John R. Carr, JRC Analytical Services, Mechanical E ngineering, PE, MSME, BSME, 6/25/2011
[email protected],, www.jrcanalyticalservices.com
[email protected] www.jrcanalyticalservices.com,, 615-218-0131 Samples of 33 Recently Solved Problems in Heat Transfer (165 total problems solved and 49 TAK 2000 models in the last last two years). References include the following:
“Heat Transfer” by J.P. Holman, 9th edition (2002) – covered 3/12 to 4/9/2011 (69 problems and 9 TAK 2000 models).
“Heat Transfer SO” by Pitts & Sissom, 2nd edition (1998) - covered 2/12 to 3/10/2011 (37 problems and 4 TAK 2000 models). “Principles of Heat Transfer” by Kreith & Bohn, 4th edition (1984) – – covered 11/16 to 12/25/2010, reviewed – 5/15 to 5/24/2011 (59 problems and 36 TAK 2000 models).
12 Sample Holman problems & solutions (from a total of 69 problems previously worked): Chapter 1 – Introduction 1-18. A small radiant radiant heater has metal strips 6 mm with a total length of 3 m. The surface emissivity of the strips strips is 0.85. To what temperature must the strips be heated if they are to dissipate 2000 W of o f heat to a room at 25 ºC?
Solution: for an enclosure q = ε1ζA1(T14 – T24) q/ε1ζA1 = T14 – T24 or T14 = T24 + q/ε1ζA1 4
4
-8
2
4
T1 = (298 K) + (2000 W)/{(0.85)*(5.669 x 10 W/m -K )*(0.006 m)*(3 m)} T1 = 1233 K = 960 ºC 1-24. One side of a plane wall is maintained at 100 ºC, while the other side is exposed to 2 a convection environment having T = 10 ºC and h = 10 W/m -ºC. The wall has k = 1.6 W/m-ºC and is 40 cm thick. thick. Calculate the heat-transfer rate through the wall. Solution: using two resistors in series, q = ∆T/∑R = ∆T/(R 1 + R 2) = ∆T/(t/k A + 1/hA) = A*∆T/(t/k + 1/h) 2 q/A = ∆T/(t/k + 1/h) = (100 – 10)ºC/{(0.40 m)/(1.6 W/m-ºC) + 1/(10 W/m -ºC) 2 2 = 90 ºC/{(0.25 + 0.1)m -ºC} = 257 W/m 1-30. A black 20-by-20 cm plate has air forced over it at a velocity of 2 m/s and a temperature of 0 ºC. The plate is placed in a large room whose walls are at 30 ºC. The back side of the plate is perfectly insulated. Calculate the temperature of the plate resulting from the convection-radiation balance. Use information from Table 1-3(for convection coefficients).
Solution: energy balance gives qin = qout 4 4 Holman uses the intro equation for net exchange of radiation q = FεFG σA(T1 – T2 ) where Fε is an emissivity function, and FG is a geometric “view factor” function. In our case, assume both are equal to 1, giving σ*A*(Twalls4 – TS4) = hc*A*(TS - T∞) 2 Table 1-3 (page 11) gives hc = 12 W/m -ºC for flow conditions and A’s cancel out, -8 2 4 4 4 4 2 giving (5.669 x 10 W/m -K )*(303 – TS ) K = (12 W/m -ºC)*(TS – 273) K 4 4 4 3 (303 – TS ) K = (211677544.54 K )*(TS – 273) Guess TS until LHS = RHS as follows: TS LHS RHS 280 1481742812 2282332481 290 3598518257 1356082481 285 2540130534 1831391856 283 2116775445 2014644560 282 1905097900 2104825905 Use TS = 282 K = 9 °C as the temperature of the plate. Chapter 2 – Steady-State Conduction – One Dimension 2-4.
Find the heat transfer per unit area through the composite wall in Figure P2-4. Assume one-dimensional heat flow.
Solution: Figure shows a wall such that the hot face (LHS) is at T = 370 ºC and the cold face (RHS) is at 66 ºC. The first layer on the left is material A with k A = 150 W/m-ºC and LA = 2.5 cm, second layer (middle) is a layer with materials B and D in parallel with k B = 30 W/m-ºC and k D = 70 W/m-ºC and LB = LD = 7.5 cm and AB = AD = 0.5*AA, and 2 the right-most layer is material C with k C = 50 W/m-ºC and AC = AA = 0.1 m . Calculating the individual resistances according to R = L/kA gives the following: 2 -3 R A = LA/k AAA = (0.025 m)/{(150 W/m-ºC)*(0.1 m )} = 1.667 x 10 ºC/W R B = LB/k BAB = (0.075)/{(30)*(0.1/2)} = 0.05 ºC/W -2 R D = LD/k DAD = (0.075)/{(70 W/m-ºC)*(0.1/2)} = 2.14286 x 10 ºC/W For the parallel resistors: -2 -2 R eq = R BR D/(R B + R D) = (0.05)*(2.14286 x 10 )/(0.5 + 2.14286 x 10 ) -3 = 2.0548 x 10 ºC/W R C = LC/k CAC = (0.05)/{(50)*(0.1)} = 0.01 ºC/W Overall heat flow is given by -3 -3 2 q = ΔT/∑R = (370 – 66)/(1.667 x 10 + 2.0548 x 10 + 0.01) = 22155 W (A = 0.1 m ) 2 gives q/A = 221550 W/m . 2-43. A plate having a thickness of 4.0 mm has an internal heat generation of 200 3 MW/m and a thermal conductivity of 25 W/m-ºC. One side of the plate is insulated and the other side is maintained at 100 ºC. Calculate the maximum temperature in the plate. 2
2
Solution: simplified differential equation is d T/dx + qdot/k = 0 (qdot = heat generation/volume as given). Two integrations give dT/dx = -(qdot/k)*x + C1
2
T(x) = -(qdot/2k)*x + C1x + C2
Applying BC’s: at x = 0, dT/dx = 0 gives 0 = 0 + C1 or C1 = 0 2
Thus T(x) = -(qdot/2k)*x + C2 2 2 Applying BC of T(L) = 100 ºC = -(qdot/2k)*L + C2 gives C2 = 100 + (qdot/2k)*L 2 2 2 2 T(x) = -(qdot/2k)*L + 100 + (qdot/2k)*L = (qdot/2k)*(L – x ) + 100 2 Max T occurs at x = 0 where dT/dx = 0 so Tmax = T(0) = (qdot/2k)*L + 100 6 3 2 = {(200 x 10 W/m )/(2*25 W/m-ºC)}*(0.004 m) + 100 = 164 ºC 2-69. A very long copper rod (k = 372 W/m-ºC) 2.5 cm in diameter has one end maintained at 90 ºC. The rod is exposed to a fluid whose temperature is 40 ºC. The heat2 transfer coefficient is 3.5 W/m -ºC. How much heat is lost by the rod? 2
θ/dx2 – (hP/kA)θ = 0 where θ = T - T∞ Boundary conditions are: at x = 0, θ = θo = To - T∞ and case 1 as x →∞, T→T∞ or θ(∞) = 0 solution to DE is θ/θo = (T - T∞)/(To - T∞) = e-mx 2 2 2 m = hP/kA = {(3.5 W/m -ºC)*π*(0.025 m)}/{(372 W/m-ºC)*(π/4)*(0.025 m) } Solution: DE is d
2
= 1.505 1/m giving m = 1.2269 1/m 1/2 q = (hPkA) θo 2 2 1/2 = {(3.5 W/m -ºC)*π*(0.025 m)*(372 W/m-ºC)*(π/4)*(0.025 m) } *(90 – 40)ºC = 11.2 W Chapter 3 – Steady-State Conduction – Multi Dimensions – I built several TAK 2000 finite difference thermal models in support of solving some of the examples and some of the homework problems in the book. I did no analytical or graphical solutions. (more information is available at www.tak2000.com). Chapter 4 – Unsteady-State Conduction – Again, several TAK 2000 models were built to simulate the example problems and compare transient numerical results. Also, a couple of lumped parameter analysis problems (among others) were worked as follows: 4-6. A piece of aluminum weighing 6 kg and initially at a temperature of 300 ºC is suddenly immersed in a fluid at 20 ºC. The convection heat-transfer coefficient is 58 2 W/m -ºC. Taking the aluminum as a sphere having the same weight as that given, estimate the time required to cool the aluminum to 90 ºC, using the lumped-capacity method of analysis (if applicable).
