\u0394y kBe \u0394x2 \u0394z T 5 T8 \u0394y So the big equation will be \u03a3q 0 kAl\u0394y2 \u0394z T 5 T 4 \u0394x k

Δy kbe δx2 δz t 5 t8 δy so the big equation will

This preview shows page 3 - 5 out of 11 pages.

) / Δy - k Be (Δx/2) Δz * (T 5 – T 8 ) / Δy So the big equation will be Σq = 0 = - k Al (Δy/2) Δz * (T 5 – T 4 ) / Δx - k Be (Δy/2) Δz * (T 5 – T 4 ) / Δx - k Al (Δx/2) Δz * (T 5 – T 2 ) / Δy - k Cu (Δx/2) Δz * (T 5 – T 2 ) / Δy - k Cu (Δy) Δz * (T 5 – T 6 ) / Δx - k Cu (Δx/2) Δz * (T 5 – T 8 ) / Δy - k Be (Δx/2) Δz * (T 5 – T 8 ) / Δy B. 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 Δ y terms since they cancel out in every term. Since Δ z = 1 m, it can also be removed as it won’t alter the calculation. This leaves: - k Al /2 * (T 5 – T 2 ) - k Be /2 * (T 5 – T 4 ) - k Al /2 * (T 5 – T 2 ) - k Cu /2 * (T 5 – T 2 ) - k Cu * (T 5 – T 6 ) - k Cu /2 * (T 5 – T 8 ) - k Be /2 * (T 5 – T 4 ) = 0 T 2 = 580 K, T 4 = 556 K, T 6 = 612 K, and T 8 = 620 K Plugging in values: - 231/2 * (T 5 – 556) – 126 /2 * (T 5 – 556) – 231/2 (T 5 -580) – 379/2 (T 5 -580) - 379 (T 5 – 612) - 379/2 * (T 5 -620) – 126/2 * (T 5 -620) = 0 - 1115 T 5 = -665000 T m,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). o o o T 2 T 4 T 5 o T 3 o T 8 Al Cu Be o T 7 o T 1 o T 6 o T 9
Image of page 3
o C 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. 4 4 8 8 8
Image of page 4
Image of page 5

You've reached the end of your free preview.

Want to read all 11 pages?

  • Fall '10
  • Brazel

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture