Prncples of Food and Boprocess Engneerng (FS 31) Solutons to Example Problems on Heat Transfer 1. We start wth Fourer s law of heat conducton: Q = k A ( T/ x) Rearrangng, we get: Q/A = k ( T/ x) Here, Q/A = 35 W/m, T = 35 C, x = 0.08 m Thus, we get: k = 0.08 W/m K. The wattage of the heater s the same as the rate at whch energy s transferred from the nsde to the outsde, across the not-so-well nsulated wall. Ths rate of energy transfer s gven by: 3. We start off by drawng the system dagram, whch s as follows: We begn by consderng transfer of heat between the nterface of the nsulaton & ppe and the nsde wall of the ppe. For ths, the energy transferred s gven by: Here, Q/L = 0 W/m, k = W/m K, r = 0.04 m, r o = 0.045 m, x = 0.005 m Thus, we get: A lm = 0.59L m Substtutng these values, we get: T = 0.1 C Thus, Temperature at nsde surface of ppe = 6 + 0.1 = 6.1 C We now consder transfer of heat between the nterface of the nsulaton & ppe and the outsde of the nsulaton. Usng the same expresson as earler for the energy transferred, we have: Q/L = 0 W/m, k = 0.0 W/m K, r = 0.045 m, r o = 0.055 m, x = 0.01 m Thus, we get: A lm = 0.97L m Substtutng these values, we get: T = 33.7 C Thus, Temperature at outsde of nsulaton = 6-33.7 = 8.3 C We now consder transfer of heat between the nsde surface of the ppe and the bulk of the product. For ths, the energy transferred s gven by: Q = h A ( T) Here, Q/L = 0 W/m, r = 0.04 m, h = 00 W/m K Thus, we get: A = 0.51L m Substtutng these values, we get: T = 0.4 C
Thus, Bulk temperature of product = 6.1 + 0.4 = 6.5 C We now consder transfer of heat between the outsde surface of the nsulaton and the bulk ar outsde. For ths, the energy transferred s gven by: Q = h A ( T) Here, Q/L = 0 W/m, r = 0.055 m, h = 5 W/m K Thus, we get: A = 0.33L m Substtutng these values, we get: T = 1 C Thus, Bulk temperature of ambent ar = 8.3-1 = 16.3 C 4. The system dagram s as shown below: We start off by consderng convecton between the wall of the heat exchanger and the bulk flud. The energy transferred s gven by: Q = h A ( T) wth A = r L Here, Q/L = 50 W/m, r = 0.03 m, T = 80-75 = 5 C Thus, A = 0.189L m Hence, h = 53 W/m K We then consder transfer of heat across the metal ppe. The energy transferred s gven by: Q = k A lm ( T/ r) Here, Q/L = 50 W/m, A lm = 0.7L (based on r = 0.03 m, r 0 = 0.06 m), r = 0.03 m Thus, T = 0.6 C Hence, T = 75-0.6 = 74.4 C 1 We then consder transfer of heat across the nsulaton. The energy transferred s gven by: Q = k A lm ( T/ r) Here, Q/L = 50 W/m, A lm = 0.5L (based on r = 0.06 m, r 0 = 1.1 m), r = 0.05 m Thus, T = 48.1 C Hence, T = 74.4-48.1 = 6.3 C 5. Ths s a forced convecton problem wth the mean temperature of water beng: (0 + 60)/ = 40 C., L = 10 m Thus, the flow s lamnar. Now, (N Re)(N Pr)(d c/l) = 16.6 Usng equaton 4.64 on page 91 (3 ed: pg. 48), we get: N Nu = 5.3 = hd/k Thus, h = 67 W/m K
