# Heat Transfer in a Nuclear Fuel Rod with Steel Slabs

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Read MoreIn summary, the problem involves a nuclear fuel surrounded by two steel slabs, with heat being generated in the fuel and removed by a fluid at a certain temperature. The steel and fuel have different thermal conductivities and the question asks for an equation for the temperature distribution in the fuel. The heat equation is used for each material and boundary conditions are set at the edges of the slabs. The heat flux at the interface between the fuel and steel is determined by the temperature gradient and thermal conductivity of both materials. The boundary conditions for the fuel are found by analyzing the behavior of the steel plates. The solution involves solving a differential equation for the temperature in the fuel.
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## Homework Statement

A nuclear fuel of thickness ##2L## has a steel slab to the left and right, each slab of thickness ##b##. Heat generates within the rod at a rate ##\dot{q}## and is removed by a fluid at ##T_{\infty}## (the question doesn't say, but I believe ##T_{\infty}## is temperature of the fluid), which is to the right of the rightmost slab of steel and is convecting by a coefficient ##h##. The other surface (to the left of the leftmost steel slab) is well insulated, and the fuel and steel have thermal conductivities of ##k_f## and ##k_s##, respectively.

Obtain an equation for the temperature distribution ##T(x)## in the nuclear fuel. Express answers in the above variables.

## Homework Equations

the heat equation: $$\frac{\partial T}{\partial t} = \alpha \frac{\partial ^2 T}{\partial x^2} + \dot{q}$$

## The Attempt at a Solution

nothing is said about time, thus we may assume ##\frac{\partial T}{\partial t} = 0##. Additionally, we are only transferring heat in 1-dimension.

Before trying to solve anything, i believe i will have to use the heat equation for each of the 3 materials (the leftmost steel slab ##T_l##, the nuclear fuel in the middle ##T_m##, and the rightmost slab of steel ##T_r##).

First, i'll try prescribing boundary conditions, assuming the center of the fuel is ##x=0##, the end of the far right steel slab is ##x=L+b## and the end of the far left steel slab is ##x=-L-b##.
$$T_l '(-L-b) = 0$$
I'm just not sure how to deal with the other boundary conditions? My professor said something about, at ##x=L+b## we have that ##-k_s T_r'(x) = hA(T_r-T_{\infty})## where I think ##A=b## since we are in 1-dimension. Can someone explain this relation?

Also, what about ##x=L,-L##? Can't we prescribe some kind of flux similarity, such as ##T_l'(-L) = T_m'(-L)## and ##T_m'(L) = T_r'(L)##?

Similarly, can't we also say ##T_l(-L) = T_m(-L)## and ##T_m(L) = T_r(L)##?

Please help! I've been working very hard and think I just need a little push in the correct direction.

Thanks everyone!

joshmccraney said:

nothing is said about time, thus we may assume ##\frac{\partial T}{\partial t} = 0##. Additionally, we are only transferring heat in 1-dimension.

First, i'll try prescribing boundary conditions, assuming the center of the fuel is ##x=0##, the end of the far right steel slab is ##x=L+b## and the end of the far left steel slab is ##x=-L-b##.
$$T_l '(-L-b) = 0$$

OK so far!

I'm just not sure how to deal with the other boundary conditions? My professor said something about, at ##x=L+b## we have that ##-k_s T_r'(x) = hA(T_r-T_{\infty})## where I think ##A=b## since we are in 1-dimension. Can someone explain this relation?

If you have a steady temperature distribution in the system, the heat flux out of the slab must be the same as the internal heat generation. You correctly said there is no heat lost at the ##x=-L-b## boundary.

##hA(T_r-T_{\infty})## looks like Newton's law of cooling, and it is the heat flux out of the steel at ##x = L+b##, where ##A## is the area of the free surface. The question seems to imply this is a "large" slab of material and you can ignore the edge effects, so you might as well consider a piece of the slab with unit area.

Also, what about ##x=L,-L##? Can't we prescribe some kind of flux similarity, such as ##T_l'(-L) = T_m'(-L)## and ##T_m'(L) = T_r'(L)##?

You are right to think about the heat flux either side of the boundary between the fuel and the steel, but the flux depends on the temperature gradient and the thermal conductivity of the two materials.

There is no heat generation in the steel plates. What does that tell you about the heat flux through the steel, and the "shape" of the temperature distribution through the steel, as a function of ##x##?

When you have got a clear idea what happens in the steel plates (and you don't really need differential equations to solve that part of the question) you will have the boundary conditions for the fuel.

You will need to write a differential equation for the temperature in the fuel, and solve it.

Note: I'm not sure what ##\dot q## means in the question. It could be the heat generated per unit volume of the fuel, or the total heat generated in the slab. It doesn't make much difference to the thought process in the question either way.

Last edited:
1 person
joshmccraney,

You did a nice job of analyzing the problem, and AlephZero correctly pointed out that you needed to include the thermal conductivities in matching the heat fluxes at the interfaces.

Regarding the equation ##-k_s T_r'(x) = hA(T_r-T_{\infty})## :

There shouldn't be an A on the right hand side of this equation. The units on both sides of the equation have to match, and they don't with the A present.

Chet

1 person
AlephZero said:
You are right to think about the heat flux either side of the boundary between the fuel and the steel, but the flux depends on the temperature gradient and the thermal conductivity of the two materials.

