, opposing the use of chemicals in growing or harvesting food and avoiding processed or canned foods. Vegetarianism probably was first practiced in connection with religious purification rituals. The idea of a fleshless diet for normal use arose in India and eastern Mediterranean lands around the 1st millennium bc. From Plato onward, many Greek and Roman philosophers and writers advocated vegetarianism as part of an ethical way of life. Hindu and Buddhist sects also considered all animal life sacred and taught that human beings should avoid harming animals. Some Jewish and Christian groups followed the principles of vegetarianism and considered eating flesh gluttonous, cruel, and wasteful. In the Roman Catholic church, Trappist monks have practiced vegetarianism since 1666. Among Protestants, the Seventh-day Adventists observe strict vegetarian guidelines. Although Muslims in general do not advocate fleshless diets, some Sufi mystics (who are among the chief guides of Muslim spiritual life) recommend a vegetarian diet for spiritual seekers. In the 19th century, members of the Bible Christian sect established the first vegetarian groups in England and in the United States. In 1847 the Vegetarian Society was founded, and in 1908 the International Vegetarian Union held the first of its regular meetings. In nutritional terms, the vegetarian diet has been recognized as having some value in reducing the risk of heart disease, certain types of cancers, and other diet-related ailments. Single vegetable products generally lack the essential amino acids found in animal products, but these amino acids can be included in a vegetarian diet by combining foods, such as beans and rice, that provide complementary amino acids (seefood and nutrition; malnutrition). Care must be exercised in a strict vegetarian diet to make sure that there is sufficient vitamin B 12 (seevitamins). Those who practice vegetarianism also claim that it is less costly to grow vegetable crops than it is to raise animals to feed a population. They maintain that converting grazing land to cropland could make more food available to the world. × It’s here: the NEW Britannica Kids website! We’ve been busy, working hard to bring you new features and an updated design. We hope you and your family enjoy the NEW Britannica Kids. Take a minute to check out all the enhancements! The same safe and trusted content for explorers of all ages. Accessible across all of today's devices: phones, tablets, and desktops. Improved homework resources designed to support a variety of curriculum subjects and standards. A new, third level of content, designed specially to meet the advanced needs of the sophisticated scholar. And so much more! Want to see it in action? 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190017	https://chem.libretexts.org/Under_Construction/Purgatory/Introductory_Chemistry_at_Solano_College_2022/13%3A_Solutions/13.11%3A_Molality	Skip to main content 13.11: Molality Last updated : Aug 26, 2022 Save as PDF 13.10: Dilution 13.12: Calculating Molar Mass Page ID : 408168 ( \newcommand{\kernel}{\mathrm{null}\,}) For many purposes, the use of molarity is very convenient. However, when we want to know the concentration of solute present in situations where there are temperature changes, molarity will not work. The volume of the solution will change somewhat with temperature, enough to affect the accuracy of data observations and calculations. Another parameter is needed, one not affected by the temperature of the material we are studying. Molality A final way to express the concentration of a solution is by its molality. The molality of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains of dissolved into of water is a "one-molal" solution of sodium chloride. The symbol for molality is a lower-case written in italics. Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying properties of solutions related to vapor pressure and temperature changes. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature. Example Determine the molality of a solution prepared by dissolving of glucose into of water. Solution Step 1: List the known quantities and plan the problem. Known Mass solute Mass solvent Molar mass Unknown Convert grams of glucose to moles and divide by the mass of the water in kilograms. Step 2: Solve. Step 3: Think about your result. The answer represents the moles of glucose per kilogram of water and has three significant figures. Molality and molarity are closely related in value for dilute aqueous solutions because the density of those solutions is relatively close to . This means that of solution has nearly a mass of . As the solution becomes more concentrated, its density will not be as close to and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem. Summary The molality of a solution is the moles of solute divided by the kilograms of solvent. The value of molality does not change with temperature. 13.10: Dilution 13.12: Calculating Molar Mass
190018	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOorgMCzP8IDlZKN8iOjqCZt1o9XoUocq9cjKBUn9bf0dx_vanfey	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190019	https://math.stackexchange.com/questions/1445584/intersection-of-line-and-circle-conditions-for-tangency	algebra precalculus - Intersection of line and circle, conditions for tangency - 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 Intersection of line and circle, conditions for tangency Ask Question Asked 10 years ago Modified10 years ago Viewed 426 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Find the quadratic equation in x x whose roots give the x x-coordinates of the points where the line y=m x y=m x intersects the circle x 2+y 2+20 x+40 y+450=0 x 2+y 2+20 x+40 y+450=0. What is the geometrical relation between the line and the circle if this quadratic equation has coincident roots? Write down the algebraic condition involving m m for this to happen and hence find the equations of the tangents from the origin to the circle. Find the coordinates of the points of contact of these tangents. Sorry, I know this is a lot of questions packed into one but I really can't crack this problem. I think that the quadratic equation the question is initially asking for is this: x 2+20 x+m 2 x 2+40 m x=−450 x 2+20 x+m 2 x 2+40 m x=−450. The answer to the second part (geometrical relation if roots are coincident), I think, is that it is a tangent to the circle as it only has one point of contact. However, for the third and fourth part I can't pull off the coefficients a a, b b and c c to substitute into b 2−4 a c b 2−4 a c as I have this m 2 m 2 and m m term. algebra-precalculus Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Sep 21, 2015 at 20:54 A confused A level studentA confused A level student 31 1 1 bronze badge 4 2 There is a double root if and only if the discriminant "b 2−4 a c b 2−4 a c", that is, (20+40 m)2−4(1+m 2)(450)(20+40 m)2−4(1+m 2)(450) is equal to 0 0.André Nicolas –André Nicolas 2015-09-21 20:59:58 +00:00 Commented Sep 21, 2015 at 20:59 Wow, I just realised that exactly as you said it. Thanks!A confused A level student –A confused A level student 2015-09-21 21:01:16 +00:00 Commented Sep 21, 2015 at 21:01 Good! You are welcome.André Nicolas –André Nicolas 2015-09-21 21:03:07 +00:00 Commented Sep 21, 2015 at 21:03 Just as a note, usually more descriptive titles are received better. Rather than "math homework help", maybe try describing the problem. Just in the future.user223391 –user223391 2015-09-21 21:19:35 +00:00 Commented Sep 21, 2015 at 21:19 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Your answers to the first two parts are fine. The idea for the third part is to think of the expression you have as quadratic in x x, with m m as a constant: (1+m 2)x 2+2(10+20 m)x+450=0,(1+m 2)x 2+2(10+20 m)x+450=0, and then stick these coefficients into the discriminant: 4(10+20 m)2−4×450(1+m 2)=0.4(10+20 m)2−4×450(1+m 2)=0. Now solve that for m m (there should be two roots, one for a tangent on both sides of the circle), and then solve the quadratic for the x x values. The y y values you can easily get from y=m x y=m x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 21, 2015 at 21:03 ChappersChappers 69.2k 11 11 gold badges 74 74 silver badges 153 153 bronze badges 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 algebra-precalculus 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 4Finding the equations of two tangents to a circle given the point of intersection. 3Finding the equation of a circle and a tangent line to the circle given two end points 3Find the possible values of p p for which the equation has coincident roots. 2Finding the intersection of a circle and a line 0Write the equation of the tangent line of a circle 1Finding the points of intersection of a circle and a line 2A line through the point P(8, -7) is a tangent to the circle C at the point T. Find the length of PT. 2y=−2 x+k y=−2 x+k and curve y=2 x−3 y=2 x−3 13Intersection of a line and a curve. 0Find the points of intersection of the line x+16=7 y x+16=7 y and the circle x 2+y 2−4 x+2 y=20 𝑥 2+𝑦 2−4 𝑥+2 𝑦=20 Hot Network Questions Checking model assumptions at cluster level vs global level? A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man The rule of necessitation seems utterly unreasonable Interpret G-code What’s the usual way to apply for a Saudi business visa from the UAE? Who is the target audience of Netanyahu's speech at the United Nations? Alternatives to Test-Driven Grading in an LLM world Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? 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190020	https://axial.substack.com/p/eulers-polyhedron-formula	Euler's polyhedron formula - Axial Axial Subscribe Sign in Discover more from Axial Agency for great inventors Over 257,000 subscribers Subscribe By subscribing, I agree to Substack's Terms of Use, and acknowledge its Information Collection Notice and Privacy Policy. Already have an account? Sign in Euler's polyhedron formula Strategy Apr 10, 2024 2 Share Axial: Axial partners with great founders and inventors. We invest in early-stage life sciences companies such as Appia Bio, Seranova Bio, Delix Therapeutics, Simcha Therapeutics, among others often when they are no more than an idea. We are fanatical about helping the rare inventor who is compelled to build their own enduring business. If you or someone you know has a great idea or company in life sciences, Axial would be excited to get to know you and possibly invest in your vision and company. We are excited to be in business with you — email us at info@axialvc.com The 5 Platonic Solids from left to tight: cube, dodecahedron, icosahedron, octahedron and tetrahedron. Euler's polyhedron formula: V - E + F = 2 or Vertices - Edges + Faces = 2 It relates the number of vertices (V), faces (F), and edges (E) of a convex polyhedron. The formula states that for any convex polyhedron, the following equation holds: V + F - E = 2 This deceptively simple equation has profound implications and has played a significant role in the development of various branches of mathematics, including graph theory, topology, and combinatorics. The study of polyhedra has fascinated mathematicians and scholars for centuries, dating back to ancient times. The Ancient Greeks, in particular, were intrigued by the regular polyhedra, also known as the Platonic solids. These are convex polyhedra with congruent regular polygonal faces, and all vertices are alike. The five Platonic solids – the tetrahedron, cube, octahedron, dodecahedron, and icosahedron – were well-known to the Greeks and held special significance in their philosophy and cosmology. Despite this early interest in polyhedra, it was not until the 18th century that the Swiss mathematician Leonhard Euler (1707-1783) formulated the polyhedron formula. In 1750, Euler mentioned the result in a letter to Christian Goldbach, and later, he published two papers providing more details and attempting to prove the formula. Euler's initial proof was incomplete and did not meet modern standards of rigor. However, his formulation of the theorem was a significant breakthrough and paved the way for further developments in the field. It was not until 1811 that Adrien-Marie Legendre provided the first accepted proof of the formula, using a geometric approach. Euler's polyhedron formula is remarkable for several reasons. First, its simplicity and elegance belie the depth of its implications. The fact that such a straightforward relationship exists between the fundamental elements of a polyhedron – vertices, faces, and edges – is truly remarkable. Secondly, the formula has far-reaching applications in various areas of mathematics. It provides a crucial link between the seemingly disparate fields of geometry, topology, and combinatorics. The formula serves as a foundation for the study of surfaces and the development of algebraic topology. In graph theory, Euler's formula finds direct applications. The graphs associated with convex polyhedra are planar graphs, meaning they can be drawn on a plane without any edges crossing. Euler's formula can be used to prove important results in planar graph theory, such as Kuratowski's theorem, which characterizes planar graphs in terms of forbidden subgraphs. The formula also plays a pivotal role in the study of surface topology. By generalizing Euler's formula to surfaces with handles (like a torus or a higher-genus surface), mathematicians were able to develop a deeper understanding of the topology of surfaces and establish a classification of orientable surfaces based on their genus. Furthermore, Euler's formula has implications in the study of coloring problems, such as the famous Four Color Theorem. This theorem states that any planar map can be colored using at most four colors, such that no two adjacent regions share the same color. The proof of this long-standing conjecture, first proposed by Francis Guthrie in 1852, relies heavily on Euler's formula and the properties of planar graphs. Over the years, numerous proofs of Euler's polyhedron formula have been developed, showcasing the richness and depth of the theorem. Some of the most elegant proofs utilize mathematical induction, either on the number of edges, faces, or vertices. Other proofs employ combinatorial techniques, such as the use of spanning trees or the flooding of "dry fields" by removing edges from the graph. One particularly insightful proof, attributed to Rademacher and Toeplitz, involves considering the polyhedron's graph as a network of dikes surrounding dry fields (the faces) and an "ocean" (the infinite region). By systematically removing edges and "flooding" the dry fields, the proof elegantly arrives at Euler's formula. Mathematicians have also explored various generalizations and extensions of Euler's formula. One notable generalization is the Euler characteristic, which extends the formula to surfaces with handles (like a torus or a higher-genus surface). This generalization played a crucial role in the development of algebraic topology and the classification of surfaces. Another generalization involves considering non-convex polyhedra or polyhedra with certain types of "holes" or "tunnels." These generalizations have led to interesting results in areas such as computational geometry and the study of polytopes. Euler's polyhedron formula has found applications in diverse fields beyond pure mathematics. In chemistry and materials science, the formula has been used to study the structure and properties of molecules and crystalline solids, which can be modeled as polyhedra or networks of polyhedra. In computer graphics and computer-aided design (CAD), the formula and related concepts from graph theory and computational geometry are employed in the representation and manipulation of three-dimensional objects and surfaces. Moreover, the formula has inspired researchers to explore related problems in combinatorics and discrete mathematics, such as counting and enumerating polyhedra with specific properties or studying the combinatorial properties of graphs associated with polyhedra. In the realm of computer science, Euler's formula has connections to graph algorithms and data structures, as well as to problems related to network analysis and optimization. Despite its age and simplicity, Euler's polyhedron formula remains an active area of research, with new applications and extensions being discovered in various fields. As our understanding of mathematics and its applications continues to evolve, the importance and relevance of this fundamental theorem will undoubtedly persist. Euler's polyhedron formula is a remarkable and enduring result that has had a profound impact on the development of mathematics and its applications. Its simplicity belies its depth and versatility, and it serves as a testament to the interconnectedness of different branches of mathematics. From its humble beginnings as a curious observation about convex polyhedra, the formula has evolved into a cornerstone of graph theory, topology, and combinatorics. It has inspired numerous proofs, generalizations, and applications, and continues to captivate mathematicians and scientists alike. As we explore new frontiers in mathematics and its applications, Euler's polyhedron formula will undoubtedly continue to reveal new insights and connections, reminding us of the timeless beauty and power of mathematical ideas. Subscribe to Axial By Joshua Elkington · Launched 6 years ago Agency for great inventors Subscribe By subscribing, I agree to Substack's Terms of Use, and acknowledge its Information Collection Notice and Privacy Policy. 2 Likes 2 Share Discussion about this post Comments Restacks Top Latest Discussions The Business of AI in Life Sciences Life sciences strategy Mar 15, 2021• Joshua Elkington 23 Axial - Recursion Pharmaceuticals Surveying great inventors and businesses May 14, 2021• Joshua Elkington 17 Drugs that are their own pipelines Life sciences strategy Feb 10, 2021• Joshua Elkington 14 See all Ready for more? 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190021	https://www.sciencedirect.com/science/article/pii/S2772963X24008287	Naxos Disease and Related Cardio-Cutaneous Syndromes - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Abstract Central Illustration Key words Abbreviations and Acronyms First part: Disease mechanisms Epidemiology: Worldwide geographical map Second Part: The Patient Pathway Management Conclusions Funding support and author disclosures Supplementary data References Show full outline Cited by (3) Figures (7) Show 1 more figure Tables (1) Table 1 Extras (1) Supplementary Data JACC: Advances Volume 4, Issue 2, February 2025, 101547 State-of-the-Art Review Naxos Disease and Related Cardio-Cutaneous Syndromes Author links open overlay panel Alexandros Protonotarios MD a, Angeliki Asimaki Ph.D b, Cristina Basso MD c, Zafeirenia Xylouri MD d, Emanuele Monda MD e, Ioannis Protonotarios MD b, Giulia Crisci MD f g, Dominic JR.Abrahms MMBS, MD, MBA h, Aris Anastasakis MD i, Loizos Antoniades MD j, Athanasios Bakalakos MD a, Andreina Carbone MD f, Aman S. Coonar MD k, Juan Ramon Gimeno MD l, George Lazaros MD m, Stamatios Lerakis MD n, Luisa Mestroni MD o, George Papadopoulos MD p, Leandro Pecchia MD q, Francesca Romana Prandi MD r…Adalena Tsatsopoulou MD v w∗ Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights •The cutaneous phenotype of Naxos disease (palmoplantar keratoderma, with or without skin erosions or bullous lesions, combined with hair shaft changes—woolly hair or occasionally absence of hair) is a red flag for thorough cardiac screening for arrhythmogenic cardiomyopathy at any age. •The diagnosis of Naxos disease or any other cardiocutaneous syndrome warrants familial and genetic evaluation. •Symptoms including palpitation, chest pain, and/or syncope typically present during adolescence. •The earliest disease manifestations are the cutaneous features and electrocardiographic abnormalities which typically precede the development of symptoms and/or arrhythmia. •Disease progression with arrhythmia and heart failure warrants comprehensive, tailored treatment and lifestyle modifications. •Global collaboration aims to better understand cardiocutaneous disease to enable development of targeted treatments to enhance patient longevity and quality of life. Abstract Naxos disease is a rare autosomal recessive condition combining arrhythmogenic right ventricular cardiomyopathy, woolly hair, and palmoplantar keratoderma. The first identified causative variant was in the gene encoding the desmosomal protein plakoglobin. Naxos disease exhibits fibro-fatty myocardial replacement with immunohistological abnormalities in cardiac protein and signaling pathways, highlighting the role of inflammation and potential anti-inflammatory treatments. Childhood cutaneous signs precede cardiac features, which are diagnosed by familial and genetic evaluation, electrocardiography and cardiac imaging. Disease progression necessitates holistic care with risk management and lifestyle adjustments, often needing treatment for arrhythmia and heart failure. Similar phenotypes have been linked to desmoplakin and rarely desmocollin2 gene variants, highlighting the importance of familial and genetic evaluation. This document summarizes current knowledge on Naxos disease and related cardiocutaneous syndromes and initiates an international endeavor to collect and study all global cases, aiming to improve understanding, treatment, and patient care through shared data and research. Central Illustration 1. Download: Download high-res image (741KB) 2. Download: Download full-size image Previous article in issue Next article in issue Key words Naxos disease cardio-cutaneous syndrome arrhythmogenic right ventricular cardiomyopathy arrhythmogenic cardiomyopathies plakoglobin desmoplakin Abbreviations and Acronyms AI artificial intelligence ARVC arrhythmogenic right ventricular cardiomyopathy CMR cardiac magnetic resonance DCM dilated cardiomyopathy DSC2 desmocollin 2 DSP desmoplakin ECG electrocardiography EMB endomyocardial biopsy ICD intracardiac cardioverter-defibrillator IDs intercalated disks JUP junctional plakoglobin LV left ventricular MDT multidisciplinary team PPK palmoplantar keratoderma RV right ventricular SCD sudden cardiac death WH woolly hair Definition and the aim of an international endeavor Naxos disease is a rare, autosomal recessive cardiocutaneous disorder characterized by arrhythmogenic right ventricular cardiomyopathy (ARVC), woolly hair (WH), and palmoplantar keratoderma (PPK) (Figure 1). It was first described by Nikos Protonotarios et al in 1986, on 9 patients across 4 families, originating from the Greek Naxos Island.1 1. Download: Download high-res image (1MB) 2. Download: Download full-size image Figure 1. Summary Features in Naxos Disease Patients (JUP2157del2) (A) Characteristic cutaneous features in a patient with Naxos disease: 1. Woolly hair, 2. Palmar and 3. Plantar keratoderma. (B) Pedigree of a family with Naxos disease (JUP 2157), showing the autosomal recessive mode of patient distribution. Solid shapes: Homozygous carriers with 100% penetrance of disease. Half-marked shapes: Heterozygous carriers without clinically detectable heart and skin phenotype. (C) 12-lead ECG from an adult patient with Naxos disease (JUP 2157): QRS prolongation, epsilon waves in V 1-V 3, T-wave inversion in V 1-V 4 and low voltage. (D) Structural myocardial changes, detected in patients with Naxos disease (JUP 2157). 1: Myocardial histology (Masson’s trichrome stain) revealing degenerating myocardial fibres (red) surrounded by fibrous (blue) and fatty (white) tissue; 2: Postmortem sample of a whole heart, anterior view, revealing a dilated RV (inflow, apex and outflow) (2 arrows) and a small part of the dilated LV (one arrow). 3: Kinetic abnormalities appearing as “accordion sign” (arrows) on the RV-free wall as shown on CMR. CMR=cardiac magnetic resonance; ECG=electrocardiography; LV=left ventricular; JUP=junctional plakoglobin; RV=right ventricular. The discovery of a junctional plakoglobin (JUP) variant in Naxos disease highlighted the role of desmosomal proteins in cardiomyopathies.2 Research has shown that desmosomal gene variants can cause both isolated cardiac and syndromic forms of ARVC and may overlap with dilated cardiomyopathy (DCM).3, 4, 5 In the following years, population screening in the Aegean islands showed that the disease occurs in about 1 in 1,000 people in certain regions.6 The discovery of the same JUP variant in a family from eastern Turkey suggested a historical connection between these populations. Other desmosomal gene variants in JUP, desmoplakin (DSP), and desmocollin 2 (DSC2) have been found in isolated cases of similar cardio-cutaneous syndromes in various parts of the world, from the Mediterranean to Argentina and Ecuador. However, no systematic study has been conducted to determine the overall prevalence of these syndromes. This document aims to summarize current knowledge on Naxos disease and similar syndromes, advocating for a global effort to collect and study cases. First part: Disease mechanisms Genetics The concurrence of hair and skin features with ARVC in 21 affected individuals across 9 Naxos families (from a cohort of 150 individuals) provided a clear diagnostic marker, effectively distinguishing those with the cardiomyopathy. Meanwhile, pedigree analysis suggested autosomal recessive inheritance. These led to homozygosity mapping and linkage analysis, which pinpointed the disease locus to 17q21.7 Subsequently, the recessive JUPc.2157del2 variant was identified.2 Shortly after, the recessive DSP7901delG variant was discovered in 6 patients from 3 Ecuadorian families, underlying a phenotype of PPK and WH similar to Naxos disease, but with a cardiomyopathy phenotype more closely resembling DCM.4 Other variants in JUP, DSP, and DCS2 were later found to cause similar cardiocutaneous disorders with various inheritance patterns. All cases exhibited both cardiac and skin features (Supplemental Table). In a study involving 6 families with dominant loss of function DSP variants, the presence of a phenotype of curly hair and minor PPK features cosegregated with arrhythmogenic cardiomyopathy phenotype.8 There is only 1 report of 2 siblings with ARVC and left ventricular (LV) involvement, associated with cutaneous features of mild PPK and WH, related to a homozygous single-base deletion in DSC2(1841delG)9 (Supplemental Table). Epidemiology: Worldwide geographical map Since the initial description of Naxos disease, additional cardiocutaneous cases have been documented worldwide. Besides Naxos Island, affected families with the same variant in JUP were found on other Aegean islands. The disease prevalence in some Aegean regions was estimated at 1:1000.6 Additionally, the same variant with a similar cardiac and skin phenotype has been reported in Turkey.10 Other homozygous JUP variants, linking arrhythmogenic cardiomyopathy with WH, alopecia, or PPK, have been reported from Argentina and Canada and have been additionally associated with lethal neonatal epidermolysis, total alopecia, and onycholysis (Supplemental Table). A decade after Naxos disease was described in 1986, a similar condition involving DCM with early arrhythmias was described in Ecuadorian families.11 This was linked to a DSP variant and named Carvajal syndrome.12,13 Other DSP variants underlying similar to Naxos disease cutaneous features with a closer to DCM phenotype were subsequently reported worldwide. One family with a recessive variant in DSC2 inherited in homozygosis has been reported from the United Kingdom.9 The global distribution and number of reported cases are detailed in the Supplemental Table, with a visual representation provided in Figure 2. 1. Download: Download high-res image (196KB) 2. Download: Download full-size image Figure 2. Map of Global Distribution of Naxos Disease and Related Cardio-Cutaneous Cases, as Reported up to 2024 Global distribution of Naxos disease and related cardio-cutaneous cases. Each circle represents the presence of reported cases within a country, color-coded by the gene involved. Circle size is proportional to the number of cases reported, with larger circles indicating a higher number of cases. Jitter is added to minimize overlap. This map provides a visual summary of the global epidemiology of Naxos-like syndrome, highlighting the geographical distribution and prevalence of genetic variants associated with the condition. Pathology and disease mechanisms Cardiac pathology-histopathology Naxos disease patients show classic features of fibro-fatty replacement of atrophic myocardium, extending from the epicardium to the endocardium, typical of autosomal dominant ARVC (Figures 1, D1 and 3, 3C).14 In 4 Naxos disease patients, 2 had typical ARVC with biventricular involvement, 1 had predominant right ventricular (RV) and focal LV involvement, and 1 had no histopathological changes, dying of leukemia before phenotypic expression at age 7 suggesting that the Naxos cardiac phenotype is acquired with age, similar to nonsyndromic ARVC. One patient exhibited extensive calcification of the RV outflow tract, possibly explaining decreased arrhythmogenicity before dying from heart failure (Figure 3, C3).15 1. Download: Download high-res image (756KB) 2. Download: Download full-size image Figure 3. Functional and Structural Myocardial Changes in Patients with Naxos Disease (JUP2157del2) (A) Diastole and (B) systole. On CMR: 1, and 2D-ECHO: 2 and 3. Arrows pointing on RV dyskinetic areas. RV free wall dyssynchronous contraction (accordion sign): 1B, RV posterior wall dyskinesia with systolic bulging (retraction sign): 2B, and apical RV aneurysm: 3A and 3B (akinesia/dyskinesia with diastolic bulging) C1: An electron micrograph shows a decreased gap junction profile length in the LV myocardium (Bar 0.5 μm). C2: Postmortem histology (hematoxylin and eosin stain; cell nuceli: blue, cytoplasm: pink, extracellular fibrosis: light pink,) from LV myocardium, showing myocyte loss with predominantly fibrous replacement of myocardial tissue. C3: Gross examination on postmortem (the patient died of heart failure waiting for heart transplantation), showing on the parasternal long-axis view of the heart, paper-thin RV walls, transmural myocardial loss (1 arrow), subepicardial left ventricular fatty infiltration (2 arrows), and calcified fibrous plaque at the level of the RV outflow tract (3 arrows). ECHO=echocardiography; other abbreviations as in Figure 1. Carvajal syndrome, a cardiocutaneous syndrome related to Naxos disease, presents with cutaneous features that are clinically indistinguishable from those of Naxos disease, along with a cardiac phenotype resembling DCM, often accompanied by early ventricular arrhythmias and heart failure.13 A whole heart from a patient with Carvajal syndrome was examined postmortem. The cardiac pathology features eccentric ventricular hypertrophy and focal aneurysms in the RV, without fatty replacement, but with fibrous tissue, often seen in autosomal dominant DSP disease with a predominant LV phenotype.16 Cardiac and extracardiac immunohistology Naxos disease patients were among the first ARVC patients whose myocardial samples were subjected to immunohistochemistry analyses, revealing that mutant JUP is expressed (as evidenced by Western immunoblotting) but fails to localize at the cardiac intercalated disks (IDs) (Figure 4, 2C).17 Immunoreactive signal for JUP is significantly depressed at the IDs in all forms of ARVC regardless of the underlying mutation.18 Furthermore, it has been shown that there is gap junction remodeling in Naxos disease (Figure 3, 1C).17 Loss of immunoreactive signal for Cx43 was also evident in myocardial samples from a 7-year-old girl homozygous for the JUP2157del2 variant who died of leukemia prior to developing any histological or functional cardiac abnormalities (Figure 4, 2C).17 These findings support the hypothesis that gap junction remodeling appears to precede structural alterations being key contributor to arrhythmogenesis. 1. Download: Download high-res image (1MB) 2. Download: Download full-size image Figure 4. Naxos Disease, Early Phenotype in JUP 2157del2 Homozygous Children Cutaneous features can be detected since early childhood: Woolly hair at 14 months (1A) and 7 years (1B). PPK mild at 14 months (2A, 3A∗∗) is potentially associated with eczematous lesions on the external foot and leg surfaces that usually resolve with age (3A∗). PPK shows a striate appearance at the age of 7 years (2B, 3B). Cardiac phenotype usually presents with an arrhythmic profile. (1C) Ventricular ectopics (14,000/day) detected on an asymptomatic child screened because of woolly hair and PPK at the age of 4 years, while she presented a normal for her age 12-lead ECG and normal 2D-ECHO. (2C) Abnormal distribution of connexin43 and plakoglobin on the right and left ventricular myocardium, detected on postmortem, after a non-cardiac death, while 2D-ECHO, gross pathology and regular histology were normal.52 ARVC in Naxos disease may present as an episode of acute myocarditis. 3C: Late contrast-enhanced image in a 14-year-old boy with Naxos disease, homozygous for JUP 2157, showing LGE (arrow) in LV (3C∗∗) as compared to (3C∗) image taken taken one year prior to (3C∗∗). The boy, followed-up yearly since the age of one year due to cutaneous features of WH and PPK, was asymptomatic till the age of 14 years, when he first presented with chest pain and troponin elevation. PPK=palmoplantar keratoderma; other abbreviations as in Figures 1 and 3. Immunohistology on postmortem examination of an Ecuadorian patient's heart revealed diminished signals for DSP, plakoglobin, and connexin43 at intercellular junctions, suggesting compromised desmosomal and gap junction integrity. Additionally, desmin was abnormally localized, failing to appear at the IDs.13 Studies have shown that buccal epithelial cells express all aforementioned protein markers and may thus act as a surrogate for the heart.19 Immunoreactive signal for JUP was lost in buccal cells from 6 patients with Naxos disease while signal for connexin43 (Cx43) was severely depressed in 5/6. These findings suggest that buccal mucosaimmunohistology may have significant potential as an easy-to-acquire disease biomarker. Intracellular pathways JUP contributes to both adhesive and signaling functionalities, as it is involved in the Wnt signaling pathway.20,21 Desmosomal perturbations can lead to nuclear accumulation of plakoglobin and downregulation of the canonical Wnt signaling and thereby enhancement of adipogenesis driven by peroxisome proliferator-activated receptor γ and CCAAT-enhancer-binding protein alpha.22,23 Furthermore, ARVC-causing mutations lead to dislocation of protein kinase C alpha and Wnt inhibition and enhanced adipogenesis via the Hippo pathway.24 GSK3β is also implicated as its pharmacological inhibition prevents and reverses the ARVC phenotype (including arrhythmias, contractile dysfunction, myocyte injury, apoptosis, exercise-induced sudden cardiac death (SCD), and redistribution of junctional proteins) in ARVC models in vitro, in vivo, and ex vivo including transgenic zebrafish and mouse models with cardiac-specific expression of the JUP2157del2 variant.23,25 Modeling of aberrations in other desmosomal proteins support the involvement of key components in Ca+2 signaling pathways that might be part of a final common pathway in all forms of ARVC.26,27 Inflammation Myocardial infiltrates are observed in >70% of ARVC myocardial samples, predominantly in areas with extensive fibrofatty replacement and contain T-lymphocytes, macrophages, neutrophils, and mast cells.14,28 Myocarditis-like “hot-phases” have been reported in Naxos disease patients, who present with ST-segment elevation and/or increased troponin levels.29,30 Several experimental models have been used to gain insights into the role of inflammation in the pathogenesis of ARVC. Neonatal rat ventricular myocytes transfected to express the JUP2157del2 variant express and release high levels of pro-inflammatory mediators in their culture medium including tumor necrosis factor-alpha, interleukin-6, macrophage inflammatory protein-1 alpha, and regulated upon activation normal T-cell expressed and secreted.23 NFκB is a master regulator of inflammation and its inhibition in disease models reduced inflammatory cytokine production and prevented key protein redistribution and apoptosis.31 Understanding the precise role inflammation plays in disease progression could aid diagnosis and risk stratification as well as provide the basis for the development of mechanism-based therapies targeting aberrant immune activation.32 Novel therapeutics Mechanistic insights from animal and cell models that aim to replicate the cardiac features of Naxos disease have the potential to uncover key molecules that can dampen disease progression, such as by modulating the innate inflammatory response.31 The most promising therapeutic pipeline, however, remains that of gene therapy, with numerous trials currently ongoing for genetic cardiomyopathies, especially where the mechanism is underlied by gene haploinsufficiency.33 Specialized funding programs and industry partnerships are crucial for developing and increasing access to novel therapeutics for Naxos disease, given its rarity and the limited viability for extensive pharmaceutical research. Second Part: The Patient Pathway Diagnostic workup Clinical presentation and natural history: Transition from pediatric to adult care Naxos disease's cutaneous features are evident from infancy, characterized by woolly, dense hair, curly eyebrows, and eyelashes (Figure 4). PPK with occasional hand and legs eczematous lesions appears within the first 2 years (Figure 4). Hyperkeratotic plaques have a distinct erythematous border. Some JUP variants may show absent or sparse hair (Supplemental Table). About half of Naxos disease patients have syncopal episodes or develop episodes of sustained ventricular tachycardia, between their teenage and 35 years. In the series studied by Protonotarios N et al, half of patients experienced arrhythmic events during the disease course.34Ventricular tachycardia on electrocardiography (ECG) consistently showed a left bundle branch block pattern.35Atrial fibrillation was documented in 3 cases.1,36 A 5-year-old girl with the JUP-2157del2 variant exhibited frequent RV extrasystoles on her ECG despite a normal echocardiogram (Figure 4, 1C). Following a noncardiac death, postmortem histopathology showed gap junction remodeling without fibrofatty replacement.17 The disease generally progresses with RV function deterioration, new wall motion abnormalities, and potential LV function decline.37 Episodes of arrhythmic storms “hot phases” may trigger structural deterioration or sudden death, sometimes resembling acute myocarditis in children or adolescents (Figure 4, 3C).29,38,39 Comprehensive cardiac evaluation is crucial for children or adults presenting with chest pain to detect ARVC progression. In Ecuadorian families, with Carvajal syndrome, all affected members showed electrocardiographic and/or echocardiographic abnormalities, with the youngest cases appearing by age 7 years. Resting 12-lead ECGs were abnormal in all, commonly showing low voltage and intraventricular conduction defects.11 T-wave inversion in V 1-V 3, sometimes extending to V 5, was frequently observed.40 All patients who underwent 24-hour Holter monitoring showed complex ventricular extrasystoles and episodes of nonsustained ventricular tachycardia. Over 90% developed LV involvement by their second decade, often severe. Though the right ventricle was also affected, detailed evaluation was limited.40 Fifty-seven percent of patients developed heart failure, with most dying during adolescence, likely from heart failure, though arrhythmic death cannot be excluded due to limited out-of-hospital data.4 The pronounced arrhythmogenicity and ECG abnormalities in these patients are now recognized as part of the nondilated LV cardiomyopathy phenotype spectrum.5 Heart failure symptoms, though rare at onset, can develop and may necessitate heart transplantation in some cases,6 particularly in those with DSP variants (Supplemental Table). In the largest longitudinal cohort with Naxos disease, annual disease-related mortality was 3%, with sudden death mortality at 2.3%.37 Sporadic reports of homozygous JUP variants include a toddler's sudden death, an 18-year-old's heart transplantation,41 and 2 cases of congenital epidermolysis bullosa and infantile death, 1 with confirmed cardiac involvement (Supplemental Table). Electrocardiography ECG abnormalities may appear early, preceding structural changes as shown in a boy with Naxos disease homozygous for JUP 2157del2 variant who was first investigated the age of 6 years due to the cutaneous features and then followed up regularly (Table 1). ECG changes might indicate right or left ventricle involvement.42,43 In the largest Naxos disease series, over 90% of patients showed a broad spectrum of ECG abnormalities as symptoms emerged (Figure 5). Electrocardiographic abnormalities included inverted T waves in leads V 1 to V 3 or across the precordial leads, epsilon waves, QRS complex prolongation in V 1 to V 3, and right bundle branch block (complete or incomplete) (Figure 5).37 Adolescents may experience T-wave reinversion in V 2 and V 3. Epsilon waves, present in 34% of patients, were associated with right ventricular outflow tract wall motion abnormalities and increased RV diameter.44 Low voltage and flat T waves in left precordial leads were seen in severe right and/or biventricular disease, with frequent ventricular extrasystoles and/or ventricular tachycardia in over 90% of cases.37 Ventricular extrasystoles of right bundle branch block configuration were less common.34,37 Table 1. Early Prestructural, ECG Evolution in a Boy With Naxos Disease \| Empty Cell \| Age, y \| \| Empty Cell \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| 12.5 \| 13 \| 14 \| 15 \| \| TWI \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| V1 \| \| TAD (V1) \| 36 \| 36 \| 44 \| 56 \| 51 \| 58 \| 60 \| 60 \| 64 \| 64 \| 60 \| \| QRSf (ms) \| 104 \| 102 \| 107 \| 115 \| 111 \| 123 \| 125 \| 127 \| 127 \| 141 \| \| \| LAS (ms) \| 32 \| 31 \| 33 \| 37 \| 35 \| 48 \| 42 \| 44 \| 54 \| 60 \| \| \| RMS (μV) \| 46 \| 62 \| 44 \| 20 \| 31 \| 11 \| 11 \| 9 \| 7 \| 9 \| \| \| VES/24h \| 0 \| 0 \| 0 \| 1 \| 2 \| 66 \| 374 \| 1684 \| 504 \| 508 \| 1814 \| \| ECHO \| N \| N \| N \| N \| N \| N \| N \| N \| N \| N \| N \| \| CMR - - - - \| N \| N ECHO=echocardiography; CMR=cardiac magnetic resonance; LAS=low-amplitude signal duration; QRSf=filtered QRS duration; N=within normal limits; TAD=terminal activation duration; TWI=T-wave inversions; RMS=root-mean-square voltage of terminal 40 ms; VES=ventricular extrasystoles; ECG=electrocardiography. 1. Download: Download high-res image (596KB) 2. Download: Download full-size image Figure 5. Electrocardiography Spectrum in Naxos Disease Patients Samples from 12-lead ECG precordial recordings in 7 different patients with Naxos disease (JUP 2157del2): A broad spectrum of QRS width, amplitude and morphology, as well as the distribution of T-wave inversions, is noted. C-denotes the case number. Abbreviations as in Figure 1. Patients with DSP cardiocutaneous disease and a DCM phenotype often exhibited T-wave inversion in inferolateral leads, low QRS voltages, and abnormal Q waves (Supplemental Table).4,42,45, 46, 47 Resting 12-lead ECG and 24-hour Holter monitoring are essential in diagnosing cardiocutaneous phenotype, as electrical abnormalities are early signs of cardiomyopathy with a high incidence of ventricular arrhythmias.4 Cardiac imaging Typical transthoracic echocardiography findings in Naxos disease resemble dominant ARVC (Figure 1), showing RV outflow tract dilation, dysfunction, and focal dyskinetic areas, often involving the RV apex and mid/apical lateral walls, with aneurysmal dilation also described.37,45,48, 49, 50, 51 In the largest Naxos disease series, RV alterations ranged from mild dilatation or regional hypokinesia (27%) to severe dilatation, diffuse hypokinesia, and aneurysms (73%), with LV abnormalities in 25% of patients.34,37,51 DSP disease typically presents with a DCM phenotype, featuring LV and/or biventricular dilation and impaired systolic function, occasionally with LV intracavitary thrombus formation (Supplemental Table).4,42,45, 46, 47 Cardiac magnetic resonance (CMR), the gold standard for noninvasive RV assessment, offers excellent spatial resolution, 3 dimensional imaging, high reproducibility, and prognostic value, with right ventricular ejection fraction<40% predicting major cardiac events in CVD patients.52 While less common in Naxos disease reports, CMR findings align with echocardiography, and late gadolinium enhancement can assess myocardial fibrosis (Figure 3).15,29,53 CMR aids arrhythmic risk stratification in ARVC, potentially guiding intracardiac cardioverter defibrillator (ICD) implantation decisions, as right ventricular ejection fraction by CMR predicts appropriate ICD shock or death in ARVC patients.54 Although cardiac CT data in Naxos disease are unavailable, CT can differentiate ARVC from other tachyarrhythmias and guide ventricular tachycardia ablation and risk stratification despite its exclusion from the 2010 ARVC task force criteria.55,56 Recommendations for screening family members: genetic testing and genetic counseling Naxos disease and related cardiocutaneous syndromes are inherited mostly in an autosomal recessive pattern.3 The most extensively studied series of JUP variants exhibit full penetrance by adolescence, with variable ARVC severity.37 In addition, early cardiac features have been observed in cases with homozygous DSP variants (Supplemental Table). Occasionally dominant DSP variants associated with a cardiomyopathy phenotype exhibit minor cutaneous abnormalities.8 Therefore, in patients with WH and palmoplantar keratosis early genotyping and cardiac assessments (ECG, echocardiogram, Holter/ECG monitoring) is mandatory. In children with normal initial cardiac investigations CMR should be performed when this is feasible without general anesthesia or sedation, usually from 8 to 10 years of age. Regular evaluations every 1 to 2 years are advised (Central Illustration). Genetic testing and counseling for children should prioritize their best interests. A confirmed pathogenic variant warrants family member screening for carrier status. Pretest and posttest counseling, crucial to the process, addresses cultural and legal factors, particularly in prenatal and preimplantation diagnostics, to support informed decisions in family planning and disease management. A likely pathogenic variant should be reassessed approximately yearly to determine if the designation has changed.5 Genetic counseling is essential for understanding inheritance risks, providing psychosocial support, and aiding informed decision-making in family planning and disease management (Central Illustration).5 1. Download: Download high-res image (741KB) 2. Download: Download full-size image Central Illustration. Naxos Disease and Related Cardio-Cutaneous Syndromes: The Patient Pathway The Patient Pathway is an opinion statement. ∗Any new P/LP variant identified either on the already involved genes of JUP, DSP, DSC2 or in other genes. DSC2=desmocollin 2; DSP=desmoplakin; other abbreviation as in Figure 1. Endomyocardial biopsy Though endomyocardial biopsy (EMB) has been a major criterion for diagnosing ARVC, since 1994,55 the use of CMR for tissue characterization increased recently, leading to a decline in EMB use, which is now primarily reserved for inconclusive noninvasive assessments or to exclude phenocopies like sarcoidosis and myocarditis, especially in sporadic cases.57 The "hot phase" of ARVC can complicate differential diagnosis with myocarditis (Figure 4, 3C). EMB is not without risks and should be performed by experienced cardiomyopathy teams, with electroanatomic voltage-guided EMB enhancing diagnostic accuracy.58,59 Multidisciplinary team, education, network organization, and artificial intelligence in managing Naxos disease and related syndromes Managing Naxos disease and related cardiocutaneous syndromes requires a comprehensive, multidisciplinary approach aligned with cardiomyopathy guidelines.5 A multidisciplinary team (MDT) is essential, including pediatric and adult cardiologists, dermatologists, geneticists, genetic counselors, pediatricians, and pathologists to provide holistic care. The MDT ensures accurate diagnosis, family screening, and ongoing management. Educating patients and health care professionals is key to raising awareness and improving understanding of these rare syndromes. Currently, no established networks exist for Naxos disease and related syndromes. Creating a network tailored to patient needs is crucial to overcoming diagnostic delays and coordinating care across primary care, specialized centers, and support services. This network should excel in diagnostics, therapy, and patient support, while considering geographic and socioeconomic diversity. Artificial intelligence (AI) offers new opportunities in managing Naxos disease by improving screening, referral, and diagnosis. AI can enhance early identification of at-risk individuals, streamline referrals to specialized centers, and support personalized management in clinical settings. The successful AI integration will rely on collaboration between clinical experts, engineers, and AI specialists to harness its full potential in advancing patient care. Management Management of cardiac disease Arrhythmias The role of anti-arrhythmics in Naxos disease and similar cardio-cutaneous syndromes has not been specifically documented, so guidance from ARVC management is typically applied.5 Medical therapy for patients with ventricular arrhythmias aims to control symptoms, reduce arrhythmia burden, and mitigate ICDshocks. Beta-blockers are commonly used in ARVC for their safety, often as the first-line therapy, sometimes combined with flecainide. Amiodarone is a second-line therapy with variable efficacy due to small sample sizes and differing disease stages among patients.60, 61, 62 Its use in children with Naxos disease requires careful consideration of potential side effects. According to the ESC guidelines for cardiomyopathies (Class IIa, Level of Evidence: C), catheter ablation, including the epicardial approach guided by 3D electroanatomical mapping, should be considered for patients with incessant ventricular tachycardia or frequent ICD interventions despite pharmacological treatment.5Atrial fibrillation, which has been observed in the largest series of Naxos disease cases,37 should be managed according to the latest guidelines for cardiomyopathies.5 Heart failure Heart failure management in ARVC involves treating LV and RV dysfunction. For LV dysfunction, the treatment in adults should align with recommendations from current international guidelines 63,64 and include beta-blockers, aldosterone antagonists, sodium-glucose cotransporter 2 inhibitors, and angiotensin receptor/neprilysin inhibitor. Conversely, in children, data on sodium-glucose transport protein 2 inhibitors are scant, and recent findings from the PANORAMA-HF trial seem to question the clinical superiority of angiotensin receptor/neprilysin inhibitor over angiotensin converting enzyme inhibitors.65,66 Treatment for RV dysfunction lacks specific trials, but common practice in experienced centers includes angiotensin converting enzyme inhibitors/angiotensin II receptor blockers, beta-blockers, and diuretics. Risk stratification Risk prediction and stratification for SCD are critical in managing patients, especially for identifying those needing ICD implantation. The first gene-specific risk study in ARVC focused on Naxos disease patients,37 revealing a high incidence of arrhythmia-related death and major events comparable to pre-ICD era ARVC literature. Despite nearly 100% penetrance of the cardiac component in Naxos disease, studies indicate similar arrhythmia and SCD risks compared to ARVC from other gene variants.51,67 Recent research highlights genotype-specific risk factors, influencing the application of ARVC risk calculators across patient subgroups.68 Although no recent studies have specifically focused on these syndromes, recommendations are extrapolated from research on nonsyndromic ARVC. According to the ESC guidelines for cardiomyopathies, ICD implantation is recommended for secondary prevention in ARVC patients who have survived cardiac arrest or experienced ventricular arrhythmia with hemodynamic instability (Class I, Level of Evidence: A) and should be considered for those with hemodynamically tolerated ventricular tachycardia (Class IIa, Level of Evidence: B). For primary prevention, high-risk features and the 2019 ARVC risk calculator guide individualized ICD decisions, especially for JUP variants, which show no major differences compared to PKP2 cases.69 In DSP variants, a gene-first risk stratification approach allows for tailored ventricular arrhythmia risk assessment, providing a more personalized approach to prevention.70 As arrhythmic risk in ARVC evolves over time, regular re-evaluation is essential. Upon identification of the specific cutaneous features, a cardiac evaluation and yearly follow-up with 12-lead resting ECG, 24-hour Holter monitoring, CMR or, if not applicable, 2D-echocardiography is required. Longitudinal studies are needed to define the optimal frequency of follow-up in Naxos disease and similar syndromes. Additionally, a dedicated risk model for these syndromes remains greatly needed and would be facilitated by developing a Registry for Cardiocutaneous Syndromes. Such a registry would be crucial, providing essential data on disease progression and treatment outcomes to improve patient care. Lifestyle modification The interplay between genotype and exercise in ARVC shows a dose-dependent relationship where more intense exercise increases disease risk and adverse outcomes in desmosomal variant carriers.71 Endurance sports are linked to earlier disease onset, pronounced structural abnormalities, higher risk of ventricular arrhythmias, and increased heart failure risk in ARVC patients.72, 73, 74 There is an exercise threshold beyond which disease phenotypes appear; thus, individuals with desmosomal variants may engage safely in low-to-moderate intensity exercise.75 Although specific guidelines for Naxos disease are unavailable, patients should avoid competitive sports and high-intensity exercise. Management of noncardiac disease Management of syndromic palmoplantar keratosis in Naxos disease focuses on symptom relief, as no specific cure exists. Treatments aim to soften and smooth thickened skin, improve function and appearance, and reduce discomfort. Methods include regular soaks and gentle scale removal. Topical keratolytics like urea, lactic acid, salicylic acid, and propylene glycol are used cautiously in children. Retinoids (eg, tazarotene, tretinoin) are reserved for severe cases due to side effects and offer temporary relief.76 Multidisciplinary team, education, and network organization leveraging artificial intelligence in managing Naxos disease and related syndromes Managing Naxos disease and related cardiocutaneous syndromes necessitates a MDT team and guideline-based approach for accurate diagnosis, family screening, and ongoing care.5 Effective networks should overcome diagnostic delays, coordinate care across various health care levels, and ensure comprehensive patient support while addressing geographic and socioeconomic diversity. AI is revolutionizing cardiac disease management77 by enhancing screening, streamlining referrals, and aiding nuanced diagnosis and tailored management. Successful AI implementation requires robust data sets and collaboration among clinical experts, biomedical engineers, and AI specialists to advance patient care for these syndromes. The unmet need: patients’ perspective A cardiovascular disease diagnosis, especially 1 with sudden death as a first symptom, is shocking, particularly for young individuals and athletes. While doctors treat the disease symptoms and prognosis, patients often struggle with the psychosocial burden. The fear, lifestyle restrictions, and work limitations cause anxiety and stress, negatively impacting mental health. Early-stage psychosocial support in clinics is essential for maintaining mental health, productivity, and quality of life. Long-term comprehensive psychosocial support is crucial to relieve the mental and emotional unrest caused by Naxos disease and similar inherited cardiocutaneous syndromes. Conclusions Naxos disease and related cardio-cutaneous syndromes are rare, globally distributed conditions caused by genetic variants affecting desmosomal proteins, leading to arrhythmogenic cardiomyopathy and skin abnormalities. The pathophysiology involves disrupted cellular adhesion and significant alterations in intracellular signaling pathways, contributing to complex cardiac and skin manifestations. Effective management requires early recognition, precise genetic identification, individualized treatment, and psychosocial support. Despite limited understanding of their epidemiology and clinical diversity, advancements in molecular technologies and targeted therapies highlight the need for an international registry to facilitate collaborative research, share best practices, and improve patient care and quality of life. Funding support and author disclosures Dr Protonotarios holds consulting agreements with Tenaya Therapeutics and Avidity Biosciences. Dr Crisci has received a research grant by the CardioPath program from Federico II University of Naples, Italy. Dr Abrams is a consultant/scientific advisory board for Thryv Therapeutics, Rockets Pharmaceuticals, Solid Biosciences. Dr Anastasakis is on the advisory board of Pfizer, Genesis, Takeda, Sanofi, and BMS. Dr Mestroni has received grant support from Tenaya Therapeutics, Pfizer, AstraZeneca, Owkin, Greenstone Bioscience, and Bristol Myers Squibb; and is a member of the scientific advisory board of Tenaya Therapeutics. Dr Saffitz is consultant for Implicit Bioscience and Rocket Pharmaceuticals. Dr McKenna is a scientific advisor to Health in Code and to Tenaya Therapeutics. Dr Elliott holds consultancies in Biomarin, Pfizer, BMS, and the Patent: US Pat No 62/903,103: GENE THERAPY COMPOSITIONS AND TREATMENT OF RIGHT VENTRICULAR ARRYTHMOGENIC CARDIOMYOPATHY. Dr Kaski has consultancy/advisory board activities with Tenaya Therapeutics, Cytokinetics, Lexeo, Avidity Biosciences, and Accord Healthcare. All other authors have reported that they have no relationships relevant to the contents of this paper to disclose. Supplementary data Download: Download Word document (17MB) Supplementary Data. Recommended articles References 1N. Protonotarios, A. Tsatsopoulou, P. Patsourakos, et al. Cardiac abnormalities in familial palmoplantar keratosis Br Heart J, 56 (1986), pp. 321-326 CrossrefView in ScopusGoogle Scholar 2G. McKoy, N. Protonotarios, A. Crosby, et al. 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Applications of artificial intelligence and machine learning in heart failure Eur Heart J Digit Health, 3 (2022), pp. 311-322 CrossrefView in ScopusGoogle Scholar Cited by (3) NAXCARE: a clinical outcome registry for Naxos disease and related cardiocutaneous syndromes 2025, Hellenic Journal of Cardiology Show abstract The NAXCARE (NAXos disease and Cardiocutaneous Assessment and Registry for Evaluation) is a global initiative designed to collect, store, and analyze clinical outcomes data on patients with Naxos disease and related cardiocutaneous syndromes (CCS). This registry aims to fill the gaps in clinical knowledge, enhance treatment approaches, and improve patient outcomes by systematically documenting disease progression, genetic profiles, and patient care pathways. The following methodology outlines the registry’s design, data collection protocols, management, security measures, and anticipated contributions to research and clinical practice. ### The history of Naxos disease: pioneering innovation and collaboration with limited means 2025, Hellenic Journal of Cardiology Show abstract In 1984, on the remote Greek island of Naxos, Drs. Nikos Protonotarios and Adalena Tsatsopoulou, during their mandatory rural service, encountered patients with a striking combination of ventricular tachycardia, woolly hair, and palmoplantar keratoderma—clinical clues suggesting a novel link between cardiac and dermatologic abnormalities. Despite the absence of advanced medical facilities and research funding, their determination turned a resource-limited setting into a crucible of discovery. Using basic diagnostic tools and self-funded research, they identified four families with nine affected individuals, culminating in the first description of Naxos disease in the British Heart Journal in 1986. Through international collaborations, their research mapped the gene locus to chromosome 17q21, revealing a plakoglobin (JUP) mutation and establishing Naxos disease as the first arrhythmogenic right ventricular cardiomyopathy (ARVC) variant linked to desmosomal defects. This pivotal discovery reframed ARVC as a disease of cell-cell adhesion, influencing broader cardiomyopathy research and enhancing risk stratification and early intervention strategies. The unique skin markers in Naxos disease enabled early identification of at-risk individuals, offering insights into disease progression rarely achievable in non-syndromic ARVC. Today, this legacy continues through NAXCARE, an international clinical outcome registry for Naxos disease and related syndromes, led by Dr. Tsatsopoulou and supported by the Hellenic Society of Cardiology. The journey from a remote island clinic to a global research network underscores how curiosity, resilience, and collaboration can transform local observations into breakthroughs that advance cardiology worldwide. ### Broad Electrocardiogram Syndromes Spectrum: From Common Emergencies to Particular Electrical Heart Disorders—Part II 2025, Diagnostics The authors attest they are in compliance with human studies committees and animal welfare regulations of the authors’ institutions and Food and Drug Administration guidelines, including patient consent where appropriate. For more information, visit the Author Center. ∗ These authors have contributed equally to the design of the manuscript. © 2025 The Authors. Published by Elsevier on behalf of the American College of Cardiology Foundation. Recommended articles ARVC/D and the dyad: A long distance relationship? 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190022	https://biz.libretexts.org/Courses/Lumen_Learning/Introduction_to_Business_(Lumen)/07%3A_Module_2-_Economic_Environment/7.17%3A_Equilibrium_Price_and_Quantity	Skip to main content 7.17: Equilibrium, Price, and Quantity Last updated : Jul 31, 2022 Save as PDF 7.16: Surpluses and Shortages 7.18: Practice- Equilibrium Page ID : 45372 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Explain equilibrium price and quantity Equilibrium: Where Supply and Demand Intersect When two lines on a diagram cross, this intersection usually means something. On a graph, the point where the supply curve (S) and the demand curve (D) intersect is the equilibrium. The equilibrium price is the only price where the desires of consumers and the desires of producers agree—that is, where the amount of the product that consumers want to buy (quantity demanded) is equal to the amount producers want to sell (quantity supplied). This mutually desired amount is called the equilibrium quantity. At any other price, the quantity demanded does not equal the quantity supplied, so the market is not in equilibrium at that price. It should be clear from the previous discussions of surpluses and shortages, that if a market is not in equilibrium, market forces will push the market to the equilibrium. If you have only the demand and supply schedules, and no graph, you can find the equilibrium by looking for the price level on the tables where the quantity demanded and the quantity supplied are equal (again, the numbers in bold in Table 1 indicate this point). Finding Equilibrium with Algebra We’ve just explained two ways of finding a market equilibrium: by looking at a table showing the quantity demanded and supplied at different prices, and by looking at a graph of demand and supply. We can also identify the equilibrium with a little algebra if we have equations for the supply and demand curves. Let’s practice solving a few equations that you will see later in the course. Right now, we are only going to focus on the math. Later you’ll learn why these models work the way they do, but let’s start by focusing on solving the equations. Suppose that the demand for soda is given by the following equation: 𝑄⁢𝑑=16−2⁢𝑃(7.17.1) where 𝑄⁢𝑑 is the amount of soda that consumers want to buy (i.e., quantity demanded), and P is the price of soda. Suppose the supply of soda is 𝑄⁢𝑠=2+5⁢𝑃(7.17.2) where Qs is the amount of soda that producers will supply (i.e., quantity supplied). (Remember, these are simple equations for lines). Finally, recall that the soda market converges to the point where supply equals demand, or 𝑄⁢𝑑=𝑄⁢𝑠(7.17.3) We now have a system of three equations and three unknowns (Qd, Qs, and P), which we can solve with algebra. Since 𝑄⁢𝑑=𝑄⁢𝑠(7.17.4), we can set the demand and supply equations equal to each other: 𝑄⁢𝑑=𝑄⁢𝑠16−2⁢𝑃=2+5⁢𝑃(7.17.5) Step 1: Isolate the variable by adding 2P to both sides of the equation, and subtracting 2 from both sides. 16−2⁢𝑃=2+5⁢𝑃−2+2⁢𝑃=−2+5⁢𝑃14=7⁢𝑃(7.17.6) Step 2: Simplify the equation by dividing both sides by 7. 147=7⁢𝑃7(7.17.7) 2=𝑃(7.17.8) The equilibrium price of soda, that is, the price where Qs = Qd will be $2. Now we want to determine the quantity amount of soda. We can do this by plugging the equilibrium price into either the equation showing the demand for soda or the equation showing the supply of soda. Let’s use demand. Remember, the formula for quantity demanded is the following: 𝑄⁢𝑑=16−2⁢𝑃(7.17.9) Taking the price of $2, and plugging it into the demand equation, we get 𝑄⁢𝑑=16−2⁢(2)𝑄⁢𝑑=16−4𝑄⁢𝑑=12(7.17.10) So, if the price is $2 each, consumers will purchase 12. How much will producers supply, or what is the quantity supplied? Taking the price of $2, and plugging it into the equation for quantity supplied, we get the following: 𝑄⁢𝑠=16+5⁢𝑃𝑄⁢𝑠=2+5⁢(2)𝑄⁢𝑠=2+10𝑄⁢𝑠=12(7.17.11) Now, if the price is $2 each, producers will supply 12 sodas. This means that we did our math correctly, since 𝑄⁢𝑑=𝑄⁢𝑠(7.17.12) and both Qd and Qs are equal to 12. That confirms that we’ve found the equilibrium quantity. Try It Watch this video for a closer look at market equilibrium: You can view the transcript for “Episode 14: Market Equilibrium” (opens in new window). Equilibrium occurs at the point where quantity supplied = quantity demanded. Equilibrium and Economic Efficiency Equilibrium is important to create both a balanced market and an efficient market. If a market is at its equilibrium price and quantity, then it has no reason to move away from that point, because it’s balancing the quantity supplied and the quantity demanded. However, if a market is not at equilibrium, then economic pressures arise to move the market toward the equilibrium price and equilibrium quantity. This happens either because there is more supply than what the market is demanding or because there is more demand than the market is supplying. This balance is a natural function of a free-market economy. Also, a competitive market that is operating at equilibrium is an efficient market. Economists typically define efficiency in this way: when it is impossible to improve the situation of one party without imposing a cost on another. Conversely, if a situation is inefficient, it becomes possible to benefit at least one party without imposing costs on others. Practice Questions Finding Equilibrium using the Four-Step Process We know that equilibrium is the place where the supply and demand curves intersect, or the point where buyers want to buy the same amount that sellers want to sell. Let’s take a closer look at how to find the equilibrium point using the four-step process. These steps explain how to first, draw the demand and supply curves on a graph and find the equilibrium. Next, consider how an economic change (e.g. a natural disaster, a change in production technology, a change in tastes and preferences, income, etc.) might affect supply or demand, then make adjustments to the graph to identify the new equilibrium point. Step 1. Draw demand and supply curves showing the market before the economic change took place. Think about the shift variables for demand, and the shift variables for supply. Using this diagram, find the initial equilibrium values for price and quantity. Step 2. Decide whether the economic change being analyzed affects demand or supply. In other words, does the event refer to something in the list of demand shift variables or supply shift variables? Step 3. Determine whether the effect on demand or supply causes the curve to shift to the right or to the left, and sketch the new demand or supply curve on the diagram. In other words, does the event increase or decrease the amount consumers want to buy or the amount producers want to sell? Step 4. Identify the new equilibrium, and then compare the original equilibrium price and quantity to the new equilibrium price and quantity. Let’s consider one example that involves a shift in supply and one that involves a shift in demand. Then we will consider an example where both supply and demand shift. Exercise 1: Good Weather for Salmon Fishing Let’s suppose that during the summer of 2015, weather conditions were excellent for commercial salmon fishing off the California coast. Heavy rains meant higher than normal levels of water in the rivers, which helps the salmon to breed. Slightly cooler ocean temperatures stimulated the growth of plankton, the microscopic organisms at the bottom of the ocean food chain, providing everything in the ocean with a hearty food supply. The ocean stayed calm during fishing season, so commercial fishing operations did not lose many days to bad weather. How did these climate conditions affect the quantity and price of salmon? Let’s consider this situation using the four-step process and the data below. Table 7.17.1:. Salmon Fishing \| Price per Pound \| Quantity Supplied in 2014 \| Quantity Supplied in 2015 \| Quantity Demanded \| --- --- \| \| $2.00 \| 80 \| 400 \| 840 \| \| $2.25 \| 120 \| 480 \| 680 \| \| $2.50 \| 160 \| 550 \| 550 \| \| $2.75 \| 200 \| 600 \| 450 \| \| $3.00 \| 230 \| 640 \| 350 \| \| $3.25 \| 250 \| 670 \| 250 \| \| $3.50 \| 270 \| 700 \| 200 \| Let’s walk through the four steps together using this example, and see how the graph changes. Use the interactive graph below (Figure 3) by clicking on the arrows at the bottom of the activity to navigate through the steps. In short, good weather conditions increased supply of the California commercial salmon. The result was a higher equilibrium quantity of salmon bought and sold in the market at a lower price. Exercise 2: Newspapers and the Internet According to the Pew Research Center for People and the Press, more and more people, especially younger people, are getting their news from online and digital sources. The majority of U.S. adults now own smartphones or tablets, and most of those Americans say they use them in part to get the news. From 2004 to 2012, the share of Americans who reported getting their news from digital sources increased from 24 percent to 39 percent. How has this trend affected consumption of print news media and radio and television news? Figure 7.17.3: and the text below illustrate the four-step analysis used to answer this question. Step 1. Draw a demand and supply model to think about what the market looked like before the event. The demand curve D0 and the supply curve S0 show the original relationships. In this case, the curves are drawn without specific numbers on the price and quantity axis. Step 2. Did the change described affect supply or demand? Exercise 7.17.1 Step 2. Did the change described affect supply or demand? Answer : A change in tastes, from traditional news sources (print, radio, and television) to digital sources, caused a change in demand for the former. x Exercise 7.17.2 Step 3. Was the effect on demand positive or negative? Answer : A shift to digital news sources will tend to mean a lower quantity demanded of traditional news sources at every given price, causing the demand curve for print and other traditional news sources to shift to the left, from D0 to D1. Exercise 7.17.3 Compare the new equilibrium price and quantity to the original equilibrium price. Answer : The decline in print news reading predates 2004. Print newspaper circulation peaked in 1973 and has declined since then due to competition from television and radio news. In 1991, 55 percent of Americans indicated that they got their news from print sources, while only 29 percent did so in 2012. Radio news has followed a similar path in recent decades, with the share of Americans getting their news from radio declining from 54 percent in 1991 to 33 percent in 2012. Television news has held its own during the last fifteen years, with the market share staying in the mid to upper fifties. What does this suggest for the future, given that two-thirds of Americans under thirty years old say they don’t get their news from television at all? Practice Questions Try It Play the simulation below multiple times to see how different choices lead to different outcomes. All simulations allow unlimited attempts so that you can gain experience applying the concepts. A link to an interactive elements can be found at the bottom of this page. Contributors and Attributions CC licensed content, Original Revision and adaptation. Provided by: Lumen Learning. License: CC BY: Attribution Food Trucks and Changes in Equilibrium. Authored by: Clark Aldrich for Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously Principles of Macroeconomics Chapter 3.3. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at 7.16: Surpluses and Shortages 7.18: Practice- Equilibrium
190023	https://en.wikipedia.org/wiki/Cell_nucleus	Jump to content Search Contents 1 Chromosomes 2 Nuclear structures and landmarks 2.1 Nuclear envelope and pores 2.2 Nuclear lamina 2.3 Nucleolus 2.4 Splicing speckles 2.5 Cajal bodies and gems 2.6 Other nuclear bodies 2.6.1 PIKA and PTF domains 2.6.2 PML-nuclear bodies 2.6.3 Paraspeckles 2.6.4 Perichromatin fibrils 2.6.5 Clastosomes 3 Function 3.1 Cell compartmentalization 3.2 Replication 3.3 Gene expression 3.4 Processing of pre-mRNA 4 Dynamics and regulation 4.1 Nuclear transport 4.2 Assembly and disassembly 4.3 Disease-related dynamics 5 Nuclei per cell 5.1 Anucleated cells 5.2 Multinucleated cells 6 Evolution 7 History 8 See also 9 References 10 Further reading 11 External links Cell nucleus Afrikaans Alemannisch العربية Aragonés Asturianu Azərbaycanca বাংলা 閩南語 / Bn-lm-gí Башҡортса Беларуская Беларуская (тарашкевіца) Български Bosanski Català Чӑвашла Čeština Cymraeg Dansk Davvisámegiella Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge Gaelg Galego ГӀалгӀай 客家語 / Hak-k-ngî 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Interlingua IsiZulu Íslenska Italiano עברית Jawa ಕನ್ನಡ ქართული Қазақша Kiswahili Kreyòl ayisyen Kurdî Кыргызча Latina Latviešu Lëtzebuergesch Lietuvių Limburgs Magyar Македонски മലയാളം მარგალური Bahasa Melayu Монгол Nederlands 日本語 Norsk bokmål Norsk nynorsk Occitan Oromoo Oʻzbekcha / ўзбекча پنجابی Plattdüütsch Polski Português Qaraqalpaqsha Romnă Русиньскый Русский Scots සිංහල Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Sunda Suomi Svenska Tagalog தமிழ் Татарча / tatarça తెలుగు ไทย Türkçe Türkmençe Українська اردو Tiếng Việt Winaray 吴语 ייִדיש 粵語中文 Edit links Article Talk Read View source View history Tools Actions Read View source 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 Eukaryotic membrane-bounded organelle containing DNA \| Cell biology \| \| Animal cell diagram \| \| Components of a typical animal cell: 1. Nucleolus 2. Nucleus 3. Ribosome (dots as part of 5) 4. Vesicle 5. Rough endoplasmic reticulum 6. Golgi apparatus (or, Golgi body) 7. Cytoskeleton 8. Smooth endoplasmic reticulum 9. Mitochondrion 10. Vacuole 11. Cytosol (fluid that contains organelles; with which, comprises cytoplasm) 12. Lysosome 13. Centrosome 14. Cell membrane \| The cell nucleus (from Latin nucleus or nuculeus 'kernel, seed'; pl.: nuclei) is a membrane-bound organelle found in eukaryotic cells. Eukaryotic cells usually have a single nucleus, but a few cell types, such as mammalian red blood cells, have no nuclei, and a few others including osteoclasts have many. The main structures making up the nucleus are the nuclear envelope, a double membrane that encloses the entire organelle and isolates its contents from the cellular cytoplasm; and the nuclear matrix, a network within the nucleus that adds mechanical support. The cell nucleus contains nearly all of the cell's genome. Nuclear DNA is often organized into multiple chromosomes – long strands of DNA dotted with various proteins, such as histones, that protect and organize the DNA. The genes within these chromosomes are structured in such a way to promote cell function. The nucleus maintains the integrity of genes and controls the activities of the cell by regulating gene expression. Because the nuclear envelope is impermeable to large molecules, nuclear pores are required to regulate nuclear transport of molecules across the envelope. The pores cross both nuclear membranes, providing a channel through which larger molecules must be actively transported by carrier proteins while allowing free movement of small molecules and ions. Movement of large molecules such as proteins and RNA through the pores is required for both gene expression and the maintenance of chromosomes. Although the interior of the nucleus does not contain any membrane-bound subcompartments, a number of nuclear bodies exist, made up of unique proteins, RNA molecules, and particular parts of the chromosomes. The best-known of these is the nucleolus, involved in the assembly of ribosomes. Chromosomes Main article: Chromosome Further information: Nuclear organization The cell nucleus contains the majority of the cell's genetic material in the form of multiple linear DNA molecules organized into structures called chromosomes. Each human cell contains roughly two meters of DNA.: 405 During most of the cell cycle these are organized in a DNA-protein complex known as chromatin, and during cell division the chromatin can be seen to form the well-defined chromosomes familiar from a karyotype. A small fraction of the cell's genes are located instead in the mitochondria.: 438 There are two types of chromatin. Euchromatin is the less compact DNA form, and contains genes that are frequently expressed by the cell. The other type, heterochromatin, is the more compact form, and contains DNA that is infrequently transcribed. This structure is further categorized into facultative heterochromatin, consisting of genes that are organized as heterochromatin only in certain cell types or at certain stages of development, and constitutive heterochromatin that consists of chromosome structural components such as telomeres and centromeres. During interphase the chromatin organizes itself into discrete individual patches, called chromosome territories. Active genes, which are generally found in the euchromatic region of the chromosome, tend to be located towards the chromosome's territory boundary. Antibodies to certain types of chromatin organization, in particular, nucleosomes, have been associated with a number of autoimmune diseases, such as systemic lupus erythematosus. These are known as anti-nuclear antibodies (ANA) and have also been observed in concert with multiple sclerosis as part of general immune system dysfunction. Nuclear structures and landmarks Further information: Nuclear equivalence The nucleus contains nearly all of the cell's DNA, surrounded by a network of fibrous intermediate filaments called the nuclear matrix, and is enveloped in a double membrane called the nuclear envelope. The nuclear envelope separates the fluid inside the nucleus, called the nucleoplasm, from the rest of the cell. The size of the nucleus is correlated to the size of the cell, and this ratio is reported across a range of cell types and species. In eukaryotes the nucleus in many cells typically occupies 10% of the cell volume.: 178 The nucleus is the largest organelle in animal cells.: 12 In human cells, the diameter of the nucleus is approximately six micrometres (μm).: 179 Nuclear envelope and pores Main articles: Nuclear envelope and Nuclear pore The nuclear envelope consists of two membranes, an inner and an outer nuclear membrane, perforated by nuclear pores.: 649 Together, these membranes serve to separate the cell's genetic material from the rest of the cell contents, and allow the nucleus to maintain an environment distinct from the rest of the cell. Despite their close apposition around much of the nucleus, the two membranes differ substantially in shape and contents. The inner membrane surrounds the nuclear content, providing its defining edge.: 14 Embedded within the inner membrane, various proteins bind the intermediate filaments that give the nucleus its structure.: 649 The outer membrane encloses the inner membrane, and is continuous with the adjacent endoplasmic reticulum membrane.: 649 As part of the endoplasmic reticulum membrane, the outer nuclear membrane is studded with ribosomes that are actively translating proteins across membrane.: 649 The space between the two membranes is called the perinuclear space, and is continuous with the endoplasmic reticulum lumen.: 649 In a mammalian nuclear envelope there are between 3000 and 4000 nuclear pore complexes (NPCs) perforating the envelope.: 650 Each NPC contains an eightfold-symmetric ring-shaped structure at a position where the inner and outer membranes fuse. The number of NPCs can vary considerably across cell types; small glial cells only have about a few hundred, with large Purkinje cells having around 20,000.: 650 The NPC provides selective transport of molecules between the nucleoplasm and the cytosol. The nuclear pore complex is composed of approximately thirty different proteins known as nucleoporins.: 649 The pores are about 60–80 million daltons in molecular mass and consist of around 50 (in yeast) to several hundred proteins (in vertebrates).: 622–4 The pores are 100 nm in total diameter; however, the gap through which molecules freely diffuse is only about 9 nm wide, due to the presence of regulatory systems within the center of the pore. This size selectively allows the passage of small water-soluble molecules while preventing larger molecules, such as nucleic acids and larger proteins, from inappropriately entering or exiting the nucleus. These large molecules must be actively transported into the nucleus instead. Attached to the ring is a structure called the nuclear basket that extends into the nucleoplasm, and a series of filamentous extensions that reach into the cytoplasm. Both structures serve to mediate binding to nuclear transport proteins.: 509–10 Most proteins, ribosomal subunits, and some RNAs are transported through the pore complexes in a process mediated by a family of transport factors known as karyopherins. Those karyopherins that mediate movement into the nucleus are also called importins, whereas those that mediate movement out of the nucleus are called exportins. Most karyopherins interact directly with their cargo, although some use adaptor proteins. Steroid hormones such as cortisol and aldosterone, as well as other small lipid-soluble molecules involved in intercellular signaling, can diffuse through the cell membrane and into the cytoplasm, where they bind nuclear receptor proteins that are trafficked into the nucleus. There they serve as transcription factors when bound to their ligand; in the absence of a ligand, many such receptors function as histone deacetylases that repress gene expression.: 488 Nuclear lamina Main article: Nuclear lamina In animal cells, two networks of intermediate filaments provide the nucleus with mechanical support: The nuclear lamina forms an organized meshwork on the internal face of the envelope, while less organized support is provided on the cytosolic face of the envelope. Both systems provide structural support for the nuclear envelope and anchoring sites for chromosomes and nuclear pores. The nuclear lamina is composed mostly of lamin proteins. Like all proteins, lamins are synthesized in the cytoplasm and later transported to the nucleus interior, where they are assembled before being incorporated into the existing network of nuclear lamina. Lamins found on the cytosolic face of the membrane, such as emerin and nesprin, bind to the cytoskeleton to provide structural support. Lamins are also found inside the nucleoplasm where they form another regular structure, known as the nucleoplasmic veil, that is visible using fluorescence microscopy. The actual function of the veil is not clear, although it is excluded from the nucleolus and is present during interphase. Lamin structures that make up the veil, such as LEM3, bind chromatin and disrupting their structure inhibits transcription of protein-coding genes. Like the components of other intermediate filaments, the lamin monomer contains an alpha-helical domain used by two monomers to coil around each other, forming a dimer structure called a coiled coil. Two of these dimer structures then join side by side, in an antiparallel arrangement, to form a tetramer called a protofilament. Eight of these protofilaments form a lateral arrangement that is twisted to form a ropelike filament. These filaments can be assembled or disassembled in a dynamic manner, meaning that changes in the length of the filament depend on the competing rates of filament addition and removal. Mutations in lamin genes leading to defects in filament assembly cause a group of rare genetic disorders known as laminopathies. The most notable laminopathy is the family of diseases known as progeria, which causes the appearance of premature aging in those with the condition. The exact mechanism by which the associated biochemical changes give rise to the aged phenotype is not well understood. Nucleolus Main article: Nucleolus Further information: Nuclear bodies The nucleolus is the largest of the discrete densely stained, membraneless structures known as nuclear bodies found in the nucleus. It forms around tandem repeats of rDNA, DNA coding for ribosomal RNA (rRNA). These regions are called nucleolar organizer regions (NOR). The main roles of the nucleolus are to synthesize rRNA and assemble ribosomes. The structural cohesion of the nucleolus depends on its activity, as ribosomal assembly in the nucleolus results in the transient association of nucleolar components, facilitating further ribosomal assembly, and hence further association. This model is supported by observations that inactivation of rDNA results in intermingling of nucleolar structures. In the first step of ribosome assembly, a protein called RNA polymerase I transcribes rDNA, which forms a large pre-rRNA precursor. This is cleaved into two large rRNA subunits – 5.8S, and 28S, and a small rRNA subunit 18S.: 328 The transcription, post-transcriptional processing, and assembly of rRNA occurs in the nucleolus, aided by small nucleolar RNA (snoRNA) molecules, some of which are derived from spliced introns from messenger RNAs encoding genes related to ribosomal function. The assembled ribosomal subunits are the largest structures passed through the nuclear pores.: 526 When observed under the electron microscope, the nucleolus can be seen to consist of three distinguishable regions: the innermost fibrillar centers (FCs), surrounded by the dense fibrillar component (DFC) (that contains fibrillarin and nucleolin), which in turn is bordered by the granular component (GC) (that contains the protein nucleophosmin). Transcription of the rDNA occurs either in the FC or at the FC-DFC boundary, and, therefore, when rDNA transcription in the cell is increased, more FCs are detected. Most of the cleavage and modification of rRNAs occurs in the DFC, while the latter steps involving protein assembly onto the ribosomal subunits occur in the GC. Splicing speckles Speckles are subnuclear structures that are enriched in pre-messenger RNA splicing factors and are located in the interchromatin regions of the nucleoplasm of mammalian cells. At the fluorescence-microscope level they appear as irregular, punctate structures, which vary in size and shape, and when examined by electron microscopy they are seen as clusters of interchromatin granules. Speckles are dynamic structures, and both their protein and RNA-protein components can cycle continuously between speckles and other nuclear locations, including active transcription sites. Speckles can work with p53 as enhancers of gene activity to directly enhance the activity of certain genes. Moreover, speckle-associating and non-associating p53 gene targets are functionally distinct. Studies on the composition, structure and behaviour of speckles have provided a model for understanding the functional compartmentalization of the nucleus and the organization of the gene-expression machinery splicing snRNPs and other splicing proteins necessary for pre-mRNA processing. Because of a cell's changing requirements, the composition and location of these bodies changes according to mRNA transcription and regulation via phosphorylation of specific proteins. The splicing speckles are also known as nuclear speckles (nuclear specks), splicing factor compartments (SF compartments), interchromatin granule clusters (IGCs), and B snurposomes. B snurposomes are found in the amphibian oocyte nuclei and in Drosophila melanogaster embryos. B snurposomes appear alone or attached to the Cajal bodies in the electron micrographs of the amphibian nuclei. While nuclear speckles were originally thought to be storage sites for the splicing factors, a more recent study demonstrated that organizing genes and pre-mRNA substrates near speckles increases the kinetic efficiency of pre-mRNA splicing, ultimately boosting protein levels by modulation of splicing. Cajal bodies and gems A nucleus typically contains between one and ten compact structures called Cajal bodies or coiled bodies (CB), whose diameter measures between 0.2 μm and 2.0 μm depending on the cell type and species. When seen under an electron microscope, they resemble balls of tangled thread and are dense foci of distribution for the protein coilin. CBs are involved in a number of different roles relating to RNA processing, specifically small nucleolar RNA (snoRNA) and small nuclear RNA (snRNA) maturation, and histone mRNA modification. Similar to Cajal bodies are Gemini of Cajal bodies, or gems, whose name is derived from the Gemini constellation in reference to their close "twin" relationship with CBs. Gems are similar in size and shape to CBs, and in fact are virtually indistinguishable under the microscope. Unlike CBs, gems do not contain small nuclear ribonucleoproteins (snRNPs), but do contain a protein called survival of motor neuron (SMN) whose function relates to snRNP biogenesis. Gems are believed to assist CBs in snRNP biogenesis, though it has also been suggested from microscopy evidence that CBs and gems are different manifestations of the same structure. Later ultrastructural studies have shown gems to be twins of Cajal bodies with the difference being in the coilin component; Cajal bodies are SMN positive and coilin positive, and gems are SMN positive and coilin negative. Other nuclear bodies Main article: Nuclear bodies Subnuclear structure sizes \| Structure name \| Structure diameter \| Ref. \| \| Cajal bodies \| 0.2–2.0 μm \| \| \| Clastosomes \| 0.2–0.5 μm \| \| \| PIKA \| 5 μm \| \| \| PML bodies \| 0.2–1.0 μm \| \| \| Paraspeckles \| 0.5–1.0 μm \| \| \| Speckles \| 20–25 nm \| \| Beyond the nuclear bodies first described by Santiago Ramón y Cajal above (e.g., nucleolus, nuclear speckles, Cajal bodies) the nucleus contains a number of other nuclear bodies. These include polymorphic interphase karyosomal association (PIKA), promyelocytic leukaemia (PML) bodies, and paraspeckles. Although little is known about a number of these domains, they are significant in that they show that the nucleoplasm is not a uniform mixture, but rather contains organized functional subdomains. Other subnuclear structures appear as part of abnormal disease processes. For example, the presence of small intranuclear rods has been reported in some cases of nemaline myopathy. This condition typically results from mutations in actin, and the rods themselves consist of mutant actin as well as other cytoskeletal proteins. PIKA and PTF domains PIKA domains, or polymorphic interphase karyosomal associations, were first described in microscopy studies in 1991. Their function remains unclear, though they were not thought to be associated with active DNA replication, transcription, or RNA processing. They have been found to often associate with discrete domains defined by dense localization of the transcription factor PTF, which promotes transcription of small nuclear RNA (snRNA). PML-nuclear bodies Promyelocytic leukemia protein (PML-nuclear bodies) are spherical bodies found scattered throughout the nucleoplasm, measuring around 0.1–1.0 μm. They are known by a number of other names, including nuclear domain 10 (ND10), Kremer bodies, and PML oncogenic domains. PML-nuclear bodies are named after one of their major components, the promyelocytic leukemia protein (PML). They are often seen in the nucleus in association with Cajal bodies and cleavage bodies. Pml-/- mice, which are unable to create PML-nuclear bodies, develop normally without obvious ill effects, showing that PML-nuclear bodies are not required for most essential biological processes. Paraspeckles Main article: Paraspeckle Discovered by Fox et al. in 2002, paraspeckles are irregularly shaped compartments in the interchromatin space of the nucleus. First documented in HeLa cells, where there are generally 10–30 per nucleus, paraspeckles are now known to also exist in all human primary cells, transformed cell lines, and tissue sections. Their name is derived from their distribution in the nucleus; the "para" is short for parallel and the "speckles" refers to the splicing speckles to which they are always in close proximity. Paraspeckles sequester nuclear proteins and RNA and thus appear to function as a molecular sponge that is involved in the regulation of gene expression. Furthermore, paraspeckles are dynamic structures that are altered in response to changes in cellular metabolic activity. They are transcription dependent and in the absence of RNA Pol II transcription, the paraspeckle disappears and all of its associated protein components (PSP1, p54nrb, PSP2, CFI(m)68, and PSF) form a crescent shaped perinucleolar cap in the nucleolus. This phenomenon is demonstrated during the cell cycle. In the cell cycle, paraspeckles are present during interphase and during all of mitosis except for telophase. During telophase, when the two daughter nuclei are formed, there is no RNA Pol II transcription so the protein components instead form a perinucleolar cap. Perichromatin fibrils Perichromatin fibrils are visible only under electron microscope. They are located next to the transcriptionally active chromatin and are hypothesized to be the sites of active pre-mRNA processing. Clastosomes Clastosomes are small nuclear bodies (0.2–0.5 μm) described as having a thick ring-shape due to the peripheral capsule around these bodies. This name is derived from the Greek klastos (κλαστός), broken and soma (σῶμα), body. Clastosomes are not typically present in normal cells, making them hard to detect. They form under high proteolytic conditions within the nucleus and degrade once there is a decrease in activity or if cells are treated with proteasome inhibitors. The scarcity of clastosomes in cells indicates that they are not required for proteasome function. Osmotic stress has also been shown to cause the formation of clastosomes. These nuclear bodies contain catalytic and regulatory subunits of the proteasome and its substrates, indicating that clastosomes are sites for degrading proteins. Function The nucleus provides a site for genetic transcription that is segregated from the location of translation in the cytoplasm, allowing levels of gene regulation that are not available to prokaryotes. The main function of the cell nucleus is to control gene expression and mediate the replication of DNA during the cell cycle.: 171 Cell compartmentalization The nuclear envelope allows control of the nuclear contents, and separates them from the rest of the cytoplasm where necessary. This is important for controlling processes on either side of the nuclear membrane: In most cases where a cytoplasmic process needs to be restricted, a key participant is removed to the nucleus, where it interacts with transcription factors to downregulate the production of certain enzymes in the pathway. This regulatory mechanism occurs in the case of glycolysis, a cellular pathway for breaking down glucose to produce energy. Hexokinase is an enzyme responsible for the first step of glycolysis, forming glucose-6-phosphate from glucose. At high concentrations of fructose-6-phosphate, a molecule made later from glucose-6-phosphate, a regulator protein removes hexokinase to the nucleus, where it forms a transcriptional repressor complex with nuclear proteins to reduce the expression of genes involved in glycolysis. In order to control which genes are being transcribed, the cell separates some transcription factor proteins responsible for regulating gene expression from physical access to the DNA until they are activated by other signaling pathways. This prevents even low levels of inappropriate gene expression. For example, in the case of NF-κB-controlled genes, which are involved in most inflammatory responses, transcription is induced in response to a signal pathway such as that initiated by the signaling molecule TNF-α, binds to a cell membrane receptor, resulting in the recruitment of signalling proteins, and eventually activating the transcription factor NF-κB. A nuclear localisation signal on the NF-κB protein allows it to be transported through the nuclear pore and into the nucleus, where it stimulates the transcription of the target genes. The compartmentalization allows the cell to prevent translation of unspliced mRNA. Eukaryotic mRNA contains introns that must be removed before being translated to produce functional proteins. The splicing is done inside the nucleus before the mRNA can be accessed by ribosomes for translation. Without the nucleus, ribosomes would translate newly transcribed (unprocessed) mRNA, resulting in malformed and nonfunctional proteins.: 108–15 Replication Main article: Eukaryotic DNA replication The main function of the cell nucleus is to control gene expression and mediate the replication of DNA during the cell cycle.: 171 It has been found that replication happens in a localised way in the cell nucleus. In the S phase of interphase of the cell cycle; replication takes place. Contrary to the traditional view of moving replication forks along stagnant DNA, a concept of replication factories emerged, which means replication forks are concentrated towards some immobilised 'factory' regions through which the template DNA strands pass like conveyor belts. Gene expression Main article: Gene expression See also: Transcription factories Gene expression first involves transcription, in which DNA is used as a template to produce RNA. In the case of genes encoding proteins, that RNA produced from this process is messenger RNA (mRNA), which then needs to be translated by ribosomes to form a protein. As ribosomes are located outside the nucleus, mRNA produced needs to be exported. Since the nucleus is the site of transcription, it also contains a variety of proteins that either directly mediate transcription or are involved in regulating the process. These proteins include helicases, which unwind the double-stranded DNA molecule to facilitate access to it, RNA polymerases, which bind to the DNA promoter to synthesize the growing RNA molecule, topoisomerases, which change the amount of supercoiling in DNA, helping it wind and unwind, as well as a large variety of transcription factors that regulate expression. Processing of pre-mRNA Main article: Post-transcriptional modification Newly synthesized mRNA molecules are known as primary transcripts or pre-mRNA. They must undergo post-transcriptional modification in the nucleus before being exported to the cytoplasm; mRNA that appears in the cytoplasm without these modifications is degraded rather than used for protein translation. The three main modifications are 5' capping, 3' polyadenylation, and RNA splicing. While in the nucleus, pre-mRNA is associated with a variety of proteins in complexes known as heterogeneous ribonucleoprotein particles (hnRNPs). Addition of the 5' cap occurs co-transcriptionally and is the first step in post-transcriptional modification. The 3' poly-adenine tail is only added after transcription is complete.: 509–18 RNA splicing, carried out by a complex called the spliceosome, is the process by which introns, or regions of DNA that do not code for protein, are removed from the pre-mRNA and the remaining exons connected to re-form a single continuous molecule. This process normally occurs after 5' capping and 3' polyadenylation but can begin before synthesis is complete in transcripts with many exons.: 494 Many pre-mRNAs can be spliced in multiple ways to produce different mature mRNAs that encode different protein sequences. This process is known as alternative splicing, and allows production of a large variety of proteins from a limited amount of DNA. Dynamics and regulation Nuclear transport Main article: Nuclear transport The entry and exit of large molecules from the nucleus is tightly controlled by the nuclear pore complexes. Although small molecules can enter the nucleus without regulation, macromolecules such as RNA and proteins require association karyopherins called importins to enter the nucleus and exportins to exit. "Cargo" proteins that must be translocated from the cytoplasm to the nucleus contain short amino acid sequences known as nuclear localization signals, which are bound by importins, while those transported from the nucleus to the cytoplasm carry nuclear export signals bound by exportins. The ability of importins and exportins to transport their cargo is regulated by GTPases, enzymes that hydrolyze the molecule guanosine triphosphate (GTP) to release energy. The key GTPase in nuclear transport is Ran, which is bound to either GTP or GDP (guanosine diphosphate), depending on whether it is located in the nucleus or the cytoplasm. Whereas importins depend on RanGTP to dissociate from their cargo, exportins require RanGTP in order to bind to their cargo. Nuclear import depends on the importin binding its cargo in the cytoplasm and carrying it through the nuclear pore into the nucleus. Inside the nucleus, RanGTP acts to separate the cargo from the importin, allowing the importin to exit the nucleus and be reused. Nuclear export is similar, as the exportin binds the cargo inside the nucleus in a process facilitated by RanGTP, exits through the nuclear pore, and separates from its cargo in the cytoplasm. Specialized export proteins exist for translocation of mature mRNA and tRNA to the cytoplasm after post-transcriptional modification is complete. This quality-control mechanism is important due to these molecules' central role in protein translation. Mis-expression of a protein due to incomplete excision of exons or mis-incorporation of amino acids could have negative consequences for the cell; thus, incompletely modified RNA that reaches the cytoplasm is degraded rather than used in translation. Assembly and disassembly During its lifetime, a nucleus may be broken down or destroyed, either in the process of cell division or as a consequence of apoptosis (the process of programmed cell death). During these events, the structural components of the nucleus — the envelope and lamina — can be systematically degraded. In most cells, the disassembly of the nuclear envelope marks the end of the prophase of mitosis. However, this disassembly of the nucleus is not a universal feature of mitosis and does not occur in all cells. Some unicellular eukaryotes (e.g., yeasts) undergo so-called closed mitosis, in which the nuclear envelope remains intact. In closed mitosis, the daughter chromosomes migrate to opposite poles of the nucleus, which then divides in two. The cells of higher eukaryotes, however, usually undergo open mitosis, which is characterized by breakdown of the nuclear envelope. The daughter chromosomes then migrate to opposite poles of the mitotic spindle, and new nuclei reassemble around them.: 854 At a certain point during the cell cycle in open mitosis, the cell divides to form two cells. In order for this process to be possible, each of the new daughter cells must have a full set of genes, a process requiring replication of the chromosomes as well as segregation of the separate sets. This occurs by the replicated chromosomes, the sister chromatids, attaching to microtubules, which in turn are attached to different centrosomes. The sister chromatids can then be pulled to separate locations in the cell. In many cells, the centrosome is located in the cytoplasm, outside the nucleus; the microtubules would be unable to attach to the chromatids in the presence of the nuclear envelope. Therefore, the early stages in the cell cycle, beginning in prophase and until around prometaphase, the nuclear membrane is dismantled. Likewise, during the same period, the nuclear lamina is also disassembled, a process regulated by phosphorylation of the lamins by protein kinases such as the CDC2 protein kinase. Towards the end of the cell cycle, the nuclear membrane is reformed, and around the same time, the nuclear lamina are reassembled by dephosphorylating the lamins. However, in dinoflagellates, the nuclear envelope remains intact, the centrosomes are located in the cytoplasm, and the microtubules come in contact with chromosomes, whose centromeric regions are incorporated into the nuclear envelope (the so-called closed mitosis with extranuclear spindle). In many other protists (e.g., ciliates, sporozoans) and fungi, the centrosomes are intranuclear, and their nuclear envelope also does not disassemble during cell division. Apoptosis is a controlled process in which the cell's structural components are destroyed, resulting in death of the cell. Changes associated with apoptosis directly affect the nucleus and its contents, for example, in the condensation of chromatin and the disintegration of the nuclear envelope and lamina. The destruction of the lamin networks is controlled by specialized apoptotic proteases called caspases, which cleave the lamin proteins and, thus, degrade the nucleus' structural integrity. Lamin cleavage is sometimes used as a laboratory indicator of caspase activity in assays for early apoptotic activity. Cells that express mutant caspase-resistant lamins are deficient in nuclear changes related to apoptosis, suggesting that lamins play a role in initiating the events that lead to apoptotic degradation of the nucleus. Inhibition of lamin assembly itself is an inducer of apoptosis. The nuclear envelope acts as a barrier that prevents both DNA and RNA viruses from entering the nucleus. Some viruses require access to proteins inside the nucleus in order to replicate and/or assemble. DNA viruses, such as herpesvirus replicate and assemble in the cell nucleus, and exit by budding through the inner nuclear membrane. This process is accompanied by disassembly of the lamina on the nuclear face of the inner membrane. Disease-related dynamics Initially, it has been suspected that immunoglobulins in general and autoantibodies in particular do not enter the nucleus. Now there is a body of evidence that under pathological conditions (e.g. lupus erythematosus) IgG can enter the nucleus. Nuclei per cell Most eukaryotic cell types usually have a single nucleus, but some have no nuclei, while others have several. This can result from normal development, as in the maturation of mammalian red blood cells, or from faulty cell division. Anucleated cells An anucleated cell contains no nucleus and is, therefore, incapable of dividing to produce daughter cells. The best-known anucleated cell is the mammalian red blood cell, or erythrocyte, which also lacks other organelles such as mitochondria, and serves primarily as a transport vessel to ferry oxygen from the lungs to the body's tissues. Erythrocytes mature through erythropoiesis in the bone marrow, where they lose their nuclei, organelles, and ribosomes. The nucleus is expelled during the process of differentiation from an erythroblast to a reticulocyte, which is the immediate precursor of the mature erythrocyte. The presence of mutagens may induce the release of some immature "micronucleated" erythrocytes into the bloodstream. Anucleated cells can also arise from flawed cell division in which one daughter lacks a nucleus and the other has two nuclei. In flowering plants, this condition occurs in sieve tube elements. Multinucleated cells Main article: Multinucleate Multinucleated cells contain multiple nuclei. Most acantharean species of protozoa and some fungi in mycorrhizae have naturally multinucleated cells. Other examples include the intestinal parasites in the genus Giardia, which have two nuclei per cell. Ciliates have two kinds of nuclei in a single cell, a somatic macronucleus and a germline micronucleus. In humans, skeletal muscle cells, also called myocytes and syncytium, become multinucleated during development; the resulting arrangement of nuclei near the periphery of the cells allows maximal intracellular space for myofibrils. Other multinucleate cells in the human are osteoclasts a type of bone cell. Multinucleated and binucleated cells can also be abnormal in humans; for example, cells arising from the fusion of monocytes and macrophages, known as giant multinucleated cells, sometimes accompany inflammation and are also implicated in tumor formation. A number of dinoflagellates are known to have two nuclei. Unlike other multinucleated cells these nuclei contain two distinct lineages of DNA: one from the dinoflagellate and the other from a symbiotic diatom. Evolution As the major defining characteristic of the eukaryotic cell, the nucleus's evolutionary origin has been the subject of much speculation. Four major hypotheses have been proposed to explain the existence of the nucleus, although none have yet earned widespread support. The first model known as the "syntrophic model" proposes that a symbiotic relationship between the archaea and bacteria created the nucleus-containing eukaryotic cell. (Organisms of the Archaeal and Bacterial domains have no cell nucleus.) It is hypothesized that the symbiosis originated when ancient archaea similar to modern methanogenic archaea, invaded and lived within bacteria similar to modern myxobacteria, eventually forming the early nucleus. This theory is analogous to the accepted theory for the origin of eukaryotic mitochondria and chloroplasts, which are thought to have developed from a similar endosymbiotic relationship between proto-eukaryotes and aerobic bacteria. One possibility is that the nuclear membrane arose as a new membrane system following the origin of mitochondria in an archaebacterial host. The nuclear membrane may have served to protect the genome from damaging reactive oxygen species produced by the protomitochondria. The archaeal origin of the nucleus is supported by observations that archaea and eukarya have similar genes for certain proteins, including histones. Observations that myxobacteria are motile, can form multicellular complexes, and possess kinases and G proteins similar to eukarya, support a bacterial origin for the eukaryotic cell. A second model proposes that proto-eukaryotic cells evolved from bacteria without an endosymbiotic stage. This model is based on the existence of modern Planctomycetota bacteria that possess a nuclear structure with primitive pores and other compartmentalized membrane structures. A similar proposal states that a eukaryote-like cell, the chronocyte, evolved first and phagocytosed archaea and bacteria to generate the nucleus and the eukaryotic cell. The most controversial model, known as viral eukaryogenesis, posits that the membrane-bound nucleus, along with other eukaryotic features, originated from the infection of a prokaryote by a virus. The suggestion is based on similarities between eukaryotes and viruses such as linear DNA strands, mRNA capping, and tight binding to proteins (analogizing histones to viral envelopes). One version of the proposal suggests that the nucleus evolved in concert with phagocytosis to form an early cellular "predator". Another variant proposes that eukaryotes originated from early archaea infected by poxviruses, on the basis of observed similarity between the DNA polymerases in modern poxviruses and eukaryotes. It has been suggested that the unresolved question of the evolution of sex could be related to the viral eukaryogenesis hypothesis. A more recent proposal, the exomembrane hypothesis, suggests that the nucleus instead originated from a single ancestral cell that evolved a second exterior cell membrane; the interior membrane enclosing the original cell then became the nuclear membrane and evolved increasingly elaborate pore structures for passage of internally synthesized cellular components such as ribosomal subunits. History The nucleus was the first organelle to be discovered. What is most likely the oldest preserved drawing dates back to the early microscopist Antonie van Leeuwenhoek (1632–1723). He observed a "lumen", the nucleus, in the red blood cells of salmon. Unlike mammalian red blood cells, those of other vertebrates still contain nuclei. The nucleus was also described by Franz Bauer in 1804 and in more detail in 1831 by Scottish botanist Robert Brown in a talk at the Linnean Society of London. Brown was studying orchids under the microscope when he observed an opaque area, which he called the "areola" or "nucleus", in the cells of the flower's outer layer. He did not suggest a potential function. In 1838, Matthias Schleiden proposed that the nucleus plays a role in generating cells, thus he introduced the name "cytoblast" ("cell builder"). He believed that he had observed new cells assembling around "cytoblasts". Franz Meyen was a strong opponent of this view, having already described cells multiplying by division and believing that many cells would have no nuclei. The idea that cells can be generated de novo, by the "cytoblast" or otherwise, contradicted work by Robert Remak (1852) and Rudolf Virchow (1855) who decisively propagated the new paradigm that cells are generated solely by cells ("Omnis cellula e cellula"). The function of the nucleus remained unclear. Between 1877 and 1878, Oscar Hertwig published several studies on the fertilization of sea urchin eggs, showing that the nucleus of the sperm enters the oocyte and fuses with its nucleus. This was the first time it was suggested that an individual develops from a (single) nucleated cell. This was in contradiction to Ernst Haeckel's theory that the complete phylogeny of a species would be repeated during embryonic development, including generation of the first nucleated cell from a "monerula", a structureless mass of primordial protoplasm ("Urschleim"). Therefore, the necessity of the sperm nucleus for fertilization was discussed for quite some time. 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"Nuclear lamins: building blocks of nuclear architecture". Genes & Development. 16 (5): 533–47. doi:10.1101/gad.960502. PMID 11877373. – a review article about nuclear lamins, explaining their structure and various roles Görlich D, Kutay U (1999). "Transport between the cell nucleus and the cytoplasm". Annual Review of Cell and Developmental Biology. 15: 607–60. doi:10.1146/annurev.cellbio.15.1.607. PMID 10611974. – a review article about nuclear transport, explains the principles of the mechanism, and the various transport pathways Lamond AI, Earnshaw WC (April 1998). "Structure and function in the nucleus" (PDF). Science. 280 (5363): 547–53. CiteSeerX 10.1.1.323.5543. doi:10.1126/science.280.5363.547. PMID 9554838. – a review article about the nucleus, explaining the structure of chromosomes within the organelle, and describing the nucleolus and other subnuclear bodies Pennisi E (August 2004). "Evolutionary biology. The birth of the nucleus". Science. 305 (5685): 766–8. doi:10.1126/science.305.5685.766. PMID 15297641. S2CID 83769250. – a review article about the evolution of the nucleus, explaining a number of different theories Pollard TD, Earnshaw WC (2004). Cell Biology. Philadelphia: Saunders. ISBN 978-0-7216-3360-2. – a university-level textbook focusing on cell biology. Contains information on nucleus structure and function, including nuclear transport, and subnuclear domains External links Wikimedia Commons has media related to Cell nucleus. Library resources about Cell nucleus Online books Resources in your library Resources in other libraries "The Nucleus". MBInfo. Archived from the original on 13 March 2014. "Learn about the Cell Nucleus". cellnucleus.com. Website covering structure and function of the nucleus from the Department of Oncology at the University of Alberta. Bickmore W. "The Nuclear Protein Database". Medical Research Council Human Genetics Unit. Information on nuclear components. "The Nucleus Collection". Image & Video Library. The American Society for Cell Biology. Archived from the original on 12 November 2006. contains peer-reviewed still images and video clips that illustrate the nucleus. Gall JG, McIntosh JR (eds.). "Nuclear Envelope and Nuclear Import Section". Landmark Papers in Cell Biology. Archived from the original on 17 November 2006. contains digitized commentaries and links to seminal research papers on the nucleus. Published online in the Image & Video Library Archived 10 June 2011 at the Wayback Machine of The American Society for Cell Biology "Cytoplasmic patterns generated by human antibodies". AntibodyPatterns.com. Archived from the original on 2 January 2007. \| v t e Structures of the cell and organelles \| \| Endomembrane system \| Cell membrane Nucleus Endoplasmic reticulum Golgi apparatus Parenthesome Autophagosome Vesicle + Exosome + Lysosome + Endosome + Phagosome + Vacuole + Acrosome Cytoplasmic granule + Melanosome + Microbody + Glyoxysome + Peroxisome + Weibel–Palade body \| \| Cytoskeleton \| Microfilament Intermediate filament Microtubule Prokaryotic cytoskeleton Microtubule organizing center + Centrosome + Centriole + Basal body + Spindle pole body Myofibril Undulipodium + Cilium + Flagellum + Axoneme + Radial spoke Pseudopodium + Lamellipodium + Filopodium \| \| Endosymbionts \| Mitochondrion Plastid + Chloroplast + Chromoplast + Gerontoplast + Leucoplast + Amyloplast + Elaioplast + Proteinoplast + Tannosome + Apicoplast Nitroplast \| \| Other internal \| Nucleolus RNA + Ribosome + Spliceosome + Vault Cytoplasm + Cytosol + Inclusions Proteasome Magnetosome \| \| External \| Cell wall Extracellular matrix \| \| v t e Structures of the cell nucleus / nuclear protein \| \| Envelope (membrane)/nuclear lamina \| Pore complex: Nucleoporin + NUP35 + NUP37 + NUP43 + NUP50 + NUP54 + NUP62 + NUP85 + NUP88 + NUP93 + NUP98 + NUP107 + NUP133 + NUP153 + NUP155 + NUP160 + NUP188 + NUP205 + NUP210 + NUP214 AAAS \| \| Nucleolus \| Cajal (coiled) body + SMN + GEMIN2 + GEMIN4 + GEMIN5 + GEMIN6 + GEMIN7 + DDX20 + COIL Perinucleolar compartment + PTBP1 + CUGBP1 TCOF ATXN7 \| \| Other \| \| \| \| Chromatin Dot (PML body) Paraspeckle \| \| SMC protein: \| Cohesin + SMC1A + SMC1B + SMC3 Condensin + NCAPD2 + NCAPD3 + NCAPG + NCAPG2 + NCAPH + NCAPH2 + SMC2 + SMC4 DNA repair + SMC5 + SMC6 \| \| Transition nuclear protein: \| + TNP1 + TNP2 \| \| Nuclear matrix (Nucleoskeleton) Nucleoplasm (Nucleosol) \| \| \| \| see also transcription factors and intracellular receptors \| \| \| see also nucleus diseases \| \| Authority control databases \| \| International \| \| \| National \| United States Japan Czech Republic Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Cell nucleus Cell anatomy Organelles Hidden categories: CS1 maint: article number as page number CS1 maint: location missing publisher CS1 Latin-language sources (la) Articles with short description Short description is different from Wikidata Wikipedia pages semi-protected against vandalism Use dmy dates from August 2018 Articles containing Latin-language text Commons category link is on Wikidata CS1: unfit URL Webarchive template wayback links Cell nucleus Add topic
190024	https://journals.lww.com/ijdr/fulltext/2012/23050/dental_caries_prevalence_and_treatment_needs_among.5.aspx	Indian Journal of Dental Research Log in or Register Get new issue alerts Get alerts;;) Submit a Manuscript Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation RegisterLogin Browsing History Home Current Issue Previous Issues For Authors Information for Authors Submit a Manuscript Published Ahead-of-Print Journal Info About the Journal Editorial Board Affiliated Society Advertising Subscription Services Reprints Rights and Permissions Articles Advanced Search Sep-Oct 2012 - Volume 23 - Issue 5 Previous Article Next Article Outline MATERIALS AND METHODS The inclusion criteria The exclusion criteria RESULTS DISCUSSION CONCLUSION Source of Support: Conflict of Interest: REFERENCES Images Slideshow Gallery Export PowerPoint file Download PDF EPUB Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Image Gallery Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection Original Research Dental caries prevalence and treatment needs among 12- and 15- Year old schoolchildren in Shimla city, Himachal Pradesh, India Shailee, Fotedar; Sogi, GM 1; Sharma, KR; Nidhi, Pruthi 1 Author Information Department of Public Health Dentistry, HP Government Dental College, Shimla, Himachal Pradesh, India 1 Department of Public Health Dentistry, M.M. College of Dental Sciences and Research, Mullana, Ambala, Haryana, India 2 Department of Public Health Dentistry, Sudha Rustagi College of Dental Sciences and Research, Faridabad, Haryana, India Address for correspondence: Dr. Fotedar Shailee E-mail: drfotedar@rediffmail.com Received July 29, 2011 Received in revised form December 20, 2011 Accepted May 05, 2012 Indian Journal of Dental Research 23(5):p 579-584, Sep–Oct 2012. \| DOI: 10.4103/0970-9290.107330 Open Abstract Context: Dental caries is one of the commonest oral diseases in children. Despite this fact, not many studies have been done on this issue among school children in Shimla. Aim: To assess the prevalence of dental caries and treatment needs among schoolchildren aged 12 years and 15 years in Shimla city, Himachal Pradesh, India. With this study we also aimed to establish reliable baseline data. Design: Cross-sectional study. Materials and Materials: This study was conducted among 12 - and 15 - year old schoolchildren in Shimla city, Himachal Pradesh-India. A sample of 1011 schoolchildren was selected by a two-stage cluster sampling method. Clinical recording of dental caries, was done according to WHO diagnostic criteria (1997). Statistical Analysis: The statistical tests used were the t- test, and the Chi-square test. Results: The prevalence of dental caries was 32.6% and 42.2% at 12 years and 15 years respectively. At 12 years of age, the mean Decayed Missing Filled Teeth was 0.62 ± 1.42 and it was 1.06 ± 2.93 at 15 years of age. Females had higher level of caries than males at both the ages. Dental caries was higher in children from government schools as compared to those from private schools. The ‘decayed’ component was the biggest contributor to the DMFT index. The highest treatment need at both ages was one surface restoration. Conclusion: The caries experience of 12- and 15- year-old children was low compared to WHO - ‘recommended’ values. Effective oral health promotion strategies need to be implemented to further improve the dental health of school children in Shimla city. Dental caries is the most prevalent dental affliction in humans. Despite creditable scientific advances and the fact that caries is preventable, the disease continues to be a major public health problem predominantly affecting children. Dental caries is often responsible for the absenteeism from school and loss of working hours. The lack of availability of dental care, postponement of treatment due to cost considerations, and under utilization of available facilities not only results in aggravation of the disease but also enhances the cost of treatment and care. Worldwide schools offer an efficient and effective way to reach children and, through them, families and community members. School age is an influential stage in people's lives, a time when lifelong sustainable oral health - related behaviors, as well as beliefs and attitudes, are being developed. Children are particularly receptive to health messages during this period and the earlier the good habits are established, the longer lasting the impact. Moreover, the messages can be reinforced regularly throughout the school years. Further, the social and economic factors for Dental Caries can be represented by the type of school. e. government and private schools. The National Oral Health Survey and Fluoride mapping - 2003 reported that 72.5% of 12 year old children and 75.4% of 15 year old children had dental caries. As it is the most common dental disease with high prevalence among children, it is important to control the disease process by rendering required treatment and by increasing awareness regarding its preventive measures. Knowledge of dental health and treatment needs of school children is important for developing appropriate preventive approaches, anticipating utilization patterns, and planning effectively for organization and financing of dental resources. As there have been no earlier studies on the prevalence of dental caries and treatment needs of schoolchildren in Shimla city,- this study was conducted with the following objectives: To assess the prevalence and severity of dental caries and the treatment needs of schoolchildren aged 12 years and 15 years in Shimla city. To compare the dental caries levels and treatment needs in government schools with that in and private schools. To establish reliable baseline data for development of national/regional oral health programs. MATERIALS AND METHODS A cross-sectional epidemiologic study was conducted among schoolgoing children aged 12 years and 15 years in Shimla city. Ethical approval to conduct the study was obtained from the Institutional Review Board of H.P. Government Dental College and Hospital, Shimla. Written consent for the participation of the children in the study was obtained from the Principals of the concerned schools. A two-stage cluster sampling technique was used for obtaining the required sample for the study. For the purpose of the study, Shimla city was arbitrarily divided into four geographical zones, which corresponded to the four administrative areas of the city: Shimla municipal and three Shimla Planning Areas (Dhalli, Tutu and New Shimla). Schools from each region were randomly selected to obtain the desired sample size. We ensured that there was equal representation from each of the four zones. Under the municipal corporation of Shimla, there were 43 schools (12 Senior Secondary, 24 Secondary, and 7 Middle), where the children in age- group of 12 and 15 were available. Among the 43 schools there were 26 government and 17 private schools as per the data available from the Director of Education, Himachal Pradesh in December, 2008. The total number of school children schoolchildren in the age groups of 12 years and 15 years was 6870. A pilot study was conducted by randomly selecting one government and one private school from the list of available schools. The results from this pilot were used to calculate the sample size (n = 985) for the main study. For obtaining this sample size, seven government and five private schools were selected randomly with proportionate representation from each category of schools, i.e., government and private schools. A total of 1011 students from these schools were examined over a period of 3 months April -June 2009. The inclusion criteria Schoolchildren (male and female) who had completed 12 and 15 years of age Children present on the day of examination The exclusion criteria Children who refused to participate were excluded. Data collection was carried out by one of the authors (SF) trained for clinical examination during several educational and clinical sessions in the Department of Public Health Dentistry, Government Dental College, Shimla. The author was assisted by an alert and co-operative recording assistant. General information about the subject and the data reagarding oral hygiene practices were obtained through interview and recorded on a modified WHO proforma. With regard to oral hygiene practices, we collected information on type of the oral hygiene aid used (i.e. tooth- brush, finger, twig of a tree and any other aid); oral hygiene material used (tooth- paste, tooth powder, charcoal, salt or any other material); and frequency of cleaning teeth (once daily, twice daily or not even once). The subjects were examined by type III clinical examination in their respective schools while seated on a comfortable chair. The diagnostic criteria and treatment codes were in accordance with those recommended by WHO. The Kappa static for intra-examiner reproducibility of caries diagnosis determined using Kappa statistic was 0.85. A referral was forwarded to the parents of the children in need of dental care. At the conclusion of the survey, an oral health education session was conducted and correct way of brushing the teeth was demonstrated in each classroom. The data collected was analyzed using Statistical Package for Social Sciences for Windows®, version 13 (SPSS Inc., Chicago, IL). The statistical tests used were the t- test for continuous variables and the Chi-square test for categorical data. A level of P ≤ .05 was considered statistically significant and P ≤ 0.001 was taken as highly significant. RESULTS The distribution of students according to type of school and gender are given in Table 1. Table 1: Distribution of subjects according to age, gender, and school Table 2 shows the distribution of subjects according to oral hygiene practices. All children (100%) in private schools used toothbrush and tooth paste whereas in government schools 95.4% used tooth brush and 93.6% used tooth paste. This difference was statistically significant, P< 0.001. The frequency of brushing twice a day was significantly higher in private schools as compared to government schools [Table 3]. Table 2: Distribution of subjects according to age, sex and oral hygiene practices Table 3: Distribution of subjects according to oral hygiene practices and school category The prevalence of dental caries at 12 years was 32.6% and at the age of 15 years it was 42.2% [Table 4]. At the age of 12 years, the mean decayed missing filled teeth was 0.62 ± 1.42 while at the age of 15 yrs it was 1.06 ± 2.93; this difference between the two age groups was statistically significant(P ≤ 0.001). In both the age groups, females showed higher mean DMFT as compared to males. The difference was statistically significant at 12 years [Table 5]. Subjects brushing their teeth once a day had higher mean DMFT as compared to those brushing twice a day; this difference was statistically significant at 12 years [Table 5]. The mean DMFT was higher in private schools as compared to government schools [Table 5]. At 12 years of age children eating vegetarian diet had significantly higher mean number of decayed teeth than children on mixed diet [Table 5]. Table 4: Prevalence of dental caries at 12 years and 15 years Table 5: Mean caries experience (DMFT) in relation to age, gender, schools, brushing frequency and Diet The largest contribution to the DMFT index was by the decayed component with 90.5% at 12 years and 84.1 % at 15 years. The mean of decayed teeth, filled teeth and missing teeth due to caries was significantly higher at 15 years as compared to 12 years. Females had higher number of mean decayed teeth (0.65) than males (0.44), and this difference was statistically significant. At both the ages, mean of decayed teeth was significantly higher in government schools as compared to private schools. Children in government schools had less number of mean filled teeth at both ages as compared to children from private schools; this difference was also statistically significant [Table 6]. Table 6: Components of DMFT according to age, gender, and school At the age of 12 years, 50.1% of children required restorative and exodontic treatment and at the age of 15 years 50.6% of children required such treatment [Table 7]. The greatest need was for single surface restorations (2.3% in 12 yrs and 2.2% in 15 yrs. Table 7: Assessment of treatment needs among study population DISCUSSION This cross-sectional study was carried out to assess the prevalence of dental caries and the treatment needs among schoolchildren in Shimla city, Himachal Pradesh,-India. Children of 12 and 15 years were chosen for this study, as these are global monitoring ages for dental caries for international comparisons and monitoring of disease trends. In the present study the sample we included schoolchildren from from both public and private schools in order to have children from all the social, economic and cultural backgrounds. In both the age groups around 97% of population used tooth brush and tooth paste for cleaning their teeth. This clearly indicates their awareness about oral hygiene. At the age of 12 years, most of the children (64%) brushed once a day which is similar with the findings of Joshi et al., but high as compared to findings of Harikaran et al. and Peng, and low as compared to Peterson et al.. The habit of brushing twice a day was more common in private schools as compared to government schools which was also reported by Mahesh Kumar et al, Tanni et al. and Peterson et al. In the present study, as the frequency of brushing increased, the prevalence of dental caries decreased. This finding is consistent with the findings of Christina SB et al, Wei S et al. and Sethi B et al. In the present study, we observed that the prevalence of dental caries was higher at the age of 15 years (42.2%) than at 12 years (32.6%); this was also reported by Rodrigues, and Damle, Singh et al. Psoter et al. Naidu et al., Goyal et al, and Mustafa et al. The reason for the higher prevalence of dental caries at 15 years as compared to 12 years is that caries being a continuous and cumulative process had obviously increased over a span of 3 years; moreover, the number of teeth is more at the age of 15 years. In the present study, the mean DMFT at 12 years and at 15 years was 0.62 and 1.06, respectively; similar values were also reported by Naidu et al., Peterson and Bajoma and Rudolph. However, our values are low compared to DMFT of 2.4 reported by National Oral Health Survey in H. Females had significantly higher mean DMFT value than males. This is in line with the findings of Al Shammery et al., Salapatal et al., Obry Musset et al., Dummer et al., Sogi and Basker., Singh et al., Mishra and Shee, Saimbi et al., This finding may be due to the fact that teeth erupt earlier in females than males which means females teeth would have been exposed to oral environment for a longer period than the male of the same age. In both the age groups there was statistically significant difference in mean decayed, filled, and missing teeth between the government and the private schools. The level of caries was higher in children attending government schools which is in line with the findings of Almedia et al., but in contrast to the results reported by Tanni, Ojofeitimi et al.,. The higher DMFT in government schools may be due to lack of availability of dental care, postponement of treatment because of cost considerations, under utilization of available facilities and lack ofawareness regarding the importance of timely dental care. Further studies are needed to assess the various barriers for utilization of services. The mean number of filled teeth and missing teeth due to other reasons were high in private schools, which may be due to positive attitude and better dental awareness, of the parents of children in private schools which is reflected in the child's oral health maintenance. The mean number of missing teeth due to other reasons was higher in private schools probably due to orthodontic interventions in these children.(“unpublished observations” with written permission from the source). In the present study, children having mixed diets had significantly lower mean decayed teeth than children having vegetarian diets. This has also been reported by Chandra and Chawla, Sarvanan et al., and Khan et al.,. It is suggested that persons who consume a mixed diet (protein rich food), will develop less amount of acid in their mouth and thereby be relatively protected from dental caries. This may be because proteins tend to putrify rather than ferment like carbohydrates, promoting alkalinity instead of acidity, and so no decalcification is usually observed. At the age of 12 years, treatment need was observed for 50.1% children, which is higher than 44.7% as reported by Abid. At the age of 15 years, treatment need was seen for 50.6% of children. On examining the treatment needs for dental diseases among children of 12 years and 15 years, we found the greatest need was for one surface restoration followed by two-surface restorations, pulp restoration, extractions and others. This is similar to the findings of Wright et al., Rodrigues and Dhamle, Kulkarniand Deshpande, Dash JK. CONCLUSION The mean DMFT in 12 - and 15- year-old children in Shimla city, as revealed by the study falls within the ‘very low’ category as per the WHO classification. To further improve the oral health of children in Shimla, we recommend the following:. Oral health promotion through well-structured oral health education program can create positive change in awareness for special groups like schoolchildren. Reinforcement of knowledge is necessary and this can be done by incorporating chapters on oral health and oral hygiene in school textbooks. Also, the teacher's training programs can ensure continuity of reinforcement. Implementation of school dental health programs, focusing on preventive programs like fluoride mouth rinse and tooth brushing programs. Preventive services should be given high priority and needs to be started at an early age to target the primary dentition and future caries in permanent dentition. Regular interval screening programs to assess the oral health and treatment needs of schoolchildren with treatment as per the need. Source of Support: Nil, Conflict of Interest: None declared. 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Prevalence of dental caries and treatment needs among children of Cuttack (Orissa) J Indian Soc Pedod Prev Dent. 2002;20:139–43 Cited Here \| PubMed View full references list Keywords: Dental caries; prevalence; schoolchildren; treatment needs © 2012 Indian Journal of Dental Research \| Published by Wolters Kluwer – Medknow View full article text Source Dental caries prevalence and treatment needs among 12- and 15- Year old schoolchildren in Shimla city, Himachal Pradesh, India Indian Journal of Dental Research23(5):579-584, Sep-Oct 2012. Full-Size Email Favorites Export View in Gallery Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. 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190025	https://pubmed.ncbi.nlm.nih.gov/3989959/	An official website of the United States government 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. Account Save citation to file Email citation Add to Collections Add to My Bibliography Your saved search Create a file for external citation management software Your RSS Feed Full text links Actions Page navigation Vibrio vulnificus. Man and the sea Vibrio vulnificus. Man and the sea Authors Abstract To identify risk factors for Vibrio vulnificus infections, we performed a regional case-control study of 19 patients identified by isolates received at a state reference laboratory. Interviews with patients or surviving relatives and with three controls for each patient were compared in a matched analysis. Patients with V vulnificus wound infection were more likely than controls to have sustained a puncture wound while handling fresh seafood or to have been exposed to salt water. More patients with primary septicemia than controls had eaten raw oysters before the onset of illness. Other risk factors for septicemia included underlying liver disease, hematopoietic disorders, chronic renal insufficiency, use of immunosuppressive agents, and heavy alcohol consumption. Although V vulnificus infection is unusual, with a regional incidence of 0.8 per 100,000 population in this study, septicemia in the immunosuppressed patient is a devastating illness that can be prevented by not eating raw seafood. PubMed Disclaimer MeSH terms LinkOut - more resources Full Text Sources NCBI Literature Resources MeSH PMC Bookshelf Disclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers
190026	https://bicycles.stackexchange.com/questions/80781/what-is-the-best-technique-to-use-when-turning-my-bicycle	riding position - What is the best technique to use when turning my bicycle? - Bicycles Stack Exchange Join Bicycles 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 Bicycles helpchat Bicycles 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 What is the best technique to use when turning my bicycle? Ask Question Asked 4 years ago Modified1 year, 9 months ago Viewed 6k times This question shows research effort; it is useful and clear 15 Save this question. Show activity on this post. Turning at low to moderate speed while road riding is unlikely to be a task of great difficulty for most riders[citation needed]. One simply rotates the handlebars, and off you go. It is when speeds get higher or turns become sharper that technique becomes more nuanced. Which foot should be down, if any? Where should one's weight be placed? Do you lean the bike? Lean the body? Lean both?? Is countersteering (the act of intentionally pointing the front wheel the opposite direction of the intended turn to initiate it) a good technique? Line choice? The question gets even more complicated in the context of off-road riding, where off-camber turns and berms (banked corners) are to be found. Recommendations will be more situation-dependent there. riding-position skills Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Sep 12, 2021 at 19:41 MaplePandaMaplePanda asked Sep 12, 2021 at 19:16 MaplePandaMaplePanda 17.5k 2 2 gold badges 43 43 silver badges 85 85 bronze badges 7 2 Citation needed, indeed.ojs –ojs 2021-09-13 06:20:56 +00:00 Commented Sep 13, 2021 at 6:20 8 One simply rotates the handlebars, and off you go. In fact you mostly change your body weight bicycles.stackexchange.com/questions/12901/…AnoE –AnoE 2021-09-13 13:57:16 +00:00 Commented Sep 13, 2021 at 13:57 3 Simply rotating the handlebars is recipe for falling straight off!Нет войне –Нет войне 2021-09-13 14:02:45 +00:00 Commented Sep 13, 2021 at 14:02 1 I'm hoping this is a canonical Q. Also, this will depend on the bicycle. Answers will be different for track bikes/fixies vs. mountain bikes, at least wrt which foot should be down.shoover –shoover 2021-09-13 20:37:43 +00:00 Commented Sep 13, 2021 at 20:37 1 @shoover I think my bike handling skills are a little better than that! 😉 Yea, answers should include that aspect of context.MaplePanda –MaplePanda 2021-09-14 15:06:07 +00:00 Commented Sep 14, 2021 at 15:06 \|Show 2 more comments 5 Answers 5 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. Which foot should be down, if any? The foot on the outside of the curve should be down and the foot on the inside curve should be up (exception: on uneven ground cranks should be horizontal so neither foot is up or down). This is because many bicycles have pedals so low that if you do this the other way around, you will hit the ground with your inside pedal. Where should one's weight be placed? On even ground, the weight distribution doesn't matter that much. Some on the handlebars, some on the pedals, some on the saddle. However, on uneven ground you will want to have absolutely no weight on the saddle. So it's only pedals (cranks being horizontal then) and handlebars. Also, on uneven ground, you will want to hold the handlebars in such a manner that your arms become "springs" / "shock absorbers". Do you lean the bike? Lean the body? Lean both?? You lean both. The body is leaned the same amount as the bike. Doing it otherwise would place side loads on the hub bearings and spokes. While angular contact cup and cone bearings can take side loads and 36-spoke 135mm/142mm OLD wheels can take reasonable side loads too, there are hubs using cartridge bearings that are really poor in taking side loads, and low-spoke-count 130mm OLD 11-speed road wheels can have so large dish and so low count of spokes that the wheel could collapse with excessive side loads. Is countersteering (the act of intentionally pointing the front wheel the opposite direction of the intended turn to initiate it) a good technique? Countersteering is the only possible way of cornering a bike. If you have cornered a bike, you have already countersteered. There's no other way possible to corner a bike. You do it subsconsiously. Line choice? Visualize the line in advance. You'll learn it if you practice it. You want to visualize the line in a path that doesn't have excessive bumps on the road, and the line should be such that at no point is the traction exceeded. Braking slightly on the corner with both brakes (equal front+rear) is possible because if your cornering g-force is 0.98g and your braking g-force is 0.1g, then the total force is sqrt((0.1g)^2+(0.98g)^2) = sqrt(0.97040g^2) = 0.985g according to the Pythagorean theorem. So moderate amount of braking increases the total traction need only very slightly. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 13, 2021 at 16:26 juhistjuhist 22.9k 3 3 gold badges 35 35 silver badges 62 62 bronze badges 7 1 It flat out shocks me how many people I see cornering (on the road, no matter the bike type) with the inside foot down! I think that this is a fear-based response to falling over, "If my foot is closer to the ground, then if I fall, I can catch myself sooner", and/or something they've seen an MX rider doing, which totally doesn't apply to cycling. Having heard horror stories from my dad who went down with a big pack because someone up front clipped a pedal, I've never done that! (Of course, most people I see doing this aren't cornering quickly but it seems so counterintuitive to me...)FreeMan –FreeMan 2021-09-13 18:00:14 +00:00 Commented Sep 13, 2021 at 18:00 Countersteering isn't just subconscious, it's created by the physics of turning - try this experiment. Remove the front wheel from the bike, hold it vertically and spin. Now lean it one way or the other - it'll also automatically rotate due to gyroscopic forces. Basically leaning is what initiates the turn and that initiates the counter steer David Waterworth –David Waterworth 2021-09-14 04:58:14 +00:00 Commented Sep 14, 2021 at 4:58 Inside foot down allows easy shifting of weight in the wanted direction. But yes, with low frame there is a risk of striking the ground.nightrider –nightrider 2021-09-14 05:04:20 +00:00 Commented Sep 14, 2021 at 5:04 @DavidWaterworth Sorry, but that's wrong: If you ride no-handed, you need to counterlean to initiate a turn. I.e. to do a left turn, you must first lean the bike to the right, allowing the gyroscopic effect to turn the front wheel right (countersteer), then you lean the bike left to use the gyroscopic effect to turn the front wheel left for the actual turn. If you don't believe me, try this: Push your bike by grabbing the saddle, only, and try to make it go round corners. The slow speed will allow you to experience the counterlean rather explicitly.cmaster - reinstate monica –cmaster - reinstate monica 2021-09-14 08:11:15 +00:00 Commented Sep 14, 2021 at 8:11 One more tip: having the outer pedal low, you can apply more pressure with the corresponding foot to help controlling the bike.clabacchio –clabacchio 2021-09-14 09:44:41 +00:00 Commented Sep 14, 2021 at 9:44 \|Show 2 more comments This answer is useful 4 Save this answer. Show activity on this post. It's an interesting question. I don't think either foot should be down, but the foot forwards for the direction of turn. Standing with knees bent, heels down, weight over back of saddle. Lean is learned intuitively as is countersteer. Once you have confidence to take a corner fast, you can analyse that in fact you do automatically countersteer and that you and the bike lean into the corner in relation to speed. Where you have a choice of line, you start wide, come as close to the apex as possible and finish wide, like a racing car. But on a narrow trail, you don't often have a choice. On a berm, the fast line is out near the top and the slow line on the inside at the bottom and that stays a constant, you don't cut the corner. Off camber turns are even harder as extreme lean angles will easily lead to wash-out especially on loose surfaces so just go slow on these until you've learned the particular bend. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 12, 2021 at 19:35 NoiseNoise 20.8k 1 1 gold badge 21 21 silver badges 64 64 bronze badges 9 1 This answer is more targeted towards mountainbiking, right?Michael –Michael 2021-09-12 19:50:19 +00:00 Commented Sep 12, 2021 at 19:50 1 @Michael yes although cornering at speed especially when descending (on a road bike) falls into a similar category, as allowance in made in body positioning for braking hard, taking corner and carrying on.Noise –Noise 2021-09-12 20:19:53 +00:00 Commented Sep 12, 2021 at 20:19 3 I don't think either foot should be down - i think whichever foot needs least clearance over the terrain if clearance is required.. which if you're leaning a lot and/or the terrain to the inside of the curve has e.g. projections that could catch a pedal, then put the outside pedal lower so your inside pedal is less likely to catch the floor..Caius Jard –Caius Jard 2021-09-13 11:28:19 +00:00 Commented Sep 13, 2021 at 11:28 Generally pedals level with ground when cornering. This means for high speed, pedals level only when on berms, but on flat, unbermed corner (with a lot of bike lean), outside pedal down. Slow speed, the bikes is leans much less, pedals more level.mattnz –mattnz 2021-09-13 22:41:48 +00:00 Commented Sep 13, 2021 at 22:41 When cornering fast on a road bike I've always felt more stable with the outside pedal down rather than level. I find it easier that way to position my body a bit like you see the TT motorbike riders, with my inside (of the corner) knee out, and the bike leant as much as possible.David Waterworth –David Waterworth 2021-09-14 04:53:02 +00:00 Commented Sep 14, 2021 at 4:53 \|Show 4 more comments This answer is useful 3 Save this answer. Show activity on this post. I watched a video on this a while ago and it showed some useful tips step by step. This and a few other videos on their channel might help you out. In this video they cover flat turns but other videos cover berms etc There are loads of tips here from the very basics to more advanced stuff but the thumbnail is a pretty good example. Outside pedal down to minimize risk that inside pedal catches the floor, and bike / body separation so you can lean the bike while retaining balance. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 15, 2021 at 8:03 answered Sep 13, 2021 at 12:50 JamesJames 31 3 3 bronze badges 2 3 Welcome to Bicycles SE. We prefer answers on this site to be self-contained. That way, the answer is still valid if the video is taken down. Please use the edit button to summarize the the tips in the video within the body of your answer. Otherwise, it is likely to be downvoted, flagged for moderator intervention, and possibly deleted.jimchristie –jimchristie♦ 2021-09-13 13:46:52 +00:00 Commented Sep 13, 2021 at 13:46 1 It would be useful to screenshot the video at relevant points and add a bit of commentary. To take a screenshot open the Windows snipping tool (on Mac use the key combo to snip a region Shift Command 4) and then you can just literally copy with Ctrl C and paste into the answer box with Ctrl V; the uploading and hosting on imgur is automatic. For example even the thumbnail shot for the video is a good example of what a couple of people are saying to do for epic trials and high speeds: outside pedal down to minimize risk that inside pedal catches the floor Caius Jard –Caius Jard 2021-09-15 05:34:17 +00:00 Commented Sep 15, 2021 at 5:34 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. On a radius turn-in, don't do a snap countersteer but smoothly and linearly leverage the handlebars. In other words don't lean as a reaction to the countersteer but lean in time with it. Now note that turning is not steering but is a leveraging of the tires against the track surface. Then since both front and rear tires have drift, there is not much apparent steering angle. (Tire drift is the fundamental action of lateral tire traction and that fundamental makes the line of the vehicle in a curve. It's a tire slide that's a loss of traction. The vehicle-dynamics source for this fundamental is tire-slip-angle which according to me should really be called tire-drift-angle.) Also, lean the body, don't stiff-arm the bicycle down and the rider up. Get the rider weight low and to the inside on the turn-in. But stay in the seat, don't move the rider's weight forward. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 13, 2021 at 23:50 answered Sep 12, 2021 at 23:46 S SpringS Spring 121 2 2 bronze badges 4 Here is the link of the development of a new sport: kbhscape.com/downhill.htm .S Spring –S Spring 2021-09-12 23:48:23 +00:00 Commented Sep 12, 2021 at 23:48 Can you define radius turn-in in the first sentence? Did you mean a turn with constant radius, rather than decreasing radius (i.e. turn gets sharper toward the end)?Weiwen Ng –Weiwen Ng 2021-09-13 15:36:47 +00:00 Commented Sep 13, 2021 at 15:36 I calculate a radius path of the vehicle based on a line within the track width. But a decreasing radius just calls for a different line and apex. In general and for all curves, a vehicle reduces speed on the turn-in, and that fundamental just due to the turn-in, but then adds-power on the drive-out. A bicycle is frustrating because it can't really begin replacing speed at mid-curve. The calculation is for a curve with tangent straights: kbhscape.com/curve.htm . And the suggested sport has track designs of curves with tangent straights.S Spring –S Spring 2021-09-13 23:38:47 +00:00 Commented Sep 13, 2021 at 23:38 As for countersteering, I hesitate to post it. To turn left, push the left bar forward and lean. But once the bicycle is leaned-over, look down at the front wheel and there is a slight pointing of the wheel to the inside even though the bars are still being pushed to the right. Now ordinary bicycle countersteering is not a correcting of a rear-wheel slide, it's just a gyroscopic action.S Spring –S Spring 2021-09-14 06:24:42 +00:00 Commented Sep 14, 2021 at 6:24 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Which foot should be down, if any? Usually your outside foot should be down, especially if you’re taking the turn at high speed (you need to lean more at higher speeds), with the intent being to avoid clipping the pedal. This is notably the exact opposite of what a lot of people seem to intuitively do, but is something that you will definately learn very quickly after the first time you clip a pedal. If the ground is especially uneven though, having both feet at the same height is generally better for controlling your weight distribution. If cornering like this, usually you want the inside pedal to be the one to the front, as that will let you capture some of the extra energy of rebalancing after the turn to help maintain speed. Where should one's weight be placed? For a lower speed on even ground, normal weight placement for the bike type is fine and generally preferable. For higher speeds or uneven ground, weight should be on your legs and arms but not the saddle, ideally with both your legs and arms slightly bent and loose so you can use them like shock absorbers. If the ground is level, you generally also want your weight back like you would position yourself while braking. The key here is to make sure you can shift your weight quickly to compensate for issues resulting from uneven ground, loss of traction, or other unforeseen complications. Do you lean the bike? Lean the body? Lean both?? Both, but unless you are doing things strangely this will happen naturally (and you should let it). From an efficiency perspective, just leaning yourself is inefficient (you have to lean further to shift enough weight to turn properly), but from a safety perspective not leaning the bike is potentially dangerous because it puts a side load on the hub bearings and the wheels (good bikes should be fine, but it’s not something you should bet on, and the types of failure this can cause are both expensive to fix and really dangerous to have happen). Is countersteering (the act of intentionally pointing the front wheel the opposite direction of the intended turn to initiate it) a good technique? In general, you should countersteer as part of turning. It’s technically possible to turn without it (leaning appropriately to produce the same effect, sometimes called ‘counter-leaning’), but you need to know what you’re doing and it’s riskier than countersteering. In practice though, if you have to think about countersteering, you’re probably approaching the process of turning incorrectly. The required motion is intuitive even to small children, and it’s instinctual enough to anyone who has ridden a bike any reasonable amount that they will generally not even notice that they’re doing it. Line choice? On level ground, a racing line is theoretically optimal but not the easiest. Outside of actual racing, you ideally want to just come from the outside, cut as close to the inside of the turn as possible, and end on the outside (this maximizes the turning radius, making it easier to safely turn at speed). On inward banked turns (that is, the inner edge is at a lower elevation than the outer edge), a similar out-in-out pattern is still safest, but it’s also relatively easy to take a constant distance from one edge if the bank angle is correct. If you have to deal with an outward banked turn (that is, the outer edge is at a lower elevation than the inner edge), a fixed distance from one edge is arguably safest. Of course, that all assumes no potholes, bumps, or other hazards. Ideal line choice obviously avoids those.and is such that you maintain proper traction throughout the turn (though that is something you kind of have to learn based on the combination of tires and ground conditions). From a practical perspective, it’s very helpful to visualize the path you intend to take ahead of time. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 19, 2023 at 16:44 answered Sep 13, 2021 at 22:28 Austin HemmelgarnAustin Hemmelgarn 2,671 7 7 silver badges 16 16 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Bicycles 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. To learn more, see our tips on writing great answers. 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190027	https://en.wiktionary.org/wiki/periodic_function	Jump to content Search Contents Beginning 1 English 1.1 Noun 1.1.1 Hypernyms 1.1.2 Hyponyms 1.1.3 Translations periodic function Bahasa Indonesia Magyar മലയാളം Русский Svenska தமிழ் 中文 Entry Discussion 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 Create a book Download as PDF Printable version In other projects Appearance From Wiktionary, the free dictionary English [edit] Noun [edit] English Wikipedia has an article on: periodic function Wikipedia periodic function (plural periodic functions) (mathematics) Any function whose value repeats after the regular addition of a period to its independent variable; i.e., a function ƒ such that, for some positive constant p, ƒ(x + p) = ƒ(x) for all x. Hypernyms [edit] function Hyponyms [edit] almost-periodic function simply periodic function Translations [edit] function whose value repeats after the regular addition of a period \| \| \| Hungarian: periodikus függvény (hu) Polish: funkcja okresowa f Romanian: funcție periodică (ro) f \| Retrieved from " Categories: English lemmas English nouns English countable nouns English multiword terms en:Mathematics en:Functions Hidden categories: Pages with entries Pages with 1 entry Entries with translation boxes Terms with Hungarian translations Terms with Polish translations Terms with Romanian translations periodic function Add topic
190028	https://www.quora.com/How-do-you-find-the-simplest-radical-form	How to find the simplest radical form - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Radicals (term) Simplifying Algebraic Exp... Basic Algebra Exponents and Radicals Simplified Radical Form Mixed Radicals Algebra Simplifying Radicals 5 How do you find the simplest radical form? All related (42) Sort Recommended James Zhou 9y Note: this answer is kind of long so jump to picture to see a visual process. In order to find the simplest radical form of a square root, you need prime factorize the number. In order to do this, you take a number and divide it by different prime numbers until all of its factors are now prime. We will use 24 as our example. 24 divided by 2 (which is prime) is 12. 12 divided by 2, is 6. 6 divided by 2 is 3. Therefor, our prime factorization of 24 is 2^3 3, which basically means 8 3. Then, take take out pairs of the same number and put on the OUTSIDE of the root. However, make sure that it i Continue Reading Note: this answer is kind of long so jump to picture to see a visual process. In order to find the simplest radical form of a square root, you need prime factorize the number. In order to do this, you take a number and divide it by different prime numbers until all of its factors are now prime. We will use 24 as our example. 24 divided by 2 (which is prime) is 12. 12 divided by 2, is 6. 6 divided by 2 is 3. Therefor, our prime factorization of 24 is 2^3 3, which basically means 8 3. Then, take take out pairs of the same number and put on the OUTSIDE of the root. However, make sure that it is NEXT to the square root and not inside the mini hook that the square root makes. This means that that number is multiplying the square root of the number. Now, the only pair in this prime factorization is 2, so take two out of the prime factorization and that leaves us with square root 6 times 2, which is the answer. Also, when taking out pairs, only take one number from the pair (in this case 2) and put it on the outside. If there is another pair, multiply that number by the other number. So if there was a pair of 3 and a pair of 2, take out 1 three and 1 two and multiply them and put on the outside. Visual: Upvote · 99 22 9 2 Ana L. Losey See how I save 90% off flights w Mighty Travels Premium. · Author has 261 answers and 16.5M answer views ·3y The simplest radical form of a number is the root of that number. To find the simplest radical form, you first need to identify the square root of the number. Once you have found the square root, you then need to find the smallest possible integer that is larger than or equal to the square root. Upvote · 999 160 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·5y Related How do you simplify radicals in Algebra 1? Here is a simple introduction… DEALING WITH SURDS (radicals). Dealing with surds is NOT a ma... Related questions More answers below How do you write in simplest radical form? How do you know if a radical expression is in the simplest form? What is the radical form of (9) 1/2? What is simplified radical form in math? When can you say that a radical is in the simplest form? Assistant Bot · 1y To find the simplest radical form: Identify the perfect squares within the radicand. A perfect square is a number that can be expressed as the product of two equal integers. Factor out the perfect squares from the radicand. The square root of a perfect square can be taken out of the radical sign. Simplify the remaining radicand. The remaining radicand should not contain any perfect squares. For example, to simplify the radical expression √72: The perfect squares within 72 are 4 and 9. Factor out the perfect squares: √72 = √(4 × 18) = √4 × √18 = 2√18. The remaining radicand, 18, cannot be factored fur Continue Reading To find the simplest radical form: Identify the perfect squares within the radicand. A perfect square is a number that can be expressed as the product of two equal integers. Factor out the perfect squares from the radicand. The square root of a perfect square can be taken out of the radical sign. Simplify the remaining radicand. The remaining radicand should not contain any perfect squares. For example, to simplify the radical expression √72: The perfect squares within 72 are 4 and 9. Factor out the perfect squares: √72 = √(4 × 18) = √4 × √18 = 2√18. The remaining radicand, 18, cannot be factored further, so the simplest radical form is 2√18. Upvote · 9 1 Related questions How do you write in simplest radical form? How do you know if a radical expression is in the simplest form? What is the radical form of (9) 1/2? What is simplified radical form in math? When can you say that a radical is in the simplest form? How do I express 16 1/3 in simplest radical form? How do I express the mixed number 9 3/2 in the simplest radical form? How do you simplify the radical of 192? How do you simplify the following radical expressions √12/32? How do you write in the simplest form? Why is a radical in a radical form? What is the simplest radical form of 58? How do you put radicals in the simplest form? What is the simplest radical form of an algebraic expression? What is a standard radical form? Related questions How do you write in simplest radical form? How do you know if a radical expression is in the simplest form? What is the radical form of (9) 1/2? What is simplified radical form in math? When can you say that a radical is in the simplest form? How do I express 16 1/3 in simplest radical form? How do I express the mixed number 9 3/2 in the simplest radical form? How do you simplify the radical of 192? How do you simplify the following radical expressions √12/32? How do you write in the simplest form? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
190029	https://www.chegg.com/homework-help/questions-and-answers/1-show-free-body-diagram-vertical-spring-mass-attached-equilibrium-show-acting-forces-q140070982	Solved 1. Show a free body diagram of a vertical spring \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Physics Physics questions and answers 1. Show a free body diagram of a vertical spring with a mass attached, while in equilibrium. Show all the acting forces. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 1. Show a free body diagram of a vertical spring with a mass attached, while in equilibrium. Show all the acting forces. Show a free body diagram of a vertical spring with a mass attached, while in equilibrium. Show all the acting forces. There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 The mass - spring system works when the mass is attached with the spring. View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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190030	https://www.inchcalculator.com/convert/cubic-centimeter-to-liter/	Inch Calculator Skip to Content Popular Searches calculate body fat what is today's date? calculate yards of concrete how long until 3:30? calculate a pay raise 30 minute timer calculate board and batten wall layout how to do long division calculate the best TV size how many days until Christmas? Cubic Centimeters to Liters Converter Enter the volume in cubic centimeters below to convert it to liters. Have a Question or Feedback? Result in Liters: 1 cm³ = 0.001 L Learn how we calculate this below Do you want to convert liters to cubic centimeters? On this page: Cubic Centimeters to Liters Converter How to Convert Cubic Centimeters to Liters What Is a Cubic Centimeter? What Is a Liter? Cubic Centimeter to Liter Conversion Table References By Joe Sexton Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications. Full bio Share on Facebook Share on X Share on Pinterest Cite This Page Sexton, J. (n.d.). Cubic Centimeters to Liters Converter. Inch Calculator. Retrieved September 22, 2025, from How to Convert Cubic Centimeters to Liters To convert a measurement in cubic centimeters to a measurement in liters, divide the volume by the following conversion ratio: 1,000 cubic centimeters/liter. Since one liter is equal to 1,000 cubic centimeters, you can use this simple formula to convert: liters = cubic centimeters ÷ 1,000 The volume in liters is equal to the volume in cubic centimeters divided by 1,000. For example, here's how to convert 500 cubic centimeters to liters using the formula above. liters = (500 cm³ ÷ 1,000) = 0.5 L Cubic centimeters and liters are both units used to measure volume. Keep reading to learn more about each unit of measure. What Is a Cubic Centimeter? The cubic centimeter is a unit that is equal to the volume of a cube with one centimeter sides. One cubic centimeter is equal to about 0.061 cubic inches. The cubic centimeter, or cubic centimetre, is a multiple of the cubic meter, which is the SI derived unit for volume. In the metric system, "centi" is the prefix for hundredths, or 10-2. Cubic centimeters can be abbreviated as cm³, and are also sometimes abbreviated as cu cm, cc, or ccm. For example, 1 cubic centimeter can be written as 1 cm³, 1 cu cm, 1 cc, or 1 ccm. Learn more about cubic centimeters. What Is a Liter? A liter is a unit of volume equal to 1,000 cubic centimeters or 0.264172 US gallons. The liter is a special name defined for the cubic decimeter and is exactly equal to the volume of one cubic decimeter (1 decimeter is 1/10 of a meter, or 10 centimeters). The liter is an SI accepted unit for volume for use with the metric system. A liter is sometimes also referred to as a litre. Liters can be abbreviated as L, and are also sometimes abbreviated as l or ℓ. For example, 1 liter can be written as 1 L, 1 l, or 1 ℓ. Learn more about liters. Cubic Centimeter to Liter Conversion Table Table showing various cubic centimeter measurements converted to liters. \| Cubic Centimeters \| Liters \| \| 1 cm³ \| 0.001 L \| \| 2 cm³ \| 0.002 L \| \| 3 cm³ \| 0.003 L \| \| 4 cm³ \| 0.004 L \| \| 5 cm³ \| 0.005 L \| \| 6 cm³ \| 0.006 L \| \| 7 cm³ \| 0.007 L \| \| 8 cm³ \| 0.008 L \| \| 9 cm³ \| 0.009 L \| \| 10 cm³ \| 0.01 L \| \| 20 cm³ \| 0.02 L \| \| 30 cm³ \| 0.03 L \| \| 40 cm³ \| 0.04 L \| \| 50 cm³ \| 0.05 L \| \| 60 cm³ \| 0.06 L \| \| 70 cm³ \| 0.07 L \| \| 80 cm³ \| 0.08 L \| \| 90 cm³ \| 0.09 L \| \| 100 cm³ \| 0.1 L \| \| 200 cm³ \| 0.2 L \| \| 300 cm³ \| 0.3 L \| \| 400 cm³ \| 0.4 L \| \| 500 cm³ \| 0.5 L \| \| 600 cm³ \| 0.6 L \| \| 700 cm³ \| 0.7 L \| \| 800 cm³ \| 0.8 L \| \| 900 cm³ \| 0.9 L \| \| 1,000 cm³ \| 1 L \| References National Institute of Standards and Technology, Specifications, Tolerances, and Other Technical Requirements for Weighing and Measuring Devices, Handbook 44 - 2019 Edition, National Institute of Standards and Technology, Units outside the SI, More Cubic Centimeter & Liter Conversions Convert from Cubic Centimeters cubic centimeters to gallons cubic centimeters to quarts cubic centimeters to pints cubic centimeters to cups cubic centimeters to fluid ounces cubic centimeters to tablespoons cubic centimeters to teaspoons cubic centimeters to cubic inches cubic centimeters to cubic feet cubic centimeters to cubic yards cubic centimeters to milliliters cubic centimeters to cubic meters Convert to Liters gallons to liters quarts to liters pints to liters cups to liters fluid ounces to liters tablespoons to liters teaspoons to liters cubic inches to liters cubic feet to liters cubic yards to liters milliliters to liters cubic meters to liters ounces to liters pounds to liters milligrams to liters grams to liters kilograms to liters
190031	https://www.reddit.com/r/math/comments/37ooad/densest_circle_packing_and_relationship_to_sphere/	Densest circle packing and relationship to sphere packing : r/math Skip to main contentDensest circle packing and relationship to sphere packing : r/math Open menu Open navigationGo to Reddit Home r/math A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to math r/math•10 yr. ago [deleted] Densest circle packing and relationship to sphere packing You'll see that my title isn't very informative. I didn't know how else to describe it but do know that this is not technically a packing problem. The following is a problem that's always been on the back of my mind and I've never been able to figure out; hope someone here can help. It's easy to see how seven equal circles can be packed most efficiently in an infinite plane. It's also interesting (and the former is implied by the fact) that 6 tangent circles fit completely around a central identical circle. In the above context, I'll define a complete fit to mean an arrangement in which a unit d-dimensional sphere is tightly covered, with no extra wiggle room, by n tangent d-spheres of radius r. For d=2, most notably, you can get a complete fit with n=6 outer circles when r=1. You can also of course get complete fits with any number of identical outer circles n > 2, and I've shown further that r is irrational for any n not equal to 6. Anyway, I'm totally at a loss attempting to move it to the third dimension, d=3. Specifically, I'd like to know whether you can get a complete fit with r=1 (all identical unit circles, I suspect not), or whether r is rational for any other complete fit. I half gave up, but did some other calculations: the solid angle of a single sphere with r=1 at the appropriate distance is about 1/19th of the total area making 19 an upper bound for n. The answer is clearly a lot lower, but I can't figure out how to prove it, and once I arrive at the answer, I also am not sure I could prove that the fit is complete. As a follow up, is there any good reason to expect a complete fit for even a single rational r? How about in higher dimensions? Reading recommendations, hints, or explanations welcome. Read more Share Related Answers Section Related Answers Applications of topology in computer science History of famous unsolved math problems Visualizing higher dimensional geometry Mathematical patterns in nature and art Role of symmetry in algebra and physics New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of May 29, 2015 Reddit reReddit: Top posts of May 2015 Reddit reReddit: Top posts of 2015 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
190032	https://www.khanacademy.org/math/ka-math-class-12/x6f66cbee9a2f805b:continuity-and-differentiability-ncert-new/x6f66cbee9a2f805b:continuity-of-different-functions/v/continuity-of-greatest-integer-function	Continuity of greatest integer function (video) \| 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. 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Skip to lesson content KA Math Class 12 Course: KA Math Class 12>Unit 5 Lesson 2: Continuity of different functions Continuity of polynomial functions Continuity of rational functions Continuity of piecewise defined functions Making a piecewise defined function continuous Making a piecewise defined function continuous Continuity of greatest integer function Algebra of continuous functions Continuity of trigonometric functions Continuity of trigonometric functions Continuity of composite functions Math> KA Math Class 12> Continuity and differentiability> Continuity of different functions © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Continuity of greatest integer function Google Classroom Microsoft Teams About About this video In this video, we'll talk about continuity of greatest integer function. We'll first plot its graph and try to figure out points of discontinuity visually. We'll then make separate cases - for integers and non-integers - and evaluate continuity for both. Finally, we'll solve a problem where we use the definition of GIF to prove discontinuity at integral points of a different function.Created by Ashish Gupta. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Video transcript Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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190033	https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar	probability - Chance of meeting in a bar - 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 Chance of meeting in a bar Ask Question Asked 13 years, 8 months ago Modified10 years, 3 months ago Viewed 23k times This question shows research effort; it is useful and clear 69 Save this question. Show activity on this post. Two people have to spend exactly 15 consecutive minutes in a bar on a given day, between 12:00 and 13:00. Assuming uniform arrival times, what is the probability they will meet? I am mainly interested to see how people would model this formally. I came up with the answer 50% (wrong!) based on the assumptions that: independent uniform arrival they will meet iff they actually overlap by some ϵ>0 ϵ>0 we can measure time continuously but my methods felt a little ad hoc to me, and I would like to learn to make it more formal. Also I'm curious whether people think the problem is formulated unambiguously. I added the assumption of independent arrival myself for instance, because I think without such an assumption the problem is not well defined. probability soft-question probability-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 27, 2012 at 22:31 Adrian Petrescu 1,305 11 11 silver badges 28 28 bronze badges asked Jan 27, 2012 at 17:11 BeltrameBeltrame 3,186 2 2 gold badges 30 30 silver badges 37 37 bronze badges 4 1 The ϵ ϵ stuff is unnecessary. If we model the process by assuming independent arrival times, each uniformly distributed in a specific interval of length 45 45 minutes, then the probability that B arrives xactly when A leaves is 0 0.André Nicolas –André Nicolas 2012-01-27 17:28:30 +00:00 Commented Jan 27, 2012 at 17:28 actually I think the 37.5% is wrong, it doesn't take into account that you cannot arrive past min 45 ! I m amending my answer to 50%Beltrame –Beltrame 2012-01-27 17:48:07 +00:00 Commented Jan 27, 2012 at 17:48 11 My quick calculation gives 5/9 5/9.André Nicolas –André Nicolas 2012-01-27 18:13:42 +00:00 Commented Jan 27, 2012 at 18:13 The probability of some of n n persons meet is 1−(1−(n−1)δ)n.1−(1−(n−1)δ)n. For more details see: Probability of no friends meeting at a coffee houseAmir –Amir 2024-04-06 21:54:03 +00:00 Commented Apr 6, 2024 at 21:54 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 97 Save this answer. Show activity on this post. This is a great question to answer graphically. First note that the two can't arrive after 12:45, since they have to spend at least 15 minutes in the bar. Second, note that they meet if their arrival times differ by less than 15 minutes. If we plot the arrival time of person 1 on the x axis, and person 2 on the y axis, then they meet if the point representing their arrival times is between the two blue lines in the figure below. So we just need to calculate that area, relative to the area of the whole box. The whole box clearly has area 45×45=2025 45×45=2025 The two triangles together have area 2×1 2×30×30=900 2×1 2×30×30=900 Therefore the probability of the friends meeting is (2025−900)/2025=1125/2025=5/9(2025−900)/2025=1125/2025=5/9 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 27, 2012 at 22:35 cambraca 103 4 4 bronze badges answered Jan 27, 2012 at 18:17 Chris TaylorChris Taylor 29.8k 7 7 gold badges 91 91 silver badges 125 125 bronze badges 2 Nice. What if both arrived and left randomly between noon and 1 pm? I can think of defining two Uniform Distributions U1, U2 for persons 1,2 respectfully in [0,60] ( 0:=noon, 60 = 1 p.m) ; assume and then computing the probability that Min U2 > MaxU1 , which entails P(MinU2 -MaxU1 >=0). But that seems top heavy, since we have to first compute the distributions of the Max, Min and the difference. Can you think of something simpler?MSIS –MSIS 2021-09-26 22:27:15 +00:00 Commented Sep 26, 2021 at 22:27 Great answer! What should we add to the graphic to show how this probability works if someone can spend less than 15 mins? There's just evidence that at least a minute in a hour every of persons will be here.. Is there any clever geometrical solution? If there are n persons, we just need to pow the time diff of 60 - minutes in bar by n, right? is it extendable to many dimensions?the_kaba –the_kaba 2023-01-02 03:46:46 +00:00 Commented Jan 2, 2023 at 3:46 Add a comment\| This answer is useful 75 Save this answer. Show activity on this post. chance they meet=green area green area + blue area=45 2−30 2 45 2 chance they meet=green area green area + blue area=45 2−30 2 45 2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 27, 2012 at 18:20 MyselfMyself 8,997 32 32 silver badges 47 47 bronze badges 0 Add a comment\| This answer is useful 18 Save this answer. Show activity on this post. I'd say that each of them arrives at some time uniformly distributed on [12:00, 12:45], where the times are independent. They meet if their arrival times differ by less than fifteen minutes. In the interest of not having annoying numbers all over the place, measure time in units of fifteen minutes, starting at noon. Let the first person's arrival time be X X and the second person's arrival time be Y Y. Then X X and Y Y are independent and uniform on [0,3][0,3] and we want P(\|X−Y\|<1)P(\|X−Y\|<1). Then this becomes a geometry problem. The area in the square [0,3]×[0,3][0,3]×[0,3] which satisfies \|x−y\|<1\|x−y\|<1 is 5 5; the area of the whole square is 9 9; the answer is 5/9 5/9. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 27, 2012 at 18:16 Michael LugoMichael Lugo 24.5k 3 3 gold badges 52 52 silver badges 98 98 bronze badges 5 Out of curiosity, is there a straightforward non-geometric proof?user2468 –user2468 2012-01-27 18:57:02 +00:00 Commented Jan 27, 2012 at 18:57 8 You could justify ∬0≤X,Y≤3;\|X−Y\|<1 d X d Y/∬0≤X,Y≤3 d X d Y∬0≤X,Y≤3;\|X−Y\|<1 d X d Y/∬0≤X,Y≤3 d X d Y on non-geometrical grounds, but the easiest way to do that first integral is geometrically. So I think trying to leave geometry out of the proof entirely is just making it more complicated than it needs to be.David Z –David Z 2012-01-27 21:03:01 +00:00 Commented Jan 27, 2012 at 21:03 6 @J.D. The answer 5 9=1−4 9=1−(2 3)2 5 9=1−4 9=1−(2 3)2 is so reminiscent of 1−P(A c)P(B c)=P(A)+P(B)−P(A)P(B)=P(A∪B)1−P(A c)P(B c)=P(A)+P(B)−P(A)P(B)=P(A∪B) where A A and B B are independent events of probability 1 3 1 3 that I too wonder if there is a simple non-geometric argument that we are all overlooking.Dilip Sarwate –Dilip Sarwate 2012-01-28 17:22:31 +00:00 Commented Jan 28, 2012 at 17:22 @user2468: Not sure if this is simple : Given two uniforms U1(0,60) and U2(0,60), assuming U1 arrives first, we need to integrate the difference of MinU2 -MaxU1 from 0 to 60. I guess we need convolution product.MSIS –MSIS 2021-10-01 21:09:13 +00:00 Commented Oct 1, 2021 at 21:09 @Michael Lugo: What if we just new X,Y arrived and left uniformly in the interval [12:00,1:00]:=[0,1]? I am thinking we set X to have arrived first, then we find the integral of Min Y- Max X , from 0 to 1. But that seems a bit clumsy. Can you think of a nicer way?MSIS –MSIS 2021-10-09 00:37:55 +00:00 Commented Oct 9, 2021 at 0:37 Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. I had a hard time envisioning both A and B times simultaneously, so I considered the probability as a function of A's arrival time. Given: both arrival times are in the range 12:00-12:45, so A can arrive at any time in those 45 minutes. If A arrives at 12:00, B is met iff B arrives in the first 15 minutes, so the probability is 1/3. If A arrives at 12:15, B can arrive any time from 0-30 minutes, so p = 2/3. This value of 2/3 remains until 12:30, because there remains a 30 minute window that their times could overlap. After 12:30, p decreases until it reaches 1/3 at 12:45 (symmetric to the condition at 12:00). Here is a graph (not as pretty as above, but I thought I'd give google docs drawing a try): not included because this is my first math post. LINK: graph of p(A meets B \| A's arrival time) The shaded portion of the first and third segments are half the area of that segment, the middle is 2/3, so the total area is p=1 3×(1 2+2 3+1 2)=1 3×5 3=5 9 p=1 3×(1 2+2 3+1 2)=1 3×5 3=5 9 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 29, 2012 at 0:46 toddkaufmanntoddkaufmann 191 2 2 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Here is a simple qbasic simulation of the situation: The answer yielded over 1 million trials was 0.5555 ~ 5/9 CLS RANDOMIZE TIMER count = 0 N = 1000000 FOR i = 1 TO N LET x = (RND 45 + 1) LET y = (RND 45 + 1) tx = x + 15 ty = y + 15 IF ABS(tx - ty) < 15 THEN count = count + 1 END IF NEXT i PRINT count / N Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 28, 2012 at 18:02 Derek KinsellaDerek Kinsella 69 1 1 bronze badge 5 8 I downvoted you. What you posted is not an acceptable within the scope & purpose for math.stackexchange. Empirical proofs are only acceptable when there are no mathematical explanations. Please read the answers above.user2468 –user2468 2012-01-28 18:24:41 +00:00 Commented Jan 28, 2012 at 18:24 thanks, I should have also used a simulation to give my initial (wrong) answer a sanity check. I was looking for the reasoning behind a rigorous answer though. For that I find empirical methods less useful.Beltrame –Beltrame 2012-01-28 20:35:35 +00:00 Commented Jan 28, 2012 at 20:35 2 @J.D.: I think your words are unnecessarily harsh. I'd certainly expect an empirical answer to receive fewer votes than an exact one (when one exists), but I don't think that makes empirical answers "not acceptable" on this site.Cam McLeman –Cam McLeman 2012-02-02 15:42:59 +00:00 Commented Feb 2, 2012 at 15:42 @CamMcLeman I was trying to leave an explanation on why I downvoted this answer. In any case, thank you for pointing this out to me, and I apologize to if my words are inappropriate.user2468 –user2468 2012-02-02 16:17:13 +00:00 Commented Feb 2, 2012 at 16:17 1 Is this BASIC? Wow, that's a blast from the past.Chris Taylor –Chris Taylor 2012-02-07 09:30:19 +00:00 Commented Feb 7, 2012 at 9:30 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Rather than measure time continuously, lets start with discrete time to try to get a handle on an approach. Divide the time into n equal minutes: d, 2d, 3d, ...,nd where nd= 1 hr. = 60 min. The first person arrives at time jd. Ignoring end points, the probability that the next person arrives within 15d of that time is (j-15)d to (j+15)d or 30d. Since j can be any of n numbers we have, ignoring end points, the probability is 30d/60d = 50%. That gives us an approach, but we have to take into account the end points, i.e. the time restriction. Well 1/n times the arrival will be at 1d and the sum will be restricted to d to 16d or just 15d. Next is arrive at 2d and sum is 16d, etc. So, instead of a constant function of 30d, we have a ramp function starting at 15d going up to 30d, staying there for awhile and going back down to 15d. Add those up with the proper weights. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 5, 2015 at 2:57 leaveswater02leaveswater02 598 5 5 silver badges 3 3 bronze badges 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 probability soft-question probability-theory See similar questions with these tags. 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He has a Bachelors Degree in Chemistry and a Master's Degree in Instructional Media with a specialization in STEM. View bio Instructor Elizabeth (Nikki) Wyman Nikki has a master's degree in teaching chemistry and has taught high school chemistry, biology and astronomy. View bio Understand the definition of a binary molecular compound. Learn the rules for naming binary molecular compounds and discover how to write their chemical formulas. Updated: 11/21/2023 Table of Contents Binary Molecular Compounds Naming Binary Molecular Compounds Writing Chemical Formulas from the Names of Binary Molecular Compounds Lesson Summary Show Frequently Asked Questions How do you write formulas for binary molecular compounds? Here is a guide to writing formulas from binary molecular compounds Step 1: Write the chemical symbol for the first of the two elements named. Step 2: Determine the subscript needed on the first element from the prefix which would come before the name of the first element. If no prefix exists, then no subscript would be needed on the first element. Step 3: Write the chemical symbol for the second element. Step 4: Determine the subscript needed on the second element by determining the prefix that is listed before the name of the second element. Is H2O a binary molecular compound? H2O is indeed a binary molecular compound. The reason that it may be slightly confusing to see that is because most people know it by its accepted common name of "Water." To sound much more scientific, however, H20 can be referred to as dihydrogen monoxide with equal accuracy. Create an account Table of Contents Binary Molecular Compounds Naming Binary Molecular Compounds Writing Chemical Formulas from the Names of Binary Molecular Compounds Lesson Summary Show Binary Molecular Compounds -------------------------- There are hundreds of binary molecular compounds that can be found on this planet. The term binary is a reference to the idea that these compounds will contain only two different elements. It is possible that there will be numerous atoms of the same element within any compound though. To be classified as molecular, a compound must contain two non-metal elements. Compounds exist when two or more elements arechemically bonded to one another. If a substance fits all three of these definitions, it can then be classified as a binary molecular compound. In addition to being called binary molecular compounds, it is also very common for these compounds to be referred to as "binary covalent compounds". This alternate name is fully acceptable and is a reference to the type of chemical bonds that two non-metals would make with each other. Examples of Binary Molecular Compounds Many compounds will fit the category of being binary molecular compounds. Most of them are not commonly known to the non-scientific community, but there are quite a few that are rather famous as well. The following list is some of the more famous binary molecular compounds as well as some of the lessons known. Take notice that there are only 2 different elements in each of the compounds: C O 2= Carbon Dioxide Fire Extinguisher C C l 4= Carbon tetrachloride N 2 O 4= Dinitrogen tetraoxide B B r 3= Boron tribromide P 3 I=triphosphorus monoiodide H 2 O= Dihydrogen monoxide Dihydrogen Monoxide puddle Click for sound 7:05 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching 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 now Try it now. Already registered? 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Your next lesson will play in 10 seconds 0:00 What Is A Binary… 1:00 Naming Binary… 4:23 Writing Formulas 6:09 Lesson Summary View Video Only Save Timeline 61K views Video Quiz Course Video Only 61K views Naming Binary Molecular Compounds --------------------------------- There is a system in place for naming binary molecular compounds that requires a bit of memorization in order to be fluent with it. As there are only two elements present in these compounds, the easiest way to communicate which element is being referred to is by the terms "first element" and "second element" when referring to the elements in the compound. There are several rules that should be followed for each of the two elements involved in the compound. Both elements will be using a prefix prior to the base element name that will match up with the subscriptof each element. The prefixes that are used for subscripts 1-10 can be found in the table: \| Subscript number \| Prefix Name \| --- \| \| 1 \| mono \| \| 2 \| di \| \| 3 \| tri \| \| 4 \| tetra \| \| 5 \| penta \| \| 6 \| hexa \| \| 7 \| Hepta \| \| 8 \| octa \| \| 9 \| nona \| \| 10 \| deca \| It is best to be familiar with the set of rules for naming these compounds below in order to gain fluency: The second element in the compound always ends in the suffix -ide. The ending of the element is usually cut off at a phonetically pleasing location. The first element will maintain its original base element's ending (no suffix added) The subscript that follows an element's symbol will be used as a prefix that is listed before the name of the element it represents. The second element ALWAYS receives a prefix (even if the subscript is an unwritten 1). The first element gets a prefix in all cases except when there is an unwritten 1 as a subscript. Examples of Naming Binary Molecular Compounds Putting these rules into action, the name for P 2 B r 5 can be determined below: Step 1 The first element's name will remain unchanged, but receive a prefix if its subscript is greater than 1. P has a subscript of 2, so it will need to use the Greek prefix of Di, standing for 2. The first word of the compound would therefore be Diphosphorus. Take notice that the entire name of the element was written out without changing the suffix. Step 2 Examining the second element, it has a subscript of 5, so the prefix from the Greek list will be penta-. Step 3 Determine where to remove the original element's ending and replace it with the suffix -ide. In this case, the element's root with a suffix would be altered to bromide. Step 4 Putting all of the information together, this compound's name would be Diphosphorus pentabromide. In a second example, the name for N O 2 will be determined. Step 1 The first element's name will remain unchanged and will NOT receive a prefix since the subscript is an unwritten 1. Nitrogen will be the first word. Step 2 The second element's prefix will be representative of that element's subscript. With a subscript of 2, the prefix for the second element will be Di-. Step 3 The base of the second element will stay the same, but the suffix -ide will replace the ending of the original element. In this case, oxygen will change to oxide. Step 4 Putting all parts of this name together, it would be called nitrogen dioxide. Writing Chemical Formulas from the Names of Binary Molecular Compounds ---------------------------------------------------------------------- To perform the reverse operation, writing the formula when provided with a name, the same naming system can be used that was mentioned above, only in the opposite direction. Here is a guide for how to approach the reverse mindset. Write the symbol for the first element's name. If the element has a prefix in front of it, write the subscript for that prefix after the symbol. Write the symbol for the element that is represented by the second element's base name. Write the subscript for the second element after the symbol that is equal to the prefix value. Example of Writing the Chemical Formula of Binary Molecular Compounds Putting the rules above into practice, given the name Tricarbon dioxide, the formula can be determined with the following steps: Write down the symbol for the first element's name, C. Add a subscript after the first symbol equal to the value of the prefix, C 3 Write the symbol for the second element, O. Determine the subscript by examining the value of the prefix, O 2 Putting those two pieces of information together, the formula would be written as C 3 O 2 Lesson Summary -------------- Binary molecular compounds have a strict naming system that must be adhered to in order to precisely describe the composition of the elements which make it up. To distinguish the naming system of a molecular compound from that of a similar compound known as ionic, a molecular compound will have prefixes written before any element that receives a subscript greater than 1 and also will also have a prefix on the second element's name. This prefix system is rather unique and is a signal that molecular compounds are being discussed. In addition to the prefix system, molecular naming also has rules when it comes to the suffix of an element as well. The first element will never receive an alteration to the original suffix of the element. The ending of the element on the Periodic Table will be the ending of the first element's name in the compound as well. The second element in the compound, however, will always have the suffix changed to -ide. To write formulas from the names of binary molecular compounds, it is important to remember the order and names of the Greek prefixes. By recognizing the prefixes, the subscript for each element can be determined. In order, the Greek Prefixes are mono, di, tri, tetra, penta, hexa, hept, octa, nona, and deca. These prefixes would match up with a subscript that is written after the symbol in a compound. Video Transcript What is a Binary Molecular Compound? Let's begin by dissecting the term binary molecular compound. Binary refers to a substance comprised of only two different kinds of materials. In this case, we're referring to atoms. Molecular refers to the sharing of electrons between non-metal atoms. Sharing of electrons is known as covalent bonding. A compound is a chemical made of at least two atoms bonded together. If we put all those terms together, we have a definition: a binary molecular compound is made of two different types of nonmetal atoms covalently bonded together through the sharing of electrons. For example, carbon dioxide (CO2) is comprised of one carbon element and two oxygen elements. There are just two types of elements present: carbon and oxygen. Both of these elements are nonmetals. Other binary molecular compounds include nitrogen monoxide (NO2) and sulfur dioxide (SO2). Naming Binary Molecular Compounds There are four main rules to naming binary molecular compounds and four easy steps to the naming process. Rule 1: If both elements in the compound are in the same period on the periodic table, the element with the smaller group number is listed first. The only exception to this rule is when oxygen and halogen, or group 7A elements like fluorine and chlorine, are in a compound. Then the halogen is listed first. Rule 2: If the elements are in two different periods, the element in the larger period will be listed first. Correct formulas should list the elements in their proper order. Thus, when naming compounds based on formulas, the elements are already in the correct order! Rule 3: Greek prefixes are used to indicate the number of each element in the compound. They are simply attached in front of name of the element that they modify. In case you're not familiar, here are the Greek prefixes for numbers 1-10. \| Number \| Greek Prefix \| --- \| \| 1 \| mono- \| \| 2 \| di- \| \| 3 \| tri- \| \| 4 \| tetra- \| \| 5 \| penta- \| \| 6 \| hexa- \| \| 7 \| hepta- \| \| 8 \| octa- \| \| 9 \| nona- \| \| 10 \| deca- \| The only exception to Rule 3 occurs when there is only one of the first element. In this case, no prefix is needed. Rule 4: The ending of the second element is dropped and replaced with '-ide'. Now let's use these steps to name P2Br5. First ensure that the elements in the compound are correctly ordered, as required by Rules 1 and 2. As phosphorus is in a lower group number (5A) than bromine (7A), the order is correct. Next, list the number and type of each element that's present in the compound: 2 phosphorus and 5 bromine. Next, replace the number with a Greek prefix, and add it to the element it modifies. So, 2 phosphorus becomes diphosphorus, and 5 bromine becomes pentabromine Now, let's remove the ending of the second element and replace it with '-ide': diphosphorus pentabromide. Here's another example: NO2. This compound follows all the naming rules with one exception. Make sure the elements in the compound are ordered correctly. As nitrogen is located in group 5A, which is lower than oxygen's group (6A), the order is correct. Then, list the number and type of each element that's present in the compound: 1 nitrogen and 2 oxygen. Next, replace the number with a Greek prefix, and add it to the element it modifies. Ahoy! Here is the exception. Remember, when there is only one of the first element in the compound, it does not get a prefix. So, 1 nitrogen is shown as nitrogen, and 2 oxygen becomes dioxygen. The last step involves removing the ending of the second element and replacing it with '-ide': nitrogen dioxide. When your prefix ends in a vowel and your element name begins with a vowel, the element's vowel trumps the prefix vowel. Thus, the vowel of the prefix is dropped and the vowel of the element is kept. For example, penta + oxide becomes pentoxide Without looking ahead, can you name H2O and CO yourself? Pause this video for a moment to determine the names for these molecules. H2O is dihydrogen monoxide (aka water) and CO is carbon monoxide. Writing Formulas Writing formulas for binary molecular compounds is like decoding a message. If you can name a compound from its formula, then it's easy to do the reverse. We will rely on some of the same rules and use essentially the same four steps. Rule 1 and 2 remain unchanged. Rule 3, which has to do with numbers, states that the number of the element will be noted as a subscript immediately following the element symbol. Elements that only occur once do not require a 1. There is no rule 4! Let's find the formula for the compound sulfur dichloride. The first step is to make sure the elements are in the correct order. Sulfur is in a lower group (6A) than chlorine (7A), so the order is correct. Next, determine the name and number of each element present: 1 sulfur and 2 chlorine. Replace the element names with their symbols: 1S and 2Cl. Lastly, turn numbers into subscripts, and place them to the bottom right of the element they modify. Since the number 1 is never used in chemical formulas, we drop it. Our formula becomes SCl2. Here's another example: carbon tetrafluoride. Ensure that the elements are in the correct order. Carbon is in a lower group (4A) than fluorine (7A), so the order is correct. Determine the name and number of each element present: 1 carbon and 4 fluorine. Next, replace the element names with their symbols: 1 C and 4 F. Turn numbers into subscripts, and place them to the bottom right of the element they modify: CF4. Lesson Summary A binary molecular compound consists of two different types of nonmetal atoms that share electrons and form a covalent bond. The proper naming of binary molecular compounds is based upon four rules and steps: Rule 1: List the element with the smaller group number first if both elements in the compound are in the same period. Rule 2 If the elements are in two different periods, list the one in the larger period first. Rule 3 Use one of the Greek prefixes to indicate the number of each element. Rule 4 Replace the ending of the second element with 'ide'. Additionally, you can use the names of binary molecular compounds to write their formulas by reversing the process and eliminating the fourth step. Some binary molecular compounds include carbon dioxide (CO2), nitrogen monoxide (NO2) and sulfur dioxide (SO2). Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? Log In Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. 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190035	https://math.stackexchange.com/questions/3666963/combinations-of-elements-from-multiple-sets-with-constraints	Skip to main content Combinations of elements from multiple sets with constraints Ask Question Asked Modified 5 years, 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. Given four sets of numbers: A={1,2,3,4} B={1,2,3,4} C={1,2,3,5} D={1,2,6} I need to find number of all possible combinations of numbers from these sets in following format: (a,b,c,d) where a∈A,b∈B,c∈C,d∈D and where each number can be present only once per combination (e.g. combination (1,2,3,4) is valid, but combination (1,1,2,3) is not (number 1 repeats) and neither is combination (2,3,5,2) (number 2 repeats)). My idea is to use inclusion-exclusion principle. First I calculate number of all possible combinations: \|A\|⋅\|B\|⋅\|C\|⋅\|D\|=192 And then I need to remove all combinations where elements are repeating 2 or more times. These sets are: \|A∩B\|⋅\|C\|⋅\|D\|=48 \|A∩C\|⋅\|B\|⋅\|D\|=36 \|A∩D\|⋅\|B\|⋅\|C\|=32 \|B∩C\|⋅\|A\|⋅\|D\|=36 \|B∩D\|⋅\|A\|⋅\|C\|=32 \|C∩D\|⋅\|A\|⋅\|B\|=32 The problem here is that I have duplicates during removal, since sum of all combinations above is 216 (some combinations that I remove - I remove multiple times). My questions is - how to find intersections between all these sets, in order to get correct number for removal (in this case it should be 142 and not 216) - so correct answer in the end should be 192−142=50. I guess I need to find following intersections but not sure how to calculate all of these: \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|A∩B\|⋅\|C\|⋅\|D\|)\|=9(1) \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|A∩C\|⋅\|B\|⋅\|D\|)\|=8 \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|A∩D\|⋅\|B\|⋅\|C\|)\|=9 \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|B∩C\|⋅\|A\|⋅\|D\|)\|=8 \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|B∩D\|⋅\|A\|⋅\|C\|)\|=8 \|(\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|C∩D\|⋅\|A\|⋅\|B\|)\|=8 ... \|(\|A∩D\|⋅\|B\|⋅\|C\|)∩(\|B∩C\|⋅\|A\|⋅\|D\|)\|=6(2) ... (1) duplicates are (1,1,1,1),(1,1,1,2),(1,1,1,6),(2,2,2,1),(2,2,2,2),(2,2,2,6),(3,3,3,1),(3,3,3,2),(3,3,3,6) (2) duplicates are (1,1,1,1),(1,2,2,1),(1,3,3,1),(2,1,1,2),(2,2,2,2),(2,3,3,2) -> what is the rule here if I have more than 4 sets like here? and then I need all 3 intersections: \|((\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|A∩C\|⋅\|B\|⋅\|D\|)∩(\|A∩D\|⋅\|B\|⋅\|C\|))\|=2 ... and then 4,5 intersections (not shown here) and finally I need remaining (6) intersection of all: \|((\|A∩B\|⋅\|C\|⋅\|D\|)∩(\|A∩C\|⋅\|B\|⋅\|D\|)∩(\|A∩D\|⋅\|B\|⋅\|C\|)∩(\|B∩C\|⋅\|A\|⋅\|D\|)∩(\|B∩D\|⋅\|A\|⋅\|C\|)∩(\|C∩D\|⋅\|A\|⋅\|B\|))\|=2 In the end I get result: 192−(216−119+65−30+12−2)=50 I tried to generalize this to more than 4 sets, but cannot find strict rule to calculate these intersections that I wrote above. Any help would be appreciated. UPDATE Based on answer below, I found a formula that makes computation much easier per partition: Part(p)=∏n=1g(−1)ln⋅ln!⋅sn Here, p is one of partitions - e.g. {{A,B},{C},{D}}; g is number of groups in that partition (in this case 3 - {A,B} ,{C} and {D}); ln is number of elements in a group minus 1 - in this case it would be 2−1=1, 1−1=0 and 1−1=0 per group; sn is number of elements in that group - in this case \|{A,B}\|=4, \|{C}\|=4 and finally \|{D}\|=3. You can see all the formula for this example in the uploaded picture below: combinatorics combinations inclusion-exclusion Share CC BY-SA 4.0 Follow this question to receive notifications edited May 11, 2020 at 8:13 Bojan Vukasovic asked May 9, 2020 at 18:39 Bojan VukasovicBojan Vukasovic 18799 bronze badges Add a comment \| 3 Answers 3 Reset to default This answer is useful 4 Save this answer. Show activity on this post. This could theoretically be done with standard inclusion–exclusion on the (n2) conditions that the elements be pairwise different (where n is the number of sets being chosen from, in your case n=4), but that would be quite cumbersome. It’s easier to apply general Möbius inversion on posets. Here the partially ordered set is the set of partitions of your set of sets, partially ordered by refinement. That is: Let S={A,B,C,D} denote the set of sets, and let P denote the set of its partitions. For each partition p∈P of S, we can count the tuples that are constant on each part of the partition. For instance, for p={{A,C},{B,D}}, the identical elements from A and C can be 1, 2 or 3, and the identical elements from B and D can be 1 or 2, so there are a total of 3⋅2=6 tuples that are constant on the parts of p. Partially order P by refinement, where p⪯q means that q is finer than p, that is, each part of q is a subset of a part of p; for instance, {{A,C},{B,D}}⪯{{A},{C},{B,D}}. Let g(p) count the tuples that are constant on the parts of p, and let f(p) count the tuples that are constant on the parts of p but not of any coarser partition. Then we have g(q)=∑p⪯qf(p). We can readily count g, and we want to invert the sum by Möbius inversion to get f(s), where s={{A},{B},{C},{D}} is the finest partition, the maximal element of P: f(s)=∑p⪯sg(p)μ(p,s). The values of the Möbius function μ that we need are determined by the initial value μ(s,s)=1 and the recurrence μ(s,p)=−∑s⪯q⪯pμ(s,q). The next-coarsest partitions after s are the ones with three parts, of the form {{A,B},{C},{D}}. There are 6 of these, one for every pair of sets. You’ve already counted the corresponding values of g. As the only finer partition is s, we have μ(s,p)=−1 for each of these, so we subtract each of their counts once, as expected. Then there are two types of partitions with two parts, of the form {{A,B},{C,D}} and of the form {{A,B,C},{D}}. There are 3 of the first type (one for each way to divide the sets into pairs) and 4 of the second type (one for each set). A partition of the first type is coarser than 2 of the partitions with three parts, and a partition of the second type is coarser than 3 of the partitions with three parts, so the Möbius function values μ(s,p) are −(1+2⋅(−1))=1 and −(1+3⋅(−1))=2, respectively. Finally, there is one partition with a single part, {{A,B,C,D}}, the minimal element of P, which is coarser than all other partitions, so the corresponding Möbius function value is μ(s,p)=−(1+6⋅(−1)+3⋅1+4⋅2)=−6. Counting the values of g as described above yields counts of 8,6,6 and 9,8,8,8 for the partitions with two parts of the first and second type, respectively, and a count of 2 for the partition with a single part. For example, the values 8, 6, 6 correspond to the partitions {{A,B},{C,D}}, {{A,C},{B,D}} and {{A,D},{B,C}}, respectively. In the first case, A and B share 4 elements (all 4) and C and D share 2 elements (1 and 2). We can combine these in 4⋅2=8 ways to form tuples that are constant on both parts. One example of such a tuple is (1,1,2,2). Analogously, A and C share 3 elements (1, 2 and 3) and B and D share 2 elements (1 and 2), so for the second partition there are 3⋅2=6 tuples. Now by Möbius inversion the desired count of tuples that are not constant on any partition coarser than the finest partition s is f(s)=192−48−36−32−36−32−32+8+6+6+2(9+8+8+8)−6⋅2=50, in agreement with your result. Share CC BY-SA 4.0 Follow this answer to receive notifications edited May 10, 2020 at 10:46 answered May 10, 2020 at 2:48 jorikijoriki 243k1414 gold badges311311 silver badges546546 bronze badges 18 1 Very nice and instructive explanation. (+1) – Markus Scheuer Commented May 10, 2020 at 4:41 Could you please write more explicit how the values of g (e.g. 8,6,6) were computed? – user Commented May 10, 2020 at 7:57 @user: The values 8,6,6 correspond to the three partitions of the form {{A,B},{C,D}}. These are {{A,B},{C,D}}, {{A,C},{B,D}} and {{A,D},{B,C}}. In the first case, A and B share 4 elements (all 4) and C and D share 2 elements (1 and 2). We can combine these in 4⋅2=8 ways to form tuples that are constant on both parts. One example of such a tuple is (1,1,2,2). Analogously, A and C share 3 elements (1, 2 and 3) and B and D share 2 elements (1 and 2), so for the second partition there are 3⋅2=6 tuples. – joriki Commented May 10, 2020 at 8:05 @joriki thanks for detailed explanation! Can you please clarify this part: "A partition of the first type is coarser than 2 of the partitions with three parts, and a partition of the second type is coarser than 3 of the partitions with three parts". How you find number of coarser partitions? – Bojan Vukasovic Commented May 10, 2020 at 9:36 1 thanks for the links. i will read all that and try to understand how to do it. I'll post my final findings here. – Bojan Vukasovic Commented May 10, 2020 at 10:30 \| Show 13 more comments This answer is useful 1 Save this answer. Show activity on this post. A selective case-based solution with four cases: d=6,c=5:1∗1∗4∗3=12d=6,c≠5:1∗3∗3∗2=18d≠5,c=5:2∗1∗3∗2=12d≠6,c≠5:2∗2∗2∗1=812+18+12+8=50 Share CC BY-SA 4.0 Follow this answer to receive notifications answered May 10, 2020 at 10:59 Daniel MathiasDaniel Mathias 6,13311 gold badge1313 silver badges2424 bronze badges 6 what is d and c ? – Bojan Vukasovic Commented May 10, 2020 at 11:02 @Bojan From your format, (a,b,c,d). The numbers in the products are options for choosing d, c, b then a – Daniel Mathias Commented May 10, 2020 at 11:06 So, if I take 2nd case d=6,c≠5, 1∗3∗3∗2 - can you please elaborate how to get 1,3,3,2? e.g. is last 2 number of elements in d without 6 ? – Bojan Vukasovic Commented May 10, 2020 at 11:11 @Bojan 1 option for d with the restriction d=6, 3 options for c≠5 (selecting from {1,2,3}), 3 options for b (the number chosen for c being excluded) and finally, 2 remaining options for a. – Daniel Mathias Commented May 10, 2020 at 11:20 Thanks. So in a,b,c,d order it is 2∗3∗3∗1. I still guess this is easier to calculate by hand only in this case, since many elements are repeating. If number of non-repeating elements increase I wold get much more possible combinations to calculate? – Bojan Vukasovic Commented May 10, 2020 at 11:25 \| Show 1 more comment This answer is useful 1 Save this answer. Show activity on this post. We calculate the wanted number with the help of rook polynomials. We consider a (6×4)-rectangular board, where the columns represent the sets A,B,C,D and the rows the elements of the sets. A valid quadruple (a,b,c,d)∈A×B×C×D can be represented by placing four non-attacking rooks on the board, with forbidden positions marked as red squares. A valid arrangement is given showing (a,b,c,d)=(2,4,1,6). Since the number of forbidden squares is smaller than the number of valid squares, we calculate the rook polynomial for the forbidden squares and subtract the corresponding number of non-attacking rook arrangements from the non-attacking rook arrangements of the complete rectangle. The rook polynomial r(x) we are looking for is r0+r1x+r2x2+r3x3+r4x4(1) with rj (0≤j≤4) denoting the number of placing exactly j non-attacking rooks on the forbidden red squares. r0: There is one arrangement placing no rooks at all. r1: We have 9 red squares where we can place a rook, giving r1=9. r2: We count all valid arrangements placing two non-attacking rooks on the forbidden squares. We start with placing a rook on (A,6) which admits 5 arrangements by placing the other rook on (B,5),(C,4),(D,5),(D,4),(D,3). We continue systematically and obtain: (A,6)→5(B,6)→4(C,6)→3(A,5)→5(B,5)→4(C,4)→2 from which r2=5+5+4+4+3+2=23 follows. r3: We count all valid arrangements placing two non-attacking rooks on the forbidden squares. We start with placing two rooks on (A,6),(B,5) which admits 3 arrangements by placing the third rook on (C,4),(D,4),(D,3). We continue systematically and obtain (A,6),(B,5)→3(A,6),(C,4)→2(A,5),(C,4)→1(B,5),(C,6)→2(A,5),(B,6)→3(A,5),(C,6)→2(B,6),(C,4)→2(B,5),(C,4)→1 from which r3=3+3+2+2+1+2+2+1=16 follows. r4: We have two valid arrangements with four rooks, namely (A,6),(B,5),(C,4),(D,3) and (A,5),(B,6),(C,4),(D,3) giving r4=2. The rook polynomial of the forbidden squares is r(x)=1+9x+23x2+16x3+2x4 Now its time to harvest. The number of all arrangements of four non-attacking rooks on the (6×4) board is 6⋅5⋅4⋅3=6!2(2) We now use the coefficients of the rook polynomial r(x) and apply the inclusion-exclusion principle. We subtract from (2) the number of arrangements with one rook on the forbidden squares, then adding the number of arrangements with two rooks on the forbidden squares. We continue this way and obtain 6!2−9⋅5!2+23⋅4!2−16⋅32+22!2=360−540+276−48+2=50 in accordance with the other answers. Share CC BY-SA 4.0 Follow this answer to receive notifications edited May 13, 2020 at 19:04 answered May 12, 2020 at 21:15 Markus ScheuerMarkus Scheuer 113k77 gold badges106106 silver badges249249 bronze badges 2 Thanks for another approach. I guess this one would not be more optimal for algorithm implementation on a PC? All these valid arrangements also take time... – Bojan Vukasovic Commented May 13, 2020 at 8:48 @BojanVukasovic: You're welcome and I agree. The main intention of this answer was to show another combinatorial approach. – Markus Scheuer Commented May 13, 2020 at 9:22 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 combinatorics combinations inclusion-exclusion See similar questions with these tags. 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190036	https://tasks.illustrativemathematics.org/content-standards/tasks/7	Engage your students with effective distance learning resources. ACCESS RESOURCES>> Ordering 3-digit numbers No Tags Alignments to Content Standards: 2.NBT.A.4 Student View Task Arrange the following numbers from least to greatest: $$ 476 \qquad \qquad 647 \qquad \qquad 74 \qquad \qquad 674 \qquad \qquad 467 $$ ______ ______ ______ ______ ______ Arrange the following numbers from greatest to least: $$ 326 \qquad \qquad 362 \qquad \qquad 63 \qquad \qquad 623 \qquad \qquad 632 $$ ______ ______ ______ ______ ______ IM Commentary Each number has at most 3 digits so that students have the opportunity to think about how digit placement affects the size of the number. Each group also contains a two-digit number so that students have to do more than just compare the first digit, the second digit, etc. Solution Least to greatest: Arranging the numbers from least to greatest means writing the numbers in an ordered list according to their values. The smallest number should be written on the left, with the next smallest number written to its immediate right. The process continues for all the given numbers, the largest of which should be on the right. First we look for the smallest number given. $74$ is the smallest number given because it is the only number with no 100s. (There is an implied zero in the hundred€™s place.) We now look for the second smallest number. There are two numbers, $476$ and $467$, with four $100$s (fours in the hundred€™s place). We must now consider the ten€™s place. $467$ is the next smallest number because it only has six $10s$ (a six in the ten€™s place), while $476$ has seven $10$s (a seven in the ten€™s place). The next smallest number is $476$ because it only has four $100$s (a four in the hundred€™s place), while all the other remaining numbers have six $100$s (sixes in the hundred€™s place). We now have two numbers remaining, $647$ and $674$. Both numbers have six $100$s (sixes in the hundred€™s place), so we must compare their $10$s. $647$ is smaller than $674$ because in has just four $10$s (a four in the ten€™s place), rather than seven $100$s (a seven in the hundred€™s place). This leaves $674$ as the largest number. $$ 74 \qquad \qquad 467 \qquad \qquad 479 \qquad \qquad 647 \qquad \qquad 674 $$ Greatest to least Arranging the numbers from greatest to least means writing the numbers in an ordered list according to their values. The largest number should be written on the left, with the next largest number written to its immediate right. The process continues for all the given numbers, the smallest of which should be on the right. First we look for the largest number given. There are three numbers that begin with six: $63$, $623$ and $632$. However, the $6$ in $63$ is in the tens place, not in the hundreds place. There are zero hundreds, making $63$ the smallest number, the only number that is less than $100$. $623$ and $632$ both have six $100$s (sixes in the hundred€™s place), so it is necessary to compare the digits in the $10$€™s place. $623$ has two $10$s (a two in the ten€™s place), while $632$ has three 10s (a tens in the ten€™s place. Since the value of the three in the $10$'s place is larger than two in the $10$'s place, $632$ is the largest number and $623$ is the second largest number. The two remaining numbers both start with three. Comparing $326$ and $362$, which both have $3$ hundreds (a three in the hundred€™s place), means comparing the numbers in the ten€™s place. Six tens (a six in the ten€™s place) is larger than three tens (a three in the ten€™s place), so $326$ is smaller than $362$. $$ 632 \qquad \qquad 623 \qquad \qquad 362 \qquad \qquad 326 \qquad \qquad 63 $$ Ordering 3-digit numbers Arrange the following numbers from least to greatest: $$ 476 \qquad \qquad 647 \qquad \qquad 74 \qquad \qquad 674 \qquad \qquad 467 $$ ______ ______ ______ ______ ______ Arrange the following numbers from greatest to least: $$ 326 \qquad \qquad 362 \qquad \qquad 63 \qquad \qquad 623 \qquad \qquad 632 $$ ______ ______ ______ ______ ______ Print Task
190037	https://math.libretexts.org/Courses/Queens_College/Preparing_for_Calculus_Bootcamp_(Gangaram)/01%3A_Day_1/1.05%3A_Rationalizing_Expressions	Skip to main content Page 1.5: Rationalizing Expressions Last updated : Aug 1, 2023 Save as PDF Page 1.4: Algebraic Expressions and Operations Page 1.6: Factoring Polynomials Page ID : 131896 The NROC Project ( \newcommand{\kernel}{\mathrm{null}\,}) Introduction An algebraic expression may sometimes have a radical in the denominator (or numerator). In calculus, a helpful technique would be to get rid of the radical in the denominator (or numerator). The technique used to rid the expression of the radical is called rationalizing. We will first discuss the idea of rationalizing the denominator, followed by rationalizing the numerator. What is Rationalizing a Denominator? The idea of rationalizing a denominator makes a bit more sense if you consider the definition of “rationalize.” Recall that the numbers 5, 12 12, and 0.75 0.75 are all known as rational numbers because they can each be expressed as a ratio of two integers ( 51 51, 12 12, and 34 34 respectively). Some radicals are irrational numbers because they cannot be represented as a ratio of two integers. As a result, the point of rationalizing a denominator is to change the expression so that the denominator becomes a rational number. Here are some examples of irrational and rational denominators. \| \| \| \| --- \| Irrational \| \| Rational \| \| 1√2 12√ \| = \| √22 2√2 \| \| 2+√3√3 2+3√3√ \| = \| 2√3+33 23√+33 \| Now let’s examine how to get from irrational to rational denominators. Rationalizing Denominators with One Term Let’s start with the fraction 1√2 12√. Its denominator is √2 2–√, an irrational number. Notice that multiplying the fraction by 1 does not change the value and so equivalently, multiplying the fraction by √2√2 2√2√ also does not change it's value. Allow us to now see what happens. 1√2⋅1=1√2⋅√2√2=√2√2⋅2=√2√4=√22 12√⋅1=12√⋅2√2√=2√2⋅2√=2√4√=2√2 The denominator of the new fraction is no longer a radical (notice, however, that the numerator is). So why choose to multiply 1√2 12√ by √2√2 2√2√? You knew that the square root of a number times itself will be a whole number. In algebraic terms, this idea is represented by √x⋅√x=x x−−√⋅x−−√=x. Look back to the denominators in the multiplication of 1√2⋅1 12√⋅1. Do you see where √2⋅√2=√4=2 2–√⋅2–√=4–√=2? Here are some more examples. Notice how the value of the fraction is not changed at all; it is simply being multiplied by another name for 1. Example Rationalize the denominator. 2+√3√3 2+3√3√ Solution \| \| \| --- \| \| 2+√3√3 2+3√3√ \| The denominator of this fraction is √3 3–√. To make it into a rational number, multiply it by √3 3–√, since √3⋅√3=3 3–√⋅3–√=3. \| \| 2+√3√3⋅√3√3 2+3√3√⋅3√3√ \| Multiply the entire fraction by another name for 1, √3√3 3√3√. \| \| √3(2+√3)√3⋅√3 3√(2+3√)3√⋅3√ \| \| \| 2√3+√3⋅√3√9 23√+3√⋅3√9√ \| Use the Distributive Property to multiply √3(2+√3) 3–√(2+3–√). \| \| 2√3+√9√9 23√+9√9√ \| Simplify the radicals, where possible. √9=3 9–√=3. \| 2+√3√3=2√3+33 2+3√3√=23√+33 You can use the same method to rationalize denominators to simplify fractions with radicals that contain a variable. As long as you multiply the original expression by another name for 1, you can eliminate a radical in the denominator without changing the value of the expression itself. Example Rationalize the denominator. √x+√y√x x√+y√x√, where x≠0 x≠0 Solution \| \| \| --- \| \| √x+√y√x⋅√x√x x√+y√x√⋅x√x√ \| The denominator is √x x−−√, so the entire expression can be multiplied by √x√x x√x√ to get rid of the radical in the denominator. \| \| √x(√x+√y)√x⋅√x x√(x√+y√)x√⋅x√ \| \| \| √x⋅√x+√x⋅√y√x⋅√x x√⋅x√+x√⋅y√x√⋅x√ \| Use the Distributive Property. Simplify the radicals, where possible. Remember that √x⋅√x=x x−−√⋅x−−√=x. \| √x+√y√x=x+√xyx x√+y√x√=x+xy√x Example Rationalize the denominator and simplify. √100x11y, where y≠0 Solution \| \| \| --- \| \| √100x√11y \| Rewrite √ab as √a√b. \| \| √100x√11y⋅√11y√11y \| The denominator is √11y, so multiplying the entire expression by √11y√11y will rationalize the denominator. \| \| √100⋅11xy√11y⋅√11y \| Multiply and simplify the radicals, where possible. \| \| √100⋅√11xy√11y⋅√11y \| 100 is a perfect square. Remember that √100=10 and √x⋅√x=x. \| √100x11y=10√11xy11y Exercise Rationalize the denominator and simplify. 3√7 3√7√7 3 7√33 3√77 Answer : 1. Incorrect. After you rationalize a denominator, the denominator should not be in radical form. Try multiplying the fraction by √7√7. The correct answer is 3√77. 2. Incorrect. The process of rationalizing does not remove the denominator, it just changes the fraction so that the denominator is no longer a radical. Try multiplying the numerator and denominator by √7√7. The correct answer is 3√77. 3. Incorrect. You may have rationalized the fraction 7√3 instead of 3√7. Try multiplying the numerator and denominator by √7√7. The correct answer is 3√77. 4. Correct. To rationalize the denominator, multiply 3√7 by √7√7. This will create an integer in the denominator, so the fraction will have been rationalized. Rationalizing Denominators with Two Terms Denominators do not always contain just one term, as shown in the previous examples. Sometimes, you will see expressions like 3√2+3 where the denominator is composed of two terms, √2 and +3. Unfortunately, you cannot rationalize these denominators the same way you rationalize single-term denominators. If you multiply √2+3 by √2, you get 2+3√2. The original √2 is gone, but now the quantity 3√2 has appeared, but this is no better! In order to rationalize this denominator, you want to square the radical term and somehow prevent the integer term from being multiplied by a radical. Is this possible? It is possible, and you have already seen how to do it! Recall that when binomials of the form (a+b)(a−b) are multiplied, the product is a2−b2. So, for example, (x+3)(x−3)=x2−3x+3x−9=x2−9; notice that the terms −3x and +3x combine to 0. Now for the connection to rationalizing denominators: what if you replaced x with √2? Look at the examples below. Just as −3x+3x combines to 0 in the top example, −3√2+3√2 combines to 0 in the bottom example. (x+3)(x−3)=x2−3x+3x−9=x2−9 (√2+3)(√2−3)=(√2)2−3√2+3√2−9=(√2)2−9=2−9=−7 There you have it! Multiplying √2+3 by √2−3 removed one radical without adding another. In this example, √2−3 is known as a conjugate, and √2+3 and √2−3 are known as a conjugate pair. To find the conjugate of a binomial that includes radicals, change the sign of the second term to its opposite as shown in the table below. \| \| \| \| --- \| Term \| Conjugate \| Product \| \| √2+3 \| √2−3 \| (√2+3)(√2−3)=(√2)2−(3)2=2−9=−7 \| \| √x−5 \| √x+5 \| (√x−5)(√x+5)=(√x)2−(5)2=x−25 \| \| 8−2√x \| 8+2√x \| (8−2√x)(8+2√x)=(8)2−(2√x)2=64−4x \| \| 1+√xy \| 1−√xy \| (1+√xy)(1−√xy)=(1)2−(√xy)2=1−xy \| More generally, if a fraction has a denominator of the form a+b√c, then the conjugate is a−b√c and vice-versa. Example Rationalize the denominator and simplify. 5−√73+√5 Solution \| \| \| --- \| \| 5−√73+√5⋅3−√53−√5 \| Find the conjugate of 3+√5. Then multiply the entire expression by 3−√53−√5. \| \| (5−√7)(3−√5)(3+√5)(3−√5) \| \| \| 5⋅3−5√5−3√7+√7⋅√53⋅3−3√5+3√5−√5⋅√5 \| Use the Distributive Property to multiply the binomials in the numerator and denominator. \| \| 15−5√5−3√7+√359−3√5+3√5−√25 \| Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0. \| \| 15−5√5−3√7+√359−√2515−5√5−3√7+√359−5 \| Simplify radicals where possible. \| 5−√73+√5=15−5√5−3√7+√354 Example Rationalize the denominator and simplify. √x√x+2 Solution \| \| \| --- \| \| √x√x+2⋅√x−2√x−2 \| Find the conjugate of √x+2. Then multiply the numerator and denominator by √x−2√x−2. \| \| √x(√x−2)(√x+2)(√x−2) \| \| \| √x⋅√x−2√x√x⋅√x−2√x+2√x−2⋅2 \| Use the Distributive Property to multiply the binomials in the numerator and denominator. \| \| √x⋅√x−2√x√x⋅√x−2√x+2√x−4 \| Simplify. Remember that √x⋅√x=x. Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0. \| √x√x+2=x−2√xx−4 One word of caution: this method will work for binomials that include a square root, but not for binomials with roots greater than 2. This is because squaring a root that has an index greater than 2 does not remove the root, as shown below. (3√10+5)(3√10−5)=(3√10)2−53√10+53√10−25=(3√10)2−25=3√100−25 3√100 cannot be simplified any further, since its prime factors are 2⋅2⋅5⋅5. There are no cubed numbers to pull out! Multiplying 3√10+5 by its conjugate does not result in a radical-free expression. Exercise Identify the conjugate of the denominator. x+2√2x−7 √2x+7 √2x−7 2x+√7 (2x)2+7 Answer : 1. Correct. The conjugate will be the binomial that, when multiplied by the denominator, eliminates the radical. The conjugate is √2x+7. 2. Incorrect. The denominator is not the conjugate. Look for the binomial that, when multiplied by the denominator, eliminates the radical. The correct answer is √2x+7. 3. Incorrect. Multiplying the denominator by 2x+√7 will not remove the radical. Look for the binomial that follows the pattern (a+b)(a−b)=a2−b2. The correct answer is √2x+7. 4. Incorrect. Multiplying the denominator by (2x)2+7 will not remove the radical. Look for the binomial that follows the pattern (a+b)(a−b)=a2−b2. The correct answer is √2x+7. Rationalizing The Numerator The process of rationalizing the numerator is extremely similar to that of rationalizing the denominator. Exercise Page1.5.1 Rationalize the numerator: √4+h−2h You may assume that h≠0. Answer : √4+h−2h=√4+h−2h×√4+h+2√4+h+2=1√4+h+2 Exercise Page1.5.1 Rationalize the numerator: √t2+9−3t You may assume that t≠0. Answer : √t2+9−3t=√t2+9−3t×√t2+9+3√t2+9+3=t√t2+9+3 Page 1.4: Algebraic Expressions and Operations Page 1.6: Factoring Polynomials
190038	https://brainly.com/question/32829073	[FREE] Two congruent squares overlap, with vertex A of one square at the intersection of the diagonals of the - brainly.com 2 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 +23k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +31,2k Ace exams faster, with practice that adapts to you Practice Worksheets +7k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Two congruent squares overlap, with vertex A of one square at the intersection of the diagonals of the other square. Each square has a side length of 12 inches. Find the area, in square inches, enclosed by the shaded region. 1 See answer Explain with Learning Companion NEW Asked by jaideeplalli564 • 05/31/2023 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 9819664 people 9M 5.0 0 Upload your school material for a more relevant answer Thus, the number of square inches enclosed by the shaded region is 72√6 square inches. Given, two congruent squares overlap, as shown, so that vertex A of one square lies at the intersection of the diagonals of the other square. The side of each square has length 12 inches. To find: The number of square inches enclosed by the shaded region. Solution: It is given that, two squares are congruent and side of each square is 12 inches. Let's find the shaded area. By Pythagorean theorem, in ΔABO, we have: OB² = AO² + AB² We know that, side of square is 12 inches. So, AO = BO = 6√2 inches AB = 12 inches Therefore, OB² = (6√2)² + 12² OB² = 72 + 144 OB² = 216 OB = 6√6 inches Area of ΔABO = 1/2 × base × height= 1/2 × AB × OB= 1/2 × 12 × 6√6= 36√6 sq. inches Area of shaded region = 2 × Area of ΔABO= 2 × 36√6= 72√6 sq. inches to know more about Pythagorean theorem visit: brainly.com/question/14930619 SPJ11 Answered by BurtonLuke •39.5K answers•9.8M people helped Thanks 0 5.0 (1 vote) 1 Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 9819664 people 9M 0.0 0 Conceptual Physics - Benjamin Crowell Ancient Philosophy Reader - Noah Levin Light and Matter - Benjamin Crowell Upload your school material for a more relevant answer The area enclosed by the shaded region formed by the overlapping squares is 72 2 square inches. Each square has a side length of 12 inches, and the area calculation involves finding triangular areas formed by the overlap. Overall, the total shaded area is doubled for the two identical triangles formed due to symmetry. Explanation To find the area enclosed by the shaded region where two congruent squares overlap, we first need to understand the configuration of these squares. Each square has a side length of 12 inches. When squares are set up such that the vertex A of one square is located at the intersection of the diagonals of the other square, it creates overlapping triangular areas around that vertex. To solve this, we can visualize the setup where each square's diagonal intersects the other square's sides. The diagonal (d) of the square can be calculated using the formula: d=s 2=12 2inches where s is the side length of the square (12 inches). The area of a triangle formed in the overlapping shaded area can be found using the base and height. The height from vertex A to the base (one side of the triangle formed) is half the diagonal due to symmetry. Thus, h=2 d=2 12 2=6 2inches The area of one triangle (formed between the intersection and sides of the square) is then given by: Area△=2 1×base×height=2 1×12×6 2=36 2square inches However, this triangle is one of two identical triangles formed by the overlapping squares. Thus, we have to account for this overlap: Total Shaded Area=2×Area△ =2×36 2=72 2square inches In conclusion, the area enclosed by the shaded region is 72 2 square inches. Examples & Evidence For further practical understanding, you can visualize or even sketch two squares of equal size overlapping in the described manner to see how the areas intersect. This will help you appreciate how the triangular sections in the overlap contribute to the total shaded area. The calculations provided use the properties of triangles and squares, adhering to geometric principles, specifically involving the Pythagorean theorem for determining side lengths and areas. Thanks 0 0.0 (0 votes) Advertisement jaideeplalli564 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Two identical squares with sides of length 10cm overlap to form a shaded region as shown. A corner of one square lies at the intersection of the diagonals of the other square. Find the area of the shaded region in square centimetres. Community Answer 5.0 131 Which statements are true of all squares? Select three options. The diagonals are perpendicular. The diagonals are congruent to each other. The diagonals bisect the vertex angles. The diagonals are congruent to the sides of the square. The diagonals are twice the length of one side of the square. Community Answer $abcd$ is a square 4 inches on a side, and each of the inside squares is formed by joining the midpoints of the outer square's sides. what is the area of the shaded region in square inches? 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? New questions in Mathematics Write an expression equivalent to 2(4 y−5)−3 y. Write an expression equivalent to 0.5(−14 a−22). Write an expression equivalent to 3 1(24 p−33). If the endpoints of A B are A(−2,3) and B(1,8), which shows the correct way to determine the coordinates of point C, the midpoint of A B? A. M=(2−2+3,2 1+8) B. M=(2−2+1,2 3+8) C. M=(2−2−3,2 1−8) D. M=(2−2−1,2 3−8) Write an expression equivalent to 3(−8+2 x). 3(−8+2 x)=□+?x 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 Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
