sm3_17 - PROBLEM ll KNOWN'I'utal tleet space and vertical...

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Unformatted text preview: PROBLEM ll? KNOWN: 'I'utal tleet' space and vertical distance between tleui‘s let a square. tlat i'net‘huilding. FINI): [a] lixpi‘essiun t‘ei' width et‘htiilding which minimises heat less. [h] Width and number ot‘tleei's which minimixe heat loss liar a prescribed tlimJ‘ space and ilistnnce between tlnnt's. tfni‘i‘espmttling heat hiss. percent heat luss reduction ti'unt 2 Hours. SCI Ilth-NI‘IC': .-’\S§il.ll\ll"l'l{)NS: Negligible heat loss te ground. .-’\N.:\l.\'$|§: [a] To minimise the heat less at. the exterior sui‘t‘ace area. As. must he minimized. From Fig. [a] A5 = w? — JAR-"H = tit-'3 + 4WN ['H [s where Nf = Alf/W3 Heneet At. = w? +4tt-"Ay H 1» / w? = W3 — 4st H {/w The optimum value et‘W eoi‘iespends to. (1W w 2 L'.‘[' '14 _. .1..- X’s-0P = (2AI~H1\_} < The competing effects ent‘W en the areas et‘the i'eet‘and sitleik'alls. and hence the basis for an optimum. is she-wit schematically in Fig. [b]. 7 [b] For Af= 32.768111" antl Hf= -'l- nt‘ - -.].-3 j W. = 2x32..7(~.2<m”><41n) =h4in < Continued PROBLEM 3.1? (Cont) Hence 1 Af 32.?63m“ NT“ 2 j: .—1 Z x W ' {\ 1’14 [11)“ and 9 9 4/12 Than? \4111 n q = LJAQAT = ] W/m‘ - K_ [(14111)‘ I 251" = 30.1300“: ‘ | Min For Nf= 3 W = [ Af.-ijll"'3 = (312.7(wa 1113.3)":2 = [IX [11 ') \ -' “" Th — -' q=1W/n'121|< [12Hm_}2 | 251‘ =512.mmw IEXm '.'-u reduction in q = [5 13mm— 31}?.3{I{}j|.-5 1 2mm = 40‘11'. (UNIMEN'I‘S: Even the minimum heat Jess is excessive and [:0qu be ['CLIUCCLI by reducing LJ. ...
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