Temperature dependent heat-conductivity

  • Thread starter Thread starter Selveste
  • Start date Start date
  • Tags Tags
    Temperature
AI Thread Summary
The discussion focuses on solving a heat conduction problem involving a cylinder with temperature-dependent thermal conductivity. The heat flow equation is derived using Fourier's law, leading to an expression for the temperature gradient. Integration yields a relationship between temperature and radius, resulting in an equation that relates the heat flow to the temperature difference across the cylinder. Participants confirm the correctness of the equations but suggest minor adjustments for clarity. To find T(r), it is advised to eliminate the heat flow term between the derived equations.
Selveste
Messages
7
Reaction score
0

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.
 
Physics news on Phys.org
k(T)dT = -\frac{\dot{Q}}{2\pi L}\frac{dr}{r}
 
Chestermiller said:
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.
 
Last edited by a moderator:
Selveste said:
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.
 
  • Like
Likes Selveste
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top