Homework Help: A Heat Conduction Problem (Final exam on Monday!)

1. Jun 5, 2010

Jamin2112

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Jun 6, 2010

LCKurtz

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?

3. Jun 6, 2010

Jamin2112

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"

4. Jun 6, 2010

LCKurtz

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

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