190039	https://www-users.cse.umn.edu/~olver/a_/bcvp.pdf	Boundary Conditions and Null Lagrangians in the Calculus of Variations and Elasticity Peter J. Olver School of Mathematics University of Minnesota Minneapolis, MN 55455 olver@umn.edu Abstract. We explicitly characterize boundary conditions that are compatible with low order variational principles. The freedom aﬀorded by adding in a null Lagrangian with-out altering the Euler–Lagrange equation signiﬁcantly expands the range of variationally admissible boundary conditions, although not all possibilities are permitted. Applications to several fundamental problems arising in elastostatics, including bars, beams, and plates, are presented. June 15, 2022 1 1. Introduction. The aim of this paper is to better understand the role of boundary conditions in the calculus of variations. Standard texts, including [12, 19, 20, 26], treat ﬁxed and natural boundary conditions, but have little to say about whether other types of boundary condi-tions, e.g., inhomogeneous Neumann boundary conditions, or Robin boundary conditions, can be incorporated into a variational framework. One reason is that, at least for ﬁrst or-der variational problems, the inhomogeneous ﬁxed boundary conditions and homogeneous natural boundary conditions are, in fact, the only possible uncoupled† boundary conditions that induce the vanishing of the boundary terms arising after integration by parts in the ﬁrst variation. However, this is not the end of the story. It is well known, [20, 32, 33, 35], that one can modify a variational problem with-out altering the corresponding Euler–Lagrange equations by adding a null Lagrangian to the integrand. As noted by Ericksen, , in the context of liquid crystal theory, this modiﬁcation (he calls null Lagrangians “nilpotent energies”) does, however, change the associated natural boundary conditions, and hence enables one to enlarge the range of boundary value problems that can be handled by variational techniques. This observation was further developed, in the context of continuum mechanics, by Edelen and Lagoudas, [13, 14], and by Lancia, Podio–Guidugli, and Vergara Caﬀarelli, [27, 37]. However, this idea has yet to be fully ﬂeshed out, even in the simple situations to be presented here. Boundary conditions that are compatible with a variational principle will be called varia-tionally admissible here; they are also termed self-adjoint or conservative, [3, 19]. I should also note that the classical text by Forsyth, , does treat boundary conditions for vari-ational problems in much greater detail than elsewhere but, as we will see, falls short in some of the analysis, and, as best as I can tell, does not consider the eﬀect of modifying the Lagrangian in this manner. The ﬁrst result is that essentially any scalar ﬁrst order variational problem can be modiﬁed by a suitable null Lagrangian in order to admit any (reasonable) uncoupled boundary conditions. On the other hand, coupled boundary conditions — also known as mixed boundary conditions, — are not always variationally admissible and we explicitly characterize those that are. Moreover, the uncoupled result is particular to this simplest case, and no longer holds for higher order scalar variational problems, variational problems involving several unknowns, or multivariate variational problems. We illustrate the various options in some of the more basic situations: ﬁrst and second order variational problems involving one and two independent variables and a single dependent variable, ﬁrst order problems involving one independent variable and two dependent variables, and ﬁrst order problems involving two or three independent and dependent variables, the latter of impor-tance in (hyper-)elastostatics, [3, 23, 28]. For second order variational problems in one independent and one dependent variable, the variationally admissible uncoupled boundary conditions are the ﬁxed conditions and generalizations of the simply supported, sliding, and † By “uncoupled” — “separated” in the terminology of Bliss, — we mean that each boundary condition only involves values of the unknowns at a single point on the boundary. Periodic, quasi-periodic, and more general coupled boundary conditions are thus excluded when making the ensuing assertion. 2 free end conditions arising in Euler–Bernoulli beam theory, [3, 40]. We further derive the natural boundary conditions for a general second order variational problem involving two independent variables and one dependent variable, and show how Timoshenko’s boundary conditions for linear plate mechanics, , are a particular case, provided one uses the physically relevant quadratic variational principle, which depends upon a particular choice of null Lagrangian. In a sense, these results can be viewed as a (partial) solution to an extended version of the inverse problem of the calculus of variations that includes boundary conditions. The basic inverse problem asks when is a given system of diﬀerential equations is the Euler–Lagrange equations of some variational principle, and is answered by the classical Helmholtz conditions, . Our concern is to characterize which boundary value problems can be assigned a variational structure as posed. We do not attempt a more general form of the inverse problem, which asks when a system is equivalent to a system coming from a variational principle; see the surveys [1, 38] for details (in the absence of boundary conditions). 2. Scalar First Order Variational Problems. We begin with a variational problem involving a scalar-valued function of a single variable. The basic minimization problem is to determine a suitable function u = f(x) that minimizes the objective functional J[u] = Z b a L(x, u, u′) dx (2.1) which we assume to be deﬁned on a compact interval [a, b] ⊂R. The integrand L(x, u, p), where p represents u′, is known as the Lagrangian for the variational problem. To avoid technicalities, we will impose the nondegeneracy condition ∂2L ∂p2 (x, u, p) ̸= 0 (2.2) although in this study, this will only be required at the endpoints a, b. All functions will be assumed to be suﬃciently smooth in order that one can perform the necessary manipulations. In most treatments of the subject, one imposes one of two possible uncoupled boundary conditions at each endpoint a, b of the interval. (The case of coupled boundary conditions will be discussed at the end of the section.) The ﬁrst is a ﬁxed boundary condition that speciﬁes the value of the minimizing function at an endpoint, e.g., u(a) = α. The second possibility is that of a free boundary, in which no conditions are imposed a priori on the minimizer at the endpoint in question. In the latter case, as we will see, the variational formulation leads to the imposition of a natural boundary condition there. The basic idea of the calculus of variations is to compare the proposed minimizer u(x) with a nearby function. We thus introduce a variation in the form Φ(ε, x) — classically written as δu — in which ε ∈R is assumed small, Φ(0, x) ≡0, and so that the function 3 u(x) + Φ(ε, x) satisﬁes the same boundary conditions as u(x) for all (suﬃciently small) ε. Thus, if u is to be a minimizer, the scalar function† h(ε) = J[u + Φ] = Z b a L(x, u + Φ, u′ + Φ′ ) dx (2.3) must have a minimum at ε = 0, and hence h′(0) = 0. Assuming suﬃcient smoothness of the integrand allows us to bring the derivative with respect to ε inside the integral and so, by the chain rule, h′(0) = d dε J[u + Φ] ε=0 = Z b a d dε L(x, u + Φ, u′ + Φ′ ) ε=0 dx = Z b a ϕ(x) ∂L ∂u (x, u, u′) + ϕ′(x) ∂L ∂p (x, u, u′) dx, (2.4) where ϕ(x) = d dε Φ(ε, x) ε=0 . (2.5) Integrating the second term in the ﬁnal integral by parts, we arrive at the basic variational formula 0 = h′(0) = ϕ(b) ∂L ∂p b, u(b), u′(b) −ϕ(a) ∂L ∂p a, u(a), u′(a) + Z b a ϕ(x) ∂L ∂u (x, u, u′) −Dx ∂L ∂p (x, u, u′) dx, (2.6) where the notation Dx refers to the total derivative, in which one diﬀerentiates with respect to x treating u as a function thereof, . We will refer to the ﬁrst two terms on the right hand side as the variational boundary terms. So far we have not explicitly speciﬁed the conditions to be imposed on our variation Φ other than setting Φ(0, x) ≡0. Suppose ﬁrst that, for each ﬁxed ε, the variation Φ(ε, x) has compact support in the open interval (a, b), which implies that it vanishes at the boundary: Φ(ε, a) = Φ(ε, b) = 0, which, by (2.5), implies the same for ϕ(a) = ϕ(b) = 0. The latter implies that the variational boundary terms in (2.6) vanish. The vanishing of the ﬁnal integral term is then governed by the Fundamental Lemma, whose proof can be found in any text on the subject. Lemma 2.1. If f(x) is continuous on [a, b], and Z b a f(x) ϕ(x) dx = 0 for every C∞ function ϕ(x) with compact support in (a, b), then f(x) ≡0 for all a ≤x ≤b. This implies that the minimizer u(x) must satisfy the usual Euler–Lagrange equation ∂L ∂u (x, u, u′) −Dx ∂L ∂p (x, u, u′) = 0, (2.7) † Primes on Φ mean derivatives with respect to x. 4 which, under the nondegeneracy condition (2.2), constitutes a second order ordinary dif-ferential equation that any suﬃciently smooth minimizer u(x) must satisfy. One word of caution: there do exist seemingly reasonable functionals whose minimizers are not, in fact, C2, and hence do not solve the Euler–Lagrange equation in the classical sense; see [7, 26, 35] for examples. Such pathologies are not considered here. Now let us look at the boundary terms in (2.6). If u satisﬁes a ﬁxed boundary condition at an endpoint, say u(a) = α, then in order that the variation continue to satisfy the same boundary condition, we must have ϕ(a) = 0, and hence the corresponding variational boundary term vanishes. On the other hand, if a is a free boundary, then there are no conditions initially imposed on either u(a) or ϕ(a). Thus, in order that the variational requirement (2.6) hold, the term multiplying ϕ(a) must vanish instead: ∂L ∂p a, u(a), u′(a) = 0. (2.8) This is known as a natural boundary condition, and imposes a constraint that any minimizer must satisfy at the free boundary x = a. An identical argument proves that at the other endpoint either u satisﬁes a ﬁxed boundary condition u(b) = β, or the corresponding natural boundary condition holds: ∂L ∂p b, u(b), u′(b) = 0. (2.9) We conclude that, in the case of uncoupled boundary conditions, any suﬃciently smooth minimizer for the variational problem (2.1) must satisfy a two-point boundary value problem for the associated second order Euler–Lagrange equation (2.7), where, at each endpoint one imposes either an (inhomogeneous) ﬁxed boundary condition or a (homo-geneous) natural boundary condition. As always, the above are just necessary conditions for local minimizers. Maximizers, if such exist, must also satisfy the same conditions, which serve to characterize the critical functions of the variational principle subject to the given boundary constraints. Determining whether a critical function is a (local) min-imizer, or maximizer, or neither, involves further criteria based on the second variation, [19, 20, 26, 35], that will not concern us here. It is also worth noting that minimizers (and maximizers) may not exist, need not be unique, and can be local or global. In continuum mechanics, under suitable assumptions, the stable equilibrium conﬁgurations are the (local and global) minimizers of the variational problem representing the stored energy in the system. Consequently, imposing any other type of boundary conditions at an endpoint, say x = a, will force the critical function to satisfy more than one boundary condition there — the natural boundary condition (2.8) plus the imposed boundary condition, thereby eﬀectively prescribing both u(a) and u′(a). In this case, the basic existence and uniqueness theorem for the initial value problem for a second order ordinary diﬀerential equation, , would imply that they are satisﬁed by one and only one solution to the Euler–Lagrange equation. This solution will probably not satisfy the additional boundary condition(s) at the other endpoint, in which case there would be no critical functions associated with the variational principle, and hence no candidate minimizers. 5 Example 2.2. Let us apply this method to the following problem: ﬁnd the shortest path between a point a = (a, α) and a vertical straight line ℓb = {x = b} in the xy plane. Assuming the solution is given by the graph of a function y = u(x), the length is given by the arc length functional J[u] = Z b a p 1 + u′(x)2 dx, (2.10) subject to the single ﬁxed boundary condition u(a) = α. The Lagrangian is L = p 1 + p2, and hence ∂L ∂u = 0, ∂L ∂p = p p 1 + p2 . (2.11) The Euler–Lagrange equation becomes 0 = −Dx u′(x) p 1 + u′(x)2 = − u′′(x) (1 + u′(x)2)3/2 , which simpliﬁes to u′′ = 0, whose solutions are straight lines. To determine the second, natural boundary condition at x = b, we apply (2.9), whence u′(b) p 1 + u′(b)2 = 0, or, simply, u′(b) = 0. This means that any critical function u(x) must have horizontal tangent at the point x = b, or, equivalently, it must be perpendicular to the vertical line ℓb. Suppose we try to impose a non-natural, non-ﬁxed boundary condition at the endpoint — for example, the Robin condition, , u′(b) = β u(b) + γ, (2.12) where β ̸= 0 and γ are constants. The solution to this boundary value problem for the Euler–Lagrange equation is easily calculated: u(x) = (αβ + γ)x + α(1 −β b) −γ a 1 −β (b −a) (2.13) provided the denominator does not vanish. However, unless αβ + γ = 0, so the graph of u is a horizontal line, this function does not provide a minimum to arc length functional among functions subject to the prescribed boundary conditions. Indeed, one can construct functions that satisfy the Robin boundary condition (2.12) whose arc length is arbitrarily close to that of the horizontal line, which is b −a. For example, we can slightly perturb the line by, say, setting uε(x) = α, a ≤x ≤b −ε, α + (αβ + γ) (x −b + ε)/(1 −ε β), b −ε ≤x ≤b, where 0 < ε < \| β \|, which does satisfy (2.12); see Figure 1. Its arc length b −a + ε s 1 + αβ + γ 1 −ε β 2 −1 ! − → b −a as ε →0. 6 a b −ε b α Figure 1. Perturbed Horizontal Line. One can even smooth oﬀits corner, which has the eﬀect of slightly decreasing its total arc length and thus does not aﬀect the convergence to b −a as ε →0. Thus, while the Robin boundary value problem has no minimizing solution, it does admit smooth functions that come arbitrarily close to the minimum possible value, which is b −a. A similar behavior is expected for other types of non-variationally admissible boundary conditions. Despite the perhaps disappointing conclusion of Example 2.2, it turns out that it is possible to impose other types of boundary conditions on a variational principle, by suitably modifying the functional. With this aim, we introduce an important notion of independent interest in the calculus of variations. Deﬁnition 2.3. A function N(x, u, p) is called a null Lagrangian if and only if its associated Euler–Lagrange expression (2.7) vanishes identically: E(N) ≡0. The key result is a special case of a general theorem characterizing higher order and multidimensional null Lagrangians; see [32; Theorem 4.7]. Proposition 2.4. A function N(x, u, p) deﬁned for all† (x, u, p) ∈R3 is a null Lagrangian if and only if it is a total derivative, so N(x, u, u′) = Dx[A(x, u)] = ∂A ∂x + u′ ∂A ∂u (2.14) for some function A that depends only on x, u. With Proposition 2.4 in hand, we can apply the Fundamental Theorem of Calculus to write the functional associated with a null Lagrangian in the following form: I[u] = Z b a N(x, u, u′) dx = Z b a Dx[A(x, u)] dx = A b, u(b) −A a, u(a) . (2.15) In other words, the value of the functional associated with a null Lagrangian depends only on the values of the function at the endpoints of the interval, and hence it is constant for any function u(x) that satisﬁes our usual ﬁxed boundary conditions u(a) = α, u(b) = β. † More generally, N can be deﬁned on a subset of R3 with trivial topology, [2, 32]. 7 The ﬂexibility aﬀorded by null Lagrangians allows us to expand the range of boundary conditions that can be handled by a variational analysis. Namely, we can modify the original variational problem by adding in a null Lagrangian, which does not alter the Euler–Lagrange equations but does change the associated natural boundary conditions. In other words, if N = DxA is a null Lagrangian, then the modiﬁed objective functional e J[u] = Z b a L(x, u, u′) + N(x, u, u′) dx = Z b a [ L(x, u, u′) + DxA(x, u) ] dx = A b, u(b) −A a, u(a) + Z b a L(x, u, u′) dx (2.16) has the same Euler–Lagrange equations as J[u] = Z b a L(x, u, u′) dx. Moreover, when sub-ject to ﬁxed boundary conditions, they have exactly the same critical functions, and hence the same minimizers, since their values diﬀer only by the initial two boundary terms in the ﬁnal expression, which depend only on the values of u at the endpoints a, b. On the other hand, as we will see, the two variational problems have diﬀerent natural boundary conditions, and this ﬂexibility allows us, at least in the present situation, to admit any uncoupled boundary conditions into the variational framework. Remark: The second expression for the modiﬁed objective functional (2.16) is, in the terminology of Bliss, , a version of the problem of Bolza, with separated end-conditions. To prove the preceding claim, and explain how to construct the required null Lag-rangian, let us assume that both boundary conditions explicitly involve the derivative of the function u(x) at the endpoint. We can already handle ﬁxed boundary conditions, and the remaining “mixed” case, in which one end is ﬁxed and the boundary condition at the other end involves the derivative, is left as an exercise for the reader. Under mild algebraic assumptions, we can solve the boundary conditions for the derivative: u′(a) = β1 u(a) , u′(b) = β2 u(b) , (2.17) for some functions β1, β2 depending only on the value of u at the endpoint in question. (They may, of course, depend on the respective endpoints a, b, but this is taken care of by allowing diﬀerent functions at each end.) Given a variational problem with Euler–Lagrange equation supplemented by the pre-scribed boundary conditions (2.17), let us add in a suitable null Lagrangian (2.14) in order that the natural boundary conditions associated with the modiﬁed Lagrangian e L(x, u, p) = L(x, u, p) + N(x, u, p) = L(x, u, p) + p ∂A ∂u (x, u) + ∂A ∂x (x, u) (2.18) are equivalent to the desired boundary conditions (2.17). Thus, at the right hand endpoint, in view of (2.9), this means 0 = ∂e L ∂p b, u(b), u′(b) = ∂L ∂p b, u(b), u′(b) + ∂A ∂u b, u(b) 8 is equivalent to the boundary equation u′(b) = β2 u(b) . For this to occur, we must require ∂A ∂u b, u(b) = ∂L ∂p b, u(b), β2 u(b) . (2.19) Similarly, at the left hand endpoint, we require ∂A ∂u a, u(a) = ∂L ∂p a, u(a), β1 u(a) . (2.20) These two conditions suﬃce to prescribe (2.17) as the natural boundary conditions for the variational problem associated with the modiﬁed Lagrangian (2.18), thus justifying our claim that by a suitable choice of A(x, u), or, equivalently, by adding in a suitable null Lagrangian N = dA/dx we can arrange for any boundary conditions of the above form to be the natural boundary conditions associated with the variational problem. We can combine the requirements (2.19–20) into a simpler form as follows. Choose an “interpolating function” B(x, u) such that B a, u(a) = β1 u(a) , B b, u(b) = β2 u(b) . (2.21) For example, we can use linear interpolation and set B(x, u) = x −a b −a β2(u) −x −b b −a β1(u). (2.22) Then (2.19–20) are implied by the interpolated equation ∂A ∂u (x, u) = −∂L ∂p x, u, B(x, u) , and thus A(x, u) = − Z ∂L ∂p x, u, B(x, u) du (2.23) is any anti-derivative of the integrand. We have thus proved a general result about varia-tional problems with speciﬁed boundary conditions. Theorem 2.5. Let J[u] = Z b a L(x, u, u′) dx be a variational problem whose mini-mizers are subject to the boundary conditions u′(a) = β a, u(a) , u′(b) = β b, u(b) , (2.24) for some function β(x, u). Let A(x, u) be deﬁned by (2.23). Then the modiﬁed variational problem e J[u] = Z b a [ L(x, u, u′) + DxA(x, u) ] dx = A b, u(b) −A a, u(a) + Z b a L(x, u, u′) dx (2.25) has the same Euler–Lagrange equations as J[u], and natural boundary conditions (2.24). Observe that the modiﬁed variational problem (2.25) diﬀers from the original only through the addition of certain “corrections” that depend only on the boundary values of u. The point is that any solution to the resulting boundary value problem will be a candidate minimizer for the modiﬁed variational problem. 9 Example 2.6. Let us consider the problem of minimizing arc length (2.10) subject to the Robin boundary conditions u′(a) = β1 u(a) + γ1, u′(b) = β2 u(b) + γ2, (2.26) in which β1, β2, γ1, γ2 are prescribed constants. We assume β1 β2 ̸= 0, leaving the cases of inhomogeneous Neumann conditions to the reader. In accordance with (2.22), set B(x, u) = β(x) u + γ(x) where β(x) = x −a b −a β2 −x −b b −a β1, γ(x) = x −a b −a γ2 −x −b b −a γ1. Substituting the second formula in (2.11) into (2.23), the required function A is obtained by integration: A(x, u) = − Z β(x) u + γ(x) p 1 + [β(x) u + γ(x)]2 du = − p 1 + [β(x) u + γ(x)]2 β(x) , provided β(x) ̸= 0. The modiﬁed variational problem (2.25) can thus be written in the form† e J[u] = − p 1 + [β2 u(b) + γ2 ]2 β2 + p 1 + [β1 u(a) + γ1 ]2 β1 + Z b a p 1 + u′(x)2 dx = − p 1 + u′(b)2 β2 + p 1 + u′(a)2 β1 + Z b a p 1 + u′(x)2 dx. (2.27) Thus, while the basic arc length functional does not, in general, admit a minimizer that satisﬁes the Robin boundary conditions, the modiﬁed arc length (2.27), which has the same Euler–Lagrange equation, (usually) does. Let us solve the Robin boundary value problem for the Euler–Lagrange equation, which, as noted above, is merely u′′ = 0, the solutions of which are straight lines u = cx+d. Substituting into the Robin boundary conditions (2.26) produces c = β1 (ca + d) + γ1 = β2 (cb + d) + γ2. Thus, if (b −a)β1β2 + β2 −β1 ̸= 0, the problem admits a unique solution, while if the left hand side is zero, then there is either a one-parameter family of solutions that all give the same value to the modiﬁed variational problem (even though they have diﬀering arc lengths), or there is no solution, depending on the values of γ1, γ2. † We avoid writing out the more complicated integral expression (2.16) involving the associated null Lagrangian. 10 What about coupled boundary conditions, which relate the values of the minimizer and its derivatives at the endpoints? The simplest are the usual periodic boundary conditions u(a) = u(b), u′(a) = u′(b). (2.28) Any variation must satisfy the same periodic conditions: ϕ(a) = ϕ(b), ϕ′(a) = ϕ′(b), and hence the diﬀerence of the two variational boundary terms in (2.6) will vanish provided ∂L/∂p is also periodic in x, meaning ∂L ∂p (a, u, p) = ∂L ∂p (b, u, p). (2.29) More generally we could try to impose a pair of coupled boundary conditions of the form F1 u(a), u′(a), u(b), u′(b) = F2 u(a), u′(a), u(b), u′(b) = 0, where each Fi(u, p, v, q) depends on 4 arguments. The variation ϕ(x) will thus satisfy the linearization of each: ∂Fi ∂u ϕ(a) + ∂Fi ∂p ϕ′(a) + ∂Fi ∂v ϕ(b) + ∂Fi ∂q ϕ′(b) = 0. (2.30) Now, if there are no algebraic constraints relating ϕ(a) and ϕ(b), meaning they can achieve independent values, then (2.6) will imply that both natural boundary conditions must be satisﬁed, and hence, unless the boundary value problem is overdetermined, variational ad-missibility implies that the boundary conditions decouple into the usual natural conditions at each end. For this not to be the case, the coupled boundary condition must relate the values of the critical function at the endpoints, say F u(a), u(b) = 0, (2.31) and hence ∂F ∂u u(a), u(b) ϕ(a) + ∂F ∂v u(a), u(b) ϕ(b) = 0. The vanishing of the variational boundary terms in (2.6) thus requires G u(a), u′(a), u(b), u′(b) = 0, (2.32) where G(u, p, v, q) = ∂F ∂v (u, v) ∂L ∂p (a, u, p) + ∂F ∂u (u, v) ∂L ∂p (b, v, q). This provides the second “natural coupled boundary condition” complementing the “ﬁxed coupled condition” (2.31). Thus, variationally admissible coupled boundary conditions necessarily take the form (2.31, 32). For example, the quasiperiodic condition u(b) = α u(a), (2.33) 11 where α is a nonzero constant, requires ∂L ∂p b, u(b), u′(b) = 1 α ∂L ∂p a, u(a), u′(a) (2.34) as its variationally admissible quasiperiodic counterpart. In the case of the arc length functional (2.10), the second quasiperiodic condition (2.34) becomes u′(b) p 1 + u′(b)2 = u′(a) α p 1 + u′(a)2 . (2.35) 3. Second Order Scalar Variational Problems. We next consider a second order scalar variational problem, with objective functional J[u] = Z b a L(x, u, u′, u′′) dx, (3.1) prescribed by the Lagrangian L(x, u, p, q), where now p, q represent the ﬁrst and second order derivatives u′, u′′. In analogy with (2.2), we will impose the nondegeneracy condition ∂2L ∂q2 x, u, p, q ̸= 0. (3.2) We introduce the variation h(ε) = J[u + Φ] = Z b a L(x, u + Φ, u′ + Φ′, u′′ + Φ′′ ) dx, where Φ(0, x) ≡0 and we set ϕ(x) = ∂Φ ∂ε (0, x). As before, any suﬃciently smooth critical function u satisﬁes 0 = h′(0) = d dε J[u + Φ] ε=0 = Z b a d dε L(x, u + Φ, u′ + Φ′, u′′ + Φ′′ ) ε=0 dx = Z b a ϕ(x) ∂L ∂u (x, u, u′, u′′) + ϕ′(x) ∂L ∂p (x, u, u′, u′′) + ϕ′′(x) ∂L ∂q (x, u, u′, u′′) dx = B(b) −B(a) + Z b a ϕ(x) E(L) dx, (3.3) where the ﬁnal formula results from integrating the second and third terms in the preceding integral by parts. Here E(L) = ∂L ∂u (x, u, u′, u′′) −Dx ∂L ∂p (x, u, u′, u′′) + D2 x ∂L ∂q (x, u, u′, u′′) , (3.4) is the associated Euler–Lagrange expression, while the variational boundary terms are obtained by evaluating B(x) = g(x) ϕ(x) + h(x) ϕ′(x), (3.5) 12 where g(x) = ∂L ∂p (x, u, u′, u′′) −Dx ∂L ∂q (x, u, u′, u′′), h(x) = ∂L ∂q (x, u, u′, u′′), at the endpoints x = a, b. If Φ(ε, x) has compact support in (a, b) for each ε, then ϕ(a) = ϕ′(a) = ϕ(b) = ϕ′(b) = 0, and the boundary terms in (3.3) both vanish: B(a) = B(b) = 0. Thus, applying the Fundamental Lemma 2.1, we conclude that any critical function u must satisfy the fourth order Euler–Lagrange equation E(L) = 0. This diﬀerential equation will be supplemented by four boundary conditions — two at each endpoint in the uncoupled case. The key question is which pairs of boundary conditions are compatible with the variational structure aﬀorded by (3.1). The variation must maintain the boundary conditions, which serves to constrain Φ(ε, x) and/or some of its derivatives at each endpoint. We can classify the possible uncoupled cases as follows. Let us designate the order 1 ≤n ≤3 of a pair of bound-ary conditions at an endpoint, either a or b, to be the highest order derivative of u that explicitly occurs in them. We will allow nonlinear boundary conditions — indeed for non-quadratic variational problems the natural boundary conditions are inevitably nonlinear — but make a mild algebraic assumption that allows us to solve one of the two boundary conditions for the highest order derivative and then, substituting this expression into the second boundary condition, solve it for the highest order remaining derivative occurring therein. We will also substitute the resulting expression back into the ﬁrst boundary condi-tion so the derivative we subsequently solved for only occurs in the second condition. (This preliminary step will become clearer in the context of speciﬁc examples.) We will focus on the left hand endpoint x = a, noting that the same analysis holds, mutatis mutandis, at x = b. Furthermore, we will modify the variational problem by adding in a null Lagrangian so as to preserve the underlying Euler–Lagrange equation. The analogue of Proposition 2.4 characterizes every second order null Lagrangians as a total derivative, so N(x, u, p, q) = DxA(x, u, p) = ∂A ∂x + p ∂A ∂u + q ∂A ∂p (3.6) for some function A that depends only on x, u, p. Replacing L by L + N in the preceding computation, we arrive at the modiﬁed boundary function (3.5) with g(x) = ∂L ∂p −Dx ∂L ∂q + ∂A ∂u = −∂2L ∂q2 u′′′(x) −∂2L ∂p∂q u′′(x) −∂2L ∂u∂q u′(x) −∂2L ∂x∂q + ∂L ∂p + ∂A ∂u , h(x) = ∂L ∂q + ∂A ∂p , (3.7) where the partial derivatives of L and A are evaluated at x, u(x), u′(x), u′′(x). 13 First order boundary conditions: If the two boundary conditions at x = a only involve u(a) and u′(a), then, under our algebraic assumption, we can solve them for u(a) = α, u′(a) = β, (3.8) where α, β ∈R, and so we are dealing with the ﬁxed boundary conditions at x = a. The corresponding variation must therefore satisfy the homogeneous ﬁxed boundary conditions ϕ(a) = 0, ϕ′(a) = 0. (3.9) In view of (3.5), this immediately implies that the boundary term B(a) = 0. Hence, as in the ﬁrst order case, ﬁxed boundary conditions are variationally admissible and no null Lagrangian modiﬁcation is required. Second order boundary conditions: These come in two subclasses. First, one boundary condition could only involve u(a), and so be of ﬁxed type. Thus, under our algebraic assumptions, we can solve for u(a) = α, u′′(a) = β u′(a) , (3.10) where α is a constant and β(p) a scalar function. Alternatively, both boundary conditions involve derivatives of u and can be solved for u′(a) = α u(a) , u′′(a) = β u(a) , (3.11) where now both α(u), β(u) are scalar functions. In the ﬁrst subcase (3.10), the ﬁrst condition again implies ϕ(a) = 0. As for the second boundary condition, the value of ϕ′(a) is no longer constrained, and so, in order that B(a) = 0, we must impose the corresponding natural boundary condition h(a) = ∂L ∂q a, u(a), u′(a), u′′(a) + ∂A ∂p a, u(a), u′(a) = 0. (3.12) Arguing as we did for ﬁrst order boundary conditions, this will be equivalent to the second boundary condition in (3.10) provided we choose A(x, u, p) such that ∂A ∂p (a, α, p) = ∂L ∂q a, α, p, β(p) . (3.13) Thus, we can always ﬁnd a null Lagrangian that makes the boundary conditions (3.10) variationally admissible. We will refer to (3.10) as generalized simply supported boundary conditions, the justiﬁcation of this choice of terminology appearing below. On the other hand, the second subcase (3.11) is not variationally admissible. Substi-tuting u + Φ into the boundary conditions, diﬀerentiating with respect to ε and setting ε = 0, we ﬁnd that ϕ satisﬁes the corresponding linearized boundary conditions ϕ′(a) = α′u(a) ϕ(a), ϕ′′(a) = β′u(a) ϕ(a). Using the ﬁrst of these, the corresponding boundary term (3.5) can thus be written in the form B(a) = g(a) + h(a) α′u(a) ϕ(a), (3.14) 14 whose vanishing requires the vanishing of the expression in brackets. However, referring back to (3.7) and the nondegeneracy condition (3.2), there is a term in g(a) that depends on u′′′(a) with non-vanishing coeﬃcient. Thus, the vanishing of the bracket expression will impose a third boundary condition involving u′′′(a) and hence, except in exceptional circumstances, we do not have a variational formulation of the boundary conditions (3.11). Third order boundary conditions come in three subclasses. Generalized free boundary conditions: u′′(a) = α u(a), u′(a) , u′′′(a) = β u(a), u′(a) ; (3.15) Generalized sliding boundary conditions: u′(a) = α u(a) , u′′′(a) = β u(a), u′′(a) ; (3.16) and a ﬁnal case: u(a) = α, u′′′(a) = β u′(a), u′′(a) . (3.17) Here α, β are smooth functions of the indicated variables except in the last case when α is constant. As we will show, in the ﬁrst two cases, variational admissibility imposes certain compatibility conditions between the functions α, β, while the third case is never variationally admissible. In all cases, the variation ϕ(x) must satisfy the corresponding linearized boundary conditions, derived in the same manner as above. Starting with the free conditions (3.15), both ϕ(a) and ϕ′(a) are unconstrained, and hence both boundary terms (3.7) must vanish when the boundary conditions are satisﬁed. Since both u(a) and u′(a) are similarly unconstrained, this requires ∂A ∂u (a, u, p) = P(u, p), ∂A ∂p (a, u, p) = Q(u, p), (3.18) for u, p ∈R, where P(u, p) = β(u, p) ∂2L ∂q2 + α(u, p) ∂2L ∂p∂q + p ∂2L ∂u∂q + ∂2L ∂x∂q −∂L ∂p , Q(u, p) = ∂L ∂q , (3.19) and the derivatives of L are all evaluated at (x, u, p, q) = a, u, p, α(u, p) . We can solve (3.18), and hence construct a null Lagrangian such that the modiﬁed variational principle is compatible with the given boundary conditions, provided the functions (3.19) satisfy the integrability constraint ∂P ∂p = ∂Q ∂u , (3.20) which eﬀectively imposes a compatibility condition on the boundary functions α, β in order that the corresponding boundary conditions be variationally admissible. Turning to the sliding conditions (3.16), linearizing the ﬁrst boundary condition as above, we deduce that (3.14) must vanish at x = a, which requires that ∂A ∂u a, u, α(u) + α′(u) ∂A ∂p a, u, α(u) = β(u, q) ∂2L ∂q2 + q ∂2L ∂p∂q + α(u) ∂2L ∂u∂q + ∂2L ∂x∂q −∂L ∂p −α′(u) ∂L ∂q , 15 where the derivatives of L are evaluated at (x, u, p, q) = a, u, α(u), q . However, here the dependence of A on p is not helpful, and we can assume the null Lagrangian (3.6) is prescribed by A(x, u), reducing the previous equation to ∂A ∂u = β(u, q) ∂2L ∂q2 + q ∂2L ∂p∂q + α(u) ∂2L ∂u∂q + ∂2L ∂x∂q −∂L ∂p −α′(u) ∂L ∂q . (3.21) Since the left hand side does not depend on q, given the nondegeneracy condition (3.2), the right hand side serves to prescribe the dependence of β on q; once this is of the proper form, the null Lagrangian can be obtained by integrating the remaining q-independent terms on the right hand side of (3.21) with respect to u to construct A(x, u). As for the third possibility, (3.17), arguing as in the case (3.10), we must impose the corresponding natural boundary condition (3.12). But this would impose an additional boundary condition constraining u′′(a). We conclude that we are unable to ﬁnd a null Lagrangian that makes (3.12) a consequence of the given boundary conditions. Example 3.1. The linearized equilibrium equations for a one-dimensional elastic beam, [3, 40], ﬁrst developed by Jacob and Daniel Bernoulli and Leonhard Euler, are obtained by minimizing the stored energy functional J[u] = Z b a 1 2 c(x) u′′2 −f(x) u dx, (3.22) prescribed by the Lagrangian L(x, u, p, q) = 1 2 c(x)q2 −f(x) u. The function c(x) > 0 is assumed positive and measures the stiﬀness of the beam at the point x, while f(x) represent an external forcing. The equilibrium conﬁguration of the beam is characterized as a solution to the fourth order Euler–Lagrange equation d2 dx2 c(x) d2u dx2 = f(x). (3.23) There are four physically important pairs of homogeneous boundary conditions that can be imposed at each endpoint. The associated names are physically motivated, and were adapted to our preceding general formulation. a) Fixed (clamped) end: u(a) = u′(a) = 0, b) Simply supported end: u(a) = u′′(a) = 0, c) Sliding end: u′(a) = u′′′(a) = 0, d) Free end: u′′(a) = u′′′(a) = 0. A second pair of boundary conditions must be imposed at the other end x = b, and can be mixed or matched in any combination. Inhomogeneous boundary conditions, both linear and nonlinear, are also allowed and used to model applied displacements or applied forces at the ends, although not all of these are variational. On the other hand, we will see that any constant inhomogeneity for the above four types of boundary conditions, i.e., setting one or both of the indicated derivatives of u at the endpoint to be a ﬁxed constant, is always admissible. 16 Let us investigate the full range of possibilities in detail based on the preceding anal-ysis. First, inhomogeneous ﬁxed boundary conditions (3.8) are always admissible. Next suppose we have inhomogeneous simply supported conditions of the form (3.11). In view of (3.13), using A(x, p) = c(x) B(p) to form our null Lagrangian, where B is a scalar func-tion of the derivative variable with β(p) = B′(p) allows us to impose any inhomogeneous simply supported conditions. For sliding boundary conditions (3.16), requiring the right hand side of (3.21) to be independent of q implies that β(u, q) = α′(u) q + γ(u) for some scalar function γ. The remaining terms give the admissibility constraints α(u) = F ′(u), β(u, q) = F ′′(u) q + G′(u), A(x, u) = c′(x) F(u) + c(x) G(u), (3.24) where F, G are arbitrary scalar functions. Finally consider the generalized free conditions (3.10). In this case, the functions (3.19) are P(u, p) = c(a) β(u, p) + c′(a) α(u, p), Q(u, p) = c′(a) α(u, p), (3.25) and the integrability constraint (3.20) required for variational admissibility reduces to c(a) ∂β ∂p + c′(a) ∂α ∂p = c′(a) ∂α ∂u . (3.26) The potential physical relevance of these constraints is not immediately clear to the author. Example 3.2. Euler’s elastica, [3, 26], governs the equilibrium conﬁgurations of a thin planar elastic rod, i.e., a nonlinear elastic beam. In the unforced case, we assume that the elastica is given by the graph† of a function y = u(x), with stored energy functional is I[u] = Z 1 2 κ2 ds = Z (u′′)2 dx 2 1 + (u′)25/2 , with Lagrangian L = q2 2(1 + p2)5/2 , (3.27) where κ = u′′ 1 + (u′)23/2 , ds = p 1 + (u′)2 dx, (3.28) are, respectively, the curvature and arc length of the elastica at the point x, u(x) . As Euler discovered, the resulting complicated, highly nonlinear fourth order Euler-Lagrange equation can be compactly written as E(L) = d2κ ds2 + 1 2 κ3 = 0, (3.29) whose solutions κ(s) can thus be expressed in terms of elliptic functions, [26, 30], whose historical developments in fact commence with Euler’s solution to the elastica. A direct derivation of (3.29) from the curvature form of the Lagrangian can be found in [2, 22, 25]. † Not all equilibrium conﬁgurations are of this form. Extending the analysis to general para-metrized curves is not diﬃcult. 17 Let us apply the preceding developments to determine which boundary conditions are variationally admissible. We will just summarize the results leaving the details of the cal-culation for the reader to ﬁll in. As always, inhomogeneous ﬁxed boundary conditions (3.8) are variationally admissible, as are inhomogeneous simply supported conditions (3.10), the latter requiring inclusion of the null Lagrangian based on A(p) = Z β(p) dp (1 + p2)5/2 . (3.30) In the case of sliding boundary conditions (3.16), using (3.21), we ﬁnd β(u, q) = 5α(u)2q2 2 1 + α(u)2 + α′(u) q + 1 + α(u)25/2 A′(u), A = A(u). (3.31) As for the free boundary conditions (3.15), setting P(u, p) = (1 + p2) β(u, p) −5 2 p α(u, p)2 (1 + p2)7/2 , Q(u, p) = α(u, p) (1 + p2)5/2 , (3.32) the integrability constraint (3.20) is required for admissibility. 4. Variational Problems Involving Several Unknowns. Next we investigate the case of ﬁrst order variational problems involving one indepen-dent and two dependent variables. We ﬁnd, unlike the scalar ﬁrst order case, compatibility constraints are required even for uncoupled boundary conditions in order that they be variationally admissible. The extension of this analysis to ﬁrst order variational problems involving more dependent variables is straightforward. We thus consider a functional J[u, v] = Z b a L(x, u, v, u′, v′) dx, (4.1) prescribed by the Lagrangian L(x, u, v, p, q) involving two unknown functions u(x), v(x); here p, q represent u′, v′, respectively. We introduce variations u(x)+Φ(ε, x), v(x)+Ψ(ε, x), so that Φ(0, x) = Ψ(0, x) ≡0, and set ϕ(x) = ∂Φ ∂ε (0, x), ψ(x) = ∂Ψ ∂ε (0, x). Arguing as before, we compute the derivative of the scalar function h(ε) = J[u + Φ, v + Ψ] = Z b a L(x, u + Φ, v + Ψ, u′ + Φ′, v′ + Ψ′ ) dx and require 0 = h′(0) = Z b a d dε L(x, u + Φ, v + Ψ, u′ + Φ′, v′ + Ψ′ ) ε=0 dx = Z b a ϕ(x) ∂L ∂u + ϕ′(x) ∂L ∂p + ψ(x) ∂L ∂v + ψ′(x) ∂L ∂q dx = B(b) −B(a) + Z b a ϕ(x) Eu(L) + ψ(x) Ev(L) dx. (4.2) 18 Here Eu(L) = ∂L ∂u (x, u, v, u′, v′) −Dx ∂L ∂p (x, u, v, u′, v′) , Ev(L) = ∂L ∂v (x, u, v, u′, v′) −Dx ∂L ∂q (x, u, v, u′, v′) , (4.3) are the Euler–Lagrange expressions associated with each of the dependent variables, while the variational boundary terms are obtained by evaluating the following function at the endpoints x = a, b: B(x) = ϕ(x) ∂L ∂p (x, u, v, u′, v′) + ψ(x) ∂L ∂q (x, u, v, u′, v′). (4.4) Taking ϕ(x), ψ(x) to have compact support in (a, b) annihilates the boundary terms in (4.2). Thus, applying the Fundamental Lemma 2.1, we conclude that any critical functions u, v must satisfy the system of second order Euler–Lagrange equations Eu(L) = 0, Ev(L) = 0. (4.5) In order that the boundary terms vanish at an endpoint, say x = a, we can either impose an inhomogeneous ﬁxed condition u(a) = α or the corresponding homogeneous natural boundary condition ∂L ∂p a, u(a), v(a), u′(a), v′(a) = 0; the same goes for the other vari-able: either v(a) = β or ∂L ∂q a, u(a), v(a), u′(a), v′(a) = 0. Another option is a mixture, imposing one ﬁxed boundary condition and letting the second be determined by naturality. In geometric terms, the problem is to ﬁnd a solution curve u(x), v(x) whose endpoint, at x = a, lies on a prescribed plane curve, say H a, u(a), v(a) = 0. (4.6) We assume the curve is nonsingular, meaning that Ha(u, v) = H(a, u, v) has non-vanishing gradient, ∇Ha = ∂Ha ∂u , ∂Ha ∂v ̸= 0, on the locus (4.6), which, by the Implicit Function Theorem, implies that one can simplify by solving for one of the boundary values in terms of the other; for example v(a) = h a, u(a) . The variations that inﬁnitesimally preserve (4.6) satisfy the linearized boundary condition: ∂H ∂u a, u(a), v(a) ϕ(a) + ∂H ∂v a, u(a), v(a) ψ(a) = 0, which implies that they be tangent to the surface. Thus, the boundary variation (4.4) will vanish provided the natural boundary condition ∂H ∂v a, u(a), v(a) ∂L ∂p a, u(a), v(a), u′(a), v′(a) −∂H ∂u a, u(a), v(a) ∂L ∂q a, u(a), v(a), u′(a), v′(a) = 0 (4.7) 19 holds. The other endpoint x = b is treated similarly. In this situation, null Lagrangians have the form N(x, u, v, p, q) = DxA(x, u, v) = ∂A ∂x + p ∂A ∂u + q ∂A ∂v (4.8) for some function A that depends only on x, u, v. Replacing L by L + N in the preceding computation, we arrive at the modiﬁed boundary function B(x) = ∂L ∂p + ∂A ∂u ϕ(x) + ∂L ∂q + ∂A ∂v ψ(x). (4.9) Thus, given a set of general ﬁrst order boundary conditions u′(a) = α a, u(a), v(a) , v′(a) = β a, u(a), v(a) , (4.10) the boundary term B(a) = 0 for all allowable variations if and only if ∂A ∂u a, u(a), v(a) = P a, u(a), v(a) , ∂A ∂v a, u(a), v(a) = Q a, u(a), v(a) , (4.11) where P(x, u, v) = −∂L ∂p x, u, v, α(x, u, v), β(x, u, v) , Q(x, u, v) = −∂L ∂q x, u, v, α(x, u, v), β(x, u, v) . (4.12) Thus, we can construct a null Lagrangian that makes the boundary conditions (4.10) variationally admissible if and only if we can solve (4.11) for A(x, u, v), which requires the compatibility constraint ∂P ∂v = ∂Q ∂u (4.13) at x = a. We conclude that, in contrast to the case of a single dependent variable, not every set of uncoupled boundary conditions is variationally admissible for a ﬁrst order problem in several dependent variables. Boundary conditions satisfying the constraint (4.13) are also known as self-adjoint boundary conditions, . Finally, in the case of a single ﬁxed boundary condition (4.6), an easy calculation shows that, with the addition of the null Lagrangian (4.8), the right hand side of the associated natural boundary condition (4.7) becomes the Jacobian determinant between A and H: ∂A ∂u a, u(a), v(a) ∂H ∂v a, u(a), v(a) −∂A ∂v a, u(a), v(a) ∂H ∂u a, u(a), v(a) , and can thus, by suitable choice of A(x, u, v), be made equal to any function of u(a), v(a). We conclude that, as in the single dependent variable problem, there are no constraints on the inhomogeneous term in the single natural boundary condition. 20 5. Multidimensional First Order Variational Problems. Next, we turn to ﬁrst order variational problems that involve several independent variables. In this section, we restrict our attention to the case of two independent variables, which we call x, y, and a single dependent variable u; later sections will deal with more general situations. Thus, we consider a functional J[u] = Z Z D L(x, y, u, ux, uy) dx dy, (5.1) given by a double integral over a prescribed bounded open domain D ⊂R2, assumed to have smooth (or, possibly, piecewise smooth) boundary, denoted by ∂D. We adopt subscript notation to denote partial derivatives, so ux = ∂u ∂x , uy = ∂u ∂y , and use the vari-ables p, q to respectively represent the ﬁrst order partial derivatives of u. The Lagrangian L(x, y, u, p, q) is assumed to be a suﬃciently smooth function of its ﬁve arguments. Our goal is to ﬁnd the function(s) u = f(x, y) that minimize the value of J[u] when subject to a set of prescribed boundary conditions on ∂D. Let Φ(ε, x, y), deﬁned for ε ∈R, (x, y) ∈D, denote a variation in the function u, where Φ(0, x, y) ≡0 and we set ϕ(x, y) = ∂Φ ∂ε (0, x, y). We thus introduce the scalar function h(ε) ≡J[u + εv ] = Z Z D L(x, y, u + Φ, ux + Φx, uy + Φy) dx dy. If u is a local minimizer, then, subject to appropriate smoothness assumptions, it must satisfy 0 = h′(0) = Z Z D ϕ(x, y) ∂L ∂u + ϕx(x, y) ∂L ∂p + ϕy(x, y) ∂L ∂q dx dy, (5.2) where the derivatives of L are all evaluated at x, y, u, ux, uy. The next step is to remove the derivatives from ϕ through an integration by parts based on the divergence form of Green’s Theorem, : Z Z D (div v ) dx dy = I ∂D v · n ds, (5.3) in which v is a C1 vector ﬁeld, n is the unit outward normal to the boundary ∂D of the domain, oriented in the counterclockwise direction, and ds is the element of arc length thereon. We set v = ϕ V, where V = ∂L ∂p , ∂L ∂q . (5.4) We conclude that h′(0) = I ∂D ϕ (V · n ) ds + Z Z D ϕ(x, y) E(L)dx dy, (5.5) 21 where E(L) = ∂L ∂u −Dx ∂L ∂p −Dy ∂L ∂q (5.6) is the Euler–Lagrange expression associated with the Lagrangian L, with Dx, Dy denoting the total derivatives with respect to the independent variables. Taking ϕ to be of compact support in D and using a two-dimensional version of the Fundamental Lemma 2.1, we con-clude that any suﬃciently smooth minimizer, or, more generally, critical function, u(x, y) must satisfy the Euler–Lagrange equation E(L) = 0. Assuming that not all the second order partial derivatives of L with respect to the variables p, q vanish, the Euler–Lagrange equation is a second order partial diﬀerential equation. Suppose ﬁrst that we have imposed a ﬁxed boundary condition u(x, y) = g(x, y) for (x, y) ∈S ⊆∂D on a subset† S of the boundary. This implies ϕ(x, y) = 0 for (x, y) ∈S, and hence that part of the boundary integral in (5.5) will vanish. This leaves an integral over the remainder T = ∂D \ S, the free boundary where the value of the minimizer u is unspeciﬁed. Since ϕ(x, y) is no longer constrained on T, a version of the Fundamental Lemma 2.1 applies to the boundary integral, and implies that V · n = 0 on T, which plays the role of the natural boundary condition. We conclude that either u(x, y) = g(x, y) or V · n = ∂L ∂p , ∂L ∂q · n = 0 at each point (x, y) ∈∂D. As before, we seek to extend the range of variationally admissible boundary conditions by including a null Lagrangian. In the present situation, ﬁrst order null Lagrangians have the form of a divergence, : N(x, y, u, p, q) = DxA(x, y, u) + DyB(x, y, u) = ∂A ∂x + p ∂A ∂u + ∂B ∂y + q ∂B ∂u , (5.7) where w = (A, B) is a general vector ﬁeld that depends on x, y and the dependent variable u, but not on derivatives of u. Applying Green’s formula (5.3), the integral of any such null Lagrangian depends only on its boundary values: Z Z D N(x, y, u, ux, uy) dx dy = I ∂D w · n ds. (5.8) Adding the null Lagrangian to the original L does not alter the Euler–Lagrange equation, but does change the natural boundary conditions to (V + w) · n = 0. (5.9) † For simplicity we assume S ⊂∂D is either empty, or all of ∂D, or a union of one or more disjoint curves contained in ∂D, including their endpoints. 22 The additional term w·n can be arranged, through a suitable choice of the functions A, B, to equal minus any desired function β(x, y, u) of a given boundary point (x, y) ∈T ⊂∂D and the value of the function u(x, y) there, i.e., we can ﬁx w · n = −β(x, y, u) for (x, y) ∈T ⊂∂D. (5.10) The corresponding null Lagrangian (5.7) can be the divergence of any vector ﬁeld w = A(x, y, u), B(x, y, u) whose normal component on T equals −β = w · n. (The explicit formula for w will depend upon the shape of the boundary.) The modiﬁed natural bound-ary conditions take the form V · n = β(x, y, u), (x, y) ∈T ⊂∂D. (5.11) these constitute the most general variationally admissible “free boundary conditions” for such a variational problem. In particular, we cannot impose any constraints on the tan-gential components of V on the boundary. Example 5.1. Consider the Dirichlet minimization problem J[u] = Z Z D 1 2 u2 x + u2 y −f(x, y) u dx dy = Z Z D 1 2 ∥∇u ∥2 −f(x, y) u dx dy, (5.12) with Lagrangian L = 1 2(p2 + q2) −f(x, y) u. The Euler–Lagrange equation (5.6) is the two-dimensional Poisson equation −∆u = −uxx −uyy = f. (5.13) Referring back to (5.4), we ﬁnd V = (ux, uy) = ∇u, and hence the natural boundary condition V · n = 0 is simply the homogeneous Neumann condition ∂u/∂n = 0. Addition of a null Lagrangian will replace this by the variationally admissible Robin-type boundary condition ∂u ∂n = β(x, y, u), (5.14) where β is any function depending on the boundary point and the value of u thereon, cf. (5.11). However, we cannot impose a variationally admissible natural boundary con-dition involving the tangential derivative of u at the boundary. (That being said, for the Dirichlet condition u = g on ∂D, the tangential derivative of u coincides with the tangential derivative of g, and so is constrained by the condition ∂u/∂t = ∂g/∂t.) Example 5.2. Minimal surfaces, [21, 29], are characterized as (local) minimizers of the surface area functional J[u] = Z Z D s 1 + ∂u ∂x 2 + ∂u ∂y 2 dx dy = Z Z D p 1 + ∥∇u ∥2 dx dy (5.15) 23 with Lagrangian L = p 1 + p2 + q2. Here we are assuming, for simplicity, that our minimal surface is non-parametric, meaning that it can be described as the graph of a function z = u(x, y) parametrized by (x, y) ∈D. (In diﬀerential geometric terms, this amounts to working on a Monge patch.) Minimal surfaces model soap ﬁlms in equilibrium, in which the surface tension serves to minimize the area, and thus forms the simplest elastic model for a two-dimensional liquid, . Note that ∂L ∂u = 0, ∂L ∂p = p p 1 + p2 + q2 , ∂L ∂q = q p 1 + p2 + q2 , and hence the Euler–Lagrange equation (5.6) becomes −∂ ∂x ux p 1 + u2 x + u2 y −∂ ∂y uy p 1 + u2 x + u2 y = −(1 + u2 y)uxx + 2uxuyuxy −(1 + u2 x)uyy (1 + u2 x + u2 y)3/2 = 0. Setting the numerator of the left hand side to 0 produces the justly famous minimal surface equation (1 + u2 y) uxx −2ux uy uxy + (1 + u2 x) uyy = 0. (5.16) On a free boundary, the minimal surface must satisfy the natural boundary conditions that the normal component V · n of the rescaled gradient vector ﬁeld V = ux p 1 + u2 x + u2 y , uy p 1 + u2 x + u2 y ! = ∇u p 1 + ∥∇u ∥2 vanishes, which is just the homogeneous Neumann boundary condition ∂u/∂n = 0. However, an inhomogeneous Neumann boundary condition is not variationally admis-sible for the surface area functional, and so a solution to the minimal surface equation that is subject to such a boundary condition cannot be characterized variationally. Indeed, according to (5.9), any variationally admissible boundary condition must, at each point on ∂D, be either a ﬁxed boundary condition u = α or of the form 0 = (V + w) · n = ∂u/∂n p 1 + ∥∇u ∥2 −β(x, y, u), where β = w · n is the normal component of w = (A, B), or, equivalently, ∂u ∂n = β(x, y, u) p 1 + ∥∇u ∥2 , (x, y) ∈∂D. (5.17) As usual, we can impose ﬁxed boundary conditions on part of the boundary and the modiﬁed natural boundary conditions on the remainder, which, when β ̸= 0, requires modiﬁcation of the surface area functional by addition of a suitable null Lagrangian. The boundary condition (5.17) appears in the theory of equilibrium conﬁgurations of capillary surfaces, [17; (7.3)], in which β is, typically, independent of u and represents the cosine of the contact angle between the ﬂuid ﬁlm and a bounding surface. Dependence of the contact angle on u would allow for patterning of the surface. 24 6. Two-Dimensional Elasticity. Let us now consider the case of a ﬁrst order variational problem involving two inde-pendent variables, x, y, and two dependent variables, u, v. Thus, we consider a functional J[u, v] = Z Z D L(x, u, ∇u) dx dy = Z Z D L(x, y, u, v, ux, uy, vx, vy) dx dy. (6.1) where D ⊂R2 is as before. The Lagrangian will be written in vectorial form L(x, u, P), where x = x y , u = u v , P = ∇u = ux uy vx vy . Of particular importance are the variational principles arising in two-dimensional hypere-lastostatics, [3, 23, 28], in which x represents the reference coordinates in a planar elastic body D ⊂R2, while u(x) represents the deformation of the body, or, in the linear approx-imation, a small displacement. The Jacobian matrix P = ∇u is known as the deformation gradient, and the Lagrangian L represents the stored energy of the deformed body (often denoted by W). Frame indiﬀerence implies that the stored energy is independent of u and has a rotational invariance, but these additional constraints will not play a role in our analysis of the boundary conditions. (See Ericksen, , for a discussion of the eﬀect of such invariances on the form of the null Lagrangians.) Applying variations ϕ(x) = ( ϕ(x, y), ψ(x, y) )T to both components of u, and per-forming the usual variational calculations leads to the ﬁrst variation integral I ∂D T ϕ · n ds + Z Z D ϕ · E(L) dx dy = I ∂D ϕ Tu + ψ Tv · n ds + Z Z D ϕ Eu(L) + ψ Ev(L) dx dy, (6.2) which must vanish for all allowable variations. Here T = (Tu, Tv) = ∂L ∂P T = ∂L/∂ux ∂L/∂vx ∂L/∂uy ∂L/∂vy (6.3) is known as the ﬁrst Piola–Kirchhoﬀstress tensor in elasticity, while the components of E(L) = ( Eu(L), Ev(L) )T are the Euler–Lagrange expressions; the corresponding second order Euler–Lagrange equations Eu(L) = ∂L ∂u −Dx ∂L ∂ux −Dy ∂L ∂uy ! = 0, Ev(L) = ∂L ∂v −Dx ∂L ∂vx −Dy ∂L ∂vy ! = 0, (6.4) are a consequence of the vanishing of the double integral in (6.2) under variations with compact support. 25 The boundary integral in (6.2) vanishes provided either the variations vanish on ∂D, or the corresponding boundary terms vanish. Thus, the variationally admissible boundary conditions include the ﬁxed (Dirichlet) conditions u(x, y) = f(x, y), v(x, y) = g(x, y), (x, y) ∈∂D, (6.5) or, alternatively, the vanishing of the boundary tractions Tu · n = Tv · n = 0, (x, y) ∈∂D. (6.6) There is also a hybrid class of boundary conditions, similar to that discussed at the end of Section 4, in which one imposes just one ﬁxed condition, say H x, y, u(x, y), v(x, y) = 0 for (x, y) ∈∂D, (6.7) and the second is the appropriate linear combination of homogeneous traction conditions (6.6) that cause the variational boundary integral to vanish, namely, ∂H ∂v Tu −∂H ∂u Tv = 0, (x, y) ∈∂D. (6.8) Geometrically, (6.7) requires that each point on the boundary of the deformed body is constrained to lie on a prescribed curve, which is allowed to vary from point to point. (A more physically realizable version of such conditions arises in the equilibrium conﬁguration of the capillary surface of a ﬂuid in a container, .) Moreover, one can decompose the boundary into subsets with diﬀerent boundary conditions (ﬁxed, traction, mixed) imposed on the sub-boundaries. As noted in , other types of boundary conditions that arise in applications, e.g., the dead load boundary conditions in which the boundary tractions are equal to a ﬁxed function of the reference coordinates, [3, 28], require the introduction of a suitable null Lagrangian into the variational functional. The most general ﬁrst order null Lagrangian is a linear combination of the basic null Lagrangians depending only on ∇u, which are all the subdeterminants thereof, [5, 15, 33]: N(x, u, ∇u) = λ + α · ∇u + β · ∇v + µ(uxvy −uyvx). (6.9) The coeﬃcients λ, µ are scalars, while α, β are vectors, all depending on (x, u). The condition that N be a null Lagrangian imposes the following constraints: ∂λ ∂u = ∇· α, ∂λ ∂v = ∇· β, ∂β ∂u −∂α ∂v = ∇⊥µ, (6.10) which are readily found by computing its Euler–Lagrange expressions. Here ∇= ∂/∂x ∂/∂y , ∇⊥= ∂/∂y −∂/∂x , (6.11) are, respectively, the gradient and skew-gradient with respect to the reference coordinates. For simplicity let us parametrize the boundary curve by arc length s, and write ∂D = { (x(s), y(s) \| 0 ≤s ≤ℓ} , (6.12) 26 where ℓis the length of ∂D and closure implies x(0) = x(ℓ), y(0) = y(ℓ). The unit tangent and unit (outwards) normal vectors are t = (xs, ys), n = (ys, −xs), x2 s + y2 s = 1. (6.13) The “null Piola–Kirchhoﬀstresses” associated with (6.9) are Su = ∂N/∂ux ∂N/∂uy = α + µ ∇⊥v, Sv = ∂N/∂vx ∂N/∂vy = β −µ ∇⊥u. (6.14) On ∂D, their normal components are the “null tractions” Su · n = α · n + µ ∇v · t = α · n + µ ∂v ∂s , Sv · n = β · n −µ ∇u · t = β · n −µ ∂u ∂s , where the second terms in each expression are multiples of the tangential (arc length) derivatives of the indicated deformations. Thus, the most general variationally admissible traction boundary conditions have the form Tu · n = α · n + µ ∇v · t, Tv · n = β · n −µ ∇u · t, (6.15) where the coeﬃcients α, β, µ satisfy the constraints (6.10). In particular, if they depend only on the reference coordinates x, y, then the third constraint implies that µ is constant, and we deduce the generalized dead load boundary conditions Tu · n = f(x, y) + µ ∇v · t, Tv · n = g(x, y) −µ ∇u · t, (6.16) where f = α · n, g = β · n, are arbitrary. The variational problem must be modiﬁed by adding in the appropriate null Lagrangian (6.9) to the stored energy. On the other hand, if the coeﬃcients in (6.15) only depend on the deformation u = (u, v), then µ(u, v) is arbitrary, but the variationally admissible “load vectors” α(u, v), β(u, v) are constrained by ∂β ∂u = ∂α ∂v . The hybrid case, in which only one ﬁxed boundary condition is imposed, is left to the reader to complete. 7. Three-Dimensional Elasticity. The fully three-dimensional case, which includes three-dimensional hyperelastostatics, [3, 23, 28], is handled similarly albeit with additional computational challenges, primarily due to the larger number of null Lagrangians. We begin with a functional J[u, v] = Z Z Z D L(x, u, ∇u) dx dy dz, (7.1) over a solid domain D ⊂R3 with smooth boundary surface ∂D, oriented using the unit out-wards normal n. The Lagrangian (stored energy) L depends on the reference coordinates x = ( x, y, z )T , the deformation (displacement) u(x) = ( u(x, y, z), v(x, y, z), w(x, y, z) )T , 27 and the deformation gradient P = ∇u, which is the 3 × 3 Jacobian matrix whose entries are the ﬁrst order partial derivatives of the dependent variables u, v, w with respect to the independent variables x, y, z. As before, frame indiﬀerence, material symmetries, and constitutive laws impose a variety of constraints on the Lagrangian, but these do not aﬀect our presentation. Applying variations ϕ(x) = ( ϕ(x, y, z), ψ(x, y, z), χ(x, y, z) )T to the three compo-nents of u, and performing the usual variational calculations leads to the ﬁrst variation integral I ∂D T ϕ · n ds + Z Z Z D ϕ · E(L) dx dy dz, (7.2) where T = (Tu, Tv, Tw) = ∂L ∂P T (7.3) is the ﬁrst Piola–Kirchhoﬀstress tensor, while E(L) = ( Eu(L), Ev(L), Ew(L) )T = 0 (7.4) is the usual system of second order Euler–Lagrange equations. The boundary integral vanishes provided either each variation vanishes on ∂D, or the corresponding traction vanishes. Thus, the variationally admissible boundary conditions include the ﬁxed (Dirichlet) conditions u(x) = f(x), x ∈∂D, (7.5) or, alternatively, the vanishing of the boundary surface tractions: T · n = ( Tu · n, Tv · n, Tw · n )T = 0, x ∈∂D. (7.6) One can also impose hybrid boundary conditions with one or two ﬁxed conditions, and two or one complementary homogeneous traction conditions, say when the boundary de-formations are restricted to prescribed curves or surfaces. As before — see also — the basic null Lagrangians depending only on ∇u are all the subdeterminants (of sizes 1 × 1, 2 × 2 and 3 × 3) of the deformation gradient†, and the general ﬁrst order null Lagrangian can be written using the vector triple product: N(x, u, ∇u) = λ + α · ∇u + β · ∇v + γ · ∇w + ρ · ∇v × ∇w + σ · ∇w × ∇u + τ · ∇u × ∇v + µ ∇u · ∇v × ∇w. (7.7) The coeﬃcients λ, µ are scalars, while α, β, γ, ρ, σ, τ are vectors, all depending on (x, u), and satisfying the following constraints: ∂λ ∂u = ∇· α, ∂λ ∂v = ∇· β, ∂λ ∂w = ∇· γ, ∂β ∂w −∂γ ∂v = ∇× ρ, ∂γ ∂u −∂α ∂w = ∇× σ, ∂α ∂v −∂β ∂u = ∇× τ, ∂ρ ∂u = ∂σ ∂v = ∂τ ∂w = ∇µ. (7.8) † See for the generalization of this result to higher order null Lagrangians. 28 The “null Piola–Kirchhoﬀstresses” Su = α + σ × ∇w −τ × ∇v + µ ∇v × ∇w, Sv = β + τ × ∇u −ρ × ∇w + µ ∇w × ∇u, Sw = γ + ρ × ∇v −σ × ∇u + µ ∇u × ∇v, (7.9) are thus found by computing the derivatives of the null Lagrangian with respect to the deformation gradient variables. The corresponding “null tractions” on the boundary are obtained by taking their dot products with the unit normal n, and these can be appended to the homogeneous traction boundary conditions without aﬀecting their variational ad-missibility. The ﬁrst terms, when α, β, γ depend only on x, produce dead load boundary conditions; the last terms are related to pressure loading on the boundary of the deformed body since the associated null Lagrangian det ∇u = ∇u·∇v×∇w measures the local change in volume. Other options appear to be less well known and would be worth investigating for their potential physical relevance. Remark: Each vector triple product arising in a null traction, say ∇u × ∇v · n, cor-responds geometrically to the volume of the tetrahedron with sides ∇u, ∇v, n or, equiv-alently, the area of the parallelogram whose sides are the orthogonal projections of the vectors ∇u, ∇v onto the tangent space of the boundary surface. 8. Multidimensional Second Order Variational Problems. While the natural boundary conditions for ﬁrst order variational problems involving several independent variables are reasonably straightforward, the same cannot be said of second and higher order problems. Forsyth, [18; Chapter 11] devotes his penultimate chapter to their study, but his presentation of the boundary conditions is incomplete, and fails to deal with the additional complications that we investigate here. On the other hand, Timoshenko, , derives the proper natural boundary conditions for the particular Lagrangian arising in linear plate mechanics, but does not extend his analysis to general variational problems. We will derive Timoshenko’s boundary conditions from our general formulas. In addition, the range of natural boundary conditions can be extended using null Lagrangians, although the details are computationally challenging and not particularly enlightening to the author. As in Section 5, for simplicity, we restrict our attention to the case of two independent variables and one dependent variable. As before, D ⊂R2 is a bounded open domain with smooth connected boundary ∂D, which we orient in the usual counterclockwise direction, although the latter assumptions can easily be weakened. We consider a functional of the form J[u] = Z Z D L(x, y, u, ux, uy, uxx, uxy, uyy) dx dy, (8.1) where the Lagrangian L depends on derivatives of order ≤2 of the scalar-valued function u(x, y). We perform the usual variational calculations, starting with the variation of the 29 Lagrangian h(ε) = J[u + Φ], with h′(0) = Z Z D ϕ ∂L ∂u + ϕx ∂L ∂ux + ϕy ∂L ∂uy + ϕxx ∂L ∂uxx + ϕxy ∂L ∂uxy + ϕyy ∂L ∂uyy ! dx dy. (8.2) We then integrate each term on the right hand side that involves a derivative of the variation ϕ by parts, leading to the Euler–Lagrange expression E(L) = ∂L ∂u −Dx ∂L ∂ux −Dy ∂L ∂uy ! + D2 x ∂L ∂uxx + DxDy ∂L ∂uxy ! + D2 y ∂L ∂uyy ! (8.3) multiplying ϕ plus a divergence, which, by Green’s formula (5.3), can be replaced by the corresponding boundary integral. Applying the Fundamental Lemma, we conclude that any suﬃciently smooth critical function must satisfy the fourth order Euler–Lagrange equation E(L) = 0. The divergence terms that result from integrating the second and third terms in (8.2) by parts are given in (5.5–6). For the fourth and sixth terms, the result is straightforward: ϕxx ∂L ∂uxx = ϕ D2 x ∂L ∂uxx + Dx ϕx ∂L ∂uxx −ϕ Dx ∂L ∂uxx , ϕyy ∂L ∂uyy = ϕ D2 y ∂L ∂uyy ! + Dy ϕy ∂L ∂uyy −ϕ Dy ∂L ∂uyy ! . (8.4) However, for the ﬁfth term, there is an ambiguity, as we can either start with the x derivative or with the y derivative, leading to the alternative expressions ϕxy ∂L ∂uxy = ϕ DxDy ∂L ∂uxy ! +            Dx ϕy ∂L ∂uxy ! + Dy −ϕ Dx ∂L ∂uxy ! , Dx −ϕ Dy ∂L ∂uxy ! + Dy ϕx ∂L ∂uxy ! . (8.5) The divergence terms produce ostensibly diﬀerent boundary integrals, and thus potentially diﬀerent natural boundary conditions! Forsyth, , takes the average of the two expres-sions in (8.5), but this is just his particular convention, and he stops his presentation before having to deal with the resolution of the problem. His ﬁnal formula, at the end of Section 337, does not properly encapsulate what the natural boundary conditions should be. Our goal is to ascertain what further analysis is needed. Observe that the diﬀerence between the two expressions in (8.5) is a null divergence, [2, 31, 33], namely a vector ﬁeld (A, B) that satisﬁes DxA + DyB ≡0, (8.6) which, assuming it is deﬁned on a topologically trivial domain, is valid if and only if the vector ﬁeld is a skew gradient: A = DyC, B = −DxC, (8.7) 30 for some function C; see [32; § 5.4]. The key point underlying the resolution of the diﬃculty is that, on the boundary, the intrinsic components of the gradient of ϕ have diﬀerent eﬀects, and there is a second integration by parts that can be employed. Let n, t denote the unit normal and unit tangent vectors at a point (x, y) ∈∂D. We can write ∇ϕ = ϕn n + ϕt t, (8.8) where the coeﬃcients ϕn, ϕt are its normal and tangential derivatives. For a general variation, the normal derivative can be speciﬁed independently of ϕ itself; however, its tangential derivative can be realized as the derivative of the restriction of ϕ to ∂D, and hence is not independent thereof. Moreover, because the boundary ∂D is a closed curve, a term involving ϕt can be integrated by parts without aﬀecting the value of the overall boundary integral. As in (6.12, 13), we parametrize the boundary curve ∂D by arc length, noting that we can identify the derivative in the tangential direction t with the arc length derivative: ∂/∂t = ∂/∂s. Substituting (6.13) into (8.8), we deduce that ϕx = ϕt xs + ϕn ys = ϕs xs + ϕn ys, ϕy = ϕt ys −ϕn xs = ϕs ys −ϕn xs, (8.9) and so ϕt = ϕs = d ds ϕ(x(s), y(s)) = ϕx xs + ϕy ys. (8.10) In this manner, referring back to (8.2), we deduce a formula of the form h′(0) = Z Z D div W + ϕ(x, y) E(L) dx dy = I ∂D (W · n ) ds + Z Z D ϕ(x, y) E(L)dx dy, where the boundary terms, resulting from the preceding integration by parts, are assembled using (5.4), (8.4), along with the ﬁrst expression in (8.5). (As the reader can check, the second expression leads to the same ﬁnal formula for the boundary integral, as does Forsyth’s average of the two.) Thus, W = (W1, W2), where W1 = ϕ ∂L ∂ux −Dx ∂L ∂uxx + ϕx ∂L ∂uxx + ϕy ∂L ∂uxy ! , W2 = ϕ ∂L ∂uy −Dx ∂L ∂uxy −Dy ∂L ∂uyy ! + ϕy ∂L ∂uyy ! . Using (6.13, 9, 10), we thus ﬁnd W · n = ϕ " ys ∂L ∂ux −Dx ∂L ∂uxx −xs ∂L ∂uy −Dx ∂L ∂uxy −Dy ∂L ∂uyy ! # + ϕs " xsys ∂L ∂uxx + y2 s ∂L ∂uxy −xsys ∂L ∂uyy # + ϕn " y2 s ∂L ∂uxx −xsys ∂L ∂uxy + x2 s ∂L ∂uyy # . 31 The term involving the tangential derivative ϕs = ϕt can be integrated by parts, using the formula I ∂D (f gs) ds = I ∂D (−fs g) ds, valid for any functions f, g on ∂D. (No boundary terms appear because ∂(∂D) = ∅.) Thus, we can rewrite I ∂D (W · n ) ds = I ∂D (F ϕ + G ϕn) ds, where F = ys ∂L ∂ux −Dx ∂L ∂uxx + xsyss + ysxss + x2 s ysDx + xsy2 s Dy ∂L ∂uyy −∂L ∂uxx ! −xs ∂L ∂uy −Dy ∂L ∂uyy ! + xsxss −ysyss + x3 s Dx −y3 s Dy ∂L ∂uxy = H · n + d ds " xsys ∂L ∂uyy −∂L ∂uxx ! + 1 2 (x2 s −y2 s) ∂L ∂uxy # , H = ∂L ∂ux −Dx ∂L ∂uxx −1 2 Dy ∂L ∂uxy , ∂L ∂uy −1 2 Dx ∂L ∂uxy −Dy ∂L ∂uyy !T , G = y2 s ∂L ∂uxx −xsys ∂L ∂uxy + x2 s ∂L ∂uyy . (8.11) where we have used the identities x2 s + y2 s = 1, xsxss + ysyss = 0, (8.12) to simplify and make the ﬁnal result evidently symmetric under a 90◦rotation of x and y, as it must be, even though the intervening calculation was not. Indeed, the ﬁnal formulas are invariant under completely general orientation-preserving changes of variables in x, y. Thus, at each point in ∂D, either ϕ = 0, which means that u must satisfy an inho-mogeneous Dirichlet boundary condition u = f, or the natural boundary condition F = 0 holds. Furthermore, either ϕn = 0, which means that u must satisfy an inhomogeneous Neumann boundary condition ∂u/∂n = g, or the natural boundary condition G = 0 holds. There are thus four possibilities, similar to what we found for a second order scalar varia-tional problem. The boundary curve ∂D could then be divided into up to four sub-curves (which need not individually be connected) on each of which one of these four possibilities is imposed. This produces the complete set of variationally admissible boundary condi-tions for the fourth order Euler–Lagrange equations before we admit the addition of a null Lagrangian, which we now investigate. The fact that the ﬁrst natural boundary condition turns out to be cubic in the components of the tangent (or the normal) to the curve and also involves its second derivatives (curvature) is striking. 32 Example 8.1. The equations of linear plate mechanics are based on the Lagrangian L = 1 2 u2 xx +ν uxxuyy + 1 2 u2 yy +(1−ν)u2 xy = 1 2 (uxx +uyy)2 +(ν −1)(uxxuyy −u2 xy), (8.13) where the constant ν denotes the Poisson ratio of the material, which is assumed to be uniform; see Timoshenko, , for details. The Euler–Lagrange equation is the biharmonic equation E(L) = uxxxx + 2uxxyy + uyyyy = ∆2u = 0, (8.14) where ∆= ∂2 x + ∂2 y is the Laplacian operator, and is independent of ν. This is because the ﬁnal term in the Lagrangian (8.13) is a multiple of the Hessian determinant H = ∇2u of the function u, which is a null Lagrangian: det H = det uxx uxy uxy uyy = uxxuyy −u2 xy = Dx(uxuyy) + Dy(−uxuxy). (8.15) Consequently the simpliﬁed Lagrangian e L = 1 2 (uxx + uyy)2 = 1 2 u2 xx + 2uxxuyy + 1 2 u2 yy (8.16) has exactly the same Euler–Lagrange expression: E(e L) = E(L) = ∆2u. However, the two Lagrangians possess quite diﬀerent natural boundary conditions, and only the former leads to physically meaningful expressions. Let us compute the natural boundary condition terms (8.11), ﬁrst for the physical plate Lagrangian (8.13). We ﬁnd F = −ys(uxxx + uxyy) + xs(uxxy + uyyy) + (ν −1) d ds xsys(uyy −uxx) + (x2 s −y2 s)uxy = −(∆u)n + (1 −ν) nT H t s, G = (ν x2 s + y2 s)uxx + 2(ν −1)xsysuxy + (x2 s + ν y2 s)uyy = ∆u + (ν −1)tT H t. (8.17) Identifying n = (ys, −xs) = (cos α, sin α), we recover the formulas in [42; eq. (110)]. Phys-ically, G represents the bending moment of the plate at the boundary point. The ﬁrst term in the second expression for F — the normal derivative of the Laplacian of the displace-ment — represents the shearing force the plate exerts on its boundary, while the second term — the arc length (tangential) derivative of the twisting moment — represents the reaction of the boundary to the twisting of the plate. On the other hand, the natural boundary conditions associated with the alternative Lagrangian (8.16) are much simpler: e F = −ys(uxxx + uxyy) + xs(uxxy + uyyy) = −(∆u)n, e G = uxx + uyy = ∆u. (8.18) However, they do not represent the proper physical forces exerted on the free boundary of an elastic plate, hence the choice of the more complicated physical Lagrangian (8.13). See [36, 39] for discussion of the role played by the null Lagrangian term in the resolution of the polygon-circle plate paradox, , in which the energy functionals of simply supported polygonal plates converge to that of a clamped circular plate. 33 Finally, as noted in (8.15), second order null Lagrangians no longer depend linearly on the second order derivatives of u. According to , every second order null Lagrangian has the form N = α + λuxx + µuxy + ν uyy + σ (uxxuyy −u2 xy) (8.19) where the coeﬃcients α, . . ., σ depend on x, y, u, ux, uy, but are not arbitrary functions. Indeed, as always, N must be a divergence: N = DxX + DyY where X = A −C uxy + Duyy, Y = B + C uxx −Duxy, (8.20) with A, B, C, D arbitrary functions of x, y, u, ux, uy, and hence α = ∂A ∂x + ux ∂A ∂u + ∂B ∂y + uy ∂B ∂u , σ = ∂C ∂uy + ∂D ∂ux , λ = ∂A ∂ux + ∂C ∂y + uy ∂C ∂u , µ = ∂A ∂uy + ∂B ∂ux −∂C ∂x −ux ∂C ∂u −∂D ∂y −uy ∂D ∂u , ν = ∂B ∂uy + ∂D ∂x + ux ∂D ∂u . (8.21) However, there is redundancy in these expressions, because one can add in a null divergence (8.6) to the vector ﬁeld (X, Y ) without aﬀecting the form of N. Indeed, if X = DyZ = ∂Z ∂y + uy ∂Z ∂u + uxy ∂Z ∂ux + uyy ∂Z ∂uy , Y = −DxZ = −∂Z ∂x −ux ∂Z ∂u −uxx ∂Z ∂ux −uxy ∂Z ∂uy , (8.22) then the corresponding N is identically zero. We can thus choose Z in order that either C or D in (8.20) is zero. As usual, adding the null Lagrangian (8.19) to the original L does not alter the Euler–Lagrange equation, but does aﬀect the natural boundary conditions. The resulting formulas can be found by replacing L by L + N in (8.11), but are quite complicated, and, at least to me, not especially enlightening. We will therefore leave this ﬁnal step to the motivated reader to try to ascertain to what extent the natural boundary conditions for a second order Lagrangian can be modiﬁed through this process. 9. Conclusions. The ﬂexibility aﬀorded by the addition of a null Lagrangian serves to enlarge the possible variationally admissible boundary conditions for the Euler–Lagrange equations associated with a variational problem. However, not every boundary condition is varia-tionally admissible; the classiﬁcation of those that are forms the subject of this paper. However, this analysis becomes increasingly complicated as the order of the Lagrangian increases, and we have therefore focussed on the simplest and most important cases for 34 applications. For variational problems arising in continuum mechanics, most of the varia-tionally admissible boundary conditions have an evident physical interpretation, and can be applied in a range of practical problems. Various further directions of research are indicated. It would be of much interest to extend this analysis to other types of elastic materials, including, for example, general rod, plate, and shell theories, and Cosserat (micropolar) continua, [3, 11, 16]. We restricted our attention to null Lagrangians that have the same order as the variational problem; higher order null Lagrangians could produce interesting higher order boundary conditions, as arising, for instance, in the modeling of plates supported on an elastic foundation, . The impact of these methods on the design of ﬁnite element and other numerical approximation schemes for such boundary value problems, [9, 41], is also worthy of de-velopment. Another interesting direction would be to extend this analysis to constrained variational problems, including incompressible and inextensible materials. Moreover, it would be worth investigating the eﬀect of null Lagrangians and the associated boundary conditions on the standard variational tests for minima and maxima (second variation, conjugate points, etc.), [19, 20, 26, 35]. Finally, we have exclusively studied the boundary value problems arising in statics; extending this analysis to initial-boundary value problems arising in dynamics would also be worth pursuing. Acknowledgments: Thanks to Aritra Lahiri for asking about the implementation of Robin boundary conditions in the calculus of variations, which served as the initial in-spiration to write this paper. Thanks also to Roger Fosdick, Eliot Fried, and Chi-Sing Man for references and insightful comments on an earlier version. I also thank the referees for their careful reading of the paper, corrections of misprints, additional references, and suggestions for improving the exposition. 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Olver, P.J., Introduction to Partial Diﬀerential Equations, Undergraduate Texts in Mathematicss, Springer, New York, 2014. Olver, P.J., The Calculus of Variations, Lecture Notes, University of Minnesota, 2021. Podio–Guidugli, P., On null-Lagrangian energy and plate paradoxes, in: Analysis of Shells, Plates, and Beams, H. Altenbach, N. Chinchaladze, R. Kienzler, and W.H. M¨ uller, eds., Springer Nature, Cham, Switzerland, 2020, pp. 367–372. Podio–Guidugli, P., and Caﬀarelli, G.V., Surface interaction potentials in elasticity, Arch. Rat. Mech. Anal. 109 (1990), 343–383. Saunders, D.J., Thirty years of the inverse problem in the calculus of variations, Rep. Math. Phys. 66 (2010), 43–53. Simmonds, J.G., A simple energetic explanation of the polygon-circle paradox for classical (Kirchhoﬀ) plate theory, J. Elasticity 99 (2010), 113–116. Strang, G., Introduction to Applied Mathematics, Wellesley Cambridge Press, Wellesley, Mass., 1986. 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190040	https://www.tutorela.com/math/question/solve-the-quadratic-equation-x2-2x80-step-by-step	Solve the Quadratic Equation: x²-2x+8=0 Step-by-Step \| Tutorela TUTORELA Learn math Learn Math by Age Practice Math Practice Math by Grade BlogAbout Features FAQTerms of ServicePrivacy Policy Login Sign up Solve the Quadratic Equation: x²-2x+8=0 Step-by-Step Home Learn math Solve the Quadratic Equation: x²-2x+8=0 Step-by-Step Solve the following equation: x 2−2 x+8=0 x^2-2x+8=0 x 2−2 x+8=0 A No solution B x 1=1,x 2=−2 x_1=1,x_2=-2 x 1=1,x 2=−2 C x=−8 x=-8 x=−8 D x 1=0,x 2=1 x_1=0,x_2=1 x 1=0,x 2=1 ❤️ Continue Your Math Journey! We have hundreds of course questions with personalized recommendations + Account 100% premium Create Free Account Step-by-step video solution Watch the teacher solve the problem with clear explanations 00:05 Let's find the value of X. 00:09 First, let's identify the coefficients. Pay attention to the numbers in front of the variables. 00:21 Next, we'll use the roots formula. This helps us find solutions to the equation. 00:50 Now, substitute the correct values based on the data we have, and let's solve the equation. 01:18 Let's calculate the square and the products step by step. 01:29 Remember! A root cannot be negative. Keep this in mind. 01:38 So, there's no solution to this question. Great try, and keep practicing! Step-by-step written solution Follow each step carefully to understand the complete solution 1 Understand the problem Solve the following equation: x 2−2 x+8=0 x^2-2x+8=0 x 2−2 x+8=0 2 Step-by-step solution To solve the quadratic equation x 2−2 x+8=0 x^2 - 2x + 8 = 0 x 2−2 x+8=0, we'll apply the quadratic formula method. The process is as follows: Step 1: Identify the coefficients from the equation: a=1 a = 1 a=1, b=−2 b = -2 b=−2, c=8 c = 8 c=8. Step 2: Calculate the discriminant using the formula Δ=b 2−4 a c \Delta = b^2 - 4ac Δ=b 2−4 a c. Step 3: For the given equation, calculate Δ=(−2)2−4(1)(8)=4−32=−28 \Delta = (-2)^2 - 4(1)(8) = 4 - 32 = -28 Δ=(−2)2−4(1)(8)=4−32=−28. Step 4: Interpret the discriminant. Since Δ=−28 \Delta = -28 Δ=−28 is negative, this indicates there are no real solutions to the equation. Without real roots, the equation has complex solutions. Since the problem requires real solutions, we conclude: The equation x 2−2 x+8=0 x^2 - 2x + 8 = 0 x 2−2 x+8=0 has no real solution. 3 Final Answer No solution Practice Quiz Test your knowledge with interactive questions Solve the following equation: 2 x 2−10 x−12=0 2x^2-10x-12=0 2 x 2−10 x−12=0 a x 1=6 x_1=6 x 1=6 x 2=−1 x_2=-1 x 2=−1 b x 1=−6 x_1=-6 x 1=−6 x 2=1 x_2=1 x 2=1 c x=6 x=6 x=6 d No solution Incorrect Correct Answer: x 1=6 x_1=6 x 1=6 x 2=−1 x_2=-1 x 2=−1 Watch Solution Continue 🌟 Unlock Your Math Potential Get unlimited access to all 18 Solving Quadratic Equations questions, detailed video solutions, and personalized progress tracking. 📹 Unlimited Video Solutions Step-by-step explanations for every problem 📊 Progress Analytics Track your mastery across all topics 🚫 Ad-Free Learning Focus on math without distractions 🚀Start Free Trial No credit card required • Cancel anytime More Questions Click on any question to see the complete solution with step-by-step explanations The Quadratic Formula ##### Solve the Quadratic Equation: 2x²-3x+5=0 Step-by-Step Click to solve##### Solve the Quadratic Equation: -x² + 6x - 10 = 0 Click to solve ##### Solve Quadratic Equation: x² + 3x + 7 = 0 in Standard Form Click to solve##### Solve the Quadratic Equation: x²+10x+25=0 Step-by-Step Click to solve [x] Table of Contents Question Video Solution Step-by-Step Solution Practice Quiz More Questions [x] Topic Navigation Main Subject: Solving Quadratic Equations 127 questions • 1 categories Articles in this Subject: Methods for Solving a Quadratic Function Specific Subject: The Quadratic Formula 127 questions • 1 question types Articles in this Subject: The quadratic equationMethods for Solving a Quadratic Function Question Types in this Subject: System of equations with no solution Pages Home About Learn math FAQ Terms of Service Privacy Policy Notice to Tutor Notice to Student Contact Us: support@tutorela.com
190041	https://math.stackexchange.com/questions/3785803/how-i-get-intervals-on-which-fx-sin-x-cos-x-increasing-or-decreasing	calculus - How I get intervals on which $f(x)=\sin x -\cos x$ increasing or decreasing? - 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 How I get intervals on which f(x)=sin x−cos x f(x)=sin⁡x−cos⁡x increasing or decreasing? Ask Question Asked 5 years, 1 month ago Modified5 years, 1 month ago Viewed 369 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Here, f′(x)=cos x+sin x f′(x)=cos⁡x+sin⁡x then f′′(x)=−sin x+cos x f″(x)=−sin⁡x+cos⁡x, f′′(x)=0 f″(x)=0 where sin x=cos x sin⁡x=cos⁡x or tan x=1 tan⁡x=1 This happens at x=π/4+π k x=π/4+π k for integer k k Next what should I do? calculus trigonometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 10, 2020 at 5:25 Andrei 40k 5 5 gold badges 31 31 silver badges 55 55 bronze badges asked Aug 10, 2020 at 5:17 Md.Atiquer RahmanMd.Atiquer Rahman 77 5 5 bronze badges 4 You just have to find x x where f′(x)f′(x) is positive or negative.SarGe –SarGe 2020-08-10 05:24:46 +00:00 Commented Aug 10, 2020 at 5:24 Wouldn't the endpoints of the intervals on which f f is increasing or decreasing be where the first derivative is zero, i.e., critical points? You seem to have found potential inflections points, points where the sign of the second derivative might change.Eric Towers –Eric Towers 2020-08-10 05:25:18 +00:00 Commented Aug 10, 2020 at 5:25 Given that f f and its derivatives are continuous, the solutions of f′(x)=0 f′(x)=0 define intervals in which f′(x)f′(x) will have a defined sign (either positive or negative, which states that f f will be increasing or decreasing). For each of those intervals, pick a point x 0 x 0 in the interval and evaluate f′(x 0)f′(x 0).cjferes –cjferes 2020-08-10 05:25:18 +00:00 Commented Aug 10, 2020 at 5:25 Arent you basically done?fleablood –fleablood 2020-08-10 05:32:26 +00:00 Commented Aug 10, 2020 at 5:32 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Hint: Before you even take the derivatives, use simple trigonometric identities: cos x=sin(π/2−x)=−sin(x−π/2)sin a+sin b=2 sin a+b 2 cos a−b 2 cos⁡x=sin⁡(π/2−x)=−sin⁡(x−π/2)sin⁡a+sin⁡b=2 sin⁡a+b 2 cos⁡a−b 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 10, 2020 at 5:24 AndreiAndrei 40k 5 5 gold badges 31 31 silver badges 55 55 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. f′(x)=cos x+sin x f′(x)=cos⁡x+sin⁡x. That means f(x)f(x) is increasing on intervals where f′(x)=cos x+sin x>0 f′(x)=cos⁡x+sin⁡x>0. .... well just look at the unit circle. In quadrant I sin,cos sin,cos are both non negative and not both 0 0 so f(x)f(x) is increasing there. In the second quadrant sin≥0 sin≥0 but cos≤0 cos≤0 but sin x>\|cos x\|sin⁡x>\|cos⁡x\| for the first 45 45 degrees. So at x=2 k π+3 π 4 x=2 k π+3 π 4, f(x)f(x) stops increasing. For the rest of the second quadrant \|cos x\|>sin x\|cos⁡x\|>sin⁡x so f(x)f(x) is decreasing. In third quadrant they are both non positive but not both zero so f(x)f(x) is decreasing. And the IV quadrant is just like the second. So on the interval (2 k π−π 4,2 k π+3 π 4)(2 k π−π 4,2 k π+3 π 4), f(x)f(x) is increasing and on the interval (2 k π+3 π 4,2 k π+7 π 4)(2 k π+3 π 4,2 k π+7 π 4), f′(x)f′(x) is decreasing. f(x)f(x) has maximum value and x=2 k π+3 π 4 x=2 k π+3 π 4 where f(x)=sin x−cos x=2√2−(−s q r t 2 2)=2–√f(x)=sin⁡x−cos⁡x=2 2−(−s q r t 2 2)=2 and a minimum value ant x=2 k π+7 π 4 x=2 k π+7 π 4 where f(x)=−2–√f(x)=−2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 10, 2020 at 5:31 fleabloodfleablood 132k 5 5 gold badges 52 52 silver badges 142 142 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. First, you would have to find out where your function reaches a maximum or a minimum by computing the derivative of f(x)f(x) you will get f′(x)=0⟹sin(x)+cos(x)=0 f′(x)=0⟹sin⁡(x)+cos⁡(x)=0. What you can do at this point is to use the fact that sin 2(x)+cos 2(x)=1 sin 2⁡(x)+cos 2⁡(x)=1 to notice that cos(x)=±1−sin 2(x)−−−−−−−−−√cos⁡(x)=±1−sin 2⁡(x) Add sin(x)sin⁡(x) to both sides you will get : sin(x)+cos(x)=±1−sin 2(x)−−−−−−−−−√+sin(x)sin⁡(x)+cos⁡(x)=±1−sin 2⁡(x)+sin⁡(x) So overall, we need to find x x for which : f′(x)=0⟹±1−sin 2(x)−−−−−−−−−√+sin(x)=0 f′(x)=0⟹±1−sin 2⁡(x)+sin⁡(x)=0 Here is a hint, if you let u=sin(x)u=sin⁡(x), you will obtain : ±1−u 2−−−−−√+u=0±1−u 2+u=0 Square both sides and you will get a quadratic equation from which you can find u u then from u u you will find the value of x x then from the value of x x you can study when f(x)f(x) increases or decreases. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 10, 2020 at 6:03 Matthew Cassell 4,364 4 4 gold badges 23 23 silver badges 31 31 bronze badges answered Aug 10, 2020 at 5:36 user815114 user815114 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. In order to find where the function is increasing or decreasing, vanishing of the second derivative is irrelevant. You need to find where the first derivative vanishes. In this case, it does where f′(x)=cos x+sin x=0 f′(x)=cos⁡x+sin⁡x=0. This means tan x=−1 tan⁡x=−1: there is no problem in dividing by cos x cos⁡x, because when cos x=0 cos⁡x=0 it holds sin x≠0 sin⁡x≠0. This yields the points 3 π/4+k π 3 π/4+k π, where k k is integer, but we're better to divide the solution set into two, for even and odd k k, because the tangent has period π π, but sine and cosine have period 2 π 2 π. Let's consider the points of the form 3 π/4+2 k π 3 π/4+2 k π: then f′′(3 π/4+2 k π)=−sin(3 π/4)+cos(3 π/4)=−2–√f″(3 π/4+2 k π)=−sin⁡(3 π/4)+cos⁡(3 π/4)=−2 so these are points of maximum. For points of the form 3 π/4+(2 k+1)π=7 π/4+2 k π 3 π/4+(2 k+1)π=7 π/4+2 k π we have f′′(7 π/4+2 k π)=−sin(7 π/4)+cos(7 π/4)=2–√f″(7 π/4+2 k π)=−sin⁡(7 π/4)+cos⁡(7 π/4)=2 so these are points of minimum. Now the intervals where the function is increasing or decreasing are easy to determine. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 10, 2020 at 7:50 egregegreg 246k 21 21 gold badges 155 155 silver badges 353 353 bronze badges 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 calculus trigonometry 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 2How to find cos A cos B−sin A sin B cos⁡A cos⁡B−sin⁡A sin⁡B? 1General values of θ θ for which (sin θ)x 2+(2 cos θ)x+cos θ+sin θ 2(sin⁡θ)x 2+(2 cos⁡θ)x+cos⁡θ+sin⁡θ 2 is the square of a linear function. 1Intervals in which f(x) is Strictly Increasing/Decreasing 1If cos(A+B+C)=cos A cos B cos C≠0 cos⁡(A+B+C)=cos⁡A cos⁡B cos⁡C≠0, then evaluate ∣∣8 sin(A+B)sin(B+C)sin(C+A)sin 2 A sin 2 B sin 2 C∣∣\|8 sin⁡(A+B)sin⁡(B+C)sin⁡(C+A)sin⁡2 A sin⁡2 B sin⁡2 C\| 3general solution beta (sin(α+β)+cos(α+2 β)sin(β))2=4 cos(α)sin(β)sin(α+β);tan(α)=3 tan(β)(sin⁡(α+β)+cos⁡(α+2 β)sin⁡(β))2=4 cos⁡(α)sin⁡(β)sin⁡(α+β);tan⁡(α)=3 tan⁡(β) 1Showing tan x=sin α sin y 1−sin α cos y tan⁡x=sin⁡α sin⁡y 1−sin⁡α cos⁡y, given sin x=sin α sin(x+y)sin⁡x=sin⁡α sin⁡(x+y) Hot Network Questions Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? 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190042	https://www.blacksacademy.net/texts/DifferenceTwoSampleMeansUnknownVariance.pdf	© blacksacademy.net 1 The Difference of Two Sample Means with Unknown Variance Prerequisites You should be familiar with (1) determining a confidence interval for a sample mean drawn from a population of known or unknown variance; (2) the biased and unbiased estimator of the population variance. Unbiased estimator of the population mean The sample mean, X , is an unbiased and consistent estimator of the population mean, . Unbiased estimator of the population variance   2 2 1 n s S n is an unbiased estimator of the population variance where     2 2 2 X S X n is the sample variance. The difference of two sample means Suppose we have two independent samples   1 2 and X X drawn one from each of two normally distributed populations of differing known variance. To say that the samples are independent is to say that values of 1 X are not in any way associated or paired with values of 2 X . That is, we are given that         2 1 1 1 2 2 2 2 , , X N X N where   2 2 1 2 and are known. Alternatively, if the variances are unknown we shall estimate them using the unbiased estimators to obtain estimates   2 2 1 2 and ˆ ˆ . Then the difference of the two sample means © blacksacademy.net 2   1 2 D X X is the linear combination of two normally distributed variables, and hence is itself normally distributed. The distribution is as follows.               2 2 1 2 1 2 1 2 1 2 ~ , D X X N n n where 1 2 and n n are the sizes of the independent samples of 1 2 and X X respectively. This result enables us to find a confidence interval for the difference of two sample means. The expression    2 2 1 2 1 2 n n is called the standard error of the difference of two means. When the population variances are not known, they must be estimated using the unbiased estimator; that is, we substitute the estimates   2 2 1 2 and ˆ ˆ for   2 2 1 2 and respectively in the above. Then, strictly, since the variances are estimated, the distribution of the standard error of the difference of the two means    2 2 1 2 1 2 ˆ ˆ n n is not normally distributed. However, if the sample sizes are large    1 2 both and 30 n n then it is approximately normally distributed and the approximation is very close. In that case we may use the normal distribution as before to determine the critical values in the confidence interval. Example (1) At a DIY store two different varieties, A and B, of sheet hardboard are sold. The management decided to measure the area, x square metres, of each of 100 randomly selected sheets of variety A. The results are summarised below.     2 302.5 916.25 x x The management also measured the area, y square metres, of each of 120 randomly selected sheets of variety B. The results are summarised below.     2 388.4 1258.90 y y Determine an approximate 95% confidence interval for the difference in the population means of the areas of the two varieties. Solution                   2 2 2 2 2 302.5 916.25 100 302.5 3.025 100 916.25 3.025 0.011875 100 x x x n x x n x S x n © blacksacademy.net 3                                2 2 2 2 2 2 2 2 2 2 2 100 0.011875 0.0119949... ˆ 1 99 388.4 1258.90 388.4 3.2366... 120 1258.90 3.2366... 0.014822... 120 120 0.014822... 0.0149467... ˆ 1 119 x x x y y y y n s S n y y y y n y S y n n s S n      The estimate for the mean difference of the areas of the two varieties is 3.2366... 3.025 0.21166... d y x           2 2 4 1 2 1 2 The standard error of the mean difference is 0.0119949... 0.0149467... ˆ ˆ SE 2.445... 10 0.015636... 100 120 n n A 95% confidence interval for the true mean difference of the areas of the two varieties is                       1.96 SE 1.96 SE 0.21166... 1.96 0.015636... 0.21166... 1.96 0.015636... 0.1810... 0.2423... 0.18 0.24 2 s.f. d d Hypothesis testing We may also be asked to test whether the two population means are equal at a given significance level.                 0 1 2 1 1 2 1 1 2 1 2 two tailed alternatively or one tailed H H H Since the difference of the two sample means is               2 2 1 2 1 2 1 2 1 2 ~ , D X X N n n the test statistic is      1 2 2 2 1 2 1 2 test X X Z n n The critical value is found using the tables for the standardised normal distribution and finding the z value corresponding to the given significance level. We may compare this test statistic with © blacksacademy.net 4 the critical value for the test depending on the significance level, or alternatively find the probability (p-value) that corresponds to the test statistic. The first example illustrates the case where the population variances are known and different. Example (2) A random sample is taken from each of two normally distributed populations. The sample size of the first population is 10 with sample mean 18.2. The sample size of the second population is 12, with sample mean 15.3. It is known that the population variances are 2.5 and 2.8 respectively. Test at the 1% level whether the means of the two populations are equal. Solution First sample        2 2 1 1 1 1 1 1 ~ , 10, 18.2, 2.5 X N n X Second sample        2 2 2 2 2 2 2 2 ~ , 12, 15.3, 2.8 X N n X The null and alternative hypotheses are formulated as follows.         0 1 2 1 1 2 two-tailed, significance level, 1% H H The test statistic is             1 2 2 2 1 2 1 2 18.2 15.3 2.5 2.8 10 12 4.17133... 4.171 3 d.p. test X X Z n n The critical z-value is    critical Z ( 0.995) since =1% and the test is 2-tailed = 2.576 P Z   test critical 1 0 Z Z Accept , reject The two population means are not equal. H H Alternatively, since     3.090 0.999 P Z , the p-value (2 tailed) associated with a test value of 4.171 test z is 0.0005 p In this next example the population variances are unknown and must be estimated. © blacksacademy.net 5 Example (3) Two random variables X and Y are normally distributed. Independently, samples of both were taken as follows.           2 1 2 2 120 250.2 1008.4 150 342.8 1278.5 n x x n y y Test the hypothesis that the mean of X is less than the mean of Y at the 1% significance level. Solution In this question the variances are unknown and therefore must be estimated using the unbiased estimator. Since the sample sizes are large we may use the z-score to evaluate the hypothesis.                                             2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 ~ , 120 250.2 1008.4 250.2 2.085 120 1008.4 2.085 4.0561... 120 120 4.0561... 4.0901... ˆ 1 119 ~ , 150 342.8 1278.5 342.8 2.2853... 150 1 x x x x y y X N n x x x x n x S x n n S n Y N n y y y y n y S y n           2 2 2 278.5 2.2853 3.30058... 150 150 3.30058... 3.3227... ˆ 1 149 x x n S n         0 1 2 1 1 2 one-tailed, significance level, 1% H H             2 2 2.085 2.2853... 4.0901... 3.3227... 120 150 0.84478... 0.845 3 d.p. test y x x y X Y Z n n      critical Z ( 0.995) 2.576 since 1% and the test is 2-tailed P Z © blacksacademy.net 6  test critical 1 0 < Reject , accept The two population means are equal. Z Z H H Alternatively, the p-value (1 tailed) associated with a test value of 0.845 test z is    1 0.6009 0.3991 p which is much greater than the critical value of 0.01. Therefore, as above reject 1 0 , accept H H . When the population variances are identical When it is given that the population variances are identical then the test is simpler, since then      1 2 test 1 2 2 Z 1 1 where is the common variance. X X n n Example (4) Two independent random samples were taken, one from each of two normally distributed populations both of which may be assumed to have the same variance, 2 =16. If the first sample had size 8 and mean 4 and the second have size 24 and mean 5, test the null hypothesis     1 2 0 against the alternative hypothesis     1 2 0 at the 1% significance level, where   1 2 and are the means of the respective populations. Solution                          2 1 1 1 1 1 2 2 2 2 2 2 2 0 1 2 1 1 2 ~ , 8 4 ~ , 24 5 16 16 4 0 0 two-tailed 1% X N n X X N n X H H            1 2 1 2 4 -5 1 1 6 0.612 (3 s.f.) 4 1 1 1 1 3 1 1 4 4 4 8 24 24 6 test X X Z n n      0 1 ( 0.995) 2.576 Accept , reject . The two means are the same. critical test critical Z P Z Z Z H H
190043	https://www.quora.com/B-takes-20-more-time-to-complete-a-job-than-A-A-and-B-work-on-alternate-days-for-the-first-6-days-then-A-alone-works-for-2-days-and-finally-B-works-alone-for-3-days-to-finish-the-job-What-percentage-of-the-total-job	B takes 20% more time to complete a job than A. A and B work on alternate days for the first 6 days, then A alone works for 2 days, and finally B works alone for 3 days to finish the job. What percentage of the total job was finished by A? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Problem Solving Work Activities Percentage Share Mathematics Word Problems Efficiency Mathematical Problems Time and Work Percentage (%) Ratio and Percentage 5 B takes 20% more time to complete a job than A. A and B work on alternate days for the first 6 days, then A alone works for 2 days, and finally B works alone for 3 days to finish the job. What percentage of the total job was finished by A? All related (38) Sort Recommended Syamal Dey Ph.D., Former Sr. Scientist, University of Maryland, USA · Author has 8.2K answers and 7.7M answer views ·3y Question, B takes 20% more time to complete a job than A. A and B work on alternate days for the first 6 days, then A alone works for 2 days, and finally B works alone for 3 days to finish the job. What percentage of the total job was finished by A? Let W denotes the whole given job. a & b denote the times (in days) required for A & B alone to complete the work W. Hence from above data we get following relation, b = (1 + 20/100)a or b = (6/5)a or b : a = 6 : 5 = k (say) …… (x1) A and B work on alternate days for the first 6 days, and the amount of work finished by A alone = 3W/a = 3W/(5k) …… (x2) [f Continue Reading Question, B takes 20% more time to complete a job than A. A and B work on alternate days for the first 6 days, then A alone works for 2 days, and finally B works alone for 3 days to finish the job. What percentage of the total job was finished by A? Let W denotes the whole given job. a & b denote the times (in days) required for A & B alone to complete the work W. Hence from above data we get following relation, b = (1 + 20/100)a or b = (6/5)a or b : a = 6 : 5 = k (say) …… (x1) A and B work on alternate days for the first 6 days, and the amount of work finished by A alone = 3W/a = 3W/(5k) …… (x2) [from (x1)] the amount of work finished by B alone = 3W/b = 3W/(6k) = W/(2k) …… (x3) [from (x1)] Next A alone works for 2 days and the amount of work finished = 2W/a = 2W(5k) …… (x4) Finally B works alone for 3 days, and the amount of work finished = 3W/b = 3W(6k) = W/(2k) …… (x5) From above we get, W = 3W/(5k) + W/(2k) + 2W(5k) + W/(2k) [from (x2) + (x3) + (x4) + (x5)] or 1 = 3/(5k) + 1/(2k) + 2/(5k) + 1/(2k) or k = 3/5 + 1/2 + 2/5 + 1/2 or k = 2 …… (x6) Therefore, the required percentage of the total job that was finished by A = 100(3W/10 + W/5)/W [from (x2), (x4) & (x6)] = 100/2 = 50% [Ans] Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 1 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 201 99 34 9 3 Related questions More answers below A is twice as efficient as B. A and B together can finish work in 5 days, then work will be finished to B alone in how many days? A can do a work in 15 days. B is 50% more efficient than A. They started work together. B left the work after some days and A does the remaining 1/3rd work. For what days did they work together? A can do a work in 21 days. B is 40% more efficient than A. The number of days required for B to finish the same work alone? A can do a job alone in 24 days. B is 50% more efficient than A. In how many days can B alone finish the job? A is 3 times efficient than B. B completes total work in 12 days. How many days will 4 B and 8 A take to finish a work? Jayanta Mukherjee B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 43.6K answers and 11.1M answer views ·3y Say, A completes the job working alone in x days. So, B can complete the job working alone in [x {1 + (20 / 100)}] = 1.2x days. Daily efficiency of A = (1 / x) Daily efficiency of B = (1 / 1.2x) A and B worked on alternate days for first 6 days. Then, A worked alone for 2 days and B worked alone for 3 days. So, (1 / x) (3 + 2) + (1 / 1.2x) (3 + 3) = 1 Or, (5 / x) + (6 / 1.2x) = 1 Or, (5 / x) + (5 / x) = 1 Or, (10 / x) = 1 Or, x = 10 So, A does (5 / 10) = (1 / 2) of the total job; that is, 50% of the total job. Upvote · 9 1 Ved Prakash Sharma Solved over a million queries · Author has 14.3K answers and 16.5M answer views ·3y Let A alone can complete a work in x days. then B alone can complete the same work in x.(1+20/100) or, 6x/5 days. Hence , A's one day work is 1/x and B's one day work is 5/6x. A's total work done = (3+2).(1/x) = 5/x. B's total work done= (3+3).(5/6x) = 5/x. (A+B)'s total work done = 5/x+5/x = 10/x . Percentage of the total job finished by A = (5/x ×100)/(10/x) = 50 %. , Answer. Upvote · 9 1 Paul Fournier Teacher · Author has 1.4K answers and 595K answer views ·3y Originally Answered: A takes 12 days to complete a work and B takes 8 days to do the same work. If they work on alternate days beginning with A, what fraction of the whole work is done by A alone? · On day 1 A does 1/12 of the work. On day 2 B does 1/8 of the work so after two days 1/12+1/8=5/24 of the work is done. So every two days they do 5/24 of the work therefore after 8 days they will have completed 4x5/24=20/24 of the work. On day 9 A does 1/12 or 2/24 of the work so that on day 10 B completes the work. A has worked on the odd number of days 1,3,5,7 and 9 therefore 5/12 of the work. Upvote · 9 2 9 1 9 1 Sponsored by JetBrains Become More Productive in Java Try IntelliJ IDEA, a JetBrains IDE, and enjoy productive Java development! Download 999 616 Related questions More answers below A and B can together finish a work in 50 days. They worked together for 20 days and then B left. After another 30 days, A finished the remaining work. In how many days can A alone finish the job? A can do a job in 9 days and B can do that job in 6 days. If B quits after working together for 3 days, in how many days will the work be completed? Two persons A and B can do a work alone in 29 days. A takes the rest of one day after every 4 days and B takes the rest of one day after every 5 days. If A and B starts working together, then the work will be completed on? A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. The number of days taken in completing the work were? A and B can finish work in 6 days and 4 days, respectively. A alone worked on it for 2 days. After that, B joined him. How much total time was taken by them to finish the work? Pradeep Hebbar Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views ·3y Originally Answered: A takes 12 days to complete a work and B takes 8 days to do the same work. If they work on alternate days beginning with A, what fraction of the whole work is done by A alone? · Part of work done by A per day =1 12=1 12 Part of work done by B per day =1 8=1 8 A A and B B work in alternate days. Hence they complete a cycle once in two days. Part of work done per cycle =1 12+1 8=5 24=1 12+1 8=5 24 Let it take n n cycles to complete the work (1 1 part). n×5 24=1⟹n=4.8 n×5 24=1⟹n=4.8 At the end of 4 4 complete cycles (8 8 days), Part of work completed by A=4×1 12=1 3 A=4×1 12=1 3 Part of work completed by B=4×1 8=1 2 B=4×1 8=1 2 Remaining work =1−(1 3+1 2)=1 6=1−(1 3+1 2)=1 6 Now, A A works on 9 9 th day an Continue Reading Part of work done by A per day =1 12=1 12 Part of work done by B per day =1 8=1 8 A A and B B work in alternate days. Hence they complete a cycle once in two days. Part of work done per cycle =1 12+1 8=5 24=1 12+1 8=5 24 Let it take n n cycles to complete the work (1 1 part). n×5 24=1⟹n=4.8 n×5 24=1⟹n=4.8 At the end of 4 4 complete cycles (8 8 days), Part of work completed by A=4×1 12=1 3 A=4×1 12=1 3 Part of work completed by B=4×1 8=1 2 B=4×1 8=1 2 Remaining work =1−(1 3+1 2)=1 6=1−(1 3+1 2)=1 6 Now, A A works on 9 9 th day and completes 1 12 1 12 part of work. Remaining work to be completed by B B on 10th day = 1 6−1 12=1 12<1 8 1 6−1 12=1 12<1 8 On 10 th day, the work gets completed. Thus, total fraction of work done by A=1 3+1 12=5 12 A=1 3+1 12=5 12 Ans: 5/12 Upvote · Sooraj M.U Studied Computer Science (Graduated 2000) · Author has 120 answers and 233.6K answer views ·5y Related A and B can finish a work, working on alternate days, in 19 days, when A works on the first. However, they can finish the work, working alternate days, in 19(5/6) days, when B works on first day. How many days does A alone take to finish the work? Let x and y be the work by A and B can do in a day If A starts first, the total work is finished in 19 days. A works for 10 days and B for 9 days So total work is W = 10x+9y —(1) If B starts first, it is completed in 19(5/6) days. B works 10 days and A works in 9(5/6) days W = 10y + 9(5/6)x —(2) Equating (1) and (2) 10x+9y = 10y+ 9(5/6)x 1/6 x = y Let's write total work in terms of x alone to calculate the number of days A would need W = 10x+ 9y = 10x + 9/6 x = 11(1/2) x A would need 11(1/2) days if he is working alone Upvote · 99 26 Sponsored by VAIZ.com Stop Fighting Your PM Tool — Switch to VAIZ VAIZ — Better Alternative to Your PM Tool. Boost Productivity with Built-in Doc Editor & Task Manager. Sign Up 9 6 Titus Ndundu Senior Lecturer at Kiriri Womens University of Science and Technology (2008–present) · Author has 160 answers and 39.5K answer views ·3y Related A and B can finish a work, working on alternate days, in 19 days, when A works on the first. However, they can finish the work, working alternate days, in 19(5/6) days, when B works on first day. How many days does A alone take to finish the work? A working alone takes 11.5 days B working alone takes 69 days Here is the method: In the first scenario, A works for 10 days and B takes 9 days In the second scenario A works for 95/6 days or 59/6 days and B works for 10 days. Let A do X fraction of work per day and let B do Y fraction of work per day. The following equations can be formed: For scenario one: 10X+9Y = 1 For scenario two: 59/6X+10Y = 1 which we simplify to 59X+60Y = 6 These are two simultaneous equations which can be solved by elimination method to get X = 2/23. Thus A does 2/23 of work per day. If 2/23 of work is done in 1 day then the wh Continue Reading A working alone takes 11.5 days B working alone takes 69 days Here is the method: In the first scenario, A works for 10 days and B takes 9 days In the second scenario A works for 95/6 days or 59/6 days and B works for 10 days. Let A do X fraction of work per day and let B do Y fraction of work per day. The following equations can be formed: For scenario one: 10X+9Y = 1 For scenario two: 59/6X+10Y = 1 which we simplify to 59X+60Y = 6 These are two simultaneous equations which can be solved by elimination method to get X = 2/23. Thus A does 2/23 of work per day. If 2/23 of work is done in 1 day then the whole work is done 1/(2/23)1=23/2= 11.5 days From the above equations we also get Y= 1/69 which implies B alone would take 69 days Upvote · 9 2 Richard Cohen Former Retired Commercial Service Rep (2012–2021) · Author has 858 answers and 1.3M answer views ·3y Originally Answered: A takes 12 days to complete a work and B takes 8 days to do the same work. If they work on alternate days beginning with A, what fraction of the whole work is done by A alone? · A = 12 days = 1 job, 1 day = 1/12 B = 8 days = 1 job , 1 day = 1/8 A = 5/12 = .4167 + ,B = 5/8 = .625 = 1.04 Upvote · 9 1 Promoted by Betterbuck Anthony Madden Writer for Betterbuck ·Updated Aug 15 What are the weirdest mistakes people make on the internet right now? Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr Continue Reading Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. 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Upvote · 999 566 99 60 9 4 Carlos Aspillera Lives in Barcelona, Spain (2018–present) · Author has 10.9K answers and 2.6M answer views ·3y Originally Answered: A takes 12 days to complete a work and B takes 8 days to do the same work. If they work on alternate days beginning with A, what fraction of the whole work is done by A alone? · ——- A — x/12 … and … B — x/8. Alternate days … (x/12 + x/8)=1 where 5x=24… x= 4.8 ——- A ((Mon) + Wed +Fri)…. (x/12)+(x/12) +0.8(x/12) = 0.233x ——- B ((Tues +Thur) … (x/8) + (x/8) = 0.250x Fraction of work done by A is (0.233/(0.233+0.250) = 0.233/0.483 = 48.2%. Upvote · 9 1 Rajendra Kumar Phophalia BE from Birla Institute of Technology and Science, Pilani · Author has 5.9K answers and 2.6M answer views ·3y Originally Answered: A takes 12 days to complete a work and B takes 8 days to do the same work. If they work on alternate days beginning with A, what fraction of the whole work is done by A alone? · per 2 day work =(1/12)+(1/8)=(2+3)/24=5/24 ,so no of days to complete the work ,(24/5)2=48/5= 9 3/5 ,so A will work for 5 days and work done by him =5(1/12)=5/12 of the work. Upvote · Vasudevan A.N.S. Former Retired Professor of Chemistry. (1977–2014) · Author has 8K answers and 13.3M answer views ·5y Related A and B can finish a work, working on alternate days, in 19 days, when A works on the first. However, they can finish the work, working alternate days, in 19(5/6) days, when B works on first day. How many days does A alone take to finish the work? If A- works for 10- days & B- works for 9- days the work will be completed in 19- days If B- works of 10- days & A- works for 9& 5/6- days the work is completed in 19& 5/6- days. Therefore 10A + 9B= 95/6 A+ 10B So 1/6 A= B. Therefore A- is 6- times more efficient than B. If A- can complete the work alone in=N- days, then B- can complete the work in = 6N - days So 10/N + 9/6N= 1 So 60 + 9= 6N So N= 69/6= 11.5- days. Therefore A- alone can complete the work in = 11.5- days. B- can complete the work alone in = 6N= (6 x 11.5)= 69- days. . Upvote · 9 3 Sushil Ganeshpure Studied at Indian Institute of Technology, Bombay (IITB) (Graduated 2022) ·7y Related A and B can complete a job in 30 and 20 days respectively. They start working together and B leaves 5 days before the work is finished. In how many days is the total work finished? Answer to this problem is 15 days. There are three methods to solve this problem. Method-1 Fraction Method A completes the work in 30 days -> In 1 day he completes 1/30 part of job. B completes the work in 20 days -> In 1 day he completes 1/20 part of job. Work done by A & B both in 1 day= 1/30 + 1/20 =(20+30)/(20x30) = 5/60 = 1/12 Let both A & B workd for 'x' days. Work done by them in x day = x/12 Remaining work = 1-x/12 ………(1) This remaining work is done by A alone in 5 days. Hence, Work done by A in 5 days = Remaining work 5/30 = 1-x/12 …. From (1) x/12 = 1–1/6 x = (5x12)/6 x=10 Total number of days to com Continue Reading Answer to this problem is 15 days. There are three methods to solve this problem. Method-1 Fraction Method A completes the work in 30 days -> In 1 day he completes 1/30 part of job. B completes the work in 20 days -> In 1 day he completes 1/20 part of job. Work done by A & B both in 1 day= 1/30 + 1/20 =(20+30)/(20x30) = 5/60 = 1/12 Let both A & B workd for 'x' days. Work done by them in x day = x/12 Remaining work = 1-x/12 ………(1) This remaining work is done by A alone in 5 days. Hence, Work done by A in 5 days = Remaining work 5/30 = 1-x/12 …. From (1) x/12 = 1–1/6 x = (5x12)/6 x=10 Total number of days to complete work = 10+5 = 15 days Method-2 LCM Method Let the total units of work be LCM(30,20)=60 A does 60/30 = 2 units per day work B does 60/2 = 3 units per day work Together they do (2+3)=5 units of per day work. Let say both of them worked for x days. Work done by them in x days = 5x Remaining work = 60–5x This remaining work done by A alone in 5 days Hence, 5(A's work per day)= 60–5x 52=60–5x x=10 Total number of days taken to complete the work = 10+5 = 15 days Method-3 Percentage Method Let the total work be 100% A does 100/30 = 3.33% work in 1 day B does 100/20 = 5% work in 1 day Both A & B do (3.33+5)=8.33% work in 1 day Let say they worked for 'x' days In x days the complete x8.33% work. Remaining work = 100%-x8.33% This remaining work completed by A alone in 5 days Work done by A in 5 days = 53.33%= 16.66% Hence, Remaining work = work done by A in 5 days 100 - x8.33%=16.66% x8.33% = 100%–16.66% x8.33% = 83.33% x = 83.33/8.33 x=10 days Total number of days taken to complete the work = 10+5 = 15 days Now, which method to use in exam? Obviously percentage method is out of question as it is only useful for very basic problems& can involve lot of unnecessary calculations. LCM & Fractions both the methods have their own merits & demerits. While LCM method is very useful to visualise the problem it involves one extra step to calculate LCM. Also It becomes difficult to use LCM method for very complex problems. Fraction method is suitable for those who have command on fractions related calculations. It is more direct method to approach the problem & can be easily used to solve any type of problem. Use the method that gives best results for you. Happy Learning! Upvote · 99 40 9 3 Abhishek Mishra 3 out of 2 people have trouble with fractions... · Author has 281 answers and 2.6M answer views ·8y Related A and B can finish a work together in 12 days, and B and C together in 16 days. If A alone works for 5 days and then B alone continues for 7 days, then remaining work is done by C in 13 days. In how many days can C alone finish the complete work? Such questions can be easily attemted by following the LCM method. You can message me, or post your Quant doubts in the Comment thread of this question. Continue Reading Such questions can be easily attemted by following the LCM method. You can message me, or post your Quant doubts in the Comment thread of this question. Upvote · 99 79 9 5 9 2 Related questions A is twice as efficient as B. A and B together can finish work in 5 days, then work will be finished to B alone in how many days? A can do a work in 15 days. B is 50% more efficient than A. They started work together. B left the work after some days and A does the remaining 1/3rd work. For what days did they work together? A can do a work in 21 days. B is 40% more efficient than A. The number of days required for B to finish the same work alone? A can do a job alone in 24 days. B is 50% more efficient than A. In how many days can B alone finish the job? A is 3 times efficient than B. B completes total work in 12 days. How many days will 4 B and 8 A take to finish a work? A and B can together finish a work in 50 days. They worked together for 20 days and then B left. After another 30 days, A finished the remaining work. In how many days can A alone finish the job? A can do a job in 9 days and B can do that job in 6 days. If B quits after working together for 3 days, in how many days will the work be completed? Two persons A and B can do a work alone in 29 days. A takes the rest of one day after every 4 days and B takes the rest of one day after every 5 days. If A and B starts working together, then the work will be completed on? A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. The number of days taken in completing the work were? A and B can finish work in 6 days and 4 days, respectively. A alone worked on it for 2 days. After that, B joined him. How much total time was taken by them to finish the work? A and B can do a job together in 12 days. A is 2 times efficient as B. In how many days can B alone complete the work? A can complete a piece of work in 10 days and B can complete it in 20 days. A starts working and B works on every alternate day. In how many days total work will be completed if B left 5 days before the completion of work? A and B together complete a work in 12 days. A alone works for 8 days and B alone can complete the remaining work in 20 days. Can B alone complete the whole work? A job can be completed by a person A alone in 24 days and by another person B alone in 40 days. A and B work on the job together and C works with them for 6 days. The three completed the job in 12 days. In how many days can C alone complete the job? A takes 6 days less than B to finish work. A and B started work and A left 3 days before the work was done. B completes 37.5% of the total work. In how many days will B complete the work? Related questions A is twice as efficient as B. A and B together can finish work in 5 days, then work will be finished to B alone in how many days? A can do a work in 15 days. B is 50% more efficient than A. They started work together. B left the work after some days and A does the remaining 1/3rd work. For what days did they work together? A can do a work in 21 days. B is 40% more efficient than A. The number of days required for B to finish the same work alone? A can do a job alone in 24 days. B is 50% more efficient than A. In how many days can B alone finish the job? A is 3 times efficient than B. B completes total work in 12 days. How many days will 4 B and 8 A take to finish a work? A and B can together finish a work in 50 days. They worked together for 20 days and then B left. After another 30 days, A finished the remaining work. In how many days can A alone finish the job? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. 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190044	https://en.wikipedia.org/wiki/Fundamental_interaction	Jump to content Search Contents 1 History 1.1 Classical theory 1.2 Standard Model 2 Overview of the fundamental interactions 3 Interactions 3.1 Gravity 3.2 Electroweak interaction 3.2.1 Electromagnetism 3.2.2 Weak interaction 3.3 Strong interaction 3.4 Higgs interaction 4 Unification 5 Beyond the Standard Model 6 See also 7 References 8 Bibliography Fundamental interaction Afrikaans Alemannisch العربية Asturianu Azərbaycanca বাংলা 閩南語 / Bn-lm-gí Беларуская Български Boarisch Bosanski Català Чӑвашла Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français 한국어 Հայերեն हिन्दी Hrvatski Ido Bahasa Indonesia Íslenska Italiano עברית ქართული Қазақша Kreyòl ayisyen Latviešu Lietuvių Lombard Magyar മലയാളം Монгол မြန်မာဘာသာ Nederlands नेपाल भाषा 日本語 Nordfriisk Norsk bokmål Norsk nynorsk Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ Plattdüütsch Polski Português Romnă Русский සිංහල Simple English Slovenčina Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் Татарча / tatarça తెలుగు ไทย Türkçe Удмурт Українська اردو Tiếng Việt 吴语粵語中文 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 Most basic type of physical force \| \| \| This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (November 2022) (Learn how and when to remove this message) \| In physics, the fundamental interactions or fundamental forces are interactions in nature that appear not to be reducible to more basic interactions. There are four fundamental interactions known to exist: gravity, electromagnetism, weak interaction, and strong interaction. The gravitational and electromagnetic interactions produce long-range forces whose effects can be seen directly in everyday life. The strong and weak interactions produce forces at subatomic scales and govern nuclear interactions inside atoms. Some scientists hypothesize that a fifth force might exist, but these hypotheses remain speculative. Each of the known fundamental interactions can be described mathematically as a field. The gravitational interaction is attributed to the curvature of spacetime, described by Einstein's general theory of relativity. The other three are discrete quantum fields, and their interactions are mediated by elementary particles described by the Standard Model of particle physics. Within the Standard Model, the strong interaction is carried by a particle called the gluon and is responsible for quarks binding together to form hadrons, such as protons and neutrons. As a residual effect, it creates the nuclear force that binds the latter particles to form atomic nuclei. The weak interaction is carried by particles called W and Z bosons, and also acts on the nucleus of atoms, mediating radioactive decay. The electromagnetic force, carried by the photon, creates electric and magnetic fields, which are responsible for the attraction between orbital electrons and atomic nuclei which holds atoms together, as well as chemical bonding and electromagnetic waves, including visible light, and forms the basis for electrical technology. Although the electromagnetic force is far stronger than gravity, it tends to cancel itself out within large objects, so over large (astronomical) distances gravity tends to be the dominant force, and is responsible for holding together the large scale structures in the universe, such as planets, stars, and galaxies. The historical success of models that show relationships between fundamental interactions have led to efforts to go beyond the Standard Model and combine all four forces in to a theory of everything. History [edit] Classical theory [edit] In his 1687 theory, Isaac Newton postulated space as an infinite and unalterable physical structure existing before, within, and around all objects while their states and relations unfold at a constant pace everywhere, thus absolute space and time. Inferring that all objects bearing mass approach at a constant rate, but collide by impact proportional to their masses, Newton inferred that matter exhibits an attractive force. His law of universal gravitation implied there to be instant interaction among all objects. As conventionally interpreted, Newton's theory of motion modelled a central force without a communicating medium. Thus Newton's theory violated the tradition, going back to Descartes, that there should be no action at a distance. Conversely, during the 1820s, when explaining magnetism, Michael Faraday inferred a field filling space and transmitting that force. Faraday conjectured that ultimately, all forces unified into one. In 1873, James Clerk Maxwell unified electricity and magnetism as effects of an electromagnetic field whose third consequence was light, travelling at constant speed in vacuum. If his electromagnetic field theory held true in all inertial frames of reference, this would contradict Newton's theory of motion, which relied on Galilean relativity. If, instead, his field theory only applied to reference frames at rest relative to a mechanical luminiferous aether—presumed to fill all space whether within matter or in vacuum and to manifest the electromagnetic field—then it could be reconciled with Galilean relativity and Newton's laws. (However, such a "Maxwell aether" was later disproven; Newton's laws did, in fact, have to be replaced.) Standard Model [edit] Main article: Standard Model See also: Standard Model (mathematical formulation) The Standard Model of particle physics was developed throughout the latter half of the 20th century. In the Standard Model, the electromagnetic, strong, and weak interactions associate with elementary particles, whose behaviours are modelled in quantum mechanics (QM). For predictive success with QM's probabilistic outcomes, particle physics conventionally models QM events across a field set to special relativity, altogether relativistic quantum field theory (QFT). Force particles, called gauge bosons—force carriers or messenger particles of underlying fields—interact with matter particles, called fermions. Everyday matter is atoms, composed of three fermion types: up-quarks and down-quarks constituting, as well as electrons orbiting, the atom's nucleus. Atoms interact, form molecules, and manifest further properties through electromagnetic interactions among their electrons absorbing and emitting photons, the electromagnetic field's force carrier, which if unimpeded traverse potentially infinite distance. Electromagnetism's QFT is quantum electrodynamics (QED). The force carriers of the weak interaction are the massive W and Z bosons. Electroweak theory (EWT) covers both electromagnetism and the weak interaction. At the high temperatures shortly after the Big Bang, the weak interaction, the electromagnetic interaction, and the Higgs boson were originally mixed components of a different set of ancient pre-symmetry-breaking fields. As the early universe cooled, these fields split into the long-range electromagnetic interaction, the short-range weak interaction, and the Higgs boson. In the Higgs mechanism, the Higgs field manifests Higgs bosons that interact with some quantum particles in a way that endows those particles with mass. The strong interaction, whose force carrier is the gluon, traversing minuscule distance among quarks, is modeled in quantum chromodynamics (QCD). EWT, QCD, and the Higgs mechanism comprise particle physics' Standard Model (SM). Predictions are usually made using calculational approximation methods, although such perturbation theory is inadequate to model some experimental observations (for instance bound states and solitons). Still, physicists widely accept the Standard Model as science's most experimentally confirmed theory. Overview of the fundamental interactions [edit] In the conceptual model of fundamental interactions, matter consists of fermions, which carry properties called charges and spin ±1⁄2 (intrinsic angular momentum ±ħ⁄2, where ħ is the reduced Planck constant). They attract or repel each other by exchanging bosons. The interaction of any pair of fermions in perturbation theory can then be modelled thus: : Two fermions go in → interaction by boson exchange → two changed fermions go out. The exchange of bosons always carries energy and momentum between the fermions, thereby changing their speed and direction. The exchange may also transport a charge between the fermions, changing the charges of the fermions in the process (e.g., turn them from one type of fermion to another). Since bosons carry one unit of angular momentum, the fermion's spin direction will flip from +1⁄2 to −1⁄2 (or vice versa) during such an exchange (in units of the reduced Planck constant). Since such interactions result in a change in momentum, they can give rise to classical Newtonian forces. In quantum mechanics, physicists often use the terms "force" and "interaction" interchangeably; for example, the weak interaction is sometimes referred to as the "weak force". According to the present understanding, there are four fundamental interactions or forces: gravitation, electromagnetism, the weak interaction, and the strong interaction. Their magnitude and behaviour vary greatly, as described in the table below. Modern physics attempts to explain every observed physical phenomenon by these fundamental interactions. The fundamental interactions can be compared using dimensionless coupling constants that characterize the intensity or "strength" of the interactions.: 102 Properties of fundamental interactions at low energy \| Interaction \| Current theory \| Mediators \| Strength \| Long-distance behavior (potential) \| Range (m) \| \| Weak \| Electroweak theory (EWT) \| W and Z bosons \| 1.027x10-5 \| \| 10−18 \| \| Strong \| Quantum chromodynamics (QCD) \| gluons \| 0.1 (short range), 1.0 (long range) \| (Color confinement, see discussion below) \| 10−15 \| \| Electromagnetic \| Quantum electrodynamics (QED) \| photons \| 1/137 \| (force) \| ∞ \| \| Gravitation \| General relativity(GR) \| gravitons (hypothetical) \| 5.9x10-39 \| (force) \| ∞ \| The modern (perturbative) quantum mechanical view of the fundamental forces other than gravity is that particles of matter (fermions) do not directly interact with each other, but rather carry a charge, and exchange virtual particles (gauge bosons), which are the interaction carriers or force mediators. For example, photons mediate the interaction of electric charges, and gluons mediate the interaction of color charges. The full theory includes perturbations beyond simply fermions exchanging bosons; these additional perturbations can involve bosons that exchange fermions, as well as the creation or destruction of particles: see Feynman diagrams for examples. Interactions [edit] Gravity [edit] Main article: Gravity Gravitation is the weakest of the four interactions at the atomic scale, where electromagnetic interactions dominate. Gravitation is the most important of the four fundamental forces for astronomical objects over astronomical distances for two reasons. First, gravitation has an infinite effective range, like electromagnetism but unlike the strong and weak interactions. Second, gravity always attracts and never repels; in contrast, astronomical bodies tend toward a near-neutral net electric charge, such that the attraction to one type of charge and the repulsion from the opposite charge mostly cancel each other out. Even though electromagnetism is far stronger than gravitation, electrostatic attraction is not relevant for large celestial bodies, such as planets, stars, and galaxies, simply because such bodies contain equal numbers of protons and electrons and so have a net electric charge of zero. Nothing "cancels" gravity, since it is only attractive, unlike electric forces which can be attractive or repulsive. On the other hand, all objects having mass are subject to the gravitational force, which only attracts. Therefore, only gravitation matters on the large-scale structure of the universe. The long range of gravitation makes it responsible for such large-scale phenomena as the structure of galaxies and black holes and, being only attractive, it slows down the expansion of the universe. Gravitation also explains astronomical phenomena on more modest scales, such as planetary orbits, as well as everyday experience: objects fall; heavy objects act as if they were glued to the ground, and animals can only jump so high. Gravitation was the first interaction to be described mathematically. In ancient times, Aristotle hypothesized that objects of different masses fall at different rates. During the Scientific Revolution, Galileo Galilei experimentally determined that this hypothesis was wrong under certain circumstances—neglecting the friction due to air resistance and buoyancy forces if an atmosphere is present (e.g. the case of a dropped air-filled balloon vs a water-filled balloon), all objects accelerate toward the Earth at the same rate. Isaac Newton's law of Universal Gravitation (1687) was a good approximation of the behaviour of gravitation. Present-day understanding of gravitation stems from Einstein's General Theory of Relativity of 1915, a more accurate (especially for cosmological masses and distances) description of gravitation in terms of the geometry of spacetime. Merging general relativity and quantum mechanics (or quantum field theory) into a more general theory of quantum gravity is an area of active research. It is hypothesized that gravitation is mediated by a massless spin-2 particle called the graviton. Although general relativity has been experimentally confirmed (at least for weak fields, i.e. not black holes) on all but the smallest scales, there are alternatives to general relativity. These theories must reduce to general relativity in some limit, and the focus of observational work is to establish limits on what deviations from general relativity are possible. Proposed extra dimensions could explain why the gravity force is so weak. Electroweak interaction [edit] Main article: Electroweak interaction Electromagnetism and weak interaction appear to be very different at everyday low energies. They can be modeled using two different theories. However, above unification energy, on the order of 100 GeV, they would merge into a single electroweak force. The electroweak theory is very important for modern cosmology, particularly on how the universe evolved. This is because shortly after the Big Bang, when the temperature was still above approximately 1015 K, the electromagnetic force and the weak force were still merged as a combined electroweak force. For contributions to the unification of the weak and electromagnetic interaction between elementary particles, Abdus Salam, Sheldon Glashow and Steven Weinberg were awarded the Nobel Prize in Physics in 1979. Electromagnetism [edit] Main article: Electromagnetism Electromagnetism is the force that acts between electrically charged particles. This phenomenon includes the electrostatic force acting between charged particles at rest, and the combined effect of electric and magnetic forces acting between charged particles moving relative to each other. Electromagnetism has an infinite range, as gravity does, but is vastly stronger. It is the force that binds electrons to atoms, and it holds molecules together. It is responsible for everyday phenomena like light, magnets, electricity, and friction. Electromagnetism fundamentally determines all macroscopic, and many atomic-level, properties of the chemical elements. In a four kilogram (~1 gallon) jug of water, there is of total electron charge. Thus, if we place two such jugs a meter apart, the electrons in one of the jugs repel those in the other jug with a force of This force is many times larger than the weight of the planet Earth. The atomic nuclei in one jug also repel those in the other with the same force. However, these repulsive forces are canceled by the attraction of the electrons in jug A with the nuclei in jug B and the attraction of the nuclei in jug A with the electrons in jug B, resulting in no net force. Electromagnetic forces are tremendously stronger than gravity, but tend to cancel out so that for astronomical-scale bodies, gravity dominates. Electrical and magnetic phenomena have been observed since ancient times, but it was only in the 19th century James Clerk Maxwell discovered that electricity and magnetism are two aspects of the same fundamental interaction. By 1864, Maxwell's equations had rigorously quantified this unified interaction. Maxwell's theory, restated using vector calculus, is the classical theory of electromagnetism, suitable for most technological purposes. The constant speed of light in vacuum (customarily denoted with a lowercase letter c) can be derived from Maxwell's equations, which are consistent with the theory of special relativity. Albert Einstein's 1905 theory of special relativity, however, which follows from the observation that the speed of light is constant no matter how fast the observer is moving, showed that the theoretical result implied by Maxwell's equations has profound implications far beyond electromagnetism on the very nature of time and space. In another work that departed from classical electro-magnetism, Einstein also explained the photoelectric effect by utilizing Max Planck's discovery that light was transmitted in 'quanta' of specific energy content based on the frequency, which we now call photons. Starting around 1927, Paul Dirac combined quantum mechanics with the relativistic theory of electromagnetism. Further work in the 1940s, by Richard Feynman, Freeman Dyson, Julian Schwinger, and Sin-Itiro Tomonaga, completed this theory, which is now called quantum electrodynamics, the revised theory of electromagnetism. Quantum electrodynamics and quantum mechanics provide a theoretical basis for electromagnetic behavior such as quantum tunneling, in which a certain percentage of electrically charged particles move in ways that would be impossible under the classical electromagnetic theory, that is necessary for everyday electronic devices such as transistors to function. Weak interaction [edit] Main article: Weak interaction The weak interaction or weak nuclear force is responsible for some nuclear phenomena such as beta decay. Electromagnetism and the weak force are now understood to be two aspects of a unified electroweak interaction — this discovery was the first step toward the unified theory known as the Standard Model. In the theory of the electroweak interaction, the carriers of the weak force are the massive gauge bosons called the W and Z bosons. The weak interaction is the only known interaction that does not conserve parity; it is left–right asymmetric. The weak interaction even violates CP symmetry but does conserve CPT. Strong interaction [edit] Main article: Strong interaction The strong interaction, or strong nuclear force, is the most complicated interaction, mainly because of the way it varies with distance. The nuclear force is powerfully attractive between nucleons at distances of about 1 femtometre (fm, or 10−15 metres), but it rapidly decreases to insignificance at distances beyond about 2.5 fm. At distances less than 0.7 fm, the nuclear force becomes repulsive. This repulsive component is responsible for the physical size of nuclei, since the nucleons can come no closer than the force allows. After the nucleus was discovered in 1908, it was clear that a new force, today known as the nuclear force, was needed to overcome the electrostatic repulsion, a manifestation of electromagnetism, of the positively charged protons. Otherwise, the nucleus could not exist. Moreover, the force had to be strong enough to squeeze the protons into a volume whose diameter is about 10−15 m, much smaller than that of the entire atom. From the short range of this force, Hideki Yukawa predicted that it was associated with a massive force particle, whose mass is approximately 100 MeV. The 1947 discovery of the pion ushered in the modern era of particle physics. Hundreds of hadrons were discovered from the 1940s to 1960s, and an extremely complicated theory of hadrons as strongly interacting particles was developed. Most notably: The pions were understood to be oscillations of vacuum condensates; Jun John Sakurai proposed the rho and omega vector bosons to be force carrying particles for approximate symmetries of isospin and hypercharge; Geoffrey Chew, Edward K. Burdett and Steven Frautschi grouped the heavier hadrons into families that could be understood as vibrational and rotational excitations of strings. While each of these approaches offered insights, no approach led directly to a fundamental theory. Murray Gell-Mann along with George Zweig first proposed fractionally charged quarks in 1961. Throughout the 1960s, different authors considered theories similar to the modern fundamental theory of quantum chromodynamics (QCD) as simple models for the interactions of quarks. The first to hypothesize the gluons of QCD were Moo-Young Han and Yoichiro Nambu, who introduced the quark color charge. Han and Nambu hypothesized that it might be associated with a force-carrying field. At that time, however, it was difficult to see how such a model could permanently confine quarks. Han and Nambu also assigned each quark color an integer electrical charge, so that the quarks were fractionally charged only on average, and they did not expect the quarks in their model to be permanently confined. In 1971, Murray Gell-Mann and Harald Fritzsch proposed that the Han/Nambu color gauge field was the correct theory of the short-distance interactions of fractionally charged quarks. A little later, David Gross, Frank Wilczek, and David Politzer discovered that this theory had the property of asymptotic freedom, allowing them to make contact with experimental evidence. They concluded that QCD was the complete theory of the strong interactions, correct at all distance scales. The discovery of asymptotic freedom led most physicists to accept QCD since it became clear that even the long-distance properties of the strong interactions could be consistent with experiment if the quarks are permanently confined: the strong force increases indefinitely with distance, trapping quarks inside the hadrons. Assuming that quarks are confined, Mikhail Shifman, Arkady Vainshtein and Valentine Zakharov were able to compute the properties of many low-lying hadrons directly from QCD, with only a few extra parameters to describe the vacuum. In 1980, Kenneth G. Wilson published computer calculations based on the first principles of QCD, establishing, to a level of confidence tantamount to certainty, that QCD will confine quarks. Since then, QCD has been the established theory of strong interactions. QCD is a theory of fractionally charged quarks interacting by means of 8 bosonic particles called gluons. The gluons also interact with each other, not just with the quarks, and at long distances the lines of force collimate into strings, loosely modeled by a linear potential, a constant attractive force. In this way, the mathematical theory of QCD not only explains how quarks interact over short distances but also the string-like behavior, discovered by Chew and Frautschi, which they manifest over longer distances. Higgs interaction [edit] Conventionally, the Higgs interaction is not counted among the four fundamental forces. Nonetheless, although not a gauge interaction nor generated by any diffeomorphism symmetry, the Higgs field's cubic Yukawa coupling produces a weakly attractive fifth interaction. After spontaneous symmetry breaking via the Higgs mechanism, Yukawa terms remain of the form : , with Yukawa coupling , particle mass (in eV), and Higgs vacuum expectation value 246.22 GeV. Hence coupled particles can exchange a virtual Higgs boson, yielding classical potentials of the form : , with Higgs mass 125.18 GeV. Because the reduced Compton wavelength of the Higgs boson is so small (1.576×10−18 m, comparable to the W and Z bosons), this potential has an effective range of a few attometers. Between two electrons, it begins roughly 1011 times weaker than the weak interaction, and grows exponentially weaker at non-zero distances. Unification [edit] The fundamental forces may become unified into a single force at very high energies and on a minuscule scale, the Planck scale. Particle accelerators cannot produce the enormous energies required to experimentally probe this regime. The weak and electromagnetic forces have already been unified with the electroweak theory of Sheldon Glashow, Abdus Salam, and Steven Weinberg, for which they received the 1979 Nobel Prize in physics. Numerous theoretical efforts have been made to systematize the existing four fundamental interactions on the model of electroweak unification. Grand Unified Theories (GUTs) are proposals to show that each of the three fundamental interactions described by the Standard Model is a different manifestation of a single interaction with symmetries that break down and create separate interactions below some extremely high level of energy. GUTs are also expected to predict some of the relationships between constants of nature that the Standard Model treats as unrelated and gauge coupling unification for the relative strengths of the electromagnetic, weak, and strong forces. (GUT). Some attempts at GUTs hypothesize "shadow" particles, such that every known matter particle associates with an undiscovered force particle, and vice versa, altogether supersymmetry (SUSY). Other theorists seek to quantize the gravitational field by the modelling behaviour of its hypothetical force carrier, the graviton and achieve quantum gravity (QG). One approach to QG is loop quantum gravity (LQG). Still other theorists seek both QG and GUT within one framework, reducing all four fundamental interactions to a Theory of Everything (ToE). The most prevalent aim at a ToE is string theory, although to model matter particles, it added SUSY to force particles—and so, strictly speaking, became superstring theory. Multiple, seemingly disparate superstring theories were unified on a backbone, M-theory. Theories beyond the Standard Model remain highly speculative, lacking great experimental support. A so-called theory of everything, which would integrate GUTs with a quantum gravity theory, faces a greater barrier because no quantum gravity theory (e.g., string theory, loop quantum gravity, and twistor theory) has secured wide acceptance. Some theories look for a graviton to complete the Standard Model list of force-carrying particles, while others, like loop quantum gravity, emphasize the possibility that time-space itself may have a quantum aspect to it. Beyond the Standard Model [edit] Main article: Physics beyond the Standard Model See also: Elementary particle § Beyond the Standard Model Some theories beyond the Standard Model include a hypothetical fifth force, and the search for such a force is an ongoing line of experimental physics research. In supersymmetric theories, some particles, known as moduli, acquire their masses only through supersymmetry breaking effects and can mediate new forces. Another reason to look for new forces is the discovery that the expansion of the universe is accelerating (also known as dark energy), creating a need to explain a nonzero cosmological constant and possibly other modifications of general relativity. Fifth forces have also been suggested to explain phenomena such as CP violations, dark matter, and dark flow. See also [edit] Physics portal Quintessence, a hypothesized fifth force Gerardus 't Hooft Edward Witten Howard Georgi References [edit] ^ a b c d e Braibant, Sylvie; Giacomelli, Giorgio; Spurio, Maurizio (2011). Particles and Fundamental Interactions: An Introduction to Particle Physics (illustrated ed.). Springer Science & Business Media. p. 109. ISBN 9789400724631. Extract of page 109 ^ "The Standard Model of Particle Physics \| symmetry magazine". www.symmetrymagazine.org. Retrieved 2018-10-30. ^ "Newton's Laws of Motion". www.grc.nasa.gov. NASA. ^ "Newton's law of gravitation \| Definition, Formula, & Facts". Encyclopedia Britannica. Retrieved 22 March 2021. ^ Nauenberg, Michael (October 2018). "Newton's graphical method for central force orbits". American Journal of Physics. 86 (10): 765–771. Bibcode:2018AmJPh..86..765N. doi:10.1119/1.5050620. S2CID 125197336. ^ Newton's absolute space was a medium, but not one transmitting gravitation. ^ Henry, John (March 2011). "Gravity and De gravitatione: the development of Newton's ideas on action at a distance" (PDF). Studies in History and Philosophy of Science Part A. 42 (1): 11–27. Bibcode:2011SHPSA..42...11H. doi:10.1016/j.shpsa.2010.11.025. hdl:20.500.11820/b84d5f3c-47b3-453a-849f-eb9add123210. ^ Faraday, Michael (2012). "Experimental Researches in Electricity". doi:10.1017/cbo9781139383165.018. {{cite journal}}: Cite journal requires \|journal= (help) ^ Goldin, Gerald A.; Shtelen, Vladimir M. (February 2001). "On Galilean invariance and nonlinearity in electrodynamics and quantum mechanics". Physics Letters A. 279 (5–6): 321–326. arXiv:quant-ph/0006067. Bibcode:2001PhLA..279..321G. doi:10.1016/S0375-9601(01)00017-2. S2CID 5398578. no fully Galilean-covariant theory of a coupled Schrödinger-Maxwell system (where the density and current of the Schrödinger field act as source of the nonrelativistic Maxwell field) is possible ^ Farhoudi, Mehrdad; Yousefian, Maysam (May 2016). "Ether and Relativity". International Journal of Theoretical Physics. 55 (5): 2436–2454. arXiv:1511.07795. Bibcode:2016IJTP...55.2436F. doi:10.1007/s10773-015-2881-y. S2CID 119258859. ^ Meinard Kuhlmann, "Physicists debate whether the world is made of particles or fields—or something else entirely", Scientific American, 24 Jul 2013. ^ Uzan, Jean-Philippe (2003-04-07). "The fundamental constants and their variation: observational and theoretical status". Reviews of Modern Physics. 75 (2): 403–455. arXiv:hep-ph/0205340. doi:10.1103/RevModPhys.75.403. ISSN 0034-6861. ^ Siegel, Ethan (2016). "What Is The Strongest Force In The Universe?". Starts With a Bang. Retrieved 22 March 2021. ^ CERN (20 January 2012). "Extra dimensions, gravitons, and tiny black holes". ^ Bais, Sander (2005), The Equations. Icons of knowledge, ISBN 978-0-674-01967-6 p.84 ^ "The Nobel Prize in Physics 1979". The Nobel Foundation. Retrieved 2008-12-16. ^ "fundamental force \| Definition, List, & Facts". Encyclopedia Britannica. Retrieved 22 March 2021. ^ "The Standard Model". CERN. Retrieved 22 March 2021. ^ Shivni, Rashmi (2016-05-16). "The Planck scale". symmetry magazine. Fermilab/SLAC. Retrieved 2018-10-30. ^ "The Nobel Prize in Physics 1979". NobelPrize.org. Retrieved 2018-10-30. ^ "The Nobel Prize in Physics 1979". NobelPrize.org. Retrieved 2018-10-30. ^ "The Nobel Prize in Physics 1979". NobelPrize.org. Retrieved 2018-10-30. ^ Krauss, Lawrence M. (2017-03-16). "A Brief History of the Grand Unified Theory of Physics". Nautilus. Bibliography [edit] Davies, Paul (1986), The Forces of Nature, Cambridge Univ. Press 2nd ed. Feynman, Richard (1967), The Character of Physical Law, MIT Press, ISBN 978-0-262-56003-0 Schumm, Bruce A. (2004), Deep Down Things, Johns Hopkins University Press While all interactions are discussed, discussion is especially thorough on the weak. Weinberg, Steven (1993), The First Three Minutes: A Modern View of the Origin of the Universe, Basic Books, ISBN 978-0-465-02437-7 Weinberg, Steven (1994), Dreams of a Final Theory, Basic Books, ISBN 978-0-679-74408-5 Padmanabhan, T. (1998), After The First Three Minutes: The Story of Our Universe, Cambridge Univ. Press, ISBN 978-0-521-62972-0 Perkins, Donald H. (2000), Introduction to High Energy Physics (4th ed.), Cambridge Univ. Press, ISBN 978-0-521-62196-0 Riazuddin (December 29, 2009). "Non-standard interactions" (PDF). NCP 5th Particle Physics Sypnoisis. 1 (1): 1–25. Archived from the original (PDF) on March 3, 2016. 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190045	https://leetcode.com/problems/expression-add-operators/	Expression Add Operators - LeetCode Problem List Problem List Debugging... Submit 00:00:00 RegisterorLog in Premium Description Description Editorial Editorial Solutions Solutions Submissions Submissions Code Code Testcase Testcase Test Result Test Result 282. Expression Add Operators Hard Topics Companies Hint Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators'+','-', and/or''between the digits ofnumso that the resultant expression evaluates to thetargetvalue. Note that operands in the returned expressions should not contain leading zeros. Note that a number can contain multiple digits. Example 1: Input: num = "123", target = 6 Output: ["123","1+2+3"] Explanation: Both "123" and "1+2+3" evaluate to 6. Example 2: Input: num = "232", target = 8 Output: ["23+2","2+32"] Explanation: Both "23+2" and "2+32" evaluate to 8. Example 3: Input: num = "3456237490", target = 9191 Output: [] Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191. Constraints: 1 <= num.length <= 10 num consists of only digits. -231 <= target <= 231 - 1 Accepted 289,693/688.8K Acceptance Rate 42.1% Topics MathStringBacktracking Companies Hint 1 Note that a number can contain multiple digits. Hint 2 Since the question asks us to find all of the valid expressions, we need a way to iterate over all of them. (Hint: Recursion!) Hint 3 We can keep track of the expression string and evaluate it at the very end. But that would take a lot of time. Can we keep track of the expression's value as well so as to avoid the evaluation at the very end of recursion? Hint 4 Think carefully about the multiply operator. It has a higher precedence than the addition and subtraction operators. 1 + 2 = 3 1 + 2 - 4 --> 3 - 4 --> -1 1 + 2 - 4 12 --> -1 12 --> -12 (WRONG!) 1 + 2 - 4 12 --> -1 - (-4) + (-4 12) --> 3 + (-48) --> -45 (CORRECT!) Hint 5 We simply need to keep track of the last operand in our expression and reverse it's effect on the expression's value while considering the multiply operator. Similar Questions Evaluate Reverse Polish Notation Medium Basic Calculator Hard Basic Calculator II Medium Different Ways to Add Parentheses Medium Target Sum Medium Discussion (76) Choose a type Comment 💡 Discussion Rules Please don't post any solutions in this discussion. The problem discussion is for asking questions about the problem or for sharing tips - anything except for solutions. If you'd like to share your solution for feedback and ideas, please head to the solutions tab and post it there. Sort by:Best No comments yet. 1 2 3 4 8 Copyright © 2025 LeetCode. All rights reserved. 3.7K 76 70 Online C++ Auto 1 2 3 4 5 6 class Solution{ public: vectoraddOperators(string num,int target){ } }; Saved Ln 1, Col 1 You need to log in / sign up to run or submit Case 1 Case 2 Case 3 num = "123" target = 6 9 1 2 3 4 5 6 › "123" 6 "232" 8 "3456237490" 9191 Source FindHeaderBarSize FindTabBarSize FindBorderBarSize Layouts Hints Apply Default Upgrade to premium to unlock this layout Subscribe Note-taking Upgrade to premium to unlock this layout Subscribe Debug Upgrade to turn on custom layouts Subscribe 🧘 Focus Mode NEW
190046	https://www.youtube.com/watch?v=gEhxBFQCYic	Sum and Difference Identities of Sine and Cosine The Organic Chemistry Tutor 9680000 subscribers 2141 likes Description 144417 views Posted: 31 Jan 2021 This trigonometry video tutorial explains how to find the exact of trigonometric expressions with angles in radians and degrees using the sum and difference identities & formulas of sine and cosine. This lesson contains plenty of examples and practice problems that will help you to get a thorough understanding of this material. Trigonometry - Free Formula Sheet: Full 1 Hour Video on YouTube: Get The Full 1 Hour Video on Patreon: Direct Link to The Full Video: 79 comments Transcript: now let's talk about the sum and difference formulas that you need to know the first one is this one cosine alpha plus or minus beta is equal to cosine alpha times cosine beta minus plus sine alpha sine beta now this equation is a way to write two equations let's say if we have alpha plus beta you need to use the negative sign so whatever sign is here on top that's the one you should use so therefore the first equation is cosine alpha plus beta is cosine alpha times cosine beta minus sine alpha sine beta so this sine is always opposite to this sign now let's say if you have the sine on the bottom you should use the sine on the bottom so therefore cosine alpha minus beta is equal to cosine alpha cosine beta plus sine alpha sine beta so make sure you understand that next we have the formula for sine sine alpha plus or minus beta that's equal to sine alpha cosine beta plus or minus cosine alpha sine beta so notice that the sine doesn't change so if you use the positive sign on top then you should use the positive sign on top as well on the right so therefore sine alpha plus beta is equal to sine alpha cosine beta plus cosine alpha sine beta and so if this was a minus sign then you would change it to a minus sign now the only other form we need to know is tangent tangent alpha plus or minus beta is equal to tangent alpha plus or minus tan beta divided by one minus plus tan alpha tan beta so if we want to use the bottom sign let's say if it's negative here then it's going to be negative on top and positive on the bottom so just keep that in mind now let's say if we want to use the upper sign let's say if it's alpha plus beta so if we're using the positive sign we need to use the upper sign in both cases so tangent alpha plus beta for example will be tangent alpha plus tangent beta divided by one minus tangent alpha times tangent beta so make sure you know these formulas now how can we use the sum and difference formulas to evaluate sine of 15 degrees so first we need to find two angles that either add or subtract to 15. two common angles that we know on a unit circle are 45 and 30. 45 minus 30 is 15. so we're going to use the formula sine alpha minus beta is equal to sine alpha cosine beta minus cosine alpha sine beta so as we can see alpha is 45 and beta is 30. so sine 45 minus 30 i'm gonna put it over here that's equal to sine 45 cosine 30 minus cosine 45 times sine 30. now we know that sine 45 is the square root of 2 over 2. cosine 30 is the square root of 3 over 2. cosine 45 is the same as sine 45 and sine 30 is one half now let's multiply the square root of two times the square root of three and that's equal to the square root of six and two times two is four and here we have the square root of two times one which is simply square root two over four so we can combine this as a single fraction if we want to so square root 6 minus square root 2 divided by 4 is equal to sine of 15 degrees now let's try another example sine 105 so go ahead and use the sum and difference formula to find the exact value of sine 105. now the first thing we need to do is find two common angles on a unit circle that adds up to 105. so this is 60 plus 45. so this time we're going to use the formula sine alpha plus beta is equal to sine alpha cosine beta plus cosine alpha times sine beta so alpha 60 beta is 45 so this is going to be sine 60 cosine 45 plus cosine 60 times sine 45 sine sixty is equal to the square root of three over two cosine forty five is the square root of two over two cosine sixty is one half and sine forty 45 is equal to the square root of two over two so now what we have is the square root of six over four plus the square root of two over four so the final answer is root six plus root two over four and so that is equal to sine of 105 degrees so that's how you can use the sum and difference formulas to evaluate certain trigonometric expressions here's another one that we can try cosine two pi divided by three minus pi over six so use the sum of difference formulas of cosine to evaluate this one so we need to use the difference one alpha minus beta so this is equal to cosine alpha times cosine beta and because we have a minus here we need to change the sign so it's going to be plus sine alpha sine beta now we can see that alpha is 2 pi over 3 and beta is pi over 6 so this is equal to cosine 2 pi divided by 3 times cosine pi over 6 plus sine of two pi over three times sine pi over six pi over six is thirty degrees two pi over three that's one twenty cosine 120 that's in quadrant two so it's negative that's equal to negative one half the reference angle is pi over three which is sixty cosine of sixty is one half cosine thirty is root 3 over 2. sine 120 is root 3 over 2 and sine 30 is simply one half so this is negative square root 3 over 4 plus positive square root three over four which really is zero and it makes sense this is one twenty minus thirty so that's cosine of at 90 degrees on the y axis the x value is zero and so that's it for this problem let's work on this example cosine 5 pi over 12. go ahead and evaluate now 5 pi over 12 is not on the unit circle so we need to use the sum and difference formulas of cosine but first let's convert 5 pi over 12 into degrees so let's multiply by 180 divided by pi so we can cancel pi 180 is 18 times 10 and 12 is 6 times 2. i'm going to break down 18 into 6 times 3 and 10 i'm going to write it as 5 times 2. so we could cancel with 6 and we could cancel a 2. five times five is twenty-five twenty-five times three is seventy-five so this is the same as cosine seventy-five degrees so we need to use this formula cosine alpha plus beta is equal to cosine alpha cosine beta and then since we have a plus sign we're going to flip it and make it minus so it's minus sine alpha times sine beta two angles that add up to 75 are 45 and 30. so this is equal to cosine 45 cosine 30 minus sine 45 times sine 30. so cosine 45 that's equal to the square root of 2 divided by 2. cosine 30 is the square root of 3 over 2. sine 45 that's root 2 over 2 and sine 30 is one half so this is going to be equal to the square root of 6 over 4 minus the square root of 2 over 4 which is root 6 minus root 2 over 4. so this is the final answer cosine 75 or cosine 5 pi over 12 is equal to that find the exact value of the expression cosine of 70 degrees multiplied by cosine of 40 plus sine 70 times sine 40. so we can't just type it in the calculator and we don't know the value of cosine 70. so what should we do in this case you need to identify the proper formula so we have cosine times cosine plus sine times sine this is associated with the sum and difference formulas of cosine now is it cosine alpha plus beta or cosine alpha minus beta with cosine you got to change the sign so we need to use this formula cosine alpha minus beta is cosine alpha cosine beta plus sine alpha sine beta so we can see that alpha is equal to 70 degrees and beta is 40 degrees so we can't evaluate the right side the right side is what we have here but we can use the left side of the equation to get the answer so therefore cosine 70 cosine 40 plus sine 70 times sine 40 all of that is equal to cosine alpha minus beta or cosine 70 minus 40. 70 minus 40 is 30. so what we need to do is find the value of cosine 30 and cosine 30 is equal to the square root of 3 divided by 2 so this is the final answer now let's work on some more examples let's say if we have cosine 20 times cosine 40 minus sine 20 times sine 40. whenever you see two cosines together and two sines together you know you're dealing with the sum and difference formulas of cosine so notice that we have a negative sign with cosine you got to reverse it so this expression is equal to cosine alpha plus beta where alpha is 20 and beta is 40. 20 plus 40 and 40 plus 20 they both equal 60. so you just got to realize one of them is 20 one of them is 40. the order really doesn't matter so cosine 20 plus 40 that's equal to cosine of 60 degrees and cosine 60 is one-half so that's the value of this entire expression it's equal to one-half here's another one sine of 110 multiplied by cosine 25 plus cosine 110 times sine 25 go ahead and evaluate that expression so we need to use this formula sine alpha cosine beta plus cosine alpha sine beta so that is equal to the sine formula and when dealing with sine the sine doesn't change so it's sine alpha plus beta so we can see that alpha is 110 and beta is 25 so then this is going to be sine 110 plus 25 which is sine of 135 135 is in quadrant two and it produces a reference angle of 45 degrees sine 45 is the square root of two over two 1 35 has the same value and it's positive in quadrant two so this is going to be equal to the square root of two divided by two and so that's the final answer for this entire expression let's try this one sine 5 pi over 12 multiplied by cosine pi over 12 minus cosine 5 pi over 12 times sine pi divided by 12. so whenever you see sine and cosine you're dealing with the sum and difference formula of sine and since we have a minus sign it's going to be sine alpha minus beta so when you're dealing with a subtraction the first one that you see here is alpha because the formula is sine alpha cosine beta minus cosine alpha sine beta so you can clearly see that alpha is 5 pi over 12. beta is pi divided by 12. so then this becomes sine 5 pi over 12 minus 1 pi over 12. 5 minus 1 is 4. so now we have sine 4 pi over 12. now we can reduce 4 pi over 12. 12 is 4 times 3 so we could cancel a 4. so therefore we're looking for the value of sine pi divided by 3. pi over 3 is 60. sine of 60 degrees is equal to the square root of 3 divided by 2. and so that's the final answer of the original expression what is the value of tangent 15 degrees so how can we get the answer for this problem so first we need to find two angles that either add or subtract to fifteen so let's use sixty minus forty five now we need the right formula tangent alpha minus beta is equal to tangent alpha minus tangent beta divided by one plus tan alpha tan beta so whatever sign you see here this sign will be the same but this sign here will change alpha is 60. beta is 45 so therefore tangent 15 which is the same as tangent 60 minus 45 that's equal to tangent of 60 degrees minus tangent of forty five degrees over one plus tangent sixty times tangent forty five now what is the value of tangent 60 tangent 60 is equal to the square root of 3. and tangent forty five is one and then we have one plus tan sixty times tan forty five so this is the square root of three minus one over one plus the square root of three so what can we do to simplify that expression whenever you have a radical on the bottom you need to get rid of it so what we need to do is multiply by the conjugate of 1 plus square root 3 which is 1 minus square root 3. so let's foil here we have the square root of 3 times 1 and that's simply square root 3 and then the square root of 3 times itself that's the square root of 9 which is 3. and then you have the negative sign next we have negative 1 times 1 that's simply negative 1 and then negative 1 times negative square root 3 which is positive square root of 3. on the bottom first we have 1 times 1 which is 1 and then 1 times negative square root 3 and then positive square root of three times one and then square root three times negative square root three which is negative square root nine in that string on the bottom these two terms will cancel on top we can add the square root of three plus the square root of three which is two square root three negative three minus one is negative four one minus three that's negative 2. so now what we're going to do is we're going to divide 2 square root 3 by negative 2 and then we'll divide negative four by negative two two divided by negative two is negative one so this is negative square root of three negative four divided by negative 2 that's positive 2. so we can rewrite this as 2 minus square root 3. and so that is the final answer that's the value of tangent of 15 degrees you
190047	https://www.youtube.com/watch?v=N9a2r01ToZk	Introduction to solubility equilibria \| Equilibrium \| AP Chemistry \| Khan Academy Khan Academy 9090000 subscribers 914 likes Description 84764 views Posted: 8 Jul 2021 The solubility product constant, Kₛₚ, is an equilibrium constant that reflects the extent to which an ionic compound dissolves in water. For compounds that dissolve to produce the same number of ions, we can directly compare their Kₛₚ values to determine their relative solubilities. If we know the solubility of a salt, we can use this information to calculate the Kₛₚ value for the compound. View more lessons or practice this subject at Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. 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Donate here: Volunteer here: 10 comments Transcript: let's say we have a beaker of distilled water at 25 degrees celsius and to the beaker we add some barium sulfate barium sulfate is a white solid a small amount of the barium sulfate dissolves in the water and forms barium 2 plus ions in solution and sulfate ions in solution so let me draw those in on our diagram so we're going to form some barium 2 plus ions and some sulfate anions but most of the barium sulfate remains undissolved and with that so we'll draw that here sitting on the bottom of the beaker so barium sulfate can dissolve to form barium two plus ions and sulfate anions in solution and it's possible for the barium two plus ion to combine with the sulfate anion to form a precipitate of barium sulfate when the rate of dissolution is equal to the rate of precipitation the system is at equilibrium these types of equilibria are referred to as solubility equilibria and when the system is at equilibrium the concentrations of barium 2 plus ion and sulfate anion in solution are constant and the amount of solid is constant too and this forms a saturated solution the balanced equation shows the dissolution of a salt barium sulfate and from the balanced equation we can write an equilibrium constant expression so we would write the equilibrium constant k is equal to the concentration of barium two plus and since there's a coefficient of one in the balanced equation it'd be the concentration raised to the first power times the concentration of sulfate also raised to the first power and since pure solids are left out of equilibrium constant expressions we would not include the solid barium sulfate for solubility equilibria we would write ksp where sp stands for solubility product the solubility product constant ksp has only one value for a given salt at a specific temperature that temperature is usually 25 degrees celsius and ksp indicates how much of that salt will dissolve for example at 25 degrees celsius the ksp value for barium sulfate is 1.1 times 10 to the negative 10th when the ksp value is much less than 1 that indicates the salt is not very soluble so barium sulfate is not a soluble salt if the ksp value is greater than one like it is for something like sodium chloride that indicates a soluble salt that dissolves easily in water the solubility of a substance refers to the amount of solid that dissolves to form a saturated solution usually the units for solubility are in grams per liter molar solubility refers to the number of moles of the solid that dissolve to form one liter of the saturated solution and therefore the units would be moles per one liter or you could just write molar ksp values can be used to predict the relative solubilities of salts that produce the same number of ions in solution for example silver chloride silver bromide and silver iodide all produce two ions in solution let's look at the dissolution equation for silver chloride to see why this is true solid silver chloride turns into a g plus and cl minus so that's one ag plus ion and one cl minus ion for a total of two ions in solution and we could write out similar equations for silver bromide and silver iodide so they all produce two ions in solution however a salt like lead to chloride produces three ions in solution so lead ii chloride would give one pb2 plus ion and two chloride anions in solution one plus two is three ions since lead two chloride produces threons three ions in solution we can't determine its solubility relative to the other three by comparing ksp values here are the ksp values for the three salts at 25 degrees celsius for silver chloride it's 1.8 times 10 to the negative 10th for silver bromide it's 5.0 times 10 to the negative 13th and for silver iodide it's 8.3 times 10 to the negative 17th when comparing salts that produce the same number of ions the higher the value of ksp the higher the solubility of the salt and since silver chloride has the highest ksp value of these three silver chloride is the most soluble salt for some insight into why this is true let's look at the ksp expression for silver chloride which we can get from the balanced equation the higher the value for ksp the higher the concentration of these ions at equilibrium which means that more of the solid must have dissolved therefore silver chloride has the highest solubility out of these three salts let's say we have some solid calcium fluoride that we add to pure water at 25 degrees celsius eventually equilibrium is reached and the equilibrium concentration of calcium 2 plus ions is measured to be 2.1 times 10 to the negative fourth molar our goal is to calculate the ksp for calcium fluoride at 25 degrees celsius the first step is to write out the dissolution equation for calcium fluoride so we would write caf2 solid and we know that calcium forms a two plus cation so we would write ca2 plus an aqueous solution and to balance everything we'd need two fluoride anions so two f minus also in aqueous solution the next step is to use the balanced equation to write the ksp expression so ksp is equal to there's a one as a coefficient in front of ca2 plus it would be the concentration of ca2 plus raised to the first power times the concentration of fluoride anion and since there's a 2 as a coefficient this is the concentration of fluoride anion squared for a ksp expression these are equilibrium concentrations and we already know the concentration of calcium 2 plus at equilibrium is 2.1 times 10 to the negative fourth so that can be plugged in for the equilibrium concentration of ca2 plus so here's our expression with 2.1 times 10 to the to the negative fourth plugged in and next we need to plug in the equilibrium concentration of fluoride anion well looking at the dissolution equation the mole ratio of calcium to plus to fluoride anion is one to two therefore at equilibrium there's twice as many fluoride ions in solution as there are calcium two plus ions therefore the equilibrium concentration of the fluoride anion would just be twice this concentration for ca2 plus so the equilibrium concentration of fluoride anion must be 4.2 times 10 to the negative fourth molar and when you do the math you get that ksp for calcium fluoride is equal to 3.7 times 10 to the negative 11th at 25 degrees celsius ksp values can be difficult to measure and therefore different sources often give different values for ksp at the same temperature for example for calcium fluoride at 25 degrees celsius one source had ksp equal to 3.5 times 10 to the negative 11th another one had 3.9 times 10 to the negative 11th since we got 3.7 times 10 to the negative 11th this sounds like a pretty good calculation for the numbers that we used for our problem finally let's think about the molar solubility of calcium fluoride so how many moles of our salt dissolve to form one liter of our saturated solution well the mole ratio of calcium two plus ions to calcium fluoride is one to one therefore the concentration of calcium two plus ions in solution 2.1 times 10 to the negative fourth molar that number must also be the molar solubility of calcium fluoride therefore for this problem we could say that we use the molar solubility of calcium fluoride to calculate the ksp value for calcium fluoride
190048	https://www.ncbi.nlm.nih.gov/books/NBK519545/	Neuroanatomy, Middle Meningeal Arteries - StatPearls - NCBI Bookshelf 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Neuroanatomy, Middle Meningeal Arteries Adriana L. Natali; Vamsi Reddy; Jonathan T. Leo. Author Information and Affiliations Authors Adriana L. Natali 1; Vamsi Reddy 2; Jonathan T. Leo 3. Affiliations 1 Lincoln Memorial Un - DeBusk College OM 2 McKinsey & Company 3 Lincoln Memorial University Last Update: September 10, 2024. Go to: Introduction The middle meningeal artery is a vital artery that plays an important clinical role. The middle meningeal artery (MMA) normally branches off the maxillary artery, which is an extension of the external carotid artery. The artery will then travel through the foramen spinosum, which is posterolateral from the foramen ovale, to supply blood to the dura mater. The middle meningeal artery arises from a complex embryological origin, which gives rise to many anatomic variations of the artery. Awareness of these anatomic variations becomes important for surgeons to reduce the risk of complications during surgical repair. The middle meningeal artery has a number of clinical implications. Rupture of the artery, which most commonly occurs at the pterion, typically leads to an epidural hematoma (see Image. Epidural Hematoma After Trauma). The resulting hematoma is described as a “lens-shaped” mass on a computed tomogram (CT) scan. Damage to the middle meningeal artery may also result in an aneurysm or arteriovenous fistulas. Due to the middle meningeal artery's attachment to the pain-sensitive dura mater, the artery also plays a role in migraine headaches. Go to: Structure and Function The middle meningeal artery most often branches off the maxillary artery and courses into the middle cranial fossa via the foramen spinosum (see Image. Middle Meningeal Artery). The middle meningeal artery provides blood to the dura mater through and through its branching arteries also supplies the periosteum of the inner aspects of the cranial bones.As the middle meningeal artery enters the dura mater, it follows a fixed course as it embeds into the groove of the inner skull face.Due to the middle meningeal artery's close contact to the inner skull, trauma to the lateral skull may result in a rupture. The consequences of rupturing the middle meningeal artery will be further explained in the clinical significance section. Go to: Embryology The development of the middle meningeal artery, similarly to other blood vessels, occurs via angiogenesis and receives guidance from vascular endothelial growth factor (VEGF) and other growth factors. During development, a series of aortic arches arise, arranged from cranial to caudal. From these arches, the third arch gives rise to the stapedial artery.The stapedial artery arises from the internal carotid artery and is later incorporated by the external carotid artery. The stapedial artery will then divide, giving rise to the middle meningeal artery. By the 10th week of development, the stapedial artery will normally degenerate. Due to the complex nature of middle meningeal artery development, many anatomic variations and anastomoses may be seen. Go to: Physiologic Variants While the middle meningeal artery primarily is derived as a branch from the maxillary artery and then passes through the foramen spinosum into the middle cranial fossa, rare anatomical variations have been observed. It is important to be aware of the possible variations of the middle meningeal artery, to reduce the risk of thromboembolism during the operative treatment of a lesion in the area. In cases where the foramen spinosum has been absent, the middle meningeal artery enters the cranial fossa through the foramen ovale alongside the mandibular nerve. The middle meningeal artery may also originate from the lateral aspect of the internal carotid artery. In this case, it will travel in the carotid canal, along with the internal carotid artery, and enter the skull through the foramen lacerum where it will then take a normal path and divide into anterior and posterior branches. The middle meningeal artery has also rarely originated from the posterior cerebellar artery, basilar artery, ascending pharyngeal artery, or ophthalmic artery.Due to the potential for variation in the middle meningeal artery, imaging should be acquired before undergoing invasive treatment. Go to: Surgical Considerations Tears of the middle meningeal artery resulting in a significant epidural hematoma are often treated surgically with overwhelming success in non-comatose patients.Studies are currently assessing the efficacy of treating small epidural hematomas non-operatively, but surgical intervention, especially in the cases of large acute epidural hematoma, is still the main treatment. Traumatic pseudoaneurysms are at high risk of secondary rupture and should be surgically managed without delay.True aneurysms of the middle meningeal artery are treated with open surgery and endovascular embolization.When treating any lesion or aneurysm of the middle meningeal artery surgically, surgeons must be aware of the variations in the artery’s course and branching to reduce the risk of complications during the operation. Go to: Clinical Significance Epidural Hematoma The meningeal vessels closely adhere to the internal aspect of the boney vault of the skull. As a result, injury to the boney vault of the skull can result in the rupture of the artery. Injury to the lateral aspect of the skull, via trauma to the pterion, may result in rupture of the vessel. Most commonly (85%), this type of injury to the middle meningeal artery will result in an epidural hematoma.Symptomatic epidural hematoma classically presents with a loss of consciousness, followed by a lucid interval, and ending with rapid deterioration in neurological status.Small epidural hematomas without mass effect may be treated conservatively,but early surgical intervention is common in symptomatic patients. Arteriovenous Fistula Arteriovenous fistulas (AVFs) are another clinical complication of the middle meningeal artery. Anatomically, the middle meningeal artery runs along with paired veins, and a traumatic tear in the skull wall may result in a traumatic AVF.The incidence of traumatic middle meningeal AVF is relatively high, but clinical complications are not seen regularly. For this reason, it is believed that many traumatic AVFs resolve spontaneously. Cases of longstanding AVF, however, have predisposed patients to complications ranging from venous congestion to intracranial hemorrhage.A middle meningeal AVF may occur without trauma, such as in the cases of craniotomy or endovascular injuries during interventional manipulation. Clinical symptoms are often seen in iatrogenic AVFs, causing intervention to be needed. Aneurysm Uncommonly, aneurysms of the middle meningeal artery may be seen. Both true aneurysms and pseudoaneurysms may be observed in the middle meningeal artery, occurring under different conditions. Pseudoaneurysms are most commonly associated with trauma to the skull and are responsible for the lucid interval before a delayed bleed.A fracture in the temporal region may cause a tear in the arterial wall. The tear is then blocked with a clot, leading to the formation of a false lumen. In contrast, true aneurysms are generally associated with increased hemodynamic stress or a pathological process affecting the middle meningeal artery.Both types of aneurysms are particularly prone to rupture, which often leads to an intracranial hemorrhage. Due to the risk of rupture, surgical resection and endovascular embolization are the most common treatment modalities. Migraine The middle meningeal artery also plays a role in the onset of migraine headaches. The dura mater that the middle meningeal artery burrows into is a pain-sensitive structure. Dilation of the middle meningeal artery, which is associated with substance P and neurokinin A, results in headaches. Even though dilation of the middle meningeal artery has been shown to cause pain consistent with migraines, it has been difficult to observe middle meningeal artery dilation during the event of a spontaneous migraine.Sumatriptan has been shown to reduce the dilation of the middle meningeal artery. This reduction in the dilation of the middle meningeal artery has been linked to decreased pain in patients experiencing migraines. While the middle meningeal artery may not be the only factor in migraine headaches, it does play a significant role. Chronic Subdural Hematoma Middle meningeal artery embolization is being proposed as a minimally invasive treatment option for the management of chronic subdural hematomas, especially the recurrent ones. Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Epidural Hematoma After Trauma, Computed Tomography. The area highlighted in red depicts an epidural hematoma posttrauma on a computed tomography scan. Contributed by S Bhimji, MD Figure Middle Meningeal Artery. This illustration portrays the middle meningeal artery in red. Illustration by E Gregory Go to: References 1. Chmielewski P, Skrzat J, Walocha J. Clinical importance of the middle meningeal artery. Folia Med Cracov. 2013;53(1):41-6. [PubMed: 24858329] 2. Yu J, Guo Y, Xu B, Xu K. Clinical importance of the middle meningeal artery: A review of the literature. Int J Med Sci. 2016;13(10):790-799. [PMC free article: PMC5069415] [PubMed: 27766029] 3. Tubbs RS, Walker AM, Demerdash A, Matusz P, Loukas M, Cohen-Gadol AA. Skull base connections between the middle meningeal and internal carotid arteries. Childs Nerv Syst. 2015 Sep;31(9):1515-20. [PubMed: 26033377] 4. Salem MM, Fusco MR, Dolati P, Reddy AS, Gross BA, Ogilvy CS, Thomas AJ. Middle meningeal artery arising from the basilar artery. J Cerebrovasc Endovasc Neurosurg. 2014 Dec;16(4):364-7. [PMC free article: PMC4296049] [PubMed: 25599045] 5. Cvetko E, Bosnjak R. Unilateral absence of foramen spinosum with bilateral ophthalmic origin of the middle meningeal artery: case report and review of the literature. Folia Morphol (Warsz). 2014 Feb;73(1):87-91. [PubMed: 24590529] 6. Maugeri R, Anderson DG, Graziano F, Meccio F, Visocchi M, Iacopino DG. Conservative vs. Surgical Management of Post-Traumatic Epidural Hematoma: A Case and Review of Literature. Am J Case Rep. 2015 Nov 14;16:811-7. [PMC free article: PMC4652627] [PubMed: 26567227] 7. Paiva WS, Andrade AF, Amorim RL, Bor-Seng-Shu E, Gattas G, Neville IS, Caldas JG, Figueiredo EG, Teixeira MJ. Computed tomography angiography for detection of middle meningeal artery lesions associated with acute epidural hematomas. Biomed Res Int. 2014;2014:413916. [PMC free article: PMC3988863] [PubMed: 24800222] 8. Suzuki M, Kominami S, Koketsu K, Mizunari T, Kobayashi S, Morita A. Endovascular Repair of a Middle Meningeal Artery Aneurysm after Cranial Surgery. NMC Case Rep J. 2014 Oct;1(1):6-8. [PMC free article: PMC5364935] [PubMed: 28663943] 9. Gutowski P, Meier U, Rohde V, Lemcke J, von der Brelie C. Clinical Outcome of Epidural Hematoma Treated Surgically in the Era of Modern Resuscitation and Trauma Care. World Neurosurg. 2018 Oct;118:e166-e174. [PubMed: 29959068] 10. Sakata H, Nishimura S, Mino M, Hori E, Fujita T, Midorikawa H, Kaimori M, Nishijima M. Serial angiography of dynamic changes of traumatic middle meningeal arteriovenous fistula: case report. Neurol Med Chir (Tokyo). 2009 Oct;49(10):462-4. [PubMed: 19855142] 11. Haldrup M, Ketharanathan B, Debrabant B, Schwartz OS, Mikkelsen R, Fugleholm K, Poulsen FR, Jensen TSR, Thaarup LV, Bergholt B. Embolization of the middle meningeal artery in patients with chronic subdural hematoma-a systematic review and meta-analysis. Acta Neurochir (Wien). 2020 Apr;162(4):777-784. [PubMed: 32086603] Disclosure:Adriana Natali declares no relevant financial relationships with ineligible companies. Disclosure:Vamsi Reddy declares no relevant financial relationships with ineligible companies. Disclosure:Jonathan Leo declares no relevant financial relationships with ineligible companies. Introduction Structure and Function Embryology Physiologic Variants Surgical Considerations Clinical Significance Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK519545 PMID: 30137830 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Introduction Structure and Function Embryology Physiologic Variants Surgical Considerations Clinical Significance Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Tranexamic acid vs. embolization of the meningeal artery as an adjunctive therapeutic regime to reduce the recurrence rate after surgical relief of chronic subdural hematomas (TABASCO)-a randomized controlled trial.[Trials. 2025]Tranexamic acid vs. embolization of the meningeal artery as an adjunctive therapeutic regime to reduce the recurrence rate after surgical relief of chronic subdural hematomas (TABASCO)-a randomized controlled trial.Scheer M, Beuchel V, Mauer UM, Efinger K, Schulz C. Trials. 2025 Jun 14; 26(1):207. Epub 2025 Jun 14. Subclavian Artery Thrombosis.[StatPearls. 2025]Subclavian Artery Thrombosis.Kimyaghalam A, Iqbal AM. StatPearls. 2025 Jan Short-Term Memory Impairment.[StatPearls. 2025]Short-Term Memory Impairment.Cascella M, Al Khalili Y. StatPearls. 2025 Jan Review The Black Book of Psychotropic Dosing and Monitoring.[Psychopharmacol Bull. 2024]Review The Black Book of Psychotropic Dosing and Monitoring.DeBattista C, Schatzberg AF. Psychopharmacol Bull. 2024 Jul 8; 54(3):8-59. Clinical importance of the middle meningeal artery: A review of the literature.[Int J Med Sci. 2016]Clinical importance of the middle meningeal artery: A review of the literature.Yu J, Guo Y, Xu B, Xu K. Int J Med Sci. 2016; 13(10):790-799. Epub 2016 Oct 17. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Neuroanatomy, Middle Meningeal Arteries - StatPearlsNeuroanatomy, Middle Meningeal Arteries - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... 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190049	https://www.ijbcp.com/index.php/ijbcp/article/view/1819	Tinea incognito with unjustified use of potent Topical Corticosteroids: a case series \| International Journal of Basic & Clinical Pharmacology Skip to main contentSkip to main navigation menuSkip to site footer Open Menu International Journal of Basic & Clinical Pharmacology Home About About the Journal Submissions Editorial Team Privacy Statement Contact Login Register Articles In Press Current Archives Author Guidelines Search Home/ Archives/ Vol. 6 No. 8 (2017): August 2017/ Case Reports Tinea incognito with unjustified use of potent Topical Corticosteroids: a case series Authors Haiya J. Sheth Department of Pharmacology, Smt. NHLMMC, Ahmedabad, Gujarat, India Santosh P. Rathod Department of Dermatology, Smt. NHLMMC, Ahmedabad, Gujarat, India Raju G. Chaudhary Department of Dermatology, Smt. NHLMMC, Ahmedabad, Gujarat, India Supriya D. Malhotra Department of Pharmacology, Smt. NHLMMC, Ahmedabad, Gujarat, India Pankaj R. Patel Department of Orthopedics, Smt. NHLMMC, Ahmedabad, Gujarat, India DOI: Keywords: Betamethasone, Clobetasol, Dermatology, Misuse, Tinea Incognito Abstract Topical corticosteroids prescribed for various dermatological conditions, if used haphazardly have their adverse effects too, like aggravation of underlying infections - tinea, herpes etc. Tinea incognito is a fungal infection of skin masked and often exacerbated by application of topical immunosuppressive or steroid agent. Tinea incognito results because local physicians/chemists might mistake tinea with a simple rash. Availability of a topical steroid cream as an over the counter (OTC) drug is also one of the causes. Once it is diagnosed (clinically with or without KOH mount preparation), it should be thoroughly treated with oral and topical antifungal agents. Four cases presenting to Dermatology OPD VSGH with chief complaints of itching and aggravation of existing lesions due to application of potent topical corticosteroids like Clobetasol and Beclomethasone have been discussed here. They were diagnosed as cases of Topical Corticosteroids induced Tinea Incognito. The patients were recovering with oral and topical antifungals prescribed by dermatologist. These were also reported to nearest ADR Monitoring Centre under PvPI with Probable causality according to WHO UMC Criteria for Causality Assessment. The wide use of topical corticosteroids is due to its immediate symptomatic relief. However, the normal presentation of superficial infection gets altered with their inappropriate use. Lack of awareness among population, steroid abuse by local physicians or no strict regulations for local chemist shops have led to its inappropriate use. Its incidence ranges from 5-15%. These figures point towards the fact that implementation of appropriate regulations for the safe use of topical corticosteroids is essential. Metrics PDF views 281 daily (first 30) \| monthly \| yearly References Wikipedia contributors, Wikipedia: The Free Encyclopedia. Topical steroid, December 15, 2016. Available at: Accessed May 4 2017. Hengge U, Ruzicka T, Schwartz R, Cork M. Adverse effects of topical glucocorticosteroids. By American Academy of Dermatology, Inc.; January 2006. DermNet, New Zealand: All About Skin. Topical Steroids, 1997. Available at: steroids/. Accessed May 4 2017. DermNet New Zealand: All About Skin. Tinea Incognito, 2003. Available at: Accessed May 4 2017. Wikipedia contributors, Wikipedia: The Free Encyclopedia. Tinea Incognito, December 14, 2015. Available at: Accessed May 4 2017. Kumar S, Goyal A, Gupta YK. Abuse of topical corticosteroids in India: Concerns and the way forward. J Pharmacol Pharmacother. 2016;7:1-5. Verma S, Madhu R. The great Indian Epidemic of superficial dermatophytosis: An appraisal. Indian J Dermatol. 2017;62:227-36. Saraswat A, Lahiri K, Chatterjee M, Barua S, Coondoo A, Mittal A, et al. Topical corticosteroid abuse on the face: A prospective, multicenter study of dermatology outpatients. Indian J Dermatol Venereol Leprol. 2011;77:160-6. Panda S, Verma S. The menace of dermatophytosis in India: The evidence that we need. Indian J Dermatol Venereol Leprol. 2017;83:281-4. Rathi SK, D’Souza P. Rational and Ethical Use of Topical Corticosteroids Based on Safety and Efficacy. Indian Journal of Dermatology. 2012;57(4):251-9. Downloads PDF Published 2017-07-22 How to Cite Sheth, H. J., Rathod, S. P., Chaudhary, R. G., Malhotra, S. D., & Patel, P. R. (2017). Tinea incognito with unjustified use of potent Topical Corticosteroids: a case series. International Journal of Basic & Clinical Pharmacology, 6(8), 2087–2090. More Citation Formats ACM ACS APA ABNT Chicago Harvard IEEE MLA Turabian Vancouver Download Citation Endnote/Zotero/Mendeley (RIS) BibTeX Issue Vol. 6 No. 8 (2017): August 2017 Section Case Reports 2 0 Rajeshwari Dabas, Manasa Shettisara Janney, Radhakrishnan Subramaniyan, Sandeep Arora, V Sandeep Lal, Navya Donaparthi (2018) Use of Over-the-Counter Topical Medications in Dermatophytosis. Indian Journal of Drugs in Dermatology, 4(1), 13. 10.4103/ijdd.ijdd_5_18 Arisha Shahid, Mohd Aleemuddin Quamri, Mohd Riyazuddin (2024) Management of Tinea incognito with Unani medicine: A case report. Journal of Ayurveda Case Reports, 7(2), 109. 10.4103/jacr.jacr_197_23 Make a Submission Make a Submission Information For Readers For Authors For Librarians Keywords Current Issue International Journal of Basic & Clinical Pharmacology. Copyright © 2025. Print ISSN: 2319-2003 \| Online ISSN: 2279-0780 medipeditor@gmail.com,editor@ijbcp.com
190050	https://www.designnews.com/industry/nist-webbook-online	Industry NIST Webbook online DN Staff March 23, 1998 1 Min Read The National Institute of Standards and Technology now offers a new compilation of chemical and physical data in the NIST Chemistry Webbook at webbook.nist.gov. The web edition contains thermochemical properties, infrared spectral data, vapor pressure data, and mass spectral data. Access is currently free and is part of the more than 50 databases available from NIST. About the Author DN Staff See more from DN Staff Sign up for Design News newsletters You May Also Like Industry SpaceX Starship Super Heavy Booster Benefits from Wind Tunnel Testing at NASA Ames March 18, 2025 Industry The Webb Space Telescope's Best Shots (So Far) March 12, 2025 Industry Trump Equity Order Kills West Point Engineering Society Chapters February 10, 2025 Industry A Super Ball, Made Especially for the Super Bowl February 04, 2025 Editors' Choice Electronics 15 Top Cities for Electrical Engineers 15 Top Cities for Electrical Engineers Sep 26, 2025 Industry The Top 12 Automation Stories That Are Rocking 2025 The Top 12 Automation Stories That Are Rocking 2025 Sep 26, 2025 Industry Which Toys & Games Will Enter the Hall of Fame for 2025? Which Toys & Games Will Enter the Hall of Fame for 2025? Sep 18, 2025 Electronics What I Found at a Maker Faire What I Found at a Maker Faire Sep 16, 2025 Automotive Engineering 2025 IAA Munich Mobility Show Cars & Concepts 2025 IAA Munich Mobility Show Cars & Concepts Sep 11, 2025 Recent Sponsored Content The Ultimate Guide to the Crane Industry Sep 26, 2025 \| 1 Min Read Sponsored Content Important success factors for dry-running bearings in construction machinery Sep 26, 2025 \| 1 Min Read Sponsored Content Digitally networked energy chains enhance STS cranes in the Port of Rotterdam Sep 26, 2025 \| 1 Min Read 15 Top Cities for Electrical Engineers Sep 26, 2025 \| 15 Slides
190051	https://www.youtube.com/watch?v=xy2VHkKmpMQ	部分分式: 怎樣找出未知數 (遮蓋法) 黑筆紅筆 75300 subscribers 851 likes Description 35027 views Posted: 10 May 2022 部分分式 (partial fraction decomposition): 怎樣找出解部分分式的未知數這是高中代數跟大學微積分裡必須知道的技巧之一! 我們會講三小例題來幫你理解怎樣快速的做好部分分式. 我們會講遮蓋法 Cover-up method 上集: 部分分式怎樣拆開 👉 0:00 [例題一] 用cover-up掩蓋法 (2x+1)/((x+1)(x+2)) 6:04 [例題二] (2x+1)/((x+1)(x^2+2)) 12:08 [例題三] (2x+1)/((x+1)(x+2)^2) 喜歡數學可訂閱 @bprptw -----感謝會員們的贊助----- 吉掰米 Dino Lin 水源ミヤゴ chuan liu ---------我是分隔線--------- 點這訂閱: 👉 贊助老師: 👉 加入會員: 👉 衣服商品: 🛍 祝你幸福黑筆紅筆微積分教學 50 comments Transcript: [例題一] 用cover-up掩蓋法 (2x+1)/((x+1)(x+2)) 微積分基礎看完這部影片你就會知道怎樣去做Partial fractions部分分式那我也教你們怎樣比較快(用遮蓋法 Cover-up method)的去找到部分分式裡的未知數
190052	https://math.stackexchange.com/questions/2669194/a-solution-using-lifting-the-exponent-lemma-to-imo-1990-p3	Skip to main content A solution using 'Lifting the exponent' lemma to IMO 1990 P3 Ask Question Asked Modified 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Question: Find all positive integers n such that n2∣2n+1 My solution: Lemma (Lifting the exponent): Let vp(n) denote the highest power of a prime p that divides n. That is, vp(n)=k such that pk∣n and pk+1∤n . We have the following relation: vp(xn−yn)=vp(x−y)+vp(n) for any odd prime p and , v2(xn−yn)=v2(x−y)+v2(x+y)+v2(n)−1 If n is odd then replacing y by −y we obtain vp(xn+yn)=vp(x+y)+vp(n) The original problem: It is easy to see that n=3 satisfies the solution. We will prove that this the only solution. Suppose for the sake of contradiction that p⩾5 is a prime divisor of n. The original problem reduces to finding n such that vp(n2)⩽vp(2n+1) Observe that n is odd and since vp(xn+yn)=vp(x+y)+vp(n), for odd n, vp(2n+1)=vp(2+1)+vp(n) Since by our assumption p⩾5 therefore vp(1+2)=vp(3)=0. And hence vp(2n+1)=vp(n) But this implies vp(n2)>vp(n) which is absurd. Therefore the only solution is n=3. Is my solution valid? I'm quite new to lifting the exponent lemma. I read it in Evan Chen's handout titled 'Orders Modulo a Prime' and was pretty fascinated at the fact that it could be used to solve at least easy-moderate Olympiad problems. Please verify and see if I missed out something obvious. Also, I have not included a proof of the lemma as I think it would be pretty useless here :) Thank you. elementary-number-theory prime-numbers contest-math divisibility Share CC BY-SA 3.0 Follow this question to receive notifications edited Feb 27, 2018 at 18:54 QFTheorist asked Feb 27, 2018 at 16:27 QFTheoristQFTheorist 78933 silver badges1414 bronze badges 8 Did you see this post, too? – Dietrich Burde Commented Feb 27, 2018 at 16:37 @DietrichBurde no I didn't, but doesn't that post ask for integers n that divide 2n+1? I mean, my question is a bit different. – QFTheorist Commented Feb 27, 2018 at 16:40 1 Don’t forget n=1… – Nemo Commented Feb 27, 2018 at 18:52 3 It seems like you're missing one of the critical features of the lifting the exponent lemma: vp(an+bn)=vp(a+b)+vp(n) requires p∣(a+b). In this case, with a=2 and b=1, that means p∣3. So any argument based on higher p can't be correct. – Steven Stadnicki Commented Feb 27, 2018 at 19:25 1 I certainly don't see a quick way to repair it, I'm afraid. – Steven Stadnicki Commented Feb 27, 2018 at 19:27 \| Show 3 more comments 2 Answers 2 Reset to default This answer is useful 3 Save this answer. Show activity on this post. You were on the right track but, as you acknowledged in the question comments, you can't use p≥5 in your case with the Lifting-the-exponent lemma (LTE) since p∤1+2=3. Instead, first use that n2∣2n+1(1) obviously requires n∣2n+1 (useful details about this are in When does n divide 2n+1?). Next, Primitive elements of the sequence of integers n such that n divides 2^n + 1 says Every element of this sequence apart from 1 and 3 is divisible either by 27 or by 171. Note that 27=33 and 171=32(19). In particular, this means that each such primitive element >3 has at least 2 factors of 3. In addition, Numbers n such that n divides 2^n + 1 states that, apart from the primitive elements, all other elements are generated among themselves by one of 3 methods, i.e., multiplication of two elements, multiple of an element having the same prime factors as the element, and the lowest common multiple of two elements. Generating any element using 3 will result in a number with at least two factors of 3. Also, note using any of the 3 methods will not reduce the number of factors of 3 in the generated element. Thus, this shows that all n other than 1 and 3 must have at least 2 factors of 3. Since n is odd, 3∤1 and 3∤2, but 3∣1+2=3, the LTE can be used to determine limits on the power of 3 in n. Using p=3, ν3(2n+1)=ν3(2+1)+ν3(n)=1+ν3(n)(2) For (1) to hold requires ν3(2n+1)≥2ν3(n)(3) As such, (2) shows that 1+ν3(n)≥2ν3(n)→1≥ν3(n)(4) Since (4) shows n can have only up to 1 factor of 3, but since all n>3 have at least 2 factors of 3, this proves n=1 and n=3 are the only solutions to (1). Share CC BY-SA 4.0 Follow this answer to receive notifications edited Sep 24, 2021 at 6:04 answered Apr 5, 2019 at 5:57 John OmielanJohn Omielan 52.9k55 gold badges3838 silver badges8787 bronze badges Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. Based somewhat on Proof Verification: Find all positive integer n for which n2\|2n+1., here's a simpler and more self-contained proof than in my other answer to determine the n which satisfy n2∣2n+1(1) For n>1, let p be the smallest prime factor of n. Since 2n≡−1(modp), then 22n≡1(modp). This means the multiplicative order of 2 modulo p must divide 2n, i.e., with ordp(2)=j, then j∣2n. Since j∣(φ(p)=p−1) (i.e., Euler's totient function of p), this means j<p. Since p is the smallest prime factor of n, then gcd(j,n)=1, so j∣2, i.e., j=2→p=3. Since n is odd, 3∤1 and 3∤2, but 3∣1+2=3, the Lifting-the-exponent lemma (LTE) can be used to get ν3(2n+1)=ν3(2+1)+ν3(n)=1+ν3(n)(2) For (1) to hold requires ν3(2n+1)≥2ν3(n)(3) As such, (2) shows that 1+ν3(n)≥2ν3(n)→1≥ν3(n)(4) This means ν3(n)=1, i.e., n has exactly one factor of 3. Thus, if n>3, then m=n3 must have other larger prime factors. Let q be the smallest among them. This means with ordq(2)=k, then k∣2(3)m, and using basically the same argument as before, we get that gcd(k,m)=1, so k∣2(3). Since 22−1=3, 23−1=7 and 26−1=32(7), this means q=7. However, 2n is congruent to just 1, 2 or 4 modulo 7. Thus, 7∤2n+1. This contradiction proves there are no n other than 1 and 3 which satisfy (1). Share CC BY-SA 4.0 Follow this answer to receive notifications edited May 17 at 17:23 answered Sep 24, 2021 at 8:37 John OmielanJohn Omielan 52.9k55 gold badges3838 silver badges8787 bronze badges 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 elementary-number-theory prime-numbers contest-math divisibility See similar questions with these tags. 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190053	https://www.youtube.com/watch?v=WB3LjyyKVNc	Binomial Expansions Practice Problems Transcended Institute 66900 subscribers 135 likes Description 15651 views Posted: 14 Jul 2023 In this Binomial Expansions Video we go over Binomial Expansions Practice Problems using the Binomial Theorem. To register and access full video lessons visit Students) Get The Full Video on Patreon: (Everyone) Direct Link to The Full Video: (Set Theory) Transcript: uh it's very important even as you basically get to do your your personal studies that you practice before you move on to a New Concept so previously we've talked about uh the binomial theorem we've also talked about how to find the N term so I want us to practice what we've learned so far so using the bomal theorem we need to expand we we can expand given expressions for example that one understand that since it is to the power 12 you start with 12 choose Z and then your a in this case is 2x so it will be the power 12 - 0 which is just 12 and then netive y will be the power zero so one thing that we we've learned so far is the fact that as you expanding the first part which in this case is 2x will always be the difference of the Coe and then the B part which is negative Y in this case will always match up with a bottom coefficient as I've shown there so when you have 12 chose two 2x will be the difference 12 - 2 and then negative y will just still be at the power two it would just always be matching up with the bottom power now certain things have already been established the fact that when you have 12 ch0 is equal to one when you have 12 choose 1 is equal to 12 and then 12 choose 12 if we add it it is also going to be equal to one so these things have already been established so that's it about the binom now I want us to answer these questions we need to write and simplify the indicated term in the expansions of the following so we also need to find the 15th term in that expansion and this will introduce us to A New Concept where we need to find the term independent of X okay so seventh term so we've already established to say when you have n and R from what I was trying to show you given that you have a plus let's say X in your expansion the basic idea there is um what you ought to do is uh your a will be the difference as I've said and then your X will always match up with the bottom power we already established that so that is what we're going to be using when it comes to finding the term so when they say find the 7eventh term 7eventh term what that means is your RoR is going to be less than by one because we start counting from zero so if you start counting from your zero as your one you end up at six as your seventh term so your seventh term the value of R is always going to be less than a one from n so therefore in this case when they say find the seventh term what we're going to have is we're going to have our value of n which is 12 we are choosing six because we start from zero so therefore our a in this case has been given to be 2x it will be the power 12 - 6 and then X which is our Y in this case negative y that is will be the power the bottom power 26 so we have 12 26 which if you have calcat is 12 combination 6 so 12 factorial 12- 6 factorial 6 factorial then you have your 2x being the^ 6 and then negative y being the power six so whenever you ra a negative number or valuable it's going to be to un even power it is going to be positive so of course for the sake of simplifications for a 12 factori there you remain remain with 12 11 10 9 8 7 the 6 factorial will just cancel out with that and then of course you still remain the 6 factorial in 12 - 6 which will be 6 5 4 3 2 1 2^ 6 I believe that should be 64 if I'm not mistaken 2^ 4 is 16 2 2 have 16 so 64 x^ 6 y^ 6 so I can try to simplify this let me just guide you 6 into 12 is 2 5 into 10 is 2 4 into 8 is two 3 into 9 is 3 2 and two one 2 it just be a one and then at Le so what you have on top is 2 3 2 7 which is easier to deal with right and then you multiply that by 64 you'll be able to get the coefficient but o in O my focus was on trying to show you how you basically get to find the term of course remind you of what we know so far so with that the concept you can answer the second one right the ninth term so the ninth term what would be the expected value of R so it would be 15 choose 8 your a is your two and then your your X or your B is x/ two so what did we say so we said for your a part to always be the difference so it will be 15us 8 then the other one will always matching up with what is on the bottom right so that's for B now let me focus on the remaining three so we need to find the middle term of this expansion so of course the half of 10 is five but if you want to make more sense if you write zero all the up to 10 0 1 2 3 4 are the first five numbers and then 10 9 8 7 6 are the last five numbers so what is on the middle is five so your value of R is five and that's what help you find the middle term so in short what you're trying to find is just 10 choose five your a is known to be two your B is known to be 3/x so 2 b^ 10 - 5 3x^ 5 we talked about that we establish that from that formula on top so we able to simplify C now very interesting are the last two on this page these are the ones I want to focus more on hopefully if you TI please feel free to pause because I did want you to get this because this is a very very important concept under binomial expansions excuse me let me drink some water talking for so long now okay um we have to find the term involving X the PA this is something very interesting so it would be easy if you of course that question was based on the first three okay so you should be able to predict by the way so for example let's say it was applying to question one I would just guess to say okay if 2x is like my first term right if I want to get a point where I will have U because what I'm saying here is want to get a point where it will be to the power where X will be to the power five where X will be to the power five so if I can substitute I would say okay I have 12 and then my RoR I don't know because that's the term I'm trying to find and then your a is 2x so in the path where is 2x now your n is 12- R so what I'm trying to do is I'm trying to find a case where the power will be equal to 5 so I'll just say when can be 12 - R be equal to 5 so 12 - 5 isal to R so your R becomes when the value is seven so in terms of any term it's the 8th term that's the one that is going to give us what that possibility let's try to see now let's try to apply that to the question at end which is uh D it's a bit interesting because we have X on the bottom and also on top so let's write it in terms of n choose R so we have seven as our n we don't know what we are trying to choose our first part is X which is a difference 7 minus r and then our second part which is our B is 1 / 2x which is also just the power R now for own information um we want to find a part where the value of x will be equal to five so it is not only determined by the top part but also the denominator part so I hope you can remember your this is so that value of r that is outside there can be distributed one the power R just R so two be the power r as well and X the power R there's a reason why I've done all that so we have X on top 7 - r on the denominator we have X being just the power R so what do your indices tell you so indices tell us if you've got the same base being divided what happens the power maintain the base then the you subtract remember when you're multiplying you basically get to so 7us R will be subtracted by that bottom power R now what we want is after we've done all that our resulting power of course it will be now be 7 - 2 R our resting power should be equal to five so therefore 7 - 5 = 2 R 7 - 5 is what two so 2 is equal 2 R so our value of R is equal to one now in terms of a term we can therefore say let me not erase that so if R is equal to one what are we talking about we are talking about second term we can also try to prove that if you want to since this is the first one we've done so far okay let me try take back my RoR outside there so if you choose your RoR to be one 71 um we're not interested in the coefficient so it would be X to reduce by one to be six and then one over so this is going to be the power one so it would be 2x the^ one will just be it remain the same so X will cancel one on where we have six it remain now x^ 5 so the answer expected will be 7 1 and then X 5 1 / 2 so you notice that this ter is one is going to have x^ 5 okay that's a basic idea or that's what the question was trying to ask us to find so it's a second term now I want you to apply whatever we've done there to e which is the next one
190054	https://www.varsitytutors.com/practice/subjects/sat-math/help/working-with-inequalities-and-absolute-value	SAT Math - Working with Inequalities and Absolute Value \| Practice Hub Skip to main content Practice Hub Search subjects AI TutorAI DiagnosticsAI FlashcardsAI WorksheetsAI SolverGamesProgress Sign In HomeSAT MathLearn by ConceptWorking with Inequalities and Absolute Value Working with Inequalities and Absolute Value Help Questions SAT Math › Working with Inequalities and Absolute Value Questions 1 - 10 1 The formula above gives the monthly cost, , of running a food truck, where the driver works hours and gallons of gasoline are used for driving and operating the truck. If, in a particular month, it costs more than $3,500 to run the truck and no more than 220 gallons of gas were used, what is the minimum number of hours that the driver worked that month? 95 96 140 480 Explanation You are given in this question that the equation C = 24h + 5y describes the cost to run a food truck and are asked to calculate the minimum number of hours the driver of the truck worked that month if the truck used no more than 220 gallons of gas. To find the minimum number of hours worked, you need to use the maximum number of gallons of gas that could have been used – 220. Substituting this value in for y gives you: C = 25h + 5(220) = 25h + 1100 You are also told that it costs more than $3500 to run the truck for the month. You can then turn the equation into an inequality and solve for a range of values for h. Remember that saying that it costs at least $3500 is the same as saying that 3500 is less than or equal to the information given on the right side of the equation: 3500 < 25h + 1100 Subtracting 1100 from both sides and dividing both sides by 25 gives you: 2400 < 25h 96 < h If h is is greater than or equal to 96, then the smallest number of hours worked must be 96. 2 A home is considered to be outside of market value in a particular market if its appraised price, "p" is more that $10,000 away from 90% of the median home value in the neighborhood, "m." Which of the following inequalities represents a scenario in which a home is outside of market value in this market? Explanation In this case, we'll want to note that it doesn't matter if the home is overvalued by over 10,000 away from 90% of the median home value, or undervalued to this extent. So, we need to utilize absolute values to address that the "distance" between p and 90% of m should be greater than 10,000, as shown in the correct answer. 3 Which of the following represents the complete solution set for the inequality above? Explanation Whenever you're working with inequalities and absolute values, you should consider two cases: 1) the expression within the absolute value directly satisfies the inequality, and 2) the expression within the absolute value satisfies the opposite of the inequality (flip the sign and multiply the other expression by -1). Let's consider both cases here: The expression directly satisfies the inequality: Subtract from and add to both sides so that you isolate the variable on the left. That yields: You may also want to quickly try a number that just barely satisfies this inequality to ensure that it works with the given information. If you try , you'll see that you have: which is which is a "close win" for the inequality, helping to confirm your work here. The expression satisfies the "opposite case" in which you flip the sign and multiply the right-hand expression by -1: Here, add and to both sides to yield: , which reduces to . Here you may again wish to do a quick test; if you choose something just less than , like perhaps , you can confirm. This way you'd have: . This means that which again is a "close win," supporting that your work is correct. The correct answer is then: 4 Which of the following provides the full solution set for the inequality above? Explanation When you're working with inequalities and absolute values, make sure that you cover both cases that would satisfy the inequality. The terms within the absolute value directly satisfy the inequality. Here that's: , so and . The "flip the sign and multiply by -1" case. Here that's: (note that the inequality sign is flipped, and the value to the right of that sign is multiplied by -1) , so 5 Which of the following provides the full solution set to the inequality above? Explanation When you're working with absolute values and inequalities, recognize that there are two cases that will satisfy the inequality: either the terms within the absolute value directly satisfy the inequality, or they satisfy the "opposite" (flip the sign and multiply by -1). For example, if , then either or . For this problem, here is how the two cases will map out: Satisfy the inequality directly: means that Flip the inequality sign and multiply the numeric value by -1: , so This can feel abstract, so it's helpful to test one number for each case to see if it really works. For case 1, try a value greater than 6, so you might try 7. because 8 > 7, so you have this right. For case 2, try a number less than -8, such as -9. because the absolute value of -8 is 8, and 8 is greater than 7. 6 Which of the following is not a possible value for x given the inequality above? 12 15 24 25 Explanation In order to simplify this inequality, we'll want to start by subtracting 12 from both sides to arrive at if we then multiply both sides of the inequality by 4 to cancel the coefficient in front of x, we get to Thus, 12 is not a possible value of x given the simplified inequality. Note - we could also solve this question by plugging each option in for x, but this route is likely somewhat more time consuming, so stay flexible in your approach on a question-to-question basis! 7 Which of the following represents the complete solution set for the inequality above? Explanation Whenever you're dealing with absolute values and inequalities, you should consider both of two cases that form the full solution set. In case 1, the expression within the absolute value directly satisfies the inequality. In case two, the expression within the absolute value satisfies the opposite, where you reverse the inequality sign and negate the expression opposite the absolute value (that explanation sounds complicated, but the example will help). Let's examine both cases for this question. Case 1 - The expression satisfies the inequality directly. Here that would mean that . You can then subtract from both sides and add to both sides to isolate the variable: It's always a good idea to check a number that just barely satisfies your new inequality to make sure that your answer works. Here if you try 14, you'll see that the given inequality becomes . That works out to , a 'close win' that tells you that your inequality works. Case 2 - The expression satisfies the opposite of the inequality. Here you need to do two things: reverse the inequality sign, and multiply the entire right side of the inequality by -1. That would give you: Adding and to both sides isolates the variable: And then finish the division: Again, try a number that just barely satisfies the inequality to check that you got it right. If you use 0, then you'll see that the given inequality becomes 8 > 5, which is another close win. Therefore, the correct answer is or . 8 Which of the following represents the complete solution set for the inequality above? Explanation Whenever you're dealing with absolute values and inequalities, you need to consider two possibilities. In case one, the expression within the absolute value directly satisfies the inequality. In case two, the expression within the absolute value will satisfy the opposite, taken by flipping the inequality sign and negating the entire other side of the inequality (think "a is greater than 1 or less than negative-one" --> case two is that "less than negative..." case). Let's examine both here. Case 1 Here you can subtract from both sides and add to both sides to isolate the variable: Then divide both sides by and you have: It's a good idea to try a number that just barely satisfies your new inequality to ensure that it fits with the given inequality. If you try here, you'll see that you have a fit: simplifies to , proving that you have a fit. Case 2 (notice that the sign is flipped, and all terms on the right are negated) Here you can add to both sides and subtract from both sides to get: So Notice here that this does not add new information. Any number less than is already less than and you've already tried a number between them (1) to ensure that it works. So the full solution set here is just any number less than , making the correct answer . 9 If , in terms of y, what is the full range of possible values for x? or or Explanation In this example, we can eliminate the absolute value signs by creating two paths, one in which we hold the sign, and one in which we flip the sign and multiply one side by -1. In this case, these two options are as follows and So, simplified, we can see that and so, 10 If an election analyst receives a bonus when her prediction "p" is within 12 votes of the actual number, "a," the scenario in which she receives her bonus is expressed by the inequality below. According to this inequality, which of the following expresses the full range of possible values for "p" if the actual number of votes, "a" for a particular candidate is 127, and the analyst received her bonus? or Explanation If the analyst receives her bonus when , and she received her bonus when the actual number of votes, "a" was 127, we can substitute 127 into the inequality as follows So, to get rid of the inequality signs, we need a situation where or where the first inequality can be simplified to and the second, to So, in order for the analyst to receive her bonus, p must be between 115 and 139, or Previous Page 1 of 2 Next Return to subject Powered by Varsity Tutors⋅ © 2025 All Rights Reserved
190055	https://math.stackexchange.com/questions/2169262/where-do-parametric-equation-for-normal-line-of-parametric-equation-come-from	calculus - Where do parametric equation for normal line of parametric equation come from? - 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 Where do parametric equation for normal line of parametric equation come from? Ask Question Asked 8 years, 7 months ago Modified8 years, 7 months ago Viewed 3k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. In my calculus II course, I am currently studying parametric curves. And they are awesome! At some point, the book states the following: _if we set parameters x = f(t), and y = g(t), the parametric equations for the tangent line at time t 0 t 0 are_ x = f (t 0 t 0) + f ' (t 0 t 0)(t - t 0 t 0) y = g (t 0 t 0) + g ' (t 0 t 0)(t - t 0 t 0) I understand that these are just the standard line equations, in which f(t 0 t 0) is the x coordinate at that time, and g (t 0 t 0) the y coordinate. The derivatives are the slope of the functions at that point. The next part is what has been puzzling me for a time now The NORMAL lines at those points are then x = f (t 0 t 0) + g ' (t 0 t 0)(t - t 0 t 0) y = g (t 0 t 0) - f ' (t 0 t 0)(t - t 0 t 0) I would have expected the slope to change from g ' (t 0 t 0) to −1 g′(t 0)−1 g′(t 0) Am I missing something obvious? Thank you very much for your time! QUESTION ANSWERED: Thank you very much to all of you, I understand now. calculus orthonormal Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 2, 2017 at 23:00 MichaelMichael asked Mar 2, 2017 at 21:57 MichaelMichael 35 1 1 silver badge 6 6 bronze badges 1 1 The slope of the tangent line isn’t g′(t 0)g′(t 0). It’s g′(t 0)/f′(t 0)g′(t 0)/f′(t 0).amd –amd 2017-03-03 03:17:22 +00:00 Commented Mar 3, 2017 at 3:17 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Note that the slope is given by d y d x=y′(t)x′(t)d y d x=y′(t)x′(t) This should fit your intuition for slope--it's change in y y divided by change in x x. Therefore, in the first formula, the slope is g′(t 0)f′(t 0)g′(t 0)f′(t 0) and in the second one, the slope is −f′(t 0)g′(t 0)−f′(t 0)g′(t 0) which is consistent with your usual formula for the slope of the perpendicular line. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 2, 2017 at 22:03 florenceflorence 13.1k 1 1 gold badge 28 28 silver badges 49 49 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. In vector terms, the tangent line has equation [x(t)y(t)]=[f(t 0)g(t 0)]+(t−t 0)[f′(t 0)g′(t 0)]=p 0→+(t−t 0)v→[x(t)y(t)]=[f(t 0)g(t 0)]+(t−t 0)[f′(t 0)g′(t 0)]=p 0→+(t−t 0)v→ where p 0→p 0→ is the point of tangency and v→v→ is the direction vector of the tangent line. Note that the vector v⊥−→=[g′(t 0)−f′(t 0)]v⊥→=[g′(t 0)−f′(t 0)] is perpendicular to the vector v→=[f′(t 0)g′(t 0)];v→=[f′(t 0)g′(t 0)]; that is to say, their dot product is equal to 0 0. Hence the equation to the line normal to [x(t)y(t)][x(t)y(t)] at the point p 0→=[f′(t 0)g′(t 0)]p 0→=[f′(t 0)g′(t 0)] is [x⊥(t)y⊥(t)]=[f(t 0)g(t 0)]+(t−t 0)[g′(t 0)−f′(t 0)]=p 0→+(t−t 0)v⊥−→[x⊥(t)y⊥(t)]=[f(t 0)g(t 0)]+(t−t 0)[g′(t 0)−f′(t 0)]=p 0→+(t−t 0)v⊥→ Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 3, 2017 at 15:14 answered Mar 2, 2017 at 22:30 Steven Alexis GregorySteven Alexis Gregory 27.7k 4 4 gold badges 49 49 silver badges 91 91 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. d y d x=d y d t d x d t d y d x=d y d t d x d t With our if we apply this to our tangent lines. d y d x=g′(t)f′(t)d y d x=g′(t)f′(t) and If we apply it to the equations of the normal lines, we get. d y d x=−f′(t)g′(t)d y d x=−f′(t)g′(t) which is indeed the negative reciprocal of the our equations for the tangent lines. However, if you wanted to, you could say: x(t)=y(t 0)+1 f′(t 0)(t−t 0)y(t)=y(t 0)−1 g′(t 0)(t−t 0)x(t)=y(t 0)+1 f′(t 0)(t−t 0)y(t)=y(t 0)−1 g′(t 0)(t−t 0) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 2, 2017 at 22:05 Doug MDoug M 58.8k 4 4 gold badges 35 35 silver badges 69 69 bronze badges 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 calculus orthonormal See similar questions with these tags. 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190056	https://www.cuemath.com/ncert-solutions/the-solution-of-the-equation-ax-b-0-is-a-x-a-b-b-x-b-c-x-b-a-d-x-b-a/	The solution of the equation ax + b = 0 is (a) x = a/b (b) x = -b (c) x = -b/a (d) x = b/a Solution: Given, the equation is ax + b = 0 We have to find the solution of the given equation. On solving the equation for x, ax + b = 0 ax = -b x = -b/a Therefore, the required solution is -b/a. ✦ Try This: The solution of the equation 5x - 10 = 0 is Given, the equation is 5x - 10 = 0 We have to find the solution of the equation. 5x = 10 x = 10/5 x = 2 Therefore, the solution of the equation is 2. ☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 2 NCERT Exemplar Class 8 Maths Chapter 4 Problem 2 The solution of the equation ax + b = 0 is (a) x = a/b, (b) x = -b, (c) x = -b/a, (d) x = b/a Summary: The solution of the equation ax + b = 0 is -b/a. ☛ Related Questions:
190057	https://web.math.princeton.edu/swim/SWIM%202010/Shi-Xie%20Presentation%20SWIM%202010.pdf	PRIME, MODULAR ARITHMETIC, AND By: Tessa Xie & Meiyi Shi OBJECTIVES  Examine Primes In Term Of Additive Properties & Modular Arithmetic  To Prove There Are Inﬁnitely Many Primes  To Prove There Are Inﬁnitely Many Primes of The Form 4n+2  To Prove There Are Inﬁnitely Many Primes of The Form 4n  To Prove There Are Inﬁnitely Many Primes of The Form 4n+3  To Prove There Are Inﬁnitely Many Primes of The Form 4n+1 Proof: Primes in Form of 4n+3 Prove By Contradiction Assumption: Assume we have a set of ﬁnitely many primes of the form 4n+3 P = {p1, p2, …,pn}. Construct a number N such that N = 4 p1 p2 … pn – 1 = 4 [ (p1 p2 … pn) – 1 ] + 3 N can either be prime or composite. If N is a prime, there’s a contradiction since N is in the form of 4n +3 but does not equal to any of the number in the set P. If N is a composite, there must exist a prime factor “a” of N such that a is in the form of 4n+3. All the primes are either in the form of 4n+1 or in the form of 4n+3. If all the prime factors are in the form of 4n+1, N should also be in the form of 4n +1. There should exist at least one prime factor of N in the form of 4n+3. 4N + 3 4N + 1 ? “a” does not belong to set P N/a = (4 p1 p2 … pn – 1) / a = (4 p1 p2 … pn ) / a - 1/a (1/a is not an integer) Conclusion: a is a prime in the form of 4n+3, but a does not belong to set P. Therefore, we proved by contradiction that there exists inﬁnitely many primes of the form 4n+3. Proof: Primes in Form of 4n+1 Prove by Fermat’s Little Theorem Let N be a positive integer Let M be a positive integer in the form: M = [ N (N-1) (N-2) … 2 1]2 + 1 (M ∈ Z+ & M is odd) = (N!) 2 +1 Let P be a prime number greater than N such that p\|M (p is odd) M ≡ 0 (mod p) Then, we can rewrite M in term of N: (N!) 2 +1 ≡ 0 (mod p) (N!) 2 ≡ -1 (mod p) Fermat’s Little Theorem: ap-1 ≡ 1 (mod p) In order to use Fermat’s Little Theorem in the proof, we would like to convert the left hand side of the equation in the form of aP-1, which can be achieved by raising the equation to the power of (p-1) / 2. (N! 2)/2 ≡ -1 (mod P)/2 We get: (N!) P-1 ≡ (-1) (P-1)/2 (mod p) Notice that the left hand side of the equation is in the form of a P-1 where N! represents a . By Fermat’s Little Theorem, we can rewrite the equation as: 1(mod p) ≡ (-1) (P-1)/2 (mod p) Since p is odd, 1 ≠ -1 (mod p). Then, 1 = (-1) (P-1)/2 The only case for this equation to hold true is when (p-1) /2 is even. If (p-1)/2 is even, it can be represented as: (p-1)/2 = 2n ( n ∈ Z) p = 4n + 1 ap = a (mod p) Since p is greater than N and N can get inﬁnitely large, as N approaches inﬁnity, p also approaches inﬁnity. Conclusion: We proved by Fermat’s Little Theorem that there exists inﬁnitely many primes in the form of 4n+1. Gratitude to Fermat!! Special Thanks To: Lillian Corina Maria & All the People who helped us
190058	https://en.wikipedia.org/wiki/Axis%E2%80%93angle_representation	Axis–angle representation - 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 Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Rotation vectorToggle Rotation vector subsection 1.1 Example 2 Uses 3 Rotating a vector 4 Relationship to other representationsToggle Relationship to other representations subsection 4.1 Exponential map from 𝔰𝔬(3) to SO(3) 4.2 Log map from SO(3) to 𝔰𝔬(3) 4.3 Unit quaternions 5 See also 6 References [x] Toggle the table of contents Axis–angle representation [x] 4 languages Español Italiano Српски / srpski 中文 Edit 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 Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Parameterization of a rotation into a unit vector and angle For broader coverage of this topic, see 3D rotation group. The angle θ and axis unit vector e define a rotation, concisely represented by the rotation vector θe. In mathematics, the axis–angle representation parameterizes a rotation in a three-dimensionalEuclidean space by two quantities: a unit vectore indicating the direction of an axis of rotation, and an angle of rotationθ describing the magnitude and sense (e.g., clockwise) of the rotation about the axis. Only two numbers, not three, are needed to define the direction of a unit vector e rooted at the origin because the magnitude of e is constrained. For example, the elevation and azimuth angles of e suffice to locate it in any particular Cartesian coordinate frame. By Rodrigues' rotation formula, the angle and axis determine a transformation that rotates three-dimensional vectors. The rotation occurs in the sense prescribed by the right-hand rule. The rotation axis is sometimes called the Euler axis. The axis–angle representation is predicated on Euler's rotation theorem, which dictates that any rotation or sequence of rotations of a rigid body in a three-dimensional space is equivalent to a pure rotation about a single fixed axis. It is one of many rotation formalisms in three dimensions. Rotation vector [edit] The axis–angle representation is equivalent to the more concise rotation vector, also called the Euler vector (not to be confused with a vector of Euler angles). In this case, both the rotation axis and the angle are represented by a vector codirectional with the rotation axis whose length is the rotation angle θ, θ=θ e.{\displaystyle {\boldsymbol {\theta }}=\theta \mathbf {e} \,.} It is used for the exponential and logarithm maps involving this representation. Many rotation vectors correspond to the same rotation. In particular, a rotation vector of length θ + 2 πM, for any integer M, encodes exactly the same rotation as a rotation vector of length θ. Thus, there are at least a countable infinity of rotation vectors corresponding to any rotation. Furthermore, all rotations by 2 πM are the same as no rotation at all, so, for a given integer M, all rotation vectors of length 2 πM, in all directions, constitute a two-parameter uncountable infinity of rotation vectors encoding the same rotation as the zero vector. These facts must be taken into account when inverting the exponential map, that is, when finding a rotation vector that corresponds to a given rotation matrix. The exponential map is onto but not one-to-one. Example [edit] Say you are standing on the ground and you pick the direction of gravity to be the negative z direction. Then if you turn to your left, you will rotate ⁠-π/2⁠ radians (or -90°) about the -z axis. Viewing the axis-angle representation as an ordered pair, this would be (a x i s,a n g l e)=([e x e y e z],θ)=([0 0−1],−π 2).{\displaystyle (\mathrm {axis} ,\mathrm {angle} )=\left({\begin{bmatrix}e_{x}\e_{y}\e_{z}\end{bmatrix}},\theta \right)=\left({\begin{bmatrix}0\0\-1\end{bmatrix}},{\frac {-\pi }{2}}\right).} The above example can be represented as a rotation vector with a magnitude of ⁠π/2⁠ pointing in the z direction, [0 0 π 2].{\displaystyle {\begin{bmatrix}0\0\{\frac {\pi }{2}}\end{bmatrix}}.} Uses [edit] The axis–angle representation is convenient when dealing with rigid-body dynamics. It is useful to both characterize rotations, and also for converting between different representations of rigid body motion, such as homogeneous transformations[clarification needed] and twists. When a rigid body rotates around a fixed axis, its axis–angle data are a constant rotation axis and the rotation angle continuously dependent on time. Plugging the three eigenvalues 1 and e±iθ and their associated three orthogonal axes in a Cartesian representation into Mercer's theorem is a convenient construction of the Cartesian representation of the Rotation Matrix in three dimensions. Rotating a vector [edit] Main article: Rodrigues' rotation formula Rodrigues' rotation formula, named after Olinde Rodrigues, is an efficient algorithm for rotating a Euclidean vector, given a rotation axis and an angle of rotation. In other words, Rodrigues' formula provides an algorithm to compute the exponential map from s o(3){\displaystyle {\mathfrak {so}}(3)} to SO(3) without computing the full matrix exponential. If v is a vector in R3 and e is a unit vector rooted at the origin describing an axis of rotation about which v is rotated by an angle θ, Rodrigues' rotation formula to obtain the rotated vector is v r o t=v+(sin⁡θ)(e×v)+(1−cos⁡θ)(e×(e×v)).{\displaystyle \mathbf {v} _{\mathrm {rot} }=\mathbf {v} +(\sin \theta )(\mathbf {e} \times \mathbf {v} )+(1-\cos \theta )(\mathbf {e} \times (\mathbf {e} \times \mathbf {v} ))\,.} For the rotation of a single vector it may be more efficient than converting e and θ into a rotation matrix to rotate the vector. Relationship to other representations [edit] Further information: Charts on SO(3) There are several ways to represent a rotation. It is useful to understand how different representations relate to one another, and how to convert between them. Here the unit vector is denoted ω instead of e. Exponential map from 𝔰𝔬(3) to SO(3) [edit] Further information: Matrix exponential, Orthogonal matrix, Lie algebra §Relation to Lie groups, and Rotation group SO(3) §Exponential map The exponential map effects a transformation from the axis-angle representation of rotations to rotation matrices, exp:s o(3)→S O(3).{\displaystyle \exp \colon {\mathfrak {so}}(3)\to \mathrm {SO} (3)\,.} Essentially, by using a Taylor expansion one derives a closed-form relation between these two representations. Given a unit vector ω∈s o(3)=R 3{\textstyle {\boldsymbol {\omega }}\in {\mathfrak {so}}(3)=\mathbb {R} ^{3}} representing the unit rotation axis, and an angle, θ ∈ R, an equivalent rotation matrix R is given as follows, where K is the cross product matrix of ω, that is, Kv = ω × v for all vectors v ∈ R3, R=exp⁡(θ K)=∑k=0∞(θ K)k k!=I+θ K+1 2!(θ K)2+1 3!(θ K)3+⋯{\displaystyle R=\exp(\theta \mathbf {K} )=\sum _{k=0}^{\infty }{\frac {(\theta \mathbf {K} )^{k}}{k!}}=I+\theta \mathbf {K} +{\frac {1}{2!}}(\theta \mathbf {K} )^{2}+{\frac {1}{3!}}(\theta \mathbf {K} )^{3}+\cdots } Because K is skew-symmetric, and the sum of the squares of its above-diagonal entries is 1, the characteristic polynomialP(t) of K is P(t) = det(K − tI) = −(t 3 + t). Since, by the Cayley–Hamilton theorem, P(K) = 0, this implies that K 3=−K.{\displaystyle \mathbf {K} ^{3}=-\mathbf {K} \,.} As a result, K4 = –K2, K5 = K, K6 = K2, K7 = –K. This cyclic pattern continues indefinitely, and so all higher powers of K can be expressed in terms of K and K2. Thus, from the above equation, it follows that R=I+(θ−θ 3 3!+θ 5 5!−⋯)K+(θ 2 2!−θ 4 4!+θ 6 6!−⋯)K 2,{\displaystyle R=I+\left(\theta -{\frac {\theta ^{3}}{3!}}+{\frac {\theta ^{5}}{5!}}-\cdots \right)\mathbf {K} +\left({\frac {\theta ^{2}}{2!}}-{\frac {\theta ^{4}}{4!}}+{\frac {\theta ^{6}}{6!}}-\cdots \right)\mathbf {K} ^{2}\,,} that is, R=I+(sin⁡θ)K+(1−cos⁡θ)K 2,{\displaystyle R=I+(\sin \theta )\mathbf {K} +(1-\cos \theta )\mathbf {K} ^{2}\,,} by the Taylor series formula for trigonometric functions. This is a Lie-algebraic derivation, in contrast to the geometric one in the article Rodrigues' rotation formula. Due to the existence of the above-mentioned exponential map, the unit vector ω representing the rotation axis, and the angle θ are sometimes called the exponential coordinates of the rotation matrix R. Log map from SO(3) to 𝔰𝔬(3) [edit] Further information: Rotation group SO(3) and Infinitesimal transformation Let K continue to denote the 3×3 matrix that effects the cross product with the rotation axis ω: K(v) = ω × v for all vectors v in what follows. To retrieve the axis–angle representation of a rotation matrix, calculate the angle of rotation from the trace of the rotation matrix: θ=arccos⁡(Tr⁡(R)−1 2){\displaystyle \theta =\arccos \left({\frac {\operatorname {Tr} (R)-1}{2}}\right)} and then use that to find the normalized axis, ω=1 2 sin⁡θ[R 32−R 23 R 13−R 31 R 21−R 12],{\displaystyle {\boldsymbol {\omega }}={\frac {1}{2\sin \theta }}{\begin{bmatrix}R_{32}-R_{23}\R_{13}-R_{31}\R_{21}-R_{12}\end{bmatrix}}~,} where R i j{\displaystyle R_{ij}} is the component of the rotation matrix, R{\displaystyle R}, in the i{\displaystyle i}-th row and j{\displaystyle j}-th column. The axis-angle representation is not unique since a rotation of −θ{\displaystyle -\theta } about −ω{\displaystyle -{\boldsymbol {\omega }}} is the same as a rotation of θ{\displaystyle \theta } about ω{\displaystyle {\boldsymbol {\omega }}}. Of course, adding any integer multiple of 2π to θ{\displaystyle \theta } will also result in the identical rotation; a better method is to constrain θ{\displaystyle \theta } to the interval [0, 2π) or (-π, π]. The above calculation of axis vector ω{\displaystyle \omega }does not work if R is symmetric. Because, this is possible only when θ{\displaystyle \theta } = π, so sin(θ{\displaystyle \theta }) = 0, causing a division by 0 in the formula. However, the limit of the formula for ω{\displaystyle \omega }, as θ{\displaystyle \theta } → π, gives the correct value for ω{\displaystyle \omega }. For the general case the ω{\displaystyle \omega } may also be found using null space of R-I, see rotation matrix#Determining the axis. The matrix logarithm of the rotation matrix R is log⁡R={0 if θ=0 θ 2 sin⁡θ(R−R T)if θ≠0 and θ∈(−π,π){\displaystyle \log R={\begin{cases}0&{\text{if }}\theta =0\{\dfrac {\theta }{2\sin \theta }}\left(R-R^{\mathsf {T}}\right)&{\text{if }}\theta \neq 0{\text{ and }}\theta \in (-\pi ,\pi )\end{cases}}} An exception occurs when R has eigenvalues equal to −1. In this case, the log is not unique. However, even in the case where θ = π the Frobenius norm of the log is ‖log⁡(R)‖F=2\|θ\|.{\displaystyle \\|\log(R)\\|{\mathrm {F} }={\sqrt {2}}\|\theta \|\,.} Given rotation matrices A and B, d g(A,B):=‖log⁡(A T B)‖F{\displaystyle d{g}(A,B):=\left\\|\log \left(A^{\mathsf {T}}B\right)\right\\|_{\mathrm {F} }} is the geodesic distance on the 3D manifold of rotation matrices. For small rotations, the above computation of θ may be numerically imprecise as the derivative of arccos goes to infinity as θ → 0. In that case, the off-axis terms will actually provide better information about θ since, for small angles, R ≈ I + θK. (This is because these are the first two terms of the Taylor series for exp(θK).) This formulation also has numerical problems at θ = π, where the off-axis terms do not give information about the rotation axis (which is still defined up to a sign ambiguity). In that case, we must reconsider the above formula. R=I+K sin⁡θ+K 2(1−cos⁡θ){\displaystyle R=I+\mathbf {K} \sin \theta +\mathbf {K} ^{2}(1-\cos \theta )} At θ = π, we have R=I+2 K 2=I+2(ω⊗ω−I)=2 ω⊗ω−I{\displaystyle R=I+2\mathbf {K} ^{2}=I+2({\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}-I)=2{\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}-I} and so let B:=ω⊗ω=1 2(R+I),{\displaystyle B:={\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}={\frac {1}{2}}(R+I)\,,} so the diagonal terms of B are the squares of the elements of ω and the signs (up to sign ambiguity) can be determined from the signs of the off-axis terms of B. Unit quaternions [edit] Main article: Quaternions and spatial rotation The following expression transforms axis–angle coordinates to versors (unit quaternions): q=(cos⁡θ 2,ω sin⁡θ 2){\displaystyle \mathbf {q} =\left(\cos {\tfrac {\theta }{2}},{\boldsymbol {\omega }}\sin {\tfrac {\theta }{2}}\right)} Given a versor q = r + v represented with its scalarr and vector v, the axis–angle coordinates can be extracted using the following: θ=2 arccos⁡r ω={v sin⁡θ 2,if θ≠0 0,otherwise.{\displaystyle {\begin{aligned}\theta &=2\arccos r\[8px]{\boldsymbol {\omega }}&={\begin{cases}{\dfrac {\mathbf {v} }{\sin {\tfrac {\theta }{2}}}},&{\text{if }}\theta \neq 0\0,&{\text{otherwise}}.\end{cases}}\end{aligned}}} A more numerically stable expression of the rotation angle uses the atan2 function: θ=2 atan2⁡(\|v\|,r),{\displaystyle \theta =2\operatorname {atan2} (\|\mathbf {v} \|,r)\,,} where \|v\| is the Euclidean norm of the 3-vector v. See also [edit] Homogeneous coordinates Pseudovector Rotations without a matrix Screw theory, a representation of rigid-body motions and velocities using the concepts of twists, screws, and wrenches References [edit] ^This holds for the triplet representation of the rotation group, i.e., spin 1. For higher dimensional representations/spins, see Curtright, T. L.; Fairlie, D. B.; Zachos, C. K. (2014). "A compact formula for rotations as spin matrix polynomials". SIGMA. 10: 084. arXiv:1402.3541. Bibcode:2014SIGMA..10..084C. doi:10.3842/SIGMA.2014.084. S2CID18776942. Retrieved from " Categories: Rotation in three dimensions Angle Hidden categories: Articles with short description Short description matches Wikidata Wikipedia articles needing clarification from January 2014 This page was last edited on 27 November 2024, at 22:30(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 Search Search [x] Toggle the table of contents Axis–angle representation 4 languagesAdd topic
190059	https://www.genome.gov/genetics-glossary/Stop-Codon	Stop Codon Scan to visit Skip to main content Due to reduction in workforce efforts, the information on this website may not be up to date, transactions submitted via the website may not be processed, and the agency may not be able to respond to inquiries. Note: Securing and protecting this website will continue. Skip to navigationSkip to searchSkip to sliderSkip to aboutSkip to subscriptionSkip to footer About Genomics About Genomics Introduction to Genomics Educational Resources Policy Issues in Genomics The Human Genome Project History of Genomics Program Research Funding Research Funding Funding Opportunities Funded Programs & Projects Division and Program Directors Scientific Program Analysts Contacts by Research Area News & Events Research at NHGRI Research at NHGRI Research Areas Research Investigators Staff Clinicians Research Projects Clinical Research Data Tools & Resources News & Events About Health About Health Genomics & Medicine Family Health History For Patients & Families For Health Professionals Careers & Training Careers & Training Jobs at NHGRI Training at NHGRI Funding for Research Training Professional Development Programs NHGRI Culture News & Events News & Events News Events Social Media Broadcast Media Image Gallery Videos Press Resources About NHGRI About NHGRI Organization Mission and Vision Policies and Guidance Budget Institute Advisors Strategic Vision Partner with NHGRI Staff Search Contact Us Breadcrumb Home About Genomics Educational Resources Talking Glossary of Genomic and Genetic Terms Stop Codon Home About Genomics Educational Resources Talking Glossary of Genomic and Genetic Terms En Español Stop Codon updated: August 28, 2025 Definition 00:00 … A stop codon is a sequence of three nucleotides (a trinucleotide) in DNA or messenger RNA (mRNA) that signals a halt to protein synthesis in the cell. There are 64 different trinucleotide codons: 61 specify amino acids and 3 are stop codons (i.e., UAA, UAG and UGA). Narration 00:00 … Stop codon. A stop codon is to protein synthesis what a period is to a sentence. Since a protein isn't infinitely long, biology created an incredibly clever way to say, “Stop! You've reached the end of this protein.” Prabarna Ganguly, Ph.D. Chief Communications Officer Nucleus Genomics Search Back to Glossary Related Amino Acids Codon Nonsense Mutation Messenger RNA (mRNA) Get Updates Enter your email address to receive updates about the latest advances in genomics research. Subscribe! Social Media Stream Footer Links Contact Accessibility Site Map Staff Search Plug-Ins Used by HHS FOIA Privacy Copyright HHS Vulnerability Disclosure
190060	https://brainly.com/question/14924566	[FREE] If a couple were planning to have three children, the sample space summarizing the gender outcomes would - brainly.com 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 +60,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +22,6k Ace exams faster, with practice that adapts to you Practice Worksheets +8,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified If a couple were planning to have three children, the sample space summarizing the gender outcomes would be: bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg. a. Construct a similar sample space for the possible gender outcomes (using b for boy and g for girl) of two children. b. Assuming that the outcomes listed in part (a) were equally likely, find the probability of getting two girl children. c. Find the probability of getting exactly one boy child and one girl child. 1 See answer Explain with Learning Companion NEW Asked by DarkPhantom2414 • 02/26/2020 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 19985847 people 19M 5.0 0 Upload your school material for a more relevant answer (a) The sample space is: S = {bb, bg, gb, gg} (b) The probability that the couple has two girl children is 0.25. (c) The probability that the couple has exactly 1 boy and 1 girl child is 0.50. Explanation A boy child is denoted by, b. A girl child is denoted by, g. (a) A couple has two children. The sample space for the possible gender of the two children are: The couple can have two boys, two girls or 1 boy and 1 girl. So the sample space is: S = {bb, bg, gb, gg} (b) It is provided that the outcomes of the sample space S are equally likely, i.e. each outcome has the same probability of success. Compute the probability that the couple has two girl children as follows: P (2 Girls) = Favorable no. of outcomes ÷ Total no. of outcomes =4 1=0.25 Thus, the probability that the couple has two girl children is 0.25. (c) Compute the probability that the couple has exactly 1 boy and 1 girl child as follows: P (1 boy & 1 girl) = Favorable no. of outcomes ÷ Total no. of outcomes =4 2=0.50 Thus, the probability that the couple has exactly 1 boy and 1 girl child is 0.50. Answered by warylucknow •2.3K answers•20M people helped Thanks 0 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 19985847 people 19M 0.0 0 Statistics - Barbara Illowsky, Susan Dean Chemical Process Dynamics and Controls - Peter J. Woolf A Practicum in Behavioral Economics - Arthur J Caplan Upload your school material for a more relevant answer The sample space for two children is {bb, bg, gb, gg}. The probability of having two girls is 0.25, and the probability of having one boy and one girl is 0.50. Explanation (a) The sample space for having two children can be represented as follows: The possible gender combinations for two children are: bb (boy, boy) bg (boy, girl) gb (girl, boy) gg (girl, girl) So the sample space is: S = {bb, bg, gb, gg} (b) To find the probability of having two girl children: In our sample space, there is only one outcome where there are two girls (gg). The total number of outcomes in our sample space is 4. Therefore, the probability P(2 Girls) can be calculated as: P(2 Girls) = Total outcomes Favorable outcomes=4 1=0.25 (c) To find the probability of getting exactly one boy and one girl child: There are two outcomes in our sample space that represent having one boy and one girl: (bg, gb). Again, the total number of outcomes is 4. So, the probability P(1 Boy & 1 Girl) can be calculated as: P(1 Boy & 1 Girl) = Total outcomes Favorable outcomes=4 2=0.50 Examples & Evidence For example, if a couple has two children, they could have two boys (bb), one boy and one girl (bg or gb), or two girls (gg). This illustrates how each combination represents a different outcome in the sample space. The calculations used to find the probabilities are based on the principle of counting the favorable outcomes and dividing by the total outcomes in the sample space. Thanks 0 0.0 (0 votes) Advertisement DarkPhantom2414 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer If a couple were planning to have three children, the sample space summarizing the gender outcomes would be: bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg. a. construct a similar sample space for the possible health outcomes (using h for healthy and s for sick) of two children. b. assuming that the outcomes listed in part (a) were equally likely, find the probability of getting two healthy children. c. find the probability of getting exactly one healthy child and one sick child. Community Answer If a couple were planning to have three children, the sample space summarizing the gender outcomes would be: bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg.A.) construct a similiar sample space for the possible weight outcomes (using o for overweight and u for underweight) of two children.B.) Assuming that the outcomes listed in part (a) were equally likely, find the probablity of getting two underweight children.C.) Find the probability of getting one overweight child and one underweight child. 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 New questions in Mathematics Compute the balance and interest earned for the following account. Round your answer to the nearest dollar. principal = 9,400,r a t e=11.9 □ Interest earned: $ □ Write the slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (1,−7) and is perpendicular to the line whose equation is x=−7. The equation of the function is □. Write the slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (9,−3) and is perpendicular to the line whose equation is x=17. For the real-valued functions g(x)=x 2+3 and h(x)=x 2−2, find the composition g∘h and specify its domain using interval notation. The rational function f is given by f(x)=−5 x 5+3 x 2+15 x+7 x−3 12. Which of the following describes the end behavior of f ?A. lim x→−∞f(x)=∞ and lim x→∞f(x)=∞.B. lim x→−∞f(x)=∞ and lim x→∞f(x)=−∞.C. lim x→−∞f(x)=−∞ and lim x→∞f(x)=∞.D. lim x→−∞f(x)=−∞ and lim x→∞f(x)=−∞. 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190061	https://brilliant.org/wiki/adjacency-matrix/	Reset password New user? Sign up Existing user? Log in Adjacency Matrix Already have an account? Log in here. An adjacency matrix is a compact way to represent the structure of a finite graph. If a graph has (n) vertices, its adjacency matrix is an (n \times n) matrix, where each entry represents the number of edges from one vertex to another. Some properties and computations in graph theory can be carried out efficiently and elegantly by using a graph's adjacency matrix. Moreover, certain questions about properties of the adjacency matrix correspond to interesting questions about the associated graph. Contents Building an Adjacency Matrix Given a graph (G) with (n) vertices, the adjacency matrix is an ( n \times n ) matrix [ A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \ a_{21} & a_{22} & \cdots & a_{2n} \ \vdots & \vdots & \ddots & \vdots \ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}, ] where the value ( a_{ij} ) equals the number of edges from the vertex (i) to the vertex (j.) If the graph is undirected, then (a_{ij} = a_{ji}) for all (i,j;) that is, the adjacency matrix is symmetric. When the graph is simple, the entries of the adjacency matrix are all (0) or (1) (there are no weights on the edges, or multiple edges) and the diagonal entries are all (0) (there are no loops). More complicated types of graphs give rise to more kinds of matrices. Create an adjacency matrix for the following weighted, undirected graph: The weights on the edges go into the entries of the adjacency matrix, as follows: [ A = \begin{pmatrix} 0&3&0&0&0&12&0 \ 3&0&5&0&0&0&4 \ 0&5&0&6&0&0&3 \ 0&0&6&0&1&0&0 \ 0&0&0&1&0&10&7 \ 12&0&0&0&10&0&2 \ 0&4&3&0&7&2&0 \end{pmatrix}.\ _\square ] This example suggests some follow-up questions, which will be addressed in a later section: Matrix Properties and Graph Properties The adjacency matrix is an array of numbers that represents all the information about the graph. Some of the properties of the graph correspond to interesting properties of its adjacency matrix, and vice versa. Here is a simple example: [m] [2n] [2m] None of these [n] Reveal the answer Let (G) be a simple graph with (n) vertices and (m) edges: that is, (G) is undirected, unweighted, and has no loops (edges from a vertex to itself). Let (A) be the adjacency matrix of (G) and let ({\bf u} = \begin{pmatrix} 1&1&\ldots&1 \end{pmatrix} ) be the (1 \times n) matrix whose entries are all equal to (1.) Then ({\bf u} A {\bf u}^T) is a (1 \times 1) matrix (\begin{pmatrix} b \end{pmatrix}). What is (b)? Notation: if (B) is a matrix, (B^T) denotes its transpose. The correct answer is: [2m] Let (G) be a simple graph with (n) vertices and (m) edges: that is, (G) is undirected, unweighted, and has no loops (edges from a vertex to itself). Let (A) be the adjacency matrix of (G) and let ({\bf u} = \begin{pmatrix} 1&1&\ldots&1 \end{pmatrix} ) be the (1 \times n) matrix whose entries are all equal to (1.) Then ({\bf u} A {\bf u}^T) is a (1 \times 1) matrix (\begin{pmatrix} b \end{pmatrix}). What is (b)? Notation: if (B) is a matrix, (B^T) denotes its transpose. Powers: One of the most well-known ways to get information about the graph from operations on the adjacency matrix is via its powers. As the following example shows, the entries of the powers of the adjacency matrix give information about paths in the graph. Let (A) be the adjacency matrix of a simple graph with (6) vertices. Suppose the entry of (A^2) in the (3^\text{rd}) row and the (4^\text{th}) column is (5.) What does this mean about the graph? Suppose (A = (a_{ij}).) Then the (3,4) entry of (A^2) is [a_{31}a_{14} + a_{32}a_{24} + a_{33}a_{34} + a_{34}a_{44} + a_{35}a_{54} + a_{36}a_{64}.] Consider (a_{31}a_{14}.) This counts the number of paths from vertex 3 to 1, multiplied by the number of paths from vertex 1 to vertex 4. This is exactly the number of two-step paths from vertex 3 to vertex 4 which pass through vertex 1. The other terms (a_{3k}a_{k4}) count the number of two-step paths from vertex 3 to vertex 4 which pass through vertex (k.) So the sum counts all of the two-step paths from vertex 3 to vertex 4. The conclusion is that there are (5) two-step paths from vertex 3 to vertex 4. (_\square) So the (i,j) entries of (A^2) count two-step walks from (i) to (j.) This generalizes as follows: Let (A) be the adjacency matrix of a graph. The (i,j) entry of (A^n) counts (n)-step walks from vertex (i) to vertex (j.) (()The proof is a straightforward induction on (n,) where the argument is essentially the same as the one given above for (A^2.)) Spectrum: The study of the eigenvalues of the adjacency matrix of a graph is called spectral graph theory. Here is a simple example of a correspondence between graph properties and eigenvectors of the adjacency matrix: An undirected graph is (k)-regular if the degree of each vertex is (k.) What can be said about the eigenvalues and eigenvectors of the adjacency matrix of a (k)-regular graph? Let (A) be the adjacency matrix of a (k)-regular graph. Let (\bf v) be the all-ones column vector in ({\mathbb R}^n.) Then the (i^\text{th}) entry of (A{\bf v}) equals the sum of the entries in the (i^\text{th}) row of (A.) This is the number of edges emanating from vertex (i,) which is exactly (k.) So [ A \begin{pmatrix} 1\1\\vdots\1 \end{pmatrix} = \begin{pmatrix} k\k\\vdots\k \end{pmatrix} = k\begin{pmatrix} 1\1\\vdots\1 \end{pmatrix}, ] i.e. (\bf v) is an eigenvector of (A) with eigenvalue (k.) (_\square) Isomorphisms: Two graphs are said to be isomorphic if one can be obtained from the other by relabeling vertices. Note that isomorphic graphs need not have the same adjacency matrix, since the adjacency matrix depends on the labeling of the vertices. But the adjacency matrices of isomorphic graphs are very closely related. Let (G) and (H) be graphs on (n) vertices with adjacency matrices (A) and (B.) Then (G) and (H) are isomorphic if and only if there is a permutation matrix (P) such that (B = PAP^{-1}.) Recall that a permutation matrix is a square matrix whose entries are all (0) or (1,) such that each row and each column contains exactly one (1.) Applications See also Graph theory Graphs - intermediate Reset password New user? Sign up Existing user? Log in
190062	https://www.reproductiveaccess.org/wp-content/uploads/2022/01/2024-US-MEC.pdf	Updated in 2024. This summary sheet only contains a subset of the recommendations from the U.S. MEC. For complete guidance, see: Most contraceptive methods do not protect against STIs. Consistent and correct use of the external (male) latex condom reduces the risk of STIs and HIV. Please see NIH guidelines for up to date recommendations on hormonal contraception and ARVs: and Condition Sub-Condition Cu-IUD LNG-IUD Implant DMPA POP CHC I C I C I C I C I C I C Age Menarche to <20 yrs:2 Menarche to <20 yrs:2 Menarche to <18 yrs:1 Menarche to <18 yrs:2 Menarche to <18 yrs:1 Menarche to <40 yrs:1 ≥20 yrs:1 ≥20 yrs:1 18–45 yrs:1 18–45 yrs:1 18–45 yrs:1 ≥40 yrs:2 >45 yrs:1 >45 yrs:2 >45 yrs:1 Anatomical abnormalities a. Distorted uterine cavity 4 4 b. Other abnormalities 2 2 Anemia, iron-deficiency 2 1 1 1 1 1 Benign ovarian tumors (including cysts) 1 1 1 1 1 1 Breast disease a. Undiagnosed mass 1 2 2 2 2 2 b. Benign breast disease 1 1 1 1 1 1 c. Family history of cancer 1 1 1 1 1 1 d. Breast cancer‡ i. Current 1 4 4 4 4 4 ii. Past and no evidence of current disease for 5 years 1 3 3 3 3 3 Breastfeeding a. <21 days postpartum 2 2 2 4 b. 21 to <30 days postpartum i. With other risk factors for VTE 2 2 2 3 ii. Without other risk factors for VTE 2 2 2 3 c. 30-42 days postpartum i. With other risk factors for VTE 1 2 1 3 ii. Without other risk factors for VTE 1 1 1 2 d. >42 days postpartum 1 1 1 2 Cervical cancer Awaiting treatment 4 2 4 2 2 2 1 2 Cervical ectropion 1 1 1 1 1 1 Cervical intraepithelial neoplasia 1 2 2 2 1 2 Chronic kidney disease‡ a. Current nephrotic syndrome 1 1 2 2 2 3 2/4 4 b. Hemodialysis 1 1 2 2 2 3 2/4 4 c. Peritoneal dialysis 2 1 2 2 2 3 2/4 4 Cirrhosis a. Compensated (normal liver function) 1 1 1 1 1 1 b. Decompensated‡ (impaired liver function) 1 2 2 3 2 4 Cystic fibrosis‡ 1 1 1 2 1 1 Deep venous thrombosis (DVT)/Pulmonary embolism (PE)‡ a. Current or history of DVT/PVE, receiving anticoagulant therapy (therapeutic dose) 2 2 2 2 2 3 b. History of DVT/PE, receiving anticoagulant therapy (prophylactic dose) i. Higher risk for recurrent DVT/PE 2 2 2 3 2 4 ii. Lower risk for recurrent DVT/PE 2 2 2 2 2 3 c. History of DVT/PE, not receiving anticoagulant therapy i. Higher risk for recurrent DVT/PE 1 2 2 3 2 4 ii. Lower risk for recurrent DVT/PE 1 2 2 2 2 3 d. Family history (first-degree relatives) 1 1 1 1 1 2 Depressive disorders 1 1 1 1 1 1 Condition Sub-Condition Cu-IUD LNG-IUD Implant DMPA POP CHC I C I C I C I C I C I C Diabetes a. History of gestational disease 1 1 1 1 1 1 b. Nonvascular disease i. Non-insulin dependent 1 2 2 2 2 2 ii. Insulin dependent‡ 1 2 2 2 2 2 c. Nephropathy, retinopathy, or neuropathy‡ 1 2 2 3 2 3/4 d. Other vascular disease or diabetes of >20 years’ duration‡ 1 2 2 3 2 3/4 Dysmenorrhea Severe 2 1 1 1 1 1 Endometrial cancer‡ 4 2 4 2 1 1 1 1 Endometrial hyperplasia 1 1 1 1 1 1 Endometriosis 2 1 1 1 1 1 Epilepsy‡ (see also Drug Interactions) 1 1 1 1 1 1 Gallbladder disease a. Asymptomatic 1 2 2 2 2 2 b. Symptomatic i. Current 1 2 2 2 2 3 ii. Treated by cholecystectomy 1 2 2 2 2 2 iii. Medically treated 1 2 2 2 2 3 Gestational trophoblastic disease (GTD)‡ a. Suspected GTD (immediate postevacuation) i. Uterine size first trimester 1 1 1 1 1 1 ii. Uterine size second trimester 2 2 1 1 1 1 b. Confirmed GTD i. Undectectable or non-pregnant β-hCG levels 1 1 1 1 1 1 1 1 ii. Decreasing ß-hCG levels 2 1 2 1 1 1 1 1 iii. Persistently elevated ß-hCG levels or malignant disease, with no evidence or suspicion of intrauterine disease 2 1 2 1 1 1 1 1 iv. Persistently elevated ß-hCG levels or malignant disease, with evidence or suspicion of intrauterine disease 4 2 4 2 1 1 1 1 Headaches a. Nonmigraine (mild or severe) 1 1 1 1 1 1 b. Migraine i. Without aura (includes menstrual migraine) 1 1 1 1 1 2 ii. With aura 1 1 1 1 1 4 History of bariatric surgery‡ a. Restrictive procedures 1 1 1 1 1 1 b. Malabsorptive procedures 1 1 1 1 3 COCs: 3 P/R: 1 History of cholestasis a. Pregnancy related 1 1 1 1 1 2 b. Past COC related 1 2 2 2 2 3 History of high blood pressure during pregnancy 1 1 1 1 1 2 History of pelvic surgery (see also Postpartum [including cesarean delivery]) 1 1 1 1 1 1 HIV a. High risk for HIV 1 1 1 1 1 1 1 1 b. HIV infection 1 1 1 1 i. Clinically well receiving ARV therapy 1 1 1 1 If on ARV, see also Drug Interactions. ii. Not clinically well or not receiving ARV therapy‡ 2 1 2 1 If on ARV, see also Drug Interactions. For accessible version, please see the summary of classifications at Summary Chart of U.S. Medical Eligibility Criteria for Contraceptive Use (U.S. MEC) KEY: Abbreviations: ARV = antiretroviral; C = continuation of contraceptive method; CHC = combined hormonal contraceptive (pill, patch, and ring); COC = combined oral contraceptive; Cu-IUD = copper intrauterine device; DMPA = depot medroxyprogesterone acetate; I = initiation of contraceptive method; LNG-IUD = levonorgestrel intrauterine device; NA = not applicable; POP = progestin-only pill; P/R = patch/ring; SSRI = selective serotonin reuptake inhibitor; STI = sexually transmitted infection; VTE = venous thromboembolism. ‡Condition associated with increased risk as a result of pregnancy. Please see the complete guidance for a clarification to this classification: CS 349509-A 1 = No restriction (method can be used) 2 = Advantages generally outweigh theoretical or proven risks 3 = Theoretical or proven risks usually outweigh the advantages 4 = Unacceptable health risk (method not to be used) This cell is empty. This cell is empty. This cell is empty. This cell is empty. Condition Sub-Condition Cu-IUD LNG-IUD Implant DMPA POP CHC I C I C I C I C I C I C Hypertension a. Adequately controlled hypertension 1 1 1 2 1 3 b. Elevated blood pressure levels (properly taken measurements) i. Systolic 140-159 or diastolic 90-99 1 1 1 2 1 3 ii. Systolic ≥160 or diastolic ≥100‡ 1 2 2 3 2 4 c. Vascular disease 1 2 2 3 2 4 Inflammatory bowel disease (ulcerative colitis or Crohn’s disease) 1 1 1 2 2 2/3 Ischemic heart disease‡ Current and history of 1 2 3 2 3 3 2 3 4 Liver tumors a. Benign i. Focal nodular hyperplasia 1 2 2 2 2 2 ii. Hepatocellular adenoma‡ 1 2 2 3 2 4 b. Malignant‡ (hepatocellular carcinoma) 1 3 3 3 3 4 Malaria 1 1 1 1 1 1 Multiple risk factors for atherosclerotic cardiovascular disease (e.g., older age, smoking, diabetes, hypertension, low HDL, high LDL, or high triglyceride levels) 1 2 2 3 2 3/4 Multiple sclerosis a. Without prolonged immobility 1 1 1 2 1 1 b. With prolonged immobility 1 1 1 2 1 3 Obesity a. Body mass index (BMI) ≥30 kg/m2 1 1 1 1 1 2 b. Menarche to <18 years and BMI ≥30 kg/m2 1 1 1 2 1 2 Ovarian cancer‡ 1 1 1 1 1 1 Parity a. Nulliparous 2 2 1 1 1 1 b. Parous 1 1 1 1 1 1 Past ectopic pregnancy 1 1 1 1 2 1 Pelvic inflammatory disease a. Current 4 2 4 2 1 1 1 1 b. Past i. With subsequent pregnancy 1 1 1 1 1 1 1 1 ii. Without subsequent pregnancy 2 2 2 2 1 1 1 1 Peripartum cardiomyopathy‡ a. Normal or mildly impaired cardiac function i. <6 months 2 2 1 2 1 4 ii. ≥6 months 2 2 1 2 1 3 b. Moderately or severely impaired cardiac function 2 2 2 3 2 4 Postabortion (spontaneous or induced) a. First trimester abortion i. Procedural (surgical) 1 1 1 1 1 1 ii. Medication 1 1 1 1/2 1 1 iii. Spontaneous abortion with no intervention 1 1 1 1 1 1 b. Second trimester abortion i. Procedural (surgical) 2 2 1 1 1 1 ii. Medication 2 2 1 1 1 1 iii. Spontaneous abortion with no intervention 2 2 1 1 1 1 c. Immediate postseptic abortion 4 4 1 1 1 1 Postpartum (nonbreastfeeding) a. <21 days 1 2 1 4 b. 21 days to 42 days i. With other risk factors for VTE 1 2 1 3 ii. Without other risk factors for VTE 1 1 1 2 c. >42 days 1 1 1 1 Postpartum (including cesarean delivery, breastfeeding, or nonbreastfeeding) a. <10 minutes after delivery of the placenta 2 2 b. 10 minutes after delivery of the placenta to <4 weeks 2 2 c. ≥4 weeks 1 1 d. Postpartum sepsis 4 4 Pregnancy 4 4 NA NA NA NA Condition Sub-Condition Cu-IUD LNG-IUD Implant DMPA POP CHC I C I C I C I C I C I C Rheumatoid arthritis a. Not on immunosuppressive therapy 1 1 1 2 1 2 b. On immunosuppressive therapy 2 1 2 1 1 2/3 1 2 Schistosomiasis a. Uncomplicated 1 1 1 1 1 1 b. Fibrosis of the liver‡ (if severe, see also Cirrhosis) 1 1 1 1 1 1 Sexually transmitted infections (STIs) a. Current purulent cervicitis or chlamydial infection or gonococcal infection 4 2 4 2 1 1 1 1 b. Vaginitis (including Trichomonas vaginalis and bacterial vaginosis) 2 2 2 2 1 1 1 1 c. Other factors related to STIs 2 2 2 2 1 1 1 1 Sickle cell disease‡ 2 1 1 2/3 1 4 Smoking a. Age <35 1 1 1 1 1 2 b. Age ≥35, <15 cigarettes/day 1 1 1 1 1 3 c. Age ≥35, ≥15 cigarettes/day 1 1 1 1 1 4 Solid organ transplantation‡ a. No graft failure 1 1 1 1 2 2/3 2 2 b. Graft failure 2 1 2 1 2 2/3 2 4 Stroke‡ History of cerebrovascular accident 1 2 2 3 3 2 3 4 Superficial venous disorders a. Varicose veins 1 1 1 1 1 1 b. Superficial venous thrombosis (acute or history) 1 1 1 2 1 3 Surgery a. Minor surgery without immobilization 1 1 1 1 1 1 b. Major surgery i. Without prolonged immobilization 1 1 1 1 1 2 ii. With prolonged immobilization 1 1 1 2 1 4 Systemic lupus erythematosus‡ a. Positive (or unknown) antiphospholipid antibodies 1 1 2 2 3 3 2 4 b. Severe thrombocytopenia 3 2 2 2 3 2 2 2 c. Immunosuppressive therapy 2 1 2 2 2 2 2 2 d. None of the above 1 1 2 2 2 2 2 2 Thalassemia 2 1 1 1 1 1 Thrombophilia‡ 1 2 2 3 2 4 Thyroid disorders Simple goiter, hyperthyroid, or hypothyroid 1 1 1 1 1 1 Tuberculosis‡ (see also Drug Interactions) a. Nonpelvic 1 1 1 1 1 1 1 1 b. Pelvic 4 3 4 3 1 1 1 1 Unexplained vaginal bleeding (suspicious for serious condition) before evaluation 4 2 4 2 3 3 2 2 Uterine fibroids 2 2 1 1 1 1 Valvular heart disease a. Uncomplicated 1 1 1 1 1 2 b. Complicated‡ 1 1 1 2 1 4 Vaginal bleeding patterns a. Irregular pattern without heavy bleeding 1 1 1 2 2 2 1 b. Heavy or prolonged bleeding 2 1 2 2 2 2 1 Viral hepatitis a. Acute or flare 1 1 1 1 1 3/4 2 b. Chronic 1 1 1 1 1 1 1 Drug Interactions Antiretrovirals (ARVs) used for prevention (PrEP) or treatment of HIV‡ Fosamprenavir (FPV) 1/2 1 1/2 1 2 2 2 3 All other ARVs are 1 or 2 for all methods Anticonvulsant therapy a. Certain anticonvulsants (phenytoin, carbamazepine, barbiturates, primidone, topiramate, oxcarbazepine) 1 1 2 1 3 3 b. Lamotrigine 1 1 1 1 1 3 Antimicrobial therapy a. Broad-spectrum antibiotics 1 1 1 1 1 1 b. Antifungals 1 1 1 1 1 1 c. Antiparasitics 1 1 1 1 1 1 d. Rifampin or rifabutin therapy 1 1 2 1 3 3 SSRIs 1 1 1 1 1 1 St. John’s wort 1 1 2 1 2 2 Summary Chart of U.S. Medical Eligibility Criteria for Contraceptive Use (U.S. MEC) CS 349509-A This cell is empty. This cell is empty. This cell is empty. This cell is empty.
