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# Suggested solutions - 10.5 From Eq(10.24 since a and the...

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Unformatted text preview: 10.5 From Eq. (10.24), since a and the unaccomplished temperature change are the same, t? varies with L3. If the film resistance of the quench liquid is negligible, {a} tT= {2.5/1}2 x 4 = 25 min {b) tT= (5x132 x 4 = 100 min 10.6 (a) Use Eq. (10.22). Quantities needed are: rm = 1.5/12 = 0.125 ft Ta = 700°F TS = 125°F a = 26/[486x0.llj = 0.486 Eb = 0.486t/0.1252 : 31.104t From Eq. (10.22}: 0.00278 0.01667 0.10 (13} (T5 ' 1],}1/(Ts _ Ta) = 0-01- From Eq. (10.25}, 2 tT = jigs—*mWﬁO-Bwﬂl) 937x0A86 = 0.0134 b or 48.2 S 10.? Use Eq. {10.40}. Total quantity of heat lost: QT / A = 2 x £J.7*(~20—5}(12/(arzxrt3.0011)ﬁn5 —2062 mm2 (inherently negative) N H {b} Assume the heated fluid is at {lUU+50)/2 75°F (535°R}. For air at T5°F: k {by interpolation) = 0.01505 Btu/h—ft-l? p = (29 x 492}/{359 x 535} = 0.0742 lb/ft3 c9: 0.25 Btu/1b—°F a = k/pcp = 0.01505/10.0742 x 0.25; = 0.811 ft2/h For water at 75°F, k = 0.349 Btu/h~ft-°F p = 42.24 1b/ft3 cp : 1.0 Btu/lb—°F a = 0.349X{62.26 x 1.0) = 0.0561 ftth From Eq. i10.3?), X? = 3.64t04t)°j For air: a. = 3.64(0.011 x 10/3600)M 0.173 ft or 2.08 in. For water: xp = 3.64(0.00561 x 10/3600)M 0.0144 ft or 0.113 in. Even though air is a poorer conductor of heat than water, it has a much lower heat capacity and the heat “penetrates” much farther than in water for the same exposure time. 4 1 1 1 3.1}. (a) —““+"—+rwa1_}2 + rinauiation 0"4. ho neglect l/hi and rmﬂl From Appendix 11, for fiberglass insulation 0.0467 W/mK ll k _ 0.027 Btu/h—ft2—°F 3 1.13073 J! 0.050»! r = =10? 0.0467 i=i+1.o7=1.12 U‘ 20 U = 0.893 W/m2°c W = 0.893(150r20} = 116 W/m2 (b) doubling the insulation thickness gives i = L+1.0'Ir'x2 = 2.19 U' 20 U I 0.45? Heat loss is reduced by a factor of 0.45?/0.893 0.51. ' 10.12 Let A refer to firebrick, B to steel, C to magnesia. From Appendix 11, kg = 0.98 Btu/hr—ft kg 26 From Appendix 10, As in Example 10.2, kg 0.034 (a) warm but not so hot as it be dangerous to the The outside surface of the magnesia can be touch. Assume its temperature is 120°F. Then the overall AT = 1200 — 120 = 1080°F The resistances are: RA = 4(12 x 0.98) = 0.34 RB: 0.5/(12 x 26} = 0.06 RC: 3/(12 x 0.034) = ?.35 R = 7 696 As in Example 10.2, Chapter 11 11.1 Caae 1: From Appendix J: x" I 0.065 x ﬂ.025£ = 0.00155 m [h - 0.75 a 25.4 - 19.1 mm Di = {L520 1: 25.4 I- 15.T mm In - {19.1 — 15.T]f1n(19.lf15.7] - 177.3 m From Eq. (11.32}, 1 l9.l 0.00165x19.l+ 1 ____._.. q. 12mmwa 121mm 1m = 130.0:101373 - 5320 mmz'uc U1 may be calculated similarly tram Eq. £11.33), Uo= but it 13 were easily found as follows: U1 ' 1.1,:ulilu:..-"l:|ul 5320 x 19.1.I’15J = 54m wrmhc Cage 2: Eu - 0.0035 In 00 = 0.025m 0i = 0.018m .0; = 0.00213m 1 0.025 0.0035x0.025 + l —+— — 0.0181720 45x0.0213 1200 = 14.2 W/m}°C Ui = 14.2 x 0.025/0.018 U0 = = 19.1 W/mec Case 3: DD = 1.315 in D1 = 1.049 in (Appendix 3) 0L = 1 17? in xw = 0.133/12 = 0.0111 ft 1 U0 - 1.315 + 0.0lllx1.315 + l L049xl30 ZﬁxLITT 14000 = 98.1 Btu/h-ft*°3 Uiz-98.l x 1.315/1.049 = 123 Btu/h—ftlmF 11.2 Use Eq. {11.35}. Case 1: AToverall = 105 A 10 = 95°C 95(19.1K15.?x12000) 115325 {AT-l = =51.3°C Wall temperature, c001 side: Twc = + = 61.3OC Wall temperature, warm side. 5T0 = 3211959991 = 35 lac 115325 10h = 105 — 36.1 = 68.9°C ...
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## This homework help was uploaded on 04/17/2008 for the course CHE 353 taught by Professor Poehl during the Fall '07 term at University of Texas.

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Suggested solutions - 10.5 From Eq(10.24 since a and the...

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