th 6. a. Lamnar flow: We need to use equaton 4.64 (4 ed: pg. 91; 3 ed: pg. 48) for ths stuaton. When the velocty of flow s doubled, Reynolds number s the only quantty that s affected n ether equaton. Let us examne equaton 4.64. It can be seen from ths equaton that as velocty s doubled, 0.33 N Re s doubled, whch n turn results n a factor of (= 1.57) ncrease n N Nu. Thus, t can be estmated that doublng the velocty results n approxmately 5.7 % ncrease n the Nusselt number or the convectve heat transfer coeffcent under lamnar flow condtons. th b. Turbulent flow: We need to use equaton 4.67 (4 ed: pg. 91; 3 ed: pg. 49) for ths. 0.8 When the flow velocty s doubled, N Re s doubled and hence N Nu ncreases by (= 1.74). Thus, t can be estmated that doublng the velocty results n approxmately 74 % ncrease n the Nusselt number or the convectve heat transfer coeffcent under turbulent flow condtons. 7. The loss n energy as water flows through the ppe s gven by: Ths s the energy that s beng transferred across the ppe. Thus, Q = U A T = 1,594 W (1) lm lm and () From equatons (1), (), and (3), we get: (3) A = r L, A = r L r = 0.05 m, r = 0.055 m, L = 100 m, A = 3.97 m x = 0.005 m, h = 7 W/m K, k = 10 W/m K o o o lm (ppe) ppe o ppe Ths yelds: Thus, (4) where R o s the outsde radus ncludng the nsulaton. We now determne the nsde heat transfer coeffcent as follows: The bulk flud temperature s: (100 + 70)/ = 85 C
Now, We then use eqn 4.67 on pg. 91 (3 ed: pg. 49) to get: Thus, h = 358 W/m K. Usng ths n equaton (4), we get: Thus, the flow s turbulent. N = 53 = (h D)/k Nu Thus, any combnaton of R o and k ns that satsfes the above equaton wll be suffcent to mnmze the heat loss as per the requrements. For example, R = 0.06 m ( x = 0.005 m), k = 0.1 W/m K o ns ns th 8. Ths s a case of flow over an mmersed object. We need to use eqn 4.69 (4 ed: pg. 9; 3 ed: pg. 49) The flm temperature n ths case s (90 + 10)/ = 50 C. The resultant velocty n ths case s 50 cm/s (= 0.5 m/s) N = 0.71 Thus, we get: N = 10 = hd/k = h (0.008)/0.07 Pr Nu h = 34 W/m K 9. The system dagram for the problem s as shown below: Consderng transfer of heat from the nsde surface of the ppe and the outsde surface of the nsulaton, we get: For ppe, x = 0.0 m, k = 15 W/m K For nsulaton, x = 0.04 m, k = 0.04 W/m K For ppe, r = 0.06 m, r = 0.04 m For nsulaton, r = 0.10 m, r = 0.06 m Ths yelds, (A ) = 0.31 L m and (A ) = 0.49 L m lm ppe o o lm ns
Thus, Q/L = 17.1 W/m a. Consderng heat transfer from the bulk of the flud to the nsde surface of the ppe, we get: Q = h (A ) (90-75) wth A = r L where r = 0.04 m Thus, 17.1 = h ( )(0.04)(15) Ths yelds h = 5 W/m K b. Consderng transfer of heat between the nsde and outsde of the ppe, we get: Q = ka( T/ x) = 15 (0.31 L) (75 - T)/(0.0) Ths yelds, T = 74.9 C c. Consderng transfer of heat between the outsde of the nsulaton and the surroundng ar, we get: Q = h (A ) (40 - T ) wth A = r L where r = 0.10 m o o a o o o Thus, 17.1 = 10 ( )(0.10)(40 - T ) a Ths yelds T = 37.3 C 10. Ths s a free convecton problem wth D = 0.3 m and T flm = (90 + 10)/ = 50 C a N Pr = 0.71 6 th N Ra = (N Gr)(N Pr) = 136.7 x 10 Thus, from Table 4. (4 ed: pg. 98-99; 3 ed: pg. 55-56), a = 0.59, m = 0.5 m Thus, N = a (N x N ) = 64 = hd/k = h(0.3)/0.07 Thus, h = 6 W/m K Nu Gr Pr 11. Ths