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Solutions Manual for Fundamentals of Heat and Mass Transfer part 2
Ryerson University
Uploaded: A year ago
Contributor: cloveb
Category: Chemical Engineering
Type: Solutions
Rating: A (1)
Filename:   ch09-heat transfer.pdf (2.49 MB)
Page Count: 178
Credit Cost: 4
Views: 300
Last Download: N/A
Transcript
PROBLEM 9.1 KNOWN: Tabulated values of density for water and definition of the volumetric thermal expansion coefficient, . FIND: Value of the volumetric expansion coefficient at 300K; compare with tabulated values. PROPERTIES: Table A-6, Water (300K): = 1/vf = 1/1.003 1 . T p The density change with temperature at constant pressure can be estimated as 12 where the subscripts (1,2) denote the property values just above and below, respectively, the condition for T = 300K denoted by the subscript (o). That is, 1 12 o . o12TT p Substituting numerical values, find 1 995.0998. 0 kg/m 300.910 K. < 997.0 kg/m 30529 5K Compare this value with the tabulation, PROBLEM 9.2 KNOWN: Object with specified characteristic length and temperature difference above ambient fluid. FIND: Grashof number for air, hydrogen, water, ethylene glycol for a pressure of 1 atm. SCHEMATIC: ASSUMPTIONS: (1) Thermophysical properties evaluated at Tf = 350K, (2) Perfect gas behavior, ( = 1/Tf). PROPERTIES: Evaluate at 1 atm, Tf = 350K: -6 2 -6 2 Table A-4, Air: = 20.92 10 m/s; Hydrogen: = 143 10 m /s Table A-6, Water (Sat. liquid): = -6 2 -3 -1 f vf = 37.5 10 m/s, f = 0.624 10 K -6 2 -3 -1 Table A-5, Ethylene glycol: = 3.17 10 m/s, = 0.65 10 K . ANALYSIS: The Grashof number is given by Eq. 9.12, gTTL 3 s Gr.L 2 Substituting numerical values for air with = 1/Tf, find 2 9.8m/s1/350 K 25K 0.25m 3 Gr L,air 2 62 20.9210 m/s 7 GrL,air2.5010. < Performing similar calculations for the other fluids, find 5 GrL,hyd5.3510 < 6 GrL,water1.7010 < 8 GrL,eth2.4810. < COMMENTS: Higher values of GrL imply increased free convection flows. However, other properties affect the value of the heat transfer coefficient. PROBLEM 9.3 KNOWN: Relation for the Rayleigh number. FIND: Rayleigh number for four fluids for prescribed conditions. SCHEMATIC: ASSUMPTIONS: (1) Perfect gas behavior for specified gases. -6 2 -6 2 PROPERTIES: Table A-4, Air (400K, 1 atm): = 26.41 10 m /s, = 38.3 10 m/s, = -3 -1 -6 2 1/T = 1/400K = 2.50 10 K ; Table A-4, Helium (400K, 1 atm): = 199 10 m /s, = 295 -6 2 -3 -1 -6 2 10 m /s, = 1/T = 2.50 10 K ; Table A-5, Glycerin (12 C = 285K): = 2830 10 m /s, = 0.964 -7 2 -3 -1 10 m /s, = 0.475 10 K ; Table A-6, Water (37 C = 310K, sat. liq.): = f vf = 695 -6 2 10 N -3 3 -6 2 -3 s/m 1.007 10 m /kg = 0.700 10 m /s, = kf vf/cp,f = 0.628 W/mK 1.007 10 3 -6 2 -6 -1 m /kg/4178 J/kgK = 0.151 10 m /s, f = 361.9 10 K . ANALYSIS: The Rayleigh number, a dimensionless parameter used in free convection analysis, is defined as the product of the Grashof and Prandtl numbers. 