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) / Δy - kBe(Δx/2) Δz * (T5– T8) / ΔySo the big equation will be Σq= 0 = - kAl(Δy/2) Δz * (T5– T4) / Δx - kBe(Δy/2) Δz * (T5– T4) / Δx - kAl(Δx/2) Δz * (T5– T2) / Δy - kCu(Δx/2) Δz * (T5– T2) / Δy- kCu(Δy) Δz * (T5– T6) / Δx - kCu(Δx/2) Δz * (T5– T8) / Δy - kBe(Δx/2) Δz * (T5– T8) / ΔyB. Here, we need to look up k values Aluminum: 231 W/m-K Beryllium: 126 W/m-K Copper: 379 W/m-K We can reduce the equation removing all Δx and Δy terms since they cancel out in every term. Since Δz = 1 m, it can also be removed as it won’t alter the calculation. This leaves: - kAl/2 * (T5– T2) - kBe/2 * (T5– T4) - kAl/2 * (T5– T2) - kCu/2 * (T5– T2) - kCu* (T5– T6) - kCu/2 * (T5– T8) - kBe/2 * (T5– T4) = 0 T2= 580 K, T4= 556 K, T6= 612 K, and T8= 620 K Plugging in values: -231/2 * (T5– 556) – 126/2 * (T5– 556) – 231/2 (T5-580) – 379/2 (T5-580) - 379 (T5– 612) -379/2 * (T5-620) – 126/2 * (T5-620) = 0 - 1115 T5= -665000 Tm,n = 596 K (if we had assumed all of the materials had the same k, we would have calculated 592 K, so it changes a little). oooT2T4T5oT3oT8AlCuBeoT7oT1oT6oT9
oC A. Find T4and T8. B. Find the total heat transfer rate leaving the right surface of the beam and entering the air. (Recall Newton’s Law of Cooling, and what area the surface temperatures represent) Note: You may assume k = constant, but determine k at 350 K as a representative temperature for this problem. 44888