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sm3_15 - PROBLEM 3 IS KNOWN Diménsiumaml materialo...

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Unformatted text preview: PROBLEM 3. IS KNOWN: Diménsiumaml materialo uwoeifitcd il-‘ilh a cqmpu'silc wait {.3 Sm-x 155m, ll] studspauh 2.5m high]. FIND: Wall lhermal rosisumce. SCH EMATIC: - Hardmd siding (Al Insulflfion T”. 40min” =L“ Hardwood {5) 32m»: = 1.: Gypsum (C) Gigs: Fibeq a. 'r faced {.0} lags/m!) ASSUB'IPTIONS: l I l Slrafiyskflccundilmns; [El 'l'mnpcmhlm qrcompusite depend: only 011 1% {surfaces normal tu'x aré: isolhcnnal). {3) Gunsmnl pl'npénits. (4.) ngllglblc contact resiolanw PROPERTIES: mm .4—3 (T o 300K}: Hardwood siding. kA = 0.094 WIm-K: Hardwood; 1. k3 = (l. | 6. W.H'n1-K;Gypsum, l-r.(- = 0. I7 Wim‘lsl'i Insulat'lfin {glass fibEI' paper faced. 28 kg‘m' l: I‘D = ern- K. ANALYSIS: Usmg lhc. isolhcnnal surface assumption. the lllcrmal Certlll assocmled wall! a Single [mil (enclosed by {lilshfid lint-s} ul'the- wall is ' LjfiAi f A LnllhAA - I Lo-fkoAs [Mlllrfil-n _ L .s'k A =-—=n.o:~24 KLFW { A A' ‘5‘) mm Wrm-Kwosmxlsm} - mam. L rk'A =—=B.]25 Kr'W { B 'B B} U.|{+'Wrm--K{0.U4mx25n1} x . {LDEkDAD} =“J—J", 4.243 KIW' 0.!le Wim‘ K'[ll.fi l 111 x 2.5m) - {Lul2m' . L Ik ;~A= :—=U.0434 Krw. { C i L) on Wu'n-Kl'll.65|11x2.5m) ' The requivalem resistance 0f the-core l5 1th = (HRH +HRD)" = (1..r3-.125+|r2.243)" :mss low and lhemmluml reststame jg Rich] = RA- + Req +-R'C =1.8-5_4-K1w; With lll such 1111115 In parallel, the total wnll resistance 15 Rm =(I'021rRmL1'f' =n.12¢54 mw. ’4 COMMENTS: 11' surfaces parallel lo the heat flflw Lilrcctlmi an: assumed adulbmlc, llu: Illcrmal' circuu and the value of Rm will'dif'fcr ...
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