333 gTLgTLgTLcc pp RaGLrPr 22 kk where = k/ cp and = / . The numerical values for the four fluids follow: Air (400K, 1 atm) 26 Ra9.8m/s 1/400262 K 30K0.01 m/26.4 110 3 L m/s38.3 10 m/s727 < Helium (400K, 1 atm) 26 Ra9.8m/s1/400 262 L K 30K0.01 m/19 910m/s 3 29510m/s 12.5 < Glycerin (285K) 2316272 Ra9.8m/L s0.47510 3 K 30K0.01m/2830 10m/s0.964 10m/s512 < Water (310K) 23162625 Ra9.8m/L s0.36210K 30K0.01m/0.70010m/s0.15 3 110m/s9.3510. < COMMENTS: (1) Note the wide variation in values of Ra for the four fluids. A large value of Ra implies enhanced free convection, however, other properties affect the value of the heat transfer coefficient. PROBLEM 9.4 KNOWN: Form of the Nusselt number correlation for natural convection and fluid properties. FIND: Expression for figure of merit FN and values for air, water and a dielectric liquid. -1 -5 2 PROPERTIES: Prescribed. Air: k = 0.026 W/mK, = 0.0035 K , = 1.5 10 m/s, Pr = 0.70. -4 -1 -6 2 Water: k = 0.600 W/mK, = 2.7 10 K , = 10 m/s, Pr = 5.0. Dielectric liquid: k = 0.064 -1 -6 2 W/mK, = 0.0014 K , = 10 m/s, Pr = 25 n ANALYSIS: With Nu~RL a, the convection coefficient may be expressed as gTL kgTLk h~~ LL The figure of merit is therefore n k FN < nn and for the three fluids, with n = 0.33 and /Pr , AirWaterDielectric FWs/mK < 5.8663209 Water is clearly the superior heat transfer fluid, while air is the least effective. COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large and small values of and . PROBLEM 9.5 KNOWN: Heat transfer rate by convection from a vertical surface, 1m high by 0.6m wide, to quiescent air that is 20K cooler. FIND: Ratio of the heat transfer rate for the above case to that for a vertical surface that is 0.6m high by 1m wide with quiescent air that is 20K warmer. SCHEMATIC: ASSUMPTIONS: (1) Thermophysical properties independent of temperature; evaluate at 300K; (2) Negligible radiation exchange with surroundings, (3) Quiescent ambient air. -6 2 -6 2 PROPERTIES: Table A-4, Air (300K, 1 atm): = 15.89 10 m /s, = 22.5 10 m /s. ANALYSIS: The rate equation for convection between the plates and quiescent air is qh AT Ls where T is either (Ts - T ) or (T - Ts); for both cases, As = Lw. The desired heat transfer ratio is then qh1L1 . (2) qh2L2 To determine the dependence of hL on geometry, first calculate the Rayleigh number, 3 Rag T L L/ (3) and substituting property values at 300K, find, 2 3 -6 2 -6 2 9 Case 1: RaL1 = 9.8 m/s (1/300K) 20K (1m) /15.89 10 m /s 22.5 10 m /s = 1.82 10 3 4 3 8 Case 2: RaL2 = RaL1 (L2/L1) = 1.82 10 (0.6m/1.0m) = 3.94 10 . Hence, Case 1 is turbulent and Case 2 is laminar. Using the correlation of Eq. 9.24, hLk NuCRa L hC Ra n n L LL kL L where for Case 1: C1 = 0.10, n1 = 1/3 and for Case 2: C2 = 0.59, n2 = 1/4. Substituting Eq. (4) into the ratio of Eq. (2) with numerical values, find n 1/3 1 9 C/LRa 0.10/1m1.8 210 q1 11 L1 0.881 < q2 n1/42 C/22LRa 0.59/0.6m3.9 8 L2 410 COMMENTS: Is this result to be expected? How do you explain this effect of plate orientation on the heat rates? PROBLEM 9.6 KNOWN: Large vertical plate with uniform surface temperature of 130 C suspended in quiescent air at 25 C and atmospheric pressure. FIND: (a) Boundary layer thickness at 0.25 m from lower edge, (b) Maximum velocity in boundary layer at this location and position of maximum, (c) Heat transfer coefficient at this