s a forced convecton problem and hence we begn by computng the Reynolds number Snce Reynolds number s greater than 10,000, the flow s turbulent N = 8.1 (at 15 C) Pr Thus, -6 wth = 308.909 x 10 Pa s w Thus, N Nu = 90 = hd/k = h (0.05)/(0.587) Thus, h = 3405 W/m K 1. The system dagram s as shown below: Ths s a forced convecton problem. So we begn by determnng the Reynolds number. Here, -6 = 1 kg/s, D = 0.054 m, = 658.06 x 10 Pa s (at 40 C) We determned the vscosty of water at the bulk temperature [T bulk = (10 + 70)/ = 40 C]
Thus, N Re = 76,179 Snce the flow s turbulent, we use the followng equaton to determne N Nu: 0.8 0.33 0.14 N Nu = 0.03 (N Re) (N Pr) ( b/ w) -6-6 Here, N Re = 76179, N Pr = 4.3 (at 40 C), b = 658.06 x 10 Pa s, w = 9.38 x 10 Pa s Thus, we get: N Nu = 335 = hd/kf wth D = 0.054 m, k f = 0.633 W/m K Solvng, we get: h = 8349 W/m K 13. The system dagram s as follows: a. Ths s a forced convecton problem. Thus, we begn by determnng the Reynolds number. Here, -6 = 0.1 kg/s, D = 0.0 m, = 9.38 x 10 Pa s (at 95 C) Thus, N Re = 1,784 Snce the flow s turbulent, we use the followng equaton to determne N Nu: 0.8 0.33 0.14 N Nu = 0.03 (N Re) (N Pr) ( b/ w) -6-6 Here, N Re = 1,784, N Pr = 1.84 (at 95 C), b = 9.38 x 10 Pa s, w = 308.909 x 10 Pa s Thus, we get: N Nu = 8.5 = hd/kf wth D = 0.0 m, k f = 0.68 W/m K Solvng, we get: h = 805 W/m K b. Q = ha (95-90) = hoa o(87-13) wth A = rl, A o = rol, r = 0.01 m, r o = 0.015 m, L = 0.9 m Solvng, we get: h = 137 W/m K o 14. In ths problem, all we need to do s determne the logarthmc mean temperature dfference. We note that T lm = ( T 1 - T )/[ln( T 1/ T)] Snce the steam s saturated steam, the temperature of the steam s the same at the nlet and outlet. Based on the pressure of 198.53 kpa, we determne the steam temperature to be 10 C (from steam tables) Thus, T 1 = 10-0 = 100 C, T = 10-70 = 50 C Thus, T = 7.1 C lm
15. Crtcal radus (r c) = k/h = 0.1/5 = 0.0 m Snce the radus of the tube s greater than the crtcal radus, addng any amount of nsulaton wll decrease the heat loss from the ppe 16. 3 U = Q/(A T lm) = [15.34 x 10 ] / [{( ) (0.05)(10)}{50.8}] = 157 W/m K 17. Consderng heat transfer from the hot water to the ambent ar, Q = 500 W, T = 85 C, h o = 0 W/m K A = rl = 1.6 m, A o = rol =.0 m ( x/ka) ppe = T/Q = 5/500 = 0.0 K/W Solvng, we get h = 49 W/m K 18. Snce we have a fnte cylnder, we consder to be a combnaton of an nfnte cylnder and an nfnte slab. For the nfnte cylnder, N B = hd/k = (40)(0.05)/50 Thus, N B = 0.04 For the nfnte slab, N B = hd/k = (40)(0.1)/50 Thus, N B = 0.08 Snce N < 0.1, the lumped parameter analyss can be used. B Here, T = 35 C, T = 0 C, T = 5 C 3 h = 40 W/m K, = 8314 kg/m, c p = 4190 J/kg K r = 0.05 m,l = 0. m 3 A = r L = 0.06 m V = r L = 0.0016 m Substtutng these values, we get: t = 16,10 s (= 4.5 hrs) a 19. Snce one of the dmensons s very large n comparson wth the other two dmensons, the product wll be assumed to nfnte n one drecton. The resultng product can be obtaned by the ntersecton of a product 4 cm thck n one dmenson (and nfnte n the other two drectons) and another product 10 cm thck n one dmenson (and nfnte n the other two dmensons). Thus, the temperature rato (TR) of the fnte object can be obtaned as the product of the TR s for the two nfnte objects. -7 = k/( c ) = 1.3 x 10 m /s,t = 3600 s p Frst slab: D = 0.0 m Thus, N B = hd/k = 4 (thus, k/hd = 0.5) and N Fo = 1. TR = 0.19 (From Hesler chart -- pg. 90)
