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**Unformatted text preview: **ChE 306: HEAT TRANSFER FALL 2014 Homework #3 ‐ SOLUTIONS (10 points each; 90 points) DUE: MONDAY, SEPTEMBER 15, 2014 in class TEAMS: You are encouraged to work in teams. A single assignment can be turned in by up to 4 team members. Include all team members’ names alphabetically on the HW set when turning in. Each team member should understand how to solve ALL problems. HOMEWORK SETS: Turn in all problems, in the order they were assigned. Staple, if possible, and fold the set in half lengthwise. Write your name (all team member names) on the outside fold, so that it will be easy to return. FORMATTING: Work neatly, draw a diagram of the problem, where appropriate, keep problems in the correct order; if the grader can’t find a problem, then it may not be graded. Write on only one side of each sheet of paper. Engineering paper is NOT required. LATE HOMEWORK: Homework will not be accepted late. Turn it in during class, send it with a classmate, or drop it in a RED folder outside my office door BEFORE the solutions post on Blackboard (~ Noon on class days). If you have an excused absence for a homework set, I will drop that set, and award a score of 0 out of 0. 1. A cylindrical pipe carries a stream of solvent from a distillation column to a storage tank. The pipe is made of plain carbon steel and is 25 meters long. It has an inside diameter of 12.5 cm and is 1.5 cm thick. The solvent cools as it goes through the pipe, entering at 75 oC and exiting at 40 oC. The convective heat transfer coefficient for the solvent is 112 W/m2‐K. Outside the pipe is surrounded by air at 26 oC, and the outside convective heat transfer coefficient is 22 W/m2‐K. A. What is UA (overall heat transfer coefficient times the surface area) for heat transfer from the solvent to air? B. Determine Ui and Uo. C. Using the definition of log mean temperature difference from homework #1, determine the rate of heat transfer from the solvent to the air. Note: You may look up k for carbon steel at 300 K and assume k is constant, the process is at steady state, heat transfer is in one direction (1‐D, radial) and that there is no heat generation (q‐dot) term. ΔTLM = {(T1,i – T2,o)‐(T1,o‐
T2,i)}/ln{(T1,i – T2,o)/(T1,o‐T2,i)} Here, the bulk air temperature does not change, so T2,0 = T2,i. A. UA = 1/sumR, so we need to find the resistance terms. Rsolvent = 1/(h*Ai) = 1/(112 * pi * .125m * 25m) = 0.0009095 K/W K carbon steel = 60.5 W/mK Rpipe = ln(r2/r1)/2*pi*L*k = ln((.0625 + .015)/.0625)/(2*pi*25m*60.5 W/mK) Be very careful with r2 and r1. Draw a diagram first to make sure you get this right! You could use D2 and D1, that would be 15.5 cm for D2 (outer) and 12.5 cm for D1 inner, since pipe thickness must be included TWICE to find the outside diameter. Rpipe = 2.26 * 10‐5 K/W Rair = 1/(h*Ao) = 1/(22 * pi * .155m * 25m) = 0.00373 K/W Rtot = 0.00467 K/W Thus, UA = 214 W/K B. UiAi = UA Ui = UA/Ai = 214 W/K / (pi * .125 * 25) = 21.8 W/m2‐K Uo = UA/Ao = 214 W/K / (pi * .155 * 25) = 17.6 W/m2‐K C. q = UA dTlm = 214 W/K * {(75‐26)‐(40‐26)}/ln{(75‐26)/(40‐26)} = 214 * 27.9 K q = 5990 W 2. The same cylindrical pipe used in problem one has fins arranged perpendicular to the cylinder surface to form multiple rings around the outside of the pipe to aid in the cooling. The fins are spaced 1.0 cm apart alongthe length of the pipe, and are also made of carbon steel. The fins extend 0.5 cm from the surface of the pipe, and can be assumed to have such a small thickness that the thin edges can be ignored in calculations. Assuming thermal conduction has a much smaller resistance than convection to the air (i.e., the surface temperature of the fin = the surface temperature of the pipe = constant), and that h is the same with and without fins, what is the expected fin efficiency based on the increased surface area exposed to air? Note: this problem can be solved primarily through geometry, and comparing relative surface areas. For a 25 meter