Heat conduction problem

[rfabdo89]

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.

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

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

dE/dt=−2Aqn

Then the author states:

dE/dt=ρ(A.2L)cdT/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:

ρcLdE/dt=−qn+u′′′L

My question is: Where did this equation come from:

dE/dt=ρ(A.2L)cdT/dt−(A.2L)u′′′

Here is an image:

 

[racookpe1978]

The problem, even as stated, makes no sense: If the temperature of the coolant remains constant (somehow, the energy from the core is removed in some other heat exchanger) , and the reactor power/volume of core plate remains constant (it is not constant in a real reactor by the way), then is the writer trying to develop the equations for the temperature of the center of the "average core plate" over time?

 

[racookpe1978]

Regardless, is this a homework assignment? It is too unrealistic to be a real world problem.

 

[rfabdo89]

It's an exercise solved by the author.
The book is Conduction Heat Transfer - Arpaci.

Actually, the author developed three formulations: lumped, integral and differential.

I agree that this problem is not real, however, the intention of the author is to explore the approximation with Kantorovich polynomial.

 

[georgeverghese]

The heat leaving the fuel plates consists of 2 components
a) The heat generated by nuclear fission which is -2L.A.u'''
b) At time t, any change in the heat content of the fuel plates, which is given by + 2L.A.ρ.c.dT/dt, where dT is the temperature change from the centre of the fuel plate to the outside surface. The author has assumed c and ρ are invariant over this dT.

 

[rfabdo89]

I think i had an insight after your answer, georgeverghese.

The variation of the energy is the same as internal energy. dE/dt=dU/dt.

Using the control volume we have: dU/dt= −2Aqn.

In other hand, we have the definition of Internal energy of multi-component systems:

dU=(∂U/∂T)dT+(∂U/∂V)dV + sum of µdN. This last therm is the Gibbs free energy.

dU=(∂U/∂T)dT+(∂U/∂V)dV + sum of µdN

For a solid body:

dU=m.cv.dT + sum of µdN

dU=2L.A.ρ.cv.dT + sum of µdN

My question is: The Gibbs free energy is negative for this reaction?

 

[georgeverghese]

Sorry there is an error in my previous reply: Correction is
b) At time t, any change in the heat content of the fuel plates, which is given by + 2L.A.ρ.c.dT/dt, where dT is the temperature change from the centre of the fuel plate to the outside surface over time dt. The author has assumed c and ρ are invariant over this dT.
The author also assumes that there is no temperature gradient from the core of the fuel plate to the external surface.. The classical laws of thermodynamics dont apply for nuclear reactions.