# Now the absolute value of eq 18 14 leads to c 2250

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Unformatted text preview: te values) of | ∆ T | = 20 C°. Now, the absolute value of Eq. 18-14 leads to c = = = = 2250 ≈ 2.3 . _______________________________________________________________________ _ Problem 18-58 ((My solution)) TH = 30°C, L2 = 0.700 L1, k3 = 0.8 k1. ℘= TH = -15°C. L3 = 0.350 L1 T −T T −T T −T dQ dT = − kA = k1 A H 12 = k 2 A 12 23 = k3 A 23 C L L L dt dx 1 2 3 There are two equations and two unknown parameters (T12 and T13). So we can solve these equations. From these two equations, we obtain (a) When k2 = 0.9 k1, T12 = 9.7°C and T23 = -6.1°C. Ak ℘ = 20.3 1 L1 ∆T2 = T12 − T23 = 15.8°C (b) When k2 = 1.1 k1, T12 = 8.3°C and T23 = -5.5°C. Ak ℘ = 21.7 1 L1 ((Mathematica)) ∆T2 = T12 − T23 = 13.8°C rule1 k2 0.9 k1, k3 0.8 k1, TC 15, TH 30, L2 0.7 L1, L3 0.350 L1 k2 0.9 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 13.125 1. T12 0.5625 T23 L1 eq2 k2 A A k1 T12 T23 L2 k3 A T23 TC L3 L1 eq1, eq2 , Simplify . rule1 Simplify 0 26.66...
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## This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton.

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