pipe, with fins every 1 cm, there are 25/.01 = 2500 fins. (possibly plus one more if you have one at the each end, but this won’t change the answer appreciably). Do = .155 m Fin Efficiency = q with fins / q without fins Since we are assuming that the temperature of the fins = the surface temperature of the pipe at all points, we Can write: q with fins = h Afinned (Ts – Tinf) q without fins = h Ano‐fins (Ts‐Tinf) Since the fin efficiency is the ratio of these two terms, the fin efficiency can be determined as Efficiency = Afinned / Ano‐fins Ano‐fins = pi * Do * L = pi * .155 m * 25 m = 12.17 m2 Afinned = Ano‐fins + 2500 * Afin The area of each fin is calculated as the area of a flat donut (big circle – small circle), but each fin has two sides, so multiply this by 2 and then by the total number of fins to get the entire finned area. Afin = 2 * {pi/4 * ((.155 + 0.010)m )2 – pi/4 (0.155)2} = 0.00503 m2 Afinned = 12.17 m2 + 2500 * .00503 = 12.17 + 6.18 = 24.7 m2 Fin efficiency = 24.7 / 12.17 = 2.03 3. A flat block of pure nickel has dimensions of 12 cm by 12 cm by 1 meter (z‐direction), and is undergoing 2‐D steady state heat transfer. The surface temperatures are known to be: @ x = 0, T = 140 oF, @ x = 12 cm, T = 230 oF, @ y = 0, T = 195 oF, and @ y = 12 cm, T = 250 oF. A. Using a finite difference grid with Δx = Δy = 6 cm, determine T(6,6) in the middle of the nickel block. B. Using a finite difference grid with Δx = Δy = 4 cm, determine T(4,4), T(4,8), T(8,4) and T(8,8), i.e. the four internal nodes‐ might be easier to write as T1, T2, T3 and T4, as long as you show on a diagram which T is for which point. A. This is simple since there is only one node internal to the block. T(6,6) = (140 + 230 + 195 + 250)/4 = 203.8 oF There is no need to know the thermal conductivity of nickel! B. This doesn’t require any new equations‐ but there are now 4 unknowns! All of the equations are for internal nodes with conduction only. To simplify writing, I will call T1 = T(4,4), T2 = T(4,8), T3 = T(8,4), T4 = T(8,8) Drawing a diagram is very helpful! 250 250 140 T2 T4 230 140 T1 T3 230 195 195 Node Balance on T1: 4* T1 = (140 + 195 + T2 + T3) Node Balance on T2: 4* T2 = (140 + 250 + T1 + T4) Node Balance on T3: 4* T3 = (230 + 195 + T1 + T4) Node Balance on T4: 4* T4 = (230 + 250 + T2 + T3) You do not need to convert to absolute temperatures! This yields 4 equations and 4 unknowns. This can be solved using linear or matrix algebra, but I will solve by substitutions. There are multiple ways to solve. I’ll choose to solve node T2 for T1 = f(T2, T4) T1 = 4*T2 ‐390 – T4 Then I’ll solve node T4 for T3 = f(T2, T4) T3 = 4*T4 – 480 – T2 I will then plug in these equations into Nodes 1 and 3 to come up with two equations with only 2 variables: T2 and T4. Node T1: 16 * T2 – 1560 – 4 * T4 = 335 + T2 + 4*T4 – 480 – T2 = 4 * T4 – 145 16 * T2 = 8 * T4 + 1415 T2 = 0.5 * T4 + 88.44 Node T3: 16 * T4 – 1920 – 4 * T2 = 425 + 4 * T2 – 390 – T4 + T4 = 35 + 4 * T2 16 * T4 = 8 * T2 + 1955 Plug in result from T1: 16 * T4 = 8 * (0.5 * T4 + 88.44) + 1955 = 4 * T4 + 2662.5 12 * T4 = 2662.5 T4 = 221.9 oF Using T2 equation from node T1: T2 = 0.5 * 221.9 + 88.44 = 199.4 oF Using the node T2 equation: T1 = 4 * 199.4 – 390 – 221.9 = 185.6 oF Using the node T4 equation: T3 = 4 * 221.9 – 480 – 199.4 = 208.1 oF Let’s put these numbers into the diagram and make sure they make sense. 250 250 Yes. Hottest number is in the hottest 140 199.4 221.9 230 Part of the material. 140 185.6 208.1 230 195 195 4. A block of lead is shaped like an L (picture at right) is experiencing 2‐D steady state heat transfer. T
1 Node temperatures T1 and T2 are on the top row, T3, T4 and T5 middle row and T6, T7 and T8 along the bottom. The surfaces between T2 & T4 and between T4 & T5 are exposed to a convection boundary with bulk fluid temperature T∞ and convective heat transfer coefficient h. The lead has thermal conductivity kPb. Grid spacings are Δx ≠ Δy. Use the normal Cartesian definition of the x and y‐axes. T2
T4 T5
T8 A. B.
