Temperature dependent heat-conductivity

[Selveste]

Homework Statement

A cylinder has length $L$, inner diameter $R_1$ and outer diameter $R_2$. The temperature on the inner cylinder surface is $T_1$ and on the outer cylinder surface $T_2$. There is no temperature variation along the cylinders lenght-axis. Assume that the heat conductivity $k$ is temperature dependent and given by

$$k = aT^{\nu}$$
where $a$ is a constant. Find $T(r), r > 0$.

Homework Equations

Fourier's law
$$\boldsymbol{j} = -k \nabla T$$
Temperature gradient
$$\nabla T = \frac{dT}{dr} \hat{e_r}$$
where $\hat{e_r}$ is a unit vector in radial direction.

The Attempt at a Solution

The stationary heat flow outwards is
$$\dot{Q} = -k\frac{dT}{dr}2\pi rL$$
rearranges to
$$dT = -\frac{\dot{Q}}{2\pi kL}\frac{dr}{r}$$
integration from $r_1$ to $r$ gives
$$T - T_1 = ?$$
Not sure what to do here when $k$ is not constant.

 

[Chestermiller]

$$k(T)dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r}$$

 

[Selveste]

$$k(T)dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r}$$

Thank you. So integrating
$$aT^{\nu}dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r}$$
from $r_1$ to $r$ I find
$$\frac{a}{\nu +1}\left(T^{\nu +1} - T_1^{\nu +1}\right) = -\frac{\dot{Q}}{2\pi L}\left(\ln r - \ln r_1\right) \tag{1}$$
which by setting $r = r_2$ gives the heat flow
$$\dot{Q} = \frac{2\pi aL}{(\nu +1)(\ln r_1 - \ln r_2)}\left(T_2 - T_1\right)^{\nu +1}\tag{2}$$
is this correct? And how would I now proceed to find $T(r)$? Thank you.

 

[Chestermiller]

Thank you. So integrating
$$aT^{\nu}dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r}$$
from $r_1$ to $r$ I find
$$\frac{a}{\nu +1}\left(T^{\nu +1} - T_1^{\nu +1}\right) = -\frac{\dot{Q}}{2\pi L}\left(\ln r - \ln r_1\right) \tag{1}$$
which by setting $r = r_2$ gives the heat flow
$$\dot{Q} = \frac{2\pi aL}{(\nu +1)(\ln r_1 - \ln r_2)}\left(T_2 - T_1\right)^{\nu +1}\tag{2}$$
is this correct?

Yes, but I would write $\ln r - \ln r_1=\ln{(r/r_1)}$. And I would correct the exponents on the T's in Eqn. 2.

And how would I now proceed to find $T(r)$? Thank you.

Just eliminate $\dot{Q}$ between Eqns. 1 and 2.