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Temperature dependent heat-conductivity

  • Thread starter Selveste
  • Start date
1. Homework Statement
A cylinder has lenght [itex] L [/itex], inner diameter [itex] R_1 [/itex] and outer diameter [itex] R_2 [/itex]. The temperature on the inner cylinder surface is [itex] T_1 [/itex] and on the outer cylinder surface [itex] T_2 [/itex]. There is no temperature variation along the cylinders lenght-axis. Assume that the heat conductivity [itex] k [/itex] is temperature dependent and given by

[tex] k = aT^{\nu} [/tex]
where [itex] a [/itex] is a constant. Find [itex] T(r), r > 0 [/itex].

2. Homework Equations

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


3. The Attempt at a Solution

The stationary heat flow outwards is
[tex] \dot{Q} = -k\frac{dT}{dr}2\pi rL [/tex]
rearranges to
[tex] dT = -\frac{\dot{Q}}{2\pi kL}\frac{dr}{r} [/tex]
integration from [itex] r_1 [/itex] to [itex] r [/itex] gives
[tex] T - T_1 = ??? [/tex]
Not sure what to do here when [itex] k [/itex] is not constant.
 
19,177
3,787
[tex] k(T)dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r} [/tex]
 
[tex] k(T)dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r} [/tex]
Thank you. So integrating
[tex] aT^{\nu}dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r} [/tex]
from [itex] r_1 [/itex] to [itex] r [/itex] I find
[tex] \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}[/tex]
which by setting [itex] r = r_2 [/itex] gives the heat flow
[tex] \dot{Q} = \frac{2\pi aL}{(\nu +1)(\ln r_1 - \ln r_2)}\left(T_2 - T_1\right)^{\nu +1}\tag{2} [/tex]
is this correct? And how would I now proceed to find [itex] T(r) [/itex]? Thank you.
 
Last edited by a moderator:
19,177
3,787
Thank you. So integrating
[tex] aT^{\nu}dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r} [/tex]
from [itex] r_1 [/itex] to [itex] r [/itex] I find
[tex] \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}[/tex]
which by setting [itex] r = r_2 [/itex] gives the heat flow
[tex] \dot{Q} = \frac{2\pi aL}{(\nu +1)(\ln r_1 - \ln r_2)}\left(T_2 - T_1\right)^{\nu +1}\tag{2} [/tex]
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 [itex] T(r) [/itex]? Thank you.
Just eliminate ##\dot{Q}## between Eqns. 1 and 2.
 

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