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Old Sep 17th 2018, 07:00 PM   #1
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Heat conduction problem

Here is the question:

Consider a pool reactor (Fig. 2-33) whose core is constructed from a number of vertical fuel plates of thickness 2L. Initially the system has the uniform temperature T; then assume that the constant nuclear internal energy u"' is uniformly generated in these plates. The heat transfer coefficient between the plates and the coolant is h. The temperature of the coolant remains constant, and the thickness of the plates is small compared with other dimensions. Thus, if the end effects are neglected, the heat transfer may be taken to be one-dimensional. We wish to formulate the unsteady temperature problem of the reactor.

Heat conduction problem-1.jpg

The author starts with the lumped formulation using this control volume:

Heat conduction problem-2.jpg

The lumped first law of thermodynamics applied to this system reduces to:

$\displaystyle \frac{dE}{dt}=-2Aq_n$

Then the author states:

$\displaystyle \frac{dE}{dt}=\rho(A.2L)c\frac{dT}{dt}-(A.2L)u'''$

u''' is the internal energy generation per unit volume.

Inserting the last equation into the first we get the lumped form:

$\displaystyle \rho c L \frac{dE}{dt}=-q_n+u'''L$

My question is: Where did this equation come from:

$\displaystyle \frac{dE}{dt}=\rho(A.2L)c\frac{dT}{dt}-(A.2L)u'''$
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Old Sep 19th 2018, 11:50 AM   #2
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Another way

I still don't understand this part, but I got the same result using another formulation.

Using the control volume surrounding the reactor:

$\displaystyle \frac{dE}{dt}=\dot{Q}$

Net heat transfer:

$\displaystyle \dot{Q}=\dot{Q}_{gen}-\dot{Q}_{out}=2LAu'''-2Aq_n$

Internal energy:

$\displaystyle \frac{dE}{dt}=\frac{dU}{dt}=m\frac{du}{dt}=mc_v \frac{dT}{dt}=\rho (2LA) c\frac{dT}{dt}$

Thus:

$\displaystyle \rho (2LA) c\frac{dT}{dt}=2LAu'''-2Aq_n$

$\displaystyle \rho L c\frac{dT}{dt}=Lu'''-q_n$
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