190063	https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_3?srsltid=AfmBOoq7LdqXkoJY2KlIEEXCntkvnTacmjeFg4yyHfNl2swFi4f2z6bp	Art of Problem Solving 2006 AIME I Problems/Problem 3 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2006 AIME I Problems/Problem 3 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2006 AIME I Problems/Problem 3 Contents [hide] 1 Problem 2 Solutions 2.1 Solution 1 2.2 Solution 2 3 Solution 3 (Quick) 4 Solution 4 5 See also Problem Find the least positiveinteger such that when its leftmost digit is deleted, the resulting integer is of the original integer. Solutions Solution 1 Suppose the original number is where the are digits and the first digit, is nonzero. Then the number we create is so But is with the digit added to the left, so Thus, The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number is never divisible by so must be divisible by But is a nonzero digit, so the only possibility is This gives or Now, we want to minimize both and so we take and Then and indeed, Solution 2 Let be the required number, and be with the first digit deleted. Now, we know that (because this is an AIME problem). Thus, has or digits. Checking the other cases, we see that it must have digits. Let , so . Thus, . By the constraints of the problem, we see that , so Now, we subtract and divide to get Clearly, must be a multiple of because both and are multiples of . Thus, . Now, we plug that into the equation: By the same line of reasoning as earlier, . We again plug that into the equation to get Now, since , , and , our number . Here's another way to finish using this solution. From the above, you have Divide by , and you get This means that has to be divisible by , and hence Now, solve for , which gives you , giving you the number Solution 3 (Quick) Note that if we let the last digit be we must have Thus we either have which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly ~Dhillonr25 (Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000). Solution 4 First we try digit numbers like . Removing the leftmost digit gives us . Now, using the information we are given we can write the equation: . Since and are relatively prime, the smallest case that works is , and , but is an digit so the number must be at least digits. We let the number be . So, using the info we know, we can conclude that . Rerraranging gives us . So, sicne and are relatively prime, the lowest possible case is that , and .So, we have . So the desired number is -jb2015007 See also Number Theory 2006 AIME I (Problems • Answer Key • Resources) Preceded by Problem 2Followed by Problem 4 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Intermediate Number Theory Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190064	https://www.ipt.ntnu.no/~kleppe/TPG4150/text.pdf	Developments in Petroleum Science, 8 fundamentals of reservoir engineering FURTHER TITLES IN THIS SERIES 1 A. GENE COLLINS GEOCHEMISTRY OF OILFIELD WATERS 2 W.H. FERTL ABNORMAL FORMATION PRESSURES 3 A.P. SZILAS PRODUCTION AND TRANSPORT OF OIL AND GAS 4 C.E.B. CONYBEARE GEOMORPHOLOGY OF OIL AND GAS FIELDS IN SANDSTONE BODIES 5 T.F. YEN and G.V. CHILINGARIAN (Editors) OIL SHALE 6 D.W. PEACEMAN FUNDAMENTALS OF NUMERICAL RESERVOIR SIMULATION 7 G.V. CHILINGARIAN and T.F. YEN (Editors) BITUMENS, ASPHALTS AND TAR SANDS 8 L.P. DAKE FUNDAMENTALS OF RESERVOIR ENGINEERING 9 K. MAGARA COMPACTION AND FLUID MIGRATION 10 M.T. SILVIA and E.A. ROBINSON DECONVOLUTION OF GEOPHYSICAL TIME SERIES IN THE EXPLORATION FOR OIL AND NATURAL GAS Developments in Petroleum Science, 8 fundamentals of reservoir engineering LP. DAKE Senior Lecturer in Reservoir Engineering, Shell Internationale Petroleum Maatschappij B. V., The Hague, The Netherlands ELSEVIER, AmsterdamLondonNew YorkTokyo ELSEVIER SCIENCE B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands First edition 1978 Second impression l979 Third impression 1980 Fourth impression 1981 Fifth impression 1982 Sixth impression 1982 Seventh impression 1983 Eighth impression 1985 Ninth impression 1986 Tenth impression 1988 Eleventh impression 1990 Twelfth impression 1991 Thirteenth impression 1993 Fourteenth impression 1994 Fifteenth impression 1995 Sixteenth impression 1997 Seventeenth impression 1998 ISBN 0-444-41830-X  1978 ELSEVIER SCIENCE B.V. All rights reserved. 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, recording or otherwise, without the prior written permission of the publisher, Elsevier Science B.V., Copyright & Permissions Department, P.O. Box 521, 1000 AM Amsterdam, The Netherlands. Special regulations for readers in the U.S.A.-This publication has been registered with the Copyright Clearance Center Inc. (CCC), 222 Rosewood Drive Danvers, MA 01923. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. This book is printed on acid-free paper Printed in The Netherlands To Grace PREFACE This teaching textbook in Hydrocarbon Reservoir Engineering is based on various lecture courses given by the author while employed in the Training Division of Shell Internationale Petroleum Maatschappij B.V. (SIPM), in the Hague, between 1974 and 1977. The primary aim of the book is to present the basic physics of reservoir engineering, using the simplest and most straightforward of mathematical techniques. It is only through having a complete understanding of the physics that the engineer can hope to appreciate and solve complex reservoir engineering problems in a practical manner. Chapters 1 through 4 serve as an introduction to the subject and contain material presented on Shell's basic training courses. They should therefore be of interest to anyone even remotely connected with the business of developing and producing hydrocarbon reserves. Chapters 5 through 8 are more specialised describing the theory and practice of well testing and pressure analysis techniques, which are probably the most important subjects in the whole of reservoir engineering. The approach is entirely general in recognising that the superposition of dimensionless pressure, or pseudo pressure functions, perm its the analysis of any rate-pressure-time record retrieved from a well test, for any type of reservoir fluid. To appreciate this generality, the reader is advised to make a cursory inspection of section 8.13 (page 295), before embarking on a more thorough reading of these chapters. The author hopes that this will serve as a useful introduction to the recently published and, as usual, excellent SPE Monograph (Advances in Well Test Analysis; by Robert C. Earlougher, Jr.), in which a knowledge is assumed of much of the theory presented in these four chapters. Chapter 9 describes the art of aquifer modelling, while Chapter 10, the final chapter, covers the subject of immiscible, incompressible displacement. The message here is-that there is but one displacement theory, that of Buckley and Leverett. Everything else is just a matter of "modifying" the relative permeability curves (known in the business as "scientific adjustment"), to account for the manner in which the fluid saturations are distributed in the dip-normal direction. These curves can then be used in conjunction with the one dimensional Buckley-Leverett equation to calculate the oil recovery. By stating the physics implicit in the generation of averaged (pseudo) relative permeabilities and illustrating their role in numerical simulation, it is hoped that this chapter will help to guide the hand of the scientific adjuster. The book also contains numerous fully worked exercises which illustrate the theory. The most notable omission, amongst the subjects covered, is the lack of any serious discussion on the complexities of hydrocarbon phase behaviour. This has al ready been made the subject of several specialist text books, most notably that of Amyx, Bass and Whiting (reference 8, page 42), which is frequently referred to throughout this text. PREFACE / ACKNOWLEDGEMENTS VIII A difficult decision to make, at the time of writing, is which set of units to employ. Although the logical decision has been made that the industry should adopt the SI (Système Internationale) units, no agreement has yet been reached concerning the extent to which "allowable" units, expressed in terms of the basic units, will be tolerated. To avoid possible error the author has therefore elected to develop the important theoretical arguments in Darcy units, while equations required for application in the field are stated in Field units. Both these systems are defined in table 4.1, in Chapter 4, which appropriately is devoted to the description of Darcy's law. This chapter also contains a section, (4.4), which describes how to convert equations expressed in one set of units to the equivalent form in any other set of units. The choice of Darcy units is based largely on tradition. Equations expressed in these units have the same form as in absolute units except in their gravity terms. Field units have been used in practical equations to enable the reader to relate to the existing AIME literature. PREFACE / ACKNOWLEDGEMENTS IX ACKNOWLEDGEMENTS The author wishes to express his thanks to SIPM for so readily granting permission to publish this work and, in particular, to H. L. Douwes Dekker, P.C. Kok and C. F.M. Heck for their sustained personal interest throughout the writing and publication, which has been a source of great encouragement. Of those who have offered technical advice, I should like to acknowledge the assistance of G.J. Harmsen; L.A. Schipper; D. Leijnse; J. van der Burgh; L. Schenk; H. van Engen and H. Brummelkamp, all sometime members of Shell's reservoir engineering staff in the Hague. My thanks for technical assistance are also due to the following members of KSEPL (Koninklijke Shell Exploratie en Productie Laboratorium) in Rijswijk, Holland: J. Offeringa; H.L. van Domseiaar; J.M. Dumore; J. van Lookeren and A.S. Williamson. Further, I am grateful to all former lecturers in reservoir engineering in Shell Training, and also to my successor A.J. de la Mar for his many helpful suggestions. Sincere thanks also to S.H. Christiansen (P.D. Oman) for his dedicated attitude while correcting the text over a period of several months, and similarly to J.M. Willetts (Shell Expro, Aberdeen) and B.J.W. Woods (NAM, Assen) for their efforts. For the preparation of the text I am indebted to G.J.W. Fransz for his co-ordinating work, and particulariy to Vera A. Kuipers-Betke for her enthusiastic hard work while composing the final copy. For the drafting of the diagrams and the layout I am grateful to J.C. Janse; C.L. Slootweg; J.H. Bor and S.O. Fraser-Mackenzie. Finally, my thanks are due to all those who suffered my lectures between 1974 and 1977 for their numerous suggestions which have helped to shape this textbook. L.P. Dake, Shell Training, The Hague, October 1977. CONTENTS PREFACE VII ACKNOWLEDGEMENTS IX CONTENTS X LIST OF FIGURES XVII LIST OF TABLES XXVII LIST OF EQUATIONS XXX NOMENCLATURE LIX CHAPTER 1 SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 1 1.1 INTRODUCTION 1 1.2 CALCULATION OF HYDROCARBON VOLUMES 1 1.3 FLUID PRESSURE REGIMES 3 1.4 OIL RECOVERY: RECOVERY FACTOR 9 1.5 VOLUMETRIC GAS RESERVOIR ENGINEERING 12 1.6 APPLICATION OF THE REAL GAS EQUATION OF STATE 20 1.7 GAS MATERIAL BALANCE: RECOVERY FACTOR 25 1.8 HYDROCARBON PHASE BEHAVIOUR 37 REFERENCES 41 CHAPTER 2 PVT ANALYSIS FOR OIL 43 2.1 INTRODUCTION 43 2.2 DEFINITION OF THE BASIC PVT PARAMETERS 43 2.3 COLLECTION OF FLUID SAMPLES 51 2.4 DETERMINATION OF THE BASIC PVT PARAMETERS IN THE LABORATORY AND CONVERSION FOR FIELD OPERATING CONDITIONS 55 2.5 ALTERNATIVE MANNER OF EXPRESSING PVT LABORATORY ANALYSIS RESULTS 65 CONTENTS XI 2.6 COMPLETE PVT ANALYSIS 69 REFERENCES 70 CHAPTER 3 MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 71 3.1 INTRODUCTION 71 3.2 GENERAL FORM OF THE MATERIAL BALANCE EQUATION FOR A HYDROCARBON RESERVOIR 71 3.3 THE MATERIAL BALANCE EXPRESSED AS A LINEAR EQUATION 76 3.4 RESERVOIR DRIVE MECHANISMS 77 3.5 SOLUTION GAS DRIVE 78 3.6 GASCAP DRIVE 86 3.7 NATURAL WATER DRIVE 91 3.8 COMPACTION DRIVE AND RELATED PORE COMPRESSIBILITY PHENOMENA 95 REFERENCES 98 CHAPTER 4 DARCY'S LAW AND APPLICATIONS 100 4.1 INTRODUCTION 100 4.2 DARCY'S LAW; FLUID POTENTIAL 100 4.3 SIGN CONVENTION 104 4.4 UNITS: UNITS CONVERSION 104 4.5 REAL GAS POTENTIAL 110 4.6 DATUM PRESSURES 111 4.7 RADIAL STEADY STATE FLOW; WELL STIMULATION 112 4.8 TWO-PHASE FLOW: EFFECTIVE AND RELATIVE PERMEABILITIES 117 4.9 THE MECHANICS OF SUPPLEMENTARY RECOVERY 121 REFERENCES 125 CHAPTER 5 THE BASIC DIFFERENTIAL EQUATION FOR RADIAL FLOW IN A POROUS MEDIUM 127 5.1 INTRODUCTION 127 5.2 DERIVATION OF THE BASIC RADIAL DIFFERENTIAL EQUATION 127 CONTENTS XII 5.3 CONDITIONS OF SOLUTION 129 5.4 THE LINEARIZATION OF EQUATION 5.1 FOR FLUIDS OF SMALL AND CONSTANT COMPRESSIBILITY 133 REFERENCES 135 CHAPTER 6 WELL INFLOW EQUATIONS FOR STABILIZED FLOW CONDITIONS 136 6.1 INTRODUCTION 136 6.2 SEMI-STEADY STATE SOLUTION 136 6.3 STEADY STATE SOLUTION 139 6.4 EXAMPLE OF THE APPLICATION OF THE STABILIZED INFLOW EQUATIONS 140 6.5 GENERALIZED FORM OF INFLOW EQUATION UNDER SEMI-STEADY STATE CONDITIONS 144 REFERENCES 146 CHAPTER 7 THE CONSTANT TERMINAL RATE SOLUTION OF THE RADIAL DIFFUSIVITY EQUATION AND ITS APPLICATION TO OILWELL TESTING 148 7.1 INTRODUCTION 148 7.2 THE CONSTANT TERMINAL RATE SOLUTION 148 7.3 THE CONSTANT TERMINAL RATE SOLUTION FOR TRANSIENT AND SEMI-STEADY STATE FLOW CONDITIONS 149 7.4 DIMENSIONLESS VARIABLES 161 7.5 SUPERPOSITION THEOREM: GENERAL THEORY OF WELL TESTING 168 7.6 THE MATTHEWS, BRONS, HAZEBROEK PRESSURE BUILDUP THEORY 173 7.7 PRESSURE BUILDUP ANALYSIS TECHNIQUES 189 7.8 MULTI-RATE DRAWDOWN TESTING 209 7.9 THE EFFECTS OF PARTIAL WELL COMPLETION 219 7.10 SOME PRACTICAL ASPECTS OF WELL SURVEYING 221 7.11 AFTERFLOW ANALYSIS 224 REFERENCES 236 CONTENTS XIII CHAPTER 8 REAL GAS FLOW: GAS WELL TESTING 239 8.1 INTRODUCTION 239 8.2 LINEARIZATION AND SOLUTION OF THE BASIC DIFFERENTIAL EQUATION FOR THE RADIAL FLOW OF A REAL GAS 239 8.3 THE RUSSELL, GOODRICH, et. al. SOLUTION TECHNIQUE 240 8.4 THE AL-HUSSAINY, RAMEY, CRAWFORD SOLUTION TECHNIQUE 243 8.5 COMPARISON OF THE PRESSURE SQUARED AND PSEUDO PRESSURE SOLUTION TECHNIQUES 251 8.6 NON-DARCY FLOW 252 8.7 DETERMINATION OF THE NON-DARCY COEFFICIENT F 255 8.8 THE CONSTANT TERMINAL RATE SOLUTION FOR THE FLOW OF A REAL GAS 257 8.9 GENERAL THEORY OF GAS WELL TESTING 260 8.10 MULTI-RATE TESTING OF GAS WELLS 262 8.11 PRESSURE BUILDUP TESTING OF GAS WELLS 278 8.12 PRESSURE BUILDUP ANALYSIS IN SOLUTION GAS DRIVE RESERVOIRS 289 8.13 SUMMARY OF PRESSURE ANALYSIS TECHNIQUES 291 REFERENCES 295 CHAPTER 9 NATURAL WATER INFLUX 297 9.1 INTRODUCTION 297 9.2 THE UNSTEADY STATE WATER INFLUX THEORY OF HURST AND VAN EVERDINGEN 298 9.3 APPLICATION OF THE HURST, VAN EVERDINGEN WATER INFLUX THEORY IN HISTORY MATCHING 308 9.4 THE APPROXIMATE WATER INFLUX THEORY OF FETKOVITCH FOR FINITE AQUIFERS 319 9.5 PREDICTING THE AMOUNT OF WATER INFLUX 328 9.6 APPLICATION OF INFLUX CALCULATION TECHNIQUES TO STEAM SOAKING 333 REFERENCES 335 CONTENTS XIV CHAPTER 10 IMMISCIBLE DISPLACEMENT 337 10.1 INTRODUCTION 337 10.2 PHYSICAL ASSUMPTIONS AND THEIR IMPLICATIONS 337 10.3 THE FRACTIONAL FLOW EQUATION 345 10.4 BUCKLEY-LEVERETT ONE DIMENSIONAL DISPLACEMENT 349 10.5 OIL RECOVERY CALCULATIONS 355 10.6 DISPLACEMENT UNDER SEGREGATED FLOW CONDITIONS 364 10.7 ALLOWANCE FOR THE EFFECT OF A FINITE CAPILLARY TRANSITION ZONE IN DISPLACEMENT CALCULATIONS 381 10.8 DISPLACEMENT IN STRATIFIED RESERVOIRS 389 10.9 DISPLACEMENT WHEN THERE IS A TOTAL LACK OF VERTICAL EQUILIBRIUM 402 10.10 THE NUMERICAL SIMULATION OF IMMISCIBLE, INCOMPRESSIBLE DISPLACEMENT 405 REFERENCES 419 AUTHOR INDEX 422 SUBJECT INDEX 424 CONTENTS XV EXERCISES EXERCISE 1.1 GAS PRESSURE GRADIENT IN THE RESERVOIR 23 EXERCISE 1.2 GAS MATERIAL BALANCE 33 EXERCISE 2.1 UNDERGROUND WITHDRAWAL 50 EXERCISE 2.2 CONVERSION OF DIFFERENTIAL LIBERATION DATA TO GIVE THE FIELD PVT PARAMETERS Bo, Rs AND Bg 63 EXERCISE 3.1 SOLUTION GAS DRIVE; UNDERSATURATED OIL RESERVOIR 79 EXERCISE 3.2 SOLUTION GAS DRIVE; BELOW BUBBLE POINT PRESSURE 80 EXERCISE 3.3 WATER INJECTION BELOW BUBBLE POINT PRESSURE 85 EXERCISE 3.4 GASCAP DRIVE 87 EXERCISE 4.1 UNITS CONVERSION 108 EXERCISE 6.1 WELLBORE DAMAGE 142 EXERCISE 7.1 ei-FUNCTION: LOGARITHMIC APPROXIMATION 154 EXERCISE 7.2 PRESSURE DRAWDOWN TESTING 157 EXERCISE 7.3 DIMENSIONLESS VARIABLES 162 EXERCISE 7.4 TRANSITION FROM TRANSIENT TO SEMI-STEADY STATE FLOW 166 EXERCISE 7.5 GENERATION OF DIMENSIONLESS PRESSURE FUNCTIONS 184 EXERCISE 7.6 HORNER PRESSURE BUILDUP ANALYSIS, INFINITE RESERVOIR CASE 200 EXERCISE 7.7 PRESSURE BUILDUP TEST ANALYSIS: BOUNDED DRAINAGE VOLUME 202 EXERCISE 7.8 MULTI-RATE FLOW TEST ANALYSIS 211 EXERCISE 7.9 AFTERFLOW ANALYSIS TECHNIQUES 231 EXERCISE 8.1 MULTI-RATE GAS WELL TEST ANALYSED ASSUMING STABILIZED FLOW CONDITIONS 265 EXERCISE 8.2 MULTI-RATE GAS WELL TEST ANALYSED ASSUMING UNSTABILIZED FLOW CONDITIONS 272 EXERCISE 8.3 PRESSURE BUILDUP ANALYSIS 282 EXERCISE 9.1 APPLICATION OF THE CONSTANT TERMINAL PRESSURE SOLUTION307 EXERCISE 9.2 AQUIFER FITTING USING THE UNSTEADY STATE THEORY OF HURST AND VAN EVERDINGEN 310 CONTENTS XVI EXERCISE 9.3 WATER INFLUX CALCULATIONS USING THE METHOD OF FETKOVITCH 324 EXERCISE 10.1 FRACTIONAL FLOW 358 EXERCISE 10.2 OIL RECOVERY PREDICTION FOR A WATERFLOOD 361 EXERCISE 10.3 DISPLACEMENT UNDER SEGREGATED FLOW CONDITIONS 375 EXERCISE 10.4 GENERATION OF AVERAGED RELATIVE PERMEABILITY CURVES FOR A LAYERED RESERVOIR (SEGREGATED FLOW) 397 LIST OF FIGURES Fig. 1.1 (a) Structural contour map of the top of the reservoir, and (b) cross section through the reservoir, along the line X−Y 2 Fig. 1.2 Overburden and hydrostatic pressure regimes (FP = fluid pressure; GP = grain pressure) 3 Fig. 1.3 Pressure regimes in the oil and gas for a typical hydrocarbon accumulation 6 Fig. 1.4 Illustrating the uncertainty in estimating the possible extent of an oil column, resulting from well testing in the gas cap 8 Fig. 1.5 Primary oil recovery resulting from oil, water and gas expansion 11 Fig. 1.6 The Z−factor correlation chart of Standing and Katz11 (Reproduced by courtesy of the SPE of the AIME) 17 Fig. 1.7 Pseudo critical properties of miscellaneous natural gases and condensate well fluids19 18 Fig. 1.8 Isothermal Z−factor as a function of pressure (gas gravity = 0.85; temperature = 200° F) 21 Fig. 1.9 Isothermal gas compressibility as a function of pressure (gas gravity = 0.85; temperature = 200° F) 24 Fig. 1.10 Graphical representations of the material balance for a volumetric depletion gas reservoir; equ. (1.35) 28 Fig. 1.11 Graphical representation of the material balance equation for a water drive gas reservoir, for various aquifer strengths; equ. (1.41) 30 Fig. 1.12 Determination of the GIIP in a water drive gas reservoir. The curved, dashed lines result from the choice of an incorrect, time dependent aquifer model; (refer Chapter 9) 31 Fig. 1.13 Gas field development rate−time schedule (Exercise 1.2) 35 Fig. 1.14 Phase diagrams for (a) pure ethane; (b) pure heptane and (c) for a 50−50 mixture of the two 37 Fig. 1.15 Schematic, multi-component, hydrocarbon phase diagrams; (a) for a natural gas; (b) for oil 38 Fig. 2.1 Production of reservoir hydrocarbons (a) above bubble point pressure, (b) below bubble point pressure 44 Fig. 2.2 Application of PVT parameters to relate surface to reservoir hydrocarbon volumes; above bubble point pressure. 45 CONTENTS XVIII Fig. 2.3 Application of PVT parameters to relate surface to reservoir hydrocarbon volumes; below bubble point pressure 47 Fig. 2.4 Producing gas oil ratio as a function of the average reservoir pressure for a typical solution gas drive reservoir 47 Fig. 2.5 PVT parameters (Bo, Rs and Bg), as functions of pressure, for the analysis presented in table 2.4; (pb = 3330 psia). 49 Fig. 2.6 Subsurface collection of PVT sample 52 Fig. 2.7 Collection of a PVT sample by surface recombination 54 Fig. 2.8 Schematic of PV cell and associated equipment 56 Fig. 2.9 Illustrating the difference between (a) flash expansion, and (b) differential liberation 56 Fig. 3.1 Volume changes in the reservoir associated with a finite pressure drop ∆p; (a) volumes at initial pressure, (b) at the reduced pressure 72 Fig. 3.2 Solution gas drive reservoir; (a) above the bubble point pressure; liquid oil, (b) below bubble point; oil plus liberated solution gas 78 Fig. 3.3 Oil recovery, at 900 psia abandonment pressure (% STOIIP), as a function of the cumulative GOR, Rp (Exercise 3.2) 82 Fig. 3.4 Schematic of the production history of a solution gas drive reservoir 84 Fig. 3.5 Illustrating two ways in which the primary recovery can be enhanced; by downdip water injection and updip injection of the separated solution gas 85 Fig. 3.6 Typical gas drive reservoir 87 Fig. 3.7 (a) Graphical method of interpretation of the material balance equation to determine the size of the gascap (Havlena and Odeh) 88 Fig. 3.7 (b) and (c); alternative graphical methods for determining m and N (according to the technique of Havlena and Odeh) 90 Fig. 3.8 Schematic of the production history of a typical gascap drive reservoir 91 Fig. 3.9 Trial and error method of determining the correct aquifer model (Havlena and Odeh) 93 Fig. 3.10 Schematic of the production history of an undersaturated oil reservoir under strong natural water drive 95 Fig. 3.11 (a) Triaxial compaction cell (Teeuw); (b) typical compaction curve 95 Fig. 3.12 Compaction curve illustrating the effect of the geological history of the reservoir on the value of the in-situ compressibility (after Merle) 97 Fig. 4.1 Schematic of Darcy's experimental equipment 101 CONTENTS XIX Fig. 4.2 Orientation of Darcy's apparatus with respect to the Earth's gravitational field 102 Fig. 4.3 Referring reservoir pressures to a datum level in the reservoir, as datum pressures (absolute units) 112 Fig. 4.4 The radial flow of oil into a well under steady state flow conditions 113 Fig. 4.5 Radial pressure profile for a damaged well 114 Fig. 4.6 (a) Typical oil and water viscosities as functions of temperature, and (b) pressure profile within the drainage radius of a steam soaked well 116 Fig. 4.7 Oil production rate as a function of time during a multi-cycle steam soak 116 Fig. 4.8 (a) Effective and (b) corresponding relative permeabilities, as functions of the water saturation. The curves are appropriate for the description of the simultaneous flow of oil and water through a porous medium 118 Fig. 4.9 Alternative manner of normalising the effective permeabilities to give relative permeability curves 119 Fig. 4.10 Water saturation distribution as a function of distance between injection and production wells for (a) ideal or piston-like displacement and (b) non-ideal displacement 121 Fig. 4.11 Illustrating two methods of mobilising the residual oil remaining after a conventional waterflood 124 Fig. 5.1 Radial flow of a single phase fluid in the vicinity of a producing well. 128 Fig. 5.2 Radial flow under semi-steady state conditions 130 Fig. 5.3 Reservoir depletion under semi-steady state conditions. 132 Fig. 5.4 Radial flow under steady state conditions 132 Fig. 6.1 Pressure distribution and geometry appropriate for the solution of the radial diffusivity equation under semi-state conditions 136 Fig. 6.2 Pressure profile during the steam soak production phase 140 Fig. 6.3 Pressure profiles and geometry (Exercise 6.1) 142 Fig. 6.4 Dietz shape factors for various geometries3 (Reproduced by courtesy of the SPE of the AIME). 146 Fig. 7.1 Constant terminal rate solution; (a) constant production rate (b) resulting decline in the bottom hole flowing pressure 148 Fig. 7.2 The exponential integral function ei(x) 153 Fig. 7.3 Graph of the ei-function for 0.001 ≤ × ≤ 5.0 154 CONTENTS XX Fig. 7.4 Single rate drawdown test; (a) wellbore flowing pressure decline during the early transient flow period, (b) during the subsequent semi-steady state decline (Exercise 7.2) 159 Fig. 7.5 Dimensionless pressure as a function of dimensionless flowing time for a well situated at the centre of a square (Exercise 7.4) 168 Fig. 7.6 Production history of a well showing both rate and bottom hole flowing pressure as functions of time 169 Fig. 7.7 Pressure buildup test; (a) rate, (b) wellbore pressure response 171 Fig. 7.8 Multi-rate flow test analysis 172 Fig. 7.9 Horner pressure buildup plot for a well draining a bounded reservoir, or part of a reservoir surrounded by a no-flow boundary 175 Fig. 7.10 Part of the infinite network of image wells required to simulate the no-flow condition across the boundary of a 2 : 1 rectangular part of a reservoir in which the real well is centrally located 176 Fig. 7.11 MBH plots for a well at the centre of a regular shaped drainage area7 (Reproduced by courtesy of the SPE of the AIME) 178 Fig. 7.12 MBH plots for a well situated within; a) a square, and b) a 2:1 rectangle7 (Reproduced by courtesy of the SPE of the AIME) 179 Fig. 7.13 MBH plots for a well situated within; a) a 4:1 rectangle, b) various rectangular geometries7 (Reproduced by courtesy of the SPE of the AIME) 180 Fig. 7.14 MBH plots for a well in a square and in rectangular 2:1 geometries7 181 Fig. 7.15 MBH plots for a well in a 2:1 rectangle and in an equilateral triangle7 (Reproduced by courtesy of the SPE of the AIME). 181 Fig. 7.16 Geometrical configurations with Dietz shape factors in the range, 4.5-5.5 184 Fig. 7.17 Plots of ∆pwf (calculated minus observed) wellbore flowing pressure as a function of the flowing time, for various geometrical configurations (Exercise 7.5) 186 Fig. 7.18 Typical Horner pressure buildup plot 190 Fig. 7.19 Illustrating the dependence of the shape of the buildup on the value of the total production time prior to the survey 192 Fig. 7.20 Analysis of a single set of buildup data using three different values of the flowing time to draw the Horner plot. A - actual flowing time; B - effective flowing time; C -time required to reach semi-steady state conditions 194 Fig. 7.21 The Dietz method applied to determine both the average pressure p and the dynamic grid block pressure d p 197 CONTENTS XXI Fig. 7.22 Numerical simulation model showing the physical no-flow boundary drained by well A and the superimposed square grid blocks used in the simulation 198 Fig. 7.23 Horner buildup plot, infinite reservoir case 201 Fig. 7.24 Position of the well with respect to its no-flow boundary; exercise 7.7 203 Fig. 7.25 Pressure buildup analysis to determine the average pressure within the no-flow boundary, and the dynamic grid block pressure (Exercise 7.7) 204 Fig. 7.26 Influence of the shape of the drainage area and degree of well asymmetry on the Horner buildup plot (Exercise 7.7) 205 Fig. 7.27 Multi-rate oilwell test (a) increasing rate sequence (b) wellbore pressure response 210 Fig. 7.28 Illustrating the dependence of multi-rate analysis on the shape of the drainage area and the degree of well asymmetry. (Exercise 7.8) 214 Fig. 7.30 Multi-rate test analysis in a partially depleted reservoir 219 Fig. 7.31 Examples of partial well completion showing; (a) well only partially penetrating the formation; (b) well producing from only the central portion of the formation; (c) well with 5 intervals open to production (After Brons and Marting19) 220 Fig. 7.32 Pseudo skin factor Sb as a function of b and h/rw (After Brons and Marting19) (Reproduced by courtesy of the SPE of the AIME) 220 Fig. 7.33 (a) Amerada pressure gauge; (b) Amerada chart for a typical pressure buildup survey in a producing well 222 Fig. 7.34 Lowering the Amerada into the hole against the flowing well stream 223 Fig. 7.35 Correction of measured pressures to datum; (a) well position in the reservoir, (b) well completion design 223 Fig. 7.36 Extreme fluid distributions in the well; (a) with water entry and no rise in the tubing head pressure, (b) without water entry and with a rise in the THP 224 Fig. 7.37 Pressure buildup plot dominated by afterflow 225 Fig. 7.38 Russell plot for analysing the effects of afterflow 226 Fig. 7.39 (a) Pressure buildup plot on transparent paper for overlay on (b) McKinley type curves, derived by computer solution of the complex afterflow problem 227 Fig. 7.40 McKinley type curves for 1 min <∆t < 1000 min. (After McKinley21) (Reproduced by courtesy of the SPE of the AIME) 228 Fig. 7.41 Buildup plot superimposed on a particular McKinley type curve for T/F = 5000 230 Fig. 7.42 Deviation of observed buildup from a McKinley type curve, indicating the presence of skin 230 CONTENTS XXII Fig. 7.43 Russell afterflow analysis Exercise 7.9) 234 Fig. 7.44 Match between McKinley type curves and superimposed observed buildup (Exercise 7.9). o − match for small ∆t (T/F = 2500) • − match for large ∆t (T/F = 5000) 235 Fig. 8.1 Radial numerical simulation model for real gas inflow 241 Fig. 8.2 Iterative calculation of pwf using the p2 formulation of the radial, semi-steady state inflow equation, (8.5) 244 Fig. 8.3 Real gas pseudo pressure, as a function of the actual pressure, as derived in table 8.1; (Gas gravity, 0.85; Temperature 200°F) 248 Fig. 8.4 2p/µZ as a function of pressure 248 Fig. 8.5 Calculation of pwf using the radial semi-steady state inflow equation expressed in terms of real gas pseudo pressures, (equ. 8.15) 250 Fig. 8.6 p/µZ as a linear function of pressure 251 Fig. 8.7 Typical plot of p/µZ as a function of pressure 252 Fig. 8.8 Laboratory determined relationship between β and the absolute permeability A relationship is usually derived of the form 256 Fig. 8.9 Gas well test analysis assuming semi-steady state conditions during each flow period. Data; table 8.3 267 Fig. 8.10 Gas well test analysis assuming semi-steady state conditions and applying equ. (8.47). Data; table 8.4 268 Fig. 8.11 MBH chart for the indicated 4:1 rectangular geometry for tDA < .01 (After Earlougher, et.al15) 273 Fig. 8.12 Essis-Thomas analysis of a multi-rate gas well test under assumed transient flow conditions. Data; table 8.6 274 Fig. 8.13 The effect of the length of the individual flow periods on the analysis of a multi-rate gas well test; (a) 4×1 hr periods, (b) 4×2 hrs, (c) 4×4 hrs 277 Fig. 8.14 (a) Rate-time schedule, and (b) corresponding wellbore pressure response during a pressure buildup test in a gas well 279 Fig. 8.15 Complete analysis of a pressure buildup test in a gas well: (a) buildup analysis (table 8.12); (b) and (c) transient flow analyses of the first and second flow periods, respectively (table 8.10) 284 Fig. 8.16 MBH plot for a well at the centre of a square, showing the deviation of mD(MBH) from pD(MBH) for large values of the dimensionless flowing time tDA 287 Fig. 8.17 Iterative determination ofp in a gas well test analysis (Kazemi10) 288 CONTENTS XXIII Fig. 8.18 Schematic of a general analysis program applicable to pressure buildup tests for any fluid system 294 Fig. 9.1 Radial aquifer geometry 300 Fig. 9.2 Linear aquifer geometry 300 Fig. 9.3 Dimensioniess water influx, constant terminal pressure case, radial flow. (After Hurst and van Everdingen, ref. 1) 302 Fig. 9.4 Dimensionless water influx, constant terminal pressure case, radial flow (After Hurst and van Everdingen, ref. 1) 303 Fig. 9.5 Dimensionless water influx, constant terminal pressure case, radial flow (After Hurst and van Everdingen, ref. 1) 304 Fig. 9.6 Dimensionless water influx, constant terminal pressure case, radial and linear flow (After Hurst and van Everdingen, ref.1) 305 Fig. 9.7 Dimensionless water influx, constant terminal pressure case, radial and linear flow (After Hurst and van Everdingen, ref.1) 306 Fig. 9.8 Water influx from a segment of a radial aquifer 307 Fig. 9.9 Matching a continuous pressure decline at the reservoir-aquifer boundary by a series of discrete pressure steps 309 Fig. 9.10 Aquifer-reservoir geometry, exercise 9.2 311 Fig. 9.11 Reservoir production and pressure history; exercise 9.2 311 Fig. 9.12 Rs and Bg as functions of pressure; exercise 9.2 312 Fig. 9.13 Bo as a function of pressure; exercise 9.2 312 Fig. 9.14 Reservoir pressure decline approximated by a series of discrete pressure steps; exercise 9.2 313 Fig. 9.15 Aquifer fitting using the interpretation technique of Havlena and Odeh 319 Fig. 9.16 Comparison between Hurst and van Everdingen and Fetkovitch for reD = 5 327 Fig. 9.17 Comparison between Hurst and van Everdingen and Fetkovitch for reD = 10 327 Fig. 9.18 Predicting the pressure decline in a water drive gas reservoir 329 Fig. 9.19 Prediction of gas reservoir pressures resulting from fluid withdrawal and water influx (Hurst and van Everdingen) 330 Fig. 9.20 Prediction of gas reservoir pressures resulting from fluid withdrawal and water influx (Fetkovitch) 332 Fig. 9.21 Conditions prior to production in a steam soak cycle 333 CONTENTS XXIV Fig. 10.1 Hysteresis in contact angle in a water wet reservoir, (a) wetting phase increasing (imbibition); (b) wetting phase decreasing (drainage) 338 Fig. 10.2 Water entrapment between two spherical sand grains in a water wet reservoir 339 Fig. 10.3 Drainage and imbibition capillary pressure functions 339 Fig. 10.4 Capillary tube experiment for an oil-water system 340 Fig. 10.5 Determination of water saturation as a function of reservoir thickness above the maximum water saturation plane, Sw = 1−Sor, of an advancing waterflood 341 Fig. 10.6 Linear prototype reservoir model, (a) plan view; (b) cross section 344 Fig. 10.7 Approximation to the diffuse flow condition for H>> h 346 Fig. 10.8 (a) Capillary pressure function and; (b) water saturation distribution as a function of distance in the displacement path 348 Fig. 10.9 Typical fractional flow curve as a function of water saturation, equ. (10.12) 349 Fig. 10.10 Mass flow rate of water through a linear volume element A dx φ 350 Fig. 10.11 (a) Saturation derivative of a typical fractional flow curve and (b) resulting water saturation distribution in the displacement path 352 Fig. 10.12 Water saturation distribution as a function of distance, prior to breakthrough in the producing well 353 Fig. 10.13 Tangent to the fractional flow curve from Sw = Swc 354 Fig. 10.14 Water saturation distributions at breakthrough and subsequently in a linear waterflood 355 Fig. 10.15 Application of the Welge graphical technique to determine the oil recovery after water breakthrough 357 Fig. 10.16 Fractional flow plots for different oil-water viscosity ratios (table 10.2) 360 Fig. 10.17 Dimensionless oil recovery (PV) as a function of dimensionless water injected (PV), and time (exercise 10.2) 364 Fig. 10.18 Displacement of oil by water under segregated flow conditions 365 Fig. 10.19 Illustrating the difference between stable and unstable displacement, under segregated flow conditions, in a dipping reservoir; (a) stable: G > M−1; M > 1; β < θ. (b) stable: G > M−1; M < 1; β > θ. (c) unstable: G < M−1. 366 Fig. 10.20 Segregated displacement of oil by water 369 Fig. 10.21 Linear, averaged relative permeability functions for describing segregated flow in a homogeneous reservoir 370 Fig. 10.22 Typical fractional flow curve for oil displacement under segregated conditions 371 CONTENTS XXV Fig. 10.23 Referring oil and water phase pressures at the interface to the centre line in the reservoir. (Unstable segregated displacement in a horizontal, homogeneous reservoir) 372 Fig. 10.24 Comparison of the oil recoveries obtained in exercises 10.2 and 10.3 for assumed diffuse and segregated flow, respectively 376 Fig. 10.25 The stable, segregated displacement of oil by water at 90% of the critical rate (exercise 10.3) 378 Fig. 10.26 Segregated downdip displacement of oil by gas at constant pressure; (a) unstable, (b) stable 380 Fig. 10.27 (a) Imbibition capillary pressure curve, and (b) laboratory measured relative permeabilities (rock curves,- table 10.1) 382 Fig. 10.28 (a) Water saturation, and (b) relative permeability distributions, with respect to thickness when the saturation at the base of the reservoir is Sw = 1 − Sor (Pc =0)383 Fig. 10.29 Water saturation and relative permeability distributions, as functions of thickness, as the maximum saturation, Sw = 1 − Sor, is allowed to rise in 10 foot increments in the reservoir 385 Fig. 10.30 Averaged relative permeability curves for a homogeneous reservoir, for diffuse and segregated flow; together with the intermediate case when the capillary transition zone is comparable to the reservoir thickness 387 Fig. 10.31 Capillary and pseudo capillary pressure curves. 387 Fig. 10.32 Comparison of oil recoveries for different assumed water saturation distributions during displacement. 387 Fig. 10.33 Variation in the pseudo capillary pressure between +2 and -2 psi as the maximum water saturation Sw = 1−Sor rises from the base to the top of the reservoir 388 Fig. 10.34 Example of a stratified, linear reservoir for which pressure communication between the layers is assumed 390 Fig. 10.35 (a)-(c) Rock relative permeabilities, and (d) laboratory measured capillary pressures for the three layered reservoir shown in fig. 10.34 391 Fig. 10.36 (a) Water saturation, and (b) relative permeability distributions, with respect to thickness, when the saturation at the base of the layered reservoir (fig. 10.34) is Sw = 1−Sor (Pc° = 2 psi) 392 Fig. 10.37 (a)-(h) Water saturation and relative permeability distributions,as functions of thickness,for various selected values of c P° (three layered reservoir, fig 10.34) 394 Fig. 10.38 Averaged relative permeability functions for the three layered reservoir, fig. 10.34: (a) high permeability layer at top, (b) at base of the reservoir 396 CONTENTS XXVI Fig. 10.39 (a) Pseudo capillary pressure, and (b) fractional flow curves for the three layered reservoir, fig. 10.34). (——high permeability at top; − − −at base of the reservoir).396 Fig. 10.40 Individual layer properties; exercise 10.4 397 Fig. 10.41 Averaged relative permeability curves; exercise 10.4 399 Fig. 10.42 (a) Pseudo capillary pressures, and (b) fractional flow curves, exercise 10.4 (—— High permeability layer at top; − − −at base of reservoir) 400 Fig. 10.43 Methods of generating averaged relative permeabilities, as functions of the thickness averaged water saturation, dependent on the homogeneity of the reservoir and the magnitude of capillary transition zone (H). The chart is only applicable when the vertical equilibrium condition pertains or when there is a total lack of vertical equilibrium 404 Fig. 10.44 Numerical simulation model for linear displacement in a homogeneous reservoir406 Fig. 10.45 Spatial linkage of the finite difference formulation of the left hand side of equ (10.83). 408 Fig. 10.46 Example of water saturation instability (oscillation) resulting from the application of the IMPES solution technique (− − − correct, and  incorrect saturations) 411 Fig. 10.47 Determination of the average, absolute permeability between grid blocks of unequal size 415 Fig. 10.48 Overshoot in relative permeability during piston-like displacement 416 Fig. 10.49 Alternative linear cross sectional models required to confirm the existence of vertical equilibrium 418 LIST OF TABLES TABLE 1.1 Physical constants of the common constituents of hydrocarbon gases12, and a typical gas composition 15 TABLE 2.1 Results of isothermal flash expansion at 200°F 57 TABLE 2.2 Results of isothermal differential liberation at 200º F 58 TABLE 2.3 Separator flash expansion experiments performed on the oil sample whose properties are listed in tables 2.1 and 2.2 62 TABLE 2.4 Field PVT parameters adjusted for single stage, surface separation at 150 psia and 80°F; f b c = .7993 (Data for pressures above 3330 psi are taken from the flash experiment, table 2.1) 64 TABLE 2.5 Differential PVT parameters as conventionally presented by laboratories, in which Bo and Rs are measured relative to the residual oil volume at 60°F 67 TABLE 3.1 88 TABLE 3.2 89 TABLE 3.3 89 TABLE 4.1 Absolute and hybrid systems of units used in Petroleum Engineering 106 TABLE 6.1 Radial inflow equations for stabilized flow conditions 139 TABLE 7.1 157 TABLE 7.2 167 TABLE 7.3 167 TABLE 7.4 187 TABLE 7.5 200 TABLE 7.6 201 TABLE 7.7 203 TABLE 7.8 207 TABLE 7.9 208 TABLE 7.10 211 TABLE 7.11 213 TABLE 7.12 213 TABLE 7.13 213 CONTENTS XXVIII TABLE 7.14 214 TABLE 7.15 218 TABLE 7.16 219 TABLE 7.17 232 TABLE 7.18 233 TABLE 8.1 Generation of the real gas pseudo pressure, as a function of the actual pressure; (Gas gravity, 0.85, temperature 200°F) 245 TABLE 8.2 265 TABLE 8.3 266 TABLE 8.4 269 TABLE 8.5 272 TABLE 8.6 274 TABLE 8.7 276 TABLE 8.8 276 TABLE 8.9 276 TABLE 8.10 282 TABLE 8.11 282 TABLE 8.12 283 TABLE 8.13 285 TABLE 8.14 292 TABLE 9.1 308 TABLE 9.2 314 TABLE 9.3 314 TABLE 9.4 316 TABLE 9.5 316 TABLE 9.6 317 TABLE 9.7 318 TABLE 9.8 322 TABLE 9.9 326 CONTENTS XXIX TABLE 9.10 328 TABLE 10.1 358 TABLE 10.2 359 TABLE 10.3 359 TABLE 10.3(a) Values of the shock front and end point relative permeabilities calculated using the data of exercise 10.1 361 TABLE 10.4 363 TABLE 10.5 363 TABLE 10.6 376 TABLE 10.7 377 TABLE 10.8 379 TABLE 10.9 Water saturation and point relative permeability distributions as functions of the reservoir thickness; fig.10.28(a) and (b). 384 TABLE 10.10 Thickness averaged saturations, relative permeabilities and pseudo capillary pressures corresponding to figs. 10.28 and 10.29 386 TABLE 10.11 Phase pressure difference, water saturation and relative permeability distributions for c P° = 2 psi; fig. 10.36 392 TABLE 10.12 Pseudo capillary pressure and averaged relative permeabilities corresponding to figs. 10.36 and 10.37 395 TABLE 10.13 398 TABLE 10.14 399 LIST OF EQUATIONS ( )( ) wc OIP V 1 S res.vol. φ = − (1.1) 1 ( ) wc oi STOIIP n v 1 S /B φ = = − (1.2) 2 OP FP GP = + (1.3) 3 ( ) ( ) d FP d GP = − (1.4) 3 w water dp p D 14.7 (psia) dD × + (1.5) 4 w water dp p D 14.7 C(psia) dD = × + + (1.6) 4 w p 0.45 D 15 (psia) = + (1.7) 5 ( ) o p 0.35D 565 psia = + (1.8) 6 ( ) o p 0.08D 1969 psia = + (1.9) 6 wc oi Ultimate Recovery (UR) (V (1 S )/B ) RF φ = − × (1.10) 9 T 1 V c V p ∂ = ∂ — (1.11) 10 dV cV p = ∆ (1.12) 10 pV nRT = (1.13) 12 ( ) 2 a (p ) V b RT V + − = (1.14) 12 pV ZnRT = (1.15) 13 pc i ci i p n p = (1.16) 14 pc i ci i T nT = (1.17) 14 pr pc p p p = (1.18) 14 pr pc T T T = (1.19) 14 2 pr 1.2(1 t) 0.06125p t e Z y − − = (1.20) 19 CONTENTS XXXI k k 1 k k dF y y F dy + = − (1.22) 19 2 3 4 2 3 4 dF 1 4y 4y 4y y (29.52t 19.52t 9.16t )y dy (1 y) + + − + = − − + − 2 3 (1.18 2.82t) (2.18 2.82t) (90.7t 242.2t 42.4t )y + + + − + (1.23) 19 sc sc sc sc V T Z p E V p T Z = = (1.24) 21 p E 35.37 (vol/ vol) ZT = (1.25) 21 wc i G V (1 S )E φ = − (1.26) 22 nM nM Mp V ZnRT /p ZRT ρ = = = (1.27) 22 gas gas g air air M M M 28.97 ρ γ ρ = = = (1.28) 22 i i i M nM = (1.29) 22 sc 0.0763 g (lbs / cu.ft) ρ γ = (1.30) 22 g 1 V 1 1 Z c V p p Z p ∂ ∂ = − = − ∂ ∂ (1.31) 23 g 1 c p = (1.32) 23 p p i Production GIIP Unproduced Gas (sc) (sc) (sc) G G (HCPV)E G G G E E = − = − = − (1.33) 25 p i G E 1 G E = − (1.34) 25 p i i G p p 1 Z Z G = − (1.35) 25 d(HCPV) = − dVw +dVf (1.36) 26 f f f f f V V 1 1 c V (FP) V p ∂ ∂ = − = ∂ ∂ (1.37) 26 CONTENTS XXXII d(HCPV) = − (cw Vw + cf Vf ) ∆p (1.38) 26 w wc f p wc i (c S c ) p G E 1 1 G 1 S E + ∆ = − − − (1.39) 27 p e i Production GIIP Unproduced Gas (sc) (sc) (sc) G G G W E E = − = − − (1.40) 29 e i p i i W E p p G 1 1 Z Z G G = − − (1.41) 29 p a i G G 1 E /E = − (1.42) 31 p e i G W E G 1 E /E = − — (1.43) 31 e a i W E G G 1 E /E = + − (1.44) 31 GWC GWC g dp p p D dD = − × ∆ (1.45) 34 2 phase ' p i i p Z p G 1 Z G − = − (1.46) 40 s g s g scf rb (R R ) B (R R ) B (rb. free gas / stb) stb scf − × = − − (2.1) 48 (Underground withdrawal)/stb = Bo + (R − Rs) Bg (rb/stb) (2.2) 48 g rb 1 B scf 5.615E = (2.3) 48 o s g y x(B ( R )B ) rb / day x + − (2.4) 50 vo (rb/ rbb) Bo f o o b v rb B c stb = (2.5) 63 F (stb/rbb) Rs f s si bf 5.615 F rb R R c stb = − (2.6) 63 E (scf/rcf) Bg g 1 rb B 5.612 E scf = (2.7) 63 CONTENTS XXXIII d d o b o b b v rb rb B c stb residual rb = − (2.8) 66 and d d d s si b 5.615 F scf R R c stb residual = − − (2.9) 66 d d si b (Maximumvalueof F) scf R 5.615 c stb residual = × − (2.10) 66 d f d f d f f b ob o o o o b b b ob c B v v B B c c c B = = = (2.11) 67 f f d d f ob s si si s ob B R R (R R ) B = − − (2.12) 68 o o o o o dv dB 1 1 c v dp B dp = − = − (2.13) 69 o oi N(B B ) (rb) − (3.1) 73 si s g N(R R ) B (rb) − (3.2) 73 g oi gi B mNB 1 (rb) B − (3.3) 73 (1+m)NBoi (rb) (3.4) 74 w wc f oi wc c S c d(HCPV) (1 m)NB p 1 S + − = + ∆ − (3.5) 74 Np (Bo + (Rp − Rs)Bg) (rb) (3.6) 74 o oi si s g p o p s g oi oi g w wc f e p w gi wc (B B ) (R R ) B N (B (R R )B ) NB B B c S c m 1 (1 m) p (W W )B B 1 S − + − + − = + + − + + ∆ + − − (3.7) 74 F = Np (Bo + (Rp Rs) Bg) + Wp Bw (rb) (3.8) 76 Eo = (Bo − Boi) + (Rsi − Rs) Bg (rb/stb) (3.9) 76 g g oi gi B E B 1 (rb / stb) B = − (3.10) 76 w wc f f,w oi wc c S c E (1 m) B p (rb / stb) 1 S + = + ∆ − (3.11) 76 CONTENTS XXXIV F = N(Eo+mEg+Ef,w) + WeBw (3.12) 76 F = NEo (3.13) 77 e o o W F n E E = + (3.14) 77 o oi w wc f p o oi oi wc (B B ) (c S c ) N B NB p B 1 S − + = + ∆ − (3.15) 78 w wc f p o oi o wc (c S c ) N B NB c p 1 S + = ∆ − (3.16) 79 o o w wc f p o oi wc c S c S c N B NB p 1 S + + = ∆ − (3.17) 79 p o oi e N B NB c p = ∆ (3.18) 79 e o o w wc f wc 1 c (c S c S c ) 1 S = + + − (3.19) 79 Np (Bo + (Rp − Rs)Bg) = N ((Bo − Boi) + (Rsi − Rs)Bg) (3.20) 81 Sg = [ N (Rsi − Rs) − Np (Rp − Rs) ] Bg (1 − Swc) / NBoi (3.21) 82 Sg = p o wc oi N B 1 1 (1 S ) N B − − − (3.22) 83 ( ) p o p s g o oi si s g g oi oi gi N B (R R )B (B B ) (R R )B B NB m 1 B B + − − + − = + − (3.23) 86 F = N (Eo + mEg) (3.24) 86 We = (cw+cf) Wi ∆p (3.25) 92 F = NEo + We (3.26) 92 2 2 e w f e o W (c c ) (r r )fh p π φ = + − ∆ (3.27) 93 e o o W F N E E = + (3.28) 93 e o g o g W F N (E mE ) (E mE ) = + + + (3.29) 94 1 2 h h h u K K I I − ∆ = = (4.1) 101 CONTENTS XXXV p hg ( gz) ρ = + (4.2) 102 dh u K dl = (4.3) 102 K d p K d(hg) u gz g dl g dl ρ = + = (4.4) 102 p 1 atm dp gz ρ − Φ = + (4.5) 103 b p b p dp g(z z ) ρ Φ = + − (4.6) 103 p gz ρ Φ = + (4.7) 103 k d u dl ρ µ Φ = (4.8) 103 k d u dl ρ µ Φ = − (4.9) 104 k d u dr ρ µ Φ = (4.10) 104 k dp u dl µ = − (4.11) 105 k d k dp dz u g dl dl dl ρ ρ µ µ Φ = = − + (4.12) 107 6 k dp g dz u dl dl 1.0133 10 ρ µ = − + × (4.13) 107 2 k(D) A(cm ) dp q(cc / sec) (atm/ cm) (cp) dl µ = − (4.14) 107 2 k(mD) A(ft ) dp q(std/ d) (constant) (psi/ ft) (cp) dl µ = − (4.15) 107 and since 2 2 2 D cm atm k mD A ft mD ft psi stb r.cc / sec rb / d dp psi q cm d rb / d stb / d (cp) dl ft ft D 1 cm atm 1 ; 30.48 and ; equ.(4.16) mD 1000 ft psi 14.7 µ × = − × = = = (4.16) 108 CONTENTS XXXVI 3 o kA dp q 1.127 10 (stb / d) dl µ − = − × (4.17) 108 3 o kA dp q 1.127 10 0.4335 sin B dl γ θ µ − = − × + (4.18) 110 b p p RT Zdp gz M p Φ = + (4.19) 111 RT Z dp d dp gdz gdz M p ρ Φ = + = + (4.20) 111 d 1 dp dz g dl dl dl ρ Φ = + (4.21) 111 kA d kA d q dl dl ρ ψ µ µ Φ = − = − (4.22) 111 kA dp q dr µ = (4.23) 113 wf w q r p p ln 2 kh r µ π − = (4.24) 113 e e wf w r q p p ln 2 kh r µ π − = (4.25) 113 skin q p S 2 kh µ π ∆ = (4.26) 114 e e wf w r q p p ln S 2 kh r µ π − = + (4.27) 114 o e e wf w q B r p p 141.2 In S kh r µ − = + (4.28) 114 e wf 3 e o w q oil rate (stb / d) PI p p pressure drawdown (psi) 7.08 10 kh r B ln S r µ − = = − × = + (4.29) 115 o w w w ro w rw w k (S ) k (S ) k (S ) and k (S ) k k = = (4.30) 118 rw rw w or k k (at S 1 S ) ′ = = − (4.31) 119 o w w w ro w rw w o w wc o w wc k (S ) k (S ) K (S ) and K (S ) k (S S ) k (S S ) = = = = (4.32) 120 CONTENTS XXXVII ko (Sw) = kkro (Sw) or ko (Sw) = ko (Sw = Swc) Kro (Sw) (4.33) 120 ro ro rw rw k K k K = (4.34) 120 r kk λ µ = (4.35) 121 rd d ro o k / Mobility of the displacing fluid M Mobility of the displaced fluid k / µ µ ′ = = ′ (4.36) 122 1 k p p r c r r r t ρ φ ρ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.1) 127 (q ) 2 rh r t ρ ρ π φ ∂ ∂ = ∂ ∂ (5.2) 128 1 k p r r r t ρ ρ ρ φ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.3) 128 m 1 c m p ρ ρ ρ ρ ρ ∂ ∂ = − = ∂ ∂ (5.4) 129 p c t t ρ ρ ∂ ∂ = ∂ ∂ (5.5) 129 e p 0 at r r r ∂ = = ∂ (5.6) 130 p t ∂ ≈ ∂ constant, for all r and t. (5.7) 130 dp dV cV q dt dt = − = − (5.8) 131 or dp q dt cV = − (5.9) 131 2 e dp q dt c r h π φ = − (5.10) 131 i i i res i i p V p V = (5.11) 131 qi ∝ Vi (5.12) 132 i i i res i i p q p q = (5.13) 132 CONTENTS XXXVIII p = pe = constant, at r = re (5.14) 132 p and t ∂ ∂ = 0 for all r and t (5.15) 132 2 2 1 k p k p k p k p p r r r c r r r r r r t r ρ ρ ρ ρ φ ρ µ µ µ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂ (5.16) 133 p c r r ρ ρ ∂ ∂ = ∂ ∂ (5.17) 133 2 2 2 1 k p k p k p k p p r c r r c r r r r r t r ρ ρ ρ ρ φ ρ µ µ µ µ ∂ ∂ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ ∂ ∂ (5.18) 133 2 2 p 1 p c p r r k t r φµ ∂ ∂ ∂ + = ∂ ∂ ∂ (5 19) 134 1 p c p r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (5.20) 134 cp << 1 (5.21) 134 ct = coSo + c wSwc + cf (5.22) 134 φ ab s ol ut e × (coSo + c wSwc + cf) (5.23) 134 φ a bs ol ut e (1 − Swc) × o o w wc f wc (c S c S c ) (1 S ) + + − (5.24) 135 cV (pi − p) = qt (6.1) 136 2 e 1 p q r r r r r kh µ π ∂ ∂ = − ∂ ∂ (6.2) 137 2 1 2 e p q r C t 2 r kh µ π ∂ = − + ∂ (6.3) 137 2 e p q 1 r r 2 kh r r µ π ∂ = − ∂ (6.4) 137 2 r 2 e wf w r p q r p lnr 2 kh 2 r p r µ π = − (6.5) 137 2 r wf 2 w e q r r p p ln 2 kh r 2r µ π − = − (6.6) 137 e e wf w r q 1 p p ln S 2 kh r 2 µ π − = − + (6.7) 137 CONTENTS XXXIX e wf e w q 2 kh PI p p r 1 ln S r 2 π µ = = − − + (6.8) 137 e w e w r r r r pdV p dV = (6.9) 138 e w r 2 e r 2 p prdr r = (6.10) 138 e w r 2 wf 2 2 w e e r 2 q r r p p . r ln dr 2 kh r r 2r µ π − = − (6.11) 138 e wf w r q 3 p p ln S 2 kh r 4 µ π − = − + (6.12) 139 1 p r 0 r r r ∂ ∂ = ∂ ∂ (6.13) 139 e wf w r q 3 p p ln 2 kh r 4 µ π − = − ′ (6.14) 140 w w r r e ′ = −s (6.15) 140 oh h h wf w q r p p ln 2 kh r µ π − = (6.16) 141 oc e e h h q r and p p ln 2 kh r µ π − = (6.17) 141 e h e wf oh oc w h oc oh e h oc w h r r q p p ln ln 2 kh r r q r r ln ln 2 kh r r µ µ π µ µ π µ − = + = + (6.18) 141 e a a a w k k r S In k r − = (6.19) 142 2 e wf 2 3/ 2 w r q 1 p p ln 2 kh 2 r e π µ π π − = (6.20) 144 2 e 3/2 2 2 2 w w w 4 r 4A 4A 4 e r 56.32r 31.6r π π γ = = (6.21) 144 CONTENTS XL wf 2 A w q 1 4A p p ln S 2 kh 2 C r µ π γ − = + (6.22) 144 i i r 0 a) p p at t 0, for all r b) p p at r , for all t p q c) lim r , for t 0 r 2 kh µ π → = = = = ∞ ∂ = > ∂ (7.1) 150 s cr t 2k t φ µ ∂ = ∂ (7.2) 150 2 2 s cr r 4k t φ µ ∂ = ∂ (7.3) 150 ( ) dp p s sp ds s 1 dp ds p s ′ ′ ′ + = − + ′ = − ′ (7.4) 151 s 2 e p C s − ′ = (7.5) 151 2 s r,t i cr x 4k t q e p p ds 4 kh s φ µ µ π − ∞ = = − (7.6) 152 2 s s x cr x 4kt e e ds ds s s φ µ − − ∞ ∞ = = (7.7) 152 ei(x) ln x 0.5772 ≈− − (7.8) 152 ( ) ei(x) ln x for x 0.01 γ ≈− < (7.9) 153 wf i 2 w q 4kt p p ln 2S 4 kh cr µ π γ φ µ = − + (7.10) 153 2 r,t i q cr p p ei 4 kh 4k t µ φ µ π = − (7.11) 154 ( ) i cAh p p qt φ − = (7.12) 156 wf i 2 A w q 4A kt p p ½ ln 2 S 2 kh cA C r µ π π φ µ γ = − + + (7.13) 156 CONTENTS XLI Cumulative Production t Effective flowing time Final flow rate = = (7.14) 156 dimensionless radius rD = w r r (7.15) 161 dimensionless time tD = 2 w kt cr φ µ (7.16) 161 and dimensionless pressure D D D i r,t 2 kh p (r ,t ) (p p ) q π µ = − (7.17) 161 D D D D D D D p p 1 r r r r t ∂ ∂ ∂ = ∂ ∂ ∂ (7.18) 161 i wf D D 2 kh (p p ) p (t ) S q π µ − = + (7.19) 161 D 2 w kt t 0.000264 t in hours cr φ µ = − (7.20) 162 D 2 w kt t 0.00634 t in days cr φ µ = − (7.21) 162 3 D i wf o kh p 7.08 10 (p p ) q B µ − = × − (7.22) 163 ( ) D 12 D D 4t p t ln γ = (7.23) 164 ( ) 12 D D D p t (ln t 0.809) = + (7.24) 164 2 w 12 D D D 2 A w r 4A p (t ) ln 2 t A C r π γ = + (7.25) 164 2 w DA D r kt t t A cA φ µ = = (7.26) 165 12 D D DA 2 A w 4A p (t ) ln 2 t C r π γ = + (7.27) 165 2 2 w D w A D r 4 t r / A C t e A π ≈ (7.28) 165 DA A DA 4 t C t e π ≈ (7.29) 165 DA kt t 0.1 cA φ µ = ≈ (7.30) 165 CONTENTS XLII ( ) ( ) j 1 n n i wf j D D D n n j 1 2 kh p p q p t t q S π µ − = − = ∆ − + (7.31) 170 i ws D D D D D 2 kh (p p ) p (t t ) p ( t ) q π µ − = + ∆ − ∆ (7.32) 171 ( ) ( ) n n j 1 n i wf j D D D j 1 n n p -p q 2 kh p t t S q q π µ − = ∆ = − + (7.33) 172 ( ) ( ) ( ) ( ) ( ) n D 2 t 2 1 n eD D D D eD 2 2 2 2 n 1 eD n 1 n eD 1 n e J r 2t 3 p t lnr 2 4 r J r J α α α α α − = ∞ = + − + − (7.34) 173 ( ) D D 1 1 2 2 i ws D D D 4 t t 2 kh t t (p p ) ln p (t t ) ln q t π µ γ + ∆ + ∆ − = + + ∆ − ∆ (7.35) 174 D 1 1 2 2 i ws D D 4t 2 kh t t (p p ) ln p (t ) ln q t π µ γ + ∆ − = + − ∆ (7.36) 174 ( ) ( ) D i ws(LIN) D D 4t 2 kh t t p p ½ ln p t ½ln q t π µ γ + ∆ − = + − ∆ (7.37) 174 ( ) i DA 2 kh 2 khqt p p 2 t q q cAh π π π µ µ φ − = = (7.38) 175 ( ) D 12 i D D 4t 2 kh p p p (t ) ln q π µ γ − = − (7.39) 176 ( ) ( ) D DA D D 4t 4 kh p p 4 t ln 2p t q π π µ γ − = + − (7.40) 176 ( ) ( ) 2 j D 1 1 2 2 i wf D D j 2 ca 4t 2 kh p p p t ln ei q 4kt φ µ π µ γ = ∞ − = = + (7.41) 177 ( ) ( ) D 1 1 2 2 D D DA D(MBH) DA 4t p t 2 t ln p t π γ = + − (7.42) 182 ( ) ( ) D(MBH) DA DA 4 kh p t p p 4 t q π π µ = − = (7.43) 182 ( ) ( ) ( ) 2 w D(MBH) DA A D A DA r 4 kh p t p p ln C t ln C t q A π µ = − = = (7.44) 182 ( ) ( ) DA D(MBH) DA A t 4 kh p t 1 p p ln C 1 q π µ = = − = = (7.45) 183 ( ) ( ) 1 1 1 2 2 2 D DA DA DA D(MBH) DA 2 w 4A p t 2 t ln t ln p t r π γ = + + − (7.46) 183 wf DA DA D(MBH) DA 0.0189(3500 p ) 2 t ½ In t 8.632 ½ p (t ) 4.5 π − = + + − + (7.46) 185 CONTENTS XLIII wf D(MBH) DA 0.0189 (3500 p ) ½ p (t ) α − = − (7.47) 185 ( ) ( ) -3 D i ws(LIN) D D o 4t kh t t 7.08 10 p p 1.151 log p t ½ ln q B t µ γ + ∆ × − = + − ∆ (7.48) 189 DA kt t 0.000264 (t-hours) cA φµ = (7.49) 189 o q B m 162.6 psi/log.cycle kh µ = (7.50) 190 -3 i wf D D o kh 7.08 10 (p p ) p (t ) S q B µ × − = + (7.51) 190 ( ) ws(LIN) 1 -hr wf 2 w p p k S 1.151 log 3.23 m cr φ µ − = − + (7.52) 191 ( ) ( ) ro w abs k k k S = × (7.53) 191 o ws i q B t t p p 162.2 log kh t µ + ∆ = − ∆ (7.54) 192 ( ) ( ) ( ) n j n j -3 i ws D D D 1 D D j 1 n o n q kh 7.08 10 p p p t t p t q B q µ − = ∆ × − = + ∆ − ∆ (7.55) 193 ( ) ( ) n n j -3 n i ws(LIN) n o n D j D D D 1 j 1 n t t kh 7.08 10 p p 1.151 log q B t 4t q p t t ½ ln q µ γ − = + ∆ × − = ∆ ∆ + + ∆ − (7.56) 193 n n t t p p m log t − = (7.57) 194 D(MBH) DA A DA n o kh p (t )0.01416 (p - p) 2.303 log (C t ) q B µ = (7.58) 195 ( ) ( ) n o A DA A DA q B p p 162.6 log C t m log C t kh µ − = = (7.59) 195 n n n t t A DA p p m log(C t ) − = (7.60) 195 ( ) ( ) n n n t t t p p p p m log t − − − = (7.61) 195 ( ) s s p p t t log t m − + ∆ = ∆ (7.62) 198 ( ) s A DA s t t log log C t t + ∆ = ∆ (7.63) 198 CONTENTS XLIV ( ) d DA d t t log log 19.1 t t + ∆ = ∆ (7.64) 199 ( ) ( ) ( ws(LIN) 1 -hr wf 2 w 6 p p k S 1.151 log 3.23 m cr 4752 4506 50 1.151 log 3.23 24.5 .2 1 20 10 .09 φµ − − = − + − = − + × × × × (7.52) 202 ( ) ws(LIN) t t 0.0144 4800 p 1.151 log 25.63 t + ∆ − = + ∆ (7.65) 206 ( ) ( ) ws(LIN) DA A DA t t 0.0144 4800 p 1.151 log 2 t ½ ln C t t t t 1.151 log t π α + ∆ − = + − ∆ + ∆ = + ∆ (7.66) 207 ( ) ( ) ( ) i ws D D D D D 0.0144 p p p t t p t − = + ∆ − ∆ (7.67) 207 ( ) ( ) ws DA D MBH t t 0.0288 p p t ln t + ∆ ∆ = ∆ − ∆ (7.68) 208 ( ) ( ) n n j 1 n i wf j 3 D D D j 1 o n n p p q kh 7.08 10 p t t S B q q µ − − = − ∆ × = − + (7.69) 209 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 1 3 2 i wf 1 2 1 3 D D D D D o 3 3 3 2 D D D 3 p p q 0 q q kh 7.08 10 p t p t t B q q q3 q q p t t S q µ − − − − × = + − + + − + (7.70) 210 ( ) ( ) n n j 1 n i wf j D D D j 1 n n p p q versus p t t q q − = − ∆ − (7.71) 210 ( ) ( ) n n i wf j 3 n j 1 2 j 1 o n n w p p q kh k 7.08 10 1.151 log t t log 3.23 0.87S B q q cr µ φµ − − = − ∆ × = − + − + (7.72) 211 ( ) 12 D D D(MDH) DA p t p (t ) α ′ ′ = − (7.73) 212 ( ) ( ) ( ) ( ) n N n j 1 n j 1 N 3 i wf j D D D D N j 1 o n j D D D n N j N 1 kh 7.08 10 p p q p t t t q S B q p t t q q S δ µ δ δ − − − = = + × − = ∆ + − + + ∆ − + − (7.74) 217 ( ) ( ) N n j 1 N j 1 N N j D D D D j D D D j 1 j 1 q p t t t q p t t δ − − = = ∆ − − ≈ ∆ − (7.75) 217 CONTENTS XLV ( ) ( ) N n n j 1 n wf wf j D D D j N 1 n N n N p p q versus p t t q q q q δ δ − = + − ∆ − − − (7.76) 217 p versus log t 1 1 C t ∆ ∆ − ∆ (7.77) 225 o 162.6q B kh m µ = (7.78) 226 wf(1 hr) wf 2 w p p k S 1.151 log 3.23 1 1/ C t cr m φµ − − = − + − ∆ (7.79) 226 pF T pT q F q ∆ ∆ × = (7.80) 229 ( ) e wf w s.cc / sec r.cc / sec Q Mscf / d r atm 3 Mscf / d s.cc / sec p p psi ln S D cm psi r 4 2 k mD h ft mD ft µ π − = − + (8.1) 241 e wf w r 711 Q ZT 3 p p ln S r 4 khp µ − = − + (8.2) 242 wf p p p 2 + = (8.3) 242 wf wf p p p p and Z Z 2 2 µ µ + + = = (8.4) 242 2 2 e wf w r 1422 Q ZT 3 p p ln S kh r 4 µ − = − + (8.5) 242 ( ) 2 2 i wf 2 w i 711 Q ZT 4 .000264kt p p ln 2S kh c r µ γ φ µ − = + (8.6) 242 ( ) b p p pdp m p 2 Z µ = (8.7) 243 Then ( ) m p 2p p r r µ ∂ ∂ = ∂ Ζ ∂ (8.8) 248 and similarly ( ) m p 2p p t t µ ∂ ∂ = ∂ Ζ ∂ (8.9) 248 CONTENTS XLVI ( ) ( ) m p m p 1 k r c r r 2p r 2p t ρ µ µ φ ρ µ ∂ ∂ ∂ Ζ Ζ = ∂ ∂ ∂ (8.10) 248 ( ) ( ) m p m p 1 c r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (8.11) 248 ( ) 2 e m p 2p p 2p q t t r h c µ µ π φ ∂ ∂ = ⋅ = − ⋅ ∂ Ζ ∂ Ζ (8.12) 249 ( ) 2 e res m p 1 2 pq r r r r r kh π ∂ ∂ = − ∂ ∂ Ζ (8.13) 249 ( ) sc sc 2 sc e m p 2p q 1 T r r r r T r kh π ∂ ∂ = − ⋅ ∂ ∂ (8.14) 249 ( ) ( ) e wf w r 1422 QT 3 m p m p ln S kh r 4 − = − + (8.15) 250 ( ) ( ) ( ) i wf 2 w i 711 QT 4 .000264kt m p m p ln 2S kh c r γ φ µ − = + (8.16) 250 or ( ) ( ) wf wf p p p p 2 µ + − Ζ equivalent to wf p p pdp µΖ (8.17) 251 dp u dr k µ = (8.18) 252 2 dp u u dr k µ βρ = + (8.19) 253 ( ) e w 2 r nD r 2p q m p dr 2 rh βρ µ π ∆ = Ζ (8.20) 254 ( ) e w 2 r g 2 2 nD r T pq m p constant dr r h γ µ β ∆ = × ΖΤ (8.21) 254 ( ) e w r 2 g sc 2 2 nD r q dr m p constant h r β γ µ Τ ∆ = × (8.22) 254 ( ) 2 g sc 2 nD w e q 1 1 m p constant r r h β γ Τ ∆ = × − (8.23) 254 ( ) 2 g 12 2 2 nD w p w Q m p 3.161 10 FQ h r β γ µ − Τ ∆ = × = (8.24) 254 CONTENTS XLVII ( ) ( ) 2 e wf w r 1422 TQ 3 m p m p ln S FQ kh r 4 − = − + + (8.25) 255 e w r 1422 TQ 3 ln S DQ kh r 4 = − + + (8.26) 255 Fkh D 1422T = (8.27) 255 constant kα β = (8.28) 256 ( ) rg constant kk α β = (8.29) 256 ( ) ( ) D i wf 4t 711QT m p m p ln 2S kh γ ′ − = + (8.30) 257 ( ) ( ) ( ) ( ) i wf D D kh m p m p m t S 1422QT ′ − = + (8.31) 257 ( ) D 12 D D 4t m t ln γ = (8.32) 258 ( ) 12 D D DA 2 A w 4A m t ln 2 t C r π γ = + (8.33) 258 D D D D D D D m m 1 r r r r t ∂ ∂ ∂ = ∂ ∂ ∂ (8.34) 258 ( ) ( ) ( ) D 1 1 2 2 D D DA DA D MBH 4t m t 2 t ln m t π γ = + − (8.35) 258 ( ) DA kt hrs t 0.000264 cA φ µ = (8.36) 258 ( ) ( ) ( ) ( ) DA D MBH kh m t m p m p 711QT = − (8.37) 259 ( ) ( ) ( ) ( ) 1 D 2 wf D 2 A w kh 4A m p m p m t S ln S 1422QT C r γ ′ ′ − = + = + (8.38) 259 ( ) ( ) ( ) ( ) n n j 1 n i wf j D D D n n j 1 kh m p m p Q m t t Q S 1422T − = ′ − = ∆ − + (8.39) 260 CONTENTS XLVIII ( ) ( ) ( ) ( ) n j 1 D 1 1 2 2 D D D D D DA DA D MBH j j j 1 n n 4t m t t m t 2 t ln m t Q Q Q and S S DQ π γ − − ′ ′ ′ ′ − = = + − ∆ = − ′ = + (8.40) 260 ( ) ( ) ( ) ( ) n n j 1 n 2 i wf n j D D D n j 1 kh m p m p FQ Q m t t Q S 1422T − = − − = ∆ − + (8.41) 260 ( ) n 2 2 i wf Q C p p = − (8.42) 262 ( ) ( ) 2 12 i wf 2 A w 1422QT 4A m p m p ln S FQ kh C r γ − = + + (8.43) 263 ( ) ( ) 2 i wf m p m p BQ FQ − = + (8.44) 263 ( ) ( ) n i wf n n m p m p versus Q Q − (8.45) 263 ( ) n 2 n i wf n j 12 j 2 j 1 n n A w i m(p ) m(p ) FQ Q 2.359T 1422T 4A t ln S Q c Ah Q kh C r φ µ γ = − − = ∆+ + (8.46) 264 n 2 n i wf n j j j 1 n n m(p ) m(p ) FQ Q versus t Q Q = − − ∆ (8.47) 264 m(p) = (0.3457p — 414.76) × 106 psia2/cp (8.48) 265 ( ) ( ) ( ) ( ) n 2 n i wf n j n j 1 2 j 1 n n w i m p m p FQ Q k m log t t m log 3.23 .87S Q Q c r φ µ − = − − ∆ = − + − + (8.49) 273 ( ) n n j 1 2 n i wf n j D D D j 1 n n m(p ) m(p ) FQ Q 1422T 1422T m t t S Q kh Q kh − = − − ∆ = − + (8.50) 275 12 D D D(MBH) DA m (t ) m (t ) α ′ ′ = − (8.51) 275 1 i ws D D D D D 1 kh (m(p ) m(p )) m (t t ) m ( t ) 1422Q T − = + ∆ − ∆ (8.52) 279 1 1 D 1 12 i ws(LIN) D D 1 4t t t kh (m(p ) m(p )) 1.151 log m (t ) ln 1422Q T t γ + ∆ − = + − ∆ (8.53) 279 1 1637Q T m kh = (8.54) 280 ws(LIN)1 hr wf 1 1 2 i w (m(p ) m(p )) k S S DQ 1.151 log 3.23 m ( c)r φ µ − − ′ = + = − + (8.55) 280 CONTENTS XLIX 1 i wf 1 2 i w 1637Q T k m(p ) m(p ) log t log 3.23 0.87S kh ( c) r φ µ ′ − = + − + (8.56) 280 i wf 1 hr 1 1 2 i w (m(p ) m(p ) ) k S S DQ 1.151 log 3.23 m ( c)r φ µ − − ′ = + = − + (8.57) 280 1 max max i wf 1 D D D D D D D 2 D D 2 2 kh (m(p ) m(p )) Q (m (t t t ) m ( t t )) 1422T Q m (t ) Q S ′ ′ − = + ∆ + − ∆ + ′ ′ + + (8.58) 281 i wf i ws 2 D D 2 2 kh kh (m(p ) m(p )) (m(p ) m(p )) Q m (t ) Q S 1422T 1422T ′ ′ ′ − = − + + (8.59) 281 ws 1 hr wf 1 hr 2 2 2 i w (m(p ) m(p ) k S S DQ 1.151 log 3.23 m ( c) r φ µ − − ′ − ′ = + = − + (8.60) 281 SSS p SSS DA SSS ( c) A t (t ) 0.000264k φ µ = (8.61) 286 b p ro o o o p k (S ) m (p) dp B µ ′ = (8.62) 289 or rg o o s g ro g k B R R k B µ µ = + (8.63) 289 D 1 1 2 2 D D DA D(MBH) DA 4t m (t ) 2 t ln m (t ) π γ ′ ′ = + − (8.64) 290 g g o s o t g w wc f o g S B S R B c B c S c B p p B p ∂ ∂ ∂ = − − + + ∂ ∂ ∂ (8.65) 290 1 c r r r k t β φµ β ∂ ∂ ∂ = ∂ ∂ ∂ (8.66) 291 D D D D D D D 1 r r r r t β β ∂ ∂ ∂ = ∂ ∂ ∂ (8.67) 292 D D f(p) (t ) S q α β = + (8.68) 292 n j 1 n n j D D D n j 1 f(p) q (t t ) q S α β − = = ∆ − + (8.69) 292 n j 1 D 1 1 2 2 o D D D D DA D(MBH) DA 4t (t t ) (t ) 2 t ln (t ) β β π β γ − ′ ′ ′ ′ − = = + − (8.70) 292 n j 1 D 12 D D D D D 4t (t t ) (t ) ln β β γ − ′ ′ − = = (8.71) 293 CONTENTS L D o r r r = (9.1) 298 D 2 0 kt t cr φµ = (9.2) 298 ( ) D D q q t 2 kh p µ π = ∆ (9.3) 299 ( ) 2 e 0 D D W 2 hcr pW t πφ = ∆ (9.4) 299 e D D W U p W (t ) = ∆ (9.5) 299 2 o U 2 f hcr π φ = (9.6) 299 D t = constant 2 0 kt cr φµ × (9.7) 300 U = 1.119 fφh 2 o cr (bbl/psi) (9.8) 300 tD = constant × 2 kt cL φµ (9.9) 301 U = .1781 wLhφc (bbl/psi) (9.10) 301 ( ) ( ) 2 12 D eD Radial W max r 1 = − (9.11) 301 ( ) D Linear W max 1 = (9.12) 301 e kct W 2hw p (ccs) φ πµ = ×∆ (9.13) 301 3 e kct W 3.26 10 hw x p (bbls) φ πµ − = × ∆ (9.14) 302 i 1 1 1 2 2 j 1 j j p p p 2 p p p 2 p p p 2 − + = + = + = (9.15) 309 CONTENTS LI i 1 i 1 1 0 i i i 1 1 2 i 2 1 2 1 2 3 1 3 1 2 2 3 2 j 1 j j j 1 j 1 j 1 j j 1 j (p p ) p p p p p p 2 2 (p p ) (p p ) p p p p p 2 2 2 (p p ) p p (p p ) p p p 2 2 2 (p p ) (p p ) p p p p p 2 2 2 − − + − + + + − ∆ = − = = + + − ∆ = − = − = + − + ∆ = − = − = + + − ∆ = − = − = (9.16) 310 ( ) j n 1 e j D D D j 0 W T U p W (T T ) − = = ∆ − (9.17) 310 e a w dW q J(p p) dt = = − (9.18) 320 We=c Wi (pi - p a) (9.19) 320 e e a i i ei i i W W p p 1 p 1 W cWp = − = − (9.20) 320 e ei a i dp dW W dt p dt = − (9.21) 320 ( ) i ei a i Jp t / W p p p p e− −= − (9.22) 320 e i ei i dW Jp t / W J(p p)e dt − = − (9.23) 321 ei i ei e i i W Jpt / W W (p p)(1 e ) p − = − − (9.24) 321 ( ) 1 ei i 1 ei 1 e i i W Jp t / W W p p (1 e ) p − ∆ ∆ = − − (9.25) 321 ( ) 1 2 ei i 2 ei a 2 e i W Jp t / W W p p (1 e ) p − ∆ ∆ = − − (9.26) 321 i 1 i e a i e W p p 1 W ∆ = − (9.27) 321 ( ) n 1 n ei i n ei a n e i W Jp t / W W p p (1 e ) p − − ∆ ∆ = − − (9.28) 321 CONTENTS LII j n 1 n 1 e j 1 a i ei W p p 1 W − − = ∆ = − (9.29) 322 khw 3 L µ (9.30) 322 khw L µ (9.31) 322 e w i dW q J(p p) dt = = − (9.32) 323 t e i 0 W J (p p)dt = − (9.33) 323 3 e o 3 7.08 10 fkh J r 3 ln r 4 7.08 10 .3889 200 100 116.5b / d/psi .55(In5 .75) µ − − × = − × × × × = = − (9.30) 325 ( ) ( ) n j n 1 n 2 e j D D D n 1 D D D j 0 W U p W T t U p W T t − − − = = ∆ − + ∆ − (9.34) 329 ( ) ( ) ( ) n j n 1 n 2 e j D D D n 2 n D D D j 0 U W U p W T t p p W T t 2 − − − = = ∆ − + − − (9.35) 329 n n p e i i i n G W E p p 1 1 Z Z G G = − − (9.36) 329 ( ) n j n 2 1 e j D D D j 0 W U p W T t − = = ∆ − (9.37) 330 ( ) n wf,n o n h oh w 2 k h p p q r ln r π µ − = (9.38) 334 ( ) 12 n n 1 n p p p − = + (9.39) 334 ( ) 12 wf,n wf,n 1 wf,n p p p − = + (9.40) 334 ( ) p,n p,n 1 n 1 n wf,n 1 wf,n N N p p p p 2 α − − − = + + − − (9.41) 334 ( ) ( ) j n 1 n 2 p,n j D D D n 1 D D D j 0 N U p W T t U p W T t − − − = = ∆ − + ∆ − (9.42) 334 CONTENTS LIII ( ) ( ) j n 2 p,n j D D D n 2 n j 0 N U p W T t p p 2 β − − = = ∆ − + − (9.43) 335 ( ) ( ) ( ) j n 2 n j D D D p,n 1 n 2 wf,n 1 wf,n n 1 j 0 1 p 2U p W T t 2N p p p p β α α β − = − − − − = ∆ − − + + + − + (9.44) 335 c o w 1 2 1 1 P p p r r σ = − = + (10.1) 338 o w c p p P gH ρ − = = ∆ (10.2) 341 o w c 2 cos p p P gh r σ ρ Θ − = = = ∆ (10.3) 341 c w o w P (S ) = p p = g cos ρ γ θ − ∆ (10.4) 342 c w 6 g y cos P (S ) (atm) 1.0133 10 ρ θ ∆ = × (10.5) 342 ( ) c w P S 0.4335 ycos (psi) γ θ = ∆ (10.6) 342 t o w i q q q q = + = (10.7) 344 o t o c rw w 6 rw ro ro q P g sin q A kk kk kk x 1.0133 10 µ µ µ ρ θ ∂ ∆ = − + = + − ∂ × (10.8) 347 ro c 6 t o w ro w rw o kk A P g sin 1 q x 1.0133 10 f k 1 k ρ θ µ µ µ ∂ ∆ + − ∂ × = + ⋅ (10.9) 347 3 ro c t o w ro w rw o kk A P 1 1.127 10 .4335 sin q x f k 1 k γ θ µ µ µ − ∂ + × − ∆ ∂ = + ⋅ (10.10) 347 c c w w P dP S x dS x ∂ ∂ = ⋅ ∂ ∂ (10.11) 348 w ro w rw o 1 f k 1 k µ µ = + ⋅ (10.12) 349 w w w w w w x dx x q q A dx ( S ) t ρ ρ φ ρ + ∂ − = ∂ (10.13) 349 w w w w (q ) A ( S ) x t ρ φ ρ ∂ ∂ = − ∂ ∂ (10.14) 350 CONTENTS LIV w w t x q S A x t φ ∂ ∂ = − ∂ ∂ (10.15) 350 w w w x t S S S dx t x dt ∂ ∂ = − ∂ ∂ (10.16) 350 t w w w t w q q S x S x ∂ ∂ ∂ = ⋅ ∂ ∂ ∂ (10.17) 350 w w t S w q dx A S dt φ ∂ = ∂ (10.18) 350 w w w t w S S S w q df dx v dt A dS φ = = (10.19) 351 w w i w S S w W df x A dS φ = (10.20) 351 ro 6 t o w ro w rw o kk A g sin 1 q 1.0133 10 f k 1 k ρ θ µ µ µ ∆ − × = + ⋅ (10.21) 352 wf i w wc 2 w w S W 1 S S x A df dS φ − = = (10.22) 353 2 1 X or 1 w X W 2 (1 S )x S dx S x − + = (10.23) 353 wf or or wf S w w or w 1 S w w 1 S w w S w df df (1 S ) S d dS dS S df dS − − − + = (10.24) 354 ( ) wf Swf w w wf w S w df S S 1 f dS = + − (10.25) 354 wf Swf w w w w S w wf we (1 f ) df 1 dS S S S S − = = − − (10.26) 354 we i id w w S W 1 W LA df dS φ = = (10.27) 355 CONTENTS LV ( ) bt bt bt wbt w pd id id bt wc w w S 1 N W q t S S df dS = = = − = (10.28) 356 bt id bt id W t q = (10.29) 356 we w we we w w S 1 S S (1 f ) df dS = + − (10.30) 356 w we we id S S (1 f )W = + − (10.31) 356 Npd = S w − Swc = (Swe − Swc) + (1 − fwe) Wid(PV) (10.32) 356 ws w o w 1 f B 1 1 1 B f = + − (10.33) 359 ro wf o rw wf w s ro o k (S )/ k (S )/ M k / µ µ µ + = ′ (10.34) 360 id t 4.39 W (years) = (10.35) 362 o w t o w 6 ro rw g sin u (p p ) kk kk x 1.0133 10 µ µ ρ θ ∂ ∆ − = − − + ′ ′ ∂ × (10.36) 366 dy 1 M 1 G 1 dx tanθ − = + (10.37) 367 rw 6 t w kk A g sin G 1.0133 10 q ρ θ µ ′ ∆ = × (10.38) 367 4 rw t w kk A sin G 4.9 10 q γ θ µ − ′ ∆ = × (10.39) 367 dy M 1 G tan tan dx G β θ −− = − = (10.40) 367 rw crit 6 w kk A g sin q r.cc / sec 1.0133 10 (M 1) ρ θ µ ′ ∆ = × − (10.41) 368 4 rw crit w 4.9 10 kk A sin q rb / d (M 1) γ θ µ − ′ × ∆ = − (10.42) 368 w wc or wc S S b 1 S S − = − − (10.43) 369 CONTENTS LVI wc rw rw or wc w w S S k (S ) k 1 S S − ′ = − − (10.44) 370 or ro ro or wc w w 1 S S k (S ) k 1 S S − − ′ = − − (10.45) 370 ro o o o (1 b)kk A p q x µ ° ′ − ∂ = − ∂ (10.46) 372 rw w w w b kk A p q x µ ° ′ ∂ = − ∂ (10.47) 372 e we e Mb f 1 (M 1)b = + − (10.48) 373 ( ) e iD 1 b W M 1 M 1 = − − (10.49) 373 ( ) we iD M f 1 W M M 1 = − − (10.50) 373 pD iD iD 1 N (2 W M W 1) M 1 = − − − (10.51) 374 bt pD 1 N M = (10.52) 374 max iD W M = (10.53) 374 iD pD iD iD W G 1 G (M 1) N 2 W M 1 1 W 1 G 1 M 1 M 1 M 1 (M 1) + = − − − − − − − − − (10.54) 374 bt pD 1 N M G = − (10.55) 374 max iD M W G 1 = + (10.56) 375 pD iD iD iD N 0.976 W (1 0.520W ) 0.535 W 0.364 = − + − (10.57) 377 2 e pD (h y ) N 1 2hL tan β − = − (10.58) 378 2 e iD pD y W N 2hL tan β = + (10.59) 379 bt bt pD iD h N W 1 2L tan β = = − (10.60) 379 CONTENTS LVII rg crit 6 t g t kk A g sin q G (M 1) q 1.0133 10 q ρ θ µ ′ ∆ = = − × (10.61) 380 dPc = .4335 (1.04−.81) dz = 0.1 dz (10.62) 382 h w 0 w S (z)dz S h = (10.63) 382 h rw w 0 rw w k (S (z))dz k (S ) h = (10.64) 383 h ro w 0 ro o k (S (z))dz k (S ) h = (10.65) 383 o w c c 6 g h p p P P z (atm) 2 1.0133 10 ρ ° ° ° ∆ − = = + − × (10.66) 388 o w c c h p p P P 0.4335 z (psi) 2 γ ° ° ° − = = + ∆ − (10.67) 388 or c 1 S 6 g h P z 2 1.0133 10 ρ ° − ∆ = − × (10.68) 388 or c 1 S h P 0.4335 z 2 γ ° − = ∆ − (10.69) 388 ( ) or c 1 S P 0.1 20 z ° − = − (10.70) 388 1 2 3 w w w 1 1 2 2 3 3 w 3 i i j 1 h S h S h S S h φ φ φ φ = + + = (10.71) 393 j j n n N j j or j j wc j 1 j n 1 w N j j j 1 h (1 S ) h S S h φ φ φ = = + = − + = (10.72) 396 n n n n N rw w j j rw j j j 1 j 1 k (S ) hk k hk = = ′ = (10.73) 396 n n n N N ro w j j ro j j j n 1 j 1 k (S ) hk k hk = + = ′ = (10.74) 397 j t w j j j w j q f v wh S φ ∆ = ∆ (10.75) 401 CONTENTS LVIII j j j j rw j j or wc k k (1 S S ) α φ ′ = − − (10.76) 401 x x dx w w w w w w w w q q A dx ( S ) Q t ρ ρ φ ρ ρ + ∂ ′ − = + ∂ (10.77) 407 rw w w w w p kk S Q x x t φ µ ∂ ∂ ∂ = + ∂ ∂ ∂ (10.78) 407 ro o o o o p kk S Q x x t φ µ ∂ ∂ ∂ = + ∂ ∂ ∂ (10.79) 407 c P° = po −pw (pseudo capillary pressure) (10.80) 408 w o S S 1 + = (10.81) 408 and w o S S 0 t t ∂ ∂ + = ∂ ∂ (10.82) 408 1 1 2 2 i i n n rw rw n 1 n 1 n 1 n 1 w,i 1 w,i w,i w,i 1 2 w w i i n 1 n w w 1 kk kk (p p ) (p p ) ( x) (S S ) t µ µ φ + + + + + − + − + − − − ∆ = − ∆ (10.83) 408 1 1 2 2 n n ro ro n 1 n 1 n 1 n 1 o,i 1 o,i o,i o,i 1 2 o o i i n 1 n o,i o,i 1 kk kk (p p ) (p p ) ( x) (S S ) t µ µ φ + + + + + − + − + − − − ∆ = − ∆ (10.84) 409 n 1 n ,n 1 n 1 n 1 ,n n n w w c o w c o w P (S ) p p P (S ) p p + ° + + + ° = − ≈ = − (10.85) 409 1 1 2 2 1 1 2 2 n n rw ro ro n 1 n 1 ,n ,n o,i w,i c,i 1 c,i 2 w o o i i n n rw ro ro n 1 n 1 w,i w,i 1 c,i 1 w o o i i ,n 1 kk kk kk (p p ) ( P P ) ( x) kk kk kk (p p ) (P ) 0 µ µ µ µ µ µ + + + + + + + − − + + ° ° ° + − − − ∆ − + − − = (10.86) 409 qw = α (pwf – pw) (10.87) 413 n w,i n 1 n n 1 w,i w,i w,i t wf,i w,i w,i d q ( S )(p p ) dS α α + + = + ∆ − (10.88) 414 NOMENCLATURE ENGLISH A area B Darcy coefficient in stabilized gas well inflow equations (ch. 8) Bg gas formation volume factor Bo oil formation volume factor Bw water formation volume factor c isothermal compressibility ce effective compressibility (applied to pore volume) cf pore compressibility ct total compressibility (applied to pore volume) c total aquifer compressibility (cw+cf) C arbitrary constant of integration C coefficient in gas well back-pressure equation (ch. 8) C pressure buildup correction factor (Russell afterflow analysis ch. 7) CA Dietz shape factor D non-Darcy flow constant appearing in rate dependent skin (ch. 8) D vertical depth e exponential ei exponential integral function E gas expansion factor Ef,w term in the material balance equation accounting for the expansion of the connate water and reduction in pore volume Eg term in the material balance equation accounting for the expansion of the gascap gas Eo term in the material balance equation accounting for the expansion of the oil and its originally dissolved gas f fraction f fractional flow of any fluid in the reservoir f function; e.g. f(p) - function of the pressure F cumulative relative gas volume in PVT differential liberation experiment (ch. 2) F non-Darcy coefficient in gas flow equation (ch. 8) F production term in the material balance equation (chs. 3,9) F wellbore parameter in McKinley afterflow analysis (ch. 7) g acceleration due to gravity g function; e.g.- g(p) function of pressure G gas initially in place – GIIP G gravity number (ch. 10) G wellbore liquid gradient (McKinley after-flow analysis, ch. 7) Ga apparent gas in place in a water drive gas reservoir (ch. 1) Gp cumulative gas production h formation thickness hp thickness of the perforated interval H total height of the capillary transition zone J Bessel function (ch. 7) J Productivity index k absolute permeability (chs. 4,9,10) k effective permeability (chs. 5,6,7,8) kr relative permeability obtained by normal-izing the effective permeability curves by dividing by the absolute permeability r k thickness averaged relative permeability k′r end point relative permeability k iteration counter Kr relative permeability obtained by normalizing the effective permeability curves by dividing by the end point permeability to oil (ch. 4) I length L length m ratio of the initial hydrocarbon pore volume of the gascap to that of the oil (material balance equation) m slope of the early, linear section of pressure analysis plot of pressure (pseudo pressure) vs. f(time), for pressure buildup, fall-off or multi-rate flow test m(p) real gas pseudo pressure m′(p) pseudo pressure for two phases (gas-oil) flow M end point mobility ratio M molecular weight Ms shock front mobility ratio n reciprocal of the slope of the gas well back pressure equation (ch. 8) n total number of moles N stock tank oil initially in place (STOIIP) Np cumulative oil production Npd dimensionless cumulative oil production (in pore volumes) NpD dimensionless cumulative oil production (in moveable oil volumes) p pressure pa average pressure in aquifer (ch. 9) pb bubble point pressure pc critical pressure pd dynamic grid block pressure pD dimensionless pressure pe pressure at the external boundary pi initial pressure NOMENCLATURE LX ppc pseudo critical pressure ppr pseudo reduced pressure psc pressure at standard conditions pwf bottom hole flowing pressure pwf(1hr) bottom hole flowing pressure rec-orded one hour after the start of flow pws bottom hole static pressure pws(LIN) (hypothetical) static pressure on the extrapolation of the early linear trend of the Horner buildup plot p average pressure p specific value of pws(LIN) at infinite closed-in time ∆p pressure drop N.B. the same subscripts/superscripts, used to distinguish between the above pressures, are also used in conjunction with pseudo pressures, hence: m(pi); m(pwf); m(pws(LIN)); etc. Pc capillary pressure c P° pseudo capillary pressure q production rate qi injection rate Q gas production rate r radial distance re external boundary radius rD dimensionless radius= r/rw (chs. 7,8) = r/ro (ch. 9) reD dimensionless radius= re/rw (chs. 7,8) = re/ro (ch. 9) rh radius of the heated zone around a steam soaked well ro reservoir radius rw wellbore radius r′w effective wellbore radius taking account of the mechanical skin (r′w = rwe-s) R producing (or instantaneous) gas oil ratio R universal gas constant Rp cumulative gas oil ratio Rs solution (or dissolved) gas oil ratio S mechanical skin factor S saturation (always expressed as a fraction of the pore volume) Sg gas saturation Sgr residual gas saturation to water So oil saturation Sor residual oil saturation to water Sw water saturation Swc connate (or irreducible) water saturation Swf water saturation at the flood front w S thickness average water saturation w S volume averaged water saturation behind an advancing flood front t reciprocal pseudo reduced temperature (Tpc/T) t time tD dimensionless time tDA dimensionless time (=tD 2 w r /A) ∆t closed-in time during a pressure buildup ∆ts closed-in time during a build-up at which pws(LIN) = p ∆td closed in time during a buildup at which pws(LIN) = pd T absolute temperature T transmissibility (McKinley afterflow analysis, ch. 7) Tc critical temperature Tpc pseudo critical temperature Tpr pseudo reduced temperature u Darcy velocity (q/A) U aquifer constant v velocity vg relative gas volume, differential liberation experiment V volume V net bulk volume of reservoir Vf pore volume (PV) Vg cumulative relative gas volume (sc), differential liberation PVT experiment w width WD dimensionless cumulative water influx (ch. 9) We cumulative water influx Wei initial amount of encroachable water in an aquifer; Wei=c Wipi (ch. 9) Wi initial volume of aquifer water (ch. 9) Wi cumulative water injected (ch. 10) Wid dimensionless cumulative water injected (pore volumes) WiD dimensionless cumulative water injected (moveable oil volumes) Wp cumulative water produced y reduced density, (Hall-Yarborough equations, ch. 1) Z Z-factor NOMENCLATURE LXI GREEK β turbulent coefficient for non-Darcy flow (ch. 8) β angle between the oil -water contact and the direction of flow, -stable segregated displacement (ch. 10) γ specific gravity (liquids,-relative to water =1 at standard conditions; gas,-relative to air=1 at standard conditions) γ exponent of Euler's constant (=1.782) ∆ difference (taken as a positive difference e.g. ∆p = pi-p) λ mobility θ dip angle of the reservoir Θ contact angle µ viscosity ρ density σ surface tension φ porosity Φ fluid potential per unit mass ψ fluid potential per unit volume (datum pressure) SUBSCRIPTS b bubble point bt breakthrough c capillary c critical d differential (PVT analysis) d dimensionless (expressed in pore volumes) d displacing phase D dimensionless (pressure, time, radius) D dimensionless (expressed in movable oil volumes) DA dimensionless (time) e effective e at the production end of a reservoir block (e.g. Swe) f flash separation (PVT) f flood front f pore (e.g. cf − pore compressibility) g gas h heated zone i cumulative injection i initial conditions n number of flow period N " " " " n (superscript) time step number o oil p cumulative production r reduced r relative r residual s steam s solution gas sc standard conditions t total w water wf wellbore flowing ws wellbore static CHAPTER 1 SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 1.1 INTRODUCTION In the process of illustrating the primary functions of a reservoir engineer, namely, the estimation of hydrocarbons in place, the calculation of a recovery factor and the attachment of a time scale to the recovery; this chapter introduces many of the fundamental concepts in reservoir engineering. The description of the calculation of oil in place concentrates largely on the determination of fluid pressure regimes and the problem of locating fluid contacts in the reservoir. Primary recovery is described in general terms by considering the significance of the isothermal compressibilities of the reservoir fluids; while the determination of the recovery factor and attachment of a time scale are illustrated by describing volumetric gas reservoir engineering. The chapter finishes with a brief quantitative account of the phase behaviour of multi-component hydrocarbon systems. 1.2 CALCULATION OF HYDROCARBON VOLUMES Consider a reservoir which is initially filled with liquid oil. The oil volume in the reservoir (oil in place) is ( )( ) wc OIP V 1 S res.vol. φ = − (1.1) where V = the net bulk volume of the reservoir rock φ = the porosity, or volume fraction of the rock which is porous and Swc = the connate or irreducible water saturation and is expressed as a fraction of the pore volume. The product Vφ is called the pore volume (PV) and is the total volume in the reservoir which can be occupied by fluids. Similarly, the product Vφ (1−Swc) is called the hydrocarbon pore volume (HCPV) and is the total reservoir volume which can be filled with hydrocarbons either oil, gas or both. The existence of the connate water saturation, which is normally 10−25% (PV), is an example of a natural phenomenon which is fundamental to the flow of fluids in porous media. That is, that when one fluid displaces another in a porous medium, the displaced fluid saturation can never be reduced to zero. This applies provided that the fluids are immiscible (do not mix) which implies that there is a finite surface tension at the interface between them. Thus oil, which is generated in deep source rock, on migrating into a water filled reservoir trap displaces some, but not all, of the water, resulting in the presence of a connate water saturation. Since the water is immobile its only influence in reservoir SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 2 engineering calculations is to reduce the reservoir volume which can be occupied by hydrocarbons. The oil volume calculated using equ. (1.1) is expressed as a reservoir volume. Since all oils, at the high prevailing pressures and temperatures in reservoirs, contain different amounts of dissolved gas per unit volume, it is more meaningful to express oil volumes at stock tank (surface) conditions, at which the oil and gas will have separated. Thus the stock tank oil initially in place is ( ) wc oi STOIIP n v 1 S /B φ = = − (1.2) where Boi is the oil formation volume factor, under initial conditions, and has the units reservoir volume/stock tank volume, usually, reservoir barrels/stock tank barrel (rb/stb). Thus a volume of Boi rb of oil will produce one stb of oil at the surface together with the volume of gas which was originally dissolved in the oil in the reservoir. The determination of the oil formation volume factor and its general application in reservoir engineering will be described in detail in Chapter 2. In equ. (1.2), the parameters φ and Swc are normally determined by petrophysical analysis and the manner of their evaluation will not be described in this text1. The net bulk volume, V, is obtained from geological and fluid pressure analysis. The geologist provides contour maps of the top and base of the reservoir, as shown in fig. 1.1. Such maps have contour lines drawn for every 50 feet, or so, of elevation 5000 5150 5250 5050 (a) X X Y Y Well WATER OWC OWC OIL (b) Fig. 1.1 (a) Structural contour map of the top of the reservoir, and (b) cross section through the reservoir, along the line X− − − −Y and the problem is to determine the level at which the oil water contact (OWC) is to be located. Measurement of the enclosed reservoir rock volume above this level will then give the net bulk volume V. For the situation depicted in fig. 1.1 (b) it would not be possible to determine this contact by inspection of logs run in the well since only the oil zone has been penetrated. Such a technique could be applied, however, if the OWC were somewhat higher in the reservoir. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 3 The manner in which the oil water contact, or fluid contacts in general, can be located requires a knowledge of fluid pressure regimes in the reservoir which will be described in the following section. 1.3 FLUID PRESSURE REGIMES The total pressure at any depth, resulting from the combined weight of the formation rock and fluids, whether water, oil or gas, is known as the overburden pressure. In the majority of sedimentary basins the overburden pressure increases linearly with depth and typically has a pressure gradient of 1 psi/ft, fig. 1.2. Pressure (psia) overburden pressure (OP) overpressure 14.7 FP GP normal hydrostatic pressure underpressure Depth (ft) Fig. 1.2 Overburden and hydrostatic pressure regimes (FP = fluid pressure; GP = grain pressure) At a given depth, the overburden pressure can be equated to the sum of the fluid pressure (FP) and the grain or matrix pressure (GP) acting between the individual rock particles, i.e. OP FP GP = + (1.3) and, in particular, since the overburden pressure remains constant at any particular depth, then ( ) ( ) d FP d GP = − (1.4) That is, a reduction in fluid pressure will lead to a corresponding increase in the grain pressure, and vice versa. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 4 Fluid pressure regimes in hydrocarbon columns are dictated by the prevailing water pressure in the vicinity of the reservoir. In a perfectly normal case the water pressure at any depth can be calculated as w water dp p D 14.7 (psia) dD × + (1.5) in which dp/dD, the water pressure gradient, is dependent on the chemical composition (salinity), and for pure water has the value of 0.4335 psi/ft. Addition of the surface pressure of one atmosphere (14.7 psia) results in the expression of the pressure in absolute rather than gauge units (psig), which are measured relative to atmospheric pressure. In many instances in reservoir engineering the main concern is with pressure differences, which are the same whether absolute or gauge pressures are employed, and are denoted simply as psi. Equation (1.5) assumes that there is both continuity of water pressure to the surface and that the salinity does not vary with depth. The former assumption is valid, in the majority of cases, even though the water bearing sands are usually interspersed with impermeable shales, since any break in the areal continuity of such apparent seals will lead to the establishment of hydrostatic pressure continuity to the surface. The latter assumption, however, is rather naive since the salinity can vary markedly with depth. Nevertheless, for the moment, a constant hydrostatic pressure gradient will be assumed, for illustrative purposes. As will be shown presently, what really matters to the engineer is the definition of the hydrostatic pressure regime in the vicinity of the hydrocarbon bearing sands. In contrast to this normal situation, abnormal hydrostatic pressure are encountered which can be defined by the equation w water dp p D 14.7 C(psia) dD = × + + (1.6) where C is a constant which is positive if the water is overpressured and negative if underpressured. For the water in any sand to be abnormally pressured, the sand must be effectively sealed off from the surrounding strata so that hydrostatic pressure continuity to the surface cannot be established. Bradley2 has listed various conditions which can cause abnormal fluid pressures in enclosed water bearing sands, which include: − temperature change; an increase in temperature of one degree-Fahrenheit can cause an increase in pressure of 125 psi in a sealed fresh water system. − geological changes such as the uplifting of the reservoir, or the equivalent, surface erosion, both of which result in the water pressure in the reservoir sand being too high for its depth of burial; the opposite effect occurs in a downthrown reservoir in which abnormally low fluid pressure can occur. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 5 − osmosis between waters having different salinity, the sealing shale acting as the semi-permeable membrane in this ionic exchange; if the water within the seal is more saline than the surrounding water the osmosis will cause an abnormally high pressure and vice versa. Some of these causes of abnormal pressuring are interactive, for instance, if a reservoir block is uplifted the resulting overpressure is partially alleviated by a decrease in reservoir temperature. The geological textbook of Chapman3 provides a comprehensive description of the mechanics of overpressuring. Reservoir engineers, however, tend to be more pragmatic about the subject of abnormal pressures than geologists, the main questions being; are the water bearing sands abnormally pressured and if so, what effect does this have on the extent of any hydrocarbon accumulations? So far only hydrostatic pressures have been considered. Hydrocarbon pressure regimes are different in that the densities of oil and gas are less than that of water and consequently, the pressure gradients are smaller, typical figures being water dp 0.45 psi/ ft dD = oil dp 0.35 psi/ ft dD = gas dp 0.08 psi/ ft dD = Thus for the reservoir containing both oil and a free gascap, shown in fig. 1.3; using the above gradients would give the pressure distribution shown on the left hand side of the diagram. At the oil-water contact, at 5500 ft, the pressure in the oil and water must be equal otherwise a static interface would not exist. The pressure in the water can be determined using equ. (1.5), rounded off to the nearest psi, as w p 0.45 D 15 (psia) = + (1.7) SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 6 2500 Exploration Well OIL GAS GOC 5200' WATER OWC 5500' TEST RESULTS at 5250 ft po = 2402 psia = 0.35psi/ft dp dD 2250 Depth (feet) 2265 5000 2369 5250 5500 2375 Pressure (psia) OWC : po = pw = 2490 GOC : po = pw = 2385 Fig. 1.3 Pressure regimes in the oil and gas for a typical hydrocarbon accumulation which assumes a normal hydrostatic pressure regime. Therefore, at the oil-water contact o w p p .45 5500 15 2490 (psia) = = × + = The linear equation for the oil pressure, above the oil water contact, is then po = 0.35D + constant and since po = 2490 psia at D = 5500 ft, the constant can be evaluated to give the equation ( ) o p 0.35D 565 psia = + (1.8) At the gas-oil contact at 5200 ft, the pressure in both fluids must be equal and can be calculated, using equ. (1.8), to be 2385 psia. The equation of the gas pressure line can then be determined as ( ) o p 0.08D 1969 psia = + (1.9) Finally, using the latter equation, the gas pressure at the very top of the structure, at 5000 ft, can be calculated as 2369 psia. The pressure lines in the hydrocarbon column are drawn in the pressure depth diagram, fig. 1.3, from which it can be seen that at the top of the structure the gas pressure exceeds the normal hydrostatic pressure by 104 psi. Thus in a well drilling through a sealing shale on the very crest of the structure there will be a sharp pressure kick from 2265 psi to 2369 psia on first penetrating the reservoir at 5000 ft. The magnitude of the pressure discontinuity on drilling into a hydrocarbon reservoir depends on the vertical distance between the point of well penetration and the hydrocarbon water contact and, for a given value of this distance, will be much greater if the reservoir contains gas alone. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 7 At the time of drilling an exploration well and discovering a new reservoir, one of the main aims is to determine the position of the fluid contacts which, as described in the previous section, will facilitate the calculation of the oil in place. Consider the exploration well, shown in fig. 1.3, which penetrates the reservoir near the top of the oil column. The gas-oil contact in the reservoir will be clearly "seen", at 5200 ft, on logs run in the well. The oil-water contact, however, will not be seen since it is some 225 ft below the point at which the well penetrates the base of the reservoir. The position of the contact can only be inferred as the result of a well test, such as a drill stem4 or wireline formation test5,6, in which the pressure and temperature are measured and an oil sample recovered. Analysis of the sample permits the calculation of the oil density at reservoir conditions and hence the oil pressure gradient (refer exercise 1, Chapter 2). Together, the pressure measurement and pressure gradient are sufficient to define the straight line which is the pressure depth relation in the oil column. If such a test were conducted at a depth of 5250 ft, in the well in fig. 1.3, then the measured pressure would be 2402 psia and the calculated oil gradient 0.35 psi/ft, which are sufficient to specify the oil pressure line as ( ) o p 0.35D 565 psia = + (1.8) and extrapolation of this line to meet the normal hydrostatic pressure line will locate the oil-water contact at 5500 ft. This type of analysis relies critically on a knowledge of the hydrostatic pressure regime. If, for instance, the water is overpressured by a mere 20 psi then the oil-water contact would be at 5300 ft instead of at 5500 ft. This fact can be checked by visual inspection of fig. 1.3 or by expressing the equation of the overpressured water line, equ. (1.6) as ( ) w p 0.45D 35 psia = + and solving simultaneously with equ. (1.8) for the condition that pw = po at the oil-water contact. The difference of 200 ft in the position of the contact can make an enormous difference to the calculated oil in place, especially if the areal extent of the reservoir is large. It is for this reason that reservoir engineers are prepared to spend a great deal of time (and therefore, money) in defining the hydrostatic pressure regime in a new field. A simple way of doing this is to run a series of wireline formation tests5,6 in the exploration well, usually after logging and prior to setting casing, in which pressures are deliberately measured in water bearing sands both above and beneath the hydrocarbon reservoir or reservoirs. The series of pressure measurements at different depths enables the hydrostatic pressure line, equ. (1.6), to be accurately defined in the vicinity of the hydrocarbon accumulation, irrespective of whether the pressure regime is normal or abnormal. Such tests are repeated in the first few wells drilled in a new field or area until the engineers are quite satisfied that there is an areal uniformity in the hydrostatic pressure. Failure to do this can lead to a significant error in the estimation of the SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 8 hydrocarbons in place which in turn can result in the formulation of woefully inaccurate field development plans. Pressure (psia) 2250 2375 2500 5000 5500 Depth (feet) GDT 5150’ DPGWC 5281’ OIL COLUMN Fig 1.3 TEST RESULTS at 5100 ft pg = 2377 psia dp dD g = .08 psi / ft GAS DPOWC 5640’ Exploration well Fig. 1.4 Illustrating the uncertainty in estimating the possible extent of an oil column, resulting from well testing in the gas cap Figure 1.4 illustrates another type of uncertainty associated with the determination of fluid contacts from pressure measurements. The reservoir is the same as depicted in fig. 1.3 but in this case the exploration well has only penetrated the gascap. A well test is conducted at a depth of 5100 ft from which it is determined that the gas pressure is 2377 psia and, from the analysis of a collected sample (refer exercise 1.1), that the gas gradient in the reservoir is 0.08 psi/ft. From these data the equation of the gas pressure line can be defined as ( ) o p 0.08D 1969 psia = + (1.9) Having seen no oil in the well the engineer may suspect that he has penetrated a gas reservoir alone, and extrapolate equ. (1.9) to meet the normal hydrostatic pressure line ( ) w p 0.45D 15 psia = + (1.7) at a depth of 5281 ft, at which pw = pg. This level is marked in fig. 1.4 as the deepest possible gas water contact (DPGWC), assuming there is no oil. Alternatively, since the deepest point at which gas has been observed in the well is 5150 ft (GDT − gas down to), there is no physical reason why an oil column should not extend from immediately beneath this point. The oil pressure at the top of such a column would be equal to the gas pressure, which can be calculated using equ. (1.9) as 2381 psia. Hence the equation of the oil pressure line, assuming the oil gradient used previously of 0.35 psi/ft, would be ( ) o p 0.35D 579 psia = + SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 9 and solving this simultaneously with equ. (1.7), for the condition that po = pw, gives the oil-water contact at a depth of 5640 ft. This is marked on fig. 1.4 as the deepest possible oil-water contact (DPOWC) and corresponds to the maximum possible oil column. Therefore, in spite of the fact that the well has been carefully tested, there remains high degree of uncertainty as to the extent of any oil column. It could indeed be zero (DPGWC − 5281 ft) or, in the most optimistic case, could extend for 490 ft (DPOWC − 5640 ft), or, alternatively, assume any value in between these limits. Also shown in fig. 1.4 is the actual oil column from fig. 1.3. Therefore, the question is always posed, on penetrating a reservoir containing only gas; is there a significant oil column, or oil rim, down-dip which could be developed? The only sure way to find out is to drill another well further down-dip on the structure or, if mechanically feasible, plug back and deviate from the original hole. When planning the drilling of an exploration well it is therefore, not always expedient to aim the well at the highest point on the structure. Doing so will tend to maximise the chance of finding hydrocarbons but will oppose one of the primary aims in drilling exploration wells, which is to gain as much information about the reservoirs and their contents as possible. Having determined the fluid contacts in the reservoir, using the methods described in this section, the engineer is then in a position to calculate the net bulk volume V required to calculate the hydrocarbons in place. In fig. 1.1 (a), for instance, this can be done by planimetering the contours above the OWC7,8. Finally, with regard to the application of equ. (1.2), the correct figure for the STOIIP will only be obtained if all the parameters in the equation are truly representative of their average values throughout the reservoir. Since it is impossible to obtain such figures it is more common to represent each parameter in the STOIIP equation by a probability distribution rather than a determinate value. For instance, there may be several different geological interpretations of the structure giving a spread in values of the net bulk volume V, which could be expressed as a probability distribution of the value of this parameter. The STOIIP equation is then evaluated using some statistical calculation procedure, commensurate with the quality of the input data, and the results expressed in terms of a probability distribution of the STOIIP. The advantage of this method is that while a mean value of the STOIIP can be extracted from the final distribution, the results can also be formulated in terms of the uncertainty attached to this figure, expressed, for instance, as a standard deviation about the mean9,10. If the uncertainty is very large it may be necessary to drill an additional well, or wells, to narrow the range before proceeding to develop the field. 1.4 OIL RECOVERY: RECOVERY FACTOR Equation (1.2), for the STOIIP, can be converted into an equation for calculating the ultimate oil recovery simply by multiplying by the recovery factor (RF), which is a number between zero and unity representing the fraction of recoverable oil, thus wc oi Ultimate Recovery (UR) (V (1 S )/B ) RF φ = − × (1.10) SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 10 And, while it is easy to say, "simply by multiplying by the recovery factor", it is much less easy to determine what the recovery factor should be for any given reservoir and, indeed, it is the determination of this figure which is the most important single task of the reservoir engineer. For a start, one can clearly distinguish between two types of recovery factor. There is one which is governed by current economic circumstances and, ever increasingly, by environmental and ecological considerations, while the second can be classed as a purely technical recovery factor depending on the physics of the reservoir-fluid system. Regrettably, the former, although possibly the more interesting, is not a subject for this book. The two main categories of hydrocarbon recovery are called primary and supplementary. Primary recovery is the volume of hydrocarbons which can be produced by virtue of utilising the natural energy available in the reservoir and its adjacent aquifer. In contrast, supplementary recovery is the oil obtained by adding energy to the reservoir-fluid system. The most common type of supplementary recovery is water flooding in which water is injected into the reservoir and displaces oil towards the producing wells, thus increasing the natural energy of the system. The mechanics of supplementary recovery will be described later, in Chapter 4, sec. 9 and in Chapter 10; for the moment only primary recovery will be considered. The entire mechanics of primary recovery relies on the expansion of fluids in the reservoir and can best be appreciated by considering the definition of isothermal compressibility. T 1 V c V p ∂ = ∂ — (1.11) The isothermal compressibility is commonly applied in the majority of reservoir engineering calculations because it is considered a reasonable approximation that as fluids are produced, and so remove heat from the reservoir by convection, the cap and base rock, which are assumed to act as heat sources of infinite extent, immediately replace this heat by conduction so that the reservoir temperature remains constant. Therefore, compressibility, when referred to in this text, should always be interpreted as the isothermal compressibility. The negative sign convention is required in equ. (1.11) because compressibility is defined as a positive number, whereas the differential, ∂V/∂p, is negative, since fluids expand when their confining pressure is decreased. When using the compressibility definition in isolation, to describe reservoir depletion, it is more illustrative to express it in the form dV cV p = ∆ (1.12) where dV is an expansion and ∆p a pressure drop, both of which are positive. This is the very basic equation underlying all forms of primary recovery mechanism. In the reservoir, if ∆p is taken as the pressure drop from initial to some lower pressure, SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 11 pi − p, then dV will be the corresponding fluid expansion, which manifests itself as production. The skill in engineering a high primary recovery factor, utilising the natural reservoir energy, is to ensure that the dV, which is the production, is the most commercially valuable fluid in the reservoir, namely, the oil. The way in which this can be done is shown schematically in fig. 1.5. aquifer gascap oil Vw dVW dV = oil production tot = dV + dV + dV o w g dVg Vg Vo Fig. 1.5 Primary oil recovery resulting from oil, water and gas expansion The diagram illustrates the fairly obvious fact that to produce an oil reservoir, wells should be drilled into the oil zone. If the reservoir is in contact with a gascap and aquifer, the oil production due to a uniform pressure drop, ∆p, in the entire system, will have components due to the separate expansion of the oil gas and water, thus dVTOT = Oil Production = dVo + dVw + dVg in which the balance is expressed in fluid volumes at reservoir conditions. Applying equ. (1.12), this may be expressed as dVTOT = co Vo ∆p + cw Vw ∆p + cg Vg ∆p Considering the following figures as typical for the compressibilities of the three components at a pressure of 2000 psia: co = 15 × 10−6/psi cw = 3 × 10−6/psi cg = 500 × 10−6/psi 1; refer sec.1.5 p ≈ it is evident that the contribution to dVTOT supplied by the oil and water expansion will only be significant if both Vo and Vw, the initial volumes of oil and water, are large. In contrast, because of its very high compressibility, even a relatively small volume of gascap gas will contribute significantly to the oil production. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 12 Therefore, while it is obvious that one would not produce an aquifer, but rather, let the water expand and displace the oil; so too, the gas in the gascap, although having commercial value, is frequently kept in the reservoir and allowed to play its very significant role in contributing to the primary recovery through its expansion. The mechanics of primary oil recovery will be considered in greater detail in Chapter 3. 1.5 VOLUMETRIC GAS RESERVOIR ENGINEERING Volumetric gas reservoir engineering is introduced at this early stage in the book because of the relative simplicity of the subject. lt will therefore be used to illustrate how a recovery factor can be determined and a time scale attached to the recovery. The reason for the simplicity is because gas is one of the few substances whose state, as defined by pressure, volume and temperature (PVT), can be described by a simple relation involving all three parameters. One other such substance is saturated steam, but for oil containing dissolved gas, for instance, no such relation exists and, as shown in Chapter 2, PVT parameters must be empirically derived which serve the purpose of defining the state of the mixture. The equation of state for an ideal gas, that is, one in which the inter-molecular attractions and the volume occupied by the molecules are both negligible, is pV nRT = (1.13) in which, for the conventional field units used in the industry p = pressure (psia); V = volume (cu.ft) T = absolute temperature − degrees Rankine (°R=460+°F) n = the number of lb. moles, where one lb. mole is the molecular weight of the gas expressed in pounds. and R = the universal gas constant which, for the above units, has the value 10.732 psia.cu.ft/lb. mole.°R. This equation results from the combined efforts of Boyle, Charles, Avogadro and Gay Lussac, and is only applicable at pressures close to atmospheric, for which it was experimentally derived, and at which gases do behave as ideal. Numerous attempts have been made in the past to account for the deviations of a real gas, from the ideal gas equation of state, under extreme conditions. One of the more celebrated of these is the equation of van der Waals which, for one Ib.mole of a gas, can be expressed as ( ) 2 a (p ) V b RT V + − = (1.14) In using this equation it is argued that the pressure p, measured at the wall of a vessel containing a real gas, is lower than it would be if the gas were ideal. This is because the momentum of a gas molecule about to strike the wall is reduced by inter-molecular attractions; and hence the pressure, which is proportional to the rate of change of SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 13 momentum, is reduced. To correct for this the term a/V2 must be added to the observed pressure, where a is a constant depending on the nature of the gas. Similarly the volume V, measured assuming the molecules occupy negligible space, must be reduced for a real gas by the factor b which again is dependent on the nature of the gas. The principal drawback in attempting to use equ. (1.14) to describe the behaviour of real gases encountered in reservoirs is that the maximum pressure for which the equation is applicable is still far below the normal range of reservoir pressures. More recent and more successful equations of state have been derived, - the Beattie-Bridgeman and Benedict-Webb-Rubin equations, for instance (which have been conveniently summarised in Chapter 3 of reference 18); but the equation most commonly used in practice by the industry is pV ZnRT = (1.15) in which the units are the same as listed for equ. (1.13) and Z, which is dimensionless, is called the Z−factor. By expressing the equation as P V = nRT Z the Z−factor can be interpreted as a term by which the pressure must be corrected to account for the departure from the ideal gas equation. The Z−factor is a function of both pressure and absolute temperature but, for reservoir engineering purposes, the main interest lies in the determination of Z, as a function of pressure, at constant reservoir temperature. The Z(p) relationship obtained is then appropriate for the description of isothermal reservoir depletion. Three ways of determining this relationship are described below. a) Experimental determination A quantity of n moles of gas are charged to a cylindrical container, the volume of which can be altered by the movement of a piston. The container is maintained at the reservoir temperature, T, throughout the experiment. If Vo is the gas volume at atmospheric pressure, then applying the real gas law, equ. (1.15), 14.7 Vo = nRT since Z=1 at atmospheric pressure. At any higher pressure p, for which the corresponding volume of the gas is V, then pV = ZnRT and dividing these equations gives SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 14 o pV Z 14.7V = By varying p and measuring V, the isothermal Z(p) function can be readily obtained. This is the most satisfactory method of determining the function but in the majority of cases the time and expense involved are not warranted since reliable methods of direct calculation are available, as described below. b) The Z-factor correlation of Standing and Katz This correlation requires a knowledge of the composition of the gas or, at least, the gas gravity. Naturally occurring hydrocarbons are composed primarily of members of the paraffin series (CnH2n+2) with an admixture of non-hydrocarbon impurities such as carbon dioxide, nitrogen and hydrogen sulphide. Natural gas differs from oil in that it predominantly consists of the lighter members of the paraffin series, methane and ethane, which usually comprise in excess of 90% of the volume. A typical gas composition is listed in table 1.1. In order to use the Standing-Katz correlation11 it is first necessary, from a knowledge of the gas composition, to determine the pseudo critical pressure and temperature of the mixture as pc i ci i p n p = (1.16) and pc i ci i T nT = (1.17) where the summation is over all the components present in the gas. The parameters pci and Tci are the critical pressure and temperature of the ith component, listed in table 1.1, while the ni are the volume fractions or, for a gas, the mole fractions of each component (Avogrado's law). The next step is to calculate the so-called pseudo reduced pressure and temperature pr pc p p p = (1.18) and pr pc T T T = (1.19) where p and T are the pressure and temperature at which it is required to determine Z. In the majority of reservoir engineering problems, which are isothermal, Tpr is constant and ppr variable. With these two parameters the Standing-Katz correlation chart, fig. 1.6, which consists of a set of isotherms giving Z as a function of the pseudo reduced pressure, can be used to determine the Z−factor. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 15 For instance, for the gas composition listed in table 1.1, and at a pressure of 2000 psia and temperature of 180° F, the reader can verify that ppc = 663.3 and Tpc = 374.1 giving pr pr 2000 640 p 3.02 and T 1.71 663.3 374.1 = = = = from which, using fig.1.6, the Z−factor can be obtained as 0.865. Component Molecular Weight Critical Pressure (psia) Constants Temp. (oR) Typical Composition (volume or mole fraction, ni) CH4 Methane 16.04 668 343 .8470 C2 H6 Ethane 30.07 708 550 .0586 C3 H8 Propane 44.10 616 666 .0220 i−C4 H10 Isobutane 58.12 529 735 .0035 n−C4 H10 Normal butane 58.12 551 765 .0058 i−C5 H12 Isopentane 72.15 490 829 .0027 n−C5 H12 Normal pentane 72.15 489 845 .0025 n−C6 H14 Normal hexane 86.18 437 913 .0028 n−C7 H14 Normal heptane 100.20 397 972 .0028 n-C8 H18 Normal octane 114.23 361 1024 .0015 n−C9 H20 Normal nonane 128.26 332 1070 .0018 n−C10 H22 Normal decane 142.29 304 1112 .0015 CO2 Carbon dioxide 44.01 1071 548 .0130 H2S Hydrogen sulphide 34.08 1306 672 .0000 N2 Nitrogen 28.01 493 227 .0345 TABLE 1.1 Physical constants of the common constituents of hydrocarbon gases12, and a typical gas composition Conventionally, the composition of natural gases is listed in terms of the individual components as far as hexane, with the heptane and heavier components being grouped together as 7 C+ (heptanes-plus). In the laboratory analysis the molecular weight and specific gravity of this group are measured, which permits the determination of the pseudo critical pressure and temperature of the 7 C+ from standard SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 16 correlations8,13. This in turn facilitates the calculation of the Z−factor using the method described above. In calculating the Z−factor it has been assumed that the non-hydrocarbon components, carbon dioxide, hydrogen sulphide and nitrogen, can be included in the summations, equs. (1.16) and (1.17), to obtain the pseudo critical pressure and temperature. This approach is only valid if the volume fractions of the non-hydrocarbon components are small, say, less than 5% vol. For larger amounts, corrections to the above calculation procedures are to be found in the text book of Amyx, Bass and Whiting8. If, however, the volume fractions of the non-hydrocarbons are very large (the carbon dioxide content of the Kapuni field, New Zealand, for instance, is 45% vol.) then it is better to determine the Z−factor experimentally as described in a), above. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 17 3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.1 1.45 1.35 1.25 1.3 1.2 1.05 1.1 1.0 0.9 0.8 0.7 0.6 Z -factor Pseudo reduced pressure 0.5 Pseudo reduced temperature 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 8 1.15 1.05 Fig. 1.6 The Z− − − −factor correlation chart of Standing and Katz11 (Reproduced by courtesy of the SPE of the AIME) If the gas composition is not available, the Standing-Katz correlation can still be used provided the gas gravity, based on the scale air = 1, at atmospheric pressure and at 60°F, is known (refer sec. 1.6). In this case fig. 1.7, is used to obtain the pseudo critical pressure and temperature; then equs. (1.18) and (1.19) can be applied to calculate the pseudo reduced parameters required to obtain the Z−factor from fig. 1.6. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 18 MISCELLANEOUS GASES CONDENSATE WELL FLUID PSEUDO CRITICAL PRESSURE, psia PSEUDO CRITICAL TEMPERATURE, degrees Rankine GAS GRAVITY (Air = 1) 700 650 600 550 500 450 400 350 300 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 Fig. 1.7 Pseudo critical properties of miscellaneous natural gases and condensate well fluids19 c) Direct calculation of Z-factors The Standing-Katz correlation is very reliable and has been used with confidence by the industry for the past thirty-five years. With the advent of computers, however, there arose the need to find some convenient technique for calculating Z−factors, for use in gas reservoir engineering programs, rather than feeding in the entire correlation chart from which Z−factors could be retrieved by table look-up. Takacs14 has compared eight different methods for calculating Z−factors which have been developed over the years. These fall into two main categories: those which attempt to analytically curve-fit the Standing-Katz isotherms and those which compute Z−factors using an equation of state. Of the latter, the method of Hall-Yarborough15 is worthy of mention because it is SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 19 both extremely accurate and very simple to program, even for small desk calculators, since it requires only five storage registers. The Hall-Yarborough equations, developed using the Starling-Carnahan equation of state, are 2 pr 1.2(1 t) 0.06125p t e Z y − − = (1.20) where ppr = the pseudo reduced pressure t = the reciprocal, pseudo reduced temperature (Tpc/T) and y = the "reduced" density which can be obtained as the solution of the equation. 2 2 3 4 2 3 2 pr 3 y y y y 1.2(1 t) 0.06125 p t e (14.76t 9.76t 4.58t )y (1 y) + + − − − − + − − + − 2 3 (2.18 2.82t) (90.7t 242.2t 42.4t )y 0 + + − + = (1.21) This non-linear equation can be conveniently solved for y using the simple Newton Raphson iterative technique. The steps involved in applying this are: 1) make an initial estimate of yk, where k is an iteration counter (which in this case is unity, e.g. y1 = 0.001) 2) substitute this value in equ. (1.21); unless the correct value of y has been initially selected, equ. (1.21) will have some small, non-zero value Fk. 3) using the first order Taylor series expansion, a better estimate of y can be determined as k k 1 k k dF y y F dy + = − (1.22) where the general expression for dF/dk can be obtained as the derivative of equ. (1.21), i.e. 2 3 4 2 3 4 dF 1 4y 4y 4y y (29.52t 19.52t 9.16t )y dy (1 y) + + − + = − − + − 2 3 (1.18 2.82t) (2.18 2.82t) (90.7t 242.2t 42.4t )y + + + − + (1.23) 4) iterate, using equs. (1.21) and (1.22), until satisfactory convergence is obtained (Fk ≈ 0). 5) substitution of the correct value of y in equ. (1.20) will give the Z−factor. (N.B. there appears to be a typographical error in the original Hall-Yarborough paper15, in that the equations presented for F (equ. 8) and dF/dy (equ. 11), contain 1−y3 and SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 20 1−y4 in the denominators of the second and first terms, respectively, instead of (1−y)3 and (1−y)4 as in equs. (1.21) and (1.23) of this text.) Takacs14 has determined that the average difference between the Standing-Katz correlation chart and the analytical Hall-Yarborough method is − 0.158% and the average absolute difference 0.518%. Figure 1.8 shows an isothermal Z−factor versus pressure relationship, obtained using the Hall-Yarborough method, for a gas with gravity 0.85 and at a reservoir temperature of 200°F. The plot coincides, within pencil thickness, with the similar relation obtained by the application of the method described in b), above. The plot shows that there is a significant deviation from the ideal gas behaviour which is particularly noticeable in the intermediate pressure range at about 2500 psia. At this pressure, use of the ideal gas equation, (1.13), would produce an error of almost 25% in calculated gas volumes. 1.6 APPLICATION OF THE REAL GAS EQUATION OF STATE The determination of the Z−factor as a function of pressure and temperature facilitates the use of the simple equation pV ZnRT = (1.15) to fully define the state of a real gas. This equation is a PVT relationship and in reservoir engineering, in general, the main use of such functions is to relate surface to reservoir volumes of hydrocarbons. For a real gas, in particular, this relation is expressed by the gas expansion factor E, where SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 21 Pressure (psia) Z - Factor 0 0.75 0.85 0.95 1000 2000 3000 4000 5000 Fig. 1.8 Isothermal Z− − − −factor as a function of pressure (gas gravity = 0.85; temperature = 200° F) sc V volume of n moles of gas at standard conditions E V volume of n moles of gas at reservoir conditions = = and applying equ. (1.15) at both standard and reservoir conditions this becomes sc sc sc sc V T Z p E V p T Z = = (1.24) For the field units defined in connection with equ. (1.13), and for standard conditions of psc = 14.7 psia, Tsc = (460+60) = 520°R and Zsc = 1, equ. (1.24) can be reduced to p E 35.37 (vol/ vol) ZT = (1.25) At a pressure of 2000 psia and reservoir temperature of 180°F the gas whose composition is detailed in table 1.1 has a Z−factor of 0.865, as already determined in sec. 1.5(b). Therefore, the corresponding gas expansion factor is 35.37 2000 E 127.8 (vol/ vol) 0.865 640 × = = × In particular, the gas initially in place (GIIP) in a reservoir can be calculated using an equation which is similar to equ. (1.2) for oil, that is SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 22 wc i G V (1 S )E φ = − (1.26) in which Ei is evaluated at the initial pressure. Other important parameters which can be conveniently expressed using the equation of state are, the real gas density, gravity and isothermal compressibility. Since the mass of n moles of gas is nM, where M is the molecular weight, then the density is nM nM Mp V ZnRT /p ZRT ρ = = = (1.27) Comparing the density of a gas, at any pressure and temperature, to the density of air at the same conditions gives gas gas air air (M/ Z) (M/ Z) ρ ρ = and, in particular, at standard conditions gas gas g air air M M M 28.97 ρ γ ρ = = = (1.28) where γg is the gas gravity relative to air at standard conditions and is conventionally expressed as, for instance, γg = 0.8 (air = 1). Therefore, if the gas gravity is known, M can be calculated using equ. (1.28) and substituted in equ. (1.27) to give the density at any pressure and temperature. Alternatively, if the gas composition is known M can be calculated as i i i M nM = (1.29) and again substituted in equ. (1.27). The molecular weights of the individual gas components, Mi, are listed in table 1.1. It is also useful to remember the density of air at standard conditions (in whichever set of units the reader employs). For the stated units this figure is (ρ air) sc = 0.0763 ib/cu.ft which permits the gas density at standard conditions to be evaluated as sc 0.0763 g (lbs / cu.ft) ρ γ = (1.30) The final application of the equation of state is to derive an expression for the isothermal compressibility of a real gas. Solving equ. (1.15) for V gives ZnRT V p = SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 23 the derivative of which, with respect to pressure, is V ZnRT 1 1 Z p p p Z p ∂ ∂ = − − ∂ ∂ and substituting these two expressions in the isothermal compressibility definition, equ. (1.11), gives g 1 V 1 1 Z c V p p Z p ∂ ∂ = − = − ∂ ∂ (1.31) In fig. 1.9, a plot of the gas compressibility defined by equ. (1.31) is compared to the approximate expression. g 1 c p = (1.32) for the 0.85 gravity gas whose isothermal Z−factor is plotted in fig. 1.8 at 200°F. As can be seen, the approximation, equ. (1.32), is valid in the intermediate pressure range between 2000−2750 psia where ∂Z/∂p is small but is less acceptable at very high or low pressures. EXERCISE 1.1 GAS PRESSURE GRADIENT IN THE RESERVOIR 1) Calculate the density of the gas, at standard conditions, whose composition is listed in table 1.1. 2) what is the gas pressure gradient in the reservoir at 2000 psia and 180° F (Z = 0.865). EXERCISE 1.1 SOLUTION 1) The molecular weight of the gas can be calculated as i i i M nM 19.91 = = and therefore, using equ. (1.28) the gravity is SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 24 Pressure (psia) GAS COMPRESSIBILITY psi-1 16 ×10-4 0 0 2 4 6 8 10 12 14 1000 2000 3000 4000 5000 cg = 1 p 1 p 1 Z -δZ δp cg = Fig. 1.9 Isothermal gas compressibility as a function of pressure (gas gravity = 0.85; temperature = 200° F) g air M 19.91 0.687 (air 1) M 28.97 γ = = = = The density at standard conditions can be evaluated using equ. (1.27) as sc sc sc sc Mp 19.91 14.7 0.0524 Ib / cu.ft Z RT 1 10.732 520 × = = = × × or, alternatively, using equ. (1.30) as sc g 0.0763 0.0524 lb / cu.ft ρ γ = = 2) The density of the gas in the reservoir can be directly calculated using equ. (1.27), or else by considering the mass conservation of a given quantity of gas as (ρV)sc = (ρV)res or ρ res = ρ scE which, using equ. (1.25), can be evaluated at 2000 psia and 180oF as sc res 35.37 p 35.37 0.0524 2000 6.696 Ib / cu.ft ZT 0.865 (180 460) ρ ρ × × = = = × + SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 25 To convert this number to a pressure gradient in psi/ft requires the following manipulation. 2 2 2 gas dp Ib 1 ft 6.696 6.696 0.0465 psi/ ft. dD ft 144 ft inch = × × = = 1.7 GAS MATERIAL BALANCE: RECOVERY FACTOR The material balance equation, for any hydrocarbon system, is simply a volume balance which equates the total production to the difference between the initial volume of hydrocarbons in the reservoir and the current volume. In gas reservoir engineering the equation is very simple and will now be considered for the separate cases in which there is no water influx into the reservoir and also when there is a significant degree of influx. a) Volumetric depletion reservoirs The term volumetric depletion, or simply depletion, applied to the performance of a reservoir means that as the pressure declines, due to production, there is an insignificant amount of water influx into the reservoir from the adjoining aquifer. This, in turn, implies that the aquifer must be small (refer sec. 1.4). Thus the reservoir volume occupied by hydrocarbons (HCPV) will not decrease during depletion. An expression for the hydrocarbon pore volume can be obtained from equ. (1.26) as HCPV = Vφ (1−Swc) = G/Ei where G is the initial gas in place expressed at standard conditions. The material balance, also expressed at standard conditions, for a given volume of production Gp, and consequent drop in the average reservoir pressure ∆p = pi−p is then, p p i Production GIIP Unproduced Gas (sc) (sc) (sc) G G (HCPV)E G G G E E = − = − = − (1.33) which can be expressed as p i G E 1 G E = − (1.34) or, using equ. (1.25), as p i i G p p 1 Z Z G = − (1.35) SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 26 The ratio Gp/G is the fractional gas recovery at any stage during depletion and, if the gas expansion factor E, in equ. (1.34), is evaluated at the proposed abandonment pressure then the corresponding value of Gp/G is the gas recovery factor. Before describing how the material balance equation is used in practice, it is worthwhile reconsidering the balance expressed by equ. (1.33) more thoroughly. Implicit in the equation is the assumption that because the water influx is negligible then the hydrocarbon pore volume remains constant during depletion. This, however, neglects two physical phenomena which are related to the pressure decline. Firstly, the connate water in the reservoir will expand and secondly, as the gas (fluid) pressure declines, the grain pressure increases in accordance with equ. (1.4). As a result of the latter, the rock particles will pack closer together and there will be a reduction in the pore volume. These two effects can be combined to give the total change in the hydrocarbon pore volume as d(HCPV) = − dVw +dVf (1.36) where Vw and Vf represent the initial connate water volume and pore volume (PV), respectively. The negative sign is necessary since an expansion of the connate water leads to a reduction in the HCPV. These volume changes can be expressed, using equ. (1.11), in terms of the water and pore compressibilities, where the latter is defined as f f f V 1 c V (GP) ∂ = − ∂ (1.4) where GP is the grain pressure which is related to the fluid pressure by d(FP) = − d(GP) therefore f f f f f V V 1 1 c V (FP) V p ∂ ∂ = − = ∂ ∂ (1.37) where p is the fluid pressure. Equation (1.36) can now be expressed as d(HCPV) = cw Vw dp + cf Vf dp or, as a reduction in hydrocarbon pore volume as d(HCPV) = − (cw Vw + cf Vf ) ∆p (1.38) where ∆p = pi − p, the drop in fluid (gas) pressure. Finally, expressing the pore and connate water volumes as f wc i wc HCPV G V PV (1 S ) E (1 S ) = = = − − and SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 27 wc w wc i wc GS V PV S E (1 S ) = × = − the reduction in hydrocarbon pore volume, equ. (1.38), can be included in equ. (1.33), to give w wc f p wc i (c S c ) p G E 1 1 G 1 S E + ∆ = − − − (1.39) as the modified material balance. Inserting the typical values of cw = 3 × 10−6/psi, cf = 10 × 10−6/psi and Swc = 0.2 in this equation, and considering a large pressure drop of ∆p = 1000 psi; the term in parenthesis becomes 6 3 (3 .2 10) 1 10 10 1 0.013 0.8 − × + − × × = − That is, the inclusion of the term accounting for the reduction in the hydrocarbon pore volume, due to the connate water expansion and pore volume reduction, only alters the material balance by 1.3% and is therefore frequently neglected. The reason for its omission is because the water and pore compressibilities are usually, although not always, insignificant in comparison to the gas compressibility, the latter being defined in sec. 1.6 as approximately the reciprocal of the pressure. As described in Chapter 3, sec. 8, however, pore compressibility can sometimes be very large in shallow unconsolidated reservoirs and values in excess of 100 × 10−6/ psi have been measured, for instance, in the Bolivar Coast fields in Venezuela. In such reservoirs it would be inadmissible to omit the pore compressibility from the gas material balance. In a reservoir which contains only liquid oil, with no free gas, allowance for the connate water and pore compressibility effects must be included in the material balance since these compressibilities have the same order of magnitude as the liquid oil itself (refer Chapter 3, sec. 5). In the majority of cases the material balance for a depletion type gas reservoir can adequately be described using equ. (1.35). This equation indicates that there is a linear relationship between p/Z and the fractional recovery Gp/G, or the cumulative production Gp, as shown in fig. 1.10(a) and (b), respectively. These diagrams illustrate one of the basic techniques in reservoir engineering which is, to try to reduce any equation, no matter how complex, to the equation of a straight line; for the simple reason that linear functions can be readily extrapolated, whereas non-linear functions, in general, cannot. Thus a plot of p versus Gp/G or Gp, would have less utility than the representations shown in fig. 1.10, since both would be non-linear. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 28 p i Zi G / G p RF Gp G ( a ) (RF) comp p Z Z p i i ( b ) p Z p Z ab Fig. 1.10 Graphical representations of the material balance for a volumetric depletion gas reservoir; equ. (1.35) Figure 1.10(a) shows how the recovery factor (RF) can be determined by entering the ordinate at the value of (p/Z)ab corresponding to the abandonment pressure. This pressure is dictated largely by the nature of the gas contract, which usually specifies that gas should be sold at some constant rate and constant surface pressure, the latter being the pressure at the delivery point, the gas pipeline. Once the pressure in the reservoir has fallen to the level at which it is less than the sum of the pressure drops required to transport the gas from the reservoir to the pipeline, then the plateau production rate can no longer be maintained. These pressure drops include the pressure drawdown in each well, which is the difference between the average reservoir and bottom hole flowing pressures, causing the gas flow into the wellbore; the pressure drop required for the vertical flow to the surface, and the pressure drop in the gas processing and transportation to the delivery point. As a result, gas reservoirs are frequently abandoned at quite high pressures. Recovery can be increased, however, by producing the gas at much lower pressures and compressing it at the surface to give the recovery (RF)comp, as shown in fig. 1.10(a). In this case the capital cost of the compressors plus their operating costs must be compensated by the increased gas recovery. Figure 1.10(b) also illustrates the important techniques in reservoir engineering, namely, "history matching" and "prediction". The circled points in the diagram, joined by the solid line, represent the observed reservoir history. That is, for recorded values of the cumulative gas production, pressures have been measured in the producing wells and an average reservoir pressure determined, as described in detail in Chapters 7 and 8. Since the plotted values of p/Z versus Gp form a straight line, the engineer may be inclined to think that the reservoir is a depletion type and proceed to extrapolate the linear trend to predict the future performance. The prediction, in this case, would be how the pressure declines as a function of production and, if the market rate is fixed, of time. In particular, extrapolation to the abscissa would give the value of the GIIP which SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 29 can be checked against the volumetric estimate obtained as described in secs. 1.2 and 1.3. This technique of matching the observed production pressure history by building a suitable mathematical model, albeit in this case a very simple one, equ. (1.35), and using the model to predict future performance is one which is fundamental to the subject of Reservoir Engineering. b) Water drive reservoirs If the reduction in reservoir pressure leads to an expansion of the adjacent aquifer water, and consequent influx into the reservoir, the material balance equation must then be modified as p e i Production GIIP Unproduced Gas (sc) (sc) (sc) G G G W E E = − = − − (1.40) where, in this case, the hydrocarbon pore volume at the lower pressure is reduced by the amount We, which is the cumulative amount of water influx resulting from the pressure drop. The equation assumes that there is no difference between surface and reservoir volumes of water and again neglects the effects of connate water expansion and pore volume reduction. If some of the water influx has been produced it can be accounted for by subtracting this volume, Wp, from the influx, We, on the right hand side of the equation. With some slight algebraic manipulation equ. (1.40) can be expressed as e i p i i W E p p G 1 1 Z Z G G = − − (1.41) where We Ei /G represents the fraction of the initial hydrocarbon pore volume flooded by water and is, therefore, always less than unity. When compared to the depletion material balance, equ. (1.35), it can be seen that the effect of the water influx is to maintain the reservoir pressure at a higher level for a given cumulative gas production. In addition, equ. (1.41) is non-linear, unlike equ. (1.35), which complicates both history matching and prediction. Typical plots of this equation, for different aquifer strengths, are shown in fig. 1.11. During the history matching phase, a separate part of the mathematical model must be designed to calculate the cumulative water influx corresponding to a given total pressure drop in the reservoir; this part of the history match being described as "aquifer fitting". For an aquifer whose dimensions are of the same order of magnitude as the reservoir itself the following simple model can be used SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 30 depletion line C B A p Z Gp G p Z ab Fig. 1.11 Graphical representation of the material balance equation for a water drive gas reservoir, for various aquifer strengths; equ. (1.41) e W cW p = ∆ where c = the total aquifer compressibility (cw + cf) W = the total volume of water, and depends primarily on the geometry of the aquifer and p ∆ = the pressure drop at the original reservoir-aquifer boundary. This model assumes that, because the aquifer is relatively small, a pressure drop in the reservoir is instantaneously transmitted throughout the entire reservoir-aquifer system. The material balance in such a case would be as shown by plot A in fig. 1.11, which is not significantly different from the depletion line. To provide the pressure response shown by lines B and C, the aquifer volume must be considerably larger than that of the reservoir and it is then inadmissible to assume instantaneous transmission of pressure throughout the system. There will now be a time lag between a pressure perturbation in the reservoir and the full aquifer response. To build an aquifer model, including this time dependence, is quite complex and the subject will be deferred until Chapter 9 where the use of such a model for both history matching and prediction will be described in detail. One of the unfortunate aspects in the delay in aquifer response is that, initially, all the material balance plots in fig. 1.11 appear to be linear and, if there is insufficient production and pressure history to show the deviation from linearity, one may be tempted to extrapolate the early trends, assuming a depletion type reservoir, which would result in the determination of too large a value of the GIIP. In such a case, a large difference between this and the volumetric estimate of the GIIP can be diagnostic in deciding whether there is an aquifer or not. It also follows that attempting to build a SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 31 mathematical model to describe the reservoir performance based on insufficient history data can produce erroneous results when used to predict future reservoir performance. If production-pressure history is available it is possible to make an estimate of the GIIP, in a water drive reservoir, using the following method as described by Bruns et.al16. The depletion material balance, equ. (1.34), is first solved to determine the apparent gas in place as p a i G G 1 E /E = − (1.42) If there is an active water drive, the value of Ga calculated using this equation, for known values of E and Gp, will not be unique. Successive, calculated values of Ga will increase as the deviation of p/Z above the depletion material balance line increases, due to the pressure maintenance provided by the aquifer. The correct value of the gas in place, however, can be obtained from equ. (1.40) as p e i G W E G 1 E /E = − — (1.43) where We is the cumulative water influx calculated, using some form of mathematic aquifer model, at the time at which both E and Gp have been measured. Ga We - too small We - correct model We - too large G We E/ (1−E / Ei) Fig. 1.12 Determination of the GIIP in a water drive gas reservoir. The curved, dashed lines result from the choice of an incorrect, time dependent aquifer model; (refer Chapter 9) Subtracting equ. (1.43) from equ. (1.42) gives e a i W E G G 1 E /E = + − (1.44) If the calculated values of Ga, equ. (1.42), are plotted as a function of WeE/(1−E/Ei) the result should be a straight line, provided the correct aquifer model has been selected, SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 32 as shown in fig. 1.12, and the proper value of G can be determined by linear extrapolation to the ordinate. Selecting the correct aquifer model (aquifer fitting) is a trial and error business which continues until a straight line is obtained. One other interesting feature shown in fig. 1.11 is that the maximum possible gas recovery, shown by the circled points, depends on the degree of pressure maintenance afforded by the aquifer, being smaller for the more responsive aquifers. The reason for this has already been mentioned in sec. 1.2; that in the immiscible displacement of one fluid by another not all of the displaced fluid can be removed from each pore space. Thus as the water advances through the reservoir a residual gas saturation is trapped behind the front. This gas saturation, Sgr, is rather high being of the order of 30−50% of the pore volume7,17, and is largely independent of the pressure at which the gas is trapped. This being the case, then applying the equation of state, equ. (1.15), to the gas trapped per cu.ft of pore volume behind the flood front, gives gr p S nRT Z = and, since Sgr is independent of pressure, then for isothermal depletion p n Z ∝ which indicates that a greater quantity of gas is trapped at high pressure than at low. The ultimate gas recovery depends both on the nature of the aquifer and the abandonment pressure. For the value of (p/Z)ab shown in fig. 1.11, the aquifer giving the pressure response corresponding to line B is the most favourable. While choice of the abandonment pressure is under the control of the engineer, the choice of the aquifer, unfortunately, is not. It is, therefore, extremely important to accurately measure both pressures and gas production to enable a reliable aquifer model to be built which, in turn, can be used for performance predictions. One of the more adventurous aspects of gas reservoir engineering is that gas sales contracts, specifying the market rate and pipeline pressure, are usually agreed between operator and purchaser very early in the life of the field, when the amount of history data is minimal. The operator is then forced to make important decisions on how long he will be able to meet the market demand, based on the rather scant data. Sensitivity studies are usually conducted at this stage, using the simple material balance equations presented in this chapter, and varying the principal parameters, i.e. − the GIIP − the aquifer model, based on the possible geometrical configurations of the aquifer − abandonment pressure; whether to apply surface compression or not − the number of producing wells and their mechanical design. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 33 (the latter point has not been discussed so far, since it requires the development of well inflow equations, which will be described in Chapter 8). The results of such a study can give initial guidance on the best way to develop the gas reserve. EXERCISE 1.2 GAS MATERIAL BALANCE The following data are available for a newly discovered gas reservoir: GWC = 9700 ft Centroid depth = 9537 ft Net bulk volume (V) = 1.776 × 1010 cu.ft φ = 0.19 Swc = 0 20 γg = 0.85 Although a gas sample was collected during a brief production test the reservoir pressure was not recorded because of tool failure. It is known, however, that the water pressure regime in the locality is pw = 0.441D + 31 psia and that the temperature gradient is 1.258°F/100 ft, with ambient surface temperature 80° F. 1) Calculate the volume of the GIIP. 2) It is intended to enter a gas sales contract in which the following points have been stipulated by the purchaser. a) during the first two years, a production rate build-up from zero −100 MMscf/d (million) must be achieved while developing the field b) the plateau rate must be continued for 15 years at a sales point delivery pressure which corresponds to a minimum reservoir pressure of 1200 psia. Can this latter requirement be fulfilled? (Assume that the aquifer is small so that the depletion material balance equation can be used). 3) Once the market requirement can no longer be satisfied the field rate will decline exponentially by 20% per annum until it is reduced to 20 MMscf/d. (This gas will either be used as fuel in the company's operations or compressed to supply part of any current market requirement). What will be the total recovery factor for the reservoir and what is the length of the entire project life? EXERCISE 1.2 SOLUTION a) In order to determine the GIIP it is first necessary to calculate the initial gas pressure at the centroid depth of the reservoir. That is, the depth at which there is SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 34 as much gas above as there is beneath, the pressures for use in the material balance equation will always be evaluated at this depth. To do this the water pressure at the gas−water contact must first be calculated as pw = 0.441 × 9700 + 31 = 4309 psia = GWC g p and the temperature as o 9700 T (1.258 ) 80 460 662 R 100 = × + + = for this 0.85 gravity gas the isothermal Z−factor plot at 200°F (660°R), fig. 1.8, can be used to determine the Z−factor at the GWC, with negligible error. Thus ZGWC = 0.888 and EGWC = 35.37 p 35.37 4309 259.3 ZT 0.888 662 × = = × The pressure gradient in the gas, at the GWC, can now be calculated, as described in exercise 1.1, as scE dp 0.0763 0.85 259.3 0.117 psi/ ft dD 144 144 ρ × × = = = The gas pressure at the centroid is therefore GWC GWC g dp p p D dD = − × ∆ (1.45) p = 4309 − 0.117 × (9700 − 9537) = 4290 psia and the absolute temperature at the centroid is o 9537 T (1.258 ) 80 460 660 R 100 = × + + = One could improve on this estimate by re−evaluating the gas gradient at the centroid, for p = 4290 psia and T = 660°R, and averaging this value with the original value at the GWC to obtain a more reliable gas gradient to use in 'equ. (1.45). Gas gradients are generally so small, however, that this correction is seldom necessary. The reader can verify that, in the present case, the correction would only alter the centroid pressure by less than half of one psi. For the centroid pressure and temperature of 4290 psia and 660°R, the GIIP can be estimated as GIIP = G = Vφ (1−Swc) Ei (1.26) SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 35 10 9 1.776x10 0.19 0.8 35.37 4290 699.70 10 scf 0.887 660 × × × × = = × × 2) The overall production schedule can be divided into three parts, the build-up, plateau production and decline periods, as shown in fig. 1.13. t 1 t 2 Qo = 100 (MMscf /d) 50 (MMscf /d) Rate Q (MMscf /d) t 3 20 (MMscf /d) Gp1 Gp2 Gp3 Fig. 1.13 Gas field development rate− − − −time schedule (Exercise 1.2) It is first required, to determine 2 P G , that is, the cumulative production when the reservoir pressure has fallen to 1200 psia and the plateau rate can no longer be maintained. When p = 1200 psia, Z = 0.832 (fig. 1.8) and using the depletion material balance, equ. (1.35), 2 2 9 p i i 9 P 1200 p 1200 Z 0.832 G G(1 ) 699.70 10 (1 ) 4290 p 0.887 Z G 491.04 10 scf = − = × = = × Since the cumulative production during the two years build-up period is 1 P G = Qavg × 2 × 365 = 50 × 106 × 2 × 365 = 36.5 × 109 scf the gas production at the plateau rate of 100 MMscf/d is 2 p G − p1 G = (491.04 – 36.50) × 109 = 454.54 × 109 scf and the time for which this rate can be maintained is 2 1 9 p p 2 6 o G G 454.54 10 t 12.45 yrs Q 100 10 365 − × = = = × × Therefore the time for which the plateau rate can be sustained will fall short of the requirement by some 2.5 years. 3) During the exponential decline period the rate at any time after the start of the decline can be calculated as SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 36 o Q Q e = −bt where Qo is the rate at t = 0, i.e. 100 MMscf/d, and b is the exponential decline factor of 0.2 p.a. Therefore, the time required for the rate to fall to 20 MMscf/d will be o Q 1 1 100 t ln ln 8.05 yrs b Q 0.2 20 = = = If gp is the cumulative gas production at time t, measured from the start of the decline, then t t p o o o bt g Qdt Q e dt − = = i.e. o p Q bt g (1 e ) b − = − and when t = 8.05 yrs. 6 9 (8.05) p 100 10 365 0.2 8.05 g (1 e ) 146.02 10 scf 0.2 × × − × = − = × Therefore, the total cumulative recovery at abandonment will be 3 2 (8.05) 9 9 p p p G G g (491.04 146.02) 10 637.06 10 scf = + = + × = × and the recovery factor 3 9 9 p G 637.06 10 RF 0.91 or 91% GIIP G 699.70 10 × = = = × which will be recovered after a total period of t1 + t2 + t3 = 2 + 12.45 + 8.05 = 22.5 years. This simple exercise covers the spectrum of reservoir engineering activity, namely, estimating the hydrocarbons in place, calculating a recovery factor and attaching a time scale to the recovery. The latter is imposed by the overall market rate required of the field, i.e. cumulative production time field rate = Later in the book, in Chapters 4, 6 and 8, the method of calculating individual well rates is described, which means that the time scale can be fixed by the more usual type of expression. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 37 ( ) 2 g 12 2 2 nD w p w Q m p 3.161 10 FQ h r β γ µ − Τ ∆ = × = 1.8 HYDROCARBON PHASE BEHAVIOUR This subject has been covered extensively in specialist books8,13,18 and is described here in a somewhat perfunctory manner simply to provide a qualitative understanding of the difference between various hydrocarbon systems as they exist in the reservoir. Consider, first of all, the simple experiment in which a cylinder containing one of the lighter members of the paraffin series, C2 H6−ethane, is subjected to a continuously increasing pressure at constant temperature. At some unique pressure (the vapour pressure) during this experiment the C2 H6, which was totally in the gas phase at low pressures will condense into a liquid. If this experiment were repeated at a series of different temperatures the resulting phase diagram, which is the pressure temperature relationship, could be drawn as shown in fig. 1.14(a). 100% - C2 H6 100% - C7 H16 50% - C2 H6 50% - C7 H16 P P P LIQUID LIQUID LIQUID + GAS LIQUID CP CP GAS GAS GAS T ( a ) T ( b ) T ( c ) CP Fig. 1.14 Phase diagrams for (a) pure ethane; (b) pure heptane and (c) for a 50− − − −50 mixture of the two The line defining the pressures at which the transition from gas to liquid occurs, at different temperatures, is known as the vapour pressure line. It terminates at the critical point (CP) at which it is no longer possible to distinguish whether the fluid is liquid or gas, the intensive properties of both phases being identical. Above the vapour pressure line the fluid is entirely liquid while below it is in the gaseous state. If the above experiment were repeated for a heavier member of the paraffin series, say, C7 H16 − heptane, the results would be as shown in fig. 1.14(b). One clear difference between (a) and (b) is that at lower temperatures and pressures there is a greater tendency for the heavier hydrocarbon, C7 H16, to be in the liquid state. For a two component system, the phase diagram for a 50% C2 H6 and 50% C7 H16 mixture would be as shown in fig. 1.14 (c). In this case, while there are regions where the fluid mixture is either entirely gas or liquid, there is now also a clearly defined region in which the gas and liquid states can coexist; the, so-called, two phase region. The shape of the envelope defining the two phase region is dependent on the composition of the mixture, being more vertically inclined if the C2H6 is the predominant component and more horizontal if it is the C7 H16. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 38 Naturally occurring hydrocarbons are more complex than the system shown in fig. 1.14 in that they contain a great many members of the paraffin series and usually some non-hydrocarbon impurities. Nevertheless, a phase diagram can similarly be defined for complex mixtures and such a diagram for a typical natural gas is shown in fig. 1.15(a). The lines defining the two phase region are described as the bubble point line, separating the liquid from the two phase region, and the dew point line, separating the gas from the two phase region. That is, on crossing the bubble point line from B U B B L E P O INT LIN E P x 90% 90% 70% 70% 50% 50% 30% 30% 10% 10% CP B A CT C LIQUID D E A B T ( a ) P LIQUID T ( b ) GAS GAS DEW POINT LI N E CP Fig. 1.15 Schematic, multi-component, hydrocarbon phase diagrams; (a) for a natural gas; (b) for oil liquid to the two phase region, the first bubbles of gas will appear while, crossing the dew point line from the gas, the first drops of liquid (dew) will appear. The lines within the two phase region represent constant liquid saturations. For a gas field, as described in secs. 1.5 − 1.8, the reservoir temperature must be such that it exceeds the so-called cricondentherm (CT), which is the maximum temperature at which the two phases can coexist for the particular hydrocarbon mixture. If the initial reservoir pressure and temperature are such that they coincide with point A in fig. 1.15(a), then for isothermal reservoir depletion, which is generally assumed, the pressure will decline from A towards point B and the dew point line will never be crossed. This means that only dry gas will exist in the reservoir at any pressure. On producing the gas to the surface, however, both pressure and temperature will decrease and the final state will be at some point X within the two phase envelope, the position of the point being dependent on the conditions of surface separation. The material balance equations presented in this chapter, equs. (1.35) and (1.41), assumed that a volume of gas in the reservoir was produced as gas at the surface. If, due to surface separation, small amounts of liquid hydrocarbon are produced, the cumulative liquid volume must be converted into an equivalent gas volume and added to the cumulative gas production to give the correct value of Gp for use in the material balance equation. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 39 Thus if n pound moles of liquid have been produced, of molecular weight M, then the total mass of liquid is nM = γ oρ w × (liquid volume) where γo is the oil gravity (water = 1), and ρw is the density of water (62.43 Ib/cu.ft). Since liquid hydrocarbon volumes are generally measured in stock tank barrels (1 bbl = 5.615 cu.ft), then the number of pound moles of liquid hydrocarbon produced in Np stb is o p N n 350.5 M γ = Expressing this number of moles of hydrocarbon as an equivalent gas volume at standard conditions, gives o p sc sc sc o p 5 sc N nRT 10.732 520 V 350.5 p M 14.7 N or V 1.33 10 M γ γ × = = × = × The correction in adding the equivalent gas volume to the cumulative gas production is generally rather small, of the order of one percent or less, and is sometimes neglected. If the initial reservoir pressure and temperature are such that the gas is at point C, fig. 1.15(a), then during isothermal depletion liquid will start to condense in the reservoir when the pressure has fallen below the dew point at D. The maximum liquid saturation deposited in the reservoir, when the pressure is between points D and E in the two phase region, is generally rather small and frequently is below the critical saturation which must be exceeded before the liquid becomes mobile. This phenomenon is analogous to the residual saturations, discussed previously, at which flow ceases. Therefore, the liquid hydrocarbons deposited in the reservoir, which are referred to as retrograde liquid condensate, are not recovered and, since the heavier components tend to condense first, this represents a loss of the most valuable part of the hydrocarbon mixture. It may be imagined that continued pressure depletion below the dew point at E would lead to re-vapourisation of the liquid condensate. This does not occur, however, because once the pressure falls below point D the overall molecular weight of the hydrocarbons remaining in the reservoir increases, since some of the heavier paraffins are left behind in the reservoir as retrograde condensate. Therefore, the composite phase envelope for the reservoir fluids tends to move downwards and to the right thus inhibiting re-vapourisation. It is sometimes economically viable to produce a gas condensate field by the process of dry gas re-cycling. That is, from the start of production at point C, fig. 1.15(a), separating the liquid condensate from the dry gas at the surface and re-injecting the latter into the reservoir in such a way that the dry gas displaces the wet gas towards the producing wells. Since only a relatively small amount of fluid is removed from the reservoir during this process, the pressure drop is small and, for a successful project, SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 40 the aim should be to operate at above dew point pressure until dry gas breakthrough occurs in the producing wells. After this, the injection is terminated, and the remaining dry gas produced. The dry gas material balance equations can also be applied to gas condensate reservoirs if the single phase Z−factor is replaced by the, so-called, two phase Z−factor. This must be experimentally determined in the laboratory by performing a constant volume depletion experiment. A volume of gas, G scf, is charged to a PVT cell at an initial pressure pi, which is above the dew point, and at reservoir temperature. The pressure is reduced in stages as gas is withdrawn from the cell, and measured as p G′ scf, without altering either the cell volume or the temperature. This simulates the production of the reservoir under volumetric depletion conditions and therefore, applying the depletion type material balance equation, (1.35), and solving explicitly for Z gives 2 phase ' p i i p Z p G 1 Z G − = − (1.46) Until the pressure has dropped to the dew point, the Z−factor measured in this experiment is identical with the Z−factor obtained using the technique described in sec. 1.5(a). Below the dew point, however, the two techniques will produce different results. The latter experiment, for determining the single phase Z−factor, implicitly assumes that a volume of reservoir fluids, below dew point pressure, is produced in its entirety to the surface. In the constant volume depletion experiment, however, allowance is made for the fact that some of the fluid remains behind in the reservoir as liquid condensate, this volume being also recorded as a function of pressure during the experiment. As a result, if a gas condensate sample is analysed using both experimental techniques, the two phase Z−factor determined during the constant volume depletion will be lower than the single phase Z−factor. This is because the retrograde liquid condensate is not included in the cumulative gas production p G′ in equ. (1.46), which is therefore lower than it would be assuming that all fluids are produced to the surface, as in the single phase experiment. Figure 1.15(b) shows a typical phase diagram for oil. As already noted, because oil contains a higher proportion of the heavier members of the paraffin series, the two phase envelope is more horizontally inclined than for gas. If the initial temperature and pressure are such that the reservoir oil is at point A in the diagram, there will only be one phase in the reservoir namely, liquid oil containing dissolved gas. Reducing the pressure isothermally will eventually bring the oil to the bubble point, B. Thereafter, further reduction in pressure will produce a two phase system in the reservoir; the liquid oil, containing an amount of dissolved gas which is commensurate with the pressure, and a volume of liberated gas. Unfortunately, when liquid oil and gas are subjected to the same pressure differential in the reservoir, the SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 41 gas, being more mobile, will travel with a much greater velocity than the oil. This leads to a certain degree of chaos in the reservoir and greatly complicates the description of fluid flow. From this point of view, it is preferable to produce the reservoir close to (or above) bubble point pressure, which greatly simplifies the mathematical description. Not only that, but as will be shown in Chapter 3, operating in such a manner enhances the oil recovery. The manner in which the reservoir pressure can be maintained at or above bubble point is conventionally by water injection, a detailed description of which is presented in Chapter 10. REFERENCES 1) Lynch, E.J.,1964. Formation Evaluation. Harper and Row, New York. 2) Bradley, J.S., 1975. Abnormal Formation Pressure. The American Ass. of Pet. Geologists Bulletin, Vol. 59, No. 6, June: 957−973. 3) Chapman, R.E., 1973. Petroleum Geology, a Concise Study. Elsevier Scientific Publishing Co., Amsterdam: 67−76. 4) Earlougher, R.C., Jr., 1977. Advances in Well Test Analysis. SPE Monograph: Chapter 8. 5) Lebourg, M., Field, R.Q. and Doh, C.A., 1957. A Method of Formation Testing on Logging Cable. Trans. AIME, 210: 260−267. 6) Schultz, A.L., Bell, W.T. and Urbanosky, H.J., 1974. Advancements in Uncased−Hole Wireline−Formation−Tester Techniques. SPE paper 5053, prepared for the Annual Fall Meeting of the SPE of AIME; Houston, Texas. October. 7) Craft, B.C. and Hawkins, M.F., Jr., 1959. Applied Petroleum Reservoir Engineering. Prentice−Hall, Inc. New Jersey. 8) Amyx, J.W., Bass, D.M. and Whiting, R. L., 1960. Petroleum Reservoir Engineering − Physical Properties. McGraw-Hill. 9) Walstrom, J.E., Mueller, T.D. and McFarlane, R.C., 1967. Evaluating Uncertainty in Engineering Calculations. J.Pet. Tech., July: 1595−1603. 10) Northern, I.G., 1967. Risk Probability and Decision-Making in Oil and Gas Development Operations. Paper presented at Petroleum Soc. of CIM. Technical Meeting, Banff, Canada. May. 11) Standing, M.B. and Katz, D.L., 1942. Density of Natural Gases. Trans. AIME, 146: 140−149. 12) Gas Processors Suppliers Association, 1972 (Revised 1974). Engineering Data Book. GPSA, Tulsa: 16−2. SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 42 13) Katz, D.L., et. al., 1959. Handbook of Natural Gas Engineering. McGraw-Hill, Inc., New York. 14) Takacs, G., 1976. Comparisons made for Computer Z−Factor Calculations. The Oil and Gas Journal, Dec. 20th: 64−66. 15) Hall, K.R. and Yarborough, L., 1974. How to Solve Equation of State for Z−Factors. The Oil and Gas Journal, Feb. 18th: 86−88. 16) Bruns, J.R., Fetkovitch, M.J. and Meitzen, V.C., 1965. The Effect of Water Influx on p/Z−Cumulative Gas Production Curves. J.Pet.Tech., March: 287−291. 17) Agarwal, R.G., Al-Hussainy, R. and Ramey, H.J., Jr., 1965. The Importance of Water Influx in Gas Reservoirs. J.Pet.Tech., November: 1336−1342. Trans. AIME. 18) McCain, W.D. Jr., 1973. The Properties of Petroleum Fluids. Petroleum Publishing Co., Tulsa. 19) Brown, G.G., Katz, D.L., Oberfell, G.B. and Alden, R.C., 1948. Natural Gasoline and Volatile Hydrocarbons. NGAA, Tulsa. CHAPTER 2 PVT ANALYSIS FOR OIL 2.1 INTRODUCTION In Chapter 1, the importance of PVT analysis was stressed for relating observed volumes of gas production at the surface to the corresponding underground withdrawal. For gas this relationship could be obtained merely by determining the single or two phase Z−factor, and using it in the equation of state. The basic PVT analysis required to relate surface production to underground withdrawal for an oil reservoir is necessarily more complex due to the presence, below the bubble point pressure, of both a liquid oil and free gas phase in the reservoir. This chapter concentrates on defining the three main parameters required to relate surface to reservoir volumes, for an oil reservoir, and then proceeds to describe how these parameters can be determined in the laboratory by controlled experiments performed on samples of the crude oil. The subject is approached from a mechanistic point of view in merely recognising that PVT parameters can be determined as functions of pressure by routine laboratory analysis. No attempt is made to describe the complex thermodynamic processes implicit in the determination of these parameters. For a more exhaustive treatment of the entire subject the reader is referred to the text of Amyx, Bass and Whiting1 . Finally, a great deal of attention is paid to the conversion of PVT data, as presented by the laboratory, to the form required in the field. The former being an absolute set of measurements while the latter depend upon the manner of surface separation of the gas and oil. 2.2 DEFINITION OF THE BASIC PVT PARAMETERS The Pressure−Volume−Temperature relation for a real gas can be uniquely defined by the simple equation of state pV = ZnRT (1.15) in which the Z−factor, which accounts for the departure from ideal gas behaviour, can be determined as described in Chapter 1, sec. 5. Using this equation, it is a relatively simple matter to determine the relationship between surface volumes of gas and volumes in the reservoir as sc sc T p 1 p E 35.37 (scf /rcf) p T Z ZT = × × = (1.25) Unfortunately, no such simple equation of state exists which will describe the PVT properties of oil. Instead, several, so-called, PVT parameters must be measured by laboratory analysis of crude oil samples. The parameters can then be used to express PVT ANALYSIS FOR OIL 44 the relationship between surface and reservoir hydrocarbon volumes, equivalent to equ. (1.25). The complexity in relating surface volumes of hydrocarbon production to their equivalent volumes in the reservoir can be appreciated by considering fig. 2.1. oil stock tank oil (a) solution gas gas oil stock tank oil (b) free gas + solution gas SURFACE RESERVOIR Fig. 2.1 Production of reservoir hydrocarbons (a) above bubble point pressure, (b) below bubble point pressure Above the bubble point only one phase exists in the reservoir − the liquid oil. If a quantity of this undersaturated oil is produced to the surface, gas will separate from the oil as shown in fig. 2.1(a), the volume of the gas being dependent on the conditions at which the surface separation is effected. In this case, it is relatively easy to relate the surface volumes of oil and gas to volumes at reservoir conditions since it is known that all the produced gas must have been dissolved in the oil in the reservoir. If the reservoir is below bubble point pressure, as depicted in fig. 2.1(b), the situation is more complicated. Now there are two hydrocarbon phases in the reservoir, gas saturated oil and liberated solution gas. During production to the surface, solution gas will be evolved from the oil phase and the total surface gas production will have two components; the gas which was free in the reservoir and the gas liberated from the oil during production. These separate components are indistinguishable at the surface and the problem is, therefore, how to divide the observed surface gas production into liberated and dissolved gas volumes in the reservoir. Below bubble point pressure there is an additional complication in that the liberated solution gas in the reservoir travels at a different velocity than the liquid oil, when both are subjected to the same pressure differential. As will be shown in Chapter 4, sec. 2, the flow velocity of a fluid in a porous medium is inversely proportional to the fluid viscosity. Typically, gas viscosity in the reservoir is about fifty times smaller than for liquid oil and consequently, the gas flow velocity is much greater. As a result, it is normal, when producing from a reservoir in which there is a free gas saturation, that gas will be produced in disproportionate amounts in comparison to the oil. That is, one PVT ANALYSIS FOR OIL 45 barrel of oil can be produced together with a volume of gas that greatly exceeds the volume originally dissolved per barrel of oil above bubble point pressure. Control in relating surface volumes of production to underground withdrawal is gained by defining the following three PVT parameters which can all be measured by laboratory experiments performed on samples of the reservoir oil, plus its originally dissolved gas. Rs − The solution (or dissolved) gas oil ratio, which is the number of standard cubic feet of gas which will dissolve in one stock tank barrel of oil when both are taken down to the reservoir at the prevailing reservoir pressure and temperature (units − scf. gas/stb oil). Bo − The oil formation volume factor, is the volume in barrels occupied in the reservoir, at the prevailing pressure and temperature, by one stock tank barrel of oil plus its dissolved gas (units – rb (oil + dissolved gas)/stb oil). Bg − The gas formation volume factor, which is the volume in barrels that one standard cubic foot of gas will occupy as free gas in the reservoir at the prevailing reservoir pressure and temperature (units − rb free gas/ssf gas). Both the standard cubic foot (scf) and the stock tank barrel (stb) referred to in the above definitions are defined at standard conditions, which in this text are taken as 60°F and one atmosphere (14.7 psia). It should also be noted that Rs and Bo are both measured relative to one stock tank barrel of oil, which is the basic unit of volume used in the field. All three parameters are strictly functions of pressure, as shown in fig. 2.5, assuming that the reservoir temperature remains constant during depletion. Precisely how these parameters can be used in relating measured surface volumes to reservoir volumes is illustrated in figs. 2.2 and 2.3. pi p T Phase diagram P Bo rb ( oil + dissolved gas) / stb 1 stb oil + R scf / stb si solution gas Fig. 2.2 Application of PVT parameters to relate surface to reservoir hydrocarbon volumes; above bubble point pressure. PVT ANALYSIS FOR OIL 46 Fig. 2.2 depicts the situation when the reservoir pressure has fallen from its initial value pi to some lower value p, which is still above the bubble point. As shown in the P−T diagram (inset) the only fluid in the reservoir is undersaturated liquid oil. When this oil is produced to the surface each stock tank barrel will yield, upon gas oil separation, Rsi standard cubic feet of gas. Since the oil is undersaturated with gas, which implies that it could dissolve more if the latter were available, then the initial value of the solution gas oil ratio must remain constant at Rsi (scf/stb) until the pressure drops to the bubble point, when the oil becomes saturated, as shown in fig. 2.5(b). Figure 2.2 also shows, in accordance with the definitions of Bo and Rs, that if Rsi scf of gas are taken down to the reservoir with one stb of oil, then the gas will totally dissolve in the oil at the reservoir pressure and temperature to give a volume of Bo rb of oil plus dissolved gas. Figure 2.5(a) shows that Bo increases slightly as the pressure is reduced from initial to the bubble point pressure. This effect is simply due to liquid expansion and, since the compressibility of the undersaturated oil in the reservoir is low, the expansion is relatively small. Typical values of Bo and Rs above the bubble point are indicated in fig. 2.5, these are the plotted results of the laboratory analysis presented in table 2.4. The initial value of the oil formation volume factor Boi is 1.2417 which increases to 1.2511 at the bubble point. Thus initially, 1.2417 reservoir barrels of oil plus its dissolved gas will produce one stb of oil. This is a rather favourable ratio indicating an oil of moderate volatility and, as would be expected in this case, the initial solution gas oil ratio is also relatively low at 510 scf/stb. Under less favourable circumstances, for more volatile oils, Boi can have much higher values. For instance, in the Statfjord field in the North Sea, Boi is 2.7 rb/stb while the value of Rsi is approximately 3000 scf/stb. Obviously the most favourable value of Boi is as close to unity as possible indicating that the oil contains hardly any dissolved gas and reservoir volumes are approximately equal to surface volumes. The small oil fields of Beykan and Kayaköy in the east of Turkey provide good examples of this latter condition having values of Boi and Rsi of 1.05 and 20 scf/stb respectively. Below the bubble point the situation is more complicated as shown in fig. 2.3. PVT ANALYSIS FOR OIL 47 pi p T p rb ( oil + dissolved gas) / stb 1 stb oil Bo (R - R ) B s g rb (free gas) / stb R = R + s (R - R ) scf / stb s Fig. 2.3 Application of PVT parameters to relate surface to reservoir hydrocarbon volumes; below bubble point pressure In this case each stock tank barrel of oil is produced in conjunction with R scf of gas, where R (scf/stb) is called the instantaneous or producing gas oil ratio and is measured daily. As already noted, some of this gas is dissolved in the oil in the reservoir and is released during production through the separator, while the remainder consists of gas which is already free in the reservoir. Furthermore, the value of R can greatly exceed Rsi, the original solution gas oil ratio, since, due to the high velocity of gas flow in comparison to oil, it is quite normal to produce a disproportionate amount of gas. This results from an effective stealing of liberated gas from all over the reservoir and its production through the relatively isolated offtake points, the wells. A typical plot of R as a function of reservoir pressure is shown as fig. 2.4. R scf / stb R = Rsi 510 scf / stb pb Reservoir pressure 4000 scf / sfb Fig. 2.4 Producing gas oil ratio as a function of the average reservoir pressure for a typical solution gas drive reservoir PVT ANALYSIS FOR OIL 48 The producing gas oil ratio can be split into two components as shown in fig. 2.3, i.e. R = Rs +(R−Rs) The first of these, Rs scf/stb, when taken down to the reservoir with the one stb of oil, will dissolve in the oil at the prevailing reservoir pressure to give Bo rb of oil plus dissolved gas. The remainder, (R − Rs) scf/stb, when taken down to the reservoir will occupy a volume s g s g scf rb (R R ) B (R R ) B (rb. free gas / stb) stb scf − × = − − (2.1) and therefore, the total underground withdrawal of hydrocarbons associated with the production of one stb of oil is (Underground withdrawal)/stb = Bo + (R − Rs) Bg (rb/stb) (2.2) The above relationship shows why the gas formation volume factor has the rather unfortunate units of rb/scf. It is simply to convert gas oil ratios, measured in scf/stb, directly to rb/stb to be compatible with the units of Bo. While Bg is used almost exclusively in oil reservoir engineering its equivalent in gas reservoir engineering is E, the gas expansion factor, which was introduced in the previous chapter and has the units scf/rcf. The relation between Bg and E is therefore, g rb 1 B scf 5.615E = (2.3) thus Bg has always very small values; for a typical value of E of, say, 150 scf/rcf the value of Bg would be .00119 rb/scf. PVT ANALYSIS FOR OIL 49 B (rb / stb) o 1.3 1.2 1.1 1.0 1000 2000 3000 4000 PRESSURE (psia) p = 3330 psia b a b c 600 400 200 R (scf / stb) s 1000 2000 3000 4000 1000 2000 3000 4000 B (rb / scf) g .010 .008 .006 .004 .002 E (scf / rcf) - 200 - 100 - 0 PRESSURE (psia) PRESSURE (psia) Fig. 2.5 PVT parameters (Bo, Rs and Bg), as functions of pressure, for the analysis presented in table 2.4; (pb = 3330 psia). PVT ANALYSIS FOR OIL 50 The shapes of the Bo and Rs curves below the bubble point, shown in fig. 2.5(a) and (b), are easily explained. As the pressure declines below pb, more and more gas is liberated from the saturated oil and thus Rs, which represents the amount of gas dissolved in a stb at the current reservoir pressure, continually decreases. Similarly, since each reservoir volume of oil contains a smaller amount of dissolved gas as the pressure declines, one stb of oil will be obtained from progressively smaller volumes of reservoir oil and Bo steadily declines with the pressure. EXERCISE 2.1 UNDERGROUND WITHDRAWAL The oil and gas rates, measured at a particular time during the producing life of a reservoir are, x stb oil/day and y scf gas/day. 1) What is the corresponding underground withdrawal rate in reservoir barrels/day. 2) If the average reservoir pressure at the time the above measurements are made is 2400 psia, calculate the daily underground withdrawal corresponding to an oil production of 2500 stb/day and a gas rate of 2.125 MMscf/day. Use the PVT relationships shown in figs. 2.5(a) − (c), which are also listed in table 2.4. 3) If the density of the oil at standard conditions is 52.8 lb/cu.ft and the gas gravity is 0.67 (air = 1) calculate the oil pressure gradient in the reservoir at 2400 psia. EXERCISE 2.1 SOLUTION 1) The instantaneous or producing gas oil ratio is R = y/x scf/stb. If, at the time the surface rates are measured, the average reservoir pressure is known, then Bo, Rs and Bg can be determined from the PVT relationships at that particular pressure. The daily volume of oil plus dissolved gas produced from the reservoir is then xBo rb, and the liberated gas volume removed daily is s y x( R ) x − Bg rb. Thus the total underground withdrawal is o s g y x(B ( R )B ) rb / day x + − (2.4) 2) At a reservoir pressure of 2400 psia, the PVT parameters obtained from table 2.4 are: Bo = 1.1822 rb/stb; Rs = 352 scf/stb and Bg = .0012 rb/ scf Therefore, evaluating equ. (2.4) for x = 2500 stb/d and y = 2.125 MMscf/d gives a total underground withdrawal rate of 2500 (1.1822 + (850 − 352) × .0012) = 4450 rb/d 3) The liquid oil gradient in the reservoir can be calculated by applying mass conservation, as demonstrated in exercise 1.1 for the calculation of the gas gradient. In the present case the mass balance is PVT ANALYSIS FOR OIL 51 Mass of 1 stb of oil Mass of Bo rb of oil + = + Rs scf dissolved gas at standard conditions dissolved gas in the reservoir or sc sc r o s g o o lb lb 1(stb) 5.615 R (scf) cu.ft cu.ft lb B (rb) 5.615 cu.ft ρ ρ ρ × × + × = × × in which the subscripts sc and r refer to standard conditions and reservoir conditions, respectively. The gas density at standard conditions is ρsc = γg × 0.0763 (refer equ. (1.30)) = 0.0511 lb/cu ft Therefore, sc sc r o s g o o ( 5.615) (R ) B 5.615 (52.8 5.615) (352 0.0511) 47.37 lb / cu ft 1.1822 5.615 ρ ρ ρ × + × = × × + × = = × and the liquid oil gradient is 47.37/144 = 0.329 psi/ft. 2.3 COLLECTION OF FLUID SAMPLES Samples of the reservoir fluid are usually collected at an early stage in the reservoir's producing life and dispatched to a laboratory for the full PVT analysis. There are basically two ways of collecting such samples, either by direct subsurface sampling or by surface recombination of the oil and gas phases. Whichever technique is used the same basic problem exists, and that is, to ensure that the proportion of gas to oil in the composite sample is the same as that existing in the reservoir. Thus, sampling a reservoir under initial conditions, each stock tank barrel of oil in the sample should be combined with Rsi standard cubic feet of gas. a) Subsurface sampling This is the more direct method of sampling and is illustrated schematically in fig. 2.6. PVT ANALYSIS FOR OIL 52 sample chamber pwf pi pb pressure r Fig. 2.6 Subsurface collection of PVT sample A special sampling bomb is run in the hole, on wireline, to the reservoir depth and the sample collected from the subsurface well stream at the prevailing bottom hole pressure. Either electrically or mechanically operated valves can be closed to trap a volume of the borehole fluids in the sampling chamber. This method will obviously yield a representative combined fluid sample providing that the oil is undersaturated with gas to such a degree that the bottom hole flowing pressure pwf at which the sample is collected, is above the bubble point pressure. In this case a single phase fluid, oil plus its dissolved gas, is flowing in the wellbore and therefore, a sample of the fluid is bound to have the oil and gas combined in the correct proportion. Many reservoirs, however, are initially at bubble point pressure and under these circumstances, irrespective of how low the producing rate is maintained during sampling, the bottom hole flowing pressure pwf will be less than the bubble point pressure pb as depicted in fig. 2.6. In this case, there will be saturated oil and a free gas phase flowing in the immediate vicinity of the wellbore, and in the wellbore itself, and consequently, there is no guarantee that the oil and gas will be collected in the correct volume proportion in the chamber. In sampling a gas saturated reservoir, two situations can arise depending on the time at which the sample is collected. If the sample is taken very early in the producing life it is possible that the fluid flowing into the wellbore is deficient in gas. This is because the initially liberated gas must build up a certain minimum gas saturation in the reservoir pores before it will start flowing under an imposed pressure differential. This, so−called, critical saturation is a phenomenon common to any fluid deposited in the reservoir, not just gas. The effect on the producing gas oil ratio, immediately below bubble point pressure, is shown in fig. 2.4 as the small dip in the value of R for a short period after the pressure has dropped below bubble point. As a result of this mechanism there will be a period during which the liberated gas remains in the reservoir and the gas oil ratio measured from a subsurface sample will be too low. Conversely, once the liberated gas saturation exceeds the critical value, then as shown in fig. 2.4 and discussed previously, the producing well will effectively steal gas from more remote parts of the reservoir and the sample is likely to have a disproportionately high gas oil ratio. PVT ANALYSIS FOR OIL 53 The problems associated with sampling an initially saturated oil reservoir, or an undersaturated reservoir in which the bottom hole flowing pressure has been allowed to fall below bubble point pressure, can be largely overcome by correct well conditioning prior to sampling. If the well has already been flowing, it should be produced at a low stabilized rate for several hours to increase the bottom hole flowing pressure and thereby re-dissolve some, if not all, of the free gas saturation in the vicinity of the well. Following this the well is closed in for a reasonable period of time during which the oil flowing into the wellbore, under an ever increasing average pressure, will hopefully re-dissolve any of the remaining free gas. If the reservoir was initially at bubble point pressure, or suspected of being so, the subsurface sample should then be collected with the well still closed in. If the reservoir is known to be initially undersaturated the sample can be collected with the well flowing at a very low rate so that the bottom hole flowing pressure is still above the bubble point. With proper well conditioning a representative combined sample can usually be obtained. One of the main drawbacks in the method is that only a small sample of the wellbore fluids is obtained, the typical sampler having a volume of only a few litres. Therefore, one of the only ways of checking whether the gas oil ratio is correct is to take several downhole samples and compare their saturation pressures at ambient temperature on the well site. This can be done using a mercury injection pump and accurate pressure gauge connected to the sampler. The chamber normally contains both oil and a free gas phase, due to the reduction in temperature between wellbore and surface. Injecting mercury increases the pressure within the chamber until at a saturation pressure corresponding to the ambient surface temperature all the gas will dissolve. This saturation pressure can be quite easily detected since there is a distinct change in compressibility between the two phase and single phase fluids. If it is experimentally determined, on the well site, that successive samples have markedly different saturation pressures, then either the tool has been malfunctioning or the well has not been conditioned properly. In addition, it is necessary to determine the static reservoir pressure and temperature by well testing, prior to collecting the samples. Further details on bottom hole sampling techniques are given in references 2 and 3 listed at the end of this chapter. b) Surface recombination sampling In collecting fluid samples at the surface, separate volumes of oil and gas are taken at separator conditions and recombined to give a composite fluid sample. The surface equipment is shown schematically in fig. 2.7. PVT ANALYSIS FOR OIL 54 gas meter gas sample oil sample p T st st stock tank oil p T sep sep well separator Fig. 2.7 Collection of a PVT sample by surface recombination The well is produced at a steady rate for a period of several hours and the gas oil ratio is measured in scf of separator gas per stock tank barrel of oil. If this ratio is steady during the period of measurement then one can feel confident that recombining the oil and gas in the same ratio will yield a representative composite sample of the reservoir fluid. In fact, a slight adjustment must be made to determine the actual ratio in which the samples should be recombined. This is because, as shown in fig. 2.7, the oil sample is collected at separator pressure and temperature whereas the gas oil ratio is measured relative to the stock tank barrel, thus the required recombination ratio is REQUIRED MEASURED SHRINKAGE sep scf scf stb R R S sep.bbl stb sep.bbl = × Dimensionally, the measured gas oil ratio must be multiplied by the shrinkage factor from separator to stock tank conditions. This factor is usually determined in the laboratory as the first stage of a PVT analysis of a surface recombination sample by placing a small volume of the oil sample in a cell at the appropriate separator conditions and discharging it (flash expansion) to a second cell maintained at the field stock tank conditions. During this process some gas will be liberated from the separator sample, due to the reduction in pressure and temperature, and the diminished stock tank oil volume is measured, thus allowing the direct calculation of S. In order to be able to perform such an experiment it is important that the engineer should accurately measure the pressure and temperature prevailing at both separator and stock tank during sampling and provide the laboratory with these data. One of the attractive features of surface recombination sampling is that statistically it gives a reliable value of the producing gas oil ratio measured over a period of hours; furthermore, it enables the collection of large fluid samples. Of course, just as for subsurface sampling, the surface recombination method will only provide the correct PVT ANALYSIS FOR OIL 55 gas oil ratio if the pressure in the vicinity of the well is at or above bubble point pressure. If not, the surface gas oil ratio will be too low or too high, depending upon whether the free gas saturation in the reservoir is below or above the critical saturation at which gas will start to flow. In this respect it should be emphasized that PVT samples should be taken as early as possible in the producing life of the field to facilitate the collection of samples in which the oil and gas are combined in the correct ratio. 2.4 DETERMINATION OF THE BASIC PVT PARAMETERS IN THE LABORATORY AND CONVERSION FOR FIELD OPERATING CONDITIONS Quite apart from the determination of the three primary PVT parameters Bo, Rs and Bg, the full laboratory analysis usually consists of the measurement or calculation of fluid densities, viscosities, composition, etc. These additional measurements will be briefly discussed in section 2.6. For the moment, the essential experiments required to determine the three basic parameters will be detailed, together with the way in which the results of a PVT analysis must be modified to match the field operating conditions. The analysis consists of three parts: − flash expansion of the fluid sample to determine the bubble point pressure; − differential expansion of the fluid sample to determine the basic parameters Bo, Rs and Bg; − flash expansion of fluid samples through various separator combinations to enable the modification of laboratory derived PVT data to match field separator conditions. The apparatus used to perform the above experiments is the PV cell, as shown in fig. 2.8. After recombining the oil and gas in the correct proportions, the fluid is charged to the PV cell which is maintained at constant temperature, the measured reservoir temperature, throughout the experiments. The cell pressure is controlled by a positive displacement mercury pump and recorded on an accurate pressure gauge. The plunger movement is calibrated in terms of volume of mercury injected or withdrawn from the PV cell so that volume changes in the cell can be measured directly. The flash and differential expansion experiments are presented schematically in figs. 2.9(a) and 2.9(b). In the flash experiment the pressure in the PV cell is initially raised to a value far in excess of the bubble point. The pressure is subsequently reduced in stages, and on each occasion the total volume vt of the cell contents is recorded. As soon as the bubble point pressure is reached, gas is liberated from the oil and the overall compressibility of the system increases significantly. Thereafter, small changes in pressure will result in large changes in the total fluid volume contained in the PV cell. In this manner, the flash expansion experiment can be used to "feel" the bubble point. Since the cell used is usually opaque the separate volumes of oil and gas, below bubble point pressure, cannot be measured in the experiment and therefore, only total fluid volumes are recorded. In the laboratory analysis the basic unit of volume, against which all others are compared, is the volume of saturated oil at the PVT ANALYSIS FOR OIL 56 bubble point, irrespective of its magnitude. In this chapter it will be assumed, for consistency, that this unit volume is one reservoir barrel of bubble point oil (1−rbb). PV cell thermal jacket Heise pressure gauge mercury reservoir mercury pump Fig. 2.8 Schematic of PV cell and associated equipment pi oil vt = vo Hg pb oil vt = 1 Hg p < pb oil vt Hg gas (a) pb oil Hg oil Hg gas vo vg p < pb vo Hg oil (b) gas vo = 1 Fig. 2.9 Illustrating the difference between (a) flash expansion, and (b) differential liberation PVT ANALYSIS FOR OIL 57 Table 2.1 lists the results of a flash expansion for an oil sample obtained by the subsurface sampling of a reservoir with an initial pressure of 4000 psia and temperature of 200°F; the experiment was conducted at this same fixed temperature. Pressure psia Relative Total Volume vt = v/vb = (rb/rbb) 5000 0.9810 4500 0.9850 4000 (pi) 0.9925 3500 0.9975 3330 (pb ) 1.0000 3290 1.0025 3000 1.0270 2700 1.0603 2400 1.1060 2100 1.1680 TABLE 2.1 Results of isothermal flash expansion at 200°F The bubble point pressure for this sample is determined from the flash expansion as 3330 psia, for which the saturated oil is assigned the unit volume. The relative total fluid volumes listed are volumes measured in relation to this bubble point volume. The flash expansion can be continued to much lower pressures although this is not usually done since the basic PVT parameters are normally obtained from the differential liberation experiment. Furthermore, the maximum volume to which the cell can expand is often a limiting factor in continuing the experiment to low pressures. The essential data obtained from the differential liberation experiment, performed on the same oil sample, are listed in table 2.2. The experiment starts at bubble point pressure since above this pressure the flash and differential experiments are identical. PVT ANALYSIS FOR OIL 58 Pressure psia Relative Gas Vol. (at p and T) vg Relative Gas Vol. (sc) Vg Cumulative Relative Gas Vol. (sc) F Gas expansion Factor E Z−factor Z Relative Oil Vol. (at p and T) vo 3330 (pb ) 1.0000 3000 .0460 8.5211 8.5211 185.24 .868 .9769 2700 .0417 6.9731 15.4942 167.22 .865 .9609 2400 .0466 6.9457 22.4399 149.05 .863 .9449 2100 .0535 6.9457 29.3856 129.83 .867 .9298 1800 .0597 6.5859 35.9715 110.32 .874 .9152 1500 .0687 6.2333 42.2048 90.73 .886 .9022 1200 .0923 6.5895 48.7943 71.39 .901 .8884 900 .1220 6.4114 55.2057 52.55 .918 .8744 600 .1818 6.2369 61.4426 34.31 .937 .8603 300 .3728 6.2297 67.6723 16.71 .962 .8459 14.7 (200°F) 74.9557 .8296 14.7 ( 60°F) 74.9557 .7794 All volumes are measured relative to the unit volume of oil at the bubble point pressure of 3330 psi TABLE 2.2 Results of isothermal differential liberation at 200º F PVT ANALYSIS FOR OIL 60 In contrast to the flash expansion, after each stage of the differential liberation, the total amount of gas liberated during the latest pressure drop is removed from the PV cell by injecting mercury at constant pressure, fig. 2.3. Thus, after the pressure drop from 2700 to 2400 psia, table 2.2, column 2, indicates that 0.0466 volumes of gas are withdrawn from the cell at the lower pressure and at 200°F. These gas volumes vg are measured relative to the unit volume of bubble point oil, as are all the relative volumes listed in table 2.2. After each stage the incremental volume of liberated gas is expanded to standard conditions and re−measured as Vg relative volumes. Column 4 is simply the cumulative amount of gas liberated below the bubble point expressed at standard conditions, in relative volumes, and is denoted by F = Σ Vg. Dividing values in column 3 by those in column 2 (Vg/vg) gives the gas expansion factor E defined in Chapter 1, sec. 6. Thus the .0466 relative volumes liberated at 2400 psia will expand to give 6.9457 relative volumes at standard conditions and the gas expansion factor is therefore 6.9457/.0466 = 149.05. Knowing E, the Z−factor of the liberated gas can be determined by explicitly solving equ . (1.25) for Z as sc sc T p 1 p Z 35.37 p T E ET = × × = and for the gas sample withdrawn at 2400 psia 2400 Z 35.37 0.863 149.05 660 = × = × These values are listed in column 6 of table 2.2. Finally, the relative oil volumes, vo, are measured at each stage of depletion after removal of the liberated gas, as listed in column 7. Before considering how the laboratory derived data presented in table 2.2 are converted to the required field parameters, Bo, Rs and Bg, it is first necessary to compare the physical difference between the flash and differential liberation experiments and decide which, if either, is suitable for describing the separation of oil and gas in the reservoir and the production of these volumes through the surface separators to the stock tank. The main difference between the two types of experiment shown in fig. 2.9(a) and (b) is that in the flash expansion no gas is removed from the PV cell but instead remains in equilibrium with the oil. As a result, the overall hydrocarbon composition in the cell remains unchanged. In the differential liberation experiment, however, at each stage of depletion the liberated gas is physically removed from contact with the oil and therefore, there is a continual compositional change in the PV cell, the remaining hydrocarbons becoming progressively richer in the heavier components, and the average molecular weight thus increasing. If both experiments are performed isothermally, in stages, through the same total pressure drop, then the resulting volumes of liquid oil remaining at the lowest pressure will, in general, be slightly different. For low volatility oils, in which the dissolved gas PVT ANALYSIS FOR OIL 61 consists mainly of methane and ethane, the resulting oil volumes from either experiment are practically the same. For higher volatility oils, containing a relatively high proportion of the intermediate hydrocarbons such as butane and pentane, the volumes can be significantly different. Generally, in this latter case, more gas escapes from solution in the flash expansion than in the differential liberation, resulting in a smaller final oil volume after the flash process. This may be explained by the fact that in the flash expansion the intermediate hydrocarbon molecules find it somewhat easier to escape into the large gas volume in contact with the oil than in the case of the differential liberation, in which the volume of liberated gas in equilibrium with the oil, at any stage in the depletion, is significantly smaller. The above description is a considerable simplification of the complex processes involved in the separation of oil and gas; also, it is not always true that the flash separation yields smaller oil volumes. What must be appreciated, however, is that the flash and differential processes will yield different oil volumes and this difference can be physically measured by experiment. The problem is, of course, which type of experiment will provide the most realistic values of Bo, Rs and Bg, required for relating measured surface volumes to volumes withdrawn from the reservoir at the current reservoir pressure and fixed temperature. The answer is that a combination of both flash and differential liberation is required for an adequate description of the overall volume changes. It is considered that the differential liberation experiment provides the better description of how the oil and gas separate in the reservoir since, because of their different flow velocities, they will not remain together in equilibrium once gas is liberated from the oil, thus corresponding to the process shown in fig. 2.9(b). The one exception to this is during the brief period after the bubble point has been reached, when the liberated gas is fairly uniformly distributed throughout the reservoir and remains immobile until the critical gas saturation is exceeded. The nature of the volume change occurring between the reservoir and stock tank is more difficult to categorise but generally, the overall effect is usually likened to a non-isothermal flash expansion. One aspect in this expansion during production is worth considering in more detail and that is, what occurs during the passage of the reservoir fluids through the surface separator or separators. Within any single separator the liberation of gas from the oil may be considered as a flash expansion in which, for a time, the gas stays in equilibrium with the oil. If two or more separators are used then the gas is physically removed from the oil leaving the first separator and the oil is again flashed in the second separator. This physical isolation of the fluids after each stage of separation corresponds to differential liberation. In fact, the overall effect of multi-stage separation corresponds to the process shown in fig. 2.9(b), which is differential liberation, only in this case it is not conducted at constant temperature. It is for this reason that multi-stage separation is commonly used in the field because, as already mentioned, differential liberation will normally yield a larger final volume of equilibrium oil than the corresponding flash expansion. PVT ANALYSIS FOR OIL 62 The conclusion reached, from the foregoing description of the effects of surface separation, is somewhat disturbing since it implies that the volume of equilibrium oil collected in the stock tank is dependent on the manner in which the oil and gas are separated. This in turn means that the basic PVT parameters Bo and Rs which are measured in terms of volume "per stock tank barrel" must also be dependent on the manner of surface separation and cannot be assigned absolute values. The only way to account for the effects of surface separation is to perform a series of separator tests on oil samples as part of the basic PVT analysis, and combine the results of these tests with differential liberation data. Samples of oil are put in the PV cell, fig. 2.8, and raised to reservoir temperature and bubble point pressure. The cell is connected to a single or multi-stage model separator system, with each separator at a fixed pressure and temperature. The bubble point oil is then flashed through the separator system to stock tank conditions and the resulting volumes of oil and gas are measured. The results of such a series of tests, using a single separator at a series of different pressures and at a fixed temperature, are listed in table 2.3 for the same oil as described previously (tables 2.1 and 2.2). Separator Stock tank Shrinkage factor GOR p T p T f si R (psia) (° F) (psia) (° F) f b c (stb/rbb) (scf/stb) 200 80 14.7 60 .7983 512 150 80 14.7 60 .7993 510 100 80 14.7 60 .7932 515 50 80 14.7 60 .7834 526 TABLE 2.3 Separator flash expansion experiments performed on the oil sample whose properties are listed in tables 2.1 and 2.2 The shrinkage factor f b c , listed in table 2.3, is the volume of oil collected in the stock tank, relative to unit volume of oil at the bubble point (stb/rbb), which is the reason for the subscript b (bubble point). The subscript f refers to the fact that these experiments are conducted under flash conditions. All such separator tests, irrespective of the number of separator stages, are described as flash although, as already mentioned, multi-stage separation is closer to a differential liberation. In any case, precisely what the overall separation process is called does not really matter since the resulting volumes of oil and gas are experimentally determined, irrespective of the title. f si R is the initial solution gas oil ratio corresponding to the separators used and is measured in the experiments in scf/stb. Using the experimental separator flash data, for a given set of separator conditions, in conjunction with the differential liberation data in table 2.2, will provide a means of obtaining the PVT parameters required for field use. It is considered that the differential liberation data can be used to describe the separation in the reservoir while the separator flash data account for the volume changes between reservoir and stock tank. PVT ANALYSIS FOR OIL 63 What is required for field use is Bo expressed in rb/stb. In the differential liberation data the corresponding parameter is vo (rb/rbb), that is, reservoir barrels of oil per unit barrel at the bubble point. But from the flash data it is known that one reservoir barrel of oil, at the bubble point, when flashed through the separators yields f b c stb. Therefore, the conversion from the differential data to give the required field parameter Bo is f o b o b b v rb rb rb B stb c stb rb = Similarly, the solution gas oil ratio required in the field is Rs (scf/stb). The parameter in the differential liberation data from which this can be obtained is F (cumulative gas vol at sc/oil vol at pb = stb/rbb). In fact, F, the cumulative gas liberated from the oil, must be proportional to f si s R R − (scf/stb), which is the initial solution gas oil ratio, as determined in the flash experiment, minus the current solution gas oil ratio at some lower pressure. The exact relationship is f f b si s b b rb scf stb scf 1 (R R ) F 5.615 stb rb stb c stb − = × × Finally, the determination of the third parameter Bg can be obtained directly from the differential parameter E as g rb 1 rcf 1 rb B scf E scf 5.615 rcf = × Thus the laboratory differential data can be transformed to give the required field PVT parameters using the following conversions Laboratory Differential Parameter Required Field Parameter Conversion vo (rb/ rbb) Bo f o o b v rb B c stb = (2.5) F (stb/rbb) Rs f s si bf 5.615 F rb R R c stb = − (2.6) E (scf/rcf) Bg g 1 rb B 5.612 E scf = (2.7) EXERCISE 2.2 CONVERSION OF DIFFERENTIAL LIBERATION DATA TO GIVE THE FIELD PVT PARAMETERS Bo, Rs AND Bg Convert the laboratory differential liberation data presented in table 2.2 to the required PVT parameters, for field use, for the optimum separator conditions listed in table 2.3. PVT ANALYSIS FOR OIL 64 EXERCISE 2.2 SOLUTION The optimum separator pressure in table 2.3 is 150 psia since this gives the largest value of the flash shrinkage factor f b c as 0.7993 (stb/rbb) and correspondingly, the lowest flash solution gas oil ratio f si R of 510 scf/stb. Using these two figures the laboratory differential data in table 2.2 can be converted to give the field parameters Bo, Rs and Bg using equs. (2.5) − (2.7), as follows Pressure (psia) f o o b v B c = (rb/stb) f f s si b 5.615 F R R c = − (scf/stb) g 1 B 5.615 E = (rb/scf) 4000 (pi) 1.2417 f oi (B ) 510 f si (R ) 3500 1.2480 510 3330 (pb) 1.2511 f f ob b 1 (B ) c = 510 .00087 3000 1.2222 450 .00096 2700 1.2022 401 .00107 2400 1.1822 352 .00119 2100 1.1633 304 .00137 1800 1.1450 257 .00161 1500 1.1287 214 .00196 1200 1.1115 167 .00249 900 1.0940 122 .00339 600 1.0763 78 .00519 300 1.0583 35 .01066 TABLE 2.4 Field PVT parameters adjusted for single stage, surface separation at 150 psia and 80°F; f b c = .7993 (Data for pressures above 3330 psi are taken from the flash experiment, table 2.1) The data in table 2.4 are plotted in fig. 2.5(a) − (c). In summary of this section, it can be stated that the laboratory differential liberation experiment, which is regarded as best simulating phase separation in the reservoir, provides an absolute set of PVT data in which all volumes are expressed relative to the unit oil volume at the bubble point, the latter being a unique volume. The PVT parameters conventionally used in the field, however, are dependent on the nature of the surface separation. The basic differential data can be modified in accordance with the surface separators employed using equs. (2.5) − (2.7) in which f b c and f b R are determined by flashing unit volume of reservoir oil through the separator system. The modified PVT parameters thus obtained approximate the process of differential PVT ANALYSIS FOR OIL 65 liberation in the reservoir and flash expansion to stock tank conditions. Therefore if, during the producing life of the reservoir, the separator conditions are changed, then the fixed differential liberation data will have to be converted to give new tables of Bo and Rs using values of f b c and f si R appropriate for the altered separator conditions. This combination of differential liberation in the reservoir and flash expansion to the surface is generally regarded as a reasonable approximation to Dodson's PVT analysis technique4. In this form of experiment a differential liberation is performed but after each pressure stage the volume of the oil remaining in the PV cell is flashed to stock tank conditions through a chosen separator combination. The ratio of stock tank oil volume to original oil volume in the PV cell prior to flashing gives a direct measure of Bo, while the gas evolved in the flash can be used directly to obtain Rs. The process is repeated taking a new oil sample for each pressure step, since the remaining oil in the PV cell is always flashed to surface conditions. This type of analysis, while more accurately representing the complex reservoir-production phase separation, is more time consuming and therefore more costly, furthermore, it requires the availability of large samples of the reservoir fluid. For low and moderately volatile crudes, the manner of deriving the PVT parameters described in this section usually provides a very good approximation to the results obtained from the Dodson analysis. For more volatile crudes, however, the more elaborate experimental technique may be justified. 2.5 ALTERNATIVE MANNER OF EXPRESSING PVT LABORATORY ANALYSIS RESULTS The results of the differential liberation experiment, as listed in table 2.2, provide an absolute set of data which can be modified, according to the surface separators used, to give the values of the PVT parameters required for field use. In table 2.2 all volumes are measured relative to the unit oil volume at the bubble point. There is, however, a more common way of representing the results of the differential liberation in which volumes are measured relative to the volume of residual oil at stock tank conditions. This volume is obtained as the final step in the differential liberation experiment by flashing the volume of oil measured at atmospheric pressure and reservoir temperature, to atmospheric pressure and 60°F. This final step is shown in table 2.2 in which 0.8296 relative oil volumes at 14.7 psia and 200°F yield 0.7794 relative oil volumes at 14.7 psia and 60°F. This value of 0.7794 is the shrinkage factor for a unit volume of bubble point oil during differential liberation to stock tank conditions and is denoted by d b c . The value of d b c ,is not dependent on any separator conditions and therefore, relating all volumes in the differential liberation to this value of d b c , which is normally referred to as the "residual oil volume", will provide an alternative means of expressing the differential liberation results. It should be noted, however, that the magnitude of d b c is dependent on the number of pressure steps taken in the differential experiment. Therefore, the differential liberation results, in which all volumes are measured relative to d b c do not provide an absolute set of data such as that obtained by relating all volumes to the unit volume of oil at the bubble point. PVT ANALYSIS FOR OIL 66 In the presentation of differential data, in which volumes are measured relative to d b c , the values of vo and F in table 2.2 are replace by d o B and d s R where d o B = Differential oil formation volume factor (rb/stb-residual oil) and d s R = Differential solution gas oil ratio (scf/stb-residual oil) Alternatively, by replacing f b c in equs. (2.5) and (2.6) by d b c , these parameters can be expressed as d d o b o b b v rb rb B c stb residual rb = − (2.8) and d d d s si b 5.615 F scf R R c stb residual = − − (2.9) where d si R is the initial dissolved gas relative to the residual barrel of oil at 60°F, and is proportional to the total gas liberated in the differential experiment, thus d d si b (Maximumvalueof F) scf R 5.615 c stb residual = × − (2.10) and for the differential data presented in table 2.2 d si 74.9557 5.615 R 540 scf stb residual oil .7794 × = = − The majority of commercial laboratories serving the industry would normally present the essential data in the differential liberation experiment (table 2.2) as shown in table 2.5. There is a danger in presenting the results of the differential liberation experiment in this way since a great many engineers are tempted to use the d o B and d s R values directly in reservoir calculations, without making the necessary corrections to allow for the surface separator conditions. In many cases, the error in directly using the data in table 2.5 is negligible, however, for moderate and high volatility oils the error can be quite significant and therefore, the reader should always make the necessary correction to the data in table 2.5 to allow for the field separator conditions, as a matter of course. PVT ANALYSIS FOR OIL 67 Pressure (psia) Formation Vol. Factor d d o o b B v / c = Solution GOR d d d s si b R R 5.615 F / c = − 4000 1.2734 540 3500 1.2798 540 3300 1.2830 d ob (B ) 540 d si (R ) 3000 1.2534 479 2700 1.2329 428 2400 1.2123 378 2100 1.1930 328 1800 1.1742 281 1500 1.1576 236 1200 1.1399 118 900 1.1219 142 600 1.1038 97 300 1.0853 52 14.7 (200°F) 1.0644 0 14.7 ( 60°F) 1.0000 0 TABLE 2.5 Differential PVT parameters as conventionally presented by laboratories, in which Bo and Rs are measured relative to the residual oil volume at 60°F The conversion can be made by expressing and Rsd, in table 2.5, in their equivalent, absolute forms of vo and F, in table 2.2, using equs. (2.8) and (2.9) and thereafter, using equs. (2.5) and (2.6) to allow for the surface separators. This will result in the required expressions for Bo and Rs. Alternatively, the required field parameters can be calculated directly as d f d f d f f b ob o o o o b b b ob c B v v B B c c c B = = = (2.11) where d o b v / c = d o B the differential oil formation volume factor measured relative to the residual oil volume as listed in table 2.5 (rb/stb-residual); f ob B = f b 1/ c is the oil formation volume factor of the bubble point oil (rbb/stb) determined by flashing the oil through the appropriate surface separators and is measured relative to the stock tank oil volume (refer tables 2.3 and 2.4); and d ob B = d b 1/ c is the oil formation volume factor of the bubble point oil determined during the differential liberation experiment and is measured relative to the residual oil volume (refer table 2.5) (rbb/stb-residual). Similarly, the required solution gas oil ratio for use under field operating conditions is, equ. (2.6) PVT ANALYSIS FOR OIL 68 d f f f d f b s si si b b b c 5.615 F 5.615 F R R R c c c = − = − which, using equ. (2.9), can be expressed as f f d d f ob s si si s ob B R R (R R ) B = − − (2.12) where f si R = solution gas oil ratio of the bubble point oil, determined by flashing the oil through the appropriate surface separators, and is measured relative to the oil volume at 60°F and 14.7 psia (refer tables 2.3 and 2.4) (scf/stb). d si R = solution gas oil ratio of the bubble point oil determined during the differential experiment and measured relative to the residual oil volume at 60°F and 14.7 psia (refer table 2.5 and equ. (2.10)) (scf/stb-residual). The differential data, as presented in table 2.5, can be directly converted to the required form, table 2.4, using the above relations. For instance, using the following data from table 2.5, at a pressure of 2400 psi d o B = 1.2123 (rb /barrel of residual oil at 60°F and 14.7 psia) d s R = 378 (scf/ —" — ) d ob B = 1.2830 (rb / — " — ) d si R = 540 (scf/ —" — ) while from the separator flash tests (table 2.3), for the optimum separator conditions of 150 psia and 80°F f f ob b B (1/ c ) 1.2511(rb / stb) = = f si R 510 (scf / stb) = Therefore, using equ. (2.11) o 1.2511 B 1.2123 1.1822 rb stb 1.2830 = × = and equ. (2.12) s 1.2511 R 510 (540 378) 352 scf / stb 1.2830 = − − × = PVT ANALYSIS FOR OIL 69 2.6 COMPLETE PVT ANALYSIS The complete PVT analysis for oil, provided by most laboratories, usually consists of the following experiments and calculations. a) Compositional analysis of the separator oil and gas, for samples collected at the surface, together with physical recombination, refer sec. 2.3(b), or; compositional analysis of the reservoir fluid collected in a subsurface sample. Such analyses usually give the mole fractions of each component up to the hexanes. The hexanes and heavier components are grouped together, and the average molecular weight and density of the latter are determined. b) Flash expansion, as described in sec. 2.4 (table 2.1), conducted at reservoir temperature. This experiment determines − the bubble point pressure − the compressibility of the undersaturated oil as o o o o o dv dB 1 1 c v dp B dp = − = − (2.13) − the total volume vt of the oil and gas at each stage of depletion. c) Differential liberation experiment as described in sec. 2.4 to determine − E, Z, F and vo (as listed in table 2.2), with F and vo measured relative to the unit volume of bubble point oil. Alternatively, by measuring d b c during the last stage of the differential liberation, the above data can be presented as − E, Z, d si R − d s R (or just d s R ) and d o B (as listed in table 2.5), with d s R and d o B measured relative to residual oil volume. In addition, the gas gravity is measured at each stage of depletion. d) Measurement of the oil viscosity at reservoir temperature (generally using the rolling ball viscometer1,3), over the entire range of pressure steps from above bubble point to atmospheric pressure. Gas viscosities are normally calculated at reservoir temperature, from a knowledge of the gas gravity, using standard correlations5. e) Separator tests to determine the shrinkage, f b c , and solution gas oil ratio, f si R , of unit volume of bubble point oil (1 barrel) when flashed through various separator combinations (refer table 2.3). Instead of actually performing these tests, in many cases the results are obtained using the phase equilibrium calculation technique1. f) Composition and gravity of the separator gas in the above separator tests. PVT ANALYSIS FOR OIL 70 REFERENCES 1) Amyx, J.W., Bass, D.M. and Whiting, R.L., 1960. Petroleum Reservoir Engineering; Physical Properties. McGraw-Hill Book Company: 359−425. 2) Reudelhuber, F.O., 1957. Sampling Procedures for Oil Reservoir Fluids. J.Pet. Tech., December. 3) Anonymous, 1966. API Recommended Practice for Sampling Petroleum Reservoir Fluids. Official publication of the American Petroleum Institute, January (API RP 44). 4) Dodson, C.R., Goodwill, D. and Mayer, E.H., 1953. Application of Laboratory PVT Data to Reservoir Engineering Problems. Trans. AIME, 198: 287−298. 5) Carr, N.L., Kobayashi, R. and Burrows, D.B., 1954. Viscosity of Hydrocarbon Gases under Pressure. Trans. AlME, 201: 264−272. CHAPTER 3 MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 3.1 INTRODUCTION The Schilthuis material balance equation has long been regarded as one of the basic tools of reservoir engineers for interpreting and predicting reservoir performance. In this chapter, the zero dimensional material balance is derived and subsequently applied, using mainly the interpretative technique of Havlena and Odeh, to gain an understanding of reservoir drive mechanisms under primary recovery conditions. Finally, some of the uncertainties attached to estimation of in-situ pore compressibility, a basic component in the material balance equation, are qualitatively discussed. Although the classical material balance techniques, once applied, have now largely been superseded by numerical simulators, which are essentially multi-dimensional, multi-phase, dynamic material balance programs, the classical approach is well worth studying since it provides a valuable insight into the behaviour of hydrocarbon reservoirs. 3.2 GENERAL FORM OF THE MATERIAL BALANCE EQUATION FOR A HYDROCARBON RESERVOIR The general form of the material balance equation was first presented by Schilthuis1 in 1941. The equation is derived as a volume balance which equates the cumulative observed production, expressed as an underground withdrawal, to the expansion of the fluids in the reservoir resulting from a finite pressure drop. The situation is depicted in fig. 3.1 in which (a) represents the fluid volume at the initial pressure pi in a reservoir which has a finite gascap. The total fluid volume in this diagram is the hydrocarbon pore volume of the reservoir (HCPV). Fig. 3.1 (b) illustrates the effect of reducing the pressure by an amount ∆p and allowing the fluid volumes to expand, in an artificial sense, in the reservoir. The original HCPV is still drawn in this diagram as the solid line. Volume A is the increase due to the expansion of the oil plus originally dissolved gas, while volume increase B is due to the expansion of the initial gascap gas. The third volume increment C is the decrease in HCPV due to the combined effects of the expansion of the connate water and reduction in reservoir pore volume as already discussed in Chapter 1, sec. 7. If the total observed surface production of oil and gas is expressed in terms of an underground withdrawal, evaluated at the lower pressure p, (which means effectively, MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 72 pi Gascap gas mNBoi (rb) Oil + originally dissolved gas NBoi (rb) ∆p p B A C (b) (a) Fig. 3.1 Volume changes in the reservoir associated with a finite pressure drop ∆ ∆ ∆ ∆p; (a) volumes at initial pressure, (b) at the reduced pressure taking all the surface production back down to the reservoir at this lower pressure) then it should fit into the volume A + B + C which is the total volume change of the original HCPV. Conversely, volume A + B + C results from expansions which are allowed to artificially occur in the reservoir. In reality, of course, these volume changes correspond to reservoir fluid which would be expelled from the reservoir as production. Thus the volume balance can be evaluated in reservoir barrels as Underground withdrawal (rb) = Expansion of oil + originally dissolved gas (rb) + Expansion of gascap gas (rb) + Reduction in HCPV due to connate water expansion and decrease in the pore volume (rb) Before evaluating the various components in the above equation it is first necessary to define the following parameters. N is the initial oil in place in stock tank barrels = V φ (1−Swc) / Boi stb m is the ratio initial hydrocarbon volume of the gascap initial hydrocarbon volume of the oil (and, being defined under initial conditions, is a constant) Np is the cumulative oil production in stock tank barrels, and Rp is the cumulative gas oil ratio MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 73 Cumulative gas production (scf) Cumulative oil production (stb) = Then the expansion terms in the material balance equation can be evaluated as follows. a) Expansion of oil plus originally dissolved gas There are two components in this term: -Liquid expansion The N stb will occupy a reservoir volume of NBoi rb, at the initial pressure, while at the lower pressure p, the reservoir volume occupied by the N stb will be NBo, where Bo is the oil formation volume factor at the lower pressure. The difference gives the liquid expansion as o oi N(B B ) (rb) − (3.1) -Liberated gas expansion Since the initial oil is in equilibrium with a gascap, the oil must be at saturation or bubble point pressure. Reducing the pressure below pi will result in the liberation of solution gas. The total amount of solution gas in the oil is NRsi scf. The amount still dissolved in the N stb of oil at the reduced pressure is NRs scf. Therefore, the gas volume liberated during the pressure drop ∆p, expressed in reservoir barrels at the lower pressure, is si s g N(R R ) B (rb) − (3.2) b) Expansion of the gascap gas The total volume of gascap gas is mNBoi rb, which in scf may be expressed as oi gi mNB G (scf) B = This amount of gas, at the reduced pressure p, will occupy a reservoir volume g oi gi B mNB (rb) B Therefore, the expansion of the gascap is g oi gi B mNB 1 (rb) B − (3.3) c) Change in the HCPV due to the connate water expansion and pore volume reduction MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 74 The total volume change due to these combined effects can be mathematically expressed as d(HCPV) = dVw + dVf (1.36) or, as a reduction in the hydrocarbon pore volume, as d (HCPV) = (cw Vw + cf Vf ) ∆p (1.38) where Vf is the total pore volume = HCPV/(1 − Swc) and Vw is the connate water volume = Vf × Swc = (HCPV)Swc/(1 − Swc). Since the total HCPV, including the gascap, is (1+m)NBoi (rb) (3.4) then the HCPV reduction can be expressed as w wc f oi wc c S c d(HCPV) (1 m)NB p 1 S + − = + ∆ − (3.5) This reduction in the volume which can be occupied by the hydrocarbons at the lower pressure, p, must correspond to an equivalent amount of fluid production expelled from the reservoir, and hence should be added to the fluid expansion terms. d) Underground withdrawal The observed surface production during the pressure drop ∆p is Np stb of oil and Np Rp scf of gas. When these volumes are taken down to the reservoir at the reduced pressure p, the volume of oil plus dissolved gas will be NpBo rb. All that is known about the total gas production is that, at the lower pressure, Np Rs scf will be dissolved in the Np stb of oil. The remaining produced gas, Np (Rp − Rs) scf is therefore, the total amount of liberated and gascap gas produced during the pressure drop ∆p and will occupy a volume N(Rp − Rs)Bg rb at the lower pressure. The total underground withdrawal term is therefore Np (Bo + (Rp − Rs)Bg) (rb) (3.6) Therefore, equating this withdrawal to the sum of the volume changes in the reservoir, equs. (3.1 ), (3.2), (3.3) and (3.5), gives the general expression for the material balance as o oi si s g p o p s g oi oi g w wc f e p w gi wc (B B ) (R R ) B N (B (R R )B ) NB B B c S c m 1 (1 m) p (W W )B B 1 S − + − + − = + + − + + ∆ + − − (3.7) MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 75 in which the final term (We − Wp)Bw is the net water influx into the reservoir. This has been intuitively added to the right hand side of the balance since any such influx must expel an equivalent amount of production from the reservoir thus increasing the left hand side of the equation by the same amount. In this influx term We = Cumulative water influx from the aquifer into the reservoir, stb. Wp = Cumulative amount of aquifer water produced, stb. and Bw = Water formation volume factor rb/stb. Bw is generally close to unity since the solubility of gas in water is rather small and this condition will be assumed throughout this text. For more detailed calculations, correlation charts for Bw are presented in references 2 and 3. The following features should be noted in connection with the expanded material balance equation − it is zero dimensional, meaning that it is evaluated at a point in the reservoir − it generally exhibits a lack of time dependence although, as will be discussed in sec. 3.7 and also in Chapter 9, the water influx has a time dependence − although the pressure only appears explicitly in the water and pore compressibility term as, ∆p = pi − p, it is implicit in all the other terms since the PVT parameters Bo, Rs and Bg are themselves functions of pressure. The water influx is also pressure dependent. − the equation is always evaluated, in the way it was derived, by comparing the current volumes at pressure p to the original volumes at pi. It is not evaluated in a step-wise or differential fashion. Although the equation appears a little intimidating, at first sight, it should be thought of as nothing more than a sophisticated version of the compressibility definition dV = c × V × ∆p Production = Expansion of reservoir fluids. and, under certain circumstances, can in fact be reduced to this simple form. In using the material balance equation, one of the main difficulties lies in the determination of the representative average reservoir pressure at which the pressure dependent parameters in the equation should be evaluated. This follows from the zero dimensional nature of the equation which implies that there should be some point in the reservoir at which a volume averaged pressure can be uniquely determined. In applying the more simple gas material balance, equ. (1.35), such a point could be defined with reasonable accuracy as the centroid point, at which pressures could be evaluated throughout the producing life of the MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 76 reservoir. In the case of an oil reservoir, however, the situation is generally more complex since below the bubble point two phases, oil and gas, will co-exist and, due to the gravity difference between the phases, will tend to segregate. As a result, the point at which the average pressure should be determined will vary with time. Precisely how the volume averaged reservoir pressure can be determined from the analysis of pressure tests in wells will be detailed in Chapter 7. 3.3 THE MATERIAL BALANCE EXPRESSED AS A LINEAR EQUATION Since the advent of sophisticated numerical reservoir simulation techniques, the Schilthuis material balance equation has been regarded by many engineers as being of historical interest only; a technique used back in the nineteen forties and fifties when people still used slide-rules. It is therefore interesting to note that as late as 1963-4, Havlena and Odeh presented two of the most interesting papers ever published on the subject of applying the material balance equation and interpreting the results. Their papers,4,5 described the technique of interpreting the material balance as the equation of a straight line, the first paper describing the technique and the second illustrating the application to reservoir case histories. To express equ. (3.7) in the way presented by Havlena and Odeh requires the definition of the following terms F = Np (Bo + (Rp Rs) Bg) + Wp Bw (rb) (3.8) which is the underground withdrawal; Eo = (Bo − Boi) + (Rsi − Rs) Bg (rb/stb) (3.9) which is the term describing the expansion of the oil and its originally dissolved gas; g g oi gi B E B 1 (rb / stb) B = − (3.10) describing the expansion of the gascap gas, and w wc f f,w oi wc c S c E (1 m) B p (rb / stb) 1 S + = + ∆ − (3.11) for the expansion of the connate water and reduction in the pore volume. Using these terms the material balance equation can be written as F = N(Eo+mEg+Ef,w) + WeBw (3.12) Havlena and Odeh have shown that in many cases equ. (3.12) can be interpreted as a linear function. For instance, in the case of a reservoir which has no initial gascap, negligible water influx and for which the connate water and rock compressibility term may be neglected; the equation can be reduced to MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 77 F = NEo (3.13) in which the observed production, evaluated as an underground withdrawal, should plot as a linear function of the expansion of the oil plus its originally dissolved gas, the latter being calculated from a knowledge of the PVT parameters at the current reservoir pressure. This interpretation technique is useful, in that, if a simple linear relationship such as equ. (3.13) is expected for a reservoir and yet the actual plot turns out to be non-linear, then this deviation can itself be diagnostic in determining the actual drive mechanisms in the reservoir. For instance, equ. (3.13) may turn out to be non-linear because there is an unsuspected water influx into the reservoir helping to maintain the pressure. In this case equ. (3.12) can still be expressed in a linear form as e o o W F n E E = + (3.14) in which F/Eo should now plot as a linear function of We/E o. Once a straight line has been achieved, based on matching observed production and pressure data, then the engineer has, in effect, built a suitable mathematical model to describe the performance of the reservoir. As previously described, in Chapter 1, sec. 7, this phase is commonly referred to as a history match. Once this has been satisfactorily achieved, the next step is to use the same mathematical model to predict how the reservoir will perform in the future, possibly for a variety of production schemes. This prediction phase is facilitated by the mathematical ease in using the simple linear expressions for the material balance equation, as presented by Havlena and Odeh. The technique will be illustrated in greater detail in the following sections. 3.4 RESERVOIR DRIVE MECHANISMS If none of the terms in the material balance equation can be neglected, then the reservoir can be described as having a combination drive in which all possible sources of energy contribute a significant part in producing the reservoir fluids and determining the primary recovery factor. In many cases, however, reservoirs can be singled out as having predominantly one main type of drive mechanism in comparison to which all other mechanisms have a negligible effect. In the following sections, such reservoirs will be described in order to isolate and study the contribution of the individual components in the material balance in influencing the recovery factor and determining the production policy of the field. The mechanisms which will be studied are: -solution gas drive -gascap drive -natural water drive -compaction drive And these individual reservoir drive mechanisms will be investigated in terms of: -reducing the material balance to a compact form, in many cases using the technique of Havlena and Odeh, in order to quantify reservoir performance MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 78 -determining the main producing characteristics, the producing gas oil ratio and watercut -determining the pressure decline in the reservoir -estimating the primary recovery factor -investigating the possibilities of increasing the primary recovery. 3.5 SOLUTION GAS DRIVE A solution gas drive reservoir is one in which the principal drive mechanism is the expansion of the oil and its originally dissolved gas. The increase in fluid volumes during the process is equivalent to the production. Two phases can be distinguished, as shown in fig. 3.2 (a) when the reservoir oil is undersaturated and (b) when the pressure has fallen below the bubble point and a free gas phase exists in the reservoir. a) Above bubble point pressure (undersaturated oil) For a solution gas drive reservoir it is assumed that there is no initial gascap, thus m = 0, and that the aquifer is relatively small in volume and the water influx is negligible. Furthermore, above the bubble point, Rs = Rsi = Rp, since all the gas produced at the surface must have been dissolved in the oil in the reservoir. Under these assumptions, the material balance equation, (3.7), can be reduced to o oi w wc f p o oi oi wc (B B ) (c S c ) N B NB p B 1 S − + = + ∆ − (3.15) OWC (a) OWC (b) Sealing fault Fig. 3.2 Solution gas drive reservoir; (a) above the bubble point pressure; liquid oil, (b) below bubble point; oil plus liberated solution gas The component describing the reduction in the hydrocarbon pore volume, due to the expansion of the connate water and reduction in pore volume, cannot be neglected for an undersaturated oil reservoir since the compressibilities cw and cf are generally of the same order of magnitude as the compressibility of the oil. The latter may be expressed as described in Chapter 2, sec. 6, as MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 79 ( ) o oi o oi B B c B p − = ∆ and substituting this in equ. (3.15) gives w wc f p o oi o wc (c S c ) N B NB c p 1 S + = ∆ − (3.16) Since there are only two fluids in the reservoir, oil and connate water, then the sum of the fluid saturations must be 100% of the pore volume, or So + Swc = 1 and substituting the latter in equ. (3.16) gives the reduced form of material balance as o o w wc f p o oi wc c S c S c N B NB p 1 S + + = ∆ − (3.17) or p o oi e N B NB c p = ∆ (3.18) in which e o o w wc f wc 1 c (c S c S c ) 1 S = + + − (3.19) is the effective, saturation-weighted compressibility of the reservoir system. Since the saturations are conventionally expressed as fractions of the pore volume, dividing by 1 − Swc expresses them as fractions of the hydrocarbon pore volume. Thus the compressibility, as defined in equ. (3.19),must be used in conjunction with the hydrocarbon pore volume. Equation (3.18) illustrates how the material balance can be reduced to nothing more than the basic definition of compressibility, equ. (1.12), in which NpBo = dV, the reservoir production expressed as an underground withdrawal, and NBoi = V the initial hydrocarbon pore volume. EXERCISE 3.1 SOLUTION GAS DRIVE; UNDERSATURATED OIL RESERVOIR Determine the fractional oil recovery, during depletion down to bubble point pressure, for the reservoir whose PVT parameters are listed in table 2.4 and for which cw = 3.0 × 10-6 / psi Swc = .20 cf = 8.6 × 10-6 / psi EXERCISE 3.1 SOLUTION The data required from table 2.4 are pi = 4000 psi Boi = 1.2417 rb/stb MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 80 pb = 3330 psi Bob = 1.2511 rb/stb Therefore, the average compressibility of the undersaturated oil between initial and bubble point pressure is 6 ob oi o oi B B 1.2511 1.2417 c 11.3 10 /psi B p 1.2417(4000 3330) − − − = = = × ∆ − The recovery at bubble point pressure can be calculated using equ. (3.18) as b p oi e ob p N B c p N B = ∆ where ( ) 6 e 6 1 c 11.3 0.8 3 0.2 8.6 10 /psi .8 22.8 10 − − = × + × + × = × and therefore, Recovery 6 1.2417 22.8 10 (4000 3330) 1.2511 − = × × × − or 1.52% of the original oil in place. Considering that the 670 psi pressure drop represents about 17% of the initial, absolute pressure, the oil recovery is extremely low. This is because the effective compressibility is small providing the reservoir contains just liquid oil and water. The situation will, however, be quite different once the pressure has fallen below bubble point. b) Below bubble point pressure (saturated oil) Below the bubble point pressure gas will be liberated from the saturated oil and a free gas saturation will develop in the reservoir. To a first order of approximation the gas compressibility is cg ≈ 1/p, as described in Chapter 1, sec. 6. Therefore, using the data of exercise 3.1, the minimum value of the free gas phase compressibility will occur at the bubble point pressure and will be equal to 1/pb = 1/3330 = 300 × 10-6/psi. This is two orders of magnitude greater than the water compressibility and 35 times greater than the pore compressibility and, as a result, the latter two are usually neglected in the material balance equation. The manner in which the reservoir will now behave is illustrated by the following exercise. EXERCISE 3.2 SOLUTION GAS DRIVE; BELOW BUBBLE POINT PRESSURE The reservoir described in exercise 3.1 will be produced down to an abandonment pressure of 900 psia. 1) Determine an expression for the recovery at abandonment as a function of the cumulative gas oil ratio Rp. MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 81 What do you conclude from the nature of this relationship? 2) Derive an expression for the free gas saturation in the reservoir at abandonment pressure. All PVT data may be taken from table 2.4. EXERCISE 3.2 SOLUTION 1) For a solution gas drive reservoir, below the bubble point, the following are assumed -m = 0; no initial gascap -negligible water influx -the term NBoi w wc f wc c S c 1 S + − ∆p is negligible once a significant free gas saturation develops in the reservoir. Under these conditions the material balance equation can be simplified as Np (Bo + (Rp − Rs)Bg) = N ((Bo − Boi) + (Rsi − Rs)Bg) (3.20) underground withdrawal = expansion of the oil plus originally dissolved gas and the recovery factor at abandonment pressure of 900 psia is o oi si s g p 900 o p s g 900 psi 900 psi (B B ) (R R ) B N (RF) N B (R R ) B − + − = = + − in which all the PVT parameters Bo, Rs and Bg are evaluated at the abandonment pressure. Using the data in table 2.4, the recovery factor can be expressed as p p 900 N (1.0940 1.2417) (510 122).00339 N 1.0940 (R 122).00339 − + − = + − which can further be reduced to p p 900 N 344 N R 201 = + This clearly demonstrates that there is an inverse relationship between the oil recovery and the cumulative gas oil ratio Rp, as illustrated in fig. 3.3. The conclusion to be drawn from the relationship is that, to obtain a high primary recovery, as much gas as possible should be kept in the reservoir, which requires that the cumulative gas oil ratio should be maintained as low as possible. By keeping the MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 82 gas in the reservoir the total reservoir system compressibility in the simple material balance dV = cV ∆p will be greatly increased by the presence of the gas and the dV, which is the production, will be large for a given pressure drop. 50 40 30 20 10 0 0 1000 2000 3000 4000 R (scf / stb) p Np N 900 % Fig. 3.3 Oil recovery, at 900 psia abandonment pressure (% STOIIP), as a function of the cumulative GOR, Rp (Exercise 3.2) 2) The free gas saturation in the reservoir may be deduced in two ways, the most obvious being to consider the overall gas balance liberated total gas gasstill gasinthe amount producedat dissolved reservoir of gas thesurface intheoil = − − which in terms of the basic PVT parameters can be evaluated at any reservoir pressure as liberated gas (rb) = (NRsi – NpRp − (N − Np) Rs) Bg and expressing this as a saturation, which is conventionally required as a fraction of the pore volume, then Sg = [ N (Rsi − Rs) − Np (Rp − Rs) ] Bg (1 − Swc) / NBoi (3.21) where NBoi / (1 − Swc) = HCPV / (1 − Swc) = the pore volume. MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 83 A simpler and more direct method is to consider that liberated gas initial total current oil in the volume of oil volume in reservoir in the reservoir the reservoir = − i.e. liberated gas = NBoi − (N − Np) Bo (rb) and therefore Sg = (NBoi − (N − Np) Bo) (1 − Swc) / NBoi or Sg = p o wc oi N B 1 1 (1 S ) N B − − − (3.22) which at abandonment pressure becomes Sg = p N 1 1 0.88 0.8 N − − × again showing that if the gas is kept in the reservoir so that Sg has a high value then Np/N will be large, and vice versa. Naturally equs. (3.21) and (3.22) are equatable through the material balance equ. (3.20). Although the lesson of the last exercise is quite clear, the practical means of keeping the gas in the ground in a solution gas drive reservoir is not obvious. Once the free gas saturation in the reservoir exceeds the critical saturation for flow, then as noted already in Chapter 2, sec. 2, the gas will start to be produced in disproportionate quantities compared to the oil and, in the majority of cases, there is little that can be done to avert this situation during the primary production phase. Under very favourable conditions the oil and gas will separate with the latter moving structurally updip in the reservoir. This process of gravity segregation relies upon a high degree of structural relief and a favourable permeability to flow in the updip direction. Under more normal circumstances, the gas is prevented from moving towards the top of the structure by inhomogeneities in the reservoir and capillary trapping forces. Reducing a well's offtake rate or closing it in temporarily to allow gas-oil separation to occur may, under these circumstances, do little to reduce the producing gas oil ratio. A typical producing history of a solution gas drive reservoir under primary producing conditions is shown in fig. 3.4. As can be seen, the instantaneous or producing gas oil ratio R will greatly exceed Rsi for pressures below bubble point and the same is true for the value of Rp. The pressure will initially decline rather sharply above bubble point because of the low compressibility of the reservoir system but this decline will be partially arrested once MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 84 free gas starts to accumulate. The primary recovery factor from such a reservoir is very low and will seldom exceed 30% of the oil in place. pi Rsi pb pressure decline R ( producing GOR ) time watercut (%) Fig. 3.4 Schematic of the production history of a solution gas drive reservoir Two ways of enhancing the primary recovery are illustrated in fig. 3.5. The first of these methods, water injection, is usually aimed at maintaining the pressure above bubble point, or above the pressure at which the gas saturation exceeds the critical value at which the gas becomes mobile. The unfortunate consequences of starting to inject water below bubble point pressure are illustrated in exercise 3.3. water treatment plant water injection OWC production well oil compressor gas injection sealing fault MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 85 Fig. 3.5 Illustrating two ways in which the primary recovery can be enhanced; by downdip water injection and updip injection of the separated solution gas EXERCISE 3.3 WATER INJECTION BELOW BUBBLE POINT PRESSURE It is planned to initiate a water injection scheme in the reservoir whose PVT properties are defined in table 2.4. The intention is to maintain pressure at the level of 2700 psia (pb = 3330 psia). If the current producing gas oil ratio of the field (R) is 3000 scf/stb, what will be the initial water injection rate required to produce 10,000 stb/d of oil. EXERCISE 3.3 SOLUTION To maintain pressure at 2700 psia the total underground withdrawal at the producing end of a reservoir block must equal the water injection rate at the injection end of the block. The total withdrawal associated with 1 stb of oil is Bo + (R − Rs)Bg rb which, evaluating at 2700 psia, using the PVT data in table 2.4, is 1.2022 + (3000 − 401) 0.00107 = 4.0 rb Therefore, to produce 10,000 stb/d oil, an initial injection rate of 40,000 rb/d of water will be required, 70% of which will be needed to displace the liberated gas. If the injection had been started at, or above bubble point pressure, a maximum injection rate of only 12,500 b/d of water would have been required. The mechanics of water injection are described in Chapter 10, including methods of calculating the recovery factor. One of the advantages in this secondary recovery process is that if the displacement is maintained at, or just below, bubble point pressure the producing gas oil ratio is constant and approximately equal to Rsi. If the gas quantities are sufficiently large it is easier, under these circumstances, to enter into a gas sales contract in which gas rates are usually specified by the customer at a plateau level. Conversely, there are obvious difficulties attached to entering such a contract with a gas oil ratio profile as shown in fig. 3.4. In such cases difficulties are frequently encountered in disposing of all the gas. Some portion of it may be sold under a fixed contract agreement but the remainder, which is frequently unpredictable in quantity, presents problems. In the "old days" (prior to the 1973−energy crisis) a lot of this excess gas, which could not conveniently be used as a local fuel supply, was flared. Even as late as the end of 1973 it was estimated that some 11% of the world's total daily gas production was flared. Today, regulations concerning gas disposal are more stringent and in many cases operators are obliged to re-inject excess gas back into the reservoir as shown in fig. 3.5. The gas is separated from the oil at high pressure and injected at a structurally high point thus forming a secondary gas cap. The oil production is taken from downdip in the reservoir thus allowing the high compressibility gas to expand and displace an equivalent amount of oil towards the MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 86 producing wells. This demonstrates a method of keeping as much gas in the reservoir as possible where it can serve its most useful purpose, as suggested in exercise 3.2. The economic success of both water and solution gas injection depends upon the additional recovery obtained as a result of the projects. The present day value of the additional oil recovery must be greater than the cost of the injection wells, surface treatment facilities (mainly for water) and compressor costs (mainly for gas). In many cases, for small reservoirs, injection of water or gas is not economically viable and the solution gas drive process must be allowed to run its full course resulting in low oil recovery factors. 3.6 GASCAP DRIVE A typical gascap drive reservoir is shown in fig. 3.6. Under initial conditions the oil at the gas oil contact must be at saturation or bubble point pressure. The oil further downdip becomes progressively less saturated at the higher pressure and temperature. Generally this effect is relatively small and reservoirs can be described using uniform PVT properties, as will be assumed in this text. There are exceptions, however, one of the most remarkable being the Brent field in the North Sea6 in which at the gas oil contact the oil has a saturation pressure of 5750 psi and a solution gas oil ratio of 2000 scf/stb, while at the oil water contact, some 500 feet deeper, the saturation pressure and solution gas oil ratio are 4000 psi and 1200 scf/stb, respectively. Such extremes are rarely encountered and in the case of the Brent field the anomaly is attributed to gravity segregation of the lighter hydrocarbon components. For a reservoir in which gascap drive is the predominant mechanism it is still assumed that the natural water influx is negligible (We = 0) and, in the presence of so much high compressibility gas, that the effect of water and pore compressibilities is also negligible. Under these circumstances, the material balance equation, (3.7), can be written as ( ) p o p s g o oi si s g g oi oi gi N B (R R )B (B B ) (R R )B B NB m 1 B B + − − + − = + − (3.23) in which the right hand side contains the term describing the expansion of the oil plus originally dissolved gas, since the solution gas drive mechanism is still active in the oil column, together with the term for the expansion of the gascap gas. Equation (3.23) is rather cumbersome and does not provide any kind of clear picture of the principles involved in the gascap drive mechanism. A better understanding of the situation can be gained by using the technique of Havlena and Odeh, described in sec. 3.3, for which the material balance, equ. (3.12), can be reduced to the form F = N (Eo + mEg) (3.24) MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 87 production wells OWC GOC OIL ( NBoi - rb) exploration well GAS ( mNBoi - rb) Fig. 3.6 Typical gas drive reservoir The way in which this equation can be used depends on the unknown quantities. For a gascap reservoir the least certain parameter in equ. (3.24) is very often m, the ratio of the initial hydrocarbon pore volume of the gascap to that of the oil column. For instance, in the reservoir depicted in fig. 3.6, the exploration well penetrated the gascap establishing the level of the gas oil contact. Thereafter, no further wells penetrated the gascap since it is not the intention to produce this gas but rather to let it expand and displace oil towards the producing wells, which are spaced in rows further downdip. As. a result there is uncertainty about the position of the sealing fault and hence in the value of m. The value of N, however, is fairly well defined from information obtained from the producing wells. Under these circumstances the best way to interpret equ. (3.24) is to plot F as a function of (Eo + mEg) for an assumed value of m. If the correct value has been chosen then the resulting plot should be a straight line passing through the origin with slope N, as shown in fig. 3.7. If the value of m selected is too small or too large, the plot will deviate above or below the line, respectively. In making this plot F can readily be calculated, at various times, as a function of the production terms Np and Rp, and the PVT parameters for the current pressure, the latter being also required to determine Eo and Eg. Alternatively, if N is unknown and m known with a greater degree of certainty, then N can be obtained as the slope of the straight line. One advantage of this particular interpretation is that the straight line must pass through the origin which therefore acts as a control point. EXERCISE 3.4 GASCAP DRIVE The gascap reservoir shown in fig. 3.6 is estimated, from volumetric calculations, to have had an initial oil volume N of 115 × 106 stb. The cumulative oil production MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 88 F (rb) m - too small correct value of m m - too large (E + mE ) (rb / stb) o g Fig. 3.7 (a) Graphical method of interpretation of the material balance equation to determine the size of the gascap (Havlena and Odeh) Np and cumulative gas oil ratio Rp are listed in table 3.1, as functions of the average reservoir pressure, over the first few years of production. (Also listed are the relevant PVT data, again taken from table 2.4, under the assumption that, for this particular application, pI = pb = 3330 psia). Pressure psia Np MMstb Rp scf/stb Bo rb/stb Rs scf/stb Bg rb/scf 3330 (pi = pb) 1.2511 510 .00087 3150 3.295 1050 1.2353 477 .00092 3000 5.903 1060 1.2222 450 .00096 2850 8.852 1160 1.2122 425 .00101 2700 11.503 1235 1.2022 401 .00107 2550 14.513 1265 1.1922 375 .00113 2400 17.730 1300 1.1822 352 .00120 TABLE 3.1 The size of the gascap is uncertain with the best estimate, based on geological information, giving the value of m = 0.4. Is this figure confirmed by the production and pressure history? If not, what is the correct value of m? EXERCISE 3.4 SOLUTION Using the technique of Havlena and Odeh the material balance for a gascap drive reservoir can be expressed as F = N (Eo + mEg ) (3.24) where F, Eo and Eg are defined in equs. (3.8 − 10). The values of these parameters, based on the production, pressure and PVT data of table 3.1, are listed in table 3.2. MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 89 Pressure psia F MM rb Eo rb/stb Eg rb/stb m = .4 Eo + mEg m = .5 m = .6 3330 (pi) 3150 5.807 .01456 .07190 .0433 .0505 .0577 3000 10.671 .02870 .12942 .0805 .0934 .1064 2850 17.302 .04695 .20133 .1275 .1476 .1677 2700 24.094 .06773 .28761 .1828 .2115 .2403 2550 31.898 .09365 .37389 .2432 .2806 .3180 2400 41.130 .12070 .47456 .3105 .3580 .4054 TABLE 3.2 The theoretical straight line for this problem can be drawn in advance as the line which, passes through the origin and has a slope of 115 × 106 stb, fig. 3.7 (b). When the plot is made of the data in table 3.2 for the value of m = 0.4, the points lie above the required line indicating that this value of m is too small. This procedure has been repeated for values of m = 0.5 and 0.6 and, as can be seen in fig. 3.7 (b), the plot for m = 0.5 coincides with the required straight line. Application of this technique relies critically upon the fact that N is known. Otherwise all three plots in fig. 3.7 (b), could be interpreted as straight lines, although the plots for m = .4 and .6 do have slight upward and downward curvature, respectively. Therefore, if there is uncertainty in the value of N, the three plots could be interpreted as m = 0.4 N = 132 × 106 stb m = 0.5 N = 114 × 106 stb m = 0.6 N = 101 × 106 stb If there is uncertainty in both the value of N and m then Havlena and Odeh suggest that equ. (3.24) should be re-expressed as g o o E F N mN E E = + a plot of F/Eo versus Eg/E o should then be linear with intercept N (when Eg/E o = 0) and slope mN. Thus for the data given in tables 3.1 and 3.2 Pressure psia F/Eo stb Eg/Eo 3330 (pi) 3150 398.8 × 106 4.938 3000 371.8 4.509 2850 368.5 4.288 2700 355.7 4.246 2550 340.6 3.992 2400 340.8 3.932 TABLE 3.3 MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 90 The plot of F/Eo versus Eg/E o is shown in fig. 3.7 (c), drawn over a limited range of each variable. The least squares fit for the six data points is the solid line which can be expressed by the equation g 6 o o E F (108.9 58.8 ) 10 stb E E = + × and therefore according to this interpretation N = 108.9 × 106 stb, and m = 0.54 F (MMrb) 40 30 20 20 10 0 0 .1 .2 .3 .4 correct straight line for N = 115 MMstb. x m = .4 o m = .5 m = .6 (b) (Eo + mEg) (rb / stb) (c) 4.0 4.5 5.0 F Eo (MMstb) 400 350 300 Eg Eo F Eg Eo Eo = (108.9 + 58.8 × 106 Fig. 3.7 (b) and (c); alternative graphical methods for determining m and N (according to the technique of Havlena and Odeh) MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 91 Both methods tend to confirm that the volumetric estimate of the oil in place is probably correct, within about 6%, and the gascap size is between m = 0.5 and 0.54. With the slight scatter in the production data it is not meaningful to try and state these figures with any greater accuracy. This estimate is made after the production of 17.7 million stb of oil or 15% recovery. As more production data become available the estimates of N and m can be revised. The pressure and production history of a typical gascap drive reservoir, under primary recovery conditions, are shown in fig. 3.8. pressure watercut time pi R = Rsi producing GOR Fig. 3.8 Schematic of the production history of a typical gascap drive reservoir Because of the gascap expansion, the pressure decline is less severe than for a solution gas drive reservoir and generally the oil recovery is greater, typically being in the range of 25−35 %, dependent on the size of the gascap. The peaks in the producing gas oil ratio curve are due to gas oil ratio (GOR) control being exercised. As the gascap expands the time will come when the updip wells start to produce gascap gas and the uppermost row of wells may have to be closed, both for the beneficial effect of keeping the gas in the reservoir and also to avoid gas disposal problems. Just as described in sec. 3.5 for a solution gas drive reservoir, if the economics are favourable water and/or gas injection will enhance the ultimate recovery. 