location, (d) Location where boundary layer becomes turbulent. SCHEMATIC: ASSUMPTIONS: (1) Isothermal, vertical surface in an extensive, quiescent medium, (2) Boundary layer assumptions valid. -6 2 PROPERTIES: Table A-4, Air T fsTT /2350K, 1 at m: = 20.92 10 m /s, k = 0.030 W/mK, Pr = 0.700. ANALYSIS: (a) From the similarity solution results, Fig. 9.4 (see above right), the boundary layer thickness corresponds to a value of 5. From Eqs. 9.13 and 9.12, yx Gr/4 1/4 x (1) m1 2 3236293 GrxsgTTx/9.8 13025K x/20.9210m/s6.71 810 x (2) 2 s 350K 1/4 y50.25 m 6.71 92 8100.25/41.74 610m 3 17.5 mm. (3) < (b) From the similarity solution shown above, the maximum velocity occurs at 1 with f0.275. From Eq.9.15, find 62 2220.9210m/s 1/2 1/29 uGr f6.71x8100.2 50.27 50.47 m/s. 3 < x0.25m The maximum velocity occurs at a value of = 1; using Eq. (3), it follows that this corresponds to a position in the boundary layer given as y1/5 17.5 m max m3.5 mm. < (c) From Eq. 9.19, the local heat transfer coefficient at x = 0.25 m is 1/4 Nuhxx/ xxkGr/4 gPr 1/43 6.718100.25/40.58 9 641.9 < The value for g(Pr) is determined from Eq. 9.20 with Pr = 0.700. (d) According to Eq. 9.23, the boundary layer becomes turbulent at xc given as 1/3 999 RaGx,rcPx,r1cc0 x10/6.718100.7000.60 m . < COMMENTS: Note that = 1/Tf is a suitable approximation for air. PROBLEM 9.7 KNOWN: Thin, vertical plates of length 0.15m at 54 C being cooled in a water bath at 20 C. FIND: Minimum spacing between plates such that no interference will occur between free- convection boundary layers. SCHEMATIC: ASSUMPTIONS: (a) Water in bath is quiescent, (b) Plates are at uniform temperature. PROPERTIES: Table A-6, Water (Tf = (Ts + T )/2 = (54 + 20) C/2 = 310K): = 1/vf = 993.05 3 -6 2 kg/m , = 695 10 Ns/m , = / = 6.998 -7 2 -6 -1 10 m /s, Pr = 4.62, = 361.9 10 K . ANALYSIS: The minimum separation distance will be twice the thickness of the boundary layer at the trailing edge where x = 0.15m. Assuming laminar, free convection boundary layer conditions, the similarity parameter, , given by Eq. 9.13, is y Gr/4 1/4 x x where y is measured normal to the plate (see Fig. 9.3). According to Fig. 9.4, the boundary layer thickness occurs at a value 5. It follows then that, y x G r/4 1/4 blx g TTx 3 s where Grx 2 2 261728 Grx9.8 m/s361.910 K5420K0.15m/6.99 810 m /s 3 8.31010. < Hence, and the minimum separation is d2 y26.3 mbl m12.6 mm. < COMMENTS: According to Eq. 9.23, the critical Grashof number for the onset of turbulent 9 8 conditions in the boundary layer is Grx,c Pr 10. For the conditions above, Grx Pr = 8.31 10 9 4.62 = 3.8 10. We conclude that the boundary layer is indeed turbulent at x = 0.15m and our calculation is only an estimate which is likely to be low. Therefore, the plate separation should be greater than 12.6 mm. PROBLEM 9.8 KNOWN: Square aluminum plate at 15 C suspended in quiescent air at 40 C. FIND: Average heat transfer coefficient by two methods – using results of boundary layer similarity and results from an empirical correlation. SCHEMATIC: ASSUMPTIONS: (1) Uniform plate surface temperature, (2) Quiescent room air, (3) Surface radiation exchange with surroundings negligible, (4) Perfect gas behavior for air, = 1/Tf. -6 PROPERTIES: Table A-4, Air (Tf = (Ts + T )/2 = (40 +15) C/2 = 300K, 1 atm): = 15.89 10 2 -6 2 m /s, k = 0.0263 W/mK, = 22.5 10 m /s, Pr = 0.707. ANALYSIS: Calculate the Rayleigh number to determine the boundary layer flow conditions, 3 Rag TL L/ 262627 Ra9.8 m/L s1/300K 4015C0.2 m/15.8910m/s 22. 