Solution: W = mg = ρVg with ρ = 2707 kg/m3 For a sphere S = A = 4πR 2 = πD2 and V = (4/3)πR 3 V = (4/3)πR 3 = m/ρ = (6 kg)/(2707 kg/m3) gives R = 0.080883 m 3 2 Biot number = Bi = hS/k = h*(V/A)/k = h*(4 πR )/{k*4πR ) = hR/3k 2
-3
= (58 W/m -ºC)*(0.080883 m)/{(3)*(204 W/m-ºC)} = 7.665 x 10 Since Bi << 0.1 lumped-parameter analysis is applicable -(hA/ρcV)τ (T - T∞)/(To - T∞) = e where c = 896 J/kg-ºC (hA/ρcV) = 1/(R rhCth) with τc = cρV/hA (time constant) 3 3 τc = (896 J/kg-ºC)*(2707 kg/m )*(4/3)*π*(0.080883 m) /
2
2
{(58 J/s-m -ºC)*(4π)*(0.088083 m) } = 1127.5 s Gives (T - T∞)/(To - T∞) = e τ/1127.5 ln{(T - T∞)/(To - T∞)} = -τ/1127.5 ln{(90 - 20)/(300 - 20)} = - τ/1127.5 τ = t = 1563 s (26 min) to cool to T = 90 ºC 4-58. A 4.0-cm cube of aluminum is initially at 450 ºC and is suddenly exposed to a 2 convection environment at 100 ºC with h = 120 W/m -ºC. How long does it take the cube to cool to 250 º? 3
Solution: for Aluminum (pure) – ρ = 2707 kg/m , c = 896 J/kg-ºC, k = 204 W/m-ºC 3 2 2 Biot number Bi = hs/k = h*(V/A)/k = h*L /(6L *k) = (120 W/m -ºC)/(6*204 W/m-ºC) -2 = (2.353/6) x 10 Since Bi ≤ 0.1 lumped parameter analysis is applicable -η/ηc where ηc = cρV/hA (time constant) Resulting solution of DE is (T - T∞)/(To - T∞) = e 3 3 ηc = (896 J/kg-ºC)*(2707 kg/m )*(0.04 m) /{(120 W/m2-ºC)*(0.04 m)2} = 808.5 s - /808.5 Time to reach T = 250 ºC is given by (250 – 100)/(450 – 100) = e τ τ = 685 s = 11.4 min Chapter 5 – Principles of Convection 5-7. Oxygen at a pressure of 2 atm and 27 ºC blows across a 50-cm-square plate at a velocity of 30 m/s. The plate temperature is maintained constant at 127 ºC. Calculate the total heat lost by the plate. Solution: film temperature is Tf = (T∞ + Tw) /2 = 77 ºC = 350 K Reynolds number ReL = u∞L/ ν = ρu∞L/μ 3 Evaluating properties of O2 at Tf = 350 K gives the following: ρ = 1.1133 kg/m , ν = 20.80 x 10-6 m2/s, k = 0.03070 W/m-ºC, and Pr = 0.702 -6 2 5 ReL = (30 m/s)*(0.50 m)/(20.80 x 10 m /s) = 721154 > 5 x 10 (turbulent at trailing edge) – use Eqn. (5-85) that takes into account laminar/turbulent flow with a transition 5 7 Reynolds number of 5 x 10 , eqn. is applicable for Re L < 10 √ 1/3 0.8 NuL = hL/k = Pr *(0.037*ReL – 871) (5-85) 1/3 0.8 = (0.702) *{(0.037)*(721154) – 871} = 823.6 2 h = (823.6)k/L = (823.6)*(0.03070)/(0.50) = 50.55 W/m -ºC q = 2hA(Tw - T∞) = 2*(50.55)*(0.5)*(0.5)*(127 – 27) = 2527.5 W (both sides of plate) 5-23. Calculate the heat transfer from a 20-cm-square plate over which air flows at 35 ºC and 14 kPa. The plate temperature is 250 ºC, and the free-stream velocity is 6 m/s. Solution: using Tf = (Ts + T∞)/2 = (250 + 35)/2 = 142.5 ºC = 415.5 K (avg film temp) Air properties at Tf = 415.5 gives the following linear interpolation 6 2 T(K) ν x 10 (m -s) k (W/m-K) Pr 400 25.90 0.03365 0.689 415.5 x1 x2 x3 450 31.71 0.03707 0.683
15.5/50 = (x1 – 25.90)/(31.71 – 25.9) = (x2 – 0.03365)/(0.03707 – 0.03365) = (x3 – 0.689)/(0.683 – 0.689) 6 -6 x1 = ν x 10 = 27.70, ν = 27.70 x 10 , x2 = k = 0.03471 W/m-K, x3 = Pr = 0.687 -6 2 5 ReL = u∞L/ν = (6 m/s)*(0.20 m)/(27.70 x 10 m /s) = 43321.3 < 5 x 10 (laminar) 1/2 1/3 NuL = hL/k = 0.664*ReL Pr equation (5.46b) 1/2 1/3 NuL = (0.664)*(43321.3) *(0.687) = 121.96 2 h = 121.96k/L = (121.96)*(0.03471)/(0.20) = 21.165 W/m -K 2 2 q = 2hA*(Ts - T∞) = 2*(21.165 W/m -K)*(0.20 m) *(250 – 35) K = 364 W (both sides of plate) Chapter 6 – Empirical and Practical Relations for Forced-Convection Heat Transfer 6-6. Water at the rate of 0.8 kg/s is heated from 35 to 40 ºC in a 2.5-cm-diameter tube whose surface is at 90 ºC. How long must the tube be to accomplish this heating? Solution T b,ave = (T b,1 + T b,2)/2 = 37.5 ºC 3 -4 Water properties – Table A-9: c p = 4174 J/kg-ºC, ρ = 993 kg/m , μ = 6.82 x 10 kg/m-s, k = 0.630 W/m-ºC, Pr = 4.53 Reynolds number Red = ρumd/μ 3 2 mdot = 0.8 kg/s = ρAum = (993 kg/m )*(π/4)*(0.025 m) *um gives um = 1.641 m/s 3 -4 Red = (993 kg/m )*(1.641 m/s)*(0.025 m)/(6.82 x 10 kg/m-s) = 59741.4 (turbulent) 6 Using Eqn. (6-4c) good for 1.5 < Pr < 500, 3000 < Re < 10 0.87 0.4 0.87 0.4 Nu = hd/k = 0.012*(Red – 280)*Pr = 0.012*(59741.4 – 280)*4.53 = 307.9 2 h = 307.9k/d = (307.9)*(0.630)/(0.025) = 7758.8 W/m -ºC q = mdot*c p*∆T b = hA*(Tw – T b)ave Computing q1 = mdot*c p*∆T b = (0.8 kg/s)*(4714 J/kg-ºC)*(40 – 35) ºC = 18856 W so 18856 W = hπdL*(Tw – T b)ave = (7758.8 W/m2-ºC)*π*(0.025 m)*L*(90 – 37.5) ºC gives L = 0.589 m. 6-42. Air at 70 kPa and 20 ºC flows across a 5-cm-diameter cylinder at a velocity of 15 m/s. Compute the drag force exerted on the cylinder. 3