3.7 NATURAL WATER DRIVE Natural water drive, as distinct from water injection, has already been qualitatively described, in Chapter 1, sec. 7, in connection with the gas material balance equation. The same principles apply when including the water influx in the general hydrocarbon reservoir material balance, equ. (3.7). A drop in the reservoir pressure, due to the production of fluids, causes the aquifer water to expand and flow into the reservoir. Applying the compressibility definition to the aquifer, then Water Influx = Aquifer Compressibility × Initial volume of water × Pressure Drop MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 92 or We = (cw+cf) Wi ∆p (3.25) in which the total aquifer compressibility is the direct sum of the water and pore compressibilities since the pore space is entirely saturated with water. The sum of cw and cf is usually very small, say 10-5/psi, therefore, unless the volume of water Wi is very large the influx into the reservoir will be relatively small and its influence as a drive mechanism will be negligible. If the aquifer is large, however, equ. (3.25) will be inadequate to describe the water influx. This is because the equation implies that the pressure drop ∆p, which is in fact the pressure drop at the reservoir boundary, is instantaneously transmitted throughout the aquifer. This will be a reasonable assumption only if the dimensions of the aquifer are of the same order of magnitude as the reservoir itself. For a very large aquifer there will be a time lag between the pressure change in the reservoir and the full response of the aquifer. In this respect natural water drive is time dependent. If the reservoir fluids are produced too quickly, the aquifer will never have a chance to "catch up" and therefore the water influx, and hence the degree of pressure maintenance, will be smaller than if the reservoir were produced at a lower rate. To account for this time dependence in water influx calculations requires a knowledge of fluid flow equations and the subject will therefore be deferred until Chapter 9, in which a full description of the phenomenon is provided. For the moment, the simple equation (3.25) will be used to illustrate the influence of water influx in the material balance. Using the technique of Havlena and Odeh (assuming that Bw = 1), the full material balance can be expressed as F = N (Eo + mEg + Ef,w) + We (3.12) in which the term Ef,w, equ. (3.11), can frequently be neglected when dealing with a water influx. This is not only for the usual reason that the water and pore compressibilities are small but also because a water influx helps to maintain the reservoir pressure and therefore, the ∆p appearing in the Ef,w term is reduced. This is a point which should be checked at the start of any material balance calculation (refer exercise 9.2). If, in addition, the reservoir has no initial gascap then equ. (3.12) can be reduced to F = NEo + We (3.26) In attempting to use this equation to match the production and pressure history of a reservoir, the greatest uncertainty is always the determination of the water influx We. In fact, in order to calculate the influx the engineer is confronted with what is inherently the greatest uncertainty in the whole subject of reservoir engineering. The reason is that the calculation of We requires a mathematical model which itself relies on the knowledge of aquifer properties. These, however, are seldom measured since wells are not deliberately drilled into the aquifer to obtain such information. For instance, suppose the influx could be described using the simple model presented as equ. (3.25). Then, if the aquifer shape is radial, the water influx can be calculated as MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 93 2 2 e w f e o W (c c ) (r r )fh p π φ = + − ∆ (3.27) in which re and ro are the radii of the aquifer and reservoir, respectively, and f is the fractional encroachment angle which is either Θ/2π or Θ/360°, depending on whether Θ is expressed in radians or degrees. It should be realised that the only term in equ. (3.27) which is known with any degree of certainty is π! The remaining terms all carry a high degree of uncertainty. For instance, what is the correct value of re? Is the aquifer continuous for 20 kilometers or is it truncated by faulting? What is the correct value of h, the average thickness of the aquifer or φ, the porosity? These can only be estimated, based on the values determined in the oil reservoir. For such reasons, building a correct aquifer model to match the production and pressure data of the reservoir is always done on a "try it and see" basis and even when a satisfactory model has been achieved it is seldom if ever, unique. Therefore, the most appropriate way of applying equ. (3.26) is by expressing it as e o o W F N E E = + (3.28) and plotting F/Eo, corresponding to the observed production, versus We/Eo, where We is calculated using an aquifer model such as equ. (3.27). we/ Eo (stb) we - too small we - correct we - too large incorrect geometry F E (stb) o 45° N Fig. 3.9 Trial and error method of determining the correct aquifer model (Havlena and Odeh) This model is linked to the reservoir by the pressure drop term ∆p which is interpreted as the pressure drop at the original reservoir-aquifer boundary, and is normally assumed to be equal to the average pressure drop in the reservoir due to the production of fluids. If the aquifer model is incorrect, the plotted data points will deviate from the theoretical straight line which has a slope of 45° and intercept N, when We/Eo = 0, as shown in fig. 3.9. The deviation labelled as being due to using the wrong geometry means that radial geometry has been assumed whereas linear geometry would probably be more appropriate. With radial geometry there is a larger body of water in close proximity to MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 94 the reservoir, for the same aquifer volume, than for a linear aquifer and, as a result, response of the radial aquifer is greater causing deviation below the theoretical straight line. Exercise 9.2 provides an example of this technique in which the aquifer model used for calculating We caters for time dependence. Once a satisfactory aquifer model has been obtained by history matching, the same model can hopefully be used in predicting reservoir performance for any scheduled offtake policy. As already mentioned, however, there are so many uncertainties involved that the aquifer model is hardly ever unique and its validity should be continually checked as fresh production and pressure data become available. If the reservoir has a gascap then equ. (3.12) has the form F = N (Eo + mEg) + We which can alternatively be expressed as e o g o g W F N (E mE ) (E mE ) = + + + (3.29) in which it is assumed that both m and N are known. By plotting F/(Eo + mEg) versus We/(Eo + mEg) the interpretation is similar to that shown in fig 3.9. Equation (3.29) demonstrates how the technique of Havlena and Odeh can be applied to a combination drive reservoir in which there are three active mechanisms, solution gas drive, gascap drive and water drive. The pressure and production history of an undersaturated reservoir under active water drive are shown in fig. 3.10. The pressure decline is relatively small due to the expansion of the aquifer water and from the producing gas oil ratio plot, it is evident that the pressure is being maintained above the saturation pressure. Recovery from water drive reservoirs can be very high, in excess of 50%, but just as in the case of the flooded out gas reservoir described in Chapter 1, sec. 7, residual oil will now be trapped behind the advancing water which can only be recovered by resorting to more advanced recovery methods, as described in Chapter 4, sec. 9. pressure watercut time pi Rsi GOR (R ≈ Rsi) MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 95 Fig. 3.10 Schematic of the production history of an undersaturated oil reservoir under strong natural water drive 3.8 COMPACTION DRIVE AND RELATED PORE COMPRESSIBILITY PHENOMENA The withdrawal of liquid or gas from a reservoir results in a reduction in the fluid pressure and consequently an increase in the effective or grain pressure, the latter being defined in Chapter 1, sec. 3, as the difference between the overburden and fluid pressures. This increased pressure between the grains will cause the reservoir to compact and this in turn can lead to subsidence at the surface. Various studies7,8,9,10 have shown that compaction depends only upon the difference between the vertically applied stress (overburden) and the internal stress (fluid pressure) and therefore, compaction can conveniently be measured in the laboratory by increasing the vertical stress on a rock sample while keeping the fluid pressure in the pores constant. If Vb is the bulk volume of a rock sample of thickness h, then the uniaxial compaction ∆Vb/Vb = ∆h/h can best be determined in the laboratory using the triaxial compaction cell described by Teeuw11, which is shown in fig. 3.11 (a). The core sample, which is completely saturated with water, is contained in a cell which has permeable cap and base plates and a cylindrical, flexible sleeve surrounding it. Vertical stress is applied by means of a piston while the fluid pressure in the pores is maintained at one atmosphere. The pressure in the fluid surrounding the flexible sleeve can be increased independently so as to maintain the condition of vertical stress lateral stress sample permeable disc elastic sleeve A B grain pressure h ∆ h (b) (a) Fig. 3.11 (a) Triaxial compaction cell (Teeuw); (b) typical compaction curve MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 96 zero lateral strain on the sample. This pressure is continually adjusted so that any change in vertical thickness of the sample ∆h is uniformly related to the measured water expelled from the porous rock. If such an experiment were performed on an uncompacted sample of sand and the compaction ∆h/h plotted as a function of the applied vertical stress which, considering the fluid pressure is maintained at one atmosphere, is equivalent to the grain pressure, then the result would be as shown in fig. 3.11 (b). The slope of this curve, at any point, is b f h / p c c h φ ∆ ∆ = ≈ refer equ. (1.37) The characteristic shape of this compaction curve is intuitively what one would expect. At low grain pressures the compressibility of the uncompacted sample is very high since it is relatively easy to effect a closer packing of the grains at this stage. As the grain pressure increases, however, it becomes progressively more difficult to compact the sample further and the compressibility decreases. What is clear from such an experiment is that the bulk or pore compressibility of a reservoir is not constant but will continually change as fluids are withdrawn and the grain pressure increases. Under normal hydrostatic conditions, since both the overburden and water pressures increase linearly with depth, then so too does the grain pressure which is the difference between the two. Thus a reservoir whose initial condition corresponds to point A will normally be buried at shallow depth, while a reservoir corresponding to point B will be buried deeper. Compaction drive is the expulsion of reservoir fluids due to the dynamic reduction of the pore volume and will only be significant as a drive mechanism if the pore compressibility cf is large. It therefore follows that such a drive mechanism will normally only provide a significant increase in the primary hydrocarbon recovery in shallow reservoirs. In parts of the Bachaquero field, Venezuela, as reported by Merle, et al12, the compaction drive mechanism accounts for more than 50% of total oil recovery. This large reservoir dips between 1000−4000 ft. and has uniaxial compressibilities in excess of 100 × 10-6/psi. If the mechanics of reservoir compaction were as simple as described above, it would appear possible to derive a relationship between uniaxial compressibility and depth, for various types of typical reservoir rock, in an attempt to apply such a correlation universally. Unfortunately, the process of compaction is frequently irreversible which in turn implies that in-situ compressibility cannot be estimated in such a simple manner. If the reservoir rock consists of well cemented grains in a rigid rock frame then the compaction, over a limited pressure range, will be approximately elastic and reversible. In loose unconsolidated sands, however, compaction is both inelastic and irreversible since upon each reloading cycle on such a sample, in a repeated loading experiment in a triaxial cell, it is possible for the individual grains to be packed in a different configuration than on the previous cycle and, in addition, some of the grains can suffer permanent mechanical deformation due to crushing. The effect of this inelastic MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 97 deformation in the reservoir is shown in fig. 3.12, which is taken from the paper of Merle et al.12 COMPACTION ∆h h unloading COMPACTION DUE TO PRODUCTION START PRODUCTION BURIAL DEPOSITION grain pressure C C′ ′ ′ ′ D B A Fig. 3.12 Compaction curve illustrating the effect of the geological history of the reservoir on the value of the in-situ compressibility (after Merle) When the reservoir sand is initially being deposited it is at point A on the compaction curve, fig. 3.12. Over geological times, as more and more material is deposited, the original sand becomes buried corresponding to point B, with grain pressure pB. Following this normal deposition, events can occur which will reduce the grain pressure below pB, such as: -uplifting of the reservoir -erosion of the surface layers above the reservoir -overpressuring of the fluid in the reservoir. As a result of one or more of these effects, in the extreme cases of either completely elastic or completely inelastic deformation of the rock during deposition, the reservoir in fig. 3.12 will be either at C or C′, respectively, corresponding to the reduced grain pressure pC In the former case, for elastic deformation, if the reservoir is produced with an initial grain pressure pC then the compaction will start immediately since the uniaxial compressibility at point C is finite. In the completely inelastic case, however, there. will be a time lag between starting to produce the reservoir and the occurrence of any significant degree of compaction. This is because the uniaxial compressibility in this latter case is the tangent to the compaction curve at point C′, which is extremely small. As shown in fig. 3.12, there will be very little compaction in the reservoir until sufficient fluids have been removed to increase the grain pressure to pB which is the maximum grain pressure experienced by the reservoir in the past. MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 98 Compaction, and its associated effect of surface subsidence, will be much more pronounced for shallow, unconsolidated reservoirs than for the deeper, more competent sands. It is therefore necessary to experimentally determine the compressibility of shallow reservoir sands in order to estimate to what degree compaction will enhance the hydrocarbon recovery, and also, to enable the prediction of the resulting surface subsidence, which can cause serious problems if the surface location of the field is adjacent to the sea or a lake13. Unfortunately, the deformation of unconsolidated sands is usually inelastic and this in turn leads to complications in relating laboratory measured compressibilities to the in-situ values in the reservoir. The nature of the problem can be appreciated by referring again to fig. 3.12. Suppose that both the grain and fluid pressures in a reservoir are normal so that under initial conditions the reservoir is at point B on the compaction curve. The process of cutting a core and raising it to the surface will cause unloading, which for a rock which deforms inelastically, will place the core at point C′, which lies off the normal compaction curve. During the re-loading the horizontal path B−C′ is not reversed, instead there is a mechanical hysteresis effect which means that the true compaction curve is not re-joined until point D, where pD > pB. As a result, the laboratory measured compressibility, determined as the slope of the line C′−D at pressure pB, will be somewhat lower than the in-situ value, which is the slope of the normal curve at pB. Thus, initial values of the in-situ compressibility are difficult to determine and usually require estimation by back extrapolation of laboratory values obtained for grain pressures in excess of pD. The above description of the various complications in estimating in-situ, uniaxial compressibility has been applied for the extreme case of a perfectly inelastic reservoir rock. Generally rock samples are neither perfectly elastic or inelastic but somewhere in between. Nevertheless, the same qualitative arguments apply and it is therefore not always meaningful to merely estimate in-situ compressibilities by reference to published charts for typical sandstones and limestones. REFERENCES 1) Schilthuis, R.J., 1936. Active Oil and Reservoir Energy. Trans., AIME, 118: 33-52. 2) Amyx, J.W., Bass, D.M., and Whiting, R.L., 1960. Petroleum Reservoir Engineering - Physical Properties. McGraw-Hill: 448-472. 3) McCain, W.D., 1973. The Properties of Petroleum Fluids. Petroleum Publishing Company, Tulsa: 268-305. 4) Havlena, D. and Odeh, A.S., 1963. The Material Balance as an Equation of a Straight Line. J.Pet.Tech. August: 896-900. Trans., AIME, 228. 5) Havlena, D. and Odeh, A.S., 1964. The Material Balance as an Equation of a Straight Line. Part II - Field Cases. J.Pet.Tech. July: 815-822. Trans., AIME., 231. MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 99 6) Kingston, P.E. and Niko, H., 1975. Development Planning of the Brent Field. J.Pet. Tech. October: 1190-1198. 7) Geertsma, J., 1966. Problems of Rock Mechanics in Petroleum Production Engineering. Proc.,1st Cong. of the Intl. Soc. of Rock Mech., Lisbon. l, 585. 8) Geertsma, J., 1957. The Effect of Fluid Pressure Decline on Volumetric Changes of Porous Rocks. Trans., AIME, 210: 331-340. 9) van der Knaap, W., 1959. Non-linear Behaviour of Elastic Porous Media. Trans., AIME, 216: 179-187. 10) Biot, M.A., 1941. General Theory of Three Dimensional Consolidation. J.Appl. Phys., Vol.12: 155. 11) Teeuw, D., 1971. Prediction of Formation Compaction from Laboratory Compressibility Data. Soc. of Pet. Eng. J., September: 263-271. 12) Merle, H.A., Kentie, C.J.P., van Opstal, G.H.C. and Schneider, G.M.G., 1976. The Bachaquero Study - A Composite Analysis of the Behaviour of a Compaction Drive/ Solution Gas Drive Reservoir. J.Pet.Tech. September: 1107-1115. 13) Geertsma, J., 1973. Land Subsidence Above Compacting Reservoirs. J.Pet.Techn. June: 734-744. CHAPTER 4 DARCY'S LAW AND APPLICATIONS 4.1 INTRODUCTION Darcy's empirical flow law was the first extension of the principles of classical fluid dynamics to the flow of fluids through porous media. This chapter contains a simple description of the law based on experimental evidence. For a more detailed theoretical treatment of the subject, the reader is referred to the classical paper by King Hubbert1 in which it is shown that Darcy's law can be derived from the Navier-Stokes equation of motion of a viscous fluid. The significance of Darcy's law is that it introduces flow rates into reservoir engineering and, since the total surface oil production rate from a reservoir is p res dN q dt = it implicitly introduces a time scale in oil recovery calculations. The practical application of this aspect of Darcy's law is demonstrated in the latter parts of the chapter in which a brief description is given of the fundamental mechanics of well stimulation and enhanced oil recovery. 4.2 DARCY'S LAW; FLUID POTENTIAL Every branch of science and engineering has its own particular heroes, one only has to think, for example, of the hallowed names of Newton and Einstein in physics or Darwin in the natural sciences. In reservoir engineering, our equivalent is the nineteenth century French engineer Henry Darcy who, although he didn't realise it, has earned himself a special place in history as the first experimental reservoir engineer. In 1856 Darcy published a detailed account of his work2 in improving the waterworks in Dijon and, in particular, on the design of a filter large enough to process the town's daily water requirements. Although fluid dynamics was a fairly advanced subject in those days, there were no published accounts of the phenomenon of fluid flow through a porous medium and so, being a practical man, Darcy designed a filter, shown schematically in fig. 4.1, in an attempt to investigate the matter. The equipment consisted of an iron cylinder containing an unconsolidated sand pack, about one metre in length, which was held between two permeable gauze screens. Manometers were connected into the cylinder immediately above and below the sand pack. By flowing water through the pack Darcy established that, for any flow rate, the velocity of flow was directly proportional to the difference in manometric heights, the relationship being DARCY'S LAW AND APPLICATIONS 101 constant rate water injection q cc/sec sand pack water collection and measurement mercury manometers h1 l h2 Fig. 4.1 Schematic of Darcy's experimental equipment 1 2 h h h u K K I I − ∆ = = (4.1) where u = flow velocity in cm/sec, which is the total measured flow rate q cc/sec, divided by the cross-sectional area of the sand pack ∆h = difference in manometric levels, cm (water equivalent) I = total length of the sand pack, cm, and K = constant. Darcy's only variation in this experiment was to change the type of sand pack, which had the effect of altering the value of the constant K; otherwise, all the experiments were carried out with water and therefore, the effects of fluid density and viscosity on the flow law were not investigated. In addition the iron cylinder was always maintained in the vertical position. Subsequently, others repeated Darcy's experiment under less restrictive conditions, and one of the first things they did was to orientate the sand pack at different angles with respect to the vertical, as shown in fig. 4.2. It was found, however, that irrespective of the orientation of the sand pack, the difference in height, ∆h, was always the same for a given flow rate. Thus Darcy's experimental law proved to be independent of the direction of flow in the earth's gravitational field. DARCY'S LAW AND APPLICATIONS 102 water manometers q cc / sec + z z l h ∆h datum plane; z = 0, p = 1 atm. Fig. 4.2 Orientation of Darcy's apparatus with respect to the Earth's gravitational field It is worthwhile considering the significance of the ∆h term appearing in Darcy's law. The pressure at any point in the flow path, fig. 4.2, which has an elevation z, relative to the datum plane, can be expressed in absolute units as p = ρg(h-z) with respect to the prevailing atmospheric pressure. In this equation h is the liquid elevation of the upper manometer, again, with respect to z = 0 and ρ is the liquid (water) density. The equation can be alternatively expressed as p hg ( gz) ρ = + (4.2) If equ. (4.1) is written in differential form as dh u K dl = (4.3) then differentiating equ. (4.2) and substituting in equ. (4.3) gives K d p K d(hg) u gz g dl g dl ρ = + = (4.4) The term ( p ρ + gz), in this latter equation, has the same units as hg which are: distance × force per unit mass, that is, potential energy per unit mass. This fluid potential is usually given the symbol Φ and defined as the work required, by a DARCY'S LAW AND APPLICATIONS 103 frictionless process, to transport a unit mass of fluid from a state of atmospheric pressure and zero elevation to the point in question, thus p 1 atm dp gz ρ − Φ = + (4.5) Although defined in this way, fluid potentials are not always measured with respect to atmospheric pressure and zero elevation, but rather, with respect to any arbitrary base pressure and elevation (pb, zb ) which modifies equ. (4.5) to b p b p dp g(z z ) ρ Φ = + − (4.6) The reason for this is that fluid flow between two points A and B is governed by the difference in potential between the points, not the absolute potentials, i.e. A B A b b B p p p A B A b B b A B p p p dp dp dp g(z z ) g(z z ) g(z z ) ρ ρ ρ Φ −Φ = + − − + − = + − It is therefore conventional, in reservoir engineering to select an arbitrary, convenient datum plane, relative to the reservoir, and express all potentials with respect to this plane. Furthermore, if it is assumed that the reservoir fluid is incompressible (ρ independent of pressure) then equ. (4.5) can be expressed as p gz ρ Φ = + (4.7) which is precisely the term appearing in equ. (4.4). It can therefore be seen that the h term in Darcy's equation is directly proportional to the difference in fluid potential between the ends of the sand pack. The constant K/g is only applicable for the flow of water, which was the liquid used exclusively in Darcy's experiments. Experiments performed with a variety of different liquids revealed that the law can be generalised as k d u dl ρ µ Φ = (4.8) in which the dependence of flow velocity on fluid density ρ and viscosity µ is fairly obvious. The new constant k has therefore been isolated as being solely dependent on the nature of the sand and is described as the permeability. It is, in fact, the absolute permeability of the sand, provided the latter is completely saturated with a fluid and, because of the manner of derivation, will have the same value irrespective of the nature of the fluid. This latter statement is largely true, under normal reservoir pressures and flow conditions, the exception being for certain circumstances encountered in real gas flow. At very low pressures there is a slippage between the gas molecules and the walls of DARCY'S LAW AND APPLICATIONS 104 each pore leading to an apparent increased permeability. This phenomenon, which is called the Klinkenberg effect3, seldom enters reservoir engineering calculations but is important in laboratory experiments in which, for convenience, rock permeabilities are determined by measuring air flow rates through core plugs at pressures close to atmospheric. This necessitates a correction to determine the absolute permeability4. Due to its very low viscosity, the flow velocity of a real gas in a reservoir is much greater than for oil or water. In a limited region around the wellbore, where the pressure drawdown is high, the gas velocity can become so large that Darcy's law does not fully describe the flow.5 This phenomenon, and the manner of its quantification in flow equations for gas, will be fully described in Chapter 8, sec. 6. 4.3 SIGN CONVENTION Darcy's empirical law was described in the previous section without regard to sign convention, it being assumed that all terms in equ. (4.8) were positive. This is adequate if the law is being used independently to calculate flow rates; however, if equ. (4.8) is used in conjunction with other mathematical equations then, just as described in connection with the definition of thermodynamic compressibility in Chapter 1, sec. 4, attention must be given to the matter of sign convention. Linear flow If distance is measured positive in the direction of flow, then the potential gradient dΦ/dl must be negative in the same direction since fluids move from high to low potential. Therefore, Darcy's law is k d u dl ρ µ Φ = − (4.9) Radial flow If production from the reservoir into the well is taken as positive, which is the convention adopted in this book, then, since the radius is measured as being positive in the direction opposite to the flow, dΦ/dr is positive and Darcy's law may be stated as k d u dr ρ µ Φ = (4.10) 4.4 UNITS: UNITS CONVERSION In any absolute set of units Darcy's equation for linear flow is k d u dl ρ µ Φ = (4.9) in which the various parameters have the following dimensions u = L/T; ρ = M/L3; µ = M/LT; I = L and Φ (potential energy/unit mass) = L2/T2. Therefore, the following dimensional analysis performed on equ. (4.9): DARCY'S LAW AND APPLICATIONS 105 [ ] [ ] [ ] 3 2 2 M/L L / T L k T M/LT L = reveals that [k] = [L2] Thus the unit of permeability should be the cm2 in cgs units, or the metre2 in Sl units. Both these units are impracticably large for the majority of reservoir rock, as will be demonstrated in exercise 4.1, and therefore, a set of units was devised in which the permeability would have a more convenient numerical size. These are the so-called "Darcy units" (refer table 4.1) in which the unit of permeability is the Darcy. The latter was defined from the statement of Darcy's law for horizontal, linear flow of an incompressible fluid k dp u dl µ = − (4.11) such that k = 1 Darcy when u = 1 cm/sec; µ = 1 cp; and dp/dl = 1 atmosphere/cm. Inspection of table 4.1 reveals that the units are a hybrid system based on the cgs units. The only difference being that pressure is expressed in atmospheres, viscosity in cp (centipoise) and, as a consequence, the permeability in Darcies. It was intended, in defining this system of units, that not only would the unit of permeability have a reasonable numerical value but also, equations expressed in these units would have the same form as equations in absolute units. That is, there would be no awkward constants involved in the equations other than multiples of π which reflect the geometry of the system. Unfortunately, this latter expectation is not always fulfilled because the Darcy, defined through the use of equ. (4.11), is based on an incomplete statement of Darcy's law. Certainly, equ. (4.11) has the same form whether expressed in absolute or Darcy units but considering the general statement of the flow law, equ. (4.9), applied to an incompressible fluid (ρ ≈ constant), then DARCY'S LAW AND APPLICATIONS 106 Absolute units Hybrid units Parameter Symbol Dimensions cgs SI Darcy Field Length I L cm metre cm ft Mass M M gm kg gm lb Time t T sec sec sec hr Velocity u L/T cm/sec metre/sec cm/sec ft/sec Rate q L3/T cc/sec metre3/sec cc/sec Pressure p (ML/T2)/L2 dyne/cm2 Newton/meter2 (Pascal) atm stb / d (liquid) Mscf / d (gas) psia Density ρ M/L3 gm/cc kg/metre3 gm/cc Ib/cu.ft Viscosity µ M/LT gm/cm.sec (Poise) kg/metre.sec cp cp Permeability k L2 cm2 metre2 Darcy mD TABLE 4.1 Absolute and hybrid systems of units used in Petroleum Engineering DARCY'S LAW AND APPLICATIONS 107 k d k dp dz u g dl dl dl ρ ρ µ µ Φ = = − + (4.12) in absolute units, while 6 k dp g dz u dl dl 1.0133 10 ρ µ = − + × (4.13) in Darcy units. The constant 1.0133 × 106 is the number of dyne/cm2 in one atmosphere and is required because both ρ and g have the same units in both the cgs and Darcy systems, and yet, the second term within the parenthesis of equ. (4.13) must have the same units as the first, namely, atm/cm. In spite of this obvious drawback, reservoir engineers tend to work theoretically using equations expressed in Darcy units. This practice will generally be adhered to in this text and, in the remaining chapters, the majority of the theoretical arguments will be developed with equations expressed in these units. When dealing with the more practical aspects of reservoir engineering, such as well test analysis described in Chapters 7 and 8, it is conventional to switch to what are called practical, or field units. The word practical is applied to such systems because all the units employed are of a convenient magnitude. There are no rules governing field units which therefore vary between countries and companies. The set of such units presented in table 4.1 is, however, probably the most widely accepted in the industry at the time of writing this book. Because of the wide variation in unit systems employed by the industry, it is very important that reservoir engineers should be adept at converting equations expressed in Darcy units to the equivalent form in field units, or for that matter, any other set of units. There is a systematic approach in making such conversions which, if rigorously applied, will exclude the possibility of error. Consider, as an example, the conversion of equ. (4.11) from Darcy to field units. Since q = u(cm/sec) × A(cm2) the equation can be expressed in more practical form, in Darcy units, as 2 k(D) A(cm ) dp q(cc / sec) (atm/ cm) (cp) dl µ = − (4.14) which, when converted to field units will have the form 2 k(mD) A(ft ) dp q(std/ d) (constant) (psi/ ft) (cp) dl µ = − (4.15) in which the same symbols are used in both equations. Making the conversion amounts to evaluating the constant in equ. (4.15) and this can be achieved simply by remembering that equations must balance. Thus, if q in DARCY'S LAW AND APPLICATIONS 108 equ. (4.14) is, say, 200 reservoir cc/sec, then the left hand side of equ. (4.15) must also have the numerical value of 200, even though q in the latter is in stb/d, i.e. conversion q(stb / d) q(r.cc / sec) factor × = which is satisfied by r.cc / sec q(stb / d) q(r.cc / sec) stb / d = This preserves the balance on the left hand side of both equations. The conversion factor can be expanded as r.cc / sec r.cc / sec rb / d stb / d rb / d stb / d = = Applying this method throughout, then and since 2 2 2 D cm atm k mD A ft mD ft psi stb r.cc / sec rb / d dp psi q cm d rb / d stb / d (cp) dl ft ft D 1 cm atm 1 ; 30.48 and ; equ.(4.16) mD 1000 ft psi 14.7 µ × = − × = = = (4.16) can be evaluated as 3 o kA dp q 1.127 10 (stb / d) dl µ − = − × (4.17) EXERCISE 4.1 UNITS CONVERSION 1) What is the conversion factor between k, expressed in Darcies, and in cm2 and metre2, respectively. 2) Convert the full equation for the linear flow of an incompressible fluid, which in Darcy units is 6 kA dp g dz q dl dl 1.0133 10 ρ µ = − + × to field units. DARCY'S LAW AND APPLICATIONS 109 EXERCISE 4.1 SOLUTION 1) For linear, horizontal flow of an incompressible fluid 2 k(D) A(cm ) dp q(cc / sec) (atm / cm) (Darcyunits) (cp) dl µ = − and 2 2 2 k(cm ) A(cm ) dp (dyne / cm ) q(cc / sec) (Absolute cgs units) (poise) dl (cm) µ = − The former equation can be converted from Darcy to cgs, absolute units by balancing both sides of the resulting equation, as follows 2 2 2 2 2 atm D (dyne / cm ) k(cm ) A(cm ) dyne / cm dp cm q(cc / sec) dl (cm) cp (poise) poise µ = − and evaluating the conversion factors [ ] 2 2 2 6 D k(cm ) A dp 1 cm q (dyne / cm ) (poise) 100 dl 1.0133 10 µ = − × or 2 2 8 D k(cm ) A dp 1 cm q dl 1.0133 10 µ = − × But the numerical constant in this equation must be unity, therefore 8 2 D 1.0133 10 cm = × so that 1 Darcy ≈ 10-8 cm2 = 10-12 metre2. It is proposed that the industry will eventually convert to Sl (Système Internationale) absolute units, (table 4.1), in which case the basic unit of permeability will be the metre2. Because this is such an impracticably large unit, it has been tentatively suggested6 that a practical unit, the micrometre2 (µm2), be "allowable" within the new system. Since 1 µm 2 = 10-12 m2 then 1 Darcy ≈ 1 µm 2 DARCY'S LAW AND APPLICATIONS 110 It is also suggested that both the Darcy and milli-Darcy be retained as allowable terms. 2) For horizontal flow, the conversion from Darcy to field units of the first part of the flow equation is 3 o kA dp q 1.127 10 B dl µ − = − × (4.17) To convert the gravity term, using the conventional manner described in the text, is rather tedious but can be easily achieved in an intuitive manner. The second term, (ρg/1.0133×106) dz/dl, must, upon conversion to field units, have the units psi/ft. The only variable involved in this latter term is ρ, the fluid density. If this is expressed as a specific gravity γ, then, since pure water has a pressure gradient of 0.4335 psi/ft, the gravity term can be expressed as 0.4335γ dz dl psi/ft Furthermore, adopting the sign convention which will be used throughout this book, that z is measured positively in the upward, vertical direction, fig. 4.2, and if θ is the dip angle of the reservoir measured counter-clockwise from the horizontal then dz dl = sin θ and the full equation, in field units, becomes 3 o kA dp q 1.127 10 0.4335 sin B dl γ θ µ − = − × + (4.18) 4.5 REAL GAS POTENTIAL The fluid potential function was defined in section 4.2, in absolute units as b p p dp gz ρ Φ = + (4.6) and for an incompressible fluid (ρ ≈ constant) as p gz ρ Φ = + (4.7) Liquids are generally considered to have a small compressibility but the same cannot be said of a real gas and therefore, it is worthwhile investigating the application of the potential function to the description of gas flow. The density of a real gas can be expressed (in absolute units) as Mp ZRT ρ = (1.27) DARCY'S LAW AND APPLICATIONS 111 and substituting this in equ. (4.6) gives the real gas potential as b p p RT Zdp gz M p Φ = + (4.19) But, since RT Z dp d dp gdz gdz M p ρ Φ = + = + (4.20) then the gradient of the gas potential in the flow direction is simply d 1 dp dz g dl dl dl ρ Φ = + (4.21) and Darcy's equation for linear flow is again k d k dp dz u g dl dl dl ρ ρ µ µ Φ = − = − + (4.12) The above merely illustrates that real gas flow can be described using precisely the same form of equations as for an incompressible liquid. 4.6 DATUM PRESSURES An alternative way of expressing the potential of any fluid is ψ = ρΦ = p + ρg z where ψ is the psi-potential and has the units-potential per unit volume. Using this function, Darcy's law becomes kA d kA d q dl dl ρ ψ µ µ Φ = − = − (4.22) The ψ potential is also frequently referred to as the "datum pressure", since the function represents the pressure at any point in the reservoir referred to the datum plane, as illustrated in fig. 4.3. DARCY'S LAW AND APPLICATIONS 112 + z arbitrary datum plane (z=z) o ψA = pA + ρg (zA − z0) ψB = pB + ρg (zB − z0) B (pB, zB) B (pA, zA) Fig. 4.3 Referring reservoir pressures to a datum level in the reservoir, as datum pressures (absolute units) Suppose pressures are measured in two wells, A and B, in a reservoir in which an arbitrary datum plane has been selected at z = z0. If the pressures are measured with respect to a datum pressure of zero, then as shown in fig. 4.3, the calculated values of ψA and ψB are simply the observed pressures in the wells referred to the datum plane, i.e. ψA = (absolute pressure)A + (gravity head)A In a practical sense it is very useful to refer, pressures measured in wells to a datum level and even to map the distribution of datum pressures throughout the reservoir. In this way the potential distribution and hence direction, of fluid movement in the reservoir can be seen at a glance since the datum pressure distribution is equivalent to the potential distribution. 4.7 RADIAL STEADY STATE FLOW; WELL STIMULATION The mathematical description of the radial flow of fluids simulates flow from a reservoir, or part of a reservoir, into the wellbore. For the radial geometry shown in fig. 4.4, flow will be described under what is called the steady state condition. This implies that, for a well producing at a constant rate q; dp/dt = 0, at all points within the radial cell. Thus the outer boundary pressure pe and the entire pressure profile remain constant with time. This condition may appear somewhat artificial but is realistic in the case of a pressure maintenance scheme, such as water injection, in which one of the aims is to keep the pressure constant. In such a case, the oil withdrawn from the radial cell is replaced by fluids crossing the outer boundary at r = re. DARCY'S LAW AND APPLICATIONS 113 pressure q = constant rw re r pwf q p = constant e Fig. 4.4 The radial flow of oil into a well under steady state flow conditions In addition, for simplicity, the reservoir will be assumed to be completely homogeneous in all reservoir parameters and the well perforated across the entire formation thickness. Under these circumstances, Darcy's law for the radial flow of single phase oil can be expressed as kA dp q dr µ = (4.23) Since the flow rate is constant, it is the same across any radial area, A = 2πrh, situated at distance r from the centre of the system. Therefore, equ. (4.23) can be expressed as 2 rkh dp q dr π µ = and separating the variables and integrating wf w p r p r q dr dp 2 kh r µ π = where pwf is the conventional symbol for the bottom hole flowing pressure. The integration results in wf w q r p p ln 2 kh r µ π − = (4.24) which shows that the pressure increases logarithmically with respect to the radius, as shown in fig. 4.4, the pressure drop being consequently much more severe close to the well than towards the outer boundary. In particular, when r = re then e e wf w r q p p ln 2 kh r µ π − = (4.25) When a well is being drilled it is always necessary to have a positive pressure differential acting from the wellbore into the formation to prevent inflow of the reservoir fluids. Because of this, some of the drilling mud will flow into the formation and the DARCY'S LAW AND APPLICATIONS 114 particles suspended in the mud can partially plug the pore spaces, reducing the permeability, and creating a damaged zone in the vicinity of the wellbore. The situation is shown in fig. 4.5, in which ra represents the radius of this zone. If pressure r w r e r a p e q r p skin ∆ Fig. 4.5 Radial pressure profile for a damaged well the well were undamaged, the pressure profile for r < ra would be as shown by the dashed line, whereas due to the reduction in permeability in the damaged zone, equ. (4.25) implies that the pressure drop will be larger than normal, or that pwf will be reduced. This additional pressure drop close to the well has been defined by van Everdingen7 as skin q p S 2 kh µ π ∆ = (4.26) in which the ∆pskin is attributed to a skin of reduced permeability around the well and S is the mechanical skin factor, which is just a dimensionless number. This definition can be included in equ. (4.25) to give the total steady state inflow equation as e e wf w r q p p ln S 2 kh r µ π − = + (4.27) in which it can be seen that if S is positive then pe - pwf the pressure drawdown, contains the additional pressure drop due to the perturbing effect of the skin. Since equ. (4.27) is frequently employed by production engineers, it is useful to express it in field units rather than the Darcy units in which it was derived. The reader should check that this will give o e e wf w q B r p p 141.2 In S kh r µ − = + (4.28) in which the geometrical factor 2π has been absorbed in the constant. This equation is frequently expressed as DARCY'S LAW AND APPLICATIONS 115 e wf 3 e o w q oil rate (stb / d) PI p p pressure drawdown (psi) 7.08 10 kh r B ln S r µ − = = − × = + (4.29) where the PI, or Productivity Index of a well, expressed in stb/d/psi, is a direct measure of the well performance. One of the aims of production engineering is to make the PI of each well as large as is practically possible, consistent with the economics of doing so. This is termed well stimulation. The ways in which a well can be stimulated can be deduced by considering how to vary the individual parameters in equ. (4.29) so as to increase the PI. The various methods are summarised below. a) Removal of Skin (S) Before making any capital expenditure to remove a positive mechanical skin, it is first necessary to check that the formation has in fact been damaged during drilling. This can best be done by performing a pressure buildup test, which is normally carried out as routine, immediately after completing the well. The manner in which S can be calculated in the analysis of such a test is detailed in Chapter 7. sec. 7. If it is determined that S is positive, the formation damage can be reduced by acid treatment. The type of acid used depends on the nature of the reservoir rock and the type of plugging materials which must be removed. If the formation is limestone, treatment with hydrochloric acid will invariably remove the skin because of the solubility of the rock itself. In sandstone reservoirs, in which the rock matrix is not soluble, special, so-called, mud acids are used. As a result of a successful acid job, the skin factor can be reduced to zero or may even become negative. b) Increasing the effective permeability (k) As noted al ready, due to the logarithmic increase of pressure with radius, the main part of the pressure drawdown occurs close to the well. Therefore, if the effective permeability in this region of high drawdown can be increased, the productivity can be considerably enhanced. This can be achieved by hydraulic fracturing, in which high fluid pressures maintained in the wellbore will induce vertical fractures in the formation. Once the fractures have been initiated, they can be propagated deep into the formation by increasing the wellbore pressure and injecting a suitable fracturing fluid, carrying granular propping agents. In carbonate reservoirs the same effect can be achieved by fracture-acidising. c) Viscosity reduction (µ) If the oil viscosity is very high, the flow rate in the reservoir will be correspondingly low, and the time scale attached to the recovery will be greatly extended. The viscosity can be siginificantly reduced by raising the temperature of the oil, a typical viscosity-temperature relation being shown in fig. 4.6(a). The thermal stimulation process DARCY'S LAW AND APPLICATIONS 116 applied to effect this viscosity reduction is steam soaking. Steam is injected into the reservoir and, in the simple model shown in fig. 4.6(b), extends to a radius rh, the magnitude of which is primarily a function of the amount of steam injected, usually several thousand tons, over a period of several days. During injection heat is lost in the wellbore and to the cap and base rock, but since steam is used, these losses are reflected as a reduction in latent heat and therefore take place without a significant change of temperature. 150 0 hot zone rw rh re p (b) (a) 150 T (°F) 400 2 cp µ (cp) µ w µ o Fig. 4.6 (a) Typical oil and water viscosities as functions of temperature, and (b) pressure profile within the drainage radius of a steam soaked well Following injection, the well is opened on production and the cold oil crossing into the heated annular region has its viscosity greatly reduced and consequently the PI is increased. A typical steam soak production rate, in comparison to the unstimulated rate, is shown in fig. 4.7. There is an initial surge in production followed by a steady decline as the temperature in the hot zone is reduced, due to the continual loss of heat to the cap and base rock, as a function of time, and the removal of heat with the produced fluids. When the production rate declines towards the unstimulated rate, the cycle is repeated. steam soak production unstimulated production time (yrs) oil rate (stb/d) 1 2 3 100 10 Fig. 4.7 Oil production rate as a function of time during a multi-cycle steam soak DARCY'S LAW AND APPLICATIONS 117 The flow equations during the initial and later part of a steam soak cycle will be described in Chapter 9, sec. 6 and Chapter 6, sec. 4, respectively, since they serve as an interesting example of the flexibility of radial flow equations. d) Reduction of the oil formation volume factor (Bo) As already described in Chapter 2, sec. 4; Bo (rb/stb) can be minimised by choosing the correct surface separator, or combination of separators. e) Reduction in the ratio re / rw Since re / rw appears as a logarithmic term, it has little influence on the PI and alteration of the ratio by, for instance, underreaming the wellbore to increase rw, is seldom considered as a means of well stimulation. f) Increasing the well penetration (h) It was assumed in deriving equ. (4.29) that the well was completed across the total formation thickness and thus the flow was entirely radial. If the well is not fully penetrating, there is a distortion of the radial flow pattern close to the well giving rise to an additional pressure drawdown. This is generally accounted for by using the full formation thickness in equ. (4.29) and including the effect of partial penetration as an additional skin factor. The method of calculating this additional skin is described in Chapter 7, sec. 9. Increasing the well penetration, if possible, will obviously increase the Pl but in many cases wells are deliberately completed over a restricted part of the reservoir to avoid excessive gas or water production from individual sands, or to prevent coning. The methods for stimulating the production of a well, described in this section, do not necessarily increase the ultimate oil recovery from the reservoir, but rather, reduce the time in which the recovery is obtained. As such, they are generally regarded as acceleration projects which speed up the production, thus having a favourable effect on the discounted cash flow. There are exceptions. For instance, if a well has stopped producing, then any stimulation which results in oil production can be regarded as increasing the recovery. These methods, however, should be distinguished from the enhanced recovery techniques, described in sec. 4.9, in which the reservoir is energised to increase the recovery. In stimulation there is frequently no net energy increase in the reservoir. In steam soaking, for instance, heat energy is supplied to the reservoir and is subsequently lost during the production cycle; as opposed to continuous steam drive, in which the aim is to keep the steam in the reservoir thus increasing the total energy of the system. 4.8 TWO-PHASE FLOW: EFFECTIVE AND RELATIVE PERMEABILITIES In describing Darcy's law, it has so far been assumed that the permeability is a rock property which is a constant, irrespective of the nature of the fluid flowing through the pores. This is correct (with the noted exception of gas flow either at low pressures or very high rates) provided that the rock is completely saturated with the fluid in question, DARCY'S LAW AND APPLICATIONS 118 and results from defining k in equ. (4.8) as the permeability, rather than the K in equ. (4.3), the latter having a dependence on the fluid properties. The permeability so defined is termed the absolute permeability. If there are two fluids, such as oil and water, flowing simultaneously through a porous medium, then each fluid has its own, so-called, effective permeability. These permeabilities are dependent on the saturations of each fluid and the sum of the effective permeabilities is always less than the absolute permeability. The saturation dependence of the effective permeabilities of oil and water is illustrated in fig. 4.8(a). It is conventional to plot both permeabilities as functions of the water saturation alone since the oil saturation is related to the former by the simple relationship So = 1−Sw. Considering the effective permeability curve for water, two points on this curve are known. When Sw = Swc, the connate or irreducible water saturation, the water will not flow and kw = 0. Also, when Sw = 1 the rock is entirely saturated with water and kw = k, the absolute permeability. Similarly for the oil, when Sw = 0 (So = 1) then ko = k and, when the oil saturation decreases to Sor, the residual saturation, there will be no oil flow and ko = 0. In between these limiting values, for both curves, the effective permeability functions assume the typical shapes shown in fig. 4.8(a). The main influence on the shapes of the curves appears to be the wettability, that is, which fluid preferentially adheres to the rock surface8. Although it is difficult to quantify this influence, the permeability curves can be measured in laboratory experiments for the wettability conditions prevailing in the reservoir9. 0 0 0 0 1 1 1 0 0 0 0 1 absolute permeability k k kw 1-Sor k′ro k′rw krw Swc Sw ko 1-Sor Swc Sw Fig. 4.8 (a) Effective and (b) corresponding relative permeabilities, as functions of the water saturation. The curves are appropriate for the description of the simultaneous flow of oil and water through a porous medium The effective permeability plots can be normalised by dividing the scales by the value of the absolute permeability k to produce the relative permeabilities o w w w ro w rw w k (S ) k (S ) k (S ) and k (S ) k k = = (4.30) DARCY'S LAW AND APPLICATIONS 119 The plots of kro and krw, corresponding to the effective permeability plots of fig. 4.8(a), are drawn in fig. 4.8(b). Both sets of curves have precisely the same shape, the only difference being that the relative permeability scales have the range zero to unity. Relative permeabilities are used as a mathematical convenience since in a great many displacement calculations the ratio of effective permeabilities appears in the equations, which can be simplified as the ratio of o w ro w ro w w w rw w rw w k (S ) k k (S ) k (S ) k (S ) k k (S ) k (S ) × = = × In figs. 4.8(a) and (b) the parts of the curves for water saturations below Sw = Swc and above Sw = 1 - Sor are drawn as dashed lines because, although these sections of the plots can be determined in laboratory experiments, they will never be encountered in fluid displacement in the reservoir, since the practical range of water saturations is Swc ≤ Sw ≤ 1—Sor The maximum relative permeabilities to oil and water that can naturally occur during displacement are called the end-point relative permeabilities and defined as (fig. 4.8(b)), ro ro w wc k k (at S S ) ′ = = and rw rw w or k k (at S 1 S ) ′ = = − (4.31) Sometimes the effective permeability curves are normalised in a different manner than described above, by dividing the scales of fig. 4.8(a) by the value of ko (Sw = Swc ) = k ro k′ , the maximum effective permeability to oil. The resulting curves are shown in fig. 4.9. S 1 1 K 0 0 0 0 ro Krw SWC or 1 S W Fig. 4.9 Alternative manner of normalising the effective permeabilities to give relative permeability curves DARCY'S LAW AND APPLICATIONS 120 In this case, the normalised relative permeability curves are defined as o w w w ro w rw w o w wc o w wc k (S ) k (S ) K (S ) and K (S ) k (S S ) k (S S ) = = = = (4.32) To describe the simultaneous flow of oil and water in the reservoir, applying Darcy's law, the absolute permeability k, implicitly used in the earlier sections of this chapter, must be replaced by the effective permeabilities ko(Sw) and kw(Sw) respectively. Using the alternative methods of normalising the effective permeability curves, the required permeabilities can be expressed as either ko (Sw) = kkro (Sw) or ko (Sw) = ko (Sw = Swc) Kro (Sw) (4.33) and kw (Sw) = kkrw (Sw) or kw (Sw) = ko (Sw = Swc) Krw (Sw) both interpretations, naturally, giving the same values of the effective permeabilities. As already mentioned, in many equations describing the displacement of one immiscible fluid by another it is the ratio of effective permeabilities which is required, and from equ. (4.33) this can be expressed as ro ro rw rw k K k K = (4.34) To complicate matters further, in the literature, it is not normal to distinguish between the two ways of presenting relative permeability curves by assigning one of them a capital letter; both interpretations are denoted by the symbol kr. In this text, the relative permeabilities used will be those obtained by normalising the effective permeability curves with the absolute permeability (fig. 4.8(b)). Relative permeabilities are measured in the laboratory by studying the displacement of oil by water (or gas) in very thin core plugs, in which it is safe to assume that the fluid saturations are uniformly distributed with respect to thickness. Therefore, these laboratory-measured, or rock-relative permeability relationships, can only be used directly to describe flow in a reservoir in which the saturations are also uniformly distributed with respect to thickness. In the majority of practical cases, however, there is a non-uniform water saturation distribution in the vertical direction which is governed by capillary and gravity forces and, therefore, there must also be a relative permeability distribution with respect to thickness. Because of this, the rock-relative permeabilities can seldom be used directly in field displacement calculations. Practically the whole of Chapter 10 is devoted to describing methods of generating averaged (or pseudo) relative permeabilities, as functions of the thickness averaged water saturation. These are used to describe the displacement of oil by water in a more realistic fashion, taking account of the manner in which the fluid saturations are distributed, with respect to thickness, as they simultaneously move through the reservoir. DARCY'S LAW AND APPLICATIONS 121 Although the above description of the concept of relative permeability has been restricted to a two phase oil-water system, the same general principle applies to any two phase system such as gas-oil or gas-water. 4.9 THE MECHANICS OF SUPPLEMENTARY RECOVERY Supplementary recovery results from increasing the natural energy of the reservoir, usually by displacing the hydrocarbons towards the producing wells with some injected fluid. By far the most common fluid injected is water because of its availability, low cost and high specific gravity which facilitates injection. The basic mechanics of oil displacement by water can be understood by considering the mobilities of the separate fluids. The mobility of any fluid is defined as r kk λ µ = (4.35) which, considering Darcy's law, can be seen to be directly proportional to the velocity of flow. Also included in this expression is the term k r/µ, which is referred to as the relative mobility. The manner in which water displaces oil is illustrated in fig. 4.10 for both an ideal and non-ideal linear horizontal waterflood. IDEAL x x NON-IDEAL Sw (a) (b) 1−Sor 1−Sor Swc Swc Sw Fig. 4.10 Water saturation distribution as a function of distance between injection and production wells for (a) ideal or piston-like displacement and (b) non-ideal displacement In the ideal case there is a sharp interface between the oil and water. Ahead of this, oil is flowing in the presence of connate water (relative mobility = kro (Sw=Swc )/µo = ro k′ /µo), while behind the interface water alone is flowing in the presence of residual oil (relative mobility = krw(Sw=1 - Sor)/µw = ro k′ /µw). This favourable type of displacement will only occur if the ratio DARCY'S LAW AND APPLICATIONS 122 rw w ro o k / M 1 k / µ µ ′ = ≤ ′ where M is known as the end point mobility ratio and, since both and ro k′ are rw k′ the end point relative permeabilities, is a constant. If M ≤ 1 it means that, under an imposed pressure differential, the oil is capable of travelling with a velocity equal to, or greater than, that of the water. Since it is the water which is pushing the oil, there is therefore, no tendency for the oil to be by-passed which results in the sharp interface between the fluids. The displacement shown in fig. 4.10(a) is, for obvious reasons, called "piston-like displacement". Its most attractive feature is that the total amount of oil that can be recovered from a linear reservoir block will be obtained by the injection of the same volume of water. This is called the movable oil volume where, 1 (MOV) = PV(1 – Sor − Swc) The non-ideal displacement depicted in fig. 4.10(b), which unfortunately is more common in nature, occurs when M > 1. In this case, the water is capable of travelling faster than the oil and, as the water pushes the oil through the reservoir, the latter will be by-passed. Water tongues develop leading to the unfavourable water saturation profile. Ahead of the water front oil is again flowing in the presence of connate water. This is followed, in many cases, by a waterflood front, or shock front, in which there is a discontinuity in the water saturation. There is then a gradual transition between the shock front saturation and the maximum saturation Sw = 1−Sor. The dashed line in fig. 4.10(b) depicts the saturation distribution at the time when the shock front breaks through into the producing well (breakthrough). In contrast to the piston-like displacement, not all of the movable oil will have been recovered at this time. As more water is injected, the plane of maximum water saturation (Sw = 1−Sor) will move slowly through the reservoir until it reaches the producing well at which time the movable oil volume has been recovered. Unfortunately, in typical cases it may take five or six MOV's of injected water to displace the one MOV of oil (as will be demonstrated in exercises 10.2 and 10.3 of Chapter 10). At a constant rate of water injection, the fact that much more water must be injected, in the unfavourable case, protracts the time scale attached to the oil recovery and this is economically unfavourable. In addition, pockets of by-passed oil are created which may never be recovered. Mobility control If the end point mobility ratio for water displacing oil is unfavourable, the injection project can be engineered to overcome this difficulty. The manner in which this is done can be appreciated by considering the general expression rd d ro o k / Mobility of the displacing fluid M Mobility of the displaced fluid k / µ µ ′ = = ′ (4.36) DARCY'S LAW AND APPLICATIONS 123 where the subscript "d" refers to the displacing fluid, which need not necessarily be water. To improve the displacement efficiency, M should be reduced to a value of unity or less which will have the effect of converting the displacement from the type shown in fig. 4.10(b), to the ideal type shown in fig. 4.10(a); this is referred to as "mobility control". The methods by which M can be reduced are. Polymer flooding (increase, µd) Polymers, such as polysaccharide, are dissolved in the injection water, this raises its viscosity, thus reducing the mobility of the water. Polymer flooding will not only accelerate the oil recovery but can also increase it, in comparison to a normal water drive, because the by-passing of oil is greatly reduced. Thermal methods (decrease, µo/µd) For very viscous crudes the ratio of µo / µd can be of the order of thousands (which means that M has the same order of magnitude) and therefore, water drive cannot be considered as a feasible project (refer Chapter 10, exercise 10.1). In such cases the viscosity ratio can be drastically reduced by increasing the temperature, as shown in fig. 4.6(a). This is achieved by one of the following methods: -hot water injection -steam injection -in-situ combustion. Although mobility control is the primary aim in applying thermal methods, there are other factors involved than merely the reduction of, µo / µd (where in this case, µd is the viscosity of the hot water or steam and differs from µw at normal reservoir temperature). In many cases distillation of the crude occurs, the lighter fractions of the oil being vapourised and providing a miscible flood in advance of the thermal front. Expansion of the oil on heating will also add to the recovery. Thermal methods can therefore be considered as basically secondary recovery processes with some tertiary side effects, such as the crude distillation, which tends to reduce the residual oil saturation. Tertiary flooding Tertiary flooding aims at recovering the oil remaining in the reservoir after a conventional secondary recovery project, such as a water drive. Oil and water are immiscible (do not mix) and as a result there is a finite surface tension at the interface between the fluids. This, in turn, leads to the trapping of oil droplets within each separate pore which is the normal state after a waterflood. From a strictly mechanical point of view, the methods commonly employed in tertiary flooding can be appreciated by considering fig. 4.11, which shows an enlargement of an oil relative permeability curve (solid line) for water-oil displacement, in the vicinity of the residual oil saturation point. After a water drive kro is zero when So = Sor, point A, and the oil will not flow. Two possibilities for improving the situation are indicated which amount to altering the oil relative permeability characteristics. The first of these is to displace the oil with a DARCY'S LAW AND APPLICATIONS 124 fluid which is soluble in it, thus increasing the oil saturation above Sor. This is equivalent to moving from point A to B on the normal relative permeability curve. As a result kro is finite and the oil becomes mobile. Alternatively, flooding can be carried out with a fluid which is miscible, or partially miscible, with the oil thus eliminating surface tension, or in some way modifying the interfacial properties, between the displacing fluid and the oil. This reduces the residual oil saturation to a very low value, or S′ in fig. 4.11, and alters the oil relative permeability curve, as shown by the dashed line. In this case, when the displacing fluid contacts the residual oil left after the waterflood, the effect is that the oil relative permeability increases from zero to point C and again the residual oil becomes mobile. B A C So Sor S’or kro Fig. 4.11 Illustrating two methods of mobilising the residual oil remaining after a conventional waterflood Obviously the second method appears the more favourable since it creates the possibility of recovering practically all of the residual oil. In the first case, only part of each swollen oil droplet is recovered. Tertiary floods generally aim at either total miscibility or else a combination of the methods described above. The ways in which such floods can be engineered are many and varied, some of the more popular being, Miscible (LPG) flooding The oil is displaced by one of the LPG (Liquid Petroleum Gas) products, ethane, propane or butane. If the reservoir conditions are such that the LPG is in the liquid phase then it is miscible with the oil and theoretically all the residual oil can be recovered. Carbon Dioxide flooding Carbon dioxide has a critical temperature of 88°F and is therefore normally injected into the reservoir as a gas. It is highly soluble in oil and this has two favourable effects. In the first place the saturation of the oil droplets, containing dissolved CO2, increases above the residual saturation, Sor, the oil permeability becomes finite and oil starts to flow. Secondly, the viscosity of the oil is reduced resulting in better mobility control. In addition the carbon dioxide, by extracting light hydrocarbons from the oil, displays miscible properties. DARCY'S LAW AND APPLICATIONS 125 Surfactant flooding (Micellar Solution flooding) Surfactants, or surface acting agents, when dissolved in minute quantities in water have a significant influence on the interfacial properties between the water and oil which it is displacing. The surfactant dissolves in the residual oil droplets thus raising the saturation above the residual level and, in addition, the surface tension between these enlarged oil droplets and the displacing water is very significantly reduced. Both these effects are active in reducing the residual oil saturation and, in laboratory tests, ninety percent residual oil recovery has been observed. The surfactants most commonly used by the industry are petroleum sulphonates. The above description of tertiary recovery mechanisms hardly "scratches the surface" of the subject. For an excellent, simplified description the reader is referred to the set of papers by Herbeck, Heintz and Hastings10, which cover all aspects of the subject including the vitally important economic considerations. The above methods are described as tertiary in that they are capable of recovering some, if not all, of the residual oil remaining after a waterflood. This does not mean, however, that they must be preceded by a waterflood. Instead, the two can be conducted simultaneously. In all tertiary recovery schemes, continuous injection of the expensive agents is unnecessary. The fluids are injected in batches and frequently the batches are followed by mobility buffers. For instance, to ensure stable displacement in a surfactant flood, the chemical slug can be displaced by water thickened with a polymer, the concentration of which is gradually decreased as the flood proceeds. REFERENCES 1) King Hubbert, M., 1956. Darcy's Law and the Field Equations of the Flow of Underground Fluids. Trans. AIME, 207: 222-239. 2) Darcy, H., 1856. Les Fontaines Publiques de la Ville de Dijon. Victor Dalmont, Paris. 3) Klinkenberg, L.J., 1941. The Permeability of Porous Media to Liquids and Gases. API Drill. and Prod. Prac.: 200. 4) Cole, F.W., 1961. Reservoir Engineering Manual. Gulf Publishing Co., Houston, Texas: 19-20. 5) Geertsma, J., 1974. Estimating the Coefficient of Inertial Resistance in Fluid Flow Through Porous Media. Soc.Pet.Eng.J., October: 445-450. 6) Campbell, J.M., 1976. Report on Tentative SPE Metrication Standards. Paper presented to the 51st Annual Fall Conference of the AIME, New Orleans, October. 7) van Everdingen, A.F., 1953. The Skin Effect and Its Impediment to Fluid Flow into a Wellbore. Trans. AIME, 198: 171-176. 8) Craig, F.F., Jr., 1971. The Reservoir Engineering Aspects of Waterflooding. SPE Monograph: DARCY'S LAW AND APPLICATIONS 126 9) Amyx, J.W., Bass, D.M. and Whiting, R. L., 1960. Petroleum Reservoir Engineering - Physical Properties. McGraw-Hill: 86-96. 10) Herbeck, E.F., Heintz, R.C. and Hastings, J.R., 1976. Fundamentals of Tertiary Oil Recovery. Series of articles appearing in the "Petroleum Engineer", January-September. CHAPTER 5 THE BASIC DIFFERENTIAL EQUATION FOR RADIAL FLOW IN A POROUS MEDIUM 5.1 INTRODUCTION In this chapter the basic equation for the radial flow of a fluid in a homogeneous porous medium is derived as 1 k p p r c r r r t ρ φ ρ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.1) This equation is non-linear since the coefficients on both sides are themselves functions of the dependent variable, the pressure. In order to obtain analytical solutions, it is first necessary to linearize the equation by expressing it in a form in which the coefficients have a negligible dependence upon the pressure and can be considered as constants. An approximate form of linearization applicable to liquid flow is presented at the end of the chapter in which equ. (5.1) is reduced to the form of the radial diffusivity equation. Solutions of this equation and their applications for the flow of oil are presented in detail in Chapters 6 and 7. For the flow of a real gas, however, a more complex linearization by integral transformation is required which will be presented separately in Chapter 8. 5.2 DERIVATION OF THE BASIC RADIAL DIFFERENTIAL EQUATION The basic differential equation will be derived in radial form thus simulating the flow of fluids in the vicinity of a well. Analytical solutions of the equation can then be obtained under various boundary and initial conditions for use in the description of well testing and well inflow, which have considerable practical application in reservoir engineering. This is considered of greater importance than deriving the basic equation in cartesian coordinates since analytical solutions of the latter are seldom used in practice by field engineers. In numerical reservoir simulation, however, cartesian geometry is more commonly used but even in this case the flow into or out of a well is controlled by equations expressed in radial form such as those presented in the next four chapters. The radial cell geometry is shown in fig. 5.1 and initially the following simplifying assumptions will be made. a) The reservoir is considered homogeneous in all rock properties and isotropic with respect to permeability. b) The producing well is completed across the entire formation thickness thus ensuring fully radial flow. c) The formation is completely saturated with a single fluid. RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 128 dr rw r h qρ \| r + dr qρ \| r re Fig. 5.1 Radial flow of a single phase fluid in the vicinity of a producing well. Consider the flow through a volume element of thickness dr situated at a distance r from the centre of the radial cell. Then applying the principle of mass conservation Mass flow rate IN − Mass flow rate OUT = Rate of change of mass in the volume element r dr qρ + − r qρ = 2 rh dr t ρ π φ ∂ ∂ where 2πrhφdr is the volume of the small element of thickness dr. The left hand side of this equation can be expanded as r r (q ) q dr q 2 rh dr r t ρ ρ ρ ρ π φ ∂ ∂ + − = ∂ ∂ which simplifies to (q ) 2 rh r t ρ ρ π φ ∂ ∂ = ∂ ∂ (5.2) By applying Darcy's Law for radial, horizontal flow it is possible to substitute for the flow rate q in equ. (5.2) since 2 khr p q r π µ ∂ = ∂ giving 2 khr p 2 rh r r t π ρ ρ π φ µ ∂ ∂ ∂ = ∂ ∂ ∂ or 1 k p r r r t ρ ρ ρ φ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.3) The time derivative of the density appearing on the right hand side of equ. (5.3) can be expressed in terms of a time derivative of the pressure by using the basic thermodynamic definition of isothermal compressibility RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 129 1 V c V p ∂ = − ∂ and since m V ρ = then the compressibility can be alternatively expressed as m 1 c m p ρ ρ ρ ρ ρ ∂ ∂ = − = ∂ ∂ (5.4) and differentiating with respect to time gives p c t t ρ ρ ∂ ∂ = ∂ ∂ (5.5) Finally, substituting equ. (5.5) in equ. (5.3) reduces the latter to 1 k p p r c r r r t ρ φ ρ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.1) This is the basic, partial differential equation for the radial flow of any single phase fluid in a porous medium. The equation is referred to as non-linear because of the implicit pressure dependence of the density, compressibility and viscosity appearing in the coefficients kρ /µ and φ cρ. Because of this, it is not possible to find simple analytical solutions of the equation without first linearizing it so that the coefficients somehow lose their pressure dependence. A simple form of linearization applicable to the flow of liquids of small and constant compressibility (undersaturated oil) will be considered in sec. 5.4, while a more rigorous method, using the Kirchhoff integral transformation, will be presented in Chapter 8 for the more complex case of linearization for the flow of a real gas. 5.3 CONDITIONS OF SOLUTION In principle, an infinite number of solutions of equ. (5.1 ) can be obtained depending on the initial and boundary conditions imposed. The most common and useful of these is called the constant terminal rate solution for which the initial condition is that at some fixed time, at which the reservoir is at equilibrium pressure pi, the well is produced at a constant rate q at the wellbore, r = rw. This type of solution will be examined in detail in Chapters 7 and 8 but it is appropriate, at this stage, to describe the three most common, although not exclusive, conditions for which the constant terminal rate solution is sought. These conditions are called transient, semi-steady state and steady state and are each applicable at different times after the start of production and for different, assumed boundary conditions. a) Transient condition RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 130 This condition is only applicable for a relatively short period after some pressure disturbance has been created in the reservoir. In terms of the radial flow model this disturbance would be typically caused by altering the well's production rate at r = rw. In the time for which the transient condition is applicable it is assumed that the pressure response in the reservoir is not affected by the presence of the outer boundary, thus the reservoir appears infinite in extent. The condition is mainly applied to the analysis of well tests in which the well's production rate is deliberately changed and the resulting pressure response in the wellbore is measured and analysed during a brief period of a few hours after the rate change has occurred. Then, unless the reservoir is extremely small, the boundary effects will not be felt and the reservoir is, mathematically, infinite. This gives rise to a complex solution of equ. (5.1) in which both the pressure and pressure derivative, with respect to time, are themselves functions of both position and time, thus and p g(r,t) p f(r,t) t = ∂ = ∂ Transient analysis techniques and their application to oil and gas well testing will be described in Chapters 7 and 8, respectively. b) Semi-Steady State condition q = constant Pressure pwf rw r re = constant = 0, at r = re pe ∂p ∂r Fig. 5.2 Radial flow under semi-steady state conditions This condition is applicable to a reservoir which has been producing for a sufficient period of time so that the effect of the outer boundary has been felt. In terms of the radial flow model, the situation is depicted in fig. 5.2. It is considered that the well is surrounded, at its outer boundary, by a solid "brick wall" which prevents the flow of fluids into the radial cell. Thus at the outer boundary, in accordance with Darcy's law e p 0 at r r r ∂ = = ∂ (5.6) Furthermore, if the well is producing at a constant flow rate then the cell pressure will decline in such a way that p t ∂ ≈ ∂ constant, for all r and t. (5.7) RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 131 The constant referred to in equ. (5.7) can be obtained from a simple material balance using the compressibility definition, thus dp dV cV q dt dt = − = − (5.8) or dp q dt cV = − (5.9) which for the drainage of a radial cell can be expressed as 2 e dp q dt c r h π φ = − (5.10) This is a condition which will be applied in Chapter 6, for oil flow, and in Chapter 8, for gas flow, to derive the well inflow equations under semi-steady state conditions, even though in the latter case the gas compressibility is not constant. One important feature of this stabilized type of solution, when applied to a depletion type reservoir, has been pointed out by Matthews, Brons and Hazebroek1 and is illustrated in fig. 5.3. This is the fact that, once the reservoir is producing under the semi-steady state condition, each well will drain from within its own no-flow boundary quite independently of the other wells. For this condition dp/dt must be approximately constant throughout the entire reservoir otherwise flow would occur across the boundaries causing a re-adjustment in their positions until stability was eventually achieved. In this case a simple technique can be applied to determine the volume averaged reservoir pressure i i i res i i p V p V = (5.11) in which th i th i V the pore volume of the i drainage volume and p the average pressure within the i drainage volume = = Equation (5.9) implies that since dp/dt is constant for the reservoir then, if the variation in the compressibility is small RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 132 p , V 4 p , V 3 q3 q4 p , V 1 q p , V 2 q2 3 4 2 1 1 Fig. 5.3 Reservoir depletion under semi-steady state conditions. qi ∝ Vi (5.12) and hence the volume average in equ. (5.11) can be replaced by a rate average, as follows i i i res i i p q p q = (5.13) and, whereas the Vi's are difficult to determine in practice, the qi's are measured on a routine basis throughout the lifetime of the field thus facilitating the calculation of res p , which is the pressure at which the reservoir material balance is evaluated. The method by which the individual i p 's can be determined will be detailed in Chapter 7. sec. 7. c) Steady State condition q = constant = 0 Pressure pe = constant fluid index re r ∂p ∂t rw pwf Fig. 5.4 Radial flow under steady state conditions The steady state condition applies, after the transient period, to a well draining a cell which has a completely open outer boundary. It is assumed that, for a constant rate of production, fluid withdrawal from the cell will be exactly balanced by fluid entry across the open boundary and therefore, p = pe = constant, at r = re (5.14) p and t ∂ ∂ = 0 for all r and t (5.15) RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 133 This condition is appropriate when pressure is being maintained in the reservoir due to either natural water influx or the injection of some displacing fluid (refer Chapter 10). It should be noted that the semi-steady state and steady state conditions may never be fully realised in the reservoir. For instance, semi-steady state flow equations are frequently applied when the rate, and consequently the position of the no-flow boundary surrounding a well, are slowly varying functions of time. Nevertheless, the defining conditions specified by equs. (5.7) and (5.15) are frequently approximated in the field since both production and injection facilities are usually designed to operate at constant rates and it makes little sense to unnecessarily alter these. If the production rate of an individual well is changed, for instance, due to closure for repair or increasing the rate to obtain a more even fluid withdrawal pattern throughout the reservoir, there will be a brief period when transient flow conditions prevail followed by stabilized flow for the new distribution of individual well rates. 5.4 THE LINEARIZATION OF EQUATION 5.1 FOR FLUIDS OF SMALL AND CONSTANT COMPRESSIBILITY A simple linearization of equ. (5.1) can be obtained by deletion of some of the terms, dependent upon making various assumptions concerning the nature of fluid for which solutions are being sought. In this section the fluid considered will be a liquid which, in a practical sense, will apply to the flow of undersaturated oil. Expanding the left hand side of equ. (5.1), using the chain rule for differentiation gives 2 2 1 k p k p k p k p p r r r c r r r r r r t r ρ ρ ρ ρ φ ρ µ µ µ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂ (5.16) and differentiating equ. (5.4) with respect to r gives p c r r ρ ρ ∂ ∂ = ∂ ∂ (5.17) which when substituted into equ. (5.16) changes the latter to 2 2 2 1 k p k p k p k p p r c r r c r r r r r t r ρ ρ ρ ρ φ ρ µ µ µ µ ∂ ∂ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ ∂ ∂ (5.18) For liquid flow, the following assumptions are conventionally made -the viscosity, µ is practically independent of pressure and may be regarded as a constant -the pressure gradient ∂p/∂r is small and therefore, terms of the order (∂p/∂r)2 can be neglected. These two assumptions eliminate the first two terms in the left hand side of equ. (5.18), reducing the latter to RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 134 2 2 p 1 p c p r r k t r φµ ∂ ∂ ∂ + = ∂ ∂ ∂ (5 19) which can be more conveniently expressed as 1 p c p r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (5.20) Making one final assumption, that the compressibility is constant, means that the coefficient φµc/k is also constant and therefore, the basic equation has been effectively linearized. For the flow of liquids the above assumptions are quite reasonable and have been frequently applied in the past. Dranchuk and Quon2, however, have shown that this simple linearization by deletion must be treated with caution and can only be applied when the product cp << 1 (5.21) This condition makes it necessary to modify the final assumption so that the compressibility is not just constant but both small and constant. The compressibility appearing in equ. (5.20) is the total, or saturation weighted, compressibility of the entire reservoir-liquid system ct = coSo + c wSwc + cf (5.22) in which the saturations are expressed as fractions of the pore volume. Using typical figures for the components of equ. (5.22) co = 10 × 10−6/psi Swc = 0.2 cw = 3 × 10−6/psi p = 3000 psi cf = 6 × 10−6/psi then ct in equ. (5.22) has the value 14.6×10−6/psi and the product expressed by equ. (5.21) has the value 0.04, which satisfies the necessary condition for this simple linearization to be valid. However, when dealing with reservoir systems which have a higher total compressibility it will be necessary to linearize equ. (5.1); using some form of integral transformation as detailed by Dranchuk and Quon. Such an approach will be required when describing the flow of a real gas since, in this case, the compressibility of the gas alone may, to a first approximation, be expressed as the reciprocal of the pressure and the cp product, equ. (5.21), will itself be unity. The linearization of equ. (5.1) under these circumstances will be described in Chapter 8, secs. 2 4. Before leaving the subject of compressibility, it should be noted that the product of φ and c in all the equations, in this and the following chapters, is conventionally expressed as φ abs ol ut e × (coSo + c wSwc + cf) (5.23) RADIAL DIFFERENTIAL EQUATION FOR FLUID FLOW 135 since it was assumed in deriving equ. (5.1) that the porous medium was completely saturated with a single fluid thus implying the use of the absolute porosity. Alternatively, allowing for the presence of a connate water saturation, the φc product can be interpreted as φ absol ute (1 − Swc) × o o w wc f wc (c S c S c ) (1 S ) + + − (5.24) in which φabsolute (1 − Swc) is the effective, hydrocarbon porosity, and the compressibility is equivalent to that derived in Chapter 3, equ. (3.19), which is used in conjunction with the hydrocarbon pore volume. In either event, the products expressed in equs. (5.23) and (5.24) have the same value, the reader must only be careful not to mix the individual terms appearing in the separate equations. Equation (5.20) is the radial diffusivity equation in which the coefficient k/φµc is called the diffusivity constant. This is an equation which is frequently applied in physics, for instance, the temperature distribution due to the conduction of heat in radial symmetry would be described by the equation 1 T 1 T r r r r K t ∂ ∂ ∂ = ∂ ∂ ∂ in which T is the absolute temperature and K the thermal diffusivity constant. Because of the general nature of equ. (5.20) it is not surprising that many reservoir engineering papers, when dealing with complex solutions of the diffusivity equation, make reference to a text book entitled "Conduction of Heat in Solids", by Carslaw and Jaeger3, which gives the solutions of the equation for a large variety of boundary and initial conditions and is regarded as a standard text in reservoir engineering. REFERENCES 1) Matthews, C.S., Brons, F. and Hazebroek, P., 1954. A Method for Determination of Average Pressure in a Bounded Reservoir. Trans. AlME. 201: 182-191. 2) Dranchuk, P.M. and Quon, D., 1967. Analysis of the Darcy Continuity Equation. Producers Monthly, October: 25-28. 3) Carslaw, H.S. and Jaeger, J.C., 1959. Conduction of Heat in Solids. Oxford at the Clarendon Press, (2nd edition). CHAPTER 6 WELL INFLOW EQUATIONS FOR STABILIZED FLOW CONDITIONS 6.1 INTRODUCTION In this chapter solutions of the radial diffusivity equation, for liquid flow, will be sought under stabilized flow conditions. These have already been defined in the previous chapter as semi-steady state and steady state for which the time derivative in equ. (5.20) is constant and zero, respectively. The solution technique for semi-steady state flow is set out in some detail since the method is a perfectly general one which can be applied for a variety of radial flow problems. Finally, the constraint that the outer boundary of the cell must be radial is removed by the introduction of Dietz shape factors. This allows a general form of inflow equation to be developed for a wide range of geometries of the drainage area and positions of the well within the boundary. 6.2 SEMI-STEADY STATE SOLUTION The radial diffusivity equation, (5.20), will be solved under semi-steady state flow conditions for the geometry and radial pressure distribution shown in fig. 6.1. r re q = constant Pressure rw pe pwf h Fig. 6.1 Pressure distribution and geometry appropriate for the solution of the radial diffusivity equation under semi-state conditions At the time when the solution is being sought the volume averaged pressure within the cell is p which can be calculated from the following simple material balance cV (pi − p) = qt (6.1) in which V is the pore volume of the radial cell, q is the constant production rate and t the total flowing time. The corresponding boundary pressures at the time of solution are pe at re and pwf at rw. For the drainage of a radial volume cell, the semi-steady state condition was derived in the previous chapter as 2 e p q r c r h π φ ∂ = − ∂ (5.10) which, when substituted in the radial diffusivity equation, (5.20), gives STABILIZED INFLOW EQUATIONS 137 2 e 1 p q r r r r r kh µ π ∂ ∂ = − ∂ ∂ (6.2) and integrating this equation 2 1 2 e p q r C t 2 r kh µ π ∂ = − + ∂ (6.3) where C1 is a constant of integration. At the outer no-flow boundary ∂p/∂r is zero and hence the constant can be evaluated as C1 = qµ/2πkh which, when substituted in equ. (6.3), gives 2 e p q 1 r r 2 kh r r µ π ∂ = − ∂ (6.4) Integrating once again 2 r 2 e wf w r p q r p lnr 2 kh 2 r p r µ π = − (6.5) or 2 r wf 2 w e q r r p p ln 2 kh r 2r µ π − = − (6.6) in which the term 2 2 w e r /r is considered to be negligible. Equation (6.6) is a general expression for the pressure as a function of the radius. In the particular case when r = re then e e wf w r q 1 p p ln S 2 kh r 2 µ π − = − + (6.7) This is the familiar well inflow equation under semi-steady state conditions and is similar to that presented as equ. (4.27) for steady state flow. It can be transposed to give the Pl relationship e wf e w q 2 kh PI p p r 1 ln S r 2 π µ = = − − + (6.8) in which the van Everdingen skin factor has been included as described in Chapter 4, sec. 7. One unfortunate aspect concerning the application of this equation is that, while both q and pwf can be measured directly, the outer boundary pressure cannot. It is therefore more common to express the pressure drawdown in terms of p − pwf instead of pe − pwf, since p, the average pressure within the drainage volume, can readily be determined from a well test as will be shown in Chapter 7, sec. 7. To express the inflow STABILIZED INFLOW EQUATIONS 138 equation in these terms requires the determination of the volume averaged pressure within the radial cell as e w e w r r r r pdV p dV = (6.9) and since dV = 2πrhφdr, equ. (6.9) can be expressed as e w r r 2 2 e w p2 rh dr p (r r )h π φ π φ = − or e w r 2 2 e w r 2 p prdr (r r ) = − and since 2 2 2 2 2 2 e w e w e e r r r (1 r /r ) r ,then − = − ≈ e w r 2 e r 2 p prdr r = (6.10) The pressure in the integrand of equ. (6.10) is obtained from equ. (6.6) which is a general expression for p as a function of r. Substituting the latter in equ. (6.10) gives e w r 2 wf 2 2 w e e r 2 q r r p p . r ln dr 2 kh r r 2r µ π − = − (6.11) The first term in the integrand is evaluated using the method of integration by parts, i.e. e e e w w w e e w w r r r 2 2 w w r r r r r 2 2 w r r 2 2 e e e w r r r 1 r r ln dr ln dr r 2 r r 2 r r r ln 2 r 4 r r r ln 2 r 4 = − = − ≈ − while the integration of the latter term in equ. (6.11) gives e e w w r r 2 3 4 e 2 2 e e r r r r r dr 8 2r 8r = ≈ STABILIZED INFLOW EQUATIONS 139 Combining these two results in equ. (6.11), and including the mechanical skin factor, results in the modified inflow equation e wf w r q 3 p p ln S 2 kh r 4 µ π − = − + (6.12) 6.3 STEADY STATE SOLUTION The steady state solution of the diffusitivy equation can be derived using precisely the same mathematical steps as for the semi-steady state solution only, in this case, since ∂p/∂t = 0, the diffusivity equation is reduced to 1 p r 0 r r r ∂ ∂ = ∂ ∂ (6.13) which is the radial form of the Laplace equation. Because of the simple form of equ. (6.13) the mathematics involved in obtaining inflow equations expressed in terms of pe and p is somewhat easier than in the previous section. The derivation of these equations will therefore be left as an exercise for the reader. The solutions of the radial diffusivity equation for both steady state and semi-steady state flow conditions are summarised in table 6.1. STEADY STATE SEMI-STEADY STATE General relationship between p and r wf w q r p p In 2 kh r µ π − = 2 wf 2 w e q r r p p ln 2 kh r 2r µ π − = − Inflow equations expressed in terms of p = pe at r = re e e wf w r q p p In 2 kh r µ π − = e e wf w r q 1 p p ln 2 kh r 2 µ π − = − Inflow equations expressed in terms of the average pressure e wf w r q 1 p p ln 2 kh r 2 µ π − = − e wf w r q 3 p p ln 2 kh r 4 µ π − = − TABLE 6.1 Radial inflow equations for stabilized flow conditions N.B. To express in field units (stb/d, psi, mD, ft.) the term q 2 kh µ π should be replaced by o 141.2q B kh µ , in each of the equations in table 6.1 In addition the mechanical skin factor can be included in the equations as shown in equs. (4.27) and (6.7). As an alternative, the skin factor can be accounted for in the inflow equations by artificially changing the wellbore radius. For example, including the skin factor, equ. (6.12) can be expressed as STABILIZED INFLOW EQUATIONS 140 e wf w r q 3 p p ln 2 kh r 4 µ π − = − ′ (6.14) in which w w r r e ′ = −s (6.15) is the effective wellbore radius due to the presence of skin. If the formation is damaged, so that the permeability close to the well is reduced, the skin factor is positive. If, however, the well has been stimulated, for instance by acidising, then the permeability close to the well can exceed the average formation permeability and the skin factor is then negative. In either case the magnitude and sign of the skin factor can be determined from pressure buildup analysis as will be described in Chapter 7, sec. 7. 6.4 EXAMPLE OF THE APPLICATION OF THE STABILIZED INFLOW EQUATIONS The solution of the diffusivity equation under semi-steady state flow conditions has been described in detail in section 6.2 since the mathematical approach is quite general and can be applied to more complex radial flow problems. Consider, for instance, the case of a well which has been stimulated by steam soaking, refer Chapter 4, sec. 7. In this type of stimulation several thousand tons of steam are injected into the well and, upon re-opening, the well will produce at a greatly increased rate. As a first approximation it will be assumed that, due to the steam injection, the temperature distribution can be described by a temperature step function1 so that, for rw < r < rh, the temperature Ts is uniform and initially equal to the condensing steam temperature at the sandface. During production, Ts will decrease due to heat losses by conduction and convection. For r > rh, the temperature is the original reservoir temperature Tr. The situation at any time during the production cycle is shown in fig. 6.2, Pressure T , r oc µ T , s oh µ pwf rw rh re pe ph Fig. 6.2 Pressure profile during the steam soak production phase where µoh and µoc are the viscosities of the oil at temperatures Ts and Tr, respectively. If the inflow equations are formulated under steady state flow conditions, the result will be as follows STABILIZED INFLOW EQUATIONS 141 oh r wf w h w oc r h h e h q r p p ln r r r 2 kh r q r and p p ln r r r 2 kh r µ π µ π − = < ≤ − = ≤ < and in particular oh h h wf w q r p p ln 2 kh r µ π − = (6.16) oc e e h h q r and p p ln 2 kh r µ π − = (6.17) Since there is continuity of pressure at r = rh, then equs. (6.16) and (6.17) will yield, upon addition e h e wf oh oc w h oc oh e h oc w h r r q p p ln ln 2 kh r r q r r ln ln 2 kh r r µ µ π µ µ π µ − = + = + (6.18) and since the inflow equation for an unstimulated well is oc e e wf w q r p p ln 2 kh r µ π − = Then the effect on the productivity index due to steam soaking can be expressed as e w oh e h oc w h PI stimulated well PI ratio increase PI unstimulated well r ln r r r ln ln r r µ µ = = + and using typical field data in the above formula, i.e. Tr = 113° F Ts = 525° F µoc = 980 cp µoh = 3.2 cp re = 382 ft rh = 65 ft rw = 0.23 ft Then 382 In .23 PI ratio increase 4.14 3.2 65 382 In In 980 .23 65 = = STABILIZED INFLOW EQUATIONS 142 This PI improvement is probably pessimistic since it has been assumed that steady state conditions prevail from the start of production. In fact, transient flow is more likely during the initial stage which will give an early boost to production, in excess of that calculated from steady state considerations. The manner in which the early transient part of the production cycle can be accounted for will be described in Chapter 9, sec. 6. EXERCISE 6.1 WELLBORE DAMAGE 1) A homogeneous formation has an average effective permeability ke. The effective permeability out to a radius ra from the well has been altered (damage/stimulation) so that its average value in this region is ka. Show that, for this situation, the skin factor may be expressed as e a a a w k k r S In k r − = (6.19) where rw is the wellbore radius. Assume that for r< ra, the flow can be approximately described under steady state conditions and that for r >ra, semi-steady state. 2) During drilling, a well is damaged out to a radius of 4 ft from the wellbore so that the permeability within the damaged zone is reduced to 1/100 th of the undamaged effective permeability. After completion the well is stimulated so that the permeability out to a distance of 10 ft from the wellbore is increased to ten times the undamaged permeability. What will be the PI ratio increase if the wellbore radius is 0.333 ft and the drainage radius 660 ft? EXERCISE 6.1 SOLUTION Pressure ke rw ra ka pa pwf pe re Fig. 6.3 Pressure profiles and geometry (Exercise 6.1) 1) The inflow equations appropriate for the pressure distribution shown in fig. 6.3 are r wf w a a w 2 r a a e 2 e a e q r p p ln r r r 2 k h r q r r p p ln r r r 2 k h r 2r µ π µ π − = < ≤ − = − ≤ < In particular STABILIZED INFLOW EQUATIONS 143 a a wf a w r q p p ln 2 k h r µ π − = and e e a a a r q 1 p p ln 2 k h r 2 µ π − = − At r = ra, there is continuity of pressure, therefore, adding the above equations e e a e wf e a a w e e e e a e w a w a w r k r q 1 p p ln ln 2 k h r 2 k r r r r k r q 1 ln ln ln ln 2 k h r 2 r r k r µ π µ π − = − + = − + − + i.e. e e a e wf e w a w r k r q 1 p p ln 1 ln 2 k h r 2 k r µ π − = − + − which must be equivalent to e e wf e w r q 1 p p ln S 2 k h r 2 µ π − = − + and therefore e a a a w k k r S ln k r − = which is an alternative expression for the skin factor presented by Craft and Hawkins2. 2) Before stimulation 1 4.333 S (100 1)ln 254.02 0.333 = − = while the skin factor after stimulation is 2 (1 10) 10.333 S ln 3.09 10 0.333 − = = − and since STABILIZED INFLOW EQUATIONS 144 e e w 2 k h PI r 1 ln S r 2 π µ = − + the Pl ratio increase is 660 1 ln 254.02 261.11 .333 2 660 1 4.00 ln 3.09 .333 2 65.3 − + = = − − = 6.5 GENERALIZED FORM OF INFLOW EQUATION UNDER SEMI-STEADY STATE CONDITIONS The semi-steady state inflow equation developed in sec. 6.2 appears to be restrictive in that it only applies for a well producing from the centre of a circular shaped drainage area. When a reservoir is producing under semi-steady state conditions each well will assume its own fixed drainage boundary, as shown in fig. 5.3, and the shapes of these may be far from circular. The inflow equation will therefore require some modification to account for this lack of symmetry. Equation (6.12) can be expressed in a generalized form by introducing the so-called Dietz shape factors3 denoted by CA, which are presented for a variety of different geometrical configurations in fig. 6.4. Precisely how these shape factors were generated, in the first place, will be explained in the appropriate place, Chapter 7, sec. 6. For the moment the reader is asked to accept the following tenuous argument for the generalization of the inflow equation. Excluding the mechanical skin factor, equ. (6.12) can be expressed as 2 e wf 2 3/ 2 w r q 1 p p ln 2 kh 2 r e π µ π π − = (6.20) in which the argument of the natural log can be modified as 2 e 3/2 2 2 2 w w w 4 r 4A 4A 4 e r 56.32r 31.6r π π γ = = (6.21) in which A is the area being drained, γ is the exponential of Euler's constant and is equal to 1.781, and 31.6 is the Dietz shape factor for a well at the centre of a circle, refer fig. 6.4. Therefore, equ. (6.20) can be expressed in the general form, including the skin factor, as wf 2 A w q 1 4A p p ln S 2 kh 2 C r µ π γ − = + (6.22) For a reservoir which is producing under semi-steady state conditions, then as already noted, the volume drained by each well is directly proportional to the well's production rate. Therefore, it is a fairly straightforward matter to estimate the volume being drained by each well and, using the average thickness in the vicinity of the well, the area. If STABILIZED INFLOW EQUATIONS 145 structural contour maps are available for the reservoir, then the areas so determined can be roughly matched to the reservoir geometry to obtain a reasonable estimate of the shape of the drainage area. Fig. 6.4 should then be consulted to determine the shape factor CA which can be seen to be dependent not only on the drainage shape but also upon the position of the well with respect to the boundary. For irregular shapes, interpolation between the geometrical configurations presented by Dietz may be necessary. Naturally it is never possible to obtain the exact shape of the drainage volume but a reasonable estimate can usually be made which, when interpreted in terms of a shape factor and used in equ. (6.22), can considerably improve the accuracy of calculations made using the inflow equation. Also listed in fig. 6.4 is the dimensionless time group tDA = kt/φµcA, in which t is the time for which the well has been producing at a reasonably steady rate of production. Unless the calculated value of tDA exceeds the figure quoted for each geometrical configuration then the well is not producing under semi-steady state conditions and the Dietz shape factors cannot be used. STABILIZED INFLOW EQUATIONS 146 60° 1 1 5 4 1 2 1 3 1 bounded reservoirs 3.45 3.43 3.45 In C A 3.45 3.32 3.30 3.12 3.12 1.68 0.86 2.56 1.52 31.6 30.9 31.6 CA 27.6 27.1 21.9 22.6 5.38 2.36 12.9 4.57 0.1 0.1 0.1 Stabilized conditions kt µ cA for > 0.2 0.2 0.4 0.2 0.7 0.7 0.6 0.5 2.38 0.73 1.00 In CA -1.46 -2.16 1.14 -0.50 -2.20 -2.32 2.95 3.22 1.58 1.22 10.8 2.07 2.72 CA 0.232 0.115 3.13 0.607 0.111 0.098 19.1 25 4.86 3.39 0.3 0.8 0.8 Stabilized conditions kt µ cA for > 2.5 3.0 0.3 1.0 1.2 0.9 0.1 0.1 1.0 0.6 1 2 1 2 1 1 2 4 1 4 1 4 2 1 4 3 In water-drive reservoirs 2 1 In reservoirs of unknown production character 1 2 Fig. 6.4 Dietz shape factors for various geometries3 (Reproduced by courtesy of the SPE of the AIME). REFERENCES 1) Boburg, T.C. and Lantz, R.B., 1966. Calculation of the Production Rate of a Thermally Stimulated Well. J.Pet.Tech., December: 1613−1623. 2) Craft, B.C. and Hawkins, M.F., Jr., 1959. Applied Petroleum Reservoir Engineering. Prentice−Hall, Inc., Englewood Cliffs, New Jersey. STABILIZED INFLOW EQUATIONS 147 3) Dietz, D.N., 1965. Determination of Average Reservoir Pressure from Build-Up Surveys. J.Pet.Tech., August: 955−959. CHAPTER 7 THE CONSTANT TERMINAL RATE SOLUTION OF THE RADIAL DIFFUSIVITY EQUATION AND ITS APPLICATION TO OILWELL TESTING 7.1 INTRODUCTION The constant terminal rate solution, which describes the pressure drop in the wellbore due to constant rate production, is the basic equation used in well test analysis. Apart from during the brief transient flow period, (infinite reservoir case) the solution depends critically on the reservoir boundary condition. In this chapter the constant terminal rate solution is presented for a well situated within a no-flow boundary for all the geometrical configurations considered by Matthews, Brons and Hazebroek and for any value of the flowing time. The solutions are expressed in dimensionless form to simplify and generalise the mathematics. Superposition of such solutions leads to a general well test equation which can be applied to the analysis of any pressure test conducted in the wellbore. In this chapter such tests are described for reservoirs containing a fluid of small and constant compressibility (undersaturated oil). In Chapter 8 the same techniques are applied to well test analysis in gas and gas saturated oil reservoirs. 7.2 THE CONSTANT TERMINAL RATE SOLUTION Starting from the static equilibrium pressure pwf = pi at t = 0, the constant terminal rate solution of the radial diffusivity equation describes how the bottom hole flowing pressure pwf varies as a function of time after imposing a rate change from 0 to q. This is illustrated in fig 7.1. time (b) Pressure pwf Transient Late Transient Semi Steady State pi Rate q time (a) Fig. 7.1 Constant terminal rate solution; (a) constant production rate (b) resulting decline in the bottom hole flowing pressure OILWELL TESTING 149 The constant terminal rate solution is therefore the equation of pwf versus t for constant rate production for any value of the flowing time. The pressure decline, fig. 7.1 (b), can normally be divided into three sections depending on the value of the flowing time and the geometry of the reservoir or part of the reservoir being drained by the well. Initially, the pressure response can be described using a transient solution of the diffusivity equation. It is assumed during this period that the pressure response at the wellbore is not affected by the drainage boundary of the well and vice versa. This is frequently referred to as the infinite reservoir case since, during the transient flow period, the reservoir appears to be infinite in extent. The transient phase is followed by the so-called late transient period during which the influence of the drainage boundary begins to be felt. For a well producing from within a no-flow boundary both the shape of the area drained and position of the well with respect to the boundary are of major importance in determining the appropriate late transient constant terminal rate solution. Eventually, stabilised flow conditions will prevail which means that for the no-flow boundary case the rate of change of wellbore pressure with respect to time is constant. This corresponds to the semi-steady state condition described in Chapter 5, sec 3(b). The constant terminal rate solution, for all values of the flowing time, was first presented to the industry by Hurst and Van Everdingen in 1949. In their classic paper on the subject1, the authors solved the radial diffusivity equation using the Laplace transform for both the constant terminal rate and constant terminal pressure cases. The latter, which is relevant to water influx calculations. will be described in Chapter 9. The full Hurst and Van Everdingen solution, equ. 7.34, is a most intimidating mathematical equation which contains as one of its components an infinite summation of Bessel functions. The complexity is due to the wellbore pressure response during the late transient period, since for transient and semi-steady state flow relatively simple solutions can be obtained which will be described in sec. 7.3. The fact that the full solution is so complex is rather unfortunate since the constant terminal rate solution of the radial diffusivity equation can be regarded as the basic equation in wellbore pressure analysis techniques. By superposition of such solutions, as will be shown in sec. 7.5, the pressure response at the wellbore can be theoretically described for any sequence of different rates acting for different periods of time, and this is the general method employed in the analysis of any form of oil or gas well test. 7.3 THE CONSTANT TERMINAL RATE SOLUTION FOR TRANSIENT AND SEMI-STEADY STATE FLOW CONDITIONS During the initial transient flow period, it has been found that the constant terminal rate solution of the radial diffusivity equation, determined using the Laplace transform, can be approximated by the so-called line source solution which assumes that in comparison to the apparently infinite reservoir the wellbore radius is negligible and the wellbore itself can be treated as a line. This leads to a considerable simplification in the OILWELL TESTING 150 mathematics and for this solution the boundary and initial conditions may be stated as follows i i r 0 a) p p at t 0, for all r b) p p at r , for all t p q c) lim r , for t 0 r 2 kh µ π → = = = = ∞ ∂ = > ∂ (7.1) Condition (a) is merely the initial condition that, before producing, the pressure everywhere within the drainage volume is equal to the initial equilibrium pressure pi. Condition (b) ensures the condition of transience, namely that the pressure at the outer, infinite boundary is not affected by the pressure disturbance at the wellbore and vice versa. Condition (c) is the line source inner boundary condition. In addition, the assumptions made in deriving the radial diffusivity equation in Chapter 5 are retained. That is, that the formation is homogeneous and isotropic, and drained by a fully penetrating well to ensure radial flow; the fluid itself must have a constant viscosity and a small and constant compressibility. The solution obtained will, therefore, be applicable to the flow of undersaturated oil. Having developed the simple theory of pressure analysis based on these assumptions, many of the restrictions will be removed by considering, for instance, the effects of partial well completion, the flow of highly compressible fluids, etc. These modifications to the basic theory will be gradually introduced in this and the following chapter. Under the above conditions the diffusivity equation 1 p c p r r r r k t φ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.20) can be solved by making use of Boltzmann's transformation 2 2 r cr s 4 (Diffusivity constant)t 4kt φ µ = = so that s cr t 2k t φ µ ∂ = ∂ (7.2) and 2 2 s cr r 4k t φ µ ∂ = ∂ (7.3) OILWELL TESTING 151 Equation (5.20) can be expressed with respect to this new variable as 1 d dp s s c dp s r r ds ds r r k ds t φ µ ∂ ∂ ∂ = ∂ ∂ ∂ and using equs. (7.2) and (7.3), this becomes 2 2 1 cr d cr dp cr dp r 2k t ds 2k t ds 2k t ds φ µ φ µ φ µ = − which can be simplified as d dp dp s s ds ds ds = − or dp d dp dp s s ds ds ds ds + = − This is an ordinary differential equation which can be solved by letting dp p ds ′ = Then ( ) dp p s sp ds s 1 dp ds p s ′ ′ ′ + = − + ′ = − ′ (7.4) Integrating equ. (7.4) gives 1 ln p ln s s C ′ = − − + or s 2 e p C s − ′ = (7.5) where C1 and C2 are constants of integration and C2 can be evaluated using the line source boundary condition r 0 p q dp s dp lim r r 2s r 2 kh ds r ds µ π → ∂ ∂ = = = ∂ ∂ therefore, OILWELL TESTING 152 s 2 q C e 4 kh µ π − = which yields, as r (and therefore s) tends to zero 2 q C 4 kh µ π = Equation (7.5) can now be integrated between the limits t = 0 (s → ∞ ) and the current value of t, for which s = x; and pi (initial pressure) and the current pressure p. i.e. i p x s p q e dp ds 4 kh s µ π − ∞ = which gives 2 s r,t i cr x 4k t q e p p ds 4 kh s φ µ µ π − ∞ = = − (7.6) Equation (7.6) is the line source solution of the diffusivity equation giving the pressure pr,t as a function of position and time. The integral 2 s s x cr x 4kt e e ds ds s s φ µ − − ∞ ∞ = = (7.7) is a standard integral, called the exponential integral, and is denoted by ei(x). Qualitatively, the nature of this integral can be understood by considering the component parts, fig. 7.2. The integral of curve (c) between x and ∞ will have the shape shown in fig. 7.2 (d). Thus ei (x) is large for small values of x, since the ei-function is the area under the graph from the particular value of x out to infinity (i.e. the shaded area in curve (c) of fig. 7.2) and, conversely, small for large values of x. The ei-function is normally plotted on a log-log scale and such a version is included as fig. 7.3. From this curve it can be seen that if x < 0.01, ei (x) can be approximated as ei(x) ln x 0.5772 ≈− − (7.8) OILWELL TESTING 153 x e-s s (a) = s (b) s = x s (c) x (d) 1 s S e s − S x e ei (x) ds s − ∞ = Fig. 7.2 The exponential integral function ei(x) where the number 0.5772 is Euler's constant, the exponential of which is denoted by 0.5772 e 1.781 γ = = and therefore equ. (7.8) can be expressed as ( ) ei(x) ln x for x 0.01 γ ≈− < (7.9) The separate plots of ei(x) and −In(γx), in Fig. 7.3, demonstrate the range of validity of equ. (7.9). The significance of this approximation; is that reservoir engineers are frequently concerned with the analysis of pressures measured in the wellbore, at r = rw. Since in this case 2 w x cr / 4kt φ µ = , it is usually found that for measurements in the wellbore, x will be less than 0.01 even for small values of t. Equation (7.6) can then be approximated as w r t wf i 2 w q 4kt p p p ln 4 kh cr µ π γ φ µ = = − Or, if the van Everdingen mechanical skin factor is included as a time independent perturbation (ref. Chapter 4, sec. 7), then wf i 2 w q 4kt p p ln 2S 4 kh cr µ π γ φ µ = − + (7.10) As expected for this transient solution there is no dependence at all upon the area drained or well position with respect to the boundary since for the short time when equ. (7.10) is applicable the reservoir appears to be infinite in extent. OILWELL TESTING 154 10 -1 8 6 4 2 10 -2 8 6 4 2 10 -3 10 -3 2 5 10 -2 2 5 10 -1 2 5 10 -1 1 10 8 6 4 2 1 8 6 4 2 2 5 5 10 ei (x) x→0 ei (x) ~ (-In γ x) - × -0.5772 ei (x) -In x ∝ ei (x) ei (x) -In γ x x 1 Fig. 7.3 Graph of the ei-function for 0.001 ≤ × × × × ≤ 5.0 Sometimes pressure tests are conducted in order to determine the degree of communication between wells, for example, pulse testing. In such cases pressure transients caused in one well are recorded in a distant well and, under these circumstances, r is large and the approximation of equ. (7.9) is no longer valid. Equation (7.6) must then be used in its full form, i.e. 2 r,t i q cr p p ei 4 kh 4k t µ φ µ π = − (7.11) in which the values of the exponential integral can be obtained from fig. 7.3. EXERCISE 7.1 ei-FUNCTION: LOGARITHMIC APPROXIMATION A well is initially producing at a rate of 400 stb/d from a reservoir which has the following rock and fluid properties k = 50 mD h = 30 ft rw = 6 inches φ = 0.3 µ = 3 cp c = 10 × 10-6 / psi Bo = 1.25 rb/stb 1) After what value of the flowing time is the approximation ei(x) = − ln(γx) valid for this system? OILWELL TESTING 155 2) What will be the pressure drop at the well after flowing at the steady rate of 400 stb/d for 3 hours, assuming transient conditions still prevail. EXERCISE 7.1 SOLUTION Converting the reservoir and fluid properties to Darcy units. k = 0.05 D h = 30 ft × 30.48 = 914.4 cm 2 w r = 0.25 sq.ft. = 0.25 × (30.48)2 cm2 = 232.3 cm2 c = 10 × 10-6/psi = 10 × 10-6 × 14.7/atm = 14.7 × 10-5/atm q = 400stb/d = 400× 1.25× 1.84 r.cc/sec = 920 r.cc/sec. The approximation ei(x) ≈ − In(γx) applies for x < 0.01 i.e. 2 w 2 w 5 cr 0.01 4k t or cr t 0.04k .3 3 14.7 10 232.3 t .04 .05 for t 15.4 seconds φ µ φ µ − < > × × × × > × > Now in a practical sense, nobody is concerned with what happens in the well during the first 15 seconds, after which the pressure decline can be calculated using the logarithmic approximation for ei(x) i.e. 2 w i wf 2 w cr q p p ei 4 kh 4k t q 4k t ln 4 kh cr φ µ µ π µ π γ φ µ − = = After producing for 3 hours at a steady rate of 400 stb/d the pressure drop at the wellbore is 5 i wf 920 3 4 .05 3 3600 10 p p ln 4 .05 914.4 1.781 .3 3 14.7 232.3 50.8atm, or 747psi π × × × × × − = × × × × × × = OILWELL TESTING 156 The constant terminal rate solution of the radial diffusivity equation during the late transient flow period is too complicated to include at this stage. A simplified method of obtaining this solution will be described in sec. 7.6. Once semi-steady state conditions prevail the solution can be determined by adding the simple material balance equation for the bounded drainage volume ( ) i cAh p p qt φ − = (7.12) to the semi-steady state inflow equation wf 2 A w q 4A p p ½ ln S 2 kh C r µ π γ − = + (6.22) to give wf i 2 A w q 4A kt p p ½ ln 2 S 2 kh cA C r µ π π φ µ γ = − + + (7.13) In this equation p is the current average pressure within the drainage boundary and CA is the Dietz shape factor introduced in Chapter 6, sec.5. The magnitude of CA depends on the shape of the area being drained and also upon the position of the well with respect to the boundary. Theoretically, for the constant terminal rate solution, the rate q in equs. (7.12) and (6.22) is the same. In practice, it is sometimes difficult to maintain the production rate of a well constant over a long period of time and therefore, the current rate in equ. (6.22) may differ from the average rate which is implicitly used in material balance, equ. (7.12). In this case the rate in equ. (7.12) is set equal to the current, or final flow rate, and the flowing time is expressed as an effective flowing time, where Cumulative Production t Effective flowing time Final flow rate = = (7.14) Use of the effective flowing time is therefore simply a method for equalising the rates and preserving the material balance and is frequently used in pressure analysis, as will be described later. Even though no equation for describing the pressure decline during the late transient flow period has yet been developed, equs. (7.10) and (7.13), which are appropriate for transient and semi-steady state flow, can be usefully employed by themselves in well test analysis. Well testing involves producing a well at a constant rate or series of rates, some of which may be zero (well closed in), while simultaneously taking a continuous recording of the changing pressure in the wellbore using some form of pressure recording device. The retrieved record of wellbore pressure as a function of time can be analysed in conjunction with the known rate sequence to determine some or all of the following reservoir parameters: OILWELL TESTING 157 − initial pressure (pi) − average pressure within the drainage boundary (p) − permeability thickness product (kh), and permeability (k) − mechanical skin factor (S) − area drained (A) − Dietz shape factor (CA) In the following example of a pressure drawdown test, a well is produced at a single constant rate from a known initial equilibrium pressure pi and pwf analysed as a function of the flowing time t. Equation (7.10) is used to determine k and S while equ. (7.13) is used for large values of t to determine A and CA. This latter part of the test is sometimes referred to as reservoir limit testing and the analysis technique used to determine the shape factor follows that presented by Earlougher2. EXERCISE 7.2 PRESSURE DRAWDOWN TESTING A well is tested by producing it at a constant rate of 1500 stb/d for a period of 100 hours. It is suspected, from seismic and geological evidence, that the well is draining an isolated reservoir block which has approximately a 2:1 rectangular geometrical shape and the extended drawdown test is intended to confirm this. The reservoir data and flowing bottom hole pressures recorded during the test are detailed below and in table 7.1 h = 20 ft c = 15 × 10-6 /psi rw = .33 ft µo = 1 cp φ = .18 Bo = 1.20 rb/stb Flowing time (hours) pwf (psia) Flowing time (hours) pwf (psia) 0 3500 (pi) 20 2762 1 2917 30 2703 2 2900 40 2650 3 2888 50 2597 4 2879 60 2545 5 2869 70 2495 7.5 2848 80 2443 10 2830 90 2392 15 2794 100 2341 TABLE 7.1 1) Calculate the effective permeability and skin factor of the well. 2) Make an estimate of the area being drained by the well and the Dietz shape factor. OILWELL TESTING 158 EXERCISE 7.2 SOLUTION 1) The permeability and skin factor can be obtained by the analysis of the transient flow period for which wf i 2 w q 4kt p p ln 2S 4 kh cr µ π γ φ µ = − + or in field units o wf i 2 w 162.6q B k p p log t log 3.23 0.87S kh cr µ φ µ = − + − + Thus for the initial period, when transient flow conditions prevail, a plot of pwf versus log t should be linear with slope m = 162.6 q µ Bo /kh, from which kh and k can be determined. Furthermore, using the value of pwf(1hr) taken from the linear trend for a flowing time of one hour and solving explicitly for S gives i wf(1hr) 2 w (p p ) k S 1.151 log 3.23 m cr φµ − = − + The plot of pwf versus log t for the first few recorded pressures, fig. 7.4, indicates that transient flow conditions last for about four hours and that the values of m and pwf(1 hr) are 61 psi/log cycle and 2917 psia, respectively. OILWELL TESTING 159 p (psi) wf 2900 2800 2700 0 .5 1.0 1.5 SLOPE m = 61 psi / log cycle p = 2917 psi wf (1hr) log t 0 20 40 60 80 100 t HOURS p (psi) wf 2900 2800 2700 2600 2500 2400 2300 dp dt = 5.08 psi / hr a b p = 2848 psi o Fig. 7.4 Single rate drawdown test; (a) wellbore flowing pressure decline during the early transient flow period, (b) during the subsequent semi-steady state decline (Exercise 7.2) OILWELL TESTING 160 Therefore, o 162.6q B 162.6 1500 1 1.2 kh 4798mD.ft m 61 µ × × × = = = and k = 240 mD. The skin factor can be evaluated ( ) 6 3500 2917 240 10 S 1.151 log 3.23 4.5 61 .18 1 15 .109 − × = − + = × × × 2) The area being drained and the shape factor can be determined from the later part of the flow test when semi-steady state flow conditions prevail. Under these circumstances dp/dt ≈ constant, and as shown in the plot of pwf versus t, fig. 7.4(b), this occurs after a flowing time of approximately 50 hours, after which dp/dt = −5.08 psi/hr. Therefore, equ. (5.9) can be used to determine the area, since dp q (atm / sec) dt cAhφ = − or, in field units o 0.2339qB dp 5.08 (psi/hr) dt cAhφ = − = − and hence 6 .2339 1500 1.2 A 35 acres 15 10 20 .18 5.08 43560 − × × = = × × × × × The equation of the linear pressure decline under semi-steady state conditions is 12 wf i 2 A w q 4A kt p p ln 2 S 2 kh cA C r µ π π φ µ γ = − + + (7.13) The linear extrapolation of this line to small values of t gives the specific value of po = 2848 psia when t = 0 and inserting this condition in equ. (7.13) gives i o 2 A w q 4A p p ln 2S 4 kh C r µ π γ − = + or in field units i o A 2 w 4A p p m log logC 0.87S r γ − = − + OILWELL TESTING 161 which can be solved to determine the shape factor as CA = 5.31, and consulting the Dietz chart, fig. 6.4, this corresponds approximately to the following geometrical configuration 2 1 7.4 DIMENSIONLESS VARIABLES For a variety of reasons, which will be duly explained, it is much more convenient to express solutions of the radial diffusivity equation in terms of the following dimensionless variables dimensionless radius rD = w r r (7.15) dimensionless time tD = 2 w kt cr φ µ (7.16) and dimensionless pressure D D D i r,t 2 kh p (r ,t ) (p p ) q π µ = − (7.17) Substitution of these variables into the radial diffusivity equ. (5.20) gives D D D D D D D p p 1 r r r r t ∂ ∂ ∂ = ∂ ∂ ∂ (7.18) the general solution of which will be for dimensionless pressure as a function of dimensionless radius and time. In particular, for analysing pressures at the wellbore, which is the main concern in this chapter, rD = 1 and D D D D i wf 2 kh p (1 ,t ) p (t ) (p p ) q π µ = = − Finally, allowing for the presence of a mechanical skin factor, the defining expression for pD (tD) may be written as i wf D D 2 kh (p p ) p (t ) S q π µ − = + (7.19) which is simply an alternative expression for the constant terminal rate solution of the radial diffusivity equation. In this text the pD functions are conventionally referred to as dimensionless pressure functions. As equ. (7.19) shows the correct term should be dimensionless pressure drop functions since pD is proportional to pi - pwf , and the latter is sometimes used in the literature. OILWELL TESTING 162 EXERCISE 7.3 DIMENSIONLESS VARIABLES 1) Show, using dimensional analysis, that both tD and pD are dimensionless. 2) Express tD in field units with the real time in hours and days, respectively. 3) Express pD (tD ) in field units. EXERCISE 7.3 SOLUTION 1) In any absolute set of units the dimensions of the parameters in the expressions for tD and pD (tD) are: [k] = L2 [p] = (ML/T2)/L2 = M/LT2 [µ] = M/LT [c] = LT2/M and therefore 2 D 2 2 2 w kt L T t cr (M/LT)(LT /M)L φ µ = = which is dimensionless, and 2 2 D i wf 3 2 kh (L )(L)(M/LT ) p (p p ) q (L / T)(M/LT) π µ = − = which is also dimensionless. 2) D 2 w kt t Darcy units cr φ µ = D 2 2 2 w 2 2 2 w D sec kmD thrs mD hrs t field units 1 psi cm c r ft psi atm ft (1/1000) (3600) kt (14.7) (30.48) cr φµ φ µ × = × × = × D 2 w kt t 0.000264 t in hours cr φ µ = − (7.20) similarly D 2 w kt t 0.00634 t in days cr φ µ = − (7.21) 3) D i wf 2 kh p (p p ) Darcy units q π µ = − OILWELL TESTING 163 i wf D i wf o D cm atm 2 kmD h ft (p p ) psi mD ft psi p fieldunits rb / d rcc / sec qstb / d stb / d rb / d 2 (1/1000) (30.48) (1/14.7) kh (p p ) B (1.84) q π µ π µ × × − = × × = − 3 D i wf o kh p 7.08 10 (p p ) q B µ − = × − (7.22) The reasons for using dimensionless variables in pressure analysis are as follows. a) The variables lead to both a simplification and generality in the mathematics. The latter is probably the more important and implies that if the radial flow of any fluid can be described by the differential equ. (7.18) then the solutions will be identical irrespective of the nature of the fluid. In this current chapter equ. (7.18) is being applied to a fluid of small and constant compressibility for which the solutions are the pD (rD,tD ) functions. In Chapter 8, however, it will be shown that an equation identical in form to equ. (7.18) can be applied to the flow of a real gas. In this case the solutions are for mD (rD, tD ) functions which are dimensionless real gas pseudo pressures. Nevertheless, solutions of equ. (7.18) expressed as pD functions will have the same form as solutions in terms of the mD functions. b) Since the variables are dimensionless then equations expressed in terms of them are invariant in form, irrespective of the units system used. The same holds true, of course, for dimensionless plots of pD as a function of tD. The scales have the same numerical value whether Darcy, field or Sl units are employed. This latter point will be referred to again in connection with the Matthews, Brons and Hazebroek plots presented in sec. 7.6. Thus suppose, for instance, a value of pD (tD) = 35.71 is determined as the result of solving an equation or reading a chart for a certain value of tD. Then if the reservoir parameters, fluid properties and rate are pi = 3500 psi (238.1 atm) Bo = 1.2 rb/stb k = 150 mD (.15D) µ = 3 cp h = 20 ft (609.6 cm) q = 100 stb/d (220.8 rcc/sec) S = 3 then equ. (7.22) (field units) can be used to determine pwf in psi as 3 wf wf wf ) 7.08 10 150 20 (3500 p 35.71 S 100 3 1.2 0.059(3500 p ) 38.71 p 2844psi − × × × − = + × × − = = OILWELL TESTING 164 or using the same value of pD (tD ) = 35.71 in conjunction with equ. (7.19) (Darcy units) to determine pwf as 193.5 atm. This example also illustrates that although equations may be developed using dimensionless pressure functions, conversion can easily be made at any stage to obtain the real pressure. c) The majority of technical papers on the subject of pressure analysis, at least those written since the late sixties, generally have all equations expressed in dimensionless form. It is therefore hoped that by introducing and using dimensionless variables in this text the engineer will be assisted in reading and understanding the current literature. To illustrate the application of dimensionless variables, the constant terminal rate solution of the radial diffusivity equation derived in sec. 7.3 for transient and semisteady state conditions, will be expressed in terms of dimensionless pressure functions. The transient solution is wf i 2 w q 4 kt p p In 2S 4 kh cr µ π γ φ µ = − + (7.10) which may be re-arranged as D 12 i wf 4t 2 kh (p p ) ln S q π µ γ − = + and therefore, from the defining equation for pD (tD ), equ. (7.19), it is evident that ( ) D 12 D D 4t p t ln γ = (7.23) which is also frequently expressed as ( ) 12 D D D p t (ln t 0.809) = + (7.24) In either case pD (tD) is strictly a function of the dimensionless time tD. For semi-steady state conditions equ. (7.13) can be expressed as 2 w 12 i wf 2 2 A w w r kh 4A kt (p p ) ln 2 S q A C r cr π π µ γ φ µ 2 − = + + or 2 w 12 i wf D 2 A w r kh 4A (p p ) ln 2 t S q A C r π π µ γ 2 − = + + and therefore, applying equ. (7.19) 2 w 12 D D D 2 A w r 4A p (t ) ln 2 t A C r π γ = + (7.25) OILWELL TESTING 165 Further, by defining a modified version of the dimensionless time as 2 w DA D r kt t t A cA φ µ = = (7.26) equ. (7.25) can be expressed in its more common form as 12 D D DA 2 A w 4A p (t ) ln 2 t C r π γ = + (7.27) The necessity for, and usefulness of, this dimensionless time tDA will be illustrated later in the chapter. No attempt can yet be made to define a pD function appropriate to describe the pressure drop at the wellbore during the late transient period. Ramey and Cobb3 have shown, however, that for a well situated at the centre of a regular shaped drainage area, for instance, a circle, square or hexagon, the late transient period is of extremely short duration and under these circumstances it is possible to equate equs. (7.23) and (7.27) to determine the approximate time at which the change from transient to semi-steady state conditions will occur, i.e. D 1 1 2 2 DA 2 A w 4t 4A ln ln 2 t C r π γ γ ≈ + which may be expressed as either 2 2 w D w A D r 4 t r / A C t e A π ≈ (7.28) or DA A DA 4 t C t e π ≈ (7.29) Solving equ. (7.28) for tD will give an approximate solution for the dimensionless transition time which is dependent both on the ratio 2 w r /A and CA. Solving equ. (7.29) for tDA, however, will give a dimensionless transition time which is only dependent on the shape factor. The solution of equ. (7.29), for CA ≈ 31, is DA kt t 0.1 cA φ µ = ≈ (7.30) so that for a well draining from the centre of one of the regular drainage area shapes mentioned, a fairly abrupt change from transient to semi-steady state flow occurs for a value of tDA ≈ 0.1, irrespective of the size of the area being drained. This in part explains the usefulness of expressing dimensionless times in terms of tDA rather than tD. The real time when the transition occurs can be determined by solving equ. (7.30) explicitly for t. OILWELL TESTING 166 EXERCISE 7.4 TRANSITION FROM TRANSIENT TO SEMI-STEADY STATE FLOW Determine the pressure response at the wellbore due to the production of a well which is situated at the centre of a square drainage area with sides measuring a) L = 100ft b) L = 500ft The relevant reservoir and fluid properties are as follows k = 50 mD φ = 0.3 µ = 1 cp c = 15 × 10-6 /psi rw = 0.3 ft CA = 30.9 When does the transition from transient to semi-steady state flow occur for both these drainage areas? EXERCISE 7.4 SOLUTION If the real time is expressed in days, then tD can be determined as D 2 w 0.00634kt t cr φ µ = (days) (7.21) with all the other parameters in field units, i.e. 5 D -6 0.00634 50 t t 7.827 10 t .3 1 15 10 .09 × × = = × × × × × The dimensionless pressure response in the well can be determined using the following functions D D D Transient flow p (t ) ½ (ln t 0.809) = + (7.24) 2 D w D D 2 A w t r 4A SSS flow p (t ) ½ ln 2 A C r π γ = + (7.25) Transient flow Equation (7.24) is independent of the reservoir geometry and will give the same value of pD (tD) irrespective of the magnitude of the area drained. Furthermore pD (tD ) is a linear function of tD when plotted on semi log paper. Using the following equations, values of pD and log tD have been calculated and the results listed in table 7.2 and plotted in fig. 7.5 OILWELL TESTING 167 pD = ½ (2.303 log tD + 0.809) = 1.151 log tD + 0.405 t (days) tD log tD pD (tD ) .05 39135 4.5926 5.69 .50 391350 5.5926 6.84 5.0 3913500 6.5926 8.00 TABLE 7.2 Semi-steady state flow Evaluating equation (7.25) for cases a) and b) (refer table 7.3) a) L = 100 ft pD (tD) = 4.4983 + 5.655 × 10-5 tD b) L = 500 ft pD (tD) = 6.1078 + 2.2619 × 10-6 tD Plots of pD (tD) versus log tD, fig. 7.5 show that, irrespective of the size of the square boundary, flow will initially be under transient conditions. Eventually, however, the boundary effects will result in a transition to semi-steady state flow. The time at which this occurs is naturally dependent upon the dimensions of the volume drained and can be read from the plots in fig. 7.5, as for L = 100 ft; log tD = 4.10; tD = 12590 and t = 0.016 days for L = 500 ft; log tD = 5.50; tD = 316230 and t = 0.404 days t (days) tD Dimensionless Pressure, pD L=100 ft. L=500 ft. .005 3914 4.72 .01 7827 4.94 .025 19568 5.60 .05 39135 6.71 .10 78270 8.92 .25 195675 15.56 6.55 .50 391350 26.63 6.99 1.00 782700 7.88 2.50 1956750 10.53 5.00 3913500 14.96 10.00 7827000 23.81 TABLE 7.3 In themselves these figures lead to no general conclusion concerning the time at which semi-steady state flow commences. However, evaluating in terms of 2 DA D w t t r / A = then for L = 100 ft; tDA = 0.113 and for L = 500 ft; tDA = 0.114 OILWELL TESTING 168 which indicate that semi-steady state conditions will occur for the same value of tDA irrespective of the size of the square. It is generally true for a well situated at the centre of -a square -a circle -a hexagon that semi-steady state conditions will prevail after a flowing time such that tDA >0.1. 32 28 24 20 16 12 8 4 0 pD 1 2 3 4 5 6 7 log tD SEMI STEADY STATE L = 100 ft L = 500 ft TRANSIENT FLOW Fig. 7.5 Dimensionless pressure as a function of dimensionless flowing time for a well situated at the centre of a square (Exercise 7.4) 7.5 SUPERPOSITION THEOREM: GENERAL THEORY OF WELL TESTING Mathematically the superposition theorem states that any sum of individual solutions of a second order linear differential equation is also a solution of the equation. In practice, this is one of the most powerful tools at the reservoir engineer's disposal for writing down solutions to complex flow problems in the reservoir without explicitly solving the full differential equation on each occasion. Applying the superposition theorem means that individual constant rate wells can be placed in any position in the reservoir at any time and an expression for the resulting pressure distribution in space and time derived by inspection. The principle will be illustrated with an example of superposition in time at a fixed location which is particularly relevant to well test analysis. OILWELL TESTING 169 Rate q1 q2 q3 q4 qn t4 t3 t2 t1 tn time t4 t3 t2 t1 tn p i p wf time Fig. 7.6 Production history of a well showing both rate and bottom hole flowing pressure as functions of time Consider the case of well producing at a series of constant rates for the different time periods shown in fig. 7.6. To determine the wellbore pressure after a total flow time tn when the current rate is qn, the superposition theorem is applied to determine a composite solution of equ. (7.18) in terms of q1 Acting for time tn + (q2 −q1 ) ” ” ” (tn−t1) + (q3 −q2) ” ” ” (tn −t2) . . + (qj −qj−1) ” ” ” (tn−tj−1) . . + (qn −qn−1) ” ” ” (tn−tn−1) That is, a solution is obtained for the initial rate q1, acting over the entire period tn. At time t1 a new well is opened to flow at precisely the same location as the original well at a rate (q2−q1) so that the net rate after t1 is q2. At time t2 a third well is opened at the same location with rate (q3−q2) which reduces the rate to q3 after time t2 ………… etc. The composite solution of equ. (7.18) for this variable rate case can then be formed by adding individual constant terminal rate solutions, equ. (7.19), for the rate-time sequence specified above, i.e. OILWELL TESTING 170 ( ) n 1 n 2 n j 1 n n 1 i wf 1 D D n n 2 1 D D D 3 2 D D D j j 1 D D D n n 1 D D D 2 kh p p (q 0) (p (t 0) S) (q q ) (p (t t ) S) (q q ) (p (t t ) S) . . (q q ) (p (t t ) S) . . (q q ) (p (t t ) S) π µ − − − − − = − − + + − − + + − − + + − − + + − − + in which n wf p is the specific value of the bottom hole flowing pressure corresponding to the total time tn which may occur at any time during the nth period of constant flow, when the rate is qn. In this summation all the skin factor terms disappear except for the last, qn S. The summation can be expressed as ( ) ( ) j 1 n n i wf j D D D n n j 1 2 kh p p q p t t q S π µ − = − = ∆ − + (7.31) in which j j j 1 q q q − ∆ = − Equation (7.31) may be regarded as the basic equation for interpreting the pressure-time-rate data collected during any well test, and with minor modifications, described in Chapter 8, can equally well be applied to gas well test analysis. The whole philosophy of well testing is to mechanically design the test with a series of different flow rates, some of which may be zero (well closed in), for different periods of time so that equ. (7.31) can be readily interpreted to yield some or all of the required reservoir parameters, pi, p, k, S, A and CA. The three most common forms of well testing are the single rate drawdown test, the pressure buildup test and the multi-rate drawdown test. The analysis of each of these tests using equ. (7.31) is briefly described below and in much greater detail in the following sections of this chapter. a) Single rate drawdown test In this type of test the well is flowed at a single constant rate for an extended period of time so that 1 1 D D n q q ; q q and t t = ∆ = = and equ. (7.31) can be reduced to ( ) i wf D D 2 kh p p p (t ) S q π µ − = + (7.19) which is simply the constant terminal rate solution expressed in dimensionless form. The flowing pressure pwf , which is recorded throughout the test, can be analysed as a OILWELL TESTING 171 function of the flowing time t to yield the basic reservoir parameters k, S, A and CA. The most common form of analysis used has already been fully described in exercise 7.2. It is assumed that the initial equilibrium pressure pi is known and this is simply the recorded pressure prior to opening the well in the first place. b) Pressure buildup testing This is probably the most common of all well test techniques for which the rate schedule and corresponding pressure response are shown in fig. 7.7. Rate q t ∆t time (a) time (b) Pressure pi pwf pws t ∆t Fig. 7.7 Pressure buildup test; (a) rate, (b) wellbore pressure response Ideally the well is flowed at a constant rate q for a total time t and then closed in. During the latter period the closed-in pressure pwf = pws is recorded as a function of the closed in time ∆t. Equation (7.31) can again be used but in this case with n n 1 1 1 D D D 2 2 D D D q q ; q q ; t t t q 0 ; q (0 q) ; t t t = ∆ = = + ∆ = ∆ = − − = ∆ the skin factor disappears by cancellation and the equation is reduced to i ws D D D D D 2 kh (p p ) p (t t ) p ( t ) q π µ − = + ∆ − ∆ (7.32) Equation (7.32) is the basic equation for pressure buildup analysis and can be interpreted in a variety of ways. The most common method of analysis is to plot the closed in pressure pws as a function of log (t t)/ t + ∆ ∆.This is called the Horner plot4 and can be used to determine pi or p, kh, and S as will be described in detail in sec. 7.7, and illustrated in exercises 7.6 and 7.7. OILWELL TESTING 172 c) Multi-rate drawdown testing In this form of test the well is flowed at a series of different rates for different periods of time and equ. (7.31) is used directly to analyse the results. The sequence is arbitrary but usually the test is conducted with either a series of increasing or decreasing rates. Providing that none of the rates is zero, the Odeh-Jones5 technique can be used to analyse the results. That is, dividing equ. (7.31) throughout by the final rate qn ( ) ( ) n n j 1 n i wf j D D D j 1 n n p -p q 2 kh p t t S q q π µ − = ∆ = − + (7.33) Values of n wf p are read from the continuos pressure record at the end of each flowing period and the corresponding values of the summation are computed on each occasion, so that each value represents a point on the graph. A plot of n i wf n (p p )/q − − n i wf p p qn − = ∆ − n j 1 n j D D D j 1 n q p (t t ) q mS m = µ 2π kh Fig. 7.8 Multi-rate flow test analysis versus ( ) n j 1 n j D D D j 1 n q p t t q − = ∆ − should be linear as shown in fig. 7.8, with slope m / 2 kh µ π = and intercept on the ordinate mS. The test yields the value of kh from the slope and S from the intercept assuming, as in the case of the single rate drawdown test, that pi is measured prior to flowing the well at the first rate. Exercise 7.8 provides an example of the traditional Odeh-Jones analysis technique. The basic oilwell test equation, (7.31), is fairly simple in form and yet it presents one major difficulty when applying it to well test analysis. The problem is, how can the pD functions, which are simply constant terminal rate solutions of the radial diffusivity equation, be evaluated for any value of the dimensionless time argument ( ) n j 1 D D t t − − ? So far in this chapter dimensionless pressure functions have only been evaluated for transient and semi-steady state flow conditions, equs. (7.23) and (7.27), respectively. For a well draining from the centre of a circular, bounded drainage area, the full constant terminal rate solution for any value of the flowing time is OILWELL TESTING 173 ( ) ( ) ( ) ( ) ( ) n D 2 t 2 1 n eD D D D eD 2 2 2 2 n 1 eD n 1 n eD 1 n e J r 2t 3 p t lnr 2 4 r J r J α α α α α − = ∞ = + − + − (7.34) in which reD = re/rw and αn are the roots of 1 n eD 1 n 1 n 1 n eD J ( r ) Y ( ) J ( ) Y ( r ) 0 α α α α − = and J1 and Y1, are Bessel functions of the first and second kind. Equation (7.34) is the full Hurst and Van Everdingen constant terminal rate solution referred to in sec. 7.2, the detailed derivation of which can be found in their original paper1, or in a concise form in Appendix A of the Matthews and Russell monograph6. One thing that can be observed immediately from this equation is that it is extremely complex, to say the least, and yet this is the expression for the case of simple radial symmetry. In fact, as already noted in sec. 7.4 and demonstrated in exercise 7.4, for a well producing from the centre of a regular shaped drainage area there is a fairly abrupt change from transient to semi-steady state flow so that equ. (7.34) need never be used in its entirity to generate pD functions. Instead, equ. (7.23) can be used for small values of the flowing time and equ. (7.27) for large values, with the transition occurring at tDA ≈ 0.1. Problems arise when trying to evaluate pD functions for wells producing from asymmetrical positions with respect to irregular shaped drainage boundaries. In this case a similar although more complex version of equ. (7.34) could be derived which again would reduce to equ. (7.23) for small tD and to equ. (7.27) for large tD. Now, however, there would be a significant late transient period during which there would be no alternative but to use the full solution to express the pD function. Due to the complexity of equations such as equ. (7.34) engineers have always tried to analyse well tests using either transient or semi-steady state analysis methods and in certain cases this approach is quite valid, such analyses having already been presented in exercise 7.2 for a single rate drawdown test. Sometimes, however, serious errors can arise through using this simplified approach and some of these will be described in detail in the following sections. It first remains, however, to describe an extremely simple method of generating pD functions for any value of the dimensionless time and for any areal geometry and well asymmetry. The method requires an understanding of the Matthews, Brons and Hazebroek pressure buildup analysis technique which is described in the following section. 7.6 THE MATTHEWS, BRONS, HAZEBROEK PRESSURE BUILDUP THEORY In this section the MBH pressure buildup analysis technique will be examined from a purely theoretical standpoint, the main aim being to illustrate a simple method of evaluating the pD function for a variety of drainage shapes and for any value of the dimensionless flowing time. The theoretical buildup equation was presented in the previous section as ( ) i ws D D D D D 2 kh (p p ) p (t t ) p t q π µ − = + ∆ − ∆ (7.32) OILWELL TESTING 174 in which tD is the dimensionless flowing time prior to closure and is therefore a constant while ∆tD is the dimensionless closed in time corresponding to the pressure pws, the latter two being variables which can be determined by interpretation of the pressure chart retrieved after the survey. For small values of ∆t, pws is a linear function of In (t+∆t)/∆t, which can be verified by adding and subtracting ½ In (tD +∆tD ) to the right hand side of equ. (7.32) and evaluating pD (∆tD) for small ∆t using equ. (7.23). Thus, D 1 1 2 2 i ws D D D D D 4 t 2 kh (p p ) p (t t ) In ln (t t ) q π µ γ ∆ − = + ∆ − ± + ∆ which can alternatively be expressed as ( ) D D 1 1 2 2 i ws D D D 4 t t 2 kh t t (p p ) ln p (t t ) ln q t π µ γ + ∆ + ∆ − = + + ∆ − ∆ (7.35) in which dimensionless time has been replaced by real time in the ratio t+∆t/∆t. Again, for small values of the closed-in time ∆t D D D ln (t t ) ln (t ) + ∆ ≈ and D D D D D p (t t ) p (t ) + ∆ ≈ and equ. (7.35) can be reduced to D 1 1 2 2 i ws D D 4t 2 kh t t (p p ) ln p (t ) ln q t π µ γ + ∆ − = + − ∆ (7.36) Since the dimensionless flowing time tD is a constant then so too are the last two terms on the right-hand side of equ. (7.36) and therefore, for small values of ∆t a plot of the observed values of pws versus In (t+∆t)/∆t should be linear with slope m q / 4 kh µ π = , from which the value of the permeability can be determined. This particular presentation of pressure buildup data is known as a Horner plot4 and is illustrated in fig. 7.9. Equation (7.36) is the equation describing the early linear buildup and due to the manner of derivation is only valid for small values of ∆t. Nevertheless, having obtained such a straight line it is perfectly valid to extrapolate the line to large values of ∆t in which case equ. (7.36) can be replaced by ( ) ( ) D i ws(LIN) D D 4t 2 kh t t p p ½ ln p t ½ln q t π µ γ + ∆ − = + − ∆ (7.37) in which pws, the actual pressure in equ. (7.36), is now replaced by pws(LIN ) which is simply the pressure for any value of ∆t on the extrapolated linear trend and while the latter may be hypothetical it is, as will be shown, mathematically very useful. The equation can be used in two ways. Firstly, drawing a straight line through the early OILWELL TESTING 175 linear trend of the observed points on the Horner buildup plot will automatically match equ. (7.37) as illustrated in fig. 7.9. Extrapolation of this line is useful in the determination of the average reservoir pressure, Alternatively, an attempt can be made to theoretically evaluate the pD function in the equation and then compare the theoretical with the actual straight line with the aim of gaining additional information about the reservoir. The application of this method will be illustrated in exercise 7.7. If the well could be closed in for an infinite period of time the initial linear buildup would typically follow the curved solid line in fig. 7.9 and could theoretically be predicted using equ. (7.32). The final buildup pressure p is the average pressure within the bounded volume being drained and is consistent with the material balance for this volume, i.e. i cAh (p p) qt φ − = (7.12) which may be expressed as ( ) i DA 2 kh 2 khqt p p 2 t q q cA h π π π µ µ φ − = = (7.38) t + ∆t ∆t In 4 3 2 1 0 A B pws equ. (7.37) equ. (7.32) large ∆t small ∆t p p Fig. 7.9 Horner pressure buildup plot for a well draining a bounded reservoir, or part of a reservoir surrounded by a no-flow boundary The closed in pressures observed during the test are plotted between points A and B. Since it is impracticable to close in a well for a sufficient period of time so that the entire buildup is obtained then it is not possible to determine p directly from the Horner plot of the observed pressures. Instead, indirect methods of calculating p are employed which rely on the linear extrapolation of the observed pressures to large values of ∆t and therefore implicitly require the use of equ. (7.37). In particular, the Matthews, Brons and Hazebroek7 method involves the extrapolation of the early linear trend to infinite closed in time. The extrapolation to In (t+∆t) / ∆t = 0 gives the value of pws(LIN) = p In the OILWELL TESTING 176 particular case of a brief initial well test in a new reservoir the amount of fluids withdrawn during the production phase will be infinitesimal and the extrapolated pressure p will be equal to the initial pressure pi which is also the average pressure p. This corresponds to the so-called infinite reservoir case for which pD (tD) in equ. (7.37) may be evaluated under transient conditions, equ. (7.23), and hence the last two terms in the former equation will cancel each other out. Apart from this special case p cannot be thought of as having any clearly defined physical meaning but is merely a mathematical device used in calculating the average reservoir pressure. Thus evaluating equ. (7.37) for infinite closed in time gives ( ) D 12 i D D 4t 2 kh p p p (t ) ln q π µ γ − = − (7.39) and subtracting this equation from the material balance for the bounded drainage volume, equ. (7.38), and multiplying throughout by 2, gives ( ) ( ) D DA D D 4t 4 kh p p 4 t ln 2p t q π π µ γ − = + − (7.40) Since p is obtained from the extrapolation of the observed pressure trend on the Horner buildup plot, then p can be calculated once the right hand side of equ. (7.40) has been correctly evaluated. This, of course, gets back to the old problem of how can pD (tD), the dimensionless pressure, be determined for any value of tD, which is the dimensionless flowing time prior to the survey? Matthews, Brons and Hazebroek derived pD (tD) functions for a variety of bounded geometrical shapes and for wells asymmetrically situated with respect to the boundary using the so-called "method of images" with which the reader who has studied electrical potential field theory will already be familiar. The method is illustrated for a 2 : 1 rectangular bounded reservoir in fig. 7.10. a j Fig. 7.10 Part of the infinite network of image wells required to simulate the no-flow condition across the boundary of a 2 : 1 rectangular part of a reservoir in which the real well is centrally located Very briefly, in order to maintain a strict no-flow condition at the outer boundary requires the placement of an infinite grid of virtual or image wells, a part of such an OILWELL TESTING 177 array being shown in fig. 7.10, each well producing at the same rate as the real well within the boundary. The constant terminal rate solution for this complex system can then be expressed as ( ) ( ) 2 j D 1 1 2 2 i wf D D j 2 ca 4t 2 kh p p p t ln ei q 4kt φ µ π µ γ = ∞ − = = + (7.41) in which the first term on the right hand side of the equation is the component of the pressure drop due to the production of the well itself, within an infinite reservoir, equ. (7.23), and the infinite summation is the contribution to the wellbore pressure drop due to the presence of the infinite array of image wells which simulate the no-flow boundary. The exponential integral function is the line source solution of the diffusivity equation introduced in sec. 7.2, equ. (7.11), for the constant terminal rate case and is necessitated by the fact that the distance aj between the producing well and the jth image well is large so that the logarithmic expression of the line source solution, equ. (7.10), is an unacceptable approximation and the full exponential integral solution must be used. The infinite summation in equ. (7.41) is therefore an example of superposition in space of the basic constant terminal rate solution of the diffusivity equation. For further details of the mathematical technique the reader should consult the appendices of the original MBH paper7. Using this method to determine pD (tD), MBH were able to evaluate equ. (7.40) for a wide variety of boundary conditions and presented their results as plots of ( ) DA 4 kh p p vs. t q π µ − where tDA is the dimensionless flowing time. These charts have been included in this text as figs. 7.11-15. The individual plots are for different geometries and different asymmetries of the producing well with respect to the no-flow boundary. The MBH charts were originally designed to facilitate the determination of p from pressure buildup data by first determining p, by the extrapolation of the Horner plot, and k from the slope of the straight line. If an estimate is made of the area being drained, tDA = kt/φµcA can be calculated for the actual flowing time t. Then, using the appropriate MBH chart the value of 4πkh (p−p)/qµ is read off the ordinate from which p can be calculated. The details of this important technique will be described in sec. 7.7. For the moment, the MBH charts will be used in a more general manner to determine pD (tD) functions for the range of geometries covered by the charts and for any value of the flowing time. OILWELL TESTING 178 7 6 5 4 3 2 1 0 0.01 0.1 1 10 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 HEXAGON AND CIRCLE SQUARE EQUILATERAL TRIANGLE RHOMBUS RIGHT TRIANGLE D(MBH) 4 kh (p p) p q π µ − = D A k t t c A φ µ = Fig. 7.11 MBH plots for a well at the centre of a regular shaped drainage area7 (Reproduced by courtesy of the SPE of the AIME) OILWELL TESTING 179 7 6 5 4 3 2 1 0 0.01 0.1 1 10 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 II I III D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA 7 6 5 4 2 2 1 0 0.01 0.1 1 10 2 3 4 6 2 3 4 6 2 3 4 6 2 1 2 1 2 1 2 1 I II III IV D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA Fig. 7.12 MBH plots for a well situated within; a) a square, and b) a 2:1 rectangle7 (Reproduced by courtesy of the SPE of the AIME) OILWELL TESTING 180 5 4 3 2 1 0 -1 0.01 0.1 1 10 2 3 4 6 2 3 4 6 2 3 4 6 -2 I II III IV 4 4 4 4 1 1 1 1 D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA 5 4 3 2 1 0 -1 0.01 0.1 1 10 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 I II III IV 1 2 4 5 -2 D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA Fig. 7.13 MBH plots for a well situated within; a) a 4:1 rectangle, b) various rectangular geometries7 (Reproduced by courtesy of the SPE of the AIME) OILWELL TESTING 181 0.01 0.1 2 3 4 5 1 2 3 4 5 10 2 3 4 5 100 2 3 4 5 -1 0 1 2 3 4 5 6 I WELL 1/8 OF HEIGHT AWAY FROM SIDE II WELL 1/8 OF HEIGHT AWAY FROM SIDE III WELL 1/8 OF HEIGHT AWAY FROM SIDE II III I 2 2 D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA Fig. 7.14 MBH plots for a well in a square and in rectangular 2:1 geometries7 0.01 0.1 2 3 4 5 1 2 3 4 5 10 2 3 4 5 100 2 3 4 5 -3 -2 -1 0 1 2 3 4 I WELL 1/16 OF LENGTH AWAY FROM SIDE II WELL 1/4 OF ALTITUDE AWAY FROM APEX 1 I 2 II D(MBH) 4 kh (p p) p q π µ − = tDA = kt φµcA Fig. 7.15 MBH plots for a well in a 2:1 rectangle and in an equilateral triangle7 (Reproduced by courtesy of the SPE of the AIME). OILWELL TESTING 182 As indicated by Cobb and Dowdle8, equ. (7.40) can be solved for pD (tD) as ( ) ( ) D 1 1 2 2 D D DA D(MBH) DA 4t p t 2 t ln p t π γ = + − (7.42) in which ( ) ( ) D(MBH) DA 4 kh p t p p q π µ = − is the dimensionless MBH pressure, which is simply the ordinate of the MBH chart evaluated for the dimensionless flowing time tDA. Equation (7.42) is extremely important since it represents the constant terminal rate solution of the diffusivity equation which, for the case of a well draining from the centre of a bounded, circular part of a reservoir, replaces the extremely complex form of equ. (7.34). It should be noted, however, that the mathematical complexity of equ. (7.34) is not being avoided since it is implicitly included in the MBH charts which were evaluated using the method of images. Furthermore, equ. (7.42) is not restricted to circular geometry and can be used for the range of the geometries and well asymmetries included in the MBH charts. For very short values of the flowing time t, when transient conditions prevail, the left hand side of equ. (7.42) can be evaluated using equ. (7.23) and the former can be reduced to ( ) ( ) D(MBH) DA DA 4 kh p t p p 4 t q π π µ = − = (7.43) Alternatively, for very long flowing times, when semi-steady state conditions prevail, the left hand side of equ. (7.42) can be expressed as equ. (7.27) and in this case equ. (7.42) becomes ( ) ( ) ( ) 2 w D(MBH) DA A D A DA r 4 kh p t p p ln C t ln C t q A π µ = − = = (7.44) Inspection of the MBH plots of fig. 7.11, for a well situated at the centre of a regular shaped bounded area, illustrates the significance of equs. (7.43) and (7.44). For small values of the dimensionless flowing time tDA the semi-log plot of pD(MBH) vs tDA is non-linear while for large tDA the plots are all linear as predicted by equ. (7.44), and have unit slope (dpD(MBH)/d (In tDA) = 1). This latter feature is common to all the MBH charts, figs. 7.11-15. that in each case there is a value of tDA, the magnitude of which depends on the geometry and well asymmetry, for which the plots become linear indicating the start of the semi-steady state flow condition. Furthermore, for the symmetry conditions of fig. 7.11 there is a fairly sharp transition between transient and semi-steady state flow at a value of tDA ≈ 0.1, which confirms the conclusion reached in sec. 7.4 and exercise 7.4. For the geometries and various degrees of well asymmetry depicted in the remaining charts, however, there is frequently a pronounced degree of curvature extending to quite large values of tDA before the start of semi-steady state flow. This OILWELL TESTING 183 part of the plots represents both the pure transient flow period, equ. (7.43), and the late transient period and it is not worthwhile trying to distinguish between the two. Equation (7.44) is interesting since it reveals how the Dietz shape factors were originally determined. Dietz, whose paper on pressure analysis9 was published some years after that of MBH, evaluated the relationship expressed in equ. (7.44) for the specific value of tDA = 1, thus ( ) ( ) DA D(MBH) DA A t 4 kh p t 1 p p ln C 1 q π µ = = − = = (7.45) Values of In CA (and hence CA ) could be determined as the ordinate of the MBH charts for each separate plot corresponding to the value of tDA = 1, and these are shown in fig. 6.4. In some cases of extreme well asymmetry, late transient flow conditions still prevail at tDA = 1 (e.g. fig 7.13) and in these cases the linear trend of pD(MBH) versus tDA must be extrapolated back to the specific value tDA = 1 to determine the correct shape factor. The usefulness of the Dietz shape factors in the formulation of equations describing semi-steady state flow, for which they were derived, has already been amply illustrated in this text. The importance of equ. (7.42) for generating dimensionless pressure functions for a variety of boundary conditions and for any value of the flowing time cannot be overemphasised. It is rather surprising that this equation has been lying dormant in the literature since 1954, the date of the original MBH paper, with its full potential being largely unrealised. It appears in disguised form in many papers and even in the classic Matthews, Russell, SPE monograph6 (equ. 10.18, p. 109), yet it was not presented in the simple form of equ. (7.42) of this text until it was highlighted in a brief J.P.T. Forum article in 1973 by Cobb and Dowdle8. The latter use a slightly modified form of the equation in which the right hand side of equ. (7.42) is expressed strictly as a function of tDA, thus ( ) ( ) 1 1 1 2 2 2 D DA DA DA D(MBH) DA 2 w 4A p t 2 t ln t ln p t r π γ = + + − (7.46) In application to general oilwell test analysis, any rate-time-pressure sequence can be analysed using the following general equations ( ) ( ) n n j 1 n i wf j D D D n j 1 2 kh p p q p t t q S π µ − = − = ∆ − + (7.31) in which ( ) ( ) n j 1 D D D D D p t t p t − ′ − = can be evaluated using either equ. (7.42) or (7.46), for dimensionless time arguments D t′ or DA t′ , respectively and as will be shown in Chapter 8, with slight modification, the same combination of equations can also be applied to gas well testing. Theoretically, at least, the use of equ. (7.42) to quantify the pD function in equ. (7.31) removes the problem of trying to decide under which flowing condition pD should be evaluated because it is valid for all flowing times. Even if tDA exceeds the maximum value on the abscissa of the MBH chart, the plots are all linear at this point, and therefore pD(MBH) can readily be calculated by linear extrapolation. For OILWELL TESTING 184 very short or very long flowing times equ. (7.42) reduces to equ. (7.23) and (7.27) respectively, which can be verified by using the argument used to derive equ. (7.43) and (7.44) in reverse, i.e. by evaluating pD(MBH) in equ. (7.42) as being equal to 4πtDA and In (CA tDA), respectively. The relative ease with which pD functions can be generated using the MBH charts is illustrated in the following exercise which is an extension of exercise 7.2. EXERCISE 7.5 GENERATION OF DIMENSIONLESS PRESSURE FUNCTIONS The analysis of the single rate drawdown test, exercise 7.2, indicated that the Dietz shape factor for the 35 acre drainage area had the value CA = 5.31. The tabulated values of CA presented in fig. 6.4 indicate that there are three geometrical configurations with shape factors in the range of 4.5 to 5.5 which are shown in fig. 7.16. 2 4 1 1 (a) (b) (c) C = 4.57 A C = 4.86 A C = 5.38 A Fig. 7.16 Geometrical configurations with Dietz shape factors in the range, 4.5-5.5 The geological evidence suggests that the 2 : 1 geometry, fig. 7.16(b), is probably correct. Using the basic data and results of exercise 7.2, confirm the geological interpretation by comparing the observed pressure decline, table 7.1, with the theoretical decline calculated for the three geometries of fig. 7.16. EXERCISE 7.5 SOLUTION The constant terminal rate solution of the radial diffusivity equation, in field units, is, equ. (7.19), ( ) ( ) 3 i wf D D o 7.08 10 kh p p p t S q B µ − × − = + in which the pD function can be determined using equ. (7.46); and evaluating for the data and results of exercise OILWELL TESTING 185 7.2 (i.e. k = 240 mD, A = 35 acres, S = 4.5). wf DA DA D(MBH) DA 0.0189(3500 p ) 2 t ½ In t 8.632 ½ p (t ) 4.5 π − = + + − + (7.46) in which DA -6 DA .000264 kt .000264 240 t (hrs) t cA .18 1 15 10 35 43560 t 0.0154 t φ µ × × = = × × × × × = For convenience, equ. (7.46) can be reduced to wf D(MBH) DA 0.0189 (3500 p ) ½ p (t ) α − = − (7.47) in which DA DA 2 t ½ ln t 13.132 α π = + + and has the same value for all three geometries shown in fig. 7.16. Values of pwf in equ. (7.47) can therefore be calculated by reading values of pD(MBH) (tDA) from the appropriate MBH plots contained in figs. 7.11-15. The values of ½ pD(MBH) (tDA) and pwf for all three geometries are listed in table 7.4 for the first 50 hours of the drawdown test. Plots of ∆pwf, which is the difference between the calculated and observed bottom hole flowing pressure, versus the flowing time, are shown in fig. 7.17. These plots tend to confirm that the geological interpretation, fig. 7.16(b), is appropriate. For the other two rectangular geometries the late transient flow period is not modelled correctly. For comparison the plot has also been made for the case of a well draining from the centre of circular bounded area, which is the simple case normally considered in the literature. As can be seen, the value of ∆pwf after 50 hours, when semi-steady state conditions prevail, is 44 psi for this latter case. OILWELL TESTING 186 = 2 = 1 TIME (hrs) 10 50 40 30 20 10 0 -20 -30 -10 20 30 40 50 = = x 4 1 wf p (psi) ∆ Fig. 7.17 Plots of ∆ ∆ ∆ ∆pwf (calculated minus observed) wellbore flowing pressure as a function of the flowing time, for various geometrical configurations (Exercise 7.5) To facilitate the calculation of dimensionless pressure functions, as illustrated in exercise 7.5, the MBH dimensionless plots of pD(MBH) versus tDA can be expressed in digitised form and used as a data bank in a simple computer program to evaluate pD functions by applying equ. (7.46) (which can always be reduced to the form of equ. (7.47)). In fact such a program could be written for the range of small desk calculators which have sufficient memory storage capacity. Digitised MBH functions have already been presented by Earlougher et al10 for all the rectangular geometrical configurations considered by Matthews, Brons and Hazebroek. OILWELL TESTING 187 t (hrs) Observed pwf (psia) tDA α 2 1 4 1 ½ pD(MBH) pwf ½ pD(MBH) pwf ½ pD(MBH) pwf ½ pD(MBH) pwf 1 2917 .0154 11.142 .093 2915 .093 2915 .093 2915 .093 2915 2 2900 .0308 11.585 .151 2895 .192 2897 .146 2895 .134 2897 3 2888 .0462 11.885 .167 2880 .267 2885 .171 2880 .285 2886 4 2879 .0616 12.126 .163 2867 .331 2876 .180 2868 .397 2879 5 2869 .0770 12.334 .148 2855 .357 2866 .168 2856 .474 2872 7.5 2848 .1155 12.779 .117 2830 .406 2845 .168 2833 .663 2859 10 2830 .1540 13.164 .117 2810 .429 2826 .194 2814 .809 2846 15 2794 .2310 13.851 .158 2776 .441 2790 .253 2781 1.008 2820 20 2762 .3080 14.478 .213 2745 .450 2758 .327 2751 1.152 2795 30 2703 .4620 15.649 .387 2692 .497 2697 .481 2698 1.357 2744 40 2650 .6160 16.760 .536 2642 .589 2644 .618 2646 1.501 2693 50 2597 .7700 17.839 .643 2590 .666 2591 .729 2595 1.602 2641 TABLE 7.4 ) DA DA 2 t ½ ln t 13.132 α π = + + OILWELL TESTING 188 In addition, the Earlougher paper describes a relatively simple method for generating MBH functions for rectangular geometries other than those included in figs. 7.11-15 and for boundary conditions other than the no-flow condition which is assumed for the MBH plots. MBH functions for a constant outer boundary pressure and for cases in between pressure maintenance and volumetric depletion, corresponding to partial water drive, can therefore be simulated. Ramey et al11 have also described the simulation of well test analysis under water drive conditions. However, while the theory exists to describe variable pressure conditions at the drainage boundary, the engineer is still faced with the perennial problem of trying to determine exactly what outer boundary condition he is trying to simulate. To use the combination of equ. (7.31) and (7.42) to describe any form of oilwell test appears at first sight to offer a simplified generalization of former analysis techniques, yet, as will be shown in the remainder of this chapter and in Chapter 8, the approach introduces certain difficulties. Providing the test is run under transient flow conditions then the pD function, equ. (7.42), can be described by the simplified form D D D 4t p (t ) ½ ln γ = (7.23) in which there is no dependence upon the magnitude or shape of the drainage boundary nor upon the degree of asymmetry of the well with respect to the boundary. Therefore, if well tests are analysed using equ. (7.23) in conjunction with equ. (7.31), the results of the test will only yield values of the permeability, k, (which is implicit in the definition of tD) and the mechanical skin factor, S. As soon as the test extends for a sufficient period of time so that either late transient or semi-steady state conditions prevail then the effect of the boundary of the drainage area begins to influence the constant terminal rate solution and the full pD function, equ. (7.42), must be used in the test analysis. In this case the interpretation can become a great deal more complex because new variables, namely, the area drained, shape and well asymmetry, are introduced which are frequently additional unknowns. Exercise 7.5 illustrated how a single rate drawdown test can be analysed to solve for these latter three parameters using pD functions expressed by equ. (7.42), thus gaining additional information from the test. Largely due to the fact that test analysis becomes more complicated when tests are run under conditions other than that of purely transient flow, the literature is permeated with transient analysis techniques. This mathematical simplification does indeed produce convenient analysis procedures but can, in some cases, lead to severe errors in determining even the basic parameters k and S from a well test, particularly in the case of multi-rate flow testing as will be illustrated in sec. 7.8. Fortunately, the pressure buildup test, if it can be applied, leads to the unambiguous determination of k and S and therefore this method will be described in considerable detail in sec. 7.7. OILWELL TESTING 189 7.7 PRESSURE BUILDUP ANALYSIS TECHNIQUES The remaining sections of this chapter concentrate on the practical application of the theory developed so far to the analysis of well tests. It is considered worthwhile at this stage to change from Darcy to field units since, in practice, tests are invariably analysed using the latter and the majority of the literature on the subject employs these units. All equations in the remainder of this chapter will therefore be formulated using the field units specified in table 4.1. Since a great many of the equations are expressed in dimensionless parameters they remain invariant, or at least partially invariant in form. For instance, the most significant equation in the present subject of pressure buildup analysis is that describing the theoretical linear buildup, which in Darcy units, is ( ) ( ) D i ws(LIN) D D 4t 2 kh t t p p ½ ln p t ½ ln q t π µ γ + ∆ − = + − ∆ (7.37) and which, on conversion to field units becomes ( ) ( ) -3 D i ws(LIN) D D o 4t kh t t 7.08 10 p p 1.151 log p t ½ ln q B t µ γ + ∆ × − = + − ∆ (7.48) The conversion of the left hand side of this equation has already been described in exercise 7.3 and is necessary to preserve this expression as dimensionless, in field units. The only change to the right hand side is that the natural log of the dimensionless time ratio has been replaced by log10, which is mainly required for plotting purposes, the remainder of the equation is invariant in form. Thus the pD function is still D D D DA D(MBH) DA 4t p (t ) 2 t ½ ln ½ p (t ) π γ = + − (7.42) which is totally invariant, although in evaluating this expression it must be remembered that now D 2 w kt t 0.000264 (t-hours) cr φµ = (7.20) and DA kt t 0.000264 (t-hours) cA φµ = (7.49) The pD(MBH) term is, in the majority of cases, just a number read from the MBH charts corresponding to tDA evaluated in field units. Only when used to calculate p using the MBH method does it require interpreting as ( ) D (MDH) o kh p 0.01416 p p q B µ = − The Horner plot for a typical buildup is shown as fig. 7.18. OILWELL TESTING 190 4 3 2 1 0 observed data equation (7.48) pws(LIN) I − hr p (psi) ws small ∆t large ∆t + ∆ ∆ t t log t + ∆ ∆ t t t 10000 1000 100 10 1 m p Fig. 7.18 Typical Horner pressure buildup plot The first part of the buildup is usually non-linear resulting from the combined effects of the skin factor and afterflow. The latter is due to the normal practice of closing in the well at the surface rather than downhole and will be described in greater detail in sec 7.11. Thereafter, a linear trend in the plotted pressures is usually observed for relatively small values of ∆t and this can be analysed to determine the effective permeability and the skin factor. The former can be obtained by measuring the slope of the straight line, m, and from equ. (7.48) it is evident that o q B m 162.6 psi/log.cycle kh µ = (7.50) Providing the well is fully penetrating and the PVT properties are known, equ. (7.50) can be solved explicitly for k. The skin factor can be determined using the API recommended procedure which consists of subtracting equ. (7.48), the theoretical equation of the linear buildup, from the constant terminal rate solution which describes the pressure drawdown prior to closure and in field units is -3 i wf D D o kh 7.08 10 (p p ) p (t ) S q B µ × − = + (7.51) where pwf is the bottom hole flowing pressure at the time of closure and t the flowing time. The subtraction results in ( ) ws(LIN) -3 D wf o 4t kh t t 7.08 10 p p ½ ln S 1.151 log q B t µ γ + ∆ × − = + − ∆ which may be solved for S to give OILWELL TESTING 191 ( ) ws(LIN) wf 2 w p p t t 0.000264 4kt S 1.151 log log t m cr γ φ µ − + ∆ × = + − ∆ in which m is the slope of the buildup. Finally, evaluating this latter equation for the specific value of ∆t = 1 hour, and assuming that t >> ∆t gives ( ) ws(LIN) 1 -hr wf 2 w p p k S 1.151 log 3.23 m cr φ µ − = − + (7.52) in which pWS(LIN) 1-hr is the hypothetical closed-in pressure read from the extrapolated linear buildup trend at ∆t = 1 hour as shown in fig. 7.18. It should be noted in connection with the determination of the permeability from the buildup plot that k is in fact, the average effective permeability of the formation being tested, thus for the simultaneous flow of oil and water in a homogeneous reservoir ( ) ( ) ro w abs k k k S = × (7.53) in which ( ) ro w k S is the average relative permeability representative for the flow of oil in the entire formation and is a function of the thickness averaged water saturation prevailing at the time of the survey. It has been assumed until now that reservoirs are perfectly homogeneous. In a test conducted in an inhomogeneous, stratified reservoir, however, providing the different layers in the reservoir are in pressure communication, the measured permeability will be representative of the average for the entire layered system for the current water saturation distribution. The concept of averaged (relative) permeability functions which account for both stratification and water saturation distribution will be described in detail in Chapter 10. The permeability measured from the buildup, or for that matter from any well test, is therefore the most useful parameter for assessing the well's productive capacity since it is measured under in-situ flow conditions. Problems occur in stratified reservoirs when the separate sands are not in pressure communication since the individual layers will be depleted at different rates. This leads to pressure differentials between the layers in the wellbore, resulting in crossflow. It is also important to note that in the subtraction of equ. (7.48) from equ. (7.51) to determine the skin factor, the pD (tD) functions in each equation disappear leading to an unambiguous determination of S. If this were not the case then one could have little confidence in the calculated value of S since the evaluation of pD (tD) at the time of closure may require a knowledge of the geometry of the drainage area and degree of well asymmetry with respect to the boundary. This point is made at this stage to contrast this method of determining the skin factor with the method which will be described in sec. 7.8, for multi-rate flow tests, in which the calculation of S does rely on the correct determination of the dimensionless pressure functions throughout the test. Figure 7.19 shows the effect of the flowing time on the Horner buildup plot. For an initial well test in a reservoir, if the flowing period prior to the buildup is short, then pD OILWELL TESTING 192 (tD) in equ. (7.48) can be approximated as ½ In(4 tD /γ) and the last two terms in the equation will cancel resulting in the simple buildup equation o ws i q B t t p p 162.2 log kh t µ + ∆ = − ∆ (7.54) which corresponds to the plot for t ≈ 0 in fig. 7.19. equation (7.48) actual buildup pws 4 3 2 1 0 p p p p p p (initial survey) 1- year 6 - months t 0 ≈ ≈ log t + ∆t ∆t Fig. 7.19 Illustrating the dependence of the shape of the buildup on the value of the total production time prior to the survey The same result can also be obtained by evaluating both pD functions in the theoretical buildup equation, (7.32), for transient flow. Equation (7.54) is the original Horner buildup equation4, for the infinite reservoir case, in which the extrapolated buildup pressure p = pi, the initial reservoir pressure, when log ( ) t t 0, t . t + ∆= ∆=∞ ∆ Furthermore, if the amount of oil withdrawn from the reservoir prior to the survey is negligible in comparison with the oil in place then the initial pressure is approximately equal to the average pressure thus, i p p p = ≈ . As the flowing time before the survey increases, so that the pD function in equ. (7.48) can no longer be evaluated under transient conditions, then the difference between the last two terms in equ. (7.48), i.e. pD (tD) − ½ In(4 tD/γ), continuously increases with the flowing time ref. exercise 7.4, fig. 7.5. Two cases are shown in fig. 7.19 for surveys conducted six months and one year after the initial survey in a well producing at a constant rate. As the flowing time increases the entire buildup is displaced downwards in fig. 7.19, resulting in ever decreasing values of p and p. This is to be expected since for long flowing times there is a significant withdrawal of oil prior to the survey and this reduces the average reservoir pressure. Such surveys correspond to the routine tests conducted in wells at regular intervals throughout the producing life of the reservoir. The main aim of these tests is to determine the average pressure within each drainage volume and hence, using equ. (5.13), the average pressure in the entire reservoir for use in the material balance equation. OILWELL TESTING 193 Since the production history of any oilwell consists of periods during which the rates vary considerably, including periods of closure for repair and testing, it may be felt by the reader that to interpret any buildup test conducted after a lengthy period of production would require the application of the superposition principle as presented in equ. (7.31) to obtain meaningful results. Fortunately, this is not necessary providing that the well is producing under semi-steady state conditions at the time of the survey. The following argument will show that, in this case, the real time can be replaced by the effective flowing time, defined by equ (7.14), without altering the value of the average pressure calculated from the buildup analysis. Suppose a well has been producing with a variable rate history prior to closure at real time tn for a buildup survey. If the final production rate is qn during the period (tn − tn-1), then the wellbore pressure at any time ∆t during the buildup can be determined using the equation ( ) ( ) ( ) n j n j -3 i ws D D D 1 D D j 1 n o n q kh 7.08 10 p p p t t p t q B q µ − = ∆ × − = + ∆ − ∆ (7.55) which is simply a direct application of equ. (7.31) for the variable rate history, including the buildup. It is analogous to the theoretical buildup equation, (7.32), which was derived for constant rate production during the entire history of the well. Therefore, repeating the steps taken in the derivation of equ. (7.37) from equ. (7.32), equ. (7.55) can be expressed as ( ) ( ) n n j -3 n i ws(LIN) n o n D j D D D 1 j 1 n t t kh 7.08 10 p p 1.151 log q B t 4t q p t t ½ ln q µ γ − = + ∆ × − = ∆ ∆ + + ∆ − (7.56) which is the theoretical linear equation which matches the actual buildup for small values of ∆t. Implicit in the derivation of equ. (7.56) is the condition that the final flow period (tn − tn- 1) >> ∆t, thus the last two terms in the equation are constants evaluated at time tn. Alternatively, if the effective flowing time t = Np/q is used in the analysis then a different buildup plot will be obtained for which the early linear trend can be matched by equ (7.48), in which the final flow rate is qn, i.e. ( ) ( ) -3 D i ws(LIN) D D o 4t kh t t 7.08 10 p p 1.151 log p t ½ ln q B t µ γ + ∆ × − = + − ∆ (7.48) The two buildup plots for real and effective flowing time are shown as lines A and B, respectively, in fig. 7.20. OILWELL TESTING 194 4 3 2 1 0 A B C m m m p p p sss equ. (7.56) equ. (7.48) pws n t m log t sss t m log t s ss t t log tn t log p tn log t + ∆t ∆t Fig. 7.20 Analysis of a single set of buildup data using three different values of the flowing time to draw the Horner plot. A - actual flowing time; B - effective flowing time; C - time required to reach semi-steady state conditions It should be noted that the difference between plots A and B is not the same as the difference between the buildups shown in fig. 7.19. The latter diagram is for three separate sets of data, pws as a function of ∆t, obtained in three different surveys. These curves are displaced downwards as a function of the flowing time, that is, as a function of the reservoir depletion. What is shown in fig. 7.20, however, is a single set of pressure-time data interpreted as Horner plots for different assumed values of the flowing time. Both the linear, extrapolated buildups, equs. (7.56) and (7.48), have the same slope m, which is dictated by the final flow rate qn. The difference between them is that a value of pws on plot A is displaced laterally by an amount n n t t t t t log log log t t t + ∆ + ∆ − ≈ ∆ ∆ with respect to the same value on plot B, providing that both t and tn>>∆t. Therefore, as shown in fig. 7.20, there is a vertical difference m log (tn/t) between the buildups for a given value of ∆t which can be interpreted as n n t t p p m log t − = (7.57) where n t p and p are the extrapolated values of pws(LIN) at ∆t = ∞ for the real and effective flowing time, respectively. In addition, if it is assumed for a routine survey that the final flow period is sufficiently long so that flow is under semi-steady state conditions, then the MBH equation, (7.44), from which p can be calculated, is OILWELL TESTING 195 D(MBH) DA A DA n o kh p (t )0.01416 (p - p) 2.303 log (C t ) q B µ = (7.58) or ( ) ( ) n o A DA A DA q B p p 162.6 log C t m log C t kh µ − = = (7.59) Equation (7.59) is appropriate for the effective flowing time while for the real time n n n t t A DA p p m log(C t ) − = (7.60) Subtracting equ. (7.59) from (7.60) gives ( ) ( ) n n n t t t p p p p m log t − − − = (7.61) which, when compared with equ. (7.57), shows that n t p p − and therefore the determination of the average pressure using the MBH method is the same whether the real or effective flowing time is employed in the analysis. Using an identical argument it can easily be demonstrated that the average pressure determined from a survey is independent of the flowing time used in the analysis. This is correct providing that the flowing time is equal to or greater than tsss, the time required for semi-steady state conditions to be established, within the drainage volume, and the final production rate is also used in the analysis. As an illustration of this statement, plot C in fig. 7.20 has been drawn for the limiting value of tsss. In this case plot C is laterally displaced with respect to plot B, for the effective flowing time, so that the equivalent equations to equ. (7.57) and (7.61) are now sss sss t p p m log t − = and ( ) ( ) sss sss sss t p p p p m log t − − − = which shows that the MBH analysis technique will yield the same values of p whether t or tsss is used to plot the buildup. This same conclusion has been presented in the literature by Pinson12 and Kazemi13. It should also be noted that the value of the skin factor determined from the analysis is also independent of the flowing time. This is because the value of pws(LIN)1-hr required for the calculation of S, (equ. (7.52), does not depend on the flowing time and is the same for plots A, B and C in fig. 7.20. It is for the above reasons that the convenient combination of final flow rate and effective flowing time is generally used in buildup analysis. The only assumption that can be regarded as restrictive is that the final flow period should be of sufficient OILWELL TESTING 196 duration so that semi-steady state flow conditions prevail at the time of closure, and even if this condition is not exactly satisfied the error introduced will be rather small. The occasion when the use of this rate-time combination may not be acceptable is for initial tests when the well may be produced for a relatively short period of time at an uneven rate. Odeh and Selig14 have described a method for buildup analysis under these conditions which can improve the accuracy of the results. In the remaining description of pressure buildup analysis the effective flowing time will be used exclusively and denoted by t, and the final production rate by q. An example of the use of tsss in buildup analysis for a gas well will be described in Chapter 8, sec. 11. Having plotted the observed pressures according to the interpretation method of Horner, the MBH method can be applied to determine p according to the following recipe. 1) Extrapolate the early linear buildup trend to t t 0 t + ∆= ∆ and determine the value of p. From the slope of the straight line calculate k using equ. (7.50). 2) Divide the reservoir into drainage volumes so that i i TOT TOT q V q V = where qi is the production rate for the ith well draining a reservoir bulk volume Vi and qTOT and VTOT are the total rate and bulk volume of the reservoir respectively. This relationship has been shown in Chapter 5, sec. 5.3, to be valid for wells draining a reservoir under semi-steady state flow conditions, however, Matthews, Brons and Hazebroek assert that the relationship can be applied with reasonable accuracy irrespective of the prevailing flow condition. This step leads to the determination of Vi and hence Ai, the area drained by the well, can be estimated by assuming that the average thickness within the area is equal to that observed in the well. With the aid of a geological structural map of the reservoir, both the shape of the drainage area and position of the well with respect to the boundary can be roughly estimated to correspond to one of the MBH geometrical configurations shown in figs. 7.11-15. 3) Evaluate the dimensionless time DA kt t 0.000264 (t hours) cA φµ = − (7.49) using values of k and A obtained from steps 1) and 2) respectively. For liquid flow the µc product is small and constant but for two phase gas-oil and for single phase gas flow this is not the case which leads to certain difficulties in interpretation which will be described in Chapter 8. 4) Enter the appropriate MBH chart, fig. 7.11-15, and, for the curve corresponding closest to the estimated geometrical configuration, read the value of the ordinate, pD(MBH) (tDA), for the calculated value of the dimensionless (effective) flowing time tDA OILWELL TESTING 197 ( ) ( ) D(MBH) DA o kh p t 0.01416 p p q B µ = − (7.58) and, since p has been determined in step 1), then pcan be directly calculated. It should be noted that the MBH charts are equally appropriate for values of tDA and pD(MBH) evaluated in either Darcy or field units since both parameters are dimensionless. An equivalent method of determining p is that presented by Dietz9 in which the aim is to calculate the value of log t t t + ∆ ∆ at which to enter the Horner plot and read off the value of p directly from the extrapolated linear buildup, as illustrated in fig. 7.21. m pd pws 4 3 2 1 0 p p (p - p) log t + ∆t ∆t log t + ∆td ∆td log t + ∆ts ∆ts Fig. 7.21 The Dietz method applied to determine both the average pressure p and the dynamic grid block pressure d p Let ∆ts be the closed-in time for which the hypothetical pressure on the extra polated linear buildup equals the average reservoir pressure. Then ws(LIN) p p = in equ. (7.48), and the latter may be expressed as ( ) ( ) 3 s D i D D o s t t 4t kh 7.08 10 p p 1.151log p t ½ ln q B t µ γ − + ∆ × − = + − ∆ but the left hand side of this equation can be evaluated using equ. (7.38), the material balance, to give ( ) s D DA D D s t t 4t 2 t 1.151log p t ½ ln t π γ + ∆ = + − ∆ If the pD function in this equation is expressed in general form using equ. (7.42), then ( ) s D(MBH) s o t t kh 2.303 log p 0.01416 p p t q B µ + ∆ = = − ∆ OILWELL TESTING 198 or ( ) s s p p t t log t m − + ∆ = ∆ (7.62) This equation, in which m = 162.6 q µ Bo/kh, the slope of the buildup, demonstrates the equivalence between the Dietz and MBH methods, which is also illustrated in fig. 7.21. In particular, Dietz concentrated on buildup analysis for wells which were producing under semi-steady state conditions at the time of survey, in which case, applying equ (7.44), in field units pD(MBH) = 2.303 log (CA tDA) and therefore ( ) s A DA s t t log log C t t + ∆ = ∆ (7.63) from which the value of log s s t t t + ∆ ∆ at which to enter the Horner plot can be calculated. An extension of Dietz method to determine pis frequently used in comparing observed well pressures with average grid block pressures calculated by numerical simulation models. A Physical no-flow boundary Grid block boundaries in the numerical simulation Fig. 7.22 Numerical simulation model showing the physical no-flow boundary drained by well A and the superimposed square grid blocks used in the simulation Suppose that a numerical simulation model is constructed so that there are several grid blocks contained within the natural no-flow boundary of the well, as shown in fig. 7.22. At the end of each time step in the simulation, the average pressure in each grid block is calculated and printed out. Therefore, by interpolation in time between the simulated pressures, it is a relatively simple matter to determine the individual grid block pressures corresponding to the time at which a buildup survey is made in well A, OILWELL TESTING 199 whether the latter time coincides with the end of a simulation time step or not. There are then two ways of comparing the observed well pressure with the simulated grid block pressures. The first of these is to calculate the average pressure within the no-flow boundary at the time of survey, using the MBH or Dietz method, and compare this with the volume averaged pressure over all the grid blocks and partial blocks within the natural no-flow boundary. This is a rather tedious business. A simpler, approximate method has been introduced by van Poollen15 and further described by Earlougher16. This consists of using the Horner buildup plot in conjunction with the Dietz method to calculate the so-called "dynamic grid block pressure" pd which is simply the average pressure in the grid block containing the well at the time of survey. The analysis seeks to determine at what value of log d d t t t + ∆ ∆ should the Horner plot be entered so that the pressure read from the hypothetical linear buildup has risen to be equal to the dynamic pressure, i.e. pws(LIN) = pd. Again, equ. (7.63) can be applied but in this case tDA must be evaluated using the grid block area rather than that of the no-flow boundary and CA takes on the fixed value of 19.1. The reasoning behind the latter choice is that the grid block boundary is not a no-flow boundary. Instead the boundary condition corresponds more closely to that of steady state flow and for such Dietz has only presented one case corresponding to a well producing from the centre of a circle for which CA = 19.1, fig 6.4. Thus the rectangular grid block shape is approximated as circular with area equal to that of the grid block. Therefore, the Horner plot is entered for a value of ( ) d DA d t t log log 19.1 t t + ∆ = ∆ (7.64) and pd read from the linear buildup as shown in fig. 7.21. Again, use of equ. (7.64) depends on the fact that the well is flowing under stabilised conditions at the time of survey. Normally, in this case t >>∆td and van Poollen, using this assumption, has presented an expression for explicitly calculating the closed in time at which pWS(LIN) = pd. This can readily be obtained from equ. (7.64), as 2 e d c r cA t 0.000264x19.1k 0.005042k φ µ π φ µ ∆ = = or 2 e d cr t 623 k φ µ ∆ = where re is the radius of the circle with area equivalent to that of the grid block. This approximate but speedy method for comparing observed with simulated pressures is very useful in performing a history match on well pressures. The following two exercises will illustrate the application of the pressure buildup techniques, described in this section, to an under saturated oil reservoir. OILWELL TESTING 200 EXERCISE 7.6 HORNER PRESSURE BUILDUP ANALYSIS, INFINITE RESERVOIR CASE A discovery well is produced for a period of approximately 100 hours prior to closure for an initial pressure buildup survey. The production data and estimated reservoir and fluid properties are listed below q = 123 stb/d φ = 0.2 Np = 500 stb µ = 1 cp h = 20 ft Boi = 1.22 rb/stb rw = 0.3 ft c = 20×10-6/psi A ≈ 300 acres = (co So +cw +Sw + f c ) and the pressures recorded during the test are listed in table 7.5. 1) What is the initial reservoir pressure? 2) If the well is completed across the entire formation thickness, calculate the effective permeability. 3) Calculate the value of the mechanical skin factor. 4) What is the additional pressure drop in the wellbore due to the skin? 5) If it is initially assumed that the well is draining from the centre of a circle, is it valid to equate pi to p? Closed-in time ∆t (hrs) Wellbore pressure (psi) Closed-in time ∆t (hrs) Wellbore pressure (psi) 0.0 4506 (pwf ) 0.5 4675 3.0 4763 0.66 4705 4.0 4766 1.0 4733 6.0 4770 1.5 4750 8.0 4773 2.0 4757 10.0 4775 2.5 4761 12.0 4777 TABLE 7.5 EXERCISE 7.6 SOLUTION 1) The effective flowing time is p final N 500 t 24 24 97.6 hrs q 123 = × = × = Points on the Horner build up plot of pws versus log t t t + ∆ ∆ are listed in table 7.6. OILWELL TESTING 201 Closed-in time ∆t (hrs) t t t + ∆ ∆ log t t t + ∆ ∆ pws (psi) 0 4506(pwf) .5 196.2 2.29 4675 .66 148.9 2.17 4705 1.0 98.6 1.99 4733 1.5 66.1 1.82 4750 2.0 49.8 1.70 4757 2.5 40.04 1.60 4761 3.0 33.53 1.52 4763 4.0 25.40 1.40 4766 6.0 17.27 1.24 4770 8.0 13.20 1.12 4773 10.0 10.76 1.03 4775 12.0 9.13 .96 4777 TABLE 7.6 The pressure buildup plot, on a linear scale, is shown as fig. 7.23. The last seven points define a straight line and the extrapolation of this trend to the value of t t t + ∆ ∆ = 0 gives p = 4800 psi and, assuming the validity of equ. (7.54) for an initial well test, pi = 4800 psi. 4800 4700 4700 4600 4600 2 1 0 p=4800 psi p (psi) ws x p = ws (1 hr) 4752 psi + ∆ ∆ t t log t Fig. 7.23 Horner buildup plot, infinite reservoir case OILWELL TESTING 202 2) The slope of the linear section of the buildup plot is m = 24.5 psi/log cycle. Therefore, since the well is fully penetrating the effective permeability of the formation is oi 162.6q B 162.6 123 1 1.22 k 50 mD mh 24.5 20 µ × × × = = = × 3) The skin factor can be evaluated using equ. (7.52) in which the hypothetical value of pws(LIN) 1-hr = 4752 psi is obtained from the extrapolation of the linear buildup trend to ∆t = 1 hour, fig. 7.23, therefore, ( ) ( ) ( ws(LIN) 1 -hr wf 2 w 6 p p k S 1.151 log 3.23 m cr 4752 4506 50 1.151 log 3.23 24.5 .2 1 20 10 .09 φµ − − = − + − = − + × × × × (7.52) 4) The additional pressure drop due to the skin, while producing, can be calculated as q p S atm. 2 kh 2m S 0.87mS psi 2.303 128 psi µ π ∆ = × = × = = 5) The assumption that pi = p relies entirely on the fact that both pD functions in the theoretical buildup equation (7.32) can be evaluated under transient flow conditions so that the equation can be reduced to the simple form equ. (7.54) for the infinite reservoir case. As already noted for a well at the centre of a circular bounded reservoir, there is a fairly sharp change from transient to semi-steady state flow for a value of tDA ≈ 0.1. Therefore, for the effective flowing time of 97.6 hours, the minimum area for which the assumption is valid is min min 6 0.000264kt 1 A 0.1 c 43560 0.000264 50 97.6 1 A 74 acres 43560 0.1 2 1 20 10 φ µ − = × × × = × ≈ × × × × and since the estimated area is 300 acres the assumption that p = pi is perfectly valid as the test is conducted entirely under transient flow conditions. EXERCISE 7.7 PRESSURE BUILDUP TEST ANALYSIS: BOUNDED DRAINAGE VOLUME A pressure buildup test is conducted in the same well described in exercise 7.6, some seven and a half months after the start of production. At the time, the well is producing 400 stb/d and the cumulative production is 74400 stb. The only change in the well data presented in the previous exercise is that Bo has increased from 1.22 to 1.23 rb/stb. The closed-in pressures as a function of time are listed in table 7.7. OILWELL TESTING 203 Closed-in time ∆t (hrs) Closed-in pressure pws (psi) Closed-in time ∆t (hrs) Closed-in pressure pws (psi) 0 1889 6 2790 0.5 2683 7.5 2795 1 2713 10 2804 1.5 2743 12 2809 2 2752 14 2813 2.5 2760 16 2817 3 2766 20 2823 3.5 2771 25 2833 4 2777 30 2840 4.5 2779 36 2844 5 2783 TABLE 7.7 At the time of the survey several wells are draining the reservoir and the well in question is estimated to be producing from a 2:1 rectangle with area 80 acres. The position of the well with respect to its no-flow boundary is shown in fig. 7.24. 80 acre drainage area Reservoir boundary Internal no-flow boundary Numerical simulation grid 2 1 Fig. 7.24 Position of the well with respect to its no-flow boundary; exercise 7.7 1) From the Horner plot determine k, S andp, the average pressure within the drainage volume. If the reservoir is modelled with grid block boundaries corresponding to the dashed lines in fig. 7.24, calculate the dynamic pressure in the grid block containing the well at the time of the survey. 2) Plot the theoretical linear buildup, equ. (7.48), and the actual buildup for this 2:1 rectangular geometry. EXERCISE 7.7 SOLUTION 1) The conventional Horner plot of the observed pressures is drawn in fig. 7.25 as the set of circled points, the plot is for an effective flowing time of t = 74400/400 × 24 OILWELL TESTING 204 = 4464 hours. The early linear trend of the observed data, for 1.5 < t < 6 hours has been extrapolated to determine t t p 3020 psi for log 0 t + ∆ = = ∆ and pws(LIN) = 2727 psi for ∆t = 1 hour The slope of buildup is m = 80 psi/log cycle from which k can be calculated as o 162.6q B 162.6 400 1 1.23 k 50 mD mh 80 20 µ × × × = = = × and using equ. (7.52), the skin factor is p ( ws psi) 3000 2900 2800 2700 2600 4 3 2 1 0 + ∆ = ∆ s s t t log .942 t + ∆ = ∆ d d t t log 2.51 t + ∆ ∆ t t log t p = 2727 ws (1 - hr) x p = 3020 psi p (DIETZ) = 2944 psi p = 2820 psi d Fig. 7.25 Pressure buildup analysis to determine the average pressure within the no-flow boundary, and the dynamic grid block pressure (Exercise 7.7) OILWELL TESTING 205 p ( ws psi) 3000 2900 2800 2700 2600 p = 2943 psi + ∆ ∆ t t log t 5 4 3 2 1 0 2 1 4 1 o-o-o ACTUAL BUILDUP (OBSERVED PRESSURES) - - - -LINEAR EXTRAPOLATION OF ACTUAL BUILDUP (OBSERVED PRESSURES) ------THEORETICAL LINEAR BUILDUPS, EQU (7.66), FOR VARIOUS GEOMETRIES  THEORETICAL BUILDUPS, EQU (7.68), FOR VARIOUS GEOMETRIES Fig. 7.26 Influence of the shape of the drainage area and degree of well asymmetry on the Horner buildup plot (Exercise 7.7) ( ) 6 2727 1889 50 S 1.151 log 3.23 6.4 80 .2 1 20 10 .09 − − = − + = × × × × both of which agree with the values obtained in the previous exercise. The average pressure within the no-flow boundary at the time of survey (t = 4464 hours) can be calculated using the MBH method for the dimensionless flowing time DA 6 kt 0.000264 50 4464 t 0.000264 4.23 cA .2 1 20 10 80 43560 φµ − × × = = = × × × × × and consulting fig. 7.12 for the appropriate geometrical configuration, the ordinate of the MBH chart for this flowing time has the value ( ) ( ) ( ) ( ) D(MBH) O kh p 4.23 .01416 p p 2.23 q B 162.6 i.e. .01416 p p .0288 p p 2.23 m and therefore p (MBH) 2943 psi µ = − = × − = − = = OILWELL TESTING 206 The MBH curve, fig. 7.12 (IV), shows that for tDA = 4.23 semi-steady state flow conditions prevail in the reservoir and therefore the method of Dietz can also be applied to calculate p, i.e. ( ) ( ) s A DA s t t log log C t t log 2.07 4.23 0.942 + ∆ = ∆ = × = (7.63) and entering the buildup plot for this value of the abscissa gives the corresponding value of pws(LIN) = p as p (Dietz) 2944 psi = To determine the dynamic pressure in the grid block containing the well at the time of survey, equ. (7.64) can be applied for ' DA t = tDA × 4 since the grid block area is only one quarter of the total drainage area. Thus ( ) t t log log 19.1 16.92 2.51 t + ∆ = × = ∆ and from the buildup plot, the corresponding dynamic pressure can be read as pws(LIN) = pd = 2820 psi. 2) The theoretical equation of the straight line which matches the observed linear buildup is ( ) ( ) -3 D i ws(LIN) D D o 4t kh t t 7.08 10 p p 1.151log p t ½ ln q B t µ γ + ∆ × − = + − ∆ (7.48) and since 2 D DA w t t A /r = × , this may be expressed as ( ) ( ) i ws(LIN) D D t t 0.0144 p p 1.151 log p t 9.862 t + ∆ − = + − ∆ Taking several points on the straight line, pD (tD) can be evaluated as pD (tD ) = 35.49 and therefore, the correct linear equation is ( ) ws(LIN) t t 0.0144 4800 p 1.151 log 25.63 t + ∆ − = + ∆ (7.65) If the geometry and well position within the bounded area have been estimated correctly, then it should be possible to match equ. (7.65) by theoretically calculating D p using equ. (7.42) or, since semi-steady state conditions prevail at the time of the survey, D p can be alternatively expressed as OILWELL TESTING 207 ( ) D D DA 2 A w 4A p t ½ ln 2 t C r π γ = + (7.27) and substituting this in equ. (7.48) reduces the latter to ( ) ( ) ws(LIN) DA A DA t t 0.0144 4800 p 1.151 log 2 t ½ ln C t t t t 1.151 log t π α + ∆ − = + − ∆ + ∆ = + ∆ (7.66) where α = 26.58 - ½ ln (CA 4.23) To investigate the effect of the geometry of the drainage area and well asymmetry, α and hence equ. (7.66), has been evaluated for the three markedly different cases shown in table 7.8. Case Geometry Shape factor α A 2 1 2.07 25.50 B 31.6 24.13 C 4 1 0.232 26.59 TABLE 7.8 The value of α for the 2:1 rectangular geometry corresponds closely to the value obtained from the plotted points, equ. (7.65), thus tending to confirm the geometrical interpretation. The linear plots of equ. (7.66) for the three cases listed in table 7.8 are shown in fig. 7.26. The actual pressure buildup, as distinct from the linear buildup, can be determined using equ. (7.32) which, in field units and for the data relevant for this exercise, is ( ) ( ) ( ) i ws D D D D D 0.0144 p p p t t p t − = + ∆ − ∆ (7.67) This function must be evaluated for all values of the closed in time ∆t. Since the well is flowing under semi-steady state conditions at the time of the buildup pD (tD + ∆tD ) can be expressed as ( ) D D D DA DA 2 A w 4A p (t t ) ½ ln 2 t t C r π γ + ∆ = + + ∆ but the second pD function must be evaluated using equ. (7.42) as ( ) D D D DA D(MBH) DA 4 t p ( t ) 2 t ½ ln ½ p t π γ ∆ ∆ = ∆ + + ∆ Substituting these functions in equ. (7.67) gives ( ) ( ) ( ) ( ) i ws DA A DA DA D MBH 0.0144 p p 2 t ½ ln c t ½ p t π − = − ∆ + ∆ OILWELL TESTING 208 and subtracting equ. (7.66), the equation of the linear buildup, from this equation gives ( ) ( ) ( ) ws DA DA DA ws LIN D MBH t t 0.0288 (p p In t -In t ln p t t + ∆ = − = ∆ − + ∆ ∆ which can be simplified as ( ) ( ) ws DA D MBH t t 0.0288 p p t ln t + ∆ ∆ = ∆ − ∆ (7.68) in which ∆pws = pws(LIN) − pws, the pressure deviation below the linear buildup trend. Values of ∆pws as a function of ∆t are listed in table 7.9 for the three geometrical configurations presented in table 7.8. The actual pressure buildups for these three cases are included in fig. 7.26 by plotting the deviations ∆pws below the linear buildups. t = 4464 hrs 2 1 4 1 ∆t (hrs) ∆tDA ln t t t + ∆ ∆ pD(MBH) ∆pws (psi) pD(MBH) ∆pws (psi) pD(MBH) ∆pws (psi) 5 .005 .001 .063 2.1 .063 2.1 .063 2.1 10 .009 .002 .106 3.6 .113 3.8 .113 3.8 20 .019 .004 .176 6.0 .232 7.9 .224 7.6 50 .047 .011 .205 6.7 .591 20.1 .334 11.2 100 .095 .022 .133 3.9 1.163 39.6 .305 9.8 250 .237 .054 .100 1.6 2.013 68.0 −.081 −4.7 500 .473 .106 .224 4.1 2.744 91.6 −.634 −25.7 1000 .947 .202 .757 19.3 3.442 112.5 −1.030 −42.8 2500 2.367 .455 1.648 41.4 4.363 135.7 −.563 −35.3 5000 4.735 .751 2.324 54.6 5.032 148.6 .134 −21.4 TABLE 7.9 Exercises 7.6 and 7.7 illustrate the common techniques applied in pressure buildup analysis. One of the most reliable features of the analysis is that the Horner plot of observed pressures pws versus t t t + ∆ ∆ can be drawn without a knowledge of the pD function at the start of the survey. Furthermore, if a linear section of the plot can be defined for small values of the closed in time this can be analysed to determine the kh value and skin factor. In partially depleted reservoirs, in which the aim is also to determine the average pressure p, the analysis is necessarily more complex. The difficulty lies in the fact that to determine p requires a knowledge of the magnitude of the area drained and the geometrical configuration, including the well position with respect to the boundary. In OILWELL TESTING 209 other words, complex boundary conditions of the differential equations implicit in the analysis are required to obtain meaningful results. As fig. 7.26 clearly demonstrates, varying the boundary conditions can have a profound influence on the shape and position of the theoretical buildup plot. One hopeful feature in this diagram is again the fact that the observed data gives an absolute buildup plot. By the appropriate choice of the boundary condition it may therefore be possible to match the observed buildup as demonstrated in exercise 7.7, in which the original geological interpretation was confirmed. With a reasonable geological map of the reservoir the technique can be diagnostic in building a model of the current drainage patterns. In addition, attempting even some crude match of the observed buildup can eliminate serious error. If it were assumed, for instance, that the well in exercise 7.7 was located at the centre of a circle, which is the conventional boundary condition assumed in the literature, then the reader can confirm by calculation, or merely by inspection of fig. 7.26, that the estimated value of p calculated in this latter exercise would be about 100 psi too low. One other feature in fig. 7.26 is of interest and that is the rather strange shape of the theoretical buildup plot for the assumed 4:1 rectangular geometry. In this case there is a pronounced increase of slope which is due to the proximity of the no-flow boundaries. This is just a more complex manifestation of the phenomenon of "doubling of the slope" due to the presence of a fault close to a well in an otherwise infinite reservoir, which has repeatedly featured in the literature4,6. References 17 and 18 of this chapter are recommended to the reader who is further interested in the subject of matching theoretical with actual pressure buildups. 7.8 MULTI-RATE DRAWDOWN TESTING Closing in a well for a pressure buildup survey is often inconvenient since it involves loss of production and sometimes it is difficult, for a variety of reasons, to start the well producing again after the survey. Therefore, multi-rate drawdown testing is sometimes practised as an alternative means of measuring the basic reservoir parameters and indeed, in some places the regulatory bodies insist that such surveys be conducted in preference to other forms of testing. This restriction is more common in the case of gas well testing which will be described separately in Chapter 8, sec. 10. The basic equation for analysing a multi-rate drawdown test for liquid flow has already been presented in sec. 7.5 as equ. (7.33). In field units this becomes ( ) ( ) n n j 1 n i wf j 3 D D D j 1 o n n p p q kh 7.08 10 p t t S B q q µ − − = − ∆ × = − + (7.69) in which n wf p is the specific value of the flowing pressure at total flowing time tn during the nth production period at rate qn. It should also be noted that throughout this section t is the actual rather than effective flowing time. Consider the typical multi-rate test shown in fig. 7.27 for four sequential flow periods. OILWELL TESTING 210 Conventionally in the analysis of such a test the pressures 1 2 wf wf p , p , . . . are read from the pressure chart at the end of each separate flow period and matched to the theoretical equation (7.69). For instance, the calculation of 3 wf p at the end of the third flow period is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 1 3 2 i wf 1 2 1 3 D D D D D o 3 3 3 2 D D D 3 p p q 0 q q kh 7.08 10 p t p t t B q q q3 q q p t t S q µ − − − − × = + − + + − + (7.70) in which, e.g. tD3 is the dimensionless flowing time evaluated for t = t3, fig. 7.27. Furthermore, it will be assumed that the test starts from some known initial equilibrium pressure pi which is a conventional although theoretically unnecessary assumption, as will be demonstrated presently. Rate t1 t2 t3 t4 Time Time pi pwf pwf1 pwf2 pwf4 pwf3 (a) (b) q1 q2 q3 q4 Fig. 7.27 Multi-rate oilwell test (a) increasing rate sequence (b) wellbore pressure response The correct way to analyse such a test, as already described in sec. 7.5, is to plot ( ) ( ) n n j 1 n i wf j D D D j 1 n n p p q versus p t t q q − = − ∆ − (7.71) which should result in a straight line with slope m = 141.2µBo/kh and intercept on the ordinate equal to mS. The main drawback to this form of analysis technique is that it pre-supposes that the engineer is able to evaluate the pD functions for all values of the dimensionless time argument during the test period, and this in turn can demand a knowledge of the drainage area, shape and degree of well asymmetry. Because of this difficulty, the literature on the subject deals exclusively with multi-rate testing under transient flow conditions thus assuming the infinite reservoir case. OILWELL TESTING 211 The original paper on the subject was presented by Odeh and Jones5 in which the analysis technique is precisely as described above except that the D p functions in equ. (7.69) were evaluated for transient flow as ( ) D 12 D D 4t p t ln γ = (7.23) This leads to the test analysis equation (with t in hours) ( ) ( ) n n i wf j 3 n j 1 2 j 1 o n n w p p q kh k 7.08 10 1.151 log t t log 3.23 0.87S B q q cr µ φµ − − = − ∆ × = − + − + (7.72) which, providing the assumption of transient flow is appropriate for the test, will give a linear plot of (pi−pwf)/qn versus Σ ∆qj/qn log(tn−tj-1), with slope m = 162.6µBo/kh and intercept m(log(k/φµc 2 w r ) −3.23 + .87S), from which k and S can be calculated. It is frequently stated in the literature that the separate flow periods should be of short duration so that transient flow conditions will prevail at each rate. While this condition is necessary, it is insufficient for the valid application of transient analysis to the test. Instead, the entire test, from start to finish, should be sufficiently short so that transience is assured throughout the whole test period. The reason for this restriction is that the largest value of the dimensionless time argument, for which the pD functions in equ. (7.69) must be evaluated, is equal to the total duration of the test. This point is illustrated in fig. 7.27 (b), which again demonstrates the basic principle of superposition and shows that in evaluating the flowing pressure at the very end of the test there is still a component of the pressure response due to the first flow rate to be included in the superposed constant terminal rate solution. The following example will illustrate the magnitude of the error that can be made by automatically assuming that a multi-rate flow test can be interpreted using transient analysis techniques. EXERCISE 7.8 MULTI-RATE FLOW TEST ANALYSIS An initial test in a discovery well is conducted by flowing the well at four different rates over a period of 12 hours as detailed in table 7.10. Flowing time (hours) Oil rate (stb/d) pwf (psia) 0 0 3000(pi) 3 500 2892 6 1000 2778 9 1500 2660 12 2000 2538 TABLE 7.10 OILWELL TESTING 212 At the end of the flow test the well is closed in for a pressure buildup from which the permeability is estimated as 610 mD. The available reservoir data and fluid properties are listed below. Drainage area A = 80 acres Geometry 2 1 φ = .22 Bo = 1.35 rb/stb h = 15 ft µ = 1 cp rw = .33 ft c = 21 × 10-6/psi 1) Analyse the test data to determine k and S using equ. (7.69) with the pD functions evaluated using equ. (7.42) or (7.46) 2) Repeat the analysis evaluating the pD function for transient flow conditions, equ. (7.23). EXERCISE 7.8 SOLUTION 1) The general multi-rate test analysis equation, (7.69) can be expressed as ( ) ( ) n n j 1 n i wf j D D D j 1 n n p p q m p t t mS q q − = − ∆ = − + where m = 141.2 µ Bo/kh and pD ( ) n j 1 D D t t − − = pD ( ' D t ) can be evaluated as ( ) 1 1 1 2 2 2 D D DA DA D(MDH) DA 2 w 4A p t 2 t ln t ln p (t ) r π γ ′ ′ ′ ′ = + + − (7.46) i.e. ( ) 12 D D D(MDH) DA p t p (t ) α ′ ′ = − (7.73) in which D 2 6 2 w .000264kt .000264 610 t t cr .22 1 21 10 (.33) φ µ − × ′ = = × × × × ( ) 5 3.2 10 t hours = × and ( ) 2 DA D w t t r / A .01t hours = × = The pD functions, equ. (7.73), are evaluated in table 7.11 for all values of the time argument required in the test, and for a variety of geometrical configurations of the 80 acre drainage area in order to investigate the sensitivity of the results to variation in shape. OILWELL TESTING 213 2 1 4 1 Time (hrs) tDA α (equ.7.73) ½ pD(MBH) pD ½ pD(MBH) pD ½ pD(MBH) pD 3 .03 7.480 .098 7.382 .189 7.291 −.069 7.549 6 .06 8.015 .098 7.917 .381 7.634 −.151 8.166 9 .09 8.407 .071 8.336 .553 7.854 −.162 8.569 12 .12 8.739 .055 8.684 .690 8.049 −.177 8.916 TABLE 7.11 The test analysis is presented in table 7.12. ( ) n j 1 n i D D D j 1 n q p t t q − = ∆ − tn hrs q (stb/d) pwf (psi) i wf n p p q − 2 1 4 1 3 500 2892 .2160 7.382 7.291 7.549 6 1000 2778 .2220 7.650 7.463 7.858 9 1500 2660 .2267 7.878 7.593 8.095 12 2000 2538 .2310 8.080 7.707 8.300 TABLE 7.12 e.g. the complex summation for n = 3 is ( ) ( ) ( ) ( ) ( ) ( ) 9 9 3 9 6 3 D D D D D j 1 D D D 500 0 1000 500 p t p t t 1500 1500 1500 1000 p t t 1500 = − − = + − − + − in which the pD functions are taken from table 7.11 for the various geometries considered. The test results in table 7.12 are plotted in fig. 7.28. The basic reservoir parameters derived from these plots are listed in table 7.13. In each case the intercept on the ordinate has been calculated by linear extrapolation. Geometry Slope m Intercept mS k (mD) S 2 1 .0124 .5280 594 2.7 .0360 −.0466 353 −1.3 4 1 0.199 .0655 639 3.3 TABLE 7.13 OILWELL TESTING 214 2) If the test is analysed assuming transient flow conditions, the evaluation would be as set out in table 7.14. .83 .81 .79 .77 .75 .73 .20 .21 .22 .23 .24 2 4 1 1 Infinite reservoir and circular geometry X X X X − n i wf n p p (psi/ stb / d) q n j 1 j D D D n q p (t t ) q − ∆ − Fig. 7.28 Illustrating the dependence of multi-rate analysis on the shape of the drainage area and the degree of well asymmetry. (Exercise 7.8) tn (hrs) tDn pD (tD) ( ) n j 1 n i wf i D D D j i n n p p q p t t q q − = − ∆ − 3 9.6×105 7.292 .2160 7.292 6 19.2 " 7.639 .2220 7.466 9 28.8 " 7.842 .2267 7.591 12 38.4 " 7.985 .2310 7.690 TABLE 7.14 To facilitate comparison with the results from the first part of this exercise, the present results have also been plotted in fig. 7.28 rather than making the more conventional Odeh-Jones semi log plot, as specified by equ. (7.72). For the infinite reservoir case the slope m = .0374 and calculated intercept mS = −.0573 which implies that k = 340 mD S = − 1.5 At first glance, the results of exercise 7.8 are somewhat alarming. Assuming that the 2:1 geometry is correct, as stated in the question, then there is an error of over forty percent in the calculated permeability, and what is in fact a damaged well (S = 2.7) appears to be stimulated (S = −1.5), merely as a result of applying transient analysis to the same set of test data. OILWELL TESTING 215 The reason for this disparity lies in the nature of the analysis technique itself. In plotting the results according to equ. (7.71) the evaluation of the abscissa, ( ) n j 1 n i D D D j 1 n q p t t q − = ∆ − , automatically involves the boundary condition in the analysis, since use of the pD function implies a knowledge of the geometrical configuration. Therefore, unlike the buildup analysis, for which a unique plot of the observed data is obtained, the multi-rate test analysis can yield a different plot for each assumed boundary condition, as shown in fig. 7.28, and all the plots appear to be approximately linear. The only time when a straight line is obtained, which has no dependence on the boundary condition, is for the infinite reservoir case. Then the Odeh-Jones plot is applicable which has as its abscissa, ( ) n j n j 1 j 1 n q log t t q − = ∆ − equ. (7.72). The problem is, of course, how can one be sure that transient analysis is valid without a knowledge of several of the basic reservoir parameters, some of which may have to be determined as results of the test analysis. As clearly shown in the MBH charts, figs. 7.11-15, the crucial parameter for deciding the flow condition is DA kt t 0.000264 cA φµ = (7.49) If tDA is extremely small when evaluated for the maximum value of t (i.e. t = total test duration) then it is probably safe to use the transient analysis technique. It is not obvious, however, just how small this limiting value of tDA should be because this too depends on the geometrical configuration. For a well positioned at the centre of a circle or square the minimum value of tDA is 0.1, at which point there is a fairly well defined change from pure transient to semi-steady state flow. For a well asymmetrically positioned within a 2:1 rectangle, e.g. curve IV of the MBH chart, fig. 7.12 (which is the correct geometrical configuration for exercise 7.8) the departure from purely transient flow, in this case to late transient flow, occurs for tDA < 0.015. Similarly for the 4:1 geometrical configuration included in the exercise the departure occurs for tDA < 0.01. In exercise 7.8, the relationship between tDA and the real time has a large coefficient of 0.01 (i.e. tDA = 0.01 t) . Th is results from the fact that the permeability is large and the area relatively small and have been deliberately chosen so to illustrate the hidden dangers in applying transient analysis techniques to multi-rate test results. After the first 3-hour flow period the corresponding value of tDA is 0.03 and therefore there is already a departure from transient flow for the 2:1 and the 4:1 geometries used in the exercise. If it is assumed that the well is at the centre of a circle, however, transient analysis can be applied throughout since the value of tDA corresponding to the entire test duration of 12-hours is tDA = 0.12 and, as already noted, the departure from transient flow for this geometry occurs for tDA = 0.1. The above points are clearly illustrated in fig. 7.28 and in tables 7.11-14. The majority of examples of multi-rate test analysis in the literature have, quite correctly, been subjected to transient analysis. For instance, there is an example of a OILWELL TESTING 216 multi-rate test in a gas well presented in the original Odeh-Jones paper5 for a well positioned at the centre of a circular shaped area of radius 3000 ft (A ≈ 650 acres) and for which the permeability is 19.2 mD. In the example tDA = 9.2×10-5, and for the geometry considered, transient analysis can be applied for a total of 1086 hours. It is in cases where reservoirs are not continuous and homogeneous over large areas but splintered into separate reservoir blocks on account of faulting that errors can occur in assuming the infinite reservoir case is applicable in the test analysis. One further, complication arises in connection with this type of analysis, and that is, that in order to apply the correct technique, using the general pD function, equ. (7.42), requires a knowledge of the permeability in order to calculate tD or tDA. In buildup analysis this presents no problem since k can be readily calculated from the slope of the linear section of the buildup plot. In multi-rate testing, however, this can prove more difficult. Sometimes it is possible to separately analyse the initial flow period by plotting pwf versus log t and applying the transient analysis technique described in exercise 7.2. Unfortunately, in high permeability reservoirs this is very difficult to apply in practice, since the pressure fall-off is initially very rapid. Under these circumstances it may be necessary, and indeed is always advisable, to conduct a buildup at the end of the flow test which tends to defeat one of the main purposes of the multi-rate test, namely, to avoid well closure. It is commonly believed that multi-rate flow tests can only be analysed if the initial equilibrium pressure within the drainage volume is known. This is an unnecessary restriction which has tended to limit the application of this technique to initial well tests for which pi can be readily determined. The following analysis shows that, with minor modifications to the method presented so far, the multi-rate test can be analysed with only a knowledge of the bottom hole pressure and surface production rate prior to the survey. Suppose that a well with the variable rate history shown in fig. 7.29 is to be tested by flowing it at a series of different rates. Prior to the test the well is produced at a constant rate qN during the Nth and final flow period before the multi-rate test commences at time tN. Then, for any value of the total time tn during the test, when the current rate is qn, the bottom hole flowing pressure n wf p can be calculated as ( ) ( ) n n j 1 n 3 i wf j D D D n j 1 o kh 7.08 10 p p q p t t q S B µ − − = × − = ∆ − + in which pi is the initial pressure at t = 0 and the summation includes all the variable rate history up to and including the test itself. This equation can be subdivided as OILWELL TESTING 217 Rate Time Start of multi-rate test t N-1 t N t n q N Fig. 7.29 Multi-rate test conducted after a variable rate production history ( ) ( ) ( ) ( ) n N n j 1 n j 1 N 3 i wf j D D D D N j 1 o n j D D D n N j N 1 kh 7.08 10 p p q p t t t q S B q p t t q q S δ µ δ δ − − − = = + × − = ∆ + − + + ∆ − + − (7.74) in which n n N t t t δ = − and j 1 j 1 N t t t for j N 1 δ − − = − ≥ + Then if the condition is imposed that (tN−tN-1)>> tn(max), i.e. the last flow period before the test commences is considerably greater than the total duration of the test itself, then ( ) ( ) N n j 1 N j 1 N N j D D D D j D D D j 1 j 1 q p t t t q p t t δ − − = = ∆ − − ≈ ∆ − (7.75) and ( ) ( ) N j 1 N N 3 j D D D N i wf j 1 o kh q p t t q S 7.08 10 p p B µ − − = ∆ − + ≈ × − where N wf p is the flowing pressure recorded immediately before the multi-rate test commences. Equation (7.74) can therefore be simplified as ( ) ( ) ( ) N n n j 1 N n 3 wf wf j D D D n n j 1 j N 1 o kh 7.08 10 p p q p t t q q S B δ δ µ − − = = + × − = ∆ − + − and therefore a plot of ( ) ( ) N n n j 1 n wf wf j D D D j N 1 n N n N p p q versus p t t q q q q δ δ − = + − ∆ − − − (7.76) should again be linear with slope m = 141.2 µBo/kh and the intercept on the ordinate equal to mS. Using this modified technique provides a useful way of applying the multi-rate flow test for routine well surveys. The only condition for its application is that the flow period before the test should be much longer than the totai test duration. This OILWELL TESTING 218 does not necessarily mean that flow should be under semi-steady state conditions at this final rate. The condition is usually satisfied since the reliable analysis of a multi-rate test, as already noted, requires that the total test duration should be brief so that transient analysis can be applied. As a demonstration of the effectiveness of this analysis technique, a test has been simulated in a well for which the following data are applicable Area drained 650 acres h = 50ft geometry re ≈3000 ft rw = .3 ft Bo = 1.2 rb/stb k = 20 mD c = 15 × 10-6 / psi φ = .23 µ = 1 cp pi = 3500 psia S = 2.0 Prior to the test the well had been producing for one year at 1000 stb/d and for a second year at 400 stb at which time a multi-rate test was conducted as detailed in table 7.15. The bottom hole flowing pressure prior to the test was N wf p = 2085 psi. Rate stb/d Cumulative time hrs Flowing pressure psia 600 4 1815 800 8 1533 1000 12 1244 1200 16 950 TABLE 7.15 For the above conditions the relationship between dimensionless and real time is tDA = 5.41 × 10-5 t (hours) and therefore, after the total test period of 16 hours tDA = 8.65 × 10-4. This means that transient analysis can be safely applied to the test since, for a well at the centre of a circle, transient conditions prevail until tDA ≈ 0.1. The test data in table 7.15 are analysed using the plotting technique of equ. (7.76), with the pD functions evaluated as ( ) D 1 1 1 2 2 2 D D DA 2 w 4t 4A p t ln t ln r γ γ = = + (7.23) The analysis is detailed in table 7.16, and the resulting plot shown as fig. 7.30. OILWELL TESTING 219 Time hrs Rate stb/d pwfn psi N n wf wf n N (p p ) q q − − pD (tD) n j 1 n j D D D j N 1 n N q p ( t t ) q q δ δ − = + ∆ − − 4 600 1815 1.350 5.968 5.968 8 800 1533 1.380 6.315 6.142 12 1000 1244 1.402 6.518 6.267 16 1200 950 1.419 6.662 6.366 TABLE 7.16 1.41 1.39 1.37 1.35 1.33 5.9 6.0 6.1 6.2 6.3 6.4 n j 1 j D D D n N q p ( t t ) q q − ∆ δ −δ − N n wf wf n N (p p ) (q q ) (psi/ stb / d) − − Fig. 7.30 Multi-rate test analysis in a partially depleted reservoir The slope and intercept of the straight line have values of 0.173 and 0.317, respectively, from which it can be calculated that k = 19.6 mD and S = 1.8. These values compare very favourably with the actual values of k = 20 mD and S = 2.0. 7.9 THE EFFECTS OF PARTIAL WELL COMPLETION In deriving the basic diffusivity equation for liquid flow, equ. (5.20), it was assumed that the well was completed across the entire producing interval thus implying fully radial flow. If for some reason the well only partially penetrates the formation, as shown in fig. 7.31 (a), then the flow can no longer be regarded as radial. Instead, in a restricted region at the base of the well, the flow could more closely be described as being spherical. OILWELL TESTING 220 (a) 30 ft 0.25 ft 150 ft (b) 150 ft 15 ft 75 ft 150 ft 6 ft 15 ft (c) Fig. 7.31 Examples of partial well completion showing; (a) well only partially penetrating the formation; (b) well producing from only the central portion of the formation; (c) well with 5 intervals open to production (After Brons and Marting19) 10,000 300 100 50 20 10 1 5 2 0 0.2 0.4 0.6 0.8 1.0 b 0 4 8 12 16 20 24 28 Sb w h r = Fig. 7.32 Pseudo skin factor Sb as a function of b and h/rw (After Brons and Marting19) (Reproduced by courtesy of the SPE of the AIME) Brons and Marting19 have shown that the deviation from radial flow due to restricted fluid entry leads to an additional pressure drop close to the wellbore which can be interpreted as an extra skin factor. This is because the deviation from radial flow only occurs in a very limited region around the well and changes in rate, for instance, will OILWELL TESTING 221 lead to an instantaneous perturbation in the wellbore pressure without any associated transient effects. This pseudo skin can be determined as a function of two parameters, the penetration ratio b and the ratio h/rw, where the total interval open to flow b the total thickness of the producing zone = and w h thickness of the producing zone r wellbore radius = The latter definition is somewhat more complex than it appears since if the well is open to flow over several sections of the total producing interval then h represents the thickness of the symmetry element within the total zone. This point is made clear in fig. 7.31 (a)-(c), which has been taken from the Brons and Marting paper and illustrates three possible types of partial well completion. In all three cases the ratio b = 30/150 = 0.2 while the ratio of h/rw is 150/.25 = 600 for case (a), 75/.25 = 300 for case (b) and 15/.25 = 60 for case (c). Having thus determined the values of b and h/rw the pseudo skin Sb can be determined using the chart presented as fig. 7.32. For the three geometric configurations shown in fig. 7.31, the pseudo skin factors are approximately 17, 15 and 9, respectively. Once the pseudo skin has been calculated it must be subtracted from the total skin measured in the well test to give the mechanical skin factor. 7.10 SOME PRACTICAL ASPECTS OF WELL SURVEYING This section deals with some of the practical aspects involved in the routine pressure testing of wells in a producing oilfield. a) Wireline Pressure Recording Instrument Due to its accuracy and ruggedness the most popular wireline pressure gauge is the Amerada (RPG), a rough schematic diagram of this tool is shown in fig. 7.33(a). The continuous trace of pressure versus time is made by the contact of a stylus with a chart which has been specially treated on one side to permit the stylus movement to be permanently recorded. The chart is held in a cylindrical chart holder which in turn is connected to a clock which drives the holder in the vertical direction. The stylus is connected to a bourdon tube and is constrained to record pressures in the perpendicular direction to the movement of the chart holder. The combined movement is such that, on removing the chart from the holder after the survey, a continuous trace of pressure versus time is obtained as shown in fig. 7.33(b), for a typical pressure buildup survey. Both the clock and pressure element can be selected to match the maximum time and pressure anticipated for the particular survey. With careful handling, regular calibration and accurate reading of the pressure chart with a magnifying instrument, an accuracy of about 0.2 can be achieved. OILWELL TESTING 222 CLOCK VERTICAL CHART MOVEMENT ∝ TIME CHART (a) STYLUS STYLUS MOVEMENT ∝ PRESSURE BOURDON PRESSURE ELEMENT PRESSURE FLOWING PRESSURE-DEPTH SURVEY STATIC PRESSURE-DEPTH SURVEY TIME t ∆ BASE LINE pwf pws (b) Fig. 7.33 (a) Amerada pressure gauge; (b) Amerada chart for a typical pressure buildup survey in a producing well b) Conducting a Pressure Buildup Survey Prior to the survey the well should be gauged to determine the gas/oil ratio and final flow rate. The Amerada is calibrated, assembled and a base pressure line recorded on the chart by disconnecting the clock and allowing the chart holder to fall slowly through its full length while in contact with the stylus at atmospheric pressure and ambient temperature. When subsequently measuring pressures after the survey, the readings are made in the direction perpendicular to this base line. The Amerada is placed in a lubricator and the latter is flanged up to the wellhead as indicated in fig. 7.34. When the gate valve beneath the lubricator is opened, the Amerada can be run in on wireline against the flowing well stream. In a flowing or gas lift well, it is common practice to stop at intervals of 1000 or 500 ft while running in with the Amerada to record a flowing pressure survey. Each stop should be made for long enough so that a series of pressure steps is discernible as shown in fig. 7.33 (b), and therefore the length of time for each stop will depend on the scale of the clock being used. The flowing pressure gradient, as a function of depth, measured in such a survey is useful for production engineers in checking the lifting efficiency of the well. Once the survey depth has been reached the bottom hole flowing pressure pwf is recorded prior to closure. The well is then closed-in, usually at the surface, and the OILWELL TESTING 223 AMERADA WITHIN LUBRICATOR GATE VALVE MASTER VALVE CASING TUBING WIRELINE WINCH Fig. 7.34 Lowering the Amerada into the hole against the flowing well stream Amerada records the increasing pressure which can be related to the closed-in time ∆t, fig. 7.33 (b). At the end of the pressure buildup survey the Amerada is pulled out, with the well still closed-in, and a static pressure survey is measured as a function of depth in a manner similar to the flowing survey made while running in. In this case, stops should be made at fairly short intervals of say 100-200 ft, close to the survey depth, and at wider spaced intervals of 500-1000 ft, higher up the hole. The information gained from such a survey can be vital in referring actual measured pressures in the well to a datum level in the reservoir, in cases where it is not possible to conduct the buildup survey adjacent to the perforated interval to be tested (refer Chapter 4, sec.6). DATUM MEASUREMENT DEPTH OIL-WATER CONTACT DATUM LEVEL AMERADA MEASUREMENT DEPTH (p ) m TOP PERFORATIONS (a) (b) h H Fig. 7.35 Correction of measured pressures to datum; (a) well position in the reservoir, (b) well completion design Consider, for instance, a survey conducted in the well shown in fig. 7.35 (a), (b). When closed-in, the distribution of fluids in the well could vary between the two extremes illustrated in fig. 7.36 (a), (b). In case (a), in which the well has been producing with a watercut, the fluid distribution may be as indicated by the solid line, which is necessary to obtain the correct pressure in the oil at the top of the perforations, the virtual oil gradient being shown by the dashed line. Alternatively, there may be no water entering OILWELL TESTING 224 the well and the fluid and pressure distribution to the surface would then be as shown in fig. 7.36(b), in which a rise in the static tubing head pressure occurs due to phase separation. THP PRESSURE GAS OIL WATER MEASUREMENT DEPTH DATUM VIRTUAL OIL GRADIENT TOP PERFS. (a) (b) THP PRESSURE GAS OIL DATUM MEASUREMENT DEPTH TOP PERFS. Fig. 7.36 Extreme fluid distributions in the well; (a) with water entry and no rise in the tubing head pressure, (b) without water entry and with a rise in the THP Between these extremes, of course, there is an infinite range of possible fluid distributions. The important thing is that the engineer should know the pressure gradient in the wellbore fluid at the survey depth, which can only be obtained from the static pressure - depth survey. If the well is completed as shown in fig. 7.3 (b), in which for mechanical reasons it is not possible to conduct the buildup survey at the actual reservoir depth, then, in order to relate the pressures measured in the borehole to the datum level in the reservoir, it is first necessary to calculate the pressure at the top of the perforations using the measured pressure gradient of the fluid in the borehole at the survey depth, thus ( ) ( ) m perfs well dp p p h dz = + × and thereafter, use the calculated oil gradient in the reservoir to correct to the datum plane. ( ) ( ) ( ) m datum well oil dp dp p p h H dz dz = + × − × 7.11 AFTERFLOW ANALYSIS When a well is closed in for a pressure survey, the closure is usually made at the surface rather than at the sandface. Because the fluids in the flow string have a higher compressibility than in the reservoir, production will continue at the sandface for some finite time after the surface production has ceased. The time lag between closing in the well at the surface and feeling the effects of closure in the reservoir is, to a large extent, dependent upon the mechanical design of the well. If the well is completed with a packed off annulus the volume of fluids in the flow string is considerably smaller than OILWELL TESTING 225 if no packer is used and the afterflow effects will be of less significance. Afterflow distorts the early part of the Horner buildup plot, as shown in fig. 7.37. pws Buildup dominated by afterflow 1000 100 10 1 p t t t + ∆ ∆ Fig. 7.37 Pressure buildup plot dominated by afterflow Several theoretical methods have been presented for analysing the pressure response during the afterflow period in order to determine kh and S. Due to the basic complexity of the problem, it should be stated from the outset that the results obtained from any of the various techniques are liable to be less accurate than those from the simple Horner analysis of the straight line part of the buildup, once the afterflow has ceased. In some cases, however, afterflow analysis provides a valuable means of obtaining information about the reservoir. For instance, in several areas in the Middle East, wells are capable of producing in excess of 50000b/d from limestone reservoirs. Because of the very high kh values, which leads to very rapid pressure buildups, and the fact that in many cases the wells produce through the casing, the afterflow period can completely dominate the pressure buildup and afterflow analysis is the only method of determining the essential reservoir parameters. The analysis methods which will be described in this section are those of Russell20 and McKinley21. a) Russell Analysis Russell developed a theoretical equation describing how the bottom hole pressure should increase as fluid accumulates in the wellbore during the buildup. As a result of this, he determined that the correct way of plotting the pressures during the part of the buildup influenced by the afterflow was as p versus log t 1 1 C t ∆ ∆ − ∆ (7.77) in which ∆p = pws(∆t) − pwf(t) (psi),and ∆t is the closed in time (hrs). The denominator of the left hand side contains a correction factor C to allow for the gradually decreasing flow into the wellbore. This constant C must be selected by trial and error so that the resulting plot is linear. This is illustrated in fig. 7.38. For very small values of ∆t the buildup is dominated by the skin factor rather than afterflow. Therefore, not all the OILWELL TESTING 226 values of ∆p and ∆t can be used in this analysis. Russell recommends that plot should be made only for values of ∆t measured after one hour of closed-in time. (psi) C - TOO SMALL C - TOO LARGE log ∆t p∆ 1 − 1 C∆t Fig. 7.38 Russell plot for analysing the effects of afterflow Having chosen the correct value of C. the slope of the straight line is measured (m−psi/log cycle) and the formation kh value can be determined using the equation o 162.6q B kh m µ = (7.78) The skin factor can be calculated using the expression wf(1 hr) wf 2 w p p k S 1.151 log 3.23 1 1/ C t cr m φµ − − = − + − ∆ (7.79) b) McKinley Analysis To apply the McKinley method it is necessary to plot the pressure buildup in a special manner and compare the resulting plot with so called "Type Curves" presented by McKinley21, as shown in fig. 7.39. OILWELL TESTING 227 1 10 100 1000 1 10 100 1000 t ∆ (mins) (psi) ∆p (a) t ∆ (mins) 1 10 100 1000 10 -4 10 -3 10 -2 10 -1 (b) T 10 000 F = T 5 000 F = T 2 500 F = q ∆pF Fig. 7.39 (a) Pressure buildup plot on transparent paper for overlay on (b) McKinley type curves, derived by computer solution of the complex afterflow problem A set of McKinley type curves is included as fig. 7.40. These curves were computed by numerical simulation of the complex afterflow process by forming a dynamic balance between the capacity of the wellbore to store fluid and the resistance of the wellbore to the flow of fluid from the reservoir. All the curves were computed for a constant value of 2 w cr /k φµ = 1.028×10-7 cp. sq ft/(mD psi), since in his original paper McKinley has demonstrated that the shape of the type curves is insensitive to variation in the value of this parameter. Furthermore, the curves were computed assuming no mechanical skin factor. If a well is damaged this fact is evident since the pressure buildup plot will deviate from the McKinley type curve and while the analysis does not explicitly determine the skin factor, it does allow a comparison to be made between the kh values in the damaged and undamaged parts of the reservoir. The abscissa of fig. 7.40 is for the parameter ∆pF/q where ∆p = pws (∆t) – pwf (t) (psi) q = oil rate in rb/d and F is the so called "wellbore parameter" Wellbore area (sq.ft) F (for partially liquid filled wells) Wellbore liquid gradient (psi/ ft) = 1 3 F Wellbore fluid compressibility (psi ) Wellbore volume (ft ) − = × (for fluid filled wells) OILWELL TESTING 228 10 -4 10 -3 10 -2 10 -1 10 0 10 1 2 3 4 5 6 7 9 2 3 4 5 6 7 9 2 3 4 5 6 7 9 2 3 4 5 6 7 9 2 3 4 5 6 7 9 2 3 4 5 6 7 9 10 2 3 4 5 6 7 9 100 2 3 4 5 6 7 9 1000 G c A V W = Wellbore liquid gradient, psi/ft = Wellbore fluid compressibility, (psi) = Wellbore cross-sectional, area, sq ft = Wellbore volume, bbls -1 = Pressure change from flowing pressure, psi = Shut-in time, minutes = Production rate, reservoir barrels per day q T F ; For partially liquid filled wells 5.6 c V; For completely fluid-filled wells W = ∆p = p - p ws wf ∆t kh µ = Wellbore Transmissibility, mD - ft cp A G SHUT-IN TIME - ∆t MINUTES T/F = 500,000 T/F = 250,000 T/F = 100,000 T/F = 50,000 T/F = 25,000 T/F = 500 T/F = 250 T/F = 100 PRESSURE BUILDUP GROUP - ∆ ∆ ∆ ∆ p f / q Fig. 7.40 McKinley type curves for 1 min <∆ ∆ ∆ ∆t < 1000 min. (After McKinley21) (Reproduced by courtesy of the SPE of the AIME) OILWELL TESTING 229 In practice, the value of F is seldom explicitly calculated in the McKinley analysis. It is a parameter which disappears by cancellation when calculating the wellbore transmissibility. Each type curve is characterised by a fixed value of T/F, where T is the transmissibility = kh/µ mD.ft/cp. The steps involved in a McKinley analysis, which is usually aimed at determining transmissibilities in the damaged/stimulated zone close to the wellbore, and the average for the formation away from the well, are as follows. a) Make a table of values of ∆t, the closed in time (minutes), and the corresponding values of ∆p = pws (∆t) - pwf (t) (psi). Unlike the Russell method it is not necessary to differentiate between the part of the buildup due to skin and that due to afterflow, all values of ∆t and ∆p can be used. b) Overlay the McKinley chart with a sheet of transparent paper and draw vertical and horizontal axes to match those of the chart. The ordinate should have the same log time scale as the McKinley chart but the abscissa, while using the same log scale, should be plotted for the most suitable range of pressure values. Plot the ∆t versus ∆p data on this transparent paper. c) The transparent paper is then moved laterally over the McKinley chart, keeping the abscissae together, until the early part of the pressure buildup coincides with one of the type curves. d) The value of the parameter T/F, characterising the type curve for which the match is obtained, is noted. e) Any one of the points on the pressure buildup, which also lies on the type curve, is now used as a match point and the corresponding value of ∆pF/q is read from the abscissa of the McKinley chart. Multiplying this value by the value of T/F gives pF T pT q F q ∆ ∆ × = (7.80) Since ∆p (psi) is known for the match point, then the transmissibility T can be calculated from this latter expression. Finally, since T = kh/µ, the values of kh and k can be determined. The procedure is illustrated in fig. 7.41. Using the figures shown in this diagram pF T pT 0.05 5000 q F q ∆ ∆ × = × = If q = 500 rb/d, then T = 5000 × 0.05 × 500 800 = 156 mD.ft/cp. OILWELL TESTING 230 X X X X X X X X X 1000 100 10 1 10 -4 10 -3 10 -2 buildup plot match point p = 800 psi ∆ McKinley type (curve (T/F = 5000) p F q ∆ = .05 p F q ∆ ∆t (mins) Fig. 7.41 Buildup plot superimposed on a particular McKinley type curve for T/F = 5000 The transmissibility obtained in this manner is for the damaged or stimulated region close to the wellbore, Tw. f) If, for large values of ∆t on the buildup plot, the pressure points trend away from the McKinley type curve, this indicates the presence of a mechanical skin factor, as shown in fig. 7.42. p F q ∆ t (mins) ∆ S (positive) S (negative) Fig. 7.42 Deviation of observed buildup from a McKinley type curve, indicating the presence of skin Since the latter part of the buildup, for large ∆t, is not influenced by the skin it should reflect the actual transmissibility of the formation beyond the damaged or stimulated zone near the well. Therefore, to obtain the formation transmissibility, Tf, the late part of the buildup plot is re-aligned with another of the type curves, for which the value of T/F is again noted. OILWELL TESTING 231 Since F is a constant, dependent upon the well design and fluid compressibility, then by simple proportionality ( ) ( ) f f w w T /F T T T /F = × or ( ) ( ) f f w w T /F k k T /F = × Apart from the two methods mentioned in this text, afterflow analysis techniques relying on the use of type curves have also been presented by Ramey22 and Earlougher and Kersch23. Which, of all these methods, is the most reliable is a question which is still unanswered. One point on which all the authors of papers on the subject agree is that afterflow analysis techniques should not be used for pressure buildup analysis when there is a clearly defined linear portion of the conventional Horner plot which can be subjected to the analysis techniques described in sec. 7.7. This is because the physical and mathematical concepts involved in the description of afterflow are vastly more complex than for the simple pressure buildup theory and therefore, the analysis results may be less reliable. Nevertheless, it is recommended that engineers should experiment with one or all of these methods in their own fields to determine which, if any, is suitable. To do this, pressure buildup tests which provide both a linear trend on the conventional Horner plot and also a significant deviation from this trend due to afterflow, for small values of ∆t, should be analysed using both the normal and afterflow analysis techniques, and the results compared. If the comparison is favourable and statistical confidence is built up in one of the afterflow methods, then the engineer can use this method for test analysis on such occasions as when buildup surveys are recovered which are dominated by the effects of afterflow. EXERCISE 7.9 AFTERFLOW ANALYSIS TECHNIQUES A twelve hour pressure buildup test was conducted in a flowing oil well from which the pressure-time record was recovered, as listed in table 7.17. The production data and reservoir and fluid properties are as follows: Np = 30,655 stb φ = 0.2 q = 231 stb/d µ = 0.6 cp h = 10 ft Bo = 1.3 rb/stb rw = 0.3 ft c = 20 × 10-6/psi OILWELL TESTING 232 ∆t (mins) pws (psi) ∆t (mins) pws (psi) 0 1600 (pwf) 20 1920 270 2879 40 2160 300 2900 60 2350 360 2935 90 2525 420 2960 120 2650 480 2980 150 2726 540 2998 180 2779 600 3011 210 2822 660 3022 240 2852 720 3035 TABLE 7.17 The conventional Horner plot of pws versus log t t t + ∆ ∆ does not become linear, even for the largest values of ∆t, and therefore the afterflow analysis techniques presented in this section must be used to analyse this test. 1) Evaluate k and S using the Russell method. 2) Determine the permeability in the vicinity of the wellbore and of the undamaged formation using the McKinley method. EXERCISE 7.9 SOLUTION 1) Russell Analysis As suggested by Russell, the analysis should only be applied for pressures measured after ∆t = 1 hour. In table 7.18 several values of the parameter C have been selected in an attempt to linearize the plot of equ. (7.77). OILWELL TESTING 233 p /(1 1/ C t) ∆ − ∆ ∆t (hrs) log ∆t ∆p (psi) C=1.7 C=2.1 C=2.5 1.5 .176 925 1522 1355 1261 2.0 .301 1050 1487 1378 1314 2.5 .398 1126 1472 1391 1340 3.0 .477 1179 1467 1401 1360 3.5 .544 1222 1469 1414 1380 4.0 .602 1252 1468 1421 1391 4.5 .653 1279 1471 1430 1404 5.0 .699 1300 1473 1437 1413 6.0 .778 1335 1480 1450 1430 7.0 .845 1360 1485 1459 1442 8.0 .903 1380 1490 1467 1453 9.0 .954 1398 1496 1476 1463 10.0 1.000 1411 1499 1482 1470 11.0 1.041 1422 1502 1486 1476 12.0 1.079 1435 1509 1494 1484 TABLE 7.18 As shown in fig. 7.43, the correct value of the parameter C to obtain a linear Russell plot is C = 2.1. Since the slope of this line is 151 psi/log cycle, then the kh product can be evaluated using equ. (7.78) as 162.6 231 0.3 1.3 kh 194 mD.ft; and k 19.4 mD 151 × × × = = = OILWELL TESTING 234 1500 1400 1300 0 .5 1.0 C= 1.7 C= 2.1 C= 2.5 ∆p 1 - 1 C t ∆ (psi) log t ∆ Fig. 7.43 Russell afterflow analysis Exercise 7.9) The value of (pws (∆t) - pwf (t))/(1−1/C∆t) at ∆t = 1 hour can be read from the linear plot as 1329 psi. Therefore the skin factor can be calculated using equ. (7.79) as 6 1329 19.4 S 1.151 log 3.23 4.7 151 .2 .6 20 10 .09 − = − + = × × × × 2) McKinley Analysis To apply the McKinley method it is necessary to plot the closed in time ∆t(minutes) versus the pressure buildup ∆p= pw, (∆t)−pwf (t) (psi) on transparent paper using the same log-log scales as on the McKinley type curve chart. When this plot is moved laterally across the McKinley chart, keeping the abscissae of both charts together, the early part of the buildup is found to match the type-curve which has the parametric value T/F = 2500, fig. 7.44. Selecting a match point ∆t = 60 minutes, ∆p = 750 psi, the corresponding value on the abscissa of the McKinley chart is ∆pF/q = 0.14. Therefore, OILWELL TESTING 235 MATCH POINT t = 60 mins p = 750 psi ∆ ∆ = 0.14 ∆pF q 10 -2 10 -1 1 10 100 ∆pF q 1000 = 5000 T F = 2500 T F 2 5 3 4 6 7 9 2 5 3 4 6 7 9 2 5 3 4 6 7 9 2 5 3 4 6 7 9 2 5 3 4 6 7 9 t (mins) ∆ Fig. 7.44 Match between McKinley type curves and superimposed observed buildup (Exercise 7.9). o − − − − match for small ∆ ∆ ∆ ∆t (T/F = 2500) • • • • − − − − match for large ∆ ∆ ∆ ∆t (T/F = 5000) pF T pF 0.14 2500 350 q F q ∆ ∆ × = = × = OILWELL TESTING 236 and since q = 300 rb/d, then the wellbore transmissibility is w w k h 3500 300 T 140 mD.ft / cp 750 µ × = = = and therefore w k 8.4 mD = For values of ∆t greater than 150 minutes the actual buildup curve breaks away from the parametric curve for T/F = 2500, indicating the presence of a positive skin factor. For large values of ∆t the buildup matches the McKinley curve with parametric value T/F = 5000, fig. 7.44. Even for this latter match, however, the buildup continues to be dominated by afterflow. Therefore, a minimum value of the formation transmissibility can be estimated as ( ) ( ) f f w w T /F T T T /F = × giving f 5000 k 8.4 16.8mD 2500 = × = Thus the comparison between the Russell and McKinley techniques is quite reasonable in this case. REFERENCES 1) van Everdingen, A.F. and Hurst, W., 1949. The Application of the Laplace Transformation to Flow Problems in Reservoirs. Trans. AIME. 1 86: 305-324. 2) Earlougher, R.C., Jr., 1971. Estimating Drainage Shapes from Reservoir Limit Tests. J. Pet. Tech., October: 1266-1268. 3) Ramey, H.J., Jr. and Cobb, W.M., 1971. A General Pressure Buildup Theory for a Well in a Closed Drainage Area. J. Pet. Tech., December: 1493-1505.Trans. AIME. 4) Horner, D.R., 1951. Pressure Build Up in Wells. Proc., Third World Petroleum Congress. E.J. Brill, Leiden. ll, 503. 5) Odeh, A.S. and Jones, L.G., 1965. Pressure Drawdown Analysis, Variable Rate Case. J. Pet. Tech., August: 960-964. Trans. AIME. 6) Matthews, C.S. and Russell, D.G., 1967. Pressure Buildup and Flow Tests in Wells. SPE Monograph: 130-133. 7) Matthews, C.S., Brons, F. and Hazebroek, P.,1954. A Method for the Determination of Average Pressure in a Bounded Reservoir. Trans. AIME.201: 182-191. OILWELL TESTING 237 8) Cobb, W.M. and Dowdle, W.L., 1973. A Simple Method for Determining Well Pressures in Closed Rectangular Reservoirs, J. Pet. Tech., November: 1305-1306. 9) Dietz, D.N., 1965. Determination of Average Reservoir Pressures from Build Up Surveys. J. Pet. Tech., August: 955-959. Trans. AIME. 10) Earlougher, R.C., Jr., Ramey, H.J., Jr., Miller, F.G. and Mueller, T.D., 1968. Pressure Distribution in Rectangular Reservoirs. J. Pet. Tech., February: 199-208. Trans. AIME. 11) Ramey, H.J., Jr., Kumar, A. and Gulati, M.S. 1973. Gas Well Test Analysis Under Water-Drive Conditions. American Gas Assn., Arlington, Va. 12) Pinson, A.E., Jr., 1972. Concerning the Value of Producing Time in Average Pressure Determinations from Pressure Buildup Analysis, J. Pet. Tech., November: 1369-1370. 13) Kazemi, H., 1974. Determining Average Reservoir Pressure from Pressure Buildup Tests. Soc. Pet. Eng. J., February: 55-62. Trans. AIME. 14) Odeh, A.S. and Selig, F., 1963. Pressure Buildup Analysis, Variable Rate Case. J. Pet. Tech., July: 790-794. Trans. AIME. 15) van Poollen, H.K., Breitenback, E.A. and Thurnau, D.H., 1968. Treatment of Individual Wells and Grids in Reservoir Modelling. Soc. Pet. Eng. J., December: 341-346. 16) Earlougher, R.C., Jr., 1972. Comparing Single-Point Pressure Buildup Data With Reservoir Simulator Results. J. Pet. Tech., June: 711-712. 17) Cobb, W.M. and Smith, J.T., 1975. An Investigation of Pressure Buildup Tests in Bounded Reservoirs. J. Pet. Tech., August: 991-996. Trans. AIME. 18) Denson, A.H., Smith, J.T. and Cobb, W.M., 1976. Determining Well Drainage Pore Volume and Porosity from Pressure Buildup Tests. Soc. Pet. Eng. J., August: 209-216. 