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anced amplifier - Can we go faster? (i.e., can we enhance its bandwidth?) (cid:131) We will look at the following - Reduction of Miller effect on Cgd - Shunt, series, and zero peaking - Distributed amplification H.-S. Lee & M.H. Perrott MIT OCW Miller Effect on Cgd Is Significant RL Id vout M1 CL Zin Cgd Cgs Rs vin Vb...
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bias Cgs M1 CL (cid:131) Consider canceling the effect of Cgd - Choose CN = Cgd - Charging of Cgd now provided by CN (cid:131) Benefit: Impact of Cgd reduced: :same as cascode H.-S. Lee & M.H. Perrott MIT OCW Neutralization, cont’d (cid:131) Issues: - What happens if CN is not precisely matched to Cgd? - Since the neu...
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Lee & M.H. Perrott MIT OCW Shunt-peaked Amplifier - Analysis Vdd Ld RL Id M1 vin Vbias Small Signal Model Zout vout Ctot Ld RL vout Cfixed M2 iin=gmvin (cid:131) Expression for gain (cid:131) Parameterize with - Corresponds to ratio of RC to LR time constants H.-S. Lee & M.H. Perrott MIT OCW The Impact of Choosing Di...
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:131) Maximum bandwidth: m = 1.41 (extension = 1.85) (cid:131) Maximally flat response: m = 2.41 (extension = 1.72) (cid:131) Best phase response: m = 3.1 (extension = 1.6) (cid:131) No inductor: m = infinity 5 0 -5 -10 -15 -20 -25 -30 -35 -40 ) B d ( i n a G d e z i l a m r o N (cid:131) Eye diagrams often ...
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ω<<ωT Transfer function can now be simplified to Adds a zero at 1/RsCs, but introduces a 2nd pole at Reduces low freq. gain to The 1st pole at 1/RLCtot can be cancelled by making RsCs=RLCtot, then the bandwidth is extended to the 2nd pole H.-S. Lee & M.H. Perrott MIT OCW Zero-peaked Amplifier Continued (cid:131) Po...
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3.185 Lectures Adam Powell Fall Semester, 2003 Abstract This represents the sum total of lecture material presented in 3.185, Transport Phenomena in Materials Engineering, in the Fall of 2003. It is not meant to be a “reader” for the course, but more of an “electronic notebook” of my own, a set of bullet points if...
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. . . . . . . . . . . 11 2.4 September 12, 2003: 9/11 remembered, ABET . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 September 15, 2003: Wrapup unsteady, boundary conditions . . . . . . . . . . . . . . . . . . . 20 . . . . . . . . . . . . . . . . . . . . . 22 2.6 September 17, 2003: Boundary conditions, l...
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3.4 October 1, 2003: Math Quiz, Graphs Wrapup, Newtonian Cooling . . . . . . . . . . . . . . . 31 3.5 October 3, 2003: Moving on... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.6 October 6, 2003: Finite Differences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.7 ...
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. . . . . . . 39 3.10 October 20, 2003: Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.11 October 22, 2003: More Radiation 4 Fluid Dynamics 43 4.1 October 24, 2003: Intro, Newtonian Fluids . . . . . . . . . ...
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5 November 3, 2003: Navier­Stokes Equations! . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 4.6 November 5, 2003: Using the Navier­Stokes Equations . . . . . . . . . . . . . . . . . . . . . . 53 4.7 November 7, 2003: Drag Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.8 No...
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bulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 4.12 November 24, 2003: Turbulence, cont’d 1 5 Coupled Fluids with Heat and Mass Transfer 66 5.1 November 26, 2003: Coupled Fluids, Heat and Mass Transfer! . . . . . . . ....
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icity . . . . . . . . . . . . 74 5.6 December 10, 2003: Bernoulli, Semester Wrapup . . . . . . . . . . . . . . . . . . . . . . . . . 76 2 Chapter 1 Introduction 1.1 September 3, 2003 Handouts: syllabus, ABET, Diffusion, PS1 due Monday 9/8. Circulate signup sheet: name, username, year, course • Introductions: me, ...
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things: low­cost high­quality processes and high performance. They don’t care about structure. Andy Groves, chairman of Intel, could care less about the electronic structure of titanium silicide­titanium aluminide diffusion barriers in aluminum interconnects, he wants cheap high­quality processes that result in high ...
