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< COMMENTS: (1) The temperature at the inner surface can be found from an energy balance on the i-node using the value found for L. ,i i ,o i rad i cv,o cd cv,i T T T T q 0 T 298.3 C R R R \u221e \u221e\u2212 \u2212 \u2032\u2032+ + = = ° \u2032\u2032 \u2032\u2032 \u2032\u2032+ It follows that Ti is close to T\u221e,i since the wall represents the dominant resistance of the system. (2) Verify that 2iq 50 W / m\u2032\u2032 = and 2oq 150 W / m .\u2032\u2032 = Is the overall energy balance on the system satisfied? PROBLEM 3.12 KNOWN: Configurations of exterior wall. Inner and outer surface conditions. FIND: Heating load for each of the three cases. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation effects. PROPERTIES: (T = 300 K): Table A.3: plaster board, kp = 0.17 W/m\u22c5K; urethane, kf = 0.026 W/m\u22c5K; wood, kw = 0.12 W/m\u22c5K; glass, kg = 1.4 W/m\u22c5K. Table A.4: air, ka = 0.0263 W/m\u22c5K. ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by the total thermal resistance. For the composite wall of unit surface area, A = 1 m2, ( ) ( ) ( ) ( ) ( ) ,i ,o i p p f f w w o T T q 1 h L k L k L k 1 h A \u221e \u221e \u2212 = + + + +\uf8ee \uf8f9\uf8f0 \uf8fb ( ) ( ) 2 2 20 C 15 C q 0.2 0.059 1.92 0.083 0.067 m K W 1m \u2212 \u2212 = + + + + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb $ $ 35 C q 15.0 W 2.33 K W = = $ < (b) For the single pane of glass, ( ) ( ) ( ) ,i ,o i g g o T T q 1 h L k 1 h A \u221e \u221e \u2212 = + +\uf8ee \uf8f9\uf8f0 \uf8fb ( ) 2 2 35 C 35 C q 130.3 W 0.269 K W0.2 0.002 0.067 m K W 1m = = = + + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb $ $ < (c) For the double pane window, ( ) ( ) ( ) ( ) ,i ,o i g g a a o T T q 1 h 2 L k L k 1 h A \u221e \u221e \u2212 = + + +\uf8ee \uf8f9\uf8f0 \uf8fb ( ) 2 2 35 C 35 C q 75.9 W 0.461K W0.2 0.004 0.190 0.067 m K W 1m = = = + + + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb $ $ < COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and the dominant contribution to its total thermal resistance (82%) is associated with the foam insulation. Even with double pane construction, heat loss through the window is significantly larger than that for the composite wall. PROBLEM 3.13 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: (a) Expression for thermal resistance of house wall, Rtot; (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: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Negligible contact resistance. PROPERTIES: Table A-3, ( ) ( )( )i oT T T / 2 20 15 C/2=2.5 C 300K := + = \u2212 \u2248$ $ Fiberglass blanket, 28 kg/m3, kb = 0.038 W/m\u22c5K; Plywood siding, ks = 0.12 W/m\u22c5K; Plasterboard, kp = 0.17 W/m\u22c5K. ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from Eq. 3.18. p b s tot i p b s o L L L1 1R . h A k A k A k A h A = + + + + < (b) The total heat loss through the house wall is ( )tot i o totq T/R T T / R .= \u2206 = \u2212 Substituting numerical values, find [ ] tot 2 2 2 2 2 2 2 5 5 tot 1 0.01m 0.10mR 30W/m K 350m 0.17W/m K 350m 0.038W/m K 350m 0.02m 1 0.12W/m K 350m 60W/m K 350m R 9.52 16.8 752 47.6 4.76 10 C/W 831 10 C/W\u2212 \u2212 = + + \u22c5 × \u22c5 × \u22c5 × + + \u22c5 × \u22c5 × = + + + + × = ×$ $ The heat loss is then, ( ) -5q= 20- -15 C/831 10 C/W=4.21 kW.