Solution: Properties of air at T = 20 ºC = 293 K, P = 70 x 10 Pa, using P = ρRT ρ = P/RT = (70 x 103 N/m2)/[(287 N-m/kg-K)*(293 K)] = 0.8324 kg/m3 -5 Table A-5 (page 602) gives μ = 1.8462 x 10 kg/m-s -5 Red = ρud/μ = (0.8324)*(15)*(0.05)/(1.8462 x 10 ) = 33815 Using Fig. 6-9 (page 282) that gives drag coefficient CD vs. Red, at Red = 33815, CD = 1.1 2 Drag force D = FD = CDA*ρu∞ /(2gc) 3 = (1.1)*(0.05 m)*(1 m)*(0.8324 kg/m )*(15 m/2)/(2*1) = 5.15 N
3 Sample Pitts & Sissom problems & solutions (from a total of 37 problems previously worked): Chapter 7 – Forced Convection: Turbulent Flow 7.26 Ethylene glycol at 0 ºC flows at the rate of 23 m/s parallel to a 0.6 m square, thin flat plate at 40 ºC, which is suspended from a balance. Assume the fluid flows over both sides of the plate and that the critical Reynolds number is 500000. (a) What drag should be indicated by the balance? (b) What is the heat transfer rate from the plate to the fluid? Solution: Tf = (0 + 40)/2 = 20 ºC = 293 K, using Table B-3(SI) to evaluate properties: ρ = 1116.65 kg/m3, c p = 2382 J/kg-K, Pr = 204, ν = 19.18 x 10-6 m2/s, k = 0.249 W/m-K -6 2 ReL = VL/ν = (23 m/s)*(0.6 m)/(19.18 x 10 m /s) = 719500 (turbulent) 1/3 Using Eqn. (7.28) given by Nu = hL/k = Pr (0.036ReL – 836) 1/3 0.8 Nu = hL/k = (204) *{0.036*(719500) – 836} = 5354 2 h = 5354*k/L = (5354)*(0.249)/0.6 = 2222 W/m -K 2 q = 2hA(Ts - T∞) = 2*(2222)*(0.6) *(40 – 0) = 63993 W 1/5 To get the drag force – use Eqn. (7.22) Cf = 0.072/ReL – (0.00334)*xc/L -6 For Rec = 500000 = Vxc/ν = (23 m/s)*xc/(19.18 x 10 ) Gives xc = 0.417 m 1/5 -3 Cf = 0.072/(719500) – (0.00334)*(0.417)/(0/6) = 2.531 x 10 -3 3 2 2 Ff = Cf *(ρV∞/2)*A = (2.531 x 10 )*(1116.65 kg/m )*(23 m/s) *(0.6 m) = 269 N D = 2Ff = 538 N 7.28 A 3 in o.d. steam pipe without insulation is exposed to a 30 mph wind blowing normal to it. The surface temperature of the pipe is 200 ºF and the air is at 40 ºF. Find the heat loss per foot of pipe. Solution: V∞ = 30 miles/h x 5280 ft/1 mile x 1 h/3600 s = 44 ft/s Tf = (T∞ - Ts)/2 = (40 + 200)/2 = 120 ºF, using Table B-4 (Eng) with interpolation gives 3 2 T (ºF) ρ (lbm/ft ) ν (ft /s) k (Btu/h-ft-ºF) Pr -5 80 0.0735 16.88 x 10 0.01516 0.708 120 x1 x2 x3 x4 -5 170 0.0623 22.38 x 10 0.01735 0.697 40/90 = (x1 – 0.0735)/(0.0623 – 0.0735) = (x2’ – 16.88)/(22.38 – 16.88) = (x3 – 0.01516)/(0.01735 – 0.01516) = (x4 – 0.708)/(0.697 – 0.708) 3 -5 -5 2 x1 = ρ = 0.06852 lbm/ft , x2’ = x2 x 10 = ν = 19.32 x 10 ft /s x3 = k = 0.01613 Btu/h-ft-ºF, x4 = Pr = 0.703 -5 2 ReD = V∞D/ν = (44 ft/s)*(3/12 ft)/(19.32 x 10 ft /s) = 56935.8 (turbulent) For our case – single cylinder in crossflow in air, use Equation (7.51a) n NuDf = hD/k f = CgReDf Our configuration gives Cg = 0.0239, n = 0.805 (for ReDf = 40000 to 250000) 0.805 NuDf = hD/k f = (0.0239)*(56935.8) = 160.87 2 h = (160.87)*(0.01613 Btu/h-ft-ºF)/(3/12 ft) = 10.38 Btu/h-ft -ºF qc = hA(Ts - T∞) = hπDL(Ts - T∞) qc/L = hπD(Ts - T∞) = π*(10.38)*(3/12)*(200 – 40) = 1304 Btu/h-ft
Chapter 8 – Natural Convection 8.20 The front panel of a dishwasher is at 95 ºF during the drying cycle. What is the rate of heat gain by the room, which is maintained at 65 ºF? The panel is 2.5 ft. square. Solution: using Equation (8.28) – Empirical correlations: isothermal surfaces, gives a hL/k = Nu = C(Gr LPr) Tf = (Ts + T∞)/2 = 80 ºF for air properties ρ = 0.0735 lbm/ft3, ν = 16.88 x 10-5 ft2/s, k = 0.01516 Btu/h-ft-ºF, Pr = 0.708
β = 1/T = 1/540 R -1 3 2 3 -5 2 Gr L = gβ(Ts + T∞)L /ν = (32.174)*(1/540)*(30)*(2.5) /(16.88 x 10 ) = 980184191 8
Gr LPr = (980184191)*(0.708) = 693970407 = 6.94 x 10 4 9 Using 10 to 10 (laminar) constants from Table 8-3: C = 0.59, a = ¼ gives 1/4 hL/k = Nu = 0.59*(693970407) = 95.76 2 h = 95.76k/L = (95.76)*(0.01516)/2.5 = 0.5807 Btu/h-ft -ºF 2 q = hA(Ts - T∞) = (0.5807)(2.5) *(30) = 108.9 Btu/h 18 Sample Kreith & Bohn problems & solutions (from a total of 59 problems previously worked): Chapter 1 – Basic Modes of Heat Transfer 1.25. A heat exchanger wall consists of a copper plate 3/8 inch thick. The surface 2 coefficients on the two sides of the plate are 48 0 and 1250 Btu/h-ft -ºF, corresponding to fluid temperatures of 200 and 90 ºF, respectively. Assuming that the thermal conductivity of the wall is 220 Btu/h-ft-ºF, (a) draw the thermal circuit, (b) compute the 2 surface temperatures in ºF, and (c) calculate the heat flux in Btu/h-ft . Solution: (a) draw the thermal circuit – drawn