[thermodynamics] spherical shell problem

In summary: I am still not sure what you did...So let's do it again.T(r) = C/r + C' with C = \frac{ab}{a-b} \DeltaT and C' an unknown to be determined.So we have a boundary condition for T(r) at r = a:T(a) = \frac{ab}{a-b} \DeltaT /a + C' = T_2so we find C' = \frac{a T_2 - b T_1}{a-b}and in the same way, at r = b, we should haveT(b) = \frac{ab}{a-b} \DeltaT /b + C' = T
  • #1
a_lawson_2k
39
0

Homework Statement



spherical shell with inner and outer radii a and b, and with temperatures of the inner and outer surfaces [tex]T_2[/tex] and [tex]T_1[/tex], the thermal conductivity of the shell is k, derive an equation for the total heat current through the shell

Homework Equations



[tex]H=\frac{dQ}{dt}=kA\frac{T_H-T_C}{L}[/tex]

The Attempt at a Solution



I am not sure how to approach the 'A' element of the above equation, as it varies and the book didn't give a hint one way or another how to approach it. Barring that,

[tex]H=kA\frac{\Delta t}{b-a}[/tex]

The book said [tex]H=k(4\pi*ab)\frac{\Delta t}{b-a}[/tex] so I can't be too far off. They have a similar problem with a cylinder I'd like to solve, so hopefully it will shed light on both of them.
 
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  • #2
[tex]H=\frac{dQ}{dt}=kA\frac{dT}{\dr}[/tex]

Then make A = 4*pi*r^2 and integrate from r1 to r2?

Doesn't seem to give me what I want
 
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  • #3
Such an integration would produce a volume, not an area.
 
  • #4
a_lawson_2k said:

Homework Statement



spherical shell with inner and outer radii a and b, and with temperatures of the inner and outer surfaces [tex]T_2[/tex] and [tex]T_1[/tex], the thermal conductivity of the shell is k, derive an equation for the total heat current through the shell

Homework Equations



[tex]H=\frac{dQ}{dt}=kA\frac{T_H-T_C}{L}[/tex]

This is true only if there is a prismatic section (that is, if A is constant between C and H, over length L). You have the problem that A is not constant from r1 to r2.
However, you can use spherical symmetry, and you can use the fact that the heat flux will be the same at all r.

So you should try to find the temperature dependence of r: T(r), with boundary conditions that T(r1) = T_2 and T(r2) = T_1. The above equation is then to be written in a differential form:

H = constant = - k A(r) dT/dr which gives you a (simple) differential equation for T(r), if you fill in the area A(r) = 4 pi r^2.
 
  • #5
[tex]\int_{a}^{b} \frac{-H}{4k\pi r^2} dr= \int_{T_1}^{T_2} dT[/tex]
[tex]H = \frac{4abk\pi \Delta T}{a-b}[/tex]

That equation wasn't even hinted at in the book; should I move to a more rigorous book if I am attempting these problems?
 
  • #6
a_lawson_2k said:
That equation wasn't even hinted at in the book; should I move to a more rigorous book if I am attempting these problems?

Different books have different purposes ; some are conceptually clear, others are encyclopedic, some are highly rigorous, still others give you good problem solving skills. So you should look at different books to get a good overall appreciation of a subject.

Concerning pure problem solving (and certainly NOT any conceptual depth), do you know about the excellent Schaum's outline series ? There is one in about any subject (including one on heat transport).
 
  • #7
Haven't heard of them, but I'll definitely look into it. Thanks for the lead.

As for the temperature variation T(R), I tried finding it by integrating the result from the previous problem H(r) divided by [tex]-k*4\pi r^2[/tex], but it didn't reconcile with the book's result. It said [tex]\frac{T_1 b(r-a)+T_2 a(b-r)}{r(b-a)}[/tex], but I kept getting a different result when I integrated [tex]\frac{H}{-4\pi r^2 k}[/tex] from a to R (a=<R<=b)...
 
  • #8
a_lawson_2k said:
Haven't heard of them, but I'll definitely look into it. Thanks for the lead.

