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There is uniform internal heat generation at $\dot{q}=5\times10^{7} \frac{W}{m^{3}}$ is occurring in a cylindrical nuclear reactor fuel rod of 50 mm diameter and under steady state conditions the temperature is distribution of the rod is $T(r)=800-4.167\times10^{7}r^{2}$. The fuel rod has a thermal conductivity of $k=30 \frac{W}{mK}$. What is the rate of heat transfer per unit length of the rod at r=0 and r = 25mm (the surface).

I did this two different ways. The first way is correct but and the second isn't so I need to understand what isn't correct about it.

So to get the rate of heat transfer at r=0 and at the surface I just used Fourier's Law $$q'=-k(2\pi r)\frac{\partial T}{\partial r}$$

I found the derivative of my temperature distribution and then plugged in the number to get....

$$q'(0)=0 \frac{W}{m}$$ $$q'(.025)=9,8182.6 \frac{W}{m}$$

Now when I looked at this question, I realized that I didn't use the internal heat generation of the rod. That made me begin to think that I had to use an energy balance on the cylinder as a whole. $$\dot{E}_{in}-\dot{E}_{out}+\dot{E}_{g}=\dot{E}_{st}$$

I know that $\dot{E}_{in}=0$ and that $\dot{E}_{out}$ is due to conduction but I am not really sure what $\dot{E}_{st}$ would be. I assumed that it would be zero. Is this true? If I were to set it equal to 0 then I wind up with... $$\dot{E}_{out}=q'(.025)=\dot{q}(\pi r^{2})=9,8174 \frac{W}{m}$$

This value does not equal to what I previously calculated above. Can someone tell me why this isnt the same? I assume that the energy storage is nonzero in this case but I dont understand why

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That would appear to contradict the claim that $T(r)$ is a steady-state solution. Of course, the difference appears in the fourth significant figure, so perhaps you are mean to neglect it. –  dmckee Jan 25 '12 at 2:36
    
@dmckee when it says "steady-state", does that mean that there is or is not energy storage? and if it doesnt explicitly say steady state, how do i determine if there is energy storage or not? –  Greg Harrington Jan 25 '12 at 3:02
    
It means the temperature distribution should not change as a function of time. The state of the system is steady, see? From that you deduce that the integrated heat flow across some closed surface must equal the generation in the enclosed volume. For a real stead-state situation your two calculations should have been equivalent. Which they are to about seven parts in $10^5$. Maybe that's good enough. –  dmckee Jan 25 '12 at 3:25
    
The equation for steady state is $-k \nabla^2 T = \dot{q}$. If $T=a-br^2$ this leads to $4kb=\dot{q}$. So, there is a typo in $\dot{q}$ or $T$. The value of $b$ temperature is approximate. It should be $4.1666\ldots\times 10^{\ldots}$. The difference of the answers is of the same order. There is no problem. –  Maksim Zholudev Jan 26 '12 at 9:32

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