sm3_19 - PROBLEM 3.19 KNOWN: Representative dimensions and...

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Unformatted text preview: PROBLEM 3.19 KNOWN: Representative dimensions and thermal conduethities tor the layers ot‘ t'ire-t'tghter's protective clothing. a turnout coat. FIND: [a] 'l'hermal cireuit representing the turnout eoat: tahuiate thermal resistances ol‘tlte layers and processes; and [h ] For a prescrihed radiant heat tlux on the lire-side surt'aee and temperature ol‘ li .l‘wll"’(.' at the inner surface. ealeulate the tire-side surface temperature. To. SC ' II lil‘t'l ..-’\'I' I (i: Shell Meleture herrler Thermal liner ("1' Lflmm)I 0.8 1.0 0551.0 3.5 RIMr'n'I-K} 0.047 0.012 0.038 .-t5."il|l\'ll"l'l()NS: [llfi'teady—state conditions I: 2] ()ne—dimensiotml conduction through the layers. [3} Heat is transt'erred hy conduetion and radiation exchange aeross the stagnant air gaps. [3] (fionstant properties. I’RUI‘ICR'I‘IICS: Edwin-Lit _.-'\ir[-—'l'.-'l} Ii. 1 atJn]: l-tah it MEN? Wfl-nt-K CLl .-\N.-\I.YSIS: [a] 'l'lte thermal cireuit is shown with laheis for the temperatures and thermal resistanees. - radah R'Iedm The conduction thermal resistances have the term Rgd = L.-' l\' while the radiation thermal resistances across the air gaps have the form 1 1 rad —w 1rad slat—I73 W The linearized radiation coefi‘ieient follows from Eqs. 1.3 and l}! with t'. = l whene Tim: represents the a werage temperature ot‘the surfaces comprising the gap hm = (7(T] + T3 )[T]2 + T33 s 4511,? For the radiation thermal resistances tabulated below. we used Tm2 = 47-1} K. lContinued PROBLEM 3.19 (Cont) Shell Air gap Barrier Air gap Liner Iium’ (s) [a-b) [nib] [e-d] [tl] {tot} shim-1K it] {llllflll use?) {LUJSXS angst! {LUU‘JJI -- iijffldljinz-K.\s] __ {LU-4246 -- mamas __ __ itgwl‘iifi-as-f] -- {Luna} -- antral -- __ Ritual -- -- —— -- —— U. am From the thermal eireuit. the resistance across the gap for the conduction and radiation processes is l _ l I l Rgap RELl i'ad and the total thermal resistance ofthe turn coat is H F! l\' n' I! IV Rtot — Rcd.s ' Rgap.a—h l Rcdanb l Rgapic—d l Reditl 1 [b] It‘ the heat flux through the eoat is “.25 \K-‘fl-cmi the fire-side surtaee temperature T0 can he ealculated from the rate equation written in terms ofthe overall thermal resistance q“ = lTo Ti lot . '1 , . a - a '~ '> . To = (aft-“C I 0.23 W sem‘ /(1H‘em 11]] xtLlU43 nt‘ - K; W T0 = amt-c ('()l\'ll\llflN'l'S: [l ] Front the tabulated results. note that the thermal resistance ofthe moisture barrier [Jnh] is nearly 3 times larger than that for the shell or air gap layers. and 45 tithes larger than the thermal liner layer. [.1] The air gap conduction and radiation iesistances were calculated based upon the average temperature otllm K. This value was determined by setting Tm L. = l To + Tit-3 and solving the equation set using HH'with kim- = led“. [Tm L.]. ...
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.

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sm3_19 - PROBLEM 3.19 KNOWN: Representative dimensions and...

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