Second slab: D = 0.05 m Thus, N B = hd/k = 10 (thus, k/hd = 0.1) and N Fo = 0.19 TR = 0.85 (From Hesler chart -- pg. 90) Thus, TR for fnte object = 0.19 x 0.85 = 0.16 = (T a - T) / (T a - T ) Thus, T = 19 C 0. a. The Bot number n ths case s: N B = hd/k = (75)(0.04)/0.4 = 7.5 k/hd = 0.13-7 Thus, we use the Hesler chart. = k/( c p) = 1.35 x 10 m /s TR = 0.17 N = 0.375 = t/d t = 4444 s (= 74 mn) Fo b. We need to use eqn. 4.159 on pg. 34 (3 ed: eqn 4.17 on pg. 84). 3 3 A = 4 r = 0.0 m, V = (4/3) r = 0.00068 m Substtutng these nto eqn. 4.159 (or eqn 4.17 n 3 ed), we get: Thus, t = 937 s (= 15.6 mn) c. The reason for the bg dfference between the answers n parts a and b are because the lumped parameter analyss s not vald. 1. a. A fnte cylnder s obtaned by the ntersecton of an nfnte cylnder and an nfnte slab. For nfnte cylnder: D = 0.03 m N B = hd/k = 1.5 Thus, k/hd = 0.67 For nfnte slab: D = 0.04 m N = hd/k =.0 Thus, k/hd = 0.5-7 Thus, we use the Hesler chart. = k/( c ) = 1.17 x 10 m /s N = t/d = 0.47 for the nfnte cylnder and N = t/d = 0.6 for the nfnte slab Fo p Fo Thus, TR = 0.45 for the nfnte cylnder and TR = 0.75 for nfnte slab Thus, TR for fnte cylnder = (0.45) (0.75) = 0.34 = (T - T) / (T - T ) B a a T = 45.4 C b. Here, the TR for fnte cylnder s gven: (T a - T) / (T a - T ) = 0.083 We know k/hd for the nfnte cylnder; but we do not know ts TR or N Fo Let us guess a value of t. Ths wll enable us to determne the TR for the nfnte cylnder. We then use ths t for the nfnte slab and thus determne the TR for the nfnte slab. We then multply the TR for the two nfnte objects and compare wth 0.083. If t s not equal to 0.083, we attempt another guess value for t and contnue the ITERATIVE procedure tll the product of the two TR s s 0.083.. The Bot number n ths case s: N B = hd/k = (1000)(0.015)/0.4 = 37.5 k/hd = 0.03 We thus use the Hesler chart. A fnte cube of sde 0.03 m s obtaned by the ntersecton of three nfnte slabs, each havng a thckness of 0.03 m 1/3 TR for fnte cube = 0. Thus, TR for nfnte slab = (0.) = 0.6-7 = k/( c p) = 1.17 x 10 m /s For one such nfnte slab: TR = 0.6 Thus, we get: N = 0. Thus, t = 385 s 3. a. The propertes requred for solvng the problem are those of the product b. h s the convectve heat transfer coeffcent for transfer of heat between the surface of the product and water c. Snce ths s an unsteady state problem, we begn by determnng the Bot number N = hd/k = (650)(0.05)/(0.5) = 3.5 B Fo