I'm not too sure what to do here. are you referring to fourier's law of heat transfer ##\vec{q} = -k \nabla T##, or in our 1-dimensional case, ##q = -k \frac{dT}{dx}##? If so, it seems that at each boundary of ##\pm L## we have that ##-k_f \frac{dT_m}{dx} = -k_s \frac{dT_i}{dx}## for the ##i^{th}## steel slab?

AlephZero said:
There is no heat generation in the steel plates. What does that tell you about the heat flux through the steel, and the "shape" of the temperature distribution through the steel, as a function of ##x##?
ahh, you're being clever, if i understand you correct. it would seem as though ##\frac{dT_l}{dx}=\frac{dT_m}{dx}=0## when ##x=-L## since we are in steady state. thus, no heat will transfer out, so ##T## is constant along ##[-L-b,-L]##. is this correct? As for the shape, it seems the temperature will decrease from left to right. do you agree?

AlephZero said:
When you have got a clear idea what happens in the steel plates (and you don't really need differential equations to solve that part of the question) you will have the boundary conditions for the fuel.

so, if the above reasoning is correct, our boundary conditions are: $$T'(-L)=0$$ $$k_f T_m'(L) = k_s T_r'(L)$$ $$-k_s T_r'(L+b) = h(T_r - T_{\infty})$$ but don't we need a B.C. for the temperature profile (not only derivatives)?

the equation we will use is first $$-T_r''(x) = \dot{q}$$ and next is $$-T_m''(x) = \dot{q}$$ and it seems we actually don't need the ##T_l## profile (if the above logic is correct)?

Chestermiller said:
Regarding the equation ##-k_s T_r'(x) = hA(T_r-T_{\infty})## :

There shouldn't be an A on the right hand side of this equation. The units on both sides of the equation have to match, and they don't with the A present.

Chet
good call! of course, what are the units associate with ##h##? i know ##A## has m^2 and ##T## has kelvins. I'm just confused why area is present at all? is it because the rate of cooling is proportional to surface area and the temperature gradient, which in this case is ambient temp minus surface temp? thus, ##hA(T_r-T_{\infty})## takes units Watts per square meter?

thanks! please let me know if this is correct.

thinking about this question more (after alephzero's and chet's comments) do you two think the heat is being generated throughout the entire fluid? if so, isn't temperature constant from ##[-L-b,L]## since the generation is taking place continuously, constantly, and everywhere through ##[-L,L]##?

joshmccraney said:
thinking about this question more (after alephzero's and chet's comments) do you two think the heat is being generated throughout the entire fluid? if so, isn't temperature constant from ##[-L-b,L]## since the generation is taking place continuously, constantly, and everywhere through ##[-L,L]##?
You are correct to say that the temperature from ##[-L-b,L]## is constant, because no heat is flowing through this region at steady state. This is going to be at the highest temperature in your system. So you might as well take the insulated boundary to be at -L; it will make no difference in the calculated temperature profiles to the right of -L.

Chet

thanks for the help guys! i think i have a solution, but can one of you comment on this brief analysis: ##\dot{q} + k_f T''(x) = 0## implies ##\dot{q}## takes units Watts per cubic meter. However, for my energy balance I have that ##\dot{q} = \frac{k_s}{b}A(T_m-T_r)##. how do i get rid of the area term? my professor said we have ##\dot{q}2L = \frac{k_s}{b}(T_m-T_r)## but i don't see how these units work out. how can we divide by area and multiply the left side by length?

nevermind, i see that the q term is my flux. thanks to you both!

## 1. What is heat transfer in a nuclear fuel rod with steel slabs?

Heat transfer is the process of thermal energy being transferred from one object or substance to another. In the case of a nuclear fuel rod with steel slabs, heat is generated by the nuclear reaction in the fuel rod and is transferred to the surrounding steel slabs.

## 2. How does heat transfer occur in a nuclear fuel rod with steel slabs?

Heat transfer in this system occurs through three main mechanisms: conduction, convection, and radiation. Conduction is the transfer of heat through direct contact between the fuel rod and steel slabs. Convection is the transfer of heat through the movement of fluids, such as coolant, around the fuel rod. Radiation is the transfer of heat through electromagnetic waves.

## 3. What factors affect heat transfer in a nuclear fuel rod with steel slabs?

The rate of heat transfer in this system can be affected by several factors, including the temperature difference between the fuel rod and steel slabs, the material properties of the fuel rod and steel slabs, the geometry of the system, and the presence of any insulating materials.

## 4. How is heat transfer in a nuclear fuel rod with steel slabs important?

Heat transfer is a crucial aspect of nuclear reactor design and operation. In a fuel rod with steel slabs, efficient heat transfer is necessary to prevent the fuel rod from overheating and potentially melting, which could lead to a nuclear meltdown. Therefore, understanding and accurately predicting heat transfer in this system is essential for ensuring the safe and efficient operation of a nuclear reactor.

## 5. How is heat transfer in a nuclear fuel rod with steel slabs studied and analyzed?

Scientists use various methods, such as mathematical modeling and experimental studies, to study and analyze heat transfer in a nuclear fuel rod with steel slabs. These methods allow for the prediction and optimization of heat transfer processes in this system, leading to advancements in nuclear reactor design and safety.

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