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a moving frame. D Dt = ∂ ∂t + � u · � Kind of like moving vector x(t), y(t), z(t): � � dC � � dx (x,y,z) = ∂C ∂t + ∂C ∂x ∂x ∂t + ∂C ∂y ∂y ∂t + ∂C ∂z . ∂z ∂t Previous feedback • Prof. Powell is cool, lectures are great, double tests are neat! (1.2) (1.3) (1.4) (1.5) (1.6) (1.7) • This cours...
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5 Chapter 2 Diffusion 2.1 September 5, 2003: 1­D Cartesian and Cylindrical Steady­ State TODO: • Check reading room to make sure texts are there. • Bring: cards, class list. • Check text to make sure chap 25 units are consistent with mine. Opener: Colleen’s facility with names... my advisee! Mechanics: • Diffusi...
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important understand how works, because from understanding flows design recommendations. ASSUME diffusion­limited, so this is the slow­step, not adsorption/desorption etc. Simple solution: unroll to a plate, with Cin on one side (equilibrium with partial pressure in natural gas) and Cout on the other (pumped away int...
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cm, this gives J = 2 × 10−8 cm2 s · g 10−5 cm 0.05cm g cm2 s· 3 = 4 × 10−12 . (2.7) Tube array with total length 10m=1000cm (e.g. 100 tubes each 10 cm long), R2 = OD/2 = 0.1cm, so throughput is J A = 2πR2LJ = 8 × 10−10π = 2.5 × 10−9 g s (2.8) Or do we use R1? That would give 1.2 × 109 . How far off is the ...
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d dr 0 = d dr � rD � . dC dr A� = rD dC dr A� Dr = dC dr C = A ln r + B where A = A�/D. From BCs: Check at R1 and R2, units. Flux= −DdC/dr: C − Cin Cout − Cin = ln(r/R1) ln(R2/R1) Jr = −D dC dr = −D d dr � Cin + (Cout − Cin) � ln(r/R1) ln(R2/R1) = D Cin − Cout 1 . ln(R2/R1) r Imp...
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sphere for a drug delivery device. Pretty cool. 2 ln(2), e.g. 3 mm OD with no change in thickness! 2 because ln( 2) = 1 √ √ 8 2.2 September 8, 2003: Steady­State with Homogeneous Chemi­ cal Reaction Mechanics: • New recitations: R12, F2, both in 8­306. • Fri: very different lecture, ABET stuff. Names. Muddy stuff...
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(2.21) (2.22) (2.23) (2.24) (2.25) (2.26) (2.27) (2.28) • Didn’t finish: timescale to steady­state τ ∼ L2/D, in this case 125,000 seconds, about a day and a half. Will explore this more rigorously on Friday. Summarize: illustrates 3.185 methodology • Problem statement: maximize throughput = flux× area • Conse...
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= A exp � � � � L k 2 D � � �� L k 2 D = A cosh � � � L k 2 D + exp − Result: C C0 = � � � cosh x � � � cosh L 2 k D k D (2.29) (2.30) (2.31) (2.32) What does this look like? Pay attention to , or more generally, L2 k . Can either be sorta uniform, D or VERY non­uniform, uniform if that num...
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∂r2 1 ∂ r ∂r + � r 1 ∂C + r ∂r � ∂C ∂r + 1 ∂2C r2 ∂θ2 ∂2C 1 r2 ∂θ2 ∂2C ∂z2 ∂2C ∂z2 + , or + , or (2.33) (2.34) (2.35) (2.36) • During derivations, important points are obscured. “What we’ve learned” summaries help. Basi­ cally following outline mentioned before, on p. 465 of W3R. Also feel free ...
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henius plot: which part is diffusion­limited, which part reaction­limited? Can’t really compare because of different units. But can sort of make a plot of log(kL2/D) vs. 1/T , look at different parts. Want: low­temperature reaction­limited case, with fast diffusion to wipe out conc gradients. Diffusion­limited means it d...
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rate also in mol/sec = V ∂C/∂t. Chapter 27 material. Resulting equation in 1­D: ∂C ∂t = D ∂2C ∂x2 + G. (2.38) Physical intuition: concentration curvature is either due to generation, or leads to time evolution, or both. Upward curvature means neighbors diffuse in, either G negative or C increases; downward mean...
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Cs − Ci)erfc ⇒ C = C0, uniform C IC t = 0 ⇒ � x √ Dt 2 � x √ Dt 2 � (2.44) (2.45) Semi­infinite. But what if your part is not semi­infinite, but thickness L? (Not many parts are semi­ infinite...) Since erf(2)=0.995 and erfc(2)=0.005, we can approximate ∞ � 2, if > 2, then erf is about 1 and erfc about 0, can...