C.
D. Write equations to determine the rate of heat transfer (q) into node 4 for a unit thickness Δz: A. from the node at T2 B. from node at T3 C. from node at T5 D. from node at T7 Express your answers in terms of Δx, Δy, Δz, kPb, h, and any of the temperatures needed. qy = kPb * ½ Δx Δz * (T2‐T4)/ Δy + h ½ Δx Δz (T∞‐T4) The temperatures are arranged so that the qy will be positive for heat transferring into node Since half of the space between T2 and T4 is solid and half fluid, both conduction and convection need to be included in the equation. qx = kPb * Δy Δz * (T3‐T4)/ Δx qx = kPb * ½ Δy Δz * (T5‐T4)/ Δy + h ½ Δy Δz (T∞‐T4) qy = kPb * Δx Δz * (T7 ‐T4)/ Δy 5. A solid piece of pure copper has known temperature and boundary conditions as shown below. Air 2 Node temperatures are given, and the grid spacing is Δx = Δy = 1.0 cm. The right and top surfaces are exposed to air with given temperature and h. 60 TA h = 740 W/m ‐K
o
T∞ = 25 C TB
54 A. Determine TA and TB. B. What is the total heat transfer rate, q, from the right edge of the block (assume 37 46 Δz = 1 m) to the air? Note: You may use k at 300 K, and assume steady state, 2‐D heat transfer, constant k, no heat generation. Temperatures to the left of the defined nodes are not known. The boundary condition on the bottom surface of the copper is not known. For part B, draw in the grids, knowing that TB, for example, represents the temperature of the solid within it’s grid, and
use Newton’s law of cooling for each portion of the grid to find total q. 1. Node A matches Case 4, external node with convection. 60 + TB + 2hΔx/k T∞ – 2(hΔx/k +1) TA = 0 In this equation h = 740 W/m2‐K dx = 1 cm (0.01 m) k (look up) = 401 W/mK 2. Node B matches Case 3, 2 * 54 + TA + 37 + 2 hΔx/k T∞ – 2(hΔx/k + 2)TB = 0 Everything in these 2 equations is known except TA and TB. 60 + TB + 2 * (740 * 0.01 / 401) * 25 – 2 * (740 * 0.01 / 401 + 1) * TA = 0 60 + TB + 0.92 – 2.037 TA = 0 108 + TA + 37 + 2 * (740 * 0.01 / 401) * 25 – 2((740 * 0.01 / 401) + 2)TB = 0 145 + TA + 0.92 – 4.037 TB = 0 Solve first equation for TB = 2.037 TA – 60.92 and plug into equation 2. 145.92 + TA – 4.037 (2.037 TA – 60.92) = 0 391.85 = 7.22 TA TA = 54.2 oC Thus, TB = 49.5 oC Part B. Use Newton’s law of cooling for three parts. Grid one around TA is ½ dy, grid two around TB is a full dy, and the grid around the 37C node is ½ dy, each with a depth of dz = 1 m. q = h A (Ts – Tinf) = h (AA (TsA – Tinf) + AB(TsA‐Tinf) + AC (TsC‐Tinf)) q = 740 W/m2‐K * (0.005m * 1m * (54.2 – 25 C) + .01m*1m * (49.5 – 25 C) + 0.005m*1m*(37‐25C)) q = 334 W 6. A solid pure aluminum sphere with a 2.0 cm diameter is initially at 350 oC when it is dropped in a vat of ice water at 4 o
C, which has a convective heat transfer coefficient of 281 W/m2‐K. (Use k, cp at 600 K, density at 300 K is OK). A. Can the lumped capacitance model be used for this problem? B. Find the temperature in the center of the sphere at t = 5 seconds and 30 seconds. C. How long will it take the sphere to cool to 25 oC? A. Bi = hLc/k Lc for a sphere = r/3 = 1/3 cm or 0.00333 m k for aluminum = 231 W/mK Bi = 281 * .00333 / 231 = 0.0041 (< 0.1), so Yes, Lumped capacitance CAN be used. B. (T‐Tinf) / (Ti‐Tinf) = exp {(‐hA/m cp) t} Ti = 350 C Tinf = 4 C h = 281 W/m2K A = 4 * pi * r2 = 4 * pi * (0.01 m)2 = 0.00126 m2 (Formula for surface area of a sphere!) m = density * volume = 2702 kg/m3 * 4/3 * pi * (0.01m)3 = 0.0113kg cp @ 600 K = 1033 J/kg‐K Plug in 5 seconds for time. Find T. (note: the temps can be entered in C or K since it is a difference. (T‐4) / (350‐4) = exp {(‐281*.00126/(.0113*1033) 5s} (careful with units‐ everything in the exponential should cancel out.) = 0.8593 (or for 30 s, this equals 0.4025) T ‐ 4 = 297.3 T ‐ 4 = 139.3 T (@ 5 s) = 301.3 C T (@ 30 s) = 143.3 C C. Use the same equation and plug in T = 25 and solve for t. (25‐4) / (350‐4) = exp {(‐281*.00126/(.0113*1033) t} 0.0607 = exp {‐.0303 t) t = 92.5 s This makes sense. More than 30 seconds (see answer in B). 7. For solid cylindrical rods made of the materials listed below, determine the maximum diameter of each so that the lumped capacitance model can be used to find T(t) for heating the rods in an oil bath with h = 159 W/m2‐K. A. AISI 316 stainless steel B. Copper C. Cement mortar, Table A.3 D. Expanded polystyrene molded beads (Styrofoam), Table A.3 Note: Assume constant k for the solids at 300 K. You may assume the rods are very long and ignore the flat ends. For each of these, we need to determine radius (diameter) of each so that the Biot number = 0.1. Since it is a cylindrical rod Lc = r/2. Bi = hLc/k A. k = 13.4 W/mK 0.1 = 159 * (r/2) / 13.4 r = 0.0169 m d = 3.37 cm B. k = 401 W/mK 0.1 = 159 * (r/2) / 401 r = 0.504 m d = 1.01 m (BIG!!!) C. k = 0.72 W/mK 0.1 = 159 * (r/2) / 0.72 r = 0.000906 m d = 1.81 mm D. k = 0.040 W/mK 0.1 = 159 * (r/2) / 0.040 r = 0.0000503 m d = 0.101 mm (TINY! 1/10000 of copper!) 8. Given the temperature vs. time data at right for a 2.5 cm diameter pure copper sphere placed in still air or with a fan turned on, determine the convective heat transfer coefficient, h, for the air in both instances. You may assume that Bi < 0.1, so the lumped capacitance model can be used. You may assume k for copper is constant, and can look up properties at 300K. The heat transfer is unsteady state, and the temperature of the sphere is measured by a thermocouple inserted at the center of the sphere (r = 0). Problem 8 Time (sec) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Data Temp, C (no fan) 185
180
175.5
171.2
167.1
163.2
159.4
155.7
152
149.1
146.4
143.8
141.3
138.9
136.6
134.4 Temp, C (fan) 185
176
168
160.5
153.6
147
140.8
134.4
128.3
122.4
116.8
111.6
106
101.3
97
93.1 ln{(T‐Tinf)/(Ti‐Tinf)} Note: You will need to use a spreadsheet to plot data, but be careful to make the plot fit an appropriate equation so that you can find h from a trendline fit to the graph. Spreadsheet has calculations, and is used to find h. no fan
fan
no fan fan
no fan fan T‐Tinf
T‐Tinf
(T‐Tinf)/(Ti‐Tinf) (T‐Tinf)/(Ti‐Tinf) Time (sec)
ln{(T‐Tinf)/(Ti‐Tinf)} ln{(T‐Tinf)/(Ti‐Tinf)} 160
160
1
1
0
0
0 0.94375
155
151
0.96875
2
‐0.031748698
‐0.057893978
150.5
143
0.940625
4
‐0.061210731
‐0.112329185 0.89375
146.2
135.5
0.91375
6
‐0.090198268
‐0.166202175 0.846875
142.1
128.6
0.888125
0.80375
8
‐0.11864278
‐0.218467003 0.7625
138.2
122
0.86375
10
‐0.146471904
‐0.271152771
‐0.32330925
134.4
115.8
0.84
12
‐0.174353387 0.72375
130.7
109.4
0.816875
0.68375
14
‐0.202269195
‐0.380162925 ‐0.230986729
127
103.3
0.79375
0.645625
16
‐0.437536439 ‐0.254086123
124.1
