A Heat Conduction Problem (Final exam on Monday)

In summary, the conversation is about finding the steady-state solution of the heat conduction equation that satisfies given boundary conditions. The approach is to use separation of variables and assume u(x,t)=X(x)T(t). The conversation also discusses the intuition behind the solution for two different problems, one with both ends fixed at certain temperatures and the other with one end insulated. The summary concludes with a good luck message for an upcoming exam.
  • #1
Jamin2112
986
12

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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  • #2
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
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"
 
  • #4
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:
 

What is a heat conduction problem?

A heat conduction problem involves the transfer of heat energy through a medium, typically a solid material. This can occur in various situations, such as when a hot object is placed in contact with a cooler object or when a temperature gradient exists within a material.

What factors affect heat conduction?

The rate of heat conduction is affected by several factors, including the thermal conductivity of the material, the temperature difference between the two ends of the material, and the cross-sectional area and length of the material through which the heat is transferring.

How is heat conduction measured?

The rate of heat conduction is typically measured in units of watts (W) or British thermal units per hour (BTU/h). These units represent the amount of heat energy transferred per unit time.

What is the equation for heat conduction?

The equation for heat conduction is Q = kAΔT / L, where Q is the amount of heat energy transferred, k is the thermal conductivity of the material, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the material.

How is heat conduction used in real-world applications?

Heat conduction is used in a variety of real-world applications, such as in cooking, heating and cooling systems, and thermal insulation. It is also important in fields such as materials science, engineering, and environmental science.

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