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ABET handout! Schedule office hours. September 11. The day meant a lot of things to a lot of people. Yesterday the occasion was commemorated in a number of ways, here in Boston, in my hometown of New York City, and around the country and the world. I can’t hope to be as profound as some of the speakers at those servi...
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of Jewish boys, and in fact, was set up to keep them safe throughout the Nazi occupation. He described the terror he felt under the occupation, and then the arrival of the American soldiers, “All of them giants,” he said, then pointed to me, “like Adam,” they had come to set the continent free. And he described the...
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free and the home of the brave.” Then after a pause, “It really is true.” His personal experience of this gave great weight to these words. During these drives along the Belt Parkway, my thoughts also turned toward the fragility of the grand edifices, and in particular to the 1993 bombing of the underground parking l...
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to the Mid­East, and this administration’s policy of deliberate neglect in the Israeli­Palestinian peace process. Then the last call, one tower had collapsed. With her voice choking from the tears, she described its fall as “like a house of cards,” and could say little more. 14 Immediately, I logged out, got on my...
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nished moving into the 89th floor of Tower 1. His staff had been told not to come in that morning until 10 AM, because their carpets, freshly washed during the 1­7 AM shift, would need to finish drying. As he drove north on the New Jersey Turnpike, he saw the first plane crash right through the windows of his new office, ...
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two in his company who didn’t make it out. To make matters worse, she was trapped in L.A. because of grounded planes, unable to get back and try to locate him, so day upon day she was not only uncertain and hurting but frustrated at being far from anyone who could help her. She is still grieving, as it’s hard to acc...
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); it was the last of these categories, serving society, which steered me into Materials Science. As a high school student, I verbalized this service as follows. As a scientist or engineer, I would be helping to solve the world’s little problems, which I listed as: • Agriculture, to feed a growing planet. • Medici...
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values, of that purpose and meaning. • Last year Ross Benson added: Tolerance of differences. An important consequence of this understanding of “little problems” and “big problems” is that being a scientist or engineer requires a lot of faith, faith that our knowledge and our inventions will be used wisely, for good...
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. I have an advisee taking this class now who took off all of last Spring for that very purpose, and ended up returning to MIT (and in fact to Materials Science) that much more focused than the previous December for the experience. Of course, you don’t have to take off a semester to do this, there are very good ways t...
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organizations I’ve given to since my undergraduate years, one even since high school. Fourth, look for opportunities to participate in the process of improving lives yourselves. Whether tutoring or mentoring, or working in a social justice organization, or writing to Congress, participating in society in a meaningf...
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faith. Human institutions, organizations, systems and even nations are terrific, but never perfect, as we learned in a powerful way on September 11. Participating in and strengthening them is an important and honorable activity, but I believe that placing all of our hope on them is not viable in the long term. At som...
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things have changed, and as citizens there are things we can do on a daily basis to improve our country’s security, and the silence from Washington has been deafening. A few months ago I saw a book provocatively titled, “When you ride alone, you ride with bin Laden.” The cover art was derived from a World War II pos...
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for a number of reasons. First, as mentioned, it was completely 17 unjustified. There was no correlation to terrorism, none to weapons of mass destruction, and there are a dozen dictators around the world whose human rights violations could similarly motivate action. If violation of U.N. resolutions was the motive,...
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potentially for many genera­ tions. The arrogance with which the administration shrugged off overwhelming international opposition was shocking, prticularly as it involved three of the five Security Council permanent members, some of our most important allies, and both of our neighbors. Even more shocking was the pathe...
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Sheet Review See the ABET sheet. Unsteady Diffusion Continued Muddy stuff: • Symbol conventions: Ci is initial, C0 I use as surface and sometimes other things; Cs always surface concentration. • Physical meaning of graphs, interpretation of these things: coming soon. • Exact criteria for each solution: will be summ...
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g(x). [Eigenfunctions...] Graph: sine wave decays with time. 19 2.5 September 15, 2003: Wrapup unsteady, boundary conditions Muddy from last time: • Appreciate 9/11 reflections, not political opinions. Apologies to those affected, particularly language. Forms a strong part of 9/11 feelings, particularly as admini...
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π/L, get: C = a exp − � π2Dt � L2 sin � �πx L . (2.60) But what use is a sine wave? Note linearity of diffusion equation: any sum of solutions is also a solution. So we can add sine waves to get something more useful, use that. Fourier transform: express any initial condition as sum of sine waves. We’ll do on...
