sm3_14 - PROBLEM 3.14 KNOWN Composite wall of a house with...

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Unformatted text preview: PROBLEM 3.14 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature. SCHEMATIC: Hoary lass blanket" (28kg fm’), to, P; a a- k VXEansﬁu/ﬂk 3 . ,"'.---r. _ Eln=273+55in6§ﬁ 051L421}, T i: D I T F = 7 ‘ f- ﬁfﬁ "" 206 =10m-l‘—"l"l.5=mmrl*—Fl-Lsa 0‘” 2 5413"“ )12 24h. mm ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 24h period), (2) Negligible contact resistance. PROPERTIES: Tobie A—j', T e 300 K: Fiberglass blanket (28 ng’mB)1 kb = 0.038 Wr’m-K; Plywood, k5 = 0.12 WEm-K: Plasterboard, kp = 0.1? WJm-K. 24h Too- T ANALYSIS: The heat loss may be approximated as Q = I %dt where 0 101' 1 [ 1 0.01m 0.1m 0.02m 1 ] —+ + + + 209m2 30 mez _ K 0.1-;Ir Wilm‘ K 0.038 Wfln- K 0.12 Win]: K 60 Wr'mz . K Rtot : 0.01454 10W. Hence the heat rate is 1 um 22: 24h 2;: Q: I 293— 273+55in—t dt+ I 293— 2?3+llsin—t dt 24 24 RTOI 0 12 12 24 Q = éssﬂﬂzows [ﬂ]cosﬂ] +[20t+1l[ﬂ]cosﬂ] 12 }K - h K 2:»? 24 III 221' 24 Q=68.8{[240+%{—1—1)]+[4SO—240+%(1+DEW-h Q = 68.8 {480-38.2+84.03} W -h Q=36.]8 kW - h=l .302x 108]. < COMMENTS: From knowledge ofthe fuel cost1 the total daily heating bill could be determined. For example1 at a cost of 0.10\$ka-h, the heating bill would be \$3.62r’day. ...
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