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Solutions to HW 5 and 6

Solutions to HW 5 and 6 - Chapter 10 10.1 Basis 1 HF Let A...

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Unformatted text preview: Chapter 10 10.1 Basis: 1 HF. Let A refer to fireolay brick, B to kaolin brick, and C to steel. From Appendixes 10 and thermal conductivities are Heat 1035: q— Thy-"Tc i’¢"+J‘:—‘i3+—Jf~C—-1-E’m k1»! k3 kc- 0.11 1150— 30 300: 21'; 0429 ooo___§ 1x.)..._.,... L7 011 45 011 ‘ (x3)a}r_eq1dv.: 0'298 m 10.2 Pipe diameters (Appendix 3): 139' = 1.315 in. 131’ = 1.049 in. 15.911212) = (1.315 — 1.0451/ 111 1.315 1' 1.049 = 1.171 in. 15.111111191151111) = (5.315 — 1.3151/ in 5.315 / 1.315 = 2.86 in. iificork) = (5.315 — 5.3151/ in 5.315 1' 5.315 = 5.30 in. Thermal conductivities, in Btu/h—ft—°F, Cork: k3 = 0.03 Magnesia: k5 = 0.034 Steel: kc = 26 Thermal resistances, based on 100 ft of pipeL RR = J32—=0.00915 0031:3630! 12)x100 RB = Lamas-47 0.034xx(2.86f12)x100 Re : 0,133/12 20.000014 26x5z'(1.177!' 12)x100 Heat loss: q = ————*—*2fi2:29———"~*— = 2130 Btufh 0.00915 + 0.06547 + 0.000014 Let TA be the temperature at the boundary between cork and magnesia and TB that at the boundary between cork and steel. The total resistance RA + R; + R; = 0.074634. Then T3; = 90 + (0.00915/0.074634} (249-90) = 109.5°F TB= 249 - (0.0000l4/0.074634) (249—901: = 248.9?°E‘ 10.3 Within the shell (r1 < r < r2): q=-k%(4m2) r. EE=EIJT fir q 72 1 1 = 40:01—13) "I": q Values of Y from Eq. (1) and Figure 10.32 are oompared in the following table. Bi 1:11 50-0") “910.3 0.5 0.2 0.761 0.77 0.5 0.5 0.506' 0.52 0.5 1.0 0.256 0.27 1.0 0.2 0.606 0.62 1.0 0.5 0.28? 0.32 1.0 1.0 0.032 0.11 For 31505, the appropriate solution, Eq. (I), gives values quite close to the exact solution. but for Biglfl, Eq. (I) is not valid. -3 10.1? Bf=hrw/k=3100'26;10 =744x10‘3 Since Bi << 0.1,the internal resistance is negligible. By heat balance, mcpdT/dr=hA(Tf —T) 4 3.06,, g- = 0415:3113 - T) 3 m J- 4517' = 3!! it Tf—T rmgrp Tf-T_ l —-3ht Tf—Y; 10 ir'pr:p —3 t: 2.3!0.6x10 !8000 K450! = 5.3 3 3x310 1n Case 2: Normal b. pt. of benzene = 80.1“C AT = 30.1 — 15 = 65.l°C Mi : 65.1(0.025m.013x20) =64 _ 2°C 1:142 T“ = 15 + 64.2 = 79.2%: M0 = m: 0.77%: 1f14.2 Twh = 80.1 — 0.77 = 79.33% Case 3: Steam temperature {Appendix 7) at 64.7r lbs/in} abs. Is 297.5% fiTflverall = 297.5 - 100 : 197.50}? 197.5(1.315f(1.049x130)) ATi = — = 186.8°F IKQSJ Tm;= 100 + 186.8 = 286.8°F ATE. = 197.5(1fl4000) = L38°F 1i98.1 Tm. = 297.5 - 1.38 = 296.12°F 11.3 Let subscript I! refer to aniline, subscript 6 refer to toluene. Assume average temperature of aniline is 175°F and that of toluene is 135°F. From Appendix 15 cm: 0.545 Btu/113°}? 0pc = 0.44 Btu/lb°E‘ (a; From Eq. {11.6) Q = 10000 x 0.545 K (200—150) = 272,500 Btu/h q ”1ch Also, Tab = + Tea = fl+ 100 = 1T2.0°F 3600x044 From Eq. (11.15) AIT = (150— 100) 253200—172) 1n—— 28 = 37.9“F Overall coefficient must be, from Eq. (11.14), U = m = 102.7 Btuxh—ftZ—w 70x319 (b) Cannot cool 10000 1b/hr of aniline to 150°F using parallel flow since T¢,would have to be 1?2.0°F and it could not exceed 150°F even in an infinitely long parallel-flow exchanger. 11.4 Assume countercurrent flow, m = 8600 lbfh C i Find an . From Eq. {11.6}: q = mkx 0.545 K {200—150) = 27.25 mh Also from Eq. (11.6) q = 8600 X 0.44 {T¢,-100} = 27.25 mh From this Tda= 0.0072nn -+ 100 From Eq. {11.15) — 0— 200—4". ATL=———-———5 (50 0) 111—.— 200—21,, 50 — (100 — 0.0072 21:3,) 50 100— 0.0072 3;“ 1n By trial to satisfy equations (A) and (B), mk is found. to be 8172 lbfh. Tch = 158.84°E‘, and ATL = 45.4402. Check: q = 0172 x 27.25 = 222,690 stuxh 0 = £15392: 2'0 Btu/h—ft2-°F as given. 70x45.44 11.5 For CClq, cp = 0.20 cal/g-°C {I 0.20 x 4.1368 = 0.837 J!g—°C (a) n1= 19000 x 1000f3600 - 5277.8 g/s q = mcp (The. - Tub) = 52?7.8 3 0.837(85~40) = 198,788 W For water, as: 13590 x lOOOOJBEOO = 3750 g/s If there is no heat loss, the exit water temperature is Td,= 20 + 198T88f{4.1868 x 3750) = 32.7°C Here 1/U = l/ho + l/hi = 1/1700 + 1/11000 = 0.0006?91 U = 1/0.0006791 = 1473 W/m2-°C .021 = 85 2 32.7 = 52.3%: ATg = 40 — 20 = 20°C ATL= (52.3-20)2’ln{52.3/20] = 33.6°C Area A = 198788/(1473 x 33.6) : 4.02 m2 (b) With parallel flow, ATl = 85 — 20 = 65°C ATZ = 40 - 32.7 7.3°C [I All: {65 _ 7.3)fln(65/7.3) = 26.3°C Factor for increase in area: 33.6/26.3 = 1.28 11.6 From Appendix 3, Di = 1.019m Do 1.315m H x“. = 0.1335: 15;. =1.115 r}: (a) U; = 1840 W/m2—°C Hi = 4800 W/m2—°C km = 45 W/m—°C D. . . x“, _, = 0133x1049 =7.05x10’5 km DL 39.37x45x1.115 From Eq. {11.33) D. A —’ =L—L—7ifix10's =2.65x10"‘ ho DE 1840 4800 ha = %x1049 = 3010 2.653510 1.315 . . . . l D. % registance :Ln OlltSlde film: m[D—‘]inx100=4-8.T% - 0 L in well: 7.05 x 10—5 x U; x 100 = 48.7% inside film: %"—x100 = 38.3% i ...
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