Find equilibrium profile T(x) Between Two Rods

[happyparticle]

Homework Statement

Find equilibrium profile T(x) Between Two Rods in contact.
$x = 0 , T= T_1$
$x = L, T=T_2$
For the first rod: $\kappa = \kappa_1$
The second rod: $\kappa = \kappa_2$

Relevant Equations

$\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial x^2}$

Knowing that we are in equilibrium $\frac{\partial}{\partial t} = 0$.

We now have a Laplace's equation $\kappa \frac{\partial^2 T}{\partial x^2} = 0$

I separated the rod in 2 halves.

The solution of this equation is $\kappa_1 \frac{\partial2 T}{\partial x2} = C_1$

Integrating both side we get $\kappa_1 T = C_1 x + C_2$

The first boundary condition: $x = 0 , T = T_1$

We now have
$\kappa_1 T = C_1 x + \kappa_1 T_1$

I'm stuck here. I can't find what is the second condition and I have the same problem with the second half of the rod.

Any help will be appreciated. I'm really stuck.

Thank you

 

[erobz]

Homework Statement: Find equilibrium profile T(x) Between Two Rods in contact.
$x = 0 , T= T_1$
$x = L, T=T_2$
For the first rod: $\kappa = \kappa_1$
The second rod: $\kappa = \kappa_2$
Relevant Equations: $\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial x^2}$

Knowing that we are in equilibrium $\frac{\partial}{\partial t} = 0$.

We now have a Laplace's equation $\kappa \frac{\partial^2 T}{\partial x^2} = 0$

I separated the rod in 2 halves.

The solution of this equation is $\kappa_1 \frac{\partial2 T}{\partial x2} = C_1$

Integrating both side we get $\kappa_1 T = C_1 x + C_2$

The first boundary condition: $x = 0 , T = T_1$

We now have
$\kappa_1 T = C_1 x + \kappa_1 T_1$

I'm stuck here. I can't find what is the second condition and I have the same problem with the second half of the rod.

Any help will be appreciated. I'm really stuck.

Thank you

How about showing a diagram that has the end conditions. Is the $x = 0$ at the meeting point, or one of the ends, or what?

 

[Chestermiller]

What are the lengths of the two rods? Are their diameters the same?

 

[happyparticle]

How about showing a diagram that has the end conditions. Is the $x = 0$ at the meeting point, or one of the ends, or what?

$x = 0 , T = T_1$ is one of the ends. I don't have any diagram.

What are the lengths of the two rods? Are their diameters the same?

Their diameters are the same and the length of each rod is L/2.

 

[Chestermiller]

$x = 0 , T = T_1$ is one of the ends. I don't have any diagram.Their diameters are the same and the length of each rod is L/2.

$$Q=k_{left}A\frac{(T_1-T_m)}{L/2}=k_{right}A\frac{(T_m-T_2)}{L/2}$$where Q is the total rate of heat flow along the rod, A is the cross sectional area, and $T_m$ is the temperature at the junction between the two materials.

 

[happyparticle]

Just to be sure, does it means that the total rate of heat flow just before the junction on the right is equal to the total rate of heat flow just after the second rod. I try to figure out what that equation means. If so, is it always the case?

Secondly, Does it means that my constant $C_1 = Q$ ?

Thank you

 

[Chestermiller]

Just to be sure, does it means that the total rate of heat flow just before the junction on the right is equal to the total rate of heat flow just after the second rod. I try to figure out what that equation means. If so, is it always the case?

Sure. The heat has nowhere else to go.

Secondly, Does it means that my constant $C_1 = Q$ ?

Thank you

Here, Q is a constant.

 

[happyparticle]

Thank you!