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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—1E’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 = Lamas47
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} (24990) = 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 1:11 500") “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 TfT_ 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 parallelflow 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= (52.320)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=48.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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 Fall '07
 Poehl

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