How Is Heat Transport in a Rod Influenced by Its Radius and Material Properties?

[Physgeek64]

Homework Statement

show $\rho c_m \frac{\partial T}{\partial t} = \kappa \frac{\partial^2T}{\partial x^2} -\frac{2}{a}R(T)$
where $R(T)=A(T-T_0)$

a) Obtain an expression for T as a function of x for the case of an infinitely long rod whose hot end has temperature $T_m$
b) Show that the heat that can be transported away by a long rod of radius a is proportional to $a^{\frac{3}{2}}$, provided A is independent of a.

Homework Equations

The Attempt at a Solution

so for part a) I got
$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{a}}x}$

Then for b) I thought that the rate of heat transfer will be $2 \pi a R(T)$ but this gives something that is proportional to a not $a^{\frac{3}{2}}$

Many thanks in advance

 

[mjc123]

a) is not right. Is your exponential term dimensionally correct?
b) 2πaR(T) is the rate per unit length. What do you get when you integrate it over x?

 

[Chestermiller]

Homework Statement

show $\rho c_m \frac{\partial T}{\partial t} = \kappa \frac{\partial^2T}{\partial x^2} -\frac{2}{a}R(T)$
where $R(T)=A(T-T_0)$

a) Obtain an expression for T as a function of x for the case of an infinitely long rod whose hot end has temperature $T_m$
b) Show that the heat that can be transported away by a long rod of radius a is proportional to $a^{\frac{3}{2}}$, provided A is independent of a.

Homework Equations

The Attempt at a Solution

so for part a) I got
$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{a}}x}$

Then for b) I thought that the rate of heat transfer will be $2 \pi a R(T)$ but this gives something that is proportional to a not $a^{\frac{3}{2}}$

Many thanks in advance

What is the temperature gradient at x = 0? If you multiply this by the cross sectional area and thermal conductivity, what do you get for the rate of heat flow into the rod?

 

[Physgeek64]

a) is not right. Is your exponential term dimensionally correct?
b) 2πaR(T) is the rate per unit length. What do you get when you integrate it over x?

Sorry, i just typed this wrong my answer is $$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{a\kappa}x}}$$

integrating over x i get the rate of heat loss is $$2\pi \sqrt{\frac{\kappa}{2A}}a^{\frac{3}{2}} (T_m - T_0)(e^{-\sqrt{\frac{2A}{a\kappa}l}}-1)$$ Does this seem right

The question then asks for what l is this approximation valid. Any ideas?

Many thanks

 

[Chestermiller]

I see that you decided not to follow the approach I suggested in post #3. So, I guess I'll answer my own questions.

The corrected equation for the temperature is: $$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{\kappa a}}x}$$
The temperature gradient at location x is $$\frac{dT}{dx}=-(T_m-T_0)\sqrt{\frac{2A}{\kappa a}}e^{-\sqrt{\frac{2A}{\kappa a}}x}$$
So the temperature gradient at x = 0 is: $$\frac{dT}{dx}=-(T_m-T_0)\sqrt{\frac{2A}{\kappa a}}$$
So the heat flux at x = 0 is: $$q(0)=-\kappa\left(\frac{dT}{dx}\right)_{x=0}=(T_m-T_0)\sqrt{\frac{2A\kappa}{ a}}$$
So the rate of heat loss is: $$Q=\pi a^2q(0)=\pi a^2(T_m-T_0)\sqrt{\frac{2A\kappa}{ a}}$$This is proportional to $a^{3/2}$

Your answer in post #4 is not consistent with this.