Order to create a temperature profile.

[mherna48]

Hi everyone,

This is a problem I need to solve in order to create a temperature profile. I have never encountered anything like this, and don't know where to start. Any suggestions?

A * \frac{\partial}{\partial x}* \left[ \kappa(T(x)) * \frac{\partial T(x)}{\partial x}\right] = 0

where
x : Position
A : Cross sectional area of the rod
T(x) : Temperature as a function of position
\kappa ( T(x) ) : the coefficient of thermal conductivity as a function of Temperature


I'm trying to solve for T(x).

 

[AiRAVATA]

Asterisks are multiplications?

 

[mherna48]

Yes, not transforms. That would be insane. =)

 

[coomast]

Hello mherna48,

Since A is a constant it is divided away with regards to the right hand side. We are left with a derivative equal to 0, giving thus a constant as solution. This means we have (ordinary derivatives because only x is the independent variable and only x):

\kappa(T) \cdot \frac{dT}{dx}=\alpha

In which \alpha the first integration constant. Can you proceed from here?

coomast

 

[mherna48]

Hello mherna48,

Since A is a constant it is divided away with regards to the right hand side. We are left with a derivative equal to 0, giving thus a constant as solution. This means we have (ordinary derivatives because only x is the independent variable and only x):

\kappa(T) \cdot \frac{dT}{dx}=\alpha

In which \alpha the first integration constant. Can you proceed from here?

coomast

Thanks for the help coomast.

I get it to be :

T(x) = \frac{\alpha}{\kappa(T(x))} \cdot x + C

Is that right?

Can I multiply both sides by \kappa(T(x)) and square root the right side? Or is that not possible because of the constant?

 

[coomast]

Mmmmm, I get the following:

\alpha \cdot x + \beta = \int \kappa(T) \cdot dT

In which \alpha and \beta are the integration constants. You can further calculate the integral once you know the dependency of the conductivity with temperature.

Does this clearify things?

coomast

 

[HallsofIvy]

I am puzzled by this "\kappa(T(x))" What is the reason for the first set of parentheses? Is \kappa a function, so this is \kappa of T(x) or is \kappa simply a constant, so this is \kappa times T (which I would write simply as \kappa T(x)).

In either case, you can't just tread "\kappa(T(x)) as a constant as you seem to be trying to do.

If it is \kappa times T(x), then \kappa T \frac{dT}{dx}= \alpha can be written as
T dT= \frac{\alpha}{\kappa} dx
and, integrating,
\frac{1}{2}T^2= \frac{\alpha}{\kappa}x+ C

If \kappa is a function of T, then
\kappa(T)dT= \alpha dx
and, integrating,
\int \kappa(T)dT= \alpha x+ C
essentially what coomast gave.

How you would solve that for T depends heavily on what function kappa is.

 

[mherna48]

I am puzzled by this "\kappa(T(x))" What is the reason for the first set of parentheses? Is \kappa a function, so this is \kappa of T(x) or is \kappa simply a constant, so this is \kappa times T (which I would write simply as \kappa T(x)).

In either case, you can't just tread "\kappa(T(x)) as a constant as you seem to be trying to do.

If it is \kappa times T(x), then \kappa T \frac{dT}{dx}= \alpha can be written as
T dT= \frac{\alpha}{\kappa} dx
and, integrating,
\frac{1}{2}T^2= \frac{\alpha}{\kappa}x+ C

If \kappa is a function of T, then
\kappa(T)dT= \alpha dx
and, integrating,
\int \kappa(T)dT= \alpha x+ C
essentially what coomast gave.

How you would solve that for T depends heavily on what function kappa is.

Thanks for the help guys.

\kappa(T(x)) is read as Thermal Conductivity as a function of Temperature, and Temperature is a function of position.

And yes, I see where I goofed on my calculation. I made a weird mistake and I don't know what I was thinking there. How would you integrate the right side of the equation, or in your case left side:
\int \kappa(T(x)) \cdot dT

Is that even possible?

 

[HallsofIvy]

So what you are saying is that \kappa(T(x)) is a function of x. Obviously, how you would integrate that depends on exactly what function that is!

 

[mherna48]

Hmm. I figured it wouldn't be easy. Maybe I can have Matlab solve it numerically. Thanks for the help everyone.