19) Brons, F. and Marting, V.E., 1961. The Effect of Restricted Fluid Entry on Well Productivity. J. Pet. Tech., February: 172-174. Trans. AIME. 20) Russell, D.G., 1966. Extensions of Pressure Buildup Analysis Methods. J. Pet. Tech., December: 1624-1636. Trans. AIME. 21) McKinley, R.M.,1971. Wellbore Transmissibility from Afterflow-Dominated Pressure Buildup Data. J. Pet. Tech. July: 863-872. Trans. AIME. 22) Ramey, H.J., Jr., 1970. Short-Time Well Test Data Interpretation in the Presence of Skin Effect and Wellbore Storage. J. Pet. Tech., January: 97-104. Trans. AIME. OILWELL TESTING 238 23) Earlougher, R.C., and Kersch, K.M., 1974. Analysis of Short-Time Transient Test Data by Type-Curve Matching. J. Pet. Tech., July: 793-800. Trans. AIME. CHAPTER 8 REAL GAS FLOW: GAS WELL TESTING 8.1 INTRODUCTION The first part of this chapter describes how the basic differential equation for radial fluid flow, equ. (5.1), can be approximately linearized for real gas flow. This is achieved using the real gas pseudo pressure function b p p pdp m(p) 2 µ = Ζ and subsequently, all equations in the chapter are expressed in terms of m(p) functions rather than real pressures. The constant terminal rate solution of the radial diffusivity equation is then presented in dimensionless form, equivalent to the pD functions for liquid flow, and the solution is applied to the analysis of gas well tests. A similar approach is used for analysing pressure buildup tests in solution gas drive reservoirs, below bubble point pressure. 8.2 LINEARIZATION AND SOLUTION OF THE BASIC DIFFERENTIAL EQUATION FOR THE RADIAL FLOW OF A REAL GAS By assuming mass conservation, Darcy's law and applying the definition of fluid compressibility, the basic equation for the radial flow of a single phase fluid in a porous medium was derived in chapter 5 as 1 k p p r c r r r t ρ φ ρ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.1) This equation was linearized for liquid flow by deletion of terms, assuming that - µ was independent of pressure - p r ∂ ∂ was small and therefore 2 p r ∂ ∂ was negligible - c was small and constant so that cp << 1 which resulted in the radial diffusivity equation 1 p c p r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (5.20) Because this equation is linear for liquid flow, simple analytical methods could be applied to describe stabilized inflow (Chapter 6) and the constant terminal rate solution (Chapter 7). The assumptions made in linearizing equ. (5.1) are inappropriate when applied to the flow of a real gas. In the first place, gas viscosity is highly pressure dependent. Secondly, the isothermal compressibility of a real gas is REAL GAS FLOW: GAS WELL TESTING 240 1 1 1 c p p p ∂Ζ = − ≈ Ζ ∂ (1.31) which again is highly pressure dependent and automatically violates the above condition that cp << 1. These problems, although severe, are not insurmountable. Nevertheless, it was not until the mid sixties that reliable analytical solutions of equ. (5.1 ) were developed. Two separate solution methods were published almost simultaneously in 1966; these are - the Russell, Goodrich et. al., p2 formulation1 - the Al-Hussainy, Ramey and Crawford, real gas pseudo pressure formulation2. Both techniques will be described in this chapter although the latter, for reasons explained in the text, is preferred. To illustrate the difference in approach the radial semi-steady state inflow equation, equivalent to equ. (6.12), will be derived in secs. 8.3 and 8.4, using both methods. Having thus established an analogy between liquid and real gas flow equations, the constant terminal rate solution for gas is stated by inference and its application described in detail in the remainder of the chapter. Because of the great disparity between gas rates measured at the surface (Q) and in the reservoir (q) it has become conventional to express gas flow equations using surface rates, at standard conditions, with all parameters expressed in field units. This practice will be adhered to in this chapter, using the following units Q -Mscf/d µ -cp(= µg) (at 60°F and 14.7 psia) Z -dimensionless t -hours p -psia k -mD T -°R (460+°F) h, r -ft In all equations µ and Z are evaluated at some defined reservoir condition. The basic derivations of the flow equations in sections 8.3 through 8.8 will still be performed in Darcy units, with conversion to field units being made upon achieving the desired form of equation. 8.3 THE RUSSELL, GOODRICH, et. al. SOLUTION TECHNIQUE The authors approached the problem by making the initial assumption that it was possible to linearize equ. (5.1) for real gas flow in precisely the same manner as for liquid flow, described in Chapter 5, sec. 4. Admittedly, this approach should yield inaccurate results. However, Russell and Goodrich also designed a numerical model of a single well draining a radial volume element which itself was subdivided into finite grid blocks as shown in fig. 8.1. REAL GAS FLOW: GAS WELL TESTING 241 rw r re Fig. 8.1 Radial numerical simulation model for real gas inflow The flow equations from block to block were solved numerically, using a finite difference approximation, making due allowance for the variation of µ and Z as functions of pressure. This is equivalent to solving the non-linear second order differential equation (5.1). The results may be expected to be in slight error due to the use of finite difference calculus, but the errors were minimized by making the grid blocks smaller in the vicinity of the wellbore, where the pressure gradients are largest, thus providing a higher resolution of solution in this region. With this model it was hoped that some correcting factor could be found which could be used to match the approximate analytical results, obtained by making the same assumptions as for a single phase liquid, with the more exact results from the numerical simulation. As an example of the approach taken by Russell and Goodrich, consideration will be given to adapting the semi-steady state inflow equation, developed in chapter 6, sec. 2, for the flow of oil, to an equivalent form which will be appropriate for the flow of gas. The equation of interest, expressed in Darcy units, is e wf w r q 3 p p ln S 2 kh r 4 µ π − = − + (6.12) which, when expressed in the field units specified in the previous section, becomes ( ) e wf w s.cc / sec r.cc / sec Q Mscf / d r atm 3 Mscf / d s.cc / sec p p psi ln S D cm psi r 4 2 k mD h ft mD ft µ π − = − + (8.1) In this conversion the ratio r.cc / sec reservoir cc / sec 1 1 s.cc / sec standard cc / sec E Gas expansion factor = = = and in field units p E 35.37 ZT = (1.25) and p, the pressure at which E is evaluated, is as yet undefined. The full conversion of the rate term in equation (8.1 ) can be expressed as REAL GAS FLOW: GAS WELL TESTING 242 Mstb / d stb / d s.cc / sec 1 Q Mscf / d q r.cc / sec Mscf / d Mstb / d stb / d E = [ ] [ ] 1 ZT Q Mscf / d 1000 1.84 q r.cc / sec 5.615 35.37p × × = QZT 9.265 Mscf / d q r.cc / sec p = Including the remaining conversion factors in equ. (8.1 ) yields e wf w r 711 Q ZT 3 p p ln S r 4 khp µ − = − + (8.2) Russell and Goodrich, comparing equ. (8.2) with the numerical simulation, found that for the same reservoir and flow conditions the two were in close agreement providing that the pressure p , at which the gas expansion factor was evaluated, was set equal to the average of the current, average reservoir pressure and the bottom hole flowing pressure i.e. wf p p p 2 + = (8.3) Furthermore, both µ and Z should also be evaluated at this same pressure so that wf wf p p p p and Z Z 2 2 µ µ + + = = (8.4) and substituting these values of p, µ and Z in equ. (8.2) gives 2 2 e wf w r 1422 Q ZT 3 p p ln S kh r 4 µ − = − + (8.5) Equ. (8.5) is the familiar p2 formulation of the well inflow equation, under semi-steady state conditions, which was tested by Russell and Goodrich and found to be applicable over a wide range of reservoir conditions and flow rates. Similarly, the transient line source solution for the same initial and boundary conditions detailed in chapter 7, sec. 2, is ( ) 2 2 i wf 2 w i 711 Q ZT 4 .000264kt p p ln 2S kh c r µ γ φ µ − = + (8.6) wf wf p p p p and Z Z 2 2 µ µ + + = = (8.4) REAL GAS FLOW: GAS WELL TESTING 243 which is the real gas equivalent of equ. (7.10) in field units. This equation was also found to compare favourably with the numerical simulation results, providing the viscosity-compressibility product was evaluated as (µc)i, at the initial pressure pi, (ref. sec. 8.8). One obvious practical disadvantage in using the p2 formulation can be appreciated by considering a frequently occurring problem in gas inflow calculations, namely, the calculation of pwf if both p and Q are known using, in this case, the semi-steady state inflow equation. A schematic of the calculation procedure is shown in fig. 8.2. If it is assumed that p has been determined in the drainage volume of the well from material balance considerations then, for a fixed offtake rate, it will be necessary to solve the inflow equation by iteration to determine pwf since both µ and Z must be evaluated at the pressure defined in equ. (8.4). In any iteration cycle k wf p is calculated using values of µk and Zk evaluated at the pressure k 1 wf (p p )/ 2 − + , where k is the iteration counter. For k = 1, both µ1 and Z1 can be evaluated at some convenient starting pressure, which in this case has been selected as p. When the difference between successive values of k wf p is less than some tolerance value (TOL) the iteration is terminated. Other disadvantages in using the p2 formulation for inflow equations will be discussed in section 8.5. 8.4 THE AL-HUSSAINY, RAMEY, CRAWFORD SOLUTION TECHNIQUE In their approach the authors attempted to linearize the basic flow equation, (5.1), using the following version of the Kirchhoff integral transformation ( ) b p p pdp m p 2 Z µ = (8.7) which was given the name, in this present context, of "the real gas pseudo pressure". REAL GAS FLOW: GAS WELL TESTING 244 p Z p k = 1 µ 1 = ( ) Z µ p 1 = Z( ) p k = 1 1/ 2 2 e wf w r 1422 Q Z T 3 p ln S p kh r 4 = − − + µk k k k k = 1 k + k > 1 -1 wf p p 2 + = µ µ k k -1 wf p p Z Z 2 + = k k k = 1 k + -+ accept pwf -1 wf wf TOL p p − − k k - from material balance Fig. 8.2 Iterative calculation of pwf using the p2 formulation of the radial, semi-steady state inflow equation, (8.5) The limits of integration are between a base pressure pb and the pressure of interest p. The value of the base pressure is arbitrary since in using the transformation only differences in pseudo pressures are considered i.e. ( ) ( ) wf b b wf p p p wf p p p pdp pdp pdp m p m p 2 2 2 Z Z Z µ µ µ − = − = As will be seen presently, it is possible, and indeed advantageous, to express all flow equations in terms of these pseudo pressures rather than in the p2 formulation of Russell and Goodrich. However, conceptually it is more difficult and generally engineers feel more comfortable dealing with p2 rather than an integral transformation. Therefore, it is worthwhile, at this stage, to examine the ease with which these functions can be generated and used. REAL GAS FLOW: GAS WELL TESTING 245 PVT data Numerical Integration Pseudo pressures p µ Z 2ρ µΖ 2p µΖ p ∆ 2p p µ × ∆ Ζ 2p m(p) p µ = Σ ∆ Ζ (psia) (cp) (psia)2/cp 400 .01286 .937 66391 33196 400 13.278×106 13.278×106 800 .01390 .882 130508 98449 " 39.380 " 52.658 " 1200 .01530 .832 188537 159522 " 63.809 " 116.467 " 1600 .01680 .794 239894 214216 " 85.686 " 202.153 " 2000 .01840 .770 282326 261110 " 104.444 " 306.597 " 2400 .02010 .763 312983 297655 " 119.062 " 425.659 " 2800 .02170 .775 332986 322985 " 129.194 " 554.853 " 3200 .02340 .797 343167 338079 " 135.231 " 690.084 " 3600 .02500 .827 348247 345707 " 138.283 " 828.367 " 4000 .02660 .860 349711 348979 " 139.592 " 967.958 " 4400 .02831 .896 346924 348318 " 139.327 " 1107.285 " TABLE 8.1 Generation of the real gas pseudo pressure, as a function of the actual pressure; (Gas gravity, 0.85, temperature 200°F) All the parameters in the integrand of equ. (8.7) are themselves functions of pressure and can be obtained directly from PVT analysis of the gas at reservoir temperature or, knowing only the gas gravity, from standard correlations of µ and Z, again at reservoir temperature. Table 8.1 lists a set of typical PVT data and shows how, using a simple graphical method for numerical integration (trapezoidal rule), a table of values of m(p) can be generated as a function of the actual pressures. A graph of the values of m(p) versus pressure, corresponding to table 8.1, is included as fig. 8.3. This plot is used in the gas well test exercises 8.1−3, (secs. 8.10−11), in which it is assumed that for high pressures, in excess of 2800 psia, the function is almost linear and can be described by m (p) = (0.3457p - 414.76)× 106 psia2/cp Having once obtained this relationship, the resulting plot should be preserved since it will be relevant for the entire lifetime of the reservoir. Using the plot, it is always quite straightforward to convert from real to pseudo pressures and vice versa. REAL GAS FLOW: GAS WELL TESTING 246 In attempting to linearize the basic radial flow equation (5.1), (using, for the moment, Darcy units), Al-Hussainy, Ramey and Crawford replaced the dependent variable p by the real gas pseudo pressure m(p) in the following manner. REAL GAS FLOW: GAS WELL TESTING 247 1200 0 800 600 400 200 1000 0 x10 6 m(p) psia /cp 2 1000 2000 3000 4000 PRESSURE (psia) REAL GAS FLOW: GAS WELL TESTING 248 Fig. 8.3 Real gas pseudo pressure, as a function of the actual pressure, as derived in table 8.1; (Gas gravity, 0.85; Temperature 200°F) m (p) 2p Z µ 2p Z µ ∆ ∆ µ (Area) = m (p) = (2p / Z) p ∆ p Pressure ∆p Fig. 8.4 2p/µZ as a function of pressure Since ( ) ( ) m p m p p r p r ∂ ∂ ∂ = ⋅ ∂ ∂ ∂ and ( ) m p 2p p µ ∂ = ∂ Ζ Then ( ) m p 2p p r r µ ∂ ∂ = ∂ Ζ ∂ (8.8) and similarly ( ) m p 2p p t t µ ∂ ∂ = ∂ Ζ ∂ (8.9) These relations are evident from fig. 8.4 and, substituting for ∂p/∂r and ∂p/∂t in equ. (5.1), using equs. (8.8) and (8.9) gives ( ) ( ) m p m p 1 k r c r r 2p r 2p t ρ µ µ φ ρ µ ∂ ∂ ∂ Ζ Ζ = ∂ ∂ ∂ (8.10) Finally, using the equation of state for a real gas Mp ZRT ρ = and substituting this expression for ρ in equ. (8.10) leads, after some cancellation of terms, to the simplified expression ( ) ( ) m p m p 1 c r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (8.11) Equation (8.11) has precisely the same form as the diffusivity equation, (5.20), except that the dependent variable has been replaced by m(p). REAL GAS FLOW: GAS WELL TESTING 249 Note that in reaching this stage it has not been necessary to make any restrictive assumptions about the viscosity being independent of pressure or that the pressure gradients are small and hence squared pressure gradient terms are negligible, as was implicit in the approach of Russell and Goodrich. Therefore, the problem has already been partially solved but it should be noted that the term φµc/k in equ. (8.11) is not a constant, as it was in the case of liquid flow, since for a real gas both µ and c are highly pressure dependent. Equation (8.11) is therefore, a non-linear form of the diffusivity equation. Continuing with the argument; in order to derive an inflow equation under semi-steady state flow conditions, then applying the simple material balance for a well draining a bounded part of the reservoir at a constant rate p V cV q t t ∂ ∂ = − = − ∂ ∂ (5.8) and for the drainage of a radial volume element 2 e p q t r h c π φ ∂ = − ∂ (5.10) Also, using equ. (8.9) ( ) 2 e m p 2p p 2p q t t r h c µ µ π φ ∂ ∂ = ⋅ = − ⋅ ∂ Ζ ∂ Ζ (8.12) and substituting equ. (8.12) in (8.11) gives ( ) 2 e m p 1 c 2p q r r r r k r h c φµ µ π φ ∂ ∂ = − ⋅ ⋅ ∂ ∂ Ζ or ( ) 2 e res m p 1 2 pq r r r r r kh π ∂ ∂ = − ∂ ∂ Ζ (8.13) Furthermore, using the real gas equation of state, sc sc sc res pq T p q T = Ζ equ. (8.13) can be expressed as ( ) sc sc 2 sc e m p 2p q 1 T r r r r T r kh π ∂ ∂ = − ⋅ ∂ ∂ (8.14) REAL GAS FLOW: GAS WELL TESTING 250 - from material balance m (p) m ( ) p p p m (p) m (p ) wf pwf p pwf m (p ) wf m (p ) = m ( ) wf p m ( ) p p p Z e w r 1422 QT 3 ln S kh r 4 − − + Fig. 8.5 Calculation of pwf using the radial semi-steady state inflow equation expressed in terms of real gas pseudo pressures, (equ. 8.15) For isothermal reservoir depletion, the right hand side of equ. (8.14) is a constant, and the differential equation has been linearized. A solution can now be obtained using precisely the same technique as applied in Chapter 6, sec. 2, for liquid flow. If, in addition, field units are employed then the resulting semi-steady state inflow equation can be expressed as ( ) ( ) e wf w r 1422 QT 3 m p m p ln S kh r 4 − = − + (8.15) Note that this equation has a similar form to the p2 formulation of equ. (8.5), except that the right hand side no longer contains the pressure dependent µZ term which is now implicit in the pseudo pressures. Because of this, the practical difficulty in having to iterate when solving the inflow equation for pwf is removed. The relevant steps corresponding to fig. 8.2 are shown in fig. 8.5. Similarly, the transient line source solution, when expressed in pseudo pressures and field units, becomes ( ) ( ) ( ) i wf 2 w i 711 QT 4 .000264kt m p m p ln 2S kh c r γ φ µ − = + (8.16) REAL GAS FLOW: GAS WELL TESTING 251 8.5 COMPARISON OF THE PRESSURE SQUARED AND PSEUDO PRESSURE SOLUTION TECHNIQUES Much has been written3,4,5 about the conditions under which the p2 and m(p) solution techniques give identical results. Comparison of the methods can best be summarised by directly comparing equ. (8.5) and equ. (8.15) i.e. when is 2 2 wf p p µ − Ζ equivalent to ( ) ( ) wf p wf p pdp m p m p 2 µ − = Ζ or ( ) ( ) wf wf p p p p 2 µ + − Ζ equivalent to wf p p pdp µΖ (8.17) where both µ and Z appearing on the left hand side are evaluated at wf (p p )/ 2 + . As shown in fig. 8.6, the equivalence expressed in equ. (8.17) is only established if pµ/Z is a linear function of the pressure. wf p p 2 Z + µ p Z µ pwf p dp Pressure Fig. 8.6 p/µZ as a linear function of pressure In this case the area under the curve between p and pwf is the integral in equ. (8.17), which is equal to ( ) ( ) wf wf p p p p 2 µ + ⋅ − Ζ . However, in general p/µZ is non-linear and has the typical shape shown in fig. 8.7. It can be seen that p/µZ versus p is only linear at high and very low pressures, the latter corresponding to the ideal gas state. In between, there is a very definite curved section in the plot where the two different solution techniques are liable to give different results. REAL GAS FLOW: GAS WELL TESTING 252 ∆p ∆p p Z µ Pressure Fig. 8.7 Typical plot of p/µZ as a function of pressure The diagram also shows that even in the non-linear part of the plot, providing the drawdown, wf p p dp, − = is small, the two methods will always give approximately the same answers. It is only when the drawdown is very large (i.e. for low kh reservoirs producing at high rates) that the results using the two methods will be significantly different. Under these circumstances the assumption implicit in the Russell Goodrich approach, namely that pressure gradients are small, is no longer valid. With the exception of the brief description of the history of gas well testing in sec. 8.10, all the equations for the flow of a real gas, in the remainder of this chapter, will be expressed in terms of real gas pseudo pressures. The reasons for adopting this approach are: - it is theoretically the better method and in using it one does not have to be concerned about the pressure ranges in which it is applicable, as is the case when using the p2 method - with a bit of practice, it is technically the more simple method to use once the basic relationship for m(p) as a function of p has been derived - the necessity for iteration in solving the inflow equation for pwf is avoided - the technique is widely used in the current literature and readers are expected to be quite familiar with its application. 8.6 NON-DARCY FLOW For the horizontal flow of fluids through a porous medium at low and moderate rates, the pressure drop in the direction of flow is proportional to the fluid velocity. The mathematical statement of this relationship is Darcy's law, which for radial flow is dp u dr k µ = (8.18) where u is the fluid velocity = q 2 rh π ⋅ REAL GAS FLOW: GAS WELL TESTING 253 At higher flow rates, in addition to the viscous force component represented by Darcy's equation, there is also an inertial force acting due to convective accelerations of the fluid particles in passing through the pore spaces6. Under these circumstances the appropriate flow equation is that of Forchheimer (1901), which is 2 dp u u dr k µ βρ = + (8.19) In this equation the first term on the right hand side is the Darcy or viscous component while the second is the non-Darcy component. In this latter term, β is the coefficient of intertial resistance and, as the following dimensional analysis shows, has the dimension (length)-1. [ ] 2 2 2 2 3 2 dp ML 1 M L u dr L T L L T β ρ = 1 L β − = The non-Darcy component in equ. (8.19) is negligible at low flow velocities and is generally omitted from liquid flow equations. For a given pressure drawdown, however, the velocity of gas is at least an order of magnitude greater than for oil, due to the low viscosity of the former, and the non-Darcy component is therefore always included in equations describing the flow of a real gas through a porous medium. Because of this it should be necessary to use the Forchheimer equation, rather than that of Darcy, in deriving the basic radial differential equation for gas flow (refer Chapter 5, sec. 2). Fortunately, even for gas, the non-Darcy component in equ. (8.19) is significant only in the restricted region of high pressure drawdown, and flow velocity, close to the wellbore. Therefore, the non-Darcy flow is conventionally included in the flow equations as an additional skin factor, that is, as a time independent perturbation affecting the solutions of the basic differential equation in the same manner as the van Everdingen skin (Chapter 4, sec. 7). Forchheimer's equation was originally derived for the flow of fluids in pipes where at high velocity there is a distinct transition from laminar to turbulent flow. In fluid flow in a porous medium, however, for most practical cases in reservoir engineering, the macroscopic flow is always laminar according to the definitions of classical fluid dynamics. What is referred to as the non-Darcy component does not correspond with classical ideas of turbulent flow but, as stated earlier, is due to the accelerations and decelerations of the fluid particles in passing through the pore spaces. Nevertheless, Forchheimer's equation can be used to describe the additional pressure drop due to this phenomenon, by integrating the second term on the right hand side of equ. (8.19), as follows. e w 2 r non Darcy r q p dr 2 rh βρ π ∆ = or expressed as a drop in the real gas pseudo pressure, using equ. (8.8) REAL GAS FLOW: GAS WELL TESTING 254 ( ) e w 2 r nD r 2p q m p dr 2 rh βρ µ π ∆ = Ζ (8.20) Also since ρ = γg × density of air at s.c. × E g p constant γ = × × ΖΤ where γg is the gas gravity (air=1) Then equ. (8.20) can be written as ( ) e w 2 r g 2 2 nD r T pq m p constant dr r h γ µ β ∆ = × ΖΤ (8.21) and since sc sc sc sc p q pq constant q T = = × ΖΤ then for isothermal reservoir depletion equ. (8.21) becomes ( ) e w r 2 g sc 2 2 nD r q dr m p constant h r β γ µ Τ ∆ = × (8.22) Since non-Darcy flow is usually confined to a localised region around the wellbore where the flow velocity is greatest, the viscosity term in the integrand of equ. (8.22) is usually evaluated at the bottom hole flowing pressure pwf and hence is not a function of position. Integrating equ. (8.22) gives ( ) 2 g sc 2 nD w e q 1 1 m p constant r r h β γ Τ ∆ = × − (8.23) If equ. (8.23) is expressed in field units (Q - Mscf/d, β - ft-1) and assuming w e 1 1 , then r r >> ( ) 2 g 12 2 2 nD w p w Q m p 3.161 10 FQ h r β γ µ − Τ ∆ = × = (8.24) where F is the non-Darcy flow coefficient psia2/cp/(Mscf/d)2. Since non-Darcy flow is only significant very close to the wellbore, two assumptions are commonly made in connection with equ. (8.24), these are -the value of the thickness h is conventionally taken as hp, the perforated interval of the well -the pseudo pressure drop ∆m(p)nD = FQ2 can be considered as a perturbation which readjusts instantaneously after a change in the production rate. REAL GAS FLOW: GAS WELL TESTING 255 Because of the latter assumption the FQ2 term can be included in equs. (8.15) and (8.16) in very much the same way as the mechanical skin factor, only in this case it is interpreted as being a rate dependent skin. Thus equ. (8.15), for instance, including the non-Darcy flow component, becomes ( ) ( ) 2 e wf w r 1422 TQ 3 m p m p ln S FQ kh r 4 − = − + + (8.25) e w r 1422 TQ 3 ln S DQ kh r 4 = − + + (8.26) where in the latter expression, which is commonly used in the literature, DQ is interpreted as the rate dependent skin factor and Fkh D 1422T = (8.27) Either F or D is used in the remainder of this chapter, to allow for non-Darcy flow, depending on which is the more convenient for the application being considered. 8.7 DETERMINATION OF THE NON-DARCY COEFFICIENT F Two methods are available for the determination of the non-Darcy flow coefficient, which are - from the analysis of well tests - by experimentally measuring the values of the coefficient of inertial resistance, β and using it in equ. (8.24) to calculate F. Of these two, the well testing method will give the more reliable result just as in the case of oil well testing in which, from the slope of the pressure buildup plot, a more meaningful value of the kh product can be obtained than by measuring values of the permeability on a selection of core samples and trying to average these results over the entire formation. Furthermore, in the well test F will be measured in the presence of any liquid saturation in the vicinity of the well. The determination of F by well testing will be described in detail in secs. 8.10 and 8.11 and will not be discussed further at this stage. To determine β experimentally, the procedure is to first measure the absolute permeability of each of the core samples and then to apply a series of increasing pressure differentials across each sample by flowing air through the core plugs at ever increasing rates. Knowing the flow rates and pressure differentials across the plugs, the coefficient of inertial resistance can be directly calculated using a linear version of the Forchheimer equation (8.19). The results are usually presented as shown in fig. 8.8 in which β is plotted as a function of the absolute permeability over the range of core samples tested. REAL GAS FLOW: GAS WELL TESTING 256 10 3 9 8 7 6 5 4 3 2 10 2 9 8 7 6 5 4 3 2 10 9 8 7 6 5 4 3 2 1 10 5 10 6 10 7 10 8 10 9 2 3 4 5 7 89 2 3 4 5 7 89 2 3 4 5 7 89 2 3 4 5 7 89 β - FACTOR (1/cm) k (md) Fig. 8.8 Laboratory determined relationship between β β β β and the absolute permeability A relationship is usually derived of the form constant kα β = (8.28) in which the exponent α is a constant. For the experimental results shown in fig. 8.8 the specific relationship is 10 1.1045 2.73 10 k β × = where k is in mD and β in ft-1. Providing that the range of porosity variation in the samples is not too great, the variation of β with φ can be neglected in comparison with the variation of β with the absolute permeability. The experimental value of β so determined is applicable to the flow of gas at 100% saturation. In the presence of a liquid saturation, e.g. connate water and immobile liquid condensate, Gewers, Nichol and Wong7,8 have experimentally determined that the permeability term in equ. (8.28) should be replaced by the effective permeability to gas at the particular liquid saturation SL, thus ( ) rg constant kk α β = (8.29) It should be noted that the experimental work of Gewers, Nichol and Wong, in directly measuring β in the presence of a liquid saturation, was conducted on microvugular carbonate rock samples for which the dry core β values are at least an order of REAL GAS FLOW: GAS WELL TESTING 257 magnitude greater than for typical sandstone samples. So far, the experiments have not been repeated on sandstone but in using equ. (8.29) it is assumed that the same physical principles will apply. Although correlation charts giving β as a function of the permeability exist in the literature9 the reader should be aware that they are not always applicable. Irregularities in the pores can greatly modify the β versus k relationship making it advisable, in many cases, to experimentally derive a relationship of the form given by equ. (8.28). 8.8 THE CONSTANT TERMINAL RATE SOLUTION FOR THE FLOW OF A REAL GAS The constant terminal rate solution of the radial diffusivity equation ( ) ( ) m p m p 1 c r r r r k t φµ ∂ ∂ ∂ = ∂ ∂ ∂ (8.11) for the flow of a real gas, describes the change in real gas pseudo pressure at the wellbore due to production at a constant rate from time t = 0. Equation (8.11) is identical in form with equ. (5.20) except that pseudo pressure replaces real pressure as the dependent variable. Therefore, the constant terminal rate solution of equ. (8.11) must, by analogy, have the same form as the solution presented for liquid flow in Chapter 7. For small flowing times, the transient, constant terminal rate solution of equ. (8.11), in Darcy units, is similar to equ. (7.10), i.e. ( ) ( ) i wf 2 w 4 kt m p m p constant ln 2S cr γ φ µ ′ − = × ⋅ + in which S' = S + DQ. The constant can be evaluated using the relationship ( ) 2p m p p µ ∆ = ∆ Ζ therefore, ( ) ( ) i wf 2 w 2p q 4 kt m p m p ln 2S 4 kh cr µ µ π γ φ µ ′ − = + Ζ and converting to field units, using the fact that pq/Z = (psc qsc)T/Tsc, gives ( ) ( ) D i wf 4t 711QT m p m p ln 2S kh γ ′ − = + (8.30) Taking the analogy with the liquid flow equations a stage further, equ. (8.30) can be expressed in dimensionless form as ( ) ( ) ( ) ( ) i wf D D kh m p m p m t S 1422QT ′ − = + (8.31) REAL GAS FLOW: GAS WELL TESTING 258 in which mD (tD) is the dimensionless real gas pseudo pressure which for transient flow conditions is simply ( ) D 12 D D 4t m t ln γ = (8.32) and is identical in form to equ. (7.23). Similarly, for long flowing times the semi-steady state constant terminal rate solution of equ. (8.11), in Darcy units, is ( ) ( ) 12 i wf 2 A w 2p q 4A kt m p m p ln 2 S 2 kh cA C r µ π µ π φ µ γ ′ − = + + Ζ which is the equivalent of equ. (7.13). In field units this becomes ( ) ( ) 12 i wf DA 2 A w 1422QT 4A m p m p ln 2 t S kh C r π γ ′ − = + + and therefore the mD (tD) function for semi-steady state flow is ( ) 12 D D DA 2 A w 4A m t ln 2 t C r π γ = + (8.33) which is equivalent to equ. (7.27). To generalise, the dimensionless real gas pseudo pressures are the constant terminal rate solutions of the equation D D D D D D D m m 1 r r r r t ∂ ∂ ∂ = ∂ ∂ ∂ (8.34) and the solution which is valid for all values of the flowing time is ( ) ( ) ( ) D 1 1 2 2 D D DA DA D MBH 4t m t 2 t ln m t π γ = + − (8.35) which is the same as equ. (7.42), for liquid flow. The mD(MBH) function, the Matthews, Brons and Hazebroek dimensionless pseudo pressure, can be read directly from the MBH charts, figs. 7.11-15, for the appropriate value of the dimensionless time argument tDA, in just the same way as the pD(MBH) function was evaluated for use in equ. (7.42). Since field units are being used in this chapter, the abscissa and ordinate of the MBH charts should be interpreted as ( ) DA kt hrs t 0.000264 cA φ µ = (8.36) and REAL GAS FLOW: GAS WELL TESTING 259 ( ) ( ) ( ) ( ) DA D MBH kh m t m p m p 711QT = − (8.37) The right hand side of equ. (8.37) is only used to calculate ( ) m p , having already determined m(p) from the extrapolated pseudo pressure buildup plot (ref. sec. 8.11). In the majority of cases, mD(MBH) is simply a number read from the MBH charts for the appropriate value of tDA, for use in equ. (8.35). Because of the equivalence in form of the constant terminal rate solution for both oil and gas, when expressed in dimensionless parameters, there is no need to elaborate further on how this solution is used in practice since the subject has been fully described in the previous chapter. The application of equ. (8.35) to gas well testing will be demonstrated in secs. 8.10 and 11. Although the form of the mD and pD functions is the same, it must always be kept in mind that the mD function applied to real gas flow is less accurate than the pD function applied to liquid flow. The reason is that the mD functions are derived as solutions of equ. (8.11), which is non-linear. The non-linearity arises from the fact that both the real gas viscosity and compressibility in the coefficient φµc/k, in equ. (8.11),are highly pressure dependent. Fortunately, the gas viscosity is directly proportional to the pressure while the compressibility, equ. (1.31), is inversely proportional to pressure and this tends to reduce the pressure dependence of the product. This favourable effect is particularly pronounced in the high pressure range where the product is fairly constant. For instance, using the PVT data presented in table 8.1, the value of µc only increases from 3.54 × 10-6 cp/psi at 4400 psia to 4.96 x 10-6 cp/psi at 3400 psia. In evaluating these figures, the isothermal gas compressibility has been calculated using equ. (1.31), treating the pore and connate water compressibilities as negligible in comparison to that of the gas. This practice is adhered to in the remainder of this chapter, thus ct ≈ cg = c. Because of this insensitivity of the µc product to pressure change it is common, when applying the mD function, to use the product (µc)i evaluated at the initial equilibrium pressure, which is also the assumption made in generating pD functions for liquid flow. Thus, tD, tDA and 1/2 mD(MBH ) (tDA ) in equ. (8.35), which are dependent on the product, are all evaluated using (µc);. Al-Hussainy, Ramey and Crawford2 have demonstrated that, using this initial value of the product, the mD functions do correlate very favourably with the pD functions for liquid flow over a wide range of conditions. While the match is very good for transient flow conditions, it is less reliable for very large values of the flowing time, once the boundary effects have been felt. Thus, equ. (8.32) correlates with the pD function better than equ. (8.33). The latter must, therefore, be used with more caution. Fortunately, the inflow equation (8.15), for semi-steady state flow, which can be expressed in dimensionless form as ( ) ( ) ( ) ( ) 1 D 2 wf D 2 A w kh 4A m p m p m t S ln S 1422QT C r γ ′ ′ − = + = + (8.38) REAL GAS FLOW: GAS WELL TESTING 260 was found to correlate almost exactly with the equivalent function for liquid flow for all values of the flowing time. This is to be expected since the, µc product is not present in equ. (8.38) as a result of the use of pseudo pressures, rather than pressures, in its formulation. The correlations between mD and pD functions were only checked for a well producing at the centre of a circular shaped reservoir, equs. (8.35) and (8.38) are generalized expressions which include the dependence of mD on geometry and well asymmetry. In practice, one is interested in applying the constant terminal rate solution (mD function) to the analysis of well tests and several examples of such usage are provided in the following sections of this chapter. All the examples considered are for initial well tests and, for this condition, the evaluation of the mD function using the (µc)i product can be expected to be quite reliable, particularly if the test duration is not excessively long and the pressure drawdowns imposed are not too large. Problems arise, however, when analysing routine pressure surveys throughout the producing lifetime of the reservoir. For instance, if a pressure test is conducted in a well several years after the start of production, at what pressure should the µc product be evaluated? This question will be dealt with in sec. 8.11 using the method presented by Kazemi10, which describes how the average reservoir pressure can be obtained from a pressure analysis using a µc product which must be iteratively determined. Once the average reservoir pressure is known, however, the inflow equation (8.38) can be used with confidence to calculate the long term deliverability of wells. 8.9 GENERAL THEORY OF GAS WELL TESTING Gas well tests can be interpreted using the following equations ( ) ( ) ( ) ( ) n n j 1 n i wf j D D D n n j 1 kh m p m p Q m t t Q S 1422T − = ′ − = ∆ − + (8.39) in which ( ) ( ) ( ) ( ) n j 1 D 1 1 2 2 D D D D D DA DA D MBH j j j 1 n n 4t m t t m t 2 t ln m t Q Q Q and S S DQ π γ − − ′ ′ ′ ′ − = = + − ∆ = − ′ = + (8.40) For convenience, equ. (8.39) is frequently expressed in the form ( ) ( ) ( ) ( ) n n j 1 n 2 i wf n j D D D n j 1 kh m p m p FQ Q m t t Q S 1422T − = − − = ∆ − + (8.41) in which F is the non-Darcy flow coefficient, equ. (8.27). These equations are analogous to equs. (7.31) and (7.42) which were used for oilwell test analysis. Equation (8.39) results from the application of the principle of superposition in time, as described in Chapter 7, sec. 5. In the summation of the REAL GAS FLOW: GAS WELL TESTING 261 individual constant terminal rate solutions it is assumed that the only skin factor term remaining which influences pwf is the value S n ′ = S + DQn. This is because the rate dependent skin factor is not, in itself, a time dependent solution of the diffusivity equation but is merely considered to have a perturbing influence on the bottom hole flowing pressure which re-adjusts instantaneously when the rate changes. Thus, even though the DQ terms in the summation leading to equ. (8.39) do not algebraically cancel, as do the mechanical skin factor components, on changing the rate from Q´n-1 to Qn the value of n wf p is only influenced by DQn. The components DQn-1, DQn-2 . . . DQn-j have no transient effects and die out immediately. The main difference between oil and gas well testing arises from the fact that the total skin factor in a gas well has two components, one of which is rate dependent. Because of this, a gas well must be tested with a minimum of two separate flow rates to be able to differentiate between these two skins. Thus, at rate Q1 the total skin 1 1 S S DQ ′ = + can be obtained from the test analysis, while at rate Q2 2 2 S S DQ ′ = + can be similarly determined. The two equations for 1 S′, and 2 S′ can then be solved simultaneously to provide S and D (or F). A further complication in applying equs. (8.39) and (8.40) to gas well test analysis lies in the fact that the superposition principle itself, which is implicit in the formulation of equ. (8.39), is strictly only valid when applied to solutions of linear differential equations. Since it is the constant terminal rate solution of the non-linear radial diffusivity equation (8.11) that is being superposed, some fundamental mathematical error might therefore be expected. Al-Hussainy, Ramey and Crawford2 have shown, by comparing superposed mD functions for gas with the superposed pD functions for liquid flow, that the error is very slight providing that the test is conducted with an increasing rather than a decreasing rate sequence. As stated in sec. 8.8, the viscosity-compressibility product used in an initial well test is (µc)i evaluated for the initial equilibrium pressure pi. In the test analysis exercises included in the following sections of the chapter, the initial pressure in each case is taken as 4290 psi and the PVT data and real gas pseudo pressures of table 8.1 are common to all. The value of (µc)i under these circumstances is 3.6 × 10-6 cp/psi and no allowance for its variation is made during the ensuing analyses. It is interesting to note that, in spite of the high compressibility, the µc product for a gas reservoir (≈ µcg) is invariably three or four times smaller than for an oil reservoir, because of the low viscosity of the gas. This implies that, for a given drainage area, permeability and porosity, the change from transient to either late transient or stabilized flow can be expected to occur much more rapidly in a gas than in an oil reservoir. This has implications which will be discussed in greater detail in secs. 8.10 and 8.11. REAL GAS FLOW: GAS WELL TESTING 262 In conclusion, it can be stated that the use of equs. (8.39) and (8.40) to analyse gas well tests is never quite as satisfactory as using the combination of equs. (7.31) and (7.42) for oilwell tests. Nevertheless, the former equations do provide what is usually described in the literature as "a reasonable engineering approximation" and their application will now be described in detail. 8.10 MULTI-RATE TESTING OF GAS WELLS This section will be presented in the form of a brief history of the subject of multirate testing of gas wells in which former analysis techniques will be compared with the method implied in the use of equ. (8.39) and (8.40). The first and perhaps the best known equation for analysing multi-rate tests is that of Schellhardt and Rawlins11 which was empirically established as the result of the analysis of some 600 gas well tests during the 1930's. The equation is ( ) n 2 2 i wf Q C p p = − (8.42) "back pressure equation". Using the observed pressures at the end of each flow period a plot can be made of log 2 2 i wf (p p ) − versus log Q, the slope of which has the value 1/n. Having thus determined n, the value of C can be calculated using equ. (8.42). It was originally believed that both C and n were constants that, once determined from the test analysis, could be used for the long term prediction of gas well deliverability. This being accomplished merely by replacing pi by the current average reservoir pressure p. Carter et al.12, have shown, however, that n is a variable with a range between .5 and 1 depending on whether the value of the non-Darcy flow component FQ2, defined by equ. (8.24), is very large (n = .5) or negligible (n = 1). Furthermore, the value of C can be shown to be dependent on k, A, CA and S and also on the pressure dependent functions, µ, Z and the flowing time, and can hardly be expected to remain constant throughout the producing life of the well. These statements will not be substantiated in this text since it is not intended to use equ. (8.42) which may be regarded, at best, as being a useful empirical approximation. Nevertheless, in spite of all the drawbacks mentioned above, it was found that the back pressure equation could be used with tolerable accuracy in analysing tests in which it was suspected that semi-steady state flow conditions prevailed during each separate flow period. As a result, it became, and still is, quite fashionable to test wells in such a way that stabilized flow is achieved at each rate. Precisely when the change from transient to semi-steady state flow occurs depends on some minimum value of tDA = .000264 kt/φ (µc)iA, (a fact which is clearly illustrated by the MBH charts, figs. 7.11-15), which in turn depends upon the geometry of the drainage area and well asymmetry. As already noted in sec. 8.8, since the µc product for gas is considerably smaller than for a liquid, there is at least some justification in attempting to analyse gas well tests assuming stabilized flow conditions, even for a test of relatively short duration. REAL GAS FLOW: GAS WELL TESTING 263 The most popular current method of analysing such a test is to use the semi-steady state inflow equation, (8.38), which, including both skin factor components and assuming that for an initial test p = pi, becomes ( ) ( ) 2 12 i wf 2 A w 1422QT 4A m p m p ln S FQ kh C r γ − = + + (8.43) which can be further simplified as ( ) ( ) 2 i wf m p m p BQ FQ − = + (8.44) where B is the Darcy coefficient and F the non-Darcy coefficient of the inflow equation. The purpose of testing is then to determine B and F and to use these values for predicting the future deliverability of the well, for which pi is replaced by the current average pressure p. The method of analysis is simply to plot ( ) ( ) n i wf n n m p m p versus Q Q − (8.45) in which the Qn are the surface production rates and pwf the values of the bottom hole flowing pressure recorded at the end of each separate flow period. The plot should be linear with slope F and intercept B, when Qn = 0; an example of such a plot is shown as fig. 8.9. This method of analysis is simple and the results obtained in terms of B and F are considered more reliable parameters for estimating well deliverabilities than the C and n determined using equ. (8.42). Nevertheless, the application of the stabilized well inflow equation (8.43), to analyse a well test is mathematically incorrect and the results obtained from such an analysis can only be considered as an approximation, although, as will be demonstrated in exercise 8.1, a perfectly acceptable approximation in many cases. The fault lies in analysing the test data using an inflow equation rather than the superposed constant terminal rate solutions, equ. (8.39). It may well be that each flow period in the test is sufficiently long so that semi-steady state conditions prevail but that does not mean that, for instance, the wellbore pressure response during the third flow period is unaffected by what happened during the first and second periods, as implied by the use of equ. (8.43) in the analysis. The rigorously correct technique is to use equ. (8.39) in which the mD functions, equ. (8.40), are evaluated for semi-steady state flow as ( ) 12 D DA DA 2 A w 4A m t ln 2 t C r π γ ′ ′ = + (8.33) The test analysis equation is then REAL GAS FLOW: GAS WELL TESTING 264 ( ) n n j 1 n 12 i wf j DA DA n n 2 j 1 A w kh 4A m(p ) m(p ) Q 2 (t t ) Q ln S DQ 1422T C r π γ − = − = ∆ − + + + or alternatively, by re-arranging the summation term ( ) n j n 12 i wf j DA n n 2 j 1 A w kh 4A m(p ) m(p ) 2 Q t Q ln S DQ 1422T C r π γ = − = ∆ + + + where j j j 1 DA DA DA t t t − ∆ = − Finally, dividing throughout by Qn, converting to real time (hours) and taking the rate dependent skin over to the left hand side of the equation gives ( ) n 2 n i wf n j 12 j 2 j 1 n n A w i m(p ) m(p ) FQ Q 2.359T 1422T 4A t ln S Q c Ah Q kh C r φ µ γ = − − = ∆+ + (8.46) It can be seen immediately, from this equation, that if the first term on the right hand side is ignored then it reduces to exactly the same form as the semi-steady state inflow equation, (8.43). The additional term ( ) n j j j 1 n i Q 2.539T t c Ah Q φ µ = ∆ can be interpreted as a material balance correction required due to the use of the initial pressure pi in the inflow equation, (8.43), rather than the current average pressure which, for t > 0, is somewhat lower and changes throughout the test. Therefore, the difference between equ. (8.46) and (8.43) is generally rather small, especially in cases of interest, that is, large gas accumulations (large Ahφ). Theoretically, a plot of n 2 n i wf n j j j 1 n n m(p ) m(p ) FQ Q versus t Q Q = − − ∆ (8.47) should be linear with slope ( ) ( ) i i 2.359T 2.359T c Ah c Pore Volume φ µ µ = × and intercept 12 2 A w 1422T 4A B ln S kh C r γ = + Of course, to draw this plot requires a knowledge of the non-Darcy flow coefficient F. As a starting point the value of F determined as the slope of the plot of equ. (8.45) can be used and the value decreased gradually until the plot of equ. (8.47) becomes linear. The slope of the latter plot will then yield the value of the pore volume and the intercept REAL GAS FLOW: GAS WELL TESTING 265 the correct value of the Darcy flow coefficient B. At first sight this analysis technique appears to be rather useful but the following exercise will reveal the great sensitivity in the plot to the chosen value of F. EXERCISE 8.1 MULTI-RATE GAS WELL TEST ANALYSED ASSUMING STABILIZED FLOW CONDITIONS A gas well is tested by producing it at four different rates over a total period of 48 hours. The rate-time sequence and pressures recorded at the end of each separate flow period are listed in table 8.2. Rate (Q) Mscf/d Cum. Flowing time (t) hrs pwf psia m (pwf) psia2/cp 10 × 103 12 4182 1030.96 × 106 20 " 24 4047 984.29 " 30 " 36 3884 927.94 " 40 " 48 3694 862.26 " TABLE 8.2 The reservoir temperature and fluid properties are the same as those listed in table 8.1 and therefore, the relationship between real pressure and pseudo pressure, fig. 8.3, can be used in this exercise. For pressures in excess of 2800 psia, this relationship is almost linear and can be matched by the equation m(p) = (0.3457p — 414.76) × 106 psia2/cp (8.48) from which the real gas pseudo pressures, m(pwf ), in table 8.2 have been calculated. The reservoir and well data for the exercise are as follows pi = 4290 psi h = 40 ft (µc)i = 3.6 × 10-6 cp/psi rw = 0.3 ft If the correct values of the flow coefficients are B = 3176 psia2/cp/Mscf/d and F = .04 psia2/cp/(Mscf/d)2 1. Calculate both B and F using the interpretation technique suggested by equ. (8.45). 2. If there is a possible error of 10 psi in the measurement of pi, determine the effect of this error on the analysis. 3. It is proposed to produce this reservoir block at a plateau production rate of 5 MMscf/d down to some fixed minimum wellhead pressure. If the plateau rate can be maintained until the average reservoir pressure has fallen to 1200 psi, calculate the error in the estimate of the cumulative gas production, at this stage, due to REAL GAS FLOW: GAS WELL TESTING 266 using the analysis results determined in the first part of this exercise, instead of the correct values of B and F. 4. Using, as an initial estimate, the value of F determined in the first part of this exercise, analyse the test data using the interpretation technique of equ. (8.47). EXERCISE 8.1 SOLUTION 1) Values of ∆m(p)/Qn required for the plot, equ. (8.45), are listed in table 8.3, for pi = 4290 psia; i.e. m(pi) = 1068.29×106 psia2/cp. (m(pI) − m(pwfn ) ) /Qn psia2/cp/Mscf/d Qn Mscf/d pi = 4290 psia: pi = 4280 psia: pi = 4300 psia: 10 × 103 3733 3388 4079 20 " 4200 4028 4373 30 " 4678 4563 4794 40 " 5151 5065 5237 TABLE 8.3 The corresponding plot of ∆m(p)/Qn versus Qn is shown as fig. 8.9. The intercept and slope of the straight line give B = 3262 psia2 /cp/Mscf/d and F = 0.047 psia2 /cp/(Mscf/d)2 2) Also listed in table 8.3 and plotted in fig. 8.9 are values of ∆m(p)/Qn as a function of Qn for initial pressures differing by ± 10 psi from the measured value of 4290 psia. i.e. pi = 4280 psia: m(pi) = 1064.84×106 psia2/cp and pi = 4300 psia: m(pi) = 1071.75×106 psia2/cp These plots exhibit quite a pronounced degree of curvature, even for the relatively small error in the measured value of pi. Therefore, if the plot of equ. (8.45) appears to be curved, rather than linear, it is first necessary to determine the correct value of pi, by trial and error, to obtain a straight line. 3) The inflow equation, using the values of B and F determined in the first part of this exercise is m( ) p − m(pwf) = 3262 Q + .047 Q2 whereas, using the correct values m( ) p − m(pwf) = 3176 Q + .040 Q2 REAL GAS FLOW: GAS WELL TESTING 267 If it is assumed that these expressions will not change throughout the producing lifetime of the well then, for a given plateau rate of 5 MMscf/d, the duration of the plateau production is determined by the value of pwf. If this pressure drops below the level 5000 4000 3000 0 10 20 30 40 Q (MMscf/d) n 4300 4000 4290 4280 n i wf m (p ) m (p ) Qn − (psia / cp / Mscf/d) 2 p (psia) i Fig. 8.9 Gas well test analysis assuming semi-steady state conditions during each flow period. Data; table 8.3 required to maintain the minimum allowable wellhead pressure then the rate must be reduced. Therefore, at this point the values of m(pwf ) in both the above equations must be equal and ( ) 2 m p 86 Q .007 Q ∆ = + and for Q = 5 MMscf/d ( ) 6 2 m p .605 10 psia / cp ∆ = × Also ( ) 2p m p p µ ∆ = ∆ Ζ and using the data in table 8.1 for p= 1200 psi REAL GAS FLOW: GAS WELL TESTING 268 6 .0153 .832 .605 10 p 3.21 psi 2 1200 × × × ∆ = = × Assuming a depletion type material balance, equ. (1.35), then 4 p i i G p G ( ) 8 10 G Z p / Z − ∆ = ∆ ≈− × Thus the use of the flow coefficients B and F, determined using the inflow equation in the analysis is only very slightly pessimistic and in general provides a perfectly acceptable approximation for highly productive wells such as the one described in this exercise. 4) For the assumed value of F = .047 psia2/cp/(Mscf/d)2, which was determined in the first part of this exercise, the data for the plot of equ. (8.47) are listed in table 8.4, and the plot itself shown as fig. 8.10. (psia / cp / Mscf/d) 2 2 n n m (p) FQ Q ∆ − 4000 3500 3000 10 20 F = 0.047; A j j n Q t Q ∆ ∆ F = 0.035; A = 60 ACRES F = 0.030; A = 43 ACRES F = 0.040; A = 100 ACRES 30 Fig. 8.10 Gas well test analysis assuming semi-steady state conditions and applying equ. (8.47). Data; table 8.4 REAL GAS FLOW: GAS WELL TESTING 269 psia2/cp/(Mscf/d) Q Mscf/d t hrs n j j j 1 n Q t Q = ∆ hrs ( ) 2 n n m p FQ Q ∆ − F = .047 .03 .035 .04 10 × 103 12 12 3263 3433 3383 3333 20 " 24 18 3260 3600 3500 3400 30 " 36 24 3268 3778 3628 3478 40 " 48 30 3271 3951 3751 3551 TABLE 8.4 These data points plot almost as a horizontal straight line which implies an infinite area. This is to be expected since the value of F determined in the first part of this exercise is under the assumption that the reservoir pressure remains constant throughout the test, which implicitly assumes that the reservoir is infinite. The correct plot should have some positive slope corresponding to the fact that a finite volume is being drained. This means that the value of F should be decreased on a trial and error basis until a straight line is obtained. The plot has been re-drawn for values of F of 0.040, 0.035 and 0.030, refer table 8.4 and fig. 8.10. For the latter two values there is a slight degree of upward curvature, whereas for the value of F = .04 the plot is linear with slope 12.2 psia2/cp/Mscf/d/hr, which implies that the pore volume drained is 35.5 × 106 cu ft. Using the values of h = 40 ft and φ = .2 as average for the drainage volume, the area is approximately 100 acres. Further, the intercept gives the value of the Darcy flow coefficient B as 3186 psia2/cp/(Mscf/d). It must be admitted that the stabilized analysis technique using the plot of equ. (8.47) is a precarious business, to say the least. In this exercise any of the plots in fig. 8.10, for different values of F, could be taken to be linear leading to estimates of the area of 60 acres, for F = .035, or 43 acres for F = .030. In a great many cases the high degree of sensitivity to the value of F precludes any reliable estimate of the pore volume or area drained being made. To summarize the foregoing remarks on stabilized gas well testing; semi-steady flow conditions will occur for some fixed value of ( ) DA i kt t 0.000264 c A φ µ = (8.36) Therefore, such tests will be appropriate in reservoirs which have a high permeability and small drainage area so that the semi-steady state condition will be reached in a relatively short period of time. The reader should be cautioned, however, that these are not the sole criteria. In exercise 8.1, for instance, the actual permeability is 100 mD and the well is draining from the centre of a square of area 100 acres. Under these circumstances the value of tDA which must be exceeded before semi-steady state conditions prevail is 0.1 (refer either MBH charts, fig. 7.11, or the Dietz chart fig.6.4). REAL GAS FLOW: GAS WELL TESTING 270 Therefore, using equ. (8.36) and the data provided in exercise 8.1, the real time before the semi-steady state condition is reached is ( ) 6 i .1 c A .1 .2 3.6 10 100 43560 t 11.9 hrs 0.000264k 0.000264 100 φ µ − × × × × × × = = = × which confirms that, in this case, the separate twelve hour flow periods are just sufficient to permit the test to be analysed using semi-steady state equations. Suppose, however, the well was not situated at the centre of a square no-flow boundary but instead, at the centre of one of the four quarters of the same square (graph III of fig. 7.12(a)). In this case the value of tDA before the semi-steady state condition is reached is 0.5 and consequently, for the same reservoir properties, each flow period would have to be about 60 hours before semi-steady state analysis could be correctly applied. In this latter case, if the flow periods were each of twelve hour duration, the application of the stabilized analysis technique would be inappropriate since the entire test would be run under the late transient flow condition for which the mD functions should be evaluated using equ. (8.40) rather than equ. (8.33). For high permeability reservoirs, such as that tested in exercise 8.1, the error made in using the incorrect mathematics in the analysis would not have too serious an effect on the values of B and F determined from the test. For lower permeability reservoirs, however, use of the incorrect mathematics could produce a more serious error. One other disadvantage of the stabilized analysis techniques described so far is that while they provide a value of the Darcy flow coefficient 12 2 A w 1422T 4A B ln S kh C r γ = + there is no way of determining the individual components in this expression. In exercise 8.1 these values happen to be k = 100 mD, A = 100 acres, CA = 30.9 and S = 6.0 but, with the exception of A, which was obtained from the rather dubious analysis in part four of the exercise, none of these can be explicitly calculated from the test. This can have unfortunate consequences if the value of B, derived from the test, is used in the inflow equation in an attempt to predict the long term deliverability of the average reservoir well. For instance, the skin factor can be expected to be high in an exploration well but this should be improved when drilling the average production well using a more compatible completion fluid. Indeed, if the skin factor could be reduced to zero in the well considered in exercise 8.1, the value of the flow coefficient B would be reduced from 3176 to 1768 psia2/cp/(Mscf/d). Furthermore, if the reservoir is large and requires more than one well for drainage the value of A and CA, which are implicit in the value of B determined for the test well, will change during the producing lifetime of the reservoir. Therefore, the use of the coefficient B in making long term predictions of well deliverability can be misleading. It would be preferable if some analysis technique could be employed which could explicitly determine k and S so that the coefficient B could be calculated from its component parts, and such methods will now be described. As deeper, less permeable gas reservoirs were discovered and tested it became evident that the stabilized analysis techniques described so far in this section were REAL GAS FLOW: GAS WELL TESTING 271 inadequate to define the B and F factors in the well inflow equation (8.44). This is because if the permeability is small then the flowing time required to reach semi-steady state conditions, which depends on tDA = 0.000264kt/φ(µc)iA, can become extremely large. Then, the use of a semi-steady state inflow equation to analyse the test is inadmissible because the pressure response at the wellbore is strongly time dependent. To cater for this time dependence, two methods have been presented in the literature for analysing tests under the assumption that the wellbore pressure response could be matched by superposing transient constant terminal rate solutions of the radial diffusivity equation. The first of these was the Odeh-Jones technique13 (1965), in which the test analysis equation was formulated using p-squared terms. This was followed in 1971 by the method of Essis and Thomas14, in which the Odeh-Jones test equation was modified by the use of real gas pseudo pressure to give ( ) ( ) ( ) ( ) n n j 1 n i wf j D D D n n j 1 kh m p m p Q m t t Q S 1422T − = ′ − = ∆ − + which is the general form of the superposed constant terminal rate solution, equ. (8.39). Both Odeh and Jones and Essis and Thomas applied their analysis techniques strictly for transient flow conditions, for which the superposed mD functions in equ. (8.39) can each be expressed as ( ) D 12 D D 4t m t ln γ = (8.32) rather than using the general expression, equ. (8.40), which assumes a knowledge of the area drained and geometry. If transient mD functions can be used then the analysis is simple and should yield values of k and S which in turn can be used to calculate a value of the Darcy flow coefficient B, equ. (8.44), for any values of the area drained and shape factor. The analysis also directly determines F (or D), the second coefficient in the inflow equation. The technique will be fully illustrated in exercise 8.2. The statements made in Chapter 7, sec. 8, about the possibility of incorrectly interpreting multi-rate test data through making an a priori judgement concerning the prevailing flow conditions are equally, if not more, valid in gas well test analysis. This is because the µc product for gas is several times smaller than for oil which implies that, for the same permeability, porosity and area drained, the boundary effects will be felt much earlier in a gas well test. To apply transient analysis techniques it is insufficient to assume that each separate flow period should be short enough so that transient conditions prevail. Instead, the entire test duration must be so brief that the maximum value of the mD function in equ. (8.39), which is ( ) ( ) max D D D m t m total dimensionless testing time ′ = can still be evaluated using the transient expression, equ. (8.32). The possible error that can be made by making the flow periods too long will be illustrated in the following exercise. REAL GAS FLOW: GAS WELL TESTING 272 EXERCISE 8.2 MULTI-RATE GAS WELL TEST ANALYSED ASSUMING UNSTABILIZED FLOW CONDITIONS A gas well is tested by producing at four different rates for periods of one hour, followed by a four hour pressure buildup. The rates, times and observed pressures during the flow test are listed in table 8.5. Rate (Q) Mscf/d Cum. Flowing Time (t) hrs pwf psia m (pwf) psia2/cp 10 × 103 1 4160 1023.35 × 106 20 × " 2 3997 967.00 " 30 × " 3 3802 899.59 " 40 × " 4 3577 821.81 " TABLE 8.5 The reservoir temperature and gas properties are again the same as detailed in table 8.1 and therefore, the pressure-pseudo pressure relationship for p > 2800 psia is given by equ. (8.48). The reservoir and well data are listed below. pi = 4290 psia (µc)i = 3.6 × 10-6 cp/psi A = 200 acres φ = .15 h = 50 ft Geometry 4 1 rw = .3 ft This exercise requires the evaluation of mD(MBH) (tDA) for values of tDA < 0.01, which is the lower limit of the dimensionless time scale in each of the MBH charts, figs. 7.11-15. For tDA < 0.01, values of mD(MBH),for the 4:1 geometry, can be obtained from fig. 8.11, which has been drawn from the tabulated values of mD(MBH) versus tDA presented by Earlougher15. (N.B. There appears to be a typographical error in the latter reference in that the data listed for the 4:1 geometries and in table 2, p. 203, should be interchanged). The pressure buildup is first analysed from which it is determined that the permeability is 50 mD. (This technique will be fully described in sec. 8.11 and illustrated in exercise 8.3). Analyse this multi-rate test to determine values of k, S and F using 1) mD evaluated for transient conditions, equ. (8.32) 2) mD evaluated using the general expression, equ. (8.40) REAL GAS FLOW: GAS WELL TESTING 273 EXERCISE 8.2 SOLUTION 1) The Essis-Thomas combination of equations (8.39) and (8.32) can be expressed in more practical terms as 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 mD (MBH) 0 1 0.005 t DA 0.010 4 Fig. 8.11 MBH chart for the indicated 4:1 rectangular geometry for tDA < .01 (After Earlougher, et.al15) ( ) ( ) ( ) ( ) n n 2 i wf n j n j 1 n 2 j 1 w i 1637T k m p m p FQ Q log t t Q log 3.23 .87S kh c r φ µ − = − − = ∆ − + − + or ( ) ( ) ( ) ( ) n 2 n i wf n j n j 1 2 j 1 n n w i m p m p FQ Q k m log t t m log 3.23 .87S Q Q c r φ µ − = − − ∆ = − + − + (8.49) Thus a plot of j n n 2 i wf n n n j 1 j 1 n Q (m(p ) m(p ) FQ )/ Q versus log Q (t t ) − = ∆ − − − should be linear with slope m = 1637 T/kh and intercept = ( ) 2 w i k m (log 3.23 .87S) c r φ µ − + from which values of k and S can be calculated. In this type of analysis the value of F must be obtained by trial and error until a straight line is achieved. The analysis is shown in table 8.6 and the plot as fig. 8.12. REAL GAS FLOW: GAS WELL TESTING 274 n 2 i wf n n m(p ) m(p ) FQ Q − − Q Mscf/d t hrs j n j 1 n Q log(t t ) Q − − pwf psia m(pwf) psia2/cp F = 0 .04 .05 .06 10×103 1 0 4160 1023.35×106 4494 4094 3994 3894 20× " 2 .1505 3997 967.00 " 5065 4265 4065 3865 30× " 3 .2594 3802 899.59 " 5623 4423 4123 3823 40× " 4 .3450 3577 821.81 " 6162 4562 4162 3762 TABLE 8.6 F = 0.0 F = 0.04 F = 0.05 F = 0.06 6000 5500 5000 4500 4000 3500 0.1 0 0.2 0.3 0.4 j n j 1 n Q log (t t ) Q − ∆ − 2 n n m(p) FQ Q ∆ − (psia / cp / Mscf / d) 2 Fig. 8.12 Essis-Thomas analysis of a multi-rate gas well test under assumed transient flow conditions. Data; table 8.6 Initially, the plot is made ignoring the effects of non-Darcy flow (F = 0) and this exhibits a marked upward curvature for large values of the abscissa, corresponding to the fact REAL GAS FLOW: GAS WELL TESTING 275 that the high flow rates occur towards the end of the test. If the well had been tested with a decreasing rate sequence, the degree of curvature would have been greater for smaller values of the abscissa14. The calculation of the left hand side of equ. (8.49) has been repeated in table 8.6 for different values of F in an attempt to linearize the plot, and from fig. 8.12 it can be seen that a value of F = .05 accomplishes this. For this value, the plot has a slope of m = 491.5 and intercept 3993, from which the permeability and skin factor can be calculated as 1637 T 1637 660 k 44 mD mh 491.5 50 × = = = × ( ) 2 w i intercept k and S 1.151 log 3.23 2.7 m c r φ µ = − + = 2) The Essis-Thomas analysis, which assumes transient flow conditions for the evaluation of mD, equ. (8.32), will now be compared with the more general case in which mD is calculated using equ. (8.40) for the 4:1 rectangular geometry. The general well test equation (8.41) can be re-arranged as ( ) n n j 1 2 n i wf n j D D D j 1 n n m(p ) m(p ) FQ Q 1422T 1422T m t t S Q kh Q kh − = − − ∆ = − + (8.50) in which mD is evaluated as 1 1 1 2 2 2 D D DA DA D(MBH) DA 2 w 4A m (t ) 2 t ln t ln m (t ) r π γ ′ ′ ′ ′ = + + − (8.40) or 12 D D D(MBH) DA m (t ) m (t ) α ′ ′ = − (8.51) Alternatively, adhering to the Essis-Thomas analysis expressed in terms of dimensionless parameters D 1 1 1 2 2 2 D D DA 2 w 4t 4A m (t ) ln ln t ln r γ γ ′ ′ = = + (8.32) A plot of the left hand side of equ. (8.50) versus n j 1 j D D D n Q m (t t ) Q − ∆ − should be linear with slope m = 1422 T/kh, and intercept = 1422 TS/kh, which will yield values of k and S, respectively. The first part of the analysis is to evaluate the mD functions, equs. (8.51) and (8.32), for all values of the dimensionless time argument n j 1 D D (t t ) − − required in the analysis. These functions are listed in table 8.7, for k = 50 mD, the permeability value obtained from the pressure buildup analysis, for which DA i .000264kt t .0028 t ( c) A φ µ = = REAL GAS FLOW: GAS WELL TESTING 276 t (hrs) tDA α equ. (8.51) ½ mD(MBH) equ. (8.51) mD equ. (8.32) mD 1 .0028 6.6772 .0174 6.6598 6.6590 2 .0056 7.0413 .0266 7.0147 7.0061 3 .0084 7.2617 .0230 7.2387 7.2080 4 .0112 7.4231 .0121 7.4110 7.3527 TABLE 8.7 Assuming the value of F = .05 determined from the Essis-Thomas analysis, equ. (8.50) can be evaluated for both mD functions as listed in table 8.8. n j 1 n j D D D j 1 n Q m (t t ) Q − = ∆ − Q Mscf/d t hrs 2 n n m(p) FQ Q ∆ − mD (equ. (8.51) ) mD (equ. (8.32) ) 10×103 1 3994 6.6598 6.6596 20 " 2 4065 6.8373 6.8329 30 " 3 4123 6.9711 6.9582 40 " 4 4162 7.0811 7.0568 TABLE 8.8 The plots of the data contained in table 8.8 are shown as fig. 8.13 (a) and it can be seen that for the total test duration of four hours the difference between them is very slight. The values of k and S calculated from the two plots are presented in table 8.9. mD (equ. (8.51) ) mD (equ. (8.32) ) Slope, m 402.8 426.5 Intercept 1312 1153 k = 1422 T/mh 46.6 mD 44.0 mD S = kh × intercept/1422 T 3.3 2.7 TABLE 8.9 As the duration of each flow period is increased, the difference between the transient and the correct analysis becomes much more pronounced. Fig. 8.13 (b) shows the difference for 4 × 2 hour flow periods while fig. 8.13 (c) is for a test of 4 × 4 hour flow REAL GAS FLOW: GAS WELL TESTING 277 2 n 2 ( m(p) FQ ) / Qn (psia / cp /Mscf / d) ∆ − 2 n 2 ( m(p) FQ ) / Qn (psia /cp/Mscf /d) ∆ − 2 n 2 ( m(p) FQ ) / Qn (psia /cp/Mscf /d) ∆ − n j 1 j D D D n Q m (t t ) Q − ∆ − n j 1 j D D D n Q m (t t ) Q − ∆ − n j 1 j D D D n Q m (t t ) Q − ∆ − 4100 4000 3900 6.5 6.75 7.0 7.25 4300 4200 4100 7.0 7.5 7.25 4400 4400 4300 4200 8.0 7.5 7.75 4500 CORRECT ANALYSIS; m EVALUATED USING EQU. (8.40) TRANSIENT D ; m (8.32) “ D “ “ “ a b c Fig. 8.13 The effect of the length of the individual flow periods on the analysis of a multi-rate gas well test; (a) 4× × × ×1 hr periods, (b) 4× × × ×2 hrs, (c) 4× × × ×4 hrs REAL GAS FLOW: GAS WELL TESTING 278 periods. In each case the solid line represents the correct analysis technique, in which mD is evaluated using equ. (8.51),while the dashed line is derived for the same observed pressure data but with the mD function evaluated under the assumption that transient flow conditions prevail throughout the test. Using the correct analysis technique always gives the same values of k and S, but the results obtained from the transient analysis are open to severe misinterpretation. One may either treat the dashed lines in figs. 8.13 (b) and (c) as being approximately linear, which would result in the calculated values of both k and S being too small. Alternatively, since both the dashed lines have a distinct upward curvature one may be tempted to linearize them either by increasing the non-Darcy flow coefficient (even though the correct value of F = .05 has been used in the analysis to produce figs. 8.13 (b) and (c) ), or by reducing the initial pressure as demonstrated in exercise 8.1. The majority of gas well test analyses described in the literature, for non-stabilized flow conditions, are for wells which have very low permeabilities; a few millidarcies or less. Under these circumstances, the transient analysis technique of Essis and Thomas, which is usually applied in one form or another, is probably quite justified. The foregoing exercise illustrates what can go wrong in applying transient analysis techniques in a moderately high permeability reservoir and, of course, at the time of planning the test the permeability is unknown. It is therefore very difficult to estimate in advance just how long the total test duration should be for the application of transient analysis to be still valid. As a safeguard, it is recommended that all sequential flow tests be followed by a pressure buildup, which under normal circumstances should provide a more reliable value of the permeability and one which can be determined independently of the boundary condition or flow condition (refer Chapter 7, sec. 7). If the permeability derived from the multi-rate test, assuming the transient flow condition, is less than that from the buildup it is likely that the multi-rate test analysis is incorrect and should be re-analysed, attempting to make allowance for the boundary condition and discarding the assumption of transient flow. The latter is easier said than done, however, for although the value of k, obtained from the buildup, facilitates the determination of tDA, required in the analysis, there always remains an ambiguity in the estimation of the correct shape of the drainage area which, as shown in exercise 7.8, cannot be resolved by conventional analysis techniques. Odeh et al.16 have also described an analysis technique for multi-rate flow tests which are followed by a buildup. The obvious drawback to this method of testing is that it negates one of the main aims of multi-rate testing which is to avoid well closure. 8.11 PRESSURE BUILDUP TESTING OF GAS WELLS Just as in the case of oilwell testing, pressure buildup tests in gas wells, if analysed correctly using the Horner buildup plot, can provide the most reliable values of the permeability and skin factor. The only difference is that a buildup in a gas well must be accompanied by two separate flow periods, one before and one after the buildup, as shown in fig. 8.14. This measure is necessary in order to determine both components of the skin factor, S and DQ. REAL GAS FLOW: GAS WELL TESTING 279 Rate Q1 Q2 t t1 ∆t ∆tmax. t ’ (a) Time (b) t t1 ∆t ∆tmax. t ’ Bottom hole Pressure pwf pws Fig. 8.14 (a) Rate-time schedule, and (b) corresponding wellbore pressure response during a pressure buildup test in a gas well The theoretical buildup equation for the rates and times shown in fig. 8.14 is just a special case of the general test equation (8.39). 1 i ws D D D D D 1 kh (m(p ) m(p )) m (t t ) m ( t ) 1422Q T − = + ∆ − ∆ (8.52) This is identical in form to equ. (7.32), the theoretical buildup equation for an oilwell test. In deriving equ. (8.52) from equ. (8.39), for superposed constant terminal rate solutions with rate changes Q1 and (0 – Q1), both the mechanical and rate dependent skin factors disappear, a fact which has been investigated by Ramey and Wattenbarger4. In analogy with the buildup theory described in Chapter 7, sec. 7, for small values of ∆t, equ. (8.52) can be expressed as a linear relationship between m(pws) and log (t1 + ∆t)/ ∆t. The equation of this straight line for any value of ∆t is 1 1 D 1 12 i ws(LIN) D D 1 4t t t kh (m(p ) m(p )) 1.151 log m (t ) ln 1422Q T t γ + ∆ − = + − ∆ (8.53) in which m(pws(LIN)) is the hypothetical pseudo pressure on the extrapolated linear trend, and 1 D D m (t ) and 1/2 ln 1 D 4t /γ , which are both evaluated for the dimensionless, effective flowing time before the buildup, are constants. For large values of ∆t the real pseudo pressure m(pws), equ. (8.52), will deviate from m(pws(LIN)) as demonstrated for the similar liquid flow equations in exercise 7.7. Therefore,. the Horner plot of REAL GAS FLOW: GAS WELL TESTING 280 1 ws t t m(p ) versus log t + ∆ ∆ for the recorded pressure data will be linear for small ∆t and the extrapolated trend can theoretically be matched by equ. (8.53). The attractive feature of the Horner buildup is that the analysis to determine k and S does not involve the specific evaluation of mD ( 1 D t ) in equ. (8.53) but merely requires that the early linear buildup trend be identified. The slope of this line is 1 1637Q T m kh = (8.54) from which kh and k can be calculated, and the total skin factor, corresponding to rate Q1, can be determined as ws(LIN)1 hr wf 1 1 2 i w (m(p ) m(p )) k S S DQ 1.151 log 3.23 m ( c)r φ µ − − ′ = + = − + (8.55) in which m(pws(LIN)1-hr) is the pseudo pressure read from the extrapolated straight line at ∆t = 1 hour. The derivation of equ. (8.55) follows the same argument leading to equ. (7.52) and therefore the calculated value of 1 S′is independent of the value of mD ( 1 D t ). The early, transient pressure response of both flow periods can be analysed to determine values of k, 1 S′ and 2 S′ (= S + DQ2). The equation describing the transient pseudo pressure drop at the wellbore at any time t during the first flow period is D 12 i wf 1 1 4t kh (m(p ) m(p )) ln S 1422Q T γ ′ − = + which can be expressed as 1 i wf 1 2 i w 1637Q T k m(p ) m(p ) log t log 3.23 0.87S kh ( c) r φ µ ′ − = + − + (8.56) Thus a plot of m(pwf) versus log t will be linear during the transient flow period with slope 1 1637Q T m kh = (8.54) again giving the value of k, while the skin factor can be calculated by evaluating equ. (8.56), for the specific value of m(pwf ) at t = 1 hr, as i wf 1 hr 1 1 2 i w (m(p ) m(p ) ) k S S DQ 1.151 log 3.23 m ( c)r φ µ − − ′ = + = − + (8.57) Only the values of m(pwf) which plot as a linear function of log t are used, which ensures that the application of transient analysis is valid. REAL GAS FLOW: GAS WELL TESTING 281 The theoretical equation describing the pseudo pressure drop during the second flow period can be derived from the basic test equation (8.39) as 1 max max i wf 1 D D D D D D D 2 D D 2 2 kh (m(p ) m(p )) Q (m (t t t ) m ( t t )) 1422T Q m (t ) Q S ′ ′ − = + ∆ + − ∆ + ′ ′ + + (8.58) where t′ is the time measured from the start of the second flow period at rate Q2, (fig. 8.14). This equation is analysed for transient conditions during the second flow period, that is, for small values of t´. In this case the expression 1 max max 1 D D D D D D D Q (m (t t t ) m ( t t )) ′ ′ + ∆ + − ∆ + in equ. (8.58) can be regarded as being constant. If both t1 and ∆tmax are short so that both the mD functions can be evaluated under transient conditions the above statement is quite correct and, in fact, the difference between the mD functions is both small and constant. For a very long initial flow period, corresponding to a routine well survey rather than an initial test, the difference between the mD functions can only be regarded as constant on the grounds that t is small, which is always the case since the wellbore pressure response at rate Q2 is only being analysed during the brief, initial, transient flow period. Therefore, equ. (8.58) implies that a plot of m(pwf) versus log t will be linear, for transient flow, with slope 2 1637Q T m kh = which leads to a re-determination of k. The skin factor can be evaluated by expressing equ. (8.58) as i wf i ws 2 D D 2 2 kh kh (m(p ) m(p )) (m(p ) m(p )) Q m (t ) Q S 1422T 1422T ′ ′ ′ − = − + + (8.59) in which ws p′ is the hypothetical static pressure that would be obtained had the buildup been continued for a time ∆tmax + t′ . The value of ws p′ will therefore increase as t´ increases. Equation (8.59) can then be solved to give 2 S′ as ws 1 hr wf 1 hr 2 2 2 i w (m(p ) m(p ) k S S DQ 1.151 log 3.23 m ( c) r φ µ − − ′ − ′ = + = − + (8.60) in which both ( ) wf m p and ( ) ws m p′ are evaluated for t′ = 1 hour. The latter can be obtained by extrapolation of the final buildup trend for one hour after the buildup has ceased. However, this correction is seldom applied and usually ( ) ws 1 hr m p − ′ is set equal to ( ) ws m p , evaluated for the final closed in pressure. The following exercise illustrates the method of buildup analysis for a well test in a new reservoir in which pi is the initial reservoir pressure. REAL GAS FLOW: GAS WELL TESTING 282 EXERCISE 8.3 PRESSURE BUILDUP ANALYSIS Instead of applying a multi-rate flow test, the reservoir described in exercise 8.2 is tested by producing it for 3 hours at a rate of 40 MMscf/d, closing in for an 8 hour buildup and finally, by producing for a further 3 hours at a rate of 60 MMscf/d. The pressures recorded during the flowing and closed in periods are listed in table 8.10 and 8.11, respectively. First Flow Period Q1 = 40 MMscf/d Second Flow Period Q2 = 60 MMscf/d Flowing time hrs pwf psia m (pwf) psia2 /cp pwf psia m (pwf) psia2 /cp .75 3602 830.45 × 106 3076 648.61 × 106 1.00 3596 828.38 " 3066 645.16 " 1.25 3591 826.65 " 3059 642.74 " 1.50 3587 825.27 " 3053 640.66 " 1.75 3583 823.88 " 3048 638.93 " 2.00 3580 822.85 " 3043 637.21 " 2.25 3577 821.81 " 3038 635.48 " 2.50 3575 821.12 " 3036 634.79 " 3.00 3570 819.39 " 3029 632.37 " TABLE 8.10 Closed in time ∆t hrs pws psia m(pws) psia2/cp Closed in time ∆t hrs pws psia m(pws) psia2/cp .5 4100 1002.61×106 3.5 4272 1062.07×106 1.0 4255 1056.19 " 4.0 4274 1062.76 " 1.5 4263 1058.96 " 5.0 4276 1063.45 " 2.0 4267 1060.34 " 6.0 4277 1063.80 " 2.5 4269 1061.03 " 7.0 4278 1064.14 " 3.0 4271 1061.72 " 8.0 4279 1064.49 " TABLE 8.11 Since the fluid properties are the same as in the two previous exercises, equ. (8.48) can be used as the relationship between real and pseudo pressures. All other data presented in exercise 8.2 can be used in the current exercise. 1) From the pressure buildup determine pi, k and 1 S′. REAL GAS FLOW: GAS WELL TESTING 283 2) From the flow tests determine k, 1 S′, 2 S′ and hence D, or F. EXERCISE 8.3 SOLUTION 1) Buildup Analysis For a flowing time of 3 hours, the data necessary to draw the Horner buildup plot are listed in table 8.12. ∆t hrs 1 t t log t + ∆ ∆ m(pws) psia2/cp ∆t hrs 1 t t log t + ∆ ∆ m(pws) psia2/cp .5 .845 1002.61×106 3.5 .269 1062.07×106 1.0 .602 1056.19 " 4.0 .243 1062.76 " 1.5 .477 1058.96 " 5.0 .204 1063.45 " 2.0 .398 1060.34 " 6.0 .176 1063.80 " 2.5 .342 1061.03 " 7.0 .155 1064.14 " 3.0 .301 1061.72 " 8.0 .138 1064.49 " (9.0) (.125) (1064.65) " TABLE 8.12 The corresponding buildup plot is shown as fig. 8.15 (a), from which the slope has been determined as 6 2 1 1637Q T m 16.17 10 psia / cp /log cycle kh = × = which for a fully penetrating well gives 3 6 1637 40 10 660 k 53.5 mD 16.17 10 50 × × × = = × × and the extrapolation to ∆t = ∞ gives m(pi) = 1066.7 × 106 psia2/cp pi = 4285 psi The value of ( ) ws 1 hr m p − ′ taken from the extrapolated linear trend is 1057 ×106 psia2/cp and therefore, using equ. (8.55) REAL GAS FLOW: GAS WELL TESTING 284 1065 1060 1055 830 820 810 650 640 630 -.1 0 .1 .2 .3 .4 .5 log t -.1 0 .1 .2 .3 .4 .5 log t 6 5 4 3 2 1 0 log t + t t ∆ ∆ m(p ) (psia /cp) 10 ws 2 6 m(p ) (psia /cp) 10 wf 2 6 m(p ) (psia /cp) 10 wf 2 6 m(p ) ws 1 - hr m(p ) ’ws 1 - hr m(p) = 1066.7 a b c × × × Fig. 8.15 Complete analysis of a pressure buildup test in a gas well: (a) buildup analysis (table 8.12); (b) and (c) transient flow analyses of the first and second flow periods, respectively (table 8.10) 6 1 1 1057 819.4 53.5 10 S S DQ 1.151 log 3.23 10.22 16.17 .15 3.6 .09 − × ′ = + = − + = × × REAL GAS FLOW: GAS WELL TESTING 285 2) Flow Analysis Plots of m(pwf) versus log t for the data listed in table 8.10, are shown as fig. 8.15 (b) and (c), from which the data presented in table 8.13 have been determined. Rate Mscf/d Slope psia2/cp/log cycle m(pwf)1-hr k mD Total Skin 40×103 16.64×106 828.38×106 45 9.9 60 " 26.08 " 645.16 " 43 11.9 TABLE 8.13 The values of the permeability in this table have been calculated using equ. (8.54) for rates Q1 and Q2 and the total skin factors using equs. (8.57) and (8.60) for the first and second flow periods respectively. In applying equ. (8.60) the value of ( ) ws 1 hr m p − ′ has been determined from the buildup plot for ∆t = 9 hours as 1064.65 × 106 psia2/cp (fig. 8.15 (a) and table 8.12). The reader can verify that irrespective of whether ( ) ws 1 hr, m p − ′ ( ) ws m p for the maximum buildup time, or m(pi) is used in equ. (8.60) makes very little difference to the calculated value of 2 S′ . Finally, S and D can be calculated by solving the equations 1 S′ = 9.9 = S + 40 × 103 D 2 S′ = 11.9 = S + 60 × 103 D to give S = 5.9; D = 1.0 × 10-4/Mscf/d and F = 1422 DT kh = 0.043 psia2/cp/(Mscf/d)2 and from these figures the Darcy flow coefficient B can be calculated for the estimated values of A and CA, equ. (8.44), for use in long term well deliverability calculations. A similar example of a pressure buildup analysis, for a low permeability reservoir (k = 5 mD),has been presented in the literature by Al-Hussainy and Ramey3. The main advantage of this type of test over the multi-rate drawdown test is the same as mentioned in Chapter 7, sec. 7. It is, that the buildup analysis will provide values of k and 1 S′ which are independent of the value of m(tD ) at the time of the survey. Furthermore, since only the transient pressure response during each flow period is interpreted, then it means that the entire test can be analysed without having to be concerned about the size or shape of the area drained or the well position with respect to the boundary. The pressure buildup test can also be used for the routine surveys conducted at regular intervals throughout the producing life of the field. Prior to the survey the well must be produced at a constant rate for a sufficient period of time so that the flow is under semi-steady state conditions. This renders transient analysis of the first flow period, at rate Q1, impossible but analysis of the buildup will yield values of k and 1 S′. Following this, k and 2 S′ can be determined from the transient analysis of the second REAL GAS FLOW: GAS WELL TESTING 286 flow period at rate Q2. The main purpose of this type of test is to determine the current average pressure within the drainage boundary of the well, p. Theoretically, this can be done by using either the method of Matthews, Brons and Hazebroek, or Dietz (refer Chapter 7, sec. 7), but the difficulty is to determine at what pressure the µc product should be evaluated which is required to calculate tDA for use with either of these methods. For the initial well tests described in exercises 8.1-3, the product (µc) i evaluated at the initial equilibrium pressure could be used but for a survey made, say, several years after the well has started to produce this can lead to serious error. The basic problem is that for very long flowing times the calculation of mD using the semi-steady state equation (8.33) with the (µc)i product does not accurately correlate with the similar pD function for liquid flow, equ. (7.27). Kazemi10 has presented an iterative method for determining the pressure at which, µc should be evaluated and hence the correct value ofp. The method is applicable for wells producing under semi-steady state conditions at the time of the survey. In this case, as shown in Chapter 7, sec. 7, the value of the flowing time used to plot the buildup is immaterial providing that t ≥ tSSS, the time required for semi-steady state conditions to be reached for the particular geometrical configuration of the drainage area. Strictly speaking, this statement is only valid when applied to a liquid, in which case the MBH plots, figs. 7.11-15, are linear functions of the dimensionless flowing time tDA. For a real gas, however, the mD(MBH) functions deviate from the linear pD(MBH) functions for large values of tDA, as shown in fig. 8.16. This implies that using the MBH charts, for a large value of the (effective) flowing time, can lead to an error in the determination ofp in the analysis of a routine buildup survey in a gas well. Kazemi argues, and substantiates his argument with detailed numerical simulation, that if the buildup is plotted for a flowing time tSSS,where SSS p SSS DA SSS ( c) A t (t ) 0.000264k φ µ = (8.61) and the MBH method applied for a dimensionless flowing time (tDA)SSS, then the portion of the MBH charts for which the liquid and gas MBH functions correlate is used and this should result in the correct determination of p. Of course, in order to calculate tSSS, using equ. (8.61), requires a knowledge of pSSS, the average pressure at a time tSSS prior to the buildup and the evaluation of the µc product at this pressure. A simple iterative scheme for calculating pSSS, tSSS and hence p is shown in fig. 8.17. REAL GAS FLOW: GAS WELL TESTING 287 m or D (MBH) p D (MBH) 01 0.1 1.0 10.0 m D (MBH) p D (MBH) (t ) DA SSS t = DA .00264 kt cA φµ Fig. 8.16 MBH plot for a well at the centre of a square, showing the deviation of mD(MBH) from pD(MBH) for large values of the dimensionless flowing time tDA In the first place an estimate is made of (tDA )SSS, for the particular geometrical configuration, fig. 6.4. Both (tDA)SSS and m(p) - m( ) p remain constant throughout the analysis. An estimate of µc is made, using either the initial pressure or some roughly estimated value of pSSS, from which an initial value of tSSS is determined using equ. (8.61). The Horner buildup is plotted using this value of tSSS instead of the effective flowing time. The MBH (or Dietz) method is then used to obtain an initial value of the average pressure at the time of the survey (refer Chapter 7, sec.7). The µc product is re-evaluated at p, and tSSS and p re-estimated. For this second, and all successive values of p, the material balance is applied to determine pSSS, the average pressure tSSS hours before the well closure, using dGp = Ql tSSS for the final flow rate Ql and the latest estimate of tSSS. A cycle is entered in which µc is evaluated for the latest determined value of pSSS and this continues until successive values of this pressure do not differ by more than some tolerance value (e.g. TOL = 5 psi). Each new estimate of tSSS leads to a new buildup plot and since m(p) – m(p) is constant, a new value of p is obtained on each occasion. In a worked example Kazemi has shown that applying the above method can lead to a correction of 100 psi in the value p compared to the normal method in which µc is evaluated as (µc)i. The method can also be used to correct the average pressure determined from a buildup test in a solution gas drive reservoir which is below bubble point pressure. REAL GAS FLOW: GAS WELL TESTING 288 k = 1 k = iteration counter ( c) = ( c) µ k i µ sss DA SSS ( c) A t (t ) .000264k φ µ = k k ( c) = ( c) µ k k-1 µ (RE)-PLOT HORNER BUILDUP EXTRAPOLATES TO m(p) k SSS ws t t m(p ) vs. log t + ∆ ∆ k MBH m(p) m( ) = constant = m log (C (t ) ) p k A DA SSS k m( ) p k p k k = 1 k > 1 k p k ( c) = ( c) µ k µ MATERIAL BALANCE k = 1 k + i 1 SSS i p p d Q t Z Z G = − k k d(p k) SSS p p d(p ) − + k k k ( c) ( c) µ k µ SSS p k k = 2 k > 2 k + -SSS SSS p p TOL − − k k-1 p p p 2 + = k k-1 TO m(p)k Fig. 8.17 Iterative determination of   p in a gas well test analysis (Kazemi10) REAL GAS FLOW: GAS WELL TESTING 289 8.12 PRESSURE BUILDUP ANALYSIS IN SOLUTION GAS DRIVE RESERVOIRS The pressure buildup theory described in sec. 7 of the previous chapter was developed for liquid flow and is therefore only appropriate for pressure surveys in undersaturated oil reservoirs. For routine pressure surveys conducted throughout the producing lifetime of the field, it is more likely that the average pressure will be below bubble point so that there will be two phases, liquid oil and free gas, in the reservoir. To analyse pressure buildup tests under these circumstances, Raghavan17 has suggested the use of the integral transformation b p ro o o o p k (S ) m (p) dp B µ ′ = (8.62) which again is referred to as a pseudo pressure, only in this case applied to the flow of oil, as denoted by the subscript "o". The kro(So ) is the oil relative permeability, which is a function of the oil saturation, while the other parameters, µo and Bo are functions of pressure. This leads to a certain degree of difficulty in determining the relation between pressure and saturation required to evaluate equ. (8.62). Raghaven has shown that this relationship can be obtained from the gas-oil ratio equation which expresses the ratio of the reservoir gas to oil rates at the time of closure of the well, i.e. s g rg o o g ro (R R )B k (rb / gas) B (rb / oil) k µ µ − = or rg o o s g ro g k B R R k B µ µ = + (8.63) In this relationship, R is the fixed value of the producing GOR at the time of closure and therefore, since krg and kro are functions of the oil saturation and Bo, Bg and Rs are functions of pressure, equ. (8.63) implicitly defines the pressure-saturation relationship. The steps in evaluating the pseudo pressure integral, equ. (8.62), are then 1) Using the value of R at the time of the survey, determine the relation krg/kro as a function of the pressure, using equ. (8.63). 2) Providing gas-oil relative permeability curves are available (kro and krg as functions of So, refer sec. 4.8) the relation between kro and pressure can be determined. 3) Using the trapezoidal rule, evaluate m (p) ′ as a function of pressure, in the same way as demonstrated in table 8.1. It should be noted that this m (p) ′ function only reflects conditions near the well at the time of the survey and must be re-calculated for each pressure survey, as R varies. Using the m (p) ′ pseudo pressure, the constant terminal rate solution of the radial diffusivity equation can be expressed in dimensionless form as REAL GAS FLOW: GAS WELL TESTING 290 3 i wf D D o kh 7.08 10 (m (p ) m (p )) m (t ) S q − ′ ′ ′ × − = + where D 1 1 2 2 D D DA D(MBH) DA 4t m (t ) 2 t ln m (t ) π γ ′ ′ = + − (8.64) in which k is the effective permeability to oil in the presence of the connate water. Raghavan has shown, using numerical simulation, that the D m′ (tD) functions correlate very well with the pD (tD) functions for liquid flow but, as in the case of real gas flow, the match is better for small values of tD, before the boundary effects are felt. It should also be noted that the compressibility used in the evaluation of tDA is the total compressibility of the system. Above the bubble point this is simply ct = coSo + cwSwc + cf (5.22) but below bubble point there must be additional components to account for the presence of free gas and for the transfer of solution gas from the oil18. Thus for the oil, a pressure drop ∆p will cause a reduction in the oil volume ∆Bo and an increase in the liberated gas volume Bg∆ Rs, and o s o g o B R 1 c B B p p ∂ ∂ = − − ∂ ∂ The total compressibility, below bubble point pressure is therefore g g o s o t g w wc f o g S B S R B c B c S c B p p B p ∂ ∂ ∂ = − − + + ∂ ∂ ∂ (8.65) in which, for a significant gas saturation, the last two terms can usually be neglected. Because of the equivalence of form of the D m′ functions, equ. (8.64), with the pD or mD functions, it is clear that the buildup theory must follow that detailed in Chapter 7, sec. 7, for oil, and sec. 8.11 of this chapter, for gas. A Horner plot of m´(pws) versus log t t t + ∆ ∆ is made and the early linear trend extrapolated to determine m´(p). The slope of the linear section is 162.6 q m kh = and the skin factor can be calculated using equ. (7.52), with pseudo pressures replacing the actual pressures in the equation. Again the MBH method can be used to determine m (p) ′ and hence the average pressure p. Furthermore, if the flowing time before the survey is very long, the correction method of Kazemi, described in the previous section, can be applied to improve the estimate of p. REAL GAS FLOW: GAS WELL TESTING 291 8.13 SUMMARY OF PRESSURE ANALYSIS TECHNIQUES For those who have struggled through this and the previous three chapters, it is worthwhile presenting a brief summary of their contents in an effort to simplify and generalise the theory of well testing. To start with, the application of the principle of mass conservation, together with Darcy's law and the definition of isothermal compressibility, lead to the non-linear, radial, second order, partial differential equation for the flow of a single phase fluid (in Darcy units) as 1 k p p r c r r r t ρ φ ρ µ ∂ ∂ ∂ = ∂ ∂ ∂ (5.1) Prior to obtaining useful solutions of this equation it must first be linearized (or partially linearized) and the method by which this can be achieved depends on the nature of the fluid under consideration, as follows. Undersaturated oil Linearization by deletion of terms, assuming that 2 ( p / r) 0;µ ∂ ∂ ≈ ≈constant and cp << 1. Real gas Partial linearization using the integral transformation b p p pdp m(p) 2 µ = Ζ (8.7) Gas-oil Partial linearization using the integral transformation b p ro o o o p k (S ) m (p) dp B µ ′ = (8.62) or, strictly speaking, in Darcy units, the correct form of this transformation should be b p ro o o o p k (S ) m (p) dp ρ µ ′ = Application of any of the above methods leads to the re-formulation of equ. (5.1) as 1 c r r r k t β φµ β ∂ ∂ ∂ = ∂ ∂ ∂ (8.66) which has the form of the radial diffusivity equation and in which: for undersaturated oil β = p REAL GAS FLOW: GAS WELL TESTING 292 for a real gas β = m(p) and for gas-oil (two phase) β = m (p) ′ Although the term linearization has been applied to the conversion of equ. (5.1) to equ. (8.66), it should be remembered that linearization is only achieved for the case of liquid flow (undersaturated oil) for which the coefficient k/φµc is a constant. For both real gas and two phase (gas-oil) systems, the µc product is pressure dependent, meaning that equ. (8.66) is still non-linear. The basic building block in well test analysis is the constant terminal rate solution of equ. (8.66), which predicts the pressure or pseudo pressure response at the wellbore, resulting from the production of a well at constant rate from a state of equilibrium pressure. Expressing equ. (8.66) in dimensionless form D D D D D D D 1 r r r r t β β ∂ ∂ ∂ = ∂ ∂ ∂ (8.67) where rD = r/rW and tD = kt/φµc 2 w r (= 0.000264 kt/φµc 2 w r in field units- t in hours), the general constant terminal rate solution, for rD = 1, can be expressed as D D f(p) (t ) S q α β = + (8.68) In this equation the various component parts are as listed in table 8.14, (in field units), again, dependent on the nature of the fluid. To interpret the majority of practical well tests requires the superposition of constant terminal rate solutions, for different constant production rates acting for different periods of time, to give the value of f(p)n at time tn during the nth flow period, as n j 1 n n j D D D n j 1 f(p) q (t t ) q S α β − = = ∆ − + (8.69) in which Undersaturated oil Real gas Two phase gas-oil (fieldunits) q α 3 o o o kh 7.08 10 q B µ − × kh 1422QT 3 o kh 7.08 10 q − × f(p) pi − pwf m (pi) − m(pwf) i wf m (p ) m (p ) ′ ′ − βD (tD) pD(tD) mD (tD) m´D(tD) S S S + DQ S TABLE 8.14 n j 1 D 1 1 2 2 o D D D D DA D(MBH) DA 4t (t t ) (t ) 2 t ln (t ) β β π β γ − ′ ′ ′ ′ − = = + − (8.70) REAL GAS FLOW: GAS WELL TESTING 293 where tDA = 0.000264kt /φµcA (field units), and βD(MBH)(tDA) is the ordinate of the MBH charts, corresponding to the value of tDA, and includes allowance for the geometry of the drainage area and the degree of well asymmetry with respect to the boundary. Well test analysis techniques presented in the literature invariably require the superposition of transient constant terminal rate solutions of equ. (8.67) which, in comparison with equ. (8.70), have the form n j 1 D 12 D D D D D 4t (t t ) (t ) ln β β γ − ′ ′ − = = (8.71) Superposition of such solutions, however, automatically assumes the infinite boundary condition and is therefore only appropriate for tests of short duration, whereas superposition of the total solution, equ. (8.70), is theoretically correct for any value of the flowing time and for any boundary condition. Examples in this and the previous chapter (exercises 7.8 and 8.2), have shown that the ad-hoc assumption of transient flow can be particularly dangerous in the interpretation of multi-rate flow tests. Furthermore, conventional multi-rate test analysis techniques do not permit the attempted matching of the drainage boundary, even using the correct βD function, equ. (8.70), in the superposed test analysis equation, (8.69), as illustrated in exercise 7.8. Multi-rate tests should only be planned when the engineer is quite confident that transient conditions apply for the duration of the entire test, which is difficult to ascertain in advance of the test itself. In this author's opinion, the safest and most useful type of test, for any fluid system, is the pressure buildup. Definition of a linear trend, for small values of the closed in time on a Horner buildup plot, can lead to the unambiguous determination of the effective permeability and the skin factor. Use of this early linear trend, in fact, imposes transience on the buildup analysis. In addition, the test can be interpreted to determine the average pressure within the drainage boundary of the well and also to gain some impression of the shape, area and well position within the boundary, as illustrated in exercise 7.7. It must be admitted that the use of equ. (8.70) instead of equ. (8.71), in the basic well test equation, (8.69), only serves to complicate well test analysis. Nevertheless, in using the correct solution one is simply recognising the basic fact that second order differential equations require the specification of both initial and boundary conditions to obtain meaningful solutions. As illustrated in the various exercises, it is not difficult to evaluate equ. (8.70) for use in test analysis. These days when engineers have access to computers or, at least, sophisticated electronic desk calculators, the evaluation is much less tedious than in the past. For instance, a perfectly general program for evaluating pressure buildup tests, for any fluid, could be structured as shown in fig. 8.18. REAL GAS FLOW: GAS WELL TESTING 294 BASIC INPUT PVT PRELIMINARY ANALYSIS KAZEMI ITERATION SPECIAL ANALYSES DIGITIZED MBH CHARTS Fig. 8.18 Schematic of a general analysis program applicable to pressure buildup tests for any fluid system The various component parts have all been detailed in the text and will only be briefly summarised here. Basic input This consists of the basic reservoir parameters, as detailed at the beginning of exercises 7.6 and 7.7, together with the specification of the flowing time to be used in the analysis and the bottom hole flowing pressure at the time of closure. A table of recorded wellbore pressures as a function of closed in time is also required and, for a gas well test, wellbore pressures as a function of time during the flowing periods before and after the buildup (exercise 8.3). PVT For a buildup in an undersaturated oil reservoir it is only necessary to specify the current oil formation volume factor, viscosity and total compressibility. For a gas well test, however, the program should be capable of generating real gas pseudo pressures as a function of the actual pressure, as shown in table 8.1. Similarly, for a test in a gas saturated reservoir the pseudo pressure function described in sec. 8.12, must be generated. Preliminary analysis The program should print a Horner plot, using closed-in pressures or pseudo pressures, as appropriate. The engineer should inspect this plot and decide which, if any, of the plotted points constitute an early, linear buildup trend. It is inadvisable to let the computer perform this task. Having defined the points on the linear trend, the computer can determine the equation of the straight line which best matches the points and subsequently determine the extrapolated pressure at infinite closed-in time, the permeability and the skin factor. REAL GAS FLOW: GAS WELL TESTING 295 For a gas well test, this part of the program should also analyse the transient pressure fall-off during the flowing periods, giving a re-determination of the permeability and two separate skin factors, at different flow rates, which facilitate the calculation of the rate dependent skin (exercise 8.3). For an initial test in either an oil or gas reservoir (exercises 7.6 and 8.3, respectively) the analysis would cease at this point. In the more common instance of routine pressure surveys made throughout the producing lifetime of the well, the following steps are necessary which incorporate the boundary condition in the analysis. Digitized MBH charts This set of data is the main-spring of the special analyses part of the program. As already mentioned, the MBH charts (figs. 7.11-15), which are equally appropriate for dimensionless pressure or pseudo pressures, have al ready been digitized by Earlougher15, for dimensionless time in the range 0.001 ≤ tDA ≤ 10. From this data bank, values of the constant terminal rate solution βD, equ. (8.70), can be extracted for any fluid type, geometry of the drainage area and degree of well asymmetry. Special analyses This part of the program should be quite flexible with the main aims being to determine the average pressure within the drainage boundary at the time of the survey and to gain some information about the magnitude and shape of the drainage area and well position with respect to the boundary. For instance, the program could be structured, as described in exercise 7.7, to match the equation of the observed linear buildup, obtained in the preliminary analysis, with a theoretical straight line for various assumed area sizes and shapes. Alternatively, an attempt could be made to match the entire buildup by evaluating both βD (tD + ∆tD) and βD (∆tD), and plotting the difference between the two, also as described in exercise 7.7. Kazemi iteration Iteration over the entire test analysis may be required for surveys conducted in gas, or in gas saturated oil reservoirs, as described in sec. 8.11. This measure is necessary to refine the estimate of the average pressure within the drainage area at the time of the survey. REFERENCES 1) Russell, D.G., Goodrich, J.H., Perry, G.E. and Bruskotter, J.F., 1966. Methods of Predicting Gas Well Performance. J.Pet.Tech., January: 99-108.Trans. AIME. 2) Al-Hussainy, R., Ramey, H.J.,Jr. and Crawford, P.B.,1966. The Flow of Real Gases Through Porous Media. J.Pet.Tech., May: 624-636. Trans. AIME. 3) Al-Hussainy, R. and Ramey, H.J., Jr., 1966. Application of Real Gas Flow Theory to Well Testing and Deliverability Forecasting. J.Pet.Tech., May: 637-642. Trans. AIME. REAL GAS FLOW: GAS WELL TESTING 296 4) Ramey, H.J., Jr. and Wattenbarger, R.A., 1968. Gas Well Testing with Turbulence, Damage and Wellbore Storage. J.Pet.Tech., August: 877-887. Trans. AIME. 5) Aziz, K., Mathar, L., Ko, S. and Brar, G.S.,1976. Use of Pressure, Pressure-Squared or Pseudo-Pressure in the Analysis of Transient Pressure Drawdown Data from Gas Wells. J.Can.Pet.Tech., April-June: 58−65. 6) Geertsma, J., 1974. Estimating the Coefficient of Inertial Resistance in Fluid Flow Through Porous Media. Soc.Pet.Eng.J., October: 445-450. 7) Gewers, C.W.W. and Nichol, L.R., 1969. Gas Turbulence Factor in a Microvugular Carbonate. J.Can.Pet.Tech., April. 8) Wong, S.W., 1970. Effects of Liquid Saturation on Turbulence Factors for Gas Liquid Systems. J.Can.Pet.Tech., October. 9) Katz, D.L., et.al., 1959. Handbook of Natural Gas Engineering. McGraw-Hill Book Company, Inc. 47−50. 10) Kazemi, H., 1974. Determining Average Reservoir Pressure from Pressure Buildup Test. Soc.Pet.Eng.J., February: 55−62. Trans. AIME. 11) Rawlins, E.L. and Schelihardt, M.A., Back Pressure Tests on Natural Gas Wellsand Their Application to Production Practices. Monograph 7, USBM. 12) Carter, R.D., Millers, S.C., Jr. and Riley, H.G., 1963. Determination of Stabilized Gas Well Performance from Short Flow Tests. J.Pet.Tech., June: 651−658. Trans. AIME. 13) Odeh, A.S. and Jones, L.G., 1965. Pressure Drawdown Analysis, Variable Rate Case. J.Pet.Tech., August: 960−964. Trans. AIME. 14) Essis, A.E. and Thomas, G.W., 1971. The Use of Open Flow Potential Test Data in Determining Formation Capacity and Skin Factor. J.Pet.Tech., July: 879−887. Trans. AIME. 15) Earlougher, R.C., Ramey, H.J., Jr., Miller, F.G. and Mueller, T.D., 1968. Pressure Distribution in Rectangular Reservoirs. J.Pet.Tech., February: 199−208. Trans. AIME. 16) Odeh, A.S., Moreland, E.E., and Schueler, S., 1975. Characterization of a Gas Well from One Flow-Test Sequence. J.Pet.Tech., December: 1500−1505. Trans. AIME. 17) Raghavan, R., 1976. Well Test Analysis: Wells Producing by Solution Gas Drive. Soc.Pet.Eng.J., August: 196−208. 18) Ramey, H.J., Jr., 1964. Rapid Method for Estimating Reservoir Compressibilities. J.Pet.Tech., April: 447−454. CHAPTER 9 NATURAL WATER INFLUX 9.1 INTRODUCTION Natural water influx into both gas and oil reservoirs has been described previously (Chapter 1, sec. 7, and Chapter 3, sec. 7) using the simple, time independent aquifer material balance We = c Wi (pi - p) for calculating the amount of influx. In this equation Wi = initial volume of water in the aquifer and is therefore dependent upon aquifer geometry pi = initial aquifer/reservoir pressure p = current reservoir pressure, which in this chapter will always be assumed to be equal to the pressure at the original oil (or gas) water contact c = total aquifer compressibility = cw + cf The equation is simply a re-statement of the basic definition of compressibility and is only applicable to very small aquifers. For large aquifers a mathematical model is required which includes time dependence, to cater for the fact that it takes a finite time for the aquifer to respond fully to a pressure change in the reservoir. Two such models will be studied in this chapter: firstly, that of Hurst and van Everdingen1 and secondly, the more recent, approximate method of Fetkovitch2. These techniques will be applied in the classical reservoir engineering manner of first building a model which will adequately match the reservoir's production and pressure history when included in the material balance equation (this is sometimes referred to as aquifer fitting) then, once a satisfactory model has been built, using it to predict the future performance of the reservoir, say, for a given offtake policy. It should be appreciated that there are more uncertainties attached to this subject, in reservoir engineering, than to any other. This is simply because one seldom drills wells into an aquifer to gain the necessary information about the porosity, permeability, thickness and fluid properties. Instead, these properties have frequently to be inferred from what has been observed in the reservoir. Even more uncertain, however, is the geometry and areal continuity of the aquifer itself. The reservoir engineer should therefore consult both production and exploration geologists, concerning the latter, rather than relying entirely upon his own judgement. Due to these inherent uncertainties the aquifer fit obtained from history matching is seldom unique and the aquifer model may require frequent updating as more production and pressure data become available, as illustrated in exercise 9.2. NATURAL WATER INFLUX 298 9.2 THE UNSTEADY STATE WATER INFLUX THEORY OF HURST AND VAN EVERDINGEN The flow equations for oil into a wellbore are identical in form to the equations describing flow from an aquifer into a cylindrical reservoir; only the radial scale is different. In the former case, when an oil well is opened up on production at a constant rate q, the pressure response at the wellbore can first of all be described under transient flow conditions, before the reservoir boundary effects are felt, followed possibly by a period of late transient flow and finally by stabilized, semi-steady state flow. Irrespective of the flow condition the general equation for calculating the pressure in the wellbore at any time was presented in Chapter 7, sec. 6, as ( ) ( ) ( ) D 12 D D DA DA D MBH 4t p t 2 t ln p t π γ = + − (7.42) where pD (tD ) = 2 kh q π µ (pi – pwf) which is the dimensionless pressure function describing the constant terminal rate case. That is, it determines the pressure drop at r = rw due to a rate change from zero to q applied at the inner boundary at time t = 0. In the description of water influx from an aquifer into a reservoir there is greater interest in calculating the influx rate rather than the pressure drop. This leads to the determination of the influx as a function of a given pressure drop at the inner boundary of the system. In this respect Hurst and van Everdingen1 solved the radial diffusivity equation for the aquifer-reservoir system by applying the Laplace transformation to the equation, expressed in terms of dimensionless variables as D D D D D D D p p 1 r r r t t ∂ ∂ ∂ = ∂ ∂ ∂ (7.18) where in this case D o r r r = (9.1) and D 2 0 kt t cr φµ = (9.2) in which ro is the outer radius of the reservoir and all the other parameters in both equs. (9.1) and (9.2) refer to aquifer rather than reservoir properties, which will be the case for all the equations in this chapter unless specifically stated otherwise. Instead of obtaining constant terminal rate solutions of equ. (7.18), Hurst and van Everdingen derived constant terminal pressure solutions of the form NATURAL WATER INFLUX 299 ( ) D D q q t 2 kh p µ π = ∆ (9.3) where qD (tD) is the dimensionless influx rate evaluated at rD = 1 and describes the change in rate from zero to q due to a pressure drop ∆p applied at the outer reservoir boundary ro at time t = 0. These functions can be generated from constant terminal rate solutions and vice-versa. It is generally more convenient to express this solution in terms of cumulative water influx rather than rate of influx. Thus integrating equ. (9.3) with respect to time D t t D D D D o o dt qdt q (t ) dt 2 kh p dt µ π = ∆ which gives ( ) 2 e 0 D D W cr W t 2 kh p k µ φµ π = ∆ and therefore ( ) 2 e 0 D D W 2 hcr pW t πφ = ∆ (9.4) in which, since Darcy units are being employed We = cumulative water influx (ccs) due to a pressure drop ∆p (atm) imposed at ro at t = 0 and WD (tD) = dimensionless, cumulative water influx function giving the dimensionless influx per unit pressure drop imposed at the reservoir aquifer boundary at t=0. Equation (9.4) is frequently expressed as e D D W U p W (t ) = ∆ (9.5) where 2 o U 2 f hcr π φ = (9.6) which is the aquifer constant for radial geometry and (encroachment angle) f 360 ° ° = which is to be used for aquifers which subtend angles of less than 360° at the centre of the reservoir-aquifer system. The dimensionless water influx WD (tD) is frequently presented in tabular form or as a set of polynomial expressions giving WD as a function of tD for a range of ratios of the NATURAL WATER INFLUX 300 aquifer to reservoir radius reD = re/ro, for radial aquifers. In contrast, plots of WD versus tD for both radial and linear geometry and, in the former case, for select values of reD are included as figs. 9.3 - 9.7. The plots are taken from the published solutions of equ. (7.18) by Hurst and van Everdingen1. Each chart has a different resolution of the dimensionless time scale. It should be noted that the graphs are valid for all values of tD and hence are applicable for calculating both the early, unstable influx (infinite aquifer case) and for the influx occurring once the aquifer boundary effects have been felt. There are differences in the way in which the dimensionless time and aquifer constant are calculated, dependent on the geometry. These are summarised in figs. 9.1 and 9.2. Water Oil ro re θ θ ° ° 360 t = θ π 2 = Radians Fig. 9.1 Radial aquifer geometry Darcy Units Field Units D 2 0 kt t cr φµ = D t = constant 2 0 kt cr φµ × (9.7) the constant = .000264 (t-hours) = .00634 (t-days) = 2.309 (t-years) U = 2π fφh 2 o cr (cc/atm) U = 1.119 fφh 2 o cr (bbl/psi) (9.8) Water Oil h L W Fig. 9.2 Linear aquifer geometry NATURAL WATER INFLUX 301 Darcy Units Field Units D 2 kt t cL φµ = (t-sec) tD = constant × 2 kt cL φµ (9.9) constant, same as for equ. (9.7) U = wLhφc (cc/atm) U = .1781 wLhφc (bbl/psi) (9.10) Other characteristic features of the plots of WD(tD) versus tD depend upon whether the aquifer is bounded or infinite in extent. Bounded Aquifers Irrespective of the geometry there is a value of tD for which the dimensionless water influx reaches a constant maximum value. This value is, however, dependent upon the geometry as follows ( ) ( ) 2 12 D eD Radial W max r 1 = − (9.11) ( ) D Linear W max 1 = (9.12) Note that if WD in equ. (9.11) is used in equ. (9.4), for a full radial aquifer (f = 1), the result is 2 2 2 e 0 12 e 0 2 0 2 2 e 0 (r r ) W 2 hcr p r (r r )h c p πφ π φ − = ×∆× = − ∆ But this latter expression is also equivalent to the total influx occurring, assuming that the ∆p is instantaneously transmitted throughout the aquifer. A similar result can be obtained using equ. (9.12) for linear geometry. Therefore, once the plateau level of WD (tD) has been reached, it means that the minimum value of tD at which this occurs has been sufficiently large for the instantaneous pressure drop ∆p to be felt throughout the aquifer. The plateau level of WD(tD) is then the maximum dimensionless water influx resulting from such a pressure drop. Infinite Aquifer Naturally, no maximum value of WD (tD) is reached in this case since the water influx is always governed by transient flow conditions. For radial geometry, values of WD (tD) can be obtained from the graphs for reD = ∞. There is no plot of WD (tD) for an infinite linear aquifer. Instead, the cumulative water influx can be calculated directly using the following equation e kct W 2hw p (ccs) φ πµ = ×∆ (9.13) NATURAL WATER INFLUX 302 which is expressed in Darcy units. The corresponding equation in field units, with t measured in hours, is 3 e kct W 3.26 10 hw x p (bbls) φ πµ − = × ∆ (9.14) Note that dimensionless time is not used in the above equations. r = 1.5 eD r = 2.0 eD r = 2.5 eD r = 3.0 eD r = 3.5 eD r = 4.0 eD r = eD 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 WD tD Fig. 9.3 Dimensioniess water influx, constant terminal pressure case, radial flow. (After Hurst and van Everdingen, ref. 1) NATURAL WATER INFLUX 303 0 0 2 4 6 8 10 12 14 16 18 20 10 20 30 40 50 60 70 r = 3.0 eD r = 3.5 eD r =5.0 D e r = 6.0 =7.0 r = 8.0 r = 4.5 D e r =4.0 eD r = 8 eD eD eD reD WD tD Fig. 9.4 Dimensionless water influx, constant terminal pressure case, radial flow (After Hurst and van Everdingen, ref. 1) NATURAL WATER INFLUX 304 110 100 90 70 80 60 50 40 30 20 10 0 0 20 40 60 80 100 120 140 160 180 200 r =5.0 eD r =6.0 eD r =7.0 eD r =8.0 eD r =9.0 eD r =10.0 eD r = eD 8 WD tD Fig. 9.5 Dimensionless water influx, constant terminal pressure case, radial flow (After Hurst and van Everdingen, ref. 1) NATURAL WATER INFLUX 305 tD Fig. 9.6 Dimensionless water influx, constant terminal pressure case, radial and linear flow (After Hurst and van Everdingen, ref.1) NATURAL WATER INFLUX 306 USE THIS SCALE FOR LINE reD = (CON’T) ONLY 8 10 2 8 6 4 2 10 8 6 4 2 1 10 2 4 6 8 10 2 4 6 8 2 10 3 2 4 6 8 10 4 10 10 2 10 3 r =15.0 eD r = eD (cont’d) 8 r = eD 8 r =2.5 eD r =3.0 eD r =3.5 eD r =4.0 eD r =4.5 eD r =5.0 eD r =7.0 eD r =7.0 eD r =8.0 eD r =9.0 eD r =10.0 eD r =6.0 eD WD tD Fig. 9.7 Dimensionless water influx, constant terminal pressure case, radial and linear flow (After Hurst and van Everdingen, ref.1) NATURAL WATER INFLUX 307 EXERCISE 9.1 APPLICATION OF THE CONSTANT TERMINAL PRESSURE SOLUTION A reservoir-aquifer system has the geometry and dimensions as shown in fig. 9.8. r = 5000’ o r = 15000’ e Oil 80 O Water Fig. 9.8 Water influx from a segment of a radial aquifer If the aquifer properties are as follows h = 50 ft µ = 0.4 cp φ = 0.25 cw = 3.0×10-6 /psi k = 50 mD cf = 6.0×10-6 /psi 1) Calculate the water influx at times t = 0.5, 1, 1.5, 2 and 3 years after an instantaneous pressure drop of ∆p = 100 psi, at the oil water contact, at time t=0. 2) What would be the corresponding water influx if it is assumed that the same pressure drop is transmitted instantaneously throughout the aquifer? EXERCISE 9.1 SOLUTION 1) Since t is measured in years, then D 6 2 2 0 D 2.309kt 2.309 50 t t .25 .4x9 10 (5000) cr t 5.131t φµ − × × = = × × × = The encroachment angle is 80°, therefore f = 80°/360° = 0.222 and We = 1.119 fφh 2 0 cr ∆pWD(tD) = 1.119×.222×.25×50×9×10-6×(5000)2×100WD (tD) We = 69868 WD (tD) bbls NATURAL WATER INFLUX 308 Using figs. 9.3 and 9.4 to obtain values of WD(tD); We can be calculated as follows: t (years) tD WD (tD) (reD = 3.00) We (bbls) .5 2.6 2.7 188644 1.0 5.1 3.5 244538 1.5 7.7 3.8 265498 2.0 10.3 3.9 272485 3.0 15.4 4.0 279472 TABLE 9.1 For dimensionless times greater than tD=15, WD(tD) = 4 and remains constant at this value indicating that the maximum amount of water influx due to the 100 psi pressure drop is 279500 bbl. 2) If the pressure drop is transmitted instantaneously throughout the aquifer, then We = ( ) 2 2 e o cf r -r π hφ ∆p/5.615 bbls = 9 × 10-6 × .222 × π (150002 − 50002) × 50 × .25 × 100/5.615 We = 279500 bbls which again is the maximum water influx due to the 100 psi pressure drop. In using the constant terminal pressure solution, a time scale has been attached to the water influx. 9.3 APPLICATION OF THE HURST, VAN EVERDINGEN WATER INFLUX THEORY IN HISTORY MATCHING In the previous section the cumulative water influx into a reservoir, due to an instantaneous pressure drop applied at the outer boundary, was expressed as ( ) e D D W U p W t = ∆ (9.5) In the more practical case of history matching the observed reservoir pressure, it is necessary to extend the theory to calculate the cumulative water influx corresponding to a continuous pressure decline at the reservoir-aquifer boundary. In order to perform such calculations it is conventional to divide the continuous decline into a series of discrete pressure steps. For the pressure drop between each step, ∆p, the corresponding water influx can be calculated using equ. (9.5). Superposition of the separate influxes, with respect to time, will give the cumulative water influx. The recommended method of approximating the continuous pressure decline, by a series of pressure steps, is that suggested by van Everdingen, Timmerman and McMahon3, which is illustrated in fig. 9.9. NATURAL WATER INFLUX 309 PRESSURE TIME p2 p1 p0 p1 p1 pi p2 0 t1 t2 t 3 p2 p3 p3 Fig. 9.9 Matching a continuous pressure decline at the reservoir-aquifer boundary by a series of discrete pressure steps Suppose that the observed reservoir pressures, which are assumed to be equal to the pressures at the original hydrocarbon-water contact, are pi, p1, p2, p3 .... etc., at times 0, t1, t2, t3 .... etc. Then the average pressure levels during the time intervals should be drawn in such a way that i 1 1 1 2 2 j 1 j j p p p 2 p p p 2 p p p 2 − + = + = + = (9.15) The pressure drops occurring at times 0, t1, t2 . . . etc. are then NATURAL WATER INFLUX 310 i 1 i 1 1 0 i i i 1 1 2 i 2 1 2 1 2 3 1 3 1 2 2 3 2 j 1 j j j 1 j 1 j 1 j j 1 j (p p ) p p p p p p 2 2 (p p ) (p p ) p p p p p 2 2 2 (p p ) p p (p p ) p p p 2 2 2 (p p ) (p p ) p p p p p 2 2 2 − − + − + + + − ∆ = − = = + + − ∆ = − = − = + − + ∆ = − = − = + + − ∆ = − = − = (9.16) Therefore, to calculate the cumulative water influx We at some arbitrary time T, which corresponds to the end of the nth time step, requires the superposition of solutions of type, equ. (9.5), to give We (T) = U [∆poWD (TD)+∆p1 WD (TD - tD1)+ ∆p2 WD (TD - tD2) +………….∆pjWD (TD - tD j)+…∆pn-1 WD (TD - tD n-1)] where ∆pj is the pressure drop at time tj, given by equ. (9.16), and WD (TD - tDj) is the dimensionless cumulative water influx, obtained from figs. 9.3 - 9.7, for the dimensionless time TD - tDj during which the effect of the pressure drop is felt. Summing the terms in the latter equation gives ( ) j n 1 e j D D D j 0 W T U p W (T T ) − = = ∆ − (9.17) In the special case of an infinite, linear aquifer for which, as noted in sec. 9.2, there is no WD function included in figs. 9.3 - 9.7, the cumulative water influx at time T due to a step-like pressure decline at the aquifer-reservoir boundary can be calculated using equ. (9.13) as ( ) n 1 e j j j 0 kc W T 2hw p T t φ πµ − = = ∆ − which, when expressed in field units has the same constant, 3.26×10-3, as equ. (9.14). The following exercise will illustrate the application of the method of superposition in history matching. EXERCISE 9.2 AQUIFER FITTING USING THE UNSTEADY STATE THEORY OF HURST AND VAN EVERDINGEN A wedge shaped reservoir is suspected of having a fairly strong natural water drive. The geometry of the reservoir-aquifer system is shown in fig. 9.10. NATURAL WATER INFLUX 311 sealing fault 140° r ? e r = 9200’ o Fig. 9.10 Aquifer-reservoir geometry, exercise 9.2 Properties common to the reservoir and aquifer Reservoir properties h = 100 ft k = 200 mD µw = 0.55 cp φ = 0.25 cw = 3.0×10-6 /psi cf = 4.0×10-6 / psi Bw = 1.0 N = 312×106 stb Swc = 0.05 (PVT data are plotted in figs. 9.12 and 9.13 and listed in table 9.3) TIME (years) 0 1 2 3 4 5 6 7 8 9 10 1000 800 600 2800 2400 2000 1600 1200 800 400 0 Psia PRESSURE Np MMstb Rp scf/stb Pressure Rp Np 100 80 60 40 20 Fig. 9.11 Reservoir production and pressure history; exercise 9.2 NATURAL WATER INFLUX 312 PRESSURE (psia) 0 1000 2000 3000 0 0 500 400 300 200 100 scf/stb Rs Bg rb/scf 0.005 0.004 0.003 0.002 0.001 Bg Rs 0.006 600 Fig. 9.12 Rs and Bg as functions of pressure; exercise 9.2 PRESSURE (psia) 0 1000 2000 3000 rb/stb 1.4 1.3 1.2 Bo 1.1 Fig. 9.13 Bo as a function of pressure; exercise 9.2 Initially the reservoir was at bubble point pressure but apparently there was no initial gas cap (m=0). The graphs of cumulative oil production (Np), cumulative GOR (Rp) and the average reservoir pressure (p), as functions of time, are shown in fig. 9.11, for the first 10 years of production. It may be assumed that the pressure decline at the oil water contact is the same as the average pressure decline in the reservoir. It was initially suspected, based on seismic and geological evidence that the value of reD = re/ro was about 10. Do you consider this to be the correct value of reD based on 10 years of production history, and if not, what is your estimate of the correct value of this ratio? NATURAL WATER INFLUX 313 PRESSURE (psia) 2700 2500 2300 2100 1900 1700 1500 1300 0 1 3 4 5 6 7 8 9 10 TIME (years) 2620 p =2500 1 2395 p =2290 2 2199 p =2109 3 2029 p =1949 4 1883 p =1818 5 1760 p =1702 6 1655 p =1608 7 1571 p =1535 8 1507 p =1480 9 1460 p =1440 10 p =2740 i Fig. 9.14 Reservoir pressure decline approximated by a series of discrete pressure steps; exercise 9.2 Interpreted Data Fig. 9.14 shows a more detailed graph of the continuous pressure decline which has been approximated by 10 annual pressure steps, as described in the text. The magnitude of the ∆p's can be calculated, using equs. (9.15) and (9.16), as shown in table 9.2. NATURAL WATER INFLUX 314 Time (years) Pressure at the O.W.C. (psia) Plateau Pressure levels (psia) ∆p (psi) 0 2740 (pi) 120 1 2500 2620 225 2 2290 2395 196 3 2109 2199 170 4 1949 2029 146 5 1818 1883 123 6 1702 1760 105 7 1608 1655 84 8 1635 1571 64 9 1480 1507 47 10 1440 1460 TABLE 9.2 The "end of the year values" of Np, the cumulative oil production, Rp, the cumulative GOR and PVT properties, corresponding to the pressure points given in column 2 of table 9.2,are listed in table 9.3. Time (years) Np (MM/stb) Rp (scf/stb) Bo (rb/stb) Rs (scf/stb) Bg (rb/scf) 0 650 (Rsi) 1.404(Boi) 650 (Rsi) .00093 (Bgi) 1 7.88 760 1.374 592 .00098 2 18.42 845 1.349 545 .00107 3 29.15 920 1.329 507 .00117 4 40.69 975 1.316 471 .00128 5 50.14 1025 1.303 442 .00139 6 58.42 1065 1.294 418 .00150 7 65.39 1095 1.287 398 .00160 8 70.74 1120 1.280 383 .00170 9 74.54 1145 1.276 381 .00176 10 77.43 1160 1.273 364 .00182 TABLE 9.3 NATURAL WATER INFLUX 315 EXERCISE 9.2 SOLUTION The procedure in tackling this type of problem is as follows. 1) Calculate the water influx using the unsteady state influx theory of Hurst and van Everdingen for reD = 10. 2) Apply the technique of Havlena and Odeh in interpreting the material balance as the equation of a straight line (refer Chapter 3, sec. 7). For a reservoir with no initial gas cap the material balance is ( ) o oi si s g w wc f p o p s g oi e oi wc (B B ) (R R )B (c S c ) N B R R B NB p W B 1 S − + − + + − = + ∆ + − where We is the cumulative water influx (rb). If the entire pressure drop over the 10 year period is considered (∆p = 1300 psi) the magnitude of the compressibility term is 6 w wc f wc (c S c ) (3 .05 4) p 10 1300 0.006 1 S .95 − + × + ∆ = × × ∼ − This is only 2% of the maximum value of the term accounting for the expansion of the oil plus its originally dissolved gas (refer table 9.6; (Eo/Boi)max = .3895/1.404 = .277) and therefore, the pore and water compressibility effects will be neglected. The material balance is then reduced to Np [Bo + (Rp - Rs)Bg] = N [(Bo - Boi) + (Rsi - Rs)Bg] + We or F = NEo+We and a plot of F/Eo versus We/Eo should be linear with intercept F/Eo = N when W/Eo = 0 and with unit slope. 3) If such a plot is non-linear for reD = 10 , vary the value of this parameter until a straight line is obtained. 1) Calculation of We for reD = 10 Since annual time steps have been selected, the dimensionless time coefficient can most conveniently be expressed, with t in years and all other parameters in field units, as D 2 o 6 2 2.309kt t cr 2.309 200 t 5.67 t .25 .55 7 10 (9200) φµ − = × × = = × × × × (9.7) Similarly, the value of the aquifer constant, in field units, is 2 o U 1.119f hcr φ = (9.8) where f = 140°/360° = 0.3889 NATURAL WATER INFLUX 316 i.e. U = 1.119×.3889×.25×100×7×10-6×(9200)2 U = 6446 bbls/psi Time (years) dimensionless time ∆pj (psi) WD (reD = 10) WD (reD = 5) 0 0 120 1 5.67 225 4.95 4.88 2 11.34 196 8.12 7.46 3 17.01 170 10.90 9.10 4 22.68 146 13.50 10.09 5 28.35 123 15.90 10.83 6 34.02 105 18.10 11.27 7 36.69 84 20.20 11.52 8 45.36 64 22.20 11.69 9 51.03 47 24.00 11.81 10 56.70 25.70 11.89 TABLE 9.4 The water influx can now be calculated, using values of ∆pj and WD listed in table 9.4, as shown in table 9.5. T (years) We = U n 1 j j 0 p − = ∆ WD (TD - tDj) We (MMrb) 1 6446 (120×4.95) 3.829 2 6446 (120×8.12 + 225×4.95) 13.460 3 6446 (120×10.90 + 225×8.12 + 196×4.95) 26.462 4 6446 (120×13.50 + 225×10.90 + 196×8.12 + 170×4.95) 41.935 5 6446 (120×15.90 + 225×13.50 + 196×10.90 + 170×8.12 + 146×4.95) 59.207 6 6446 (120×18.10 + 225×15.90 + 196×13.50 + 170×10.90 + 146×8.12 + 123×4.95) 77.628 7 6446 (120×20.20 + 225×18.10 + 196×15.90 + 170×13.50 + 146×10.90 + 123×8.12 + 105×4.95) 96.805 8 6446 (120×22.20 + 225×20.20 + 196×18.10 + 170×15.90 + 146×13.50 + 123×10.90 + 105×8.12 + 84×4.95) 116.284 9 6446 (120×24.00 + 225×22.20 + 196×20.20 + 170×18.10 + 146×15.90 + 123×13.50 + 105×10.90 + 84×8.12 + 64×4.95) 135.601 10 6446 (120×25.70 + 225×24.00 + 196×22.20 + 170×20.20 + 146×18.10 + 123×15.90 + 105×13.50 + 84×10.90 + 64×8.12 + 47×4.95) 154.401 TABLE 9.5 NATURAL WATER INFLUX 317 2) Material Balance Calculation F = Np (Bo + (Rp - Rs)Bg) Eo = ((Bo - Boi) + (Rsi - Rs)Bg) Time (years) F (MMrb) Eo (rb/stb) F/Eo (MMstb) We/Eo (reD = 10) We/Eo (reD = 5) 1 12.124 .0268 452.4 142.9 140.9 2 30.761 .0574 535.9 234.5 223.8 3 52.826 .0.923 5.72.3 286.7 260.3 4 79.798 .1411 565.5 297.2 253.5 5 105.964 .1881 563.3 314.8 251.3 6 132.292 .2380 555.8 326.2 243.8 7 157.080 .2862 548.8 338.2 236.8 8 179.177 .3299 543.1 352.5 231.2 9 196.654 .3630 541.7 373.6 229.7 10 210.743 .3895 541.1 396.4 229.1 TABLE 9.6 The plot of F/Eo versus We/Eo, fig. 9.15, shows that the trend deviates below the theoretical 45° line after the first year. This indicates that the correct value of reD should be somewhat smaller. The calculation has therefore been repeated with reD=5. 3) Calculation of We for reD = 5. Values of We for reD=5 are listed in table 9.7. These have been calculated using the WD values listed in column 5 of table 9.4. NATURAL WATER INFLUX 318 T (years) We = U n 1 j j 0 p − = ∆ WD (TD - tDj) We (MMrb) 1 6446 (120×4.88) 3.775 2 6446 (120×7.46 + 225×4.88) 12.848 3 6446 (120×9.10 + 225×7.46 + 196×4.88) 24.024 4 6446 (120×10.09 + 225×9.10 + 196×7.46 + 170×4.88) 35.775 5 6446 (120×10.83 + 225×10.09 + 196×9.10 + 170×7.46 + 146×4.88) 47.276 6 6446 (120×11.27 + 225×10.83 + 196×10.09 + 170×9.10 + 146×7.46 + 123×4.88) 58.035 7 6446 (120×11.52 + 225×11.27 + 196×10.83 + 170×10.09 + 146×9.10 + 123×7.46 + 105×4.88) 67.778 8 6446 (120×11.69 + 225×11.52 + 196×11.27 + 170×10.83 + 146×10.09 + 123×9.10 + 105×7.46 + 84×4.88) 76.259 9 6446 (120×11.81 + 225×11.69 + 196×11.52 + 170×11.27 + 146×10.83 + 123×10.09 + 105×9.10 + 84×7.46 + 64×4.88) 83.398 10 6446 (120×11.89 + 225×11.81 + 196×11.69 + 170×11.52 + 146×11.27 + 123×10.83 + 105×10.09 + 84×9.10 + 64×7.46 + 47×4.88) 89.225 TABLE 9.7 The corresponding values of We/Eo, required for the material balance calculations, are listed in table 9.6 and plotted in fig. 9.15. As can be seen, in this case, all the points lie on the straight line of unit slope which intercepts the ordinate at Np = 312 MMstb, thus confirming that reD = 5 is the correct value of the ratio of the aquifer/reservoir radius. Also apparent from this plot is the fact that after the third year there occurs a reversal in the trend of plotted points with both We/ Eo and F/Eo decreasing with time. Havlena and Odeh have pointed out4,5 that this reversal is to be expected for a finite acting aquifer but nevertheless all points, for the correct value of reD, must lie on the linear trend. For the incorrect aquifer model (reD = 10), however, the calculated values of We/Eo continually increase as F/Eo decrease, as illustrated in fig. 9.15. NATURAL WATER INFLUX 319 1 2 34 5 6 7 8 9 9,10 600 550 500 450 400 350 N =312 p 300 F/E (MMstb) o 100 200 300 400 W /E (MMstb) e o 1,2,3...etc. Time (years) r =5 eD r =10 eD 10 8 3 4 5 6 7 2 1 Fig. 9.15 Aquifer fitting using the interpretation technique of Havlena and Odeh 9.4 THE APPROXIMATE WATER INFLUX THEORY OF FETKOVITCH FOR FINITE AQUIFERS The unsteady state influx theory of Hurst and van Everdingen provides the correct method for calculating the cumulative water influx, under practically all circumstances, for radial and linear aquifers. Unfortunately, it has the disadvantage that calculations performed using the method are rather tedious, due to the complexity of superposing solutions for each time step. This drawback is exaggerated by the fact that influx calculations, when history matching, usually require a trial and error approach. Because of this, many attempts have been made in the past to find a more direct computational method of performing water influx calculations which would duplicate results obtained with the Hurst and van Everdingen method and remove the necessity of superposition. The most successful of the methods is that proposed by Fetkovitch in 19712. In this approach the flow of aquifer water into a hydrocarbon reservoir is modelled in precisely NATURAL WATER INFLUX 320 the same way as the flow of oil from a reservoir into a well. An inflow equation of the form e a w dW q J(p p) dt = = − (9.18) is used where, qw = water influx rate J = aquifer productivity index p = reservoir pressure, i.e. pressure at the oil or gas water contact. a p = average pressure in the aquifer. The latter is evaluated using the simple aquifer material balance We=c Wi (pi - p a) (9.19) in which pi is the initial pressure in the aquifer and reservoir. This balance can be alternatively expressed as e e a i i ei i i W W p p 1 p 1 W cWp = − = − (9.20) where Wei = c Wipi is defined as the initial amount of encroachable water and represents the maximum possible expansion of the aquifer. Differentiating equ. (9.20) with respect to time gives e ei a i dp dW W dt p dt = − (9.21) and substituting equ. (9.21) into equ. (9.18) and separating the variables gives a i ei a dp Jp dt W p p = − − This equation will now be integrated for the initial condition that at t=0 (We = 0, a p = pi) a pressure drop ∆p = pi - p is imposed at the reservoir boundary. Furthermore, the boundary pressure p remains constant during the period of interest so that ( ) i a ei Jp t ln p p C W − = − + where C is an arbitrary constant of integration which can be evaluated from the initial conditions as C = In(pi - p), and therefore ( ) i ei a i Jp t / W p p p p e− −= − (9.22) NATURAL WATER INFLUX 321 which on substituting in the inflow equation (9.18) gives e i ei i dW Jp t / W J(p p)e dt − = − (9.23) Finally, integrating equ. (9.23) for the stated initial conditions yields the following expression for the cumulative water influx ei i ei e i i W Jpt / W W (p p)(1 e ) p − = − − (9.24) What can be observed immediately from this expression is that as t tends to infinity, then ei e i i i i W W (p p) p cW (p p) − = = − which is the maximum amount of water influx that could occur once the pressure drop pi - p has been transmitted throughout the aquifer. As it stands, equ. (9.24) is not particularly useful since it was derived for a constant inner boundary pressure. To use this solution in the practical case, in which the boundary pressure is varying continuously as a function of time, it should again be necessary to apply the superposition theorem. Fetkovitch has shown, however, that a difference form of equ. (9.24) can be used which eliminates the need for superposition. That is, for the influx during the first time step ∆t1, equ. (9.24) can be expressed as ( ) 1 ei i 1 ei 1 e i i W Jp t / W W p p (1 e ) p − ∆ ∆ = − − (9.25) where 1 p is the average reservoir boundary pressure during the first time interval. For the second interval ∆t2 ( ) 1 2 ei i 2 ei a 2 e i W Jp t / W W p p (1 e ) p − ∆ ∆ = − − (9.26) where 1 a p is the average aquifer pressure at the end of the first time interval and is evaluated using equ. (9.20) as i 1 i e a i e W p p 1 W ∆ = − (9.27) In general for the nth time period, ( ) n 1 n ei i n ei a n e i W Jp t / W W p p (1 e ) p − − ∆ ∆ = − − (9.28) where NATURAL WATER INFLUX 322 j n 1 n 1 e j 1 a i ei W p p 1 W − − = ∆ = − (9.29) The values of n p , the average reservoir boundary pressure, are calculated, as described in section 9.3, as n 1 n n p p p 2 −+ = (9.15) Fetkovitch has demonstrated that using equs. (9.28) and (9.29), in a stepwise fashion, the water influx calculated for a variety of different aquifer geometries matches closely the results obtained using the unsteady state influx theory of Hurst and van Everdingen for finite aquifers. Values of the aquifer productivity index J, which depend both on the geometry and flowing conditions, are listed in table 9.8, in Darcy units. Multiplying the radial Pl functions by 7.08×10-3 and the linear by 1.127×10-3 will convert these expressions to field units. The radial values of J for semi-steady state and steady state influx will be recognised as identical in form to the productivity indices listed in Chapter 6, table 6.1, for the flow of a liquid into a wellbore. The only difference is that ro, the reservoir radius, now replaces rw, the wellbore radius. Note also that, while semi-steady state expressions for J, equs. (9.30), are used in conjunction with the Fetkovitch equations (9.28) and (9.29), the steady state expressions, (9.31), are used in a different manner. In applying these values it is assumed that the water influx from the aquifer into the reservoir is replaced by water from an external source, such as an artesian water supply, so that the pressure Flowing Condition Radial Aquifers J (cc/sec/atm) Linear Aquifers J (cc/sec/atm) Semi-steady state (used with drawdown expressed as a p p − ) e o 2 fkh r 3 ln r 4 π µ − khw 3 L µ (9.30) Semi-steady state (used with drawdown expressed as i p p − ) e o 2 fkh r ln r π µ khw L µ (9.31) TABLE 9.8 at the external boundary of the aquifer remains constant at its initial value pi. In this case it is unnecessary to keep evaluating the average pressure in the aquifer since it remains unchanged. The J values expressed in equ. (9.31) are now used in conjunction with the drawdown measured as pi - p. Referring to equ. (9.23), NATURAL WATER INFLUX 323 e i ei w i dW Jp t / W q J(p p)e dt − = = − (9.23) the steady state case implies that Wei, the encroachable water, is infinite and therefore e w i dW q J(p p) dt = = − (9.32) which, upon integration, gives the cumulative water influx as t e i 0 W J (p p)dt = − (9.33) Equs. (9.32) and (9.33), which are a special case of Fetkovitch's theory, were first presented in 1936 by Schilthuis6 and described as steady state influx equations. Equation (9.33) can be evaluated in stepwise fashion in which values of n p , the inner boundary pressure during the nth time period, are calculated using equ. (9.15). The reader should also be aware that the PI expressions presented in table 9.8 were derived in similar form in Chapter 6, sec. 2, under the assumption that (rw/re)2 was approximately zero. For small radial aquifers, the equivalent assumption that (ro/re)2 is negligible may not always be applicable and the correct PI expression should then be obtained by solving the radial diffusivity equation, using exactly the same steps as shown in Chapter 6 but, without neglecting such terms. Considering the inherent uncertainties in aquifer fitting this approach is generally unnecessary and, in fact, Fetkovitch has demonstrated an almost perfect match between his results and those of Hurst and van Everdingen for values of reD as small as three. For the case of a reservoir asymmetrically situated within a non-circular shaped aquifer it should, with tolerable accuracy, be possible to use the Dietz shape factors presented in fig. 6.4, and described in Chapter 7, sec. 7, to modify the semi-steady state PI expressions. Thus the radial PI in equ. (9.30) can be generalised as 2 A o 2 fkh J 4A ln 2 C r π µ γ = which has precisely the same form as equ. (6.22). For very large aquifers, the initial flow of water into the reservoir will be governed by transient flow conditions. In this case, it takes a finite time for the initial pressure disturbance at the reservoir-aquifer boundary to feel the effect of the outer boundary of the aquifer. Unfortunately, during this transient flow period it is no longer possible, in analogy with wellbore equations, to derive a simple expression for the productivity index J. This is because for inflow into a reservoir it is incorrect to use the approximate line source solution of the radial diffusivity equation, in an attempt to evaluate a PI under transient conditions, since ro is always finite and the boundary conditions for this type of solution, expressed in equ. (7.1), can no longer be justified. Therefore, the method of Fetkovitch cannot be used for the description of influx from an infinite aquifer NATURAL WATER INFLUX 324 and, when dealing with very large, finite aquifers, it is initially still necessary to apply the unsteady state influx theory of Hurst and van Everdingen for the first few time steps. The following example will illustrate the speed and accuracy in using the method of Fetkovitch in comparison to that of Hurst and van Everdingen. In addition, it will demonstrate how the two methods can be combined when dealing with a large aquifer, reD = 10, in which, for the first few years, the influx occurs under transient flow conditions. EXERCISE 9.3 WATER INFLUX CALCULATIONS USING THE METHOD OF FETKOVITCH Recalculate the cumulative water influx as a function of time, using all the reservoir and aquifer data presented in exercise 9.2, but applying the method of Fetkovitch. Perform the calculations for both reD = 5 and 10. EXERCISE 9.3 SOLUTION Using the method of Fetkovitch the following two equations are required j n 1 n 1 e j 1 a i ei W p p 1 W − − = ∆ = − (9.29) and n 1 n ei i n ei a n e i W Jp t / W W (p p )(1 e ) p − − ∆ ∆ = − − (9.28) where n-1 a p is the average pressure in the aquifer at the end of the (n - 1)th time interval and n p is the average reservoir-aquifer boundary pressure during the nth time interval. Since in this present application a history match is being sought for available reservoir pressures, that is, values of n p which are listed in column 3 of table 9.2 in the previous exercise, the manner of solving the above equations, to explicitly calculate the cumulative water influx, is as follows -having obtained ∆Wen-1 1for the n - 1th time step then Wen-1= j n 1 e j 1 W − = ∆ -using equ. (9.29), evaluate n 1 a p − -insert n 1 a p −in equ. (9.28) and solve for ∆Wen NATURAL WATER INFLUX 325 For the first time step n-1 a p = pi, the initial reservoir and aquifer pressure. The constant terms in equs. (9.28) and (9.29) can be evaluated (in field units), for the correct aquifer size, i.e. reD = 5, as follows: ( ) ei i i 2 2 e o i 6 6 ei W cWp cf r r h p / 5.615 rb 7 10 .3889 (2116 85) 10 100 .25 2740 5.615 W 211.9 MMrb π φ π − = = − × × × − × × × × = = For a finite radial aquifer 3 e o 3 7.08 10 fkh J r 3 ln r 4 7.08 10 .3889 200 100 116.5b / d/psi .55(In5 .75) µ − − × = − × × × × = = − (9.30) Therefore, Jpi/Wei = 116.5×2740/211.9×106 = 1.506×10-3 /day Since the ∆t in equ. (9.28) is in days, then for time steps of one year 3 i ei Jp t / W 1.506 10 365 0.4229 1 e 1 e − − ∆ − × × = − = − Equ. (9.28) can therefore be reduced to ( ) ( ) n n 1 n n 1 6 e a n e a n 211.9 10 W p p 0.4229 2740 W 32705 p p rb − − × ∆ = − × ∆ = − NATURAL WATER INFLUX 326 The calculation of the water influx is as shown in table 9.9. Time (years) n p (psia) n 1 a n p p −− (psi) n e W ∆ (MMrb) n e W (MMrb) n a p (psia) 0 2740(pi) 1 2620 120 3.925 3.925 2689 2 2395 294 9.615 13.540 2565 3 2199 366 11.970 25.510 2410 4 2029 381 12.461 37.971 2249 5 1883 366 11.970 49.941 2094 6 1760 334 10.924 60.865 1953 7 1655 298 9.746 70.611 1827 8 1571 256 8.373 78.984 1719 9 1507 212 6.934 85.918 1629 10 1460 169 5.527 91.445 1558 TABLE 9.9 The cumulative water influx is shown in fig. 9.16, in comparison to that calculated in exercise 9.2, for reD = 5, (table 9.7). As can be seen, the agreement between the two is excellent. The water influx has been recalculated, using the method of Fetkovitch, for reD = 10, in which case: Wei = 874.2 MMrb; J = 64.5 b/ d/psi and the influx equation becomes ∆Wen = 22685 n 1 a n p p −− This influx, as a function of time, is shown in fig. 9.17, in which it can be seen that there is a disparity between this and the unsteady state influx calculated in table 9.5. This is because the method of Fetkovitch cannot correctly model the early, transient influx from a large aquifer. The figures can be improved by applying the Hurst and van Everdingen method for the first few years, which is computationally simple, and then switching to Fetkovitch's method. This procedure is illustrated in table 9.10, using values of We from table 9.5, for the first four years. NATURAL WATER INFLUX 327 r = 5 eD We (MMrb) 100 80 60 40 20 1 2 3 4 Hurst, van Everdingen (table 9.7) 5 6 7 8 9 10 Fetkovitch (table 9.9) Time (years) Fig. 9.16 Comparison between Hurst and van Everdingen and Fetkovitch for reD = 5 160 120 100 80 60 40 20 1 140 2 3 4 5 6 8 10 9 We (MMrb) r = 10 eD Hurst, van Everdingen (table 9.5) Fetkovitch Fetkovitch (modified) (table 9.10) Time (years) Fig. 9.17 Comparison between Hurst and van Everdingen and Fetkovitch for reD = 10 NATURAL WATER INFLUX 328 Time (years) n p (psia) n-1 a n p -p (psi) n e ∆W (MMrb) n e W (MMrb) n a p (psia) 0 2740 2740 1 2620 3.829 2728 2 2395 13.460 2698 3 2199 26.462 2657 4 2029 41.935 2609 5 1883 726 16.469 58.404 2557 6 1760 797 18.080 76.484 2500 7 1655 845 19.169 95.653 2440 8 1571 869 19.713 115.366 2378 9 1507 871 19.759 135.125 2316 10 1460 856 19.418 154.543 2256 TABLE 9.10 Using this combined method (Fetkovitch-modified) it can be seen that the results are almost identical with those obtained using the unsteady state influx theory throughout, as shown in fig. 9.17. 9.5 PREDICTING THE AMOUNT OF WATER INFLUX Sections 9.2 through 9.4 considered the ways in which a mathematical aquifer model is constructed and matched to the reservoir production and pressure history. If it is felt confident that the model so developed is satisfactory in matching the history, then the next step is to use it in predicting the future reservoir performance. The aim here is usually to determine how the reservoir pressure will decline for a given offtake rate of reservoir fluids. A knowledge of this decline will assist in calculating the recovery factor, consistent with production engineering and economic constraints. All the mathematical tools necessary to perform such an exercise have already been presented; all that is necessary to consider is how to solve the various equations to explicitly determine the pressure. The basic equations are the reservoir material balance and the water influx equation. These can be solved simultaneously, by an iterative process, to give the reservoir pressure. To illustrate the method of solution the case of water influx into a gas reservoir will be considered for which the material balance is very simple and, as shown in Chapter 1, sec. 7, can be expressed as p e i i i G W E p p 1 1 Z Z G G = − − (1.41) NATURAL WATER INFLUX 329 in which the cumulative gas production Gp is constrained to meet a fixed market offtake rate. The methods of Hurst and van Everdingen and Fetkovitch will be described separately. a) Hurst and van Everdingen pressure pn-2 pn-1 n-2 n-1 n T time pn pi Fig. 9.18 Predicting the pressure decline in a water drive gas reservoir Fig. 9.18 illustrates the situation. Up to time level n - 1 everything has been determined and the water influx up to this point has been correctly included in the material balance. The next step is the determination of pn, the current reservoir pressure at the end of the nth time interval, that is at time T. The water influx is then ( ) n j n 1 e j D D D j 0 W U p W T t − = = ∆ − (9.17) which may be expanded as ( ) ( ) n j n 1 n 2 e j D D D n 1 D D D j 0 W U p W T t U p W T t − − − = = ∆ − + ∆ − (9.34) and, using equ. (9.16) n 2 n n 1 p p p 2 − − − ∆ = equ. (9.34) may be written as ( ) ( ) ( ) n j n 1 n 2 e j D D D n 2 n D D D j 0 U W U p W T t p p W T t 2 − − − = = ∆ − + − − (9.35) In this equation there are only two unknowns n e W and n p . These two are also related through the material balance n n p e i i i n G W E p p 1 1 Z Z G G = − − (9.36) A convenient way of solving equs. (9.35) and (9.36) is by the iterative method shown in fig. 9.19. The sequence of steps during any time period may be described as follows. NATURAL WATER INFLUX 330 k = 1 Time step = n n i n 2 p i i j D D D i n j G p E p 1 1 U p W (T t ) Z Z G G − =α = − − ∆ − k n pk n j n 1 n 2 e j D D D n 2 n D D D j 0 U W U p W (T t ) (p p )W (T t ) 2 − − − = = ∆ − + − − k k n e Wk k= 1 k+ -1 n n p p = k k n=n+1 n n p e i i i n G W E p p 1 1 Z Z G G = − − k k n pk k = 1 k 1 n n p p TOL − − − k k -+ k = iteration counter TOL = tolerance pressure difference (psi) 0 Fig. 9.19 Prediction of gas reservoir pressures resulting from fluid withdrawal and water influx (Hurst and van Everdingen) Make an initial estimate of the reservoir pressure pn k = pn 1 at the end of time step n by evaluating the material balance equation with the water influx initially set equal to ( ) n j n 2 1 e j D D D j 0 W U p W T t − = = ∆ − (9.37) i.e. neglecting the final term U∆pn-1 WD (TD – tDn-1) in equ. (9-35). Note: one could also use the value of n 1 e W −, the water influx after time step n-1, in the material balance but equ. (9.37) is usually closer to the actual influx and will lead to a smaller number of iterations. NATURAL WATER INFLUX 331 − Insert this initial estimate of pn 1, which is admittedly too small, in the water influx equation thus calculating a new value of Wen 1, which will now be too large. − Rework the material balance with this new value of n 1 e W , giving 1 n p which must now be too large. − Iterate finding n 2 2 e n (W ,p ), n 3 3 e n (W ,p ) . . . etc. until the difference between successive values of pn k is less than some tolerance level, at which point the procedure is repeated for the next time step. b) Fetkovitch The iterative solution technique can again be applied but in this case the equations used are (9.28) and (9.29). Aquifer material balance n-1 n 1 e a i ei W p p 1 W − = − Water influx ( ) ( ) n 1 n k k ei i n ei 12 a e n 1 n i W Jp t / W W p p p (1 e ) p − − − ∆ ∆ = − + − n j n n 1 k k e e e j 1 W W W − = = ∆ + ∆ Reservoir material balance n n k k p e i i i n G W E p p 1 1 Z Z G G = − − A flow chart showing the sequence of steps is included as fig. 9.20. An example illustrating this technique is presented in Fetkovitch's paper.2 NATURAL WATER INFLUX 332 Time step = n n 1 n 1 e i a ei W p p 1 W − − = − k = 1 n n 1 p e i i i G W E p p 1 1 Z Z G G − = − − k n n n 1 ei 1 e n 1 n a 2 i Jp t /W n i ei W W (p (p p )) 1 e p − − − ∆ ∆ = − + − k k n n 1 n e e e W W W − = + ∆ k k pn k = pn k-1 n=n 1 + pn k pn k k k = 1 + k = 1 k n n p e i i i G W E p p 1 1 Z Z G G = − − k k n -1 n n p p TOL − − k k k = iteration counter TOL = tolerance pressure difference (psi) -+ Fig. 9.20 Prediction of gas reservoir pressures resulting from fluid withdrawal and water influx (Fetkovitch) NATURAL WATER INFLUX 333 9.6 APPLICATION OF INFLUX CALCULATION TECHNIQUES TO STEAM SOAKING The method of predicting aquifer performance, using the unsteady state theory of Hurst and van Everdingen presented in the previous section, is not necessarily restricted in use to the description of reservoir-aquifer systems. The same technique can be used to predict fluid influx in any system which has the same geometry as the reservoir-aquifer model described in this chapter. As an example, the method will be used to determine the oil producing rate during the early, transient phase of a steam soak cycle. This subject was discussed in Chapter 6, sec. 4, in which an expression was derived for the PI ratio increase during the later stabilized flow part of the cycle. Immediately upon opening the well on production, however, there will be a period when transient flow conditions prevail, that is, before the effect of the boundary of the drainage volume has been felt. The initial situation is shown in fig. 9.21, in which, after injecting several thousand tons of steam, a hot zone of radius rh is created around the well in which the temperature is Ts, the condensing steam temperature at the prevailing reservoir pressure. cold hot Ts µoh T , r µoc rw rh r Fig. 9.21 Conditions prior to production in a steam soak cycle The radius of the hot zone can be calculated using the technique of Marx and Langenheim7, which allows for heat losses to both cap and base rock during the steam injection. This simplified description of the temperature distribution is justified by the experimental findings of Niko and Troost8, which indicate that the dominant factor in the production cycle is the total amount of heat injected into the formation, the additional oil production being largely independent of the way in which the temperature is distributed. The overall geometry of this system is the same as for an aquifer producing into a reservoir, only on a smaller scale. During the transient flow period there will be an influx of cold oil into the heated region which can be described using the water influx calculation procedures al ready developed in this chapter. If the transient period is divided into small, equal time steps of length ∆t, then it is required to calculate the average oil rate n q during the nth time step and, assuming steady state flow at all times within the small volume of the hot zone close to the well, then NATURAL WATER INFLUX 334 ( ) n wf,n o n h oh w 2 k h p p q r ln r π µ − = (9.38) which is a re-formulation of the steady state inflow equation presented in table 6.1. The oil viscosity, µoh is a function of the average temperature in the hot zone during the nth time interval. The temperature will decline continuously during production as heat is lost by conduction to the cap and base rock and by convection through the produced fluids. A simple method for predicting the temperature decline allowing for both effects has been presented by Boburg and Lantz9. The pressure n p in equ. (9.38) is the average pressure during the time step at rh, the outer boundary of the hot zone, and wf,n p the average wellbore pressure during the same period. If it is assumed that the pressure declines at rh and rw can be approximated by a series of discrete pressure steps then, in accordance with equ. (9.15), n p and wf,n p can be expressed as ( ) 12 n n 1 n p p p − = + (9.39) and ( ) 12 wf,n wf,n 1 wf,n p p p − = + (9.40) where pn and pwf,n are the pressures at rh and rw respectively at the end of the nth time step. The cumulative oil influx across the boundary rh by the end of the time step will be n p,n p,n 1 N N q . t − = + ∆ in which Np,n-l is the known influx at the end of the (n - 1)th time step. Using equs. (9.39) and (9.40), the influx can be expressed as ( ) p,n p,n 1 n 1 n wf,n 1 wf,n N N p p p p 2 α − − − = + + − − (9.41) where o h oh w 2 k h t r ln r π α µ ∆ = An expression for the cumulative oil influx can also be obtained using the unsteady state influx theory of Hurst and van Everdingen in the manner similar to that of equ. (9.34) ( ) ( ) j n 1 n 2 p,n j D D D n 1 D D D j 0 N U p W T t U p W T t − − − = = ∆ − + ∆ − (9.42) in which T = n. ∆t and WD is the dimensionless influx function of Hurst and van Everdingen, figs. 9.3-9.7. U is the aquifer constant defined in equ. (9.6), only in this case applied to the oil reservoir for which f = 1, and the radius ro is replaced by rh. NATURAL WATER INFLUX 335 Values of the pressure drop ∆pj at the hot zone boundary can be evaluated using equ. (9.16) as ( ) 12 i j 1 j 1 p p p − + ∆ = − (9.16) and in particular ( ) 12 n 1 n 2 n p p p − − ∆ = − therefore, equ. (9.42) can be written as ( ) ( ) j n 2 p,n j D D D n 2 n j 0 N U p W T t p p 2 β − − = = ∆ − + − (9.43) where β = U WD (TD−tDn -1) Equations (9.41) and (9.43) can now be equated and solved explicitly for the boundary pressure pn at rh at the end of the nth time step, i.e. ( ) ( ) ( ) j n 2 n j D D D p,n 1 n 2 wf,n 1 wf,n n 1 j 0 1 p 2U p W T t 2N p p p p β α α β − = − − − − = ∆ − − + + + − + (9.44) Providing that the manner in which the bottom hole flowing pressure will be allowed to decline is specified, then everything on the right hand side of equ. (9.44) is determined, thus pn can be calculated and subsequently n p using equ. (9.39). Finally, substituting this value of n p in equ.(9.38) will lead to the determination of n q . In fact, it is necessary to iterate on this oil rate to correctly model the heat loss due to convection, since, as mentioned already, the temperature in the hot zone, which determines the oil viscosity in equ. (9.38), is reduced due to the removal of heat by the produced oil. This convective heat loss is directly proportional to the production rate and therefore, the oil rate and viscosity are dependent upon one another necessitating the iterative solution technique. Bentsen and Donohue10 have reported the use of the above technique incorporated in a dynamic programming model for optimising steam soak operations. REFERENCES 1) van Everdingen, A.F. and Hurst, W., 1949.TheApplication of the Laplace Transformation to Flow Problems in Reservoirs. Trans. AIME. 186: 305-324. 2) Fetkovitch, M.J., 1971. A Simplified Approach to Water Influx Calculations-Finite Aquifer Systems. J.Pet.Tech., July: 814-828. 3) van Everdingen, A.F., Timmerman, E.H. and McMahon, J.J., 1953. Application of the Material Balance Equation to a Partial Water-Drive Reservoir. Trans. AIME. 198:51. 4) Havlena, D. and Odeh, A.S., 1963. The Material Balance as an Equation of a Straight Line. J.Pet.Techn, August: 896-900. Trans. AIME. NATURAL WATER INFLUX 336 5) Havlena, D. and Odeh, A.S.,1964. The Material Balance as an Equation of a Straight Line-Part II, Field Cases. J.Pet.Tech., July: 815-822. Trans. AIME. 6) Schilthuis, R.J., 1936. Active Oil and Reservoir Energy. Trans. AIME. 118: 37. 7) Marx, J.W. and Langenheim, R.H., 1959. Reservoir Heating by Hot Fluid Injection. Trans. AIME: 118:37. 8) Niko, H. and Troost, P.J.P.M., 1971. Experimental Investigation of Steam Soaking in a Depletion-Type Reservoir. J.Pet.Tech., August: 1006-1014. Trans. AIME. 9) Boberg, T.C. and Lantz, R.B., Jr., 1966. Calculation of the Production Rate of a Thermally Stimulated Well. J.Pet.Tech., December: 1613-1623. 10) Bentsen, R.G. and Donohue, D.A.T., 1969. A Dynamic Programming Model of the Cyclic Steam Injection Process. J.Pet.Tech., December: 1582-1596. CHAPTER 10 IMMISCIBLE DISPLACEMENT 10.1 INTRODUCTION This chapter describes how to calculate the oil recovery resulting from displacement by an immiscible (non-mixing) fluid which is primarily taken to be water. After considering several basic assumptions, the subject is introduced in the conventional manner by describing fractional flow and the Buckley-Leverett equation. Since the latter is one dimensional its direct application, in calculating oil recovery, is restricted to cases in which the water saturation distribution is uniform with respect to thickness. In more practical cases where there is a non-uniform saturation distribution, defined, for instance, by the assumption of vertical equilibrium, then it is necessary to generate relative permeabilities, which are functions of the thickness averaged water saturation, for use in conjunction with the Buckley-Leverett theory. This has the effect of reducing two dimensional problems to one dimension. The remainder of the chapter concentrates on the theme of generating these averaged functions, for various assumed flow conditions in homogeneous and layered reservoirs, and describes their application in numerical reservoir simulation. 10.2 PHYSICAL ASSUMPTIONS AND THEIR IMPLICATIONS Before undertaking to describe the mechanics of displacement, it is first necessary to consider some of the basic physical assumptions which will later be incorporated in the simple mathematical description of the process. The implications of each assumption are described in detail. a) Water is displacing oil in a water wet reservoir When two immiscible fluids, such as oil and water, are together in contact with a rock face the situation is as depicted in fig. 10.1. The angle Θ, measured through the water, is called the contact angle. If Θ < 90° the reservoir rock is described as being water wet, whereas if Θ > 90° it is oil wet. The wettability, as defined by the angle Θ, is a measure of which fluid preferentially adheres to the rock. The two dynamic situations shown in fig. 10.1 (a) and (b) are described as (a) Imbibition; in which the wetting phase saturation is Increasing and (b) Drainage; in which the wetting phase saturation is Decreasing. It has been determined experimentally that the contact angle is larger when the wetting phase is advancing over the rock face than when retreating and this difference is described as the hysteresis of the contact angle. IMMISCIBLE DISPLACEMENT 338 OIL OIL WATER (a) WATER (b) Θ Θ Fig. 10.1 Hysteresis in contact angle in a water wet reservoir, (a) wetting phase increasing (imbibition); (b) wetting phase decreasing (drainage) Whether the majority of reservoirs are water wet, oil wet, or of intermediate wettability (Θ ≈ 90°) is still very much a matter of debate and research. It is argued by many that since all sands were initially saturated with water (water wet), prior to the migration of hydrocarbons into the reservoir traps, then this initial wettability should be retained. In this chapter it will also be assumed that the reservoirs described are water wet. This is not necessarily because this author is convinced by the above argument but more for the sake of uniformity. The displacement of oil by water in a water wet reservoir is, therefore, an imbibition process. As such, the capillary pressure curve and relative permeabilities used in the description of the displacement must be measured under imbibition conditions. Conversely, the displacement of oil by water in an oil wet reservoir would be a drainage process and require capillary pressures and relative permeabilities measured under drainage conditions. There is a basic difference between the two due to the hysteresis of contact angle1,2. The fact that the oil and water are immiscible is also important. When such fluids are in contact a clearly defined interface exists between them. The molecules near the interface are unevenly attracted by their neighbours and this gives rise to a free surface energy per unit area or interfacial tension. If the interface is curved the pressure on the concave side exceeds that on the convex and this difference is known as the capillary pressure. The general expression for calculating the capillary pressure at any point on an interface between oil and water is given by the Laplace equation c o w 1 2 1 1 P p p r r σ = − = + (10.1) where Pc = the capillary pressure (absolute units) σ = the interfacial tension and r1 and r2 = the principal radii of curvature at any point on the interface where the pressures in the oil and water are po and pw, respectively. There is also a sign convention that the radii are positive if measured in the oil and negative if measured in the water, in a water wet system. IMMISCIBLE DISPLACEMENT 339 Consider a volume of water contained between two spherical sand grains, in a water wet reservoir, as shown in fig. 10.2. r1 r2 ROCK WATER OIL x Fig. 10.2 Water entrapment between two spherical sand grains in a water wet reservoir In applying equ. (10.1) to calculate the capillary or phase pressure difference at point X on the interface, one of the principal radii of curvature, say, r1, is positive, since it is measured through the oil, while the other, r2, which is measured through the water, is negative. Since r1 < r2, however, the capillary pressure is positive. What is also evident from fig. 10.2 is the fact that as the volume of water (water saturation) decreases, the radii decrease, and therefore there must be some inverse relationship between Pc and Sw. This relationship is called the capillary pressure curve (function) and is measured routinely in the laboratory. Typically, such experiments are conducted, for convenience, using air-brine or air-mercury fluid combinations and the resulting capillary pressure curve converted for the oil-water system in the reservoir3,4. For the sake of consistency, however, a hypothetical experiment will be considered, firstly for the (non-commercial) case of oil displacing water, in a water wet core sample, and then the displacement process will be reversed. The results of such an experiment are shown in fig. 10.3. Pc B C A DRAINAGE IMBIBTION Swc 1 - Sor 100% C S (% OF PORE VOLUME) w Fig. 10.3 Drainage and imbibition capillary pressure functions IMMISCIBLE DISPLACEMENT 340 Starting at point A, with the core sample 100% saturated with water, the water is displaced by the oil, which is a drainage process. If the difference in phase pressures (imposed pressure differential) is plotted as a function of the decreasing water saturation the result would be the dashed line shown in fig. 10.3, the capillary pressure drainage curve. At the connate water saturation (point B) there is an apparent discontinuity at which the water saturation cannot be reduced further, irrespective of the imposed difference in phase (capillary) pressure. If the experiment is reversed, by displacing the oil with water, the result would be the imbibition curve shown as the solid line in fig. 10.3. The drainage and imbibition plots differ due to the hysteresis in contact angle, the latter being the one required for the displacement described in this chapter. When the water saturation has risen to its maximum value Sw = 1 − Sor the capillary pressure is zero (point C). At this point the residual oil saturation, Sor, cannot be reduced, irrespective of the pressure difference applied between the water and oil (Pc−negative). The capillary pressure curve can also be interpreted in terms of the elevation of a plane of constant water saturation above the level at which Pc = 0. The analogy is usually drawn between capillary rise in the reservoir and the simple laboratory experiment, shown in fig. 10.4, performed with oil and water, the latter being the wetting phase. At the level interface, application of equ. (10.1) for infinite r1 and r2 indicates that Pc = 0 and therefore at this point po = pw = p. The water will rise in the capillary tube until it reaches a height H, above the level interface, when capillary-gravity (hydrostatic) equilibrium is achieved. If po and pw are the oil and water pressures on opposite sides of the curved interface then, in absolute units po + ρogH = p and Oil Water oil water Pressure elevation Pc po pw r R H p = p o w c = p(P = 0) capillary tube Θ Fig. 10.4 Capillary tube experiment for an oil-water system po + ρwgH = p which, on subtraction, give IMMISCIBLE DISPLACEMENT 341 o w c p p P gH ρ − = = ∆ (10.2) where ∆ρ = ρw - ρo. Furthermore, considering in detail the geometry at the interface in the capillary tube, fig. 10.4. If the curvature is approximately spherical with radius R, then in applying the Laplace equation, (10.1), r1 = r2 = R at all points on the interface. Also if r is the radius of the capillary tube, then r = RcosΘ and therefore o w c 2 cos p p P gh r σ ρ Θ − = = = ∆ (10.3) This equation is frequently used to draw a comparison between simple capillary rise experiments, as described above, and capillary rise in the reservoir, the porous tracts in the latter being likened to a collection of capillary tubes with different radii. In this comparison it can be seen that the capillary rise of water will be greater for small r, equ. (10.3), and will decrease as r increases. The decrease in capillary rise will be apparently a continuous function due to the continuous range of pore capillaries in the reservoir and will define, in fact, the capillary pressure-saturation relation. This argument is frequently applied to consider the static case of water distribution above the 100% water saturation level in the reservoir, under initial conditions, for which the drainage capillary pressure curve is required. There will generally be no sharp interface between the water and oil but, rather, a zone in which saturations decrease with height above the 100% water saturation level, at which Pc = 0, in accordance with the capillary pressure (capillary rise)-saturation relationship. The vertical distance between the point at which Sw = 100%, Pc = 0 and Sw = Swc is called the capillary transition zone and is denoted by H. This chapter is more concerned with the effect of capillary pressure on the displacement of oil by water, for which the imbibition capillary pressure curve is relevant (Pc = 0 at Sw = 1 − Sor). Consider the static situation during a water drive shown in fig. 10.5. WATER z A S = 1 - S w or y X θ z = y cos θ Pc Swc Swa Sw 1 - Sor ∆ρgz OIL Fig. 10.5 Determination of water saturation as a function of reservoir thickness above the maximum water saturation plane, Sw = 1− − − −Sor, of an advancing waterflood IMMISCIBLE DISPLACEMENT 342 By static it is meant that the downdip water injection is stopped when the maximum water saturation plane, Sw = 1 − Sor, at which Pc = 0, has just reached point X in the linear displacement path. If the imbibition capillary pressure curve has a distinct capillary transition zone, as shown on the right hand side of the diagram, then above point X the water saturation will be distributed in accordance with the capillary pressure (capillary rise)-saturation relationship. In particular, the capillary pressure at point A, a distance y above the base of the reservoir, in the dip-normal direction (normal to the flow direction), can be calculated as c w o w P (S ) = p p = g cos ρ γ θ − ∆ (10.4) from which the saturation at point A can be determined from the capillary pressure curve, as shown in fig. 10.5. Equation (10.4) is referred to, in this text, as the capillary pressure equation expressed, in this case, in absolute units. In Darcy units, which are used in this chapter to develop theoretical arguments, the general equation becomes c w 6 g y cos P (S ) (atm) 1.0133 10 ρ θ ∆ = × (10.5) while in the field units defined in table 4.1, which are employed in the exercises ( ) c w P S 0.4335 ycos (psi) γ θ = ∆ (10.6) where ∆γ is the difference between the water and oil specific gravities in the reservoir. Allowing the Sw = 1 − Sor plane to rise incrementally in fig. 10.5 will result in a different water saturation distribution, in the dip-normal direction, at point X in the displacement path. This concept is applied in sec. 10.7 in which dynamic displacement is viewed as a series of static positions of the Sw = 1 − Sor plane as the flood moves through the reservoir, each position leading to a new water saturation distribution dictated by the capillary pressure-saturation relationship. b) Displacement generally occurs under conditions of vertical equilibrium Coats5 has qualitatively explained the concept of vertical equilibrium by drawing an analogy with a simple problem in heat conduction. If one were trying to mathematically describe heat conduction in a thin metal plate, say, 1/8th of an inch thick and with an area of several square feet, no allowance would be made for the heat distribution across the thickness of the metal, in which direction thermal equilibrium would be assumed. Since reservoirs, typically, have dimensions in proportion to those described for the metal plate, displacement problems can be frequently tackled in a similar manner. In this case, however, the assumption made is that of fluid potential equilibrium across the thickness of the reservoir. The condition for fluid potential equilibrium is simply that of hydrostatic equilibrium, discussed previously, for which the saturation distribution can be determined as a function of capillary pressure and, therefore, height, as IMMISCIBLE DISPLACEMENT 343 ( ) c w 6 g y cos P S 1.0133 10 ρ θ ∆ = × (10.5) that is, the fluids are distributed in accordance with capillary-gravity equilibrium. The vertical equilibrium condition can therefore be interpreted in the following manner. When, during the displacement of oil by water, the water saturation at any point in the reservoir increases by a small amount, the new water saturation is instantaneously redistributed as indicated by equ. (10.5). This means that the vertical velocities of oil and water, as the two are redistributed in accordance with capillary-gravity equilibrium, appear to be infinite in comparison with the velocity of fluid movement parallel to the reservoir bedding planes resulting from the Darcy or viscous forces. The condition of vertical equilibrium will be promoted by -a large vertical permeability (kv) -small reservoir thickness (h) -large density difference between the fluids (∆ρ) -high capillary forces (large capillary transition zone H) -low fluid viscosities -low injection rates. Coats5,6 has presented two dimensionless groups, relating the above terms, the magnitudes of which can be used as "rough rules of thumb" for deciding whether vertical equilibrium conditions prevail in the reservoir. The two cases considered can be applied when the capillary transition zone is large and also when it is negligible. These dimensionless groups are not presented in this text since, irrespective of their magnitude, the only way to check the validity of the vertical equilibrium is by using the numerical simulation techniques described in sec. 10.10. In any case, when applying simple, analytical techniques to describe the displacement process one is obliged to assume that either vertical equilibrium is valid or else the complete opposite, that there is a total lack of vertical equilibrium. The latter case will apply when, for instance, the injection rate is so high that the water and oil velocities, parallel to the bedding planes are much greater than their velocity components in the dip-normal direction. Under these circumstances the water saturation will be uniformly distributed with respect to thickness. These two extremes both represent conditions under which the saturation profile in the dip-normal direction is definable and this facilitates the application of the analytical techniques described in this chapter. For in-between cases the engineer must resort to numerical simulation techniques (refer sec. 10.10). It has been found that the vertical equilibrium condition is approximately satisfied in a great many reservoirs. Nevertheless, it will be repeatedly stated throughout the remainder of the chapter precisely when this condition is being assumed, and when not. IMMISCIBLE DISPLACEMENT 344 c) The displacement is considered as incompressible This assumption implies that steady state conditions prevail in the reservoir with the pressure at any point remaining constant. There must, of course, be a pressure differential between injection and production wells but the variation in the pressure dependent variables, viscosities and densities, resulting from this differential is ignored. This type of displacement will occur if t o w i q q q q = + = (10.7) where qt = total flow rate (r.vol/time) qo = oil " " " qw = water " " " qI = water injection rate The assumption is quite realistic since the engineer has control over the displacement process to a much greater extent than, say, during the volumetric depletion of a reservoir. Therefore, the wells and surface facilities are usually designed for constant rate injection/production for it makes little sense to do otherwise. In addition, from a reservoir engineering point of view, there are definite advantages in maintaining the pressure at a constant level above that at which the solution gas first becomes mobile (refer Chapter 3, sec. 5). Because of this assumption the methods described in this chapter will be equally appropriate for the description of the displacement of oil by gas at constant pressure with no mass transfer between the phases. d) The displacement is considered to be linear Throughout the chapter displacement will be considered exclusively in a linear prototype reservoir model, as shown in fig. 10.6. Injection Production w L (a) (b) Z θ x y h qi qt Fig. 10.6 Linear prototype reservoir model, (a) plan view; (b) cross section IMMISCIBLE DISPLACEMENT 345 The model represents a symmetry element taken from a line drive pattern. The co-ordinates used in describing displacement in the linear cross section are shown in fig. 10.6(b). Both the injection and production wells are considered to be perforated across the entire formation thickness, in the dip-normal direction. No account is taken of the distortion of the linear flow streamlines (lines of constant fluid potential) in the vicinity of the wells and saturations are assumed to be uniformly distributed across the entire width of the block, that is, normal to plane shown in fig. 10.6(b). The main concern in this chapter is, therefore, to account for the fluid saturation distributions in the dip-normal direction (y-direction) as the flood moves through the linear reservoir block. No analytical methods are presented to account for the areal distribution of saturations in the reservoir. Such methods do exist for regular grid spacings of injection and production wells and are described in the Craig monograph1. For irregular well spacing, however, the analytical methods are extremely complex and have largely been superseded by numerical simulation techniques. In fact, one of the main purposes in using simulators is to determine the areal distribution of oil and water (or displacing fluid in general) resulting from a flood. This knowledge enables the engineer to place injection and production wells to gain the maximum recovery. To provide such results, however, it is necessary that the simulator be informed of what can be expected to occur in the dip-normal direction, that is, how the fluids will be distributed in this direction. This information is generally provided as input to the simulation. The whole intent of this chapter is, therefore, to describe the physics governing displacement, as viewed through a linear cross section of the reservoir, which is of vital concern if the areal distributions of fluids are to be correctly modelled. The chapter describes, in the first place, displacement in a homogeneous linear section, secs. 10.3-7, and subsequently extends the methods developed to the description of displacement in inhomogeneous (layered) reservoirs. 10.3 THE FRACTIONAL FLOW EQUATION In this, and the following two sections, oil displacement will be assumed to take place under the so-called diffuse flow condition. This means that fluid saturations at any point in the linear displacement path are uniformly distributed with respect to thickness. The sole reason for making this assumption is that it permits the displacement to be described, mathematically, in one dimension and this provides the simplest possible model of the displacement process. The one dimensional description follows from the fact that since the water saturation is uniformly distributed in the dip-normal direction then so too are the relative permeabilities to oil and water, which are themselves functions of the water saturation at any point. This means that the simultaneous flow of oil and water can be modelled using thickness averaged relative permeabilities, along the centre line of the reservoir, which are also equivalent to relative permeabilities at any point throughout the thickness. IMMISCIBLE DISPLACEMENT 346 The diffuse flow condition can be encountered under two extreme physical conditions: a) when displacement occurs at very high injection rates so that, as described in sec. 10.2, the condition of vertical equilibrium is not satisfied and the effects of the capillary and gravity forces are negligible, and b) for displacement at low injection rates in reservoirs for which the measured capillary transition zone greatly exceeds the reservoir thickness (H >> h) and the vertical equilibrium condition applies. The latter case can be visualized by considering the capillary pressure curve, fig. 10.7. Since, H >> h then it will appear that the water saturation is, to a first approximation, uniformly distributed with respect to thickness in the reservoir. H Pc Swc Sw 1 - Sor RESERVOIR THICKNESS h Fig. 10.7 Approximation to the diffuse flow condition for H>> h It should also be noted that relative permeabilities are measured in the laboratory under the diffuse flow condition. This normally results from displacing one fluid by another, in thin core plugs, at high flow rates3. As such, the laboratory, or rock relative permeabilities, must be regarded as point relative permeabilities which are functions of the point water saturation in the reservoir. It is, therefore, only when describing displacement, under the diffuse flow condition, that rock relative permeabilities can be used directly in calculations since, in this case, they also represent the thickness averaged relative permeabilities. Consider then, oil displacement in a tilted reservoir block, as shown in fig. 10.6(b), which has a uniform cross sectional area A. Applying Darcy's law, for linear flow, the one dimensional equations for the simultaneous flow of oil and water are ro o o ro o o o 6 o o kk A kk A p g sin q x x 1.0133 10 ρ ρ θ µ µ ∂Φ ∂ = − = − + ∂ ∂ × and rw w w rw w w w 6 w w kk A kk A p g sin q x x 1.0133 10 ρ ρ θ µ µ ∂Φ ∂ = − = − + ∂ ∂ × IMMISCIBLE DISPLACEMENT 347 By expressing the oil rate as qo = qt - qw the subtraction of the above equations gives o t o c rw w 6 rw ro ro q P g sin q A kk kk kk x 1.0133 10 µ µ µ ρ θ ∂ ∆ = − + = + − ∂ × (10.8) in which c o w P p p x x x ∂ ∂ ∂ = − ∂ ∂ ∂ the capillary pressure gradient in the direction of flow, and ∆ρ = ρw - ρo The fractional flow of water, at any point in the reservoir, is defined as w w w o w t q q f q q q = = + and substitution of this in equ. (10.8) gives ro c 6 t o w ro w rw o kk A P g sin 1 q x 1.0133 10 f k 1 k ρ θ µ µ µ ∂ ∆ + − ∂ × = + ⋅ (10.9) while, by analogy with equ. (4.18), this equation can be expressed in field units as 3 ro c t o w ro w rw o kk A P 1 1.127 10 .4335 sin q x f k 1 k γ θ µ µ µ − ∂ + × − ∆ ∂ = + ⋅ (10.10) both of these being fractional flow equations for the displacement of oil by water, in one dimension. It is worthwhile considering the influence of the various component parts of this expression. According to the convention adopted in this text θ is the angle measured from the horizontal to the line indicating the direction of flow. Therefore, the gravity term ∆ρ g sinθ /1.0133×106 will be positive for oil displacement in the updip direction (0 < θ < π), as shown in fig. 10.6(b), and negative for displacement downdip (π < θ < 2π). As a result, provided all the other terms in equ. (10.9) are the same, the fractional flow of water for displacement updip is lower than for displacement downdip since in the former case gravity tends to suppress the flow of water. IMMISCIBLE DISPLACEMENT 348 The effect of the capillary pressure gradient term is less obvious but can be qualitatively understood by expressing the gradient as c c w w P dP S x dS x ∂ ∂ = ⋅ ∂ ∂ (10.11) The first term on the right hand side is the slope of the capillary pressure curve, fig. 10.8(a), and is always negative. The second term is the slope of the water saturation profile in the direction of flow, a typical profile being shown in fig. 10.8(b). Swc Sw 1 - Sor Swc Sw 1 - Sor +dSw -dPc Pc (a) -dSw +dx Swf (b) x Fig. 10.8 (a) Capillary pressure function and; (b) water saturation distribution as a function of distance in the displacement path From this it can be seen that ∂Sw/∂x is also negative. Therefore ∂Pc/∂x is always positive and consequently the presence of the capillary pressure gradient term tends to increase the fractional flow of water. Quantitatively, it is difficult to allow for the capillary pressure gradient for, although the capillary pressure curve may be available, the water saturation profile, fig. 10.8(b), is unknown and, as will be shown presently, is the required result of displacement calculations. The water saturation distribution shown in fig. 10.8(b), corresponding to the situation after injecting a given volume of water, may be regarded as typical for the displacement of oil by water. The diagram shows that there is a distinct flood front, or shock front, at which point there is a discontinuity in the water saturation which increases abruptly from Swc to Swf, the flood front saturation. It is at this shock front where both derivatives on the right hand side of equ. (10.11) have their maximum value, which is evident by inspection of fig. 10.8(a) and (b), and therefore ∂Pc/∂x is also maximum. behind the flood front there is gradual increase in saturations from Swf up to the maximum value 1 - Sor. In this region it is normally considered that both dPc/dSw and ∂Sw/∂x are small and therefore ∂Pc/∂x can be neglected in the fractional flow equation. For displacement in a horizontal reservoir (sinθ = 0), and neglecting, for the moment, the capillary pressure gradient, the fractional flow equation is reduced to IMMISCIBLE DISPLACEMENT 349 w ro w rw o 1 f k 1 k µ µ = + ⋅ (10.12) Provided the oil displacement occurs at a constant temperature then the oil and water viscosities have fixed values and equ. (10.12) is strictly a function of the water saturation, as related through the relative permeabilities. For a typical set of relative permeabilities, as shown in fig. 4.8, the fractional flow equation, (10.12), usually has the shape indicated in fig. 10.9, with saturation limits Swc and 1 − Sor, between which the fractional flow increases from zero to unity. The manner in which the shape of this curve is influenced by the viscosity ratio of oil to water will be studied in exercise 10.1. Swc Sw 1 - Sor fw f = 1 w Fig. 10.9 Typical fractional flow curve as a function of water saturation, equ. (10.12) The fractional flow equation is used to calculate the fraction of the total flow which is water, at any point in the reservoir, assuming the water saturation at that point is known. Precisely how to determine when a given water saturation plane reaches a particular point in the linear system requires the application of the displacement theory presented in the following section. 10.4 BUCKLEY-LEVERETT ONE DIMENSIONAL DISPLACEMENT In 1942 Buckley and Leverett presented what is recognised as the basic equation for describing immiscible displacement in one dimension7. For water displacing oil, the equation determines the velocity of a plane of constant water saturation travelling through a linear system. Assuming the diffuse flow condition, the conservation of mass of water flowing through volume element A dx, φ fig. 10.10, may be expressed as Mass flow rate = Rate of increase of mass In - Out in the volume element w w w w w w x dx x q q A dx ( S ) t ρ ρ φ ρ + ∂ − = ∂ (10.13) IMMISCIBLE DISPLACEMENT 350 q \| wρw x q \| ρ w w x+dx x dx Fig. 10.10 Mass flow rate of water through a linear volume element A dx φ or w w w w w w w w x x q q (q )dx A dx ( S ) x t ρ ρ ρ φ ρ ∂ ∂ − + = ∂ ∂ which can be reduced to w w w w (q ) A ( S ) x t ρ φ ρ ∂ ∂ = − ∂ ∂ (10.14) and for the assumption of incompressible displacement (ρw ≈ constant) w w t x q S A x t φ ∂ ∂ = − ∂ ∂ (10.15) The full differential of the water saturation is x w w w t S S dS dx dt x t ∂ ∂ = + ∂ ∂ and since it is the intention to study the movement of a plane of constant water saturation, that is, dSw = 0, then w w w x t S S S dx t x dt ∂ ∂ = − ∂ ∂ (10.16) Furthermore, t w w w t w q q S x S x ∂ ∂ ∂ = ⋅ ∂ ∂ ∂ (10.17) and substituting equs. (10.16) and (10.17) in equ. (10.15) gives w w t S w q dx A S dt φ ∂ = ∂ (10.18) Again, for incompressible displacement, qt is constant and, since qw = qtfw, equ. (10.18) may be expressed as IMMISCIBLE DISPLACEMENT 351 w w w t w S S S w q df dx v dt A dS φ = = (10.19) This is the equation of Buckley-Leverett which implies that, for a constant rate of water injection (qt = qi), the velocity of a plane of constant water saturation is directly proportional to the derivative of the fractional flow equation evaluated for that saturation. If the capillary pressure gradient term is neglected in equ. (10.9) then the fractional flow is strictly a function of the water saturation, irrespective of whether the gravity term is included or not, hence the use of the total differential of fw in the Buckley-Leverett equation. Integrating for the total time since the start of injection gives w t w S t w 0 df 1 x q dt A dS φ = or w w i w S S w W df x A dS φ = (10.20) where Wi is the cumulative water injected and it is assumed, as an initial condition, that Wi = 0 when t = 0 . Therefore, at a given time after the start of injection (Wi = constant) the position of different water saturation planes can be plotted, using equ. (10.20), merely by determining the slope of the fractional flow curve for the particular value of each saturation. There is a mathematical difficulty encountered in applying this technique which can be appreciated by considering the typical fractional flow curve shown in fig. 10.9 in conjunction with equ. (10.20). Since there is frequently a point of inflexion in the fractional flow curve then the plot of dfw/dSw versus Sw will have a maximum point, as shown in fig. 10.11 (a). Using equ. (10.20) to plot the saturation distribution at a particular time will therefore result in the solid line shown in fig. 10.111b). This bulbous saturation profile is physically impossible since it indicates that multiple water saturations can co-exist at a given point in the reservoir. What actually occurs is that the intermediate values of the water saturation, which as shown in fig. 10.11 (a) have the maximum velocity, will initially tend to overtake the lower saturations resulting in the formation of a saturation discontinuity or shock front. IMMISCIBLE DISPLACEMENT 352 1 - Sor Sw Swc B A x Swc Sw 1 - Sor w w s w df v dS (a) (b) Swf ∝ Fig. 10.11 (a) Saturation derivative of a typical fractional flow curve and (b) resulting water saturation distribution in the displacement path Because of this discontinuity the mathematical approach of Buckley-Leverett, which assumes that Sw is continuous and differentiable, will be inappropriate to describe the situation at the front itself. Behind the front, however, in the saturation range Swf < Sw < 1−Sor where Swf is the shock front saturation, equs. (10.19) and (10.20) can be applied to determine the water saturation velocity and position. Furthermore, in this saturation range the capillary pressure gradient is usually negligible, as noted in the previous section, and the fractional flow equation to be used in equs. (10.19) and (10.20) is simply w ro w rw o 1 f k 1 k µ µ = + ⋅ (10.12) in a horizontal reservoir, or ro 6 t o w ro w rw o kk A g sin 1 q 1.0133 10 f k 1 k ρ θ µ µ µ ∆ − × = + ⋅ (10.21) in a dipping reservoir. To draw the correct water saturation profile using the Buckley-Leverett technique requires the determination of the vertical dashed line, shown in fig. 10.11(b), such that the shaded areas A and B are equal. The dashed line then represents the shock front saturation discontinuity. A more elegant method of achieving the same result was presented by Welge in 19528, This consists of integrating the saturation distribution over the distance from the injection point to the front, thus obtaining the average water saturation behind the front w S , as shown in fig. 10.12. IMMISCIBLE DISPLACEMENT 353 x1 x x2 Swf 1 - Sor Sw Swc Sw Fig. 10.12 Water saturation distribution as a function of distance, prior to breakthrough in the producing well The situation depicted is at a fixed time, before water breakthrough in the producing well, corresponding to an amount of water injection Wi. At this time the maximum water saturation, Sw = 1 - Sor, has moved a distance x1, its velocity being proportional to the slope of the fractional flow curve evaluated for the maximum saturation which, as shown in figs. 10.9 and 10.11 (a), is small but finite. The flood front saturation Swf is located at position x2 measured from the injection point. Applying the simple material balance w i 2 wc W x A (S S ) φ = − or i w wc 2 W S S x Aφ − = and using equ. (10.20) which is applicable up to the flood front at x2, then wf i w wc 2 w w S W 1 S S x A df dS φ − = = (10.22) An expression for the average water saturation behind the front can also be obtained by direct integration of the saturation profile as 2 1 X or 1 w X W 2 (1 S )x S dx S x − + = (10.23) and again since w w w S S w df x dS ∝ for a given volume of injected water, and for Sw ≥ Swf, then equ. (10.23) can be expressed as IMMISCIBLE DISPLACEMENT 354 wf or or wf S w w or w 1 S w w 1 S w w S w df df (1 S ) S d dS dS S df dS − − − + = (10.24) The integral in the numerator of this equation can be evaluated using the method of integration by parts, i.e. ∫ udv = uv − ∫ vdu to give [ ] S S wf wf wf 1 S 1 S or or or S w w w w w w w 1 S df df S d S f dS dS − − − = − and substituting this in equ. (10.24) and cancelling terms gives ( ) wf Swf w w wf w S w df S S 1 f dS = + − (10.25) in which both fw and its derivative are evaluated for the shock front saturation Swf. Finally, equating (10.22) and (10.25) gives wf Swf w w w w S w wf we (1 f ) df 1 dS S S S S − = = − − (10.26) The significance of this result is illustrated in fig.10.13. To satisfy equ. (10.26) the tangent to the fractional flow curve, from the point Sw = Swc; fw = 0, must have a point of tangency with co-ordinates wf wf w w w S S S ;f f = = , and the extrapolated tangent must intercept the line fw = 1 at the point w w w S S ;f 1. = = 1 - Sor Swc Sw Swf, fwSwf fw fw = 1 Sw Fig. 10.13 Tangent to the fractional flow curve from Sw = Swc IMMISCIBLE DISPLACEMENT 355 This method of determining Swf, Swf w w f and S , requires that the fractional flow curve be plotted, using either equ. (10.12) or equ. (10.21), for the entire water saturation range Swc < Sw < 1 – Sor As noted previously, the use of either of these equations ignores the effect of the capillary pressure gradient, ∂Pc/∂x. This neglect, however, is only admissible behind the flood front for Swf < Sw < 1 – Sor The part of the fractional flow curve for saturations less than Swf is, therefore, virtual and the first real point on the curve has the co-ordinates Swf, wf S w f , corresponding to the shock front. This simple graphical technique of Welge has much wider application in the field of oil recovery calculations which will be described in the following section. 10.5 OIL RECOVERY CALCULATIONS Before water breakthrough (bt) in the producing well, equ. (10.20) can be applied to determine the positions of planes of constant water saturation, for Swf < Sw < 1 – Sor, as the flood moves through the reservoir, and hence the water saturation profile. At the time of breakthrough and subsequently, this equation is used in a different manner, to study the effect of increasing the water saturation at the producing well. In this case x = L, the length of the reservoir block, which is a constant, and equ. (10.20) can be expressed as we i id w w S W 1 W LA df dS φ = = (10.27) in which Swe is the current value of the water saturation in the producing well, fig. 10.14, and Wid the dimensionless number of pore volumes of water injected (1PV= LAφ). 1 - Sor Sw Swc Sw bt Sw Swe 0 x L bt Fig. 10.14 Water saturation distributions at breakthrough and subsequently in a linear waterflood IMMISCIBLE DISPLACEMENT 356 Before breakthrough occurs the oil recovery calculations are trivial. For incompressible displacement the oil recovered is simply equal to the volume of water injected, there being no water production during this phase. At the time of breakthrough the flood front saturation, bt wf w S S , = reaches the producing well and the reservoir watercut increases suddenly from zero to bt wf w w s f f = a phenomenon frequently observed in the field and one which confirms the existence of a shock front. At this time equ. (10.22) can be interpreted in terms of equ. (10.27) to give ( ) bt bt bt wbt w pd id id bt wc w w S 1 N W q t S S df dS = = = − = (10.28) in which all volumes are expressed, for convenience, as dimensionless pore volumes. In particular, the dimensionless injection rate is qi/(LAφ) (PV/unit of time) which facilitates the calculation of the time at which breakthrough occurs as bt id bt id W t q = (10.29) After breakthrough, L remains constant in equ. (10.27) and Swe and fwe, the water saturation and fractional flow at the producing well, gradually increase as the flood moves through the reservoir, as shown in fig. 10.14. During this phase the calculation of the oil recovery is somewhat more complex and requires application of the Welge equation, (10.25), as we w we we w w S 1 S S (1 f ) df dS = + − (10.30) which, using equ. (10.27), can also be expressed as w we we id S S (1 f )W = + − (10.31) Finally, subtracting Swc from both sides of equ. (10.31) gives the oil recovery equation Npd = S w − Swc = (Swe − Swc) + (1 − fwe) Wid(PV) (10.32) The manner in which equs. (10.28) and (10.32) can be used in practice is described below. a) Draw the fractional flow curve, equ. (10.12) or (10.21), allowing for gravity effects, if necessary, but neglecting the capillary pressure gradient ∂Pc/∂x. b) Draw the tangent to this curve from the point Sw = Swc, fw = 0. As described in the previous section, the point of tangency has the co-ordinates Sw = Swf = bt bt wf w w w w s S , f f f = = and the extrapolation of this line to fw = 1 gives the value of the average saturation behind the front at breakthrough bt w w S S . = IMMISCIBLE DISPLACEMENT 357 Equations (10.28) and (10.29) can then be applied to calculate the oil recovery and time at which breakthrough occurs. c) Choosing Swe as the independent variable; allow its value to increase in increments of, say, 5% above the saturation at breakthrough. Each point on the fractional flow curve, for bt we w S S > , has co-ordinates Sw = Swe, fw = fwe and, applying equ. (10.30), fig. 10.15 demonstrates that the tangent to fractional flow curve intersects the line fw = 1 to give the current value of the average water saturation in the reservoir block, w S . For each new value of Swe the corresponding value of w S is determined graphically and the oil recovery calculated as Npd = w S − Swc (PV) The reciprocal of the slope of the fractional flow curve, for each value of Swe, gives Wid, the number of pore volumes of water injected, equ. (10.27). This allows a time scale to be attached to the recovery since Wid = qid t f = 1 w Sw Sw 1 - Sor (1 - f ) we ( - S ) Sw we (S , f we we) bt fw 1 - fwe Sw we - S = dfw dSw Swe Fig. 10.15 Application of the Welge graphical technique to determine the oil recovery after water breakthrough Alternatively, equ. (10.32) can be used directly to calculate the oil recovery by determining fwe and Wid from the fractional flow curve for each chosen value of Swe. This latter method is illustrated in exercise (10.2) in which Npd and Wid are evaluated numerically. The Welge technique for calculating oil recovery, as a function of water injection and time, has been described in detail because it is the very basic method of performing such calculations. It should be emphasized, however, that the theory has been developed under the assumption of diffuse flow which implies a one dimensional mathematical description of the displacement process. In the remaining sections of this IMMISCIBLE DISPLACEMENT 358 chapter, oil displacement will be considered for conditions which apparently require a two dimensional description to account for the vertical distributions of fluid saturations with respect to thickness, e.g. segregated flow and displacement in stratified reservoirs. Nevertheless, by averaging the saturations, and saturation dependent relative permeabilities, in the direction normal to the flow the majority of two dimensional problems can be reduced to one dimension. A fractional flow curve can be drawn using the averaged relative permeability curves, instead of the rock curves, and the Buckley-Leverett/Welge technique applied to oil recovery calculations. It is worth keeping the above comments in mind when reading the remainder of this chapter. EXERCISE 10.1 FRACTIONAL FLOW Oil is being displaced by water in a horizontal, direct line drive under the diffuse flow condition. The rock relative permeability functions for water and oil are listed in table 10.1. Sw krw kro Sw krw kro .20 0 .800 .50 .075 .163 .25 .002 .610 .55 .100 .120 .30 .009 .470 .60 .132 .081 .35 .020 .370 .65 .170 .050 .40 .033 .285 .70 .208 .027 .45 .051 .220 .75 .251 .010 .80 .300 0 TABLE 10.1 Pressure is being maintained at its initial value for which Bo = 1.3 rb/stb and Bw = 1.0 rb/stb Compare the values of the producing watercut (at surface conditions) and the cumulative oil recovery at breakthrough for the following fluid combinations. Case oil viscosity water viscosity 1 50 cp .5 cp 2 5 " .5 " 3 .4 " 1.0 " Assume that the relative permeability and PVT data are relevant for all three cases. EXERCISE 10.1 SOLUTION 1) For horizontal flow the fractional flow in the reservoir is w ro w rw o 1 f k 1 k µ µ = + ⋅ (10.12) IMMISCIBLE DISPLACEMENT 359 while the producing watercut at the surface, fws, is w w ws w w o o q /B f q /B q /B = + where the rates are expressed in rb/d. Combining the above two equations leads to an expression for the surface watercut as ws w o w 1 f B 1 1 1 B f = + − (10.33) The fractional flow in the reservoir for the three cases can be calculated as follows. Fractional Flow (fw) Case 1 Case 2 Case3 Sw krw kro kro/krw µw/µo = .01 µw/µo = .1 µw/µo = 2.5 .2 0 .800 ∞ 0 0 0 .25 .002 .610 305.000 .247 .032 .001 .30 .009 .470 52.222 .657 .161 .008 .35 .020 .370 18.500 .844 .351 .021 .40 .033 .285 8.636 .921 .537 .044 .45 .051 .220 4.314 .959 .699 .085 .50 .075 .163 2.173 .979 .821 .155 .55 .100 .120 1.200 .988 .893 .250 .60 .132 .081 .614 .994 .942 .394 .65 .170 .050 .294 .997 .971 .576 .70 .208 .027 .130 .999 .987 .755 .75 .251 .010 .040 .999 .996 .909 .80 .300 0 0 1.000 1.000 1.000 TABLE 10.2 Fractional flow plots for the three cases are shown in fig. 10.16, and the results obtained by applying Welge's graphical technique, at breakthrough, are listed in table 10.3. Case bt w S bt w f (reservoir) bt ws f (surface) bt w S bt pd N (PV) 1 .28 .55 .61 .34 .14 2 .45 .70 .75 .55 .35 3 .80 1.00 1.00 .80 .60 TABLE 10.3 An important parameter in determining the effectiveness of a waterflood is the end point mobility ratio defined in Chapter 4, sec. 9, as IMMISCIBLE DISPLACEMENT 360 rw w ro o k / M k / µ µ ′ = ′ And, for horizontal flow, stable, piston-like displacement will occur for M ≤ 1. An even more significant parameter for characterising the stability of Buckley Leverett displacement is the shock front mobility ratio, Ms, defined as ro wf o rw wf w s ro o k (S )/ k (S )/ M k / µ µ µ + = ′ (10.34) CASE 2 CASE 3 CASE 1 1.0 0.9 1.0 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 = .01 w o µ µ = .1 w o µ µ = 2.5 w o µ µ Sw f (rb/rb) w Fig. 10.16 Fractional flow plots for different oil-water viscosity ratios (table 10.2) in which the relative permeabilities in the numerator are evaluated for the shock front water saturation, Swf. Hagoort has shown9, using a theoretical argument backed by experiment, that Buckley-Leverett displacement can be regarded as stable for the less restrictive condition that Ms < 1. If this condition is not satisfied there will be severe viscous channelling of water through the oil and breakthrough will occur even earlier than predicted using the Welge technique10. Values of M and Ms for the three cases defined in exercise 10.1 are listed in table 10.3(a). Using these data IMMISCIBLE DISPLACEMENT 361 Case No. (exercise 10.1) o w µ µ Swf krw(Swf) kro(Swf) Ms M 1 100 .28 .006 .520 1.40 37.50 2 10 .45 .051 .220 .91 3.75 3 .4 .80 .300 0 .15 0.15 TABLE 10.3(a) Values of the shock front and end point relative permeabilities calculated using the data of exercise 10.1 the results of exercise 10.1 can be analysed as follows: a) Case 1 - this displacement is unstable due to the very high value of the oil/water viscosity ratio. This results in the by-passing of oil and consequently the premature breakthrough of water. The oil recovery at breakthrough is very small and a great many pore volumes of water will have to be injected to recover all the movable oil. Under these circumstances oil recovery by water injection is hardly feasible and consideration should be given to the application of thermal recovery methods with the aim of reducing the viscosity ratio. b) Case 2 - the oil/water viscosity ratio is an order of magnitude lower than in case 1 which leads to a stable and much more favourable type of displacement (Ms < 1). This case will be analysed in greater detail in exercise 10.2, in which the oil recovery after breakthrough is determined as a function of the cumulative water injected and time. c) Case 3 - for the displacement of this very low viscosity oil (µo = .4 cp) both the end point and shock front mobility ratios are less than unity and piston-like displacement occurs. The tangent to the fractional flow curve, from Sw = Swc, fw = 0, meets the curve at the point bt bt w or w S 1 S , f 1 = − = and therefore bt wbt w or S S 1 S = = − . The total oil recovery at breakthrough is wbt wc or wc S S 1 S S = = − − , which is the total movable oil volume. EXERCISE 10.2 OIL RECOVERY PREDICTION FOR A WATERFLOOD Water is being injected at a constant rate of 1000 b/d/well in a direct line drive in a reservoir which has the following rock and fluid properties. φ = 0.18 Swc = 0.20 Sor = 0.20 µo = 5 cp µw = 0.5 cp IMMISCIBLE DISPLACEMENT 362 The relative permeabilities for oil and water are presented in table 10.1 and the flood pattern geometry is as follows Dip angle = 0° Reservoir thickness = 40 ft Distance between injection wells = 625 ft Distance between injectors and producers = 2000 ft Assuming that diffuse flow conditions prevail and that the injection project starts simultaneously with oil production from the reservoir 1) determine the time when breakthrough occurs 2) determine the cumulative oil production as a function of both the cumulative water injected and the time. EXERCISE 10.2 SOLUTION The relative permeabilities and viscosities of the oil and water are identical with those of "Case 2" in exercise 10.1. Therefore, the fractional flow curve is the same as drawn in fig. 10.16, for, which the breakthrough occurs when bt bt bt bt w w id pd S 0.45 f 0.70 and W N 0.35 = = = = 1) Calculation of the breakthrough time For a constant rate of water injection the time is related to the dimensionless water influx by the general expression id i W (one pore volume) (cu.ft) t q 5.615 365 (cu.ft / year) × = × × id W 625 40 2000 .18 t (years) 1000 5.615 365 × × × × = × × id t 4.39 W (years) = (10.35) Therefore breakthrough will occur after a time tbt = 4.39 × 0.35 = 1.54 years 2) Cumulative oil recovery The oil recovery after breakthrough, expressed in pore volumes, can be calculated using the equation Npd = (Swe – Swc) + (1 – fwe) Wid (10.32) IMMISCIBLE DISPLACEMENT 363 where we id w w S 1 W df dS = (10.27) Allowing Swe, the water saturation at the producing end of the block, to rise in increments of 5% (for Swe ≥ bt w S ) the corresponding values of Wid are calculated in table 10.4 using the data listed in table 10.2 for Case 2. Swe fwe ∆Swe ∆fwe ∆fwe/∆Swe we S∗ Wid .45 (bt) .699 .05 .122 2.440 .475 .410 .50 .821 .05 .072 1.440 .525 .694 .55 .893 .05 .049 .980 .575 1.020 .60 .942 .05 .029 .580 .625 1.724 .65 .971 .05 .016 .320 .675 3.125 .70 .987 .05 .009 .180 .725 5.556 .75 .996 .05 .004 .080 .775 12.500 .80 1.000 TABLE 10.4 In this table, values of ∆fwe/∆Swe have been calculated rather than determined graphically as suggested in the text. The values of we S∗ in column 6 are the mid points of each saturation increment, at which discrete values of Wid have been calculated using equ. (10.27). The oil recovery as a function of both Wid and time can now be determined using equ. (10.32), as listed in table 10.5. we S∗ we wc S S ∗− we f∗ we 1 f∗ − ( ) id W PV ( ) pd N PV time (yrs) equ (10.35) .475 .275 .765 .235 .410 .371 1.80 .525 .325 .870 .130 .694 .415 3.05 .575 .375 .925 .075 1.020 .452 4.48 .625 .425 .962 .038 1.724 .491 7.57 .675 .475 .982 .018 3.125 .531 13.72 .725 .525 .993 .007 5.556 .564 24.39 TABLE 10.5 IMMISCIBLE DISPLACEMENT 364 The values of we f∗ in column 3 of table 10.5 have been obtained from fig. 10.16 (Case 2), for the corresponding values of we S∗. The oil recovery, in reservoir pore volumes, is plotted as a function of Wid and time in fig. 10.17. The maximum possible recovery is one movable oil volume, i.e. (1 − Swc − Sor) = 0.6 PV. In the general case in which the displacement takes place at a fixed pressure, which is above the bubble point pressure, then .6 .5 .4 .3 .2 .1 0 0 1 2 3 4 5 6 7 0 5 10 15 20 25 30 W (PV) id time (yrs) q = 1000 rb/d i N (PV) pd Fig. 10.17 Dimensionless oil recovery (PV) as a function of dimensionless water injected (PV), and time (exercise 10.2) p o pd wc oi N B oil production (rb) N (1 S ) one pore volume (rb) N B = = − and the conventional expression for the oil recovery is p pd oi o wc N N B (stb.oil) N B (1 S ) STOIIP (stb) = − In exercise 10.2, Bo = Boi, since displacement occurs at the initial reservoir pressure, Np / N = Npd / (1−Swc) 10.6 DISPLACEMENT UNDER SEGREGATED FLOW CONDITIONS In the previous two sections a one dimensional displacement theory was presented which relied on the assumption of diffuse flow. Now, precisely the opposite will be assumed, namely, that displacement occurs under the segregated flow condition shown in fig.10.18. In the flooded part of the reservoir water alone is flowing, in the presence of residual oil, with effective permeability kw = rw kk′ , where rw k′ is the end point relative permeability IMMISCIBLE DISPLACEMENT 365 to water. Similarly, in the unflooded zone oil is flowing in the presence of connate water with effective permeability ko = ro kk′ , where ro k′ is the end point relative permeability to oil. Furthermore, at any point on the interface between the fluids the pressures in the oil and water are assumed to be equal. This means that there is a distinct interface with no capillary transition zone. Segregated flow also assumes that the displacement is governed by vertical equilibrium, as discussed in WATER S = 1 - S S = S w or o or OIL S = S S = 1 - S w wc o wc Fig. 10.18 Displacement of oil by water under segregated flow conditions sec. 10.2. In this case, since there is no capillary transition zone, gravity forces alone are responsible for the instantaneous distribution of the fluids in the dip-normal direction5. Dietz has investigated this type of displacement11 and, in particular, the conditions under which it can be regarded as stable; the difference between stable and unstable displacement in a dipping reservoir being illustrated in fig.10.19. The condition for stable displacement is that the angle between the fluids' interface and the direction of flow should remain constant throughout the displacement, fig.10.19(a), (b), such that dy tan dx β = − = constant This is only satisfied at relatively low injection rates when the gravity force, arising from difference in density between the fluids, will act to try and maintain the interface horizontal and, in the limiting case in which the rate is reduced to zero, a horizontal interface would result. At high injection rates the viscous forces, driving the fluids through the reservoir, will prevail over the component of the gravity force, acting in the downdip direction, resulting in the unstable displacement shown in fig. 10.19(c). Due to the density difference the water will underrun the oil in the form of a water tongue, leading to premature water breakthrough. Unstable displacement will occur for the limiting condition that dy tan 0 dx β = − = IMMISCIBLE DISPLACEMENT 366 y x y x y x θ θ θ WATER WATER WATER dx -dy OIL OIL OIL dx -dy β (a) (b) (c) β Fig. 10.19 Illustrating the difference between stable and unstable displacement, under segregated flow conditions, in a dipping reservoir; (a) stable: G > M− − − −1; M > 1; β β β β < θ θ θ θ. (b) stable: G > M− − − −1; M < 1; β β β β > θ θ θ θ. (c) unstable: G < M− − − −1. If the incompressible displacement is stable then, at all points on the interface, the oil and water must have the same velocity and, applying Darcy's law at any point on the interface for displacement in the x direction ro o o o t 6 o kk p g sin u u x 1.0133 10 ρ θ µ ′ ∂ = = − + ∂ × and rw w w w t 6 w kk p g sin u u x 1.0133 10 ρ θ µ ′ ∂ = = − + ∂ × where uo, uw and ut are the oil, water and total flow velocities, respectively. These equations can be combined to give o w t o w 6 ro rw g sin u (p p ) kk kk x 1.0133 10 µ µ ρ θ ∂ ∆ − = − − + ′ ′ ∂ × (10.36) IMMISCIBLE DISPLACEMENT 367 where ∆ρ = ρw − ρo. Also, applying the capillary pressure equation, (10.5), c o w 6 g cos dP d(p p ) dy 1.0133 10 ρ θ ∆ = − = × and for stable displacement (dy/dx—negative) c 6 P g cos dy x dx 1.0133 10 ρ θ ∂ ∆ = − ∂ × which, when substituted in equ. (10.36) gives o w t 6 ro rw g dy u cos sin kk kk dx 1.0133 10 µ µ ρ θ θ ∆ − = + ′ ′ × This equation can be alternatively expressed in terms of the total flow rate, qt, as ro rw rw 6 w o w t k k kk A g sin dy 1 1 1 dx tan 1.0133 10 q ρ θ µ µ θ µ ′ ′ ′ ∆ − = + × or dy 1 M 1 G 1 dx tanθ − = + (10.37) where M is the end point mobility ratio and G the dimensionless gravity number, which in Darcy units is rw 6 t w kk A g sin G 1.0133 10 q ρ θ µ ′ ∆ = × (10.38) and in field units can be deduced from equ. (10.10) to be 4 rw t w kk A sin G 4.9 10 q γ θ µ − ′ ∆ = × (10.39) Equation (10.37) can be solved to give the slope of the interface for stable flow as dy M 1 G tan tan dx G β θ −− = − = (10.40) In this equation M is a constant and, when displacing oil by water at a fixed rate in the updip direction, G is a positive constant. Therefore, the inclination of the interface dy/dx assumes a fixed value. For stable displacement, as already mentioned, dy/dx must be a negative constant and this imposes the condition for stability that G > M−1 The limiting case is when dy/dx = 0 then, as shown in fig. 10.19(c), the water will underrun the oil in the form of a water tongue. This will occur when IMMISCIBLE DISPLACEMENT 368 G = M−1 which, using equ. (10.38), can be solved to determine the so-called critical rate for by-passing as rw crit 6 w kk A g sin q r.cc / sec 1.0133 10 (M 1) ρ θ µ ′ ∆ = × − (10.41) or in field units 4 rw crit w 4.9 10 kk A sin q rb / d (M 1) γ θ µ − ′ × ∆ = − (10.42) Provided the injection rate is maintained below qcrit gravity forces will stabilize the displacement. The magnitude of the mobility ratio also influences the displacement. This can be appreciated from equ. (10.40), as detailed below. M > 1 This is the most common physical condition. The displacement is stable if G > M−1, in which case β < θ, fig. 10.19(a), and unstable if G < M−1. M = 1 This is a very favourable mobility ratio for which there is no tendency for the water to by-pass the oil (refer Chapter 4, sec. 9). For M = 1 the displacement is unconditionally stable. Furthermore, β = θ, and the interface rises horizontally in the reservoir. M < 1 This mobility ratio also leads to unconditionally stable displacement but in this case, β > θ, fig. 10.19(b). If the displacement is stable the oil recovery, as a function of cumulative water injected and time, can be calculated from simple geometrical considerations, as will be demonstrated in exercise 10.3. An alternative approach is to attempt to reduce the description of segregated displacement to one dimension and then calculate the oil recovery using the Buckley-Leverett displacement theory. It is worthwhile pursuing this idea because it is quite general and can be applied whether the displacement is stable or not. Consider then the general segregated displacement in a linear, homogeneous reservoir as shown in fig.10.20. IMMISCIBLE DISPLACEMENT 369 WATER S = 1 - S S = S w or o or OIL S = S S = 1 - S w wc o wc x y h 1 - b b (x, y) Fig. 10.20 Segregated displacement of oil by water Unlike the oil displacement under the assumed diffuse flow condition, described in the previous section, which could be accounted for using one dimensional mathematics; the segregated flow depicted in fig. 10.20 is very definitely a two dimensional problem. In attempting to reduce the mathematical description to one dimension it is necessary to average the saturations and saturation dependent relative permeabilities over the reservoir thickness. Flow can then be described as occurring along the centre line of the reservoir. At any point x in the linear displacement path, let b be the fractional thickness of the water, fig. 10.20, thus b = y/h. The thickness averaged water saturation at x is then Sw = b(1−Sor) + (1−b)Swc which can be solved for b to give w wc or wc S S b 1 S S − = − − (10.43) and, since Sor and Swc are constants, equ. (10.43) indicates that b is directly proportional to the average saturation. The thickness averaged relative permeability to water can be similarly derived as krw ( ) w S = b krw (Sw = 1 − Sor) + (1 − b) krw (Sw = Swc) and since krw (Sw = Swc) is zero and krw (Sw = 1 − Sor) = k´rw this can be reduced to krw ( ) w S = b rw k′ where rw k′ is the end point relative permeability to water. For the oil, the same argument gives the thickness averaged relative permeability as kro ( ) w S = b kro (Sw = 1 − Sor) + (1 − b) kro (Sw = Swc) IMMISCIBLE DISPLACEMENT 370 or kro ( ) w S = b ro k′ where ro k′ is the end point relative permeability to oil. Substituting for b in these expressions, using equ. (10.43), gives wc rw rw or wc w w S S k (S ) k 1 S S − ′ = − − (10.44) and or ro ro or wc w w 1 S S k (S ) k 1 S S − − ′ = − − (10.45) These equations indicate that the thickness averaged relative permeabilities, for segregated flow, are simply linear functions of the thickness averaged water saturation, as shown by the solid lines in fig. 10.21. k'ro k S ro w ( ) equ. (10.45) k S rw w ( ) equ. (10.44) k'rw Swc Sw 1 - Sor Fig. 10.21 Linear, averaged relative permeability functions for describing segregated flow in a homogeneous reservoir Also shown in fig. 10.21, as dashed lines, are the rock relative permeability curves obtained by measuring relative permeabilities in the laboratory using thin core plugs. They are measured under conditions which correspond to diffuse flow and represent point relative permeabilities in the reservoir. As mentioned already, they can only be used directly in displacement calculations if the water saturation is the same at all points throughout the thickness. In this unique case point relative permeabilities are equal to thickness averaged relative permeabilities. In contrast, the linear functions shown in fig. 10.21, result from the thickness averaging process required to facilitate IMMISCIBLE DISPLACEMENT 371 the description of two dimensional, segregated flow using one dimensional equations. Therefore oil recovery calculations, for either stable or unstable, segregated flow, can be performed using the linear relative permeabilities in conjunction with the Buckley-Leverett displacement theory. This is because the theory was based simply on the conservation of water mass, in one dimension, equ. (10.13). Therefore, whether the water is uniformly distributed with respect to thickness or segregated from the oil does not matter provided the resulting displacement can be described using one dimensional mathematics; the same basic principle of mass conservation still applies. The fractional flow equation can be plotted using the linear relative permeability functions and the Welge graphical technique applied as illustrated in exercise 10.2. In this case, the fractional flow curve will have no inflexion point, as shown in fig. 10.22, since there is no shock front for segregated flow. All points on the fractional flow curve are used in the recovery calculations after breakthrough. f = 1 w Swc 1 - Sor Fig. 10.22 Typical fractional flow curve for oil displacement under segregated conditions Because the thickness averaged relative permeabilities are linear for segregated flow, it is also possible to derive a simple analytical expression for the oil recovery as a function of the cumulative water injected. As mentioned al ready, this is unnecessary for stable displacement but for unstable displacement it provides a rapid means of predicting recovery. The following argument will, for simplicity, be developed for the unstable displacement of oil by water in a horizontal reservoir. As described in Chapter 4, sec. 9 and illustrated in exercise 10.1, this will occur if M > 1. An analytical expression for the fractional flow of water will be derived and used in the oil recovery formula of Welge, equ. (10.32). The one dimensional equations for the separate flow of oil and water, under segregated conditions in a horizontal reservoir, are IMMISCIBLE DISPLACEMENT 372 ro o o o (1 b)kk A p q x µ ° ′ − ∂ = − ∂ (10.46) and rw w w w b kk A p q x µ ° ′ ∂ = − ∂ (10.47) in which A is the area of cross-section and o p° and w p° are the oil and water phase pressures referred to the centre line of the reservoir, as shown in fig. 10.23. Centre line oil - water interface h 2 y x po pw } } h 2 ( -y) y h Fig. 10.23 Referring oil and water phase pressures at the interface to the centre line in the reservoir. (Unstable segregated displacement in a horizontal, homogeneous reservoir) From this diagram it can be seen that o o o 6 g h p p y atm 2 1.0133 10 ρ = − − × and w w w 6 g h p p y atm 2 1.0133 10 ρ ° = − − × where y is the actual thickness of the water; i.e. y = bh. Since the pressures at the interface, po and pw, are equal for segregated flow then the phase pressure gradient, resulting from the differentiation and subtraction of the above equations, is o w 6 p p g dy x x dx 1.0133 10 ρ ° ° ∂ ∂ ∆ − = − ∂ ∂ × For unstable, horizontal displacement the approximation is usually made that the angle of inclination of the interface, dy/dx, is small and therefore the gradient of the phase pressure difference can be neglected. In this case, using equs. (10.46) and (10.47), and following the argument used in sec. 10.3, in deriving the fractional flow equation, results in ro rw ro w w o rw ro w k k f k 1 b b k µ µ µ µ ′ ⋅ ′ = ′ − + ⋅ ′ IMMISCIBLE DISPLACEMENT 373 which can be simplified as w Mb f 1 (M 1)b = + − in which M is the end point mobility ratio. Until the moment of breakthrough the oil recovery is simply equal to the cumulative water injected. After breakthrough, let be be the fractional thickness of the water at the producing end of the reservoir block. Then, for a fully penetrating well, the fractional flow of water into the well is e we e Mb f 1 (M 1)b = + − (10.48) Applying equ. (10.27) at the producing end of the block we we id df 1 W dS = and using equ. (10.43) for the thickness averaged water saturation, Swe we we e we we w id e e or wc df df db df 1 1 W db db (1 S S ) dS dS = = ⋅ = ⋅ − − and therefore we or wc e id iD df (1 S S ) 1 db W W − − = = in which WiD is the cumulative water injection expressed in movable oil volumes where 1 MOV = PV(1 − Swc − Sor) Differentiating equ. (10.48) with respect to be gives we 2 e iD e df 1 M db W (1 (M 1)b ) = = + − from which it can be determined that ( ) e iD 1 b W M 1 M 1 = − − (10.49) and substituting for be in equ. (10.48) gives ( ) we iD M f 1 W M M 1 = − − (10.50) The oil recovery equation, (10.32), can be expressed in MOV's as we wc pD we iD or wc S S N (1 f )W 1 S S − = − − − IMMISCIBLE DISPLACEMENT 374 or pD e we iD N b (1 f )W = + − and substituting in this equation for be and fwe, using equs. (10.49) and (10.50), results in the simple recovery formula pD iD iD 1 N (2 W M W 1) M 1 = − − − (10.51) in which all volumes are expressed in MOV's. It should again be stressed that equ. (10.51) is only applicable for horizontal displacement under segregated conditions and for unstable flow (M > 1). At the time of water breakthrough NpD = WiD and solving equ. (10.51) for this condition gives bt pD 1 N M = (10.52) which shows that in the limiting case of M = 1, stable, piston-like displacement occurs for which bt pD N 1 = . Similarly, when the total amount of oil has been recovered, NpD = 1 (MOV), and substituting this condition in equ. (10.51) gives max iD W M = (10.53) Equations (10.52) and (10.53) clearly demonstrate the significance of the mobility ratio in characterising oil recovery under segregated flow conditions. For the more general case of unstable displacement in a dipping reservoir (G<M−1), the fractional flow equation, equivalent to equ. (10.48), is e e e we e Mb b (1 b )G f 1 (M 1) b − − = + − and repeating the steps leading to equ. (10.51 ) will, in this more complex case, yield the recovery formula iD pD iD iD W G 1 G (M 1) N 2 W M 1 1 W 1 G 1 M 1 M 1 M 1 (M 1) + = − − − − − − − − − (10.54) which if G = 0 (horizontal reservoir) reduces to the form of equ. (10.51). Equation (10.54) can readily be solved for the breakthrough condition (NpD = WiD) to give bt pD 1 N M G = − (10.55) while for the maximum recovery (NpD = 1) IMMISCIBLE DISPLACEMENT 375 max iD M W G 1 = + (10.56) EXERCISE 10.3 DISPLACEMENT UNDER SEGREGATED FLOW CONDITIONS 1) Re-work Exercise 10.2, for the same water injection rate and using precisely the same reservoir and fluid property data but assuming that displacement takes place under segregated flow conditions. Compare the results with those obtained in exercise 10.2. 2) If the same reservoir had a dip angle of 25°, what would be the critical rate for water displacing oil updip? Compare the breakthrough times and the oil recovery at breakthrough when injecting at 1000 rb/d and also at 90% of the critical rate. Additional information k = 2.0 Darcy γw = 1.04 specific gravity in the reservoir γo = .81 " " " " EXERCISE 10.3 SOLUTION 1) Using the relative permeability relations given in exercise 10.1, table 10.1, and for µo = 5 cp, µw = 0.5 cp (Case 2 of exercise 10.1), the end point mobility ratio can be evaluated as rw w ro o k / .3 .8 M 3.750 k / .5 5 µ µ ′ = = = ′ Therefore, using equs. (10.52) and (10.53), breakthrough will occur when bt bt iD pD 1 W N .267 (MOV) .160 (PV) M = = = = since 1 MOV = PV(1 − Sor − Swc) = PV(1−0.2−0.2) = 0.6 PV; and the maximum oil recovery when WiD = M = 3.75 (MOV) = 2.25 (PV) Between breakthrough and maximum recovery, the oil recovery as a function of WiD can be calculated using equ. (10.51), with WiD as the independent variable. The results are listed in table 10.6. IMMISCIBLE DISPLACEMENT 376 WiD NpD Wid Npd t = 4.39 Wid (yrs) (MOV) (MOV) (PV) (PV) (equ. (10.35) ) .267 (bt) .267 .160 .160 .702 .300 .299 .180 .179 .790 .500 .450 .300 .270 1.317 1.000 .681 .600 .409 2.634 1.500 .816 .900 .489 3.951 2.000 .901 1.200 .540 5.268 3.000 .985 1.800 .591 7.902 3.750 1.000 2.250 .600 9.878 TABLE 10.6 In fig. 10.24 plots of the oil recovery, assuming the segregated flow condition, are compared with the results from exercise 10.2, assuming diffuse flow. The comparison shows that, although the breakthrough occurs much earlier for segregated flow, the ultimate recovery is obtained sooner and for a much smaller throughput of water. N (PV) pd .6 .4 .5 .2 .3 0 .1 0 1 2 3 4 5 6 7 0 5 10 15 20 25 30 W (PV) id time (yrs) q = 1000 rb / d i DIFFUSE FLOW ( EXERCISE 10.2) SEGREGATED FLOW (EXERCISE 10.3) Fig. 10.24 Comparison of the oil recoveries obtained in exercises 10.2 and 10.3 for assumed diffuse and segregated flow, respectively 2) For the data given in exercise 10.2, the critical rate for by-passing can be calculated, in field units, as 4 rw crit w 4.9 10 kk A sin q (M 1) γ θ µ − ′ × ∆ = − (10.42) IMMISCIBLE DISPLACEMENT 377 4 o crit 4.9 10 2000 .3 625 40 (1.04 0.81) sin25 q (rb / d) .5(3.75 1) 520 rb.water / d − × × × × × × − × = − = While for injection at this critical rate G = M − 1 = 2.75 Comparison of equs. (10.39) and (10.42) indicates that qcrit (M − 1) = qtG and therefore, at an injection rate of qi = qt = 1000 rb/d 520 G 2.75 1.430 1000 = × = Substituting this value of G and the value of M = 3.75 in equ. (10.54), for this unstable displacement (G < M−1), reduces the latter to pD iD iD iD N 0.976 W (1 0.520W ) 0.535 W 0.364 = − + − (10.57) At the time of water breakthrough NpD = WiD and equ. (10.55) can be applied to determine the breakthrough recovery as bt pD 1 1 N 0.431 (MOV) 0.259 (PV) M G 3.75 1.43 = = = = − − which, when injecting at 1000 rb/d, will occur after 4.39 bt iD W = 1.137 years, equ. (10.35). Similarly, the maximum cumulative water injection to recover the one movable oil volume can be determined using equ. (10.56) as max iD M 3.75 W 1.543 (MOV) 0.926 (PV) G 1 2.43 = = = = + Between breakthrough and total recovery equ. (10.57) can be used to calculate the oil recovery, as listed in table 10.7. WiD (MOV) NpD (MOV) Wid (PV) Npd (PV) t = 4.39 Wid (yrs) (equ. (10.35)) .431 (bt) .431 .259 .259 1.137 .500 .497 .300 .298 1.317 .750 .697 .450 .418 1.976 1.000 .847 .600 .508 2.634 1.250 .950 .750 .570 3.293 1.543 1.000 .926 .600 4.064 TABLE 10.7 When compared to the water drive performance, at an injection rate of 1000 rb/d in a horizontal reservoir (table 10.6), it can be seen that, even though the displacement is IMMISCIBLE DISPLACEMENT 378 unstable in both cases, the gravity force in a reservoir with a 25° dip has a very favourable effect on the water drive performance, the maximum recovery being obtained in less than half the time required in a horizontal reservoir. For stable displacement, at 90% of the critical rate (468 rb/d), the gravity number is crit t q 1 G (M 1) 2.75 3.056 q 0.9 = × − = × = The angle of inclination of the oil water interface to the direction of flow can be determined using equ. (10.40) as dy M 1 G 3.750 1 3.056 tan tan 0.4663 0.0467 dx G 3.056 β θ −− −− = − = = × = − and therefore β = 2.673°. The situation after breakthrough, when the water has risen a height ye at the level of the producing well, is shown in fig. 10.25. h - y tan e β β OIL WATER θ ye y / tan e β h Fig. 10.25 The stable, segregated displacement of oil by water at 90% of the critical rate (exercise 10.3) If the width, thickness and length of the reservoir are denoted by w, h and L, respectively, then the total movable volume of oil is whLφ (1−Sor−Swc). For the situation depicted in fig.10.25, the volume of oil uncontacted by the water is e e or wc (h y ) (h y ) w (1 S S ) 2 tan φ β − − × − − and therefore the oil recovery at this stage, expressed in MOV's, is 2 e pD (h y ) N 1 2hL tan β − = − (10.58) The amount of water injected at this stage can be estimated by ignoring the presence of the producing well. The volume of water by-passing the well, as shown in fig. 10.25 is IMMISCIBLE DISPLACEMENT 379 e e or wc y y w (1 S S ) 2 tan φ β ⋅ − − and therefore the number of MOV's of water injected is 2 e iD pD y W N 2hL tan β = + (10.59) In particular, at breakthrough, ye = 0 and bt bt pD iD h N W 1 2L tan β = = − (10.60) For the data relevant to this exercise, the recovery at breakthrough can be calculated using equ. (10.60) as bt pD 40 N 1 0.786 MOV .472 PV 2 2000 .0467 = − = = × × At an injection rate of 468 rb/d the relation between water injected and time is now (refer equ. (10.35) id id 1000 t 4.39 W 9.38 W (years) 468 = × = and therefore breakthrough occurs after 9.38 × .472 = 4.43 years. Thereafter the oil recovery and cumulative water injection, as functions of time, are listed in table 10.8. These have been calculated using equs. (10.58) and (10.59) for increasing values of ye. ye ft NpD (MOV) WiD (MOV) Npd (PV) Wid (PV) t = 9.38Wid years 0 (bt) .786 .786 .472 .472 4.427 10 .880 .893 .528 .536 5.028 20 .946 1.000 .568 .600 5.628 30 .987 1.107 .592 .664 6.228 40 1.000 1.214 .600 .728 6.829 TABLE 10.8 When compared to the results presented in table 10.7, which were obtained at an injection rate of 1000 rb/d, the effect of displacing the oil at a rate which is below critical (468 rb/d) is quite naturally to increase both the breakthrough time (by nearly 300% ) and the total recovery time (by almost 70% ). The main advantage in displacement at the lower rate is that the total water requirements are reduced from 1.543 to 1.214 (MOV). If there are problems concerning either the availability of injection water or disposal of produced water it may be considered expedient to inject at a rate which is just below critical. IMMISCIBLE DISPLACEMENT 380 The same analytical methods described in this section can be applied to the case of gas displacing oil downdip, at constant pressure. As shown in fig.10.26,for unstable displacement the gas will tend to override the oil causing premature gas breakthrough in the downdip producing wells. For stable displacement the angle of inclination of the gas-oil interface remains constant. GAS OIL y x y x (a) GAS OIL β (b) Fig. 10.26 Segregated downdip displacement of oil by gas at constant pressure; (a) unstable, (b) stable Applying the method described in this section for determining the critical rate (in this case it is algebraically convenient to keep the same sign convention, x-positive in the updip direction, and reverse the signs of the flow velocities) will again give the condition for stable displacement as G > M−1 This leads to the critical rate formula rg crit 6 g kk A g sin q 1.0133 10 (M 1) ρ θ µ ′ ∆ = × − in which, for gas displacing oil o g ρ ρ ρ ∆ = − rg crit 6 t g t kk A g sin q G (M 1) q 1.0133 10 q ρ θ µ ′ ∆ = = − × (10.61) and rg g ro o k / M k / µ µ ′ = ′ Since µg is very small compared to µo, the mobility ratio for this displacement is large and the requirement for unconditional stability (M ≤ 1) is never satisfied. Stability then depends on the magnitude of G and hence on the dip angle. An interesting extension of the methods described in this section has been presented by van Daalen and van Domselaar12 who considered segregated displacement in reservoirs in which there is a defined (absolute) permeability distribution in the dip normal direction. In addition, Richardson and Blackwell13 have analysed some quite IMMISCIBLE DISPLACEMENT 381 complex displacement problems using the assumption of segregated flow, these include gravity drainage and bottom water coning. In this section a great deal of attention has been focussed on the presentation of approximate analytical methods for predicting oil recovery resulting from segregated displacement. In reading the remainder of this chapter, however, the main point to keep in mind is that the description of segregated flow in one dimension necessitates the use of linear, averaged relative permeability functions, irrespective of whether the displacement is stable or not. It is this fact which facilitated the derivation of such simple recovery formulae. 10.7 ALLOWANCE FOR THE EFFECT OF A FINITE CAPILLARY TRANSITION ZONE IN DISPLACEMENT CALCULATIONS For the displacement of oil by water, exercises 10.2 and 10.3 clearly demonstrate the sensitivity of the calculated oil recovery, as a function of time, to the assumed water saturation distribution in the dip-normal direction. So far, too extreme notions of this distribution have been considered; the uniform distribution (diffuse flow) and that due to fluid segregation. From the information provided in the two exercises it is not possible for the engineer to decide which, if either, of the assumed saturation distributions is appropriate to describe the displacement. One vital piece of data has been omitted and that is the capillary pressure curve and, in particular, the thickness of the capillary transition zone. If h is the reservoir thickness and H the thickness of the capillary transition zone then the water saturation distribution can be approximated as uniform or segregated depending on whether H >> h (uniform) H << h (segregated) If the reservoir is very thin in comparison to the capillary transition zone the saturation in the advancing water flood will appear to be uniformly distributed with respect to thickness, fig. 10.7. On the other hand, if the transition zone is of negligible thickness compared to the reservoir then it will appear that the oil and water are segregated. Linear relative permeability curves can be used to describe such displacement. The question then arises of how to describe oil displacement in a homogeneous reservoir when the capillary transition zone is of the same order of magnitude as the reservoir thickness (H ≈ h). Consider, for instance, the capillary pressure curve shown in fig. 10.27(a) in which the capillary pressure difference across the transition zone is 3 psi. IMMISCIBLE DISPLACEMENT 382 P (psi) c 5 4 3 2 1 0 0 .2 .4 .6 .8 1.0 50 40 30 20 10 0 z (ft) a k rw 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 k ro b OIL WATER Sw Sw Fig. 10.27 (a) Imbibition capillary pressure curve, and (b) laboratory measured relative permeabilities (rock curves,- table 10.1) If oil displacement in the horizontal reservoir described in exercises 10.2 and 10.3 is re-examined using this capillary pressure curve, then, since γw = 1.04 and γo = 0.81; the capillary pressure versus capillary rise equation, (10.6), applied in the differential form dPc = .4335 ∆γ dz gives the relationship dPc = .4335 (1.04−.81) dz = 0.1 dz (10.62) Thus for dPc = 3 psi the height of the capillary transition zone is 30 ft which, since the reservoir is 40 ft thick, means that neither diffuse nor segregated flow conditions can be assumed to govern the displacement. Instead, it is necessary to generate averaged relative permeability curves, which are functions of the thickness averaged water saturation, and use these in the oil recovery calculations. The manner in which this is done is illustrated graphically in fig. 10.28. Consider first of all fig. 10.28(a). This represents the water saturation distribution, at any point in the displacement path, as a function of thickness. For this initial case the maximum water saturation, Sw = 1−Sor (Pc = 0), is assumed to be at the base of the reservoir, above which the saturation is distributed in accordance with the saturation-capillary rise function, fig. 10.27(a). Since the reservoir is homogeneous, the thickness averaged water saturation can be expressed mathematically as h w 0 w S (z)dz S h = (10.63) IMMISCIBLE DISPLACEMENT 383 40 30 20 10 0 0 .2 .4 .6 .8 1.0 Sw a Sw = .357 z (ft) 40 30 20 10 0 0 .2 .4 .6 .8 1.0 k r b krw = .047; kro = .481 z (ft) kro(z) krw (z) Fig. 10.28 (a) Water saturation, and (b) relative permeability distributions, with respect to thickness when the saturation at the base of the reservoir is Sw = 1 − − − − Sor (Pc =0) which can be evaluated graphically by measuring the shaded area to the left and beneath the saturation-thickness function, fig. 10.28(a), and dividing this by the total height, h = 40 ft. For this initial saturation distribution Sw = 0.357. The rock relative permeabilities presented in table 10.1, which have been used in exercises 10.2 and 10.3, are plotted as fig. 10.27(b). As already mentioned, these represent point relative permeabilities in the reservoir and are dependent on the water saturation at the point in question. Therefore, since there is a defined water saturation distribution in the reservoir, fig. 10.28(a), there must also be a relative permeability distribution, with respect to thickness, for both oil and water. These distributions can be determined by selecting a particular height in the reservoir and reading the water saturation at that point from fig. 10.28(a). For this saturation the corresponding point relative permeabilities are read from fig. 10.27 (b). These relative permeability distributions are listed in table 10.9 and plotted in fig. 10.28(b). Mathematically, the thickness averaged relative permeabilities are h rw w 0 rw w k (S (z))dz k (S ) h = (10.64) and h ro w 0 ro o k (S (z))dz k (S ) h = (10.65) IMMISCIBLE DISPLACEMENT 384 z ft Sw fig.10.28(a) krw fig.10.28(b) kro fig.10.28(b) 0 .800 .300 0 5 .650 .170 .055 10 .470 .060 .195 15 .350 .020 .370 20 .275 .006 .540 25 .225 .002 .690 30 .200 0 .800 40 .200 0 .800 TABLE 10.9 Water saturation and point relative permeability distributions as functions of the reservoir thickness; fig.10.28(a) and (b). Graphically these values can be determined by measuring the area to the left of each curve in fig. 10.28(b) and dividing by the total thickness. For this initial saturation distribution, ro w k (S ) = 0.047 and ro w k (S ) = 0.481 After all this work all that has been obtained is one value of the thickness averaged saturation w S and the corresponding values of thickness averaged water and oil relative permeabilities rw w k (S ) and ro w k (S ) . These three values are only relevant for the initial assumption that the maximum water saturation has reached the base of the reservoir at the point in question. The next stage in generating averaged relative permeability curves is to allow the maximum water saturation, Sw = 1−Sor, to rise by an arbitrary amount and recalculate the new thickness averaged water saturation and relative permeabilities corresponding to the new water saturation distribution. This process is illustrated in fig. 10.29 in which the maximum water saturation is allowed to rise in 10 foot increments. Physically this corresponds to a series of saturation distributions observed at a fixed point in the reservoir as the flood passes that point. The main assumption which permits the drawing of the saturation distributions, shown in fig.10.29 (a)−(h), is that of vertical equilibrium described in sec. 10.2. This implies that as the average water saturation at the point of observation increases, the water and oil are instantaneously distributed in accordance with capillary-gravity equilibrium. Thus in fig. 10.29(a), above the height of 10 feet, the capillary pressure curve, fig. 10.27(a), is retraced to give the appropriate water saturation distribution. For each height rise of the saturation, Sw = 1−Sor, relative permeability distributions have also been plotted in fig. 10.29 and in each case values of Sw,krw w S and ro w k (S ) have been calculated by graphical integration, as described previously. These average values are listed in table 10.10 and plotted in fig. 10.30 as the circled points. IMMISCIBLE DISPLACEMENT 385 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 Sw Sw Sw Sw kr kr kr kr a Sw = .504 c Sw = .648 e Sw = .756 g Sw = .800 z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) b krw = .130 ; kro = .280 d k k rw ro = .203 ; = .120 h k k rw ro = .300 ; = 0 k (z) ro k (z) rw k (z) ro k (z) rw k (z) rw f k k rw ro = .269 ; = .025 k (z) rw k (z) ro Fig. 10.29 Water saturation and relative permeability distributions, as functions of thickness, as the maximum saturation, Sw = 1 − − − − Sor, is allowed to rise in 10 foot increments in the reservoir IMMISCIBLE DISPLACEMENT 386 w S rw w k (S ) ro w k (S ) c P (psi) ° .20 (Swc) 0 .8 5.0 .357 .047 .481 2.0 .504 .130 .280 1.0 .648 .203 .120 0 .756 .269 .025 -1.0 .800 .300 0 -2.0 TABLE 10.10 Thickness averaged saturations, relative permeabilities and pseudo capillary pressures corresponding to figs. 10.28 and 10.29 The rw w k (S )and ro w k (S ) relationships are the thickness averaged relative permeabilities and will be referred to as such in the remainder of this chapter. In the literature they are frequently called pseudo relative permeabilities, or simply "pseudo-curves". This author feels, however, that there are already a sufficient number of "pseudos" in reservoir engineering without introducing another to describe parameters derived from a simple averaging process. As might be expected the averaged relative permeability curves are intermediate between the rock curves (H >> h) and the linear relations for segregated flow (H << h). Using the averaged curves in effect reduces the description of the displacement process from two to one dimension, along the centre line of the reservoir. The curves can therefore be used in conjunction with the one dimensional Buckley Leverett theory by drawing the corresponding fractional flow curve and applying the practical graphical technique of Welge, described in sec. 10.5 and illustrated in exercise 10.2, to determine oil recovery as a function of cumulative water injected and time. The results obtained from such calculations, for the same linear reservoir model described in exercises 10.2 and 10.3 and at the same injection rate of 1000 rb/d (µo = 5 cp; µw = .5 cp), are shown in fig. 10.32 and again are intermediate between those obtained assuming the diffuse flow condition (table 10.5; fig. 10.17) and segregated flow (table 10.6; fig. 10.24). The graphical technique for determining averaged relative permeabilities is rather laborious and has only been included in this section for illustrative purposes. In practice it is very easy to compute the functions using a simple computer program for which the necessary input data include rock relative permeabilities and the capillary pressure function. Average saturations and permeabilities can then be obtained by numerically evaluating the integrals equs. (10.63-65) for different elevations of the maximum saturation Sw = 1−Sor. Also listed in table 10.10 are values of c P° the pseudo capillary pressure (a real "pseudo"!). This is simply the phase pressure difference between the oil and water at the centre of the reservoir and the relationship between pseudo capillary pressure and thickness averaged water saturation, fig. 10.31, is called the pseudo capillary pressure curve. IMMISCIBLE DISPLACEMENT 387 DIFFUSE FLOW (ROCK CURVES) SEGREGATED FLOW INTERMEDIATE (FINITE CAPILLARY TRANSITION ZONE) 1.0 .9 .8 .7 .6 .5 .4 .3 .2 .1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 krw k'ro kro k'rw Fig. 10.30 Averaged relative permeability curves for a homogeneous reservoir, for diffuse and segregated flow; together with the intermediate case when the capillary transition zone is comparable to the reservoir thickness PSEUDO CAPILLARY PRESSURE LABORATORY MEASURED CURVE 5 4 3 2 1 0 -1 -2 .2 .4 .6 1.0 .8 Sw Sw P P ° (psi) c c 0 5 10 15 20 25 30 time (yrs) 0 0 1 2 3 4 5 6 7 .6 .5 .4 .3 .2 .1 q; = 1000 rb / d DIFFUSE FLOW (EX. 10.2) SEGREGATED FLOW (EX. 10.3) INTERMEDIATE (H h) N (PV) pd W (PV) id Fig. 10.31 Capillary and pseudo capillary pressure curves. Fig. 10.32 Comparison of oil recoveries for different assumed water saturation distributions during displacement. Let po and pw be the oil and water pressures at any point in a horizontal reservoir at an elevation z above the base. If o p° and w p° are the corresponding pressures referred to the centre line of the reservoir then, if the reservoir has a total thickness h, the relations between po and o p° , and pw and, w p° , under conditions of hydrostatic equilibrium, are o o o 6 g h p p z (Darcy units) 2 1.0133 10 ρ ° = − − × and IMMISCIBLE DISPLACEMENT 388 w w w 6 g h p p z ( " " ) 2 1.0133 10 ρ ° = − − × Subtraction of these equations gives o w c c 6 g h p p P P z (atm) 2 1.0133 10 ρ ° ° ° ∆ − = = + − × (10.66) or, converting to field units o w c c h p p P P 0.4335 z (psi) 2 γ ° ° ° − = = + ∆ − (10.67) It is convenient to choose the value of z to coincide with the position of the maximum water saturation, Sw = 1−Sor, in the reservoir (i.e. or 1 S z z − = ). At this point po−pw = Pc = 0 and equations (10.66) and (10.67) can be reduced to or c 1 S 6 g h P z 2 1.0133 10 ρ ° − ∆ = − × (10.68) and or c 1 S h P 0.4335 z 2 γ ° − = ∆ − (10.69) respectively. Thus, in the reservoir being currently described (∆γ = 0.230; h = 40 ft.) equ. (10.69) becomes ( ) or c 1 S P 0.1 20 z ° − = − (10.70) From equ. (10.70) it can be seen that the pseudo capillary pressure will vary between 2.0 psi and -2.0 psi as z, the position of the saturation Sw = 1 − Sor varies between 0 and 40 feet, as shown in fig. 10.33. Pressure Pressure Pressure Z Z = 0 1 -S P = 2 psi ° c P = -2 psi ° c P = 0 ° c water oil water oil water oil Z = 20ft 1 -S Z = 40ft Z = 20ft Z = 0 Z = 40ft 1 -S or or or Fig. 10.33 Variation in the pseudo capillary pressure between +2 and -2 psi as the maximum water saturation Sw = 1− − − −Sor rises from the base to the top of the reservoir IMMISCIBLE DISPLACEMENT 389 Values of c P° as a function ofSw are listed in table 10.10, and the relationship plotted in fig. 10.31. In particular, the maximum value of c P° is included for Sw = Swc = .2, the connate water saturation. In this case, the water saturation is also 0.2 at the base of reservoir which, as shown in fig. 27(a), corresponds to a capillary pressure of at least 3 psi. Therefore the phase pressure difference at the centre of the reservoir must be, at least, c P° = 5 psi. Using the combination of averaged relative permeabilities and pseudo capillary pressure a one dimensional fractional flow equation, representing the average flow along the centre line of the reservoir, can be developed which is analogous to equ. (10.9). The only difference will be that the rock relative permeabilities will be replaced by averaged permeabilities and the capillary pressure gradient term c P x ∂ ∂ by c P x ° ∂ ∂ . And, just as the capillary pressure gradient was neglected in the oil recovery calculations described in sec. 10.5, so too, the c P x ° ∂ ∂ term is neglected when drawing the fractional flow curve for the present recovery calculations. The pseudo capillary pressure-saturation relationship, however, plays an important role in numerical reservoir simulation which will be described in sec. 10.10. The methods presented in this section can also be applied in a dipping reservoir. The averaging is again carried out in the dip normal direction with the result that the capillary pressures, evaluated in this section as dPc ∝ dz are replaced throughout by expressions of the form dPc ∝ cosθ dy where z is measured vertically upwards and y in the dip-normal direction from the base of the reservoir. 