by hand with 3 resistors in series, heat 2 flow from left to right, with T1 = 200 ºF (left-most fluid temp), h c1 = 480 Btu/h-ft -ºF, R 1 = 1/hc1A, Tx unknown (left-most surface temp), R 2 = t/kA (copper wall), Ty unknown 2 (right-most surface temp), T2 = 90 ºF (right-most fluid temp), hc2 = 1250 Btu/h-ft -ºF, and R 3 = 1/hc2A. (b) compute the surface temperatures – overall heat flux is given by q/A = ΔT/∑RA = (200 – 90)/{1/480 + (0.375/12)/220 + 1/1250} = 36359 Btu/h-ft2 which is the same through each of the series resistors, so 36359 = (200 – Tx)/(1/480), Tx = 124 ºF 36359 = (Ty – 90), Ty = 119 ºF 2 2 (c) calculate the heat flux in Btu/h-ft – is given above as q/A = 36359 Btu/h-ft 1.31. A simple solar heater consists of a flat plate of glass below which is located a shallow pan filled with water, so that the water is i n contact with the glass plate above it. 2 Solar radiation is passing through the glass at the rate of 156 Btu/h-ft . The water is at 200 ºF and the surrounding air is 80 ºF. If the heat transfer coefficients between the 2 2 water and the glass and the glass and the air are 5 Btu/h-ft -ºF, and 1.2 Btu/h-ft -ºF
2
respectively, determine the time required to transfer 100 Btu/ft of surface to the water in the pan. The lower surface of the pan may be assumed insulated. Solution: Assuming hci and hco act in series and neglecting resistance of the glass, gives 2 qc/A = ΔT/∑RA = (200 – 80)/{(1/5) + (1/1.2)} = 116.129 Btu/h-ft (heat loss) 2 Qnet/A = 100 Btu/ft = {(qs – qc)/A}*Δt = (156 – 116.129)*Δt
gives Δt = 2.5 h
Chapter 2 – Conduction 5
3
2.5. A plane wall, 7.5 cm thick, generates heat internally at the rate of 10 W/m . One side of the wall is insulated, and the other side is exposed to an environment at 93 ºF. 2 The convection coefficient between the wall and the environment is 567 W/m -K. If the thermal conductivity of the wall is 0.12 W/m-K, calculate the maximum temperature in the wall.
Solution: Using Fourier’s law of heat conduction (1D, SS with heat generation) gives k ∂2T/∂x2 = - qGdot (heat generated/volume) or ∂2T/∂x2 = - qGdot/k integration gives ∂T/∂x = (- qGdot/k)*x + C1 and again
2
T(x) = (- qGdot/2k)*x + C1x + C2 apply BC’s: at x = 0, ∂T/∂x = 0 = 0 + C1 gives C1 = 0, thus T(x) = (- q Gdot/k)*x + C2 at x = 0, T = Tmax gives Tmax = C2 and ∂T/∂x = (- qGdot/k)*x at x = L, -kA*(∂T/∂x)x=L = hcA*(T2 - T∞) qGdot*A*L = hcA*(T2 - T∞) qGdot*A*L = hcA*(T2 - T∞) equation (1) 2 at x = L, T(L) = (- q Gdot/2k)*L + Tmax = T2 equation (2) 2 substitute equation (2) into equation (1) giving qGdot*L/hc = qGdot*L /2k + Tmax - T∞ 5 Tmax = T∞ + qGL*(L/2k + 1/hc) = 93 + (10 )*(0.075)*{(0.075)/(2*0.12) + (1/567)} = 2450 ºC A TAK 2000 finite difference thermal model (problem2_5.ou t) gives Tmax = 2450 ºC (run as a check) 5 Solving equation (1) for T2 = qGdot*L/hc + T∞ = (10 )*(0.075)/(567) + 93 = 106.2 ºC 2.15. Estimate the rate of heat loss per unit length from a 2 in-ID, 2.375 in-OD steel pipe covered with asbestos insulation (3.375 in-OD). Steam flows in the pipe. It has a quality 2 of 99% and is at 300 ºF. The unit thermal resistance at the inner wall is 0.015 h-ft ºF/Btu, the heat transfer coefficient at the outer surface is 3.0 Btu/h-ft2-ºF, and the ambient temperature is 60 ºF. Solution: The thermal circuit (drawn by hand) shows 4 resistors in series with Ti = 300 ºF as the internal fluid temperature, resistor R ci = 1/hciAi (internal fluid resistor), R 1 = ln(r 2/r 1)/(2πLk 1) (conduction resistor through the steel), R 2 = ln(r 3/r 2)/(2πLk 2) (conduction resistor through the asbestos), R o = 1/hcoAo (outer fluid resistor), and T∞ = 60 ºF (outer fluid temperature). q = ΔT/(R i + R 1 + R 2 + R o)
steel – use k = 43 W/m-K = 24.84 Btu/h-ft-ºF asbestos – k = 0.113 W/m-K = 0.06528 Btu/h-ft-ºF 2 for hci – 1/R i = 1/0.015 = 66.67 Btu/h-ft -ºF = hci q = (300 – 60)/{(1)/(66.67*π*2*1/12) + ln(2.375/2)/(2*π*1*24.84)
+ ln(3.375/2.375)/(2*π*1*0.06528) + 1/(3*π*3.375*1/12)} 2