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at that time? 4/π × exp(−π2) = 6.5 × 10−5, quite close to zero! 20 Other application: finite system with thickness L, uniform IC, constant C boundary conditions. Graph, show the “virtual” wave outside. Note that at short times, erf is easier; long times, one­term sine wave is easy. TA wanted to do 2­D and 3­D sep...
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. • Semiconductor devices: initial treatment with phosphorous­containing gas, initially no phosphorous so t = 0 CP = 0; fixed concentration at surface C = Ceq or Cs or C0. Erf. Then seal the top, no more gas, drive­in diffusion gives shrinking Gaussian. ⇒ • Galvanizing steel: thin layer of zinc on iron. Initial: x < ...
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galvanizing situation, why erf for small times? Isn’t it a finite amount of zinc? At very small times, can consider even the zinc as semi­infinite. Also, at small times t < δ2/D, shrinking Gaussian doesn’t work, goes to infinitely tall and thin. • What to use in example 3 between δ2/16D < t < δ2/D? Nothing covered here...
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equilibrium here would be a brick of SiO2. Local equilibrium gives concentrations at interface sometimes. Can be out of local equilibrium and at steady state too. Misconception 2: kinetic reaction rate order is NOT thermodynamic order. Layer growth Motivating example: silicon oxidation (example W3R pp. 487­489). El...
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graph needed to transform that much Si to SiO2. For equilibrium, set J = ΔC/Y , solve for Y : C3 − C1 Y = Y dY = Y 2 − Y 2 0 2 = 2 moles O 1 mole SiO2 1 mole SiO2 2 moles O 1 mole SiO2 2 moles O dY ρSiO2 dt MSiO2 MSiO2 (C3 − C1) ρSiO2 MSiO2 (C3 − C1) ρSiO2 dt (t − t0). (2.68) (2.69) (2.70) ...
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W3R pp. 487­489). How much does it grow, how long to leave it there? Back to phase diagram: C0, C1 and C3 as three equilibrium concentrations at operating temperature. Out of equilibrium, but interfaces at local equilibrium: pseudo­steady­state. For equilibrium, set J = ΔC/Y , solve for Y : Y 2 − Y0 2 2 = 1 mol...
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Y D Y kY D C1 + C3 kY D + 1 � D Y D Y C3 − C1 � C1 D Y D Y k(C3 − C1) kY D + 1 J = k � D Y C3 − k + C3 − C1 = 1+ k Y D J = Resistances in series. What dominates? Biot number! kY mean small Biot, reaction­limited. Long times and large Y mean large Biot, diffusion­limited. D Ratio of resistance...
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independent! Very often there are multiple “right answers” (fluid dynamics), choose the one which is most convenient. Step 5 Form the π groups from what’s left, which are unitless versions of the parameters. Dimensionless J, called πJ , is J · [C3 − C1]a [D] b · [Y ] . Easy way: make a table with base units across t...
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= 0, so πJ = πk, JO = Purpose: simplify down to an easier expression, single graph. If couldn’t solve equation, single graph could be obtained from one experiment, generalized to any other reaction­diffusion problem of the same nature. Physical modeling, e.g. wind tunnel: get the dimensionless numbers right, every det...
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this term... • How to form dimensionless quantities? If counted i = n − r correctly, and chose dimensionally­ independent parameters to eliminate, then like simultaneous equations: units of J * (units of ΔC)a ... etc. Table as an easier way of doing that. Will do an example today with πk . • (Multiple people) How is...
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D = 1 + 1 D kY = D kY D 1 + kY − D + 1 kY = 1 − 1 1 + kY D So, πJ = 1 − 1 1 + πk D 1 Limiting cases: large πk means πJ = 1 − 0 = 1, so JO = Y (C3 − C1). For small πk, use 1+x � 1 − x near x = 0, so πJ = πk, JO = Purpose: simplify down to an easier expression, single graph. If couldn’t solve equation, s...
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­D heat equation, same as diffusion in section 2.5 (p. 20). Simplify constant k, 1­D, so: ρcp ∂T ∂t = k ∂2T ∂x2 + ˙q. (3.2) Define thermal diffusivity α = k/ρcp, with no gen reduces to diffusion equation, and give 1­D solutions: • 1­D steady­state: linear temperature. • Cylindrical steady­state: T = A ln r + B; ...
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PS3 due today, PS4 due Monday 10/6. • Last test 1 material next Wednesday, following math quiz in 2­143! • Tests 1, 2 (10/10, 11/19) first part in 2­143. • Final Mon 12/15 1:30­4:30 “2­105”... • Zeiss materials microscopy truck at Chapel Turnaround 10/2 9­4. Muddy from last time: • Why is πJ = πk at small πk ? Oka...