3 510m/s1.827 10 where 9 = 1/Tf and T = T - Ts. Since RaL < 10 , the flow is laminar and the similarity solution of Section 9.4 is applicable. From Eqs. 9.21 and 9.20, h LL4 NuGr/4 gPr 1/4 L L k3 1/2 0.75 Pr gPr 1/4 1/2 0.6091.221 Pr1.238 Pr and substituting numerical values with GrL = RaL/Pr, find 1/4 gP r0.7 50.707 1/21/2 /0.6091.220.7071.2380.7070.5 01 < The appropriate empirical correlation for estimating hL is given by Eq. 9.27, 1/4 h LL 0.670 Ra Nu0.68 L L k 4/9 10.492/Pr 9/16 1/4 4/9 h0.0263 W/ mK/0.20m 0.680.67 7 9/16 L 01.82710/10.4 92/0.707 2 h4.42 W/ L mK. < COMMENTS: The agreement of hL calculated by these two methods is excellent. Using the Churchill-Chu correlation, Eq. 9.26, find h4.87 W/ L mK. This relation is not the most accurate for the laminar regime, but is suitable for both laminar and turbulent regions. PROBLEM 9.9 KNOWN: Dimensions of vertical rectangular fins. Temperature of fins and quiescent air. FIND: (a) Optimum fin spacing, (b) Rate of heat transfer from an array of fins at the optimal spacing. SCHEMATIC: ASSUMPTIONS: (1) Fins are isothermal, (2) Radiation effects are negligible, (3) Air is quiescent. -6 2 PROPERTIES: Table A-4, Air (Tf = 325K, 1 atm): = 18.41 10 m /s, k = 0.0282 W/mK, Pr = 0.703. ANALYSIS: (a) If fins are too close, boundary layers on adjoining surfaces will coalesce and heat transfer will decrease. If fins are too far apart, the surface area becomes too small and heat transfer decreases. Sop x=H. From Fig. 9.4, the edge of boundary layer corresponds to /H G r/45. 1/4 H gTT H9.8 m/ 32s1/325K 50K 0.15m 3 s 7 Hence, GrH1.510 H2 2 62 18.411 0m/s 1/4 H5 0.15 m/1.5 7 10/40.017m17m mS34mm. op < (b) The number of fins N can be found as NW/St355/35. op 510 and the rate is q2 N hHL TT. s For laminar flow conditions 4/9 1/4 Nu0.6H 80.67 Ra/10.492/Pr 9/16 L 1/4 4/9 7 9/16 Nu0.6H 80.671.5100.70 3/10.492/0.703 30 hk Nu/H0.0282 W/H mK30/0.15 m 5.6 W/ 2 mK q2105.6 W/ 2 mK0.15m0.02 m 350300K16.8 W. < COMMENTS: Part (a) result is a conservative estimate of the optimum spacing. The increase in area resulting from a further reduction in S would more than compensate for the effect of fluid entrapment due to boundary layer merger. From a more rigorous treatment (see Section 9.7.1), Sop 10 mm is obtained for the prescribed conditions. PROBLEM 9.10 KNOWN: Interior air and wall temperatures; wall height. FIND: (a) Average heat transfer coefficient when T = 20 C and Ts = 10 C, (b) Average heat transfer coefficient when T = 27 C and Ts = 37 C. SCHEMATIC: ASSUMPTIONS: (a) Wall is at a uniform temperature, (b) Room air is quiescent. -3 -1 -6 PROPERTIES: Table A-4, Air (Tf = 298K, 1 atm): = 1/Tf = 3.472 10 K , = 14.82 10 2 m /s, k = 0.0253 W/mK, = 20.9 -6 2 10 m /s, Pr = 0.710; (Tf = 305K, 1 atm): = 1/Tf = 3.279 -3 -1 -6 2 10 K , = 16.39 -6 2 10 m /s, k = 0.0267 W/mK, = 23.2 10 m /s, Pr = 0.706. ANALYSIS: The appropriate correlation for the average heat transfer coefficient for free convection on a vertical wall is Eq. 9.26. 