Heat conduction in nuclear reactor (introductory question)

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Homework Help Overview

The discussion revolves around heat conduction in a boiling water nuclear reactor, specifically focusing on the thermal behavior of uranium dioxide rods and their aluminum cladding under certain operational conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore modeling the heat transfer situation, questioning the appropriateness of using a slab model and the equations applied. There is an attempt to calculate the heat generated and the heat carried away, with some participants expressing confusion about the calculations leading to the expected temperature results.

Discussion Status

The discussion includes attempts to clarify the heat generation and transfer equations, with hints provided to guide the original poster towards understanding the relationship between heat generation and heat removal. Some participants have expressed satisfaction with their progress, while others are still seeking clarification on specific calculations.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the thermal dynamics without providing direct solutions.

*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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