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2 h + ... + Ln + kn 1 h2 Same qx everywhere implies that layers with higher k have lower ∂T /∂x. Cylindrical is slightly different, uses flux­area product, based on log solution: Q = qA = 2πL(T1 − T4) 1 ln R2 + 1 ln R3 + k2 k1 R2 R1 1 hR3 Temperature trick: use Biot number equivalent: T0 − T2 T2 − Tn = res...
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Ti, outside fluid at Tf l. Boundary conditions: r = R ⇒ qr = h(T −Tf l). Want to know temperature distribution through time, or temperature history. This requires a Bessel function series!! How to do understand? • Dimensional analysis! • Qualitative description of behavior. • Graphs in text. • Simpified low Biot nu...
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end there, continuing after the Math Quiz on Wednesday... 30 3.4 October 1, 2003: Math Quiz, Graphs Wrapup, Newtonian Cooling Mechanics: • Zeiss Materials Microscopy Truck scheduled tomorrow: cancelled! Muddy stuff: • Mass transfer: diffusion/reaction­limited. Heat transfer: conduction/convection­limited. Mass tr...
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.5 October 3, 2003: Moving on... Mechanics: • Test 1 next Friday 2­143; handout, answer any questions. • Regular office hours; zephyrable (instance) most of next Tuesday. • PS4 due next Monday 10/6, correction: #2a in BTU/hr not kW. Corrected version on Stellar. • PS2#3c solution error: “at t = 1 second, x = 9.6 × 1...
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∞ − T T∞ − T0 � = f X = αt 2 x1 , n = x x1 , m = � k hx1 Wrapup Newtonian cooling Last time we did accum = − − out for the whole shape, got to: T − Tf l = exp Ti − Tf l � − � Aht V ρcp First, examine terms, timescale, larger/smaller h, rho cp, V /A. Plug in V /A: • Sphere: R/3 • Cylinder: R/2 • Pla...
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2 1.77 × 10−3 (both 300K) Influence of porosity and humidity/water absorption. Gases are very bad conductors, water not quite as bad but has very high specific heat! (PS4 #1d, water has four times cp of aluminum which is highest there.) Typical conductivity values: 0.1 to 300 m·K . Porous→less, metals high, gases real...
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at a finite number of “timesteps” at times tn, so with both, we have Ti,n. For simplicity, Δt uniform. • Make some approximations about derivatives: ∂T ∂t � � � � � xi ,tn+1/2 � � ∂T � � ∂x xi+1/2 ,tn � � xi−1/2 ,tn ∂T − ∂x � � � xi+1/2 ,tn Ti,n+1 − Ti,n Δt Ti+1 − Ti Δx Δx � Ti−1,n − 2Ti,n...
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: Ti,n+1 = Ti,n(1 − 2FoM ) + 2FoM Ti−1,n + Ti+1,n + 2 Δt q˙ ρcp 1 So, it’s like a weighted average between Ti,n and the average of the two (show graphically). When FoM > 2 , the Ti,n part is negative, so we shoot past it! So, the criterion is that it must be ≤ 2 , larger timestep means less work, so use 1 .2 ...
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T3,n+1 = T2,n + ρcp T3,BC T4,n+1 = T0,BC 1 −FoM ⎛ ⎜ ⎜ ⎝ (1 + 2FoM ) −FoM −FoM (1 + 2FoM ) ⎛⎞ ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ T0 T1 T2 T3 −FoM 1 ⎛⎞ ⎜ ⎜ ⎜ ⎝ ⎟ ⎟ ⎠ = T0,BC T1,n + q˙1 Δt ρcp q˙1 Δt T2,n + ρcp T3,BC ⎞ ⎟ ⎟ ⎟ ⎠ Now just use 18.06 matrix techniques: Gaussian elimination, LU decomposition, eigenvalues, etc. ...
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. “You said you were going to start each lecture with a ‘motivating factor’—a real example to tie things to so the lecture isn’t just so many symbols and numbers—where was today’s motivating factor? “I’m hoping to at least be able to see a problem being solved where all this is useful. Otherwise, this makes no sens...