10.8 DISPLACEMENT IN STRATIFIED RESERVOIRS So far, displacement has only been considered in homogeneous, linear reservoirs. In stratified reservoirs in which there is a defined variation in reservoir parameters with thickness, in the dip-normal direction, the description of displacement is necessarily more complex. Nevertheless, the same basic method is used, as described in the previous section, namely, to generate thickness averaged relative permeabilities as functions of the thickness averaged water saturation. This will again reduce the mathematical description to one dimension permitting the use of the Buckley Leverett theory and the Welge graphical technique (sec. 10.5) for approximate calculations of the oil recovery. Two cases can be distinguished, which will both be dealt with in this section. Firstly, when there is pressure communication between the individual sand layers and vertical equilibrium pertains across the entire formation thickness and, secondly, when the IMMISCIBLE DISPLACEMENT 390 individual sands are isolated from one another by impermeable shale layers so that there is a total lack of pressure communication. a) With pressure communication between the layers Consider the case of a 40 ft. thick horizontal reservoir which can be subdivided into three homogeneous layers each with different thickness, porosity and permeability, as shown in fig.10.34. layer 3 layer 2 layer 1 k = 200mD; 3 k = 100mD; k = 50mD; 2 1 φ3 = 0.20; φ φ 2 1 = 0.17; = 0.15; h = 10 ft. 3 h h 2 1 = 20 ft. = 10 ft. h = 40 ft. Fig. 10.34 Example of a stratified, linear reservoir for which pressure communication between the layers is assumed If the water and oil again have specific gravities of 1.04 and 0.81, respectively then the capillary pressure is related to capillary rise by the equation dPc = 0.1 dz (psi) (10.62) Similarly, the pseudo capillary pressure equation for calculating the phase pressure difference at the centre of the reservoir is still or c 1 S P 0.1 (20 z ) − ° = − (10.70) where or 1 S z − is the elevation of the maximum water saturation, Sw = 1 − Sor, in the reservoir, at which point the phase pressure Pc is always zero. Laboratory measured relative permeabilities and capillary pressures for the three layers are shown in fig. 10.35. IMMISCIBLE DISPLACEMENT 391 k rw 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 k ro a OIL WATER Sw LAYER 1 (psi) Pc 5 4 3 2 1 0 0 .2 .4 .6 .8 1.0 z (ft) d Sw LAYER NUMBER 50 40 30 20 10 0 1 2 3 k rw 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 k ro c OIL WATER Sw LAYER 3 k rw 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 k ro b OIL WATER Sw LAYER 2 Fig. 10.35 (a)-(c) Rock relative permeabilities, and (d) laboratory measured capillary pressures for the three layered reservoir shown in fig. 10.34 Averaged relative permeability curves can be generated by allowing the maximum saturation, Sw = 1 − Sor, to rise in the reservoir and, assuming vertical equilibrium, determining for each selected value of or 1 S z , − values of Sw, rw w k (S )and ro w k (S ) as described in the previous section. In the current example, as will be shown presently, it is more convenient to use the pseudo capillary pressure as the independent variable, since the pressures in the oil and water are continuous across the reservoir thickness, whereas saturations are not. The values of Pc ° will vary between + 2 psi, when or 1 S z − is zero, and − 2 psi, when or 1 S z − = 40 ft, the latter corresponding to complete flood-out of the reservoir at the point of observation. In total, five values of c P° have been selected in the range between ± 2 psi for which the corresponding saturation and relative permeability distributions have been plotted in figs.10.36 and 10.37. IMMISCIBLE DISPLACEMENT 392 40 30 20 10 0 0 .2 .4 .6 .8 1.0 z (ft) Sw Sw = .396 a 40 30 20 10 0 0 .2 .4 .6 .8 1.0 z (ft) kr krw = .042 ; kro = .510 b P = 2 psi o c Fig. 10.36 (a) Water saturation, and (b) relative permeability distributions, with respect to thickness, when the saturation at the base of the layered reservoir (fig. 10.34) is Sw = 1− − − −Sor (Pc° = 2 psi) Consider the initial situation when c P° = 2 psi or or 1 S z − = 0. Table 10.11 lists the corresponding phase pressure difference across the reservoir together with the water saturation and relative permeability distributions which have been obtained from fig. 10.35(a) - (d). The oil and water pressures, and hence the phase pressure difference, are continuous across the reservoir. The water saturation, however, is discontinuous at the layer boundaries, fig. 10.36(a), since the saturation values in table 10.11 are obtained from the three capillary pressure curves shown in fig. 10.35(d). Thus at the interface between layers 1 and 2, where Pc = 1.0 psi, the water saturation in layer 1 is 0.69 and in layer 2 is 0.63. Similarly, since there are three sets of relative permeability curves there are also discontinuities in the relative permeability distributions, to oil and water, at the layer boundaries. Layer z (ft) Pc (psi) Sw krw kro 40 4.0 .2 0 .8 35 3.5 .2 0 .8 3 30 3.0 .2 0 .8 30 3.0 .22 .001 .55 25 2.5 .24 .003 .50 20 2.0 c (P ) ° .29 .02 .40 15 1.5 .45 .07 .18 2 10 1.0 .63 .17 .05 10 1.0 .69 .18 .02 5 .5 .78 .23 .002 1 0 0 .80 .24 0 TABLE 10.11 Phase pressure difference, water saturation and relative permeability distributions for c P° = 2 psi; fig. 10.36 IMMISCIBLE DISPLACEMENT 393 For the saturation distribution depicted in fig.10.36(a) the average water saturation can be evaluated mathematically as h h w w 0 0 S (z)S (z)dz (z)dz φ φ = which, since there are three distinct, homogeneous layers, can be expressed as 1 2 3 w w w 1 1 2 2 3 3 w 3 i i j 1 h S h S h S S h φ φ φ φ = + + = (10.71) in which, for instance, 1 1 h w w 1 0 S S (z)dz h = The average saturations, 1 2 3 w w w S ,S and S ,can be evaluated either graphically or numerically, as described in the previous section. Similarly, rw w k (S )can be evaluated as h rw w 0 rw w h 0 k(z)k (S (z))dz k (S ) k(z)dz = or 1 2 3 1 2 3 rw rw rw w w w 1 1 2 2 3 3 rw w 3 j j j 1 h k k (S ) h k k (S ) h k k (S ) k (S ) hk = + + = where, for instance 1 1 1 h rw w 0 rw w 1 k (S (z))dz k (S ) h = and similarly for ro w k (S ) . IMMISCIBLE DISPLACEMENT 394 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 40 30 20 10 0 0 .2 .4 .6 .8 1.0 Sw = .524 Sw = .634 Sw = .765 Sw = .800 z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) z (ft) b krw = .100 ; kro = .395 d k k rw ro = .158 ; = .257 h k k rw ro = .338 ; = 0 f k k rw ro = .280 ; = .026 a Sw Sw Sw Sw kr kr kr kr c e g P = -2 psi ° c P = -1 psi ° c P = 0 psi ° c P = 1 psi ° c kro krw kro krw kro krw kro krw Fig. 10.37 (a)-(h) Water saturation and relative permeability distributions,as functions of thickness,for various selected values of c P° (three layered reservoir, fig 10.34) IMMISCIBLE DISPLACEMENT 395 Values of c P , ° rw w w S ,k (S ) and ro w k (S ) are listed in table 10.12 and plotted in figs. 10.39(a) and 10.38(a). The latter three have been obtained by the graphical integration of the distributions shown in fig. 10.36(a) and (b) and fig. 10.37(a)-(h) for the five selected values of c P° c P° (psi) Sw krw kro 7.0 .200 0 .678 2.0 .396 .042 .510 1.0 .524 .100 .395 0 .634 .158 .257 -1.0 .765 .280 .026 -2.0 .800 .338 0 TABLE 10.12 Pseudo capillary pressure and averaged relative permeabilities corresponding to figs. 10.36 and 10.37 In particular, when Sw = 0.2, the connate water saturation, then using the capillary pressure curve for layer 1, fig. 10.35(d), the phase pressure at the base of the reservoir must be at least 5 psi and hence c P° = 7 psi. The method described above is perfectly general and can be applied equally well when the end point saturations also vary from layer to layer. The case studied is for the gradation of permeabilities from high to low from the top to the bottom of the reservoir. If the reservoir shown in fig. 10.34 were inverted, repetition of the above exercise would yield the averaged relative permeability and pseudo capillary pressure curves shown in figs. 10.38(b) and 10.39(a) (dashed curve). The corresponding fractional flow curves for the two cases are plotted in fig. 10.39(b) assuming that µo = 5 cp and µw = .5 cp, as in exercises 10.2 and 10.3. The latter curves indicate that the more favourable displacement occurs when the high permeability layer is uppermost in the reservoir. In this case the injected water will preferentially travel through the top layer and in doing so will be pulled downwards, due to the gravity difference between the water and oil, resulting in fairly even saturation distribution. If, however, the high permeability layer is at the base of the reservoir, capillary forces will tend to suck up the water but this is not as effective in creating a uniform saturation distribution. A special, simple case of the method of averaging presented in this section occurs when the capillary transition zone in each layer is negligible so that flow occurs under segregated conditions, as described in sec. 10.6. Suppose there are a total of N layers in the reservoir. When the water-oil interface has risen so that it coincides with the top of the nth layer, the thickness averaged water saturation is, in analogy with equ. (10.71) IMMISCIBLE DISPLACEMENT 396 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 b OIL WATER Sw 1.0 .8 .6 .4 .2 0 0 .2 .4 .6 .8 1.0 OIL WATER Sw krw kro kro krw a Fig. 10.38 Averaged relative permeability functions for the three layered reservoir, fig. 10.34: (a) high permeability layer at top, (b) at base of the reservoir 0 .2 .4 .6 .8 1.0 b Sw 3 2 1 0 -1 -2 .2 .4 .6 .8 fw 1.0 .8 .6 .4 .2 0 a P° c Sw Fig. 10.39 (a) Pseudo capillary pressure, and (b) fractional flow curves for the three layered reservoir, fig. 10.34). (——high permeability at top; − − − − − − − − − − − −at base of the reservoir). j j n n N j j or j j wc j 1 j n 1 w N j j j 1 h (1 S ) h S S h φ φ φ = = + = − + = (10.72) while the corresponding averaged relative permeabilities to water and oil are n n n n N rw w j j rw j j j 1 j 1 k (S ) hk k hk = = ′ = (10.73) and IMMISCIBLE DISPLACEMENT 397 n n n N N ro w j j ro j j j n 1 j 1 k (S ) hk k hk = + = ′ = (10.74) where j rw k′ and j ro k′ are the end point relative permeabilities to water and oil in the jth layer, for water saturations j or (1 S ) − and j wc S , respectively. Furthermore, as demonstrated in sec. 10.6, as the water-oil interface rises in any given layer the water saturation and relative permeabilities for the layer increase in direct proportion to the fractional thickness of the water. Thus the discrete values of n n n rw ro w S ,k and k calculated using equs. (10.72-74) are connected to the values calculated for n - 1 and n + 1 by straight lines. The averaged relative permeability curves are therefore made up of straight line segments as illustrated in the following exercise. EXERCISE 10.4 GENERATION OF AVERAGED RELATIVE PERMEABILITY CURVES FOR A LAYERED RESERVOIR (SEGREGATED FLOW) Generate and plot the averaged relative permeabilities and the pseudo capillary pressure relationship, as functions of the thickness averaged water saturation, for the 40 ft. thick reservoir described in this section, fig.10.34, assuming that flow will take place under segregated conditions (no capillary transition zone). The end point relative permeabilities and saturations for each layer may be taken from fig. 10.35 (a)-(c) and the specific gravities of the water and oil are 1.04 and 0.81, respectively. Repeat the exercise with the reservoir model inverted so that the high permeability layer is at the base of the reservoir. EXERCISE 10.4 SOLUTION Figure 10.40 includes all the data for the three-layered reservoir specified in the question. layer 3 h3 = 10ft.; k3 = 200mD; φ3 = .20; 3 rw k .4; ′ = 3 ro k .8; ′ = layer 2 h2 = 20ft.; k2 = 100mD; φ2 = .17; 2 rw k .6; ′ = 2 ro k .6; ′ = layer 1 h1 = 10ft.; k1 = 50mD; φ1 = .15; 1 rw k .25; ′ = 1 ro k .5; ′ = Fig. 10.40 Individual layer properties; exercise 10.4 In addition, the end point saturations in each layer are Sor = Swc = 0.2 and 3 j j j 1 h 6.9 ft φ = = 3 j j j 1 h k 4500 mD.ft = = IMMISCIBLE DISPLACEMENT 398 When there is only connate water in each layer the initial average saturation and relative permeabilities can be determined as (n=0) 0 3 3 w j j wc j j j 1 j 1 S h S h 0.2 φ φ = = = = 0 0 rw w k (S ) 0 = and 0 0 j 3 3 ro w j j ro j j j 1 j 1 k (S ) h k k hk 0.678 = = ′ = = As the water-oil interface rises in the reservoir, values of n n n rw w w S , k (S ) and n n ro w k (S )can be computed using equs. (10.72-74), as n varies from 1 to 3, giving the values listed in table 10.13. n n w S n rw k n ro k c P° 0 .200 0 .678 2.0 1 .330 .028 .622 1.0 2 .626 .161 .356 -1.0 3 .800 .339 0 -2.0 TABLE 10.13 For instance, when n=2, so that the water-oil interface has risen to the top of the middle layer, 2 1 2 3 3 w 1 1 or 2 2 or 3 3 wc j 1 j 1 S (h (1 S ) h (1 S ) h S ) / h φ φ φ φ = = − + − + = (10× .15×.8 + 20×.17×.8 + 10× .20×.2) /6.9 = 0.626 2 2 1 2 3 rw w 1 1 rw 2 2 rw j j j 1 k (S ) (h k k h k k ) / hk = ′ ′ = + = (10×50×.25 + 20×100×.3) /4500 = 0.161 and 2 2 3 3 ro w 3 3 ro j j j 1 k (S ) h k k / hk = ′ = = 10×200×.8 / 4500 = 0.356 The pseudo capillary pressures have again been calculated using the equation or c 1 S P 0.1 (20 z ) − ° = − (10.70) where or 1 S z − is the elevation of the water-oil interface above the base of the reservoir. IMMISCIBLE DISPLACEMENT 399 If the reservoir were inverted, the reader can verify that the average saturations, relative permeabilities and pseudo capillary pressures would be as listed in table 10.14, in which the layer ordering has now been reversed. The averaged relative permeabilities and pseudo capillary pressure functions are plotted in figs. 10.41 and 42(a), respectively, each consisting of a series of linear segments. n n w S n rw k n ro k c P° 0 .200 0 .678 2.0 1 .374 .178 .322 1.0 2 .670 .311 .056 -1.0 3 .800 .339 0 -2.0 TABLE 10.14 .9 .8 .7 .6 .5 .4 .3 .2 .1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 krw kro HIGH PERMEABILITY - TOP HIGH PERMEABILITY - BASE Sw Fig. 10.41 Averaged relative permeability curves; exercise 10.4 IMMISCIBLE DISPLACEMENT 400 b fw 1.0 .8 .6 .4 .2 0 .2 .3 .4 .5 .6 .7 .8 Sw .3 .4 .6 Sw 2.0 1.0 0 -2.0 P (psi) ° c a .8 .7 Fig. 10.42 (a) Pseudo capillary pressures, and (b) fractional flow curves, exercise 10.4 (—— High permeability layer at top; − − − − − − − − − − − −at base of reservoir) Inversion of the reservoir, so that the highest permeability layer is at the base, results in the functions represented by the dashed lines, which are simply reflections of the originals. Also shown in fig. 10.42(b) are the fractional flow curves for the two cases which again indicate that the displacement is much more favourable if the high permeability layer is at the top of the reservoir, since vertical fluid movement is dependent on gravity alone. While the fractional flow curve for the favourable case (solid line), increases continuously with increasing water saturation, the derivative of the curve, which is required in oil recovery calculations, does not. Therefore to use the curve, in conjunction with the Welge graphical technique for determining oil recovery, requires that it first be smoothed, as shown in fig. 10.42(b). This smoothing is unnecessary in the case of unfavourable displacement, with the high permeability layer at the base of the reservoir (dashed line). b) No pressure communication between the layers If the individual layers are isolated from one another by impermeable barriers so that there is a lack of pressure communication between them, there will be no flow of fluids in the dip normal direction. Even though injection and production wells are completed across all the layers the problem is one of dealing with a set of isolated reservoirs between which pressure communication can only exist in the wellbore. Stiles has presented a simple, approximate method14 for calculating oil recovery under these circumstances and further examples of its application have been presented by Craft and Hawkins15 and Colel6. Fundamental to the Stiles method is the assumption that piston-like displacement occurs in each separate layer, which is equivalent to assuming that the end point mobility ratio is unity. Furthermore, for this stable displacement the oil and water must have the same velocity which, applying Darcy's law, implies that IMMISCIBLE DISPLACEMENT 401 o w p p M x x ∂ ∂ = ∂ ∂ and since M = 1, the pressure gradients in the oil and water must be the same. Therefore, if a constant pressure difference ∆p is applied between injection and production wells, then o w p p p x x L ∂ ∂ ∆ = = ∂ ∂ in all layers, where L is the length of the reservoir block. Applying the Buckley Leverett theory to calculate the velocity of the water in the jth layer, equ. (10.19) j t w j j j w j q f v wh S φ ∆ = ∆ (10.75) where w is the reservoir width and, at the front, for piston-like displacement, j j w w or wc j f 1 S (1 S S ) ∆ = ∆ − − Finally, since j j j rw w t j w k k p q wh x µ ′ ∂ = − ∂ then equ. (10.75) can be expressed as j j j j rw j j w or wc k k p v (1 S S ) L φ µ ′ ∆ = − − One additional imposition, inherent in the assumption that M = 1 in all layers, is that the ratio of end point relative permeabilities, j j rw ro k /k ′ ′ , is constant. The layer order in which water will break through to the producing well will therefore be in the sequence of decreasing values of j j j j rw j j or wc k k (1 S S ) α φ ′ = − − (10.76) Application of the Stiles method to generate averaged relative permeability curves requires, as a preliminary, the re-ordering of the layer numbers in the sequence in which water breakthrough will occur. Thus the sand having the highest value of ∝, equ. (10.76), becomes layer number one; the sand having the second largest value of a becomes layer number two, etc. If there is a total of N layers in the reservoir then, observing the water breakthrough at the producing well, the average saturation, n w S , after breakthrough in the nth layer, can be calculated using an equation which is identical in form to equ. (10.72). Similarly, the IMMISCIBLE DISPLACEMENT 402 averaged relative permeabilities, n rw k and n ro k , can be calculated using equs. (10.73) and (10.74), respectively. The averaged relative permeabilities so generated are, in fact, independent of the position in the linear reservoir at which the flood is observed and are not restricted to application at the wellbore. This is because, although there is a difference in water velocity between any two sands, the difference remains constant as the flood advances; this being implicit in the assumption of piston-like displacement. When plotting the relative permeability curves, the separate points are not joined by line segments but, instead, by step functions since after breakthrough in layer n the relative permeabilities remain constant until breakthrough occurs in layer n+1 when there is a discrete jump to the new values of n 1 rw k + and n 1 ro k + . If there are a large number of separate layers in the reservoir the step functions can be smoothed into continuous curves. The main assumption in using the Stiles method, that M=1, should theoretically limit its application to reservoirs in which this condition is approximately satisfied. Nevertheless, it is frequently used as a starting point even when M > 1 and the resulting relative permeability curves adjusted so that calculated oil recovery matches that observed in pilot floods. 10.9 DISPLACEMENT WHEN THERE IS A TOTAL LACK OF VERTICAL EQUILIBRIUM The generation of averaged relative permeability curves, described in secs. 10.6-8, relies on the assumption that the oil displacement occurs under the vertical equilibrium condition (refer sec. 10.2). Because of this it is possible to define both saturation and relative permeability distributions with respect to thickness and averaging these facilitates the description of two dimensional displacement using simple, one dimensional equations. A similar procedure can also be adopted for precisely the opposite flow condition, that is, when there is an extreme lack of vertical equilibrium. This will occur when the fluid velocity parallel to the bedding-greatly exceeds the velocity in the dipnormal direction. In extreme cases the fluid saturations will be uniformly distributed with respect to thickness, in a homogeneous reservoir, and the displacement can be described using the rock relative permeability curves with the one dimensional Buckley-Leverett theory. The corresponding pseudo capillary pressure relationship for such displacement is simply c P° (Sw) = o p° − w p° = 0 In a stratified reservoir, a total lack of vertical equilibrium implies that the fluids will move through each layer in isolation from fluids in adjacent layers. Since there is negligible fluid movement in the dip-normal direction such displacement can be described in one dimension using the method of Stiles for generating averaged relative permeabilities17. The methods described in this chapter for reducing the description of oil displacement to one dimension are summarised in fig. 10.43. The starting point in this flow chart is IMMISCIBLE DISPLACEMENT 403 the decision concerning the validity of the assumption of vertical equilibrium. To attempt to describe oil displacement using ,simple analytical methods requires either the assumption that vertical equilibrium predominates, or alternatively, that it does not exist at all. In both cases averaged relative permeability curves can be generated for use with the Buckley-Leverett, one dimensional, displacement theory. Alternatively, the averaged relative permeability functions are used as basic input for the numerical simulation of the displacement process. The following section will describe the importance of using averaged relative permeabilities and pseudo capillary pressures in numerical simulation. In addition, the manner in which the assumption of vertical equilibrium can be verified, using numerical simulation, and the manner in which displacement can be described under conditions intermediate between total and non-existent vertical equilibrium, will also be described. IMMISCIBLE DISPLACEMENT 404 No Homogeneous Yes No VERTICAL EQUILIBRIUM Yes No Homogeneous Yes Capillary Transition Zone H << h H >> h No Yes Pressure Communication Yes No Capillary Transition Zone Stiles Method k S rw w ( ) as step functions (sec. 10.8) k S ro w ( ) Direct cal-culation of as straight line segments (sec. 10.8; ex. 10.4) k S k S rw w ro w ( ), ( ) (sec. 10.8) k S k S rw w ro w ( ), ( ) k k S ro rw w and are linear functions of (sec. 10.6) (sec. 10.7) k S k S rw w ro w ( ) ( ) Rock relative permeabilties Buckly-Leverett (application Welge sec. 10.5; ex. 10.2) OR Input Numerical Simulation (sec. 10.10) H h h w w 0 (z)S (z)dz S h φ = φ h w 0 S(z)dz S h = ˜ ˜ Fig. 10.43 Methods of generating averaged relative permeabilities, as functions of the thickness averaged water saturation, dependent on the homogeneity of the reservoir and the magnitude of capillary transition zone (H). The chart is only applicable when the vertical equilibrium condition pertains or when there is a total lack of vertical equilibrium . IMMISCIBLE DISPLACEMENT 405 10.10 THE NUMERICAL SIMULATION OF IMMISCIBLE, INCOMPRESSIBLE DISPLACEMENT So far, this text has concentrated on developing simple mathematical models to describe the physics of reservoir depletion and fluid flow. Examples of these are the zero dimensional material balance equation (Chapter 3) and the analytical solutions of linear second order differential equations for radial flow (Chapters 5-8). Sometimes, however, these simple models are totally inadequate to provide the required solutions to reservoir engineering problems. In a strong natural water drive reservoir, for instance, the zero dimensional material balance can be used to predict the volume of water influx, but it can never predict where the water will preferentially move in the reservoir and such knowledge may be required to determine the location of additional production or injection wells. Furthermore, not all reservoir problems can be formulated in terms of linear differential equations for which standard solutions can be sought. Nobody, for instance, would attempt to solve simultaneously the equations for three phase (oil — water—gas) flow in three dimensions with irregular boundary conditions, using an analytical approach. To deal with complex reservoir problems the engineer must resort to numerical simulation methods and particularly so for displacement problems, in which one of the main aims is to determine the areal distribution of fluids in the reservoir resulting from a flood. A numerical simulator is a computer program which permits the user to divide the reservoir into discrete grid blocks which may each be characterised by having different reservoir properties. The flow of fluids from block to block is governed by the principle of mass conservation coupled with Darcy's law. Flow into or out of a block, due to presence of an injection or production well, is also catered for. Most simulators are capable of solving large sets of second order differential equations for the simultaneous flow of oil, gas and water, in three dimensions. In addition the effects of natural water influx, fluid compressibility, mass transfer between gas and liquid phases and the variation of such parameters as porosity and permeability, as functions of pressure, can all be modelled. The differential equations themselves are generally formulated using the finite difference analogue for first and second order differentials and solved simultaneously using numerical techniques with some acceptable, small error attached to each solution. It is not intended, in this text, to describe numerical simulation in any great detail. Indeed, at the time of writing, such a description would require not merely another text book but a small encyclopaedia. Instead, since this chapter has concentrated on the generation of averaged relative permeabilities and pseudo capillary pressures, as functions of the thickness averaged water saturation; the description of simulation will be confined to the way in which these vitally important parameters are handled by the simulator. To accomplish this, the simple case of linear, incompressible displacement of oil by water will be considered, for which the appropriate mass conservation equations for reservoir fluids will be presented. This simplifies and concentrates the description since the pressure dependent terms, densities and viscosities, can safely be regarded as being independent of pressure. Thus only the variation in time and IMMISCIBLE DISPLACEMENT 406 space of the water saturation and saturation dependent functions is considered, but this description is equally applicable to more complex simulations. For those unfamiliar with numerical simulation methods, references 18-20, at the end of this chapter, provide a more general introduction to the subject, while the selected AIME papers21 give details of the more successful mathematical techniques employed, together with a selection of field case histories which have been interpreted using numerical simulation methods. Consider the displacement of oil by water in the linear, horizontal, homogeneous reservoir element shown in fig.10.44. y x z 1 2 i − 1 2 i + Fig. 10.44 Numerical simulation model for linear displacement in a homogeneous reservoir For simulation purposes the element has been subdivided into a row of discrete, regular grid blocks. The geometrical centre of each grid block is called a node and these points are numbered in increasing sequence from left to right. At first glance it may appear that this simple model is three dimensional since the geometry is defined by regular Cartesian coordinates. What the reservoir simulator deals with internally, however, are sets of grid block averaged data concentrated at each node and, since these are discrete points in space, this has the effect of reducing the flow description to one dimension. At any time the data used by the simulator, for each node, consists of the average oil and water pressures in each block together with the corresponding pressure dependent functions, the fluid densities and viscosities. In the case of incompressible displacement, these can be considered as constants. Also used are the average water saturation and the saturation dependent relative permeabilities and capillary pressures. Since no allowance is made for the variation of any parameter across the width of a grid block (y-direction), the average saturations used are the thickness averaged values,Sw, and the relative permeabilities are the values rw w k (S ) and ro w k (S ) , the generation of which, for a homogeneous reservoir, has been described in detail in sec. 10.7. Finally, the saturation dependent capillary pressure used is the pseudo capillary pressure or phase pressure difference at the centre of the grid block, which is a function ofSw. It is therefore necessary, even for such a simple, one dimensional problem, to provide as basic input rw w k (S ), ro w k (S ) and c P° w (S ), as functions ofSw (e.g. table 10.10), IMMISCIBLE DISPLACEMENT 407 rather than the rock relative permeability and capillary pressure functions measured on thin core plugs in the laboratory. For this, or any other problem, the basic physical principle employed by the simulator is that of mass conservation. Usually fluid quantities are conserved at stock tank conditions and related to reservoir fluid quantities through the pressure dependent PVT parameters. When studying the immiscible displacement of oil by water, however, there- is no loss in generality in applying mass conservation directly to both fluids in the reservoir. Considering the one dimensional flow of water through a small volume element of thickness dx, which may also contain a fluid source or sink (injection or production well), the mass conservation of water can be expressed as x x dx w w w w w w w w q q A dx ( S ) Q t ρ ρ φ ρ ρ + ∂ ′ − = + ∂ (10.77) This has the same form as equ. (10.13), which was used as the starting point of the derivation of the Buckley-Leverett equation, except that now a mass source term w w Q ρ ′ has been added; where w Q′ is the reservoir rate of water injection or production, for which it conventionally has a positive or negative sign, respectively. Reducing the equation to the form w w w w w w Q (q ) A ( S ) x t dx ρ ρ φ ρ ′ ∂ ∂ − = + ∂ ∂ and substituting for the reservoir flow rate qw using Darcy's equation for horizontal flow rw w w w p kk A q x µ ∂ = − ∂ gives rw w w w w p kk S Q x x t φ µ ∂ ∂ ∂ = + ∂ ∂ ∂ (10.78) in which it has been assumed that, for this incompressible displacement both ρw and φ are constant, and Qw = w Q / Adx ′ which is the injection or production rate per unit of reservoir bulk volume. Furthermore, in this, and in subsequent equations, the pressure should be interpreted as the average value at the centre of the reservoir and the viscosity evaluated at this same pressure. The analogous equation to equ. (10.78) for the mass conservation of oil is ro o o o o p kk S Q x x t φ µ ∂ ∂ ∂ = + ∂ ∂ ∂ (10.79) Instead of attempting to analytically solve either equ. (10.78) or (10.79), which was the approach employed in developing the Buckley-Leverett displacement theory, the IMMISCIBLE DISPLACEMENT 408 equations are expressed in finite difference form in the simulator and solved for the oil and water pressures and saturations, using the ancillary equations c P° = po −pw (pseudo capillary pressure) (10.80) w o S S 1 + = (10.81) and w o S S 0 t t ∂ ∂ + = ∂ ∂ (10.82) In expressing equs. (10.78) and (10.79) in finite difference form, the simulator considers time and space to be discrete rather than continuous as for analytical solutions. The time intervals, or time steps, are denoted by ∆t and, for this one dimensional problem, the space increments by ∆x. For grid blocks of equal length, ∆x is the length of a block, which is also the distance between successive nodes. Equation (10.78) is commonly formulated in finite difference terms using the so-called central difference approximation for the left hand side which, for flow through grid block i during the timestep ∆t, gives 1 1 2 2 i i n n rw rw n 1 n 1 n 1 n 1 w,i 1 w,i w,i w,i 1 2 w w i i n 1 n w w 1 kk kk (p p ) (p p ) ( x) (S S ) t µ µ φ + + + + + − + − + − − − ∆ = − ∆ (10.83) in which it has been assumed, for the moment, that the grid block does not contain a well. The superscript n is attached to parameters evaluated at time tn and n+1 to parameters evaluated at time tn+1, where tn+1 − tn = ∆t, the current time step. Similarly the subscripts i, i+½ and i-½ refer to parameters specified at the grid block nodes and grid block faces, fig. (10.44). Flow between grid block i − 1 and i and between i and i+1 is governed by the values of rw w kk / µ evaluated at the grid block faces i − ½ and i+½ , respectively. The manner of the spatial linkage of the left hand side of equ. (10.83) can be appreciated from fig.10.45. ∆x (pw,i − pw,i−1) ∆x i−½ k krw µw i - 1 i + 1 (pw,i+1 − pw,i) ∆x i+½ k krw µw Fig. 10.45 Spatial linkage of the finite difference formulation of the left hand side of equ (10.83). IMMISCIBLE DISPLACEMENT 409 Similarly, equ. (10.79), for oil, can be expressed in finite difference terms as 1 1 2 2 n n ro ro n 1 n 1 n 1 n 1 o,i 1 o,i o,i o,i 1 2 o o i i n 1 n o,i o,i 1 kk kk (p p ) (p p ) ( x) (S S ) t µ µ φ + + + + + − + − + − − − ∆ = − ∆ (10.84) and adding equs. (10.83) and (10.84), using equ. (10.82) to eliminate the sum of the time derivatives of the saturations, and equ. (10.80) expressed as n 1 n ,n 1 n 1 n 1 ,n n n w w c o w c o w P (S ) p p P (S ) p p + ° + + + ° = − ≈ = − (10.85) to substitute for po n+1, gives 1 1 2 2 1 1 2 2 n n rw ro ro n 1 n 1 ,n ,n o,i w,i c,i 1 c,i 2 w o o i i n n rw ro ro n 1 n 1 w,i w,i 1 c,i 1 w o o i i ,n 1 kk kk kk (p p ) ( P P ) ( x) kk kk kk (p p ) (P ) 0 µ µ µ µ µ µ + + + + + + + − − + + ° ° ° + − − − ∆ − + − − = (10.86) in which it is assumed, equ. (10.85), that while oil and water pressures may individually change during a time step, the corresponding change in the pseudo capillary pressure can be neglected. The finite difference expression of equ. (10.86) is not unique, but is the one which has been most commonly applied in reservoir simulation. In the equation the water pressures are all dated at the new time level n+1, at which they are unknown, whereas the kr/µ term, which is both saturation and pressure dependent, and the pseudo capillary pressures, equ. (10.85), which are saturation dependent, are all dated at the old time level n at which their values are known. Therefore, equ. (10.86), which was specifically formulated for the flow of water through grid block i, contains only three unknowns n 1 n 1 n 1 w,i 1 w,i w,i 1 p , p and p + + + − + but, when the equation is linked with the similar equations for all the grid blocks in the one dimensional model (i = 1........n), the set of equations can be solved simultaneously to determine the value of n 1 w p + in each block. This simultaneous solution for pressures at the new time level is referred to as being IMPLICIT in pressure. Having obtained the values of n 1 w p + , for each grid block, the remainder of the solution at time level n+1, and in the jth grid block, proceeds as follows: − substitute the calculated values of n 1 n 1 n 1 w,i 1 w,i w,i 1 p , p and p + + + − + in equ. (10.83) and solve for n 1 w,i S + . (This is referred to as an EXPLICIT solution for n 1 w,i S + , since it is the only unknown in the equation). − the value of n 1 w,i S + is used to calculate n 1 o,i S + , since n 1 w,i S + + n 1 o,i S + = 1, equ. (10.81) IMMISCIBLE DISPLACEMENT 410 − the value of n 1 w,i S + is used to determine ,n 1 c,i P + ° n 1 w,i (S ) + since the relationship between pseudo capillary pressure and average water saturation is part of the input data − the updated value of the pseudo capillary pressure is used to calculate n 1 o,i p + , since n 1 o,i p + = ,n 1 c,i P + ° + n 1 w,i p + , equ. (10.85). − and finally, the values of all the pressure and saturation dependent variables (densities, viscosities, and relative permeabilities) are all updated at time level n+1 using the newly determined pressures and saturations and the input tables of these variables, as functions of pressure and saturation. The procedure described above for the simulation of immiscible, incompressible displacement of oil by water is one that can be readily extended to three phase flow. The finite difference conservation equations for oil, gas and water, including the possibility of mass transfer between the gas and liquid phases, are again summed to eliminate the total time change in saturation during the time step g o w S S S 0 t t t ∂ ∂ ∂ + + = ∂ ∂ ∂ This results in a more complex form of the water pressure equation, (10.86), in which the oil and gas phase pressures can be eliminated using the pseudo capillary pressure functions n 1 n n c oil o w (P ) p p + ° ≈ − and n 1 n n c gas g o (P ) p p + ° ≈ − Thereafter, water pressures are solved for implicitly, and the remainder of the solution proceeds in very much the same way as described above. The method, in which pressures are determined IMPLICITLY and saturations EXPLCITLY, is known as the IMPES solution technique. One of the main disadvantages in its application is that in solving the water pressure equation, (10.86), at time level n+1, the saturation dependent relative permeabilities and pseudo capillary pressures are evaluated for the known water saturation at time level n. For problems in which the water saturation change in a grid block can be large during a time step, such as when sharp saturation fronts move through the reservoir, the evaluation o saturation dependent functions at the old time level can lead to severe instabilities in the IMPES solution. IMMISCIBLE DISPLACEMENT 411 w S w S w S (a) TIME LEVEL n (c) TIME LEVEL n + 2 (b) TIME LEVEL n + 1 i i i Fig. 10.46 Example of water saturation instability (oscillation) resulting from the application of the IMPES solution technique (− − − − − − − − − − − − correct, and     incorrect saturations) Consider, as an example, a water drive from left to right in the two blocks of a linear model, shown in fig. 10.46(a) - (c), in which the average saturations depicted by the dashed lines are correct, and by the solid lines, incorrect. During the time step from n to n+1 the water saturation in block i must increase due to the advancing water from block i − 1 (not shown). Since the water relative permeability in block i is held constant, commensurate with the saturation shown in fig. 10.46(a), then not enough water will flow out of block i during the time step, and consequently the water saturation will be artificially high in the block at time n+1. For the time step from n+1 to n+2, the water relative permeability is evaluated for this latter saturation and too much water will leave the block resulting, at time n+2, in too low a saturation in block i and too high a value in block i+1. The overall result will be an oscillation in solution for the water saturation. Instabilities of this sort, which are inherent in the IMPES solution method, can be overcome by drastically reducing the length of the time step, but this results in excessively long computing times. Blair and Weinaug22 developed a numerical simulator which employed implicitly evaluated permeabilities. This so-called fully implicit procedure greatly improved the stability of solution of the finite difference equations and consequently permitted the use of much larger time steps in the simulation. Unfortunately, this benefit was largely offset by the increased computational work per time step due to the difficulty of solving the sets of fully implicit finite difference equations. An alternative and computationally simpler method is that of semi-implicit formulation of the finite difference equations23,24,25, in which the saturation dependent relative permeabilities and pseudo IMMISCIBLE DISPLACEMENT 412 capillary pressures are evaluated at time level n+1 using the first order Taylor series expansion, i.e. n n 1 n r w r r t w dk k k S dS + = + ∆ and n ,n 1 ,n c w r c t w dP P P S dS + ° ° ° = + ∆ where ∆tS is the time change in saturation, n 1 n w w S S , + − and the first order derivatives are evaluated at time level n. Peaceman has demonstrated26 that this approach produces perfectly acceptable stability in the majority of numerical simulations, while Chappelear27 has illustrated some of the practical difficulties associated with its use and how they can be overcome. On substituting such expressions for n 1 ,n 1 r c k and P , + + ° in the finite difference equations (10.83) and (10.84); the latter equations can be algebraically manipulated in such a way that the saturation changes again disappear on addition since w o t t S S 0 ∆ + ∆ = The resulting water pressure equation, which is similar although more complex than equ. (10.86), can then be solved, following which the saturations can be determined as described previously for the IMPES method. The time required to solve the finite difference equations using the semi-implicit method is longer than when using explicit saturation dependent parameters, but this is more than compensated by the additional length of the time step which can be tolerated using the method. In describing the solution of equ. (10.86) the fact that a grid block could contain an injection or production well was temporarily overlooked. In such a case there will be an additional term, Qo + Qw, on the right hand side of the equation, which is the total oil and water rate per unit of reservoir bulk volume and is positive for injection and negative for production. Precisely how these rates are incorporated into the solution of equations such as equ. (10.86) depends on the way the simulator is being operated. There is a method of running the simulator in what is described as the material balance mode. When using this, one attempts to match the observed reservoir pressure history with the numerical simulation. In this mode the desired time steps are usually specified as input to the simulation and the observed injection and production rates of each fluid are assigned to each grid block, containing a well, as a function of time. In this case, since all the rates are specified they do not add any complication to the method of solution of equ. (10.86), described previously. As the outcome, the pressures calculated by the simulator, for each grid block containing a well, should be compared with the pressures measured in the field, using the dynamic grid block pressure method described in Chapter 7, sec. 7. This aspect of history matching is tantamount to calibrating the simulation model and will be recognised as being nothing more than a IMMISCIBLE DISPLACEMENT 413 sophisticated version of the history matching, using the volumetric material balance equation, described in Chapters 1, 3 and 9. In more normal application, however, it is necessary for the simulator to forecast the injection or production rate of each well. Considering a single well situated at the centre of a grid block the simulator must be able to determine, for the currently evaluated average fluid pressures and saturations, how much of each fluid can flow into or out of the grid block, through the well. Since pressure and saturation gradients do not exist within each individual grid block, but only between blocks, the physical modelling of well performance presents difficulties. Pressure-gradients can be very high in the proximity of the well which should require the subdivision of the grid block into a fine mesh of blocks close to the well in order to correctly account for these gradients. Such an approach would result in very high computer costs due to the large number of grid blocks involved in the simulation. An alternative, approximate method of modelling well performance is to design an analytical well model to account for either injection or production from a grid block. For instance, the total production of water from a block, qw = Qwx (net bulk volume of block), could be predicted as qw = α (pwf – pw) (10.87) in which pwf and pw are the current bottom hole flowing pressure and average grid block pressure for the water, respectively . The order of subtraction of these pressures ensures that water production is negative and injection Positive, in accordance with the normal convention. The term a is the productivity index of the well and is normally evaluated for steady state flow (Chapter 6, table 1) as rw e w w 2 kk h (Darcy units) r 1 ln S r 2 π α µ = − + or 3 rw e w w 7.08 10 kk h (field units) r 1 ln S r 2 α µ − × = − + where re is the equivalent grid block radius e x y r π ∆∆ = for a well draining from the centre of a block. The choice of a steady state well model is usually justified by the fact that transients, within a grid block, normally die out in a relatively short time compared to the average time step used by the simulator; and the condition at the block boundary is probably closer to steady state (open boundary — constant pressure) than semi-steady state (closed boundary — pressure depletion). IMMISCIBLE DISPLACEMENT 414 It has been found that, even for a simulation which can be satisfactorily run using the IMPES solution technique, the well model will be unstable unless extremely small time steps are employed. This is because of the saturation dependence of the relative permeability term in the productivity index α. For this reason the PI is always evaluated semi implicitly as n n 1 n t d S dS α α α + = + ∆ for each liquid. Thus the equation for the water rate, (10.87), at time n+1, in block i, which contains a well, is n w,i n 1 n n 1 w,i w,i w,i t wf,i w,i w,i d q ( S )(p p ) dS α α + + = + ∆ − (10.88) and a similar expression can be formulated for the oil phase. Such expressions are added to the right hand side of equs. (10.83) and (10.84), in grid blocks containing wells and addition of the equations, again with much algebraic manipulation, will lead to a modified form of the pressure equation, (10.86), in which the tS ∆ terms cancel on addition. The subsequent solution for phase saturations and pressures permits the calculation of the rates by back substitution of these values in the well model equation, (10.88), for both oil and water. It is intuitively obvious that since the term r kk / µ controls the flow of fluids from one grid block to the next, it should be evaluated at the grid block faces. What is not so obvious, however, is precisely how the values of the three parameters in this expression are determined at the block faces since they are only defined at the grid block nodes. The viscosities, which are pressure dependent, are considered to be slowly varying functions of position, especially in the case of incompressible displacement, and are therefore evaluated as the average of their values at adjacent grid block nodes; thus for regular grid block spacing µi+½ = (µi + µi+1) / 2 The absolute permeability, which normally remains constant throughout the simulation, is. determined as the harmonic mean of its value at adjacent grid nodes. Consider the general case of flow between grid blocks of unequal length, as shown in fig. 10.47. IMMISCIBLE DISPLACEMENT 415 i k i k 1 + 1 i 2 k + i p ∆ i p 1 ∆ + i x 1 ∆ + i x ∆ p ∆ Fig. 10.47 Determination of the average, absolute permeability between grid blocks of unequal size Assuming, for the calculation of absolute permeability, that the reservoir is completely saturated with a single fluid and that the flow rate in each block is the same (qi = qi+l = q). Then applying Darcy's law in each grid block i i i x q p A k µ ∆ ∆ = and i 1 i 1 i 1 x q p A k µ + + + ∆ ∆ = But the total pressure drop across the two blocks is i´ 1 i i i 1 i i 1 i i 1 k x k x q p p p A kk µ + + + + ∆ + ∆ ∆ = ∆ + ∆ = Also 12 i i 1 i ( x x ) q p A k µ + + ∆ + ∆ ∆ = where ki+½ is the average permeability for flow through the blocks and is the value assigned at the grid block face i+½. Therefore, equating these two expressions for ∆p gives 12 i i 1 i i 1 i i 1 i i i 1 kk ( x x ) k (k x k x ) + + + + + ∆ + ∆ = ∆ + ∆ or, in the more normal case, for equal grid block sizes 12 i i 1 i i 1 i 2 kk k k k + + + = + It is more difficult to extrapolate the averaged relative permeabilities from the grid block nodes, at which they are specified, to the grid block faces. This is due to the saturation dependence of this parameter which means that it can vary substantially between one grid block node and the next. Todd et.al.28 have demonstrated that a satisfactory IMMISCIBLE DISPLACEMENT 416 means of extrapolation is by the method of two point upstream weighting. Suppose flow in the one dimensional reservoir is in the direction of increasing values of i, then the opposite direction is "upstream" and 12 i r,i r,i x kr k k x 2 + ∆ ∂ = + ∂ in which, for unequal grid block lengths, r k / x ∂ ∂is established from the two upstream nodes to give 12 r,i r,i 1 r,i r,i i i i 1 (k k ) k k x ( x x ) − + − − = + ∆ ∆ + ∆ or for equal grid block lengths 12 r,i r,i r,i-1 3 1 k k k 2 2 + = − Similarly if flow is in the opposite direction, equ. (10.89) becomes 12 r,i r,i 1 r,i 2 3 1 k k k 2 2 + + + = − The direction of flow at the time of evaluation is decided by the potential difference (pressure difference in a horizontal reservoir) between adjacent grid blocks. Todd, et.al.28 have also shown that a method such as two point upstream weighting cannot be inserted in a numerical simulator without careful screening. Consider, for instance, the very simple problem of piston-like displacement of oil by water as shown in fig. 10.48. i - 1 i w S wc S or I S − x rw k 0 x l l rw rw rw rw k k k k = = 1 2 i ∆ + l 1 rw rw 2 1 k , i k 2 + = − rw k 0 = Fig. 10.48 Overshoot in relative permeability during piston-like displacement IMMISCIBLE DISPLACEMENT 417 When the sharp water front is about to enter block i the value of the relative permeability to water at i+½ is − ½ k´rw, equ. (10.89). This negative relative permeability is referred to as overshoot and can obviously occur any time rw,i 1 rw,i k 3k −> . This situation can arise when a shock front water saturation discontinuity moves through the reservoir. Such overshoot is forbidden and is overcome by imposing the condition that rw k 0 ≥ . By considering the oil saturation and relative permeabilities corresponding to the situation shown in fig. 10.48, the reader can verify that another condition which must be imposed is that 12 12 ro,i ro,i ro,i k k the greater of k + + ≤ The above two rules controlling overshoot are generally applicable for any type of displacement; that 12 rw,i k + is not allowed to be negative and 12 ro,i k + must be less than or equal to the greater of the values at adjacent grid nodes. Although more difficult to code than single point upstream weighting 12 r,i r,i (k k ) + = , the two point method is more accurate and, using it, larger grid block sizes can be tolerated. This reduces the program running time and therefore, simulation costs. Actual reservoir simulation studies are usually undertaken to determine the areal distribution of fluids in the reservoir, resulting either from natural water influx or water injection . As such they require the building of a two or three dimensional model of the reservoir in which variation in the basic reservoir parameters, such as porosity, permeability etc., can be allowed for by assigning different values of these parameters to each grid block. Once a model has been constructed, which closely resembles the reservoir, it is then necessary to select relative permeabilities and a pseudo capillary pressure function which characterise the manner in which the water saturation is distributed in the dip-normal direction, as the flood moves through the reservoir. The manner in which this can be done is described below. a) Generate a set of averaged relative permeabilities and the pseudo capillary pressure for a typical linear cross section through the reservoir, between the injection and production wells, assuming vertical equilibrium. b) Set up two simulator models of the same cross section through the reservoir. Both models should have only one grid block in the y direction but, in the dip-normal direction, one of them should be one grid block thick, fig. 10.49(a) IMMISCIBLE DISPLACEMENT 418 (a) (b) Y Z X MODEL A MODEL B Fig. 10.49 Alternative linear cross sectional models required to confirm the existence of vertical equilibrium (model A), using thickness averaged porosity and absolute permeability, while the second, fig. 10.49(b) (model B) should be constructed with a large number of layers, irrespective of whether the reservoir is layered or homogeneous. c) Simulate the displacement of oil by water in both models using the thickness averaged relative permeabilities and pseudo capillary pressure functions in model A, and the relevant rock relative permeabilities and actual capillary pressures, in each separate layer, in model B. If the breakthrough times and subsequent oil recovery as a function of time are the same in both models then it means that the assumption of vertical equilibrium, implicit in the generation of the averaged relative permeability curves, is valid. Since vertical equilibrium is less likely to occur at high flow rates the comparison between the simulations should be made using the maximum water injection rate to be used in the study. d) If the agreement between the cross sectional simulations is poor, so that vertical equilibrium does not pertain, re-run model A using averaged relative permeabilities generated using the Stiles method, if the reservoir is layered, or the rock curves if the reservoir is homogeneous in the dip-normal direction. Agreement between simulations with models A and B means, in this case, that there is an extreme lack of vertical equilibrium. For either case c) or d), if agreement is achieved, the cross sectional model can be reduced to one dimension and the total three dimensional simulation to an areal, two dimensional model, with consequent reduction in program running time and considerable cost saving. In addition, it is worthwhile comparing the results of model A with those obtained using the analytical method of Buckley-Leverett, using the averaged relative permeability curves. Favourable comparison, in this latter case, gives the engineer added confidence in using the simulator. The comparison between the analytical and simulated results is never exact for, although both methods are based on the conservation of (water) mass, there are differences. The Buckley-Leverett theory, for instance, ignores the effect of the capillary pressure gradient behind the flood front, whereas capillary effects are included in the simulation. The analytical solution, however, uses continuous IMMISCIBLE DISPLACEMENT 419 space and time, whereas these assume discrete values in the simulator. Because of this, shock front displacement is never accurately modelled with a finite difference simulator. Examples of the comparison between analytical and simulated shock front displacement have been presented by Todd, et.al.28 e) If agreement is not achieved between the cross sectional simulations in either c) or d) above, then no definite assumption can be made concerning the vertical equilibrium condition. In this case model A should be re-structured with two layers of grid blocks in the dip-normal direction, each layer having its own averaged porosity and permeability. Step a) should be repeated, generating averaged relative permeabilities and pseudo capillary pressures for each layer. Simulations with this two layer model should again be compared with the results from model B. Agreement, in this case, means that while vertical equilibrium may not apply across the entire thickness of the reservoir, it is appropriate across half the reservoir. If there is still no agreement between the cross sectional simulations, model A should be re-structured as a three or four layered model and the process repeated until satisfactory agreement is achieved. In this manner it is usually possible to significantly reduce the number of layers. The procedure outlined above should be undertaken at the commencement of any large numerical simulation study. It can usually be conducted very quickly and at only a small fraction of the total study cost. If the reservoir is not areally homogeneous, so that a cross section between injection and production wells which is representative of the entire reservoir cannot be selected, then several cross sections must be chosen. It will then be necessary to repeat step a) — the generation of averaged relative permeabilities and pseudo capillary pressure for each cross section. If, however, vertical equilibrium has been clearly validated in one cross section, it is fairly safe to assume its validity throughout the reservoir, unless there are extremes of thickness or vertical permeability. The relative permeability curves obtained from the above study are input to the reservoir simulation model which has, hopefully, by this stage, been reduced to two dimensions. Simulations are then conducted to try and match the available history data. Following a successful history match, the simulator can be used to predict the future reservoir performance for different proposed production policies, well positions etc.; and, as Coats18 has observed, although a reservoir can only be produced once, its performance can be simulated in many different ways and at comparatively low cost. REFERENCES 1) Craig, F.F., Jr., 1971. The Reservoir Engineering Aspects of Waterflooding. SPE Monograph: Chapter 1. 2) Morrow, N.R., 1976. Capillary Pressure Correlations for Uniformly Wetted Porous Media. J.Can.Pet.Tech., October-December: 49-69. 3) Amyx, J.W., Bass, D.M. and Whiting, R.L., 1960. Petroleum Reservoir Engineering - Physical Properties, McGraw-Hill: 176-196. IMMISCIBLE DISPLACEMENT 420 4) Dumore, J.M., 1974 Drainage Capillary Pressure Functions and their Computation from One Another. Soc.Pet.Eng.J., October: 440 5) Coats, K.H., Dempsey, J.R. and Henderson, J.H.,1971. The Use of Vertical Equilibrium in Two Dimensional Simulation of Three Dimensional Reservoir Performance. Soc. Pet.Eng.J., March: 63-71. Trans. AIME. 6) Coats, K.H., Nielsen, R.L., Terhune, Mary H. and Weber, A.G., 1967. Simulation of Three Dimensional, Two Phase Flow in Oil and Gas Reservoirs. Soc.Pet.Eng.J., December: 377-388. Trans. AIME. 7) Buckley, S.E. and Leverett, M.C.,1942. Mechanism of Fluid Displacement in Sands. Trans. AIME. 146: 107-116. 8) Welge, H.J., 1952. A Simplified Method for Computing Oil Recovery by Gas or Water Drive. Trans. AIME. 195: 91-98. 9) Hagoort, J., 1974. Displacement Stability of Water Drives in Water Wet Connate Water Bearing Reservoirs. Soc.Pet.Eng.J., February: 63-74. Trans. AI ME. 10) Jacquard, P. and Seguier, P., 1962. Mouvement de Deux Fluides en Contact dans un Milieu Poreux. J. de Mechanique, Vol. 1:25. 11) Dietz, D.N., 1953. A Theoretical Approach to the Problem of Encroaching and By-Passing Edge Water. Akad. van Wetenschappen, Amsterdam. Proc. V.56-B: 83. 12) van Daalen, F. and van Domselaar, H.R., 1972. Water Drive in Inhomogeneous Reservoirs - Permeability Variations Perpendicular to the Layer. Soc.Pet.Eng.J., June: 211 -219. Trans. AIME. 13) Richardson, J.G. and Blackwell, R.J., 1971. Use of Simple Mathematical Models for Predicting Reservoir Behaviour. J.Pet.Tech., September: 1145-1154. Trans. AIME. 14) Stiles, W.E., 1 949. Use of Permeability Distribution in Water Flood Calculations. Trans. AIME,186:9. 15) Craft, B.C. and Hawkins, M.F., Jr., 1959. Applied Petroleum Reservoir Engineering. Prentice-Hall, Inc. New Jersey: 393-406. 16) Cole, F.W., 1961. Reservoir Engineering Manual. Gulf Publishing Company, Houston, Texas: 200-213. 17) Hearn, C.L., 1971. Simulation of Stratified Waterflooding by Pseudo Relative Permeability Curves. J.Pet.Tech., July: 805. 18) Coats, K.H., 1969. Use and Misuse of Reservoir Simulation Models. J.Pet.Tech., November: 1391-1398. Trans. AIME. 19) Staggs, H.M. and Herbeck, E.F., 1971. Reservoir Simulation Models - An Engineering Overview. J.Pet.Tech., December: 1428-1435. Trans. AIME. IMMISCIBLE DISPLACEMENT 421 20) O'Dell, P.M., 1974. Numerical Reservoir Simulation: Review and State of the Art. Paper presented at 76th National AlChE Meeting, Tulsa, Oklahoma. March. 21) 1973. Numerical Simulation. SPE Reprint Series No. 11. Society of Petroleum Engineers of AIME, Dallas, Texas. 22) Blair, P.M. and Weinaug, C.F., 1969. Solution of Two Phase Flow Problems Using Implicit Difference Equations. Soc.Pet.Eng.J., December: 417-424. Trans. AIME. 23) MacDonald, R.C. and Coats, K.H., 1970. Methods for Numerical Simulation of Water and Gas Coning. Soc.Pet.Eng.J., December: 425-436. Trans. AIME. 24) Letkeman, J.P. and Ridings, R.L.,1970. A Numerical Coning Model. Soc.Pet.Eng.J., December: 418-424. Trans. AIME. 25) Nolen, J.S. and Berry, D.W., 1972. Tests of Stability and Time-Step Sensitivity of Semi-lmplicit Reservoir Simulation Techniques. Soc.Pet.Eng.J., June: 253-266. Trans. AIME. 26) Peaceman, D.W., 1977. A Nonlinear Stability Analysis for Difference Equations Using Semi-lmplicit Mobility. Soc.Pet.Eng.J., February: 79-91. 27) Chappelear, J.E. and Rogers, W.L., 1974. Some Practical Considerations in the Construction of a Semi-lmplicit Simulator. Soc.Pet.Eng.J., June: 216-220. 28) Todd, M.R., O'Dell, P.M. and Hirasaki, G.J., 1971. Methods for Increased Accuracy in Numerical Reservoir Simulators. Soc.Pet.Eng.J., December: 515-530. Trans. AIME. AUTHOR INDEX Agarwal, R.G. Alden, R.C. Al-Hussainy, R. Amyx, J.W. Aziz, K. Bass, D.M. Bell, W.T. Bentsen, R.G. Berry, D.W. Biot, M.A. Blackwell, R.J. Blair, P.M. Boburg, T.C. Bradley, J.S. Brar, G.S. Breitenback, E.A. Brons, F Brown, G.G. Bruns, J.R. Bruskotter, J.F. Buckley, S.E. Burrows, D.B. Campbell, J.M. Carr, N.L. Carslaw, H.S. Carter, R.D. Chapman, R.E. Chappelear, J.E. Coats, K.H. Cobb, W.M. Cole, F.W. Craft, B.C. Craig, F.F. Crawford, P.B. Daalen, F. van Darcy, H. Dempsey, J.R. Denson, A.H. Dietz, D.N. Dodson, C.R. Doh, C.A. Domselaar, H.R. van Donohue, D.A.T. Dowdle, W.L. Dranchuk, P.M. Dumoré, J.M. Earlougher, R.C., Jr. Essis, A.E. Everdingen, A.F. van Fetovitch, M.J. Field, R.Q. Geertsma, J. Gewers, C.W.W. Goodrich, J.H. Goodwill, D. Gulati, M.S. Hagoort, J. Hall, K.R. Hastings, J.R. Havlena, D. Hawkins, M.F., Jr. Hazebroek, P. Hearn, C.L. Heintz, R.C. Henderson, J.H. Herbeck, E.F. Hirasaki, G.J. Horner, D.R. Hurst, W. Jacquard, P. Jaeger, J.C. Jones, L.G. Katz, D.L. Kazemi, H., Jr. Kentie, C.J.P. Kersch, K.M. King Hubbert, M., Kingston, P.E. Klinkenberg, L.J. AUTHOR INDEX 423 Knaap, W. van der Ko, S. Kobayashi, R. Kumar, A. Langenheim, R.H. Lantz, R.B. Lebourg, M. Letkeman, J.P. Leverett, M.C. Lynch, E.J. McCain, W.D. MacDonald, R.C. McFarlane, R.C. McKinley, R.M. McMahon, J.J. Marting, V.E. Marx, J.W. Marthar, L. Matthews, C.S. Mayer, E.H. Meitzen, V.C. Merle, H.A. Miller, F.G. Millers, S.C., Jr. Moreland, E.E. Morrow, N.R. Mueller, T.D. Nichol, L.R. Nielsen, R.L. Niko, H. Nolen, J.S. Northern, I.G. Oberfell, G.B. Odeh, A.S. O'Dell, P.M. Opstal, G.H.C. van Peaceman, D.W. Perry, G.E. Pinson, A.E., Jr. Poollen, H.K. van Quon, D. Raghavan, R. Ramey, H.J., Jr. Rawlins, E.L. Reudelhuber, F.O. Richardon, J.G. Ridings, R.L. Riley, H.G. Rogers, W.L. Russell, D.G. Schellhardt, M.A. Schilthuis, R.J. Schueler, S. Schultz, A.L. Seguier, P. Selig, F. Smith, J.T. Staggs, H.M. Standing, M.B. Stiles, W.E. Takacs, G. Teeuw, D. Terhune, Mary H. Thomas, G.W. Thurnau, D.H. Timmerman, E.L. Todd, M.R. Troost, P.J.P.M. Urbanosky, H.J. Walstrom, J.E. Wattenbarger, R.A. Weber, A.G. Weinaug, C.F. Welge, H.J. Whiting, R.L. Wong, S.W. Yarbourough, L. SUBJECT INDEX Abnormal fluid pressure Absolute permeability measurement of Acceleration project Acid treatment Afterflow Afterflow analysis Earlougher and Kersch McKinley Ramey Russell Al-Hussainy, Ramey and Crawford: real gas flow Amerada pressure gauge Aquifer average pressure, bounded, compressibility, total, constant, encroachable water encroachable angle fitting infinite acting model pressure decline productivity index steady state Aquifer expansion time dependence of Aquifer geometry: general, linear, radial, Average reservoir pressure, Averaged relative permeability and saturation (see Thickness averaged relative permeability) Bachaquero field (Venezuela), Back pressure testing, Basic differential equation, radial flow, conditions of solution, derivation of, linearization, real gas, linearization, general, linearization, liquid, Basic well test analysis equation: general, for liquid, for real gas, Beykan field (Turkey), Bolivar Coast fields (Venezuela), Boltzmann's transformation, Boundary conditions, constant pressure, infinite, no-flow, Breakthrough: of water in a producing well, premature, saturation, time, Brent field (North Sea), Brons and Marting, partial well penetration, Bubble point line, pressure, Buckley-Leverett one dimensional displacement, stability of, Buckley-Leverett equation, derivation of, mathematical difficulty with, Buildup (see Pressure buildup) SUBJECT INDEX 425 By-passing: critical rate to prevent, oil by gas, oil by water, Capillary: force, -gravity equilibrium, rise, capillary tube experiment, rise, in the reservoir, tube, Capillary pressure, equation, Laplace equation, negative value, psuedo (see Psuedo capillary pressure) Capillary pressure curve (function): laboratory determination, oil-water, drainage, oil-water, imbibition, Capillary pressure gradient, neglected in description of segregated flow, neglect in fractional flow equation, Capillary transition zone, allowance for in displacement calculations, (homogeneous reservoir), (layered reservoir), compared to reservoir thickness, lack of, Carbon dioxide flooding, Coats, K.H., Coefficient of inertial resistance, dependence on liquid saturation, experimental determination of units of, Combination drive, Compaction, cell, derive, Compaction curve, hysteresis of, Comparison between p2 and m(p) solution techniques, real gas flow, Compressibility, isothermal, application of basic definition, aquifer, total effective, saturation weighted, (undersaturated reservoir) fluids in the well, oil, pore, total, gas saturated reservoir, total, saturation weighted (undersaturated reservoir), water, Compressibility program for generating thickness averaged relative permeabilities, numerical simulation, well test analysis, Condensate, retrograde, Conduction heat transfer, to cap and base rock, reservoir from cap and base rock, Coning, Connate water saturation, expansion of, Constant terminal pressure solution (radial diffusivity equation), Constant terminal rate solution (radial diffusivity equation): liquid, real gas, semi-steady state flow (liquid, (real gas) transient flow (liquid), (real gas) Constant volume depletion experiment, Contact angle, hysteresis of, SUBJECT INDEX 426 Convection heat loss, Correction of measured wellbore pressure to a datum level, Correction of PVT differential data to allow for surface separation, Cricodentherm, Critical: point, pressure, saturation (see Saturation, critical) temperature, Critical rate for by-passing: oil by gas, oil by water, Damage to formation, Darcy: Henry, the, units, Darcy's experiment, flow coefficient, gas, law, linear flow, law radial flow, Datum plane, pressure, Depletion drive, gas reservoir, development planning: gas field, general, Dew point line, Dietz: determination of average pressure in a buildup survey, segregated displacement, shape factors, Differential liberation experiment, Diffuse flow conditions, Diffusivity constant, Diffusivity equation (see Radial diffusivity equation) Dimensionless cumulative oil production: moveable oil volumes, pore volumes, Dimensionless cumulative water influx function, maximum value of, bounded aquifer, plots of , superposition of, Dimensionless cumulative water injected: moveable oil volumes, pore volumes, Dimensionless pressure functions: definition of difficulty in application, general expression for any value of the flowing time, generation of, semi-steady state, superposition of, transient, Dimensionless pressure/pseudo pressure function for any fluid, superposition of, transient, Dimensionless pseudo pressure, gas-oil, Dimensionless radius: aquifer/reservoir, reservoir/wellbore, Dimensionless real gas pseudo pressure function, difficulty in application, general expression for any value of the flowing time, generation of, semi-steady state, superposition of, transient, SUBJECT INDEX 427 Dimensionless time, tD: aquifer, reservoir, Dimensionless time, tDA, difficulty in evaluation for a real gas, Dimensionless time for semi-steady state flow, (tDA)SSS, Dimensionless variable, pressure analysis, reasons for use, Dip-angle of reservoir, effect on the fractional flow equation, Displacement: immiscible (see Immiscible displacement) miscible, Dissolved gas (see Gas, disolved) Distillation of crude oil in the reservoir, Distortion of radial flow in the vicinity of a well, Dodson's PVT experiment, Drainage (capillary), Drill stem test, Drilling mud, Drive mechanisms (see Reservoir drive mechanisms) Dry gas re-cycling, Dynamic grid block pressure, Earlougher, R.C., Jr: afterflow anlaysis, digitized MBH charts, dynamic grid block pressure, generation of MBH charts, pressure drawdown analysis, Effective flowing time, Effective permeability, curves, oil-water, saturation dependence, Effective permeability, measurement of: afterflow analysis buildup analysis, gas-oil, " " , oil, 9 " " , oil, real gas, multi-rate drawdown analysis, oil multi-rate drawdown analysis, real gas, single rate drawdown analysis, oil, " " " " , real gas, Effective well bore raduis, End point: mobility ratio, gas/oil, mobility ratio, water/oil, relative permeabilities, gas-oil, relative permeabilities, water-oil, Equations of state: ideal gas, real gas, van der Waals, Equivalent gas volume, Erosion of surface, Essis-Thomas analysis, Euler's constant, Everdingen, A.F. van, mechanical skin factor (see Mechanical skin factor) Expansion of reservoir fluids, Exploration well, Exponential integral, logarithmic approximation, Fetovictch, M.J., comparison between calculated water influx and that of Hurst and van Everdingen, equations for water influx modified method for large aquifers, Final flow rate, Finite difference approximation Flash expansion of: oil sample to determine the bubble point pressure, separator oil sample, SUBJECT INDEX 428 unit volume of bubble point oil through model separators, Flow charts: calculation of flowing pressure in a gas well, Kazemi iteration, predication of pressure decline in a water drive gas reservoir, Flowing pressure survey, Flowing time Fluid: contacts in the reservoir, distributions in the well, potential equilibrium, Fluid samples: subsurface collection, surface recombination, Forchheimer equation, Formation volume factor: gas (see Gas formation volume factor) oil (see Oil " " ") water (see Water formation volume factor) Fractional flow, calculation of, surface conditions, Fractional flow curve tangent to, Fractional flow equation: dependence on water saturation, derivation of, derivation of horizontal, diffuse flow, horizontal, segregated, unstable displacement, influence of capillary pressure gradient, influence of dip angle (gravity), Fracturing fluid, Gas composition, compressibility (isothermal), condensate, constant, universal, density, disposal, dissolved (solution), equation of state (see Equation of state) expansion factor, expansion in the reservoir, flow, (see Real gas flow) flow velocity in the reservoir formation volume factor, gravity, ideal, injection, liberated, solution, material balance (see Material balance, gas) pressure gradient, PVT experiment, slippage, viscosity-compressibility product, volume equivalent, Z-factor (see Z-factor) Gascap, drive, expansion, secondary, Gas in place: apparent, in water drive gas reservoir, initial, GIIP, Gas-oil contact, Gas-oil ratio: control, cumulative, equation, producing (instantaneous), solution, solution, initial, of bubble point oil (differential), SUBJECT INDEX 429 solution, initial, of bubble point oil (flash), Gas recovery, Gas saturation, critical, residual, to water, Gas-water contact, Gas well testing: general analysis theory, buildup, multi-rate drawdown, Grain (matrix) pressure, Gravity: force, segregation, Gravity number: gas displacing oil, water displacing oil, Hall-Yarborough, calculation of Z-factors, Havlena-Odeh, interpretation of material balance, gascap drive, natural water drive, solution gas drive, History matching: reservoir performance, pressure, Horner buildup plot: gas-oil general (any fluid), oil, real gas, Hot water injection, Hurst and van Everdingen constant terminal pressure solution (radial diffusivity equation), constant terminal rate solution (radial diffusivity equation), unsteady state water influx theory, Hydrocarbon: accumulations, columns, migration, phase behaviour, recovery (see Recovery of hydrocarbons) Hydrocarbon pore volume (HCPV) gascap, reduction of during depletion, Hydrostatic: equilibrium, pressure regime (abnormal), " " (normal), Image wells, Imbibition, Immisible fluids, oil by gas, oil by water, physical assumptions, water by oil, with total lack of vertical equilibrium, Incompressible fluids, Increasing: effective permeability, well penetration, productivity index, Infinite: acting aquifer, " reservoir, closed in time, Inflow equations: aquifer, semi-steady state (oil), (gas), steady state (oil), " " , steam soaked well In-situ combustion, Interface between immiscible fluids, Interfacial tension, Kapuni field (New Zealand), Kayaköy field (Turkey), Kazemi, H., Jr., SUBJECT INDEX 430 Klinkenberg effect, Kirchhoff integral transformation, Laplace, capillary pressure equation, steady state equation, transform Late transient flow, Line transient flow, 153-54, 304 Late transient flow, Line source solution (radial diffusivity equation) liquid, real gas, Linear prototype reservoir, Linearization (see Basic differential equation) Logs, petrophysical, McKinley afterflow analysis, transmissiblity, type curves, wellbore parameter, Maps: datum pressure, geological contour, Mass conservation, fluid, radial flow, oil, linear flow, water, linear displacement, Mass transfer, Material balance: aquifer, gas reservoir, depletion, " " , water drive, general hydrocarbon reservoir, " , linear function, linear incompressible water injection, production from a closed radial cell, reduced equation, gascap drive reservoir, " ", natural water drive reservoir, solution gas drive reservoir, undersaturated, solution gas drive reservoir, gas saturated, Matthews, Brons, Hazebroek (MBH): charts, determination of average pressure, buildup test, gas-oil, general, oil, real gas, dimensionless pressure, dimensionless pseudo pressure, theory, buildup testing, volume (rate) averaged reservoir pressure, Mechanical skin factor, Mechanical skin factor, determination of afterflow analysis, oil buildup testing gas-oil, oil, real gas, multi-rate drawdown oil, real gas, single rate drawdown oil, Mercury injection pump, Method of images, Micellar solution flooding, Mobilituy: buffer, control, fluid, relative, Mobility ratio: end point, gas/oil, end point, general, " ", water/oil, flood front, SUBJECT INDEX 431 reduction of, significance of, Mobilising residual oil, Movable oil volume, Multi-rate testing, gas wells, back pressure testing, Essis-Thomas analysis, Odeh-Jones analysis, non-stabilized analysis, validity of transient analysis, Multi-rate testing, oilwells, 97 ambiguity in interpretation, analysis, basic analysis equation, Odeh-Jones analysis, routine surveys, partially depleted reservoirs, validity of transient analysis, Natural water influx (see Water influx) Net bulk volume, Non-Darcy flow, Non-Darcy flow coefficient (F), Normalization of effective permeability curves, Numerical simulation, Numerical simulation of immiscible, incompressible displacement, comparison with analytical methods, conducting a study, evaluation of viscosity at grid block faces, finite difference formulation of 1-D conservation equations, harmonic averaging of absolute permeability, IMPES solution technique, instabilities in solution, material balance mode, one-dimensional model, purpose of, reduction of dimensions, role of thickness averaged relative permeabilities and pseudo capillary pressures, semi implicit rates, well model, semi implicit relative permeabilities; pseudo capillary pressure, two point upstream weighting, relative permeabilities, well model, Odeh-Jones analysis, Oil column, compressibility, density, reservoir, " , standard conditions, displacement (see Immiscible, incompressible displacement) expansion in reservoir, " on heating, formation volume factor of bubble point oil, differential, of bubble point oil, flash, differential, relative to residual oil volume, gravity, in place, pressure gradient, saturation, " , residual, undersaturated, velosity, during displacement, viscosity, volume, residual, Oil relative permeability, altering the characteristics, Oil viscosity reduction: carbon dioxide flooding, thermal, Oil-water contract, 9 Oil wet, Osmosis, Overburden: SUBJECT INDEX 432 pressure, " , gradient, Overpressure, Paraffin series, Partical penetration, of well, Permeability: absolute, dimensions of, effective (see Effective permeability) increasing by well stimulation, reduction in vicinity of well, relative (see Relative permeability) units of, vertical, Phase diagram, Phase equilibrium calculations, Piston-like displacement, Polymer flooding, Pore compressibility, experimental determination, Pore volume (PV), determination of, gas well test " ", oil well test, reduction during depletion, Porosity, Potential: dimensions, fluid, gradient, real gas, Practical aspects of well surveying, Predicting pressure decline, water drive gas reservoir, reservoir performance water influx, Pressure: absolute, average, aquifer, " , grid block, " , reservoir, " , within well's drainage boundary, bottom hole flowing, critical, drawdown, drop, across skin, dynamic grid block, extrapolated, infinite closed-in time, gauge, grain (matrix), initial, determination of, overburden, pseudo critical, pseudo reduced, squared, real gas flow, water abnormal, " , normal Pressure buildup: computer program, dominated by afterflow, Matthews, Brons, Hazebroek theory, plot, Horner (see Horner buildup plot) practical aspects, theoretical equation, theoretical linear equation, Pressure buildup analysis: gas-oil, general, oil, real gas, Pressure drawdown testing, multi-rate, (see Multi-rate testing) Pressure drawdown testing, single rate: analysis of, oil, " ", real gas, purpose of, Pressure gradient: capillary (see Capillary pressure gradient) fluids in the well, gas, overburden, SUBJECT INDEX 433 water, Pressure maintenance, partial: gas reservoir, oil reservoir, Pressure maintenance, total, Primary recovery (see Recovery of hydrocarbons) Producing gas-oil ratio (see Gas-oil ratio) Productivity index(PI): aquifer, increase, well stimulation, numerical simulator, well model, ratio increase, skin removal, " " , steam soaking, semi-steady state, liquid, Pseudo: critical pressure, " temperature, pressure, real gas (see Real gas pseudo pressure) pressure, two phase, gas-oil, reduced pressure, reduced temperature, relative permeabilities (see Thickness averaged relative permeabilities) skin factor, Pseudo capillary pressure, function , in dipping reservoir, role in numerical simulation, Psi potential, PV cell, PVT: gas, oil, PVT analysis Dodson's experiment, gas, gas condensate, oil, basic analysis, " , complete analysis, PVT oil, presentation of results: corrected for surface separation, uncorrected for surface separation, Radial diffusivity equation: constant terminal rate solution (see Constant terminal rate) difficulty in solving, real gas,general, line source solution (see Line source solution) oil, real gas, Radial diffusivity equation, dimensionless general, liquid, real gas, Ramey, H.J., Jr., Real gas pseudo pressure, generation of, reasons for using, Recombination, fluid samples, Recovery factor Recovery of hydrocarbons, primary, compaction drive, gas, gascap drive, solution gas drive Recovery of hydrocarbons, tertiary, Reduction of displacement problems to one dimension, layered reservoirs, numerical simulation, segregated displacement, homogeneous, Relative permeability, curves, rock, end point values, gas-oil, end point values, water-oil, gas, linear, measurement of , oil, SUBJECT INDEX 434 thickness averaged (see Thickness averaged relative permeabilities) Relative volumes, PVT cumulative gas, gas, oil, total, Removal of mechanical skin, Reservoir drive mechanisms, combination, compaction, solution, gas, water, Reservoir limit testing, Reservoir simulation (see Numerical simulation) Residual: gas saturation, oil saturation, oil volume, PVT, Retrograde liquid condensate, Russell, afterflow analysis, Russell-Goodrich, real gas flow, difficulty in application, Sample collection (see Fluid samples) Saturation: critical, gas, " , gas condensate, gas (see Gas saturation) oil (see Oil saturation) water (see Water saturation) Schilthuis, R.J.: material balance equation, steady state water influx. Segregated displacement, oil by gas, " " , oil by water (see Water drive) Semi-steady state: condition, time, Separation, gas-oil: reservoir, surface, Shrinkage factor: bubble point oil, differential, " " ", flash, oil separator-stock tank, Sign convention: Darcy's equation, isothermal compressibility, Laplace, capillary pressure equation, source-sink terms, numerical, Skin factor: mechanical (see Mechanical skin) pseudo, rate dependent, Solution gas drive: above bubble point, below bubble point, Solution gas-oil ration (see Gas-oil ratio) Standard conditions, Standing-Katz, Z-factor correlation, application of 14-18 Statfjörd field (North Sea), Static pressure survey, Steady state condition, Steam: drive, injection, Stiles method, Stock tank oil, initially in place, STOIIP, probablistic determination, Subsidence, surface, Superposition: inspace, in time, Summary: chart, generation of averaged relative permeabilities, pf pressure analysis techniques, SUBJECT INDEX 435 Surface tension, Surfactant flooding, Temperature: ambient, surface, distribution, steam soaked well, effect, sealed fresh water system, reservoir, seperator, Tertiary flooding, Thermal recovery, Thickness averaged relative permeabilities/water saturations, diffuse flow, graphical determination, homogeneous reservoir, finite capillary transition zone, homogeneous reservoir, segregated flow, layered reservoir, lack of vertical equilibrium, layered reservoir, pressure communication, role in numerical simulation, summary chart, with total lack of vertical equilibrium, Time scale for hydrocarbon recovery, Time to reach semi-steady state Transient flow condition, Transient well test analysis buildup, general, " , oil, " , real gas, multi-rate drawdown, general, " " " , oil, multi-rate drawdown, real gas single rate drawdown, oil " " " , real gas, Transition from transient to semi-steady state flow, well at centre of regular shaped drainage area, Trapezoidal rule for integration, Ultimate recovery, Underground withdrawal, Underreaming, Undersaturated oil, Units: absolute, cgs, Darcy, field, for real gas flow, Systeme Internationale (SI), Units conversion, Darcy's ;linear equation, dimensionless variables, real gas flow, Universal gas constant, Uplifting, of reservoir, Variable rate history: prior to buildup test, prior to multi-rate flow test, Velocity of gas, with respect to oil, maximum water saturation plane, plane of constant water saturation, Vertical equilibrium, influences on, total lack of, Viscosity: gas, oil, ratio, oil/displacing fluid, water, Viscosity-compressibility product, real gas, Viscosity-compressibility product, real gas, Viscosity-temperature relation, oil Viscosity-Z factor product, real gas, SUBJECT INDEX 436 Volume: hydrocarbon (see Hydrocarbon pore volume) moveable oil, pore (see Pore volume) Water: bearing sand, breakthrough (see Breakthrough) compressibility, expansion, formation volume factor, injection, pressure, " , equation, " , gradient, salinity, tongue, viscosity, Water drive, Buckley-Leverett theory, homogeneous reservoir, diffuse flow, " " , finite capillary transition zone, homogeneous reservoir, segregated flow, ideal (piston-like), layered reservoir, with vertical equilibrium, layered reservoir, without vertical equilibrium, mechanics of , non-ideal, numerical simulation of, Water drive, oil recovery equations: stable, segregated flow, unstable, segregated flow, " " ", inclined, Welge, Water influx: dimensionless cumulative influx function, dimensionless influx rate, gas reservoir, oil reservoir, time dependance of, Water influx calculations: applied to steam soaking, Fetkovitch, history matching, " , prediction, Hurst-van Everdingen, history matching, Hurst-van Everdingen, prediction, uncertainty in, Water saturation: average, behind flood front, breakthrough, connate (see Connate water saturation) flood front (shock front), influence, effective and relative permeabilities, Water saturation distribution: areal, definable, diffuse (uniform), dip normal (see Thickness averaged relative permeability/water saturation Water wet reservoir, Welge's equation, graphical technique, Well: asymmetry, conditioning, exploration, model, numerical simulation, partial penetration, stimulation, Well testing: collection of fluid samples, initial, practical aspects, purpose of, routine, SUBJECT INDEX 437 Wettability, Wireline formation test, Z-factor: application of, as function of pressure, average value, Russell-Goodrich inflow equations, two phase, Z-factor, determination of: correlation charts, direct calculation, experiment, The author of "Fundamentals of Reservoir Engineering" is especially well-qualified to write on this subject, having spent the past few years with the Training Division of Shell International Petroleum Company, lecturing on reservoir engineering. In this book he gives a coherent account of the basic physics of reservoir engineering, a thorough knowledge of which is essential in the petroleum industry for the efficient recovery of hydrocarbons. Much of the text is based on various lecture courses given by the author and throughout the book only the simplest and most straightforward mathematical techniques are used. The first four chapters serve as an introduction to the subject and will interest all those connected in any way with development and production of hydrocarbon reserves. The next four chapters are more specialised and describe the most important aspects of the field, i.e. the theory and practice of oil and gas well testing and pressure analysis techniques. This is dealt with in a concise, unified and applied manner not readily found in other texts. Water influx and fluid displacement are described in the last two chapters with the aim of helping the reader understand the numerical simulation of reservoir performance. Numerous modern computational techniques are described with the aid of flow charts and illustrated with examples. This is an ideal text for the student as it assumes no prior knowledge of reservoir engineering, but it is also of value for the practising reservoir engineer, production technologist and production engineer.
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Open Middle Team Submit English English Español Français Home>Grade 7>Unit Rates with Fractions 1 Unit Rates with Fractions 1 Directions: Using the digits 0 to 9 at most one time each, fill in the boxes to create two unit rates. You may reuse all the digits each equation. Hint How will the digits you choose affect whether you can have a single digit numerator in the unit rate? 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190066	https://saestatsteaching.tech/analysis-of-variance	Chapter 17 Analysis of Variance The ANOVA (ANalysis Of VAriance) test is a classical statistical test associated with R. Fisher. The test is suited to situations where we have observations grouped by a categorical variable and we want to understand whether or not there are differences in the group means. The mathematical notation for ANOVA is horrible, but the intuition is relatively easy to understand. The ANOVA test statistic follows the F-distribution, and large values are evidence against the null hypothesis that the group means are all the same. In practice we can think of the test as dividing the observed variation in the data into: The variation between the group means and the grand mean (the average across all the groups); and The variation within each group. If the variation between the group means and the grand mean is high, and the variation within groups is small, this suggests there are true differences in the group means. On the other hand, if the variation between the group means and the grand mean is small, and the variation within groups is large, this suggests there are no real differences in the group means i.e. the variations we observe is just sampling variation. For the ANOVA test, if we observe a p-value of less than 0.05 we say we have sufficient evidence to reject the null hypothesis that all the group means are the same. If we reject the null we then move on to pair-wise t-tests to try and determine which specific group means are different. For the pair-wise t-tests we use an adjustment method to control the Type I error rate. The need to adjust p-values to control the Type I error rate is one of the reasons we need specialist software. If we observe a p-value that is greater than 0.05 we say we have insufficient evidence to reject the null hypothesis that all the group means are the same. If we do not reject the null hypothesis we do not proceed to pair-wise t-tests. We just stop our analysis at the ANOVA test. We have concluded that there are no significant differences so we stop.\ Similar to the case for the t-test, for the ANOVA test there are two possible formulas that we can use: one formula when the group variances are the same, and one formula for when the group variances are not the same. So, before we conduct an ANOVA test we have to conduct a pre-test to establish whether the group variances are the same or not. The test we use for this in the Bartlett test. If we reject the null for the ANOVA test we then go on to look for the specific differences between groups. This means that when we conduct an ANOVA test we usually have three tests to conduct: A Bartlett test to check whether the group variances are the same The actual ANOVA test (with either equal or unequal variance) Pair-wise t-tests, with an adjustment for multiple comparisons (if we reject the null for the ANOVA test). We will work through these steps below. As with all previous statistical tests, we will set the alpha level at 0.05. 17.1 ANOVA - One Factor For this example we will use the base R data set chickwts, which has the weights of chickens fed six different types of feed. The data is in long format. This means we have one continuous variable (weight) and one factor variable (feed) that has six levels. When working with multiple groups it is easiest to work with data in long format. Remark. This is the same data set used in the Equal Variance Testing in R reference guide. What we are trying to do is help a farmer identify which feed(s) promote the greatest weight gain in chickens. Let’s start by looking at our data and creating a box plot: ``` get summary & structure and create a boxplot of our differences str(chickwts) > 'data.frame': 71 obs. of 2 variables: > $ weight: num 179 160 136 227 217 168 108 124 143 140 ... > $ feed : Factor w/ 6 levels "casein","horsebean",..: 2 2 2 2 2 2 2 2 2 2 ... summary(chickwts) > weight feed > Min. :108 casein :12 > 1st Qu.:204 horsebean:10 > Median :258 linseed :12 > Mean :261 meatmeal :11 > 3rd Qu.:324 soybean :14 > Max. :423 sunflower:12 with(chickwts,boxplot(weight~feed, col= "lightgray", main= "", xlab= "Feed type", ylab= "Weight (g)", ylim= c(100,450), las= 1)) ``` Figure 17.1: Chicken weight distributions for six different feed types Based on the boxplot there look to be clear differences between at least some of the means, but it is not easy to see what the actual values are. Let’s use a new function, tapply(), to calculate the mean and standard deviation for each of our groups. Here you are applying the function mean() and sd() to the vector weight, grouped by feed. The tapply() function works when we have data in long format. The function generates results similar to the way results are generated using the pivot table function in MS Excel. Personally, I prefer to use MS Excel to work through data management issues, but you can do similar things with R. ``` with(chickwts, tapply(weight, feed, mean)) > casein horsebean linseed meatmeal soybean sunflower > 324 160 219 277 246 329 with(chickwts, tapply(weight, feed, sd)) > casein horsebean linseed meatmeal soybean sunflower > 64.4 38.6 52.2 64.9 54.1 48.8 ``` As we saw in our boxplot, there is quite a difference between some of our group means, particularly horse bean on the lower end, and sunflower and casein on the higher end. What about our standard deviations? Remember, the standard deviation is the square root of the variance so the group standard deviations gives us an idea of whether the group variances are equal or not. The units of measurement for the standard deviation are just a bit easier to interpret than the units of measurement for the variance. Look carefully at the reported standard deviation values and also at the boxplot. Can you see the link between the visual display of the data and the group standard deviation values? Now let’s formally test for equal variance across the groups. Because we have multiple groups the test we use is the Bartlett Test of Homogeneity of Variances. The Null hypothesis for the test is that the group variances are all equal; the Alternate hypothesis is that the group variances are not equal. The implementation of the test follows the standard format we use for most tests. First we use the with() function to direct R to the dataset we want to use. Next we use the bartlett.test() function to tell R what function to apply. Finally we specify the column names that contain the data we are working with. Although this explanation is quite brief, there is a separate reference guide on the Bartlett test that explains the steps in greater detail. ``` with(chickwts, bartlett.test(weight ~ feed)) # long data format, use ~ not , > > Bartlett test of homogeneity of variances > > data: weight by feed > Bartlett's K-squared = 3, df = 5, p-value = 0.7 ``` As the test p-value = 0.66, which is greater than 0.05, we do not reject the null hypothesis that the group variances are all equal: we have insufficient evidence to reject the null. As such we set the parameter var.equal to TRUE in our ANOVA test of whether the groups means are the same. Note: If we reject the null for the Bartlett test (i.e. we have a test p-value that is less than 0.05) when we conduct the ANOVA test we set the parameter var.equal to FALSE. To implement the ANOVA test we use the function oneway.test(), and so for our formal test we have: Step 1: Set the Null and Alternate Hypotheses Null hypothesis: The group means are all equal Alternate hypothesis: At least one group mean is different from the other groups Step 2: Implement Analysis of Variance Test with(chickwts, oneway.test(weight ~ feed, var.equal = TRUE)) #> #> One-way analysis of means #> #> data: weight and feed #> F = 15, num df = 5, denom df = 65, p-value = 6e-10 Step 3: Interpret the Results From the output we see the p-value = 5.936e-10; which we report as < 0.001. As the test p-value is smaller than 0.05 we: Reject the null hypothesis that all the means are equal. In our output we can also see the F-statistic (15.36). This is the statistic that our p-value is based on. For large numbers we reject the null. So what now? If you were a farmer wanting to know which feed to give your chickens, which feed would you choose? You could decide from your boxplot, likely choosing casein or sunflower, but the figure doesn’t tell you if there is a feed which is statistically the highest, and the ANOVA test only tells you that the feeds are not ALL the same. This is where pair-wise comparisons can be helpful. 17.1.1 Pair-wise Comparisons Let’s come back to our research question - which feed(s) should the farmer consider based on the weights of chickens eating that feed. Here we will use the function pairwise.t.test() to compare all our individual group means. With this function the parameter pool.sd is the equivalent of var.equal for the t.test(). We use the Bartlett test result to decide whether we set this parameter to TRUE or FALSE. If we do not reject the null for the Bartlett test we set the parameter to TRUE. If we reject the null for the Bartlett test we set the parameter to FALSE. So, for this example we will set it to TRUE. The pairwise.t.test() function does two things that make our life easy. First, it calculates all the different possible pair-wise combinations for us at once. Rather than having to type the t.test() command 15 times (one for each pair-wise combination) we just need to write one command. The second thing that the pairwise.t.test() command does for us is to automatically adjust the reported p-values to control the Type I error rate. The automatic adjustment makes life easy for us as we can still apply our standard p-value decision rule to the results, where we know that the values have been adjusted to address the multiple comparison issue. This is something that only specialist software such as R is able to do, and it really saves a lot of time compared to the alternative of working through the adjustments manually. It is possible to use a variety of different adjustments for multiple comparisons. For example, by changing the parameter p.adjust.method and providing “none” for no adjustment (lower p-values), or “bon” for a more restrictive adjustment (higher p-values). The default adjustment used by R is the holm adjustment, and this is a good choice. As such we will not be changing this parameter value. So conducting all the required comparisons is quite easy. ``` with(chickwts, pairwise.t.test(weight, feed, pool.sd = TRUE)) > > Pairwise comparisons using t tests with pooled SD > > data: weight and feed > > casein horsebean linseed meatmeal soybean > horsebean 3e-08 - - - - > linseed 2e-04 0.094 - - - > meatmeal 0.182 9e-05 0.094 - - > soybean 0.005 0.003 0.518 0.518 - > sunflower 0.812 1e-08 8e-05 0.132 0.003 > > P value adjustment method: holm ``` As you probably suspected based on the boxplot, we have some significant differences between groups (e.g. horsebean and casein) and some non-significant differences (e.g meat meal and linseed). The feeds with the two highest means, casein and sunflower, are not significantly different. We also see that the feed with the third highest mean, meat meal, is also not significantly different from either casein or sunflower. So, it looks like the farmer has some options in terms of the feed they select. To make a final decision the farmer might now look to other factors such as cost or availability to decide what feed to provide her chickens. It is this final step that is important. The statistical test is just one part of the process of working out what to do. You then typically need to look at other information to workout what you actually need to do. 17.1.2 Reporting Test Results How you present numerical results in a table is just as important as to how you present figures. You must include information in a clear and concise way, without causing distraction with unnecessary information. Table 17.1 presents the results of our pair-wise comparisons. This is a good format for you to follow. Table 17.1: Pair-wise t-test results (Holm adjusted p-values): effect of feed on chicken weights \| \| Casein \| Horsebean \| Linseed \| Meatmeal \| Soybean \| \| \| Horsebean \| <0.001 - - \| \| \| Linseed \| <0.001 \| 0.094 - \| \| Meatmeal \| 0.182 \| <0.001 \| 0.094 - \| \| \| Soybean \| 0.005 \| 0.003 \| 0.518 \| 0.518 \| \| Sunflower \| 0.812 \| <0.001 \| <0.001 \| 0.132 \| 0.003 \| \| 17.2 The notation of the ANOVA model The ANOVA test can be motivated by saying that when looking at grouped data the observed variation can be separated into two parts: The variation due to differences in the group means; and The variation of individual observations around each individual group mean. The variation between group means is called the between group variation. The variation of individual observations around each individual group mean is called the within group variation. The ANOVA test can be summarised with two propositions: If the variation due to differences in the group means is relatively large, while the variation of observations around each group mean is relatively small, it is likely there are true differences in the group means. For this scenario the test will generate a small p-value. If the variation due to differences in the group means is relatively small, while the variation of observations around the group mean is relatively large, it is unlikely there are true differences in the group means. For this scenario the test will generate a relatively large p-value. Figure 17.2: A comparison of where differences are likely and unlikely If you understand the interpretation of the ANOVA test presented above you understand the fundamental element of ANOVA test. Unfortunately, the formal notation used to set out the ANOVA model is relatively intimidating. The complexity of the notation is one of the reasons a great many people refuse to come to terms with ANOVA. For the vast majority of people formal notation serves to obscure rather than highlight what is going on. The ANOVA model is especially bad in this respect. The type of notation used for ANOVA analysis will appear again and again throughout your studies. Because of this, a relatively formal introduction to ANOVA notation is presented below. 17.2.1 Introduction to notation Consider the case where we have three groups. In group 1 we have 4 observations, in group 2 we have 5 observations, and in Group 3 we have 4 observations. That means we have something like the following: For group 1 we have four observations: X1,X2,X3, and X4 which we summarise as Xj,j=1,…,4. For group 2 we have five observations: X1,X2,X3,X4, and X5,which we summarise as Xj, j=1,...,5. For group 3 we have four observations: X1,X2,X3, and X4 which we summarise as Xj, j=1,...,4. What you can see is that when we have groups of observations things get confusing if we use a single sub-script to index observations. In the above expressions j was used to keep track of the number of observations in each group. To keep track of which group an individual observation belongs to we need to introduce a second subscript. This looks complex, but it we take things slowly it makes sense. When we have groups we use the notation Xij to keep track of all the observations. The i tells us which group the observation belongs to and the j keeps track of the individual observations within each group. So for the specific case we have: For group 1 we have: X11, X12, X13, and X14 For group 2 we have: X21, X22, X23, X24, and X25; For group 3 we have: X31, X32, X33, and X34. The number of observations in each group can be different. In our example there are 4 observations in group 1 and group 3, but 5 observations in group 2. The indexation for j (the number of observations within each group) must reflect this, and so we write j=1,…,ni. This says that the number of observations in group i runs from 1 through to the maximum number for group i , and so reflects the fact that the number of observations in each group can vary. In our example we have n1=4, n2=5, and n3=4. We can denote the individual group means as ¯X1, ¯¯¯¯¯X2 , and ¯¯¯¯¯X3, and the average across all observations as ¯¯¯¯¯XG, which we call the grand mean. We know how to calculate these values. Formally, we write the group mean for group i as: ¯¯¯¯¯Xi=1ni∑jXij. This expression says, to find the mean for any given group, take all the observations for that group, add them up, and divide through by the total number of observations for that group. The expression ∑j is read j. So, if we take group 1 to illustrate we would have: ¯¯¯¯¯X1=1n1∑jX1j=14(X11+X12+X13+X14)=average for group 1. To find the overall grand mean for the data we know that we simply have to add up all the observations and divide through by the total number of observations. While in practice this is a simple operation in Excel or R, the formal notation is a little involved. First we have to implement a rule that will give us the total number of observations, which we will denote with N . The formal notation we use to denote the total number of observations is as follows: N=∑ini=n1+n2+n3=4+5+4=13. Now, we need a way to say add up in a sequence all the observations in group 1, group 2, and group 3. If we had a single group we could use the expression ∑jXj , which say add up all the observations indexed by j. However, we need to say add up all the observations in group 1, group 2, and group 3. To see how this works let us start by writing out in full what we want to do. We want to write something that will say add up all 13 observations: X11+X12+X13+X14+X21+X22+X23+X24+X25+X31+X32+X33+X44 We know how to say add up all the values within a group. For group 1 we write this as ∑jX1j, for group 2 we write ∑jX2j, and for group 3 we write ∑jX3j. Now we want to add all these values together so we have something that looks like: ∑jX1j+∑jX2j+∑jX3j. What we need is a second summation sign that says add up across the groups. Formally we write ∑i∑jXij if we want to say add up all the observations from all groups. This expression says add up all the observations in all groups; where the groups are indexed by i and the individual observations within a group are indexed by j. For the three groups we have considered the term ∑i∑jXij translates as follows: ∑i∑jXij=X11+X12+X13+X14+X21+X22+X23+X24+X25+X31+X32+X33+X44. To find the average of the observations so we need to divide through by the total number of observation, which is N=∑ini. So, formally we denote the grand mean as: ¯¯¯¯¯XG=1N∑i∑jXij=113(X11+X12+X13+X14+X21+X22+X23+X24+X25+X31+X32+X33+X44). In terms of notation this is as complicated as things get, and we now have all the elements we need to consider a formal ANOVA model. 17.2.2 The ANOVA model The proposition presented regarding the ANOVA test is that if the variation due to differences in the group means is relatively large, while the variation of observations around each group mean is relatively small it is likely there are true differences in the group means. The alternative contrasting scenario is one where the variation due to differences in the group means is relatively small, while the variation of observations around each group mean is relatively large it is unlikely there are true differences in the group means.\ The ANOVA test statistic is: standardisied measure of between group variationstandardisied measure of within group variation. This test statistic follows the F-distribution, and large values for F (p-value < 0.05) lead us to reject the null hypothesis of no differences between the groups. When adding up the differences between individual observations and the group mean, or differences between the group mean and the grand mean, positive and negative values will cancel out. To ensure that we have a positive value for the variation we therefore square each difference calculation. 17.2.3 Between group variation We find the total between group variation for each group as the difference between the group mean and the grand mean multiplied by the number of observations in each group. This measure can be written as: n1(¯¯¯¯¯X1−¯¯¯¯¯XG)2+n2(¯¯¯¯¯X2−¯¯¯¯¯XG)2+n3(¯¯¯¯¯X3−¯¯¯¯¯XG)2. In our example we have 4 observations in group 1, 5 observations in group 2, and 4 observations in group 3, so we have n1=4, n2=5, n3=4 or 4×(¯¯¯¯¯X1−¯¯¯¯¯XG)2+5×(¯¯¯¯¯X2−¯¯¯¯¯XG)2+4×(¯¯¯¯¯X3−¯¯¯¯¯XG)2. We can however also use the summation operator to write this in a formal way as: ∑ini(¯¯¯¯¯Xi−¯¯¯¯¯XG)2 = the total between group variation, and we call this measure the between group sum of squares. Note that the indexation subscript uses i as we are comparing group means to the grand mean. The standardisation we use is to divide through by the degrees of freedom. In a statistics context degrees of freedom works as follows. When we consider the sum of squares calculations there is a constraint that means if we have all but the last piece of information we are able to retrieve the final missing value. The final value is not free to take any value but rather is constrained to the value required by the formula we are using. For our between group variation, if we know any k−1 values we can retrieve the final value. The standardisation we use for the between group sum of squares is therefore k−1. So, to summarise we have:\ ∑ini(¯¯¯¯¯Xi−¯¯¯¯¯XG)2=the between group sum of squares. We then have as our standardised measure of the between group variation: ∑ini(¯¯¯¯¯Xi−¯¯¯¯¯XG)2k−1=the standardised between group sum of squares=the between group mean squares. 17.2.4 Within group variation For the within group variation we have the same problem with positive and negative values that would cancel out, so again we square each deviation around the individual group means. The variation we want to add up is the variation of the individual observations around their respective group means. Specifically, the values that we want to add up are the following values: Sum of squared deviationsfrom group means=(X11−¯¯¯¯¯X1)2+⋯+(X14−¯¯¯¯¯X1)2+(X21−¯¯¯¯¯X2)2+⋯+(X25−¯¯¯¯¯X2)2+(X31−¯¯¯¯¯X3)2+⋯+(X34−¯¯¯¯¯X3)2 We know that we can write the sum of the squared deviations from the mean for the groups individually as: within group variation for Group 1 =∑j(X1j−¯¯¯¯¯X1)2 within group variation for Group 2 =∑j(X2j−¯¯¯¯¯X2)2 within group variation for Group 3 =∑j(X3j−¯¯¯¯¯X3)2 This means that we can, in the first instance pull together the information that describes the within group variation as: ∑i∑j(Xij−¯¯¯¯¯Xi)2=(X11−¯¯¯¯¯X1)2+⋯+(X14−¯¯¯¯¯X1)2+(X21−¯¯¯¯¯X2)2+⋯+(X25−¯¯¯¯¯X2)2+(X31−¯¯¯¯¯X3)2+⋯+(X34−¯¯¯¯¯X3)2 We call this the within group sum of squares. Within each group there is one value that is not free to vary, and so the degrees of freedom is equal to N−k; the total number of observations minus the number of groups. In the example we have been working with there are 19 observations and 3 groups so we have 16 degrees of freedom. We then have as our standardised measure of the within group variation: ∑i∑j(Xij−¯¯¯¯¯Xi)2N−k=the standardised within group sum of squares=within group mean squares. Again, technically this value is a variance estimate which follows a chi-squared distribution. The ANOVA test is then conducted as: F=Between Group MSWithin Group MS, and we reject the null hypothesis of no significant difference in the group means if this value is large, where large is defined by the F-distribution (We use the fact that F-value, as the ratio of two chi-squared variables, follows the F-distribution). In practice we need to know how to implement the test, and we can implement the test in R using the oneway.test() function.
190067	http://www.cpdee.ufmg.br/~palhares/Zeros-poles-rational-function.pdf	(1) (2) zeros and poles of rational function A rational function of a complex variable may be presented by the equation where the numerator and the denominator are mutually irreducible polynomials with complex coefﬁcients and ( ). If ( ), then the real and imaginary parts of are rational functions of and . When we factorize the numerator and the denominator in the ring , we can write where for all . The form (2) of the rational function expresses the zeros and the inﬁnity places of the function. One can write (2) as where is a rational function which in gets a ﬁnite non-zero value. Accordingly one says that the point is a zero of with the order ( ). One can also write (2) as where is a rational function getting in the point a ﬁnite non-zero value. As , the modulus increases unboundedly in such a manner that tends to a ﬁnite non-zero limit. So one says that has in the point a pole with the order ( ). Behaviour at inﬁnity Now let increase unboundedly. When we write z R (z) = , a0zm + a1zm−1 + … + am b0zn + b1zn−1 + … + bn aj bk a0b0 ≠0 z = x+iy x, y ∈R R (z) x y C [z] R (z) = , a0(z −α1)µ1(z −α2)µ2…(z −αr)µr b0(z −β1)ν1(z −β2)ν2…(z −βs)νs αj ≠βk j, k αj βk R (z) = (z−αj)µjSj (z) Sj (z) z = αj αj R (z) µj j = 1, 2, …, r R (z) = Tk (z) 1 (z−βk)νk Tk (z) βk z →βk \|R (z)\| \|z −βk\|νk \|R (z)\| R (z) βk νk k = 1, 2, …, s \|z\| zeros and poles of rational function 1 of 3 08/05/2024, 08:33 (3) we get three cases: • If , then . Since is ﬁnite and non-zero, the point is the pole of with the order . • If , we have and thus has in the inﬁnity a ﬁnite non-zero value. • If , we have in such a manner that . This means that has in inﬁnity a zero with the order . In any case, has equally many zeros and poles, provided that each zero and pole is counted so many times as its order says. The common number of the zeros and poles is called the order of the rational function. It is the greatest of the degrees ( PolynomialRing) and of the numerator and denominator. -places Denote by any non-zero complex number. The -place of means such a point for which . If is a -place of where the polynomials and have no common factor ( DivisibilityInRings), then is a zero of If this zero is of order , then one says that is of order as the -place of . The numerator and denominator of (3) cannot have common factor (otherwise any common factor would be also a factor of ). This implies that the order of the rational function deﬁned by (3) is the same as the order of . Because (3) gets times the value , also gets times the value . Thus we have derived the Theorem. A rational function attains any complex value so many times as its order is. References 1 R. NEVANLINNA & V. PAATERO: Funktioteoria. Kustannusosakeyhtiö Otava. Helsinki (1963). R (z) = zm−n ⋅ , a0 + + … + a1 z am zm b0 + + … + b1 z bn zn m > n limz→∞R (z) = ∞ limz→∞ = R(z) zm−n a0 b0 z = ∞ R (z) m−n m = n limz→∞R (z) = a0 b0 R (z) m < n limz→∞R (z) = 0 limz→∞zn−mR (z) = a0 b0 R (z) n−m R (z) m n c c c R (z) z R (z) = c z0 c R (z) = P (z) Q (z) P (z) Q (z) z0 R (z) −c = . P (z) −c Q (z) Q (z) µ z0 µ c R (z) P (z) k R (z) k 0 R (z) k c zeros and poles of rational function 2 of 3 08/05/2024, 08:33
190068	https://www.engineeringtoolbox.com/gross-net-heating-values-d_420.html	Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Gases - Gross and Net Heat Values Gross heat and net heat values for gases like hydrogen, methane and more. Gross heating and net heating values for some common gases can be found in the table below: Fuel Gases - Gross and Net Heat Values \| Gas \| (kcal/kg) \| (kcal/nm3)1) \| \| Gross Heating Values \| Net Heating Values \| Gross Heating Values \| Net Heating Values \| \| Hydrogen \| 33889 \| 28555 \| 3050 \| 2570 \| \| Methane \| 13284 \| 11946 \| 9530 \| 8570 \| \| Ethane \| 12400 \| 11350 \| 16700 \| 15300 \| \| Ethylene \| 12020 \| 11270 \| 15100 \| 14200 \| \| Natural Gas - approximately \| 12000 \| 11000 \| 9000 \| 8000 \| \| Propane \| 12030 \| 11080 \| 24200 \| 22250 \| \| Propylene \| 11700 \| 10940 \| 22400 \| 20900 \| \| n-Butane \| 11830 \| 10930 \| 31900 \| 29400 \| \| Iso-Butane \| 11810 \| 10900 \| 31700 \| 29200 \| \| Butylene-1 \| 11580 \| 10830 \| 29900 \| 27900 \| \| Iso-Pentane (liquid) \| 11600 \| 10730 \| \| LPG (average) \| 11920 \| 10997 \| 28000 \| 25775 \| \| Acetylene \| 11932 \| 11514 \| 13980 \| 13490 \| \| Carbon Monoxide \| 2411 \| 2411 \| 3014 \| 3014 \| 1) nm3 = normal cubic metre 1 kJ/kg = 1 J/g = 0.4299 Btu/ lbm = 0.23884 kcal/kg Unit Converter Unit Converter . Make Shortcut to Home Screen? Cookie Settings
190069	https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/02%3A_Molecules_Ions_and_Chemical_Formulas/2.02%3A_Chemical_Formulas	Skip to main content 2.2: Chemical Formulas Last updated : Jul 4, 2022 Save as PDF 2.1: Chemical Compounds 2.3: Naming Ionic Compounds Buy Print CopyView on Commons Donate Page ID : 24740 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To describe the composition of a chemical compound. When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C4H10. The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C2H5. The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane, for example, has the empirical formula C2H5, but it contains two C2H5 formula units, giving a molecular formula of C4H10. Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na+ and Cl−, then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg2+ and Cl−, for example, two Cl− ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of MgCl2. Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O. Note the Pattern Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions. Binary Ionic Compounds An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges. When crossing charges, you will sometimes find it necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2: This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion. Example Ga3+ and As3− Eu3+ and O2− calcium and chlorine Given: ions or elements Asked for: empirical formula for binary ionic compound Strategy: A If not given, determine the ionic charges based on the location of the elements in the periodic table. B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers to write the empirical formula. Check to make sure the empirical formula is electrically neutral. Solution B Using the absolute values of the charges on the ions as the subscripts gives Ga3As3: Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3− have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3− ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices. 2. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion: The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens. 3. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl−. B Two Cl− ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl− as the subscript for Ca: The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter. Exercise Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. Li+ and N3− Al3+ and O2− lithium and oxygen Answer : 1. Li3N 2. Al2O3 3. Li2O Polyatomic Ions Polyatomic ions are groups of atoms that bear a net electrical charge, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in Table . Table Common Polyatomic Ions and Their Names \| Formula \| Name of Ion \| --- \| \| NH4+ \| ammonium \| \| CH3NH3+ \| methylammonium \| \| OH− \| hydroxide \| \| O22− \| peroxide \| \| CN− \| cyanide \| \| SCN− \| thiocyanate \| \| NO2− \| nitrite \| \| NO3− \| nitrate \| \| CO32− \| carbonate \| \| HCO3− \| hydrogen carbonate, or bicarbonate \| \| SO32− \| sulfite \| \| SO42− \| sulfate \| \| HSO4− \| hydrogen sulfate, or bisulfate \| \| PO43− \| phosphate \| \| HPO42− \| hydrogen phosphate \| \| H2PO4− \| dihydrogen phosphate \| \| ClO− \| hypochlorite \| \| ClO2− \| chlorite \| \| ClO3− \| chlorate \| \| ClO4− \| perchlorate \| \| MnO4− \| permanganate \| \| CrO42− \| chromate \| \| Cr2O72− \| dichromate \| \| C2O42− \| oxalate \| \| HCO2− \| formate \| \| CH3CO2− \| acetate \| \| C6H5CO2− \| benzoate \| The method we used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus K+ and NO3− ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42− form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3− form NH4NO3 (ammonium nitrate), which is a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43− ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43− ions. Example Write the empirical formula for the compound formed from each ion pair. Na+ and HPO42− potassium cation and cyanide anion calcium cation and hypochlorite anion Given: ions Asked for: empirical formula for ionic compound Strategy: A If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use Table 2.4 to find the charge on a polyatomic ion. B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula. Solution B Because HPO42− has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts. A The potassium cation is K+, and the cyanide anion is CN−. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally. A The calcium cation is Ca2+, and the hypochlorite anion is ClO−. B Two ClO− ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools. Exercise Write the empirical formula for the compound formed from each ion pair. Ca2+ and H2PO4− sodium cation and bicarbonate anion ammonium cation and sulfate anion Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder. NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda. (NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers. Answer Hydrates Many ionic compounds occur as hydrates, compounds that contain specific ratios of loosely bound water molecules, called waters of hydration. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per Ca(H2PO4)2 unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is Ca(H2PO4)2·H2O. In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula CuSO4·5H2O. When heated, all five water molecules are lost, giving a white solid with the empirical formula CuSO4 (Figure 2.9). Loss of Water from a Hydrate with Heating When blue CuSO4·5H2O is heated, two molecules of water are lost at 30°C, two more at 110°C, and the last at 250°C to give white CuSO4. Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, CaSO4·½H2O is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas CaSO4·2H2O is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid CaSO4·2H2O. Similar processes are used in the setting of cement and concrete. Summary An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. KEY TAKEAWAY The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit. CONCEPTUAL PROBLEMS What are the differences and similarities between a polyatomic ion and a molecule? Classify each compound as ionic or covalent. Zn3(PO4)2 C6H5CO2H K2Cr2O7 CH3CH2SH NH4Br CCl2F2 Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds? CH3CH2CO2H CaCl2 Y(NO3)3 H2S NaC2H3O2 Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed? Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH3OH rather than CH4O. Explain why the ROH notation is preferred. The compound dimethyl sulfide has the empirical formula C2H6S and the structural formula CH3SCH3. What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound. What is the correct formula for magnesium hydroxide—MgOH2 or Mg(OH)2? Why? Magnesium cyanide is written as Mg(CN)2, not MgCN2. Why? Does a given hydrate always contain the same number of waters of hydration? Answer The structural formula gives us the connectivity of the atoms in the molecule or ion, as well as a schematic representation of their arrangement in space. Empirical formulas tell us only the ratios of the atoms present. The condensed structural formula of dimethylsulfide is (CH3)2S. NUMERICAL PROBLEMS Write the formula for each compound. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms sodium hypochlorite pentahydrate, which has 1 chlorine atom, 1 sodium atom, 6 oxygen atoms, and 10 hydrogen atoms Write the formula for each compound. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair. \| Ion \| K+ \| Fe3+ \| NH4+ \| Ba2+ \| --- --- \| Cl− \| KCl \| \| \| \| \| SO42− \| \| \| \| \| \| PO43− \| \| \| \| \| \| NO3− \| \| \| \| \| \| OH− \| \| \| \| \| 4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. 1. zinc and sulfur 2. barium and iodine 3. magnesium and chlorine 4. silicon and oxygen 5. sodium and sulfur 5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. 1. lithium and nitrogen 2. cesium and chlorine 3. germanium and oxygen 4. rubidium and sulfur 5. arsenic and sodium 6. Write the empirical formula for each compound. 1. Na2S2O4 2. B2H6 3. C6H12O6 4. P4O10 5. KMnO4 7. Write the empirical formula for each compound. 1. Al2Cl6 2. K2Cr2O7 3. C2H4 4. (NH2)2CNH 5. CH3COOH AnswerS MgSO4 C2H6O2 C2H4O2 KClO3 NaOCl·5H2O \| Ion \| K + \| Fe 3+ \| NH 4 + \| Ba 2+ \| --- --- \| Cl − \| KCl \| FeCl3 \| NH4Cl \| BaCl2 \| \| SO 4 2− \| K2SO4 \| Fe2(SO4)3 \| (NH4)2SO4 \| BaSO4 \| \| PO 4 3− \| K3PO4 \| FePO4 \| (NH4)3PO4 \| Ba3(PO4)2 \| \| NO 3 − \| KNO3 \| Fe(NO3)3 \| NH4NO3 \| Ba(NO3)2 \| \| OH − \| KOH \| Fe(OH)3 \| NH4OH \| Ba(OH)2 \| Li3N CsCl GeO2 Rb2S Na3As AlCl3 K2Cr2O7 CH2 CH5N3 CH2O 2.1: Chemical Compounds 2.3: Naming Ionic Compounds
190070	https://math.stackexchange.com/questions/2776291/a-geometry-problem-with-the-reflection-of-the-incenter	contest math - A geometry problem with the reflection of the incenter - 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 A geometry problem with the reflection of the incenter Ask Question Asked 7 years, 4 months ago Modified4 years, 10 months ago Viewed 1k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Consider △A B C△A B C and its incenter I I. Let M M be the midpoint of arc B C B C (not containing A A) and P P be the second intersection point of the circumcircle of △A B C△A B C and the circle with center I I and radius I A I A. Prove that if I′I′ is the reflection of I I with respect to B C B C, then I′,P,M I′,P,M are collinear (lie on the same line). Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur. For help, here is a figure: geometry contest-math euclidean-geometry reflection geometric-transformation Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Nov 13, 2020 at 16:05 nonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges asked May 11, 2018 at 7:46 Pet123Pet123 1,262 7 7 silver badges 15 15 bronze badges Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 5 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by nonuser Show activity on this post. @Alex Zhao's proof is incorrect, but I think I've managed to make the idea work. As in the original, this is just a big angle chase. Let K K be the intersection of (I)(I) and B C B C closest to C C. The triangle A B K A B K is symmetric about B I B I.(∗)(∗) Therefore ∡I K B=∡B A I=∡I A C⟹∡I A C+∡C K I=π⟹C K I A is cyclic⟹∡K I A=π−∡A C B.(∗∗)∡I K B=∡B A I=∡I A C⟹∡I A C+∡C K I=π⟹C K I A is cyclic⟹∡K I A=π−∡A C B.(∗∗) We claim ∡K P B+∡K I′B=π∡K P B+∡K I′B=π, because ∡K P B=∡A P B−∡A P K=π−∡A C B−∡K I A/2=(π−∡A C B)/2(by(∗∗))=(∡C A B+∡C B A)/2=∡I A B+∡I B A=π−∡A I B=π−∡B I K(by(∗))=π−∡K I′B.∡K P B=∡A P B−∡A P K=π−∡A C B−∡K I A/2=(π−∡A C B)/2(by(∗∗))=(∡C A B+∡C B A)/2=∡I A B+∡I B A=π−∡A I B=π−∡B I K(by(∗))=π−∡K I′B. Thus P B I′K P B I′K is cyclic. Therefore ∡B P I′=∡B K I′=∡I K B=∡B A I=∡B A M=∡B P M.∡B P I′=∡B K I′=∡I K B=∡B A I=∡B A M=∡B P M. Thus P,I′,M P,I′,M are collinear and we're done. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 17, 2020 at 6:19 brainjambrainjam 9,212 1 1 gold badge 23 23 silver badges 45 45 bronze badges Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Denote circumcircle of △A B C△A B C as Ω Ω and circle with center M M and radius M I M I as Γ Γ. It's well-known that Γ Γ is the circumcircle of △B I C△B I C. Let Φ Φ be the inversion with respect to the circle Γ Γ. Then, for any point X≠M X≠M of the plane denote its image under the Φ Φ as X∗=Φ(X)X∗=Φ(X). Since I I and J J (here J=I′J=I′) are symmetric with respect to the line B C B C their images I∗I∗ and J∗J∗ are symmetric with respect to Φ(B C)=Ω Φ(B C)=Ω (because inversion preserves symmetry eith respect to generalized circles; Φ({I,B,C})={A,B,C}Φ({I,B,C})={A,B,C}). However, Φ(I)=I Φ(I)=I, so J∗J∗ is the image of I I under the inversion with respect to the circle Ω Ω. In particular, J∗J∗ lies on the line O I O I. It's clear that A A and P P are symmetric with respect to O I O I. Hence, if D D is the point symmetric to M M with respect to O I O I, then D∈Ω D∈Ω, D,I,P D,I,P are collinear and lines M P M P, A D A D and O I O I are concurrent. Denote the intersection of M P M P, A D A D and O I O I as K K. Now note that the collinearity of P P, M M and J J is equivalent to the collinearity of P P, M M and J∗J∗ (since M M is the center of Γ Γ). Thus, we need to prove that J∗=K J∗=K. Indeed, we have cyclic quadrilateral A D M P A D M P with A P∥M D A P∥M D, where K=A D∩P M K=A D∩P M, I=A M∩P D I=A M∩P D. It means that I I and K K are inverse images of each other with respect to Ω Ω. Therefore, J∗=K J∗=K, as desired. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Nov 13, 2020 at 16:57 answered Nov 13, 2020 at 16:49 richrowrichrow 4,727 2 2 gold badges 15 15 silver badges 34 34 bronze badges 2 +1. This is really elegant.brainjam –brainjam 2020-11-13 17:36:52 +00:00 Commented Nov 13, 2020 at 17:36 This a solution from other world. Very nice. But now it seem to me very artificial problem. Someone just compose two inversion and get the problem.nonuser –nonuser 2020-11-13 19:37:05 +00:00 Commented Nov 13, 2020 at 19:37 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. After a suggestion by @brainjam I found an offical solution: This was posted initial and it is not a solution, just partial result if someone find it usefull. Let P M P M meet smaller circle at D D. Then A D⊥B C A D⊥B C. Proof: It is easy to see that angle between altitude from A A and angle bisector from is \|β−γ\|\|β−γ\|. Now we see that ∠M B C=∠M A C=α∠M B C=∠M A C=α so ∠D P A=∠M P A=∠M B A=α+2 β∠D P A=∠M P A=∠M B A=α+2 β From here we get (notice: α+β+γ=90∘α+β+γ=90∘ and that A D I A D I is isosceles) ∠D J A=360∘−2(α+2 β)⟹∠D A I=β−γ∠D J A=360∘−2(α+2 β)⟹∠D A I=β−γ and we are done. Let perpendicular from I I to B C B C meet P M P M at J J. If we prove that B C B C bisect I J I J we are done, i.e. J=I′J=I′. But no idea how to finish. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Nov 15, 2020 at 13:49 answered Nov 12, 2020 at 16:29 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges 6 1 Another observation (possibly irrelevant): if you repeat the construction for the other two sides (e.g. midpoint of other two arcs, two other circles centered at I I, etc), you get two more lines. All three lines are concurrent, presumably at an obscure triangle center.brainjam –brainjam 2020-11-12 17:02:09 +00:00 Commented Nov 12, 2020 at 17:02 Are you working on this problem? @brainjam nonuser –nonuser 2020-11-13 16:11:30 +00:00 Commented Nov 13, 2020 at 16:11 Yes, but happy to stop if the new answer is correct. The concurrency point referred to in my earlier comment is X(36), by the way. It shows up in @richrow's answer as K K.brainjam –brainjam 2020-11-13 17:13:15 +00:00 Commented Nov 13, 2020 at 17:13 I think we can all stop now, new answer looks really good. FWIW, there is another (less elegant IMO) answer at AoPSbrainjam –brainjam 2020-11-13 17:40:43 +00:00 Commented Nov 13, 2020 at 17:40 1 OP mentioned Math Excalibur, so I found mention of problem G2 of 2018 Croatian Math Olympiad here. (That issue was after OP posted, so there must have been an earlier reference). AoPS has a lot of olympiad solutions, so wasn't too hard to track down.brainjam –brainjam 2020-11-13 17:57:03 +00:00 Commented Nov 13, 2020 at 17:57 \|Show 1 more comment This answer is useful 0 Save this answer. Show activity on this post. Hint: The line through P P and I I must intersect the perpendicular bisectors of the triangle; in particular it must intersect the bisector of the side B C B C. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 11, 2018 at 10:50 AllawonderAllawonder 13.6k 1 1 gold badge 22 22 silver badges 28 28 bronze badges 0 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let the circle centered at I intersect BC again at K. I claim that PBI'K is cyclic. Note that ∡K P B=∡A P B−∡A P K=∡A C B−∡A I K 2=∡A C B−∡A I B=∡A C B−(∡A C B+∡I A C+∡C B I)=∡C A I+∡I B C=∡I A B+∡A B I=∡A I B=∡B I K=∡K I′B.∡K P B=∡A P B−∡A P K=∡A C B−∡A I K 2=∡A C B−∡A I B=∡A C B−(∡A C B+∡I A C+∡C B I)=∡C A I+∡I B C=∡I A B+∡A B I=∡A I B=∡B I K=∡K I′B. Therefore ∡B P I′=∡B K I′=∡I K B=∡B A I=∡B A M=∡B P M.∡B P I′=∡B K I′=∡I K B=∡B A I=∡B A M=∡B P M. Thus BPI'M are collinear and we're done. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 28, 2020 at 20:06 Alex ZhaoAlex Zhao 9 3 3 bronze badges 0 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 geometry contest-math euclidean-geometry reflection geometric-transformation See similar questions with these tags. 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190071	https://brainly.com/question/29594745	[FREE] Use the discriminant to determine how many and what kind of solutions the quadratic equation 2x^2 - 4x = -2 - brainly.com 5 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 +76,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +44,5k Ace exams faster, with practice that adapts to you Practice Worksheets +6,6k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Use the discriminant to determine how many and what kind of solutions the quadratic equation 2x^2 - 4x = -2 has. Please explain how you got the solution. 20 Points! Thanks! 2 See answers Explain with Learning Companion NEW Asked by kayliexkaylie33 • 11/26/2022 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer 5.0 7 Upload your school material for a more relevant answer In this case, we first need to simplify the givenquadratic equation 2x² - 4x = -2, and to do so,we perform the following simplification steps: For a quadratic equation of the form ax² + bx + c = 0, the discriminant is calculated using the following formula: Δ = b² - 4ac In this case, we have that theequation that we must solve is 2x²-4=-2, which implies that thecorresponding coefficients are: a = 2 b = -4 c = 2 Byintroducing these values into the formulawe get: Δ = b² - 4ac = (-4)² - 4·(2)·(2) = 0 Therefore, the discriminant for the given quadratic equation is Δ=0=0, which is zero, and indicates that the equation2x²-4=-2 has only one real root. We can say: It has a Real solution. Answered by Brainly User Thanks 7 5.0 (4 votes) 1 Textbook &Expert-Verified⬈(opens in a new tab) 5.0 7 Celestial Mechanics - Jeremy Tatum Calculus-Based Physics - Jeffrey W. Schnick Fundamentals of Calculus - Joel Robbin, Sigurd Angenent Upload your school material for a more relevant answer The quadratic equation 2 x 2−4 x+2=0 has exactly one real solution, as indicated by the discriminant being zero. This means the graph touches the x-axis at one point. Thus, the equation has one repeated root. Explanation To determine how many and what type of solutions the quadratic equation 2 x 2−4 x=−2 has, we first need to get the equation into standard form, which is a x 2+b x+c=0. We start by rearranging the equation: 2 x 2−4 x+2=0 Now, we can identify the coefficients: a=2 b=−4 c=2 Next, we calculate the discriminant (Δ) using the formula: Δ=b 2−4 a c Substituting in the values we found: Δ=(−4)2−4(2)(2) Δ=16−16=0 Since the discriminant is equal to zero (Δ=0), it indicates that the quadratic equation has exactly one real solution (or one repeated root). This means the graph of the equation touches the x-axis at one point. In summary, the quadratic equation 2 x 2−4 x+2=0 contains one real solution, which can also be referred to as a double root. Examples & Evidence For example, if we encountered a quadratic equation like x 2−4 x+4=0, we would also find that it has one real solution since its discriminant would also equal zero, confirming the concept that a zero discriminant leads to one solution. The properties of the discriminant are well-established in algebra. If Δ>0, there are two distinct real roots; if Δ=0, there is one real root; and if Δ<0, there are no real roots. This can be found in algebra textbooks discussing quadratic equations. Thanks 7 5.0 (4 votes) Advertisement Community Answer This answer helped 12615 people 12K 5.0 1 Answer: One Real Solution We can Identify the coefficients a, b, and c from your equation because as it says, it is in the format of the quadratic equation ( ax^2 + bx + c = 0 ) a=2,b=-4,c=2 Now we can use the discriminant to determine how many solutions this has using the equation b^2 - 4ac Once we plug in our coefficients: (-4)^2 - 4 x 2 x 2 Once we simplify, we get 0. One real solution. I hope this helps. Answered by Flamestone •41 answers•12.6K people helped Thanks 1 5.0 (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 Multiply or divide as indicated. Be sure to reduce all answers to lowest terms. x 2−16 5 x−15⋅x−3 x 2−2 x−8= A bubble solution contains 4 3 cups of corn starch for every 2 cups of water. Which fraction represents the cups of corn starch per cup of water? A. 2 4 3 B. 4 2 C. 4 3 D. 4 32 The p-value was found to be approximately 0.2063. Using a 5% significance level, we should: A. accept the null hypothesis, because the observed statistic value is larger than the significance level B. accept the null hypothesis, because the p-value is larger than the significance level C. reject the null hypothesis, because the p-value is larger than the significance level D. fail to reject the null hypothesis, because the p-value is larger than the significance level E. reject the null hypothesis, because the observed statistic value is larger than the significance level F. fail to reject the null hypothesis, because the observed statistic value is larger than the significance level Multiply or divide as indicated. Be sure to reduce all answers to lowest terms. 4 y−20 y 2−7 y+10÷y−5 4 y−8=□ Solve for n and s: 2 s−2 n−3 5 n+3=s 0 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
190072	https://www.supersummary.com/non-sequitur/	Non-sequitur in Literature: Definition & Examples \| SuperSummary Opt-out Preferences We use third-party cookies that help us analyse how you use this website, store your preferences, and provide the content and advertisements that are relevant to you. However, you can opt out of these cookies by checking "Do Not Sell or Share My Personal Information" and clicking the "Save My Preferences" button. Once you opt out, you can opt in again at any time by unchecking "Do Not Sell or Share My Personal Information" and clicking the "Save My Preferences" button. [x] Do Not Sell or Share My Personal Information Cancel Save My Preferences Powered by Log InSign Up Home / Literary Devices / Rhetoric: Persuasive / Non Sequitur non sequitur What Is a Non Sequitur? Definition, Usage, and Literary Examples Non Sequitur Definition Uses and Types of Non Sequitur Writers Known for Non Sequiturs Examples of Non Sequiturs in Literature Further Resources on Non Sequiturs Related Terms Non Sequitur Definition In literature, a non sequitur(nahn SEK-wit-ur) is a statement or conclusion that does not logically arise from the thought that precedes it. Writers commonly utilize non sequiturs to heighten the comedic elements of a literary work, especially intheatrical playsand humorous writing. They commonly serve as comedic transitions, acting as purposefully awkward ways to change the subject or mood or signal a shift in theplotor conversation. Outside of literature, non sequiturs are mostly known as statements that seem to come unexpectedly. The word non sequitur stems from the Latin phrase meaning “it does not follow.” The first recognized use of a non sequitur dates to 1540. Uses and Types of Non Sequitur Non sequiturs appear frequently in literature and written dialogue. They’re an easy way to inject humor into a text, change the topic of conversation or the trajectory of a plot, or make a bold statement that requires anything from a creative deduction to a massive leap in logic. Non sequiturs were especially popular among playwrights ofthe theatre of the absurd, an era in dramatic writing that explored life’s idiosyncrasies, contradictions, and complexities through irrational and illogical language. For instance, in Tom Stoppard’s frenetic playRosencrantz and Guildenstern Are Dead, the title characters ramble in non sequitur after non sequitur. Here is a brief passage: This made for a kind of harmony and a kind of confidence. It related the fortuitous and the ordained into a reassuring union, which we recognized as nature. The sun came up about as often as it went down, in the long run, and a coin showed heads about as often as it showed tails. Then a messenger arrived. We had been sent for. Nothing else happened. Non sequiturs also appear in casual, everyday conversation. For instance, a person who experienced a bad reaction to a medication might say, “I had nasty side effects from that medicine, so doctors shouldn’t prescribe it to anybody.” This conclusion is obviously illogical, since not everyone will experience the same side effects. In the legal system, a non sequitur is an illogical conclusion drawn based on previously presented information. One famous example comes from the O.J. Simpson murder trial and a pair of gloves: “If it doesn’t fit, you must acquit.” Extemporaneous speech comes with a higher risk of non sequiturs, as a speaker making off-the-cuff comments attempts to explain the rationale behind their thoughts, even if such rationale is haphazard. In this regard, non sequiturs are prevalent among politicians. For instance, Donald Trump gave a non sequitur-laden interview with The New York Times in 2017, at one point saying: “They’ve won five wars where the armies that went against them froze to death. It’s pretty amazing. So, we’re having a good time. The economy is doing great.” Non Sequiturs as Logical Fallacies In logic and philosophy, a logical fallacy is an argument that employs faulty reasoning. Non sequiturs are classified as logical fallacies because faulty reasoning is applied to reach conclusions that are inconsistent with the supplied information. There are several different types of non sequiturs that also qualify as logical fallacies. Affirming the consequent is when a true statement is followed by an invalid conclusion, and the statement’s converse is also true. For example: “The street in front of her house is wet, so it must have rained earlier.” By swapping the two pieces of this statement, it would read: “If it rained earlier, the street in front of her house would be wet.” This makes a conditionally true statement. Leaving it as is, however, represents a logical fallacy because there are things other than rain that can make a street wet. Therefore, the conclusion is not valid. Affirming the disjunct is a situation when the speaker/writer assumes that because one statement is true, another must be false—even if both are correct. For example: “To be a movie star, you have to be beautiful or well-connected. Julia Roberts is beautiful. Therefore, she is not well-connected.” Denying the antecedent usually occurs when an if/then (or therefore) statement erroneously concludes that when the if part of the statement is untrue, the then/therefore part must also be untrue. For instance: “If it meows, it is a cat. It does not meow. Therefore, it is not a cat.” Denying the conjunct is a statement in which the first premise says that at least one of the two conjuncts is false and concludes that the other conjunct must then be true. For example: “He is not both a Republican and a Democrat. He is not a Republican. Therefore, he is a Democrat.” A fallacy of the undistributed middle is when the middle term of a syllogism—a logical argument that relies on deductive reasoning—is not distributed in at least one of the statement’s other premises, such as: “All human beings are mortal. All dogs are mortal. Therefore, all dogs are human beings.” Non Sequiturs in Comedy Non sequiturs are abundant in comedic works because their innately nonsensical nature elicits laughs or, at the least, a spirit of lightheartedness. Take, for example, Ralph Wiggum, a supporting character on The Simpsons. Ralph is practically a walking non sequitur: always stringing together disjointed thoughts, replying to questions with nonsensical answers, and speaking in hilarious word salads. In one episode, Lisa Simpson and her rival-turned-friend Allison invite Ralph to play anagrams with them; in lieu of an answer, Ralph replies, “My cat’s breath smells like cat food.” In another episode, Ralph recounts his summer vacation: “…then, the doctor told me that both my eyes were lazy! And that’s why it was the best summer ever!” Non Sequitur is also the name of a long-running comic strip. Created by Wiley Miller, it follows the escapades of the Pyle family and the other citizens of Whatchacallit, Maine. Wiley likely chose the title because of the humor and satire associated with the term—two elements that are front and center in the comic. Generational humor, specificallymillennial humor, often makes use of non sequiturs. For instance, millennial humor tends to utilize a lot of randomness, which is just another way of saying it relies on a lot of non sequiturs. As a result, some consider the millennial sense of humor to be increasingly surreal and borderline absurd. You can see examples of this humor on the shows Family Guy,Rick and Morty, and The Eric Andre Show. Writers Known for Non Sequiturs Edward Albee,The American Dream,Three Tall Women Samuel Beckett,Waiting for Godot,Endgame,Happy Days Eugene Ionesco,The Bald Soprano Harold Pinter,The Room,The Birthday Party Tom Stoppard,Rosencrantz and Guildenstern Are Dead Examples of Non Sequiturs in Literature 1. Samuel Beckett,Waiting for Godot Beckett wrote his classic play almost exclusively in non sequiturs. The characters of Vladimir and Estragon are waiting for a man named Godot to appear. As they pass the time, they have a series of nonsensical conversations: VLADIMIR: “Consult his family…” ESTRAGON: (anxious) “And we?” ESTRAGON: “And why would he shout?” VLADIMIR: “At his horse. Silence.” ESTRAGON: (violently) “I’m hungry!” VLADIMIR: “Do you want a carrot…” VLADIMIR: “I might have some turnips…” VLADIMIR: “Oh pardon! I could have sworn it was a carrot…” ESTRAGON: (chewing) “I asked you a question.” VLADIMIR: “Ah.” ESTRAGON: “Did you reply?” VLADIMIR: “How’s the carrot?” The play is a cornerstone of the theatre of the absurd, and while it is certainly open to interpretation, it seems to reflect the inherent emptiness and meaninglessness of life. In the above exchange, the two characters bat non sequiturs back and forth: ignoring questions, answering questions with unrelated questions, and making comments that don’t logically stem from anything that came prior. 2. Susanna Kaysen,Girl, Interrupted In Kaysen’s memoir, she documents the time she spent in a psychiatric hospital as a young woman: It was a spring day, the sort that gives people hope: all soft winds and delicate smells of warm earth. Suicide weather. Daisy had killed herself the week before. This passage has a glaring non sequitur that forces a narrative transition from hopeful to bleak. After describing the advent of spring and the loveliness of the wind and earth, Kaysen abruptly calls it “suicide weather.” Logically, there’s no such thing as suicide weather, but it provides the right change of mood to describe Daisy’s death and the subsequent atmosphere in the hospital. 3. Joseph Heller,Catch-22 Heller’s novel is a satirical look at World War II through the eyes of an antihero Army captain named John Yossarian. Another character, a bomber pilot named Orr, speaks mainly in non sequiturs, much to the confusion of his fellow soldiers: “There’s a leak in here,” Orr said. “I’m trying to fix it.” “Please stop it,” said Yossarian. “You’re making me nervous.” “When I was a kid,” Orr replied, “I used to walk around all day with crab apples in my cheeks. One in each cheek.” In this passage, Orr attempts to fix a leaky tent, and when Yossarian asks him to stop, he responds with the non sequitur about stuffing crab apples in his cheeks as a child. Further Resources on Non Sequiturs TV Tropes has a detailed list ofnon sequiturs in tv, movies, animation, and popular culture. The Internet Encyclopedia of Philosophy offers an in-depth look at varioustypes of logical fallacies, including non sequiturs. The Hartford Courant discusseshow Donald Trump’s speech “flummoxes” linguists, especially his rampant use of non sequiturs, repetition, and asides. A CSUN website provides a primer onhow to use logical fallacies, like non sequiturs, in debates. Colleen Abel delves intothe “major moment” non sequiturs are having in modern poetry. Related Terms Antecedent Aside Conjunct Consequent Disjunct Dramatic Monologue Syllogism 7-day Money-Back Guarantee company About Us Our Literary Experts Wall of Love Work With Us library Study Guides Teaching Guides Plot Summaries Study Guide Collections New This Week Book Briefs resources Literary Devices Resource Guides Discussion Questions Tool readers Student Teacher Book Club Member Parent contact Help Feedback Suggest a Title Copyright ® 2025 SuperSummary/All Rights Reserved Privacy Policy \| Terms of Service \| Do not Share my Personal Information
190073	https://arxiv.org/abs/1307.1065	[1307.1065] The Combinatorics of Flat Folds: a Survey Skip to main content We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate >math> arXiv:1307.1065 Help \| Advanced Search Search GO quick links Login Help Pages About Mathematics > Metric Geometry arXiv:1307.1065 (math) [Submitted on 3 Jul 2013] Title:The Combinatorics of Flat Folds: a Survey Authors:Thomas C. Hull View a PDF of the paper titled The Combinatorics of Flat Folds: a Survey, by Thomas C. Hull View PDF Abstract:We survey results on the foldability of flat origami models. The main topics are the question of when a given crease pattern can fold flat, the combinatorics of mountain and valley creases, and counting how many ways a given crease pattern can be folded. In particular, we explore generalizations of Maekawa's and Kawasaki's Theorems, develop a necessary and sufficient condition for a given assignment of mountains and valleys to fold up in a special case of single vertex folds, and describe recursive formulas to enumerate the number of ways that single vertex in a crease pattern can be folded. Comments:10 pages, 4 figures Subjects:Metric Geometry (math.MG); Combinatorics (math.CO) MSC classes:05A02 Cite as:arXiv:1307.1065 [math.MG] (or arXiv:1307.1065v1 [math.MG] for this version) Focus to learn more arXiv-issued DOI via DataCite Journal reference:Origami3: The Third International Meeting of Origami Science, Mathematics, and Education, T. Hull ed., AK Peters, 2002 Submission history From: Thomas Hull [view email] [v1] Wed, 3 Jul 2013 16:22:24 UTC (21 KB) Full-text links: Access Paper: View a PDF of the paper titled The Combinatorics of Flat Folds: a Survey, by Thomas C. Hull View PDF TeX Source Other Formats view license Current browse context: math.MG <prev \| next> new \| recent \| 2013-07 Change to browse by: math math.CO References & Citations NASA ADS Google Scholar Semantic Scholar export BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic Tools Bibliographic and Citation Tools [x] Bibliographic Explorer Toggle Bibliographic Explorer (What is the Explorer?) [x] Connected Papers Toggle Connected Papers (What is Connected Papers?) [x] Litmaps Toggle Litmaps (What is Litmaps?) [x] scite.ai Toggle scite Smart Citations (What are Smart Citations?) Code, Data, Media Code, Data and Media Associated with this Article [x] alphaXiv Toggle alphaXiv (What is alphaXiv?) [x] Links to Code Toggle CatalyzeX Code Finder for Papers (What is CatalyzeX?) [x] DagsHub Toggle DagsHub (What is DagsHub?) [x] GotitPub Toggle Gotit.pub (What is GotitPub?) [x] Huggingface Toggle Hugging Face (What is Huggingface?) [x] Links to Code Toggle Papers with Code (What is Papers with Code?) [x] ScienceCast Toggle ScienceCast (What is ScienceCast?) Demos Demos [x] Replicate Toggle Replicate (What is Replicate?) [x] Spaces Toggle Hugging Face Spaces (What is Spaces?) [x] Spaces Toggle TXYZ.AI (What is TXYZ.AI?) Related Papers Recommenders and Search Tools [x] Link to Influence Flower Influence Flower (What are Influence Flowers?) [x] Core recommender toggle CORE Recommender (What is CORE?) Author Venue Institution Topic About arXivLabs arXivLabs: experimental projects with community collaborators arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs. Which authors of this paper are endorsers? \| Disable MathJax (What is MathJax?) About Help Contact Subscribe Copyright Privacy Policy Web Accessibility Assistance arXiv Operational Status Get status notifications via email or slack
190074	https://www.youtube.com/watch?v=UhyaTVofNXw	Symmetry in Charge Configurations and Electric Fields Maxwell Ross 427 subscribers 9 likes Description 740 views Posted: 12 Apr 2019 A qualitative discussion of charge symmetries and how they can be used to learn things about the electric field at specific locations. Transcript: hey everybody I want to talk about electric fields and symmetry and using symmetry to simplify some well what might be more complex problems so what's important is that if we have a charge configuration the symmetries of the charge configuration are the same as the symmetries of the electric field so we can use the charge configuration symmetry to figure out some things about the electric field so let's do some examples just to see how this goes let's look at this first set of charges right here we have a plate a positive charge at the top and a plate of positive charge down at the bottom now pardon that the drawings not perfect these are supposed to be pretty much equal and opposite about the y-axis or about the x axis flipped so we have two symmetries here we have reflective symmetry against the R cross I should say the x axis if we flipped it our charge would be the same that means if we flip it our electric field has to be the same so anywhere along this line if I were to sketch out the electric field I were to guess that the electric field were this well that can't be right because if we flip our charge symmetry that would have to flip the electric field which means it would have to be this and that can't be the case because when we reverse it it's the same the only way that we can have reflective symmetry here is if our electric field points somewhere along this axis so our electric field has to either point to the right or maybe a points to the left but the vertical components have to cancel so there's one little bit of symmetry we also have symmetry about the y-axis we can make the same argument if we reflect about the y-axis the electric fields or sorry we get the same configuration which means that our electric fields have to well basically it means that there can't be any horizontal component of the electric field on this axis which means our field has to point up or directly down on this axis and we can combine those all together at this point in the center oops just deleted my Banksy's at this point in the center at this point in the center we have horizontal and vertical symmetry and the only way that our electric field can be the same when it flipped both horizontally and vertically is if the electric field is zero so that means we have no B field let's look at this next charge configuration we have a ring of negative charge now if we look at this we have rotational symmetry about our center so the ohm so the only way what sorry that means that our electric field has to be rotationally symmetric about the center if we rotate the electric if we rotate the charge we get the same charge so we'll get the same electric field on rotation that means our field everywhere has to be radially symmetric rotationally symmetric so it's going to either point out or point in but we know that since its negative charge their field lines are going to point in towards it any time we have a radially symmetric charge we have radially symmetric electric field that also means right at the center here our electric field again has to be zero because if it pointed in any direction it wouldn't be able they wouldn't preserve that symmetry under rotation let's take a look at this X of positive charge over on the right we have we have a lot of different symmetries here we have symmetry about the this line here symmetry about this line here symmetry about this line here reflective symmetry about this line here which means our electric field has to be has to be the same on any reflection there that again tells us that components of the electric field have to cancel along these lines the only way we can get an electric field that's the same under a reflection as if it points along those lines it's a positive charge we know it's going to point away the other components have to cancel or the components along along these lines of symmetry have to cancel in this case you could do that anywhere along those lines so here we have something that is symmetric and only one way it's only symmetric about this x equals zero line or Y axis which means our electric field is only symmetric about this x equals 0 or Y axis again means along this y axis our electric field has to point either straight up or straight down I think it's going to point down when it's above our charges because electric field points towards negative and up when it's below our charges so I expect it will be something like that and the fields shape if I were to actually sketch out the field lines I think are going to be something like like this let's look at a few more interesting examples let's go down here here we have a positive plate on the left and a negative charge on the right this isn't symmetric about a reflection but it is what we call anti-symmetric if we reflect it and we switch the charge if we were to reflect this positive plate and reverse the charge it would be the same so we can't say exactly the same things that we could about the electric field is before but because it's anti-symmetric we can still figure some things out we know that this positive plate is going to produce an electric field that points away from the positive plate I want to draw it as a red arrow here and if we were to ignore this negative plate for now let's pretend that we flipped our positive plate across that axis of symmetry and ended up with a positive plate over here well then we know from symmetry that the electric field from that right positive plate would have to point to the left with the same magnitude as this arrow here but since this is anti-symmetric since we actually have a negative plate over here the field produced from this negative plate would be the same as this except reversed so our net electric field is just going to be the field from one of our two plates but doubled we have a rightward field from our leftward play and we also have an equal right word field from our rightward plate so our net field passed a point to the right our net field is going to be something like that in green pardon my messy arrows my green arrow and we can use this to figure out the electric fields for more complicated arrangements here we have a positive weirdly shaped electric field and a negative weirdly shaped electric field but there is some symmetry about well rotational symmetry here if we rotate 180 degrees about our center point so since it's anti-symmetric and rotation by 180 degrees we can use that to figure out the electric field we know that the positive plate is going to produce an electric field that goes this way and because it's anti-symmetric about rotation the negative plate has to produce an equal magnitude field in the same direction so our net electric field at the origin is going to be something like that that green arrow we can also combine these symmetries to figure out more complicated electric fields here in this rightmost diagram over here we have two negative plate here in this rightmost diagram over here we have two negative plates that are symmetric this one and this one and a positive and a negative plate that are anti-symmetric if we wanted to find out the direction of the electric field at the origin well we can think about these images imagery's that are involved in cabina the two negative plates are going to cancel out at the origin because there's that rotational symmetry this are the two large negative plates that is the smaller negative plate and a smaller positive plate our anti symmetric about rotation which means they'll produce equal electric fields so since my negative plate produces a field that goes towards it my positive plate producing produces it equal field that goes towards that negative plate so the net electric field is going to be in that direction it's just a little bit on using symmetries to solve for electric fields that's all
190075	https://www.mathportal.org/calculators/popular-problems/complex-simplify.php?val1=(-1-i)%5E2&val2=	Simplify (-1-i)^2 Site map Math Tests Math Lessons Math Formulas Calculators Math Calculators, Lessons and Formulas It is time to solve your math problem mathportal.org HW Help (paid service) Math Lessons Math Formulas Calculators ◀ back to index Question Answer (−1−i)2=2 i Explanation Discover more Calculator Mathematics calculator Math calculators math Calculators Statistics Tap the blue circles to see an explanation. (−1−i)21◯1+2 i+i 2 2◯1+2 i−1 3◯2 i ① Find (−1−i)2 in two steps. S1: Change all signs inside bracket. S2: Apply formula (A+B)2=A 2+2⋅A⋅B+B 2 where A=1 and B=i. (−1−i)2S 1(1+i)2 S 2 1 2+2⋅1⋅i+i 2==1+2 i+i 2 ② i 2=−1 ③ Combine like terms: 2 i+1−1=2 i This page was created using Complex Expression calculator About the Author Welcome to MathPortal. This website's owner is mathematician Miloš Petrović. I designed this website and wrote all the calculators, lessons, and formulas. If you want to contact me, probably have some questions, write me using the contact form or email me on mathhelp@mathportal.org. Send Me A Comment Comment: Email (optional): Main Navigation Online Math Calculators Math Formulas Math Lessons Math Tests Site Map Privacy Policy
190076	https://www.quora.com/What-is-the-significance-of-sign-changes-in-a-quadratic-polynomial	Something went wrong. Wait a moment and try again. Types of Polynomials Roots (mathematics) Quadratic Forms Algebraic Functions Polynomial Theory Degrees (polynomials) Roots of Polynomials 5 What is the significance of sign changes in a quadratic polynomial? · The significance of sign changes in a quadratic polynomial can be understood in terms of the polynomial's roots and its graph. A quadratic polynomial typically has the form: f(x)=ax2+bx+c where a, b, and c are constants and a≠0. Key Points of Sign Changes: Roots and Intervals: The roots (or zeros) of the polynomial are the values of x for which f(x)=0. The sign of f(x) changes at these roots. If the quadratic has two distinct real roots, the polynomial will change its sign at each root. For example, if the roots are r1 and r2 (with r1<r2), then: f(x)>0 for x<r1 f(x The significance of sign changes in a quadratic polynomial can be understood in terms of the polynomial's roots and its graph. A quadratic polynomial typically has the form: f(x)=ax2+bx+c where a, b, and c are constants and a≠0. Key Points of Sign Changes: Roots and Intervals: The roots (or zeros) of the polynomial are the values of x for which f(x)=0. The sign of f(x) changes at these roots. If the quadratic has two distinct real roots, the polynomial will change its sign at each root. For example, if the roots are r1 and r2 (with r1<r2), then: f(x)>0 for x<r1 f(x)<0 for r1<x<r2 f(x)>0 for x>r2 Discriminant: The discriminant D=b2−4ac determines the nature of the roots: If D>0: Two distinct real roots (two sign changes). If D=0: One real root (the polynomial touches the x-axis and does not change sign). If D<0: No real roots (the polynomial does not change sign; it is either always positive or always negative). Graphical Interpretation: The graph of a quadratic polynomial is a parabola. The direction of the parabola (opening upwards or downwards) is determined by the coefficient a: If a>0: The parabola opens upwards, and it will be positive outside the roots (if they exist). If a<0: The parabola opens downwards, and it will be negative outside the roots. Behavior at Infinity: The sign of the leading coefficient a also tells us about the behavior of the polynomial as x approaches positive or negative infinity. For example: If a>0, as x→∞ or x→−∞, f(x)→∞. If a<0, as x→∞ or x→−∞, f(x)→−∞. In summary, sign changes in a quadratic polynomial are significant as they indicate the location of the roots, the intervals where the polynomial is positive or negative, and the overall shape and behavior of the graph. Understanding these sign changes helps in analyzing the polynomial's properties and its graph effectively. Vaibhav Sharma Physics Grad Student in IIT Bombay · Author has 567 answers and 2.8M answer views · 11y For the quadratic polynomial, ax2+bx+c, if all the coefficients are real, then both the roots are either real or both are imaginary & complex conjugate of each other, that is, they will be of the form x+iy & x−iy Sum of the roots = −b/a Product of the roots = c/a If both the roots are real and positive, then clearly −b/a and c/a have to be positive. Also if the roots are imaginary with positive real parts, the sum will be positive as well as the product. This means that a, −b and c must be of the same sign. If both the roots are real and negative, then sum of the For the quadratic polynomial, ax2+bx+c, if all the coefficients are real, then both the roots are either real or both are imaginary & complex conjugate of each other, that is, they will be of the form x+iy & x−iy Sum of the roots = −b/a Product of the roots = c/a If both the roots are real and positive, then clearly −b/a and c/a have to be positive. Also if the roots are imaginary with positive real parts, the sum will be positive as well as the product. This means that a, −b and c must be of the same sign. If both the roots are real and negative, then sum of the roots −b/a will be negative while the product c/a will be positive. If the roots are imaginary with negative real parts, then also the sum will be negative while the product will be positive. Thus in this case, a and c must be of the same sign while −b should be of a different sign. If one root is positive and one negative, then the product c/a will always be negative. Remember the roots cannot be imaginary in this case as the product of complex conjugate imaginary numbers is always positive ( (x+iy)∗(x−iy)=x2+y2>0. The sum can be both positive or negative depending upon whether the negative root is bigger in magnitude or the positive one. That requires c and a to be of opposite sign while −b can either have the same or different sign from a. In short : If a, −b & c are of the same sign, then either the roots are real and positive or imaginary roots with positive real parts. If a and c are of the same sign while −b is of a different sign, then either the roots are real and negative or they are imaginary with negative real parts. If a and c are of a different sign and −b is of the same sign as a, then both the roots are real with one positive and one negative where the magnitude of the positive root is bigger than the negative one. If a and c are of opposite signs and −b is of the opposite sign of a as well, then both the roots are again real with one positive and one negative but this time the magnitude of the negative root is bigger than the magnitude of the positive root. Nic Banks MMath in Pure Mathematics, University of Waterloo (Graduated 2021) · Upvoted by Babita Gupta , M.Sc Mathematics · Author has 535 answers and 3.1M answer views · 6y Related Why is the quadratic formula so significant? First I’ll type out the formula. It’s worth doing at least once. If a, b, and c are real numbers then: ax2+bx+c=0⟹x=−b±√b2−4ac2a One reason that it’s significant is because of the simplicity. It’s hard to think of it as simple when you’re first learning and working with it, but think about what it gives us: no matter which quadratic function you give me, I can always find where it crosses the x-axis. ALWAYS. This is an extremely hard problem for other types of functions. Unless x is a very particular number, then sin(x)=0 is probably impossible to express using radical First I’ll type out the formula. It’s worth doing at least once. If a, , and are real numbers then: One reason that it’s significant is because of the simplicity. It’s hard to think of it as simple when you’re first learning and working with it, but think about what it gives us: no matter which quadratic function you give me, I can always find where it crosses the -axis. ALWAYS. This is an extremely hard problem for other types of functions. Unless is a very particular number, then is probably impossible to express using radicals and other simple operations. Even for cubic polynomials there is a so-called cubic formula, which is a formula that finds the root of a cubic function. But it’s VASTLY more complex than the quadratic formula; just writing it on a whiteboard would probably take a whole board. The quartic formula is, again, enormously complex compared to the cubic and quadratic versions. And for a quintic, AKA degree , function, it is impossible to find such a formula using roots and other algebraic operations. It’s not just that we haven’t been clever enough to find one - it was proven that such a formula which works for every quintic is impossible (called the Abel-Ruffini theorem). It’s incredible that a simple polynomial can be unsolvable like that. So, one reason that the quadratic formula is significant is its existence and simplicity. Another significant aspect of the formula is the form of it. For one thing, we know of quadratics that don’t cross the -axis; is such a quadratic. That means that no real values of satisfy . The quadratic formula in fact predicts this: using the formula we encounter the number . What real number squares to ? Of course the answer is “none”. This is true for any negative number. For this reason the quantity is very important when studying quadratic functions. This is called the discriminant (a capital D in Greek) because it discriminates (tells the difference between) different types of quadratics. Namely, if , then the square root in the formula is not a real number, meaning that the quadratic has no real number roots (although it does have complex roots). The discriminant tells us even more. If , then where the formula would normally have two answers due to the , we actually only have one. This is called a double root, and it’s immediately clear from the formula when a quadratic has a double root. Of course, if then we are in the “nice” case where the quadratic has two distinct real roots. Right away, the formula has allowed us to break ALL quadratics into three important categories. That’s significant to me. Another minor observation - let’s break the formula into two parts: The first part might be familiar: this is the -coordinate for the vertex of the graph of the quadratic. Then the formula tells us that the two roots are equally spaced from the vertex - one lower, and one greater. This is an amazing insight that you might not have ever had without the formula. So, why is it significant? It gives us insight into the three types of quadratic functions and tells us something fascinating about their shape. Related questions What is a real life example of a quadratic polynomial? What is the significance of the quadratic formula? How can you find the signs of the coefficients of a quadratic polynomial by just looking at the graph? What is the need to change signs while performing a division of polynomials? Why does a polynomial with real, simple roots change its sign between its roots? Elizabeth Jean Stapel tutor (1989 to present), instructor (1991+), author (1998+) · Author has 5.9K answers and 3.3M answer views · 1y Related How does changing the sign of an equation affect the quadratic formula? The Quadratic Formula applies to quadratic equations in the following form: To “change the sign” on the equation, multiply though by : This will have absolutely no effect on the values of which solve the quadratic equation, as these -values are where the quadratic crosses the -axis. And, by multiplying through by , all you’ve done to the associated quadratic function, , is flip it upside-down. The -intercepts remain the same. Max Gretinski Studied Mathematics · Author has 6.5K answers and 2.4M answer views · Jul 29 Related How do double roots affect the sign changes in a polynomial inequality like (x+7) (x-3) (x-5) < 0, and what are they exactly? If there is a factor raised to an even power (2, 4, 6, etc.), then instead of changing sign at the point where that factor is zero, the function on the left-hand side keeps the same sign. Example: f(x) = math^2(x - 3) [/math] The function changes sign where x - 3 = 0; that is, when x = 3. But the graph “bounces off” the x-axis when x + 4 = 0; that is, when x = -4. It does not change sign. The zeros of the function break the real line up into intervals. For example, in the above case, we have intervals (-[math]\infty[/math], -4) (-4, 3) (3,[math] \infty)[/math] Start your sign chart on the right side. If the lead coefficient is If there is a factor raised to an even power (2, 4, 6, etc.), then instead of changing sign at the point where that factor is zero, the function on the left-hand side keeps the same sign. Example: f(x) = math^2(x - 3) [/math] The function changes sign where x - 3 = 0; that is, when x = 3. But the graph “bounces off” the x-axis when x + 4 = 0; that is, when x = -4. It does not change sign. The zeros of the function break the real line up into intervals. For example, in the above case, we have intervals (-[math]\infty[/math], -4) (-4, 3) (3,[math] \infty)[/math] Start your sign chart on the right side. If the lead coefficient is positive, the sign of the function on the rightmost interval will always be positive (+). So I write: (-[math]\infty[/math], -4) (-4, 3) (3,[math] \infty)[/math] + Now, the sign alternates when we cross “3” into the next interval. Write a - there instead of a +. (-[math]\infty[/math], -4) (-4, 3) - (3,[math] \infty)[/math] + The sign does NOT alternate when crossing the “even” root at x = -4. Keep the sign negative. (-[math]\infty[/math], -4) - (-4, 3) - (3,[math] \infty)[/math] + You now have all of the information that you need to solve any polynomial inequality involving the function. Graph courtesy desmos reveals that the method works: Let’s try another. g(x) =[math] (x + 3)^3(x - 1)^2(x - 5)[/math] The roots of -3, 1, and 5 give us these intervals (-[math]\infty[/math], -3) ; (-3, 1) ; (1, 5) ; (5, [math]\infty[/math]) The sign is positive on (5, [math]\infty[/math]). Since x - 5 is not raised to an even power, the sign alternates when moving from the interval (5, [math]\infty[/math]) to the interval (1, 5). The function is negative on (1, 5). The graph crosses the x-axis at (5, 0). Since x - 1 is raised to an even power, the sign remains the same when moving from (1, 5) to the interval (-3, 1). The function is still negative on (-3, 1), and the graph “bounces off” of the x-axis at (1, 0). Since x + 3 is not raised to an even power, the sign alternates when moving from (-3, 1) to (-[math]\infty[/math], -3). The function is positive on (-[math]\infty[/math], -3). The graph crosses the x-axis at (-3, 0). Graph courtesy of desmos again. Therefore, if you were asked to solve the inequality math^3(x - 1)^2(x - 5) \ge 0[/math] We would see that the function is greater than 0 (positive) on (-[math]\infty[/math], -3) U (5, [math]\infty[/math]) It is equal to zero at -3, 1, and 5. Shaded above, the solution is (-[math]\infty[/math], -30 U {1} U [5, [math]\infty[/math]) Jörg Straube M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.2K answers and 1.6M answer views · 2y Related Why does the concavity of a quadratic function f(x) =ax^2+bx+c depend on the sign of a? First of all: what is the concavity of a curve? Let’s have a look at this graph: The blue parabola has “a” = 1 = positive. The orange parabola has “a” = -1 = negative. If you follow the blue curve from left to right, its arc turns counter-clockwise. See blue arrow If you follow the orange curve from left to right, its arc turns clockwise. See orange arrow The orientation of this arrow (clockwise vs counter-clockwise) is called “curvature”. Mathematically, the curvature is given by the second derivative. If the second derivative > 0, the curvature is counter-clockwise. If the second derivative < First of all: what is the concavity of a curve? Let’s have a look at this graph: The blue parabola has “a” = 1 = positive. The orange parabola has “a” = -1 = negative. If you follow the blue curve from left to right, its arc turns counter-clockwise. See blue arrow If you follow the orange curve from left to right, its arc turns clockwise. See orange arrow The orientation of this arrow (clockwise vs counter-clockwise) is called “curvature”. Mathematically, the curvature is given by the second derivative. If the second derivative > 0, the curvature is counter-clockwise. If the second derivative < 0, the curvature is clockwise. f(x) = ax^2 + bx + c. f’(x) = 2ax + b. f”(x) = 2a. So, the sign of “a” decides if the arrow turns clockwise or counter-clockwise. The following graph shows a cubic curve. The orientation of the arrow changes at x = 0 Related questions What is the significance of quadratic equations in graphing functions? What is the definition of a real root of a quadratic polynomial? Do all real roots of a quadratic polynomial lie within its roots? What is the difference between a quadratic polynomial and any other polynomial? What is special about quadratic polynomials in particular (mathematics)? What is the significance of constant polynomial? What is a second order polynomial or quadratic equation called in math terms? How many different types of second order polynomials or quadratic equations are there? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.5M answer views · 7y Related What is the significance of studying quadratic equations? Mathematicians study things because they want to find out all there is to know about those things! Let me give a simple example: Suppose a very young mathematician has only been familiar with linear equations and so expects problems to have just one solution but then he/she meets a new type of problem like this: he/she would be very surprised that there are 2 different solutions to this problem! (The answer could be 2 or 6 of course.) The young mathematician would start investigating (or be lead by a good teacher) and find out that the basic key to these equations lies in this logic: If a × b = 0 t Mathematicians study things because they want to find out all there is to know about those things! Let me give a simple example: Suppose a very young mathematician has only been familiar with linear equations and so expects problems to have just one solution but then he/she meets a new type of problem like this: he/she would be very surprised that there are 2 different solutions to this problem! (The answer could be 2 or 6 of course.) The young mathematician would start investigating (or be lead by a good teacher) and find out that the basic key to these equations lies in this logic: If a × b = 0 then a = 0 or b = 0 leading to : if (x – 3)(x + 5) = 0 then x = 3 or x = –5 When the young mathematician is ready, equations such as: can be studied leading to “completing the square” method and, assuming he/she has good algebra skills the quadratic formula can be proved by “completing the square” on the general form: This of course leads to other investigations because these new equations seem to have different types of solutions. This of course leads to other investigations because these new equations seem to have different types of solutions. He/she would find that one type, like x^2 – 8x + 12 = 0 has 2 rational solutions then a slight change such as x^2 – 8x + 16 = 0 means the equation would only have 1 rational solution. Another slight change produces an equation like x^2 – 8x + 14 = 0 which would have 2 irrational solutions and some equations such as x^2 – 8x + 17 = 0 seem to have no real solutions at all. (This would later lead the young mathematician off into the realm of complex numbers.) The next step would be to find out WHY. It is actually a huge mental leap forward for many young students to find out that the solutions of f(x) = 0 are where the graph of y = f(x) crosses the x axis! Lots of mathematics is done for its own sake simply because it is interesting and enjoyable. The young mathematician wants to know all about this new type of equation and all the different cases. At the time, the young mathematician is not very concerned about the applications of these equations to some real life problem. Of course, subsequently it will be found there are many applications! This is where the mathematician would get this feeling of great self satisfaction again. Doug Dillon Ph.D. Mathematics · Author has 12.3K answers and 11.4M answer views · 4y Related How would the remainder theorem change if you divided by a quadratic polynomial instead of a linear polynomial? The remainder theorem is only described when the divisor is linear. That is, the remainder when p(x) is divided by x-a is p(a). However, with care a version of this can be used which the divisor is of higher degree. An example will serve a better purpose than a general statement. Just so that the question will be done in finite time, I will chose the divisor and dividend to be near in terms of degree. Example: Find the remainder when [math]x^4+3x^2+2x+1[/math] is divided by [math]x^2–4x-2[/math], In [math]x^4+3x^2+2x+1[/math], replace every occurence of [math]x^2[/math] with [math]4x+2[/math] and write math^2+3(4x+2)+2x+1[/math] [math]=16x^2+30x+11[/math] [math]\equiv 16(4x+2)+30x+11[/math] [math]=94[/math] The remainder theorem is only described when the divisor is linear. That is, the remainder when p(x) is divided by x-a is p(a). However, with care a version of this can be used which the divisor is of higher degree. An example will serve a better purpose than a general statement. Just so that the question will be done in finite time, I will chose the divisor and dividend to be near in terms of degree. Example: Find the remainder when [math]x^4+3x^2+2x+1[/math] is divided by [math]x^2–4x-2[/math], In [math]x^4+3x^2+2x+1[/math], replace every occurence of [math]x^2[/math] with [math]4x+2[/math] and write math^2+3(4x+2)+2x+1[/math] [math]=16x^2+30x+11[/math] [math]\equiv 16(4x+2)+30x+11[/math] [math]=94x+43[/math] and that, if no arithmetic mistakes, is the remainder. I used [math]\equiv[/math] loosely. Chahat Deep Singh Designeer • Geek • Soporific • Roboticist • Sapiosexual · Author has 62 answers and 412.5K answer views · 9y Related What's the difference between polynomial and quadratic functions? In mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. So, basically a polynomial is of the form: If [math]n=2[/math], the equation is a quadractic polynomial equation commonly known as quadratic equation. In mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. So, basically a polynomial is of the form: If [math]n=2[/math], the equation is a quadractic polynomial equation commonly known as quadratic equation. Soumya Sharma Studied Interior Design at Amity University (Graduated 2021) · 8y Related What is the difference between quadratic and polynomial? In mathematics, the term Quadratic pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms. Whereas, Polynomial is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). (Above given, were only the definition to the terms) On the basis of application,both have different entity : QUADRATIC Algebra (elementary and abstract) Quadratic function (or Quadratic polynomial), a polynomial function that contains terms of at most second degree Complex qua In mathematics, the term Quadratic pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms. Whereas, Polynomial is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s). (Above given, were only the definition to the terms) On the basis of application,both have different entity : QUADRATIC Algebra (elementary and abstract) Quadratic function (or Quadratic polynomial), a polynomial function that contains terms of at most second degree Complex quadratic polynomials are particularly interesting for their sometimes chaotic properties under iteration Quadratic equation, a polynomial equation of degree 2 (reducible to 0 = ax2 + bx + c) Quadratic formula, calculation to solve a quadratric equation for the independent variable (x) Quadratic field, an algebraic number field of degree two over the field of rational numbers Quadratic irrational or "quadratic surd", an irrational number that is a root of a quadratic polynomial Calculus Quadratic integral, the integral of the reciprocal of a second-degree polynomial. POLYNOMIAL A polynomial function is of the form- f(x) = anx n + an−1x n−1 + . . . + a2x 2 + a1x + a0 . The degree of a polynomial is the highest power of x in its expression. Constant (non-zero) polynomials, linear polynomials, quadratics, cubics and quartics are polynomials of degree 0, 1, 2 , 3 and 4 respectively. The function f(x) = 0 is also a polynomial, but we say that its degree is ‘undefined’. Polynomial Equation- is simply a polynomial that has been set equal to zero in an equation. Calculus polynomial : Power rule - Wikipedia Rijul Saini Math Enthusiast · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Anurag Bishnoi , Ph.D. Mathematics, Ghent University (2016) · 12y Related What is the significance of the quadratic formula? Most of the times in Mathematics, it's not the formula itself, but the idea behind it, which carries the most value. So, what significance does the quadratic formula hold for me? Nothing apart from giving me roots of quadratic equations. On the other hand, the quadratic formula has the idea of "Completing the square" behind it, which is quite useful in Number Theory (more specifically diophantine equations). Another nice thing this gives is that the discriminant must be a perfect square if you want to have integer solutions. Buddha Buck Studied at University at Buffalo · Author has 5.7K answers and 16.8M answer views · 7y Related What is the significance of studying quadratic equations? There are a few reasons to study quadratic equations: Quadratics are a step up in complexity from linear equations, and thus provide practice for learning how to deal with complicated equations without being too difficult. Physics has a lot of 2nd-order systems (Newtonian physics is centered around forces, which are virtually defined by the 2nd-order system [math]\mathbf{F} = m\mathbf{a}[/math] ). 2nd-order systems can often be transformed into quadratics to solve, yielding to a lot of uses for quadratics. Quadratic equations define conic sections — [math]x^2 - 3y^2 = 5[/math] is an equation describing a hyperbola, [math]57x^2 +[/math] There are a few reasons to study quadratic equations: Quadratics are a step up in complexity from linear equations, and thus provide practice for learning how to deal with complicated equations without being too difficult. Physics has a lot of 2nd-order systems (Newtonian physics is centered around forces, which are virtually defined by the 2nd-order system [math]\mathbf{F} = m\mathbf{a}[/math]). 2nd-order systems can often be transformed into quadratics to solve, yielding to a lot of uses for quadratics. Quadratic equations define conic sections — [math]x^2 - 3y^2 = 5[/math] is an equation describing a hyperbola, [math]57x^2 + 43y^2 = 48xy + 25[/math] is an equation describing an ellipse, and [math]x = 4y^2 + 3[/math] is an equation describing a parabola. Historically, conic sections have been the source of immense study and have loads of amazing properties. The study of quadratic equations is intimately tied to the study of conic sections. I’m sure more reasons could be listed. One main observation in mathematics is that things are connected in surprising ways across disciplines, and so studying one discipline can yield fruitful results in another. Two of the reasons above show connections between traditional aspects of physics and geometry to quadratics in ways which come about naturally given the subjects. I’m sure there are plenty of others. One example that doesn’t have to do with quadratics directly is the extremely fruitful field of study of elliptical curves (which came about from attempts to measure the perimeter of an ellipse), a particular class of cubic equations. One end result of the study of elliptical curves was a proof of Fermat’s Last Theorem. Another field of elliptical curves is used in cryptography. While they are not quadratics, the practice you get in manipulating equations from working with quadratics helps you understand what’s going on with the cubics. Karthik S Bharadwaj Electronics Engineer · 7y Related What is the importance of quadratic equations in engineering? Quadratic equations are kind of cradle to all Engineers around dealing with non-linearity. The places they are important are: When you deal with parabolic curves (trajectories and other geometric problems.) When you deal with harmonic oscillators and other passive circuits where charge and voltage, or mechanical elements where displacement and force or rotational dynamics where displacement and torque are considered. And sometimes with Maxwell equations. In the last two points, the importance can be seen when you try to solve second order ordinary and partial differential equations. Enjoy the essenc Quadratic equations are kind of cradle to all Engineers around dealing with non-linearity. The places they are important are: When you deal with parabolic curves (trajectories and other geometric problems.) When you deal with harmonic oscillators and other passive circuits where charge and voltage, or mechanical elements where displacement and force or rotational dynamics where displacement and torque are considered. And sometimes with Maxwell equations. In the last two points, the importance can be seen when you try to solve second order ordinary and partial differential equations. Enjoy the essence of mathematics given in engineering courses because in real world you would just feed the numbers to systems where there are softwares which solve them. Related questions What is a real life example of a quadratic polynomial? What is the significance of the quadratic formula? How can you find the signs of the coefficients of a quadratic polynomial by just looking at the graph? What is the need to change signs while performing a division of polynomials? Why does a polynomial with real, simple roots change its sign between its roots? What is the significance of quadratic equations in graphing functions? What is the definition of a real root of a quadratic polynomial? Do all real roots of a quadratic polynomial lie within its roots? What is the difference between a quadratic polynomial and any other polynomial? What is special about quadratic polynomials in particular (mathematics)? What is the significance of constant polynomial? What is a second order polynomial or quadratic equation called in math terms? How many different types of second order polynomials or quadratic equations are there? What are the zeroes of the quadratic polynomial 3x^2-75? What is the significance of [latex] \Delta\geq 0 [/latex] in determining the nature of roots in quadratic polynomial equations? How does changing the sign of an equation affect the quadratic formula? What is the method for determining the number of real solutions to an equation represented by a quadratic polynomial? How can you use the quadratic equation to solve for the value of x in a polynomial equation? Related questions What is a real life example of a quadratic polynomial? What is the significance of the quadratic formula? How can you find the signs of the coefficients of a quadratic polynomial by just looking at the graph? What is the need to change signs while performing a division of polynomials? Why does a polynomial with real, simple roots change its sign between its roots? What is the significance of quadratic equations in graphing functions? What is the definition of a real root of a quadratic polynomial? Do all real roots of a quadratic polynomial lie within its roots? What is the difference between a quadratic polynomial and any other polynomial? What is special about quadratic polynomials in particular (mathematics)? What is the significance of constant polynomial? What is a second order polynomial or quadratic equation called in math terms? How many different types of second order polynomials or quadratic equations are there? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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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 How to calculate cumulative normal distribution? Ask Question Asked 16 years, 5 months ago Modified8 months ago Viewed 271k times This question shows research effort; it is useful and clear 146 Save this question. Show activity on this post. I am looking for a function in Numpy or Scipy (or any rigorous Python library) that will give me the cumulative normal distribution function in Python. python numpy scipy statistics Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Oct 13, 2019 at 19:59 martineau 124k 29 29 gold badges 180 180 silver badges 317 317 bronze badges asked Apr 30, 2009 at 22:13 toma toma Add a comment\| 8 Answers 8 Sorted by: Reset to default This answer is useful 175 Save this answer. Show activity on this post. Here's an example: ```python from scipy.stats import norm norm.cdf(1.96) 0.9750021048517795 norm.cdf(-1.96) 0.024997895148220435 ``` In other words, approximately 95% of the standard normal interval lies within two standard deviations, centered on a standard mean of zero (0.975 - 0.0250 ~= 0.95). If you need the inverse CDF: ```python norm.ppf(norm.cdf(1.96)) array(1.9599999999999991) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 29 at 21:07 answered Apr 30, 2009 at 22:24 Alex ReynoldsAlex Reynolds 97.1k 59 59 gold badges 251 251 silver badges 354 354 bronze badges 7 Comments Add a comment Irvan IrvanOver a year ago Also, you can specify the mean (loc) and variance (scale) as parameters. e.g, d = norm(loc=10.0, scale=2.0); d.cdf(12.0); Details here: docs.scipy.org/doc/scipy-0.14.0/reference/generated/… 2014-10-31T13:41:14.207Z+00:00 16 Reply Copy link qkhhly qkhhlyOver a year ago @Irvan, the scale parameter is actually the standard deviation, NOT the variance. 2015-06-02T19:08:35.703Z+00:00 10 Reply Copy link WestCoastProjects WestCoastProjectsOver a year ago Why does scipy name these as loc and scale ? I used the help(norm.ppf) but then what the heck are loc and scale - need a help for the help.. 2016-12-22T20:31:28.233Z+00:00 2 Reply Copy link Michael Ohlrogge Michael OhlroggeOver a year ago @javadba - location and scale are more general terms in statistics that are used to parameterize a wide range of distributions. For the normal distribution, they line up with mean and sd, but not so for other distributions. 2017-08-25T17:59:27.763Z+00:00 4 Reply Copy link WestCoastProjects WestCoastProjectsOver a year ago @MichaelOhlrogge . Thx! Here is a page from NIST explaining further itl.nist.gov/div898/handbook/eda/section3/eda364.htm 2017-08-25T18:03:24.203Z+00:00 1 Reply Copy link Add a comment\|Show 2 more comments This answer is useful 61 Save this answer. Show activity on this post. It may be too late to answer the question but since Google still leads people here, I decide to write my solution here. That is, since Python 2.7, the math library has integrated the error function math.erf(x) The erf() function can be used to compute traditional statistical functions such as the cumulative standard normal distribution: python from math import def phi(x): #'Cumulative distribution function for the standard normal distribution' return (1.0 + erf(x / sqrt(2.0))) / 2.0 Ref: How are the Error Function and Standard Normal distribution function related? Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 7, 2018 at 20:01 gibbone 2,720 23 23 silver badges 22 22 bronze badges answered Mar 26, 2015 at 7:40 WTIFSWTIFS 1,030 9 9 silver badges 13 13 bronze badges 2 Comments Add a comment Hannes Landeholm Hannes LandeholmOver a year ago This was exactly what I was looking for. If someone else than me wonders how this can be used to calculate "percentage of data that lies within the standard distribution", well: 1 - (1 - phi(1)) 2 = 0.6827 ("68% of data within 1 standard deviation") 2017-07-10T18:30:48.377Z+00:00 3 Reply Copy link Bernhard Barker Bernhard BarkerOver a year ago For a general normal distribution, it would be def phi(x, mu, sigma): return (1 + erf((x - mu) / sigma / sqrt(2))) / 2. 2020-03-15T19:18:02.793Z+00:00 4 Reply Copy link This answer is useful 61 Save this answer. Show activity on this post. Starting Python 3.8, the standard library provides the NormalDist object as part of the statistics module. It can be used to get the cumulative distribution function (cdf - probability that a random sample X will be less than or equal to x) for a given mean (mu) and standard deviation (sigma): ```python from statistics import NormalDist NormalDist(mu=0, sigma=1).cdf(1.96) 0.9750021048517796 ``` Which can be simplified for the standard normal distribution (mu = 0 and sigma = 1): ```python NormalDist().cdf(1.96) 0.9750021048517796 NormalDist().cdf(-1.96) 0.024997895148220428 ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 28, 2019 at 19:50 Xavier GuihotXavier Guihot 62.6k 26 26 gold badges 319 319 silver badges 202 202 bronze badges 4 Comments Add a comment dcl dclOver a year ago Based on some quick checks, this is significantly faster than norm.cdf from scipy.stats and a fair bit faster than both scipy and math implementations of erf. 2021-03-15T04:33:50.66Z+00:00 9 Reply Copy link hasManyStupidQuestions hasManyStupidQuestionsOver a year ago Does this vectorize? Or should someone use the scipy implementation if they need to compute the CDF evaluated at all points in an array? 2021-05-16T14:33:47.297Z+00:00 2 Reply Copy link Juozas JuozasOver a year ago Awesome. Maybe you know how to get inverse (normsinv)? Edit: OK, it is inv_cdf(). Thank you! 2022-08-28T13:39:38.097Z+00:00 1 Reply Copy link zeit zeitDec 5, 2024 at 12:09 A quick look at NormalDist().cdf() shows that it depends on erf(). Ref: github.com/python/cpython/blob/3.8/Lib/statistics.py#L941 2024-12-05T12:09:12.273Z+00:00 0 Reply Copy link Add a comment This answer is useful 21 Save this answer. Show activity on this post. Adapted from here ```python from math import def erfcc(x): """Complementary error function.""" z = abs(x) t = 1. / (1. + 0.5z) r = t exp(-zz-1.26551223+t(1.00002368+t(.37409196+ t(.09678418+t(-.18628806+t(.27886807+ t(-1.13520398+t(1.48851587+t(-.82215223+ t.17087277))))))))) if (x >= 0.): return r else: return 2. - r def ncdf(x): return 1. - 0.5erfcc(x/(20.5)) ``` Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Apr 30, 2009 at 22:23 UnknownUnknown 46.9k 29 29 gold badges 142 142 silver badges 183 183 bronze badges 3 Comments Add a comment Marc MarcOver a year ago Since the std lib implements math.erf(), there is no need for a sep implementation. 2016-02-25T20:10:14.793Z+00:00 5 Reply Copy link TmSmth TmSmthOver a year ago i was not able to find an answer, where do those numbers come from ? 2020-01-15T23:31:51.2Z+00:00 1 Reply Copy link tbrugere tbrugereOver a year ago @TmSmth If I had to guess this looks like some kind of approximation of what is inside the exponential, so you probably can calculate them with some kind of taylor expansion after fiddling with your function a bit (changing vars, then say r = t exp( - z2 -f(t)) and do a taylor expansion of f (which can be found numerically 2021-06-01T07:29:55.8Z+00:00 2 Reply Copy link Add a comment This answer is useful 19 Save this answer. Show activity on this post. To build upon Unknown's example, the Python equivalent of the function normdist() implemented in a lot of libraries would be: ```python def normcdf(x, mu, sigma): t = x-mu; y = 0.5erfcc(-t/(sigmasqrt(2.0))); if y>1.0: y = 1.0; return y def normpdf(x, mu, sigma): u = (x-mu)/abs(sigma) y = (1/(sqrt(2pi)abs(sigma)))exp(-uu/2) return y def normdist(x, mu, sigma, f): if f: y = normcdf(x,mu,sigma) else: y = normpdf(x,mu,sigma) return y ``` Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Aug 19, 2010 at 20:14 answered Aug 19, 2010 at 19:35 CerinCerin 65.4k 106 106 gold badges 346 346 silver badges 560 560 bronze badges Comments Add a comment This answer is useful 16 Save this answer. Show activity on this post. Alex's answer shows you a solution for standard normal distribution (mean = 0, standard deviation = 1). If you have normal distribution with mean and std (which is sqr(var)) and you want to calculate: ```python from scipy.stats import norm cdf(x < val) print norm.cdf(val, m, s) cdf(x > val) print 1 - norm.cdf(val, m, s) cdf(v1 < x < v2) print norm.cdf(v2, m, s) - norm.cdf(v1, m, s) ``` Read more about cdf here and scipy implementation of normal distribution with many formulas here. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 20, 2015 at 10:24 Salvador DaliSalvador Dali 224k 151 151 gold badges 724 724 silver badges 766 766 bronze badges Comments Add a comment This answer is useful 2 Save this answer. Show activity on this post. Taken from above: ```python from scipy.stats import norm norm.cdf(1.96) 0.9750021048517795 norm.cdf(-1.96) 0.024997895148220435 ``` For a two-tailed test: python Import numpy as np z = 1.96 p_value = 2 norm.cdf(-np.abs(z)) 0.04999579029644087 Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 6, 2019 at 22:17 David MillerDavid Miller 517 5 5 silver badges 5 5 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Simple like this: python import math def my_cdf(x): return 0.5(1+math.erf(x/math.sqrt(2))) I found the formula in this page Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered May 30, 2020 at 14:55 Samuel CorradiSamuel Corradi 49 3 3 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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190078	https://arxiv.org/pdf/1906.02145	Cubic-Spline Flows Conor Durkan 1 Artur Bekasov 1 Iain Murray 1 George Papamakarios 1 Abstract A normalizing flow models a complex probability density as an invertible transformation of a simple density. The invertibility means that we can eval-uate densities and generate samples from a flow. In practice, autoregressive flow-based models are slow to invert, making either density estimation or sample generation slow. Flows based on cou-pling transforms are fast for both tasks, but have previously performed less well at density estima-tion than autoregressive flows. We stack a new coupling transform, based on monotonic cubic splines, with LU-decomposed linear layers. The resulting cubic-spline flow retains an exact one-pass inverse, can be used to generate high-quality images, and closes the gap with autoregressive flows on a suite of density-estimation tasks. 1. Introduction Normalizing flows are flexible probabilistic models of con-tinuous data. A normalizing flow models data x as the output of an invertible smooth transformation f of noise u: x = f (u) where u ∼ π(u). (1) Typically, the noise distribution π(u) is taken to be sim-ple (a standard normal is a common choice), whereas the transformation f and its inverse f −1 are implemented by an invertible neural network. The probability density of x under the flow is obtained by a change of variables: p(x) = π(f −1(x)) ∣∣∣∣det ( ∂f −1 ∂x )∣ ∣∣∣. (2) Given training data {x1, . . . , xN }, the flow is trained with backpropagation, by maximizing the total log likelihood ∑Nn=1 log p(xn) with respect to the parameters of the Equal contribution 1School of Informatics, Univer-sity of Edinburgh, United Kingdom. Correspondence to: Conor Durkan <conor.durkan@ed.ac.uk >, Artur Bekasov <ar-tur.bekasov@ed.ac.uk >.First workshop on Invertible Neural Networks and Normalizing Flows (ICML 2019), Long Beach, CA, USA transformation f . Normalizing flows have been success-fully used for density estimation [ 5, 23 ], variational in-ference [ 17 , 27 , 29 ], image, audio and video generation [14, 16, 18, 26], and likelihood-free inference . In practice, the Jacobian determinant of f must be tractable, so that the density p(x) can be computed efficiently. Several invertible transformations with a tractable Jacobian deter-minant have been proposed, including coupling transforms [ 4, 5], autoregressive transforms [ 3, 13 , 17 , 23 ], invertible convolutions [ 11 , 16 ], transformations based on the matrix-determinant lemma [ 27 , 29 ], and continuous transforma-tions based on differential equations [2, 8]. Even though the above transformations are invertible in theory , they are not always efficiently invertible in practice. For instance, the affine autoregressive transforms used by MAF [ 23 ] and IAF [ 17 ] are D times slower to invert than to evaluate, where D is the dimensionality of x. Even worse, inverting the non-affine transformations used by NAF [ 13 ]and block-NAF [ 3] would require numerical optimization. In practice, the forward transformation f is needed for gener-ating data, whereas the inverse transformation f −1 is needed for computing p(x); for a flow to be applicable in both situ-ations, both directions need to be fast. Transformations that are equally fast to invert and evaluate do exist. Affine coupling transforms, used by Real NVP [ 5]and Glow [ 16 ], only require a single neural-network pass in either direction; however, more flexible autoregressive trans-forms have obtained better density-estimation results. More flexible coupling transforms have been proposed, based on piecewise-quadratic splines [ 20 ], although these have not previously been evaluated as density estimators. Continuous flows such as Neural ODEs [ 2] and FFJORD [ 8] are also equally fast in both directions, but they require numerically integrating a differential equation in each direction, which can be slower than a single neural-network pass. In this paper we introduce cubic-spline flows , which, like Real NVP and Glow, only require a single neural-network pass in either the forward or the inverse direction, but in practice are as flexible as state-of-the-art autoregressive flows. Cubic-spline flows alternate between non-affine coupling transforms that use flexible monotonic cubic splines to transform their input, and LU-decomposed linear layers that combine their input across dimensions. arXiv:1906.02145v1 [stat.ML] 5 Jun 2019 Cubic-Spline Flows 0.0 0.5 1.0 x 0.0 0.5 1.0 y = P(x′ x) Cubic Spline Inverse Knots 0.0 0.5 1.0 x p(x) Figure 1: Left : Illustration of a random monotonic cubic spline with K = 10 bins which maps [0 , 1] to [0 , 1] . Monotonic splines act as drop-in replacements for affine or additive transformations in coupling layers. Right : Since the cubic spline can be inter-preted as a cumulative distribution function, its derivative defines a probability density function which is piecewise quadratic. Experimentally, we show that cubic-spline flows match or exceed state-of-the-art performance in density estimation, and that they are capable of generating high-quality images with only a single neural-network pass. 2. Method 2.1. Coupling transforms A coupling transform φ maps an input x to an output y in the following way: 1. Split the input x into two parts, x = [ x1: d−1, xd:D ].2. Compute parameters θ = NN (x1: d−1), where NN is an arbitrary neural network. 3. Compute yi = gθi (xi) for i = d : D, where gθi is an invertible function parameterized by θi.4. Set y1: d−1 = x1: d−1, and return y = [ y1: d−1, yd:D ].The Jacobian matrix of a coupling transform is lower trian-gular, since yd:D is given by transforming xd:D elementwise as a function of x1: d−1, and y1: d−1 is equal to x1: d−1. Thus, the Jacobian determinant of the coupling transform φ is det ( ∂φ ∂x ) = D ∏ i=d ∂g θi ∂x i . (3) Coupling transforms solve two important problems for nor-malizing flows: first, they have a tractable Jacobian determi-nant; second, they can be inverted exactly in a single pass. The inverse of a coupling transform can be easily computed by running steps 1–4 above, this time inputting y, and using g−1 θi to compute xd:D in step 3. 2.2. Monotonic-spline coupling transforms Typically, the function gθi takes the form of an affine or additive transformation for computational ease. The affine transformation is given by: gθi (xi) = αixi + βi where θi = {αi, β i}. (4) The additive transformation corresponds to the special case αi = 1 . Both the affine and the additive transformation are easy to invert, but they lack flexibility. Nonetheless, we are free to choose any function gθi as long as it is differentiable and we can easily invert it. Recently, M ¨uller et al. proposed a powerful generalization of the above forms, based on monotonic piecewise polynomi-als. The idea is to restrict the input domain of gθi between 0 and 1, partition the input domain into K bins, and de-fine gθi to be a simple polynomial segment within each bin (see fig. 1). M ¨uller et al. restrict themselves to monotonically-increasing linear and quadratic polynomial segments, whose coefficients are parameterized by θi. More-over, the polynomial segments are restricted to match at the boundaries so that gθi is continuous. Functions of this form are known as polynomial splines . 2.3. Monotonic cubic-spline coupling transforms Inspired by this direction of research, we introduce cubic-spline flows , a natural extension to the framework of M ¨uller et al. . We propose to implement gθi as a mono-tonic cubic spline [ 7], where each segment is defined by a monotonically-increasing cubic polynomial (fig. 1). Cubic splines are well-known across science, finding use in many applications, and their theory is well-explored [6, 12]. We employ Steffen’s method [ 28 ] to parameterize amonotonically-increasing cubic spline that maps [0 , 1] to [0 , 1] . Given a fixed number of bins K, Steffen’s method takes K + 1 coordinates {(xk, y k)}Kk=0 , known as knots ,and two values for the derivatives at inputs 0 and 1. Then, the method constructs a continuously-differentiable cubic spline that passes through the knots, has the given boundary derivatives, and is monotonic on each segment (for details, see appendix A.1). To ensure the spline is monotonically-increasing and maps [0 , 1] to [0 , 1] , we require: (x0, y 0) = (0 , 0) , (xK , y K ) = (1 , 1) , (5) xk < x k+1 and yk < y k+1 for k = 0 : K − 1. (6) Our implementation of the cubic-spline coupling transform is as follows: 1. A neural network NN takes x1: d−1 and outputs an un-constrained parameter vector θi of length 2K + 2 for each i = d : D.2. Vector θi is partitioned as θi = [θwi , θhi , θdi ], where θwi and θhi have length K, and θdi has length 2.3. Vectors θwi and θhi are each passed through a softmax; the outputs are interpreted as the widths and heights of the K bins, which must be positive and sum to 1.Cubic-Spline Flows A cumulative sum of the K bin widths and heights yields the K + 1 knots {(xk, y k)}Kk=0 . Vector θdi is interpreted as the two boundary derivatives. Evaluating a cubic spline at location x requires finding the bin in which x lies, which can be done efficiently with binary search, since the bins are sorted. The Jacobian determinant of the cubic-spline coupling transform can be computed in closed form as a product of quadratic-polynomial derivatives (see appendix A.2). For the inverse, we need to compute the roots of a cubic polynomial whose constant term depends on the desired value to invert, which can be done analytically. However, in early experiments, we found that na ¨ıve root finding using trigonometric and hyperbolic methods was unstable for certain values of the coefficients. To address this issue, we follow a modified version, provided by Peters , of the scheme outlined by Blinn , which stabilizes many of the potentially difficult cases (see appendix A.3). Unlike the affine and additive transformations which have limited flexibility, a monotonic spline can approximate any continuous monotonic function arbitrarily well, given suf-ficiently many bins. Yet, cubic-spline coupling transforms have a closed-form Jacobian determinant and can be in-verted in one pass. Finally, our parameterization of cubic splines using Steffen’s method is fully differentiable, so the transform can be trained by gradient methods. In addition to transforming xd:D by monotonic cubic splines whose parameters are a function of x1: d−1, we introduce a set of monotonic cubic splines which act elementwise on x1: d−1, whose parameters are optimized directly with stochastic gradient descent. This means that our coupling layer transforms all input variables at once, rather than being restricted to a subset, as follows: θ1: d−1 = Trainable parameters (7) θd:D = NN (x1: d−1) (8) yi = gθi (xi) for each i = 1 : D. (9) 2.4. Linear transforms with an LU decomposition To ensure all input variables can interact with each other, it is common to permute the elements of intermediate layers in a normalizing flow. Permutation is an invertible linear trans-formation, with absolute determinant equal to 1. Kingma & Dhariwal generalize this approach using an efficient parameterization of a linear transformation, demonstrating improvements on a range of image tasks. In particular, they parameterize the lower-upper or LU decomposition of a square matrix as W = PLU , where P is a fixed permuta-tion matrix, L is lower-triangular with ones on the diagonal, and U is upper triangular. Written in this form, the determi-nant of W can be calculated in O(D) time as the product of the diagonal elements of U. Also, by parameterizing the diagonal elements of U to be positive, W is guaranteed to Training data Flow density Flow samples Figure 2: Qualitative results for two-dimensional synthetic datasets using cubic-spline flows with two coupling layers. Some previous flows struggle to model such fine details, as demonstrated by, e.g., Nash & Durkan . be invertible. Given that we have the LU factorization of W, we can efficiently apply the inverse transformation, and sample from the model in one pass. 2.5. Cubic-spline flows We propose a general-purpose flow based on alternating LU-decomposed linear layers and monotonic cubic-spline coupling transforms. Since the splines map [0 , 1] to [0 , 1] ,we apply a sigmoid, and its inverse, the logit, before and after each coupling layer, so the overall transform is com-patible with unconstrained data and linear layers. We clip the inputs of the logit to [10 −6, 1−10 −6] to prevent satura-tion with 32-bit floating-point precision. The function is no longer strictly reversible for all inputs due to this clipping. Like Real NVP or Glow, cubic-spline flows can represent either a transformation from data to noise, or from noise to data. In both cases, the transformation requires only a single pass of the neural network defining the flow. Overall, our proposed cubic-spline flow resembles a traditional feed-forward neural network architecture, alternating between linear transformations and piecewise nonlinearities, while retaining exact invertibility. 3. Experiments For our experiments, NN in eq. (8) is a fully-connected or convolutional neural network with two pre-activation resid-ual blocks [ 9, 10 ]. We use the Adam optimizer [ 15 ], and a cosine schedule for annealing the learning rate [ 19 ]. All experiments use a standard normal for the noise distribution π(u), except for the two-dimensional experiments, which use a uniform distribution. Cubic-Spline Flows Table 1: Test log likelihood (in nats) for UCI datasets and BSDS300; higher is better. Error bars correspond to two standard deviations (FFJORD do not report error bars). Apart from quadratic and cubic splines, all results are taken from existing literature. NAF † report error bars across five repeated runs rather than across the test set. MODEL POWER GAS HEPMASS MINIBOONE BSDS300 ONE -PASS FLOWS FFJORD 0.46 8.59 −14 .92 −10 .43 157 .40 QUADRATIC -SPLINE 0.65 ±0.01 13 .13 ±0.02 −14 .95 ±0.02 −9.18 ±0.43 157 .49 ±0.28 CUBIC -SPLINE 0.65 ±0.01 13 .14 ±0.02 −14 .59 ±0.02 −9.06 ±0.44 157 .24 ±0.28 AUTO - REGRESSIVE FLOWS MAF 0.30 ±0.01 10 .08 ±0.02 −17 .39 ±0.02 −11 .68 ±0.44 156 .36 ±0.28 NAF † 0.62 ±0.01 11 .96 ±0.33 −15 .09 ±0.40 −8.86 ±0.15 157 .73 ±0.04 BLOCK -NAF 0.61 ±0.01 12 .06 ±0.09 −14 .71 ±0.38 −8.95 ±0.07 157 .36 ±0.03 TAN VARIOUS 0.60 ±0.01 12 .06 ±0.02 −13 .78 ±0.02 −11 .01 ±0.48 159 .80 ±0.07 3.1. Synthetic datasets We first demonstrate the flexibility of cubic-spline flows on synthetic two-dimensional datasets (fig. 2). The mixture of 225 Gaussians and Einstein are taken from [ 21 ], and we generate another dataset using an image of Gauss. For each task, we train on one-million data points, use a cubic-spline flow with two coupling layers, no linear layers, and K = 128 bins. As we can see, the model is flexible enough to fit complex, multimodal densities that other flows would struggle with. In each case, the model consists of fewer than 30% of the parameters than the 512 × 512 histograms which are used to display the data. 3.2. Density estimation on tabular data We next consider the UCI and BSDS300 datasets, a stan-dard suite of benchmarks for density estimation of tabular data. Our spline flows alternate coupling layers with LU-decomposed linear layers, and we use K = 10 bins in all experiments. For all datasets, we stack ten of these com-posite linear and coupling layers, apart from MINIBOONE, where we use five. Hyperparameters such as the number of training steps and the hidden dimension used by NN in cou-pling layers are tuned separately for each cubic model. As an ablation study, we compare with a flow using quadratic instead of cubic splines, which extends [ 20 ] with LU layers and elementwise transformations as in section 2. Table 1 shows our results. The cubic and quadratic mod-els are close in performance, with the error bars on test log likelihood overlapping. However, a paired comparison computing the mean and standard error of the difference be-tween models on individual examples shows that the cubic-spline flow is statistically significantly better on three of the five datasets, while being indistinguishable on POWER and slightly worse than the quadratic-spline flow on BSDS300. On the other hand, the quadratic-spline flow is slightly faster and less prone to numerical issues. Appendix A.4 further compares the cubic and quadratic models. Spline flows achieve state-of-the-art performance on all tasks against FFJORD [ 8], the previously strongest flow-based model with a one-pass inverse. Moreover, the spline flows are competitive against the best-performing autore-gressive flows, achieving state of the art for any flow model on POWER and GAS. These results demonstrate that it is possible for neural density estimators based on coupling layers to challenge autoregressive models, and that it may not be necessary to sacrifice one-pass sampling for density-estimation performance. 3.3. Image generation Finally, we demonstrate that cubic-spline flows scale to high-dimensional data using the Fashion-MNIST dataset [ 30 ]. We use a Glow-like model [ 16 ] with cubic-spline cou-pling transforms, which includes a multi-scale architecture, actnorm layers with data-dependent initialization, and in-vertible 1×1 convolutions. In line with Kingma & Dhariwal , we illustrate the effect of scaling the standard deviation of the noise distribution π(u) by a temperature T ∈ [0 , 1] .Qualitative results are shown in fig. 3. We leave the com-parison with state-of-the-art flow-based image models on higher-dimensional image datasets for future work. Figure 3: Qualitative results for Fashion-MNIST dataset. Left :Training data. Right : Unconditional samples from the cubic-spline flow. Samples for three temperatures are shown in separate blocks, top-to-bottom: 0.5, 0.75 , 1.0. 4. Conclusion The cubic-spline flow is the first model that combines the density-estimation performance of state-of-the-art autore-gressive models like NAF [ 13 ] and TAN [ 22 ], while still Cubic-Spline Flows being able to generate realistic samples in a single pass like Real NVP [5 ] and Glow [ 16 ]. More generally, we have shown that flexible monotonic splines are a useful differ-entiable module, which could be included in many models that can be trained end-to-end. We intend to implement autoregressive flows with these transformations, which may further improve the state of the art in density estimation. Acknowledgements This work was supported in part by the EPSRC Centre for Doctoral Training in Data Science, funded by the UK Engineering and Physical Sciences Research Council (grant EP/L016427/1) and the University of Edinburgh. George Papamakarios was also supported by Microsoft Research through its PhD Scholarship Programme. A. More details on monotonic cubic splines A.1. Parameterization of the spline We here include an outline of the method of Steffen which closely matches the original paper. Let {(xk, y k)}Kk=0 be a given set of knot points in the plane which satisfy (x0, y 0) = (0 , 0) , (xK , y K ) = (1 , 1) , (10) xk < x k+1 and yk < y k+1 for k = 0 : K − 1. (11) If additionally the derivatives {dk}Kk=0 at the knot points are known, a unique cubic polynomial fk(ξ) = αk0 + αk1ξ + αk2ξ2 + αk3ξ3, (12) where ξ = x − xk, is determined on each bin. The coeffi-cients are given by αk0 = yk (13) αk1 = dk (14) αk2 = 3sk − 2dk − dk+1 wk (15) αk3 = dk + dk+1 − 2sk (wk)2 , (16) where wk = xk+1 − xk is the width of each bin, and sk = ( yk+1 − yk)/w k is the slope of the line joining con-secutive knots. The resulting cubic spline is continuously differentiable on the interval [0 , 1] , and it remains only to determine the derivative values {dk}Kk=0 so that the overall spline is monotonic on this interval. To this end, Steffen uses the unique quadratic func-tion passing through the points (xk−1, y k−1), (xk, y k) and (xk+1 , y k+1 ) to determine the derivative at location xk.Monotonicity of this quadratic on the interval defined by these points is sufficient to specify an appropriate derivative, which is given by pk = sk−1wk + skwk−1 wk−1 + wk . (17) However, it is possible that the quadratic in question is not monotonic on the given interval, and in this case the derivative value needs to be modified. Steffen proposes using the smallest value for the derivative at xk such that the quadratic is monotonic on the given interval, showing that dk = { 2 min( sk−1, s k) if pk > 2 min( sk−1, s k) pk otherwise (18) is sufficient. A.2. Computing the derivative The derivative of eq. (12) is straightforwardly given by dfk(ξ)dx = αk1 + 2 αk2ξ + 3 αk3ξ2. (19) The logarithm of the absolute value of the determinant of a cubic-spline coupling transform is thus given by a sum of the logarithm of eq. (19) for each transformed x. A.3. Computing the inverse Computing the inverse of a cubic polynomial requires solv-ing for the roots of a cubic polynomial whose constant term depends on the value to invert. That is, for a given y, we wish to find x which satisfies (αk0 − y) + αk1ξ + αk2ξ2 + αk3ξ3 = 0 , (20) where ξ is defined as before. Monotonicity means there is at most one root x in each bin, and it is possible to determine this value analytically, using either Cardano’s formula, or trigonometric or hyperbolic methods. In practice, careful numerical consideration is necessary to ensure stable root-finding. We found the comprehensive treatment of Blinn particularly useful, ultimately settling on a slightly modified version of a procedure outlined in this latter work [ 25 ]. We refer readers to Blinn for full details. A.4. Comparison with monotonic quadratic splines Following M ¨uller et al. , we construct a quadratic spline by integrating an unconstrained linear spline. The result is a monotonic piecewise-quadratic function, where the values and derivatives of the function are continuous at the knots. The derivatives of the quadratic spline at the knots, corre-sponding to the density, are set directly by the linear spline. However, some monotonically-increasing knot locations (xk, y k) cannot be expressed through this construction. Cubic-Spline Flows In contrast, Steffen’s cubic-spline construction [ 28 ] can in-terpolate arbitrary monotonic settings of the knot locations (xk, y k), so unlike the quadratic spline can set arbitrary quantiles of the implied distribution at the knot locations. However, the method sets the derivatives of the function based on prior smoothness assumptions, rather than allow-ing the user to set them directly as in the quadratic spline. As a result, the number of parameters for the two spline constructions are similar. The cubic spline could be made strictly more flexible by adding parameters for the deriva-tives at the knots, constrained to maintain monotonicity. As they are, the cubic and quadratic splines give different functions for knots specifying the same quantiles, corre-sponding to different inductive biases. Steffen dis-cusses several nice properties of the interpolants chosen by his cubic-spline method. References Blinn, J. F. How to solve a cubic equation, part 5: Back to numerics. IEEE Computer Graphics and Applications , 27(3):78–89, 2007. Chen, R. T. Q., Rubanova, Y., Bettencourt, J., and Du-venaud, D. K. Neural ordinary differential equations. Advances in Neural Information Processing Systems ,2018. De Cao, N., Titov, I., and Aziz, W. Block neural autoregressive flow. Conference on Uncertainty in Artificial Intelligence , 2019. Dinh, L., Krueger, D., and Bengio, Y. NICE: Non-linear independent components estimation. Interna-tional Conference on Learning Representations, Work-shop track , 2015. Dinh, L., Sohl-Dickstein, J., and Bengio, S. Density estimation using Real NVP. International Conference on Learning Representations , 2017. Durrleman, S. and Simon, R. Flexible regression mod-els with cubic splines. Statistics in medicine , 8(5): 551–561, 1989. Fritsch, F. N. and Carlson, R. E. Monotone piece-wise cubic interpolation. SIAM Journal on Numerical Analysis , 17(2):238–246, 1980. Grathwohl, W., Chen, R. T. Q., Bettencourt, J., Sutskever, I., and Duvenaud, D. K. FFJORD: Free-form continuous dynamics for scalable reversible gen-erative models. International Conference on Learning Representations , 2018. He, K., Zhang, X., Ren, S., and Sun, J. Deep residual learning for image recognition. IEEE Conference on Computer Vision and Pattern Recognition , 2016. He, K., Zhang, X., Ren, S., and Sun, J. Identity map-pings in deep residual networks. European Conference on Computer Vision , 2016. Hoogeboom, E., van den Berg, R., and Welling, M. Emerging convolutions for generative normalizing flows. arXiv:1901.11137 , 2019. Hou, H. and Andrews, H. Cubic splines for image interpolation and digital filtering. IEEE Transactions on acoustics, speech, and signal processing , 26(6): 508–517, 1978. Huang, C.-W., Krueger, D., Lacoste, A., and Courville, A. Neural autoregressive flows. International Confer-ence on Machine Learning , 2018. Kim, S., Lee, S.-g., Song, J., and Yoon, S. FloWaveNet: A generative flow for raw audio. arXiv:1811.02155 , 2018. Kingma, D. P. and Ba, J. Adam: A method for stochas-tic optimization. International Conference on Learn-ing Representations , 2014. Kingma, D. P. and Dhariwal, P. Glow: Generative flow with invertible 1 × 1 convolutions. Advances in Neural Information Processing Systems , 2018. Kingma, D. P., Salimans, T., Jozefowicz, R., Chen, X., Sutskever, I., and Welling, M. Improved variational inference with inverse autoregressive flow. Advances in Neural Information Processing Systems , 2016. Kumar, M., Babaeizadeh, M., Erhan, D., Finn, C., Levine, S., Dinh, L., and Kingma, D. Vide-oFlow: A flow-based generative model for video. arXiv:1903.01434 , 2019. Loshchilov, I. and Hutter, F. SGDR: Stochastic gra-dient descent with warm restarts. arXiv:1608.03983 ,2016. M ¨uller, T., McWilliams, B., Rousselle, F., Gross, M., and Nov ´ak, J. Neural importance sampling. arXiv:1808.03856 , 2018. Nash, C. and Durkan, C. Autoregressive energy ma-chines. International Conference on Machine Learn-ing , 2019. Oliva, J., Dubey, A., Zaheer, M., Poczos, B., Salakhut-dinov, R., Xing, E., and Schneider, J. Transformation autoregressive networks. International Conference on Machine Learning , 2018. Cubic-Spline Flows Papamakarios, G., Pavlakou, T., and Murray, I. Masked autoregressive flow for density estimation. Advances in Neural Information Processing Systems ,2017. Papamakarios, G., Sterratt, D. C., and Murray, I. Se-quential neural likelihood: Fast likelihood-free infer-ence with autoregressive flows. International Confer-ence on Artificial Intelligence and Statistics , 2019. Peters, C. How to solve a cubic equation, revis-ited. 105 , 2016. Accessed: 2019-04-17. Prenger, R., Valle, R., and Catanzaro, B. WaveGlow: A flow-based generative network for speech synthesis. arXiv:1811.00002 , 2018. Rezende, D. J. and Mohamed, S. Variational inference with normalizing flows. International Conference on Machine Learning , 2015. Steffen, M. A simple method for monotonic interpola-tion in one dimension. Astronomy and Astrophysics ,239:443, 1990. van den Berg, R., Hasenclever, L., Tomczak, J. M., and Welling, M. Sylvester normalizing flows for vari-ational inference. Conference on Uncertainty in Artifi-cial Intelligence , 2018. Xiao, H., Rasul, K., and Vollgraf, R. Fashion-MNIST: a novel image dataset for benchmarking machine learn-ing algorithms. arXiv:1708.07747 , 2017.
190079	https://damek.github.io/teaching/orie6300/Rec5.pdf	ORIE 6300 Mathematical Programming I September 21, 2016 Recitation 5 Lecturer: Calvin Wylie Scribe: Xueyu Tian The Transportation Problem 1 Suppose there are a set, S, of suppliers, each with supply si and a set, D, of customers each with demand dj that must be met. Further, suppose that each unit shipped from supply node i to demand node j incurs a cost cij. Let xij equal the amount to ship from i ∈S to j ∈D, of cij per unit. Find a minimum-cost shipping scheme that satisﬁes all the demand and supply restrictions. Clearly, this problem is only feasible if P i si ≥P j dj. Without loss of generality, suppose P i si = P j dj since otherwise we can set a dummy demand node k with dk = P i si −P j dj. Then we can formulate this as an linear program by: min P i∈S,j∈D cijxij s.t. P j∈D xij = si ∀i ∈S P i∈S xij = dj ∀j ∈D xij ≥0. In some cases, we might wish to restrict the value of xij to be integral. First, note that the rows of the constraint matrix are linearly dependent because: X i∈S X j∈D xij = X i∈S si = X j∈D dj = X j∈D X i∈S xij i.e., adding all the rows in the ﬁrst constraint set gives the same result as adding all the rows of the second constraint set. Hence, without loss of generality, we can remove one constraint arbitrarily. 1Based on previous notes of Chaoxu Tong 5-1 Let \|S\| = m, \|D\| = n. Rank of constraint matrix is n+m−1. Consider a graphical representation of this problem where G is a bipartite graph with vertex sets corresponding to S and D and an edge set E = S × D, i.e., {i, j} ∈E for every i ∈S, j ∈D, i.e., G = (S ∪D, EB), EB = {(i, j) ∈ E\|xij > 0}. Note that each decision variable corresponds to an edge in the graph. Hence, a basic solution to the LP corresponds to some subgraph. e.g. Claim 1 The subgraph corresponding to a basic solution to the LP doesn’t contain cycles. Cycle = {(i1, j1), (j1, i2), ..., (jk, i1)} Proof: Let C be a cycle in the corresponding subgraph induced by the basic solution. Then C = i1 →j1 →. . . →jk →i1 where il ∈S, jl ∈D. i.e., C = {(i1, j1), (j1, i2), ..., (jk, i1)}. Consider the vector yij which equals 1 if yij corresponds a cycle edge from S to D, −1 if yij corresponds to a cycle edge from D to S, and 0 otherwise. Note that for every node in the cycle, there is exactly one edge ’entering’ and one ’leaving’. Hence, it must be that Ay = 0, where A is our constraint matrix. So, consider any basis B and suppose C is a cycle using only our basis variables. Let AB be the corresponding basis matrix and yB be the matching components of y. Since yi = 0 for i ̸∈B. 0 = Ay = AB AN yB yN = AByB + ANyN = AByB + AN0 constraints AB having full rank. Noting that our graph has n + m vertices, and each basis has n + m −1 edges (since there are that many constraints), and each basis cannot contain a cycle, it follows that each basic solution corresponds to a spanning tree of our graph. 2 Dual: max Pm i=1 ui + Pn j=1 vi s.t. ui + vj ≤cij, ∀i, j Since one constraint in the primal was redundant, one variable in the dual can be set arbitrarily. 5-2 Algorithm Idea Assume we start with an initial basic feasible solution, (xB, xN) Solve ui + vj = cij, ∀(i, j) ∈B (Easy since one can be set arbitrarily, then solve for all others since basis induces a spanning tree) (complementary slackness condition) Check if ui + vi ≤cij, ∀(i, j) ∈N (Dual feasibility). If so, it’s optimal. Otherwise, uˆ i + vˆ j > cˆ iˆ j, for some (ˆ i, ˆ j) ∈N. We’ll see in lecture that putting (ˆ i, ˆ j) in the basis will improve solution. Add edge (ˆ i, ˆ j) to the subgraph. This will create a cycle. Alternatingly increase (from i →j) and decrease (from j →k) ﬂow around edges in the cycle to preserve 0 net ﬂow. Let δ = min(i,j){xij : xij is a basic variable , (i, j) is a decreasing edge}. Increase all ’forward’ edges by δ, decrease all ’backward’ edges by δ. Remove from the basis an edge which achieves the minimum. 5-3
190080	https://unacademy.com/content/question-answer/chemistry/what-is-the-molar-mass-of-methane-ch4/	Access free live classes and tests on the app Login Join for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd Question & Answer » Chemistry Questions » What is the Molar Mass of Methane, CH4 What is the Molar Mass of Methane, CH4 What is the molar mass of methane, CH4? Find the answer to this question and access a vast question bank that is customised for the student. Share Answer: The mass of one mole of a compound and the number of grams per mole of a compound is defined by its molar mass. To put it another way, the molar mass is the total mass in grams of all the atoms that make up a mole of a specific molecule. The molar mass is a mass property of a substance, not a molecular property. The molecular mass is usually expressed in grams per mole. Methane has the chemical formula CH4 indicating that between the C and H atoms, there are a total of four bonds. Its structure is: The steps to be followed to find Methane’s molar mass are: Methane’s molecular formula is CH4 according to which we can find the number of atoms in a molecule. Next, using the mass of carbon, and hydrogen given in the Periodic Table, calculate the atomic mass of each element. Then, multiply the subscript representing the number of atoms by the atomic mass of hydrogen. Add the atomic masses of all the atoms in the molecule to get the molar mass. Thus, when the atomic masses of carbon and hydrogen are rounded to four significant figures, the molar mass will be: CH4=( Atomic mass of C)+4( Atomic mass of H) CH4=(12.01)+4(1.008) CH4=16.042g/mol . Some applications of methane are: Methane is utilized as a fuel in autos, furnaces, and water heaters since it is an energy powerhouse. Used to generate energy, which is a common application. As a rocket fuel when processed into a liquid state. As an antifreeze element in industry. As a fertilizer. Methane is present in several sanitization products. In the testing of gas appliances, methane is employed. Share via
190081	https://actascientific.com/ASCS/pdf/ASCS-06-0525.pdf	Acta Scientific COMPUTER SCIENCES Volume 6 Issue 6 June 2024 Boolean Algebras and their Applications Review Article Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam Department of Mathematics, University College in Addair, Jazan University, K.S.A Corresponding Author: Rasha Ahmed Hamid Ahmed Department of Mathematics, University College in Addair, Jazan University, K.S.A. Received: April 29, 2024 Published: May 07, 2024 © All rights are reserved by Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. Expression Boolean algebra expressions are statements that make use of logical operators such as AND, OR, NOT, XOR, etc. These logical statements can only have two outputs, either true or false. In digital circuits and logic gates "1" and "0" are used to denote the input and output conditions. For example, if we write A OR B it becomes a Boolean. Abstract Keywords: Boolean Algebra; Logical Operators; Binary Operations; Unary Operation; Digital Circuits; Logic Gates; Duality; Boolean Expression; Isomorphic; Simplification; Complement; Boolean Variables; Literal; Truth Table Introduction In the middle of the nineteenth century the English mathemati cian George Boole introduced the algebras which are now named after him. These algebras give a mathematical basis to logical rea soning they are also the basis of electronic computer design via the physical implementation of Boolean operations. In this paper we introduce Boolean algebras and their operations, and we de fine Boolean functions which specify the operation performed by a Boolean circuit, or the truth value of a logical formula. Finally, we show how to represent a Boolean function by a polynomial expres sion built up using the basic operations of Boolean algebras. Boolean algebra is a branch of algebra dealing with logical operations on variables. There can be only two possible values of variables in boolean algebra, i.e. either 1 or 0. In other words, the variables can only denote two options, true or false. The three main logical operations of boolean algebra are conjunction, dis junction, and negation. In elementary algebra, mathematical expressions are used to mainly denote numbers whereas, in boolean algebra, expressions represent truth values. The truth values use binary variables or bits “1” and “0” to represent the status of the input as well as the output. The logical operators AND, OR, and NOT form the three ba sic boolean operators. In this article, we will learn more about the definition, laws, operations, and theorems of boolean algebra . Very special Boolean algebra, denoted by IB, is the Boolean al gebra containing only the two elements 0 and 1. (It is usually re ferred to as The Boolean algebra.) The sum, product and comple ment operations on this two element algebra are described in the 0 and 1, not forgetting that 1 + 1 = 1! Boolean Algebras Boolean algebra is defined as a set contain- ing two distinct ele ments 0 and 1, together with binary operations +, ·, and a unary operation, having the following properties: The definition of an ab stract Boolean algebra gives the axioms for an 1.1 . Definition (2-1): The Boolean algebras so far have all been concrete, consisting of bit vectors or equivalently of subsets of some set. Such a Boolean algebra consists of a set and operations on that set which can be shown to satisfy the laws of Boolean algebra. Instead of showing that the Boolean laws are satisfied, we can instead postulate a set X, two binary operations on X, and one unary operation, and require that those operations satisfy the laws of Boolean algebra. The elements of X need not be bit vectors or subsets but can be anything at all. This leads to the more general abstract definition. A Boolean algebra is any set with binary operations 𝖠 and ∨ and a unary operation ¬ thereon satisfying the Boolean laws. Definition (2-2) We say the system mathematics (B, +, ., , 0,1) Such that B is set contains at least two different elements. de fined two binary operations (+,.) and a unary operation (). 0≠1 two elements. B is Boolean Algebra if is satisfy Boolean lows Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 Boolean expression A logical statement that results in a Boolean value, either be True or False, is a Boolean expression. Sometimes, synonyms are used to express the statement such as ‘Yes’ for ‘True’ and ‘No’ for ‘False’. Also, 1 and 0 are used for digital circuits for True and False, respectively . Boolean expressions are the statements that use logical opera tors, i.e., AND, OR, XOR and NOT. Thus, if we write X AND Y = True, then it is a Boolean expression. Boolean algebra terminologies Now, let us discuss the important terminologies covered in Bool ean algebra. • Boolean Algebra: Boolean algebra is the branch of algebra that deals with logical operations and binary variables . • Boolean Variables: A Boolean variable is defined as a vari able or a symbol defined as a variable or a symbol, generally an alphabet that represents the logical quantities such as 0 or 1 . • Boolean Function: A Boolean function consists of binary variables, logical operators, constants such as 0 and 1, equal to the operator, and the parenthesis symbols . • Literal: A literal may be a variable or a complement of a vari able . • Complement: The complement is defined as the inverse of a variable, which is represented by a bar over the variable . • Truth Table: The truth table is a table that gives all the pos sible values of logical variables and the combination of the variables. It is possible to convert the Boolean equation into a truth table. The number of rows in the truth table should be equal to 2n, where “n” is the number of variables in the equa tion. For example, if a Boolean equation consists of 3 vari ables, then the number of rows in the truth table is 8. (i.e.,) 23 = 8. Definition (2-5) Duality Duality is the quality or state of having two different or opposite parts or elements, or a difference between two opposite things. If E is Boolean expression the duality expressions 𝐸 𝑑is expres sion find from change all adding operation of multiplying operation and O of I, I of O. Example 2-1 Solution 𝐸𝑑 = (x + 𝑧́) +( x + 1) + (𝑥́ + o) Theorem 2-1: If B Boolean Algebra and a, b ∈ B such that • (i) a+ a = a, (ii) a. a = a • (i) a + 1 = a, (ii) a. 0 = 0 • (i) a + a b = a, (ii) a (a + b) = a • (i) (𝑎) = a, (ii) a + 𝑎́ = 1 • (i) 0́ = 1, (ii) 1́ = 0 • (i) (a +́ 𝑏) = 𝑎́𝑏́, (ii) (𝑎𝑏)́= 𝑎́ + 𝑏́ Theorem (2-2):If B is Boolean Algebra and a ∈B → 𝑎́ is ungue ness. Definition (2-6): If B is Boolean Algebra a≠0 ∈B We say a is a tom if ∀ 𝑏 ∈B (0 ≤ 𝑏 ≤ 𝑎 → 𝑏 = 0 𝑣 𝑏 = 𝑎) Definition (2-6): If 𝐵1and 𝐵2are Boolean Algebra say 𝐵1and 𝐵2 are isomorphic if we find Then say is isomorphic between 𝐵1and 𝐵2. Boolean Functions Definition (3-1) If n ∈ 𝑧+and 𝐵2 𝑛 = {𝑎1, 𝑎2, … … …., 𝑎𝑛: 𝑎𝑖 ∈ 𝐵2} called application 𝐵2 𝑛 → 𝐵2 𝑖𝑠 Boolean Function. Example (3-1) Solution Definition (3-2) Example (3-2) Prove that the Boolean functions 04 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 Solution g (x, y, z) = x y (x𝑧́ + yz) = x y x 𝑧́ + xyy = x y 𝑧́ + xyz = x y (𝑧́ + z) = x y = f (x, y, z) then f (x, y, z) = x y, g (x, y, z) = x y (x 𝑧́ + yz) are equivalent. Definition (3-3) If 𝑥1, 𝑥1, 𝑥2,......................................, 𝑥𝑛 → Boolean function (i) ∀ 1 ≤ 𝑖 ≤ 𝑛, 𝑥𝑖, 𝑥́ 𝑖 called literal. (ii) 𝑦1, … … …., 𝑦𝑛, = 1 ∶ 𝑥𝑖, = 𝑦𝑖 or 𝑦𝑖 = 𝑥́ 𝑖 ∀ 1 ≤ 𝑖 ≤ 𝑛 Called min term. (iii) If f written the form sum minterms. This from is called (com plete sum of products) denoted (C S P). Algorithm (3 − 1) If 𝑥1, 𝑥1, 𝑥2, ……………., 𝑥𝑛 → Boolean function we put f (x, y, z) in the from complete sum of products C S P per from the following steps. • Find truth table off. • Find where value of f = 1. • Find the minterm of 𝑦1𝑦2, ………. 𝑦𝑛 where 𝑥𝑖, = 𝑦𝑖 = 1 (Or) 𝑥́ 𝑖 = 𝑦𝑖 = 0. • Find C S P is sum minterms of step (3) Example (3-3) Write the function, f (x, y, z) = (x + y) 𝑧́ in the from complete sum of products C S P. Solution f (x, y, z) = (x + y) 𝑧́ in the from C S P = x y 𝑧́ + x 𝑦́𝑧́ + 𝑥́ y 𝑧́ Logarithm (3 − 2) If 𝑥1, 𝑥1, 𝑥2, ……………., 𝑥𝑛 → Boolean function we put f ( x, y, z ) in the from C S P perform the following steps Find truth table of f . 2.Find where value of f = 0. Find the maxterm of 𝑦1 + 𝑦2 +, . ............................................+𝑦𝑛 ∀ 𝑥𝑖, = 𝑦𝑖, 𝑖𝑓 𝑥𝑖, = 0 Or 𝑥́ 𝑖 = 𝑦𝑖, 𝑖𝑓 𝑥𝑖, = 1. Find C P S is product maxterm of step (3) Example (3-4) Write the function, f (x, y, z) = (x + y) 𝑧́ in the from complete products of sum C P S Solution. f (x, y, z) = (x + y) 𝑧́ in the from C P S = (𝑥́ + 𝑦́ + 𝑧́) (𝑥́ + 𝑦 + 𝑧́) (𝑥 + 𝑦́ + 𝑧́) (𝑥 + 𝑦 + 𝑧́) (𝑥 + 𝑦 +𝑧) The minimal Boolean algebra (3-2) A very special Boolean algebra, denoted by IB, is the Boolean algebra containing only the two elements 0 and 1. (It is usually re ferred to as The Boolean algebra.) The sum, product and comple ment operations on this two element algebra are described in the following table. The function f (x, y) = 𝑥̅𝑦 + 𝑦̅ = is a polynomial function. Its values are given by the following table Laws of Boolean Algebra There are six types of Boolean algebra laws. They are • Commutative law • Associative law • Distributive law • AND law • OR law • Inversion law Those six laws are explained in detail here. Commutative law Any binary operation which satisfies the following expression is referred to as a commutative operation. Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit . • A. B = B. A • A + B = B + A Associative law It states that the order in which the logic operations are per formed is irrelevant as their effect is the same . 05 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 • (A. B). C = A. (B. C) • (A + B) + C = A + (B + C) Distributive law Distributive law states the following conditions • A. (B + C) = (A. B) + (A. C) • A + (B. C) = (A + B). (A + C) and law These laws use the AND operation. Therefore they are called AND laws. • A. 0 = 0 • A. 1 = A • A. A = A Simplification boolean function Boolean Algebra Simplification and how to simplify Boolean al gebra expressions using some basic rules applied to their variables, literals and terms. Boolean Algebra simplification is not that diffi cult to understand if you realise that the use of the symbols or signs of: “+” and “.” represent the operation of logical functions. Logical functions test whether a condition or state is either TRUE or FALSE but not both at the same time. So depending on the result of that test, a digital circuit can then decide to do one thing or another . Definition (5-1) Let f, g two Boolean functions w-ritten in the form product of sums say that f is simplest of g if • Number of term f less than number of term g • Number of literal f less than number of literal g Definition (5-2) Let S the set of all equivalent functions each of which in the form of sum products . If f is simplest function say that f in the form (min imial sum of products) . Having established the switching operation of the AND, OR, and NOT functions. We can now look at simplifying some basic Boolean Algebra expressions to obtain a final expression that has the mini mum number of terms . Example (5-1) First let us start with something simple such as The Boolean Expression: A. (A + B) Multiplying out the brackets gives us Example (5-2) This time we will use three Boolean terms, A, B, and C and use the same Boolean algebra simplification rules as before. Boolean Expression: (A + B) (A + C) Again, multiplying out the brackets gives us Boolean algebra theorems. The two important theorems which are extremely used in Bool ean algebra are De Morgan’s First law and De Morgan’s second law. These two theorems are used to change the Boolean expression. This theorem basically helps to reduce the given Boolean expres sion in the simplified form. These two De Morgan’s laws are used to change the expression from one form to another form. Now, let us discuss these two theorems in detail. De Morgan’s First Law (6-1) De Morgan’s First Law states that (A.B)’ = A’+B’. 06 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 The first law states that the complement of the product of the variables is equal to the sum of their individual complements of a variable. The truth table that shows the verification of De Morgan’s First law is given as follows The last two columns show that (A+B)’ = A’. B’. Hence, De Mor gan’s second law is proved. The other theorems in Boolean algebra are complementary the orem, duality theorem, transposition theorem, redundancy theo rem and so on. All these theorems are used to simplify the given Boolean expression. The reduced Boolean expression should be equivalent to the given Boolean expression. Logic network Logic network is switching network but we use gates. We find some gates: • NOT gate • OR gate • AND gate • NOR gate • NAND gate Core boolean operators (7 − 1) Universal gates (7-2) Universality of nor truth table • NOT: A nor A • AND: (A nor A) nor (B nor B) • OR: (A nor B) nor (A nor B) Universality of NAND Truth table 07 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 • NOT: A nand A • AND: (A nand B) nand (A nand B) • OR: (A nand A) nand (B nand B) From Boolean expressions to circuits (4-4) • Operation: not ( A or B ) Circuit : • 4 inputs (A, B, C, D), output 1 iff all inputs are 1 • Operation: (A and B) and (C and D) • Circuit: Example (7-1) Solution Frequently Asked Questions on Boolean Algebra . Q1 What is meant by Boolean algebra? In Mathematics, Boolean algebra is called logical algebra con sisting of binary variables that hold the values 0 or 1, and logical operations . Q2 What are some applications of Boolean algebra? In electrical and electronic circuits, Boolean algebra is used to simplify and analyze the logical or digital circuits. Q3 What are the three main Boolean operators? The three important Boolean operators are: AND (Conjunction) OR (Disjunction) NOT (Negation) Q4 Is the value 0 true or false? In Boolean logic, zero (0) represents false and one (1) repre sents true. In many applications, zero is interpreted as false and a non-zero value is interpreted as true. Q5 Mention the six important laws of Boolean algebra. The six im portant laws of Boolean algebra are: Commutative law Associative law Distributive law Inversion law AND law OR law 08 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 Main results • Boolean algebra is a branch of algebra. It differs from elemen tary algebra in two ways. First, the values of the variables are the truth values true and false, usually denoted 1 and 0, whereas in elementary algebra the values of the variables are numbers. • Second, Boolean algebra uses logical operators such as con junction (and) denoted as 𝖠, disjunction (or) denoted as ∨, and the negation (not) denoted as ¬. Elementary algebra, on the other hand • Uses arithmetic operators such as addition, multiplication, subtraction, and division. Boolean algebra is therefore a for mal way of describing logical operations, in the same way that elementary algebra describes numerical operations. Conclusion In this scientific paper, I discussed the concept of Boolean Alge bras and its characteristics and some of its applications. And, I have sought to provide assistance to students through the explanation in the text and examples to understand the definition of Boolean algebra, its laws, and some of its applications. 09 Boolean Algebras and their Applications Citation: Rasha Ahmed Hamid Ahmed and Um Salama Ahmed Abd Alla Alemam. “Boolean Algebras and their Applications". Acta Scientific Computer Sciences 6.6 (2024): 03-09 Bibliography 1. KH Rosen. “Discrete Mathematics And Its Applications, 7th Edi tion, McGraw-Hill (2007). 2. RP Grimaldi. “Discrete and Combinatorial Mathematics: Anap plied Introduction”. Addison-Waesly (1989). 3. Web sites dedicated to Graph Theory And Its Applications on the intern 4- Discrete Mathematics, written by Marouf Samhan and Ahmed Sharari . 4. J Eldon Whitesitt. “Boolean Algebra and Its Applications. 5. Stone Marshall H. “Boolean algebras and their application to topology”. Proceedings of the National Academy of Sciences 20.3 (1934): 197- 202. 6. Nies, André. “Effectively dense Boolean algebras and their ap plications”. Transactions of the American Mathematical Society 352.11 (2000): 4989-5012. 7. Whitesitt J Eldon. “Boolean algebra and its applications”. Cou rier Corporation (2012).
