1 dimensional heat transfer question

In summary, the homework statement is trying to solve for q_cond and q_gen, but it seems to give the wrong answer. I believe to get the right answer, they need to use the first equation they wrote in the relevant equations section.
  • #1
theBEAST
364
0

Homework Statement


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Homework Equations


d/dx(kdT/dx) + S = 0 (for steady state)
q_cond = -kAdT/dx
q_gen = S*Vol

The Attempt at a Solution


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So I tried to equate q_cond to q_gen but it seemed to give me the wrong answer. Does anyone know why? I believe to get the right answer I have to use the first equation I wrote in the relevant equations section.

Does my method not work because S is not uniform? But then it tells me that at x = 0, S = So so it shouldn't matter since we have a boundary condition?
 
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  • #2
The quick reply would be that there is no reason to equate q_cond to q_gen: there are two mechanisms that cause temperature change, namely conduction and heat generation. The sum of the contributions to dT/dt has to cause a steady state, hence the = 0.
 
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  • #3
Could you also check q_gen = S * Vol ? This looks too much as if S is constant all over the place.
 
  • #4
Hmm, I am not too sure about the equations you use for S, but I think you need to do something about the constants, C2 and C3 as well.

I would suggest looking at the boundary conditions. You have already been given an expression for the temperature, which you solved correctly. When T=T2 you get 1 (as you also did).

But the deriviative of the equation also has to be continuous over the boundary. (Note: can you explain why?)

You then have 2 equations with 2 unknowns. Remember that at this boundary x=L! (which you also wrote down).

So you could then express the constants C2 and C3 in terms of L. I think that is what you should start by doing.
EDIT: I solved it now (phew), but the equation you need would be:

$$ \frac{d^2T}{dx^2}+\frac{S}{k}=0 $$

Are you familiar with this expression, or?
 
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  • #5
This is exactly the same equation you wrote down (correctly) in your original post. So you shouldn't be too surprised that it gives you the right answer.
 
  • #6
Chestermiller said:
This is exactly the same equation you wrote down (correctly) in your original post. So you shouldn't be too surprised that it gives you the right answer.

But not the one used in the attached paper :-) but yeah, then you must be familiar with it!
 

Related to 1 dimensional heat transfer question

1. What is 1 dimensional heat transfer?

1 dimensional heat transfer is the process of heat transfer occurring in a single direction, typically from a higher temperature region to a lower temperature region. It is a simplified model used to study heat transfer in systems that are long and narrow, such as a rod or pipe.

2. What are the factors that affect 1 dimensional heat transfer?

The factors that affect 1 dimensional heat transfer include the temperature difference between the two regions, the thermal conductivity of the material, the length and cross-sectional area of the system, and the surface area available for heat transfer.

3. How is 1 dimensional heat transfer calculated?

1 dimensional heat transfer is calculated using the Fourier's Law of Heat Conduction, which states that the rate of heat transfer is directly proportional to the temperature difference, thermal conductivity, and surface area, and inversely proportional to the length of the system.

4. What are the applications of 1 dimensional heat transfer?

1 dimensional heat transfer has various applications in engineering and physics, such as in the design of heating and cooling systems, thermal insulation, and in studying heat transfer in long and narrow objects such as wires or pipes.

5. How does 1 dimensional heat transfer differ from 2 or 3 dimensional heat transfer?

1 dimensional heat transfer only considers the transfer of heat in one direction, while 2 and 3 dimensional heat transfer take into account heat transfer in multiple directions. This makes 1 dimensional heat transfer a simpler model that is often used to approximate more complex systems.

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