q = 240/(0.0286 + 0.00110 + 0.857 + 0.377) = 190 Btu/h-ft (heat loss) Chapter 4 – Analysis of Convection Heat and Mass Transfer 4.17. Hydrogen at 15 ºC and a pressure of 1 atm is flowing along a flat plate at a velocity of 3 m/s. If the plate is 0.3 m wide and at 71 ºC, calculate the following quantities at x = 5 0.3 m and at the distance corresponding to the transition point, i.e., Rex = 5 x 10 (take properties at 43 ºC): (a) Hydrodynamic boundary layer thickness, in cm, (b) Local friction coefficient, (c) Average friction coefficient, (d) Drag force, in N, (e) Thickness of 2 thermal boundary layer, in cm, (f) Local convective-heat-transfer coefficient, in W/m 2 ºC, (g) Average convective-heat-transfer coefficient, in W/m -ºC, and (h) Rate of heat transfer, in W. Solution: Table 31 (page 669) gives properties for H2 at p = 1 atm – linear interpolation k μ x 106 T(ºC) c p Pr 27 14314 0.182 8.963 0.706 43 x1 x2 x3 x4 77 14436 0.206 9.954 0.697 16/50 = (x1 – 14314)/(14436 – 14314) = (x2 – 0.182)/0.024 = (x3 – 8.963)/(9.954 – 8.963) = (x4 – 0.706)/(0.697 – 0.706) 6 x1 = c p = 14353 J/kg-K, x2 = k = 0.190 W/m-K, x3 = μ x 10 = 9.280, μ = 9.280 x 10-6 N-s/m2, x4 = Pr = 0.703, ρ = 0.07811 kg/m3 5 Transition occurs at Recr = 5 x 10 , solving for xc with Re = ρVL/μ gives 5 3 -6 2 2 5 x 10 = (0.07811 kg/m )*(3 m/s)*(xc)/(9.280 x 10 kg-m-s/s -m ), gives xx = 19.8 m -6 at x = 0.3, Rex=0.3 = (0.07811)*(3)*(0.3)/(9.280 x 10 ) = 7575 (a) Hydrodynamic boundary layer thickness in cm x = 0.3 m, δ = 5x/(Rex)1/2 = (5)*(0.3 m)/(7575)1/2 = 1.7234 x 10-2 m = 1.7234 cm x = 19.8 m, δ = 5x/(Rex)1/2 = (5)*(19.8 m)/(500000)1/2 = 0.140 m = 14 cm 2 1/2 (b) Local friction coefficient, Cfx = ηs/(ρU∞ /2) = 0.664/(Rex) 1/2 x = 0.3, Cfx = 0.664/(7575) = 0.007629 1/2 x = 19.8, Cfx = 0.664/(500000) = 0.000939 (c) Average friction coefficient, Cf = (1/L)*∫ Cfx dx (integrate from 0 to L) = 1.33*(μ/ρU∞L) = 1.33/(ReL)1/2, thus for a given x or L, C f = 2Cfx x = 0.3, Cf = 2*(0.007629) = 0.01526 x = 19.8, Cf = 2*(0.000939) = 0.001878 (d) Drag force, D = ηsA where ηs = Cf *(ρU∞2/2), thus D = Cf A*(ρU∞2/2) 2 2 for x = 0.3, D = (0.01528)*(0.07811)*(3) *(0.30) /2 = 0.0004827 N 2 for x = 19.8, D = (0.001878)*(0.07811)*(3) *(0.30)*(19.8)/2 = 0.003921 N
(e) Thickness of thermal boundary layer, in cm, using Eqn. (4.47) δrh = δ/Pr 1/3 for x = 0.3, δth = (1.7234 x 10-2)/(0.703)1/3 = 1.938 x 10-2 m = 1.938 cm for x = 19.8, δth = (0.140)/(0.703)1/3 = 0.1575 m = 15.75 cm
2
(f) Local convective-heat-transfer coefficient, in W/m -ºC 0.5 0.33 Nux = hcxx/k = 0.332*Re *Pr 0.5 0.33 for x = 0.3, Nux = (0.332)*(7575) *(0.703) = 25.693 2 hcx = 25.693*k/x = (25.693)*(0.190)/0.3 = 16.27 W/m -ºC 0.5 0.33 for x = 19.8, Nux = (0.332)*(500000) *(0.703) = 208.7 2 hcx = 208.7*k/x = (208.7)*(0.190)/19.8 = 2.003 W/m -ºC 2 (g) Average convective-heat-transfer coefficient, in W/m -ºC 0.5 0.33 NuL = 0.664*ReL Pr = 2*Nux and hc = 2*hcx for x = L = 0.3, NuL = 2*25.693 = 51.386 2 hc = 2*16.27 = 32.54 W/m -ºC for x = L = 19.8, NuL = 2*208.7 = 417.4 2 hc = 2*2.003 = 4.006 W/m -ºC What does the following equation give for x = xc = 19.8 m: 0.33 0.8 0.33 0.8 NuL = 0.036*Pr *(Re – 23200) = 0.036*(0.703) (500000 – 23200) 2 = 417.87 = hcL/k so hc = 4.01 W/m -ºC 2 which matches well with laminar value of hc = 4.006 W/m -ºC at x = xcr = 19.8 m. 4.19. Determine the rate of heat loss in Btu/hr from the wall of a building in a 10-mph wind blowing parallel to its surface. The wall is 80 ft long, 20 ft high, its surface temperature is 80 ºF, and the temperature of the ambient air is 40 ºF. Solution: Tmean = Tf = (Ts - T∞)/2 = 60 ºF Table 27 (page 665) – air properties (use values at 68 ºF) ρ = 0.07267 lbm/ft3, c p = 0.2417 Btu/lbm-ºF, k = 0.01450 Btu/h-ft-ºF, Pr = 0.71 μ = 12.257 x 10-6 lbm/ft-s ReL = ρVL/μ V = (10 miles/h) * (5280 ft/1 mile) * (1 h/3600 s) = 14.67 ft/s -6 5 ReL = (0.07267)*(14.67)*)*(80)/(12.257 x 10 ) = 6956367 = 69.6 x 10 (turbulent) 0.33 0.8 0.33 0.8 Use NuL = hcL/k = 0.036*Pr *(ReL – 23200) = 0.036*(0.71) *(6956367 – 23200) 2 NuL = hcL/k = 8829, hc = 8829*(0.01450)/80 = 1.6002 Btu/h-ft -ºF Heat loss qc = hcA*(Ts - T∞) = (1.6002)*(80)*(20)*(80 – 40) = 102413 Btu/h Chapter 5 – Natural Convection 5.4 Compare the rate of heat loss from a human body with the typical energy intake from consumption of food (1300 kcal/day). Model the body as a vertical cylinder 30 cm in diameter and 1.8 m high in still air. Assume the skin temperature is 2 ºC below normal body temperature. Neglect radiation, transpiration cooling (sweating), and the effects of clothing. Solution: assume T∞ = 20 ºF (not given) 3 Ein = 1300 x 10 cal/day x 4.186 J/1 cal x 1 day/24 h x 1 h/3600 s = 62.9838 W T b = 98.6 ºF, T(ºC) = {T(ºF) – 32}*(5/9) = {98.6 – 32}*(5/9) = 37 ºC -2 ºC below normal body temperature gives Ts = 35 ºC
