# Rule1 simplify 0 266667 1 t12 277778 t23 eq3 solve

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Unformatted text preview: 67 1. T12 2.77778 T23 eq3 Solve . rule1 L2 0 T12, T23 T12 9.68652, T23 6.11285 T12 T23 . eq3 1 15.7994 P k1 A TH T12 . eq3 L1 1 . rule1 20.3135 A k1 L1 rule2 k2 1.1 k1, k3 0.8 k1, TC 15, TH 30, L2 0.7 L1, L3 0.350 L1 k2 1.1 k1, k3 0.8 k1, TC 15, TH 30, L2 0.7 L1, L3 0.35 L1 eq1 k1 A A k1 TH T12 k2 A L1 T12 T23 L2 11.6667 1. T12 0.611111 T23 L1 eq2 k2 A A k1 T12 T23 L2 k3 A T23 TC L3 21.8182 1. T12 2.45455 T23 L1 eq3 Solve eq1, eq2 , T12, T23 T12 8.30137, T23 5.50685 T12 T23 . eq3 1 13.8082 P k1 A TH T12 L1 21.6986 A k1 L1 . eq3 1 . rule2 . rule2 Simplify 0 . rule2 0 Simplify ((WileyPlus)) 58. (a) As in Sample Problem 18-6, we take the rate of conductive heat transfer through each layer to be the same. Thus, the rate of heat transfer across the entire wall Pw is equal to the rate across layer 2 (P2 ). Using Eq. 18-37 and canceling out the common factor of area A, we obtain = ⇒ = which leads to ∆ T2 = 15.8 °C. (b) We expect (and this is supported by the result in the next part) that greater conductivity should mean a large...
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