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, yj , tn; explicit form: Ti,n+1 − Ti,n Δt = α � Ti−1,j,n − 2Ti,j,n + Ti+1,j,n (Δx)2 + Ti,j−1,n − 2Ti,j,n + Ti,j+1,n (Δy)2 � With Δx = Δy, FoM = αΔt/(Δx)2, we have: Ti,n+1 = (1 − 4FoM )Ti,j + 4FoM Ti−1,j,n + Ti+1,j,n + Ti,j−1,n + Ti,j+1,n 4 So the stability criterion is: Fom ≤ 1 4 ⇒ Δt ≤ 2 Δx 4α . In...
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. Sorry! • Office hours: Today 2:30­3:30. • SOFCs and energy today 12:15 Marlar Lounge (37­252), Ashley Predith, MIT. • Magnetic nanodots Monday 3­4 Chipman, Igor Roshchin, UCSD. Moving body Example: VAR of titanium alloys, nickel superalloys. Start, during operation. Nickel: 6­8 kA, 17→20”; Ti around 30 kA, 30→36”....
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˙q Divide by ρcp: ∂T ∂t + ux ∂T ∂x = α ∂2T ∂x2 + q˙ ρcp Discuss terms: why proportional to ∂T /∂x, competing effects of positive ∂2T /∂x2 and negative −∂T /∂x. Graphical explanation. What introductory math concept does this remind us of? The substantial derivative! Rewrite: Note that’s the time derivative in ...
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� = −ρcpux(Tm − Ti) Note ρcp(Tm −Ti) is the enthalpy per unit volume to heat metal to its melting point. Mult by ux for enthalpy per unit area to heat metal coming at a rate of ux, which is a cool result. Next time: heat flux required to melt... 38 3.9 October 17, 2003: Phase Change Ask Andy re retake... Mechani...
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the time rate of change at a fixed point (in a certain frame). • What’s the significance of qx = −ρcpux(Tm − Ti)? Well, ρcpΔT is the heat per unit volume. How much heat to raise Ti from the initial temp to the melting point. Times ux gives the heat/area/time, the flux required to raise titanium coming in at that speed....
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conversion factor. If not pure, then mult by activity. Either way, multiply material flux J by ΔHvap for heat flux influence. 39 3.10 October 20, 2003: Radiation Mechanics: • Test stats first time around: 62­86 within a std dev. But significant clustering, low 80s and low 60s. Problem Mean before Std. Dev. Max 1. 2...
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radiation, theoretical construct with some practical application. Also emits maximum possible radiation. Handwaving explanation: zero reflection at the interface. Defs: e is power emitted per unit area, eb is power emitted by black body per unit area, eλ is power per unit wavelength per unit area, eb,λ is power by bl...
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α = f (incident spectrum). Example: global warming, CO2 absorbs in the infrared, admits sun in visible. 40 3.11 October 22, 2003: More Radiation Mechanics: • CONGRATS TO ALBERT! • Test stats first time around: 62­86 within a std dev. But significant clustering, low 80s and low 60s. Problem Mean before Std. Dev. ...
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be pretty hot to peak in the visible spectrum. Little table: BB Emission Actual emission Emissivity Absorptivity Wavelength eb,λ eλ �λ = eλ/eb,λ αλ ≡ �λ Total/average � eb = ∞ eb,λdλ 0 � ∞ eλdλ e(= q) = 0 �(T ) = e/eb α(incident) Fortunately eb is quite simple: eb = � ∞ 0 eb,λdλ = σT 4, σ = 5.67 × ...
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1 i=1 if they form an enclosure. Simple thing. With these two, can do complex stuff. Simple geom graphs on pp. 396–398. Note: F11 = 0 if concave. For coaxial disks of same radius, graph F12 vs. d/r, values below. Example: disk and cylinder section height d/4 to d/2 above, viewfactor for disks d/4 is 0.6, for d/2 is ...
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A1 F1R 1 A2 F2R Substitute that in instead of A1F12 in Q12,net equation above. Done with radiation, with heat transfer, on to fluids! 42 � Chapter 4 Fluid Dynamics 4.1 October 24, 2003: Intro, Newtonian Fluids TODO: look up Wiedmann­Franz, falling film in new textbook. Mechanics: • PS5: Get the spreadsheet fro...
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“momentum diffusion” tensor as shear stress. Show this using units: momentum per unit area per unit time: kg m N s = m2 s m · s2 m2 · kg = Two parallel plates, fluid between, zero and constant velocity. x­momentum diffusing in z­direction, call it τzx, one component of 2nd­rank tensor. Some conservation of math: ...