190082	https://www.biointeractive.org/sites/default/files/SickleCell_Genetics_Teacher.pdf	Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics Published July 2012 Revised October 2013 www.BioInteractive.org Page 1 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans MENDELIAN GENETICS, PROBABILITY, PEDIGREES, AND CHI-SQUARE STATISTICS OVERVIEW This classroom lesson uses the information presented in the short film The Making of the Fittest: Natural Selection in Humans ( to take students through a series of questions pertaining to the genetics of sickle cell disease and its relationship to malaria resistance. We divided the questions into sections: “Mendelian Genetics and Probability,” “Pedigrees,” and “Chi-Square Statistics.” Within each section, the questions sequentially move from a basic level to a more advanced level in order to develop the skills of the students. KEY CONCEPTS AND LEARNING OBJECTIVES • Having two of the same alleles of a given gene means an individual is homozygous; if the alleles are not identical, then the individual is heterozygous. • The sickle cell allele arose as a random mutation in the hemoglobin gene. • Individuals who are homozygous for the sickle cell allele have sickle cell disease; individuals who are homozygous for the normal hemoglobin allele do not have sickle cell disease but can contract malaria. • Individuals who are heterozygous for the sickle cell allele are protected against malaria but do not get sickle cell disease. Heterozygous individuals have a mix of normal and abnormal hemoglobin molecules. Students will be able to • use Punnett squares to predict the frequencies of genotypes in the next generation based on the genotypes of the parents; • understand the rules of probability as they relate to genetics problems; • analyze pedigrees to deduce genotypes, phenotypes, and probabilities; and • use the chi-square statistical analysis test to determine the significance of genetics data. CURRICULUM CONNECTIONS Curriculum Standards NGSS (April 2013) HS-LS3-1, HS-LS3-2, HS-LS3-3, HS-LS4-2, HS-LS4-3, HS-LS4-4 LS1.A, LS2.A, LS3.A, LS3.B, LS4.B, LS4.C Common Core (2010) ELA-Literacy.RST.9-10.7, ELA-Literacy.RST.11-12.4, ELA-Literacy.RST.11-12.7 Math.Content.HSS-IC.A.1, Math.Content.HSS-MD.B.6 AP Biology (2012–13) 1.A.2, 3.A.3, 3.C.1, 3.C.2, 4.C.1 IB Biology (2009) Topic 1, 4.1, 4.3, 5.4, 10.2 KEY TERMS chi-square statistical analysis, codominance, dominant, genotype, hemoglobin, heterozygous, homozygous, incomplete dominance, independent assortment, malaria, Mendelian genetics, pedigree, phenotype, probability, recessive, red blood cells, sickle cell anemia, sickle cell disease TIME REQUIREMENTS This lesson was designed to be completed within one to two 50-minute class periods if the chi-square statistics section is not included; if the chi-square statistics section is included, additional time may be required depending on the pace and background of the students. Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 2 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans SUGGESTED AUDIENCE This lesson is appropriate for high school biology (all levels including AP and IB) and undergraduate introductory biology. PRIOR KNOWLEDGE Students should have prior knowledge of the basics of Mendelian genetics (genotype, phenotype, homozygous, heterozygous, incomplete dominance, and codominance) and the rules of probability. They should also be familiar with how to draw and interpret pedigrees (including standard symbols used therein), use pedigrees to show family relationships, and analyze the pattern of inheritance of a particular trait. More-advanced students should have a working knowledge of the chi-square statistical analysis test. MATERIALS critical values table TEACHING TIPS • You could use this lesson and the story line of the short film as a culminating unit classroom assignment on genetics that ties together all levels of genetic analysis: Punnett squares, probability, pedigrees, and chi-square analysis. • You may discuss with the class how sickle cell disease provides an interesting example of the arbitrary nature of dominance, incomplete dominance, and codominance. Sickle cell disease, at an organismal level, is defined as an autosomal recessive disorder because one copy of HbA produces enough normal hemoglobin to prevent anemia. At the cellular level, regarding blood cell shape, the phenotype of the sickled red blood cell is incompletely dominant because heterozygotes can display some sickled red blood cells (RBCs) in low-oxygen environments. Finally, regarding hemoglobin at the molecular level, there is codominance. In heterozygotes, both HbA and HbS alleles are expressed. • The chi-square statistics portion of this lesson is optional. If you teach a course in which chi-square analysis is not required, you may remove this section from this lesson; we have placed it on separate pages for this reason. ANSWER KEY MENDELIAN GENETICS AND PROBABILITY i. If two people with sickle cell trait have children, what is the chance that a child will have normal RBCs in both high- and low-oxygen environments? What is the chance that a child will have sickle cell disease? Write the possible genotypes in the Punnett square. Normal RBCs in high- and low-oxygen environments 1/4 (25%) Sickle cell disease 1/4 (25%) a. What is the chance that a child will carry the HbS gene but not have sickle cell disease? 1/2 (50%) b. What are the chances that these parents will have three children who are homozygous for normal RBCs? (Show your work.) 1/4 × 1/4 × 1/4 = 1/64 (1.56%) A S A S AA AS AS SS Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 3 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans A A A S A S S S AS AS SS SS c. What are the chances that these parents will have three children who have both normal and mutant hemoglobin beta chains? (Show your work.) 1/2 × 1/2 × 1/2 = 1/8 (12.5%) d. What are the chances that all three of their children will show the disease phenotype? (Show your work.) 1/4 × 1/4 × 1/4 = 1/64 (1.56%) e. What are the chances that these parents will have two children with sickle cell trait and one with sickle cell disease? (Show your work.) 1/2 × 1/2 × 1/4 = 1/16 (6.25%) f. In the cross above, if you know that the child does not have sickle cell disease, what is the chance that the child has sickle cell trait? 2/3 (66.67%) (Note: Because you know the child does not have sickle cell disease, the child cannot have the SS genotype; thus, you can eliminate it from the Punnett square. The individual must be either AA or AS. There are 2 out of 3 chances that the individual will have the AS genotype.) ii. An individual who has sickle cell trait has children with an individual who does not have the HbS allele. a. What are the genotypes of the parents? AA and AS b. In a Punnett square, show all the possible genotypes of their children. State the genotype and phenotype ratios of the offspring. Genotype Ratio 50% (1/2) AA: 50% (1/2) AS Phenotype Ratio 50% (1/2) normal hemoglobin (normal RBCs):50% (1/2) normal and mutant hemoglobin (sickle cell trait) c. What are the chances that any one of this couple’s children will have sickle cell disease? 0% d. If this couple lives in the lowlands of East Africa, what are the chances that one of their children would be resistant to malaria if exposed to the malaria parasite? 1/2 (50%) iii. A woman with sickle cell disease has children with a man who has sickle cell trait. Answer the following questions. a. What are the genotypes of the parents? AS and SS b. What is the genetic makeup of the gametes the mother can produce? S c. What is the genetic makeup of the gametes the father can produce? A or S d. In the Punnett square, show all the possible genotypes of their children. Then summarize the genotype and phenotype ratios of the possible offspring. Genotype Ratio 50% (1/2) AS: 50% (1/2) SS Phenotype Ratio 50% (1/2) normal and mutant hemoglobin (sickle cell trait): 50% (1/2) mutant hemoglobin (sickle cell disease) AS AS AA AA Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 4 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans e. What are the chances that any one of this couple’s children will have sickle cell disease? 1/2 (50%) f. If this couple moves to the lowlands of East Africa and has children, which of their children would be more likely to survive? Explain your answer. If this couple moves to the moist lowlands of East Africa, the family would be exposed to the Anopheles mosquito that transmits the Plasmodium parasite, which causes malaria. Children who have sickle cell disease (SS) have a lethal disease and will be less likely to survive regardless of where they live. Children with sickle cell trait (AS) have two advantages: they have a greater resistance to malaria, and they normally do not show symptoms of sickle cell disease. Therefore, heterozygotes are more likely to survive. iv. In humans, blood type is a result of multiple alleles: IA, IB, and iO. A few simple rules of blood type genetics are that • IA is dominant over iO, • IB is dominant over iO, and • IAIB are codominant. Two parents who are heterozygous for type A blood and have sickle cell trait have children. Answer the following questions. a. What is the genotype of the parents? IAiOAS b. What are the genetic makeups of all the possible gametes they can produce? IAA, IAS, iOA, or iOS c. Complete the dihybrid Punnett square to determine the frequency of the different phenotypes in the offspring. (Note: Consider blood type and normal versus mutant hemoglobin in the various phenotypes.) IAA IAS ioA ioS IAA IAIAAA IAIAAS IAioAA IAioAS IAS IAIAAS IAIASS IAioAS IAioSS ioA IAioAA IAioAS ioioAA ioioAS ioS IAioAS IAioSS ioioAS ioioSS 3/16 (18.75%) Blood type A, normal hemoglobin (normal RBCs) 3/8 (6/16, or 37.5%) Blood type A, normal and mutant hemoglobin (sickle cell trait) 3/16 (18.75%) Blood type A, mutant hemoglobin (sickle cell anemia) 1/16 (6.25%) Blood type O, normal hemoglobin (normal RBCs) 1/8 (2/16, or 12.5%) Blood type O, normal and mutant hemoglobin (sickle cell trait) 1/16 (6.25%) Blood type O, mutant hemoglobin (sickle cell anemia) v. Now try a different way of solving a dihybrid cross. Because of Mendel’s (second) law of independent assortment, you can work with the blood type gene and the hemoglobin gene separately. Set up two monohybrid crosses with the following parents: the mother is heterozygous for type B blood and has sickle cell trait, while the father has type AB blood and also has sickle cell trait. IB iO A S Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 5 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans IA IAIB IAiO A AA AS IB IBIB IBiO S AS SS a. What are the chances that a child of this couple will have type B blood and sickle cell trait? (Show your work.) 1/2 × 1/2 = 1/4 (25%) b. What are the chances that a child will have type AB blood and will not have sickle cell disease? (Show your work.) 1/4 × 3/4 = 3/16 (18.75%) c. What are the chances that a child will have type B blood and sickle cell disease? (Show your work.) 1/2 × 1/4 = 1/8 (12.5%) d. What are the chances that a child will have type B blood and at least some normal hemoglobin? (Show your work.) 1/2 × 3/4 = 3/8 (37.5%) PEDIGREES vi. The following pedigree traces sickle cell disease through three generations of a family. Use the pedigree to answer the following questions. a. What is the genotype of the father in the first generation? AS b. What is the genotype of the daughter in the second generation? SS c. What is the genotype of individual 3 in the second generation? How do you know? AS; he and his son do not have sickle cell anemia, so he has at least one normal hemoglobin gene (A). He also has a son with sickle cell disease (SS). Therefore, he must carry one mutant hemoglobin gene (S) in order to have passed it on to his son. d. If the couple in the second generation has another child, what are the chances that the child will have the following? Sickle cell disease 1/2 (50%) Sickle cell trait 1/2 (50%) Completely normal hemoglobin 0% e. If the entire family moves to the lowlands of East Africa, four of the five males in the pedigree will have two genetic advantages over the other individuals in the family. Explain the two advantages. Moving to the moist lowlands of East Africa exposes this family to mosquitoes carrying the Plasmodium parasite. Therefore, the four males who are heterozygous (AS) for the sickle cell allele have two distinct genetic advantages. First, they do not suffer from sickle cell disease. Second, they are more resistant to malaria infection due to their heterozygous genotype. vii. The following pedigree traces sickle cell disease through four generations of a family living in New York City. Use the pedigree to answer the following questions. a. What is the genotype of the mother in the first generation? SS b. What are the possible genotypes of the father in the first generation? AA or AS c. What can you say about the genotype of all the children of the couple in the first generation? Explain your answer. All the children in the second generation are heterozygous (AS) for the sickle cell allele. None of the children have sickle cell disease, so they possess at least one normal hemoglobin gene (A). Each child would have inherited the mutant hemoglobin gene (S) from the mother, because she has sickle cell anemia (SS). Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 6 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans d. Regarding the answer to Question 7c, based on where the family resides, why would this genotype be considered a disadvantage? This family lives in New York City, which has a very low prevalence of malaria infection, so their AS genotype confers no genetic advantage. For the most part, the heterozygous genotype in New York City confers no distinct advantage or disadvantage. However, if any of these individuals mated with another heterozygous (AS) individual, they would have a 25% chance of having children with sickle cell anemia, which can be a deadly disease. The disadvantage of the AS genotype is in the possibility of future generations having sickle cell disease. e. What are the genotypes of the parents in the third generation? Explain how you know. Mother AS Father AS Neither parent has sickle cell anemia, so each parent possesses at least one normal hemoglobin gene (A). They do have children with sickle cell anemia (SS), so each must possess at least one mutant hemoglobin gene to pass on to their children. f. What is the possible genotype or genotypes of the mother in the second generation? AA or AS g. If the couple in the third generation has another child, what are the child’s chances of the following? Having sickle cell disease 1/4 (25%) Having sickle cell trait 1/2 (50%) Being homozygous for normal RBCs 1/4 (25%) Being resistant to malaria and not having sickle cell disease 1/2 (50%) viii. The following pedigree traces sickle cell disease through four generations of a family living in the highlands of eastern Africa. Use the pedigree to answer the following questions. a. What are the genotypes of the following individuals? (If more than one genotype pertains, include all possibilities.) Individual 1 AS Individual 10 AA or AS Individual 2 SS Individual 13 AS Individual 7 AS Individual 17 AA or AS b. If individuals 13 and 14 have another child, what are the chances that the child will have sickle cell disease? 1/4 (25%) c. If the same couple has three more children, what are the chances that the three children will have sickle cell trait? (Show your work.) 1/2 × 1/2 × 1/2 = 1/8 (12.5%) d. Based on where this family lives, is the sickle cell trait genotype a genetic advantage? Explain. Since this family lives in the relatively dry highlands of eastern Africa, there is a low incidence of malaria; therefore, the heterozygous genotype confers no significant genetic advantage. e. If individuals 8 and 9 have four more children, what are the chances that two of the children will be homozygous for normal RBCs? Explain why. There is a 0% chance this couple will have a child who is homozygous (AA) for normal hemoglobin. This is because the father has sickle cell disease (SS); therefore, he can only pass on the sickle cell allele (mutant hemoglobin [HbS]). Each child of this father has will either have sickle cell disease (SS) or be a carrier for the sickle cell allele (AS). Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 7 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans ix. Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Jerome was married before, and he and his first wife have a daughter with sickle cell disease. The brother of his current wife, Michaela, died of complications from sickle cell disease, but neither of her parents has the disease. a. Draw a pedigree representing this family. Be sure to clearly label Jerome and Michaela. Key J = Jerome M = Michaela b. What is the probability that Jerome and Michaela will have a baby with sickle cell disease? Note that neither Jerome nor Michaela has sickle cell disease. (Show your work.) 2/3 × 1/4 = 2/12 = 1/6 (16.67%) Based on his child from his first marriage, Jerome is heterozygous (AS) for the sickle cell allele. In order for Michaela and Jerome to have a child with sickle cell anemia, they must both be heterozygous, which would confer a 1/4 chance of having a child with the disease. However, based on the information available, Michaela has a 1/3 chance of being homozygous (AA) normal and a 2/3 chance of being heterozygous (AS). Therefore, to calculate the probability of this “combined” event, the rule of multiplication must be applied (2/3 chance of Michaela being AS × 1/4 chance of having an SS child). x. Natasha and Demarcus are planning on having children. Each has a sister with sickle cell disease. Neither Natasha nor Demarcus nor any of their parents have the disease, and none of them has been tested to see if they have sickle cell trait. a. Draw a pedigree representing this family. Be sure to clearly label Natasha and Demarcus. Key D = Demarcus N = Natasha b. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle cell disease. 2/3 × 2/3 × 1/4 = 4/36 = 1/9 (11.11%) Similar to Question 9, each parent has a 2/3 chance of being heterozygous (AS), and there is a 1/4 chance of having a child with the disease. Therefore, this is a combined event concerning its probability, so all values must be multiplied together. Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 8 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans CHI-SQUARE STATISTICS xi. Multiple couples living in a small village in the eastern African lowlands, all of whom are heterozygous for the HbS allele, have 500 children among them. Of these children, 139 are homozygous for HbA, 279 are heterozygous for HbS, and 82 suffer from sickle cell disease. Are these data statistically significant? Explain using a chi-square statistical analysis test. Chi-Square Data Table Phenotype/Genotype Observed (o) Expected (e) (o − e) (o − e)2/e Normal RBC/AA 139 125 14 1.57 Sickle cell trait/AS 279 250 29 3.36 Sickle cell disease/SS 82 125 −43 14.79 a. What is the chi-square value (χ2)? χ2 = 19.72 b. Calculate the degrees of freedom (df). df = 3 – 1 = 2 c. Using the critical values table (see page 12), determine the P value. P < 0.01 d. Interpret the P value as it relates to these data. Explain the significance. Since P < 0.05, the null hypothesis is rejected, which means that there is a statistically significant difference between the observed and expected data. Therefore, the difference between the observed and expected data is not solely due to chance. e. Which of the children have the greatest chance of survival? Explain why. Because these families live in a malaria “hot spot” in Africa, namely the moist eastern African lowlands, the children who are heterozygous for the sickle cell allele (AS) have a selective advantage over both the homozygous normal hemoglobin (AA) and the homozygous sickle cell anemia (SS) genotypes. xii. Suppose there are 50 couples with the same blood type and hemoglobin genotypes. They live on a small, isolated Pacific island on which very few mosquitoes have been identified. All the individuals are heterozygous for both type A blood and have sickle cell trait. The 50 couples had 224 children over the years. The children were all tested for blood type and for the presence of the sickle cell allele. Are these data significant? Explain using a chi-square statistical analysis test. (Use the table below if you need assistance.) Chi-Square Data Table Phenotype Observed (o) Expected (e) (o − e) (o− e)2/e Type A, normal RBCs 48 42 6 0.86 Type O, normal RBCs 18 14 4 1.14 Type A, sickle cell trait 92 84 8 0.76 Type O, sickle cell trait 33 28 5 0.89 Type A, sickle cell disease 27 42 −15 5.36 Type O, sickle cell disease 6 14 −8 4.57 Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 9 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans a. What is χ2? χ2 = 13.58 b. Calculate df. df = 6 – 1 = 5 c. Using the critical values table, determine the P value. P < 0.05 (0.025 < P < 0.01) d. Interpret the P value as it relates to these data. Explain the significance. Since P < 0.05, the null hypothesis is rejected, which suggests that there is a statistically significant difference between the observed and expected data. Therefore, the difference between the observed and expected data is not solely due to chance. e. From what you know about hemoglobin, sickle cell disease, and blood type, what selection pressure is acting on this population of children and causing the null hypothesis to be rejected? Explain your answer. (Hint: Look at the actual differences between the observed and expected numbers.) This population of individuals is isolated on a small Pacific island where very little quality health care is available. Therefore, there is selection against the children with sickle cell disease. The presence of this selection pressure skews the observed numbers from the expected values, causing the null hypothesis to be rejected, which suggests that something other than chance is acting on the population. In this case, that “thing” is the selection against the SS genotype. f. Due to the increase in global travel and the prevalence of invasive species, the Anopheles mosquito carrying the malaria parasite was inadvertently introduced to this isolated Pacific island. A researcher, 100 years from the present day, decides to complete a follow-up study and monitors another 50 couples who are all heterozygous for type A blood and have sickle cell trait. These couples had 136 children. Based on the introduction of the Anopheles mosquito carrying the malaria parasite, predict scientifically logical observed numbers of children for each genotype possibility and complete a chi-square statistical analysis test. Sample data: Student predictions for the observed numbers (o) will vary. The predicted observed numbers should show lower numbers for offspring with both normal RBCs and with sickle cell disease than the expected numbers, as well as higher numbers of offspring with sickle cell trait than expected. Additionally, the 1:4 type O:type A blood ratios should be maintained. The expected numbers should be exactly as written in the data table below. Chi-Square Data Table: Sample Data Phenotype Predicted Observed (o) Expected (e) (o − e) (o − e)2/e Type A, normal RBCs 19 25.5 −6.5 1.66 Type O, normal RBCs 5 8.5 −3.5 1.44 Type A, sickle cell trait 70 51 21 8.65 Type O, sickle cell trait 25 17 8 3.76 Type A, sickle cell disease 13 25.5 −12.5 6.13 Type O, sickle cell disease 4 8.5 −4.5 2.38 i. What is your predicted chi-square value (χ2)? Answers will vary depending on the values the students choose. Sample data: χ 2 = 24.02 ( χ2 should be greater than the value in Question 12a. ii. Calculate df. df = 6 – 1 = 5 iii. Using the critical values table, determine the predicted P value. P < 0.01 iv. From your predicted numbers, do you accept or reject the null hypothesis? Reject Mendelian Genetics, Probability, Pedigrees, and Chi-Square Statistics www.BioInteractive.org Page 10 of 10 LESSON TEACHER MATERIALS The Making of the Fittest: Natural Selection in Humans v. Based on what you know about hemoglobin, sickle cell disease, blood type, and malaria, what selection pressures are acting on this population of children? Explain your answer. Answers will vary; however, the predicted answers should be along the following lines: The isolated population sampled here is facing selection pressures caused by both limited health care as it relates to sickle cell disease and the introduction of malaria through the Anopheles mosquito, which carries the malaria-causing parasite. Therefore, the deviation of the observed numbers from the expected numbers is larger than it was from the first scenario due to the presence of three selection pressures. First, the sickle cell disease genotype (SS) is strongly selected against; second, the normal (AA) genotype is selected against due to the higher susceptibility to malaria infection. Finally, the data suggest that there is selection for the sickle cell trait (AS) genotype because these individuals do not have sickle cell disease and are more resistant to malaria. AUTHOR Ann Brokaw, Rocky River High School, Ohio FIELD TESTERS Marjorie Davis, Mount Saint Joseph Academy; David Knuffke, Deer Park High School; Mark Little, Broomfield High School; Dawn Norton, Minnetonka High School
190083	https://www.wolframalpha.com/input/?i=inequality+system+solves&a=FSelect_	inequality system solves - Wolfram\|Alpha UPGRADE TO PRO APPS TOUR Sign in Step-by-Step Solutions with Pro Get a step ahead with your homework Go Pro Now inequality system solves Natural Language Math Input Extended KeyboardExamplesUpload Random Computational Inputs: Assuming a system of four inequalities \| Use a system of three inequalities or a system of two inequalities instead » inequality 1: » inequality 2: » inequality 3: » inequality 4: Compute Input interpretation Download Page POWERED BY THE WOLFRAM LANGUAGE
190084	https://www.youtube.com/watch?v=XdZqk8BXPwg	The easy way to convert decimal numbers to binary (base 2) MindYourDecisions 3230000 subscribers 15256 likes Description 730091 views Posted: 30 Jul 2014 This video gives a method to convert decimal numbers to binary numbers quickly. This is a variation of the remainder system that is typically taught in courses. I like this method because it gives the answer in the correct order from left to right. Proof of why it works: I also give a proof in this post on my blog (skip down to the section with the proof and the discussion of dividing by 2 and ignoring the remainder) If you like my videos, you can support me at Patreon: Connect on social media. I update each site when I have a new video or blog post, so you can follow me on whichever method is most convenient for you. My Blog: Twitter: Facebook: Google+: Pinterest: Tumblr: Instagram: Patreon: Newsletter (sent about 2 times a year): My Books "The Joy of Game Theory" shows how you can use math to out-think your competition. (rated 4/5 stars on 23 reviews) "The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 5/5 stars on 1 review) "Math Puzzles Volume 1" features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.5/5 stars on 11 reviews. "Math Puzzles Volume 2" is a sequel book with more great problems. "Math Puzzles Volume 3" is the third in the series. "40 Paradoxes in Logic, Probability, and Game Theory" contains thought-provoking and counter-intuitive results. (rated 4.9/5 stars on 7 reviews) "The Best Mental Math Tricks" teaches how you can look like a math genius by solving problems in your head (rated 4.7/5 stars on 3 reviews) "Multiply Numbers By Drawing Lines" This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 5/5 stars on 1 review) 883 comments Transcript: Intro hi this is PR toer in this video I'm going to teach you how to convert numbers into binary or base 2 which is the number system used in virtually all modern computers we'll start out with an Example example of 13 the process we're going to use is we're going to divide the number in half and ignore the remainder so 13 / 2 is 6 with the remainder of one which we ignore and we're going to keep repeating this proc process until we get to the number one so dividing 6 and2 we get three dividing 3 and2 we get one now what we're going to do is we're going to write a one below any of the numbers that are odd and a zero below any of the numbers that are even so the number one is odd three is also odd six is an even number and 13 is an odd number and that's actually going to give us our answer 13 will equal 1 01 in binary you can do this process for other numbers we'll do another example like 244 and the step we're going to do is we're going to keep repeatedly dividing the number in half and ignoring the remainder so we get 122 dividing in half again we get 61 1 then we get 30 15 7 3 and finally one and once again all we do now is we write a one below any number that's odd and a zero below any number that's even so we have one one one one we have an even number and then we just fill out the rest and once again that gives us our answer we can convert 244 into binary from these numbers a final example we'll do 2014 we're going to keep on dividing this number in half ignoring the remainder and it's going to be a few steps for this and now once we're here we simply write a one below any of the odd numbers and a zero below any of the even numbers and once again this will give us the answer of 2014 in binary I hope you like this video please Outro subscribe to my channel I make videos on math and Game Theory you can catch me on my blog Mindy decisions or on Twitter at press toall Walker
190085	https://philosophy.lander.edu/logic/middle_fall.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Homepage » Logic » Syllogism Topics » Undistributed Middle \| \| \| \| \| \| \| \| \| \| \| \| \| Quizzes \| \| Tests \| \| Links \| \| Search \| \| Readings \| \| Syllabus \| \| \| Syllogistic Fallacy 2: The Fallacy of the Undistributed Middle Term Abstract: The syllogistic Fallacy of the Undistributed Middle Term is explained and illustrated with examples. 1. Fallacy of the undistributed middle term is a formal fallacy committed in standard form syllogisms whenever the term appearing in both premises is undistributed. (I.e., each of middle terms refers to some but not all of the individuals in the premise statements.) 1. Example of a distributed and undistributed term: In the statement … All [persons]D are [primates].U …the subject term “persons” is said to be a distributed term since it refers to, or denotes, each and every person. 1. The predicate term “primates” is said to be undistributed since it does not refer to each and every primate — “primates” in that sentence only makes reference to those individuals who are “persons” and not to other kinds of individuals such as as lemurs or gorillas. [If you want to review the topic of the distribution of subject and predicate terms in categorical statements, here's the link: Distribution of a Term.] 2. The “undistributed middle” is the second fallacy in our list of syllogistic fallacies. 2. Example of the Fallacy of an Undistributed Middle Term: In the following argument, the only persons likely to approve of the reasoning are those who are focusing only on the truth of the premises and not the structure of the argument: All [modern feminists]D are [social equality seekers].U All [protofeminists]D are [social equality seekers].U All [protofeminists]D are [modern feminists].U The Venn Diagram is sketched like this: 1. The blank space in the class of “protofeminists” which is outside of the class of “modern feminists” (indicated by the arrow) represents nothing is known about possible individuals existing in this area. Consequently, the premises allow for the possibility that some protofeminists are not in the class of “modern feminists.” Thus, the conclusion … “All protofeminists are modern feminists” … has not been deductively proved. 2. Intuitively, it is fairly evident that for the conclusion to follow validly, one would have to falsely presuppose the converse major premise … “All social equality seekers are modern feminists” … (not the originally given “All modern feminists are are social equality seekers”). So if the given major premise were to be converted in this manner, the subject class “social equality seekers” would become distributed. close ×Distribution: A term is distributed if in the context of the statement it refers to each and every member of the class it denotes; otherwise, the term is said to be undistributed. 3. So, with the major premise converted, the validity of the argument can be illustrated by visualizing how the new argument nests its terms: protofeminists ⊳ social equality seekers ⊳ modern feminists However, we know from the meaning of the terms used, not all social equality seekers are feminists. So, since this premise if false, the revised argument, although valid, is not a sound close ×Sound Argument A sound argument is a valid argument with true premises. (And, therefore, of course, the conclusion is necessarily true also.) argument. 4. In conclusion, in the original argument given above in the lightly shaded box, the class “social equality seekers” is undistributed in both premises since each appears in the predicate of an A statement. Since the middle term is undistributed in both premises, the argument commits the undistributed middle fallacy. 2. The Fallacy of the Undistributed Middle Term occurs when the middle term is undistributed in both premises. 1. Corresponding Rule: In a valid standard form categorical syllogism, the middle term must be distributed in at least one premise. 2. Reason: for the two terms of the conclusion to be connected through the third, as in the mechanism shown here, at least one term must be related to the whole of the class designated by the middle term. Otherwise, the connection might be with different parts of the middle term, as illustrated in the diagram above, and no connection can be made from S to P since the middle term could be referring to nothing in common between the two premises. John Buridan, a fourteenth century logician, explained the reason for this rule in this manner: Since the middle is not distributed in either [premise] it is possible that its conjunction with the major extreme is true for one thing and its conjunction with the minor is true for another; and from this no conjunction of the extremes with one another can be inferred … The premises can related to each other only if at least one of the middle terms is distributed. 3. In the history of logic, occasionally, an undistributed middle term has been considered an example of equivocation (fallacia æquivocationis) presumably because the denotation of the terms are different since the references of the terms are not necessarily identical in the context of their use. 4. Note: Remember for the Fallacy of the Undistributed Middle Term to occur, the middle term must be undistributed in both premises, not just one premise. 3. Optional Exercises: Try these challenging exercises. I think you'll find the mental mastery obtained from the effort well worthwhile. Try to picture in your mind the position and resultant distribution of the middle terms from the following syllogistic mood and figures without writing them down. --- Reminder: The position of the middle term is described by the figure of the syllogism. The figures are named “1,” “2,” “3,” and “4.” They are easily remembered because they form the shape of a flying brick or very wide “W”. Think of the middle four Ms on the corners of the orange square representing the empty space between figures 2 and 3. --- (1) Visualize the position of the middle term in the following syllogistic forms without writing the three statements in standard form syllogisms. (2) Then picture in your mind the distribution status of the two statements by the kind of statements used (i.e., an A, E, I, or O,). (3) State whether or not the fallacy of the undistributed middle term occurs. At first, picturing in your mind the distribution statuses to determine the fallacy occurs will be challenging, but working through a few problems will presently become rewarding. 1. AAA-2 All P is MU All Sis MU All S is P Since both premise predicate classes are undistributed, the fallacy of undistributed middle term occurs. 2. EAE-1 No MD is P All S is MU No SD is P The E major premise distributes all terms, so the undistributed middle falllacy does not occur. (The EAE-1 syllogism is the second most common valid syllogism found in argumentation.) 3. AAI-2 All P is MU All S is MU Some S is P Since both premise predicate classes are undistributed, the fallacy of undistributed middle term occurs. 4. AII-4 All P is MU Some MU is P Some S is P The middle term is undistributed in both premises, so the fallacy of the undistributed middle occurs. 5. OAO-4 Some P is not MU All MD is S Some S is not PD The middle term is distributed in the minor premise so the fallacy of the undistributed middle does not occur. (Later we will learn since the P term is undistributed in a premise but distributed in the conclusion, the fallacy of the illicit major term occurs since only some Ps are referred to in the premises but all Ps are referred to in the conclusion. Reasoning from “some” to “all” is not deductive reasoning.) 6. III-1 Some MU is P Some S is MU Some S is P The middle term is undistributed in both premises, so the fallacy of the undistributed middle occurs. 7. AAA-1 All MD is P All S is MU All S is P The middle term is distributed in the major premise, so the fallacy of the undistributed middle term does not occur. The AAA-1 syllogism is the most common occurring valid syllogism used in deductive argumentation. 8. OII-3 Some MU is not P Some MU is S Some S is P The middle term is undistributed in both premises, so the fallacy of the undistributed middle occurs. 9. OIO-2 Some PU is not MD Some S is MU Some S is not PD The middle term is distributed in the major premise, so the fallacy of the undistributed term does not occur. (Later we will study the fallacy of the illicit major term where the major term P is undistributed in a premise but distributed in the conclusion. Again, reasoning from “some” to “all” is not deductive reasoning.) 10. EOI-1 No MD is P Some S is not MD Some S is P The middle term is distributed in both premises so the fallacy of the undistributed middle does not occur. (Later we will learn that all syllogisms with two negative premises are invalid arguments.) Evaluate the following two syllogisms stated by Biron, a lord attending the king, in Shakespeare's Love's Labor Lost: “Well, ‘set thee down, sorrow!” for so they say the fool said, and so say I, and I the fool. Well proved, wit. By the Lord, this love is as mad as Ajax: it kills sheep; it kills me — I am sheep. Well proved again o’ my side.” Hint: the arguments can be setup as follows: Argument 1: The fool said “Well, set thee down sorrow!” I said “Well, set thee down sorrow!” ∴ I am a fool. Argument 2: Sheep are killed by a love as mad as Ajax. I am killed by a love as mad as Ajax. ∴ I am a sheep. Translate the arguments into standard order and form and test for validity. Argument 1: A fool is a person who said “Well … sorrow!”U I am a person who said “Well, … sorrow!”U I am a fool. The first argument commits the fallacy of the undistributed middle term. Argument 2: Sheep are mammals killed by a love as mad as Ajax.U I am a mammal killed by a love as mad as Ajax.U I am a sheep. The second argument commits the fallacy of four terms since in the major premise “killed” denotes “caused to die” and in the minor premise denotes “caused pain.” --- Notes 1. To say the revised argument is valid does not imply the revised argument is sound unless both of the premises presented are known to be true. This argument would be sound if all “social equality seekers” were understood to be narrowly classified as “feminists.”↩ 2. John Buridan, Treatise on Consequences trans. Stephen Read (New York: Fordham University Press, 2015), 121. doi: 0.5422/fordham/9780823257188.001.0001↩ 3. E.g., see Henry Joseph Turrell, A Manual of Logic, Or, A Statement and Explanation of the Laws of Formal Thought (London: Rivingtons, 1870), 105.)↩ 4. Shakespeare, Love's Labor Lost IV iii.4. (In Homer's Illiad, Ajax lost his senses when he was not awarded Achilles’ armor by the Acheans, and thinking their sheep and cattle were Acheans, began killing the sheep and cattle.)↩ \| \| \| \| \| Relay corrections or suggestions to philhelp@philosophy.lander.edu Read the disclaimer concerning this page. 1997-2021 Licensed under GFDL and Creative Commons 3.0 The “Copyleft” copyright assures the user the freedom to use, copy, redistribute, make modifications with the same terms. Works for sale must link to a free copy. The “Creative Commons” copyright assures the user the freedom to copy, distribute, display, and modify on the same terms. Works for sale must link to a free copy. --- Arguments \| Language \| Fallacies \| Propositions \| Syllogisms \| Ordinary Language \| Symbolic \| \|
190086	https://pubs.acs.org/doi/10.1021/acsomega.9b01581	Molecular and Dissociative Adsorption of Oxygen on Au–Pd Bimetallic Clusters: Role of Composition and Spin State of the Cluster \| ACS Omega 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 Manage Preferences Recently Viewedclose modal ACS ACS Publications C&EN CAS Access through institution Log In Molecular and Dissociative Adsorption of Oxygen on Au–Pd Bimetallic Clusters: Role of Composition and Spin State of the Cluster Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X Wechat LinkedIn Reddit Email Bluesky Jump to Abstract 1. Introduction 2. Computational Details 3. Results and Discussion 4. 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SI ACS Omega Vol 4/Issue 7 Article Get e-Alerts This publication is Open Access under the license indicated. Learn More Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X WeChat LinkedIn Reddit Email Bluesky Jump to Abstract 1. Introduction 2. Computational Details 3. Results and Discussion 4. Conclusions Supporting Information Author Information Acknowledgments References Cited By Expand Collapse Article July 25, 2019 Molecular and Dissociative Adsorption of Oxygen on Au–Pd Bimetallic Clusters: Role of Composition and Spin State of the Cluster Click to copy article link Article link copied! Manzoor Ahmad DarManzoor Ahmad Dar Department of Chemistry, Islamic University of Science and Technology, Awantipora, Jammu and Kashmir 192122, India E-mail: manzoor.dar@islamicuniversity.edu.inMore by Manzoor Ahmad Dar Sailaja Krishnamurty Sailaja Krishnamurty Physical Chemistry Division, CSIR-National Chemical Laboratory, Pune 411008, India More by Sailaja Krishnamurty Open PDFSupporting Information (1) ACS Omega Cite this: ACS Omega 2019, 4, 7, 12687–12695 Click to copy citation Citation copied! Published July 25, 2019 Publication History Received 30 May 2019 Accepted 26 June 2019 Published online 25 July 2019 Published in issue 31 July 2019 research-article Copyright © 2019 American Chemical Society. This publication is licensed under these Terms of Use. Request reuse permissions Article Views 1533 Altmetric - Citations 18 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts. The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated. Abstract Click to copy section link Section link copied! High Resolution Image Download MS PowerPoint Slide Utilization of molecular oxygen as an oxidizing agent in industrially important reactions is the ultimate goal to design environmentally benign processes under ambient conditions. However, the high thermal stability and a large O–O dissociation barrier in O 2 molecule pose a great challenge toward its successful application in the oxidative chemistry. To achieve this goal, different catalysts based on monometallic and bimetallic clusters have been developed over the years to promote binding and dissociation of molecular oxygen. The successful design of efficient metal cluster catalysis needs an in-depth knowledge of synergistic effects between different metal atoms and intrinsic catalytic mechanisms for O 2 adsorption and dissociation. Here, we present a systematic theoretical investigation of reaction pathways for O 2 adsorption and dissociation on Au 8, Pd 8, and Au 8–n Pd n (n = 1–7) nanoclusters in different spin states. The density functional calculations point out that the O 2 dissociation barriers can be significantly reduced with the help of certain bimetallic clusters along specific spin channels. Our results particularly indicate that Au 5 Pd 3 and Au 1 Pd 7 show very large O 2 binding energies of 1.76 and 1.69 eV, respectively. The enhanced O 2 binding subsequently leads to low activation barriers of 0.98 and 1.19 eV along the doublet and quartet spin channels, respectively, without the involvement of any spin flip-over for O 2 dissociation. Furthermore, the computed O 2 dissociation barriers are significantly low as compared to the already reported barriers (1.95–3.65 eV) on monometallic and bimetallic Au–Ag clusters. The results provide key mechanistic insights into the interaction and dissociation of molecular oxygen with Au–Pd clusters, which can prove informative for the design of efficient catalysts for oxidative chemistry involving molecular oxygen as a reactant. ACS Publications Copyright © 2019 American Chemical Society 1. Introduction Click to copy section link Section link copied! Activation and dissociation of molecular oxygen is one of the primary and key steps for driving environmentally and industrially important reactions such as CO oxidation, ethylene epoxidation, hydrocarbon oxidation, and so on. (1−14) Because of its triplet ground-state and strong oxygen–oxygen bond, molecular oxygen is kinetically very stable under ambient conditions. Owing to these challenges, a number of catalysts based on different metal nanoclusters (15−28) have been explored toward activation and dissociation of molecular oxygen for its successful utilization in oxidative chemistry. Among the various nanoclusters, gold-based nanoclusters have been at the forefront of both experimental and theoretical research for understanding the intrinsic mechanistic pathways of O 2 activation and dissociation on metal nanoclusters. Early experimental investigations were carried out by the research groups of Kaldor (29,30) and Whettan (31) on O 2 adsorption on a series of anionic gold nanoclusters. The authors found that O 2 exhibits profound adsorption with even sized anionic clusters and O 2 reactivity strongly depends on the cluster size and charge. Subsequently, Wallace et al. (32,33) performed detailed experimental studies to discern the role of CO and OH coadsorption on the reactivity of molecular oxygen with small gold clusters. Huang et al. (34) using combined experimental and theoretical methods further demonstrated molecular chemisorption and physisorption of O 2 on small sized anionic gold clusters with even and odd number of atoms, respectively. Woodham et al. (35) using infrared multiple photon dissociation spectra of anionic O 2–gold cluster complexes experimentally provided direct evidence for the formation of a superoxo moiety upon O 2 complexation. Recent investigations by Zeng and co-workers (36) unraveled that a superoxo to peroxo binding transition of O 2 molecule on anionic gold clusters occurs at Au 8–. The above experimental findings of O 2 interaction with gold clusters were explained by numerous theoretical studies. For example, Mills et al. (37) using density functional theory (DFT) found that O 2 binds more strongly with neutral and anionic gold clusters having odd number of electrons. Landman and co-workers (38) suggested that small sized anionic gold clusters (n = 1–3) prefer molecular adsorption and large sized clusters (n = 4–8) prefer dissociative adsorption. The authors also revealed that O 2 dissociation entails very high barriers. Similar results were found by Wang and Gong (39) while investigating O 2 chemisorption on Au 32 as a model system. Jena et al. (40) investigated the promotional effect of hydrogen atom doping on the reactivity of Au 8 cluster toward molecular oxygen. Zeng and Gao (41) and Lyalin and Taketsugu (42) in two independent studies discerned the effect of water and ethylene adsorption on the O 2 activation and dissociation on small sized gold clusters. Several other theoretical works were carried out to examine the effect of factors such as charge state, (43−46) doping, (47−52) coadsorption, (53,54) and support (55−58) on O 2 interactions with gold clusters. Although O 2 adsorption and dissociation are studied comprehensively on monometallic gold clusters both experimentally and theoretically, very few investigations are available wherein the role of composition and spin state vis-a-vis O 2 adsorption and dissociation on bimetallic clusters, particularly on Au–Pd clusters, has been elucidated thus far. Recently, García-Cruz et al. (59) studied the O 2 adsorption and dissociation on small bimetallic Au m Ag n clusters (m + n< 6). The authors concluded that the O 2 binding energies on the pure and bimetallic systems show a strong dependence on the spin state of the cluster. Similarly, Joshi et al. (60) revealed that Pt 6 and Au n Pt m (n + m = 6) with a double-planar shape are catalytically more efficient toward O–O bond activation. Bimetallic clusters hold a special position in the field of catalysis as they possess strikingly distinct properties compared to the parent metals, thereby acting as suitable candidates for design of superior catalysts with enhanced stability, activity, and selectivity. (61,62) Among the various bimetallic systems, the Au–Pd cluster catalysts are known to display a promising efficiency for catalyzing a wide range of oxidation reactions, namely, CO oxidation, direct synthesis of hydrogen peroxide from hydrogen and oxygen, and selective oxidation of alcohols to aldehydes. (63−67) Despite their superior and promising catalytic behavior, a comprehensive knowledge of the synergic effects between Au and Pd and the intrinsic factors controlling the adsorption and dissociation of molecular oxygen on these bimetallic systems are still elusive. Thus, an in-depth molecular-level understanding of the factors controlling the O 2 adsorption and dissociation to produce atomic oxygen species on the bimetallic Au–Pd clusters is highly imperative for their optimization for the aforementioned oxidation reactions involving oxygen. In this context, in the current work, DFT calculations were carried out to systematically investigate the influence of spin state and composition of Au–Pd bimetallic clusters on the adsorption and dissociation of molecular oxygen. The calculations were performed on selected ground-state conformations of monometallic Au 8,/Pd 8, and bimetallic Au8 8–n Pd n (n = 1–7) clusters according to the data available in the literature. 2. Computational Details Click to copy section link Section link copied! All the calculations were performed by DFT using the Perdew–Burke–Ernzerhof (PBE) (68) functional as implemented in the Gaussian 09 package. (69) The coordinates of the starting geometries for Au 8, Pd 8, and Au–Pd clusters were taken from the available literature (70) and fully optimized in the singlet/triplet and doublet/quartet spin states depending on whether the cluster has even or odd number of electrons. Geometry optimizations of the different structures were carried out using the default convergence cutoff of 10–4 with the help of Berny algorithm. The LANL2DZ basis set and the corresponding Los Alamos effective core potential were used for gold and palladium atoms, respectively. For oxygen, the TZVP basis set was used. The energetically most stable configurations of O 2 adsorbed complexes of Au 8, Pd 8, and Au–Pd clusters were found by adsorbing the O 2 molecule both via atop/superoxo (with one metal–oxygen bond) and bridged/peroxo (with two metal–oxygen bonds) modes on various distinct sites in these clusters. The oxygen adsorption was studied in two different spin states (singlet/triplet or doublet/quartet) depending on the cluster. It is important to highlight here that O 2 interaction and dissociation on closed shell metal clusters may involve spin crossovers, and thus, computation of reaction pathways for O 2 dissociation quantitatively requires advanced wave-function-based methods. However, because of computational restrictions, DFT in the framework of PBE has been widely used to study the spin and structural details of O 2 adsorption and dissociation (44,59,71,72) on monometallic and bimetallic clusters. Moreover, in a recent study, Dononelli and Klüner (72) showed that the DFT/PBE and high-levelcoupled cluster singles, doubles, and triples method results for the O 2 and CO adsorption energies on group 11 nanoclusters were in close agreement with each other. Vibrational frequency calculations were carried out to guarantee that the optimized structures are local minima. The O 2 binding energies (E b(O 2)) were computed with the help of the given equation where E O 2 is the energy of the O 2 molecule in the triplet sate, E C is the energy of the cluster in the lowest-energy spin state, and E complex is the energy of the O 2–cluster complexes in the lowest-energy spin state. The transition states were located by the linear synchronous transit method and were characterized by the presence of single imaginary frequency corresponding to the O–O scission. 3. Results and Discussion Click to copy section link Section link copied! The geometries of the monometallic (Au 8 and Pd 8) and bimetallic clusters (Au 8–n Pd n) are shown in Figure1. As reported earlier, Au 8 has a planar structure and prefers to be in the singlet spin state, whereas the Pd 8 cluster possesses a three-dimensional structure with a triplet spin state. All the bimetallic clusters exhibit three-dimensional structures. The bimetallic clusters Au 7 Pd 1 and Au 6 Pd 2 with high percentage of Au atoms prefer to be in the low spin singlet or doublet states as compared to the triplet and quartet spin states. However, Au–Pd clusters with three or more than three Pd atoms favor higher spin states of triplet or quartet multiplicity depending on whether the cluster has even or odd number of electrons. Thus, a spin transition is involved as the composition of the clusters is varied with Pd, playing a dominant role in deciding the overall spin state of the bimetallic clusters. Figure 1 Figure 1. Lowest-energy structures of the Au 8, Pd 8, and Au n–1 Pd n (n = 1–7) clusters and their corresponding spin states. High Resolution Image Download MS PowerPoint Slide We next look at the mode of the O 2 adsorption and binding on the monometallic Au 8 and Pd 8 clusters. Figure2 represents the equilibrium structures of O 2 adsorbed complexes of the Au 8, Pd 8, and Au 8–n Pd n clusters in their lowest-energy spin states. The structures corresponding to the higher energy spin states and their relative energies with respect to the lowest energy spin state structures are given in Figure S1 of the Supporting Information. It can be seen from Figure2 that O 2 adsorbs via atop or superoxo mode of bonding on the pristine Au 8 cluster with a single Au–O bond in both singlet and triplet spin states. Among the singlet and triplet spin states, the Au 8–O 2 complex in the triplet state has 1.06 eV lower energy than the singlet state. The calculated O 2 binding energy on Au 8 in the triplet state was found to be 0.31 eV. Moreover, interestingly, it was found that the Au 8–O 2 complex in the singlet state has higher energy than the isolated Au 8 cluster and O 2 molecule. Thus, the binding of O 2 molecule on Au 8 cluster is not favored from a thermodynamic point of view in the singlet spin state. No bridged or peroxo type of bonding was seen for O 2 adsorption on the Au 8 cluster in both spin states. This is in line with the earlier reported results of O 2 adsorption on Au 8 cluster. (38,40,51,55) Figure 2 Figure 2. Most stable geometries of the O 2-adsorbed complexes of Au 8, Pd 8, and Au n–1 Pd n (n = 1–7) clusters. The spin state of the system is given below each O 2–cluster complex. All the geometrical parameters marked are in angstroms. High Resolution Image Download MS PowerPoint Slide In contrast to Au 8, pristine Pd 8 cluster prefers a peroxo/bridged type of bonding with two Pd–O bonds with O 2 molecule both in the singlet and in the triplet spin state. The triplet state of the Pd 8–O 2 complex is found to be 0.05 eV lower in energy than the singlet state. The Pd 8 cluster shows considerably higher O 2 binding energies of 1.43 eV in the triplet state. To further corroborate the enhanced O 2 binding energy in the case of the Pd 8 cluster, we calculated the O–O bond lengths, O–O stretching frequencies, and net natural bond order charges on the O 2 molecule in the lowest energy spin states of Au 8/Pd 8–O 2 complexes. The abovementioned parameters are known to qualitatively describe the interaction of the O 2 molecule with metal nanoclusters. It is well known that the O 2 molecule acts as an electron acceptor and interacts strongly with clusters which can donate electrons easily into the antibonding orbitals of the O 2 molecule. This results in the increase in the O–O bond length and red shift in the O–O stretching frequencies. As can be seen from Table1, there is a notable increase in the O–O bond length and decrease in the O–O stretching frequency on the Pd 8 cluster as compared to the Au 8 cluster. We further note a significant charge transfer of −0.41 electrons from the Pd 8 cluster to the O 2 molecule as compared to the −0.05 electrons in the case of Au 8 cluster. Table 1. Calculated O 2 Binding Energy (E b(O 2)), O–O Bond Distances (R o–o), O–O Stretching Frequencies, and Net Natural Bond Orbital Charges (Q o–o) on the Au 8–O 2, Pd 8–O 2, and Au n–1 Pd n–O 2 Complexes in Their Lowest-Energy Spin States \| system \| spin state \| E b(O 2) (eV) \| R o–o (Å) \| ν o–o (cm–1) \| Q o–o \| :---: :---: :---: \| \| Au 8–O 2 \| triplet \| 0.31 \| 1.23 \| 1420 \| –0.05 \| \| Pd 8–O 2 \| triplet \| 1.43 \| 1.33 \| 1002 \| –0.41 \| \| Au 7 Pd 1–O 2 \| doublet \| 0.64 \| 1.26 \| 1279 \| –0.14 \| \| Au 6 Pd 2–O 2 \| triplet \| 0.61 \| 1.26 \| 1292 \| –0.12 \| \| Au 5 Pd 3–O 2 \| doublet \| 1.76 \| 1.26 \| 1271 \| –0.16 \| \| Au 4 Pd 4–O 2 \| triplet \| 1.37 \| 1.31 \| 1248 \| –0.16 \| \| Au 3 Pd 5–O 2 \| quartet \| 1.26 \| 1.31 \| 1073 \| –0.28 \| \| Au 2 Pd 6–O 2 \| triplet \| 1.34 \| 1.32 \| 1032 \| –0.34 \| \| Au 1 Pd 7–O 2 \| doublet \| 1.69 \| 1.33 \| 1016 \| –0.36 \| The lowest-energy structures of the O 2 adsorbed complexes of Au–Pd bimetallic clusters are also presented in Figure2. As can be seen, O 2 prefers a superoxo/atop mode of binding on Au–Pd clusters such as Au 7 Pd, Au 6 Pd 2, and Au 5 Pd 3 with a high percentage of Au atoms, whereas peroxo/bridged mode of bonding is preferred by O 2 on Au 4 Pd 4, Au 3 Pd 5, Au 2 Pd 6, and Au 1 Pd 7 clusters with more than three Pd atoms. Thus, we note a superoxo to peroxo transition in the bonding of oxygen molecule with respect to the composition of the cluster. A similar type of superoxo to peroxo transition of O 2 binding on anionic gold clusters as a function of size was revealed by Zeng and co-workers (36) in a recent study using photoelectron spectroscopy and DFT calculations. Coming to the spin state, the O 2-adsorbed complexes of Au 7 Pd 1, Au 5 Pd 3, and Au 1 Pd 7 clusters having odd number of electrons prefer a doublet spin state, whereas the O 2-adsorbed complexes of Au 6 Pd 2 and Au 4 Pd 4 having even number of electrons prefer to be in the triplet spin state. The lowest-energy doublet and quartet states of O 2–Au 3 Pd 5 complexes were found to have an energy difference of 0.02 eV, whereas the singlet and triplet spin states of O 2–Au 2 Pd 6 complexes were found to have the same energy. The equilibrium structures of the O 2-adsorbed complexes of Au–Pd clusters in their high energy spin states and their relative energies are given in Figure S2 of the Supporting Information. A careful analysis of the energetics of the O 2-adsorbed complexes reveals that the energy difference between the triplet and singlet spin states of even electron clusters and doublet and quartet spin states of odd electron clusters decreases as the number of Pd atoms increase in the cluster. This is particularly important as O 2 adsorption is known to involve a spin transition from singlet to triplet state in the case of less reactive monometallic closed shell gold clusters. Thus, incorporation of heteroatoms such as Pd in closed shell gold clusters can be used to overcome the spin activation energy and thereby leading to facile channels for O 2 dissociation. We next calculated the O 2 binding energies on the Au–Pd clusters in the lowest energy spin states which are given in Table1. O 2 adsorbs in atop mode on Au 7 Pd, Au 6 Pd 2, and Au 5 Pd 3 clusters, with binding energy values of 0.64, 0.61, and 1.76 eV, respectively. For clusters such as Au 4 Pd 4, Au 3 Pd 5, Au 2 Pd 6, and Au 1 Pd 7 where O 2 adsorbs in a bridged manner, the calculated binding energies are found to be 1.37, 1.26, 1.34, and 1.69 eV, respectively. Thus, importantly, the binding energies of O 2 with Au–Pd clusters are seen to increase significantly with the increase in the Pd content irrespective of whether the cluster is an odd or even electron one. Thus, changing the composition by incorporating atomic species such as Pd can be an efficient strategy to improve the binding and reactivity of O 2 with the known less reactive even electron gold clusters. We further looked at the O–O stretching frequencies and net negative charge on the O 2 molecule on the O 2–Au–Pd clusters in the atop mode as well as the bridged mode. The calculated O–O stretching frequencies as given in Table1 show a larger red shift for O 2 adsorption in bridged mode than in the atop mode. The large red shift in the O–O stretching frequencies in the case of bridged O 2 adsorption is also supported by a considerable charge transfer in to the antibonding orbitals of O 2 during O 2–cluster interaction as compared to the atop mode of O 2 adsorption. We finally investigated the O 2 dissociation pathways in different spin states (singlet and triplet states for even electron clusters and doublet and quartet spin states for odd electron clusters) on the Au 8, Pd 8, and Au–Pd clusters. The computed reaction energy profiles of O 2 dissociation on the Au 8 and Pd 8 clusters along with the reactants, transition states, and products are given in Figure3. The energies of the reactants, transition states, and products in the reaction profiles are given with respect to the energy of the cluster in the lowest energy spin state and oxygen in the triplet spin state. As can be seen from the reaction profiles, the singlet 1 R → 1 TS → 1 P and triplet 3 R → 3 TS → 3 P pathways of O 2 dissociation on Au 8 have very large activation barriers of 2.59 and 3.83 eV, respectively. On the other hand, in the case of Pd 8, the activation barriers for the singlet and triplet O 2 dissociation pathways are 2.03 and 1.56 eV, respectively. The product 3 P is less stable than the reactant 3 R in the case of Au 8 and thus O 2 dissociation is more unlikely to occur on Au 8. Moreover, based on the energy analysis of the O 2 dissociation pathway, it can be concluded that the process starts with triplet 3 Au 8–O 2 complex, passes through the singlet 1 Au 8–O 2 transition state, and ends with the triplet O–Au 8–O product. Thus, there are two spin crossover steps, one before the transition state and the other after the transition state in the O 2 dissociation process on the Au 8 cluster. The transition states corresponding to O 2 dissociation on Au 8 both in the singlet and triplet pathways connect the reactants with the transition states as can be seen in Figure S3 of the Supporting Information. For the Pd 8 cluster, the O 2 dissociation starts with the triplet 3 Pd 8–O 2 complex, passes through the triplet 3 Pd 8–O 2 transition state, and ends with the singlet O–Pd 8–O product. Hence, only a single spin crossover step after the transition state occurs in the case of O 2 dissociation on the Pd 8 cluster. The reaction profile diagrams for O 2 dissociation on the bimetallic clusters are presented in Figures4 and 5. As can be seen from the figures, the calculated activation barriers for O 2 dissociation on the Au–Pd clusters range from 0.98 to 1.94 eV depending on the composition of the cluster and spin state of the dissociation pathway. The activation barriers for O 2 dissociation along certain specific spin channels on the Au–Pd clusters are found to be very low as compared to the parent Au 8 and Pd 8 clusters. Among the bimetallic clusters, Au 4 Pd 4 shows high activation barriers of 1.94 and 1.86 eV for the singlet and triplet pathways, respectively. Importantly, very low activation barriers of 0.98 and 1.19 eV are seen for O 2 dissociation along the doublet and quartet spin channels on Au 5 Pd 3 and Au 1 Pd 7 clusters. It is important to mention that Au 5 Pd 3 and Au 1 Pd 7 were found to adsorb O 2 strongly with very large binding energies. Moreover, the obtained O 2 dissociation barriers are significantly lower than the already reported activation barriers (1.95–3.65 eV) on pure gold and bimetallic Au–Ag clusters. (59) The activation barriers along the quartet and doublet O 2 dissociation pathways on these clusters were found to be 1.45 and 1.55 eV, respectively. From the reaction profile diagrams of O 2 dissociation on Au 5 Pd 3 and Au 1 Pd 7, it can be seen that the process starts with doublet 2 Au 5 Pd 3–O 2 and quartet 4 Au 1 Pd 7–O 2 complexes, passes through doublet and quartet transition states, and ends with doublet and quartet products, respectively, on these clusters. Thus, not only do Au 5 Pd 3 and Au 1 Pd 7 clusters reveal significantly low activation barriers for O 2 dissociation but also do not involve any spin crossing during the O 2 dissociation. Further, the reaction energies for O 2 dissociation as presented in Table2 are thermodynamically more favorable on the abovementioned clusters than the other cases studied. The binding energy, activation barrier, and reaction energy analysis of O 2 dissociation on various bimetallic clusters confirm that certain compositions can lead to facile reaction pathways with no spin flipping involved during the reaction coordinate. Thus, the current study highlights that the composition of a binary cluster and spin state of the various species during a reaction coordinate on metal clusters can play a pivotal role in deciding the kinetic and thermodynamic feasibility of the reaction. Figure 3 Figure 3. (A) Calculated reaction pathways (in eV) for O–O bond rupture on the Au 8 and Pd 8 clusters in singlet (red) and triplet (green) spin states. The energies of the reactants, transition states, and products are with respect to the energy of oxygen molecules in the triplet spin state and the isolated cluster in the lowest energy spin state. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide Figure 4 Figure 4. (A) Calculated reaction pathways (in eV) for O–O bond rupture on Au 7 Pd 1, Au 5 Pd 3, Au 3 Pd 5, and Au 1 Pd 7 clusters in doublet (red) and quartet (green) spin states. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide Figure 5 Figure 5. (A) Calculated reaction pathways (in eV) for O–O bond rupture on Au 6 Pd 2, Au 4 Pd 4, and Au 2 Pd 6 clusters in singlet (red) and triplet (green) spin states. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide Table 2. Reaction Energiesa (Δ E) and Activation Energiesb (E a) in eV for the O–O Bond Dissociation on the Au 8, Pd 8, and Au n–1 Pd n Clusters in Various Spin States \| system \| reaction pathway \| Δ E (eV) \| E a (eV) \| :---: :---: \| \| Au 8 \| 1 R→1 TS→1 P \| –0.11 \| 2.59 \| \| \| 3 R→3 TS→3 P \| 0.59 \| 3.83 \| \| Pd 8 \| 1 R→1 TS→1 P \| –0.42 \| 2.03 \| \| \| 3 R→3 TS→3 P \| –0.20 \| 1.56 \| \| Au 7 Pd 1 \| 2 R→2 TS→2 P \| 0.45 \| 1.42 \| \| \| 4 R→4 TS→4 P \| 0.44 \| 1.83 \| \| Au 6 Pd 2 \| 1 R→1 TS→1 P \| 0.30 \| 1.49 \| \| \| 3 R→3 TS→3 P \| 0.75 \| 1.78 \| \| Au 5 Pd 3 \| 2 R→2 TS→2 P \| –0.16 \| 0.98 \| \| \| 4 R→4 TS→4 P \| 0.29 \| 1.45 \| \| Au 4 Pd 4 \| 1 R→1 TS→1 P \| 0.37 \| 1.94 \| \| \| 3 R→3 TS→3 P \| 0.15 \| 1.86 \| \| Au 3 Pd 5 \| 2 R→2 TS→2 P \| 0.36 \| 1.09 \| \| \| 4 R→4 TS→4 P \| 0.48 \| 1.59 \| \| Au 2 Pd 6 \| 1 R→1 TS→1 P \| 0.32 \| 1.42 \| \| \| 3 R→3 TS→3 P \| 0.43 \| 1.43 \| \| Au 1 Pd 7 \| 2 R→2 TS→2 P \| –0.01 \| 1.55 \| \| \| 4 R→4 TS→4 P \| –0.40 \| 1.19 \| a The reaction energy is calculated as Δ E = E(i P) – E(i R). b The activation energy is calculated as E a = E(i TS) – E(i R). 4. Conclusions Click to copy section link Section link copied! In summary, we have carried out DFT-based calculations to investigate the effect of composition and spin state of bimetallic Au–Pd clusters on the O 2 adsorption and dissociation. Our results indicate that the O 2 adsorption energy for both atop and bridged mode of binding increases significantly with the increase in the number of Pd atoms in the bimetallic clusters. Thus, the prominent odd/even effect of O 2 adsorption prevalent in pristine metal clusters is seen to disappear in the bimetallic Au–Pd clusters. Moreover, among the bimetallic Au–Pd clusters Au 5 Pd 3 and Au 1 Pd 7 are seen to show very low activation barriers of 0.98 and 1.19 eV along the doublet and quartet spin channels, respectively, without any spin flipping for O 2 dissociation. The considerably low activation barriers for the above clusters are attributed to their significantly large O 2 adsorption energies. In addition to low activation barriers, O–O bond rupture is found to be thermodynamically favorable with negative reaction energies on Au 5 Pd 3 and Au 1 Pd 7 clusters along the doublet and quartet reaction pathways. The current theoretical study thus points out the critical role of composition and spin state of a cluster to exploit alternate and efficient ways of nanocatalyst design. Supporting Information Click to copy section link Section link copied! The Supporting Information is available free of charge on the ACS Publications website at DOI: 10.1021/acsomega.9b01581. Optimized geometries of Au 8, Pd 8, Au 8–n Pd n, and their O 2-adsorbed complexes in the higher energy spin states and an internal reaction coordinate plot for O 2 dissociation on the Au 8 cluster (PDF) ao9b01581_si_001.pdf (376.02 kb) Molecular and Dissociative Adsorption of Oxygen on Au–Pd Bimetallic Clusters: Role of Composition and Spin State of the Cluster 517 views 0 shares 0 downloads Skip to figshare navigation ShareDownload figshare Terms & Conditions Most electronic Supporting Information files are available without a subscription to ACS Web Editions. Such files may be downloaded by article for research use (if there is a public use license linked to the relevant article, that license may permit other uses). Permission may be obtained from ACS for other uses through requests via the RightsLink permission system: Author Information Click to copy section link Section link copied! Corresponding Author Manzoor Ahmad Dar - Department of Chemistry, Islamic University of Science and Technology, Awantipora, Jammu and Kashmir 192122, India; Email: manzoor.dar@islamicuniversity.edu.in Author Sailaja Krishnamurty - Physical Chemistry Division, CSIR-National Chemical Laboratory, Pune 411008, India Notes The authors declare no competing financial interest. Acknowledgments Click to copy section link Section link copied! M.A.D. acknowledges the Head, Department of Chemistry and Dean School of Sciences, Islamic University of Science and Technology, for their constant support and encouragement. S.K. acknowledges MLP035426 for funding. References Click to copy section link Section link copied! This article references 72 other publications. 1Guo, Y.; Gu, D.; Jin, Z.; Du, P.-P.; Si, R.; Tao, J.; Xu, W.-Q.; Huang, Y.-Y.; Senanayake, S.; Song, Q.-S.; Jia, C.-J.; Schüth, F.Uniform 2 nm gold nanoparticles supported on iron oxides as active catalysts for CO oxidation reaction: structure–activity relationship. Nanoscale 2015, 7, 4920– 4928, DOI: 10.1039/c4nr06967f Google Scholar1 Uniform 2 nm gold nanoparticles supported on iron oxides as active catalysts for CO oxidation reaction: structure-activity relationship Guo, Yu; Gu, Dong; Jin, Zhao; Du, Pei-Pei; Si, Rui; Tao, Jing; Xu, Wen-Qian; Huang, Yu-Ying; Senanayake, Sanjaya; Song, Qi-Sheng; Jia, Chun-Jiang; Schuth, Ferdi Nanoscale (2015), 7 (11), 4920-4928 CODEN: NANOHL; ISSN:2040-3372. (Royal Society of Chemistry) Uniform Au nanoparticles (∼2 nm) with narrow size-distribution (std. deviation: 0.5-0.6 nm) supported on both hydroxylated (Fe_OH) and dehydrated iron oxide (Fe_O) have been prepd. by either deposition-pptn. (DP) or colloidal-deposition (CD) methods. Different structural and textural characterizations were applied to the dried, calcined and used gold-iron oxide samples. Transmission electron microscopy (TEM) and high-resoln. TEM (HRTEM) showed high homogeneity in the supported Au nanoparticles. The ex situ and in situ X-ray absorption fine structure (XAFS) characterization monitored the electronic and short-range local structure of active gold species. The synchrotron-based in situ X-ray diffraction (XRD), together with the corresponding temp.-programmed redn. by hydrogen (H2-TPR), indicated a structural evolution of the iron-oxide supports, correlating to their reducibility. An inverse order of catalytic activity between DP (Au/Fe_OH < Au/Fe_O) and CD (Au/Fe_OH > Au/Fe_O) was obsd. Effective gold-support interaction results in a high activity for gold nanoparticles, locally generated by the sintering of dispersed Au atoms on the oxide support in the DP synthesis, while a hydroxylated surface favors the reactivity of externally introduced Au nanoparticles on Fe_OH support for the CD approach. This work reveals why differences in the synthetic protocol translate to differences in the catalytic performance of Au/FeOx catalysts with very similar structural characteristics in CO oxidn. 2Liu, J.-X.; Filot, I. A. W.; Su, Y.; Zijlstra, B.; Hensen, E. J. M.Optimum Particle Size for Gold-Catalyzed CO Oxidation. J. Phys. Chem. C 2018, 122, 8327– 8340, DOI: 10.1021/acs.jpcc.7b12711 Google Scholar2 Optimum Particle Size for Gold-Catalyzed CO Oxidation Liu, Jin-Xun; Filot, Ivo A. W.; Su, Yaqiong; Zijlstra, Bart; Hensen, Emiel J. M. Journal of Physical Chemistry C (2018), 122 (15), 8327-8340 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) The structure sensitivity of Au-catalyzed CO oxidn. is presented by analyzing the dependence of CO oxidn. rate on particle size. Clusters with <14 Au atoms adopt a planar structure, whereas larger ones adopt a 3-dimensional structure. The CO and O2 adsorption properties depend strongly on particle structure and size. All of the reaction barriers relevant to CO oxidn. display linear scaling relations with CO and O2 binding strengths as main reactivity descriptors. Planar and 3-dimensional Au clusters exhibit different linear scaling relation due to different surface topols. and different coordination nos. of the surface atoms. From these linear scaling relations, 1st-principles microkinetics simulations were conducted to det. CO oxidn. rates and possible rate-detg. step of Au particles. Planar Au9 and 3-dimensional Au79 clusters present the highest CO oxidn. rates for planar and 3-dimensional clusters, resp. The planar Au9 cluster is much more active than the optimum Au79 cluster. A common feature of optimum CO oxidn. performance is the intermediate binding strengths of CO and O2, resulting in intermediate coverages of CO, O2, and O. Both these optimum particles present lower performance than max. Sabatier performance, indicating that there is sufficient room for improvement of Au catalysts for CO oxidn. 3Adnan, R. H.; Golovko, V. B.; Golovko, B. V.Benzyl Alcohol Oxidation Using Gold Catalysts Derived from Au8 Clusters on TiO2. Catal. Lett. 2019, 149, 449– 455, DOI: 10.1007/s10562-018-2625-8 Google Scholar3 Benzyl Alcohol Oxidation Using Gold Catalysts Derived from Au8 Clusters on TiO2 Adnan, Rohul H.; Golovko, Vladimir B. Catalysis Letters (2019), 149 (2), 449-455 CODEN: CALEER; ISSN:1011-372X. (Springer) Atomically-precise gold clusters have gained attraction in catalysis due to high fraction of low-coordinated gold atoms, unique structural geometry and ligand effect. Phosphine-ligated gold clusters offer an advantage in the light of the labile gold-phosphorous bond for easy ligand removal. Here, heterogeneous gold catalysts were prepd. by depositing atomically-precise phosphine-ligated gold clusters, Au8(PPh3)8(NO3)2 onto anatase-phase TiO2 nanoparticles. The catalysts were then calcined under two different conditions: O2 (Au8/TiO2:O2) and O2 followed by H2 (Au8/TiO2:O2-H2) at 200 °C, to dislodge phosphine ligands from the Au core. It was found that Au8/TiO2:O2-H2 catalyst showed a decent catalytic activity in benzyl alc. oxidn. while Au8/TiO2 and Au8/TiO2:O2 were completely inactive. Such results imply that small-size gold clusters (2-3 nm) alone do not always contribute to high catalytic activity of gold catalysts. It is suggested that the presence of NO3- species defines the catalytic activity of supported gold clusters in benzyl alc. oxidn. in the case of these catalysts and reinforces our initial claim in the previous work. 4Coquet, R.; Howard, K. L.; Willock, D. J.Theory and simulation in heterogeneous gold catalysis. Chem. Soc. Rev. 2008, 37, 2046– 2076, DOI: 10.1039/b707385m Google Scholar4 Theory and simulation in heterogeneous gold catalysis Coquet, Rudy; Howard, Kara L.; Willock, David J. Chemical Society Reviews (2008), 37 (9), 2046-2076 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) This crit. review covers the application of quantum chem. to the burgeoning area of the heterogeneous oxidn. by Au. We focus on the most established reaction, the oxidn. of CO at low temp. The review begins with an overview of the methods available comparing the treatment of the electron-electron interaction and relativistic effects. The structure of Au particles and their interaction with oxide reviews is then discussed in detail. Calcns. of the adsorption and reaction of CO and O2 are then considered and results from isolated and supported Au clusters compared (155 refs.). 5Turner, M.; Golovko, V. B.; Vaughan, O. P. H.; Abdulkin, P.; Berenguer-Murcia, A.; Tikhov, M. S.; Johnson, B. F. G.; Lambert, R. M.Selective oxidation with dioxygen by gold nanoparticle catalysts derived from 55-atom clusters. Nature 2008, 454, 981– 983, DOI: 10.1038/nature07194 Google Scholar5 Selective oxidation with dioxygen by gold nanoparticle catalysts derived from 55-atom clusters Turner, Mark; Golovko, Vladimir B.; Vaughan, Owain P. H.; Abdulkin, Pavel; Berenguer-Murcia, Angel; Tikhov, Mintcho S.; Johnson, Brian F. G.; Lambert, Richard M. Nature (London, United Kingdom) (2008), 454 (7207), 981-983 CODEN: NATUAS; ISSN:0028-0836. (Nature Publishing Group) Supported gold nanoparticles have excited much interest owing to their unusual and somewhat unexpected catalytic properties, but the origin of the catalytic activity is still not fully understood. Exptl. work on gold particles supported on a titanium dioxide (110) single-crystal surface has established a striking size threshold effect assocd. with a metal-to-insulator transition, with gold particles catalytically active only if their diams. fall below ∼3.5 nm. However, the remarkable catalytic behavior might also in part arise from strong electronic interaction between the gold and the titanium dioxide support. In the case of industrially important selective oxidn. reactions, explanation of the effectiveness of gold nanoparticle catalysts is complicated by the need for additives to drive the reaction, and/or the presence of strong support interactions and incomplete understanding of their possible catalytic role. Here we show that very small gold entities (∼1.4 nm) derived from 55-atom gold clusters and supported on inert materials are efficient and robust catalysts for the selective oxidn. of styrene by dioxygen. We find a sharp size threshold in catalytic activity, in that particles with diams. of ∼2 nm and above are completely inactive. Our observations suggest that catalytic activity arises from the altered electronic structure intrinsic to small gold nanoparticles, and that the use of 55-atom gold clusters may prove a viable route to the synthesis of robust gold catalysts suited to practical application. 6Herzing, A. A.; Kiely, C. J.; Carley, A. F.; Landon, P.; Hutchings, G. J.Identification of active gold nanoclusters on iron oxide supports for CO oxidation. Science 2008, 321, 1331– 1335, DOI: 10.1126/science.1159639 Google Scholar6 Identification of Active Gold Nanoclusters on Iron Oxide Supports for CO Oxidation Herzing, Andrew A.; Kiely, Christopher J.; Carley, Albert F.; Landon, Philip; Hutchings, Graham J. Science (Washington, DC, United States) (2008), 321 (5894), 1331-1335 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Gold nanocrystals absorbed on metal oxides have exceptional properties in oxidn. catalysis, including the oxidn. of carbon monoxide at ambient temps., but the identification of the active catalytic gold species among the many present on real catalysts is challenging. We have used aberration-cor. scanning transmission electron microscopy to analyze several iron oxide-supported catalyst samples, ranging from those with little or no activity to others with high activities. High catalytic activity for carbon monoxide oxidn. is correlated with the presence of bilayer clusters that are ~0.5 nm in diam. and contain only ~10 gold atoms. The activity of these bilayer clusters is consistent with that demonstrated previously with the use of model catalyst systems. 7Gao, Y.; Shao, N.; Bulusu, S.; Zeng, X. C.Effective CO Oxidation on Endohedral Gold-Cage Nanoclusters. J. Phys. Chem. C 2008, 112, 8234– 8238, DOI: 10.1021/jp801262v Google Scholar7 Effective CO Oxidation on Endohedral Gold-Cage Nanoclusters Gao, Yi; Shao, Nan; Bulusu, Satya; Zeng, X. C. Journal of Physical Chemistry C (2008), 112 (22), 8234-8238 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Ab initio calcns. are performed to study the CO oxidn. on six endohedral gold-cage clusters (W@Au12, Nb@Au13, Zr@Au14, Sc@Au15, Ca@Au16, and Na@Au17). The calcns. suggest that three clusters (Nb@Au13, Zr@Au14, and Sc@Au15) can be very effective nanocatalysts. The reaction energy barriers are lower than those assocd. with either bare gold clusters or gold surfaces and are comparable to those assocd. with the support Au clusters. The effective CO oxidn. can be attributed to the charge transfer from the dopant to the gold cage, the low coordination no. for gold atoms on the cages, as well as the fluxionality of the cage. 8Gao, Y.; Shao, N.; Pei, Y.; Zeng, X. C.Icosahedral Crown Gold Nanocluster Au 43 Cu 12 with High Catalytic Activity. Nano Lett. 2010, 10, 1055– 1062, DOI: 10.1021/nl100017u Google Scholar8 Icosahedral Crown Gold Nanocluster Au43Cu12 with High Catalytic Activity Gao, Yi; Shao, Nan; Pei, Yong; Zeng, Xiao Cheng Nano Letters (2010), 10 (3), 1055-1062 CODEN: NALEFD; ISSN:1530-6984. (American Chemical Society) Structural and catalytic properties of the gold alloy nanocluster, Au43Cu12, were assessed using a d.-functional method. In contrast to a pure Au55 nanocluster which exhibited a low-symmetry C1 structure, the 55-atom crown Au nanocluster exhibited a multi-shell structure, denoted by Au@Cu12@Au42, with the highest icosahedral group-symmetry. Also, d. functional calcns. suggested this geometric magic-no. nanocluster possessed comparable catalytic capability as a small-sized Au10 cluster for CO oxidn. due, in part, to their low-coordinated Au atoms on the vertexes. The gold alloy nanocluster also exhibited higher selectivity for styrene oxidn. than a bare Au(111) surface. 9Gao, Y.; Shao, N.; Pei, Y.; Chen, Z.; Zeng, X. C.Catalytic Activities of Subnanometer Gold Clusters (Au 16-Au 18, Au 20 and Au 27-Au 35) for CO Oxidation. ACS Nano 2011, 5, 7818– 7829, DOI: 10.1021/nn201817b Google Scholar9 Catalytic Activities of Subnanometer Gold Clusters (Au16-Au18, Au20, and Au27-Au35) for CO Oxidation Gao, Yi; Shao, Nan; Pei, Yong; Chen, Zhongfang; Zeng, Xiao Cheng ACS Nano (2011), 5 (10), 7818-7829 CODEN: ANCAC3; ISSN:1936-0851. (American Chemical Society) Using the CO oxidn. as a chem. probe, we perform a comprehensive ab initio study of catalytic activities of subnanometer gold clusters. Particular attention is placed on 12 different clusters in the size range of Au16-Au35, whose at. structures in the anionic state have been resolved from previous expts. Adsorption energies of a single CO or O2 mol. as well as coadsorption energies of both CO and O2 mols. on various distinctive surface sites of each anionic cluster and their neutral counterpart are computed. In general, the anionic clusters can adsorb CO and O2 more strongly than their neutral counterparts. The coadsorption energies of both CO and O2 mols. decrease as the size of gold clusters increases with the exception of Au34 (an electronic "magic-no." cluster). Besides the known factor of low coordination site, we find that a relatively small cone angle (<110°) assocd. with each surface site is another key geometric factor that can enhance the binding strength of CO and O2. For the subnanometer clusters, although the size effect can be important to the strength of CO adsorption, it is less important to the activation energy. Using Au34 as a prototype model, we show that strong CO and O2 adsorption sites tend to yield a lower reaction barrier for the CO oxidn., but they have little effect on the stability of the reaction intermediate. Our calcns. support the notion that CO and O2 adsorption energies on the gold clusters can be an effective indicator to assess catalytic activities of subnanometer gold clusters. This systematic study of the site- and size-dependent adsorption energies and reaction pathways enables a quant. assessment of the site-size-activity relationship for the CO oxidn. on subnanometer gold clusters. 10Patil, N. S.; Uphade, B. S.; Jana, P.; Sonawane, R. S.; Bhargava, S. K.; Choudhary, V. R.Oxidation of Benzyl Alcohol to Benzaldehyde by tert-Butyl Hydroperoxide over Nanogold Supported on TiO2 and other Transition and Rare-Earth Metal Oxides. Catal. Lett. 2004, 94, 89– 93, DOI: 10.1023/b:catl.0000019336.34270.5d Google Scholar10 Epoxidation of styrene by anhydrous t-butyl hydroperoxide over Au/TiO2 catalysts Patil, N. S.; Uphade, B. S.; Jana, P.; Sonawane, R. S.; Bhargava, S. K.; Choudhary, V. R. Catalysis Letters (2004), 94 (1-2), 89-93 CODEN: CALEER; ISSN:1011-372X. (Kluwer Academic/Plenum Publishers) Nanosize gold deposited on TiO2, by homogeneous deposition-pptn. (HDP) using urea as the pptn. agent is an active/selective and reusable catalyst for the epoxidn. of styrene (at 82°) by anhyd. t-Bu hydroperoxide (TBHP). The activity and epoxide selectivity of the Au/TiO2 catalyst in the epoxidn. is increased with increasing the Au loading on the TiO2 support. The Au/TiO2catalyst prepd. by the deposition pptn. using sodium hydroxide as the pptn. agent (DP method) has much lower gold loading and also lower Au dispersion and consequently possesses lower epoxidn. activity as compared to that prepd. by the HDP method. The styrene oxide selectivity is not influenced very significantly by the catalyst prepn. method or by the Au loading (except at the very low Au loading). In the Au/TiO2 catalysts prepd. by both the methods, Au is found to exist in both the metallic (Au°) and cationic (Au3+) forms. 11Deng, X.; Friend, C. M.Selective Oxidation of Styrene on an Oxygen-Covered Au(111). J. Am. Chem. Soc. 2005, 127, 17178– 17179, DOI: 10.1021/ja0557031 Google Scholar11 Selective Oxidation of Styrene on an Oxygen-Covered Au(111) Deng, Xingyi; Friend, Cynthia M. Journal of the American Chemical Society (2005), 127 (49), 17178-17179 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) For the first time, we demonstrate olefin epoxidn. promoted by an extended Au surface. The oxidn. of styrene to styrene epoxide, benzoic acid, and benzeneacetic acid is promoted on Au(111) covered with 0.2 ML of oxygen atoms. The estd. selectivity for styrene epoxide formation is ∼53%. Total combustion to CO2 accounts for ∼20% of the styrene reaction. We propose that styrene epoxide, benzoic acid, and benzeneacetic acid are produced via two possible oxametallacycle intermediates. Our work demonstrates that extended Au is an effective material for olefin oxidn., which has implications for understanding the activity of nanoscale Au catalysts. 12Hashmi, A. S. K.Homogeneous Gold Catalysts and Alkynes: A Successful Liaison. Gold Bull. 2003, 36, 3– 9, DOI: 10.1007/bf03214859 Google Scholar12 Homogeneous gold catalysts and alkynes: a successful liaison Hashmi, A. Stephen K. Gold Bulletin (London, United Kingdom) (2003), 36 (1), 3-9 CODEN: GOBUFW; ISSN:1027-8591. (World Gold Council) A review. The use of the combination of homogeneous gold-catalysts and alkynes in org. synthesis is reviewed from its beginnings in C-N-bond formation to the newest developments in C-C-bond formation. The common basic principle of these reactions is discussed. Special attention is devoted to the question where the gold catalysts are superior to either other catalysts or more traditional synthetic approaches to the product mols. 13Hughes, M. D.; Xu, Y.-J.; Jenkins, P.; McMorn, P.; Landon, P.; Enache, D. I.; Carley, A. F.; Attard, G. A.; Hutchings, G. J.; King, F.; Stitt, E. H.; Johnston, P.; Griffin, K.; Kiely, C. J.Tunable Gold Catalysts For Selective Hydrocarbon Oxidation Under Mild Conditions. Nature 2005, 437, 1132– 1135, DOI: 10.1038/nature04190 Google Scholar13 Tunable gold catalysts for selective hydrocarbon oxidation under mild conditions Hughes, Mathew D.; Xu, Yi-Jun; Jenkins, Patrick; McMorn, Paul; Landon, Philip; Enache, Dan I.; Carley, Albert F.; Attard, Gary A.; Hutchings, Graham J.; King, Frank; Stitt, E. Hugh; Johnston, Peter; Griffin, Ken; Kiely, Christopher J. Nature (London, United Kingdom) (2005), 437 (7062), 1132-1135 CODEN: NATUAS; ISSN:0028-0836. (Nature Publishing Group) Oxidn. is an important method for the synthesis of chem. intermediates in the manuf. of high-tonnage commodities, high-value fine chems., agrochems. and pharmaceuticals: but oxidns. are often inefficient. The introduction of catalytic systems using oxygen from air is preferred for 'green' processing. Gold catalysis is now showing potential in selective redox processes, particularly for alc. oxidn. and the direct synthesis of hydrogen peroxide. However, a major challenge that persists is the synthesis of an epoxide by the direct electrophilic addn. of oxygen to an alkene. Although ethene is epoxidized efficiently using mol. oxygen with silver catalysts in a large-scale industrial process, this is unique because higher alkenes can only be effectively epoxidized using hydrogen peroxide, hydroperoxides or stoichiometric oxygen donors. Here we show that nanocryst. gold catalysts can provide tunable active catalysts for the oxidn. of alkenes using air, with exceptionally high selectivity to partial oxidn. products (∼98%) and significant conversions. Our finding significantly extends the discovery by Haruta that nanocryst. gold can epoxidize alkenes when hydrogen is used to activate the mol. oxygen; in our case, no sacrificial reductant is needed. We anticipate that our finding will initiate attempts to understand more fully the mechanism of oxygen activation at gold surfaces, which might lead to com. exploitation of the high redox activity of gold nanocrystals. 14Corma, A.; Garcia, H.Supported Gold Nanoparticles as Catalysts for Organic Reactions. Chem. Soc. Rev. 2008, 37, 2096– 2126, DOI: 10.1039/b707314n Google Scholar14 Supported gold nanoparticles as catalysts for organic reactions Corma, Avelino; Garcia, Hermenegildo Chemical Society Reviews (2008), 37 (9), 2096-2126 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) A review. This crit. review is intended to attract the interest of org. chemists and researchers on green and sustainable chem. on the catalytic activity of supported gold nanoparticles in org. transformations. In the general part of this crit. review, emphasis is given to the different procedures to form supported gold nanoparticles and to the importance of the support cooperating in the catalysis. Also the convergence of homogeneous and heterogeneous catalysis in the study of gold nanoparticles has been discussed. The core part of this review is constituted by sections in which the reactions catalyzed by supported gold nanoparticles are described. Special emphasis is made on the unique ability of gold catalysts to promote addns. to multiple C-C bonds, benzannulations, and alc. oxidn. by oxygen. 15Oemry, F.; Padama, A. A. B.; Kishi, H.; Kunikata, S.; Nakanishi, H.; Kasai, H.; Maekawa, H.; Osumi, K.; Sato, K.Effects of Cluster Size on Platinum-Oxygen Bonds Formation in Small Platinum Clusters. J. Appl. Phys. 2012, 51, 035002, DOI: 10.7567/jjap.51.035002 Google Scholar15 Effects of cluster size on platinum-oxygen bonds formation in small platinum clusters Oemry, Ferensa; Padama, Allan Abraham B.; Kishi, Hirofumi; Kunikata, Shinichi; Nakanishi, Hiroshi; Kasai, Hideaki; Maekawa, Hiroyoshi; Osumi, Kazuo; Sato, Kaoru Japanese Journal of Applied Physics (2012), 51 (3, Pt. 1), 035002/1-035002/8 CODEN: JJAPB6 ISSN:. (Japan Society of Applied Physics) We present the results of d. functional theory calcn. in oxygen dissociative adsorption process on two types of isolated platinum (Pt) clusters: Pt4 and Pt10, by taking into account the effect of cluster reconstruction. The strength of Pt-Pt bonds in the clusters is mainly defined by d-d hybridization and interstitial bonding orbitals (IBO). Oxygen that adsorbed on the clusters is weakening the IBO and thus inducing geometry reconstruction as occurred in Pt10 cluster. However, cluster that could undergo structural deformation is found to promote oxygen dissocn. with no energy barrier. The details show that maintaining well-balanced of attractive and repulsive (Hellmann-Feynman) forces between atoms is considered to be the main key to avoid any considerable rise of energy barrier. Furthermore, a modest energy barrier that gained in Pt4 cluster is presumed to be originate from inequality of intramol. forces between atoms. 16Lv, P.; Lu, Z.; Li, S.; Ma, D.; Zhang, W.; Zhang, Y.; Yang, Z.Tuning Metal Cluster Catalytic Activity with Morphology and Composition: A DFT Study of O 2 Dissociation at the Global Minimum of Pt m Pd n (m + n = 5) Clusters. RSC Adv. 2016, 6, 104388– 104397, DOI: 10.1039/c6ra23266c Google Scholar16 Tuning metal cluster catalytic activity with morphology and composition: a DFT study of O2 dissociation at the global minimum of PtmPdn (m + n = 5) clusters Lv, Peng; Lu, Zhansheng; Li, Shuo; Ma, Dongwei; Zhang, Wenjin; Zhang, Yi; Yang, Zongxian RSC Advances (2016), 6 (106), 104388-104397 CODEN: RSCACL; ISSN:2046-2069. (Royal Society of Chemistry) The Pt-based alloyed clusters are the important catalysts in the chem. industry. The global min. structures of PtmPdn (m + n = 5) clusters were searched for based on the CALYPSO algorithm, and the various properties of the lowest-energy PtmPdn clusters have been studied. The results show that introducing the Pd element can modify the morphol. and compn. of the pure Pt5 cluster, i.e., from a 2D planar to a 3D trigonal bipyramid structure, except for the Pt4Pd1 (square pyramid). According to the av. cohesive energy, the Pt4Pd1 cluster possesses the highest stability among these Pt-Pd alloyed clusters. The O2 mol. prefers to anchor on the clusters by the Yeager mode. The catalytic property for O2 dissocn. of the pure Pt5 cluster can be further improved by introducing the Pd atoms. Moreover, the Pt4Pd1 cluster with the high Pt compn. and the structure of square pyramid shows the highest catalytic activity. More importantly, the Bronsted-Evans-Polanyi relationship is found to be applicable to the current mini alloyed metal clusters. This study will help to understand the prediction of the global min. structure of metal cluster by CALYPSO and shed light on the design of the less expensive and more effective alloyed cluster catalysts by controlling the morphol. and compn. in the fuel cells. 17Xu, Y.; Shelton, W. A.; Schneider, W. F.Effect of Particle Size on the Oxidizability of Platinum Clusters. J. Phys. Chem. A 2006, 110, 5839– 5846, DOI: 10.1021/jp0547111 Google Scholar17 Effect of Particle Size on the Oxidizability of Platinum Clusters Xu, Ye; Shelton, William A.; Schneider, William F. Journal of Physical Chemistry A (2006), 110 (17), 5839-5846 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The catalytic properties of transition metal particles often depend crucially on their chem. environment, but so far, little is known about how the effects of the environment vary with particle size, esp. for clusters consisting of only a few atoms. To gain insight into this topic, we have studied the oxygen affinity of free Ptx clusters as a function of cluster size (x = 1, 2, 3, 4, 5, and 10) using d. functional theory (DFT) calcns. (GGA-PW91). DFT-based Nose-Hoover mol. dynamics has been used to explore the configuration space of the PtxOx and PtxO2x clusters, leading to the discovery of several novel Pt-oxide structures. The formation of small Pt-oxide clusters by oxidizing the corresponding Ptx clusters is found to be significantly more exothermic than the formation of bulk Pt-oxides from Pt metal. The exothermicity generally increases as cluster size decreases but exhibits strongly nonlinear dependence on the cluster size. The nanoclusters are also structurally distinct from the bulk oxides and prefer one- and two-dimensional chain and ringlike shapes. These findings help elucidate the oxidn. behavior of Pt nanoclusters and lay the foundation for understanding the reactivity of Pt nanoclusters in oxidizing chem. environments. 18Kerpal, C.; Harding, D. J.; Hermes, A. C.; Meijer, G.; Mackenzie, S. R.; Fielicke, A.Structures of Platinum Oxide Clusters in the Gas Phase. J. Phys. Chem. A 2013, 117, 1233– 1239, DOI: 10.1021/jp3055137 Google Scholar18 Structures of Platinum Oxide Clusters in the Gas Phase Kerpal, Christian; Harding, Dan J.; Hermes, Alexander C.; Meijer, Gerard; Mackenzie, Stuart R.; Fielicke, Andre Journal of Physical Chemistry A (2013), 117 (6), 1233-1239 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The structures of small gas-phase PtnO2m+ (n = 1-6, m = 1, 2) cluster cations have been investigated in a combined IR multiple photon dissocn. (IRMPD) spectroscopy and d. functional theory (DFT) study. On the basis of the IR spectra obtained, it is concluded that in most clusters oxygen is bound dissociatively, preferring 2-fold bridge binding motifs, sometimes combined with singly coordinated terminal binding. Comparison of the oxide cluster structures with those of bare cationic platinum clusters reported previously reveals major structural changes induced in the platinum core upon oxygen binding. For some cluster sizes the presence of the Ar messenger atom(s) is found to induce a significant change in the obsd. cluster structure. 19Reber, A. C.; Khanna, S. N.; Roach, P. J.; Woodward, W. H.; Castleman, A. W.Spin Accommodation and Reactivity of Aluminum Based Clusters with O2. J. Am. Chem. Soc. 2007, 129, 16098– 16101, DOI: 10.1021/ja075998d Google Scholar19 Spin accommodation and reactivity of aluminum based clusters with O2 Reber, Arthur C.; Khanna, Shiv N.; Roach, Patrick J.; Woodward, William H.; Castleman, A. Welford, Jr. Journal of the American Chemical Society (2007), 129 (51), 16098-16101 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) It is shown that spin accommodation plays a detg. role in the reactivity of aluminum based anion clusters with oxygen. Exptl. reactivity studies on aluminum and aluminum-hydrogen clusters show variable reactivity in even electron systems and rapid etching in odd electron systems. The reactivity of even electron clusters is governed by a spin transfer to the singlet cluster through filling of the spin down antibonding orbitals on triplet oxygen. Theor. investigations show that when the spin transfer cannot occur, the species is unreactive. When spin accommodation is possible, more subtle effects appear, such as the required spin excitation energy, which raises the total energy of the system, and the filling of the antibonding levels of the O2 mol., which is stabilized by becoming an aluminum oxygen π bond. This explanation is consistent with obsd. behavior in oxygen etching reactions with a variety of clusters including AlnHm-, Aln-, AlnIm- and AlnC-. The proposed reaction mechanism lends a phys. interpretation as to why the HOMO-LUMO gap successfully predicts oxygen etching behavior of the considered systems. 20Samanta, B.; Sengupta, T.; Pal, S.Specificity of Amino Acid-Aluminum Cluster Interaction and Subsequent Oxygen Activation by the above Complex. J. Phys. Chem. C 2018, 122, 28310– 28323, DOI: 10.1021/acs.jpcc.8b08396 Google Scholar20 Specificity of Amino Acid-Aluminum Cluster Interaction and Subsequent Oxygen Activation by the above Complex Samanta, Bipasa; Sengupta, Turbasu; Pal, Sourav Journal of Physical Chemistry C (2018), 122 (49), 28310-28323 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Amino acid-aluminum cluster complex has been shown to be a promising oxygen mol. activator and has the potential to dissoc. O-O bond on the cluster surface. The study of the complex also reveals specificity of amino acid to aluminum cluster in the above complex, which can be used in the detection of amino acid. The results have been obtained by the d. functional theory at the B3LYP/6-311++g(d,p) level of calcn. 21Samanta, B.; Sengupta, T.; Pal, S.Aluminum cluster for CO and O 2 adsorption. J. Mol. Model. 2019, 25, 2, DOI: 10.1007/s00894-018-3869-3 Google ScholarThere is no corresponding record for this reference. 22Burgert, R.; Schnockel, H.; Grubisic, A.; Li, X.; Stokes, S. T.; Bowen, K. H.; Gantefor, G. F.; Kiran, B.; Jena, P.Spin Conservation Accounts for Aluminum Cluster Anion Reactivity Pattern with O 2. Science 2008, 319, 438– 442, DOI: 10.1126/science.1148643 Google Scholar22 Spin Conservation Accounts for Aluminum Cluster Anion Reactivity Pattern with O2 Burgert, R.; Schnoeckel, H.; Grubisic, A.; Li, X.; Stokes, S. T.; Bowen, K. H.; Gantefoer, G. F.; Kiran, B.; Jena, P. Science (Washington, DC, United States) (2008), 319 (5862), 438-442 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) The reactivity pattern of small (~10 to 20 atoms) anionic aluminum clusters with oxygen has posed a long-standing puzzle. Those clusters with an odd no. of atoms tend to react much more slowly than their even-numbered counterparts. We used Fourier transform ion cyclotron resonance mass spectrometry to show that spin conservation straightforwardly accounts for this trend. The reaction rate of odd-numbered clusters increased appreciably when singlet oxygen was used in place of ground-state (triplet) oxygen. Conversely, monohydride clusters AlnH-, in which addn. of the hydrogen atom shifts the spin state by converting formerly open-shell structures to closed-shell ones (and vice versa), exhibited an opposing trend: The odd-n hydride clusters reacted more rapidly with triplet oxygen. These findings are supported by theor. simulations and highlight the general importance of spin selection rules in mediating cluster reactivity. 23Neumaier, M.; Olzmann, M.; Kiran, B.; Bowen, K. H.; Eichhorn, B.; Stokes, S. T.; Buonaugurio, A.; Burgert, R.; Schnöckel, H.The Reaction Rates of O2 with Closed-Shell and Open-Shell Al x– and Ga x– Clusters under Single-Collision Conditions: Experimental and Theoretical Investigations toward a Generally Valid Model for the Hindered Reactions of O2 with Metal Atom Clusters. J. Am. Chem. Soc. 2014, 136, 3607– 3616, DOI: 10.1021/ja4125548 Google Scholar23 The Reaction Rates of O2 with Closed-Shell and Open-Shell Alx- and Gax- Clusters under Single-Collision Conditions: Experimental and Theoretical Investigations toward a Generally Valid Model for the Hindered Reactions of O2 with Metal Atom Clusters Neumaier, Marco; Olzmann, Matthias; Kiran, Boggavarapu; Bowen, Kit H.; Eichhorn, Bryan; Stokes, Sarah T.; Buonaugurio, Angela; Burgert, Ralf; Schnoeckel, Hansgeorg Journal of the American Chemical Society (2014), 136 (9), 3607-3616 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) In order to characterize the oxidn. of metallic surfaces, the reactions of O2 with a no. of Alx- and, for the first time, Gax- clusters as mol. models have been investigated, and the results are presented here for x = 9-14. The rate coeffs. were detd. with FT-ICR mass spectrometry under single-collision conditions at O2 pressures of ∼10-8 mbar. In this way, the qual. known differences in the reactivities of the even- and odd-numbered clusters toward O2 could be quantified exptl. To obtain information about the elementary steps, we addnl. performed d. functional theory calcns. The results show that for both even- and odd-numbered clusters the formation of the most stable dioxide species, [MxO2]-, proceeds via the less stable peroxo species, [Mx+···O22-]-, which contains M-O-O-M moieties. We conclude that the formation of these peroxo intermediates may be a reason for the decreased reactivity of the metal clusters toward O2. This could be one of the main reasons why O2 reactions with metal surfaces proceed more slowly than Cl2 reactions with such surfaces, even though O2 reactions with both Al metal and Al clusters are more exothermic than are reactions of Cl2 with them. Furthermore, our results indicate that the spin-forbidden reactions of 3O2 with closed-shell clusters and the spin-allowed reactions with open-shell clusters to give singlet [Mx+···O22-]- are the root cause for the obsd. even/odd differences in reactivity. 24Hagen, J.; Socaciu, L. D.; Le Roux, J.; Popolan, D.; Bernhardt, T. M.; Wöste, L.; Mitrić, R.; Noack, H.; Bonačić-Koutecký, V.Cooperative Effects in the Activation of Molecular Oxygen by Anionic Silver Clusters. J. Am. Chem. Soc. 2004, 126, 3442– 3443, DOI: 10.1021/ja038948r Google Scholar24 Cooperative Effects in the Activation of Molecular Oxygen by Anionic Silver Clusters Hagen, Jan; Socaciu, Liana D.; Le Roux, Jerome; Popolan, Denisia; Bernhardt, Thorsten M.; Woeste, Ludger; Mitric, Roland; Noack, Holger; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2004), 126 (11), 3442-3443 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) A novel size dependence in the adsorption reaction of multiple O2 mols. onto anionic silver clusters Ag-n (n = 1-5) is revealed by gas-phase reaction studies in an rf-ion trap. Ab initio theor. modeling based on DFT method provides insight into the reaction mechanism and finds cooperative electronic and structural effects to be responsible for the size selective reactivity of Agn- clusters toward one or more O2. In particular, Ag-n clusters with odd n have paired electrons and therefore bind one O2 only weakly, but they are simultaneously activated to adsorb a strongly bound second oxygen mol. For the clusters Ag3O4- and Ag5O4-, this cooperative effect results in a superoxo-like, doubly bound O2 subunit with potentially high activity in catalytic silver cluster oxidn. processes. 25Socaciu, L. D.; Hagen, J.; Le Roux, J.; Popolan, D.; Bernhardt, T. M.; Wöste, L.; Vajda, Š.Strongly cluster size dependent reaction behavior of CO with O2 on free silver cluster anions. J. Chem. Phys. 2004, 120, 2078, DOI: 10.1063/1.1644103 Google Scholar25 Strongly cluster size dependent reaction behavior of CO with O2 on free silver cluster anions Socaciu, Liana D.; Hagen, Jan; Le Roux, Jerome; Popolan, Denisia; Bernhardt, Thorsten M.; Woste, Ludger; Vajda, Stefan Journal of Chemical Physics (2004), 120 (5), 2078-2081 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) Reactions of free silver anions Agn- (n=1-13) with O2, CO, and their mixts. are investigated in a temp. controlled radio frequency ion trap setup. Cluster anions Agn- (n=1-11) readily react with mol. oxygen to yield AgnOm- (m=2, 4, or 6) oxide products. In contrast, no reaction of the silver cluster anions with carbon monoxide is detected. However, if silver cluster anions are exposed to the mixt. of O2 and CO, new reaction products and a pronounced, discontinuous size dependence in the reaction behavior is obsd. In particular, coadsorption complexes Agn(CO)O2- are detected for cluster sizes with n=4 and 6 and, the most striking observation, in the case of the larger odd atom no. clusters Ag7-, Ag9-, and Ag11-, the oxide product concn. decreases while a reappearance of the bare metal cluster signal is obsd. This leads to the conclusion that carbon monoxide reacts with the activated oxygen on these silver clusters and indicates the prevalence of a catalytic reaction cycle. 26Bernhardt, T. M.; Hagen, J.; Lang, S. M.; Popolan, D. M.; Socaciu-Siebert, L. D.; Wöste, L.Binding Energies of O2 and CO to Small Gold, Silver, and Binary Silver–Gold Cluster Anions from Temperature Dependent Reaction Kinetics Measurements. J. Phys. Chem. A 2009, 113, 2724– 2733, DOI: 10.1021/jp810055q Google Scholar26 Binding Energies of O2 and CO to Small Gold, Silver, and Binary Silver-Gold Cluster Anions from Temperature Dependent Reaction Kinetics Measurements Bernhardt, Thorsten M.; Hagen, Jan; Lang, Sandra M.; Popolan, Denisia M.; Socaciu-Siebert, Liana D.; Woste, Ludger Journal of Physical Chemistry A (2009), 113 (12), 2724-2733 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) A detailed anal. of exptl. obtained temp.-dependent gas-phase kinetic data for the oxygen and carbon monoxide adsorption on small anionic gold (Aun-, n = 1-3), silver (Agn-, n = 1-5), and binary silver-gold (AgnAum-, n + m = 2, 3) clusters is presented. The Lindemann energy transfer model in conjunction with statistical unimol. reaction rate theory is employed to det. the bond dissocn. energies E of the obsd. metal cluster complexes with O2 and CO. The accuracy limits of the obtained binding energies are evaluated by applying different transition-state models. The assumptions involved in the data evaluation procedure as well as possible sources of error are discussed. The thus-derived binding energies of O2 to pure silver and binary silver-gold cluster anions are generally in excellent agreement with previously reported theor. values. In marked contrast, the binding energies of O2 and CO to Au2- and Au3- detd. via temp.-dependent reaction kinetics are consistently lower than the theor. predicted values. 27Luo, Z.; Gamboa, G. U.; Smith, J. C.; Reber, A. C.; Reveles, J. U.; Khanna, S. N.; Castleman, A. W.Spin Accommodation and Reactivity of Silver Clusters with Oxygen: The Enhanced Stability of Ag13–. J. Am. Chem. Soc. 2012, 134, 18973– 18978, DOI: 10.1021/ja303268w Google Scholar27 Spin Accommodation and Reactivity of Silver Clusters with Oxygen: The Enhanced Stability of Ag13- Luo, Zhixun; Gamboa, Gabriel U.; Smith, Jordan C.; Reber, Arthur C.; Reveles, J. Ulises; Khanna, Shiv N.; Castleman, A. W., Jr. Journal of the American Chemical Society (2012), 134 (46), 18973-18978 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Spin accommodation is demonstrated to play a detg. role in the reactivity of silver cluster anions with oxygen. Odd-electron silver clusters are found to be esp. reactive, while the anionic 13-atom cluster exhibits unexpected stability against reactivity with oxygen. Theor. studies show that the odd-even selective behavior is correlated with the excitation needed to activate the O-O bond in O2. Furthermore, by comparison with the reactivity of proximate even-electron clusters, we demonstrate that the inactivity of Ag13- is assocd. with its large spin excitation energy, ascribed to a crystal-field-like splitting of the orbitals caused by the bilayer at. structure, which induces a large gap despite not having a magic no. of valence electrons. 28Ma, L.; Melander, M.; Laasonen, K.; Akola, J.CO Oxidation Catalyzed by Neutral and Anionic Cu20 Clusters: Relationship between Charge and Activity. Phys. Chem. Chem. Phys. 2015, 17, 7067– 7076, DOI: 10.1039/c5cp00365b Google Scholar28 CO oxidation catalyzed by neutral and anionic Cu20 clusters: relationship between charge and activity Ma, Li; Melander, Marko; Laasonen, Kari; Akola, Jaakko Physical Chemistry Chemical Physics (2015), 17 (10), 7067-7076 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) Reactions of CO and O2 on neutral and anionic Cu20 clusters have been investigated by spin-polarized d. functional theory. Three reaction mechanisms of CO oxidn. are explored: reactions with at. oxygen (dissocd. O2) as well as reactions with mol. oxygen, including Langmuir-Hinshelwood (LH) and Eley-Rideal (ER) mechanisms. The adsorption energies, reaction pathways, and reaction barriers for CO oxidn. are calcd. systematically. The anionic Cu20- cluster can adsorb CO and O2 more strongly than the neutral counterpart due to the superat. shell closing of 20 valence electrons which leaves one electron above the band gap. The activation of O2 mol. upon adsorption is crucial to det. the rate of CO oxidn. The CO oxidn. proceeds efficiently on both Cu20 and Cu20- clusters, when O2 is pre-adsorbed dissociatively. The ER mechanism has a lower reaction barrier than the LH mechanism on the neutral Cu20. In general, CO oxidn. occurs more readily on the anionic Cu20- (effective reaction barriers 0.1-0.3 eV) than on the neutral Cu20 cluster (0.3-0.5 eV). Moreover, Cu20- exhibits enhanced binding for CO2. From the anal. of the reverse direction of CO oxidn., it is obsd. that the transition of CO2 to CO + O can occur on the Cu20- cluster, which demonstrates that Cu clusters may serve as good catalyst for CO2 chem. 29Cox, D. M.; Brickman, R.; Creegan, K.; Kaldor, A.Gold Clusters: Reactions and Deuterium Uptake. Z. Phys. D: At., Mol. Clusters 1991, 19, 353– 355, DOI: 10.1007/bf01448327 Google Scholar29 Gold clusters: reactions and deuterium uptake Cox, D. M.; Brickman, R.; Creegan, K.; Kaldor, A. Zeitschrift fuer Physik D: Atoms, Molecules and Clusters (1991), 19 (1-4), 353-5 CODEN: ZDACE2; ISSN:0178-7683. The reactivity and satn. of small gold clusters (cations, neutrals and anions) towards several mols. was examd. Specific small gold clusters exhibit a pronounced variation in their reactivity towards H, CH4 and O. The reactivity not only depends strongly on cluster size but also on the cluster charge state. For example, Aun+ (n < 15) react readily with D2, but no evidence of reaction is obsd. for the anions under our exptl. conditions. Similar behavior is seen for CH4. With O only even atom (odd electron) anions are reactive, and Au10+ is the only cation which exhibits evidence of reaction. The global features (small cluster cations reactive towards H2, CH4, but large ones not reactive, odd electron anions reactive towards O2) are qual. explained by a simple frontier orbital picture. The uptake of D and CH4 on gold clusters also exhibits a pronounced size dependence with D/Au varying from a high of 3 for the dimer to zero for clusters contg. more than 15 Au atoms. Comparison of the CH4 and D satn. behavior indicated that CH4 is dissocd. and bound as CH3 and H. 30Cox, D. M.; Brickman, R. O.; Creegan, K.; Kaldor, A.Studies of the Chemical Properties of Size Selected Metal Clusters: Kinetics and Saturation. Mater. Res. Soc. Symp. Proc. 1991, 206, 43, DOI: 10.1557/proc-206-43 Google Scholar30 Studies of the chemical properties of size selected metal clusters: kinetics and saturation Cox, Donald M.; Brickman, Robert O.; Creegan, Kathleen; Kaldor, Andrew Materials Research Society Symposium Proceedings (1991), 206 (Clusters Cluster-Assem. Mater.), 43-8 CODEN: MRSPDH; ISSN:0272-9172. Activated dissociative mol. chemisorption is studied of hydrogen (deuterium) and small alkanes on cationic, neutral and anionic platinum and gold clusters. Methane or ethane chemisorption occurs most readily only on small Pt and Au cluster cations and least rapidly, if at all, on cluster anions. The larger Pt and Au clusters are found to be non-reactive towards the alkanes under our exptl. conditions. At the other extreme, at steady state, small clusters of Pt, Rh, Ni, Pd, V, Nb, and Ta are hydrogen "rich" exhibiting H/M stoichiometries » 1. These results are consistent with a limited data base on supported cluster obtained via EXAFS measurements and have important implications for catalytic reactions involving hydrogen and light hydrocarbons. 31Salisbury, B. E.; Wallace, W. T.; Whetten, R. L.Low-Temperature Activation of Molecular Oxygen by Gold Clusters: a Stoichiometric Process Correlated to Electron Affinity. Chem. Phys. 2000, 262, 131– 141, DOI: 10.1016/s0301-0104(00)00272-x Google Scholar31 Low-temperature activation of molecular oxygen by gold clusters: a stoichiometric process correlated to electron affinity Salisbury, B. E.; Wallace, W. T.; Whetten, R. L. Chemical Physics (2000), 262 (1), 131-141 CODEN: CMPHC2; ISSN:0301-0104. (Elsevier Science B.V.) The adsorption-desorption equil. of di-oxygen (O2) on Au cluster anions (AuN-, N = 2-22) were measured in the high-pressure limit by pulsed flow-reactor methods at ambient temp. Only certain clusters (N = 2,4,6,8,10,12,14,18,20) exhibit measurable adsorption, and for most of these the reaction can be driven to completion (>90% conversion to AuNO2-). In no case could secondary adsorption be detected, i.e. a 1:1 reaction stoichiometry is strictly obeyed. The size (N) variation in the adsorption free energies is correlated with the known electron affinities of the Au clusters. Taken with the pos. electron affinity of O2, these results suggest that O2 acts as a single-electron acceptor, yielding adsorbed superoxide (O2-), while the Au cluster strives to pair electrons (or close electron shells). This species could serve as an activated form for low-temp. oxygen-atom transfer reactions catalyzed by supported Au clusters, wherein the activation of mol. oxygen is regarded as a crucial step. 32Wallace, W. T.; Whetten, R. L.Coadsorption of CO and O 2 on Selected Gold Clusters: Evidence for Efficient Room-Temperature CO2 Generation. J. Am. Chem. Soc. 2002, 124, 7499– 7505, DOI: 10.1021/ja0175439 Google Scholar32 Coadsorption of CO and O2 on Selected Gold Clusters: Evidence for Efficient Room-Temperature CO2 Generation Wallace, William T.; Whetten, Robert L. Journal of the American Chemical Society (2002), 124 (25), 7499-7505 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Spurred by the recent demonstrations of the size- and support-dependent reactivity of supported gold clusters, here we present results on the coadsorption of CO and O2 on selected anionic gold clusters, AuN-, in the gas phase. O2 adsorbs in a binary (0,1) fashion as a one-electron acceptor on the AuN- clusters, with even-N clusters showing varying reactivity toward O2 adsorption, while odd-N clusters show no evidence of reactivity. CO shows a highly size-dependent reactivity for AuN- sizes from N = 4 to 19, but no adsorption on the gold dimer or trimer. When the gold clusters are exposed to both reactants, either simultaneously or sequentially, interesting effects have been obsd. While the same rules pertaining to individual O2 or CO adsorption continue to apply, the preadsorption of one reactant on a cluster may lead to the increased reactivity of the cluster to the other reactant. Thus, the adsorbates are not competing for bonding sites (competitive coadsorption), but, instead, aid in the adsorption of one another (cooperative coadsorption). New peaks also arise in the mass spectrum of Au6- under CO and O2 coadsorption conditions, which can be attributed to the loss of a CO2 mol. (or mols.). By studying the relative amt. of reaction, and relating it to the reaction time, it is found that the gas-phase Au6 anion is capable of oxidizing CO at a rate 100 times that reported for com. or model gold catalysts. 33Wallace, W. T.; Wyrwas, R. B.; Whetten, R. L.; Mitrić, R.; Bonačić-Koutecký, V.Oxygen Adsorption on Hydrated Gold Cluster Anions: Experiment and Theory. J. Am. Chem. Soc. 2003, 125, 8408– 8414, DOI: 10.1021/ja034905z Google Scholar33 Oxygen Adsorption on Hydrated Gold Cluster Anions: Experiment and Theory Wallace, William T.; Wyrwas, Richard B.; Whetten, Robert L.; Mitric, Roland; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2003), 125 (27), 8408-8414 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) The discovery that supported gold clusters act as highly efficient catalysts for low-temp. oxidn. reactions has led to a great deal of work aimed at understanding the origins of the catalytic activity. Several studies have shown that the presence of trace moisture is required for the catalysts to function. Using near-atm. pressure flow reactor techniques, we have studied humidity and temp. effects on the reactivity of gas-phase gold cluster anions with O2. Near room temp., the humid source produces abundant gold-hydroxy cluster anions, AuNOH-, and these have a reversed O2 adsorption activity: nonreactive bare gold clusters become active when in the form AuNOH-, while active bare clusters are inactive when -OH is bound. The binding energies for the stable structures obtained from d. functional calcns. confirm fully these findings. Moreover, the theory provides evidence that electron-transfer induced by the binding of a OH group enhances the reactivity toward mol. oxygen for odd anionic gold clusters and suppresses the reactivity for the even ones. The temp. dependence of O2 addn. to Au3OH- and Au-4 indicates deviations from equil. control at temps. below room temp. The effects of humidity on gold cluster adsorption activity support the conclusion drawn for the mechanism of O2 adsorption on "dry" gold cluster anions and provides insight into the possible role of water in the enhanced activity of supported gold cluster catalysts. 34Huang, W.; Zhai, H.-J.; Wang, L.-S.Probing the Interactions of O 2 with Small Gold Cluster Anions (Au n–, n = 1–7): Chemisorption vs Physisorption. J. Am. Chem. Soc. 2010, 132, 4344– 4351, DOI: 10.1021/ja910401x Google Scholar34 Probing the Interactions of O2 with Small Gold Cluster Anions (Aun-, n = 1-7): Chemisorption vs. Physisorption Huang, Wei; Zhai, Hua-Jin; Wang, Lai-Sheng Journal of the American Chemical Society (2010), 132 (12), 4344-4351 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Activation of O2 is the most crit. step in catalytic oxidn. reactions involving gold and remains poorly understood. Here the authors report a systematic study of the interactions between O2 and small gold cluster anions Aun- (n = 1-7) using photoelectron spectroscopy. Higher resoln. photoelectron spectra are obtained for the molecularly chemisorbed even-sized AunO2- (n = 2, 4, 6) complexes. Well-resolved vibrational structures due to O-O stretching are obsd. and can be readily distinguished from the Au-derived PES bands. The adiabatic detachment energies and O-O vibrational frequencies are 3.03 ± 0.04, 3.53 ± 0.05, and 3.17 ± 0.05 eV, and 1360 ± 80, 1360 ± 80, and 1330 ± 80 cm-1 for n = 2, 4, 6, resp. Physisorbed Aun-(O2) complexes for n = 1, 3, 5, 7 are obsd. for the 1st time, providing direct evidence for the inertness of the closed-shell odd-sized Aun- clusters toward O2. Neutral even-sized Aun clusters are closed-shell and are expected to be inert toward O2, which is not consistent with the reduced O-O vibrational frequencies obsd. in the photoelectron spectra relative to free O2. Probably the photodetachment transitions can only access excited states of the neutral even-sized AunO2 complexes; a double-well potential is proposed consisting of the ground-state van der Walls well at long Aun-O2 distances and a higher energy deeper well at short Aun-O2 distances derived from singlet O2 (1Δg). The current study provides further insight into O2 interactions with small gold clusters, as well as accurate exptl. data to benchmark theor. studies. 35Woodham, A. P.; Meijer, G.; Fielicke, A.Activation of Molecular Oxygen by Anionic Gold Clusters. Angew. Chem., Int. Ed. 2012, 51, 4444– 4447, DOI: 10.1002/anie.201108958 Google Scholar35 Activation of Molecular Oxygen by Anionic Gold Clusters Woodham, Alex P.; Meijer, Gerard; Fielicke, Andre Angewandte Chemie, International Edition (2012), 51 (18), 4444-4447, S4444/1-S4444/11 CODEN: ACIEF5; ISSN:1433-7851. (Wiley-VCH Verlag GmbH & Co. KGaA) Activation of mol. oxygen by anionic gold clusters is discussed. 36Pal, R.; Wang, L.-M.; Pei, Y.; Wang, L.-S.; Zeng, X. C.Unraveling the Mechanisms of O 2 Activation by Size-Selected Gold Clusters: Transition from Superoxo to Peroxo Chemisorption. J. Am. Chem. Soc. 2012, 134, 9438– 9445, DOI: 10.1021/ja302902p Google Scholar36 Unraveling the Mechanisms of O2 Activation by Size-Selected Gold Clusters: Transition from Superoxo to Peroxo Chemisorption Pal, Rhitankar; Wang, Lei-Ming; Pei, Yong; Wang, Lai-Sheng; Zeng, Xiao Cheng Journal of the American Chemical Society (2012), 134 (22), 9438-9445 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) The activation of dioxygen is a key step in CO oxidn. catalyzed by gold nanoparticles. It is known that small gold cluster anions with even-numbered atoms can molecularly chemisorb O2 via one-electron transfer from Aun- to O2, whereas clusters with odd-numbered atoms are inert toward O2. Here we report spectroscopic evidence of two modes of O2 activation by the small even-sized Aun- clusters: superoxo and peroxo chemisorption. Photoelectron spectroscopy of O2Au8- revealed two distinct isomers, which can be converted from one to the other depending on the reaction time. Ab initio calcns. show that there are two close-lying mol. O2-chemisorbed isomers for O2Au8-: the lower energy isomer involves a peroxo-type binding of O2 onto Au8-, while the superoxo chemisorption is a slightly higher energy isomer. The computed detachment transitions of the superoxo and peroxo species are in good agreement with the exptl. observation. The current work shows that there is a superoxo to peroxo chemisorption transition of O2 on gold clusters at Au8-: O2Aun- (n = 2, 4, 6) involves superoxo binding and n = 10, 12, 14, 18 involves peroxo binding, whereas the superoxo binding re-emerges at n = 20 due to the high symmetry tetrahedral structure of Au20, which has a very low electron affinity. Hence, the two-dimensional (2D) Au8- is the smallest anionic gold nanoparticle that prefers peroxo binding with O2. At Au12-, although both 2D and 3D isomers coexist in the cluster beam, the 3D isomer prefers the peroxo binding with O2. 37Mills, G.; Gordon, M. S.; Metiu, H.The Adsorption of Molecular Oxygen on Neutral and Negative Au n Clusters (n = 2–5). Chem. Phys. Lett. 2002, 359, 493– 499, DOI: 10.1016/s0009-2614(02)00746-7 Google Scholar37 The adsorption of molecular oxygen on neutral and negative Aun clusters (n=2-5) Mills, Greg; Gordon, Mark S.; Metiu, Horia Chemical Physics Letters (2002), 359 (5,6), 493-499 CODEN: CHPLBC; ISSN:0009-2614. (Elsevier Science B.V.) We use d. functional theory to examine the binding of O2 to Aun and Aun- clusters. O2 binds more strongly to clusters having an odd no. of electrons than to those with an even no. A second O2 mol. binds more weakly than the first. 38Yoon, B.; Häkkinen, H.; Landman, U.Interaction of O 2 with Gold Clusters: Molecular and Dissociative Adsorption. J. Phys. Chem. A 2003, 107, 4066– 4071, DOI: 10.1021/jp027596s Google Scholar38 Interaction of O2 with Gold Clusters: Molecular and Dissociative Adsorption Yoon, Bokwon; Haekkinen, Hannu; Landman, Uzi Journal of Physical Chemistry A (2003), 107 (20), 4066-4071 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The interaction of O2 with charged and neutral gold clusters, AuN- and AuN contg. up to eight atoms, investigated through the use of d. functional theory with generalized gradient corrections and scalar relativistic pseudopotentials, exhibits a pronounced sensitivity to the cluster size and to its charge state. Mol. adsorption is found to be favorable regardless of the charge state for clusters with N ≤ 3, while dissociative adsorption is favored for larger clusters. The dissocn. process involves a significant barrier (1 eV or more), and it is predicted to result in a large structural distortion in the host gold cluster. The interaction energy is largest for the anionic gold clusters where it exhibits an odd-even alternation as a function of the no. of gold atoms, with the maxima occurring for AuNO2- complexes with even N. The mol. bonding mechanism to these complexes involves charge transfer to the oxygen mol. with a concomitant activation of the O-O bond to a superoxo-like state. 39Wang, Y.; Gong, X. G.First-principles study of interaction of cluster Au32 with CO, H2, and O2. J. Chem. Phys. 2006, 125, 124703, DOI: 10.1063/1.2352749 Google Scholar39 First-principles study of interaction of cluster Au32 with CO, H2, and O2 Wang, Yao; Gong, X. G. Journal of Chemical Physics (2006), 125 (12), 124703/1-124703/12 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) First-principles calcns. are performed to study the interaction of cluster Au32 with small mols., such as CO, H2, and O2. The cagelike Au32(Ih) shows a higher chem. inertness than the amorphous Au32(C1) with respect to the interaction with small mols. CO, H2, and O2. H2 can only be phys. adsorbed on Au32(Ih), while it can be dissociatively chemisorbed on Au32(C1). Although CO can be chem. adsorbed on Au32(Ih) and Au32(C1) with one electron transferred from Au32 to the antibonding π orbit of CO, it is bound more strongly on Au32(C1) than on Au32(Ih). Spin polarized and spin nonpolarized calcns. result almost identical ground state structures of Au32(Ih)-O2 and Au32(C1)-O2, in which O2 is dissociatively chemisorbed. 40Jena, N. K.; Chandrakumar, K. R. S.; Ghosh, S. K.Beyond the Gold-Hydrogen Analogy: Doping Gold Cluster with H-atom-O 2 Activation and Reduction of the Reaction Barrier for CO Oxidation. J. Phys. Chem. Lett. 2011, 2, 1476– 1480, DOI: 10.1021/jz2006435 Google Scholar40 Beyond the Gold-Hydrogen Analogy: Doping Gold Cluster with H-atom-O2 Activation and Reduction of the Reaction Barrier for CO Oxidation Jena, Naresh K.; Chandrakumar, K. R. S.; Ghosh, Swapan K. Journal of Physical Chemistry Letters (2011), 2 (12), 1476-1480 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) The gold-hydrogen analogy is well established in terms of analogous chem. exhibited by gold and hydrogen atoms. A step beyond this analogy is demonstrated here by showing that the replacement of a Au atom with a H-atom in a gold cluster can lead to considerable enhancement in its reactivity. The present computational study reveals that while the closed-shell neutral pristine gold cluster Au8 can bind only weakly with an O2 mol., its doped counterpart Au7H shows remarkable enhancement in its binding with mol. oxygen. It also shows an increment in O-O bond length and a red shift in vibrational frequency, indicating the activation of the O2 mol. Furthermore, the H-doped gold cluster reduces the barrier height for the environmentally important CO oxidn. reaction as compared to the pristine cluster. These observations may spur further studies in the design of a cost-effective and efficient catalyst by doping gold clusters with a cheaper element like hydrogen. 41Gao, Y.; Zeng, X. C.Water-Promoted O 2 Dissociation on Small-Sized Anionic Gold Clusters. ACS Catal. 2012, 2, 2614– 2621, DOI: 10.1021/cs300383p Google Scholar41 Water-Promoted O2 Dissociation on Small-Sized Anionic Gold Clusters Gao, Yi; Zeng, Xiao Cheng ACS Catalysis (2012), 2 (12), 2614-2621 CODEN: ACCACS; ISSN:2155-5435. (American Chemical Society) Although thermodynamically O2 favors dissociative adsorption over mol. adsorption on small-sized anionic gold clusters (except Au2-), O2 dissocn. is unlikely to proceed under ambient conditions because of the high activation energy barrier (>2.0 eV). Here, we present a systematic theor. study of reaction pathways for the O2 dissocn. on small-sized anionic gold nanoclusters Aun- (n = 1-6) with and without involvement of a water mol. The d. functional theory calcns. indicate that the activation barriers from the mol. adsorption state of O2 to dissociative adsorption can be significantly lowered with the involvement of a H2O mol. Once the O2 dissocs. on small-size gold clusters, at. oxygen is readily available for other reactions, such as the CO oxidn., on the surface of gold clusters. This theor. study supports previous exptl. evidence that H2O can be used to activate O2, which suggests an alternative way to exploit catalytic capability of gold clusters for oxidn. applications. 42Lyalin, A.; Taketsugu, T.Reactant-Promoted Oxygen Dissociation on Gold Clusters. J. Phys. Chem. Lett. 2010, 1, 1752– 1757, DOI: 10.1021/jz100503j Google Scholar42 Reactant-Promoted Oxygen Dissociation on Gold Clusters Lyalin, Andrey; Taketsugu, Tetsuya Journal of Physical Chemistry Letters (2010), 1 (12), 1752-1757 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) The mol. and dissociative adsorption of O2 on pure odd-size Aun (n = 1, 3, 5, 7, 9) clusters and Aun with a coadsorbed C2H4 mol. has been studied theor. using d. functional theory. It is demonstrated that catalytic activation of the adsorbed O2 on pure gold clusters in the considered size range cannot lead to O2 dissocn. Coadsorption of C2H4, however, involves an extra charge transfer from the gold cluster to O2, energetically promoting oxygen dissocn. Our finding indicates that O2 dissocn. on the surface of small gold clusters is sensitive to the presence of other adsorbates, including the reactant mol. itself. This effect can be particularly important for understanding the mechanism of catalytic oxidn. on gold clusters. 43Bürgel, C.; Reilly, N. M.; Johnson, G. E.; Mitric, R.; Kimble, M. L.; Castleman, A. W., Jr.; Bonačić-Koutecký, V.Influence of Charge State on the Mechanism of CO Oxidation on Gold Clusters. J. Am. Chem. Soc. 2008, 130, 1694– 1698, DOI: 10.1021/ja0768542 Google Scholar43 Influence of Charge State on the Mechanism of CO Oxidation on Gold Clusters Buergel, Christian; Reilly, Nelly M.; Johnson, Grant E.; Mitric, Roland; Kimble, Michele L.; Castleman, A. W., Jr.; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2008), 130 (5), 1694-1698 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) We present results from our joint exptl. and theor. study of the reactivity of anionic and cationic gold oxide clusters toward CO, focusing on the role of at. oxygen, different charge states, and mechanisms for oxidn. We show that anionic clusters react by an Eley-Rideal-like mechanism involving the preferential attack of CO on oxygen rather than gold. In contrast, the oxidn. of CO on cationic gold oxide clusters can occur by both an Eley-Rideal-like and a Langmuir-Hinshelwood-like mechanism at multiple collision conditions as a result of the high adsorption energy of two CO mols. This large energy of CO adsorption on cationic gold oxide clusters is the driving force for the CO oxidn. Therefore, in the presence of cationic gold species at high pressures of CO, the oxidn. reaction is self-promoting (i.e., the oxidn. of one CO mol. is promoted by the binding of a second CO). Our findings provide new insight into the role of charge state in gold-cluster-based nanocatalysis. 44Tang, D.; Hu, C.DFT Insight into CO Oxidation Catalyzed by Gold Nanoclusters: Charge Effect and Multi-State Reactivity. J. Phys. Chem. Lett. 2011, 2, 2972– 2977, DOI: 10.1021/jz201290x Google Scholar44 DFT Insight into CO Oxidation Catalyzed by Gold Nanoclusters: Charge Effect and Multi-State Reactivity Tang, Dianyong; Hu, Changwei Journal of Physical Chemistry Letters (2011), 2 (23), 2972-2977 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) All-electron relativistic d. functional theory (DFT) calcns. on CO oxidn. catalyzed by 55-atom gold nanoclusters (Au55) have revealed that all of neutral, anionic, and cationic Au55 clusters exhibit stronger catalytic efficiency toward CO oxidn. at low temps. High spin energy profiles and spin crossing play a noticeably important role in CO oxidn. catalyzed by gold nanoclusters. Our findings provide new insight into the charge and spin-state effects of gold nanoclusters within gold nanocatalysts. 45Yoon, B.; Häkkinen, H.; Landman, U.; Wörz, A.; Antonietti, J.; Abbet, S.; Judai, K.; Heiz, U.Charging Effects on Bonding and Catalyzed Oxidation of CO on Au 8 Clusters on MgO. Science 2005, 307, 403– 407, DOI: 10.1126/science.1104168 Google Scholar45 Charging Effects on Bonding and Catalyzed Oxidation of CO on Au8 Clusters on MgO Yoon, Bokwon; Haekkinen, Hannu; Landman, Uzi; Woerz, Anke S.; Antonietti, Jean-Marie; Abbet, Stephane; Judai, Ken; Heiz, Ueli Science (Washington, DC, United States) (2005), 307 (5708), 403-407 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Gold octamers (Au8) bound to oxygen-vacancy F-center defects on Mg(001) are the smallest clusters to catalyze the low-temp. oxidn. of CO to CO2, whereas clusters deposited on close-to-perfect magnesia surfaces remain chem. inert. Charging of the supported clusters plays a key role in promoting their chem. activity. IR measurements of the stretch vibration of CO adsorbed on mass-selected gold octamers soft-landed on MgO(001) with coadsorbed O2 show a red shift on an F-center-rich surface with respect to the perfect surface. The expts. agree with quantum ab initio calcns. that predict that a red shift of the C-O vibration should arise via electron back-donation to the CO antibonding orbital. 46Baishya, S.; Deka, R. C.Catalytic Activities of Au 6, Au 6-, and Au 6+ Clusters for CO Oxidation: A density functional study. Int. J. Quantum Chem. 2014, 114, 1559– 1566, DOI: 10.1002/qua.24725 Google Scholar46 Catalytic Activities of Au6, Au6-, and Au6+ Clusters for CO oxidation: A density functional study Baishya, Subhi; Deka, Ramesh C. International Journal of Quantum Chemistry (2014), 114 (22), 1559-1566 CODEN: IJQCB2; ISSN:0020-7608. (John Wiley & Sons, Inc.) We have studied the CO oxidn. over neutral, anionic, and cationic gold hexamer clusters using d. functional theory which elucidates the effect of cluster charge state on the catalytic activity. Herein, we have considered the conventional bimol. Langmuir-Hinshelwood mechanism with coadsorbed CO and O2 at the neighboring sites in all the clusters. Among the three clusters, Au 6- entails lower barriers during the various steps of the oxidn. mechanism. The stability of all the species including the transition states with respect to the interacting species in Au 6- indicates no thermal activation. Our study suggests better catalytic activity of Au 6- as compared to the neutral and cationic counterparts. 47Torres, M. B.; Fernández, E. M.; Balbás, L. C.Theoretical Study of Oxygen Adsorption on Pure Au n+1+ and Doped MAu n+ Cationic Gold Clusters for M = Ti, Fe and n = 3–7. J. Phys. Chem. A 2008, 112, 6678– 6689, DOI: 10.1021/jp800247n Google Scholar47 Theoretical Study of Oxygen Adsorption on Pure Aun+1+ and Doped MAun+ Cationic Gold Clusters for M = Ti, Fe and n = 3-7 Torres, M. Begona; Fernandez, Eva M.; Balbas, Luis C. Journal of Physical Chemistry A (2008), 112 (29), 6678-6689 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) A comparative study of the adsorption of an O2 mol. on pure Aun+1+ and doped MAun+ cationic Au clusters for n = 3-7 and M = Ti, Fe is presented. The simultaneous adsorption of 2 oxygen atoms also was studied. This work was performed by 1st principles calcns. based on norm-conserving pseudo-potentials and numerical basis sets. For pure Au4+, Au6+, and Au7+ clusters, the O2 mol. is adsorbed preferably on top of low coordinated Au atoms, with an adsorption energy smaller than 0.5 eV. Instead, for Au5+ and Au8+, bridge adsorption sites are preferred with adsorption energies of 0.56 and 0.69 eV, resp. The ground-state geometry of Aun+ is almost unperturbed after O2 adsorption. The electronic charge flows towards O2 when the mol. is adsorbed in bridge positions and towards the Au cluster when O2 is adsorbed on top of Au atoms, and both the adsorption energy and the O-O bond length of adsorbed oxygen increase when the amt. of electronic charge on O2 increases. However, the authors studied the adsorption of an O2 mol. on doped MAun+ clusters, giving (MAunO2+)ad complexes with different equil. configurations. The highest adsorption energy was obtained when both atoms of O2 bind on top of the M impurity, and it is larger for Ti doped clusters than for Fe doped clusters, showing an odd-even effect trend with size n, which is opposite for Ti as compared to Fe complexes. For those adsorption configurations of (MAunO2+)ad involving only Au sites, the adsorption energy is similar to or smaller than that for similar configurations of Aun+1O2+ complexes. However, the highest adsorption energy of (MAunO2+)ad is higher than that for (Aun+1O2+)ad by a factor of ∼4.0 (1.2) for M = Ti (M = Fe). The trends with size n are rationalized in terms of O-O and O-M bond distances, as well as charge transfer between oxygen and cluster substrates. The spin multiplicity of those (MAunO2+)ad complexes with the highest O2 adsorption energy is a max. (min.) for M = Fe (Ti), corresponding to parallel (anti-parallel) spin coupling of MAun+ clusters and O2 mols. Finally, the authors obtained the min. energy equil. structure of (AunO2+)dis and (MAunO2+)dis contg. 2 sepd. O atoms bonded at different sites of Aun+ and MAun+ clusters, resp. For (MAunO2+)dis, the equil. configuration with the highest adsorption energy is stable against sepn. in MAun+ and O2 fragments, resp. Instead, for (AunO2+)dis, only the complex n = 6 is stable against sepn. in Aun+ and O2 fragments. The max. sepn. energy of (MAunO2+)dis is higher than the O2 adsorption energy of (MAunO2+)ad complexes by factors of ∼1.6 (2.5), 1.6 (1.7), 1.5 (2.4), 1.5 (1.3), and 1.6 (1.8) for M = Ti (Fe) complexes in the range n = 3-7, resp. 48Fernández, E. M.; Torres, M. B.; Balbás, L. C.Theoretical Study of the Coadsorption of CO and O 2 on Doped Cationic Gold Clusters MAu n+ (M = Ti, Fe, Au; n = 1, 6, 7). Eur. Phys. J. D 2009, 52, 135– 138, DOI: 10.1140/epjd/e2009-00010-4 Google Scholar48 Theoretical study of the coadsorption of CO and O2 on doped cationic gold clusters MAu+n (M = Ti, Fe, Au; n = 1, 6, 7) Fernandez, E. M.; Torres, M. B.; Balbas, L. C. European Physical Journal D: Atomic, Molecular, Optical and Plasma Physics (2009), 52 (1-3), 135-138 CODEN: EPJDF6; ISSN:1434-6060. (EDP Sciences) We present a study based on first-principles calcns. of the adsorption of CO on selected equil. configurations of MAunO+2 (M = Ti, Fe; n = 1, 6, 7) complexes resulting from the adsorption of O2 on doped cationic gold clusters MAu+n. Empirical rules for the formation of CO2-MAunO+ complexes are outlined. The desorption energy of CO2 is calcd. The adsorption of a second CO mol. on the residual MAunO+ complex leads in some cases to the formation of CO2-MAu+n. The desorption of a second CO2 mol. brings back to the initial doped gold cluster MAu+n. 49Manzoor, D.; Krishnamurty, S.; Pal, S.Contriving a Catalytically Active Structure from an Inert Conformation: A Density Functional Investigation of Al, Hf, and Ge Doping of Au20Tetrahedral Clusters. J. Phys. Chem. C 2016, 120, 19636– 19641, DOI: 10.1021/acs.jpcc.6b04305 Google Scholar49 Contriving a Catalytically Active Structure from an Inert Conformation: A Density Functional Investigation of Al, Hf, and Ge Doping of Au20 Tetrahedral Clusters Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Journal of Physical Chemistry C (2016), 120 (35), 19636-19641 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) The transformation of an inert structure into a catalytically active conformation has immense potential from a structural engineering point of view. In the present work, we explored methods of achieving such a transformation through a d.-functional-based study. Au20 is known be one of the most stable and catalytically inert gold clusters, with a well-known tetrahedral structure. We demonstrate the transformation of the inert Au20 conformation into a highly active catalytic cluster through selective doping with Hf and Ge atoms. Depending on the dopant, the inert tetrahedral Au20 cluster evolves into either an endohedral or a hollow-cage-like conformation. The structural changes are manifested in the catalytic properties as well, with the result that the transformed doped cages exhibit extremely low activation barriers for the environmentally important CO oxidn. reaction as compared to reported inert Au20 cluster. The activation barriers for CO oxidn. are particularly low ( < 0.12 eV) when germanium is directly involved in the CO oxidn. Thus, the current work highlights the importance of engineering structural properties of metal nanoclusters with the help of heteroatom dopants for future applications in efficient catalysis. 50Manzoor, D.; Krishnamurty, S.; Pal, S.Endohedrelly Doped Gold Cages: Efficient Catalysts for Oxygen Activation and CO Oxidation. Phys. Chem. Chem. Phys. 2016, 18, 7068– 7074, DOI: 10.1039/c5cp05624a Google Scholar50 Endohedrally doped gold nanocages: efficient catalysts for O2 activation and CO oxidation Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Physical Chemistry Chemical Physics (2016), 18 (10), 7068-7074 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) Gold nanocages are the most attractive catalytic materials as all the atoms in the cage type clusters reside on the surface, making them available for chemisorption by reacting mols. Due to a hollow space at the center, their chem. and catalytic properties can be tuned effectively and easily by endohedral doping. While a significant exptl. and theor. understanding is currently available on the structural and electronic properties of doped gold cages, very little information is available on their reactivity and catalytic behavior. In the present work, with the help of d. functional theory calcns. we demonstrate that endohedral doping leads to a notable increase in the binding energy of mol. oxygen on the gold nanocages. The enhancement in the O2 binding energy on the doped gold cages is also confirmed by a significant decrease in the Au-O and an increase in the O-O bond lengths, corroborated by a red shift (∼250 cm-1) in the O-O stretching frequency as compared to the pristine cage. Furthermore, interestingly, the doped gold cages show very low activation barriers for the environmentally important CO oxidn. reaction as compared to the pristine gold cage. Importantly, the decrease in the barrier height is comparatively greater for the rate limiting step of O-O-C-O intermediate formation and as a result the CO oxidn. is expected to be more facile on the doped gold cages. Thus, the current study highlights the role of heteroatom doping in imparting new chem. and catalytic properties to gold cages and is expected to spur further research in the design of efficient gold nanocatalysts. 51Manzoor, D.; Krishnamurty, S.; Pal, S.Effect of Silicon Doping on the Reactivity and Catalytic Activity of Gold Clusters. J. Phys. Chem. C 2014, 118, 7501– 7507, DOI: 10.1021/jp501611t Google Scholar51 Effect of Silicon Doping on the Reactivity and Catalytic Activity of Gold Clusters Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Journal of Physical Chemistry C (2014), 118 (14), 7501-7507 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Doping is known to be an excellent and simple way of catalyst design. Although notable progress has been made in understanding the reactivity and catalytic activity of gas-phase and supported gold clusters, very few studies have been carried out on the doped gold clusters. In the present work, we have carried out d. functional theory calcns. to investigate the effect of silicon doping on the reactivity and catalytic activity of gold nanoclusters. The present work particularly focuses on the adsorption and activation of mol. oxygen on the pristine and silicon-doped gold clusters. The results confirm that the silicon-doped Au7Si cluster shows considerable binding and activation of the O2 mol. in comparison to the pristine Au8 cluster as reflected in the relevant geometrical parameters (O-O and Au-O bond lengths) and O-O stretching frequency. However, silicon doping has no contrasting effect on the reactivity and catalytic activity of the Au7 cluster. In addn. to the stronger binding and activation of the O2 mol., the doped Au7Si cluster leads to a significant redn. in the activation barrier (0.57 eV) for the environmentally important CO oxidn. reaction in contrast to the catalytically inactive pristine Au8 cluster (1.22 eV). Thus, our results highlight the crit. role of doping foreign impurities for future endeavors in the field of gold nanocatalysis. 52Molina, L. M.; Hammer, B.Some Recent Theoretical Advances in the Understanding of the Catalytic Activity of Au. Appl. Catal., A 2005, 291, 21– 31, DOI: 10.1016/j.apcata.2005.01.050 Google Scholar52 Some recent theoretical advances in the understanding of the catalytic activity of Au Molina, L. M.; Hammer, B. Applied Catalysis, A: General (2005), 291 (1-2), 21-31 CODEN: ACAGE4; ISSN:0926-860X. (Elsevier B.V.) A review; we present a small review of recent d.-functional-theory (DFT) studies of the reactivity towards CO oxidn. of supported Au nano-particles. The possible structure of the periphery of the interface between a Au particle and an oxide support is discussed. A certain structure, in which low coordinated Au atoms are overhanging the support without binding directly to the oxide atoms, is argued to be prototypical of medium-sized Au particles. This structure is shown to be particularly active both at the edges and at the corners of Au particles. Examples from the literature of Au systems supported on MgO(1 0 0) and rutile-TiO2(1 1 0) are reviewed and new data are given for the reactivity of facet, edge, and corner sites of a Au34 cluster supported on MgO(1 0 0). On the non-reducible oxide support, MgO(1 0 0), the CO oxidn. is found to occur via CO adsorption to the Au particles and subsequent CO-promoted O2 capture and formation of a CO·O2 reaction intermediate complex. On the reducible oxide support, TiO2(1 1 0), the O2 is found to adsorb independently of the CO. However, on this support, the reaction still proceeds via CO·O2 formation rather than via O2 dissocn. 53Liu, C.; Tan, Y.; Lin, S.; Li, H.; Wu, X.; Li, L.; Pei, Y.; Zeng, X. C.CO Self-Promoting Oxidation on Nanosized Gold Clusters: Triangular Au3 Active Site and CO Induced O-O Scission. J. Am. Chem. Soc. 2013, 135, 2583– 2595, DOI: 10.1021/ja309460v Google Scholar53 CO Self-Promoting Oxidation on Nanosized Gold Clusters: Triangular Au3 Active Site and CO Induced O-O Scission Liu, Chunyan; Tan, Yingzi; Lin, Sisi; Li, Hui; Wu, Xiaojun; Li, Lei; Pei, Yong; Zeng, Xiao Cheng Journal of the American Chemical Society (2013), 135 (7), 2583-2595 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) We have systematically studied the CO oxidn. on various nanosized gold clusters with sizes ranging from 0.3 to 0.8 nm on the basis of d. functional theory (DFT) calcns. A hitherto unreported trimol. Langmuir-Hinshelwood (LH) mechanism is proposed, which offers new insights into the fundamental mechanism for CO oxidn. on nanosized gold clusters. Specifically, we find that the coadsorbed CO mol. at a unique triangular Au3 active site can act as a promoter for the scission of an O-O bond, leading to the spontaneous formation (due to extremely low energy barrier) of two CO2 mols. as product. The key step to the O-O bond scission in the OCOO intermediate is significantly accelerated due to the electrophilic attack of the coadsorbed neighboring CO mol. on the triangular Au3 site. This new mechanism is called CO self-promoting oxidn., which can be visualized in real time from the trajectory of a Born-Oppenheimer mol. dynamics (BOMD) simulation. We also find that such CO self-promoting oxidn. is quite universal, as long as the triangular Au3 reaction site is available. This is demonstrated in two prototype metal oxide supported gold nanostructure systems: namely, Aun/MgO and bilayer-Au/TiO2. The coadsorbed CO can not only serve as a promoter for its own oxidn. but also promote other oxidn. reactions such as styrene oxidn. through expediting O-O scission on gold nanostructures. 54Manzoor, D.; Pal, S.Hydrogen Atom Chemisorbed Gold Clusters as Highly Active Catalysts for Oxygen Activation and CO Oxidation. J. Phys. Chem. C 2014, 118, 30057– 30062, DOI: 10.1021/jp510488v Google Scholar54 Hydrogen Atom Chemisorbed Gold Clusters as Highly Active Catalysts for Oxygen Activation and CO Oxidation Manzoor, Dar; Pal, Sourav Journal of Physical Chemistry C (2014), 118 (51), 30057-30062 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Using oxygen activation and carbon monoxide oxidn. as a probe, the authors have carried out d. functional theory (DFT) calcns. to study the effect of hydrogen atom chemisorption on the reactivity and catalytic activity of gold nanoclusters. The authors particularly focus on the closed shell neutral Aun (n = 2, 4, 6, 8) gold clusters which are reported to be chem. inert and catalytically less active. The hydrogen atom chemisorption activates the inert clusters, thereby leading to enhanced binding and activation of the O2 mol. A significant amt. of increase in the O-O and decrease in the Au-O bond lengths together with a pertinent red shift in the O-O stretching frequencies clearly indicates the stronger binding and activation of oxygen mol. in the case of AunH clusters. Further, results reveal that the hydrogen atom chemisorbed gold clusters are catalytically more active with low activation barriers for the CO oxidn. reaction compared to the parent less active closed shell gold clusters. Thus, in short the current theor. observation provides key inputs for enhancing the reactivity and catalytic activity of inert gold clusters through the adsorption of a small atom or a mol. 55Zhang, C.; Yoon, B.; Landman, U.Predicted Oxidation of CO Catalyzed by Au Nanoclusters on a Thin Defect-Free MgO Film Supported on a Mo(100) Surface. J. Am. Chem. Soc. 2007, 129, 2228– 2229, DOI: 10.1021/ja0684545 Google Scholar55 Predicted Oxidation of CO Catalyzed by Au Nanoclusters on a Thin Defect-Free MgO Film Supported on a Mo(100) Surface Zhang, Chun; Yoon, Bokwon; Landman, Uzi Journal of the American Chemical Society (2007), 129 (8), 2228-2229 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) With the use of first-principles simulations, low barrier CO oxidn. reactions are predicted to occur via a Langmuir-Hinshelwood or an Eley-Rideal mechanism, on two-dimensional gold nanocluster islands adsorbed on two-layer MgO films that are supported on Mo(100). Underlying the emergent catalytic activity, predicted to occur even in the absence of oxygen vacancy F-center defects, is the excess electronic charge at the gold cluster/magnesia interface originating from the penetration of metal states through the thin magnesia film. This excess charge stabilizes the planar structure of the gold cluster and activates the oxygen mols. adsorbed at the interfacial periphery of the gold islands. 56Molina, L. M.; Hammer, B.Active Role of Oxide Support during CO Oxidation at Au/MgO. Phys. Rev. Lett. 2003, 90, 206102, DOI: 10.1103/physrevlett.90.206102 Google Scholar56 Active Role of Oxide Support during CO Oxidation at Au/MgO Molina, L. M.; Hammer, B. Physical Review Letters (2003), 90 (20), 206102/1-206102/4 CODEN: PRLTAO; ISSN:0031-9007. (American Physical Society) The oxidn. of CO at MgO supported gold aggregates is studied by means of d. functional theory calcns. In addn. to serving as a structural promoter holding the gold particles, the supporting oxide also takes an active role in the bonding and activation of adsorbates bound to the gold. The oxide stabilizes a peroxolike reaction intermediate, CO·O2, and causes steric repulsion to CO. The most reactive site at Au/MgO appears where the gold shelters the MgO thereby creating a cavity where several low-coordinated Au atoms and Mg2+ cations from the substrate can interact simultaneously with an adsorbate. 57Harding, C.; Habibpour, V.; Kunz, S.; Farnbacher, A. N.-S.; Heiz, U.; Yoon, B.; Landman, U.Control and Manipulation of Gold Nanocatalysis: Effects of Metal Oxide Support Thickness and Composition. J. Am. Chem. Soc. 2009, 131, 538– 548, DOI: 10.1021/ja804893b Google Scholar57 Control and Manipulation of Gold Nanocatalysis: Effects of Metal Oxide Support Thickness and Composition Harding, Chris; Habibpour, Vahideh; Kunz, Sebastian; Farnbacher, Adrian Nam-Su; Heiz, Ueli; Yoon, Bokwon; Landman, Uzi Journal of the American Chemical Society (2009), 131 (2), 538-548 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Control and tunability of the catalytic oxidn. of CO by gold clusters deposited on MgO surfaces grown on molybdenum, Mo(100), to various thicknesses are explored through temp.-programmed reaction measurements on mass-selected 20-atom gold clusters and via first-principles d. functional theory calcns. Au20 was chosen because in the gas phase it is characterized as an extraordinarily stable tetrahedral-pyramidal structure. Dependencies of the catalytic activities and microscopic reaction mechanisms on the thickness and stoichiometry of the MgO films and on the dimensionalities and structures of the adsorbed gold clusters are demonstrated and elucidated. Langmuir-Hinshelwood mechanisms and reaction barriers corresponding to obsd. low- and high-temp. CO oxidn. reactions are calcd. and analyzed. These reactions involve adsorbed O2 mols. that are activated to a superoxo- or peroxo-like state through partial occupation of the antibonding orbitals. In some cases, we find activated, dissociative adsorption of O2 mols., adsorbing at the cluster peripheral interface with the MgO surface. The reactant CO mols. either adsorb on the MgO surface in the cluster proximity or bind directly to the gold cluster. Along with the oxidn. reactions on stoichiometric ultrathin MgO films, we also study reactions catalyzed by Au20 nanoclusters adsorbed on relatively thick defect-poor MgO films supported on Mo and on defect-rich thick MgO surfaces contg. oxygen vacancy defects. 58Baishya, S.; Deka, R. C.Hybrid Density Functional/Molecular Mechanics Studies on Activated Adsorption of Oxygen on Zeolite Supported Gold Monomer. J. Chem. Phys. 2011, 135, 244703, DOI: 10.1063/1.3667206 Google Scholar58 Hybrid density functional/molecular mechanics studies on activated adsorption of oxygen on zeolite supported gold monomer Baishya, Subhi; Deka, Ramesh C. Journal of Chemical Physics (2011), 135 (24), 244703/1-244703/7 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) D. functional theory calcns. on oxygen adsorption over gas phase and faujasite supported Au monomer has been studied using hybrid quantum mechanics/mol. mechanics method, surface integrated MO mol. mechanics implemented in GAMESS package. Three different oxidn. states of Au (0, +1, +3) and three different adsorption modes viz., top, bridge, and dissociative adsorption of oxygen have been considered in our calcns. Red shift in the νO-O value from that in gas phase O2 indicates activation of O2 upon adsorption over faujasite supported gold monomer. The activation of O2 is an important step in the catalytic oxidn. of CO. The presence of adsorbed O2 increases the interaction of the Au monomer with the faujasite support. In faujasite supported cationic Au monomer, O2 preferably remains bridge bonded to Au rather than being dissocd. (c) 2011 American Institute of Physics. 59García-Cruz, R.; Poulain, E.; Hernández-Pérez, I.; Reyes-Nava, J. A.; González-Torres, J. C.; Rubio-Ponce, A.; Olvera-Neria, O.Effect of Spin Multiplicity in O 2 Adsorption and Dissociation on Small Bimetallic AuAg Clusters. J. Phys. Chem. A 2017, 121, 6079– 6089, DOI: 10.1021/acs.jpca.7b01968 Google Scholar59 Effect of Spin Multiplicity in O2 Adsorption and Dissociation on Small Bimetallic AuAg Clusters Garcia-Cruz, Raul; Poulain, Enrique; Hernandez-Perez, Isaias; Reyes-Nava, Juan A.; Gonzalez-Torres, Julio C.; Rubio-Ponce, A.; Olvera-Neria, Oscar Journal of Physical Chemistry A (2017), 121 (32), 6079-6089 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) To dispose of at. oxygen, it is necessary the O2 activation; however, an energy barrier must be overcome to break the O-O bond. This work presents theor. calcns. of the O2 adsorption and dissocn. on small pure Aun and Agm and bimetallic AunAgm (n + m ≤ 6) clusters using the d. functional theory (DFT) and the zeroth-order regular approxn. (ZORA) to explicitly include scalar relativistic effects. The most stable AunAgm clusters contain a higher concn. of Au with Ag atoms located in the center of the cluster. The O2 adsorption energy on pure and bimetallic clusters and the ensuing geometries depend on the spin multiplicity of the system. For a doublet multiplicity, O2 is adsorbed in a bridge configuration, whereas for a triplet only one O-metal bond is formed. The charge transfer from metal toward O2 occupies the σO-O antibonding natural bond orbital, which weakens the oxygen bond. The Au3 (2A) cluster presents the lowest activation energy to dissoc. O2, whereas the opposite applies to the AuAg (3A) system. In the O2 activation, bimetallic clusters are not as active as pure Aun clusters due to the charge donated by Ag atoms being shared between O2 and Au atoms. 60Joshi, K.; Krishnamurty, S.Tailoring the structure and electronic properties of platinum and gold-platinum nanocatalysts towards enhanced O2 activation. New J. Chem. 2016, 40, 1336– 1346, DOI: 10.1039/c5nj02213d Google Scholar60 Tailoring the structure and electronic properties of platinum and gold-platinum nanocatalysts towards enhanced O2 activation Joshi, Krati; Krishnamurty, Sailaja New Journal of Chemistry (2016), 40 (2), 1336-1346 CODEN: NJCHE5; ISSN:1144-0546. (Royal Society of Chemistry) Geometric structure plays a crucial role in enhancing the catalytic activity of a material towards reactions such as O redn., MeOH oxidn. etc. Pt and Pt based bimetallic alloys of different morphologies such as nanorods, core-shell etc. were synthesized so far contributing to varying catalytic activity in fuel cells. However, the electronic factors contributing to enhanced catalytic activity of particular structure/shape/morphol. were not understood so far. In a 1st such computational study, the authors demonstrate this complex structure-property correlation on model six-atom Pt clusters in the context of O2 activation at the cathode of direct MeOH fuel cells. Since recent studies showed that Au-Pt bimetallic alloys exhibit superior catalytic activity towards O redn. reaction with respect to other bimetallic clusters, the authors have identified 4 distinct six-atom Pt cluster morphologies, viz., trigonal planar, distorted octahedral, normal octahedral and double square planar, and doped them sequentially with Au atoms until a ratio of 1:1 was obtained. Thus, clusters with varying conformations/shapes and bimetallic compns. are considered for potential O2 activation. The O-O bond stretching frequencies evaluate the degree of bond activation in O2 mols. The authors' study reveals that the Pt6 and AunPtm (n + m = 6) clusters with a double square planar shape exhibit superior catalytic activity towards O-O bond activation with respect to other six-atom conformations. Anal. of their electronic properties demonstrates that the amt. of charge transferred by the cluster to the O2 mol. and the effective overlap between the d-orbitals of the metal atom and the pi-MOs of O2 can be directly correlated with their catalytic activities. To validate these conclusions, the authors have further extended such anal. to various Ptm (m = 3-13) conformations, wherein, once again the frontier MO compn. is seen to play a predominant role in O activation. Thus, the present study unveils the electronic properties detg. the catalytic activity of a catalytic cluster towards O-O bond activation. 61Campbell, C.Bimetallic Surface Chemistry. Annu. Rev. Phys. Chem. 1990, 41, 775– 837, DOI: 10.1146/annurev.pc.41.100190.004015 Google Scholar61 Bimetallic surface chemistry Campbell, Charles T. Annual Review of Physical Chemistry (1990), 41 (), 775-837 CODEN: ARPLAP; ISSN:0066-426X. A review on adsorption, chemisorption, catalysis, alloy surface properties, and bimetallic catalysts. 283 Refs. 62Gao, F.; Goodman, D. W.Pd-Au Bimetallic Catalysts: Understanding Alloy Effects from Planar Models and (Supported) Nanoparticles. Chem. Soc. Rev. 2012, 41, 8009– 8020, DOI: 10.1039/c2cs35160a Google Scholar62 Pd-Au bimetallic catalysts: understanding alloy effects from planar models and (supported) nanoparticles Gao, Feng; Goodman, D. Wayne Chemical Society Reviews (2012), 41 (24), 8009-8020 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) A review. Pd-Au bimetallic catalysts often display enhanced catalytic activities and selectivities compared with Pd-alone catalysts. This enhancement is often caused by two alloy effects, i.e., ensemble and ligand effects. The ensemble effect is a diln. of surface Pd by Au. With increasing surface Au coverage, contiguous Pd ensembles disappear and isolated Pd ensembles form. For certain reactions, for example vinyl acetate synthesis, this effect is responsible for reaction rate enhancement via the formation of highly active surface sites, e.g., isolated Pd pairs. The disappearance of contiguous Pd ensembles also switches off side reactions catalyzed by these sites. This explains the selectivity increase of certain reactions, for example direct H2O2 synthesis. The ligand effects are electronic perturbation of Pd by Au. Via direct charge transfer or by affecting bond lengths, the ligand effects cause the Pd d band to be more filled, moving the d-band center away from the Fermi level. Both changes make Pd more "at. like" therefore binding reactants and products more weakly. For certain reactions, this eliminates a so-called "self-poisoning" effect and enhances activity/selectivity. 63Gao, F.; Wang, Y.; Goodman, D. W.Reaction Kinetics and Polarization-Modulation Infrared Reflection Absorption Spectroscopy (PM-IRAS) Investigation of CO Oxidation over Supported Pd–Au Alloy Catalysts. J. Phys. Chem. C 2010, 114, 4036– 4043, DOI: 10.1021/jp910896k Google Scholar63 Reaction Kinetics and Polarization-Modulation Infrared Reflection Absorption Spectroscopy (PM-IRAS) Investigation of CO Oxidation over Supported Pd-Au Alloy Catalysts Gao, Feng; Wang, Yilin; Goodman, D. Wayne Journal of Physical Chemistry C (2010), 114 (9), 4036-4043 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Pd-Au bimetallic model catalysts, synthesized either as thin films on Mo(110) or as nanoparticles on a TiO2 thin film, were used in CO oxidn. at both low (1 × 10-7 Torr) and elevated (8-16 Torr) CO pressures. Alloying with Au forms isolated Pd sites that are incapable of dissocg. O2. This inability causes reactivity loss at low reactant pressures. Moreover, alloying also electronically modifies surface sites and affects the reaction activation energy. At elevated pressures, Pd preferentially segregates to the surface to form contiguous Pd sites and CO oxidn. reactivity is regained. Under stoichiometric conditions and relatively low temps., Pd-Au alloys show superior reactivity compared with pure Pd due to more facile CO desorption (less CO inhibition). Under net oxidizing conditions, metallic Pd displays superior reactivity due to its higher capability of dissocg. O2. However, pure Pd oxidizes and loses reactivity more readily than do Pd-Au alloys. 64Xu, J.; White, T.; Li, P.; He, C.; Yu, J.; Yuan, W.; Han, Y.-F.Biphasic Pd–Au Alloy Catalyst for Low-Temperature CO Oxidation. J. Am. Chem. Soc. 2010, 132, 10398– 10406, DOI: 10.1021/ja102617r Google Scholar64 Biphasic Pd-Au Alloy Catalyst for Low-Temperature CO Oxidation Xu, Jing; White, Tim; Li, Ping; He, Chongheng; Yu, Jianguo; Yuan, Weikang; Han, Yi-Fan Journal of the American Chemical Society (2010), 132 (30), 10398-10406 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Low-temp. CO oxidn. over a compositional series of SiO2 fume-supported Pd/Au nano-alloy catalysts was assessed. Except for the pure metals, these materials invariably displayed biphasic sepn. into Pd- and Au-rich components. Performance was optimal for a bulk compn. catalyst, Pd4Au1, a mixt. of Pd90Au10 (72.5 at. percent) and Pd31Au69 (27.5 at. percent), which was remarkably active at 300° K and more stable than a pure Au catalyst. For bulk materials dominated by Pd (Pd:Au = 16:1; 8:1; 4:1), the Pd-rich alloy fraction frequently adopted hollow sphere or annular morphol.; Au-rich crystals were often multiply twinned. Quant. powder x-ray diffraction showed that under the synthesis conditions used, the Au soly. limit in Pd crystals was ∼12 at. percent; Pd was more sol. in Au (∼31 at. percent). This was consistent with XPS, which showed that Pd-rich alloy surfaces were enriched in Au relative to the bulk compn. In-situ Fourier transform IR spectra collected during CO oxidn. contained a new band at 2114 cm-1 (attributed to linear CO-Au/Au-Pd bonds) and reduced intensity of a band at 2090 cm-1 (from a linear CO-Pd bond) with escalating Au content, indicating the Pd sites became increasingly obscured by Au. High resoln. electron micrographs of Pd-rich alloys showed at. scale surface defects consistent with this interpretation. Results demonstrated that Au-contg. biphasic Pd nano-alloys may be highly durable alternatives for a range of catalytic reactions. 65Han, Y.; Wang, J.; Kumar, D.; Yan, Z.; Goodman, D.A Kinetic Study of Vinyl Acetate Synthesis over Pd-Based Catalysts: Kinetics of Vinyl Acetate Synthesis over Pd-Au/SiO 2 and Pd/SiO 2 Catalysts. J. Catal. 2005, 232, 467– 475, DOI: 10.1016/j.jcat.2005.04.001 Google Scholar65 A kinetic study of vinyl acetate synthesis over Pd-based catalysts: kinetics of vinyl acetate synthesis over Pd-Au/SiO2 and Pd/SiO2 catalysts Han, Y.-F.; Wang, J.-H.; Kumar, D.; Yan, Z.; Goodman, D. W. Journal of Catalysis (2005), 232 (2), 467-475 CODEN: JCTLA5; ISSN:0021-9517. (Elsevier) Kinetic studies of vinyl acetate (VA) synthesis were carried out for Pd (1.0 wt%)/SiO2 and Pd (1.0 wt%)-Au (0.5 wt%)/SiO2 catalysts, with highly dispersed metal particles (Pd and Pd-Au) characterized by X-ray diffraction and transmission electron microscopy-energy dispersive spectroscopy (TEM-EDS). The kinetics of the related CO oxidn. reaction were also explored. The kinetics of VA synthesis and CO oxidn. reactions proceed via a Langmuir-Hinshelwood mechanism. For Pd-only catalysts, dissociative adsorption of O2 is believed to be the rate-detg. step. This step is suppressed by adsorbed CO/C2H4 and gives rise to a neg. reaction order with respect to CO/C2H4 and a pos. order with respect to O2. However, the reaction mechanism was significantly modified by the addn. of Au to Pd, as indicated by the change in the reaction order with respect to CO/C2H4; in particular, the order with respect to C2H4 becomes pos. The modification of the mechanism for the Pd-Au catalyst correlates with the redn. in the concn. of Pd-Pd ensembles upon alloying with Au. By TEM-EDS, Au surface enrichment was detected for the Pd-Au alloy, and the interaction between Au and Pd leads to a formation of more active Pd ensembles, such as PdAu5 and PdAu6. The surface properties of Pd vs. Pd-Au catalysts were explored by CO/C2D4-TPD on thick films of Pd and Pd-Au. These studies indicate that the adsorption sites on Pd are significantly modified by Au; concurrently, the bonding of CO and C2D4 to Pd in the Pd-Au alloy also varied. As a result, the coverage of CO and C2H4 on the Pd-Au surface decreased markedly. The enhanced capacity of the Pd-Au surface for oxygen and/or the enhanced mobility of adsorbed oxygen under the reaction conditions are likely responsible for the unusually high reactivity of Pd-Au alloy catalysts for VA synthesis. 66Edwards, J. K.; Solsona, B.; N, E. N.; Carley, A. F.; Herzing, A. A.; Kiely, C. J.; Hutchings, G. J.Switching Off Hydrogen Peroxide Hydrogenation in the Direct Synthesis Process. Science 2009, 323, 1037– 1041, DOI: 10.1126/science.1168980 Google Scholar66 Switching Off Hydrogen Peroxide Hydrogenation in the Direct Synthesis Process Edwards, Jennifer K.; Solsona, Benjamin; Ntainjua N, Edwin; Carley, Albert F.; Herzing, Andrew A.; Kiely, Christopher J.; Hutchings, Graham J. Science (Washington, DC, United States) (2009), 323 (5917), 1037-1041 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Hydrogen peroxide (H2O2) is an important disinfectant and bleach and is currently manufd. from an indirect process involving sequential hydrogenation/oxidn. of anthaquinones. However, a direct process in which H2 and O2 are reacted would be preferable. Unfortunately, catalysts for the direct synthesis of H2O2 are also effective for its subsequent decompn., and this has limited their development. We show that acid pretreatment of a carbon support for gold-palladium alloy catalysts switches off the decompn. of H2O2. This treatment decreases the size of the alloy nanoparticles, and these smaller nanoparticles presumably decorate and inhibit the sites for the decompn. reaction. Hence, when used in the direct synthesis of H2O2, the acid-pretreated catalysts give high yields of H2O2 with hydrogen selectivities greater than 95%. 67Chen, M.; Kumar, D.; Yi, C. W.; Goodman, D. W.The Promotional Effect of Gold in Catalysis by Palladium-Gold. Science 2005, 310, 291– 293, DOI: 10.1126/science.1115800 Google Scholar67 The Promotional Effect of Gold in Catalysis by Palladium-Gold Chen, Mingshu; Kumar, Dheeraj; Yi, Cheol-Woo; Goodman, D. Wayne Science (Washington, DC, United States) (2005), 310 (5746), 291-293 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Acetoxylation of ethylene to vinyl acetate (VA) was used to investigate the mechanism of the promotional effect of Au in a Pd-Au alloy catalyst. The enhanced rates of VA formation for low Pd coverage relative to high Pd coverage on Au single-crystal surfaces demonstrate that the crit. reaction site for VA synthesis consists of two noncontiguous, suitably spaced, Pd monomers. The role of Au is to isolate single Pd sites that facilitate the coupling of crit. surface species to product, while inhibiting the formation of undesirable reaction byproducts. 68Perdew, J. P.; Burke, K.; Ernzerhof, M.Generalized Gradient Approximation Made Simple. Phys. Rev. Lett. 1996, 77, 3865, DOI: 10.1103/physrevlett.77.3865 Google Scholar68 Generalized gradient approximation made simple Perdew, John P.; Burke, Kieron; Ernzerhof, Matthias Physical Review Letters (1996), 77 (18), 3865-3868 CODEN: PRLTAO; ISSN:0031-9007. (American Physical Society) Generalized gradient approxns. (GGA's) for the exchange-correlation energy improve upon the local spin d. (LSD) description of atoms, mols., and solids. We present a simple derivation of a simple GGA, in which all parameters (other than those in LSD) are fundamental consts. Only general features of the detailed construction underlying the Perdew-Wang 1991 (PW91) GGA are invoked. Improvements over PW91 include an accurate description of the linear response of the uniform electron gas, correct behavior under uniform scaling, and a smoother potential. 69Frisch, M. J.; Trucks, G. W.; Schlegel, H. B.; Scuseria, G. E.; Robb, M. A.; Cheeseman, J. R.; Scalmani, G.; Barone, V.; Mennucci, B.; Petersson, G. A.; Gaussian 09, Revision A.1; Gaussian, Inc.: Wallingford, CT, 2009.Google ScholarThere is no corresponding record for this reference. 70Zanti, G.; Peeters, D.DFT Study of Bimetallic Palladium-Gold Clusters Pd n Au m of Low Nuclearities (n + m ≤ 14). J. Phys. Chem. A 2010, 114, 10345– 10356, DOI: 10.1021/jp1041298 Google Scholar70 DFT Study of Bimetallic Palladium-Gold Clusters PdnAum of Low Nuclearities (n + m ≤ 14) Zanti, Giuseppe; Peeters, Daniel Journal of Physical Chemistry A (2010), 114 (38), 10345-10356 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) Bimetallic palladium-gold PdnAum clusters of low nuclearity (n + m ≤ 14) are studied using the d. functional theory at B3LYP level with a Lanl2DZ pseudopotential to understand the evolution of various structural, electronic, and energetic properties as a function of size (n + m) and compn. (n/m) of the system. The potential energy surfaces have been explored for many different structures, and the min. obtained were then collected and used as a starting point for comparing the selected properties. Theor. results show a logical evolution of the properties depending on the size and the compn. of the system. Pdn clusters clearly prefer 3D structures while Aum clusters favor planar configurations. The geometry of the bimetallic PdnAum clusters mainly depends on their compn., i.e., clusters enriched in palladium atoms prefer 3D structures while increasing gold contents promotes planar configurations with deviation from planarity near Pd centers. Regarding the electronic properties, NBO anal. reveals that the unique closed-shell electronic structure of Pd atoms (4d10) requires a (4d → 5s) promotion to form stable bonds. In contrast, the half-occupied Au 6s AO implies effective Au-Au interaction and the electronic structure of Au atoms remains almost unchanged upon formation of bimetallic bonds. Consequently, clusters enriched in palladium atoms have spin multiplicities that increase with the cluster size while clusters enriched in gold atoms maintain the lowest possible spin multiplicity. Finally, the stability of these systems shows a synergic gain in cohesion for mixed PdnAum clusters compared to their monometallic Pdn and Aum counterparts. The maximal stabilization effect corresponds to n ≈ m, compns. for which the no. of mixed Pd-Au bonds is maximized. 71Pichugina, D. A.; Polynskaya, Y. G.; Kuz’menko, N. E.Spin and Structural Features in Oxygen Dissociation on Tetrahedral Ag 20 and Ag 19 Au Clusters. Phys. Chem. Chem. Phys. 2016, 18, 18033– 18044, DOI: 10.1039/c6cp01630h Google Scholar71 Spin and structural features of oxygen dissociation on tetrahedral Ag20 and Ag19Au clusters Pichugina, D. A.; Polynskaya, Y. G.; Kuz'menko, N. E. Physical Chemistry Chemical Physics (2016), 18 (27), 18033-18044 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) The spin-crossing mechanism of oxygen dissocn. on Ag20 and monodoped Ag19Au clusters was investigated via spin-polarized scalar-relativistic DFT calcns. using the PBE, TPSSh, M06L, mPBE, BLYP, OLYP, and B3LYP functionals. In particular, the singlet and triplet O-O bond rupture pathways at vertex-edge and facet-edge sites on the tetrahedral clusters were studied. The calcns. reveal that for the Ag20O2 and Ag19AuO2 complexes, the spin inversion from the triplet to singlet state occurs during the last step, which involves O-O bond rupture through a singlet transition state (TS). When spin crossing during oxygen dissocn. on the clusters is considered, the activation energies decrease by 10-29 kJ mol-1; however, they are still high due to the magic nature of the clusters and high vertical spin excitation energies. For these silver clusters, size effects based on the relationship between the TS structure and oxidn. activation energy were predicted. 72Dononelli, W.; Klüner, T.CO Adsorption and Oxygen Activation on Group 11 Nanoparticles–A Combined DFT and High Level CCSD(T) Study About Size Effects and Activation Processes. Faraday Discuss. 2018, 208, 105– 121, DOI: 10.1039/c7fd00225d Google Scholar72 CO adsorption and oxygen activation on group 11 nanoparticles - a combined DFT and high level CCSD(T) study about size effects and activation processes Dononelli, Wilke; Kluener, Thorsten Faraday Discussions (2018), 208 (Designing Nanoparticle Systems for Catalysis), 105-121 CODEN: FDISE6; ISSN:1359-6640. (Royal Society of Chemistry) The focus of this study lies in the activation of mol. oxygen and reaction with CO within d. functional theory (DFT) and high level CCSD(T) calcns. Therefore, we use M13 and M55 nanoparticles (NPs) and periodic M(321) surfaces as model systems and compare the catalytic activity of gold substrates to Ag and Cu based NP catalysts. In the first step, the adsorption energies of CO were compared for nanoparticles of different sizes for Au, Ag and Cu. The adsorption energies on M(321) and M55 NPs (M = Au, Ag, Cu) are virtually identical. For smaller M13 NPs the adsorption energies differ by ∼0.2 eV for Ag, ∼0.4 eV for Au, and ∼0.6 eV for Cu at the PBE level of theory. This can be explained by size effects, as the M13 NPs show a more mol.-like character. Presumably, CO binds more strongly to these very small NPs at the PBE level of theory. However, a benchmark calcn. in the framework of CCSD(T)-theory reveals an adsorption energy of CO on Au13 of -0.88 eV, comparable to the adsorption energies calcd. at the PBE level for Au55 and Au(321). For Au55, the adsorption energy calcd. at the CCSD(T) level is -0.85 eV. This is in perfect agreement with the PBE result. In addn. to adsorption energies, dissocn. barriers have been calcd. on M(321) surfaces. The dissocn. energies of O2 on coinage metal catalysts are high, so that direct CO-oxidn. reactions with mol. oxygen should be the dominant reaction mechanism compared to the dissocn. and reaction of CO and at. oxygen at least for silver and copper catalysts. Cited By Click to copy section link Section link copied! Citation Statements beta Smart citations byscite.aiinclude citation statements extracted from the full text of the citing article. The number of the statements may be higher than the number of citations provided by ACS Publications if one paper cites another multiple times or lower if scite has not yet processed some of the citing articles. Supporting Supporting 1 Mentioning Mentioning 21 Contrasting Contrasting 0 Explore this article's citation statements onscite.ai powered by This article is cited by 18 publications. Insha Anis, Mohd. 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Published July 25, 2019 Publication History Received 30 May 2019 Accepted 26 June 2019 Published online 25 July 2019 Published in issue 31 July 2019 Copyright © 2019 American Chemical Society. This publication is licensed under these Terms of Use. Request reuse permissions Article Views 1533 Altmetric - Citations 18 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts. The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated. Recommended Articles ### O 2 Activation by Metal Surfaces: Implications for Bonding and Reactivity on Heterogeneous Catalysts November 8, 2017 Chemical Reviews Matthew M. Montemore , Matthijs A. van Spronsen , Robert J. Madix ,and Cynthia M. Friend ADVERTISEMENT Figures References Supporting Info Abstract High Resolution Image Download MS PowerPoint Slide Figure 1 Figure 1. Lowest-energy structures of the Au 8, Pd 8, and Au n–1 Pd n (n = 1–7) clusters and their corresponding spin states. High Resolution Image Download MS PowerPoint Slide Figure 2 Figure 2. Most stable geometries of the O 2-adsorbed complexes of Au 8, Pd 8, and Au n–1 Pd n (n = 1–7) clusters. The spin state of the system is given below each O 2–cluster complex. All the geometrical parameters marked are in angstroms. High Resolution Image Download MS PowerPoint Slide Figure 3 Figure 3. (A) Calculated reaction pathways (in eV) for O–O bond rupture on the Au 8 and Pd 8 clusters in singlet (red) and triplet (green) spin states. The energies of the reactants, transition states, and products are with respect to the energy of oxygen molecules in the triplet spin state and the isolated cluster in the lowest energy spin state. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide Figure 4 Figure 4. (A) Calculated reaction pathways (in eV) for O–O bond rupture on Au 7 Pd 1, Au 5 Pd 3, Au 3 Pd 5, and Au 1 Pd 7 clusters in doublet (red) and quartet (green) spin states. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide Figure 5 Figure 5. (A) Calculated reaction pathways (in eV) for O–O bond rupture on Au 6 Pd 2, Au 4 Pd 4, and Au 2 Pd 6 clusters in singlet (red) and triplet (green) spin states. (B) Structures of the reactants, transition states, and products along the reaction coordinate in different spin states. High Resolution Image Download MS PowerPoint Slide References This article references 72 other publications. 1. 1Guo, Y.; Gu, D.; Jin, Z.; Du, P.-P.; Si, R.; Tao, J.; Xu, W.-Q.; Huang, Y.-Y.; Senanayake, S.; Song, Q.-S.; Jia, C.-J.; Schüth, F.Uniform 2 nm gold nanoparticles supported on iron oxides as active catalysts for CO oxidation reaction: structure–activity relationship. _Nanoscale_ 2015, _7_, 4920– 4928, DOI: 10.1039/c4nr06967f 1 Uniform 2 nm gold nanoparticles supported on iron oxides as active catalysts for CO oxidation reaction: structure-activity relationship Guo, Yu; Gu, Dong; Jin, Zhao; Du, Pei-Pei; Si, Rui; Tao, Jing; Xu, Wen-Qian; Huang, Yu-Ying; Senanayake, Sanjaya; Song, Qi-Sheng; Jia, Chun-Jiang; Schuth, Ferdi Nanoscale (2015), 7 (11), 4920-4928 CODEN: NANOHL; ISSN:2040-3372. (Royal Society of Chemistry) Uniform Au nanoparticles (∼2 nm) with narrow size-distribution (std. deviation: 0.5-0.6 nm) supported on both hydroxylated (Fe_OH) and dehydrated iron oxide (Fe_O) have been prepd. by either deposition-pptn. (DP) or colloidal-deposition (CD) methods. Different structural and textural characterizations were applied to the dried, calcined and used gold-iron oxide samples. Transmission electron microscopy (TEM) and high-resoln. TEM (HRTEM) showed high homogeneity in the supported Au nanoparticles. The ex situ and in situ X-ray absorption fine structure (XAFS) characterization monitored the electronic and short-range local structure of active gold species. The synchrotron-based in situ X-ray diffraction (XRD), together with the corresponding temp.-programmed redn. by hydrogen (H2-TPR), indicated a structural evolution of the iron-oxide supports, correlating to their reducibility. An inverse order of catalytic activity between DP (Au/Fe_OH < Au/Fe_O) and CD (Au/Fe_OH > Au/Fe_O) was obsd. Effective gold-support interaction results in a high activity for gold nanoparticles, locally generated by the sintering of dispersed Au atoms on the oxide support in the DP synthesis, while a hydroxylated surface favors the reactivity of externally introduced Au nanoparticles on Fe_OH support for the CD approach. This work reveals why differences in the synthetic protocol translate to differences in the catalytic performance of Au/FeOx catalysts with very similar structural characteristics in CO oxidn. 2. 2Liu, J.-X.; Filot, I. A. W.; Su, Y.; Zijlstra, B.; Hensen, E. J. M.Optimum Particle Size for Gold-Catalyzed CO Oxidation. _J. Phys. Chem. C_ 2018, _122_, 8327– 8340, DOI: 10.1021/acs.jpcc.7b12711 2 Optimum Particle Size for Gold-Catalyzed CO Oxidation Liu, Jin-Xun; Filot, Ivo A. W.; Su, Yaqiong; Zijlstra, Bart; Hensen, Emiel J. M. Journal of Physical Chemistry C (2018), 122 (15), 8327-8340 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) The structure sensitivity of Au-catalyzed CO oxidn. is presented by analyzing the dependence of CO oxidn. rate on particle size. Clusters with <14 Au atoms adopt a planar structure, whereas larger ones adopt a 3-dimensional structure. The CO and O2 adsorption properties depend strongly on particle structure and size. All of the reaction barriers relevant to CO oxidn. display linear scaling relations with CO and O2 binding strengths as main reactivity descriptors. Planar and 3-dimensional Au clusters exhibit different linear scaling relation due to different surface topols. and different coordination nos. of the surface atoms. From these linear scaling relations, 1st-principles microkinetics simulations were conducted to det. CO oxidn. rates and possible rate-detg. step of Au particles. Planar Au9 and 3-dimensional Au79 clusters present the highest CO oxidn. rates for planar and 3-dimensional clusters, resp. The planar Au9 cluster is much more active than the optimum Au79 cluster. A common feature of optimum CO oxidn. performance is the intermediate binding strengths of CO and O2, resulting in intermediate coverages of CO, O2, and O. Both these optimum particles present lower performance than max. Sabatier performance, indicating that there is sufficient room for improvement of Au catalysts for CO oxidn. 3. 3Adnan, R. H.; Golovko, V. B.; Golovko, B. V.Benzyl Alcohol Oxidation Using Gold Catalysts Derived from Au8 Clusters on TiO2. _Catal. Lett._ 2019, _149_, 449– 455, DOI: 10.1007/s10562-018-2625-8 3 Benzyl Alcohol Oxidation Using Gold Catalysts Derived from Au8 Clusters on TiO2 Adnan, Rohul H.; Golovko, Vladimir B. Catalysis Letters (2019), 149 (2), 449-455 CODEN: CALEER; ISSN:1011-372X. (Springer) Atomically-precise gold clusters have gained attraction in catalysis due to high fraction of low-coordinated gold atoms, unique structural geometry and ligand effect. Phosphine-ligated gold clusters offer an advantage in the light of the labile gold-phosphorous bond for easy ligand removal. Here, heterogeneous gold catalysts were prepd. by depositing atomically-precise phosphine-ligated gold clusters, Au8(PPh3)8(NO3)2 onto anatase-phase TiO2 nanoparticles. The catalysts were then calcined under two different conditions: O2 (Au8/TiO2:O2) and O2 followed by H2 (Au8/TiO2:O2-H2) at 200 °C, to dislodge phosphine ligands from the Au core. It was found that Au8/TiO2:O2-H2 catalyst showed a decent catalytic activity in benzyl alc. oxidn. while Au8/TiO2 and Au8/TiO2:O2 were completely inactive. Such results imply that small-size gold clusters (2-3 nm) alone do not always contribute to high catalytic activity of gold catalysts. It is suggested that the presence of NO3- species defines the catalytic activity of supported gold clusters in benzyl alc. oxidn. in the case of these catalysts and reinforces our initial claim in the previous work. 4. 4Coquet, R.; Howard, K. L.; Willock, D. J.Theory and simulation in heterogeneous gold catalysis. _Chem. Soc. Rev._ 2008, _37_, 2046– 2076, DOI: 10.1039/b707385m 4 Theory and simulation in heterogeneous gold catalysis Coquet, Rudy; Howard, Kara L.; Willock, David J. Chemical Society Reviews (2008), 37 (9), 2046-2076 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) This crit. review covers the application of quantum chem. to the burgeoning area of the heterogeneous oxidn. by Au. We focus on the most established reaction, the oxidn. of CO at low temp. The review begins with an overview of the methods available comparing the treatment of the electron-electron interaction and relativistic effects. The structure of Au particles and their interaction with oxide reviews is then discussed in detail. Calcns. of the adsorption and reaction of CO and O2 are then considered and results from isolated and supported Au clusters compared (155 refs.). 5. 5Turner, M.; Golovko, V. B.; Vaughan, O. P. H.; Abdulkin, P.; Berenguer-Murcia, A.; Tikhov, M. S.; Johnson, B. F. G.; Lambert, R. M.Selective oxidation with dioxygen by gold nanoparticle catalysts derived from 55-atom clusters. _Nature_ 2008, _454_, 981– 983, DOI: 10.1038/nature07194 5 Selective oxidation with dioxygen by gold nanoparticle catalysts derived from 55-atom clusters Turner, Mark; Golovko, Vladimir B.; Vaughan, Owain P. H.; Abdulkin, Pavel; Berenguer-Murcia, Angel; Tikhov, Mintcho S.; Johnson, Brian F. G.; Lambert, Richard M. Nature (London, United Kingdom) (2008), 454 (7207), 981-983 CODEN: NATUAS; ISSN:0028-0836. (Nature Publishing Group) Supported gold nanoparticles have excited much interest owing to their unusual and somewhat unexpected catalytic properties, but the origin of the catalytic activity is still not fully understood. Exptl. work on gold particles supported on a titanium dioxide (110) single-crystal surface has established a striking size threshold effect assocd. with a metal-to-insulator transition, with gold particles catalytically active only if their diams. fall below ∼3.5 nm. However, the remarkable catalytic behavior might also in part arise from strong electronic interaction between the gold and the titanium dioxide support. In the case of industrially important selective oxidn. reactions, explanation of the effectiveness of gold nanoparticle catalysts is complicated by the need for additives to drive the reaction, and/or the presence of strong support interactions and incomplete understanding of their possible catalytic role. Here we show that very small gold entities (∼1.4 nm) derived from 55-atom gold clusters and supported on inert materials are efficient and robust catalysts for the selective oxidn. of styrene by dioxygen. We find a sharp size threshold in catalytic activity, in that particles with diams. of ∼2 nm and above are completely inactive. Our observations suggest that catalytic activity arises from the altered electronic structure intrinsic to small gold nanoparticles, and that the use of 55-atom gold clusters may prove a viable route to the synthesis of robust gold catalysts suited to practical application. 6. 6Herzing, A. A.; Kiely, C. J.; Carley, A. F.; Landon, P.; Hutchings, G. J.Identification of active gold nanoclusters on iron oxide supports for CO oxidation. _Science_ 2008, _321_, 1331– 1335, DOI: 10.1126/science.1159639 6 Identification of Active Gold Nanoclusters on Iron Oxide Supports for CO Oxidation Herzing, Andrew A.; Kiely, Christopher J.; Carley, Albert F.; Landon, Philip; Hutchings, Graham J. Science (Washington, DC, United States) (2008), 321 (5894), 1331-1335 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Gold nanocrystals absorbed on metal oxides have exceptional properties in oxidn. catalysis, including the oxidn. of carbon monoxide at ambient temps., but the identification of the active catalytic gold species among the many present on real catalysts is challenging. We have used aberration-cor. scanning transmission electron microscopy to analyze several iron oxide-supported catalyst samples, ranging from those with little or no activity to others with high activities. High catalytic activity for carbon monoxide oxidn. is correlated with the presence of bilayer clusters that are ~0.5 nm in diam. and contain only ~10 gold atoms. The activity of these bilayer clusters is consistent with that demonstrated previously with the use of model catalyst systems. 7. 7Gao, Y.; Shao, N.; Bulusu, S.; Zeng, X. C.Effective CO Oxidation on Endohedral Gold-Cage Nanoclusters. _J. Phys. Chem. C_ 2008, _112_, 8234– 8238, DOI: 10.1021/jp801262v 7 Effective CO Oxidation on Endohedral Gold-Cage Nanoclusters Gao, Yi; Shao, Nan; Bulusu, Satya; Zeng, X. C. Journal of Physical Chemistry C (2008), 112 (22), 8234-8238 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Ab initio calcns. are performed to study the CO oxidn. on six endohedral gold-cage clusters (W@Au12, Nb@Au13, Zr@Au14, Sc@Au15, Ca@Au16, and Na@Au17). The calcns. suggest that three clusters (Nb@Au13, Zr@Au14, and Sc@Au15) can be very effective nanocatalysts. The reaction energy barriers are lower than those assocd. with either bare gold clusters or gold surfaces and are comparable to those assocd. with the support Au clusters. The effective CO oxidn. can be attributed to the charge transfer from the dopant to the gold cage, the low coordination no. for gold atoms on the cages, as well as the fluxionality of the cage. 8. 8Gao, Y.; Shao, N.; Pei, Y.; Zeng, X. C.Icosahedral Crown Gold Nanocluster Au 43 Cu 12 with High Catalytic Activity. _Nano Lett._ 2010, _10_, 1055– 1062, DOI: 10.1021/nl100017u 8 Icosahedral Crown Gold Nanocluster Au43Cu12 with High Catalytic Activity Gao, Yi; Shao, Nan; Pei, Yong; Zeng, Xiao Cheng Nano Letters (2010), 10 (3), 1055-1062 CODEN: NALEFD; ISSN:1530-6984. (American Chemical Society) Structural and catalytic properties of the gold alloy nanocluster, Au43Cu12, were assessed using a d.-functional method. In contrast to a pure Au55 nanocluster which exhibited a low-symmetry C1 structure, the 55-atom crown Au nanocluster exhibited a multi-shell structure, denoted by Au@Cu12@Au42, with the highest icosahedral group-symmetry. Also, d. functional calcns. suggested this geometric magic-no. nanocluster possessed comparable catalytic capability as a small-sized Au10 cluster for CO oxidn. due, in part, to their low-coordinated Au atoms on the vertexes. The gold alloy nanocluster also exhibited higher selectivity for styrene oxidn. than a bare Au(111) surface. 9. 9Gao, Y.; Shao, N.; Pei, Y.; Chen, Z.; Zeng, X. C.Catalytic Activities of Subnanometer Gold Clusters (Au 16-Au 18, Au 20 and Au 27-Au 35) for CO Oxidation. _ACS Nano_ 2011, _5_, 7818– 7829, DOI: 10.1021/nn201817b 9 Catalytic Activities of Subnanometer Gold Clusters (Au16-Au18, Au20, and Au27-Au35) for CO Oxidation Gao, Yi; Shao, Nan; Pei, Yong; Chen, Zhongfang; Zeng, Xiao Cheng ACS Nano (2011), 5 (10), 7818-7829 CODEN: ANCAC3; ISSN:1936-0851. (American Chemical Society) Using the CO oxidn. as a chem. probe, we perform a comprehensive ab initio study of catalytic activities of subnanometer gold clusters. Particular attention is placed on 12 different clusters in the size range of Au16-Au35, whose at. structures in the anionic state have been resolved from previous expts. Adsorption energies of a single CO or O2 mol. as well as coadsorption energies of both CO and O2 mols. on various distinctive surface sites of each anionic cluster and their neutral counterpart are computed. In general, the anionic clusters can adsorb CO and O2 more strongly than their neutral counterparts. The coadsorption energies of both CO and O2 mols. decrease as the size of gold clusters increases with the exception of Au34 (an electronic "magic-no." cluster). Besides the known factor of low coordination site, we find that a relatively small cone angle (<110°) assocd. with each surface site is another key geometric factor that can enhance the binding strength of CO and O2. For the subnanometer clusters, although the size effect can be important to the strength of CO adsorption, it is less important to the activation energy. Using Au34 as a prototype model, we show that strong CO and O2 adsorption sites tend to yield a lower reaction barrier for the CO oxidn., but they have little effect on the stability of the reaction intermediate. Our calcns. support the notion that CO and O2 adsorption energies on the gold clusters can be an effective indicator to assess catalytic activities of subnanometer gold clusters. This systematic study of the site- and size-dependent adsorption energies and reaction pathways enables a quant. assessment of the site-size-activity relationship for the CO oxidn. on subnanometer gold clusters. 10. 10Patil, N. S.; Uphade, B. S.; Jana, P.; Sonawane, R. S.; Bhargava, S. K.; Choudhary, V. R.Oxidation of Benzyl Alcohol to Benzaldehyde by _tert_-Butyl Hydroperoxide over Nanogold Supported on TiO2 and other Transition and Rare-Earth Metal Oxides. _Catal. Lett._ 2004, _94_, 89– 93, DOI: 10.1023/b:catl.0000019336.34270.5d 10 Epoxidation of styrene by anhydrous t-butyl hydroperoxide over Au/TiO2 catalysts Patil, N. S.; Uphade, B. S.; Jana, P.; Sonawane, R. S.; Bhargava, S. K.; Choudhary, V. R. Catalysis Letters (2004), 94 (1-2), 89-93 CODEN: CALEER; ISSN:1011-372X. (Kluwer Academic/Plenum Publishers) Nanosize gold deposited on TiO2, by homogeneous deposition-pptn. (HDP) using urea as the pptn. agent is an active/selective and reusable catalyst for the epoxidn. of styrene (at 82°) by anhyd. t-Bu hydroperoxide (TBHP). The activity and epoxide selectivity of the Au/TiO2 catalyst in the epoxidn. is increased with increasing the Au loading on the TiO2 support. The Au/TiO2catalyst prepd. by the deposition pptn. using sodium hydroxide as the pptn. agent (DP method) has much lower gold loading and also lower Au dispersion and consequently possesses lower epoxidn. activity as compared to that prepd. by the HDP method. The styrene oxide selectivity is not influenced very significantly by the catalyst prepn. method or by the Au loading (except at the very low Au loading). In the Au/TiO2 catalysts prepd. by both the methods, Au is found to exist in both the metallic (Au°) and cationic (Au3+) forms. 11. 11Deng, X.; Friend, C. M.Selective Oxidation of Styrene on an Oxygen-Covered Au(111). _J. Am. Chem. Soc._ 2005, _127_, 17178– 17179, DOI: 10.1021/ja0557031 11 Selective Oxidation of Styrene on an Oxygen-Covered Au(111) Deng, Xingyi; Friend, Cynthia M. Journal of the American Chemical Society (2005), 127 (49), 17178-17179 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) For the first time, we demonstrate olefin epoxidn. promoted by an extended Au surface. The oxidn. of styrene to styrene epoxide, benzoic acid, and benzeneacetic acid is promoted on Au(111) covered with 0.2 ML of oxygen atoms. The estd. selectivity for styrene epoxide formation is ∼53%. Total combustion to CO2 accounts for ∼20% of the styrene reaction. We propose that styrene epoxide, benzoic acid, and benzeneacetic acid are produced via two possible oxametallacycle intermediates. Our work demonstrates that extended Au is an effective material for olefin oxidn., which has implications for understanding the activity of nanoscale Au catalysts. 12. 12Hashmi, A. S. K.Homogeneous Gold Catalysts and Alkynes: A Successful Liaison. _Gold Bull._ 2003, _36_, 3– 9, DOI: 10.1007/bf03214859 12 Homogeneous gold catalysts and alkynes: a successful liaison Hashmi, A. Stephen K. Gold Bulletin (London, United Kingdom) (2003), 36 (1), 3-9 CODEN: GOBUFW; ISSN:1027-8591. (World Gold Council) A review. The use of the combination of homogeneous gold-catalysts and alkynes in org. synthesis is reviewed from its beginnings in C-N-bond formation to the newest developments in C-C-bond formation. The common basic principle of these reactions is discussed. Special attention is devoted to the question where the gold catalysts are superior to either other catalysts or more traditional synthetic approaches to the product mols. 13. 13Hughes, M. D.; Xu, Y.-J.; Jenkins, P.; McMorn, P.; Landon, P.; Enache, D. I.; Carley, A. F.; Attard, G. A.; Hutchings, G. J.; King, F.; Stitt, E. H.; Johnston, P.; Griffin, K.; Kiely, C. J.Tunable Gold Catalysts For Selective Hydrocarbon Oxidation Under Mild Conditions. _Nature_ 2005, _437_, 1132– 1135, DOI: 10.1038/nature04190 13 Tunable gold catalysts for selective hydrocarbon oxidation under mild conditions Hughes, Mathew D.; Xu, Yi-Jun; Jenkins, Patrick; McMorn, Paul; Landon, Philip; Enache, Dan I.; Carley, Albert F.; Attard, Gary A.; Hutchings, Graham J.; King, Frank; Stitt, E. Hugh; Johnston, Peter; Griffin, Ken; Kiely, Christopher J. Nature (London, United Kingdom) (2005), 437 (7062), 1132-1135 CODEN: NATUAS; ISSN:0028-0836. (Nature Publishing Group) Oxidn. is an important method for the synthesis of chem. intermediates in the manuf. of high-tonnage commodities, high-value fine chems., agrochems. and pharmaceuticals: but oxidns. are often inefficient. The introduction of catalytic systems using oxygen from air is preferred for 'green' processing. Gold catalysis is now showing potential in selective redox processes, particularly for alc. oxidn. and the direct synthesis of hydrogen peroxide. However, a major challenge that persists is the synthesis of an epoxide by the direct electrophilic addn. of oxygen to an alkene. Although ethene is epoxidized efficiently using mol. oxygen with silver catalysts in a large-scale industrial process, this is unique because higher alkenes can only be effectively epoxidized using hydrogen peroxide, hydroperoxides or stoichiometric oxygen donors. Here we show that nanocryst. gold catalysts can provide tunable active catalysts for the oxidn. of alkenes using air, with exceptionally high selectivity to partial oxidn. products (∼98%) and significant conversions. Our finding significantly extends the discovery by Haruta that nanocryst. gold can epoxidize alkenes when hydrogen is used to activate the mol. oxygen; in our case, no sacrificial reductant is needed. We anticipate that our finding will initiate attempts to understand more fully the mechanism of oxygen activation at gold surfaces, which might lead to com. exploitation of the high redox activity of gold nanocrystals. 14. 14Corma, A.; Garcia, H.Supported Gold Nanoparticles as Catalysts for Organic Reactions. _Chem. Soc. Rev._ 2008, _37_, 2096– 2126, DOI: 10.1039/b707314n 14 Supported gold nanoparticles as catalysts for organic reactions Corma, Avelino; Garcia, Hermenegildo Chemical Society Reviews (2008), 37 (9), 2096-2126 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) A review. This crit. review is intended to attract the interest of org. chemists and researchers on green and sustainable chem. on the catalytic activity of supported gold nanoparticles in org. transformations. In the general part of this crit. review, emphasis is given to the different procedures to form supported gold nanoparticles and to the importance of the support cooperating in the catalysis. Also the convergence of homogeneous and heterogeneous catalysis in the study of gold nanoparticles has been discussed. The core part of this review is constituted by sections in which the reactions catalyzed by supported gold nanoparticles are described. Special emphasis is made on the unique ability of gold catalysts to promote addns. to multiple C-C bonds, benzannulations, and alc. oxidn. by oxygen. 15. 15Oemry, F.; Padama, A. A. B.; Kishi, H.; Kunikata, S.; Nakanishi, H.; Kasai, H.; Maekawa, H.; Osumi, K.; Sato, K.Effects of Cluster Size on Platinum-Oxygen Bonds Formation in Small Platinum Clusters. _J. Appl. Phys._ 2012, _51_, 035002, DOI: 10.7567/jjap.51.035002 15 Effects of cluster size on platinum-oxygen bonds formation in small platinum clusters Oemry, Ferensa; Padama, Allan Abraham B.; Kishi, Hirofumi; Kunikata, Shinichi; Nakanishi, Hiroshi; Kasai, Hideaki; Maekawa, Hiroyoshi; Osumi, Kazuo; Sato, Kaoru Japanese Journal of Applied Physics (2012), 51 (3, Pt. 1), 035002/1-035002/8 CODEN: JJAPB6 ISSN:. (Japan Society of Applied Physics) We present the results of d. functional theory calcn. in oxygen dissociative adsorption process on two types of isolated platinum (Pt) clusters: Pt4 and Pt10, by taking into account the effect of cluster reconstruction. The strength of Pt-Pt bonds in the clusters is mainly defined by d-d hybridization and interstitial bonding orbitals (IBO). Oxygen that adsorbed on the clusters is weakening the IBO and thus inducing geometry reconstruction as occurred in Pt10 cluster. However, cluster that could undergo structural deformation is found to promote oxygen dissocn. with no energy barrier. The details show that maintaining well-balanced of attractive and repulsive (Hellmann-Feynman) forces between atoms is considered to be the main key to avoid any considerable rise of energy barrier. Furthermore, a modest energy barrier that gained in Pt4 cluster is presumed to be originate from inequality of intramol. forces between atoms. 16. 16Lv, P.; Lu, Z.; Li, S.; Ma, D.; Zhang, W.; Zhang, Y.; Yang, Z.Tuning Metal Cluster Catalytic Activity with Morphology and Composition: A DFT Study of O 2 Dissociation at the Global Minimum of Pt m Pd n (m + n = 5) Clusters. _RSC Adv._ 2016, _6_, 104388– 104397, DOI: 10.1039/c6ra23266c 16 Tuning metal cluster catalytic activity with morphology and composition: a DFT study of O2 dissociation at the global minimum of PtmPdn (m + n = 5) clusters Lv, Peng; Lu, Zhansheng; Li, Shuo; Ma, Dongwei; Zhang, Wenjin; Zhang, Yi; Yang, Zongxian RSC Advances (2016), 6 (106), 104388-104397 CODEN: RSCACL; ISSN:2046-2069. (Royal Society of Chemistry) The Pt-based alloyed clusters are the important catalysts in the chem. industry. The global min. structures of PtmPdn (m + n = 5) clusters were searched for based on the CALYPSO algorithm, and the various properties of the lowest-energy PtmPdn clusters have been studied. The results show that introducing the Pd element can modify the morphol. and compn. of the pure Pt5 cluster, i.e., from a 2D planar to a 3D trigonal bipyramid structure, except for the Pt4Pd1 (square pyramid). According to the av. cohesive energy, the Pt4Pd1 cluster possesses the highest stability among these Pt-Pd alloyed clusters. The O2 mol. prefers to anchor on the clusters by the Yeager mode. The catalytic property for O2 dissocn. of the pure Pt5 cluster can be further improved by introducing the Pd atoms. Moreover, the Pt4Pd1 cluster with the high Pt compn. and the structure of square pyramid shows the highest catalytic activity. More importantly, the Bronsted-Evans-Polanyi relationship is found to be applicable to the current mini alloyed metal clusters. This study will help to understand the prediction of the global min. structure of metal cluster by CALYPSO and shed light on the design of the less expensive and more effective alloyed cluster catalysts by controlling the morphol. and compn. in the fuel cells. 17. 17Xu, Y.; Shelton, W. A.; Schneider, W. F.Effect of Particle Size on the Oxidizability of Platinum Clusters. _J. Phys. Chem. A_ 2006, _110_, 5839– 5846, DOI: 10.1021/jp0547111 17 Effect of Particle Size on the Oxidizability of Platinum Clusters Xu, Ye; Shelton, William A.; Schneider, William F. Journal of Physical Chemistry A (2006), 110 (17), 5839-5846 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The catalytic properties of transition metal particles often depend crucially on their chem. environment, but so far, little is known about how the effects of the environment vary with particle size, esp. for clusters consisting of only a few atoms. To gain insight into this topic, we have studied the oxygen affinity of free Ptx clusters as a function of cluster size (x = 1, 2, 3, 4, 5, and 10) using d. functional theory (DFT) calcns. (GGA-PW91). DFT-based Nose-Hoover mol. dynamics has been used to explore the configuration space of the PtxOx and PtxO2x clusters, leading to the discovery of several novel Pt-oxide structures. The formation of small Pt-oxide clusters by oxidizing the corresponding Ptx clusters is found to be significantly more exothermic than the formation of bulk Pt-oxides from Pt metal. The exothermicity generally increases as cluster size decreases but exhibits strongly nonlinear dependence on the cluster size. The nanoclusters are also structurally distinct from the bulk oxides and prefer one- and two-dimensional chain and ringlike shapes. These findings help elucidate the oxidn. behavior of Pt nanoclusters and lay the foundation for understanding the reactivity of Pt nanoclusters in oxidizing chem. environments. 18. 18Kerpal, C.; Harding, D. J.; Hermes, A. C.; Meijer, G.; Mackenzie, S. R.; Fielicke, A.Structures of Platinum Oxide Clusters in the Gas Phase. _J. Phys. Chem. A_ 2013, _117_, 1233– 1239, DOI: 10.1021/jp3055137 18 Structures of Platinum Oxide Clusters in the Gas Phase Kerpal, Christian; Harding, Dan J.; Hermes, Alexander C.; Meijer, Gerard; Mackenzie, Stuart R.; Fielicke, Andre Journal of Physical Chemistry A (2013), 117 (6), 1233-1239 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The structures of small gas-phase PtnO2m+ (n = 1-6, m = 1, 2) cluster cations have been investigated in a combined IR multiple photon dissocn. (IRMPD) spectroscopy and d. functional theory (DFT) study. On the basis of the IR spectra obtained, it is concluded that in most clusters oxygen is bound dissociatively, preferring 2-fold bridge binding motifs, sometimes combined with singly coordinated terminal binding. Comparison of the oxide cluster structures with those of bare cationic platinum clusters reported previously reveals major structural changes induced in the platinum core upon oxygen binding. For some cluster sizes the presence of the Ar messenger atom(s) is found to induce a significant change in the obsd. cluster structure. 19. 19Reber, A. C.; Khanna, S. N.; Roach, P. J.; Woodward, W. H.; Castleman, A. W.Spin Accommodation and Reactivity of Aluminum Based Clusters with O2. _J. Am. Chem. Soc._ 2007, _129_, 16098– 16101, DOI: 10.1021/ja075998d 19 Spin accommodation and reactivity of aluminum based clusters with O2 Reber, Arthur C.; Khanna, Shiv N.; Roach, Patrick J.; Woodward, William H.; Castleman, A. Welford, Jr. Journal of the American Chemical Society (2007), 129 (51), 16098-16101 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) It is shown that spin accommodation plays a detg. role in the reactivity of aluminum based anion clusters with oxygen. Exptl. reactivity studies on aluminum and aluminum-hydrogen clusters show variable reactivity in even electron systems and rapid etching in odd electron systems. The reactivity of even electron clusters is governed by a spin transfer to the singlet cluster through filling of the spin down antibonding orbitals on triplet oxygen. Theor. investigations show that when the spin transfer cannot occur, the species is unreactive. When spin accommodation is possible, more subtle effects appear, such as the required spin excitation energy, which raises the total energy of the system, and the filling of the antibonding levels of the O2 mol., which is stabilized by becoming an aluminum oxygen π bond. This explanation is consistent with obsd. behavior in oxygen etching reactions with a variety of clusters including AlnHm-, Aln-, AlnIm- and AlnC-. The proposed reaction mechanism lends a phys. interpretation as to why the HOMO-LUMO gap successfully predicts oxygen etching behavior of the considered systems. 20. 20Samanta, B.; Sengupta, T.; Pal, S.Specificity of Amino Acid-Aluminum Cluster Interaction and Subsequent Oxygen Activation by the above Complex. _J. Phys. Chem. C_ 2018, _122_, 28310– 28323, DOI: 10.1021/acs.jpcc.8b08396 20 Specificity of Amino Acid-Aluminum Cluster Interaction and Subsequent Oxygen Activation by the above Complex Samanta, Bipasa; Sengupta, Turbasu; Pal, Sourav Journal of Physical Chemistry C (2018), 122 (49), 28310-28323 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Amino acid-aluminum cluster complex has been shown to be a promising oxygen mol. activator and has the potential to dissoc. O-O bond on the cluster surface. The study of the complex also reveals specificity of amino acid to aluminum cluster in the above complex, which can be used in the detection of amino acid. The results have been obtained by the d. functional theory at the B3LYP/6-311++g(d,p) level of calcn. 21. 21Samanta, B.; Sengupta, T.; Pal, S.Aluminum cluster for CO and O 2 adsorption. _J. Mol. Model._ 2019, _25_, 2, DOI: 10.1007/s00894-018-3869-3 There is no corresponding record for this reference. 22. 22Burgert, R.; Schnockel, H.; Grubisic, A.; Li, X.; Stokes, S. T.; Bowen, K. H.; Gantefor, G. F.; Kiran, B.; Jena, P.Spin Conservation Accounts for Aluminum Cluster Anion Reactivity Pattern with O 2. _Science_ 2008, _319_, 438– 442, DOI: 10.1126/science.1148643 22 Spin Conservation Accounts for Aluminum Cluster Anion Reactivity Pattern with O2 Burgert, R.; Schnoeckel, H.; Grubisic, A.; Li, X.; Stokes, S. T.; Bowen, K. H.; Gantefoer, G. F.; Kiran, B.; Jena, P. Science (Washington, DC, United States) (2008), 319 (5862), 438-442 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) The reactivity pattern of small (~10 to 20 atoms) anionic aluminum clusters with oxygen has posed a long-standing puzzle. Those clusters with an odd no. of atoms tend to react much more slowly than their even-numbered counterparts. We used Fourier transform ion cyclotron resonance mass spectrometry to show that spin conservation straightforwardly accounts for this trend. The reaction rate of odd-numbered clusters increased appreciably when singlet oxygen was used in place of ground-state (triplet) oxygen. Conversely, monohydride clusters AlnH-, in which addn. of the hydrogen atom shifts the spin state by converting formerly open-shell structures to closed-shell ones (and vice versa), exhibited an opposing trend: The odd-n hydride clusters reacted more rapidly with triplet oxygen. These findings are supported by theor. simulations and highlight the general importance of spin selection rules in mediating cluster reactivity. 23. 23Neumaier, M.; Olzmann, M.; Kiran, B.; Bowen, K. H.; Eichhorn, B.; Stokes, S. T.; Buonaugurio, A.; Burgert, R.; Schnöckel, H.The Reaction Rates of O2 with Closed-Shell and Open-Shell Al _x_– and Ga _x_– Clusters under Single-Collision Conditions: Experimental and Theoretical Investigations toward a Generally Valid Model for the Hindered Reactions of O2 with Metal Atom Clusters. _J. Am. Chem. Soc._ 2014, _136_, 3607– 3616, DOI: 10.1021/ja4125548 23 The Reaction Rates of O2 with Closed-Shell and Open-Shell Alx- and Gax- Clusters under Single-Collision Conditions: Experimental and Theoretical Investigations toward a Generally Valid Model for the Hindered Reactions of O2 with Metal Atom Clusters Neumaier, Marco; Olzmann, Matthias; Kiran, Boggavarapu; Bowen, Kit H.; Eichhorn, Bryan; Stokes, Sarah T.; Buonaugurio, Angela; Burgert, Ralf; Schnoeckel, Hansgeorg Journal of the American Chemical Society (2014), 136 (9), 3607-3616 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) In order to characterize the oxidn. of metallic surfaces, the reactions of O2 with a no. of Alx- and, for the first time, Gax- clusters as mol. models have been investigated, and the results are presented here for x = 9-14. The rate coeffs. were detd. with FT-ICR mass spectrometry under single-collision conditions at O2 pressures of ∼10-8 mbar. In this way, the qual. known differences in the reactivities of the even- and odd-numbered clusters toward O2 could be quantified exptl. To obtain information about the elementary steps, we addnl. performed d. functional theory calcns. The results show that for both even- and odd-numbered clusters the formation of the most stable dioxide species, [MxO2]-, proceeds via the less stable peroxo species, [Mx+···O22-]-, which contains M-O-O-M moieties. We conclude that the formation of these peroxo intermediates may be a reason for the decreased reactivity of the metal clusters toward O2. This could be one of the main reasons why O2 reactions with metal surfaces proceed more slowly than Cl2 reactions with such surfaces, even though O2 reactions with both Al metal and Al clusters are more exothermic than are reactions of Cl2 with them. Furthermore, our results indicate that the spin-forbidden reactions of 3O2 with closed-shell clusters and the spin-allowed reactions with open-shell clusters to give singlet [Mx+···O22-]- are the root cause for the obsd. even/odd differences in reactivity. 24. 24Hagen, J.; Socaciu, L. D.; Le Roux, J.; Popolan, D.; Bernhardt, T. M.; Wöste, L.; Mitrić, R.; Noack, H.; Bonačić-Koutecký, V.Cooperative Effects in the Activation of Molecular Oxygen by Anionic Silver Clusters. _J. Am. Chem. Soc._ 2004, _126_, 3442– 3443, DOI: 10.1021/ja038948r 24 Cooperative Effects in the Activation of Molecular Oxygen by Anionic Silver Clusters Hagen, Jan; Socaciu, Liana D.; Le Roux, Jerome; Popolan, Denisia; Bernhardt, Thorsten M.; Woeste, Ludger; Mitric, Roland; Noack, Holger; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2004), 126 (11), 3442-3443 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) A novel size dependence in the adsorption reaction of multiple O2 mols. onto anionic silver clusters Ag-n (n = 1-5) is revealed by gas-phase reaction studies in an rf-ion trap. Ab initio theor. modeling based on DFT method provides insight into the reaction mechanism and finds cooperative electronic and structural effects to be responsible for the size selective reactivity of Agn- clusters toward one or more O2. In particular, Ag-n clusters with odd n have paired electrons and therefore bind one O2 only weakly, but they are simultaneously activated to adsorb a strongly bound second oxygen mol. For the clusters Ag3O4- and Ag5O4-, this cooperative effect results in a superoxo-like, doubly bound O2 subunit with potentially high activity in catalytic silver cluster oxidn. processes. 25. 25Socaciu, L. D.; Hagen, J.; Le Roux, J.; Popolan, D.; Bernhardt, T. M.; Wöste, L.; Vajda, Š.Strongly cluster size dependent reaction behavior of CO with O2 on free silver cluster anions. _J. Chem. Phys._ 2004, _120_, 2078, DOI: 10.1063/1.1644103 25 Strongly cluster size dependent reaction behavior of CO with O2 on free silver cluster anions Socaciu, Liana D.; Hagen, Jan; Le Roux, Jerome; Popolan, Denisia; Bernhardt, Thorsten M.; Woste, Ludger; Vajda, Stefan Journal of Chemical Physics (2004), 120 (5), 2078-2081 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) Reactions of free silver anions Agn- (n=1-13) with O2, CO, and their mixts. are investigated in a temp. controlled radio frequency ion trap setup. Cluster anions Agn- (n=1-11) readily react with mol. oxygen to yield AgnOm- (m=2, 4, or 6) oxide products. In contrast, no reaction of the silver cluster anions with carbon monoxide is detected. However, if silver cluster anions are exposed to the mixt. of O2 and CO, new reaction products and a pronounced, discontinuous size dependence in the reaction behavior is obsd. In particular, coadsorption complexes Agn(CO)O2- are detected for cluster sizes with n=4 and 6 and, the most striking observation, in the case of the larger odd atom no. clusters Ag7-, Ag9-, and Ag11-, the oxide product concn. decreases while a reappearance of the bare metal cluster signal is obsd. This leads to the conclusion that carbon monoxide reacts with the activated oxygen on these silver clusters and indicates the prevalence of a catalytic reaction cycle. 26. 26Bernhardt, T. M.; Hagen, J.; Lang, S. M.; Popolan, D. M.; Socaciu-Siebert, L. D.; Wöste, L.Binding Energies of O2 and CO to Small Gold, Silver, and Binary Silver–Gold Cluster Anions from Temperature Dependent Reaction Kinetics Measurements. _J. Phys. Chem. A_ 2009, _113_, 2724– 2733, DOI: 10.1021/jp810055q 26 Binding Energies of O2 and CO to Small Gold, Silver, and Binary Silver-Gold Cluster Anions from Temperature Dependent Reaction Kinetics Measurements Bernhardt, Thorsten M.; Hagen, Jan; Lang, Sandra M.; Popolan, Denisia M.; Socaciu-Siebert, Liana D.; Woste, Ludger Journal of Physical Chemistry A (2009), 113 (12), 2724-2733 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) A detailed anal. of exptl. obtained temp.-dependent gas-phase kinetic data for the oxygen and carbon monoxide adsorption on small anionic gold (Aun-, n = 1-3), silver (Agn-, n = 1-5), and binary silver-gold (AgnAum-, n + m = 2, 3) clusters is presented. The Lindemann energy transfer model in conjunction with statistical unimol. reaction rate theory is employed to det. the bond dissocn. energies E of the obsd. metal cluster complexes with O2 and CO. The accuracy limits of the obtained binding energies are evaluated by applying different transition-state models. The assumptions involved in the data evaluation procedure as well as possible sources of error are discussed. The thus-derived binding energies of O2 to pure silver and binary silver-gold cluster anions are generally in excellent agreement with previously reported theor. values. In marked contrast, the binding energies of O2 and CO to Au2- and Au3- detd. via temp.-dependent reaction kinetics are consistently lower than the theor. predicted values. 27. 27Luo, Z.; Gamboa, G. U.; Smith, J. C.; Reber, A. C.; Reveles, J. U.; Khanna, S. N.; Castleman, A. W.Spin Accommodation and Reactivity of Silver Clusters with Oxygen: The Enhanced Stability of Ag13–. _J. Am. Chem. Soc._ 2012, _134_, 18973– 18978, DOI: 10.1021/ja303268w 27 Spin Accommodation and Reactivity of Silver Clusters with Oxygen: The Enhanced Stability of Ag13- Luo, Zhixun; Gamboa, Gabriel U.; Smith, Jordan C.; Reber, Arthur C.; Reveles, J. Ulises; Khanna, Shiv N.; Castleman, A. W., Jr. Journal of the American Chemical Society (2012), 134 (46), 18973-18978 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Spin accommodation is demonstrated to play a detg. role in the reactivity of silver cluster anions with oxygen. Odd-electron silver clusters are found to be esp. reactive, while the anionic 13-atom cluster exhibits unexpected stability against reactivity with oxygen. Theor. studies show that the odd-even selective behavior is correlated with the excitation needed to activate the O-O bond in O2. Furthermore, by comparison with the reactivity of proximate even-electron clusters, we demonstrate that the inactivity of Ag13- is assocd. with its large spin excitation energy, ascribed to a crystal-field-like splitting of the orbitals caused by the bilayer at. structure, which induces a large gap despite not having a magic no. of valence electrons. 28. 28Ma, L.; Melander, M.; Laasonen, K.; Akola, J.CO Oxidation Catalyzed by Neutral and Anionic Cu20 Clusters: Relationship between Charge and Activity. _Phys. Chem. Chem. Phys._ 2015, _17_, 7067– 7076, DOI: 10.1039/c5cp00365b 28 CO oxidation catalyzed by neutral and anionic Cu20 clusters: relationship between charge and activity Ma, Li; Melander, Marko; Laasonen, Kari; Akola, Jaakko Physical Chemistry Chemical Physics (2015), 17 (10), 7067-7076 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) Reactions of CO and O2 on neutral and anionic Cu20 clusters have been investigated by spin-polarized d. functional theory. Three reaction mechanisms of CO oxidn. are explored: reactions with at. oxygen (dissocd. O2) as well as reactions with mol. oxygen, including Langmuir-Hinshelwood (LH) and Eley-Rideal (ER) mechanisms. The adsorption energies, reaction pathways, and reaction barriers for CO oxidn. are calcd. systematically. The anionic Cu20- cluster can adsorb CO and O2 more strongly than the neutral counterpart due to the superat. shell closing of 20 valence electrons which leaves one electron above the band gap. The activation of O2 mol. upon adsorption is crucial to det. the rate of CO oxidn. The CO oxidn. proceeds efficiently on both Cu20 and Cu20- clusters, when O2 is pre-adsorbed dissociatively. The ER mechanism has a lower reaction barrier than the LH mechanism on the neutral Cu20. In general, CO oxidn. occurs more readily on the anionic Cu20- (effective reaction barriers 0.1-0.3 eV) than on the neutral Cu20 cluster (0.3-0.5 eV). Moreover, Cu20- exhibits enhanced binding for CO2. From the anal. of the reverse direction of CO oxidn., it is obsd. that the transition of CO2 to CO + O can occur on the Cu20- cluster, which demonstrates that Cu clusters may serve as good catalyst for CO2 chem. 29. 29Cox, D. M.; Brickman, R.; Creegan, K.; Kaldor, A.Gold Clusters: Reactions and Deuterium Uptake. _Z. Phys. D: At., Mol. Clusters_ 1991, _19_, 353– 355, DOI: 10.1007/bf01448327 29 Gold clusters: reactions and deuterium uptake Cox, D. M.; Brickman, R.; Creegan, K.; Kaldor, A. Zeitschrift fuer Physik D: Atoms, Molecules and Clusters (1991), 19 (1-4), 353-5 CODEN: ZDACE2; ISSN:0178-7683. The reactivity and satn. of small gold clusters (cations, neutrals and anions) towards several mols. was examd. Specific small gold clusters exhibit a pronounced variation in their reactivity towards H, CH4 and O. The reactivity not only depends strongly on cluster size but also on the cluster charge state. For example, Aun+ (n < 15) react readily with D2, but no evidence of reaction is obsd. for the anions under our exptl. conditions. Similar behavior is seen for CH4. With O only even atom (odd electron) anions are reactive, and Au10+ is the only cation which exhibits evidence of reaction. The global features (small cluster cations reactive towards H2, CH4, but large ones not reactive, odd electron anions reactive towards O2) are qual. explained by a simple frontier orbital picture. The uptake of D and CH4 on gold clusters also exhibits a pronounced size dependence with D/Au varying from a high of 3 for the dimer to zero for clusters contg. more than 15 Au atoms. Comparison of the CH4 and D satn. behavior indicated that CH4 is dissocd. and bound as CH3 and H. 30. 30Cox, D. M.; Brickman, R. O.; Creegan, K.; Kaldor, A.Studies of the Chemical Properties of Size Selected Metal Clusters: Kinetics and Saturation. _Mater. Res. Soc. Symp. Proc._ 1991, _206_, 43, DOI: 10.1557/proc-206-43 30 Studies of the chemical properties of size selected metal clusters: kinetics and saturation Cox, Donald M.; Brickman, Robert O.; Creegan, Kathleen; Kaldor, Andrew Materials Research Society Symposium Proceedings (1991), 206 (Clusters Cluster-Assem. Mater.), 43-8 CODEN: MRSPDH; ISSN:0272-9172. Activated dissociative mol. chemisorption is studied of hydrogen (deuterium) and small alkanes on cationic, neutral and anionic platinum and gold clusters. Methane or ethane chemisorption occurs most readily only on small Pt and Au cluster cations and least rapidly, if at all, on cluster anions. The larger Pt and Au clusters are found to be non-reactive towards the alkanes under our exptl. conditions. At the other extreme, at steady state, small clusters of Pt, Rh, Ni, Pd, V, Nb, and Ta are hydrogen "rich" exhibiting H/M stoichiometries » 1. These results are consistent with a limited data base on supported cluster obtained via EXAFS measurements and have important implications for catalytic reactions involving hydrogen and light hydrocarbons. 31. 31Salisbury, B. E.; Wallace, W. T.; Whetten, R. L.Low-Temperature Activation of Molecular Oxygen by Gold Clusters: a Stoichiometric Process Correlated to Electron Affinity. _Chem. Phys._ 2000, _262_, 131– 141, DOI: 10.1016/s0301-0104(00)00272-x 31 Low-temperature activation of molecular oxygen by gold clusters: a stoichiometric process correlated to electron affinity Salisbury, B. E.; Wallace, W. T.; Whetten, R. L. Chemical Physics (2000), 262 (1), 131-141 CODEN: CMPHC2; ISSN:0301-0104. (Elsevier Science B.V.) The adsorption-desorption equil. of di-oxygen (O2) on Au cluster anions (AuN-, N = 2-22) were measured in the high-pressure limit by pulsed flow-reactor methods at ambient temp. Only certain clusters (N = 2,4,6,8,10,12,14,18,20) exhibit measurable adsorption, and for most of these the reaction can be driven to completion (>90% conversion to AuNO2-). In no case could secondary adsorption be detected, i.e. a 1:1 reaction stoichiometry is strictly obeyed. The size (N) variation in the adsorption free energies is correlated with the known electron affinities of the Au clusters. Taken with the pos. electron affinity of O2, these results suggest that O2 acts as a single-electron acceptor, yielding adsorbed superoxide (O2-), while the Au cluster strives to pair electrons (or close electron shells). This species could serve as an activated form for low-temp. oxygen-atom transfer reactions catalyzed by supported Au clusters, wherein the activation of mol. oxygen is regarded as a crucial step. 32. 32Wallace, W. T.; Whetten, R. L.Coadsorption of CO and O 2 on Selected Gold Clusters: Evidence for Efficient Room-Temperature CO2 Generation. _J. Am. Chem. Soc._ 2002, _124_, 7499– 7505, DOI: 10.1021/ja0175439 32 Coadsorption of CO and O2 on Selected Gold Clusters: Evidence for Efficient Room-Temperature CO2 Generation Wallace, William T.; Whetten, Robert L. Journal of the American Chemical Society (2002), 124 (25), 7499-7505 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Spurred by the recent demonstrations of the size- and support-dependent reactivity of supported gold clusters, here we present results on the coadsorption of CO and O2 on selected anionic gold clusters, AuN-, in the gas phase. O2 adsorbs in a binary (0,1) fashion as a one-electron acceptor on the AuN- clusters, with even-N clusters showing varying reactivity toward O2 adsorption, while odd-N clusters show no evidence of reactivity. CO shows a highly size-dependent reactivity for AuN- sizes from N = 4 to 19, but no adsorption on the gold dimer or trimer. When the gold clusters are exposed to both reactants, either simultaneously or sequentially, interesting effects have been obsd. While the same rules pertaining to individual O2 or CO adsorption continue to apply, the preadsorption of one reactant on a cluster may lead to the increased reactivity of the cluster to the other reactant. Thus, the adsorbates are not competing for bonding sites (competitive coadsorption), but, instead, aid in the adsorption of one another (cooperative coadsorption). New peaks also arise in the mass spectrum of Au6- under CO and O2 coadsorption conditions, which can be attributed to the loss of a CO2 mol. (or mols.). By studying the relative amt. of reaction, and relating it to the reaction time, it is found that the gas-phase Au6 anion is capable of oxidizing CO at a rate 100 times that reported for com. or model gold catalysts. 33. 33Wallace, W. T.; Wyrwas, R. B.; Whetten, R. L.; Mitrić, R.; Bonačić-Koutecký, V.Oxygen Adsorption on Hydrated Gold Cluster Anions: Experiment and Theory. _J. Am. Chem. Soc._ 2003, _125_, 8408– 8414, DOI: 10.1021/ja034905z 33 Oxygen Adsorption on Hydrated Gold Cluster Anions: Experiment and Theory Wallace, William T.; Wyrwas, Richard B.; Whetten, Robert L.; Mitric, Roland; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2003), 125 (27), 8408-8414 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) The discovery that supported gold clusters act as highly efficient catalysts for low-temp. oxidn. reactions has led to a great deal of work aimed at understanding the origins of the catalytic activity. Several studies have shown that the presence of trace moisture is required for the catalysts to function. Using near-atm. pressure flow reactor techniques, we have studied humidity and temp. effects on the reactivity of gas-phase gold cluster anions with O2. Near room temp., the humid source produces abundant gold-hydroxy cluster anions, AuNOH-, and these have a reversed O2 adsorption activity: nonreactive bare gold clusters become active when in the form AuNOH-, while active bare clusters are inactive when -OH is bound. The binding energies for the stable structures obtained from d. functional calcns. confirm fully these findings. Moreover, the theory provides evidence that electron-transfer induced by the binding of a OH group enhances the reactivity toward mol. oxygen for odd anionic gold clusters and suppresses the reactivity for the even ones. The temp. dependence of O2 addn. to Au3OH- and Au-4 indicates deviations from equil. control at temps. below room temp. The effects of humidity on gold cluster adsorption activity support the conclusion drawn for the mechanism of O2 adsorption on "dry" gold cluster anions and provides insight into the possible role of water in the enhanced activity of supported gold cluster catalysts. 34. 34Huang, W.; Zhai, H.-J.; Wang, L.-S.Probing the Interactions of O 2 with Small Gold Cluster Anions (Au n–, n = 1–7): Chemisorption vs Physisorption. _J. Am. Chem. Soc._ 2010, _132_, 4344– 4351, DOI: 10.1021/ja910401x 34 Probing the Interactions of O2 with Small Gold Cluster Anions (Aun-, n = 1-7): Chemisorption vs. Physisorption Huang, Wei; Zhai, Hua-Jin; Wang, Lai-Sheng Journal of the American Chemical Society (2010), 132 (12), 4344-4351 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Activation of O2 is the most crit. step in catalytic oxidn. reactions involving gold and remains poorly understood. Here the authors report a systematic study of the interactions between O2 and small gold cluster anions Aun- (n = 1-7) using photoelectron spectroscopy. Higher resoln. photoelectron spectra are obtained for the molecularly chemisorbed even-sized AunO2- (n = 2, 4, 6) complexes. Well-resolved vibrational structures due to O-O stretching are obsd. and can be readily distinguished from the Au-derived PES bands. The adiabatic detachment energies and O-O vibrational frequencies are 3.03 ± 0.04, 3.53 ± 0.05, and 3.17 ± 0.05 eV, and 1360 ± 80, 1360 ± 80, and 1330 ± 80 cm-1 for n = 2, 4, 6, resp. Physisorbed Aun-(O2) complexes for n = 1, 3, 5, 7 are obsd. for the 1st time, providing direct evidence for the inertness of the closed-shell odd-sized Aun- clusters toward O2. Neutral even-sized Aun clusters are closed-shell and are expected to be inert toward O2, which is not consistent with the reduced O-O vibrational frequencies obsd. in the photoelectron spectra relative to free O2. Probably the photodetachment transitions can only access excited states of the neutral even-sized AunO2 complexes; a double-well potential is proposed consisting of the ground-state van der Walls well at long Aun-O2 distances and a higher energy deeper well at short Aun-O2 distances derived from singlet O2 (1Δg). The current study provides further insight into O2 interactions with small gold clusters, as well as accurate exptl. data to benchmark theor. studies. 35. 35Woodham, A. P.; Meijer, G.; Fielicke, A.Activation of Molecular Oxygen by Anionic Gold Clusters. _Angew. Chem., Int. Ed._ 2012, _51_, 4444– 4447, DOI: 10.1002/anie.201108958 35 Activation of Molecular Oxygen by Anionic Gold Clusters Woodham, Alex P.; Meijer, Gerard; Fielicke, Andre Angewandte Chemie, International Edition (2012), 51 (18), 4444-4447, S4444/1-S4444/11 CODEN: ACIEF5; ISSN:1433-7851. (Wiley-VCH Verlag GmbH & Co. KGaA) Activation of mol. oxygen by anionic gold clusters is discussed. 36. 36Pal, R.; Wang, L.-M.; Pei, Y.; Wang, L.-S.; Zeng, X. C.Unraveling the Mechanisms of O 2 Activation by Size-Selected Gold Clusters: Transition from Superoxo to Peroxo Chemisorption. _J. Am. Chem. Soc._ 2012, _134_, 9438– 9445, DOI: 10.1021/ja302902p 36 Unraveling the Mechanisms of O2 Activation by Size-Selected Gold Clusters: Transition from Superoxo to Peroxo Chemisorption Pal, Rhitankar; Wang, Lei-Ming; Pei, Yong; Wang, Lai-Sheng; Zeng, Xiao Cheng Journal of the American Chemical Society (2012), 134 (22), 9438-9445 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) The activation of dioxygen is a key step in CO oxidn. catalyzed by gold nanoparticles. It is known that small gold cluster anions with even-numbered atoms can molecularly chemisorb O2 via one-electron transfer from Aun- to O2, whereas clusters with odd-numbered atoms are inert toward O2. Here we report spectroscopic evidence of two modes of O2 activation by the small even-sized Aun- clusters: superoxo and peroxo chemisorption. Photoelectron spectroscopy of O2Au8- revealed two distinct isomers, which can be converted from one to the other depending on the reaction time. Ab initio calcns. show that there are two close-lying mol. O2-chemisorbed isomers for O2Au8-: the lower energy isomer involves a peroxo-type binding of O2 onto Au8-, while the superoxo chemisorption is a slightly higher energy isomer. The computed detachment transitions of the superoxo and peroxo species are in good agreement with the exptl. observation. The current work shows that there is a superoxo to peroxo chemisorption transition of O2 on gold clusters at Au8-: O2Aun- (n = 2, 4, 6) involves superoxo binding and n = 10, 12, 14, 18 involves peroxo binding, whereas the superoxo binding re-emerges at n = 20 due to the high symmetry tetrahedral structure of Au20, which has a very low electron affinity. Hence, the two-dimensional (2D) Au8- is the smallest anionic gold nanoparticle that prefers peroxo binding with O2. At Au12-, although both 2D and 3D isomers coexist in the cluster beam, the 3D isomer prefers the peroxo binding with O2. 37. 37Mills, G.; Gordon, M. S.; Metiu, H.The Adsorption of Molecular Oxygen on Neutral and Negative Au n Clusters (n = 2–5). _Chem. Phys. Lett._ 2002, _359_, 493– 499, DOI: 10.1016/s0009-2614(02)00746-7 37 The adsorption of molecular oxygen on neutral and negative Aun clusters (n=2-5) Mills, Greg; Gordon, Mark S.; Metiu, Horia Chemical Physics Letters (2002), 359 (5,6), 493-499 CODEN: CHPLBC; ISSN:0009-2614. (Elsevier Science B.V.) We use d. functional theory to examine the binding of O2 to Aun and Aun- clusters. O2 binds more strongly to clusters having an odd no. of electrons than to those with an even no. A second O2 mol. binds more weakly than the first. 38. 38Yoon, B.; Häkkinen, H.; Landman, U.Interaction of O 2 with Gold Clusters: Molecular and Dissociative Adsorption. _J. Phys. Chem. A_ 2003, _107_, 4066– 4071, DOI: 10.1021/jp027596s 38 Interaction of O2 with Gold Clusters: Molecular and Dissociative Adsorption Yoon, Bokwon; Haekkinen, Hannu; Landman, Uzi Journal of Physical Chemistry A (2003), 107 (20), 4066-4071 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) The interaction of O2 with charged and neutral gold clusters, AuN- and AuN contg. up to eight atoms, investigated through the use of d. functional theory with generalized gradient corrections and scalar relativistic pseudopotentials, exhibits a pronounced sensitivity to the cluster size and to its charge state. Mol. adsorption is found to be favorable regardless of the charge state for clusters with N ≤ 3, while dissociative adsorption is favored for larger clusters. The dissocn. process involves a significant barrier (1 eV or more), and it is predicted to result in a large structural distortion in the host gold cluster. The interaction energy is largest for the anionic gold clusters where it exhibits an odd-even alternation as a function of the no. of gold atoms, with the maxima occurring for AuNO2- complexes with even N. The mol. bonding mechanism to these complexes involves charge transfer to the oxygen mol. with a concomitant activation of the O-O bond to a superoxo-like state. 39. 39Wang, Y.; Gong, X. G.First-principles study of interaction of cluster Au32 with CO, H2, and O2. _J. Chem. Phys._ 2006, _125_, 124703, DOI: 10.1063/1.2352749 39 First-principles study of interaction of cluster Au32 with CO, H2, and O2 Wang, Yao; Gong, X. G. Journal of Chemical Physics (2006), 125 (12), 124703/1-124703/12 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) First-principles calcns. are performed to study the interaction of cluster Au32 with small mols., such as CO, H2, and O2. The cagelike Au32(Ih) shows a higher chem. inertness than the amorphous Au32(C1) with respect to the interaction with small mols. CO, H2, and O2. H2 can only be phys. adsorbed on Au32(Ih), while it can be dissociatively chemisorbed on Au32(C1). Although CO can be chem. adsorbed on Au32(Ih) and Au32(C1) with one electron transferred from Au32 to the antibonding π orbit of CO, it is bound more strongly on Au32(C1) than on Au32(Ih). Spin polarized and spin nonpolarized calcns. result almost identical ground state structures of Au32(Ih)-O2 and Au32(C1)-O2, in which O2 is dissociatively chemisorbed. 40. 40Jena, N. K.; Chandrakumar, K. R. S.; Ghosh, S. K.Beyond the Gold-Hydrogen Analogy: Doping Gold Cluster with H-atom-O 2 Activation and Reduction of the Reaction Barrier for CO Oxidation. _J. Phys. Chem. Lett._ 2011, _2_, 1476– 1480, DOI: 10.1021/jz2006435 40 Beyond the Gold-Hydrogen Analogy: Doping Gold Cluster with H-atom-O2 Activation and Reduction of the Reaction Barrier for CO Oxidation Jena, Naresh K.; Chandrakumar, K. R. S.; Ghosh, Swapan K. Journal of Physical Chemistry Letters (2011), 2 (12), 1476-1480 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) The gold-hydrogen analogy is well established in terms of analogous chem. exhibited by gold and hydrogen atoms. A step beyond this analogy is demonstrated here by showing that the replacement of a Au atom with a H-atom in a gold cluster can lead to considerable enhancement in its reactivity. The present computational study reveals that while the closed-shell neutral pristine gold cluster Au8 can bind only weakly with an O2 mol., its doped counterpart Au7H shows remarkable enhancement in its binding with mol. oxygen. It also shows an increment in O-O bond length and a red shift in vibrational frequency, indicating the activation of the O2 mol. Furthermore, the H-doped gold cluster reduces the barrier height for the environmentally important CO oxidn. reaction as compared to the pristine cluster. These observations may spur further studies in the design of a cost-effective and efficient catalyst by doping gold clusters with a cheaper element like hydrogen. 41. 41Gao, Y.; Zeng, X. C.Water-Promoted O 2 Dissociation on Small-Sized Anionic Gold Clusters. _ACS Catal._ 2012, _2_, 2614– 2621, DOI: 10.1021/cs300383p 41 Water-Promoted O2 Dissociation on Small-Sized Anionic Gold Clusters Gao, Yi; Zeng, Xiao Cheng ACS Catalysis (2012), 2 (12), 2614-2621 CODEN: ACCACS; ISSN:2155-5435. (American Chemical Society) Although thermodynamically O2 favors dissociative adsorption over mol. adsorption on small-sized anionic gold clusters (except Au2-), O2 dissocn. is unlikely to proceed under ambient conditions because of the high activation energy barrier (>2.0 eV). Here, we present a systematic theor. study of reaction pathways for the O2 dissocn. on small-sized anionic gold nanoclusters Aun- (n = 1-6) with and without involvement of a water mol. The d. functional theory calcns. indicate that the activation barriers from the mol. adsorption state of O2 to dissociative adsorption can be significantly lowered with the involvement of a H2O mol. Once the O2 dissocs. on small-size gold clusters, at. oxygen is readily available for other reactions, such as the CO oxidn., on the surface of gold clusters. This theor. study supports previous exptl. evidence that H2O can be used to activate O2, which suggests an alternative way to exploit catalytic capability of gold clusters for oxidn. applications. 42. 42Lyalin, A.; Taketsugu, T.Reactant-Promoted Oxygen Dissociation on Gold Clusters. _J. Phys. Chem. Lett._ 2010, _1_, 1752– 1757, DOI: 10.1021/jz100503j 42 Reactant-Promoted Oxygen Dissociation on Gold Clusters Lyalin, Andrey; Taketsugu, Tetsuya Journal of Physical Chemistry Letters (2010), 1 (12), 1752-1757 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) The mol. and dissociative adsorption of O2 on pure odd-size Aun (n = 1, 3, 5, 7, 9) clusters and Aun with a coadsorbed C2H4 mol. has been studied theor. using d. functional theory. It is demonstrated that catalytic activation of the adsorbed O2 on pure gold clusters in the considered size range cannot lead to O2 dissocn. Coadsorption of C2H4, however, involves an extra charge transfer from the gold cluster to O2, energetically promoting oxygen dissocn. Our finding indicates that O2 dissocn. on the surface of small gold clusters is sensitive to the presence of other adsorbates, including the reactant mol. itself. This effect can be particularly important for understanding the mechanism of catalytic oxidn. on gold clusters. 43. 43Bürgel, C.; Reilly, N. M.; Johnson, G. E.; Mitric, R.; Kimble, M. L.; Castleman, A. W., Jr.; Bonačić-Koutecký, V.Influence of Charge State on the Mechanism of CO Oxidation on Gold Clusters. _J. Am. Chem. Soc._ 2008, _130_, 1694– 1698, DOI: 10.1021/ja0768542 43 Influence of Charge State on the Mechanism of CO Oxidation on Gold Clusters Buergel, Christian; Reilly, Nelly M.; Johnson, Grant E.; Mitric, Roland; Kimble, Michele L.; Castleman, A. W., Jr.; Bonacic-Koutecky, Vlasta Journal of the American Chemical Society (2008), 130 (5), 1694-1698 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) We present results from our joint exptl. and theor. study of the reactivity of anionic and cationic gold oxide clusters toward CO, focusing on the role of at. oxygen, different charge states, and mechanisms for oxidn. We show that anionic clusters react by an Eley-Rideal-like mechanism involving the preferential attack of CO on oxygen rather than gold. In contrast, the oxidn. of CO on cationic gold oxide clusters can occur by both an Eley-Rideal-like and a Langmuir-Hinshelwood-like mechanism at multiple collision conditions as a result of the high adsorption energy of two CO mols. This large energy of CO adsorption on cationic gold oxide clusters is the driving force for the CO oxidn. Therefore, in the presence of cationic gold species at high pressures of CO, the oxidn. reaction is self-promoting (i.e., the oxidn. of one CO mol. is promoted by the binding of a second CO). Our findings provide new insight into the role of charge state in gold-cluster-based nanocatalysis. 44. 44Tang, D.; Hu, C.DFT Insight into CO Oxidation Catalyzed by Gold Nanoclusters: Charge Effect and Multi-State Reactivity. _J. Phys. Chem. Lett._ 2011, _2_, 2972– 2977, DOI: 10.1021/jz201290x 44 DFT Insight into CO Oxidation Catalyzed by Gold Nanoclusters: Charge Effect and Multi-State Reactivity Tang, Dianyong; Hu, Changwei Journal of Physical Chemistry Letters (2011), 2 (23), 2972-2977 CODEN: JPCLCD; ISSN:1948-7185. (American Chemical Society) All-electron relativistic d. functional theory (DFT) calcns. on CO oxidn. catalyzed by 55-atom gold nanoclusters (Au55) have revealed that all of neutral, anionic, and cationic Au55 clusters exhibit stronger catalytic efficiency toward CO oxidn. at low temps. High spin energy profiles and spin crossing play a noticeably important role in CO oxidn. catalyzed by gold nanoclusters. Our findings provide new insight into the charge and spin-state effects of gold nanoclusters within gold nanocatalysts. 45. 45Yoon, B.; Häkkinen, H.; Landman, U.; Wörz, A.; Antonietti, J.; Abbet, S.; Judai, K.; Heiz, U.Charging Effects on Bonding and Catalyzed Oxidation of CO on Au 8 Clusters on MgO. _Science_ 2005, _307_, 403– 407, DOI: 10.1126/science.1104168 45 Charging Effects on Bonding and Catalyzed Oxidation of CO on Au8 Clusters on MgO Yoon, Bokwon; Haekkinen, Hannu; Landman, Uzi; Woerz, Anke S.; Antonietti, Jean-Marie; Abbet, Stephane; Judai, Ken; Heiz, Ueli Science (Washington, DC, United States) (2005), 307 (5708), 403-407 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Gold octamers (Au8) bound to oxygen-vacancy F-center defects on Mg(001) are the smallest clusters to catalyze the low-temp. oxidn. of CO to CO2, whereas clusters deposited on close-to-perfect magnesia surfaces remain chem. inert. Charging of the supported clusters plays a key role in promoting their chem. activity. IR measurements of the stretch vibration of CO adsorbed on mass-selected gold octamers soft-landed on MgO(001) with coadsorbed O2 show a red shift on an F-center-rich surface with respect to the perfect surface. The expts. agree with quantum ab initio calcns. that predict that a red shift of the C-O vibration should arise via electron back-donation to the CO antibonding orbital. 46. 46Baishya, S.; Deka, R. C.Catalytic Activities of Au 6, Au 6-, and Au 6+ Clusters for CO Oxidation: A density functional study. _Int. J. Quantum Chem._ 2014, _114_, 1559– 1566, DOI: 10.1002/qua.24725 46 Catalytic Activities of Au6, Au6-, and Au6+ Clusters for CO oxidation: A density functional study Baishya, Subhi; Deka, Ramesh C. International Journal of Quantum Chemistry (2014), 114 (22), 1559-1566 CODEN: IJQCB2; ISSN:0020-7608. (John Wiley & Sons, Inc.) We have studied the CO oxidn. over neutral, anionic, and cationic gold hexamer clusters using d. functional theory which elucidates the effect of cluster charge state on the catalytic activity. Herein, we have considered the conventional bimol. Langmuir-Hinshelwood mechanism with coadsorbed CO and O2 at the neighboring sites in all the clusters. Among the three clusters, Au 6- entails lower barriers during the various steps of the oxidn. mechanism. The stability of all the species including the transition states with respect to the interacting species in Au 6- indicates no thermal activation. Our study suggests better catalytic activity of Au 6- as compared to the neutral and cationic counterparts. 