- #1

womfalcs3

- 62

- 5

Thank you very much.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter womfalcs3
- Start date

- #1

womfalcs3

- 62

- 5

Thank you very much.

- #2

Chestermiller

Mentor

- 22,129

- 5,062

Chet

- #3

womfalcs3

- 62

- 5

The actual boundary conditions of interest specify the heat fluxes; however, the heat fluxes are dependent on the temperatures on the boundaries, which are also unknown.

I just also include T(0,t) and T(L,t) being non-zero and positive. Is it wrong of me to impose those Dirichlet conditions in the solution process?

Thanks.

EDIT: I have tried to resolve it using only the derivative conditions: What irks me is that G(t) disappears, which means the thermal diffusivity disappears. That doesn't give me confidence in the solution.

I just also include T(0,t) and T(L,t) being non-zero and positive. Is it wrong of me to impose those Dirichlet conditions in the solution process?

Thanks.

EDIT: I have tried to resolve it using only the derivative conditions: What irks me is that G(t) disappears, which means the thermal diffusivity disappears. That doesn't give me confidence in the solution.

Last edited:

- #4

Chestermiller

Mentor

- 22,129

- 5,062

Chet

- #5

womfalcs3

- 62

- 5

Chet

Correct. The heat fluxes at the boundaries are dependent on temperatures at the boundaries, and they vary with time. This is an example of what the heat flux would like:

q''(t)=constant*(T(0,t)^4-Toutdoor(t)^4)... where T(0,t) is unknown and Toutdoor(t) is known.

Last edited:

- #6

Chestermiller

Mentor

- 22,129

- 5,062

Correct. The heat fluxes at the boundaries are dependent on temperatures at the boundaries, and they vary with time. This is an example of what the heat flux would like:

q'(t)'=constant*(T(0,t)^4-Tindoor(t)^4)... where T(t) is unknown and Tindoor(t) is known.

OK. Some questions: Is this radiation boundary condition the same on both sides of the slab? How are T

As the problem stands now, it is non-linear in T because of the boundary condition. But, if we could linearize the boundary condition, that would make things much simpler.

Is this an actual real world problem, or is it a homework problem that is asking for some kind of analytic solution procedure? If it is a real world problem, we may be able to make some significant simplifications.

Chet

- #7

womfalcs3

- 62

- 5

Solving for the general solution of the linear governing equation should be straight forward. The BCs just complicate getting expressions for the coefficients. Do you think what I have done in post #3 is wrong?

- #8

Chestermiller

Mentor

- 22,129

- 5,062

Solving for the general solution of the linear governing equation should be straight forward. The BCs just complicate getting expressions for the coefficients. Do you think what I have done in post #3 is wrong?

I haven't looked over your detailed analysis, but the final result in post #3 does not look correct. The temperature profile at time t should involve integrals over the history of the flux variations at the boundary, not just the present values of the fluxes. The boundary conditions are going to be bears to deal with if you are going to try to solve the problem by your present method.

As an experienced heat transfer guy, I can tell you that this is not the way I would ever approach this problem. I would start out by "doping things out," which means getting an idea of the kinds of approximations I can make to simplify the solution (possibly getting an analytic solution), while at the same time not sacrificing accuracy. This would depend on the specific parameters for the system. My next best choice would be to solve the problem numerically, which should be a snap.

Please let me help you dope out the problem. I don't think you will be wasting your time. Please tell me the diffusivity and thickness L of the slab. Also please tell me the time scale for the variations in "inside" and "outside" temperatures: are they going to be varying on a time scale of hours, minutes, or seconds (roughly)? These temperatures are the drivers for the whole problem. My guess is that they are varying on a time scale of hours. How far out in time do you want to describe the temperature variations? Is the initial temperature going to matter on your time scale of interest? What is the amplitude of the "inside" and "outside" temperature variations over time? How different is the initial temperature going to be from the driving temperatures?

Hope you are willing to let me help.

Chet

- #9

womfalcs3

- 62

- 5

I've just solved the problem where I have 1-D space with one end's temperature fixed to zero and the other has convection term equal to the conduction flux. But I seem to have come to the same issue, how does one analytically solve a problem when both ends have an undefined heat flux. It also seems doable if a we have a homogeneous Nuemann condition on one end and a non-zero Robin condition on the other.

The boundary conditions I would resort to would then be:

-k*(dT/dx)=h1*(Tsol-air(t)-T(0,t)) at x=0

and

-k*(dT/dx)=h2*(T(L,t)-Tindoor(t)) at x=L

The coefficients are exogenously defined.

I would just like some direction rather than a solution.

- #10

Chestermiller

Mentor

- 22,129

- 5,062

I've just solved the problem where I have 1-D space with one end's temperature fixed to zero and the other has convection term equal to the conduction flux. But I seem to have come to the same issue, how does one analytically solve a problem when both ends have an undefined heat flux. It also seems doable if a we have a homogeneous Nuemann condition on one end and a non-zero Robin condition on the other.

The boundary conditions I would resort to would then be:

-k*(dT/dx)=h1*(Tsol-air(t)-T(0,t)) at x=0

and

-k*(dT/dx)=h2*(T(L,t)-Tindoor(t)) at x=L

The coefficients are exogenously defined.

I would just like some direction rather than a solution.

OK. I can help you work through the solution to this version of the formulation, which is linearized with respect to the external heat transfer.

The way I would solve this problem would be to use both linear superposition and time convoluation.

The drivers for this problem are Tsol-air(t) and Tindoor(t). Both of these are specified functions of time. Step 1 in what we are going to do is to solve the problem with Tsol-air and Tindoor being constant with time. We will then show how to extend the solution to include time varying Tsol-air and Tindoor. Anyway, if we can't solve the problem for this case, we certainly won't be able to solve it for the time varying case.

We will represent the solution to the problem in step 1 as the linear sum of the solutions to two other problems:

Problem 1: Same differential equation, but with Tsol-air specified, and Tindoor = T

Problem 2: Same differential equation, but with Tindoor specified, and Tsol-air = T

Because of the linearity of the equations, the sum of the solutions to these problems will be the solution for Step 1.

To solve Problem 1, the first thing we will do will be to determine the steady state solution for long times. This will be the solution to the differential equation and boundary conditions with the time derivative set equal to zero. Call this steady state solution T

T* = T

-k*(dT*/dx)=-h1*(T*(0,t)) at x=0

and

-k*(dT*/dx)=h2*(T*(L,t)) at x=L

This problem can be readily solved by T*=F(t)G(x).

See if you can get the analytic solution to this problem and then get back with me. We will then be able to immediately get the solution to Problem 2 from this result. This will then give us the general solution for Step 1.

Chet

Share:

- Last Post

- Replies
- 6

- Views
- 2K

- Replies
- 3

- Views
- 4K

- Replies
- 4

- Views
- 1K

- Replies
- 3

- Views
- 3K

- Replies
- 0

- Views
- 4K

- Replies
- 3

- Views
- 5K

- Replies
- 1

- Views
- 2K

- Replies
- 0

- Views
- 3K

- Replies
- 4

- Views
- 5K

- Replies
- 2

- Views
- 2K