Objective: to compare the rate of heat loss (due to free convection) from a human body to the heat gain, using vertical cylinder correlation – see Figure 4. Air properties – use Tmean = (35 + 20)/2 = 27.5 ºC 2 3 2 Dimensionless parameters: Gr L = gβ*(Ts - T∞)*ρ L /μ , Pr = c pμ/k = ν/α Linear interpolation of air properties table: (without interpolation Pr = 0.71) β x 103 c p k μ x 106 T(ºC) ρ 20 1.164 3.41 1012 0.0251 18.24 27.5 x1 x2 x3 x4 x5 40 1.092 3.19 1014 0.0265 19.123 7.5/20 = (x1 – 1.164)/(1.092 – 1.164) = (x2 – 3.41)/(3.19 – 3.41) = (x3 – 1012)/(1014 – 1012) = (x4 – 0.0251)/(0.0265 – 0.0251) = (x5 – 18.24)/(19.123 – 18.24) 3 3 x1 = ρ = 1.137 kg/m , x2 = β x 10 = 3.3275 1/K, x3 = c p = 1012.75 J/kg-K 6 2 x4 = k = 0.025625 W/m-K, x5 = μ x 10 = 18.571125 N-s/m -3 3 3 Gr L = (9.81 m/s)*(3.3275 x 10 1/K)*(15 K)*(1.137 kg/m )*(1.8 m) -6 2 2 2 /(18.571125 x 10 kg-m-s/s -m ) = 10703858243.6 9 Gr LPr = 7599739353 = 7.599737353 x 10 Figure 5.4 gives 2 appropriate correlation curves – one for laminar and one for transition/turbulent region 1/4 Using the laminar correlation: NuL = 0.555*(Gr LPr) = 163.867 = hcL/k f 2 hc = (163.867)*(0.025625)/1.8 = 2.333 W/m -K Eout = qc = hcA*(Ts - T∞) = (2.333)*π*(0.30)*(1.8)*(15) = 59.36 W 2/5 Using the transition/turbulent region correlation: NuL = 0.0210*(Gr LPr) = 188.165 = hcL/k f 2 hc = (188.165)*(0.025625)/1.8 = 2.6787 W/m -K Eout = qc = hcA*(Ts - T∞) = (2.6787)*π*(0.30)*(1.8)*(15) = 68.17 W Both calculated Eout values obtained from an idealized geometric model and simplified thermal model (sensible convection only) compare well (energ y-balance wise) with Ein = 62.98 W. 5.16 Estimate the rate of heat transfer across a 1-m tall double-pane window assembly in which the outside pane is at 0 ºC and the inside pane is at 20 ºC. The panes are spaced 1 cm apart. What is the thermal resistance (“R” value) of the window? 2
3
2
Solution: Gr δ = gβρ *(T1 – T2)*δ /μ , use Tmean = (T1 + T2)/2 = 10 ºC to evaluate props. 3 3 -1 Table A27 gives the following: ρ = 1.208 kg/m , β x 10 = 3.535 K , -6 2 k = 0.0244 W/m-K, μ x 10 = 17.848 N-s/m , Pr = 0.71 -3 2 3 -6 2 Gr δ = (9.81)*(3.535 x 10 )*(1.208) *(20)*(0.01) /(17.848 x 10 ) = 3177.19 Gr δPr = 2255.8, L/δ = 1/0.01 = 100 Because Gr δ ≤ 8000, the heat transfer mechanism is equivalent to conduction across the enclosure, giving qk = kA*(T1 – T2) = (0.0244)*(1)*(1)*(20)/0.01 = 48.8 W 2 R-value: R k = δ/kA = (0.01 m)/{(0.0244 W/m-K)*(1 x 1 m ) = 0.4098 K/W
Chapter 6 – Forced Convection Inside Tubes and Ducts 6.3 For water at a bulk temperature of 32 ºC flowing at a velocity of 1.5 m/s through a 2.54 cm ID duct with a wall temperature of 43 ºC, calculate the Nusselt number and the convection heat transfer coefficient by three different methods and compare the results. Solution: For water at T b = 32 ºC – Table 13 (page 651) – linear interpolation gives ρ c p k μ x 106 Pr T(ºC) 30 995.7 4176 0.615 792.4 5.4 32 x1 x2 x3 x4 x5 35 994.1 4175 0.624 719.8 4.8 2/5 = (x1 - 995.7)/(994.1 – 995.7) = (x2 – 4.176)/(-1) = (x3 – 0.615)/(0.009) = (x4 – 792.4)/(719.8 - 792.4) = (x5 – 5.4)/(-0.6) 3 x1 = ρ = 995.06 kg/m , x2 = c p = 4175.6, x3 = k = 0.6186 W/m-K, 6 2 x4 = μ x 10 = 763.36 N-s/m , x5 = Pr = 5.16 -6 ReD = ρVD/μ = (995.06)*(1.5)*(0.0254)/(763.36 x 10 ) = 49664 (turbulent) Method (1): using Table 6.3 (page 324) and first equation 0.8 0.4 0.8 0.4 NuD = 0.023*ReD Pr = 0.023*(49664) *(5.16) = 253.29 2 hc = (253.29)*k f/D = (253.29)*(0.6186)/(0.0254) = 6169 W/m -K 0.8 1/3 0.2 Method (2): using Table 6.3 and second equation NuD = 0.027*ReD Pr *(μ b/μs) At Tw = Ts = 43 ºC linear interpolation gives 6 T(ºC) μ x 10 3/5 = (y1 – 658)/(605.1 – 658) -6 2 40 658.0 y1= μs = 626.26 x 10 N-s/m 43 y1 45 605.1 0.8 1/3 0.2 NuD = 0.027*(49664) *(5.16) *(763.36/626.26) = 273 = hcD/k f 2 hc = (273)*(0.6186)/(0.0254) = 6647 W/m -K Method (3): using Table 6.3 and third equation 1/2 2/3 NuD = (f/8)*ReDPr/{K 1 + K 2*(f/8) *(Pr – 1)} -2 where f = (1.82*log10ReD – 1.64) and log10x = (ln x)/(ln 10), gives 1/3 f = 0.020963, K 1 = 1 + 3.4f = 1.07127, K 2 = 11.7 + 1.8/Pr = 12.74165 NuD = (0.020963/8)*(49664)*(5.16)/ 1/2 2/3 {1.07127 + 12.74165*(0.020963/8) *(5.16 – 1)} = 284 = hcD/k f 2 hc = (284)*(0.6186)/(0.0254) = 6910 W/m -K Summary: (1) NuD = 253, (2) NuD = 273, and (3) NuD = 284 gives (NuD)avg = 270 (+6% ,-6%) 6.16 Water at 82.2 ºC is flowing through a thin copper tube (15.2 cm ID) at a velocity of 7.6 m/s. The duct is located in a room at 15.6 ºC and the unit-surface-conductance at the 2 outer surface of the duct is 14.1 W/m -K. (a) Determine the heat transfer coefficient at the inner surface. (b) Estimate the length of duct in which the water temperature drops (5/9) ºC.