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the momentum diffusivity. Units: N · s/m2 or kg/m · s, Poiseuille. CGS units: g/cm · s, Poise = 0.1 Poiseuille. Water: .01 Poise = .001 Poiseuille. So, sub constitutive equation in the conservation equation, with uy = 0: With constant ρ and µ: It’s a diffusion equation! ∂(ρux) ∂t = − ∂ ∂z � −µ � ∂ux + Fx ∂z ...
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quite confusing, τzx is flux of x­momentum in z­direction. Momentum is a vector, so we have three conservation equations: conservation of x­momentum, y­momentum, z­momentum. This vector field thing is a bit tricky, especially the vector gradient. • I left out: a Newtonian fluid exhibits linear stress­strain rate behav...
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from t = 0. ux = U erfc � � L − z √ νt 2 , τzx = − 2 1 √ νt 2 √ π exp � − (L − z)2 4νt � • New: steady­state, generation, like book’s falling film in problem 4.15 of W3R, with θ the inclination angle off­normal so gx = g sin θ, z is the distance from the plane. The steady­state equation reduces to: 0 = ...
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g sin θ Lz2 2 ν − �3 z 6 = W g sin θ L3 3 ν uav = Q LW = g sin θL2 ; 3ν umax = ux|z=L = g sin θL2 . 2ν So average velocity is 2/3 of maximum for falling film, channel flow, etc. 46 4.3 October 29: 1­D Laminar Newtonian Wrapup, Summary Mechanics: • Next Weds 11/5: 3B Symposium! Muddy from last time: ...
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� = (L − z)ρg sin θ. In z­direction, this is balanced by pressure: P = Patm + ρg cos θ(L − z). Nice segue into pressure­driven flows. Suppose fluid in a cylinder, a pipe for example of length L and radius R, P1 on one end, P2 on other. Net force: (P1−P2)Axs, force per unit volume is (P1−P2)V /Axs = (P1−P2)/L. Can sh...
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� R uz 2πrdr = P1 − P2 (R2 − r 2)2πrdr 4µL Q = � π(P1 − P2) R2r 2 2µL − 4 r 4 �R 0 = π(P1 − P2)R4 . 8µL 0 2 H¨agen­Poisseuille equation, note 4th­power relation is extremely strong! 3/4” vs. 1/2” pipe... 47 Summary Summary of the three phenomena thus far: What’s conserved? Local density Units of flux C...
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p is (σxx + σyy + σzz)/3, so τxx + τyy + τzz = 0, τxy = τyx = −σxy = −σyx. Note that τyx = τxy almost always, otherwise infinite rotation... Also: mechanics uses displacement for �u, acceleration is its second derivative with time. Simple shear: σ = −τ − P I ρ ∂2�u ∂t2 = � · σ + � ⇒ ρ F ∂2ux = ∂t2 ∂σxx ∂x2 + ...
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curvature of the ram and cylinder. The ram motion is also vastly powerful in terms of driving flow than the weight of the fluid, so we can neglect rho*g in the fluid. With these two simplifications, velocity profile between the cylinder and ram is linear, making the problem a lot simpler. • In problem 4a, Arrhenius mean...
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nonzero A in A ln r gives infinite → • Mechanics analogy: very rushed, deserves better treatment, even though not a part of this class. Reynolds number Like other dimensionless numbers, a ratio, this time of convective/diffusive momentum transfer, a.k.a. inertial/viscous forces. Formula: Re = ρU L η Describes dim...
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Dilatant (shear­thickening), example: fluid with high­aspect ratio solid bits; blood. More mixing, momentum mixing, acts like viscosity. Platelet diffusivity, concentration near walls... Model: power­law, n > 1. • Pseudoplastic (shear­thinning), examples: heavily­loaded semi­solid, many polymers get oriented then she...
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difference between a liquid and a Newtonian fluid? Liquids are fluids, as are gases, and plasmas. Definition: finite (nonzero) shear strain rate for small stresses. Bingham plastic not tech­ nically a fluid, nor in a sense is a strict power­law pseudoplastic (though tends to break down near 0). The Navier­Stokes equations...
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? In convective mass transfer it was ρcp� uT . Now it’s ρ�u. Wierd outer product second­rank tensor! u� But what is momentum convection? Those uy ∂y terms. We’ll come back to those later. Recall heat ∂ux transfer with convection: With fluids, it’s the same: ρcp ∂T ∂t + � · (ρcpuT ) = −� · q� + ˙q � ∂(ρ�u) ∂t + ...
