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.