Heat flux through a material with variable heat conductance

[shivamchouhan5077]

Homework Statement

Can anyone explain how temperature at any point on the rod (or any material in general) varies with distance when conductivity is variable. This question came to my mind from my physics test last week, I got the following question in it.

*A rod of length $l$ and cross-sectional area $A$ has a variable conductivity given by $K = aT$, where $a$ is a positive constant and $T$ is temperature in kelvin. Two ends of the rod are maintained at temperature $T_1$ and $T_2$ $(T_1 > T_2)$. Heat current flowing through the rod will be*:

The answer given in the answer key was $$\frac{Aa\left(T_1^2-T_2^2\right)}{2l}$$ I could not solve the question during the test. But now the test is over and I want to find what mistakes I made.So as I was analyzing the question, trying to solve this on my own at home, I spent hours but somehow my answer didn't match with the answer key.

Relevant Equations

$$I_H = \frac{T_1-T_2}{R_{total}}$$

My first thought was that as the heat current $I_H$ is going to be constant at any cross section of the rod, by the equation of continuity, as the area of cross section is constant throughout the rod.
So using: (where $R_{total}$ is the total thermal resistance of the rod) $$I_H = \frac{T_1-T_2}{R_{total}}$$
If I can find net thermal resistance of the rod, I can solve the question.
We know that thermal resistance is given by:$$R = \frac{l}{KA}$$
For a small thermal resistance of an element of rod of length $\mathrm dx$ at a distance x from $T_1$ $$\mathrm dR = \frac{\mathrm dx}{\left(aT\right)A}$$
Now we need to find a relation between $T$ and $x$.
By using the concept of temperature gradient, We know that,$$T_1-T = \frac{x}{l}(T_1-T_2)$$
Therefore, $$T = T_1- \frac{x}{l}(T_1-T_2)$$
On put the value of $T$, $$\int_0^R dR = \frac{1}{aA}\int_0^l \frac{1}{T_1-\frac{x}{l}(T_1-T_2)} dx$$
On simplification, $$R = \frac{l}{aA(T_1-T_2)} ln\left(\frac{T_1}{T_2}\right)$$

On put $R$ in the equation $1$, $$I_H = \frac{aA(T_1-T_2)^2}{l \times ln\left(\frac{T_1}{T_2}\right)}$$

So this was my answer, but as you know my answer is somehow wrong according of the answer key. Although I have now understood that the mistake in my solution was that temperature gradient is not constant throughout the rod. But still, I don't the concept about how can we find temperature at any point on the rod (or any material in general)

 

[Chestermiller]

Homework Statement: Can anyone explain how temperature at any point on the rod (or any material in general) varies with distance when conductivity is variable. This question came to my mind from my physics test last week, I got the following question in it.

*A rod of length $l$ and cross-sectional area $A$ has a variable conductivity given by $K = aT$, where $a$ is a positive constant and $T$ is temperature in kelvin. Two ends of the rod are maintained at temperature $T_1$ and $T_2$ $(T_1 > T_2)$. Heat current flowing through the rod will be*:

The answer given in the answer key was $$\frac{Aa\left(T_1^2-T_2^2\right)}{2l}$$ I could not solve the question during the test. But now the test is over and I want to find what mistakes I made.So as I was analyzing the question, trying to solve this on my own at home, I spent hours but somehow my answer didn't match with the answer key.
Relevant Equations: $$I_H = \frac{T_1-T_2}{R_{total}}$$

My first thought was that as the heat current $I_H$ is going to be constant at any cross section of the rod, by the equation of continuity, as the area of cross section is constant throughout the rod.
So using: (where $R_{total}$ is the total thermal resistance of the rod) $$I_H = \frac{T_1-T_2}{R_{total}}$$
If I can find net thermal resistance of the rod, I can solve the question.
We know that thermal resistance is given by:$$R = \frac{l}{KA}$$
For a small thermal resistance of an element of rod of length $\mathrm dx$ at a distance x from $T_1$ $$\mathrm dR = \frac{\mathrm dx}{\left(aT\right)A}$$
Now we need to find a relation between $T$ and $x$.
By using the concept of temperature gradient, We know that,$$T_1-T = \frac{x}{l}(T_1-T_2)$$

This assumes that temperature is a linear function of x, which, for this material is incorrect.

 

[erobz]

I think you can start with the (reduced) steady state heat equation:

$$ \frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) = 0 $$

Work forward from there? With regards to the partial derivatives $k$ is a function of temperature $T$, which is a function of $x$. I think you have to keep that in mind.