47. 47Torres, M. B.; Fernández, E. M.; Balbás, L. C.Theoretical Study of Oxygen Adsorption on Pure Au _n_+1+ and Doped MAu _n_+ Cationic Gold Clusters for M = Ti, Fe and n = 3–7. _J. Phys. Chem. A_ 2008, _112_, 6678– 6689, DOI: 10.1021/jp800247n 47 Theoretical Study of Oxygen Adsorption on Pure Aun+1+ and Doped MAun+ Cationic Gold Clusters for M = Ti, Fe and n = 3-7 Torres, M. Begona; Fernandez, Eva M.; Balbas, Luis C. Journal of Physical Chemistry A (2008), 112 (29), 6678-6689 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) A comparative study of the adsorption of an O2 mol. on pure Aun+1+ and doped MAun+ cationic Au clusters for n = 3-7 and M = Ti, Fe is presented. The simultaneous adsorption of 2 oxygen atoms also was studied. This work was performed by 1st principles calcns. based on norm-conserving pseudo-potentials and numerical basis sets. For pure Au4+, Au6+, and Au7+ clusters, the O2 mol. is adsorbed preferably on top of low coordinated Au atoms, with an adsorption energy smaller than 0.5 eV. Instead, for Au5+ and Au8+, bridge adsorption sites are preferred with adsorption energies of 0.56 and 0.69 eV, resp. The ground-state geometry of Aun+ is almost unperturbed after O2 adsorption. The electronic charge flows towards O2 when the mol. is adsorbed in bridge positions and towards the Au cluster when O2 is adsorbed on top of Au atoms, and both the adsorption energy and the O-O bond length of adsorbed oxygen increase when the amt. of electronic charge on O2 increases. However, the authors studied the adsorption of an O2 mol. on doped MAun+ clusters, giving (MAunO2+)ad complexes with different equil. configurations. The highest adsorption energy was obtained when both atoms of O2 bind on top of the M impurity, and it is larger for Ti doped clusters than for Fe doped clusters, showing an odd-even effect trend with size n, which is opposite for Ti as compared to Fe complexes. For those adsorption configurations of (MAunO2+)ad involving only Au sites, the adsorption energy is similar to or smaller than that for similar configurations of Aun+1O2+ complexes. However, the highest adsorption energy of (MAunO2+)ad is higher than that for (Aun+1O2+)ad by a factor of ∼4.0 (1.2) for M = Ti (M = Fe). The trends with size n are rationalized in terms of O-O and O-M bond distances, as well as charge transfer between oxygen and cluster substrates. The spin multiplicity of those (MAunO2+)ad complexes with the highest O2 adsorption energy is a max. (min.) for M = Fe (Ti), corresponding to parallel (anti-parallel) spin coupling of MAun+ clusters and O2 mols. Finally, the authors obtained the min. energy equil. structure of (AunO2+)dis and (MAunO2+)dis contg. 2 sepd. O atoms bonded at different sites of Aun+ and MAun+ clusters, resp. For (MAunO2+)dis, the equil. configuration with the highest adsorption energy is stable against sepn. in MAun+ and O2 fragments, resp. Instead, for (AunO2+)dis, only the complex n = 6 is stable against sepn. in Aun+ and O2 fragments. The max. sepn. energy of (MAunO2+)dis is higher than the O2 adsorption energy of (MAunO2+)ad complexes by factors of ∼1.6 (2.5), 1.6 (1.7), 1.5 (2.4), 1.5 (1.3), and 1.6 (1.8) for M = Ti (Fe) complexes in the range n = 3-7, resp. 48. 48Fernández, E. M.; Torres, M. B.; Balbás, L. C.Theoretical Study of the Coadsorption of CO and O 2 on Doped Cationic Gold Clusters MAu n+ (M = Ti, Fe, Au; n = 1, 6, 7). _Eur. Phys. J. D_ 2009, _52_, 135– 138, DOI: 10.1140/epjd/e2009-00010-4 48 Theoretical study of the coadsorption of CO and O2 on doped cationic gold clusters MAu+n (M = Ti, Fe, Au; n = 1, 6, 7) Fernandez, E. M.; Torres, M. B.; Balbas, L. C. European Physical Journal D: Atomic, Molecular, Optical and Plasma Physics (2009), 52 (1-3), 135-138 CODEN: EPJDF6; ISSN:1434-6060. (EDP Sciences) We present a study based on first-principles calcns. of the adsorption of CO on selected equil. configurations of MAunO+2 (M = Ti, Fe; n = 1, 6, 7) complexes resulting from the adsorption of O2 on doped cationic gold clusters MAu+n. Empirical rules for the formation of CO2-MAunO+ complexes are outlined. The desorption energy of CO2 is calcd. The adsorption of a second CO mol. on the residual MAunO+ complex leads in some cases to the formation of CO2-MAu+n. The desorption of a second CO2 mol. brings back to the initial doped gold cluster MAu+n. 49. 49Manzoor, D.; Krishnamurty, S.; Pal, S.Contriving a Catalytically Active Structure from an Inert Conformation: A Density Functional Investigation of Al, Hf, and Ge Doping of Au20Tetrahedral Clusters. _J. Phys. Chem. C_ 2016, _120_, 19636– 19641, DOI: 10.1021/acs.jpcc.6b04305 49 Contriving a Catalytically Active Structure from an Inert Conformation: A Density Functional Investigation of Al, Hf, and Ge Doping of Au20 Tetrahedral Clusters Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Journal of Physical Chemistry C (2016), 120 (35), 19636-19641 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) The transformation of an inert structure into a catalytically active conformation has immense potential from a structural engineering point of view. In the present work, we explored methods of achieving such a transformation through a d.-functional-based study. Au20 is known be one of the most stable and catalytically inert gold clusters, with a well-known tetrahedral structure. We demonstrate the transformation of the inert Au20 conformation into a highly active catalytic cluster through selective doping with Hf and Ge atoms. Depending on the dopant, the inert tetrahedral Au20 cluster evolves into either an endohedral or a hollow-cage-like conformation. The structural changes are manifested in the catalytic properties as well, with the result that the transformed doped cages exhibit extremely low activation barriers for the environmentally important CO oxidn. reaction as compared to reported inert Au20 cluster. The activation barriers for CO oxidn. are particularly low ( < 0.12 eV) when germanium is directly involved in the CO oxidn. Thus, the current work highlights the importance of engineering structural properties of metal nanoclusters with the help of heteroatom dopants for future applications in efficient catalysis. 50. 50Manzoor, D.; Krishnamurty, S.; Pal, S.Endohedrelly Doped Gold Cages: Efficient Catalysts for Oxygen Activation and CO Oxidation. _Phys. Chem. Chem. Phys._ 2016, _18_, 7068– 7074, DOI: 10.1039/c5cp05624a 50 Endohedrally doped gold nanocages: efficient catalysts for O2 activation and CO oxidation Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Physical Chemistry Chemical Physics (2016), 18 (10), 7068-7074 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) Gold nanocages are the most attractive catalytic materials as all the atoms in the cage type clusters reside on the surface, making them available for chemisorption by reacting mols. Due to a hollow space at the center, their chem. and catalytic properties can be tuned effectively and easily by endohedral doping. While a significant exptl. and theor. understanding is currently available on the structural and electronic properties of doped gold cages, very little information is available on their reactivity and catalytic behavior. In the present work, with the help of d. functional theory calcns. we demonstrate that endohedral doping leads to a notable increase in the binding energy of mol. oxygen on the gold nanocages. The enhancement in the O2 binding energy on the doped gold cages is also confirmed by a significant decrease in the Au-O and an increase in the O-O bond lengths, corroborated by a red shift (∼250 cm-1) in the O-O stretching frequency as compared to the pristine cage. Furthermore, interestingly, the doped gold cages show very low activation barriers for the environmentally important CO oxidn. reaction as compared to the pristine gold cage. Importantly, the decrease in the barrier height is comparatively greater for the rate limiting step of O-O-C-O intermediate formation and as a result the CO oxidn. is expected to be more facile on the doped gold cages. Thus, the current study highlights the role of heteroatom doping in imparting new chem. and catalytic properties to gold cages and is expected to spur further research in the design of efficient gold nanocatalysts. 51. 51Manzoor, D.; Krishnamurty, S.; Pal, S.Effect of Silicon Doping on the Reactivity and Catalytic Activity of Gold Clusters. _J. Phys. Chem. C_ 2014, _118_, 7501– 7507, DOI: 10.1021/jp501611t 51 Effect of Silicon Doping on the Reactivity and Catalytic Activity of Gold Clusters Manzoor, Dar; Krishnamurty, Sailaja; Pal, Sourav Journal of Physical Chemistry C (2014), 118 (14), 7501-7507 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Doping is known to be an excellent and simple way of catalyst design. Although notable progress has been made in understanding the reactivity and catalytic activity of gas-phase and supported gold clusters, very few studies have been carried out on the doped gold clusters. In the present work, we have carried out d. functional theory calcns. to investigate the effect of silicon doping on the reactivity and catalytic activity of gold nanoclusters. The present work particularly focuses on the adsorption and activation of mol. oxygen on the pristine and silicon-doped gold clusters. The results confirm that the silicon-doped Au7Si cluster shows considerable binding and activation of the O2 mol. in comparison to the pristine Au8 cluster as reflected in the relevant geometrical parameters (O-O and Au-O bond lengths) and O-O stretching frequency. However, silicon doping has no contrasting effect on the reactivity and catalytic activity of the Au7 cluster. In addn. to the stronger binding and activation of the O2 mol., the doped Au7Si cluster leads to a significant redn. in the activation barrier (0.57 eV) for the environmentally important CO oxidn. reaction in contrast to the catalytically inactive pristine Au8 cluster (1.22 eV). Thus, our results highlight the crit. role of doping foreign impurities for future endeavors in the field of gold nanocatalysis. 52. 52Molina, L. M.; Hammer, B.Some Recent Theoretical Advances in the Understanding of the Catalytic Activity of Au. _Appl. Catal., A_ 2005, _291_, 21– 31, DOI: 10.1016/j.apcata.2005.01.050 52 Some recent theoretical advances in the understanding of the catalytic activity of Au Molina, L. M.; Hammer, B. Applied Catalysis, A: General (2005), 291 (1-2), 21-31 CODEN: ACAGE4; ISSN:0926-860X. (Elsevier B.V.) A review; we present a small review of recent d.-functional-theory (DFT) studies of the reactivity towards CO oxidn. of supported Au nano-particles. The possible structure of the periphery of the interface between a Au particle and an oxide support is discussed. A certain structure, in which low coordinated Au atoms are overhanging the support without binding directly to the oxide atoms, is argued to be prototypical of medium-sized Au particles. This structure is shown to be particularly active both at the edges and at the corners of Au particles. Examples from the literature of Au systems supported on MgO(1 0 0) and rutile-TiO2(1 1 0) are reviewed and new data are given for the reactivity of facet, edge, and corner sites of a Au34 cluster supported on MgO(1 0 0). On the non-reducible oxide support, MgO(1 0 0), the CO oxidn. is found to occur via CO adsorption to the Au particles and subsequent CO-promoted O2 capture and formation of a CO·O2 reaction intermediate complex. On the reducible oxide support, TiO2(1 1 0), the O2 is found to adsorb independently of the CO. However, on this support, the reaction still proceeds via CO·O2 formation rather than via O2 dissocn. 53. 53Liu, C.; Tan, Y.; Lin, S.; Li, H.; Wu, X.; Li, L.; Pei, Y.; Zeng, X. C.CO Self-Promoting Oxidation on Nanosized Gold Clusters: Triangular Au3 Active Site and CO Induced O-O Scission. _J. Am. Chem. Soc._ 2013, _135_, 2583– 2595, DOI: 10.1021/ja309460v 53 CO Self-Promoting Oxidation on Nanosized Gold Clusters: Triangular Au3 Active Site and CO Induced O-O Scission Liu, Chunyan; Tan, Yingzi; Lin, Sisi; Li, Hui; Wu, Xiaojun; Li, Lei; Pei, Yong; Zeng, Xiao Cheng Journal of the American Chemical Society (2013), 135 (7), 2583-2595 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) We have systematically studied the CO oxidn. on various nanosized gold clusters with sizes ranging from 0.3 to 0.8 nm on the basis of d. functional theory (DFT) calcns. A hitherto unreported trimol. Langmuir-Hinshelwood (LH) mechanism is proposed, which offers new insights into the fundamental mechanism for CO oxidn. on nanosized gold clusters. Specifically, we find that the coadsorbed CO mol. at a unique triangular Au3 active site can act as a promoter for the scission of an O-O bond, leading to the spontaneous formation (due to extremely low energy barrier) of two CO2 mols. as product. The key step to the O-O bond scission in the OCOO intermediate is significantly accelerated due to the electrophilic attack of the coadsorbed neighboring CO mol. on the triangular Au3 site. This new mechanism is called CO self-promoting oxidn., which can be visualized in real time from the trajectory of a Born-Oppenheimer mol. dynamics (BOMD) simulation. We also find that such CO self-promoting oxidn. is quite universal, as long as the triangular Au3 reaction site is available. This is demonstrated in two prototype metal oxide supported gold nanostructure systems: namely, Aun/MgO and bilayer-Au/TiO2. The coadsorbed CO can not only serve as a promoter for its own oxidn. but also promote other oxidn. reactions such as styrene oxidn. through expediting O-O scission on gold nanostructures. 54. 54Manzoor, D.; Pal, S.Hydrogen Atom Chemisorbed Gold Clusters as Highly Active Catalysts for Oxygen Activation and CO Oxidation. _J. Phys. Chem. C_ 2014, _118_, 30057– 30062, DOI: 10.1021/jp510488v 54 Hydrogen Atom Chemisorbed Gold Clusters as Highly Active Catalysts for Oxygen Activation and CO Oxidation Manzoor, Dar; Pal, Sourav Journal of Physical Chemistry C (2014), 118 (51), 30057-30062 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Using oxygen activation and carbon monoxide oxidn. as a probe, the authors have carried out d. functional theory (DFT) calcns. to study the effect of hydrogen atom chemisorption on the reactivity and catalytic activity of gold nanoclusters. The authors particularly focus on the closed shell neutral Aun (n = 2, 4, 6, 8) gold clusters which are reported to be chem. inert and catalytically less active. The hydrogen atom chemisorption activates the inert clusters, thereby leading to enhanced binding and activation of the O2 mol. A significant amt. of increase in the O-O and decrease in the Au-O bond lengths together with a pertinent red shift in the O-O stretching frequencies clearly indicates the stronger binding and activation of oxygen mol. in the case of AunH clusters. Further, results reveal that the hydrogen atom chemisorbed gold clusters are catalytically more active with low activation barriers for the CO oxidn. reaction compared to the parent less active closed shell gold clusters. Thus, in short the current theor. observation provides key inputs for enhancing the reactivity and catalytic activity of inert gold clusters through the adsorption of a small atom or a mol. 55. 55Zhang, C.; Yoon, B.; Landman, U.Predicted Oxidation of CO Catalyzed by Au Nanoclusters on a Thin Defect-Free MgO Film Supported on a Mo(100) Surface. _J. Am. Chem. Soc._ 2007, _129_, 2228– 2229, DOI: 10.1021/ja0684545 55 Predicted Oxidation of CO Catalyzed by Au Nanoclusters on a Thin Defect-Free MgO Film Supported on a Mo(100) Surface Zhang, Chun; Yoon, Bokwon; Landman, Uzi Journal of the American Chemical Society (2007), 129 (8), 2228-2229 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) With the use of first-principles simulations, low barrier CO oxidn. reactions are predicted to occur via a Langmuir-Hinshelwood or an Eley-Rideal mechanism, on two-dimensional gold nanocluster islands adsorbed on two-layer MgO films that are supported on Mo(100). Underlying the emergent catalytic activity, predicted to occur even in the absence of oxygen vacancy F-center defects, is the excess electronic charge at the gold cluster/magnesia interface originating from the penetration of metal states through the thin magnesia film. This excess charge stabilizes the planar structure of the gold cluster and activates the oxygen mols. adsorbed at the interfacial periphery of the gold islands. 56. 56Molina, L. M.; Hammer, B.Active Role of Oxide Support during CO Oxidation at Au/MgO. _Phys. Rev. Lett._ 2003, _90_, 206102, DOI: 10.1103/physrevlett.90.206102 56 Active Role of Oxide Support during CO Oxidation at Au/MgO Molina, L. M.; Hammer, B. Physical Review Letters (2003), 90 (20), 206102/1-206102/4 CODEN: PRLTAO; ISSN:0031-9007. (American Physical Society) The oxidn. of CO at MgO supported gold aggregates is studied by means of d. functional theory calcns. In addn. to serving as a structural promoter holding the gold particles, the supporting oxide also takes an active role in the bonding and activation of adsorbates bound to the gold. The oxide stabilizes a peroxolike reaction intermediate, CO·O2, and causes steric repulsion to CO. The most reactive site at Au/MgO appears where the gold shelters the MgO thereby creating a cavity where several low-coordinated Au atoms and Mg2+ cations from the substrate can interact simultaneously with an adsorbate. 57. 57Harding, C.; Habibpour, V.; Kunz, S.; Farnbacher, A. N.-S.; Heiz, U.; Yoon, B.; Landman, U.Control and Manipulation of Gold Nanocatalysis: Effects of Metal Oxide Support Thickness and Composition. _J. Am. Chem. Soc._ 2009, _131_, 538– 548, DOI: 10.1021/ja804893b 57 Control and Manipulation of Gold Nanocatalysis: Effects of Metal Oxide Support Thickness and Composition Harding, Chris; Habibpour, Vahideh; Kunz, Sebastian; Farnbacher, Adrian Nam-Su; Heiz, Ueli; Yoon, Bokwon; Landman, Uzi Journal of the American Chemical Society (2009), 131 (2), 538-548 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Control and tunability of the catalytic oxidn. of CO by gold clusters deposited on MgO surfaces grown on molybdenum, Mo(100), to various thicknesses are explored through temp.-programmed reaction measurements on mass-selected 20-atom gold clusters and via first-principles d. functional theory calcns. Au20 was chosen because in the gas phase it is characterized as an extraordinarily stable tetrahedral-pyramidal structure. Dependencies of the catalytic activities and microscopic reaction mechanisms on the thickness and stoichiometry of the MgO films and on the dimensionalities and structures of the adsorbed gold clusters are demonstrated and elucidated. Langmuir-Hinshelwood mechanisms and reaction barriers corresponding to obsd. low- and high-temp. CO oxidn. reactions are calcd. and analyzed. These reactions involve adsorbed O2 mols. that are activated to a superoxo- or peroxo-like state through partial occupation of the antibonding orbitals. In some cases, we find activated, dissociative adsorption of O2 mols., adsorbing at the cluster peripheral interface with the MgO surface. The reactant CO mols. either adsorb on the MgO surface in the cluster proximity or bind directly to the gold cluster. Along with the oxidn. reactions on stoichiometric ultrathin MgO films, we also study reactions catalyzed by Au20 nanoclusters adsorbed on relatively thick defect-poor MgO films supported on Mo and on defect-rich thick MgO surfaces contg. oxygen vacancy defects. 58. 58Baishya, S.; Deka, R. C.Hybrid Density Functional/Molecular Mechanics Studies on Activated Adsorption of Oxygen on Zeolite Supported Gold Monomer. _J. Chem. Phys._ 2011, _135_, 244703, DOI: 10.1063/1.3667206 58 Hybrid density functional/molecular mechanics studies on activated adsorption of oxygen on zeolite supported gold monomer Baishya, Subhi; Deka, Ramesh C. Journal of Chemical Physics (2011), 135 (24), 244703/1-244703/7 CODEN: JCPSA6; ISSN:0021-9606. (American Institute of Physics) D. functional theory calcns. on oxygen adsorption over gas phase and faujasite supported Au monomer has been studied using hybrid quantum mechanics/mol. mechanics method, surface integrated MO mol. mechanics implemented in GAMESS package. Three different oxidn. states of Au (0, +1, +3) and three different adsorption modes viz., top, bridge, and dissociative adsorption of oxygen have been considered in our calcns. Red shift in the νO-O value from that in gas phase O2 indicates activation of O2 upon adsorption over faujasite supported gold monomer. The activation of O2 is an important step in the catalytic oxidn. of CO. The presence of adsorbed O2 increases the interaction of the Au monomer with the faujasite support. In faujasite supported cationic Au monomer, O2 preferably remains bridge bonded to Au rather than being dissocd. (c) 2011 American Institute of Physics. 59. 59García-Cruz, R.; Poulain, E.; Hernández-Pérez, I.; Reyes-Nava, J. A.; González-Torres, J. C.; Rubio-Ponce, A.; Olvera-Neria, O.Effect of Spin Multiplicity in O 2 Adsorption and Dissociation on Small Bimetallic AuAg Clusters. _J. Phys. Chem. A_ 2017, _121_, 6079– 6089, DOI: 10.1021/acs.jpca.7b01968 59 Effect of Spin Multiplicity in O2 Adsorption and Dissociation on Small Bimetallic AuAg Clusters Garcia-Cruz, Raul; Poulain, Enrique; Hernandez-Perez, Isaias; Reyes-Nava, Juan A.; Gonzalez-Torres, Julio C.; Rubio-Ponce, A.; Olvera-Neria, Oscar Journal of Physical Chemistry A (2017), 121 (32), 6079-6089 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) To dispose of at. oxygen, it is necessary the O2 activation; however, an energy barrier must be overcome to break the O-O bond. This work presents theor. calcns. of the O2 adsorption and dissocn. on small pure Aun and Agm and bimetallic AunAgm (n + m ≤ 6) clusters using the d. functional theory (DFT) and the zeroth-order regular approxn. (ZORA) to explicitly include scalar relativistic effects. The most stable AunAgm clusters contain a higher concn. of Au with Ag atoms located in the center of the cluster. The O2 adsorption energy on pure and bimetallic clusters and the ensuing geometries depend on the spin multiplicity of the system. For a doublet multiplicity, O2 is adsorbed in a bridge configuration, whereas for a triplet only one O-metal bond is formed. The charge transfer from metal toward O2 occupies the σO-O antibonding natural bond orbital, which weakens the oxygen bond. The Au3 (2A) cluster presents the lowest activation energy to dissoc. O2, whereas the opposite applies to the AuAg (3A) system. In the O2 activation, bimetallic clusters are not as active as pure Aun clusters due to the charge donated by Ag atoms being shared between O2 and Au atoms. 60. 60Joshi, K.; Krishnamurty, S.Tailoring the structure and electronic properties of platinum and gold-platinum nanocatalysts towards enhanced O2 activation. _New J. Chem._ 2016, _40_, 1336– 1346, DOI: 10.1039/c5nj02213d 60 Tailoring the structure and electronic properties of platinum and gold-platinum nanocatalysts towards enhanced O2 activation Joshi, Krati; Krishnamurty, Sailaja New Journal of Chemistry (2016), 40 (2), 1336-1346 CODEN: NJCHE5; ISSN:1144-0546. (Royal Society of Chemistry) Geometric structure plays a crucial role in enhancing the catalytic activity of a material towards reactions such as O redn., MeOH oxidn. etc. Pt and Pt based bimetallic alloys of different morphologies such as nanorods, core-shell etc. were synthesized so far contributing to varying catalytic activity in fuel cells. However, the electronic factors contributing to enhanced catalytic activity of particular structure/shape/morphol. were not understood so far. In a 1st such computational study, the authors demonstrate this complex structure-property correlation on model six-atom Pt clusters in the context of O2 activation at the cathode of direct MeOH fuel cells. Since recent studies showed that Au-Pt bimetallic alloys exhibit superior catalytic activity towards O redn. reaction with respect to other bimetallic clusters, the authors have identified 4 distinct six-atom Pt cluster morphologies, viz., trigonal planar, distorted octahedral, normal octahedral and double square planar, and doped them sequentially with Au atoms until a ratio of 1:1 was obtained. Thus, clusters with varying conformations/shapes and bimetallic compns. are considered for potential O2 activation. The O-O bond stretching frequencies evaluate the degree of bond activation in O2 mols. The authors' study reveals that the Pt6 and AunPtm (n + m = 6) clusters with a double square planar shape exhibit superior catalytic activity towards O-O bond activation with respect to other six-atom conformations. Anal. of their electronic properties demonstrates that the amt. of charge transferred by the cluster to the O2 mol. and the effective overlap between the d-orbitals of the metal atom and the pi-MOs of O2 can be directly correlated with their catalytic activities. To validate these conclusions, the authors have further extended such anal. to various Ptm (m = 3-13) conformations, wherein, once again the frontier MO compn. is seen to play a predominant role in O activation. Thus, the present study unveils the electronic properties detg. the catalytic activity of a catalytic cluster towards O-O bond activation. 61. 61Campbell, C.Bimetallic Surface Chemistry. _Annu. Rev. Phys. Chem._ 1990, _41_, 775– 837, DOI: 10.1146/annurev.pc.41.100190.004015 61 Bimetallic surface chemistry Campbell, Charles T. Annual Review of Physical Chemistry (1990), 41 (), 775-837 CODEN: ARPLAP; ISSN:0066-426X. A review on adsorption, chemisorption, catalysis, alloy surface properties, and bimetallic catalysts. 283 Refs. 62. 62Gao, F.; Goodman, D. W.Pd-Au Bimetallic Catalysts: Understanding Alloy Effects from Planar Models and (Supported) Nanoparticles. _Chem. Soc. Rev._ 2012, _41_, 8009– 8020, DOI: 10.1039/c2cs35160a 62 Pd-Au bimetallic catalysts: understanding alloy effects from planar models and (supported) nanoparticles Gao, Feng; Goodman, D. Wayne Chemical Society Reviews (2012), 41 (24), 8009-8020 CODEN: CSRVBR; ISSN:0306-0012. (Royal Society of Chemistry) A review. Pd-Au bimetallic catalysts often display enhanced catalytic activities and selectivities compared with Pd-alone catalysts. This enhancement is often caused by two alloy effects, i.e., ensemble and ligand effects. The ensemble effect is a diln. of surface Pd by Au. With increasing surface Au coverage, contiguous Pd ensembles disappear and isolated Pd ensembles form. For certain reactions, for example vinyl acetate synthesis, this effect is responsible for reaction rate enhancement via the formation of highly active surface sites, e.g., isolated Pd pairs. The disappearance of contiguous Pd ensembles also switches off side reactions catalyzed by these sites. This explains the selectivity increase of certain reactions, for example direct H2O2 synthesis. The ligand effects are electronic perturbation of Pd by Au. Via direct charge transfer or by affecting bond lengths, the ligand effects cause the Pd d band to be more filled, moving the d-band center away from the Fermi level. Both changes make Pd more "at. like" therefore binding reactants and products more weakly. For certain reactions, this eliminates a so-called "self-poisoning" effect and enhances activity/selectivity. 63. 63Gao, F.; Wang, Y.; Goodman, D. W.Reaction Kinetics and Polarization-Modulation Infrared Reflection Absorption Spectroscopy (PM-IRAS) Investigation of CO Oxidation over Supported Pd–Au Alloy Catalysts. _J. Phys. Chem. C_ 2010, _114_, 4036– 4043, DOI: 10.1021/jp910896k 63 Reaction Kinetics and Polarization-Modulation Infrared Reflection Absorption Spectroscopy (PM-IRAS) Investigation of CO Oxidation over Supported Pd-Au Alloy Catalysts Gao, Feng; Wang, Yilin; Goodman, D. Wayne Journal of Physical Chemistry C (2010), 114 (9), 4036-4043 CODEN: JPCCCK; ISSN:1932-7447. (American Chemical Society) Pd-Au bimetallic model catalysts, synthesized either as thin films on Mo(110) or as nanoparticles on a TiO2 thin film, were used in CO oxidn. at both low (1 × 10-7 Torr) and elevated (8-16 Torr) CO pressures. Alloying with Au forms isolated Pd sites that are incapable of dissocg. O2. This inability causes reactivity loss at low reactant pressures. Moreover, alloying also electronically modifies surface sites and affects the reaction activation energy. At elevated pressures, Pd preferentially segregates to the surface to form contiguous Pd sites and CO oxidn. reactivity is regained. Under stoichiometric conditions and relatively low temps., Pd-Au alloys show superior reactivity compared with pure Pd due to more facile CO desorption (less CO inhibition). Under net oxidizing conditions, metallic Pd displays superior reactivity due to its higher capability of dissocg. O2. However, pure Pd oxidizes and loses reactivity more readily than do Pd-Au alloys. 64. 64Xu, J.; White, T.; Li, P.; He, C.; Yu, J.; Yuan, W.; Han, Y.-F.Biphasic Pd–Au Alloy Catalyst for Low-Temperature CO Oxidation. _J. Am. Chem. Soc._ 2010, _132_, 10398– 10406, DOI: 10.1021/ja102617r 64 Biphasic Pd-Au Alloy Catalyst for Low-Temperature CO Oxidation Xu, Jing; White, Tim; Li, Ping; He, Chongheng; Yu, Jianguo; Yuan, Weikang; Han, Yi-Fan Journal of the American Chemical Society (2010), 132 (30), 10398-10406 CODEN: JACSAT; ISSN:0002-7863. (American Chemical Society) Low-temp. CO oxidn. over a compositional series of SiO2 fume-supported Pd/Au nano-alloy catalysts was assessed. Except for the pure metals, these materials invariably displayed biphasic sepn. into Pd- and Au-rich components. Performance was optimal for a bulk compn. catalyst, Pd4Au1, a mixt. of Pd90Au10 (72.5 at. percent) and Pd31Au69 (27.5 at. percent), which was remarkably active at 300° K and more stable than a pure Au catalyst. For bulk materials dominated by Pd (Pd:Au = 16:1; 8:1; 4:1), the Pd-rich alloy fraction frequently adopted hollow sphere or annular morphol.; Au-rich crystals were often multiply twinned. Quant. powder x-ray diffraction showed that under the synthesis conditions used, the Au soly. limit in Pd crystals was ∼12 at. percent; Pd was more sol. in Au (∼31 at. percent). This was consistent with XPS, which showed that Pd-rich alloy surfaces were enriched in Au relative to the bulk compn. In-situ Fourier transform IR spectra collected during CO oxidn. contained a new band at 2114 cm-1 (attributed to linear CO-Au/Au-Pd bonds) and reduced intensity of a band at 2090 cm-1 (from a linear CO-Pd bond) with escalating Au content, indicating the Pd sites became increasingly obscured by Au. High resoln. electron micrographs of Pd-rich alloys showed at. scale surface defects consistent with this interpretation. Results demonstrated that Au-contg. biphasic Pd nano-alloys may be highly durable alternatives for a range of catalytic reactions. 65. 65Han, Y.; Wang, J.; Kumar, D.; Yan, Z.; Goodman, D.A Kinetic Study of Vinyl Acetate Synthesis over Pd-Based Catalysts: Kinetics of Vinyl Acetate Synthesis over Pd-Au/SiO 2 and Pd/SiO 2 Catalysts. _J. Catal._ 2005, _232_, 467– 475, DOI: 10.1016/j.jcat.2005.04.001 65 A kinetic study of vinyl acetate synthesis over Pd-based catalysts: kinetics of vinyl acetate synthesis over Pd-Au/SiO2 and Pd/SiO2 catalysts Han, Y.-F.; Wang, J.-H.; Kumar, D.; Yan, Z.; Goodman, D. W. Journal of Catalysis (2005), 232 (2), 467-475 CODEN: JCTLA5; ISSN:0021-9517. (Elsevier) Kinetic studies of vinyl acetate (VA) synthesis were carried out for Pd (1.0 wt%)/SiO2 and Pd (1.0 wt%)-Au (0.5 wt%)/SiO2 catalysts, with highly dispersed metal particles (Pd and Pd-Au) characterized by X-ray diffraction and transmission electron microscopy-energy dispersive spectroscopy (TEM-EDS). The kinetics of the related CO oxidn. reaction were also explored. The kinetics of VA synthesis and CO oxidn. reactions proceed via a Langmuir-Hinshelwood mechanism. For Pd-only catalysts, dissociative adsorption of O2 is believed to be the rate-detg. step. This step is suppressed by adsorbed CO/C2H4 and gives rise to a neg. reaction order with respect to CO/C2H4 and a pos. order with respect to O2. However, the reaction mechanism was significantly modified by the addn. of Au to Pd, as indicated by the change in the reaction order with respect to CO/C2H4; in particular, the order with respect to C2H4 becomes pos. The modification of the mechanism for the Pd-Au catalyst correlates with the redn. in the concn. of Pd-Pd ensembles upon alloying with Au. By TEM-EDS, Au surface enrichment was detected for the Pd-Au alloy, and the interaction between Au and Pd leads to a formation of more active Pd ensembles, such as PdAu5 and PdAu6. The surface properties of Pd vs. Pd-Au catalysts were explored by CO/C2D4-TPD on thick films of Pd and Pd-Au. These studies indicate that the adsorption sites on Pd are significantly modified by Au; concurrently, the bonding of CO and C2D4 to Pd in the Pd-Au alloy also varied. As a result, the coverage of CO and C2H4 on the Pd-Au surface decreased markedly. The enhanced capacity of the Pd-Au surface for oxygen and/or the enhanced mobility of adsorbed oxygen under the reaction conditions are likely responsible for the unusually high reactivity of Pd-Au alloy catalysts for VA synthesis. 66. 66Edwards, J. K.; Solsona, B.; N, E. N.; Carley, A. F.; Herzing, A. A.; Kiely, C. J.; Hutchings, G. J.Switching Off Hydrogen Peroxide Hydrogenation in the Direct Synthesis Process. _Science_ 2009, _323_, 1037– 1041, DOI: 10.1126/science.1168980 66 Switching Off Hydrogen Peroxide Hydrogenation in the Direct Synthesis Process Edwards, Jennifer K.; Solsona, Benjamin; Ntainjua N, Edwin; Carley, Albert F.; Herzing, Andrew A.; Kiely, Christopher J.; Hutchings, Graham J. Science (Washington, DC, United States) (2009), 323 (5917), 1037-1041 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Hydrogen peroxide (H2O2) is an important disinfectant and bleach and is currently manufd. from an indirect process involving sequential hydrogenation/oxidn. of anthaquinones. However, a direct process in which H2 and O2 are reacted would be preferable. Unfortunately, catalysts for the direct synthesis of H2O2 are also effective for its subsequent decompn., and this has limited their development. We show that acid pretreatment of a carbon support for gold-palladium alloy catalysts switches off the decompn. of H2O2. This treatment decreases the size of the alloy nanoparticles, and these smaller nanoparticles presumably decorate and inhibit the sites for the decompn. reaction. Hence, when used in the direct synthesis of H2O2, the acid-pretreated catalysts give high yields of H2O2 with hydrogen selectivities greater than 95%. 67. 67Chen, M.; Kumar, D.; Yi, C. W.; Goodman, D. W.The Promotional Effect of Gold in Catalysis by Palladium-Gold. _Science_ 2005, _310_, 291– 293, DOI: 10.1126/science.1115800 67 The Promotional Effect of Gold in Catalysis by Palladium-Gold Chen, Mingshu; Kumar, Dheeraj; Yi, Cheol-Woo; Goodman, D. Wayne Science (Washington, DC, United States) (2005), 310 (5746), 291-293 CODEN: SCIEAS; ISSN:0036-8075. (American Association for the Advancement of Science) Acetoxylation of ethylene to vinyl acetate (VA) was used to investigate the mechanism of the promotional effect of Au in a Pd-Au alloy catalyst. The enhanced rates of VA formation for low Pd coverage relative to high Pd coverage on Au single-crystal surfaces demonstrate that the crit. reaction site for VA synthesis consists of two noncontiguous, suitably spaced, Pd monomers. The role of Au is to isolate single Pd sites that facilitate the coupling of crit. surface species to product, while inhibiting the formation of undesirable reaction byproducts. 68. 68Perdew, J. P.; Burke, K.; Ernzerhof, M.Generalized Gradient Approximation Made Simple. _Phys. Rev. Lett._ 1996, _77_, 3865, DOI: 10.1103/physrevlett.77.3865 68 Generalized gradient approximation made simple Perdew, John P.; Burke, Kieron; Ernzerhof, Matthias Physical Review Letters (1996), 77 (18), 3865-3868 CODEN: PRLTAO; ISSN:0031-9007. (American Physical Society) Generalized gradient approxns. (GGA's) for the exchange-correlation energy improve upon the local spin d. (LSD) description of atoms, mols., and solids. We present a simple derivation of a simple GGA, in which all parameters (other than those in LSD) are fundamental consts. Only general features of the detailed construction underlying the Perdew-Wang 1991 (PW91) GGA are invoked. Improvements over PW91 include an accurate description of the linear response of the uniform electron gas, correct behavior under uniform scaling, and a smoother potential. 69. 69Frisch, M. J.; Trucks, G. W.; Schlegel, H. B.; Scuseria, G. E.; Robb, M. A.; Cheeseman, J. R.; Scalmani, G.; Barone, V.; Mennucci, B.; Petersson, G. A.; _Gaussian 09_, Revision A.1; Gaussian, Inc.: Wallingford, CT, 2009.There is no corresponding record for this reference. 70. 70Zanti, G.; Peeters, D.DFT Study of Bimetallic Palladium-Gold Clusters Pd n Au m of Low Nuclearities (_n + m_ ≤ 14). _J. Phys. Chem. A_ 2010, _114_, 10345– 10356, DOI: 10.1021/jp1041298 70 DFT Study of Bimetallic Palladium-Gold Clusters PdnAum of Low Nuclearities (n + m ≤ 14) Zanti, Giuseppe; Peeters, Daniel Journal of Physical Chemistry A (2010), 114 (38), 10345-10356 CODEN: JPCAFH; ISSN:1089-5639. (American Chemical Society) Bimetallic palladium-gold PdnAum clusters of low nuclearity (n + m ≤ 14) are studied using the d. functional theory at B3LYP level with a Lanl2DZ pseudopotential to understand the evolution of various structural, electronic, and energetic properties as a function of size (n + m) and compn. (n/m) of the system. The potential energy surfaces have been explored for many different structures, and the min. obtained were then collected and used as a starting point for comparing the selected properties. Theor. results show a logical evolution of the properties depending on the size and the compn. of the system. Pdn clusters clearly prefer 3D structures while Aum clusters favor planar configurations. The geometry of the bimetallic PdnAum clusters mainly depends on their compn., i.e., clusters enriched in palladium atoms prefer 3D structures while increasing gold contents promotes planar configurations with deviation from planarity near Pd centers. Regarding the electronic properties, NBO anal. reveals that the unique closed-shell electronic structure of Pd atoms (4d10) requires a (4d → 5s) promotion to form stable bonds. In contrast, the half-occupied Au 6s AO implies effective Au-Au interaction and the electronic structure of Au atoms remains almost unchanged upon formation of bimetallic bonds. Consequently, clusters enriched in palladium atoms have spin multiplicities that increase with the cluster size while clusters enriched in gold atoms maintain the lowest possible spin multiplicity. Finally, the stability of these systems shows a synergic gain in cohesion for mixed PdnAum clusters compared to their monometallic Pdn and Aum counterparts. The maximal stabilization effect corresponds to n ≈ m, compns. for which the no. of mixed Pd-Au bonds is maximized. 71. 71Pichugina, D. A.; Polynskaya, Y. G.; Kuz’menko, N. E.Spin and Structural Features in Oxygen Dissociation on Tetrahedral Ag 20 and Ag 19 Au Clusters. _Phys. Chem. Chem. Phys._ 2016, _18_, 18033– 18044, DOI: 10.1039/c6cp01630h 71 Spin and structural features of oxygen dissociation on tetrahedral Ag20 and Ag19Au clusters Pichugina, D. A.; Polynskaya, Y. G.; Kuz'menko, N. E. Physical Chemistry Chemical Physics (2016), 18 (27), 18033-18044 CODEN: PPCPFQ; ISSN:1463-9076. (Royal Society of Chemistry) The spin-crossing mechanism of oxygen dissocn. on Ag20 and monodoped Ag19Au clusters was investigated via spin-polarized scalar-relativistic DFT calcns. using the PBE, TPSSh, M06L, mPBE, BLYP, OLYP, and B3LYP functionals. In particular, the singlet and triplet O-O bond rupture pathways at vertex-edge and facet-edge sites on the tetrahedral clusters were studied. The calcns. reveal that for the Ag20O2 and Ag19AuO2 complexes, the spin inversion from the triplet to singlet state occurs during the last step, which involves O-O bond rupture through a singlet transition state (TS). When spin crossing during oxygen dissocn. on the clusters is considered, the activation energies decrease by 10-29 kJ mol-1; however, they are still high due to the magic nature of the clusters and high vertical spin excitation energies. For these silver clusters, size effects based on the relationship between the TS structure and oxidn. activation energy were predicted. 72. 72Dononelli, W.; Klüner, T.CO Adsorption and Oxygen Activation on Group 11 Nanoparticles–A Combined DFT and High Level CCSD(T) Study About Size Effects and Activation Processes. _Faraday Discuss._ 2018, _208_, 105– 121, DOI: 10.1039/c7fd00225d 72 CO adsorption and oxygen activation on group 11 nanoparticles - a combined DFT and high level CCSD(T) study about size effects and activation processes Dononelli, Wilke; Kluener, Thorsten Faraday Discussions (2018), 208 (Designing Nanoparticle Systems for Catalysis), 105-121 CODEN: FDISE6; ISSN:1359-6640. (Royal Society of Chemistry) The focus of this study lies in the activation of mol. oxygen and reaction with CO within d. functional theory (DFT) and high level CCSD(T) calcns. Therefore, we use M13 and M55 nanoparticles (NPs) and periodic M(321) surfaces as model systems and compare the catalytic activity of gold substrates to Ag and Cu based NP catalysts. In the first step, the adsorption energies of CO were compared for nanoparticles of different sizes for Au, Ag and Cu. The adsorption energies on M(321) and M55 NPs (M = Au, Ag, Cu) are virtually identical. For smaller M13 NPs the adsorption energies differ by ∼0.2 eV for Ag, ∼0.4 eV for Au, and ∼0.6 eV for Cu at the PBE level of theory. This can be explained by size effects, as the M13 NPs show a more mol.-like character. Presumably, CO binds more strongly to these very small NPs at the PBE level of theory. However, a benchmark calcn. in the framework of CCSD(T)-theory reveals an adsorption energy of CO on Au13 of -0.88 eV, comparable to the adsorption energies calcd. at the PBE level for Au55 and Au(321). For Au55, the adsorption energy calcd. at the CCSD(T) level is -0.85 eV. This is in perfect agreement with the PBE result. In addn. to adsorption energies, dissocn. barriers have been calcd. on M(321) surfaces. The dissocn. energies of O2 on coinage metal catalysts are high, so that direct CO-oxidn. reactions with mol. oxygen should be the dominant reaction mechanism compared to the dissocn. and reaction of CO and at. oxygen at least for silver and copper catalysts. Supporting Information Supporting Information The Supporting Information is available free of charge on the ACS Publications website at DOI: 10.1021/acsomega.9b01581. Optimized geometries of Au 8, Pd 8, Au 8–_n_ Pd _n_, and their O 2-adsorbed complexes in the higher energy spin states and an internal reaction coordinate plot for O 2 dissociation on the Au 8 cluster (PDF) ao9b01581_si_001.pdf (376.02 kb) Terms & Conditions Most electronic Supporting Information files are available without a subscription to ACS Web Editions. 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190087	https://cs.stackexchange.com/questions/80865/greedy-algorithms-minimum-sum-number-pairing	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 Greedy algorithms: Minimum sum number pairing Ask Question Asked Modified 8 years ago Viewed 2k times 5 $\begingroup$ Given $n$ real numbers (where $n$ is even) find a pairing which minimizes the maximum sum of a pair. I think the optimal pairing is obtained by sorting the original set, pairing the first element with the last one, and so on. But I get stuck trying to prove it. I tried use the "difference reduction method", examining differences between an optimal answer and the "greedy" answer, and trying to get the greedy from the optimal without losing optimality, but I didn't reach anything. Any idea? correctness-proof greedy-algorithms Share Improve this question edited Sep 5, 2017 at 9:02 Yuval Filmus 281k2727 gold badges317317 silver badges515515 bronze badges asked Sep 5, 2017 at 3:58 Álvaro G. TenorioÁlvaro G. Tenorio 34522 silver badges1515 bronze badges $\endgroup$ 0 Add a comment \| 1 Answer 1 Reset to default 5 $\begingroup$ Suppose that numbers are $x_1, \ldots x_{2n}$, and let us rename them as $a_1, \ldots a_n, b_1, \ldots, b_n$, where $a_i \geq b_j$ for any $i, j$, $a_1 \geq a_2 \geq \ldots \geq a_n$, and $b_1 \leq b_2 \leq \ldots \leq b_n$. In this notation, the suggested optimum solution is $(a_1, b_1), \ldots (a_n, b_n)$. Given some arbitrary pairing, let us show we can reduce it to the right pairing without increasing the maximum sum of a pair. This shows that the expected correct solution is at least as good as any other. Equality might hold, eg. if all the values are the same, then any pairing is the joint best. First suppose that there is some pair $(a_u, a_v)$. Then since there are an equal number of $a$'s and $b$'s, there must also be some pair $(b_s, b_t)$. Clearly $a_u + a_v \geq b_s + b_t$, furthermore $a_u + a_v \geq a_u + b_s$ and similarly $a_u + a_v \geq a_v + b_t$. So if we replace the pairs $(a_u, a_v), (b_s, b_t)$ by $(a_u, b_s), (a_v, b_t)$, we have reduced the maximum sum of a pair, or at worst kept it the same. So now assume that our pairing is of the form $(a_{i_1}, b_{i_1}), \ldots (a_{i_n}, b_{i_n})$, but that it's not $(a_1, b_1), \ldots (a_n, b_n)$. There must be at least one pair of pairs, $(a_j, b_k)$ and $(a_l, b_m)$, where $j < l$ but $k > m$. (Otherwise, the $b$'s are in the same order as the $a$'s, which means it's the suggested solution.) This means that $a_j \leq a_l$ and $b_k \leq b_m$. So $a_l + b_m \geq a_j + b_k$, also $a_l + b_m \geq a_j + b_m$, and similarly $a_l + b_m \geq a_l + b_k$. So $(a_l, b_m)$ must be the bigger of $(a_j, b_k), (a_l, b_m)$, and replacing those two pairs with $(a_l, b_k), (a_j, b_m)$ can only reduce the maximum (or keep it the same). This shows that whatever pairing we start out with, we can swap pairs until we reach the suggested solution, without making the maximum sum any bigger. So the suggested solution must be the best. I believe you could collapse the two steps into a single argument since both are very similar and rely on showing that two pairs can be 'flipped' without making the maximum sum any bigger. However I find it easier to imagine in two stages. Share Improve this answer edited Sep 6, 2017 at 7:38 answered Sep 5, 2017 at 9:55 jwgjwg 20511 silver badge66 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions correctness-proof greedy-algorithms 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 1 Proof of a greedy algorithm concerning Buy and Resell Problem Related 0 Greedy algorithms tutorial Greedy algorithm correctness proof for "Elegant Permuted Sum" (UVa 11158) N numbers, N/2 pairs. Minimizing the maximum sum of a pairing. 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190088	https://math.stackexchange.com/questions/1695211/prove-a-sum-of-fractions-less-than-a-value	riemann sum - Prove a sum of fractions less than a value - 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 Prove a sum of fractions less than a value Ask Question Asked 9 years, 6 months ago Modified9 years, 6 months ago Viewed 399 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I happened to see this problem from an elementary school textbook, but cannot solve it: C=1 5+1 6+1 7+...+1 15+1 16+1 17 C=1 5+1 6+1 7+...+1 15+1 16+1 17 Prove C<2 C<2 Very much appreciate someone to enlighten me, especially with an elementary solution. fractions riemann-sum Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Mar 13, 2016 at 4:57 user1390686user1390686 1 2 1 sum of first five terms < 1, sum of last ten terms < 1 user44092 –user44092 2016-03-13 05:00:35 +00:00 Commented Mar 13, 2016 at 5:00 awesome! I got it. Thanks!user1390686 –user1390686 2016-03-13 05:18:01 +00:00 Commented Mar 13, 2016 at 5:18 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. they are usually comparing denominators to powers of two; sum(5,6,7) <3/4<3/4 sum(8,9,10,11,12,13,14,15) <8/8=1<8/8=1 sum(16,17) <2/16=1/8<2/16=1/8 C<15 8<2 C<15 8<2 In the other direction: sum(5,6,7,8) >4/8=1/2>4/8=1/2 sum(9,10,11,12,13,14,15,16) >8/16=1/2>8/16=1/2 C>1 C>1 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 13, 2016 at 5:05 Will JagyWill Jagy 147k 8 8 gold badges 156 156 silver badges 285 285 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. C=1 5+1 6+1 7+...+1 15+1 16+1 17 C=1 5+1 6+1 7+...+1 15+1 16+1 17 C=1 5+1 6+1 7+1 8+1 9+1 10+1 11+1 12+1 13+1 14+1 15+1 16+1 17≤(1 5×6)+(1 10×7)=19 10<2 C=1 5+1 6+1 7+1 8+1 9+1 10+1 11+1 12+1 13+1 14+1 15+1 16+1 17≤(1 5×6)+(1 10×7)=19 10<2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 13, 2016 at 5:02 GoodDeedsGoodDeeds 11.4k 3 3 gold badges 23 23 silver badges 43 43 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. It is easily seen that the terms 1 5,1 6,1 7,1 8,1 9 all are less than or equal to 1 5, so the sum of them is less than 1 and similarly the sum of other is less than 1 also all of them is less than or equal to 1 10 then C<2. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 13, 2016 at 5:08 AliAli 2,340 2 2 gold badges 14 14 silver badges 23 23 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. C≤3×1 5+10×1 8⟹C≤3 5+5 4⟹C≤12 20+25 20⟹C≤37 20≤40 20=2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 13, 2016 at 6:43 user249332 answered Mar 13, 2016 at 6:34 Olive StemfornOlive Stemforn 200 2 2 silver badges 8 8 bronze badges 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 fractions riemann-sum See similar questions with these tags. 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190089	https://smallfarms.cornell.edu/guide/guia-para-granjas-en-nueva-york/como-establecer-los-precios-de-los-productos-agricolas/?lang=es	24: Cómo Establecer los Precios de los Productos Agrícolas - Cornell Small Farms Skip to content SEARCH: This Site Cornell Menú Acerca de Proyectos Agroforestería Farm Ops Futuro en Ag Labranza Reducida en Hortalizas Hongos de Especialidad Agricultura Urbana Cursos Recursos Contacto Contactos locales Noticias 24: Cómo Establecer los Precios de los Productos Agrícolas Por Hannah Rae Warren \| abril 18, 2022 Home » Guía para la Agricultura en Nueva York » 24: Cómo Establecer los Precios de los Productos Agrícolas 24: Cómo Establecer los Precios de los Productos Agrícolas Conozca sus Costos y Precios para Obtener Beneficios El precio que usted pide por un producto o un servicio es una de las cuatro "P" de la comercialización: Precio, Producto, Posición y Promoción. El precio es de importancia crítica para determinar beneficio de la operación agrícola, pero las otras "P" de la comercialización contribuyen de gran manera al precio que se puede cobrar. El beneficio (profit en inglés) es la quinta "P" que influye en el negocio. Muchos agricultores principiantes se limitan a mirar lo que cobran otras operaciones agrícolas y ponen un precio similar a sus productos. Aunque conocer los precios de sus competidores es importante, es sólo una parte del panorama, y utilizar simplemente esto como estrategia de precios puede significar que su negocio no sea rentable. También debería analizar cuidadosamente sus costos, para estar seguro de que el precio que cobra le ayudará a lograr el beneficio que desea. Hay varios gastos que intervienen a la hora de decidir el precio que cobrará por su producto. Comience con los costos de los insumos = Costos Variables (CV), es decir, fertilizantes, semillas, gasolina, mano de obra. Si no los cubre, tendrá que cerrar en poco tiempo, a menos que subvencione la operación agrícola con otros ingresos. Agregue los costos de propiedad = Costos Fijos (CF), es decir, depreciación, intereses, mantenimiento, impuestos, seguros. Si apenas los cubre, alcanzará el umbral de costos del negocio, pero no le quedará nada adicional para usted. Cada elemento debe contribuir a los costos de propiedad. Si no cubre los costos de propiedad, tal vez no pueda operar durante mucho tiempo antes de tener que cerrar (a menos que cubra estos costos con los ingresos de fuera de la operación agrícola, pero esto significa que su negocio no es financieramente sostenible y depende de sus ingresos de fuera de la operación agrícola). Agregue un retorno para usted = Precio Rentable -este es el precio que necesita para sobrevivir a largo plazo. Asignar los Gastos por Rubro Para seguir los costos de la mano de obra y el equipo por producto se necesitan registros detallados. Puede hacer un seguimiento de las tareas y los gastos, como el tiempo de labranza y el fertilizante para toda la operación agrícola y asignar por pies cuadrados usados por un producto en particular. Lleve un registro del tiempo diario dedicado a los esfuerzos o gastos especiales que requieren los productos específicos, como el trasplante por separado. Sume todo esto para determinar los costos por producto. Es importante dar seguimiento al rendimiento cosechado o de la cantidad de producto que realmente se ha vendido, ya que esto influye en el precio por unidad, de forma significativa. Valor vs. Precio Muchos productores del mercado directo tienen miedo de cobrar lo necesario para tener algún beneficio para ellos mismos. Tenga en cuenta que usted está proporcionando más valor al estar más cerca del cliente. Pregúntese quiénes son sus competidores. ¿Quiere ser un "fijador" de precios o un "tomador" de precios? Valor = Calidad + Servicio + Precio Sus compradores quieren un producto de calidad, que ofrece excelente sabor en lugar de la durabilidad y la vida útil Sus compradores quieren saber cómo se han cultivado sus alimentos. Les gusta tener una relación con usted. Esto requiere tiempo por su parte, pero están dispuestos a pagar por ello. Puede darles a conocer nuevos productos y formas de cocinar artículos especiales. Se trata de una enseñanza por la que están dispuestos a pagar Los productos frescos sin encerar, el menor uso de combustible y el apoyo de la comunidad también se citan como razones por las cuales muchos consumidores están dispuestos a pagar más por los productos locales Puede cobrar más por los productos de principios de temporada, cuando los clientes están ansiosos por probar las primeras fresas locales frescas o el maíz dulce, así que esfuércese por hacer las ventas tempranas Resumen de la Ficha Conozca sus Costos y Precios para Obtener Beneficios Asignar los Gastos por Rubro Valor vs. Precio Cálculos para Determinar el Precio Planifique para Obtener Beneficios: No Rebaje los Precios Cálculos para Determinar el Precio Método de Costos y Beneficios Sume sus costos variables + sus costos fijos + la ganancia necesaria para el producto en particular = Ingresos Divida por el número de unidades producidas = precio/unidad Por Ejemplo: Si le cuesta 3.000 dólares en costos variables totales y 2.000 dólares en costos fijos totales y quiere obtener 2.000 dólares de beneficio por un producto específico, entonces sus ingresos totales por ese producto tienen que ser de 7.000 dólares. Divida esto por el número de unidades producidas, y tendrá el precio por unidad: 7.000 dólares / 950 unidades = 7,38 dólares/unidad Método del Margen Bruto Este método se basa en el conjunto de las ventas del negocio, los costos y el beneficio previsto. Este método suele ser utilizado por los negocios minoristas que revenden productos. Por Ejemplo: El método del margen bruto en un negocio de verduras podría calcularse así: Conozca el total de ventas de verduras previstas = 10.000 dólares Conozca sus costos fijos totales + el beneficio deseado = 3.000 dólares: éste es el margen bruto necesario. Divida su margen bruto por el total de ventas: 3.000 $ / 10.000 $ = 30%. Conozca su costo variable por unidad = 5,00 dólares Divida el precio por unidad por 1 menos el 30% del costo variable por unidad para determinar el precio 5,00 dólares / (1-30%) = 5,00 dólares / 0,7 = 7,14 dólares por unidad Planifique para Obtener Beneficios: No Rebaje los Precios ¿Qué pasa si usted tiene el maíz a 3,50 dólares/docena según sus cálculos y su vecino lo tiene a 3,00 dólares/docena? ¿Puede seguir obteniendo ganancias bajando el precio? A veces es mejor vender menos al precio más alto que vender más al precio más bajo. Por ejemplo, si su margen sobre los 3,50 dólares es de 0,50 dólares de beneficio, si vende 300 docenas, eso le dará 150 dólares de beneficio. Usted tendría que vender 600 docenas si vendiera a 3,25 dólares para obtener el mismo beneficio. Para una disminución del 7% en el precio hay que vender el doble de producto. ¡No ponga el precio de su producto agrícola por debajo del mercado sólo porque los ingresos de la operación agrícolas sean intrascendentes para usted! Si su operación es muy pequeña y tiene ingresos ajenos a la misma, quizá pueda permitirse vender una docena de huevos marrones frescos por 2 dólares sólo para cubrir el costo de los piensos, pero otros agricultores locales que dependen de los ingresos de sus negocios agrícolas para mantener su hogar no pueden hacerlo: podrían necesitar el precio completo de más de 4 dólares la docena para cubrir todos sus gastos. Podrían perder ventas injustamente debido a su indiscreción. En aras de la cooperación con su comunidad agrícola local, mantenga sus precios en línea con los precios de mercado para cualquier producto agrícola, incluso si puede permitirse no hacerlo. Si el beneficio económico no forma parte de sus motivaciones de practicar la agricultura, considere la posibilidad de asociarse con un banco de alimentos local, un comedor social o una red de donaciones para enviar los buenos alimentos que produce a personas que no pueden obtenerlos de otro modo. Los Precios Vigentes en la Zona del Mercado Muchos agricultores principiantes empiezan con una estrategia de precios que refleja lo que cobran todos los demás. Aunque este es un buen punto de partida, no es el lugar en el que se quiere estar siempre. Es importante conocer los costos y el precio para obtener beneficios. El sitio web de Cornell Small Farms Program tiene manuales y guías sobre mercadeo directo, CSA (Agricultura Apoyada por la Comunidad), Food Hubs (Centros de Alimentos), así como el mercadeo de productos específicos como orgánicos, de valor agregado y de venta al por mayor. El Small Farms Program también ofrece un curso en línea de 6 semanas para agricultores principiantes que recién comienzan a explorar sus mercados potenciales: Small Farms Program Curso En Línea: BF 102: Explorando Mercados y Beneficios PRÓXIMAMENTE EN ESPAÑOL ← 23: Cómo Evaluar Su Potencial de Mercado25: Cómo Encontrar Información sobre Precios → Proyectos Menú Proyectos Proyectos Anteriores Summit Cursos Menú Cursos y capacitación Elegir un Curso Preguntas Frecuentes Recursos Menú Centro de Recursos para Agricultores Quarterly Menú Latest Issue Subscribe Contribute Apoyar a Small Farms Si tiene alguna discapacidad y tiene problemas para acceder a la información de este sitio web o necesita materiales en un formato alternativo, póngase en contacto con la Universidad de Cornell a través del correo electrónico web-accessibility@cornell.edu para solicitar ayuda. El Programa de Pequeñas Explotaciones Agrícolas de Cornell tiene su sede en la Facultad de Agricultura y Ciencias de la Vida (CALS) de la Universidad de Cornell, en Ithaca (Nueva York), y trabaja en colaboración con la Extensión Cooperativa de Cornell en todo el Estado de Nueva York. Reconocimiento de la tierra: La Universidad de Cornell está situada en las tierras natales tradicionales de los Gayogo̱hó:nǫɁ (la Nación Cayuga). Los Gayogo̱hó:nǫɁ son miembros de la Confederación Haudenosaunee, una alianza de seis Naciones soberanas con presencia histórica y contemporánea en esta tierra. La Confederación precede a la creación de la Universidad de Cornell, el estado de Nueva York y los Estados Unidos de América. Reconocemos la dolorosa historia del despojo de Gayogo̱hó:nǫɁ, y honramos la conexión actual del pueblo Gayogo̱hó:nǫɁ, pasado y presente, con estas tierras y aguas. © 2022 Cornell Small Farms. 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190090	https://commons.wikimedia.org/wiki/Category:Young_diagrams	Category:Young diagrams - Wikimedia Commons Jump to content [x] Main menu Main menu move to sidebar hide Navigate Main page Welcome Community portal Village pump Help center Language select Participate Upload file Recent changes Latest files Random file Contact us Special pages In Wikipedia Add links Search Search English [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size 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 Category:Young diagrams Good pictures Advanced... All images Featured pictures Featured videos Quality images Valued images In this category and in... In this category but not in... About FastCCI... Help Category Discussion [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Permanent link Page information Get shortened URL Download QR code RSS feed Nominate category for discussion Print/export Create a book Download as PDF Printable version In other projects From Wikimedia Commons, the free media repository A Young diagram (also called Ferrers diagram, particularly when represented using dots) is a graphical representation of an integer partition. Subcategories This category has the following 2 subcategories, out of 2 total. S Single 3-bit Walsh permutations; SVG inversions matrix; Young diagrams(3 C) W Weight partitions of seals(4 F) Category Slideshow Media in category "Young diagrams" The following 33 files are in this category, out of 33 total. Corners of young.png 190 × 191; 5 KB Diagrama de Young-dot.svg 800 × 700; 17 KB Diagrama de Young-fr-dot.svg 800 × 700; 16 KB Diagrama de Young-fr.svg 800 × 700; 16 KB Diagrama de Young.svg 800 × 700; 16 KB Diagramme partition 331.svg 512 × 307; 3 KB Dominance order partitions of 6.png 321 × 859; 19 KB Ferrer 331.png 219 × 220; 1 KB Ferrer 5441-432.png 265 × 222; 2 KB Ferrer partitioning diagrams.svg 430 × 430; 11 KB Mors.png 215 × 140; 843 bytes MurnaghanNakayama1.svg 519 × 125; 39 KB MurnaghanNakayama2.svg 818 × 83; 61 KB MurnaghanNakayama3.svg 266 × 83; 13 KB Partition aleatoire.svg 600 × 600; 1 KB Partition.png 144 × 89; 2 KB Partition2.png 87 × 140; 2 KB Partitions of n with biggest addend k.svg 1,350 × 1,108; 124 KB Partitions of n with k addends.svg 1,350 × 1,108; 173 KB Rank of a partition es.svg 297 × 198; 25 KB Rank of a partition.svg 297 × 198; 25 KB Schur functions rectangular example.png 683 × 300; 10 KB Young 456 French.svg 126 × 72; 2 KB Young diagram 5 4 1.svg 1,134 × 720; 1 KB Young diagram for 541 partition french.svg 180 × 108; 945 bytes Young diagram for 541 partition-French.svg 205 × 125; 6 KB Young diagram for 541 partition.jpg 197 × 122; 3 KB Young diagram for 541 partition.svg 180 × 108; 950 bytes Young tableau.svg 264 × 35; 6 KB Young's lattice.svg 514 × 379; 6 KB Youngdiag.png 512 × 493; 2 KB 杨表12223.jpg 126 × 208; 4 KB 杨表145.jpg 208 × 126; 4 KB Retrieved from " Categories: Group diagrams by type Young tableaux Integer partitions Representation theory This page was last edited on 13 April 2024, at 09:01. Files are available under licenses specified on their description page. All structured data from the file namespace is available under the Creative Commons CC0 License; all unstructured text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and the Privacy Policy. Privacy policy About Wikimedia Commons Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Search Search Category:Young diagrams Add topic
190091	https://en.wikipedia.org/wiki/Monte_Carlo_methods_for_electron_transport	Jump to content Search Contents 1 Background 1.1 Boltzmann transport equation 1.2 Monte Carlo method 1.3 Hydrodynamic and drift diffusion method 1.4 Comparison of semiclassical models 2 Monte Carlo for semiconductor transport 2.1 Band structure 2.1.1 Parabolic band structure 2.1.2 Non-parabolic band structure 2.1.3 Full band structure 2.2 Types of Monte Carlo simulation 2.2.1 One-particle Monte Carlo 2.2.2 Ensemble Monte Carlo 2.2.3 Self-consistent ensemble Monte Carlo 2.3 Random flight selection 3 Scattering mechanisms 3.1 General background in solid-state physics 3.2 Inclusion of scattering mechanisms in Monte Carlo 3.3 Selection of scattering mode and scattered trajectory 4 Quantum corrections for Monte Carlo simulation 4.1 Wigner-based correction 4.2 Effective potential correction 4.3 Schrödinger-based correction 5 See also 6 References Monte Carlo methods for electron transport Add 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 Wikidata item Appearance From Wikipedia, the free encyclopedia The Monte Carlo method for electron transport is a semiclassical Monte Carlo (MC) approach of modeling semiconductor transport. Assuming the carrier motion consists of free flights interrupted by scattering mechanisms, a computer is utilized to simulate the trajectories of particles as they move across the device under the influence of an electric field using classical mechanics. The scattering events and the duration of particle flight is determined through the use of random numbers. Background [edit] Boltzmann transport equation [edit] The Boltzmann transport equation model has been the main tool used in the analysis of transport in semiconductors. The BTE equation is given by[citation needed]: The distribution function, f, is a dimensionless function which is used to extract all observable of interest and gives a full depiction of electron distribution in both real and k-space. Further, it physically represents the probability of particle occupation of energy k at position r and time t. In addition, due to being a seven-dimensional integro-differential equation (six dimensions in the phase space and one in time) the solution to the BTE is cumbersome and can be solved in closed analytical form under very special restrictions. Numerically, solution to the BTE is employed using either a deterministic method or a stochastic method. Deterministic method solution is based on a grid-based numerical method such as the spherical harmonics approach, whereas the Monte Carlo is the stochastic approach used to solve the BTE. Monte Carlo method [edit] The semiclassical Monte Carlo method is a statistical method used to yield exact solution to the Boltzmann transport equation which includes complex band structure and scattering processes. This approach is semiclassical for the reason that scattering mechanisms are treated quantum mechanically using the Fermi's Golden Rule, whereas the transport between scattering events is treated using the classical particle notion. The Monte Carlo model in essence tracks the particle trajectory at each free flight and chooses a corresponding scattering mechanism stochastically. Two of the great advantages of semiclassical Monte Carlo are its capability to provide accurate quantum mechanical treatment of various distinct scattering mechanisms within the scattering terms, and the absence of assumption about the form of carrier distribution in energy or k-space. The semiclassical equation describing the motion of an electron is where F is the electric field, E(k) is the energy dispersion relation, and k is the momentum wave vector. To solve the above equation, one needs strong knowledge of the band structure (E(k)). The E(k) relation describes how the particle moves inside the device, in addition to depicting useful information necessary for transport such as the density of states (DOS) and the particle velocity. A Full-band E(K) relation can be obtained using the semi-empirical pseudopotential method. Hydrodynamic and drift diffusion method [edit] Both drift diffusion (DD) and the hydrodynamic (HD) models can be derived from the moments of the Boltzmann transport equation (BTE) using simplified approximation valid for long channel devices. The DD scheme is the most classical approach and usually solves the Poisson equation and the continuity equations for carriers considering the drift and diffusion components. In this approach, the charge transit time is assumed to be very large in comparison to the energy relaxation time. On the other hand, the HD method solves the DD scheme with the energy balance equations obtained from the moments of BTE. Thus, one may capture and calculate physical details such as carrier heating and the velocity overshoot effect. Needless to say, an accurate discretization method is required in HD simulation, since the governing equations are strongly coupled and one has to deal with larger number of variables compared to the DD scheme. Comparison of semiclassical models [edit] The accuracy of semiclassical models are compared based on the BTE by investigating how they treat the classical velocity overshoot problem, a key short channel effect (SCE) in transistor structures. Essentially, velocity overshoot is a nonlocal effects of scaled devices, which is related to the experimentally observed increase in current drive and transconductance. As the channel length becomes smaller, the velocity is no longer saturated in the high field region, but it overshoots the predicted saturation velocity. The cause of this phenomenon is that the carrier transit time becomes comparable to the energy relaxation time, and therefore the mobile carriers do not have enough time to reach equilibrium with the applied electric field by scattering in the short channel devices. The summary of simulation results (Illinois Tool: MOCA) with DD and HD model is shown in figure beside. In the figure (a), the case when the field is not high enough to cause the velocity overshoot effect in the whole channel region is shown. Note that at such limit, the data from the DD model fit well to the MC model in the non-overshoot region, but the HD model overestimate the velocity in that region. The velocity overshoot is observed only near the drain junction in the MC data and the HD model fits well in that region. From the MC data, it can be noticed that the velocity overshoot effect is abrupt in the high-field region, which is not properly included in the HD model. For high field conditions as shown in the figure (b) the velocity overshoot effect almost all over the channel and the HD results and the MC results are very close in the channel region. Monte Carlo for semiconductor transport [edit] Band structure [edit] Band structure describes the relationship between energy(E) and wave vector(k). The band structure is used to compute the movement of carriers under the action of the electric field, scattering rate, and final state after the collision. Silicon band structure and its Brillouin zone are shown in figure below, but there is no analytical expression which satisfies entire Brillouin zone. By using some approximation, there are two analytical models for band structure, namely the parabolic and the non-parabolic modes. Parabolic band structure [edit] For the concept of band structure, parabolic energy bands are generally assumed for simplicity. Electrons reside, at least when close to equilibrium, close to the minima of the E(k) relation. Then the E(k) relation can be extended in a Taylor series as Because the first derivative vanishes at the band minimum, so the gradient of E(k) is zero at k = 0. Thus, which yields the definition of the effective mass tensor This expression is true for semiconductor which has isotropic effective mass, for instance GaAs. In case of silicon, conduction band minima does not lie at k = 0 and the effective mass depends on the crystallographic orientation of the minimum as where describe longitudinal and transverse effective mass, respectively. Non-parabolic band structure [edit] For higher applied fields, carriers reside above the minimum and the dispersion relation, E(k), does not satisfy the simple parabolic expression described above. This non-parabolicity is generally described by where is a coefficient of non-parabolicity given by where is the electron mass in vacuum, and Eg is the energy gap. Full band structure [edit] For many applications, non-parabolic band structure provides reasonable approximation. However, in case of very high field transport, which requires the better physical model of the full band structure. For full band approach, numerically generated table of E(k) is used. Full band approach for Monte Carlo simulation was first used by Karl Hess at the University of Illinois at Urbana-Champaign. This approach is based on empirical pseudopotential method suggested by Cohen and Bergstresser . Full band approach is computationally expensive, however, following the advancement of the computational power, it can be used as a more general approach. Types of Monte Carlo simulation [edit] One-particle Monte Carlo [edit] For this type of simulation, one carrier is injected and the motion is tracked in the domain, until it exits through contact. Another carrier is then injected and the process repeated to simulate an ensemble of trajectories. This approach is mostly useful to study bulk properties, like the steady state drift velocity as a function of field. Ensemble Monte Carlo [edit] Instead of single carrier, a large ensemble of carriers is simulated at the same time. This procedure is obviously a good candidate for super-computation, since one may apply parallelization and vectorization. Also, it is now possible to perform ensemble averages directly. This approach is suitable for transient simulations. Self-consistent ensemble Monte Carlo [edit] This method couples the ensemble Monte Carlo procedure to Poisson's equation, and is the most suitable for device simulation. Typically, Poisson's equation is solved at fixed intervals to update the internal field, to reflect the internal redistribution of charge, due to the movement of carriers. Random flight selection [edit] The probability that the electron will suffer its next collision during dt around t is given by where P[k(t)]dt is the probability that an electron in the state k suffers a collision during the time dt. Because of the complexity of the integral at the exponent, it is impractical to generate stochastic free flights with the distribution of the equation above. In order to overcome this difficulty, people use a fictitious “self-scattering” scheme. By doing this, the total scattering rate, including this self-scattering, is constant and equal to, say, . By random selection, if self-scattering is selected, k′ after the collision is the same as k and the carrier continues its flight without perturbation. Introducing a constant , the above equation reduces to Random numbers r can be used very simply to generate stochastic free flights, which duration will then be given by . The computer time used for self-scattering is more than compensated for by the simplification of the calculation of the free-flight duration. To enhance the speed of free flight time calculation, several schemes such as “Constant Technique”, and “Piecewise Technique” are used to minimize the self-scattering events. Scattering mechanisms [edit] General background in solid-state physics [edit] Important charge transport properties of semiconductor devices such as the deviance from Ohm's law and the saturation of carriers mobility are a direct consequence of scattering mechanisms. It is thus of great importance for a semiconductor device simulation to capture the physics of such mechanisms. The semiconductor Monte Carlo simulation, in this scope, is a very powerful tool for the ease and the precision with which an almost exhaustive array of scattering mechanisms can be included. The duration of the free flights is determined from the scattering rates. At the end of each flight, the appropriate scattering mechanism must be chosen in order to determine the final energy of the scattered carrier, or equivalently, its new momentum and scattering angle. In this sense, one will distinguish two broad types of scattering mechanisms which naturally derive form the classic kinetic theory of collision between two bodies: Elastic scattering, where the energy of the particle is conserved after being scattered. Elastic scattering will hence only change the direction of the particle's momentum. Impurity scattering and surface scattering are, with a fair approximation, two good examples of elastic scattering processes. Inelastic scattering, where energy is transferred between the scattered particle and the scattering center. Electronphonon interactions are essentially inelastic since a phonon of definite energy is either emitted or absorbed by the scattered particle. Before characterizing scattering mechanisms in greater mathematical details, it is important to note that when running semiconductor Monte Carlo simulations, one has to deal mainly with the following types of scattering events: Acoustic Phonon: The charge carrier exchanges energy with an acoustic mode of the vibration of atoms in the crystal lattice. Acoustic Phonons mainly arise from thermal excitation of the crystal lattice. Polar Optical: The charge carrier exchanges energy with one of the polar optical modes of the crystal lattice. These modes are not present in covalent semiconductors. Optical phonons arise from the vibration against each other of atoms of different types when there is more than one atom in the smallest unit cell, and are usually excited by light. Non-Polar Optical: Energy is exchanged with an optical mode. Non-polar optical phonons must generally be considered in covalent semiconductors and the L-valley of GaAs. Equivalent Intervalley Phonon: Due to the interaction with a phonon, the charge carrier transitions from initial states to final states which belong to different but equivalent valleys. Typically, this type of scattering mechanism describes the transition of an electron from one X-valley to another X-valley, or from one L-valley to another L-valley. Non Equivalent Intervalley Phonon: Involves the transition of a charge carrier between valleys of different types. Piezoelectric Phonon: For low temperatures. Ionized Impurity: Reflects the deviation of a particle from it ballistic trajectory due to Coulomb interaction with an ionized impurity in the crystal lattice. Because the mass of an electron is relatively small in comparison to the one of an impurity, the Coulomb cross section decreases rapidly with the difference of the modulus of momentum between the initial and final state. Therefore, impurity scattering events are mostly considered for intravalley scattering, intraband scattering and, to a minor extent, interband scattering. Carrier-Carrier: (electron-electron, hole-hole and electron-hole interactions). When carrier concentration is high, this type of scattering reflects the electrostatic interaction between charge carriers. This problem becomes very quickly computationally intensive with an increasing number of particles in an ensemble simulation. In this scope, Particle-Particle–Particle-Mesh (P3M) algorithms, which distinguish short range and long range interaction of a particle with its surrounding charge gas, have proved efficient in including carrier-carrier interaction in the semiconductor Monte Carlo simulation. Very often, the charge of the carriers is assigned to a grid using a Cloud-in-Cell method, where part of the charge of a given particle is assigned to a given number of closest grid points with a certain weight factor. Plasmon: Reflects the effect of the collective oscillation of the charge carriers on a given particle. Inclusion of scattering mechanisms in Monte Carlo [edit] A computationally efficient approach to including scattering in Monte Carlo simulation consists in storing the scattering rates of the individual mechanisms in tables. Given the different scattering rates for a precise particle state, one may then randomly select the scattering process at the end of the free flight. These scattering rates are very often derived using the Born approximation, in which a scattering event is merely a transition between two momentum states of the carrier involved. As discussed in section II-I, the quantum many-body problem arising from the interaction of a carrier with its surrounding environment (phonons, electrons, holes, plasmons, impurities,...) can be reduced to a two-body problem using the quasiparticle approximation, which separates the carrier of interest from the rest of the crystal. Within these approximations, Fermi's Golden Rule gives, to the first order, the transition probability per unit time for a scattering mechanism from a state to a state : where H' is the perturbation Hamiltonian representing the collision and E and E′ are respectively the initial and final energies of the system constituted of both the carrier and the electron and phonon gas. The Dirac -function stands for the conservation of energy. In addition, the term , generally referred to as the matrix element, mathematically represents an inner product of the initial and final wave functions of the carrier: In a crystal lattice, the wavefunctions and are simply Bloch waves. When it is possible, analytic expression of the Matrix elements are commonly found by Fourier expanding the Hamiltonian H', as in the case of Impurity scattering or acoustic phonon scattering. In the important case of a transition from an energy state E to an energy state E' due to a phonon of wave vector q and frequency , the energy and momentum change is: where R is a reciprocal lattice vector. Umklapp processes (or U-processes) change the momentum of the particle after scattering and are therefore limiting the conduction in semiconductor crystals. Physically, U-processes occur when the final momentum of the particle points out of the first Brillouin zone. Once one knows the scattering probability per unit time from a state k to a state k', it is interesting to determine the scattering rate for a given scattering process. The scattering rate gives the probability per unit time to scatter from a state k to any other state in the reciprocal space. Therefore, the scattering rate is which can be readily used to determine the free flight time and the scattering process as discussed in section 3-3. It is important to note that this scattering rate will be dependent on the band structure of the material (the dependence arises from the matrix elements). Selection of scattering mode and scattered trajectory [edit] At the end of a free flight, a scattering mode and angle must be randomly chosen. In order to determine the scattering mechanism, one has to consider all the scattering rates of the mechanisms relevant to the simulation as well as the total scattering rate at the time of scattering Selecting a scattering mechanism then simply results in generating a uniformly distributed random number 0 < r < 1 and referring to the following rules A computationally efficient approach to selecting the scattering mechanism consists in adding a “void” scattering mechanism so that remains constant over time. If a particle is scattered according to this mechanism, it will keep its ballistic trajectory after scattering takes place. In order to choose a new trajectory, one must first derive the energy (or momentum) of the particle after scattering where the term accounts for phonon emission or absorption and the term is non-null for inter-valley scattering. The final energy (and the band structure) directly yield the modulus of the new momentum k'. At this point one only needs to choose a new direction (or angle) for the scattered particle. In some simple cases as phonon scattering and a parabolic dispersion relation, the scattering angle is random and evenly distributed on the sphere of radius k'. Using spherical coordinates, the process of choosing the angle is equivalent to randomly picking two angles and . If the angle is distributed with a distribution , then for a uniform distribution of angles, the probability to pick a point of the sphere is It is possible, in this case, to separate the two variables. Integrating over then over , one finds The two spherical angles can then be chosen, in the uniform case, by generating two random numbers 0 < r1, r2 < 1 such that Quantum corrections for Monte Carlo simulation [edit] The current trend of scaling down semiconductor devices has forced physicists to incorporate quantum mechanical issues in order to acquire a thorough understanding of device behavior. Simulating the behavior of nano-scale devices necessitates the use of a full quantum transport model especially for cases when the quantum effects cannot be ignored. This complication, however, can be avoided in the case of practical devices like the modern day MOSFET, by employing quantum corrections within a semi-classical framework. The semi-classical Monte Carlo model can then be employed to simulate the device characteristics. The quantum corrections can be incorporated into a Monte Carlo simulator by simply introducing a quantum potential term which is superimposed onto the classical electrostatic potential seen by the simulated particles. Figure beside pictorially depicts the essential features of this technique. The various quantum approaches available for implementation are described in the following subsections. Wigner-based correction [edit] The Wigner transport equation forms the basis for the Wigner-based quantum correction.[citation needed] where, k is the crystal momentum, V is the classical potential, the term on the RHS is the effect of collision, the fourth term on the LHS represents non-local quantum mechanical effects. The standard Boltzmann Transport Equation is obtained when the non-local terms on the LHS disappear in the limit of slow spatial variations. The simplified (for ) quantum corrected BTE then becomes where the quantum potential is contained in the term (must be an error: was never mentioned). Effective potential correction [edit] This method for quantum correction was developed by Feynman and Hibbs in 1965.[citation needed] In this method the effective potential is derived by calculating the contribution to the path integral of a particle's quantum fluctuations around its classical path. This calculation is undertaken by a variational method using a trial potential to first order. The effective classical potential in the average point on each path then becomes Schrödinger-based correction [edit] This approach involves periodical solving of a Schrödinger equation in a simulation with the input being the self-consistent electrostatic potential. The exact energy levels and wavefunctions relating to the electrostatic potential solution are employed to calculate the quantum potential. The quantum correction obtained on the basis of this method can be visualised by the following equation where Vschr is the quantum correction potential, z is the direction perpendicular to the interface, nq is the quantum density from the Schrödinger equation which is equivalent to the converged Monte Carlo concentration, Vp is the potential from the Poisson solution, V0 is the arbitrary reference potential far away from the quantum region such that the correction goes to null in the region of semi-classical behavior. Even though the above-mentioned potentials for quantum correction differ in their method of calculation and their basic assumptions, yet when it comes to their inclusion into Monte Carlo simulation they are all incorporated the same way. See also [edit] Monte Carlo method Semiconductor device Monte Carlo method for photon transport Electronic band structure Method of quantum characteristics Quantum Monte Carlo Quasi-Monte Carlo method References [edit] ^ Karl Hess, ed. (1991). Monte Carlo Device Simulation: Full Band and Beyond. Springer US. doi:10.1007/978-1-4615-4026-7. ISBN 978-1-4615-4026-7. ^ S. M. Sze; Kwok K. Ng (2007). Physics of Semiconductor Devices (third ed.). John Wiley and Sons, Inc. ISBN 978-0-471-14323-9. ^ Choi, W.S.; Ahn, J.-K.; Park, Y.-J.; Min, H.-S.; Hwang., C.-G. (1994). "A time dependent hydrodynamic device simulator SNU-2D with new discretization scheme and algorithm". IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems. 13 (7). Institute of Electrical and Electronics Engineers (IEEE): 899–908. doi:10.1109/43.293947. ISSN 0278-0070. ^ Forghieri, A.; Guerrieri, R.; Ciampolini, P.; Gnudi, A.; Rudan, M.; Baccarani, G. (1988). "A new discretization strategy of the semiconductor equations comprising momentum and energy balance". IEEE Transactions on Computer-Aided Design of Integrated Circuits and Systems. 7 (2). Institute of Electrical and Electronics Engineers (IEEE): 231–242. doi:10.1109/43.3153. ISSN 0278-0070. ^ Sai-Halasz, G.A.; Wordeman, M.R.; Kern, D.P.; Rishton, S.; Ganin, E. (1988). "High transconductance and velocity overshoot in NMOS devices at the 0.1-μm gate-length level". IEEE Electron Device Letters. 9 (9). Institute of Electrical and Electronics Engineers (IEEE): 464–466. Bibcode:1988IEDL....9..464S. doi:10.1109/55.6946. ISSN 0741-3106. S2CID 43748586. ^ Song, J.H.; Park, Y.J.; Min, H.S. (1996). "Drain current enhancement due to velocity overshoot effects and its analytic modeling". IEEE Transactions on Electron Devices. 43 (11). Institute of Electrical and Electronics Engineers (IEEE): 1870–1875. Bibcode:1996ITED...43.1870S. doi:10.1109/16.543021. ISSN 0018-9383. ^ "6.3 Silicon Band Structure Models". ^ Cohen, Marvin L.; Bergstresser, T. K. (1966-01-14). "Band Structures and Pseudopotential Form Factors for Fourteen Semiconductors of the Diamond and Zinc-blende Structures". Physical Review. 141 (2). American Physical Society (APS): 789–796. Bibcode:1966PhRv..141..789C. doi:10.1103/physrev.141.789. ISSN 0031-899X. ^ a b c d Jacoboni, Carlo; Reggiani, Lino (1983-07-01). "The Monte Carlo method for the solution of charge transport in semiconductors with applications to covalent materials". Reviews of Modern Physics. 55 (3). American Physical Society (APS): 645–705. Bibcode:1983RvMP...55..645J. doi:10.1103/revmodphys.55.645. ISSN 0034-6861. ^ "2.5.2.4 Intervalley Phonon Scattering". ^ R. Hockney, J. Eastwood, “Computer Simulations Using Particles” McGraw Hill, Ch. 10 (1981) ^ D.K. Ferry, “Quantum Mechanics: An Introduction for Device Physicist and Electrical Engineer” Institute of Physics, ed. 1, p.186 (1995) ^ K. Hess, “Advanced Theory of Semiconductor Devices” Wiley, ed. 1, pp.94–95 (1999) ^ K. Hess, “Advanced Theory of Semiconductor Devices” Wiley, ed. 1, pp.97–99(1999) \| v t e Statistics \| \| Outline Index \| \| \| Descriptive statistics \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Continuous data \| \| \| \| --- \| \| Center \| Mean + Arithmetic + Arithmetic-Geometric + Contraharmonic + Cubic + Generalized/power + Geometric + Harmonic + Heronian + Heinz + Lehmer Median Mode \| \| Dispersion \| Average absolute deviation Coefficient of variation Interquartile range Percentile Range Standard deviation Variance \| \| Shape \| Central limit theorem Moments + Kurtosis + L-moments + Skewness \| \| \| Count data \| Index of dispersion \| \| Summary tables \| Contingency table Frequency distribution Grouped data \| \| Dependence \| Partial correlation Pearson product-moment correlation Rank correlation + Kendall's τ + Spearman's ρ Scatter plot \| \| Graphics \| Bar chart Biplot Box plot Control chart Correlogram Fan chart Forest plot Histogram Pie chart Q–Q plot Radar chart Run chart Scatter plot Stem-and-leaf display Violin plot \| \| \| \| \| Data collection \| \| \| \| \| --- \| \| Study design \| Effect size Missing data Optimal design Population Replication Sample size determination Statistic Statistical power \| \| Survey methodology \| Sampling + Cluster + Stratified Opinion poll Questionnaire Standard error \| \| Controlled experiments \| Blocking Factorial experiment Interaction Random assignment Randomized controlled trial Randomized experiment Scientific control \| \| Adaptive designs \| Adaptive clinical trial Stochastic approximation Up-and-down designs \| \| Observational studies \| Cohort study Cross-sectional study Natural experiment Quasi-experiment \| \| \| \| \| Statistical inference \| \| \| \| \| --- \| \| Statistical theory \| Population Statistic Probability distribution Sampling distribution + Order statistic Empirical distribution + Density estimation Statistical model + Model specification + Lp space Parameter + location + scale + shape Parametric family + Likelihood (monotone) + Location–scale family + Exponential family Completeness Sufficiency Statistical functional + Bootstrap + U + V Optimal decision + loss function Efficiency Statistical distance + divergence Asymptotics Robustness \| \| Frequentist inference \| \| \| \| --- \| \| Point estimation \| Estimating equations + Maximum likelihood + Method of moments + M-estimator + Minimum distance Unbiased estimators + Mean-unbiased minimum-variance - 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190092	http://www.numericana.com/answer/primes.htm	\| \| \| home \| index \| units \| counting \| geometry \| algebra \| trigonometry \| calculus \| functions analysis \| sets & logic \| number theory \| recreational \| misc \| nomenclature & history \| physics \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| --- \| \| \| \| \| \| \| Final Answers © 2000-2023 Gérard P. Michon, Ph.D.The Set of Primes God may not play dice with the universe, but something strange is going on with the prime numbers. Paul Erdös (1913-1996) We have reason to believe that [the sequence of the primes] is a mystery into which the mind will never penetrate. Leonhard Euler (1707-1783) Divisors can be defined in any additive semigroup. Relatively prime integers: The GCD of two coprime integers is 1. A prime number is a positive integer with 2 distinct divisors (1 and itself). Lemma: Every integer larger than 1 has at least one prime factor. Fundamental theorem of arithmetic. The factorization into primes is unique. Euclid's proof: There are infinitely many primes. Dirichlet's theorem: There are infinitely many primes of the form kN+a. Green-Tao theorem: Arbitrarily long arithmetic progressions of primes. Von Mangoldt's function is Log p for a power of a prime p, 0 otherwise. Prime Number Theorem: The probability that N is prime is roughly 1/ln(N). Riemann's power-prime counting function (J). Riemann's explicit formula. The average number of factors of a large number N is Log N. The average number of distinct prime factors of N is Log Log N. The largest known prime: Historical records, old and new. The Lucas-Lehmer Test checks the primality of a Mersenne number fast. Proth primes are almost as fast to check as Mersenne primes. Sierpinski numbers are moduli for which no Proth number is prime. Pratt primality certificates give quick automated proofs of primality. The AKS test determines the primality of any number in polynomial time. Formulas giving only primes may not help with new primes. Ulam's Lucky Numbers and other sequences generated by sieves. Related articles on this site: Ring theory \| Category theory. Unique factorization of Gaussian integers into Gaussian primes. The Riemann Hypothesis: {Re(z) > 0 & z(z) = 0} Þ {Re(z) = ½}. The Twin-Prime Conjecture: Infinitely many pairs of primes differ by 2. Number of squarefree numbers (or cubefree numbers) not exceeding N. Carmichael Numbers: An absolute pseudoprime n always divides (an-a). Pseudoprimes & primality tests \| Modular arithmetic \| p-adic arithmetic. Prime factorizations: Resolving composite integers into their prime factors. 1 isn't a prime number (or else the fundamental th. of arithmetic would fail). Wieferich primes & base 68. Factoring Wendt numbers, M708 & M1860. \| \| \| Related Links (Outside this Site)Primality Tests by Evangelos Kranakis (Yale, Dec. 1984). The n-1 and n+1 primality tests by Curtis Bright, INTP (2013-10-09). Structure and randomness in the prime numbers. by Terence Chi-Shen Tao. How Many Primes Are There? by Chris K. Caldwell. Proving Primality in Polynomial Time by Chris Caldwell. Dirichlet's Theorem: A real variable approach. by Robin Chapman. Sloane's On-Line Encyclopedia of Integer Sequences Records for prime numbers by Paul Zimmermann. \| 1964: C.P. Willans' formula for the nth prime (14:46) by Eric Rowland (2022-09-22). The Mathematical Gazette, 48, 366, 413-415 (DBradford ec. 1964) Christopher Paul Willans (1942-1971) was born in Bradford (1942-01-17) and attended Bradford Grammar School from 1953 to 1959 (O-levels in 1958). He was at the University of Birmingham in 1964 and died in Bradford in 1971. \| \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Introduction to Prime Numbers With our order of presentation, fundamental concepts aren't used before they're introduced. The edifice rests on greatest common divisors, Euclid's algorithm and Bézout's lemma. The basic structure of the prime numbers can then be freely examined in the second part. (2016-11-07) Divisors (or Factors ) and divisibility relation : Divisors are well-defined within any additive semigroup. In an additive semigroup (or, more generally, a magma merely endowed with a power-associative addition) an element x is said to be a divisor (or factor ) of another element y (we also say that "x divides y") when the latter is the sum of one or more terms equal to the former: x + x + ... + x = y The divisibility relation "x divides y" is a preorder relation, since: It's reflexive (any element, including zero, divides itself ). It's transitive (i.e., if x divides y and y divides z, then x divides z) because addition is associative. (Power-associativity is enough.) As usual, the qualifier proper can be used to disallow reflexivity. Thus, x is never a proper divisor of itself. It's paramount to define a concept as general as this one in the poorest of cases (to avoid inferior definitions in richer contexts). Here for example, we must disallow empty sums at the outset because such things are only defined when there's a neutral element for addition. Thus, when an additive element (zero) does exist, the existence of proper divisors of zero isn't trivial (it's an important consideration with the generalized notion of divisor used in Ring Theory). For the monoid of nonnegative integers, a third property is true, which makes divisibility a full-fledged partial ordering relation : It's antisymmetric (i.e., if x divides y and y divides x, then x = y). Proof : We first remark that, among nonnegative integers, a number is always greater than or equal to any of its divisors. Then, the antisymmetry of divisibility is obtained from the antisymmetry of the usual ordering. Note that zero (the neutral element for addition) divides only itself. Although zero is a multiple of any other integer, none is a divisor of zero. (2015-01-25) Relatively prime (or coprime ) integers. Two integers are coprime when their greatest common divisor is 1. The greatest common divisor (GCD) of two integers is defined as the largest positive integer that divides them both. The GCD is efficiently computed using Euclid's algorithm (or, far less efficiently, with the subtractive version which can be more convenient for theoretical proofs). Likewise, several integers are coprime when their only common positive divisor is 1 (French: nombres premiers entre eux, nombres étrangers ). Such numbers need not be pairwise coprime (e.g., 6, 10 and 15). Theorem : Arguably, the most important fact about coprime integers is: If the number d divides m n and is coprime with n, then d divides m. Proof : Using Bézout's lemma, (itself a consequence of Euclid's algorithm, quoted in the above definition of coprime numbers) we know that there exists two integers u and v such that: u d + v n = 1 Since d divides m n it also divides the following quantity: u d m + v m n = m ( u d + v n ) = m Note that this result is crucial in the proof of the fundamental theorem of arithmetic given below. (In other words, the above establishes the unicity of prime factorizations; it's not the other way around.) Corollary : If k is coprime with n, then k Ù m n = k Ù m Proof : The main theorem says that any divisor d of m n divides m. So d = k Ù m n divides m and k. Therefore, this d divides their GCD (k Ù m). The converse is trivially true, so the two are equal. (2006-11-25) Prime Integers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 ... (A000040) The symbol denotes the set of all primes. A positive integer p is said to be prime when it has just two distinct divisors among positive integers (1 and p). Besides the number 1 (one) itself, which is not considered prime, any positive integer is thus either a prime or a composite number (namely, the product of two lesser factors). Two distinct prime numbers p and q are clearly coprime (HINT: the set of their common divisors is the intersection of {1,p} and {1,q}). That fact, known as the coprimality of primes, implies that a prime which divides a product of distinct primes must be equal to one of them (HINT: Prove this by induction on the number of factors). That's needed to prove the fundamental theorem of arithmetic. The First 360 Primes ( p0 = 2 to p359 = 2423 ) \| 0 \| 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 \| \| 10 \| 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 \| \| 20 \| 73, 79, 83, 89, 97, 101, 103, 107, 109, 113 \| \| 30 \| 127, 131, 137, 139, 149, 151, 157, 163, 167, 173 \| \| 40 \| 179, 181, 191, 193, 197, 199, 211, 223, 227, 229 \| \| 50 \| 233, 239, 241, 251, 257, 263, 269, 271, 277, 281 \| \| 60 \| 283, 293, 307, 311, 313, 317, 331, 337, 347, 349 \| \| 70 \| 353, 359, 367, 373, 379, 383, 389, 397, 401, 409 \| \| 80 \| 419, 421, 431, 433, 439, 443, 449, 457, 461, 463 \| \| 90 \| 467, 479, 487, 491, 499, 503, 509, 521, 523, 541 \| \| 100 \| 547, 557, 563, 569, 571, 577, 587, 593, 599, 601 \| \| 110 \| 607, 613, 617, 619, 631, 641, 643, 647, 653, 659 \| \| 120 \| 661, 673, 677, 683, 691, 701, 709, 719, 727, 733 \| \| 130 \| 739, 743, 751, 757, 761, 769, 773, 787, 797, 809 \| \| 140 \| 811, 821, 823, 827, 829, 839, 853, 857, 859, 863 \| \| 150 \| 877, 881, 883, 887, 907, 911, 919, 929, 937, 941 \| \| 160 \| 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013 \| \| 170 \| 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069 \| \| 180 \| 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151 \| \| 190 \| 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223 \| \| 200 \| 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291 \| \| 210 \| 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373 \| \| 220 \| 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451 \| \| 230 \| 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511 \| \| 240 \| 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583 \| \| 250 \| 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657 \| \| 260 \| 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733 \| \| 270 \| 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811 \| \| 280 \| 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889 \| \| 290 \| 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987 \| \| 300 \| 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053 \| \| 310 \| 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129 \| \| 320 \| 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213 \| \| 330 \| 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287 \| \| 340 \| 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357 \| \| 350 \| 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423 \| (2006-11-25) Lemma [ Euclid's Elements VII, 31 ] Any integer greater than 1 has at least one prime factor. This humble statement is often overlooked. Yet, it's a required preliminary to two otherwise unrelated proofs for two fundamental theorems : The fundamental theorem of arithmetic: Every positive integer has a unique factorization into primes. Euclid's theorem: There are infinitely many primes. Proof : We proceed by induction, using the fact that any integer greater than 1 is either prime or composite. If such an integer is prime, we're done (the prime divisor we're after is the number itself). Otherwise, it has a lesser divisor greater than 1. Using the induction hypothesis, that divisor has at least one prime factor. By transitivity, that prime factor also divides the original number. (2006-11-25) Fundamental Theorem of Arithmetic Every positive integer has a unique factorization into primes. For example: 168 = 2 3 3 7 Such a factorization can be specified by the sequence of the exponents to which the successive primes should be raised. In the example of 168, this is: ( 3, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ... ) We say that "2 has multiplicity 3 in 168". The multiplicities of 3 and 7 in 168 are equal to 1. Any other prime has multiplicity 0. When discussing the prime divisors of a certain integer, it's sometimes more elegant to consider the multiplicity of a prime without assuming it's nonzero. (On 2015-12-31, I found it convenient to formulate this way a conjecture about the prime divisors of the form 6n+1 of the Wendt determinant of odd order n). Conversely, any sequence of nonnegative integers with finitely many nonzero elements is associated with a unique positive integer. In the factorization of the integer 1, all exponents are zero: ( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... ) The sequence of exponents associated with the product of two positive integers is the direct sum of the sequences associated with those factors. Such sequences can be represented by lists ending at the last nonzero term. That would be the list (3,1,0,1) for 168 and the empty list ( ) for 1. Proof of the Fundamental Theorem of Arithmetic : Using the above lemma, we first prove (by induction) that any positive integer n has a prime factorization (i.e., it's a product of prime numbers). Gauss (1798) didn't think such a proof was needed. I beg to differ: If n = 1, the factorization we seek is simply empty. Otherwise, the lemma says that n has a prime divisor p. The induction hypothesis tells us that the positive integer n/p has a prime factorization. Combining p with that factorization, we obtain a prime factorization of n. To establish that the prime factorization of an integer is unique, we remark that dividing repeatedly two such factorizations by any prime they have in common, must yield an empty product. Otherwise, we'd have a prime dividing a product of primes without being equal to any of them (which contradicts the coprimality of distinct primes). Unique factorization domain (2006-05-25) Euclid's Theorem (c. 300 BC) There are infinitely many prime numbers. All we have to prove is the existence of a prime above any given prime P. This can be established in just two lines (one of the greatest proofs ever): If Q is the product of all primes less than or equal to P, then any prime factor of Q+1 can't divide Q and must, therefore, be a prime greater than P. Euclid's original proof isn't as simple but it's arguably more elementary because it doesn't involve the product of an unbounded number of factors, which ancient Greeks may have found repugnant. It appears as proposition 20 in Book IX of The Elements. This great proof is often needlessly presented as a proof by contradiction (as Euclid himself did). It's also often obscured by preliminary discussions about such things as the nature of infinity, which aren't needed either. Finally, the need for a not-so-obvious lemma (duly noted by Euclid) is rarely stated, if ever. \| \| \| \| \| \| \| \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| The Set of Prime Numbers (2007-04-30) Dirichlet's theorem on primes in arithmetic progressions: If a and N are coprime, infinitely many primes are of the form kN+a. This statement was conjectured by Gauss (Euler had previously stated the special case a = 1). It was proved by Dirichlet in 1837, using the Dirichlet characters and related L-series which he introduced for that very purpose. There are infinitely many primes in all arithmetic progressions with coprime elements (Dirichlet, 1837-07-27). (2007-04-30) Green-Tao Theorem (2004) There are arbitrarily long prime arithmetic progressions (PAP). This was proved in 2004 by Terence Tao (1975-) and Ben Green (1977-). In 1975, Endre Szemerédi (Abel prize 2012) proved that any set of integers of positive upper density contains arbitrarily long arithmetic progressions (Szemerédi's theorem). This had been conjectured by Erdös (1913-1996) and Turán (1910-1976) in 1936. That result doesn't apply to the set of primes (which has zero upper density) but Erdös strengthened the Erdös-Turán conjecture in 1973 to state that arbitrarily long arithmetic progressions exist in any set of positive integers when the series of its reciprocals diverges. That's a very strong statement which would imply the Green-Tao theorem of 2004. In 2006 (pdf), Tao and Tamar Ziegler generalized that, by showing that there are arbitrarily long polynomial progressions of primes. More precisely: For any sequence of k integer-valued polynomials (Q1, Q2 ... Qk) and any positive e, there are infinitely many choices of integers x and m < xe which make all expressions x+mQi(m) prime. The original Green-Tao theorem corresponds to the special case Qi(m) = i. The existence of infinitely many arithmetic progressions of length 3 among primes was established in 1939, by the Dutch mathematician Johannes van der Corput (1890-1971). Originally, Green and Tao wanted to prove that there are infinitely many equally spaced sequences of 4 primes, but they saw that their methods proved the existence of such sequences of any length... Smallest Prime Arithmetic Progressions (PAP) of Given Length \| Author \| Length \| N \| Prime Numbers a + kN \| \| \| 1, 2 \| 1 \| 2, 3. \| \| 3 \| 2 \| 3, 5, 7. \| \| 4, 5 \| 6 \| 5, 11, 17, 23, 29. \| \| G. Lemaire (1909) \| 6 \| 30 = 5# \| 7, 37, 67, 97, 127, 157. \| \| 7 \| 150 \| 7, 157, 307, 457, 607, 757, 907. \| \| Edward B. Escott (1910) \| 8, 9 10 \| 210 = 7# \| 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089. \| \| Edgar Karst (1967) \| 11 12 \| 13860 = 6 . 11# \| 110437, 124297, 138157, 152017, 165877, 179737, 193597, 207457, 221317, 235177, 249037, 262897. \| \| V. N. Seredinskij (1963) \| 13 \| 60060 = 2 . 13# \| 4943, 65003, 125063, 185123, 245183, 305243, 365303, 425363, 485423, 545483, 605543, 665603, 725663. \| \| Paul A. Pritchard (1983) \| 14 \| 14 . 13# \| 31385539, 31805959 ... 36850999. \| \| 15 \| 138 . 13# \| 115453391, 119597531 ... 173471351. \| \| Sol Weintraub (1976-1977) \| 16 \| 323 . 13# \| 53297929, 62997619 ... 198793279. \| \| 17 \| 171 . 17# \| 3430751869, 3518049079 ... 4827507229 \| \| Paul A. Pritchard (1984) \| 18 \| 1406 . 17# \| 4808316343 ... 17010526363 \| \| 19 \| 431 . 19# \| 8297644387 ... 83547839407 \| \| Jeff Young & James Fry (1987) \| 20 \| 1943 . 19# \| 214861583621 ... 72945039351 \| \| Pritchard (1992) \| 21 \| 2681 . 19# \| 5749146449311 ... 6269243827111 \| If an arithmetic progression (AP) of k primes starts above k, then its common difference (N) must be a multiple of all the primes less than or equal to k (or else, one such prime would be a proper factor of a term in the progression). This is made explicit in the above, using the (fairly standard) notation p# to denote the primorial of p, namely the product of all primes between 2 and p (A002110). Recent Results and New Records : Since the record breakers are rarely found in strict order of size, we can't reliably extend the above table. Instead, we'll just mention some newer results, starting with the first known example of 22 primes in arithmetic progression, discovered by Andrew Moran, Paul A. Pritchard and Anthony Thyssen on 1993-03-17: 11410337850553 + 4609098694200 k (k = 0 to 21). On 2004-07-24, Markus Frind, Paul Jobling and Paul Underwood found an arithmetic progression of 23 primes (ending with 449924511422857) : 56211383760397 + 44546738095860 k (k = 0 to 22). A smaller instance (ending with 1036239621869317) was later found by Frind (2006-04-01) featuring a common difference N = 9523 . 23#. 403185216600637 + 2124513401010 k (k = 0 to 22). The first arithmetic progression of 24 primes was discovered by Jaroslaw "Jarek" Wróblewski on 2007-01-18: 468395662504823 + k . 205619 . 23# (k = 0 to 23). Wróblewski and Raanan Chermoni found a PAP of length 25 on 2008-05-17: 6171054912832631 + k . 366384 . 23# (k = 0 to 24). In a distributed PrimeGrid project using software written by Jaroslaw Wroblewski and Geoff Reynolds, the Playstation 3 of Benoît Périchon found 26 primes in arithmetic progression on 2010-04-12: 43142746595714191 + k . 23681770 . 23# (k = 0 to 25). A204189 The second example, discovered on 2012-03-16 by James Fry, is smaller: 3486107472997423 + k . 1666981 . 23# (k = 0 to 25). The first example of 27 primes in arithmetic progression was discovered by Rob Gahan with PrimeGrid, on 2019-09-23: 224584605939537911 + k . 81292139 . 23# (k = 0 to 26). Consecutive Primes in Arithmetic Progression : It's much more difficult to find consecutive primes in arithmetic progression. A sequence of length 10 was first found on 1998-03-02. Namely, P + 210 k (for k = 0 to 9) with the following 93-digit value for P. 100996972469714247637786655587969840329509324689190041803603417758904341703348882159067229719 It's highly unlikely that a longer sequence (length 11) will be found any time soon, although it's conjectured that there are infinitely many instances of k consecutive primes in arithmetic progression, for any k... Van der Waerden theorem (1927) Green-Tao theorem \| PAP \| Ten consecutive primes in arithmetic progression Primes in Arithmetic Progression Records (page created and maintained by Jens Kruse Andersen). (2009-06-26) The von Mangoldt function L(n) L(n) = Log p if n is a power of the prime p. Otherwise, L(n) = 0 The fundamental theorem of arithmetic is essentially equivalent to the following relation, which states that the logarithm of an integer is equal to the sum of the values of the von Mangoldt function over all its divisors: \| \| \| \| --- \| Log n = \| å \| L(d) \| \| \| d \| n \| \| That Dirichlet convolution can be inverted, using the Möbius function: \| \| \| \| \| \| --- --- \| L (n) = \| å \| m(d) Log (n/d) = \| å \| - m(d) Log d \| \| \| d \| n \| \| d \| n \| The function L is named after Hans von Mangoldt (1854-1925) who got his doctorate at Berlin under Weierstrass and Kummer. He paved the way for the two proofs of the Prime number theorem (1896) by providing rigorous proofs for two statements which had only partial proofs in Riemann's 1859 seminal paper on the Zeta function. In 1895, von Mangoldt proved a conjecture stated by Bernhard Riemann (1826-1866) in 1859. The number of roots of the zeta function with a positive imaginary part less than 2kp is indeed: k Log k - k + O(Log k) Mangoldt Function (Mathworld) von Mangoldt function (Monday Math 95) by Twisted One 151 (2006-11-25) PNT: The Prime Number Theorem (1896) A random integer N is prime with a probability roughly equal to 1/ln(N). In 1792, at age 15, Gauss made the above statement as a private conjecture in his notebook. This is usually recast in terms of various (equivalent) approximations to the prime-counting function which lets p(x) denote the number of primes below x : There's no controversy about the value of p(x) between prime values of x. At those points of discontinuity however, the modern convention is used (consistent with the results of Fourier analysis) which makes the value at a jump discontinuity equal to the half-sum of the left and right limits. Bizarre as it may look at first, this normalization is also adopted for other counting functions, including the prime-power counting function (J) discussed below. p(1) = 0 , p(2) = 1/2 , p(3) = 3/2 , p(4) = 2 , p(5) = 5/2 , p(6) = 3 , ... In terms of Kronecker's signum and/or Heaviside's step function : \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| --- --- \| p(x) \| = \| \| å \| ½ [ 1 + sgn (x-p) ] \| \| \| \| \| p prime \| \| \| = \| \| å \| H ( x-p ) \| \| \| \| \| p prime \| \| This flavor of the prime counting function is what Riemann described when he introduced the concept in 1859 (in his paper, he used the symbol F for this). That's sometimes called the normalized prime counting function, using a nought subscript to single it out. However, since there's no agreement on what the unnormalized version should be, I recommend dropping both the qualifier and the subscript ! In 1808, Legendre proposed the following approximation, involving a parameter B (about -1.08366) sometimes known as Legendre's constant. (For extremely large values of N, the best value of B would be 1.) p(x) ~ x / (B + ln x) The aforementioned conjectural remark of the young Gauss is equivalent to equating p(x) and either flavor of the logarithmic integral li(x) or Li(x). Gauss put it in this form in 1849 (although his remark appeared in print only posthumously, in 1863). The resulting statement became known as the prime number theorem (PNT). \| \| \| \| --- \| p(x) \| ~ \| li(x) ~ Li(x) \| \| \| ~ \| x / ln(x) + x / (ln x)2 + 2x / (ln x)3 + ... + k! x / (ln x)k+1 + ... \| Although it would be better to retain the first 3 terms in the above asymptotic expansion of the logarithm integral, the PNT is usually stated by retaining only the leading term x/ln(x). Namely: Prime-Number Theorem (PNT) \| \| \| \| \| \| \| --- --- \| \| lim x® ¥ \| p(x) \| = 1 \| \| \| \| \| x / Log x \| \| In 1848, well before this relation was unconditionally proved, Chebyshev showed that if the limit exists, then it must be 1. (video by Ram Murty). Two independent proofs of this famous theorem were given in 1896, by Hadamard and Vallée-Poussin. In 1951, Wiener made clear that both of those proofs rely on the established fact that the Riemann Zeta Function z does not have any zeroes of the form 1+it. A similar remark holds for a third PNT proof given in 1903 by Edmund Landau (1877-1938) as a prelude to his proof of the prime ideal theorem for algebraic number fields. In 1949, a beautiful elementary proof of the PNT was again found by two mathematicians simultaneously: Paul Erdös and Atle Selberg. The celebrated Riemann Hypothesis (which states that the nontrivial zeroes of z are all of the form ½+it ) is equivalent to the following statement: p(x) = Li(x) + O(Öx ln x) = li(x) + O(Öx ln x) Against all numerical evidence, which never shows p(x) above li(x), John E. Littlewood (1885-1977) proved, in 1914, that the sign of p(x)-li(x) changes infinitely many times. It's now known that the first such reversal of sign must happen for some number x with 370 digits or fewer. \| n \| p(n) \| li(n) \| n / ln(n) \| --- --- \| \| 2 \| 0.5 \| 1.045 \| 2.885 \| \| 3 \| 1.5 \| 2.164 \| 2.730 \| \| 4 \| 2 \| 2.968 \| 2.885 \| \| 5 \| 2.5 \| 3.635 \| 3.107 \| \| 10 \| 4 \| 5.120 \| 4.343 \| \| 100 \| 25 \| 30.126 \| 21.715 \| \| 1000 \| 168 \| 177.610 \| 144.765 \| \| 10000 \| 1229 \| 1246.137 \| 1085.736 \| \| 100000 \| 9592 \| 9629.809 \| 8685.890 \| \| 1000000 \| 78498 \| 78627.549 \| 72382.414 \| \| 10000000 \| 664579 \| 664918.405 \| 620420.688 \| \| 100000000 \| 5761455 \| 5762209.365 \| 5428681.024 \| \| 1000000000 \| 50847534 \| 50849234.957 \| 48254942.434 \| \| 10000000000 \| 455052511 \| 455055614.587 \| 434294481.904 \| Prime Number Theorem (Theorem of the Day #33) by Robin Whitty (2014-07-24) Riemann's weighted prime-power counting function J Giving partial credit to the powers of primes. It turns out that Euler's Zeta function, the undisputed Queen of analytic number theory, is directly tied with a special counting function closely related to the prime-counting function: Instead of just counting the primes below x, like the function p does, the function J also gives a fractional score of 1/n to the n-th power of a prime. Since there are clearly p ( x1/n) such numbers below x, we have: J ( x ) = p ( x ) + p ( x1/2) / 2 + p ( x1/3) / 3 + p ( x1/4) / 4 + ... J has jump discontinuities at prime-powers, where its value is defined as the half-sum of its lower and upper limits (to be consistent with Fourier analysis). The above formula can be inverted using the Möbius function m : p ( x ) = J ( x ) - J( x1/2) / 2 - J( x1/3) / 3 + ... + m(n) J( x1/n) / n + ... Proof : To check this, use the above definition of J to expand the RHS and then, for every integer k, regroup all terms where x is raised to 1/k : å m å n m(n) p( x1/mn) / mn = å k [ å n\|k m(n) ] p( x1/k ) / k The bracketed sum is a known function of k which vanishes unless k=1. J is sometimes termed Riemann's Jump Function. Bernhard Riemann introduced it (using the symbol f at the time) in his famous 1859 paper on the Zeta function, as part of the following construction... The paper entitled "Über die Anzahl der Primzahlen unter einer gegebenen Grösse" (Nov. 1859) is the only one Riemann ever wrote on the subject of the Zeta function (z) and/or Number Theory. In it, he formulates the wonderful conjecture now known as the Riemann hypothesis (RH) but the rest of the paper doesn't depend on that. Riemann's own words were: "It's very probable that all roots are real. One would surely wish for a rigorous proof here. I have temporarily put aside the search for one, after some futile attempts, because it's not necessary for the immediate goal of my study." When Re(z) > 1, the series are absolutely convergent and we may write: Log z(z) = å - Log ( 1-p-z ) = åp-z + åp-2z/2 + åp-3z/3 + åp-4z/4 + ... Riemann then substitutes every occurrence of p-kz with a definite integral: \| \| \| \| \| --- --- \| \| p-kz = z \| ò \| ¥ \| x-z-1 dx \| \| pk \| This turns the whole sum into an integral involving the above function J : \| \| \| \| \| --- --- \| \| Log z(z) = z \| ò \| ¥ \| J(x) x-z-1 dx \| \| 0 \| Riemann's Explicit Formula (1859) : Now comes the great part: The above integral transformation would later be called a Mellin transform (Hjalmar Mellin was only 5 years old at the time). Riemann realized that it could be inverted to retrieve J from z. As we already know how the prime-counting function is derived from J, this will give us an explicit way to obtain it from z... The sum is understood to be performed over all the zeroes of the Zeta function, taken in increasing order of magnitude (the value of the sum depends on this order because the series isn't absolutely convergent). Riemann Prime-Counting Function(s) by Eric W. Weisstein ("J" is called "f" in that article) Wikipedia : Explicit formulae for L-functions \| Mellin inversion theorem \| Chebyshev function (2009-06-26) Number of divisors of an integer N A large number N is expected to have about Log N divisors. In 1838, Dirichlet evaluated the average number of divisors of all positive numbers not exceeding N. This involves the Euler-Mascheroni constant g : Log N + 2 g - 1 » Log N + 0.15443... (2009-06-26) On the number w(N) of prime factors of an integer N A large number N has about Log Log N prime factors (1917) The number of distinct prime factors of n is traditionally denoted w(n). The function w (see A001221) is an additive function. That's to say: w( ab ) = w( a ) + w( b ) whenever a and b are coprime. In 1917, G.H. Hardy and S. Ramanujan proved w(n) to be asymptotically equal to Log Log n . In 1933, Paul Turán (1910-1976) gave an innovative one-page proof of that fact, also featured in his doctoral dissertation (1934). (2022-09-17) The Erdös-Kac Theorem (1940) Fundamental law of probabilistic number theory. w( N ) is normally distributed about its mean. Erdös-Kac Theorem (1940) \| Paul Erdös (1913-1996) \| Mark Kac (1914-1984) Paul Erdö et l'anatomie des entiers (1:30:59) by Gérald Tenenbaum, X1972 (BNF, 2017-02-22) (2006-05-24) The Largest Known Prime Until a fast formula is found, the record will be broken again and again. The largest known prime has very often been of the form 2n-1. Such numbers are called Mersenne numbers and their prime values are known as Mersenne primes (we discuss elsewhere their history, the special form of their factors and the connection with perfect numbers). It's easy to see that a Mersenne number can't be prime unless the exponent (n) is itself prime. (This happens to be also a consequence of a nice general property of integer sequences which start with 0 and 1 and obey a second-order recurrence, as we demonstrate elsewhere.) The primality of the exponent is not sufficient. For example, the 11th Mersenne number 2047 is the product of 23 and 89, whereas the 23rd is divisible by 47... Nevertheless, the primality of Mersenne primes is (currently) significantly easier to establish than that of all other integers of similar magnitudes. The gap between the Mersenne primes 2127-1 and 2521-1 was sufficiently large to allow other approaches to break the record, as documented in the table below. This happened again between the discovery of the primality of 2216091-1 (Slowinski, 1985) and that of 2756839-1 (Slowinski, Gage et al., 1992) when an "Amdahl 1200" computer was used to prove the primality of the following number (J. Brown, C. Noll, B. Parady, G. Smith, J. Smith and S. Zarantonello, 1989). 391581 ´ 2 216193 - 1 Since 1996, the scene has been dominated by the "Great Internet Mersenne Prime Search" (GIMPS) which has harnessed thousands of microcomputers and found all the latest record primes (see GIMPS for an update). The "Largest Known Prime", by Date (until the dawn of the Computer Era) \| When \| Who \| How \| Expression \| Digits \| \| January 30, 1952 \| Raphael M. Robinson \| SWAC \| 2607 - 1 \| 183 \| \| 2521 - 1 \| 157 \| \| Early July 1951 (see note below) \| J.C.P. Miller D. J. Wheeler \| EDSAC \| 180 (2127 - 1 )2 + 1 \| 79 \| \| Aimé Ferrier \| Mechanical Desk Calculator \| (2148+1) / 17 \| 44 \| \| June 1951 \| J.C.P. Miller D. J. Wheeler \| EDSAC \| 978 (2127 - 1) + 1 \| 42 \| \| June 7, 1951 \| 934 (2127 - 1) + 1 \| \| May-June, 1951 \| k (2127 - 1) + 1 for k = 696, 738, 774, 780 \| \| \| k (2127 - 1) + 1 for k = 114, 124, 388, 408, 498 \| \| 41 \| \| 1876 \| Edouard Lucas \| Lucas Test \| 2127-1 \| 39 \| \| 1867 \| Fortuné Landry \| Trial Division (Optimized) \| (259-1) / 179951 \| 13 \| \| 1851 \| W. Looff \| 1012 - 106 + 1 \| 12 \| \| 1772 \| Leonhard Euler \| 231-1 \| 10 \| \| 1732 \| (232+1) / 641 \| 7 \| \| 1588 \| Pietro Cataldi \| Trial Division \| 219-1 \| 6 \| \| 217-1 \| Le Lasseur's Number : The 17-digit number 13373763765986881 divides the 360th Fibonacci number. It was the second-largest known prime when it was discovered in 1879 by the noted French amateur Henri Le Lasseur (Henri Le Lasseur de Ranzay, 1821-1894; also spelled "de Ranzey"). It has been argued that this number was (briefly) the largest "known" prime, as the prime status of M127 (39 digits) was supposedly not firmly established at the time. This ain't so, unfortunately. The heroic proof of the primality of a "general" number as large as Le Lasseur's number would have deserved a bright spot in the record books, instead of a mere footnote... Le Lasseur's number was then the largest prime whose primality had been established by general-purpose methods. It's more than 4000 times larger than the runner-up 3203431780337 (Landry's number, 1867). By contrast, the Frenchman Aimé Ferrier officially reported his 44-digit record-breaking prime on Bastille day, July 14, 1951. He had been working on this since May and Jeff Miller may have been aware of Ferrier's ongoing work. According to the 1997 recollections of family member David Miller as reported by Chris Caldwell : "Jeff Miller went to some length to make sure Ferrier's result was not overlooked ". Miller may well have changed his original strategy (leading to his 79-digit record) specifically to beat Ferrier's upcoming result which would have overshadowed Miller's other results (the first of which broke the 75-year old record of Lucas). However, Miller deliberately reported both his own 79-digit number and Ferrier's 44-digit prime as having been discovered "in early July". This may have been a professional courtesy to Ferrier, although a deeper inquiry (which probably never took place) may or may not have revealed that Ferrier's results came a few hours too late to enter the record book. Giving priority to Ferrier puts both numbers in the record book and still gives credit to Miller and Wheeler for having broken the long-standing record of Lucas with the earliest of their 41-digit numbers. Ferrier's number itself stands out as the largest prime ever discovered without the help of an electronic computer. This healthy ambiguity ought to be strictly respected now. This is just what D.H. Lehmer did when he summarized the "Recent Discoveries of Large Primes" very shortly after those events [MTAC, 5, 36, Oct. 1951]. A machine printed the primality of 24253-1 before that of 24423-1. However, a human being (Alexander Hurwitz) read about the latter before anybody knew about the former, which was thus never largest anong "known primes" (for more details, see our presentation of Mersenne primes and perfect numbers). Largest Known Prime by Year (Chris Caldwell) (2007-05-08) The Lucas-Lehmer Test (1930) A fast way to check the primality of the pth Mersenne number 2p-1. The Lucas-Lehmer test is a special case of the modern way to check the primality of n when all the prime factors of n+1 are known. It boils down to a procedure devised in 1878 by Edouard Lucas (1842-1991) and streamlined by D.H. Lehmer (1905-1991) in 1930: Consider the following recursively-defined sequence, modulo 2p-1 L0 = 4 Ln+1 = Ln2 - 2 [mod 2p-1] For an odd prime p, 2p-1 is prime if and only if Lp-2 is zero. That's all! So, the primality of 2p-1 can be determined with just p-2 multiplications. The Lucas-Lehmer Test (Theorem of the Day #127) by Robin Whitty (2017-11-13) Proth Numbers & Proth Primes (Proth's theorem, 1878) Proth numbers are integers of the form k 2n + 1 with k < 2n Proth's theorem (1878): A Proth number N = k 2n + 1 [ with k < 2n ] is prime if and only if a(N-1)/2 is congruent to -1 modulo N, for some a. Proof : The condition is clearly necessary because it's satisfied for any odd prime N, by Euler's criterion, for exactly 50% of the possible positive choices for a (namely, the so-called quadratic nonresidues). It's more delicate to establish that the stated condition is sufficient. Anachronistically, that's a corollary of the Pocklington-Lehmer theorem. Thus, the primality of a Proth number N can be easily checked by exponentiation modulo a suitable base a. The test itself does tell if the base is suitable and 50% of the random choices are. François Proth (1852-1879) \| Pépin's test (1877) \| Théophile Pépin (1826-1904) A primality test for k.pn+1 by José Maria Grau & Antonio M. Oller-Marcén (2011-04-26). Deterministic Primality Proving on Proth Numbers by Tsz-Wo (2011-07-15). Toward a proof of Proth's theorem (AoPS, 2008-01-22 & 2013-11-12) n-1 tests and Pepin's tests for Fermats by Chris Caldwell. 78557 and Proth Primes (8:39) by James Grime (Brady Haran's Numberphile, 2017-11-13). (2017-11-13) Sierpinski Numbers [of the second kind] Odd numbers k for which k 2n + 1 is never prime. A Sierpinski number is an odd modulus k for which the Proth number k 2n + 1 is never prime, for any integer n. There are infinitely many Sierpinski numbers. At this writing, the smallest known one is 78557. It is due to John Selfridge (1927-2010) who showed, in 1962, that: 78557 . 2n + 1 is divisible by 3, 5, 7, 13, 19, 37 or 73 In 1967, Selfridge and Sierpinski conjectured that 78557 was the smallest Sierpinski number. In 2002, only 17 lesser possibilities remained open and the Seventeen or Bust online collaboration was formed to tackle them all. Between March 2002 and April 2016, they managed to eliminate eleven of those candidates, before merging with the PrimeGrid project, which soon eliminated one more (10223) by using Proth's theorem to prove the primality of the following 9383761-digit number: 10223 . 231172165 + 1 [ Halloween Prime, discovered on 2016-10-31 ] As of November 2017, the above is the seventh-largest known prime (all of the larger ones are Mersenne primes). Only 5 remaining numbers could falsify the Selfridge-Sierpinski conjecture: 21181, 22699, 24737, 55459, 67607 Waclaw Sierpinski (1882-1969) \| John Selfridge (1927-2010) "Sierpinski numbers of the first kind" nn+1 (A014566) are unrelated to the above Sierpinski numbers. 78557 and Proth Primes (8:39) by James Grime (Brady Haran's Numberphile, 2017-11-13). (2017-08-03) Pratt's theorem & Pratt primality certificates (1975) Keys to a quick proof that a given number is prime. The converse of Fermat's little theorem can be used to design a nondeterministic algorithm which can prove the primality of a prime in polynomial time. This establishes that PRIMES (the set of all primes numbers, expressed as strings of digits) is in NP (Pratt's theorem, 1975). The lucky guesses in that nondeterministic algorithm can be construed as a standardized proof that a given number is prime; they form what is known as a Pratt primality certificate. The validity of such a certificate can be checked deterministically rather quickly (in polynomial time). A Pratt certificate for a prime p is recursively defined as consisting of: A generator (witness) for the multiplicative group (/p) The prime factorization of p-1. A Pratt certificate for each prime in that factorization. Pratt certificates are typically recorded in databases of very large primes, including The Prime Pages of Chris Caldwell. Pratt's theorem and primality certificates by Robin Whitty (Theorem of the Day). Wikipedia : Vaughan Pratt (2017-08-03) From Pratt's theorem (1975) to the AKS test (2002) : Telling whether any number is prime or composite, in polynomial time. The AKS algorithm was a great theoretical breakthrough which proved unconditionally that the primality of an integer can be checked deterministically in polynomial time. However, for all practical purposes, the randomized Rabin-Miller test remains unsurpassed. PRIMES is in P by Manindra Agrawal, Neeraj Kayal & Nitin Saxena (2002). Fool-Proof Test for Primes (3:42) by James Grime (Brady Haran's Numberphile, 2014-02-06). (2006-05-24) Is there a formula which gives only primes? What's a "formula" anyway? As of this writing, checking the primality of 2p-1 for increasingly large values of the exponent p is the most efficient way to name large primes (see GIMPS). Anything faster than that would be major news. Something vastly faster could essentially end the above record-breaking game by making it easy to "name" explicitly primes with so many digits that they could not possibly be written down. This would not necessarily end the search for primes of a specific type (like "Mersenne primes") but it would make the title of "largest known prime" as meaningless as that of "largest known integer" (whatever integer is named, something like "two to the power of that" will name something vastly larger). It's not enough to have a "formula" for larger primes. It must be an effective formula... The formula discussed next isn't an effective one! Mills' Formula : There are uncountably many values of x for which x3n will always round down to a prime integer. This was shown in 1947 by a student of Emil Artin at Princeton: William H. Mills (1921-2007 ?). Mills used deep results about prime gaps which had been obtained by Guido Hoheisel (1894-1968) in 1930 and by Albert Ingham (1900-1967) in 1937. Assuming the Riemann Hypothesis to be true, the lowest such x may be explicitly computed, by bracketing its powers: x1, rounded down, is the lowest prime, namely 2 x3, rounded down, is the lowest prime between 23 and 33, namely 11 x9, rounded down, is the lowest prime between 113 and 123, i.e., 1361 x27, rounded down, is the lowest prime between 13613 and 13623, etc. Thus, raising x = 2.229494772491595235722852237656- to the power of 3n (and rounding down) only yields prime numbers, namely: 2, 11, 1361, 2521008887, 16022236204009818131831320183, etc. One fallacy with that "etc." is that we had to prove the primality of the last value to obtain x with this much precision (conversely, the precision given is barely enough to obtain this last prime number correctly). The whole thing is merely a way to encode an infinite sequence of prime values into the infinitely many decimals of a real number. It doesn't help in constructing such a sequence. The other fallacy is that the iteration of the process depends on the "obvious fact" that there's always a prime between consecutive cubes. Although the Hoheisel-Ingham theorem does guarantees that for sufficiently large cubes, it doesn't say exactly how large is large. Therefore, we cannot proceed "by inspection" for the above sequence until a putative lower bound is reached. Currently, our only option is to assume the validity of Riemann's hypothesis and conclude on that conditional basis (otherwise no satisfactory value of x can be established). The above sequence is usually expressed as A3, A9... A3n... where n is a nonzero integer and A (the cube root of the above x) is now known as Mills' constant (cf. A051021). Thanks to T.D. Noe for clarifying this, on 2008-09-23 (reminding me that we attended graduate school together). A = x1/3 = 1.306377883863080690468614492602605712916784585156713644368... Using squares instead of cubes in the above would be fine if we knew that there's always a prime between consecutive squares. If that's true, then the lowest number that truncates down to a prime when raised to the power of a 2-power is 2.3247099696648664983923017- The first primes so obtained are 2, 5, 29, 853, 727613, 529420677791 and 280286254072681840639693. See A059784. Mills' Theorem \| Mills' Constant \| Prime Formulas \| A143935 (Noe's conjecture) (2009-01-23) Ulam's Lucky Numbers and Ludic Numbers Prime-likes sequences obtained by modifying the Sieve of Erathostenes. The prime numbers not exceeding some integer n can be obtained by crossing out some of the positive integers not exceeding n, according to the following sieving procedure, known as the Sieve of Eratosthenes which was devised by Eratosthenes of Cyrene (276-194 BC). Cross out the number 1 (which is not prime). Circle the smallest number p not yet crossed out (or circled) and cross out every pth number thereafter (i.e., cross out 2p, 3p, 4p, 5p, etc.) Repeat the above step until all numbers in your table (up to n) is either circled or crossed. The circled numbers are the prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499 (A000040) In the above, we could instead cross out every pth number among the subsequent numbers not yet crossed out. In this case, we do not obtain the sequence of primes, but the so-called ludic numbers 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107, 115, 119, 121, 127, 131, 143, 149, 157, 161, 173, 175, 179, 181, 193, 209, 211, 221, 223, 227, 233, 235, 239, 247, 257, 265, 277, 283, 287, 301, 307, 313, 329, 331, 337, 341, 353, 359, 361, 377, 383, 389, 397, 407, 415, 419, 421, 431, 433, 437, 445, 463, 467, 475, 481, 493, 497 (A003309) If we start with the odd numbers and repeatedly remove every pth number (counting from the very beginning of the surviving sequence) we obtain the [odd] lucky numbers (French: nombres fastes ): 1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, 51, 63, 67, 69, 73, 75, 79, 87, 93, 99, 105, 111, 115, 127, 129, 133, 135, 141, 151, 159, 163, 169, 171, 189, 193, 195, 201, 205, 211, 219, 223, 231, 235, 237, 241, 259, 261, 267, 273, 283, 285, 289, 297, 303, 307, 319, 321, 327, 331, 339, 349, 357, 361, 367, 385, 391, 393, 399, 409, 415, 421, 427, 429, 433, 451, 463, 475, 477, 483, 487, 489, 495, 511, 517, 519, 529, 535, 537, 541, 553, 559, 577, 579, 583, 591, 601, 613, 615, 619, 621, 631, 639, 643, 645, 651, 655, 673, 679, 685, 693, 699, 717, 723, 727, 729, 735, 739, 741, 745, 769, 777, 781, 787, 801, 805, 819, 823, 831, 841, 855, 867, 873, 883, 885, 895, 897, 903, 925, 927, 931, 933, 937, 957, 961, 975, 979, 981, 991, 993, 997 (A000959) This latest process applied to the even numbers yields the even lucky numbers : 2, 4, 6, 10, 12, 18, 20, 22, 26, 34, 36, 42, 44, 50, 52, 54, 58, 68, 70, 76, 84, 90, 98, 100, 102, 108, 114, 116, 118, 130, 132, 138, 140, 148, 150, 164, 170, 172, 178, 182, 186, 196, 198, 212, 214, 218, 228, 230, 234, 244, 246, 260, 262, 268, 278, 282, 290, 298, 300, 308, 310, 314, 324, 326, 332, 346, 354, 358, 362, 372, 374, 386, 388, 390, 394, 406, 418, 420, 426, 434, 436, 438, 442, 452, 458, 470, 474, 482, 490, 498, 502, 516, 518, 522, 524, 532, 534, 546, 548, 570, 578, 586, 588, 596, 598, 602, 614, 628, 630, 642, 644, 646, 660, 666, 674, 684, 690, 706, 708, 710, 714, 716, 724, 738, 740, 742, 746, 754, 772, 778, 780, 790, 794, 804, 818, 822, 826, 838, 852, 868, 870, 874, 882, 884, 886, 900, 906, 914, 916, 922, 938, 946, 954, 962, 964, 966, 972, 982, 986, 994, 996, 998 (A045954) With even numbers, we may also start the sieving directly with p=2, which yields: 2, 6, 10, 14, 18, 26, 30, 34, 38, 50, 54, 58, 62, 74, 78, 82, 86, 102, 106, 110, 114, 122, 126, 130, 134, 154, 158, 162, 170, 178, 182, 194, 202, 210, 222, 226, 230, 246, 250, 254, 258, 266, 270, 274, 278, 290, 298, 314, 318, 326, 338, 342, 346, 354, 370, 374, 386, 394, 398, 410, 414, 434, 438, 446, 450, 458, 466, 470, 482, 486, 494, 498, 510, 530, 534, 538, 542, 558, 562, 566, 578, 586, 594, 602, 606, 630, 634, 638, 642, 654, 658, 678, 682, 686, 690, 698, 702, 706, 722, 726, 734, 758, 770, 774, 782, 794, 798, 806, 826, 834, 842, 846, 850, 866, 878, 882, 894, 898, 902, 914, 918, 942, 946, 950, 962, 974, 978, 986, 990 ... (A194282) \| \| \| \| \| \| \| www.numericana.com/answer/primes.htm updated 2023-10-07 05:10 ln
190093	https://www.math.aau.at/heuberger/publications/pdf/dnaf-1.pdf	ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS CLEMENS HEUBERGER‡ AND HELMUT PRODINGER∗ Abstract. It is known that every positive integer n can be represented as a ﬁnite sum of the form P i ai2i, where ai ∈{0, 1, −1} and no two consecutive ai’s are non-zero (“nonadjacent form”, NAF). Recently, Muir and Stinson [12, 13] investigated other digit sets of the form {0, 1, x}, such that each integer has a nonadjacent representation (such a number x is called admissible). The present paper continues this line of research. The topics covered include transducers that translate the standard binary representa-tion into such a NAF and a careful topological study of the (exceptional) set (which is of fractal nature) of those numbers where no ﬁnite look-ahead is suﬃcient to construct the NAF from left-to-right, counting the number of digits 1 (resp. x) in a (random) represen-tation, and the non-optimality of the representations if x is diﬀerent from 3 or −1. 1. Introduction Redundant number systems have been studied by many people, and for various reasons. Probably the most famous one uses the base 2, but instead of the traditional digits 0, 1, it allows a third digit −1. One can make this system unique by superimposing a condition: no two adjacent digits diﬀerent from zero are permitted. This non-adjacent-form (“NAF”) goes back (at least) to Reitwiesner . For pointers to the earlier literature we refer the reader to and . Recently, Muir and Stinson [12, 13] asked for other sets of digits, in particular of the form {0, 1, x}, such that every (positive) integer has a unique representation using these digits, base 2, and obeying again the condition that two adjacent digits cannot both be diﬀerent from zero. The present paper analyzes their number systems with respect to frequencies of digits, description of characteristic sets and similar questions. Our main focus is to provide results for general x as much as possible; we give explicit results as well as algorithms to compute some characteristic quantities for speciﬁc x. We summarize here what follows in the later sections. In order to describe the digits x that work, we deﬁne a suitable graph. The answer depends on whether each node can be reached from the starting node. We propose ‡ This paper was written while the ﬁrst author was a visitor at the John Knopfmacher Centre for Applicable Analysis and Number Theory, School of Mathematics, University of the Witwatersrand, Jo-hannesburg. He thanks the centre for its hospitality. He was also supported by the grant S8307-MAT of the Austrian Science Fund. ∗This author is supported by the grant NRF 2053748 of the South African National Research Foundation. 1 2 C. HEUBERGER AND H. PRODINGER an O(\|x\|)-time algorithm for this decision. The list of admissible numbers x starts like 3, −1, −5, −13, . . . (they must all be ≡3 mod 4). We then deﬁne transducers (ﬁnite automata with output) that translate the standard binary representation into a NAF using digits {0, 1, x} by reading the input word from right to left (thus starting with the least signiﬁcant digit). Such a transducer has at most \|x\| + 4 states. It can be constructed even for such x’s that don’t lead to NAFs; in such cases, not every number has a representation, and one can “see” from the transducer when this happens. (For the standard NAF, such a transducer is of course well known.) One of the advantages of the (standard) NAF, especially for applications in cryptogra-phy, is, that many digits are zero “on average.” To be more precise, if one considers all admissible words of length N, with digits (letters) from the set {0, 1, −1}, then a random word has on average N 6 digits ‘1’ resp. ‘−1’ and 2N 3 zeros. Interestingly, these frequencies persist even in the general case, with error terms depending on the extra digit x; in prin-ciple (for any particular x), they can be computed. The variance follows a similar pattern, with the constant 1 6 being replaced by 11 108. From this one can conclude that the underlying distribution is Gaussian. For various reasons (on-line computations!) it would be nice to generate the NAF when reading the digits of the binary expansion from left-to-right. It is known for the standard case x = −1 that this is not possible, but there exist equivalent representations (with respect to the large number of zeros, i.e., small “Hamming weight”). There are always situations where a ﬁnite look-ahead will not be suﬃcient to decide which digit should be taken. We scale this exceptional set B down to the unit interval and study it as a function of the length of the look-ahead. While this set is just { 1 6, 2 6, 4 6, 5 6} in the standard instance, we enter the realm of fractals for x = −5, −13, . . . . The (Lebesgue-) measure of B is zero, but its Hausdorﬀdimension is positive and strictly less than 1. A similar situation occurred in the study of joint expansions in Grabner, Heuberger, and Prodinger . For particular x, the Hausdorﬀdimension can be explicitly computed from the dominant eigenvalue of the adjacency matrix of a certain auxiliary automaton, but it is quite challenging to say something in general. We manage, however, to derive nontrivial lower and upper bounds. As already mentioned, for x = −1, the NAF is optimal, as there is no representation of any integer with fewer nonzero digits. We show in an algorithmic fashion that this is also true for x = 3. This is no longer true for the other possible values of x. In most cases, the number 3 serves as a counter example; in a few exceptional cases one must take another one. As a byproduct of our analysis, we can describe (for x = 1 resp. x = 3) all optimal representations; recall that in these cases a representation of an integer m is optimal if it has the same number on nonzero digits as the NAF of m. These representations can be described by ﬁnite automata. It is always beneﬁcial in order to understand how the arithmetic of a number system works to understand ﬁrst how addition of 1 (or any other ﬁxed integer) works. Knuth’s book has several such examples. Here, we conﬁne ourselves to the transducer describing the addition of 1 in the instance x = −5. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 3 For the standard NAF case (x = −1) it is possible to give an explicit formula for the digits of the expansion, cf. Prodinger . This is somehow a reﬂection of the fact that the above mentioned exceptional set is so simple. In the instance x = −5, there is still an explicit formula for the digits. However, it sometimes predicts a nonzero digit when it actually is zero, but is correct otherwise. This corresponds to Figure 9, where one can see that the intervals that contain green resp. blue points, are separated. For other values of x nothing like that exists anymore. Before entering into the details, we will ﬁx a minimum amount of notation. Let n be an integer and D ⊂Z. A (binary) D-expansion of n is a sequence (. . . , d2, d1, d0) ∈D 0 such that only ﬁnitely many digits are nonzero and such that n = P j≥0 dj2j. The standard binary expansion of a nonnegative integer n is its unique {0, 1}-expansion. We will usually denote sequences by boldface letters. Where appropriate, we will also think about D-expansions as ﬁnite or inﬁnite words over the alphabet D. By value(. . . , d2, d1, d0), we denote P j≥0 dj2j. The position of the most signiﬁcant digit MSB(d) is max{j : dj ̸= 0}. 2. Nonadjacent Digit Sets We recall some deﬁnitions of Muir and Stinson [12, 13]. Deﬁnition 1. (1) Let 0 ∈D ⊂Z and n ∈Z. Then a D-expansion η of n is called a D-nonadjacent form (D-NAF) of n, if ηjηj+1 = 0 for all j ≥0. (2) Let 0 ∈D ⊂Z. If there is a D-NAF for every positive integer n, then D is called a nonadjacent digit set (NADS). Muir and Stinson [12, 13] study digit sets D = {0, 1, x} for integers x. The following results have been proved in their paper: Proposition 2. Let D = {0, 1, x}. (1) If D is a NADS, then x ≡3 (mod 4). (2) If x ≡3 (mod 4), then each positive integer has at most one D-NAF. (3) The only NADS D = {0, 1, x} with positive x is {0, 1, 3}. Deﬁnition 3. Let D = {0, 1, x} with x ≡3 (mod 4). Then we deﬁne η0 : Z →D and r : Z →Z by setting η0(n) :=      0, if n ≡0 (mod 2), 1, if n ≡1 (mod 4), x, if n ≡3 (mod 4), r(n) := n −η0(n) 2 . It is easily seen (cf. Muir and Stinson [12, 13]) that n ∈N admits a D-NAF if and only if r(n) admits a D-NAF and that if (. . . , η′ 1, η′ 0) is the D-NAF of r(n) then (. . . , η′ 1, η′ 0, η0(n)) is the D-NAF of n. (Note that r(n) is even if n is odd). Since r(n) < n for all positive n ̸≡3 (mod 4) and r2(n) := r(r(n)) < n for all positive n ≡3 (mod 4) with n > \|x\| /3, the set D is a NADS if and only if all 0 < n ≤\|x\| /3 with n ≡3 (mod 4) have a D-NAF. 4 C. HEUBERGER AND H. PRODINGER Proposition 4. Let D = {0, 1, x} with x < 0 and x ≡3 (mod 4). Deﬁne the directed graph G := (V, A) by V = {0, . . . , ⌊\|x\| /3⌋} and A := {(m, n) ∈V 2 : n ∈{2m, 4m + 1, 4m + x}}. Then D is a NADS if and only if every n ∈V is reachable from 0. Proof. We observe that if 0 ̸= n ∈V , then there is exactly one edge with head n, namely (ri(n), n), where i = (n mod 2) + 1. Therefore, there is a path1 from 0 to n if and only if n has a D-NAF. □ Therefore, we may check whether a set D is a NADS by simple breadth-ﬁrst search (cf. Algorithm 1). Algorithm 1 Check for NADS Input: D = {0, 1, x} with x < 0 and x ≡3 (mod 4) Output: Decision, whether D is a NADS S ←(0) c ←1 while S ̸= () do m ←First element of S Remove m from S T ←{2m, 4m + 1, 4m + x} ∩{0, . . . , ⌊\|x\| /3⌋} c ←c + #T Append T to S end while if c = ⌊\|x\| /3)⌋+ 1 then Return(True) else Return(False) end if It is clear that the run-time of this algorithm is O(\|x\|). Muir and Stinson [12, 13] give a list of all NADS {0, 1, x} with \|x\| ≤10 000. The list starts with 3, −1, −5, −13, −17, −25, −29, −37, −53, −61, −65, . . . They also gave some necessary and some suﬃcient conditions on x such that {0, 1, x} is a NADS. We remark that for negative integers n, we always have r(n) > n if x < 0. This implies that for some ﬁnite positive k, we have rk(n) ≥0. Proposition 5. Let D := {0, 1, x} with x < 0 be a NADS. Then every integer n ∈Z has a D-NAF. 1In this paper, a path in a transducer, in an automaton or in a directed graph is not required to have distinct vertices. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 5 3. Calculating a D-NAF from right to left Let D = {0, 1, x} with x ≡3 (mod 4) be ﬁxed throughout this section. Our aim is to give a transducer which transforms the binary expansion of n into its D-NAF from right to left. Since the least signiﬁcant digit η0(n) depends on n mod 4, a transducer will need a look-ahead of 1 in order to be able to make a decision. The transducer T0 over the input alphabet {0, 1} and the output alphabet D is deﬁned as follows: It has the set of states Q0 = {0, . . ., 2+\|x\|}∪{I}, where the states 0 ≤m ≤2+\|x\| represent carries and I denotes the initial state. The sets of initial and terminal states are {I} and {0}, respectively. The set of transitions is (3.1) E0 = {I 0\|ε − →0, I 1\|ε − →1} ∪ m d\|η0(2d+m) − − − − − − →r(2d + m) : 0 ≤m ≤2 + \|x\| , d ∈{0, 1} , where ε denotes the empty word. Since r(2d + m) = 2d + m −η0(2d + m) 2 ≤2 + (2 + \|x\|) + \|x\| 2 = 2 + \|x\| , this ﬁnite transducer is well deﬁned. Of course, we only have to consider the accessible states in T0; we therefore deﬁne T to be the subgraph of T0 spanned by the accessible states. The sets of states and transitions of T will be denoted by Q and E, respectively. Note that this is a right-to-left transducer, i.e., a path m dJ−1···d0\|ηJ−1···η0 − − − − − − − − − − →m′ is m d0\|η0 − − − → m1 d1\|η1 − − − →m2 d2\|η2 − − − →· · · dJ−1\|ηJ−1 − − − − − − →m′. To prove that the transducer indeed calculates a D-NAF of an integer when reading its standard binary expansion, the following lemma is useful. Lemma 6. (1) Let I iJ···i0\|oJ−1···o0 − − − − − − − − →m be a path in T . Then (oJ−1, . . . , o0) is the D-NAF of value(iJ, . . . , i0) −m2J. (2) If there is a path m dJ−1···d0\|ηJ−1···η0 − − − − − − − − − − →m′ in T for some m ̸= I, then we have (3.2) 2Jm′ + value(ηJ−1, . . ., η0) = 2 value(dJ−1, . . ., d0) + m. Proof. The lemma is easily proved by using the deﬁnition of the transducer and induction. □ The following theorem states that the transducer T does what we promised. Theorem 7. Let D = {0, 1, x} with x ≡3 (mod 4) and T = (Q, E, {I}, {0}) the trans-ducer constructed above. Then the following holds: (1) #Q ≤\|x\| + 4. (2) An integer n with binary expansion d has a D-NAF if and only if there is a suc-cessful path with input label (dMSB(d)+#Q−2, . . . , d0) in T . In this case, the output label of this successful path is the D-NAF of n. (3) The set D is a NADS if and only if the only cycle in T with input label 0 · · ·0 is 0 0\|0 − →0. 6 C. HEUBERGER AND H. PRODINGER Proof. (1) Follows from the deﬁnition. (2) Assume that n > 0 has a D-NAF η and a standard binary expansion d. Let J := max{MSB(η), MSB(d)}. Consider the path I dJ+2dJ+1dJ···d0\|oJ+1oJ···o0 − − − − − − − − − − − − − − − − →m in T . From Lemma 6 we conclude that n −m2J+2 has a D-NAF (oJ+1, oJ, . . . , o0). This implies that value(ηJ, . . . , η0) ≡value(oJ+1, . . ., o0) (mod 2J+2). Since both expansions are D-NAFs, we infer that ηj = oj for 0 ≤j ≤J. This yields oJ+12J+1 = −m2J+2 from which we conclude that oJ+1 = m = 0. This implies that I dJ+2dJ+1dJ···d0\|0ηJ···η0 − − − − − − − − − − − − − − →0 is a successful path in T . We will prove that the length of a successful path is at most MSB(d) + #Q −1 after proving the third part of the theorem. On the other hand, if there is some successful path, Lemma 6 shows that it corresponds to a D-NAF of the value of its input. (3) When processing the binary expansion of n on the transducer, we can distinguish between two phases: in the ﬁrst phase, we read “signiﬁcant” input, in the second phase, we just read leading zeros of the binary expansion. If we reach the terminal state 0 in this second phase, we are successful and got a D-NAF of our input. However, if we enter a cycle in the second phase apart from the trivial cycle 0 0\|0 − →0, it is clear that we will never reach the terminal state, i.e., there is no D-NAF. This implies that after reading dMSB(d), we will reach each of the states Q{I, 0} at most once, hence we need at most #Q −2 leading zeros to reach the terminal state 0. □ For x = 3, x = −1, x = −5, x = −9, and x = −13, the transducers T are shown in Figures 1, 2, 3, 4, and 5, respectively. Note that for x = −9, there is a cycle 3 0\|¯ 9 − →6 0\|0 − →3, therefore, {0, 1, −9} is not a NADS. I 0 1 0\|ε 1\|ε 0\|0 1\|0 0\|1, 1\|3 Figure 1. Transducer T for calculating the {0, 1, 3}-NAF of n from its standard binary expansion from right to left. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 7 I 0 1 2 0\|ε 1\|ε 0\|0 1\|0 0\|1 1\|1 0\|0 1\|0 Figure 2. Transducer T for calculating the {0, 1, −1}-NAF of n from its standard binary expansion from right to left. I 0 1 4 2 3 0\|ε 1\|ε 0\|0 1\|0 0\|1 1\|5 0\|0 1\|0 0\|0 1\|0 0\|5 1\|1 Figure 3. Transducer T for calculating the {0, 1, −5}-NAF of n from its standard binary expansion from right to left. In some parts of this paper, we will study the input automaton A of T , i.e., we only consider the input labels of the transitions in T . By construction, the automaton A is deterministic. We will use the notation m′ = (dJ−1 · · ·d0) · m for the transition function in this automaton, which means that there is a path in A from m to m′ with (input) label (dJ−1 · · · d0). Furthermore, we will apply these transitions to sets of states also, i.e. (dJ−1 · · · d0) · M := {(dJ−1 · · · d0) · m : m ∈M} for M ⊆Q. The following lemma will be used several times: Lemma 8. Let x < 0, (3.3) k0 := 2 + max{MSB(η(n)) : −2 ≤n ≤2 + \|x\|} 8 C. HEUBERGER AND H. PRODINGER I 0 1 6 3 4 2 0\|ε 1\|ε 0\|0 1\|0 0\|1 1\|9 0\|0 1\|0 0\|9 1\|1 1\|0 0\|0 0\|0 1\|0 Figure 4. Transducer T for calculating the {0, 1, −9}-NAF of n from its standard binary expansion from right to left. and m ∈Q \ {I}. Then for any k ≥k0, we have (0k) · m = 0, (1k) · m = 2, where dk means the word consisting of k repetitions of the letter d. Proof. Let d ∈{0, 1}. We consider the path m dk\|ηk−1···η0 − − − − − − →m′ in T . By (3.2), we have (3.4) 2km′ + value(ηk−1 · · · η0) = 2d(2k −1) + m = d2k+1 + m −2d. By deﬁnition of k0, the D-NAF η′ of m−2d satisﬁes MSB(η′) ≤k −2. Then (3.4) implies that value(ηk−1 · · · η0) ≡value(η′ k−1 · · · η′ 0) (mod 2k), which yields ηj = η′ j for 0 ≤j ≤k−2. Inserting this in (3.4) we immediately see that ηk−1 = 0 and m′ = 2d. □ Furthermore, we can also construct a transducer e T0 which takes an arbitrary binary {0, 1, x}-expansion and transforms it to the D-NAF: The set of states e Q0 is {I} ∪{−2 \|x\|, ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 9 I 0 1 8 4 5 2 3 10 6 0\|ε 1\|ε 0\|0 1\|0 0\|1 1\|13 0\|0 1\|0 0\|0 1\|0 0\|1 1\|13 0\|0 1\|0 0\|13 1\|1 0\|0 1\|0 1\|0 0\|0 Figure 5. Transducer T for calculating the {0, 1, −13}-NAF of n from its standard binary expansion from right to left. . . . , \|x\| + 2}, the set of transitions e E0 is (3.5) e E0 = {I 0\|ε − →0, I 1\|ε − →1, I x\|ε − →x} ∪ m d\|η0(2d+m) − − − − − − →r(2d + m) : −2 \|x\| ≤m ≤2 + \|x\| , d ∈{0, 1, x} . The transducer e T is obtained by removing inaccessible states. The transducers e T for x = 3 and x = −1 are shown in Figures 13 and 15, respectively. 4. Frequency of digits Let D = {0, 1, x} be a ﬁxed NADS. We denote the D-NAF of a nonnegative integer n by η(n). For d ∈{1, x}, we denote the number of occurrences of the digit d in the D-NAF of n ∈N0 by (4.1) fd(n) := X j≥0 [ηj(n) = d] , 10 C. HEUBERGER AND H. PRODINGER where we use Iverson’s notation [expr] = 1 if expr is true and [expr] = 0 otherwise, cf. . Let XN be a uniformly distributed random variable on {0, . . ., 2N −1}. We are interested in the distribution of the random variable Fd,N := fd(XN). We will ﬁrst calculate the main terms of mean and variance using a recurrence approach. In a second step, we study the probability generating function and derive a central limit law using Hwang’s “quasi power theorem.” For concrete values of x, the generating function approach gives also second order terms for the mean and the variance. We derive recursive formulæ for the ﬁrst two moments Ed(N) := E(Fd,N) = 1 2N 2N−1 X n=0 fd(n), Sd(N) := E((Fd,N)2) = 1 2N 2N−1 X n=0 f 2 d(n) for N ≥N0, where N0 := max{MSB(η(n)) : 0 ≤n < ⌈−x/4⌉} + 4. We split the sum into three parts, depending on the residue class of n modulo 4, writing n = 2m, n = 4m + 1, and n = 4m + x, respectively: Ed(N) = 1 2N 2N−1−1 X m=0 fd(m) + 1 2N 2N−2−1 X m=0 fd(m) + 1 2N 2N−2+y−1 X m=y fd(m) + 1 4 = 1 2Ed(N −1) + 1 2Ed(N −2) + 1 4 + 1 2N y−1 X m=0 X j≥0 ηj(2N−2 + m) = d −[ηj(m) = d] , where y = ⌈−x/4⌉. By our choice of N0, we conclude that for 0 ≤m < y, the D-NAF of 2N−2 + m is simply the D-NAF of m with the digit at position N −2 set to 1, i.e., ηj(2N−2 + m) = ηj(m) + [j = N −2] . This yields Ed(N) = 1 2Ed(N −1) + 1 2Ed(N −2) + 1 4 + y [d = 1] 2N , N ≥N0. Solving this linear recurrence with constant coeﬃcients, we get (4.2) Ed(N) = E(Fd,N) = 1 6N + ed + O 1 2N , where ed is some constant. A recurrence relation for the second moment can be derived similarly, we get Sd(N) = 1 2Sd(N −1) + 1 2Sd(N −2) + 1 2Ed(N −2) + 1 4 + C 2N , N ≥N0, ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 11 where C = [d = 1] Py−1 m=0(2fd(m) + 1) is a constant. Inserting (4.2) (with undetermined constants), we again get a linear recurrence with constant coeﬃcients. Solving it, we get Sd(N) = E((Fd,N)2) = 1 36N2 + 11 108 + ed 3 N + (vd + e2 d) + O N 2N for some constant vd. Subtracting (Ed(N))2, we get (4.3) V(Fd,N) = E((Fd,N)2) −(E(Fd,N))2 = 11 108N + vd + O N 2N . We now consider the probability generating function Gd(Y, Z) = X N≥0 X m≥0 P(Fd,N = m)Y mZN. We deﬁne the (#Q×#Q)-matrices Ad := Ad(Y ) = (aij)i,j∈Q and Bd = Bd(Y ) = (bij)i,j∈Q by aij = X v∈{0,1} i v\|η − →j∈E 1 2Y [η=d], bij = ( Y [η=d], if there is a transition i 0\|η − →j ∈E, 0, otherwise, where the rows and columns of Ad and Bd are ordered as I, 0, 1, . . .. Then Gd(Y, Z) can be expressed as Gd(Y, Z) = (1, 0, . . ., 0)(I −AdZ)−1B#Q−2 d (0, 1, 0, . . ., 0)t, where the factor B#Q−2 d ensures that we come back to the terminal state 0. For concrete x, Gd can be calculated explicitly, for instance, we have G1(Y, Z) = 16 −8Z + 8Y Z −4Z2 −Y Z4 + Y 2Z4 2(−2 + Z)(−4 + 2Z + Z2 + Y Z2) , G−5(Y, Z) = −(8 + 4Z −Y Z3 + Y 2Z3) 2(−4 + 2Z + Z2 + Y Z2) for x = −5. Then we have E(Fd,N) = [ZN] ∂Gd ∂Y Y =1 , V(Fd,N) = [ZN] ∂2Gd ∂Y 2 Y =1 + E(Fd,N) −(E(Fd,N))2. For the ﬁrst ten values of x, we calculated means and variances by this approach. Com-paring with (4.2) and (4.3) for the ﬁrst few values of x gives the constants ed and vd in Table 1. From the deﬁnition (3.1) of the transducer T it is clear that Ad(Y ) = 0 1 1 0 . . . 0 0 e Ad(Y ) 12 C. HEUBERGER AND H. PRODINGER x e1 v1 ex vx 3 2 9 ≈0.2222 8 81 ≈ 0.0987 −1 9 ≈−0.1111 −4 81 ≈−0.0493 −1 5 9 ≈0.5555 −16 81 ≈−0.1975 −1 9 ≈−0.1111 −4 81 ≈−0.0493 −5 23 36 ≈0.6388 −253 1296 ≈−0.1952 −1 36 ≈−0.0277 155 1296 ≈ 0.1195 −13 467 576 ≈0.8107 −61609 331776 ≈−0.1856 −25 576 ≈−0.0434 35279 331776 ≈ 0.1063 −17 319 288 ≈1.1076 −13825 82944 ≈−0.1666 71 576 ≈ 0.1232 176783 331776 ≈ 0.5328 −25 7553 9216 ≈0.8195 −20629249 84934656 ≈−0.2428 637 4608 ≈ 0.1382 8343287 21233664 ≈ 0.3929 −29 2183 2304 ≈0.9474 −1241137 5308416 ≈−0.2338 −37 2304 ≈−0.0160 893351 5308416 ≈ 0.1682 −37 47939 36864 ≈1.3004 −197600137 1358954496 ≈−0.1454 26143 73728 ≈ 0.3545 6190893119 5435817984 ≈ 1.1389 −53 22009 18432 ≈1.1940 −54592561 339738624 ≈−0.1606 14561 36864 ≈ 0.3949 1292596799 1358954496 ≈ 0.9511 −61 10289 9216 ≈1.1164 −19291489 84934656 ≈−0.2271 −187 9216 ≈−0.0202 13417319 84934656 ≈ 0.1579 Table 1. Constant terms in means and variances of Fd,N for some (#Q−1) ×(#Q−1)-matrix e Ad. By deﬁnition of T and Lemma 8, the matrix ˜ Ad is the adjacency matrix of a strongly connected directed graph. Moreover, since there is a loop 0 0\|0 − →0 in T , the matrix e Aℓ d(1) is positive for some positive integer ℓ. Therefore, e Ad(1) is primitive (cf. for instance [11, §15.6]) and has a unique simple eigenvalue of maximal modulus by the Perron-Frobenius theorem. Since the spectrum σ(Ad(1)) = σ( e Ad(1))∪{0}, the same holds for Ad(1) and, by continuity, for Ad(Y ) for Y in some neighbourhood of 1. Thus, the generating function Gd is a rational function whose numerator has a unique simple pole of minimal modulus for Y suﬃciently close to 1. We obtain E(eFd,Ns) = eu(s)N+v(s) 1 + O ρN for some analytic functions u(s) and v(s) and some 0 < ρ < 1 for \|s\| suﬃciently small. By Hwang’s “quasi power theorem” we conclude the central limit theorem (4.4). Theorem 9. Let D = {0, 1, x} be a NADS, d ∈{1, x} and Fd,N the number of occurrences of the digit d in the D-NAF of a randomly chosen integer in the interval {0, . . ., 2N −1}. Then we have E(Fd,N) = 1 6N + ed + O 1 2N , V(Fd,N) = 11 108N + vd + O N 2N for some constants ed and vd depending on x, which can be computed explicitly. For \|x\| ≤ 61, they are given in Table 1. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 13 Furthermore, we have (4.4) P Fd,N ≤1 6N + z r 11 108N ! = 1 √ 2π Z z −∞ e−t2/2 dt + O 1 √ N uniformly with respect to z, z ∈R. 5. “Calculating” Digits from Left To Right The aim of this section is to give a description of arbitrary digits in a NADS D = {0, 1, x}. If n has standard binary expansion (. . . dℓ+1dℓdℓ−1 · · ·d0) and D-NAF (. . . ηℓ+1ηℓηℓ−1 · · ·η0), then ηℓdepends on n mod 2ℓ+2 only. Alternatively, ηℓdepends on n/2ℓ+2 , where {z} denotes the fractional part of z. In Figures 7, 8, 9, and 10, we draw the second digit from the left for x = 3, −1, −5, and −13 respectively: For k = 3, . . . , 12 and n = 0, . . . , 2k −1, we calculated the ℓth digit of the D-NAF of all integers m such that n 2k ≤ n m 2ℓ+2 o < n + 1 2k ; these are those integers whose ﬁrst (from the left!) digits in the standard binary expansion agree with the ﬁrst digits of n. In some cases, the ℓth digit was the same for all integers m which lie in the given interval, in some cases, several digits could occur. We mapped the sets of possible digits to colors according to Figure 6 and ﬁlled a rectangle of width 1/2k and height 1 with this color at position (n/2k, 12 −k). We remark that for x = −1, the picture is rather regular, whereas for x = −13, many decisions are still open after 12 digits. x 1 0 {1, x} {0, 1} {0, x} D Figure 6. Colors for Figures 7–10. In what follows, we will describe this phenomenon in terms of the states of the transducer T . Since the digit ηℓdepends on the state mℓ+1 = (dℓ· · · d0) · I and on dℓ+1, we will try to predict the state mℓ+1 from the knowledge of dℓ· · · dℓ−r+1 for small r, wherever possible. Since we do not know in which state we are after reading the unknown digits dℓ−r · · · d0, we have to assume that we are in any state (apart from the initial state). So we denote the set Q \ {I} by Q∗. 14 C. HEUBERGER AND H. PRODINGER Figure 7. Second digit from the left for x = 3. Figure 8. Second digit from the left for x = −1. Figure 9. Second digit from the left for x = −5. We deﬁne the map Φr : {0, 1}r →Q∗by Φr(x1x2 · · ·xr) := (x1x2 · · · xr) · Q∗. It is clear that Φr+1(x1x2 · · · xrxr+1) ⊆Φr(x1x2 · · · xr). Therefore, for any x = (x1x2 · · · ) ∈ {0, 1} , the limit Φ(x1x2 · · · ) := lim r→∞Φr(x1 . . . xr) ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 15 Figure 10. Second digit from the left for x = −13. exists. We note that if Φ(x) is a singleton {m} for some x ∈{0, 1} , we have Φr(x1x2 · · · xr) = {m} for some r ∈N. In this case, we have Φ(y) = Φ(x) for any y ∈{0, 1} with the property that the preﬁxes of length r of x and y coincide. This means that Φ is continuous in all points where it has a singleton image. We now assume x < 0. From Lemma 8, we see that for any (x1, . . . , xj) ∈{0, 1}j and k ≥k0, we have Φk+j+1(x1 · · · xj10k) = (x1 · · · xj1) · {0} = (x1 · · · xj) · {1} = (x1 · · · xj0) · {2} = Φk+j+1(x1 · · · xj01k). This implies that Φ(x1x2 · · · xj0111 · · ·) = Φ(x1x2 · · · xj1000 · · ·). Therefore, Φ descends to a function on the unit interval [0, 1] (which we call Φ again) by Φ X j≥1 xj2−j ! := Φ(x1x2 · · · ). This function is continuous at all points z ∈[0, 1] where Φ(z) is a singleton. This implies that the sets Aj := Φ−1({j}) are open for j ∈Q∗. The set A = S j∈Q∗Aj is therefore open. Since the set of real numbers in [0, 1] whose binary expansion contains a block (0k0) has Lebesgue measure 1 and is a subset of A by Lemma 8, we conclude that λ(A) = 1, too. We translate our results back to the ℓth digit: We have ηℓ(n) = η0(2dℓ+1 +mℓ+1) = η for mℓ+1 = (dℓ· · · d0)·I and the standard binary expansion d. The relation mℓ+1 = (dℓ· · · d0)·I is equivalent to n/2ℓ+1 ∈Amℓ+1, since Φk0+ℓ+1( n/2ℓ+1 ) is a singleton by Lemma 8. To include dℓ+1 in our description, we deﬁne ϕd(z) := (d + z)/2 for d ∈{0, 1} and see that n n 2ℓ+2 o ∈ϕdℓ+1(Amℓ+1). We now collect all sets ϕd(Am) which lead to the same digit η ∈D in the set Wη := int [ d∈{0,1} m∈Q∗ η0(2d+m)=η ϕd(Am) , 16 C. HEUBERGER AND H. PRODINGER where the bar denotes closure and int denotes the interior. We take the interior of the closure for “aesthetical reasons”: We want to avoid “holes” in the sets which would only be meaningful at the level of the transducer, but not on the level of the digits. Since n/2ℓ+1 is in the interior of some Am anyway, the dyadic points are not aﬀected by this operation. This yields the following theorem. Theorem 10. Let D = {0, 1, x} be a NADS with x < 0. There are disjoint open subsets Wη, η ∈D, of the unit interval [0, 1] such that ηℓ(n) = η ⇐ ⇒ n/2ℓ+2 ∈Wη for ℓ≥0. The sum of the Lebesgue measures of the Wη, η ∈D, equals 1. 6. Dimension of the boundary Theorem 10 shows that [0, 1] \ (W0 ∪W1 ∪Wx) = ∂(W0 ∪W1 ∪Wx) = ∂W0 ∪∂W1 ∪∂Wx has Lebesgue measure zero. However, Figures 9 and 10 demonstrate that this “exceptional set” is quite “irregular.” The aim of this section is to quantify this “irregularity” in terms of Hausdorﬀdimension. We will calculate the Hausdorﬀdimension of ∂(W0 ∪W1 ∪Wx) as the spectral radius of the adjacency matrix of an auxiliary automaton as follows. We construct auxiliary automata Am for m ≥2. The set of states is the set Qm := {I ⊆Q∗: #I = m} of m element subsets of Q∗. The alphabet is still {0, 1}. The set of transitions Em := {I d − →J : J = d · I = {d · i : i ∈I}, d ∈{0, 1}, I, J ∈Qm} is deﬁned in terms of the transitions in the input automaton underlying T . All states are initial and terminal states. The adjacency matrix of Am will be denoted by Mm. For x = −5 and m = 2, this automaton is shown in Figure 11. We will mainly work with A2, which is the only automaton to occur in the statement of our result, however in some cases, we may be forced to use Am for some m > 2 instead. The remaining part of this section is devoted to the proof of the following theorem. Theorem 11. Let D = {0, 1, x} with x < 0 be a NADS and Wη, η ∈D, be the sets described in Theorem 10. Let ρ(M2) denote the spectral radius of the adjacency matrix M2 of A2. Then we have dimH(∂(W0 ∪W1 ∪Wx)) = log2 ρ(M2) and 0 < Hlog2 ρ(M2)(∂(W0 ∪W1 ∪Wx)) < ∞. For x = −1, we have log2 ρ(M2) = 0, for \|x\| ≥5, we have 0 < 0.347121 −1.04136 ℓ+ 2 ≤log2 ρ(M2) ≤1 − 1 2k0 log 2 + 1 22k0+1 log 2 < 1, ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 17 0, 1 0, 4 0, 2 0, 3 1, 4 1, 2 1, 3 4, 2 4, 3 2, 3 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 Figure 11. Automaton A2 for x = −5. where k0 has been deﬁned in (3.3) and ℓ= ⌊log2(\|x\| + 3)⌋−1. For \|x\| ≤29, the values log2 ρ(M2) are shown in Table 2. The transitions in Am will again be denoted by J = d · I, since this coincides with the earlier deﬁnition for subsets of Q∗. However, this transition is not deﬁned in Am for every d and I, since it might be the case that I is mapped to some J with #J < m. We will also reuse the notation d · I for strings d ∈{0, 1}n. As in the previous section, we ﬁrst study the problem at the level of the states of the transducer and will translate them back to the level of output digits afterwards. We consider the set B := [0, 1]\A of those real numbers which do not allow us to decide about the ﬁnal state from the knowledge of a ﬁnite number of digits. To calculate the Hausdorﬀ dimension of B for general x, we proceed as follows: We ﬁrst derive an upper bound for the box dimension using a suitable covering of B. In a second step, we construct a lower bound for the box dimension. Finally we will show that a subset of B can be interpreted as a ﬁnite union of graph directed sets (cf. Falconer ) satisfying an open set condition, which implies (cf. Edgar ) that the Hausdorﬀ, box and similarity dimensions are equal. Finally, we use this fact for calculating the HausdorﬀDimension of B explicitly for small values of \|x\|. 18 C. HEUBERGER AND H. PRODINGER x #Q #Q2 r(A2) dimH(∂(W0 ∪W1 ∪Wx)) −1 3 3 2 0 −5 5 10 3 0.7618 −13 9 36 7 0.9003 −17 10 45 6 0.9037 −25 14 91 17 0.9364 −29 14 91 8 0.9357 −37 18 153 34 0.9492 −53 22 231 25 0.9606 −61 25 300 55 0.9547 −65 25 300 33 0.9477 −113 37 666 79 0.9733 −121 41 820 147 0.9729 −125 39 741 83 0.9649 −137 42 861 98 0.9757 −145 48 1128 228 0.9763 −149 46 1035 130 0.9790 Table 2. Order of T and of A2, number r(A2) of strongly connected com-ponents of A2, and Hausdorﬀdimension of ∂(W0 ∪W1 ∪Wx). So we start by giving an upper bound for the upper box dimension dimBB = lim sup n→∞ log2 N2−n(B) n , where N2−n(B) denotes the number of 2−n-mesh intervals that intersect B. We have (6.1) N2−n(B) = #{d ∈{0, 1}n : Φn(d) is not a singleton}, since if Φn(d) is not a singleton, there are two points z1 and z2 in the interval [P j dj2−j, P j dj2−j+ 2−n] lying in disjoint open sets Am1 and Am2. Therefore, there must be a point of B between z1 and z2. Lemma 12. For k0 deﬁned in (3.3), we have (6.2) dimB(B) ≤1 − 1 2k0 log 2 + 1 22k0+1 log 2. Proof. By Lemma 8, we have N2−n(B) ≤#Un, where Un = (x1, . . . , xn) ∈{0, 1}n : (xj, . . . , xj+k0−1) / ∈{(0k0), (1k0)} for all 1 ≤j ≤n −k0 + 1 . The strings in Un can be described by a regular expression (ε + 0 + · · · + 0k0−1) (1 + 12 + · · · + 1k0−1)(0 + 02 + · · · + 0k0−1) ∗(ε + 1 + · · · + 1k0−1), ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 19 which can be translated to the generating function G(z) := X n≥0 #Unzn = (1 + · · · + zk0−1) · 1 1 −(z + · · · + zk0−1)2 · (1 + · · · + zk0−1) = 1 −zk0 1 −2z + zk0 . Let q(z) := 1 −2z + zk0. We note that we have k0 ≥4 by (3.3). Then \|q(z) −(1 −2z)\| = \|zk0\| < 2 \|z\| −1 ≤\|1 −2z\| for 1 −δ < \|z\| < 1 for a suitable δ. Therefore, q(z) has exactly one zero with modulus less than 1, say ρ, by Rouch´ e’s Theorem. Since q(1/2) > 0 and q(1/2 + 1/(2k0)) < 0, we know that ρ is real and ρ = 1/2 + O(k0 −1). By bootstrapping, we obtain ρ = 1 + ρk0 2 = 1 2 + O(2−k0), ρ = 1 2 + 1 2k0+1 + O k0 4k0 . In fact, we have (6.3) ρ ≥1 2 + 1 2k0+1 for k0 ≥4, since q(2−1 + 2−(k0+1)) > 0. Deﬁning the constant ck0 := lim z→ρ G(z) 1 −z ρ , we have #Un = [zn]G(z) = ck0 ρn + O(1). Therefore, we conclude that the upper box dimension of B can be bounded from above by dimB(B) ≤−log2 ρ, hence (6.3) yields (6.2). □ We now derive lower bounds for the box dimension. Lemma 13. Let \|x\| ≥5. Then we have (6.4) dimB(B) = lim inf n→∞ log2 N2−n(B) n ≥ 1 2 − 3 2ℓ+ 4 log2 1 + √ 5 2 ≈0.347121 −1.04136 ℓ+ 2 > 0, where ℓ= ⌊log2(\|x\| + 3)⌋−1. Proof. From the deﬁnition of T , we deduce that 0 and 1 always have the neighbourhood shown in Figure 12, where s := (3 + \|x\|)/2. We further notice that any path from s to 1 has at least length ℓ= ⌊log2 s⌋, since m ≥2i implies r(2d + m) ≥2i−1 for d ∈{0, 1}. We consider the set of sequences (6.5) Ln := {d = (dn, . . . , d1) ∈{0, 1}n : didi+1 = 0 for 1 ≤i < n and if (di · · ·d1) · s = 1 then di+1 = 1 for 1 ≤i < n}. We claim that d · s ̸= d · 0 for all d ∈Ln. To prove this, we observe that d · 0 = dn. For any 1 ≤i ≤n, we have (di · · · d1) · s ≥1 by deﬁnition (and the known neighbourhood of 0 20 C. HEUBERGER AND H. PRODINGER I 0 1 s 2 other states 0\|ε 1\|ε 0\|0 1\|0 0\|1 1\|x 0\|0 1\|0 Figure 12. Structure of T for general x (Neighbourhoud of I, 0, 1 is correct.) and 1). If (di · · · d1) · s = 1, we have di = 0, which implies (di · · · d1) · 0 = 0. This proves the claim. By (6.1), this implies that N2−n(B) ≥#Ln. We now derive a lower bound for #Ln. For 1 ≤t ≤ℓwe have Lt := {(dt, . . . , d1) ∈{0, 1}t : didi+1 = 0 for 1 ≤i < t}, since the second condition is no restriction for sequences of that length. The regular expression for sequences which avoid adjacent 1s is 0∗(100∗)∗(ε + 1), the corresponding generating function equals 1 1 −z 1 1 − z2 1−z (1 + z) = 1 + z 1 −z −z2 = X t≥0 Ft+2zt, where Ft denotes the Fibonacci sequence Ft+2 = Ft+1 + Ft with F0 = 0 and F1 = 1. Therefore, we get #Lt = Ft+2 = αt+2 √ 5 −O(α−t) ≥αt. for α = (1 + √ 5)/2 and 1 ≤t ≤ℓ. For d = (dℓ−2, . . . , d0) ∈Lℓ−1 we deﬁne ψ(d) := 1h((d0) · s)d0, where h(m) is the input label of a shortest path from state m to state 1 whose input label does not start (at the right) with 1 and does not contain adjacent ones. We note that if (d0) · s = 1, we have dℓ−2 = 0, h(1) = ε, and ψ(d) does not contain adjacent ones. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 21 If m ̸= 0, we write h(m) = h′(0 · m), where h′(m′) is the input label of a shortest path from state m′ to state 1 whose input label does not contain adjacent ones. We claim that (6.6) ⌊log2 m′⌋≤\|h′(m′)\| ≤⌊log2 m′⌋+ 1, where \|h′(m′)\| denotes the length of h′(m′). The left inequality follows from the argument given before (6.5). For the right inequality, we choose the input digit d to be 0 unless m′ ≡3 (mod 4), in which case we take d = 1. Thus η0(2d + m′) ∈{0, 1} and r(2d + m′) = d+⌊m′/2⌋. If we choose d = 1, it is clear that r(2d+m′) is even so that we will not choose input 1 in the next step. Furthermore, we have ⌊log2(r(2d + m′))⌋= ⌊log2(m′)⌋−1 unless m′ = 2g −1 for some g. But in the latter case, we have r(2d+m′) = 2g−1 with \|h′(m′)\| = g. Therefore, (6.6) follows by induction. We conclude that \|h′(m′)\| ≤⌊log2 m′⌋+ 1 ≤⌊log2(\|x\| + 2)⌋+ 1 ≤ℓ+ 2 and therefore \|ψ(d)\| ≤2ℓ+ 4. We now deﬁne an injective map L⌊n/(2ℓ+4)⌋ ℓ−1 →Ln by (d⌊n/(2ℓ+4)⌋, . . . , d1) 7→d0 n − ⌊n/(2ℓ+4)⌋ X j=1 \|ψ(dj)\| ψ(d⌊n/(2ℓ+4)⌋) · · · ψ(d1), where d0(m) ∈Lm is some ﬁxed admissible string of length m which is used to obtain the required length n. This yields #Ln ≥α(ℓ−1)⌊n/(2ℓ+4)⌋≥ 1 αℓ−1 · α(ℓ−1)/(2ℓ+4)n . For \|x\| ≥5, the bound (6.4) follows. □ We want to prove that the lower box, upper box, Hausdorﬀand similarity dimensions of B agree. To this aim, we ﬁrst collect a few simple consequences of Perron-Frobenius theory. We denote the spectral radius of a (n × n)-matrix M by ρ(M). If I, J ⊆{1, . . ., n}, we denote the submatrix of M with rows I and columns J by M(I, J). If M = (mij)1≤i,j≤n is a nonnegative (n × n)-matrix, we will consider the digraph G induced by M, i.e., the directed graph with set of vertices {1, . . . , n} and set of arcs {(i, j) : mij > 0}. We will call any strongly connected component C ⊆{1, . . . , n} of G which fulﬁlls (6.7) ρ(M(C, C)) = ρ(M) a dominating component of M. We will identify subsets of {1, . . ., n} and the subgraph of G induced by these vertices when speaking about strongly connected components. Lemma 14. Let M be a nonnegative (n × n)-matrix. Then there is a nonnegative vector 0 ̸= x ∈Rn such that Mx = ρ(M)x. Let G be the directed graph induced by M. Then M has a dominating component. 22 C. HEUBERGER AND H. PRODINGER Proof. By continuity, the assertion on the nonnegative eigenvector follows from the Perron-Frobenius theorem, cf. [11, Theorem 15.5.1], for instance. Let C1, . . . , Cr be the strongly connected components of G. We consider the auxiliary digraph G′ with set of vertices {C1, . . . , Cr} and an arc from Ci to Cj if there is an arc from some vertex in Ci to some vertex in Cj in the original digraph G. By construction, the auxiliary digraph G′ has no directed cycle. This implies that we may sort the components in such a way that the adjacency matrix of G′ is upper triagonal. If we permute the rows and columns of M according to the strongly connected components, we get a matrix which is block upper triagonal. Therefore, the spectrum σ(M) of M equals σ(M) = r [ i=1 σ(M(Ci, Ci)). This implies that ρ(M) = maxi=1,...,r ρ(M(Ci, Ci)). This proves the existence of a dominat-ing component. □ Lemma 15. Let M be a nonnegative (n × n)-matrix and e = (1, . . ., 1)t ∈Rn. Then lim m→∞ log2(etMme) m = log2 ρ(M), which may equal −∞. Proof. Each entry of M m is bounded by a constant times mn−1ρ(M)m, so the same holds for the nonnegative number etMme. On the other hand, let x be a nonnegative eigenvector of M for ρ(M), which exists by Lemma 14. Without loss of generality, 0 ≤xj ≤1 for j = 1, . . . , n. Therefore, etMme ≥xtMmx = ρ(M)mxtx. Taking logarithms and limits yields the result. □ In general, we cannot expect the automaton Am to be strongly connected. Therefore, we cannot apply the results on graph directed sets directly, but we have to apply it on the strongly connected components. To cover the remaining parts of B, we give an upper bound for the upper box dimension (and therefore for the Hausdorﬀdimension) using the spectral radius ρ(M2) of M2 at this point. Lemma 16. We have dimH(B) ≤dimB(B) ≤log2 ρ(M2). For \|x\| ≥5, we have ρ(M2) > 1. Proof. Consider d ∈{0, 1}n such that Φn(d) is not a singleton. Then there are u ̸= v ∈Q∗ such that d · u ̸= d · v, i.e., there is at least one path in A2 from some {u, v} ∈Q2 to some {u′, v′} ∈Q2 with label d. Thus the number N2−n(B) can be bounded from above by the number of paths of length n in A2, which equals etMn 2 e. From Lemma 15 we conclude that dimB(B) = lim sup n→∞ log2 N2−n(B) n ≤lim sup n→∞ log2 etMn 2 e n = log2 ρ(M2). ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 23 It is well known that the Hausdorﬀdimension is always less or equal the lower (and therefore the upper) box dimension. Since dimB(B) > 0 for \|x\| ≥5 by Lemma 13, we conclude that ρ(M2) > 1 in that case. □ Proof of Theorem 11. We assume that \|x\| ≥5. Since 0 / ∈1 · Q∗(cf. Figure 12), the automaton A#Q∗has one state and at most one transition, hence ρ(M#Q∗) ≤1 < ρ(M2) by Lemma 16. We now choose m ≥2 maximal such that ρ(Mm) ≥ρ(M2). The preceeding observation implies that m < #Q∗. We deﬁne sets VI := {x ∈[0, 1] : I ⊆Φ(x)} for I ∈Qm. By deﬁnition of Φ, these sets are compact. We consider the contracting similarities ϕd(z) := (d + z)/2 for d ∈{0, 1}. We clearly have Φ(ϕd(z)) = d · Φ(z) for d ∈{0, 1} and z ∈[0, 1]. Then it is easily proved that (6.8) VI = [ d∈{0,1} J∈Qm I=d·J ϕd(VJ), I ∈Qm. Let now C ⊆Qm be a dominant component of Mm, i.e., ρ(Mm(C, C)) ≥ρ(M2) > 1. First, we claim that (6.9) VI ̸= ∅for every I ∈C. To this aim, we note that there is a directed cycle I d − →I in C, where d ∈{0, 1}r for some 1 ≤r ≤#C. It follows that Φsr(ds) = ds · Q∗⊇ds · I = I, which implies that Φ(dω) ⊇I, where dω denotes the inﬁnite word ddd · · · . The real number z with binary expansion 0.ddd · · · is therefore an element of VI. This proves (6.9). Next, we claim that (6.10) UI := {z ∈[0, 1] : I = Φ(z)} ̸= ∅for every I ∈C. To prove this, we assume that UI = ∅for some I ∈C. Let now d ∈{0, 1}n for some n ∈N be the label of a path of length n in C, i.e., there are K1, K2 ∈C such that K1 d − →K2 is a path in C. We take some z ∈VK1 and a path K2 d′ − →I in C. We have Φ(ϕd′d(z)) = d′d · Φ(z) ⊇d′d · K1 = I, where ϕ is deﬁned by ϕδ1···δn := ϕδ1 ◦· · · ◦ϕδn. Since UI = ∅by assumption, this implies that # (d · Q∗) ≥# (d′d · Q∗) ≥#Φ(ϕd′d(z)) > #I = m, which yields # (d · Q∗) ≥m + 1. This means that there is a path K3 d − →K4 of length n in Am+1. The above construction shows that there is an injective map from the set of labels of paths of length n in C to the set of paths of length n in Am+1. We obtain etMn m+1e ≥#{d ∈{0, 1}n : d is label of a path of length n in C} ≥ 1 #C2etMm(C, C)ne. 24 C. HEUBERGER AND H. PRODINGER By Lemma 15, we conclude that log2 ρ(Mm+1) = lim n→∞ log2(etMn m+1e) n ≥lim n→∞ log2 1 #C2etMm(C, C)ne n = log2 ρ(Mm(C, C)) = log2(ρ(M2)). This yields ρ(Mm+1) ≥ρ(M2), a contradiction to the choice of m. Hence our claim (6.10) is proved. We now restrict (6.8) to the component C: There is a unique collection of nonempty compact sets (V C I )I∈C, such that (6.11) V C I = [ d∈{0,1} J∈C I=d·J ϕd(V C J ), I ∈C, cf. for instance Edgar [3, Theorem 4.3.5]. These sets V C I , I ∈Qm, are the graph directed sets deﬁned by C and the contractions ϕd, cf. Falconer . It is clear that V C I ⊆VI for each I ∈C, since the ﬁxed point (V C I )I∈C of (6.11) can be obtained by iterating the right hand side of (6.11) starting with the collection (VI)I∈C, which yields a sequence of tuples of compact sets which is nonincreasing in each component by (6.8). For I ∈C, we deﬁne OI := {z ∈[0, 1] : d(z, VI) < d(z, VJ) for all J ∈Qm ∪{∂} with J ̸= I}, where V∂:= {0, 1} is the boundary of the unit interval, d(z, K) := inf{d(z, y) : y ∈K} for compact sets K and the Euclidean distance d(z, y). The sets OI, I ∈C, are open. Since Φ(0) = {0} and Φ(1) = {2} by Lemma 8, we have UI ⊆OI, hence OI ̸= ∅for I ∈C by (6.10). It is easily checked that for ﬁxed I ∈C, we have (6.12) ϕd(OJ) ⊆OI, for all d ∈{0, 1}, J ∈C with I = d · J, the sets ϕd(OJ) for d ∈{0, 1}, J ∈C with I = d · J are disjoint. This is the open set condition, cf. Edgar [3, Deﬁnition 6.4.7]. Therefore, by Edgar [3, Theorem 6.4.8], we have dimH(V C I ) = s and Hs(V C I ) > 0 for all I ∈C, where s is the unique similarity dimension such that ρ(2−sMm(C, C)) = 1. Thus we have s = log2 ρ(Mm). Since V C I ⊆VI ⊆B and 0 < Hlog2 ρ(Mm)(V C I ) ≤Hlog2 ρ(Mm)(B) ≤Hlog2 ρ(M2)(B) < ∞ by deﬁnition of m and Lemma 16, we have dimH(B) = dimB(B) = log2 ρ(M2). Finally, we translate this result to the sets Wη deﬁned in Theorem 10. From the deﬁnition of Wη we conclude that for η ∈{0, 1, x}, we have (6.13) Hlog2 ρ(M2)(∂Wη) ≤ X d∈{0,1} m∈Q∗ η0(2d+m)=η 1 ρ(M2)Hlog2 ρ(M2)(∂Am) < ∞. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 25 We now choose I ∈C and {u, v} ⊆I in such a way that u ̸= v and v2(u −v) is minimal, where v2(n) denotes the maximal t such that 2t divides n. Since ρ(M2) > 1 by Lemma 16, there is a transition J = d·I in C. We set u′ = d·u and v′ = d·v. As #I = #J = m, we have u′ ̸= v′. If u ≡v (mod 4), then u′−v′ = 1/2(u−η0(2d+u))−1/2(v−η0(2d+v)) = 1/2(u−v), so v2(u′ −v′) < v2(u −v), a contradiction. If u and v are both even, we similarly get u′−v′ = 1/2(u−v), which is also a contradiction. Therefore, we have η0(2d+u) ̸= η0(2d+v). Let now z ∈VI. By deﬁnition of Φ, every neighborhood of z contains a point z1 ∈Au and a point z2 ∈Av. We clearly have ϕd(z1) ∈Wη0(2d+u) and ϕd(z2) ∈Wη0(2d+v), which implies that ϕd(z) ∈∂(W0 ∪W1 ∪Wx). We conclude that (6.14) 0 < 1 ρ(M2)Hlog2 ρ(M2)(V C I ) = Hlog2 ρ(M2)(ϕd(V C I )) ≤Hlog2 ρ(M2)(∂(W0 ∪W1 ∪Wx)). This concludes the proof of the theorem for \|x\| ≥5. For x = −1, we note that ρ(M2) = 1, which implies that dimH(∂(W0 ∪W1 ∪Wx)) = 0 by Lemma 16. It can easily be checked that in this case, we have W0 = [0, 1/6) ∪(2/6, 4/6) ∪(5/6, 1], W1 = (1/6, 2/6), W−1 = (4/6, 5/6), and therefore H0(∂(W0 ∪W1 ∪Wx)) = 4. The remaining dimensions in Table 2 have been computed using Mathematica r ⃝. □ 7. Geometric Approach for Calculating the Frequency of Digits We give a geometric approach (going back to Delange ) using the results in the pre-ceeding sections to compute the summatory function of the frequency of digits. In contrast to the results in Section 4, we are now able to compute the summatory function up to some integer N instead of considering the full block length {0, . . ., 2L −1} as in Section 4. However, we will need the results of that section to compute the Lebesgue measures of the sets Wd, d ∈{0, 1, x}. For d ∈{1, x} and positive N ∈Z, let Hd(N) := N−1 X n=0 fd(n) = N−1 X n=0 X k≥0 [ηk(n) = d] , where fd(n) has already been deﬁned in (4.1). Since ηk(n) = 0 for k > ⌊log2(n)⌋+ #Q −2 by Theorem 7, Theorem 10 implies that Hd(N) = N−1 X n=0 K+1 X k=0 hn n 2k+2 o ∈Wd i , where K = ⌊log2 N⌋+ #Q −2. We proceed as in Section 4 of : We replace Wd by an appropriate approximation Wd,k, replace the sum by an integral and pull out the main 26 C. HEUBERGER AND H. PRODINGER term given by the Lebesgue measure of Wd. Since the technical modiﬁcations are straight forward, we skip the details. We get (7.1) Hd(N) = λ(Wd)N log2 N + Nψd({log2 N}) + O(N log2 ρ(M2)), where ρ(M2) is the dominant eigenvalue of the auxiliary automaton as in Theorem 11, ψd(z) := λ(Wd)(#Q −{z}) + 2{z}+#Q−2Ψd(2{z}+2−#Q) + ∞ X k=−1 βk, Ψd(t) := ∞ X k=0 Z t 0 x2k−2 ∈Wd −λ(Wd) dx, βk := Z 1 0 [{x} ∈Wd,k] −[{x} ∈Wd] dx, Wd,k := [ m∈2k+2Wd∩ m 2k+2, m + 1 2k+2 . As in we see that ψd(z) is continous in [0, 1) and that limz→1−ψd(z) exists. Of course, ψd is 1-periodic. We cannot conclude that ψd is continuous at z = 1 by direct computation as in . Instead, we get continuity by considering O(L) = Hd(2L) −Hd(2L −1) = 2L ψd(0) −ψd n log2 1 −1 2L o + O(L). Comparing with Theorem 9, we see that (7.1) implies that λ(Wd) = 1/6 for d ∈{1, x}. Since λ(W0 ∪W1 ∪Wx) = 1 we must have λ(W0) = 2/3. We summarize this result in the following theorem. Theorem 17. Let D = {0, 1, x} be a NADS, d ∈{1, x} and N be a positive integer. Then the number of occurrences of the digit d in the D-NAFS of the integers 0, . . . , N −1 equals Hd(N) = 1 6N log2 N + Nψd({log2 N}) + O(N log2 ρ(M2)), where ψd is a 1-periodic continuous function and ρ(M2) < 2 is the spectral radius of the adjacency matrix of the auxiliary automaton described in Section 6. Moreover, the Lebesgue measures of the sets Wd described in Theorem 10 equal λ(W1) = λ(Wx) = 1 6, λ(W0) = 2 3. 8. Non-Optimality For D = {0, 1, −1}, the D-NAF of n has minimal Hamming weight amongst all {0, 1, −1}-expansions of n, cf. Reitwiesner , where the Hamming weight c(η) of an expansion η ∈D 0 equals the number of nonzero digits #{j : ηj ̸= 0}. For x ≤−5, this is no longer the case: Theorem 18. Let D = {0, 1, x} be a NADS. Then the following two conditions are equiv-alent: ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 27 • For every positive integer n, the Hamming weight of the D-NAF of n is minimum amongst all D-expansions of n. • x ∈{−1, 3}. We ﬁrst consider the case x = 3 and give an algorithmic proof, which can also be used in more general situations. For instance, the case x = −1 follows along the same lines. Lemma 19. The {0, 1, 3}-NAF is the {0, 1, 3}-expansion of minimal Hamming weight. Proof of the lemma. We consider the transducer e T (cf. Figure 13). We introduce weights I 0 1 3 2 4 5 0\|ε 1\|ε 3\|ε 0\|0 1\|0 3\|0 0\|1, 1\|3 3\|3 0\|3 1\|1 3\|1 0\|0 1\|0 3\|0 1\|0 0\|0 3\|0 0\|1, 1\|3 3\|3 Figure 13. Transducer e T calculating a {0, 1, 3}-NAF of n from any {0, 1, 3}-expansion of n from right to left. for the transitions: w(i d\|o − →j) := c(d) −c(o), where c denotes the Hamming weight and c(ε) = 0. If we have a successful path I d\| − − →0, the weight of this path equals c(d) −c(η). We calculate the shortest path from I to 0 in this weighted digraph using the Ford-Bellman algorithm (cf. Cook et al. ). It turns out that the shortest path has weight 0, i.e. c(d) −c(η) ≥0, as requested. □ Remark 20. We note that the above proof of Lemma 19 also yields a complete description of all {0, 1, 3}-expansions of minimal weight: The Ford-Bellman algorithm calculates a feasible potential π : V →Z, where π(i) is the weight of a shortest path from I to i. In our case, we have i I 0 1 2 3 4 5, π(i) 0 0 1 1 1 1 2. Thus if we have π(i) + w(i d\|o − →j) > π(j) for some transition, the transition corresponds to an actual gain when modifying the given representation d to the D-NAF. Therefore, 28 C. HEUBERGER AND H. PRODINGER we remove all those transitions and all output labels and get an automaton which accepts minimal {0, 1, 3}-expansions, see Figure 14. In this ﬁgure, we also identiﬁed states 0 and I, since they only diﬀered in the output labels of the transitions leaving them. 0 1 3 2 4 5 0 1 3 0 3 0 1 3 0 0 3 0 Figure 14. Automaton accepting all binary {0, 1, 3}-expansions of minimal Hamming weight. If we use the same approach for x = −1, we get the transducer in Figure 15 and the automaton in Figure 16, respectively. The transducer corresponds to the algorithm due to Jedwab and Mitchell , the automaton to the syntactical rules described in Heuberger [?]. Proof of Theorem 18. By Lemma 19 and Reitwiesner , we only have to consider the case x ≤−5. If \|x\| + 3 = 2g for some g ≥4, we consider n = 2g+1 + 7. We have n = 1 · 22g + x · 2g + 1 · 22 + x · 20, (8.1) n = 1 · 2g+2 + x · 2 + 1. (8.2) Both are {0, 1, x}-expansions, (8.1) is the D-NAF with Hamming weight 4, whereas (8.2) is an expansion of Hamming weight 3. If x = −5, we have 23 = value(100¯ 50¯ 50¯ 5), (8.3) 23 = value( 1000¯ 51). (8.4) The ﬁrst expansion (8.3) is the D-NAF, again with Hamming weight 4, the second expan-sion has Hamming weight 3. Next, if \|x\| + 3 is not a power of 2, we consider n = 3. We have η0(3) = x, η1(3) = 0 and r2(3) = (3 + \|x\|)/4. By assumption, (3 + \|x\|)/4 is not a power of 2. Since (3 + \|x\|)/4 < \|x\|, we conclude that any D-expansion of (3+\|x\|)/4 has Hamming weight at least 2, therefore, ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 29 I 0 1 −1 2 −2 0\|ε 1\|ε 1\|ε 0\|0 1\|0 1\|0 0\|1, 1\|1 1\|1 0\|1, 1\|1 1\|1 1\|0 0\|0 1\|0 1\|0 0\|0 1\|0 Figure 15. Transducer e T calculating a {0, 1, −1}-NAF of n from any {0, 1, −1}-expansion of n from right to left. 0 1 −1 2 −2 0 1 1 0 1 0 1 0 0 Figure 16. Automaton accepting all binary {0, 1, −1}-expansions of mini-mal Hamming weight. the D-NAF of 3 has Hamming weight at least 3. However, the standard binary expansion 3 = value(11) has lower Hamming weight. □ 9. Addition of 1 The transducer for calculating the addition of 1 for x = −5 is shown in Figure 17. 30 C. HEUBERGER AND H. PRODINGER 1 2 3 4 5 6 0¯ 5\|00 0\|ε, 01\|0 0\|ε 01\|00 0¯ 5\|0 01\|00¯ 5 0¯ 5\|0¯ 5 0\|01 0\|0¯ 5 01\|001 0¯ 5\|01 0\|0, 01\|01, 0¯ 5\|0¯ 5 0\|01, 0¯ 5\|00¯ 5 01\|0¯ 5 Figure 17. Add +1 or −1 in {0, 1, −5}-NAF from right to left. Start node is 1 for addition of +1 and 2 for addition of −1 10. Digit formulæ It was proved in [14, 7] that for the digit set {0, 1, −1}, the formula n = X k≥0 n 2k+2 + 5 6 − n 2k+2 + 4 6 − n 2k+2 + 2 6 + n 2k+2 + 1 6 2k produces the NAF, i. e., that if we write the NAF of n as . . . a2(n)a1(n)a0(n), the digits are given by ak(n) = n 2k+2 + 5 6 − n 2k+2 + 4 6 − n 2k+2 + 2 6 + n 2k+2 + 1 6 . In the instance {0, 1, −5}, there is a similar formula, viz. ak(n) = n 2k+2 + 23 24 − n 2k+2 + 1 24 −6 n 2k+2 + 16 24 + 6 n 2k+2 + 4 24 . If ak(n) ∈{1, −5}, this formula produces the right result, but it sometimes happens that if ak(n) = 0, the formula gives either 1 or −5. For other systems like {1, 0, −13} etc. not even such a formula exists. ANALYSIS OF ALTERNATIVE DIGIT SETS FOR NONADJACENT REPRESENTATIONS 31 References 1. W. J. Cook, W. H. Cunningham, W. R. Pulleyblank, and A. Schrijver, Combinatorial optimization, Wiley-Interscience Series in Discrete Mathematics and Optimization, John Wiley & Sons Inc., New York, 1998. MR 99b:90098 2. H. Delange, Sur la fonction sommatoire de la fonction“somme des chiﬀres”, Enseignement Math. (2) 21 (1975), no. 1, 31–47. MR 52 #319 3. G. A. Edgar, Measure, topology, and fractal geometry, Undergraduate Texts in Mathematics, Springer-Verlag, New York, 1990. MR 92a:54001 4. K. Falconer, Techniques in fractal geometry, John Wiley & Sons Ltd., Chichester, 1997. MR 99f:28013 5. P. J. Grabner, C. Heuberger, and H. Prodinger, Distribution results for low-weight binary representa-tions for pairs of integers, Theoret. Comput. Sci. 319 (2004), 307–331. 6. R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete mathematics. A foundation for computer science, second ed., Addison-Wesley, 1994. MR 97d:68003 7. C. Heuberger and H. Prodinger, On minimal expansions in redundant number systems: Algorithms and quantitative analysis, Computing 66 (2001), 377–393. 8. H.-K. Hwang, On convergence rates in the central limit theorems for combinatorial structures, Euro-pean J. Combin. 19 (1998), 329–343. MR 99c:60014 9. J. Jedwab and C. J. Mitchell, Minimum weight modiﬁed signed-digit representations and fast exponen-tiation, Electron. Lett. 25 (1989), 1171–1172. 10. D. E. Knuth, Seminumerical algorithms, third ed., The Art of Computer Programming, vol. 2, Addison-Wesley, 1998. 11. P. Lancaster and M. Tismenetsky, The theory of matrices, second ed., Computer Science and Applied Mathematics, Academic Press Inc., Orlando, FL, 1985. MR 87a:15001 12. J. A. Muir and D. R. Stinson, Alternative digit sets for nonadjacent representations, to appear in Lecture Notes in Computer Science (Selected Areas in Cryptography, 2003), Preprint available at 13. , Alternative digit sets for nonadjacent representations, Journal Version of ; Preprint avail-able at 14. H. Prodinger, On binary representations of integers with digits −1, 0, 1, Integers 0 (2000), A08, avail-able at 15. G. W. Reitwiesner, Binary arithmetic, Advances in computers, vol. 1, Academic Press, New York, 1960, pp. 231–308. (C. Heuberger) Institut f¨ ur Mathematik B, Technische Universit¨ at Graz, Steyrergasse 30, 8010 Graz, Austria E-mail address: clemens.heuberger@tugraz.at (H. Prodinger) The John Knopfmacher Centre for Applicable Analysis and Number The-ory, School of Mathematics, University of the Witwatersrand, P. O. Wits, 2050 Johan-nesburg, South Africa E-mail address: helmut@maths.wits.ac.za
190094	https://www.youtube.com/watch?v=PylNyRNOe1Q	Using the Pythagorean Theorem in the Coordinate Plane Math and Science 1630000 subscribers 81 likes Description 1762 views Posted: 7 Feb 2024 This tutorial delves into the practical application of the Pythagorean Theorem in the coordinate plane, a crucial concept for students and professionals in mathematics, engineering, and related fields. This video is designed to provide a clear understanding of how this fundamental theorem can be used to solve problems involving distance and geometry within the Cartesian coordinate system. The video begins with a brief refresher on the Pythagorean Theorem, outlining its basic principle that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We then explore how this theorem applies to the coordinate plane, where the sides of a right triangle can be formed by the coordinates of points. Through step-by-step examples, viewers will learn how to use the Pythagorean Theorem to calculate distances between points on a grid. This includes determining the lengths of horizontal and vertical lines and then applying the theorem to find the hypotenuse, which represents the distance between two points. The video also addresses common misconceptions and pitfalls in applying the Pythagorean Theorem in coordinate geometry. By the end of this tutorial, viewers will have a solid grasp of using the Pythagorean Theorem in the coordinate plane, enhancing their problem-solving skills in geometry and beyond. Join us to master this essential mathematical technique. More Lessons: Twitter: 6 comments Transcript: so once again as in the last lesson we have a coordinate plane with using a map of the city the units are in kilometers use the Pythagorean theorem to find the distance from the school s to the library l so here is the school here is the library here is City Hall in the center of town and you can see I have this grid system set up these are kilometer one kilometer increments and so we're trying to figure out the straight line distance as the bird flies from the school to the library so how do we proceed well we can see that we have a right triangle of course we're trying to find this length but we have a right triangle here and so here is a 90 degree angle so the uh the hypotenuse C is listed right here so the distances of the other side sides are one two three four five six seven and eight and the other side is one two three four five six so the distances are eight and six kilometers so what we have is c squared is equal to a squared plus b squared and we're trying to find this distance here we know that one side is eight and the other side is six so we have to square them and so 8 times 8 is 64 and 6 times 6 is 36. when we add these numbers together we get a nice even number of 100. so c squared is 100. so how do we get rid of this Square term right here we take the square root of both sides and since we're applying our own square root we put a plus or minus cancel the root with the square and then what we have is the square root of 100 which is 10. we're going to throw away the negative because we don't have negative distances so 10 what 10 kilometers so if we were to construct a grid system on a map where City Hall was in the center and everything was measured relative to it right and you would go you know this many distance units to the left and this many distance units up to get to the school and similar thing down to the library then the straight line distance between them would actually be 10 kilometers as we've shown mathematically here let's take this one down and solve our next problem all right let's take a look at problem number two we're going to use the Pythagorean theorem to find the distance from the ice cream parlor I to the movie theater in so the ice cream parlor is here and the movie theater is here so we're trying to find this straight line distance so let's draw that distance and find a triangle that fits here's the straight line distance now we could draw a triangle down here but it's fine to draw a triangle like this either way it's going to work because the distance here is the same as the distance here and the distance here is the same as the distance here so you're going to get the same answer the hypotenuse C is what we're trying to find here so c squared is equal to a squared plus b squared now c squared is then equal to one side here is one two three four five six seven eight nine ten so it's ten squared and then the other side is one two three four squared so we have 10 squared is 100 and then 4 squared 4 times 4 is 16 so we have c squared is equal to the whoops let me get too far ahead of myself is uh 100 plus 16 116. and in order to solve for that we're going to do the opposite of the square we're going to apply the square root and add our plus or minus here and so we have the distance C is equal to this so we have to figure out what the square root of 116 is so we go over here and say what is the square root of 116. now you know it's an even number so it's divisible by 2. but it turns out that it's also divisible by four turns out that 29 times 4 is exactly equal to 116. now this is 2 times 2 and this is not really able to be broken up into anything that's going to have pairs so you just leave that there so what you end up with is a pair of twos which comes out as 2 times the square root of 29 so the 2 comes out the this stays under the radical so it's 2 times the square root of 29 which is approximately equal to 10.77 and it's rounded of course so 29 square root of 29 times 2 is this and also square root of 16 is this these are the same thing so this is the final answer in the unit here is in kilometers all right so just a little bit over 10 kilometers close to 11 kilometers is that distance all right we'll take this one down and do our last problem and here's our last problem we're trying to find the distance between the Delhi D and the grocery store G Deli to the or is it grocery store so this straight line distance let's draw that distance here it's from the deli to the grocery store we can form a right triangle of course right here and so the red line is the hypotenuse C all right so c squared is a squared plus b squared so we're trying to find C I have two distance units here so 2 squared and then we have one two three four five six seven eight in the other direction so we have eight squared up here and so 2 times 2 is 4 and 8 times 8 is 64. so c squared is this becomes 68 so c squared is 68. now we want to solve for C the distance C here so we're going to take the square root of both sides cancel these guys we have to add our plus or minus and the distance C is going to be the square root of 68. I don't know what that is so what times what is 68 well it appears that it's divisible by 2 but it turns out this is divisible by 4 as well 17 times 4. works out to be 68 and then the 4 is 2 times 2. I'm looking for pairs so the 2 comes out the 17 stays under the radical so 2 times the square root of 17. I discard the negative length because I can't have negative lengths of a triangle and this is the final answer now this also works out to be approximately equal to 8.25 kilometers so the square root of 68 or 2 times the square root of 17 work out to 8.25 and the units here are kilometers so as I said we could do more problems here but it kind of like gets very repetitive after a while there's not much more we can squeeze out of this just know that when you have grid systems of any type you can use the Pythagorean theorem the entire Earth is covered in a grid system called latitude and longitude and so even though the Earth is curved as long as we're not talking about distances all the way around the planet if we're talking about distances maybe within a city where it's not so far that the curvature of the earth comes into play then we can still use the Pythagorean theorem but as I mentioned early on if the surface is curved over which you're doing the stuff in the Pythagorean theorem doesn't hold anyway so if you try to find the distance using the Pythagorean theorem from Paris to you know Antarctica or something then the distance is so far that the the curvature of the Earth really makes the Pythagorean theorem not work really anymore because that curvature changes the quality and makes it not equal anymore but on a local situation in a city or a town or even maybe even a broad section that's not too big it it works just fine there are other applications as well we'll get to later but for now once you understand this we're going to shift gears and start talking about volume and surface area of geometric objects learn anything at mathandscience.com
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190096	https://study.com/academy/lesson/how-to-calculate-sales-price.html	Sale Price Formula, Calculation & Examples - Lesson \| 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 Copyright Math Courses 3rd Grade Math Sale Price Formula, Calculation & Examples Contributors: Christina Harvey, Bethany Calderwood Author Author: Christina Harvey Show more Instructor Instructor: Bethany Calderwood Show more Learn how to calculate sale price and selling point. Discover the sales price and selling price formulas and work through examples of calculating these prices. Updated: 11/21/2023 Table of Contents What are Sale Price and Selling Point? Sale Price and Selling Price Formula How to Calculate Selling Price from Cost and Margin How to Find Sale Price with Discounts How to Calculate Selling Price of a Product Lesson Summary Show FAQ How do you calculate sale price? A discount dollar amount is needed to calculate a sale price. This is done by multiplying the original amount by the sale percentage. This product is then subtracted from the original price to generate the sale price. How do you calculate buying and selling price? A selling price can be calculated in multiple ways. One way is to calculate the selling price based on a predetermined markup using the following equation: M a r k u p=P r i c e−C o s t C o s t∗100 For consumers, the buying price is the selling price, minus any sales or discounts. What is the sale price? A sale price is the pride an item is sold at minus any discounts. Sale prices may be calculated with a percentage off or dollar amount off. In some instance, sale prices may need to be calculated based off special promotions, such as "buy-one-get-one" events. Create an account LessonTranscript VideoQuizCourse An error occurred trying to load this video. Try refreshing the page, or contact customer support. You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching 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 now Already registered? 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Your next lesson will play in 10 seconds 0:04 Sales Price 0:33 Calculating the Discount 1:32 Calculating the Sales Price 1:53 Another Example 2:48 Lesson Summary QuizCourseView Video OnlySaveTimeline 25K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Percents: Definition, Application & Examples 6:20 ##### Percent Equation \| Definition, Formula & Calculation 7:21 ##### Calculating a 20 Percent Discount: How-to & Steps 3:27 ##### Calculating a 10 Percent Discount: How-to & Steps 3:08 ##### Calculating Percentage Discount \| Equation & Formula 6:23 ##### Percentage Increase & Decrease \| Formula & Examples 5:06 ##### Percentage Increase \| Definition, Formula & Calculation 5:16 ##### Percentages - Comparisons: Study.com SAT® Math Exam Prep 10:06 ##### What's 20 Percent of 1000? - How-to & Steps 3:29 ##### Percentages \| Study.com ACT® Math Test Prep 8:12 ##### Percentages - Combinations: Study.com SAT® Math Exam Prep 11:33 ##### Solve Problems Using Percents 7:50 ##### How to Estimate the Percent of a Number 5:25 ##### Percent Error \| Definition, Formula & Examples 3:48 ##### Fractional Percents \| Uses, Conversions, & Examples 4:54 ##### How to Use Percents Greater Than 100 5:12 ##### Finding the Whole From a Percent \| Overview & Applications 5:21 ##### Percentage Definition, Formula & Examples 4:32 ##### MEGA Elementary Education Mathematics Subtest: Practice & Study Guide ##### MEGA Elementary Education Study Guide and Test Prep ##### Basic Geometry: Help & Review ##### Student Resources ##### Math for Kids ##### ICAS Mathematics - Paper A: Test Prep & Practice ##### ICAS Mathematics - Paper B: Test Prep & Practice ##### ICAS Mathematics - Paper C: Test Prep & Practice ##### EQAO Grade 3 Math: Test Prep & Practice ##### GED Math: Quantitative, Arithmetic & Algebraic Problem Solving ##### AP Calculus AB & BC: Exam Prep ##### ELM: CSU Math Study Guide ##### Study.com SAT Study Guide and Test Prep ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - Number & Quantity: High School Standards ##### Common Core Math - Algebra: High School Standards ##### Common Core Math - Statistics & Probability: High School Standards ##### Common Core Math - Geometry: High School Standards ##### Common Core Math - Functions: High School Standards ##### Praxis 5165 Study Guide - Mathematics Exam Prep What are Sale Price and Selling Point? -------------------------------------- Maria is shopping for a new dress and wants to keep her total under $100. After a few minutes of browsing, Maria comes across a rack of dresses that suit her style. However, a quick glance at the price tags indicates that all the dresses are above Maria's budget. Feeling discouraged, Maria turns to leave when a store clerk stops her. The clerk points out that all dresses in the store are 20% off and that the price tags do not reflect the sale price. A sale price is the price of an item after all discounts have been deducted. After hearing this, Maria finds that the sale prices of many of the dresses are within her budget. Understanding how to calculate a sale price can help one become a savvy shopper. Not only will this ensure that one is selecting items that are the best deal, but understanding how to interpret sales and calculate discounts will ensure that the price paid reflects current store promotions and coupons. To unlock this lesson you must be a Study.com memberCreate an account Sale Price and Selling Price Formula ------------------------------------ Below, the sale price formula explores how to calculate a sale price from an original price with a known discount. In the following equation, "P" represents the sale price, "O" represents the original price, and "D" represents a known discount: P=O−D When sales offer a dollar amount off, then D is replaced with this known value. However, discounts are often presented as a particular percentage of the original price. When this is the case, the discount can be calculated from the percentage off using the following formula: D=O∗(P e r c e n t a g e) Markup and Gross Margin Understanding how prices work is not only important for consumers. Retailers who want to better understand the relationship between the sale price of an item and its cost of production will explore markups and gross margins. A markup is the percentage increase in price between the selling price and the cost of an item's production. A gross margin demonstrates the relationship between an item's sale price and cost reflected as a percentage of its revenue. Revenue is the amount of money that is acquired from selling an item. For instance, an item that sells for $6 generates $6 of revenue, no matter its cost of production. Because the cost must be lower than the selling price in order to generate a profit, an item's markup will always be higher than its gross margin. To unlock this lesson you must be a Study.com memberCreate an account How to Calculate Selling Price from Cost and Margin --------------------------------------------------- Businesses need to know what price to sell products for, to earn their desired margin. This margin needs to be set to cover desired earnings and overhead expenses, such as the cost to pay employees and lease store space. Once the margin has been determined, it can be applied to each product to determine the selling price, based on the cost of the product. The information needed to calculate the selling price from cost and margin are: Cost of the product Desired margin percentage Once this information is known the selling price can be calculated by: Converting the desired margin into a decimal Adding 1 to the margin decimal Multiplying the cost of the product by the previous sum Example of Calculating Selling Price from Cost and Margin A restaurant has determined that its margin is 25%. The restaurant has several new menu items: Salmon and rice: cost of ingredients is $15 Mushroom risotto: cost of ingredients is $13 Tomato soup: cost of ingredients is $10 To calculate the selling price for each of these new items, the margin needs to be considered. First convert the percentage into a decimal: 25% is equal to 0.25. Next, add one to the decimal: 1+0.25=1.25. Finally, multiply the cost of each product by 1.25: Salmon and rice: 15∗1.25=18.75 Mushroom risotto: 13∗1.25=16.25 Tomato soup: 10∗1.25=12.50 Using this method, the selling price for the salmon and rice should be $18.75, the mushroom risotto should be $16.25, and the tomato soup should be $12.50. To unlock this lesson you must be a Study.com memberCreate an account How to Find Sale Price with Discounts ------------------------------------- Finding a sale price can be done utilizing the following steps: Convert the discount into a percentage if it is not already given as one Determine a discount's dollar amount Subtract the dollar amount from the original selling price The first step is to determine the dollar amount of a discount. If the discount is already provided in a dollar amount, no conversion is needed. At times, discounts will be displayed in fractions. For instance, a discount may be described as "half-off." This means that the discount is represented by a fraction (1 2). The numerator must be divided by the denominator to generate a decimal. In this instance, "half-off" represents a discount of 0.5 (or 50%). More complicated fractions such as 3 4 of 7 8 are rarely used to display sales but follow the same procedure to generate a sales percentage if needed. The next step is to determine the discount's dollar amount. This can be done by multiplying the discount percentage by the original selling price. Once the discount amount is known, it is subtracted from the original selling price to determine the sale price. Examples of How to Calculate Sale Price The following examples illustrate how to calculate a sale price when given different types of sales. Continue reading to explore how to generate a sale price from a known: Half-off sale Percentage-off discount Half-Off Sale Katelyn is looking to buy a new set of tires and wants to know which of the two retailers is providing a better deal. The first retailer is selling tires for $100 each. The second retailer is selling tires for $150 each. However, the second retailer is running a special, where if a buyer purchases two tires, the next two tires are half off. For the first option, the total price for the tires is $400, as reflected by the equation below: 400.00=(4 T i r e s)100.00 1 T i r e The second retailer's total price must be calculated with the sale in mind. The first two tires are being sold at their normal price: 300.00=(2 T i r e s)150.00 1 T i r e The next two tires are being sold at half-off. To determine their sale price, the discount dollar amount must be calculated. A "half-off" sale has a discount of 0.5 (50%). Thus, the discounted price is 50% of $150, or $75. The sale price is thus calculated with the formula provided earlier in this lesson: P=O−D 75.00=150.00−75.00 At $75 per tire, the total cost of the two sale tires is $150. Thus, the total cost of the four tires with the two sale tires is $450. Even though the second retailer is having a sale, the first retailer is still a better deal. Percentage-Off Discount Jeremy's employees are given a 20% employee discount to eat at his restaurant. One of his waiters is bringing his family to the restaurant for dinner. Their total tab comes up to $50. Jeremy begins to calculate their total with the employee's discount by determining the dollar amount to take off of the total price. First, calculate 20% off of $50.00: 10=(50)∗(0.20) Now that the dollar amount is known, the discounted price for his employee's meal can be calculated: 40=50−10 With his 20% discount, Jeremy's employee paid a total discounted price of $40 from an original bill of $50. To unlock this lesson you must be a Study.com memberCreate an account How to Calculate Selling Price of a Product ------------------------------------------- A seller may want to know how much to sell their product for given a particular markup. The markup and selling price formula is as follows: M a r k u p=P r i c e−C o s t C o s t∗100 Knowing the amount of a markup is needed to know how to calculate the selling price of a product. If the seller knows how much they would like to markup their item, it can be plugged into the above equation to determine the item's cost. Similarly, gross margin (GM) can be calculated with the following equation: G M=R e v e n u e−C o s t R e v e n u e∗100 Explore these examples below to better understand how to price a product with calculations. Examples of How to Find Selling Price Tiffany wants to sell her new fudge at the farmer's market. After doing some research, she has found that competitor fudge makers sell their product with a markup of between 10-20%. Tiffany has decided that she wants to use a markup of 15% to start and wants to know what price her product must be sold at to meet this markup. Each piece of fudge she makes costs $3.15. With this, Tiffany's markup equation appears as such: 0.20=P r i c e−3.15 3.15∗100 3.15∗0.20=P r i c e−3.15 (3.15∗0.20)+3.15=P r i c e 3.78=P r i c e To meet a 20% markup, Tiffany must sell her fudge for $3.78 per piece. Now that she knows her selling price, Tiffany wants to know what her gross margin is. She calculates her gross margin as follows: G M=3.78−3.15 3.78∗100 G M=0.167 At a markup of 20% and a cost of $3.15 per piece, Tiffany's gross margin is 16.7%. To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- A sale price is the price of an item, minus any discounts. The sale price can be calculated by subtracting the dollar amount of any discount from the original price. A discount can be calculated by multiplying the percentage of the discount by the original price. In some cases, discounts are given as a dollar amount and don't need to be calculated. Sellers may want to determine the price of the product based on a markup, which is the difference between the sale price and the cost of production divided by the cost of production. Gross margin is like a markup but is calculated by subtracting the cost of production from the revenue and then dividing it by the revenue. Revenue is the amount of money obtained from selling a product before subtracting costs. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript Sales Price Mona wants to buy a new outfit, so she goes to ''Fabulous Frank's Clothing Store.'' There is a sign saying ''BIG SALE: 10% OFF SHIRTS.'' Mona wants to know how much each shirt will cost. To do that, Mona has to calculate sales price, the final price that needs to be paid after a discount is taken. Here is an equation you can use to calculate sales price: Original Price - Discount = Sales Price We will use Mona's shopping trip as an example to help us learn how to calculate sales price. Calculating the Discount To find out the sales price, you have to first calculate the discount. The discount is the percent off—it is the amount from the original price that you don't have to pay. If the sale is 10% off, the discount is 10% of the original price. The shirt Mona is looking at has a discount of 10% off. There is more than one way to calculate the discount. We will learn a way that, once you practice, you could do in your head if you were at the store with no calculator. Here is some information that will help us figure out how much money to subtract for the discount: Percent (%) means ''out of one hundred.'' One dollar is equal to 100 cents. This means a 10% discount is equal to 10 cents ''out of one hundred,'' or 10 cents taken off for every dollar in the original price. The original price on Mona's shirt is $5.00. A 10% discount for a 5 dollar shirt is 10 cents for every dollar, or 10 cents multiplied by 5. The equation can be written like this: $0.10 × 5 = $0.50 The discount is 50 cents. Calculating the Sales Price Remember the equation we mentioned earlier? Now it is time to use it: Original Price - Discount = Sales Price The original price of the shirt is $5.00. The discount we just calculated is $0.50. So: $5.00 - $0.50 = $4.50 The sales price for Mona's shirt is $4.50. Another Example Because the shirt is on sale, Mona decides to buy a pair of pants, too. She finds a rack of pants that says ''CLEARANCE: 50% OFF.'' The pants Mona likes have a price tag that says the original price is $20.00. Let's calculate the sales price. First, we have to find the discount. Mona's discount is 50% off. Remember, percent means ''out of one hundred.'' A 10% discount is 10 cents taken off for every dollar in the original price, and a 50% discount is 50 cents taken off for every dollar in the original price. For Mona's pants, 50 cents is multiplied by 20 because the original price was $20.00: $0.50 × 20 = $10.00 The discount is 10 dollars. Now we use the equation for sales price: Original Price - Discount = Sales Price $20.00 - $10.00= $10.00 The sales price on Mona's pants is $10.00. Lesson Summary To find sales price, the amount a person needs to pay for an item after a discount has been taken, you must first calculate the discount, the amount of the original price that you don't have to pay. Remember, percent (%) means ''out of one hundred,'' so ''10% off'' is 10 cents taken off for every dollar of the original price, ''50% off'' is 50 cents taken off for every dollar of the original price, and so on. Once you know the discount, subtract it from the original price. Then you know your sales price. Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now 3rd Grade Math 10 chapters 86 lessons Chapter 1 3rd Grade Math: Number Sense How to Read & Write Numbers Up to Six Digits: Lesson for Kids 4:21 min Recognizing Place Values to Hundred Thousands: Lesson for Kids 3:23 min Put Numbers in Order from Least to Greatest \| Overview & Purpose 6:49 min Comparison Symbols in Math \| Definition & Examples 3:01 min Number Line Definition, Use & Examples 5:16 min Introduction to Negative Numbers 6:27 min Rounding Whole Numbers \| Overview & Examples 6:03 min Rounding Numbers to the Nearest 1000, 10,000 & 100,000 7:04 min Expanded Form: Lesson for Kids 3:26 min Ordinal Numbers \| Definition, List & Examples 2:46 min How to Identify Even & Odd Numbers 4:00 min What is Skip Counting? 5:18 min Roman Numerals 1 to 20 \| Overview, List & Rules 6:41 min Chapter 2 3rd Grade Math: Creating & Interpreting Graphs Reading & Interpreting Bar Graphs 3:33 min Practice with Bar Graphs 4:12 min Line Graphs: Lesson for Kids 3:39 min Recording & Graphing Outcomes of a Simple Event: Lesson for Kids 3:14 min Chapter 3 3rd Grade Math: Addition & Subtraction What is Addition? \| Definition, Operation & Examples 4:07 min Regrouping in Addition \| Definition & Examples 3:09 min How to Add Two Numbers up to Three Digits 2:47 min How to Perform Subtraction: Steps & Examples 3:46 min Subtraction with Regrouping \| Overview & Examples 3:01 min How to Subtract Numbers Up to Three Digits 4:13 min Finding the Missing Addend 3:06 min Mental Math: Addition and Subtraction 4:34 min Chapter 4 3rd Grade Math: Multiplication & Division How to Perform Multiplication: Steps & Examples 5:22 min Learning Multiplication Facts to 10 Using Skip Counting 2:53 min Learning Multiplication Facts for 6s-9s Using Finger Tricks 3:52 min Working with Multiplication Input-Output Tables 3:32 min How to Multiply by 10 4:06 min Multiplying Whole Numbers With & Without Regrouping: Lesson for Kids 3:52 min How to Multiply Three or More Numbers 2:48 min Expanded Notation Method for Multiplication 3:23 min Using Mental Math for Multiplication 4:44 min The Relationship Between Multiplication & Division 3:50 min Division Lesson for Kids: Definition & Method 3:04 min How to Divide by 1 3:13 min How to Perform Division: Steps & Examples 3:56 min Long Division Steps: Lesson for Kids 3:44 min Division with Remainders \| Overview & Examples 3:19 min Chapter 5 3rd Grade Math: Understanding Fractions & Decimals Fractions: Lesson for Kids 2:39 min Numerator & Denominator Lesson for Kids 3:16 min Mixed Number to Improper Fraction \| Definition & Conversion 3:56 min Equivalent Fractions \| Overview, Chart & Calculation 5:12 min Comparing Fractions With Like Denominators: Lesson for Kids 3:18 min How to Read & Write Decimals to the Hundredths Place: Lesson for Kids 2:44 min Fractions to Decimals: Lesson for Kids 4:13 min Chapter 6 3rd Grade Math: Arithmetic With Money How to Write Amounts of Money: Lesson for Kids 3:14 min How to Find the Missing Coin 3:35 min Making Change from a Purchase 4:44 min Adding & Subtracting Money Amounts 4:29 min Multiplying Money by Whole Numbers & Decimals 5:57 min Dividing Money Amounts 5:14 min Estimation: Lesson for Kids 3:44 min Chapter 7 3rd Grade Math: Word Problems & Equations Multiplication Sentence \| Parts, Types & Examples 3:00 min Missing Quantities in Multiplication & Division Problems: Lesson for Kids 3:31 min Two-Step Math Word Problems 4:03 min Three Digit Multiplication Word Problems 5:19 min Order of Operations in Math \| Steps & Examples 5:50 min Chapter 8 3rd Grade Math: Measuring Physical Quantities Linear Measurements in Metric & English Units: Lesson for Kids 4:00 min How to Draw & Measure Line Segments: Lesson for Kids 3:24 min How to Estimate Measurements of Distance: Lesson for Kids 3:08 min How to Measure & Compare Weights: Lesson for Kids 4:16 min How to Estimate the Weight of Objects: Lesson for Kids 4:03 min How to Measure & Compare Liquid Volume: Lesson for Kids 3:31 min How to Estimate Liquid Volume: Lesson for Kids 3:31 min Temperature Definition: Lesson for Kids 3:59 min Measuring Temperature: Lesson for Kids 4:10 min Chapter 9 3rd Grade Math: Reading & Recording Time How to Tell Time on an Analog Clock 6:17 min Time Measurement \| Definition, History & Elapsed Time Calculation 3:21 min Reading a Calendar 4:31 min How to Write the Date with Words & Numbers: Lesson for Kids 2:35 min Chapter 10 3rd Grade Math: Geometric Shapes & Figures Identifying Types of Lines: Lesson for Kids 3:30 min Line Segment: Lesson for Kids 2:54 min Regular Polygon \| Definition, Sides & Types 6:01 min Quadrilateral Lesson for Kids: Definition & Shapes 3:18 min Pentagon Shape Facts: Lesson for Kids 2:39 min Hexagon Lesson for Kids: Definition & Facts 2:29 min Octagon Lesson for Kids: Definition & Facts 2:38 min Facts About Right Angles 3:09 min What is a Line of Symmetry in Geometric Shapes? 3:19 min Congruent in Geometry \| Definition & Examples 2:13 min Solid Figures \| Definition, Shapes & Examples 4:20 min How to Find the Area of a Triangle: Lesson for Kids 3:01 min Surface Area Lesson for Kids 3:58 min Perimeter Lesson for Kids: Definition & Examples 3:42 min Related Study Materials Sale Price Formula, Calculation & Examples LessonsCoursesTopics ##### Percents: Definition, Application & Examples 6:20 ##### Percent Equation \| Definition, Formula & Calculation 7:21 ##### Calculating a 20 Percent Discount: How-to & Steps 3:27 ##### Calculating a 10 Percent Discount: How-to & Steps 3:08 ##### Calculating Percentage Discount \| Equation & Formula 6:23 ##### Percentage Increase & Decrease \| Formula & Examples 5:06 ##### Percentage Increase \| Definition, Formula & Calculation 5:16 ##### Percentages - Comparisons: Study.com SAT® Math Exam Prep 10:06 ##### What's 20 Percent of 1000? - How-to & Steps 3:29 ##### Percentages \| Study.com ACT® Math Test Prep 8:12 ##### Percentages - Combinations: Study.com SAT® Math Exam Prep 11:33 ##### Solve Problems Using 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190097	https://artofproblemsolving.com/wiki/index.php/Geometric_sequence?srsltid=AfmBOorYjT4QoNYXQG9dBqmzGHh-JPk7nQEa2IgODpkHA4BnzuN-YIU0	Art of Problem Solving Geometric sequence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Geometric sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Geometric sequence In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence. For example, is a geometric sequence with common ratio and is a geometric sequence with common ratio ; however, and are not geometric sequences, as the ratio between consecutive terms varies. More formally, the sequence is a geometric progression if and only if . A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants , , and are in geometric progression if and only if . Contents 1 Properties 2 Sum 2.1 Finite 2.2 Infinite 3 Problems 3.1 Introductory 3.2 Intermediate 4 See also Properties Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let be the first term, be the th term, and be the common ratio of any geometric sequence; then, . A common lemma is that a sequence is in geometric progression if and only if is the geometric mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences. Sum A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms. Finite A finite geometric series with first term , common ratio not equal to one, and total terms has a value equal to . Proof: Let the geometric series have value . Then Factoring out , mulltiplying both sides by , and using the difference of powers factorization yields Dividing both sides by yields , as desired. Infinite An infinite geometric series converges if and only if ; if this condition is satisfied, the series converges to . Proof: The proof that the series convergence if and only if is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, Multiplying both sides by and adding , we find that Thus, , and so . Problems Here are some problems with solutions that utilize geometric sequences and series. Introductory 2025 AMC 8 Problem 20 Intermediate 1965 AHSME Problem 36 2005 AIME II Problem 3 2007 AIME II Problem 12 See also Arithmetic sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190098	https://ahdictionary.com/word/search.html?q=alluringly	American Heritage Dictionary Entry: alluringly HOW TO USE THE DICTIONARY To look up an entry in The American Heritage Dictionary of the English Language, use the search window above. For best results, after typing in the word, click on the “Search” button instead of using the “enter” key. Some compound words (like bus rapid transit, dog whistle, or identity theft) don’t appear on the drop-down list when you type them in the search bar. For best results with compound words, place a quotation mark before the compound word in the search window. guide to the dictionary THE USAGE PANEL The Usage Panel is a group of nearly 200 prominent scholars, creative writers, journalists, diplomats, and others in occupations requiring mastery of language. Annual surveys have gauged the acceptability of particular usages and grammatical constructions. The Panelists AMERICAN HERITAGE DICTIONARY APP The new American Heritage Dictionary app is now available for iOS and Android. THE AMERICAN HERITAGE DICTIONARY BLOG The articles in our blog examine new words, revised definitions, interesting images from the fifth edition, discussions of usage, and more. See word lists from the best-selling 100 Words Series! Find out more! INTERESTED IN DICTIONARIES? Check out the Dictionary Society of North America at al·lure (ə-lr) Share: Tweet v.al·lured, al·lur·ing, al·lures v. tr. To attract with something desirable; entice:Promises of quick profits allure the unwary investor. v. intr. To be highly, often subtly attractive:charms that still allure. n. The power to attract; enticement. [Middle English aluren, from Old French alurer : a-, to (from Latin ad-; see AD-) + loirre, bait (of Germanic origin).] al·lurementn. al·lurern. al·luring·lyadv. The American Heritage® Dictionary of the English Language, Fifth Edition copyright ©2022 by HarperCollins Publishers. All rights reserved. Indo-European & Semitic Roots Appendices Thousands of entries in the dictionary include etymologies that trace their origins back to reconstructed proto-languages. You can obtain more information about these forms in our online appendices: Indo-European Roots Semitic Roots The Indo-European appendix covers nearly half of the Indo-European roots that have left their mark on English words. A more complete treatment of Indo-European roots and the English words derived from them is available in our Dictionary of Indo-European Roots. American Heritage Dictionary Products The American Heritage Dictionary, 5th Edition The American Heritage Dictionary of Idioms The American Heritage Roget's Thesaurus Curious George's Dictionary The American Heritage Children's Dictionary CONTACT US Customer Service Make Me An Author Ebooks Help with Glose Reader ABOUT US Company Profile Leadership Team Corporate Social Responsibility HarperCollins Careers HarperCollins Imprints HarperGreen Social Media Directory Accessibility FOR READERS Browse Reading Guides FOR AUTHORS Submit a Manuscript Report Piracy Agent Portal MEDIA Publicity Contacts Press Room SERVICES HarperCollins Speakers Bureau Library Services Academic Services Desk & Exam Copies Review Copies OpenBook API Marketing Partnerships COVID-19 RESOURCES & PERMISSIONS Permissions for Adult Online Readings Permissions for Kids Online Readings SALES & RIGHTS Booksellers & Retailer Ordering HarperCollins Catalogs Permissions Subsidiary Rights Media Rights and Content Development GLOSE APP iPhone Android GLOBAL DIVISIONS HarperCollins US HarperCollins Canada HarperCollins Christian HarperCollins Australia HarperCollins India HarperCollins UK Terms of Use•Terms of Sale•Your Ad Choices•Privacy Policy•California Privacy Policy Do Not Sell My Personal Information Copyright 2022 HarperCollins Publishers All rights reserved. This website is best viewed in Chrome, Firefox, Microsoft Edge, or Safari. Some characters in pronunciations and etymologies cannot be displayed properly in Internet Explorer.
190099	https://mathoverflow.net/questions/3332/two-commuting-mappings-in-the-disk	gn.general topology - Two commuting mappings in the disk - MathOverflow Join MathOverflow 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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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 Two commuting mappings in the disk Ask Question Asked 15 years, 11 months ago Modified10 days ago Viewed 13k times This question shows research effort; it is useful and clear 188 Save this question. Show activity on this post. Suppose that f f and g g are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in R n R n) to itself. Does there always exist a point x x such that f(x)=g(x)f(x)=g(x)? If one of the mappings is invertible, then it is just a restatement of Brouwer's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension 1 1. gn.general-topology open-problems Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited May 31, 2015 at 11:34 Ricardo Andrade 6,291 5 5 gold badges 43 43 silver badges 69 69 bronze badges asked Oct 29, 2009 at 20:16 fedjafedja 63.8k 11 11 gold badges 164 164 silver badges 308 308 bronze badges 5 2 Could more be true? Would f and g necessarily have a common fixed point? This might perhaps be easier to prove if true.Harald Hanche-Olsen –Harald Hanche-Olsen 2009-10-29 23:35:35 +00:00 Commented Oct 29, 2009 at 23:35 12 Apparently f and g may not have a common fixed point even in the dimension 1 case. This is mentioned in the first paragraph of "Equivalent Conditions involving Common Fixed Points in the Unit Interval" by Jachymski. Unfortunately I can't follow the reference given. @fedja, what an amazing problem! Alon Amit –Alon Amit 2009-10-30 16:58:33 +00:00 Commented Oct 30, 2009 at 16:58 9 Ah, indeed: Jachymski refers to an abstract in an old Notices. But searching MR reveals J.P. Huneke: On common fixed points of commuting continuous functions on an interval. Trans. Amer. Math. Soc. 139 1969 371--381. See jstor.org/stable/1995330 if you have JSTOR access. From the abstract: “This paper offers two methods of constructing commuting pairs of continuous functions [...] which map [0,1] to itself without common fixed points”. Jachymski also notes that if the iterates of one function forms an equicontinuous family, there is a common fixed point.Harald Hanche-Olsen –Harald Hanche-Olsen 2009-10-30 18:02:53 +00:00 Commented Oct 30, 2009 at 18:02 9 If I understand well, you don't ask whether f f and g g have a common fixed point. Yet, the answers given so far speak of fixed points ...Denis Serre –Denis Serre 2011-03-31 14:28:29 +00:00 Commented Mar 31, 2011 at 14:28 A paper of Christian Bonatti is vaguely related: "Un point fixe commun pour des difféomorphismes commutants de S^2", Annals of Maths 1989.Benoît Kloeckner –Benoît Kloeckner 2014-01-10 20:14:43 +00:00 Commented Jan 10, 2014 at 20:14 Add a comment\| 8 Answers 8 Sorted by: Reset to default This answer is useful 28 Save this answer. Show activity on this post. A positive answer to this question would imply a positive answer to the open problem on existence of a "coincidence point" of two commuting maps f 1,f 2:T→T f 1,f 2:T→T, where T T is the triod (tripod), see Question 1 in McDowell's survey The point is that if f 1,f 2:T→T f 1,f 2:T→T are commuting maps as above, one can define maps f i~=f i∘R:D 2→D 2 f i~=f i∘R:D 2→D 2 where R:D 2→T R:D 2→T is a retraction. (I am assuming that T T is embedded in D 2 D 2.) Then f 1~,f 2~f 1~,f 2~ commute if and only if f 1 f 1 and f 2 f 2 commute, furthermore, f 1 f 1 and f 2 f 2 have a coincidence point x,f 1(x)=f 2(x),x,f 1(x)=f 2(x), if and only if f 1~,f 2~f 1~,f 2~ do. Of course, if one were to look for counter-examples, it would be easier to find ones among maps of the 2-disk. Incidentally, it appears that coincidence problem (in the general setting of compact manifolds) was first addressed by Lefschetz, see . Lefschetz formula for coincidence was extended to compact manifolds with boundary in 1980s, see references in Saveliev's paper Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 18 at 10:20 Martin Sleziak 4,783 4 4 gold badges 38 38 silver badges 42 42 bronze badges answered Apr 26, 2012 at 18:30 MishaMisha 31.8k 1 1 gold badge 97 97 silver badges 166 166 bronze badges Add a comment\| This answer is useful 22 Save this answer. Show activity on this post. I'm sure what I write has been thought of by many, but it's a starting point that I thought should be written down. First by the Brouwer fixed point theorem f f has at least one fixed point, say x¯=f(x¯)x¯=f(x¯). If that fixed point is unique (contraction mappings spring to mind for a bunch of examples of this) we're done since g(x¯)=g(f(x¯))=f(g(x¯))g(x¯)=g(f(x¯))=f(g(x¯)) and we see that g(x¯)g(x¯) is "another" fixed point of f f, since the fixed point was unique g(x¯)=x¯=f(x¯)g(x¯)=x¯=f(x¯). For "less nice" f f we still have that f(g(x¯))=g(x¯)f(g(x¯))=g(x¯)... in fact for any n∈N n∈N we have f(g n(x¯))=g n(x¯)f(g n(x¯))=g n(x¯). If (without resorting to sequences) g n(x¯)→y g n(x¯)→y, we can again claim success since we'll have g(y)=y g(y)=y and f(y)=y f(y)=y. Unless there's another "obvious" easy case I missed it seems like the interesting cases will be when g n(x¯)g n(x¯) does not converge. Two sub-cases spring to mind: when g n(x¯)g n(x¯) has finitely many accumulation points (like when g n(x¯)g n(x¯) is a periodic point of g g), or ... it has lots. Intuition (really thinking about rational and then irrational rotations about the origin as one way to generate those two cases) tells me that in either of these cases what we really need to do drop the x¯x¯ as a "starting point". It "would be nice" if we can show g g conjugate to a rotation in the above two cases. Any thought on if that is true or not? I suspect not else g g would have a unique fixed point and we'd be done (as above)... Maybe semi-conjugate... but would that help? New minds, any thoughts? Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications edited Mar 31, 2011 at 14:42 Denis Serre 53k 10 10 gold badges 152 152 silver badges 315 315 bronze badges answered Jun 4, 2010 at 6:14 MarkMark 281 2 2 silver badges 6 6 bronze badges 1 g g could be a reflection instead of a rotation. Draw a vector field in the disc which is symmetric across the y y-axis and which has exactly two zeros, and let f f be a small flow along this vector field. Let g g be reflection in the y y-axis. There are exactly two places f f and g g agree: (0,1)(0,1) and (0,−1)(0,−1).Sean Eberhard –Sean Eberhard 2015-04-13 16:16:39 +00:00 Commented Apr 13, 2015 at 16:16 Add a comment\| This answer is useful 13 Save this answer. Show activity on this post. There is also a result due to Shields which implies a common fixed point (and hence a coincidence point) assuming that f f and g g are both analytic on the disk. A. L. Shields, On fixed points of commuting analytic functions, Proc. Amer. Math. Soc. 15(1964), 703-706. MR 29 #2790. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 20, 2015 at 19:41 Jeff NordenJeff Norden 563 4 4 silver badges 9 9 bronze badges Add a comment\| This answer is useful 13 Save this answer. Show activity on this post. I just ran across some information on especially the one-dimensional version of this problem, which appears in Victor Klee's unpublished Unsolved Problems in Intuitive Geometry. His formulation is: "Suppose I I is a closed interval of real numbers, and f f and g g are commuting continuous maps of I I into itself. Must they have a common fixed point?" He attributes the problem to J.R. Isbell from 1957 ["Research Problem #7: Commuting mappings of trees," Bull. Amer. Math. Soc. 63 (1957), 419]. Negative answers by Boyce and Huneke were published (in Notices Amer. Math. Soc.) in the late 1960's. (Harald Hanche-Olsen cites the Huneke paper in comments above.) Branko Grünbaum cites a recent survey on the topic: Eric L. McDowell, "Coincidence Values of Commuting Functions," Topology Proceedings 34 (2009) pp. 365-384. I cannot easily access this paper, but Grünbaum says it contains new results and an extensive bibliography. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 18 at 10:21 Martin Sleziak 4,783 4 4 gold badges 38 38 silver badges 42 42 bronze badges answered Jul 14, 2010 at 15:53 Joseph O'RourkeJoseph O'Rourke 153k 36 36 gold badges 371 371 silver badges 994 994 bronze badges Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. The following is not a solution but rather a reformulation of the original problem. To make f:B→B f:B→B invertible let us pass from the ball B B to the solenoid S f S f, S f={{x n;n∈Z}:x n+1=f(x n)}S f={{x n;n∈Z}:x n+1=f(x n)} Map f f induces the shift f∗:S f→S f f∗:S f→S f and g induces g∗({x n})={g(x n)}g∗({x n})={g(x n)}. They commute and f∗f∗ is invertible. If Brower's fixed point theorem were true for S f S f the result would follow. Indeed, a fixed point of f−1∗∘g∗f∗−1∘g∗ gives the desired orbit. Solenoid S f S f is a non-empty closed subset of B Z B Z eqiupped with the product topology. Hence S f S f is compact. In fact, S f⊂K Z S f⊂K Z, where K=∩n>0 f n(B)K=∩n>0 f n(B). And I think it is plausible (can anybody give a proof?) that S f S f contracts to the orbit of the fixed point of f f. (later: I am not that sure now) Unfortunately, Brower's fixed point theorem doesn't hold for compact contractible spaces. But this seem to be quite old and active research area as I learned from the introduction to this paper People prove positive results under various additional assumptions. So, maybe it's useful to look at the problem from this point of view. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 18 at 10:21 Martin Sleziak 4,783 4 4 gold badges 38 38 silver badges 42 42 bronze badges answered Jun 13, 2010 at 7:55 Andrey GogolevAndrey Gogolev 4,278 1 1 gold badge 22 22 silver badges 26 26 bronze badges 2 1 Could you put that pdf somewhere for those of us who aren't at utexas? Well, or at least the title of the paper so we can find it through our own institutions? At the moment, that link only works for utexas people.Harry Gindi –Harry Gindi 2010-06-13 08:59:55 +00:00 Commented Jun 13, 2010 at 8:59 Well, I give up trying to fix the formula at the end of the first paragraph.Andrey Gogolev –Andrey Gogolev 2010-06-15 17:47:57 +00:00 Commented Jun 15, 2010 at 17:47 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. Off on a tangent ... Kakutani's Fixed-Point Theorem A commuting family of continuous affine maps of a weakly compact convex set into itself has a common fixed point. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Nov 19, 2009 at 13:51 Gerald EdgarGerald Edgar 41.7k 5 5 gold badges 126 126 silver badges 220 220 bronze badges 1 A related tangent: applying this other Kakutani Fixed-point theorem (en.wikipedia.org/wiki/Kakutani_fixed_point_theorem) to the multi-valued function g−1∘f g−1∘f looks like it solves the problem... unfortunately the sets f−1(g(x))f−1(g(x)) aren't convex (which is a pretty significant assumption!)Fabrizio Polo –Fabrizio Polo 2010-05-02 11:12:45 +00:00 Commented May 2, 2010 at 11:12 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. May be a possible way to solve the problem is to set up it in the context of closed relations instead of functions, by putting R=g−1∘f:B n→B n R=g−1∘f:B n→B n, and assuming for instance f(B n)⊂g(B n)f(B n)⊂g(B n). So, we are looking for x∈B n x∈B n such that x∈R(x)x∈R(x). In shape theory there are defined the so-called multihomotopy groups, which generalize the classical homotopy groups by considering multi-valued loops instead of the single-valued ones. See Sanjurjo, José M. R., Multihomotopy, Čech spaces of loops and shape groups, Proc. Lond. Math. Soc., III. Ser. 69, No. 2, 330–344 (1994). Zbl 0826.55004. and Giraldo, Antonio; Sanjurjo, Jose M. R., Strong multihomotopy and Steenrod loop spaces, J. Math. Soc. Japan 47, No. 3, 475–489 (1995). Zbl 0842.55006. Project Euclid. Under nice circumstances the multihomotopy groups of spheres are π~n(S n)≅Z π~n(S n)≅Z, while for the n n-disk are trivial. One can try to adapt the proof of Brouwer's fixed point theorem in this setting. So, assuming by contradicition that f(x)≠g(x)f(x)≠g(x) for all x∈B n x∈B n, one can consider the relation T:B n→S n−1 T:B n→S n−1 where T(x)T(x) is the intersection between S n−1 S n−1 and the half-lines starting from points of R(x)R(x), and passing through x x. It follows that T T is the identity on S n−1 S n−1, so T T is a multi-valued retraction B n→S n−1 B n→S n−1, and this contradicts the fact that π~n−1(S n−1)π~n−1(S n−1) is not trivial. I think that there are many details to fill, and that one should give the "nice" conditions under this approach works (for instance, assuming that g−1(y)g−1(y) is countable for all y∈B n y∈B n does suffice?) but this could be an interesting application of multihomotopy groups. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 18 at 10:12 The Amplitwist 1,661 4 4 gold badges 14 14 silver badges 28 28 bronze badges answered May 14, 2013 at 10:30 Daniele ZuddasDaniele Zuddas 2,323 13 13 silver badges 20 20 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I made a little google search and found this paper, that mentions that the conjecture holds true for polynomials and special functions Cohen calls "full functions". I'm not able to download the paper though. Cohen's paper is: H. Cohen, On fixed points of commuting functions, Proc. Amer. Math. Soc. 15. (1964), An idea that occured to me, since apparently this is true for polynomials. Can we not use some form of approximation theorem (Stone-Weierstrass or whatever there is) to conclude the result for other continuous functions between the closed unit balls? Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications edited Nov 6, 2009 at 18:03 answered Nov 6, 2009 at 12:55 Jose CapcoJose Capco 2,646 5 5 gold badges 32 32 silver badges 38 38 bronze badges 5 By the Brouwer fixed point theorem, for every f there is an x such that f(x) = x, so the answer is yes when g is the identity.Autumn Kent –Autumn Kent 2009-11-06 12:58:26 +00:00 Commented Nov 6, 2009 at 12:58 Ok sorry.. Kent is correct.. Brouwers fixed point theorem, my argument does not work because the function that I "drew" below the diagonal must have values in -0.5 an 0.5 respectively so it will necessarily cross the diagonal.. Jose Capco –Jose Capco 2009-11-06 13:48:44 +00:00 Commented Nov 6, 2009 at 13:48 8 The trouble with proving this via an approximation scheme is that you would have to prove that if f f and g g commute, then you can approximate f f and g g by polynomials f′f′ and g′g′ that commute. This sounds unlikely.Andy Putman –Andy Putman 2009-11-06 18:10:14 +00:00 Commented Nov 6, 2009 at 18:10 1 I think that you should avoid editing your answers the way you did (erasing everything and writing a new answer). It's better to delete the answer and write a new one.Ori Gurel-Gurevich –Ori Gurel-Gurevich 2009-11-07 07:47:59 +00:00 Commented Nov 7, 2009 at 7:47 I wasn't able to delete after several posts.. and things get being added to the features I am able to do here. I think this is a overlflow.net system, it's a bit confusing and annoying. I wasn't even able to make comments before. But now I can at least delete :) Jose Capco –Jose Capco 2009-11-07 09:34:19 +00:00 Commented Nov 7, 2009 at 9:34 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 gn.general-topology open-problems See similar questions with these tags. 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