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

) / 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

o C
A. Find T
4
and T
8
.
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