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.