Solution: water at T bi = 82.2 ºC – Table 13 (page 651) – linear interpolation ρ c p k μ x 106 Pr T(ºC) 75 974.9 4190 0.671 376.6 2.23 82.2 x1 x2 x3 x4 x5 100 958.4 4211 0.682 277.5 1.75 0.288 = (x1 – 974.9)/(958.4 – 974.9) = (x2 – 4190)/(4211 – 4190) = (x3 – 0.671)/0.011 = (x4 – 376.6)/(277.5 – 376.6) = (x5 – 2.23)/(1.75 – 2.23) 3 x1 = ρ = 970.148 kg/m , x2 = c p = 9196.048 J/kg-K, x3 = k = 0.674168 W/m-K, 6 2 x4 = μ x 10 = 348.0592 N-s/m , x5 = Pr = 2.09176 Re = ρVD/μ = ( 970.148)*(7.6)*(0.152)/(348.0592 x 10-6) = 3219898 = 32.19898 x 105 (turbulent) 0.8 0.3 0.8 0.3 Using NuD 0.023*(ReD) *(Pr) = 0.023*(3219898) (2.09176) = 4614.8 = hcD/k f 2 hc = (4614.8)*(0.674168)/(0.152) = 20468 W/m -K
Neglecting copper wall ΔT or R th hciA*(T b – Ts) = hcoA*(Ts - T∞) or hci*(T b – Ts) = hco*(Ts - T∞)
Ts*(hco + hci) = hciT b + hcoT∞ Ts = (hciT b + hcoT∞)/(hco + hci) = {(204.68)*(82.2) + (14.1)*(15.6)}/(204.68 + 14.1) = 82.155 ºC (b) hciA*(T b – Ts) = mdot*c pΔT b mdot = ρAV = (970.148 kg/m3)*(π/4)*(0.152 m)2*(7.6 m/s) = 133.79 kg/s hciπDL*(T b – Ts) = mdot*c pΔT b L = (mdot*c pΔT b)/{hciπD*(T b – Ts)} = (133.79 kg/s)*(4196.048 J/kg-K)*(5/9 ºC)/ 2 {(20468 W/m -K)*π*(0.152 m)*(82.2 – 82.155) ºC} = 2228 m (for ΔT b = 5/9 ºC drop) Chapter 7 – Forced Convection over Exterior Surfaces 7.4 Steam at 1 atm and 100 ºC is flowing across a 5-cm-OD tube at a velo city of 6 m/s. Estimate the Nusselt number, the heat transfer coefficient, and the rate of heat transfer per meter length of pipe if the pipe is at 200 ºC. m
n
0.25
Solution: Use correlation NuD = hcD/k = C*(ρU∞D/μ) *Pr *(Pr/Pr s) (7.3) 3 Steam at 1 atm and 100 ºC – Table 34 (page 672) gives ρ = 0.5977 kg/m , 6 2 c p = 2034 J/kg-K, k = 0.0249 W/m-K, μ x 10 = 12.10 N-s/m , Pr = 0.987 To get Pr s at Ts = 200 ºC linear interpolation gives T(ºC) Pr s 23/50 = (x1 – 1.010)/(0.996 – 1.010) 177 1.010 x1 = Pr s = 1.00356 200 x1 227 0.996 Re = ρU∞D/μ = (0.5977)*(6)*(0.05)/(12.10 x 10-6) = 14819 = 1.4819 x 104 Gives for Equation (7.3) the following constants: C = 0.26, m = 0.6, n = 0.36 0.6 0.36 0.25 NuD = hcD/k = 0.26*(14819) *(0.987) *(0.987/1.00356) = 81.96 2 hc = (81.96)*(0.0249)/(0.05) = 40.82 W/m -K qc = hcA*(Ts - T∞) = hcπDL*(Ts - T∞) qc/L = hcπD*(Ts - T∞) = (40.82)*π*(0.05)*(100) = 256 W/m
7.6 Determine the average unit-surface conductance for air at 60 ºC flowing at a velocity of 1 m/s over a bank of 6-cm-OD tubes arranged as shown in the accompanying sketch (shows a staggered tube bank). The tube-wall temperature is 117 ºC. Solution: Reference Figure 7.18 (page 372) for a typical staggered tube arrangement. In our case, D = 6 cm, SL = 7.6 cm, ST = 5.1 cm, 2 2 2 2 2 SL’ = ST + SL = 5.1 + 7.6 gives SL’ = 9.1526 cm, ST/SL = 5.1/7.6 ≤ 2 Air properties at Tf = (Ts + T∞)/2 = (117 + 60)/2 = 88.5 ºC using linear interpolation gives ρ c p k μ x 106 Pr = 0.71 T(ºC) 80 0.968 10.19 0.0293 20.790 88.5 x1 x2 x3 x4 100 0.916 10.22 0.0307 21.673 0.425 = (x1 – 0.968)/(0.916 – 0.968) = (x2 – 1019)/3 = (x3 – 0.0293)/(0.0307 – 0.0293) = (x4 – 20.790)/(21.673 – 20.790) 3 x1 = ρ = 0.9459 kg/m , x2 = c p = 1020.275 J/kg-K, x3 = k = 0.029895 W/m-K, 6 2 x4 = μ x 10 = 21.165275 N-s.m -6 ReD = ρVD/μ = (0.9459)*(1)*(0.06)/(21.165275 x 10 ) = 2681.5 3 5 Using a transition regime equation (for 10 ≤ ReD ≤ 2 x 10 ) 0.2 0.6 0.36 0.25 NuD = hcD/k = 0.35*(ST/SL) ReD Pr (Pr/Pr s) 0.2 0.6 0.36 0.25 = 0.35*(5.1/7.6) *(2681.5) (0.71) (1) = 32.576 2 hc = (32.576)*(0.029895)/(0.06) = 16.23 W/m -K Chapter 8 – Heat Exchangers 8.3 A light oil flows through a copper tube of 2.6 cm ID and 3.2 cm OD. Air is flowing over the exterior of the tube. The convective heat transfer coefficient for the oil is 120 2 2 W/m -K and for the air is 35 W/m -K. Calculate the overall heat transfer coefficient based on the outside area of the tube (a) considering the thermal resistance of the tube, (b) neglecting the resistance of the tube. Solution: Using q = UA*(Th – Tc) where UA = 1/∑R = 1/{(1/hciAi) + ln(r o/r i)/(2πkL) + (1/hcoAo)} 2 2 hci = 120 W/m -K, hco = 35 W/m -K, Ai = πDiL = π(0.026)*(1), Ao = πDoL =
π(0.032)(1)
A0/Ai = D0/Di, for copper k = 399 W/m-K (a) including R th for the tube and based upon the outer area Uo = 1/{(D0/Di)*(1/hci) + πDoL*ln(r o/r i)/(2πkL) + (1/hco)}
= 1/{(3.2/2.6)/(1/120) + π(0.032)*(1)*ln(3.2/2.6)/(2*π*399*1) + (1/35)} -2
-6
-2
2
= 1/{(1.0256 x 10 ) + (8.3264 x 10 ) + (2.857 x 10 )} = 25.75 W/m -K 2 (b) without including the thermal resistance of the copper w all gives U0 = 25.76 W/m -K. 8.9 A shell-and-tube heat exchanger has one shell pass and four tube passes. The fluid in the tubes enters at 200 ºC and leaves at 100 ºC. The temperature of the fluid entering the shell is 20 ºC and is 90 ºC as it leaves the shell. The overall heat transfer coefficient 2 2 based on a surface area of 12 m is 300 W/m -K. Calculate the heat transfer rate between fluids.
