# sm3_13 - PROBLEM 3.13 KNOWN Composite wall ofa house with...

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Unformatted text preview: PROBLEM 3.13 KNOWN: Composite wall ofa house with prescribed convection processes at inner and outer surfaces. FIND: (a) Expression for thermal resistance of house wall, Rm: (b) Total heat loss, q(W),‘ (c) Effect on heat loss due to increase in outside heat transfer convection coefﬁcient, ho: and (d) Controlling resistance for heat loss from house. SCHEMATIC: Iriéergfass blankcfaskgjm’lkb aka-350ml P; d‘ 'ch' k h,=30W/m1-K yw‘ao .5: n9. 3 1! f 1%: soWImz-K , 7;= -15'c ASSUMPTIONS: (I) One-dimensional conduction, (2) Steady-state conditions, (3) Negligible contact reSJ stance. PROPERTIES: TableA—j',(T=(Ti+To)i2=[20—15)°Ci2=2.5°Cm300K): Fiberglass 'i blanket, 28 kgfm' , kb = 0.038 “7me; Plywood siding, k5 = 0.12 me-K; Plasterboard, kp = 0. I? me-K. ANALYSIS: (a) The expression for the total thermal resistance ofthe house wall follows from Eq. 3.18. L Riot=;+—p+i+i+;' < hiA kPA kbA ksA hoA (b) The total heat loss through the house wall is cl = AT’lRtot = [Ti ‘Tolll Rtot- Substituting numerical values, find 1 0.01m 0.10m Rtot = + + 30am2 - 5633mm2 0.17me- K>i350m2 0.038me- Kx350m2 . m + —+— 0.lzwmn-szsom2 sowxngxgjomz R101=[9.52+l6.8+T52+4T.6+4.?6]><10—5°CM=331X10—5 DOW The heat loss is then, q=[20—(—15)]°cx831x10'5 °cxw=4.21 kW. < (c) If he changes from 60 to 300 Wimz-K, R0 = lr‘hoA changes from 4.76 x 10-5 “’wa to 0.95 x 10—5 °C!W. This reduces Rm to 826 x 10—5 °CHW, which is a 0.6% decrease and hence a 0.6% increase in q. (d) From the expression for Rm in part (b), note that the insulation resistance, LbfkbA, is 752;”830 a 90% of the total resistance. Hence, this material layer controls the resistance of the wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an increase in the wind velocity has a negligible effect on the heat loss. ...
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