They are really worth it: they're pretty cheap, easily available (on amazon for instance) and very efficient to learn to solve problems in an efficient way.
(and no, I have no stock in them :-)

As for the temperature variation T(R), I tried finding it by integrating the result from the previous problem H(r) divided by [tex]-k*4\pi r^2[/tex], but it didn't reconcile with the book's result. It said [tex]\frac{T_1 b(r-a)+T_2 a(b-r)}{r(b-a)}[/tex], but I kept getting a different result when I integrated [tex]\frac{H}{-4\pi r^2 k}[/tex] from a to R (a=<R<=b)...

This should come out right...(I mean, the approach is correct).

Let's check: we have
C/r^2 = - dT/dr, so after integration, T is of the form T(r) = C/r + C'
Note that [tex]C = \frac{ab}{a-b} \DeltaT[/tex] in here.

Now, you can check that the expression of your book does comply to this T(r) form (check the coefficient of 1/r, it is C all right, plus a constant).

In order to check the constant C', we only have to verify that the expression of the book has the right value at r = a (and should then also have the right expression at r = b, because that's how we fixed C - and hence H - in the first place).
Fill it in: it is all right.

So this should come out all right...
 
  • #9
[tex]\int_{a}{R}-\frac{\frac{4abk\pi (T_1-T_2)}{b-a}}{4\pi r^2 k}[/tex]

Maple spit out the same thing, I finally found the first part of the solutions to University Physics (set me back $20, still grumbling about that, I'm switching over to 11th once I find a good price) so this may be resolved sooner than later thankfully.

I'm still not sure why it didn't reconcile. To shed light, I used Maple to expand the results:
Book's:
[tex]{\frac {{\it T1}\,b}{b-a}}-{\frac {ab{\it T1}}{R \left( b-a \right) }}
+{\frac {ab{\it T2}}{R \left( b-a \right) }}-{\frac {{\it T2}\,a}{b-a}}[/tex]

mine:

[tex]-{\frac {ab{\it T1}}{ \left( -b+a \right) R}}+{\frac {ab{\it T2}}{ \left( -b+a \right) R}}[/tex]

I appear to be off by the far left and right terms and then by a factor of -1.
 
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  • #10
Still haven't figured it out; further insight would be appreciated.

Thanks
 
  • #11
I can only repeat what I wrote already...



vanesch said:
Let's check: we have
C/r^2 = - dT/dr, so after integration, T is of the form T(r) = C/r + C'
Note that [tex]C = \frac{ab}{a-b} \DeltaT[/tex] in here.

Now, you can check that the expression of your book does comply to this T(r) form (check the coefficient of 1/r, it is C all right, plus a constant).

In order to check the constant C', we only have to verify that the expression of the book has the right value at r = a (and should then also have the right expression at r = b, because that's how we fixed C - and hence H - in the first place).
Fill it in: it is all right.
 

Related to [thermodynamics] spherical shell problem

1. What is the spherical shell problem in thermodynamics?

The spherical shell problem in thermodynamics refers to a theoretical scenario where a gas is enclosed in a perfectly spherical container. This problem is often used to illustrate the concept of entropy and the second law of thermodynamics.

2. How is the spherical shell problem related to thermodynamics?

The spherical shell problem is related to thermodynamics because it helps to explain the relationship between energy, work, and heat in a confined system. It also demonstrates the principle of maximum entropy, which states that a system will tend towards a state of maximum disorder or entropy.

3. What is the significance of the spherical shell problem in thermodynamics?

The significance of the spherical shell problem lies in its ability to demonstrate important principles of thermodynamics, such as the second law and the concept of entropy. It also helps to illustrate the limitations of the first law of thermodynamics, which only considers the conservation of energy.

4. How does the spherical shell problem help us understand real-world systems?

The spherical shell problem serves as a simplified model for understanding the behavior of real-world systems. It allows us to make predictions and draw conclusions about the behavior of gases in confined spaces, which has practical applications in fields such as engineering and environmental science.

5. What are some limitations of the spherical shell problem?

One limitation of the spherical shell problem is that it assumes a perfect, isolated system, which is not always the case in the real world. It also does not account for factors such as external influences and non-ideal behavior of gases. Additionally, it is a simplified model and may not accurately represent the complexity of thermodynamic systems in the real world.

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