Snce the Bot number les between 0.1 and 40, we use the Hesler chart The temperature rato for the fnte cube s (85-50)/(85-5) = 0.44 We now note that a fnte cube can be obtaned as the ntersecton of 3 nfnte slabs, each of thckness equal to that of the length of the cube. 3 1/3 Thus, (TR nfnte slab) = TRfnte cube Thus, TR nfnte slab = (0.44) = 0.76 We thus use the Hesler chart for an nfnte slab wth TR = 0.76 and k/hd = 0.03 to obtan a Fourer number of 0. -7 Thus, ( t/d ) = 0. = k/( c p) = 0.5/(950)(3700) = 1.4 x 10 m /s D = 0.05 Thus, we get t = 880 s = 14.7 mn 4. Ths s an unsteady state heat transfer problem and hence we begn by computng Bot number. Snce ths s a fnte cylnder, we consder t to be a combnaton of an nfnte cylnder and an nfnte slab. For nfnte cylnder, N B = hd/k = (5)(0.04)/(10) = 0.0 For nfnte slab, N B = hd/k = (5)(0.05)/(10) = 0.05 th Snce Bot number s less than 0.1, we use the lumped parameter analyss (4 ed: eqn. 4.159 on pg. 34; 3 ed: eqn. 4.17 on pg. 84). Thus, A = rl + r = 0.035 m 3 V = r L = 0.005 m t = 1800 s Solvng, we get: (T - 5)/(90-5) = 0.85 Thus, T = 77.3 C 5. A fnte brck-shaped object can be obtaned by the ntersecton of 3 nfnte slabs. Neglect the convectve resstance mples that k/hd = 0 (h 0; n fact h s very hgh) d = 0.03 m, d = 0.035 m, d = 0.04 m 1 3 Thus, (T - 5) / (-5-5) = 0.048 T = 3.6 C 6. N = (0 x 0.0) / (5) = 0.08. Snce N < 0.1, use the lumped parameter analyss. B B Thus, t = 7 s = 1 mn 7. a. A fnte cylnder s obtaned by the ntersecton of an nfnte cylnder and an nfnte slab. For nfnte cylnder: D = 0.04 m N B = hd/k =.0 Thus, k/hd = 0.5 For nfnte slab: D = 0.05 m N B = hd/k =.5 Thus, k/hd = 0.4-7 Thus, we use the Hesler chart. = k/( c p) = 1.3 x 10 m /s N Fo = t/d = 0.5 for the nfnte cylnder and N Fo = t/d = 0.33 for the nfnte slab Thus, TR = 0.4 for the nfnte cylnder and TR = 0.9 for nfnte slab Thus, TR for fnte cylnder = (0.4) (0.9) = 0.36 = (T - T ) / (T - T ) T = 45. C b. Here, the TR for fnte cylnder s gven: (T - T ) / (T - T ) = (40-0)/(90-0) = 0.9 We know k/hd for the nfnte cylnder; but we do not know ts TR or N. Let us guess a Fo
value of t. Ths wll enable us to determne the TR for the nfnte cylnder. We then use ths t for the nfnte slab and determne the TR for the nfnte slab. We then multply the TR for the two nfnte objects. If ths s not equal to 0.9, we attempt another guess value for t and contnue the ITERATIVE procedure tll the product of the two TR s s 0.9. 8. The system dagram s as follows: Snce ths s a forced convecton problem, we begn by determnng the Reynolds number. Here, = 0.5 kg/s, D = 0.06 m, = 0.001 Pa s (at 85 C) Thus, N Re = 10,610 Snce the flow s turbulent, we use the followng equaton to determne N Nu: 0.8 0.33 0.14 N Nu = 0.03 (N Re) (N Pr) ( b/ w) Here, N Re = 10,610, N Pr = cp /k = 8 Snce we do not know the wall temperature, we assume that the vscosty correcton factor ( b/ w) s neglgble. Thus, we get: N Nu = 76 = hd/kf wth D = 0.06 m, k f = 0.5 W/m K Solvng, we get: h = 633 W/m K The heat loss n the un-nsulated case s gven by: = 0.5 (4000) (10) = 0,000 Thus heat loss n the nsulated case = 10,000 W (Ths means that the ext temperature of the water wll be 85 C) For the nsulated case, Q = 10000 = UA lm Tlm Thus, 1/UA lm = T lm/q = (1/hA ) + ( r/ka lm) + (1/hoA o) Here, T lm = 77.5 C (based on T 1 = 90-10 = 80 C and T = 85-10 = 75 C) A = rl = 3.77 m (based on r = 0.03 m) A o = rol = 4.4 m (based on r o = 0.035 m) A lm = 4.08 m (based on A and A o) Also, h = 633 W/m K, h o = 100 W/m K Thus, 77.5/10000 = [1/(633)(3.77)] + [0.005/(k)(4.08)] + [1/(100)(4.4)] Smplfyng, we get: 198 = (4.08 k)/0.005 Solvng, we get: k = 0.4 W/m K 9. The Bot number n ths stuaton s gven by: N B = hd c/k Here, h = 10 W/m K, d = 0.05 m, k = 3.5 W/m K c
