A Heat Conduction Problem (Final exam on Monday)

AI Thread Summary
The discussion focuses on finding steady-state solutions to heat conduction problems using separation of variables. For the first problem, the steady-state temperature distribution is determined to be v(x) = x/5 + 10, based on the boundary conditions u(0,t)=10 and u(50,t)=40. In the third problem, with an insulated end and the other end held at 0 degrees, the solution is v(x) = 0, indicating that the temperature throughout the rod will be zero. The participants emphasize the importance of understanding the physical implications of boundary conditions in solving these problems. Overall, the discussion prepares participants for potential exam questions on heat conduction.
Jamin2112
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Homework Statement



I need to go into this test with great aplomb.


In each of Problems 1 through 8 find the steady-state solution of the heat conduction equation a2uxx=ut that satisfies the given boundary conditions.

1. u(0,t)=10, u(50,t)=40

...


3. ux(0,t)=0, u(L,t)=0

Homework Equations



Will be using separation of variables; so assume u(x,t)=X(x)T(t)

The Attempt at a Solution



After a long time---that is, as t approaches ∞---I anticipate that a steady state temperature distribution v(x) will be reached.

Then u(x,t) will just be a2v''(x)=v'(t). Since v is not a function of t, v'(t) = 0 and v''(x)=0.

v(x) must be a 1st degree polynomial: v(x) = Ax + B.

Notice that the initial conditions to problem one imply that

X(0)T(t)=10, X(50)T(t)=40.

I don't want T(t) to be something trivial, so

X(0)=10, X(50)=40

----> 10= 0 + B ---> B = 10

40=A*50 + 10 ----> A = (40-30) / 50 = 1/5

-----> v(x) = x/5 + 10

That's the steady state heat distribution, which is all the problem asked for.

Now problem 3. Notice the sub x.

Hmmmm... need to put the pieces of this puzzle together.

ux(0,t)=0 is saying that the rod is insulated at x=0.

So X'(0)=0.

The other initial condition says X(50)=40.

Where do I go from here?
 
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Well, as you concluded above, you have v''(x) = 0. This gives:

v(x) = Ax + B

0 = X'(0) = v'(0) = A

so v(x) = B. Now v(L) = 0 = B so

v(x) = 0x + 0 = 0.

Doesn't that agree with your intuition if you insulate one end and hold the other at 0?
 
LCKurtz said:
Well, as you concluded above, you have v''(x) = 0. This gives:

v(x) = Ax + B

0 = X'(0) = v'(0) = A

so v(x) = B. Now v(L) = 0 = B so

v(x) = 0x + 0 = 0.

Doesn't that agree with your intuition if you insulate one end and hold the other at 0?

In other words, the temperature will be zero throughout the tube because only 0-degree air can come in?

If a question like this comes up on the final tomorrow, I'll just write: "Steady-state solution is zero because 0-degree air comes in one end"
 
Jamin2112 said:
In other words, the temperature will be zero throughout the tube because only 0-degree air can come in?

If a question like this comes up on the final tomorrow, I'll just write: "Steady-state solution is zero because 0-degree air comes in one end"

It could be a solid metal rod insulated laterally and on one end with the other end in ice water. :smile:

Good luck on your exam. Just be ready to solve the problem completely if you are asked to do so. :cry:
 
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