Heat conduction in nuclear reactor (introductory question)

AI Thread Summary
The discussion focuses on calculating the temperature of the cladding and the maximum temperature within a boiling water nuclear reactor's uranium dioxide rods. The initial approach modeled the heat transfer as a slab, but the calculations did not yield the expected temperature of 328°C. Participants suggested calculating the heat generated per unit length of the rod and the heat carried away by the water, emphasizing the need to equate these values for equilibrium. A hint was provided to adjust the equation to account for the surface area in the heat transfer calculations. The conversation highlights the importance of accurately modeling heat conduction in reactor components.
*Alice*
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Homework Statement



The elements of a boiling water nuclear reactor consist of long cylindrical rods of uranium dioxide (U02) of diameter 8mm surrounded by a thin layer of aluminium cladding. In the reactor core the elements are cooled by boiling water at 285°C with a heat transfer coefficient of 35kW/m^2K. If heat is generated uniformly within the rod at a rate of 760 MW/m^3, calculate the temperature of the cladding and the maximum temperature within the rod. The mean thermal conductivity of U=2 is 2.3 W/m K

Homework Equations



first part: model it as a slab with equation Q= h(T*-285) ?

second part : T-T* = (Q/4k)r^2


The Attempt at a Solution



first part: I just assumed that one can model the situation at the wall as a slab with very thin walls and using the equation above with the values of

h= 35kW/m^2K
and
Q=760 MW/m^3*2*pi*0.004 = 19.1 MW/m^3

does not give the required solution of 328°C. I don't really see what exactly is wrong with this calculation and would therefore appreciate if anyone could give me a hint.

second part: completely fine

thanks a lot
 
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Hint:
First, you have to calculate the rate of heat gererated per unit length of the rod...
Then, calculate the rate of heat carry away by the water, (it depends on the temperature different between the rod and the water and the surface area)
At the equalibrium point, The heat generated is equal to the heat carry away... set them equal to get the \Delta T
 
Last edited:
Thanks a lot - I now have the answer!
 
Last edited:
Q = h * (T-285)
The Q on the LHS is the heat created per unit length,
but h*(T-285) is heat carry away per unit area

You have to multiple something on the RHS to make this equation works... can you tell me what you have missed?
 
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