\uf8ee \uf8f9 ×\uf8f0 \uf8fb$ $ < (c) If ho changes from 60 to 300 W/m2\u22c5K, Ro = 1/hoA changes from 4.76 × 10-5 °C/W to 0.95 × 10-5 °C/W. This reduces Rtot to 826 × 10 -5 °C/W, which is a 0.5% decrease and hence a 0.5% increase in q. (d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is 752/830 \u2248 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 3.14 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: ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 24h period), (2) Negligible contact resistance. PROPERTIES: Table A-3, T \u2248 300 K: Fiberglass blanket (28 kg/m3), kb = 0.038 W/m\u22c5K; Plywood, ks = 0.12 W/m\u22c5K; Plasterboard, kp = 0.17 W/m\u22c5K. ANALYSIS: The heat loss may be approximated as 24h ,i ,o tot0 T TQ dt where R \u221e \u221e\u2212 = \u222b p b s tot i p b s o tot 2 2 2 tot L L L1 1 1 R A h k k k h 1 1 0.01m 0.1m 0.02m 1 R 0.17 W/m K 0.038 W/m K 0.12 W/m K200m 30 W/m K 60 W/m K R 0.01454 K/W. = + + + + = + + + + \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 = \uf8ee \uf8f9\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb Hence the heat rate is 12h 24h tot 0 12 1 2 2Q 293 273 5 sin t dt 293 273 11 sin t dt R 24 24 \u3c0 \u3c0\uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9\uf8f4 \uf8f4\uf8ee \uf8f9 \uf8ee \uf8f9 = \u2212 + + \u2212 +\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f4 \uf8f4\uf8f3 \uf8fe \u222b \u222b 12 24 0 12 W 24 2 t 24 2 tQ 68.8 20t+5 cos 20t+11 cos K h K 2 24 2 24 \u3c0 \u3c0 \u3c0 \u3c0 \uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9\uf8f4 \uf8f4\uf8ee \uf8f9 \uf8ee \uf8f9 = + \u22c5\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f4 \uf8f4\uf8f3 \uf8fe ( ) ( )60 132Q 68.8 240 1 1 480 240 1 1 W h \u3c0 \u3c0 \uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9 = + \u2212 \u2212 + \u2212 + + \u22c5\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f3 \uf8fe { }Q 68.8 480-38.2+84.03 W h= \u22c5 8Q=36.18 kW h=1.302 10 J.\u22c5 × < COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be determined. For example, at a cost of 0.10$/kW\u22c5h, the heating bill would be $3.62/day. PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (2.5m × 6.5m, 10 studs each 2.5m high). FIND: Wall thermal resistance. SCHEMATIC: 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 \u2248 300K): Hardwood siding, kA = 0.094 W/m\u22c5K; Hardwood, kB = 0.16 W/m\u22c5K; Gypsum, kC = 0.17 W/m\u22c5K; Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038 W/m\u22c5K. ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single unit (enclosed by dashed lines) of the wall is ( ) ( )A A A 0.008mL / k A 0.0524 K/W 0.094 W/m K 0.65m 2.5m = = \u22c5 × ( ) ( )B B B 0.13mL / k A 8.125 K/W 0.16 W/m K 0.04m 2.5m = = \u22c5 × ( ) ( )D D D 0.13mL /k A 2.243 K/W 0.038 W/m K 0.61m 2.5m = = \u22c5 × ( ) ( )C C C 0.012mL / k A 0.0434 K/W. 0.17 W/m K 0.65m 2.5m = = \u22c5 × The equivalent resistance of the core is ( ) ( )1 1eq B DR 1/ R 1/ R 1/ 8.125 1/ 2.243 1.758 K/W\u2212 \u2212= + = + = and the total unit resistance is tot,1 A eq CR R R R 1.854 K/W.= + + = With 10 such units in parallel, the total wall resistance is ( ) 1tot tot,1R 10 1/ R 0.1854 K/W.\u2212= × = < COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value of Rtot will differ. PROBLEM 3.16 KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator compartment. FIND: Coefficient of performance (COP). SCHEMATIC: ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartment completely sealed from ambient air. ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heat transfer between the interior of the refrigerator and the ambient air. Applying an energy balance to a control surface about the refrigerator, it follows from Eq. 1.11a that, at any instant, g outE E 0\u2212 =\ufffd \ufffd Hence, elec outq q 0\u2212 = where ( )out ,i ,o tq T T R\u221e \u221e= \u2212 . It follows that ( ),i ,o t elec T T 90 25 C R 3.25 C/W q 20 W \u221e \u221e \u2212 \u2212 = = = $ $ For Case (b), heat transfer from the ambient air to