Thus, N B = 0.07 Snce N < 0.1, the lumped parameter analyss can be used. B Here, T a = 150 C, T = 5 C, T = 80 C, h = 10 W/m K, A = 6 (0.05) = 0.015 m = 950 kg/m, c = 3700 J/kg K, V = (0.05) = 15 x 10 m Thus, TR = 0.483, ha/ cpv = 0.000341 s -1 Solvng, we get: t =,134 s 3 3-6 3 p 30. a. The system dagram s as follows: Snce ths s a forced convecton problem, we begn by determnng the Reynolds number. Here, = 0. kg/s, D = 0.04 m, = 0.005 Pa s Thus, N Re = 173 Snce the flow s lamnar, we use the followng equaton to determne N Nu: 0.33 0.14 N Nu = 1.86 [N Re N Pr (d c/l)] ( b/ w) Here, N Re = 173, N Pr = cp /k = 41 Snce we do not know the nsde wall temperature, we assume that the vscosty correcton factor ( b/ w) s neglgble. Thus, we get: N Nu = 10.8 = hd/kf wth D = 0.04 m, k f = 0.5 W/m K Solvng, we get: h = 135 W/m K Consderng heat transfer between the outsde wall and the bulk of the flud, we get: Here, T = 85-80 = 5 C h = 135 W/m K, k = 8 W/m K A = rl, A o = rol, r = 0.0 m, r o = 0.05 m, L = 10 m, x = 0.05-0.0 = 0.005 m Thus, A = 1.6 m, A o = 1.57 m, A lm = (A o - A )/[ln(a o/a )] = 1.41 m Solvng, we get: Q = 791 W
Now, Q = hoa o (T surface - T a) Here, Q = 791 W, A o = 1.57 m, T surface = 80 C, T a = 10 C Solvng, we get: h = 7. W/m K o Note: You can also solve ths problem consderng free convecton at the surface b. Here, r c = k/h o = 0.1/7. = 0.014 m In ths case, r = 0.05 m; Snce r > r, addng any amount of nsulaton wll dec. heat loss o o c 31. a. The overall heat transfer coeffcent can be determned usng the followng equaton: Q = UA T (1) lm In ths case, lm = 0. (400)(70-5) = 54,600 W Snce there are no heat losses, ths amount of energy s the energy lost by hot water. Thus, 54,600 = = 0.4 (400) (90 - T ) Thus, T h(o) = 57.5 C Now, T lm = ( T 1 - T )/[ln( T 1/ T)] Here, T 1 = 90-70 = 0 C, T = 57.5-5 = 5.5 C Thus, T lm = 33.7 C Now, A lm = [ L (r o - r )]/ln (r o/r )] Here, L = 10 m, r o = 0.04 m, r = 0.05 m Substtutng, we get: A = 1.46 m lm Substtutng these values n equaton 1, we get: U = 1,110 W/m K b. When a rapd ntal rate of coolng s desred, a co-current confguraton s preferable snce the drvng force (temperature dfference) s hgher for ths confguraton. 3. When the lumped parameter analyss s vald, we can use the followng equaton to determne the temperature as a functon of tme: h(o) If the ntal and ambent condtons are the same n both cases, the temperature after a gven amount of tme would depend on the value of A/V. 3 For the cube, A = 6D, V = D ; thus, A/V = 6/D 3 For the sphere, A = 4 R, V = (4/3) R ; thus, A/V = 3/R = 6/D Thus, t would take the same amount of tme for the cube and the sphere to heat up (or cool down) to the same temperature.
