# Heat Flow

1. Apr 2, 2006

### bemigh

Hey everyone...
I think im not picking up on something here...
The rate of heat flow across a slab is:
P = (k*A*T)/D
where k is the thermal conductivity of the medium,
A is the cross sectional medium
and T is the temperature difference
and P is power....

Now.. .for my lab, im using to concentric cylinders...

and I have to derive this expression for the heat flow between two concentric cylinders:
P = [(2*pi)*L*k*T] / ln(b/a)
where L is the length of the cylinders, and b is the radius of the outer cyliner, and a is the radius of the inner cylinder...

Now the lab is saying that the mathematics is essentially the same as for the electric field between two cylinders, and the heat flow is analogous to the electric flux... but i cant see how that helps me...

My thinking is this... the area A becomes the area of a cylinder, 2*pi*r*L, and D becomes r... but how can I possibly integrate from b to a, because now my r's will cancel?? any help is appreciated...
Cheers

2. Apr 2, 2006

### gulsen

I'm assuming you have a constant J.
$$J = -\kappa A \nabla T$$
$$J = -\kappa (2 \pi r L) \frac{\partial T}{\partial r}$$
$$\frac{dr}{r} = -\kappa (2 \pi L) dT / J$$
$$\ln(b/A) = -\kappa (2 \pi L) \Delta T / J$$
$$J = -\kappa (2 \pi L) \Delta T / \ln(b/A)$$

3. Apr 3, 2006

### bemigh

thanks a lot for your help,
PS, how do i type that script you use for your reply?

4. Apr 3, 2006

### Astronuc

Staff Emeritus
Last edited: Apr 3, 2006