2 0.1667 hL0.387 Ra Nu0.825L L k 0.296 10.492/Pr 0.563 3 where RaL = g T L / , Eq. 9.25, with T = Ts - T or T - Ts. (a) Substituting numerical values typical of winter conditions gives 231 9.8 m/s3.47210 K2010 K 2.5m 3 10 Ra1.71 L 110 62-62 14.8210m/s20.9610m/s 0.1667 2 10 0.3871.71 110 Nu0.825299.6. L 0.296 10.492/0.710 0.563 2 Hence, hNu k/LL299.60.0253 W/ mK/2.5m3.03 W/ mK. < (b) Substituting numerical values typical of summer conditions gives 231 9.8 m/s3.27910 K3727 K 2.5 m 3 10 Ra1.32 L 010 6262 23.210m/s16.3910m/s 2 0.1667 10 0.3871.32 010 Nu0.82 L 5275.8. 0.296 10.492/0.706 0.563 2 Hence, hNu k/LL275.80.0267 W/mK/2.5m2.94 W/mK. < COMMENTS: There is a small influence due to Tf on h for these conditions. We should expect radiation effects to be important with such low values of h. PROBLEM 9.11 KNOWN: Vertical plate experiencing free convection with quiescent air at atmospheric pressure and film temperature 400 K. FIND: Form of correlation for average heat transfer coefficient in terms of T and characteristic length. SCHEMATIC: ASSUMPTIONS: (1) Air is extensive, quiescent medium, (2) Perfect gas behavior. -6 2 PROPERTIES: Table A-6, Air (Tf = 400K, 1 atm): = 26.41 10 m/s, k = 0.0338 W/mK, -6 2 = 38.3 10 m /s. ANALYSIS: Consider the correlation having the form of Eq. 9.24 with RaL defined by Eq. 9.25. n NuhL/kCRa L L L gTT L9.8 m/ 323 s s1/400 KTL 73 Ra2.42L 210TL. (2) 6262 26.4110m/s38.310m/s Combining Eqs. (1) and (2), 0.0338 W/mK n hk/LCR aC n73 L 2.42210TL. L (3) L 4 9 From Fig. 9.6, note that for laminar boundary layer conditions, 10 < RaL < 10, C = 0.59 and n = 1/4. Using Eq. (3), 1/4 1/4 13 T h1.40LTL1.40 < L 9 13 For turbulent conditions in the range 10 < RaL < 10 , C = 0.10 and n = 1/3. Using Eq. (3), < COMMENTS: Note the dependence of the average heat transfer coefficient on T and L for laminar and turbulent conditions. The characteristic length L does not influence hL for turbulent conditions. PROBLEM 9.12 1/4 KNOWN: Temperature dependence of free convection coefficient, hCT,=∆ for a solid suddenly submerged in a quiescent fluid. FIND: (a) Expression for cooling time, tf, (b) Considering a plate of prescribed geometry and thermal conditions, the time required to reach 80° C using the appropriate correlation from Problem 9.10 and (c) Plot the temperature-time history obtained from part (b) and compare with results using a constant ho from an appropriate correlation based upon an average surface temperature TTT2=+()if . SCHEMATIC: ASSUMPTIONS: (1) Lumped capacitance approximation is valid, (2) Negligible radiation, (3) Constant properties. 3 PROPERTIES: Table A.1, Aluminum alloy 2024 ()TTT2400K=+≈() ρ if : = 2770 kg/m, cp = -5 2 925 J/kg⋅K, k = 186 W/m⋅K; Table A.4, Air (T ν × ⋅ film= 362 K): = 2.221 10 m/s, k = 0.03069 W/mK, α -5 2 = 3.187 × 10 m/s, Pr = 0.6976, β = 1/Tfilm.
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