33. The system dagram s as follows: In oer to determne T, we consder heat transfer between the nterface of the two layers of nsulaton and the outsde ar and wrte down the expresson for heat transfer between these two ponts as follows: Solvng, we get T = 8.7 C 34. Ths s a free convecton problem. Hence we begn by determnng N Gr. Assumng that the entre object s hot and that t s placed wth the 0 cm face on the counter-top, d = 0. m and T = (40 + 4)/ = C c flm N Pr = 0.71 N Ra = (N Gr)(N Pr) = 3.1 x 10 7 th Thus, from Table 4. (page 55 of 3 ed; pg. 99 of 4 ed), a = 0.59, m = 0.5 m Thus, N Nu = a (N Gr x N Pr) = 44.1 = hd c/k = h(0.)/0.05 Thus, h = 5.6 W/m K 35. Ths s a forced convecton problem. So, we begn by computng the Reynolds number. = (4)(0.35)/[( )(0.05)(0.0015)] = 11,883.6 Snce N Re > 10,000, the flow s turbulent and we use the followng equaton to determne h: 0.8 0.33 0.14 N Nu = 0.03 (N Re) (N Pr) ( b/ w) Here, N Re = 11,883.6, N Pr = cp /k = (4100)(0.0015)/(0.55) = 11., b = w = 0.0015 Pa s Thus, we get: N Nu = 9.9 = hd c/kf wth d c = 0.05 m, k f = 0.55 W/m K Solvng, we get: h =,044 W/m K
36. A fnte cylnder s obtaned by the ntersecton of an nfnte cylnder and an nfnte slab. For nfnte cylnder: d c = 0.04 m, k = 0.54 W/m K N B = hd c/k = 1.85 Thus, k/hd c = 0.54 For nfnte slab: d c = 0.05 m, k = 0.54 W/m K N B = hd c/k =.31 Thus, k/hd c = 0.43-7 Thus, we use the Hesler chart. = k/( c p) = 1.54 x 10 m /s t = 90 mn = 5400 s N Fo = t/d c = 0.5 for the nfnte cylnder and N Fo = t/d c = 0.33 for the nfnte slab Thus, TR = 0.4 for the nfnte cylnder and TR = 0.9 for nfnte slab Thus, TR for fnte cylnder = (0.4) (0.9) = 0.36 = (T - T ) / (T - T ) T = 45. C 37. a. Snce the tubular heat exchanger s well-nsulated, all the heat lost by hot water s ganed by the product. Thus, Solvng, we get: The energy transferred from hot water to the product s gven by: Q = 0. (400) (90-70) = 16800 W Also, Q = U A lm T lm (1) T lm = ( T 1 - T )/[ln( T 1/ T )] () Here, T 1 = 70-10 = 60 C, T = 90-60 = 30 C Substtutng these n equaton, we get: T lm = 43.3 C Also, A lm = (A o - A )/[ln(a o/a )] (3) Here, A = rl, A o = rol wth r = 0.013 m, r o = 0.014 m, L = 50 m Thus, A = 4.4 m, A o = 6.6 m Substtutng these n equaton 3, we get: A lm = 5.4 m Substtutng ths n equaton (1), we get: U = 7 W/m K b. Ths s a forced convecton problem. So, we begn be determnng the Reynolds number., L = 0 m Thus, the flow s lamnar. So, we use equaton 4.64 wth the vscosty correcton factor assumed to be 1.0. Thus, equaton 4.64 reduced to: N Nu = 1.86 {(N Re) (N Pr) (d c/l)} 0.33 Here, N Pr = (c p)( )/k = (4000) (0.005)/0.4 = 50 Also, d c = 0.06 m, L = 0 m Thus, N Nu = 6.9 Also, N Nu = hd c/kf Substtutng N = 6.9, d = 0.06 m, k = 0.4 W/m K yelds: h = 106 W/m K Nu c f 38. Ths s an unsteady state heat transfer problem. So, we have to determne the Bot number. However, we do not know the convectve heat transfer coeffcent. We determne that by performng a free convecton analyss.
d c = ½ D wth D = 0.01 m Thus, we get: d c = 0.0157 m We begn by determnng the N as follows: Gr We determned prop. of water at the flm temperature (= avg. of 15 C and 45 C = 30 C) Also, N Pr = 5.4 N Ra = (N Gr)(N Pr) =.91 x 10 6 Substtutng ths n the Nusselt number equaton n Table 4. (pg. 99), we get: N Nu = 4 = hd c/kf wth d c = 0.0157 m and k f = 0.615 W/m K Thus, h = 940 W/m K We now begn the unsteady state heat transfer analyss as follows: The Bot number n ths case s: N B = hd c/k s = (940)(0.005)/0.4 = 11.75 Snce N B > 0.1, we use the Hesler chart for a sphere. 1/N B = 1/11.75 = 0.09 TR = (T a - T)/ (T a - T ) = (45-30)/(45-15) = 0.5 From the Hesler chart, we get: N Fo = 0.15 N Fo = t/d c = 0.15-7 Here, = k/( c p) = 1.5 x 10 m /s and d c = 0.005 m Solvng, we get: t = 30 s