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−� · τx = µ � 2 ∂2ux ∂x2 + ∂2ux ∂y2 + � ∂2uy ∂x∂y � �2 ux + = µ � ∂ux ∂x ∂ ∂x + ∂uy ∂y �� So, that simplifies things quite a bit. Incompressible Newtonian result: ρ Dux Dt = − ∂p ∂x + µ�2 ux + Fx Written out in 3­D: � ρ ∂ux ∂t + ux ∂ux ∂x + uy ∂ux ∂y + uz � ∂ux ∂z = −∂p/∂x +...
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takes too long); negatives: 1/3 too fast, 1/2 math­ intense; need concept summaries. – TA: split, most comfortable, 1/3 unhelpful, many: recs need more PS help. “Makes the class not hurt so badly.” – PSes: most like, old PSes a prob, need more probs for test studying. – Test: like policy, but too long. – Text: few...
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= µ ∂ r ∂r ∂r∂z ∂2uz = 0! P = A(r, θ)z + B(r, θ) ⇒ uz = ∂P/∂z 4µ r 2 + A ln r + B = − P1 − P2 z 2 + A ln r + B. 4µL ∂2P ∂z2 = � ∂ µ ∂ ∂z r ∂r � � ∂P ∂z Lateral pressure: if θ = 0 points up: µ ∂ r ∂r ∂uz ∂r 0 = − + Resulting pressure: ρgr = −ρg cos θ = ∂P/∂r, ρgθ = ρg sin θ = (1/r)∂P/∂θ. P = −ρ...
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before due! This time: option Weds/Fri, okay? – Test: like policy, but too long, too long delay to retake. – Text: few helpful different approach, most useless! But better... – Time: from 2­5 or 3+ to 12­18; “4 PS + 2­4 banging head against wall.” • Test 2 in less than two weeks... Shorter wait for retake this time...
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(stress) times area: Fd = τ · A. Here, looking at τrz : Integrated traction (force per unit area) in one direction. For tubes, it’s in the z­direction, uz = P1 − P2 (R2 − r 2) ⇒ τrz = −µ 4µL ∂uz ∂r = P1 − P2 r. 2L Fd = 2πRL · τrz r=R = 2πRL | P1 − P2 R = πR2(P1 − P2). 2L Neat result: this is just the net pre...
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� 2µ τ ρU 2 πother πµ = µ ρU d πρ = ρU d µ πU = ρU d µ πd = ρU d µ First and last are essentially the same, though last is more familiar (Reynolds number), so ignore the first. Third is just a mess, so throw it out. Second is a great fit to what’s above. So use last, or second? With turbulence, there are di...
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• Test 2 11/19 in 2­143, preview handout today. • PS7 extended to Friday. Muddy from last time: • How is f = 16µ/ρU d? τrz = 8µ d uav ⇒ τrz f = 1 ρu2 av 2 = 8µuav /d 1 ρu2 av 2 = 16 16µ . ρuav d Re = • What’s the point of defining f ? We just need τ , right? Well, f is easy for laminar flow, for tur...
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ynolds Number revisited Low velocity: shear stress; high velocity: braking kinetic energy. Ratio of forces: Re = convective momentum transfer shear momentum transfer = inertial forces viscous forces � ∂ux ρuy ∂y ∂2 ux µ ∂y2 � ρU U/L µU/L2 = ρU L . µ Flow past a sphere Motivating process: Electron bea...
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ignore all convective terms; analytical result in 3.21 notes, drag force: Fd = 3πµU d = f · ρU 2 1 2 · 1 4 πd2 ⇒ f = 24µ 24 . = ρU d Re 56 If faster, though not turbulent, f becomes a constant. Graph on W3R p. 153 of f (they call cD ) vs. Re. Constant: about 0.44, that’s what I’ve known as the drag coeffici...
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. • PS7 solution error: #1 replace L with δ, “length” with L. Correction on Stellar. Muddy from last time: • Which are the convective and viscous terms? In vector Navier­Stokes momentum: ρ u D� Dt � ∂� u ∂t = ρ + � u · �� � u = −�P + µ�2� u + � F . Convective terms are the � such a pain to solve and create...
https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6e459228e4e97c9c29e50c481b262c5c_lectures.pdf
surface; no generation. Full equation: DT Dt = α�2T Define boundary layer thickness δx where temperature deviates at least 1% from far­field. If δ � x, then ∂2T ∂y2 >> ∂2T ∂x2 ux ∂T ∂x = α ∂2T ∂y2 Transform: τ = x/ux, becomes diffusion equation, erf solution: T − Ts = erf � Ti − Ts y 2 αx/ux If we defin...
https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6e459228e4e97c9c29e50c481b262c5c_lectures.pdf