Solution: ΔTmean = (LMTD)*(F)
(eqn 8.18) Using Fig. 8.12 for a HX with one shell pass and four tube passes P = (Tt,out – Tt,in)/(Ts,in – Tt,in) = (100 – 200)/(20 – 200) = 100/180 = 0.18 Z = (mdottc pt)/(mdotsc ps) = (Ts,in – Ts,out)/(Tt,out – Tt,in) = (20 – 90)/(100 – 200) = 0.70 From Fig. 8.12, gives F = 0.98 F * LMTD = F * (ΔTa – ΔT b)/ln(ΔTa/ΔT b) ΔTa = Th,in – Tc,out = 200 – 90 = 110 ºC ΔT b = Th,out – Tc,in = 100 – 20 = 80 ºC F * LMTD = (0.98)*(110 – 80)/ln(110/80) = 92.32 ºC Heat transfer rate is given by 2 2 q = UA*F*LMTD = (300 W/m -K)*(12 m )*(92.32 C) = 332356 W (ΔK = ΔC) (counterflow HX was assumed) 8.12 Water entering a shell-and-tube heat exchanger is at 35 ºC is to be heated to 75 ºC by an oil. The oil enters at 110 ºC and leaves at 75 ºC. The heat exchanger is arranged for counterflow with water making one shell pass and the oil two tube passes. If the water flow rate is 68 kg/min and the overall heat transfer coefficient is estimated from 2 Table 8.1 to be 320 W/m -K, calculate the required heat exchanger area. Solution: q = UA*F*LMTD Using Figure 8.12 for two tube passes P = (Tt,out – Tt,in)/(Ts,in – Tt,in) = (75 – 110)/(35 – 110) = 35/75 = 0.4667 Z = (mdottc pt)/(mdotsc ps) = (Ts,in – Ts,out)/(Tt,out – Tt,in) = (35 – 75)/(75 – 110) = 1.143 Figure 8.12 gives F = 0.8 ΔTa = Th,in – Tc,out = 110 – 75 = 35 ºC ΔT b = Th,out – Tc,in = 75 – 35 = 40 ºC F * LMTD = F * (ΔTa – ΔT b)/ln(ΔTa/ΔT b) = 0.80*(35 – 40)/ln(35/40) = 29.96 ºC Heat transfer – multiple pass – need to solve for A with q = UA*F*LMTD (equation 1) For the water mdot = 68 kg/min, at Tmean = (35 + 75)/2 = 55 ºC (use 50 ºC) c p = 4178 J/kg-K qwater = mdot*c p*ΔT = (68 kg/min)*(1 min/60 s)*(4178 J/kg-K)*(40 K) = 189402.67 W 2 equate to equation 1: 189402.67 W = (320 W/m -K)*A*(29.96 K) 2 gives A = 19.8 m Chapter 9 – Heat Transfer by Radiation 9.3. Determine the total average hemispherical emittance and the emissive power of a surface which has a spectral hemispherical emittance of 0.8 at wavelengths less than 1.5
μm, 0.6 from 1.5 to 2.5 μm, and 0.4 at wavelengths longer than 2.5 μm. The surface temperature is 1111 K.
Solution: at λ = 1.5 μm, λT = (1.5 μm)*(1111 K) = 1666.5 μm-K = 1.6665 x 10-3 m-K 4 Table 9.1 (Blackbody Radiation Functions) gives E b(0→λT)/ζT = 0.0262455 (ε = 0.8) at λ = 2.5 μm, λT = (2.5 μm)*(1111 K) = 2.7775 x 10-3 m-K 4 table 9.1 gives E b(0→λT)/ζT = 0.222871 giving
ε = ∫0λ1 ελ(λ)*E bλ(λT)*dλ*{(∫λ1∞ ελ(λ)*E bλ(λT)*dλ)/E b} ε = 0.8*(0.0262455) + 0.6*(0.222871 – 0.0262455) + 0.4*(1 – 0.222871) = 0.4498 4 -8 4 2 Emissive power is given by Eg = εζT = (0.4498)*(5.67 x 10 *(1111) = 38858 W/m 9.23. A black sphere (1 inch diameter) is placed in a large infrared heating oven whose walls are maintained at 700 ºF. The temperature of the air in the oven is 200 ºF and the heat-transfer coefficient for convection between the surface of the sphere and air is 5 2 Btu/h-ft -ºF. Estimate the net rate of heat flow to the sphere when its surface temperature is 100 ºF.
Solution: sphere surface area S = 4πR 2 = πD2 = π in2 x ft2/(144 in2) = π/144 ft2 Energy balance on sphere – assume steady state qx-1 = (E b2 – E b1)*A1F1-2 + hcAs*(T∞ – T1) -8 2 4 4 4 4 2 = (0.1714 x 10 Btu/h-ft -R )*(1160 – 560 )R (π/144 ft ) 2 2 + (5 Btu/h-ft -ºF)*(π/144 ft )*(200 – 100) = 64.03 + 10.91 = 74.9 Btu/h (radiation) (convection) 9.30. Three thin sheets of polished aluminum are placed parallel to each other so that the distance between them is very small compared to the size of the sheets. If on of the outer sheets is at 540 ºF, whereas the other outer sheet is at 140 ºF, calculate the net rate of heat flow by radiation and the temperature of the intermediate sheet. Convection may be ignored. Solution: outer sheet 1 T1 = 540 ºF = 1000 ºR, outer sheet 2 T2 = 140 ºF = 600 ºR Intermediate or middle sheet 3, known are geometric view factors F1-3 = F3-2 = 1, equal
areas, use ε = 0.05, ρ = 1 – ε = 0.95 Using q1-3 = A1Ḟ1-3*(E b1 = E b3) = q1-3 = A1Ḟ1-3*ζ(T b1 - T b3) Configuration factor Ḟ1-3 = 1/(1/ε1 + 1/ε3 + 1) = 1/(1/0.05 + 1/0.05 + 1) = 0.02439 4 4 q1-3 = ζA1*(0.02439)*(1000 – T3 ) 4 4 also q3-2 = A3Ḟ3-2*ζ(T b3 - T b2) = ζA3*(0.02439)*(T3 – 600 )
equate for energy balance q1-3 = q3-2 ζA1*(0.02439)*(10004 – T34) = ζA3*(0.02439)*(T34 – 6004) 4 4 4 4 1000 – T3 = T3 – 600 gives T3 = 867 ºR = 407 ºF -8 4 4 2 q1-3 = A1Ḟ1-3*ζ(T b1 - T b3) = A*(0.02439)*(0.1718 x 10 )*(1000 - 867 ) = 18.2 Btu/h-ft
