chap3assign13-15 - PROBLEM 3.13 KNOWN Composite wall of a...

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Unformatted text preview: PROBLEM 3.13 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: (3) Ex ression for thermal resistance of house wall, Rm; (b) Total heat loss, q(W); (0) Effect on heat 055 due to increase in outside heat transfer convection coefficient, ho; and (d) Controlling resistance for heat loss from house. SCHEMATIC: Plasrer board- kP Fiberglass blankcfggkg/makb Plywood siding, ks ASSUMPTIONS: (l) One~dimensional conduction, (2) Steady—state conditions, (3) Negligible contact resistance. PROPERTIES: Table A-3, (T : (Ti + TO )I 2 z (20 —15)° C/2=2.5°C = 300K): Fiberglass blanket, 28 kg/m3, kb = 0.038 W/m-K; Plywood siding, kS = 0.12 W/m-K; Plasterboard, kp = 0.17 W/m-K. ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from Eq. 3.18. L L Rtot :;+_p_+__I:b_+ S + 1 - hiA kpA kbA ksA hOA (b) The total heat loss through the house wall is C1 = ATI'Rtot : (Ti ‘To)/Rtot~ < Substituting numerical values, find 1 0.01m 010m Rtot _ T—fiw“ ___.___—+ +—.___ 30W/m2 16653501112 0.17W/m- K T 350m2 0.038W/n1~K><350m2 . m + ———— + w—ww 0.12W/m . K><350m2 60W/rn2 -K><350m2 Rtot = [952+16.s+752+47.6+4.76]x10“5 °C/w = 831x10“5 °C/w The heat loss is then, q=[20—(-15)]° C/831x10'5 °C/W=4.21 kW. < (c) If h0 changes from 60 to 300 W/mZ-K, R0 = 1/h0A changes from 4.76 X 10-5 °CIW to 0.95 X 10-5 °C/W. This reduces Rm to 826 x 10-5 °C/W, which is a 05% decrease and hence a 0.5% increase in q. (d) From the expression for Rm: in part (b), note that the insulation resistance, Lb/kbA, is 752/830 = 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. PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (2.5m x 6.5m, 10 studs each 2.5m high). FIND: Wall thermal resistance. SCHEMATIC: .. l'“—“ __ __9_-€1~’>Lm_::’l Hardwood Siding (A) m E— _ Insu/afion 40min,” _ LA Glass fibec I Hardwood (3) paper faced (D) /me=Lc (281:9/7713) ————————————— —- Gypsum (C) ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of composite depends only on x (surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resistance. PROPERTIES: Table A-3 (T : 300K): Hardwood siding, kA = 0.094 W/m-K; Hardwood, kB : 0.16 W/rn‘K; Gypsum, kg = 0.17 W/m-K; Insulation (glass fiber paper faced, 28 kg/rn3), kD = 0.038 W/m-K. ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single unit (enclosed by dashed lines) of the wall is L [k A M,“ A“ talks As (LA lkAAA) =——00083L~——=0.0524 KfW 0.094 W/rn - K (0.65mx 2.5m) (LB/kBAB)=——0—‘13m—m= 8.125 W 6.16 W/m-K (0.04mx2.5m) (LD/kDAD ) = A = 2.243 WW 0038 W/m-K(0.61m>< 2.5m) (LC IkCAC ) = $ = 0.0434 KfW. 0.17 W/m-K(0.65m><2.5m) The equivalent resistance of the core is Req =(1/RB +1/RD)~1=(1/8.125+1/2.243)_121.758 W and the total unit resistance is Rtot,1: RA + Req + RC = 1.854 KIW. With 10 such units in parallel, the total wall resistance is Rtot =(10X1/Rt0[,1)h1 20.1854 KfW. < COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value of Rtot will differ. ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at University of Wisconsin.

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chap3assign13-15 - PROBLEM 3.13 KNOWN Composite wall of a...

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