97.4
0.775625
0.60875
18
‐0.496347605
121.4
91.8
0.75875
0.57375
20
‐0.555561518 ‐0.276082937
118.8
86.6
0.7425
0.54125
22
‐0.297732408
‐0.613874 ‐0.319000756
116.3
81
0.726875
0.50625
24
‐0.680724661 ‐0.339852945
113.9
76.3
0.711875
0.476875
26
‐0.740500877
111.6
72
0.6975
0.45
28
‐0.798507696 ‐0.360252765
109.4
68.1
0.68375
0.425625
30
‐0.380162925
‐0.854196602 0.1 0
10
20
30
40
‐0.1 0 y = ‐0.0127x ‐ 0.0147
Series1
‐0.2 R² = 0.9945
Series2
‐0.3
y = ‐0.0134x R² = 0.9902
Linear (Series1)
‐0.4 Linear (Series2)
‐0.5 Linear (Series1)
‐0.6
y = ‐0.0281x y = ‐0.0285x + 0.0082
Linear (Series2)
‐0.7
R² = 0.9989
R² = 0.9992 ‐0.8 ‐0.9
Time (sec) The blue data are for no fan, the red for with a fan. The equations give slopes which must match ‐hA/mcp Both equations should have been forced to go through the origin, since the theoretical equation has no intercept when plotted this way. (Note: I converted the data into a linear plot; it is possible to leave with a ln or exp, as long as a different fit is used). A = 4 * pi * r2 = 4 * pi * 0.0125^2 = 0.00196 m2 m = 4/3 * pi * r3 * density = 4/3 * pi * (0.0125)^3 * 8933 = 0.0731 kg density = 8933 kg/m3 cp = 385 J/kg‐K With no fan, hA/mcp = 0.0134 s‐1= h* 0.00196m2 / {0.0731 kg) * (385 J/kg‐K) h (no fan) = 192 W/m2K With the fan, hA/mcp = 0.0281 s‐1= h* 0.00196m2 / {0.0731 kg) * (385 J/kg‐K) h (fan)= 403 W/m2K 9. A large solid sphere with a 10 inch diameter initially has a uniform temperature of 67 oF until it is placed in an oven at 400 oF (for the purposes of this problem, you can assume that the sphere is surrounded by air on all sides does not touch any surface). A. Draw a qualitative sketch of T(r) from –r to r for a yellow pine sphere for four time points, including t = 0, t = ∞ and two mes in between. Assume Bi > 0.1. B. Draw a similar qualitative sketch of T(r) from –r to r for a 10 in diameter aluminum sphere, assuming Bi < 0.1. A. here it is probably obvious that the pine sphere will not meet the requirements for lumped capacitance. Since the problem says that we can assume Bi > 0.1, that means that the temperature of the sphere will be different on the edge and in the middle. A qualitative graph simply means to draw an estimate of the profile at different times, showing how you expect the T(r) profile to look. Here, the sphere will be initially at 67, then the edges will warm up slightly, while the core stays near 67, then the edges will reach close to 400, while the center lags behind, finally at infinite time, the whole sphere will reach 400 F. Long time 400F Temp Initial 67F ‐r r B. Since h is not given, we can only follow the assumption given that Bi < 0.1. So, the temperature of the aluminum sphere will be the same at all radii for any given time. That means that the aluminum sphere will start with a uniform temperature of 67 F, then at a later time it will have a uniform temperature of, say, 100 F. After more time, the sphere will have a flat profile at a higher temperature, and finally with enough time, the whole sphere will reach 400 F, but the T(r) profile will be flat for all times since heat conducts so easily to re‐equilibrate the sphere as heat enters from the surface. Long time 400F Temp t3 350F t2 100F Initial 67F ‐r r ...

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