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Homework Help: Heat Transfer within a Cone

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Sketched below is a solid, truncated cone, with a side profile of y=Cx. Based on the geometry, the area (in m2) of the left (truncated, x = 1m) and the right face of the cone is 4∏
    and 36∏, respectively. The temperature of the left face is T1=50°C and T2=30°C. Assuming that the system is one-dimensional, steady state, and has constant thermal conductivity k=1/∏, answer the following questions:

    1) What is the actual value of "C"? What is the location of the right face?
    2) Show that the temperature profile in the cone can be described as T(x)=a(1/x)+b. Determine the values of a and b.
    3) Determine the expression of heat flux q"(x) and the heat transfer rate rate q(x), as a function of x.


    2. Relevant equations

    Heat Equation:

    [itex]\frac{∂}{∂x}[/itex](k[itex]\frac{∂T}{∂x}[/itex])+[itex]\frac{∂}{∂y}[/itex](k[itex]\frac{∂T}{∂y}[/itex])+[itex]\frac{∂}{∂z}[/itex](k[itex]\frac{∂T}{∂z}[/itex])+q = ρcp[itex]\frac{∂T}{∂t}[/itex]

    3. The attempt at a solution

    The main issue I'm having is with Part 2.

    Under the assumptions of constant k, no internal heat generation, and steady-state one-dimensional heat flow along the axial direction x, the heat equation simplifies to:

    [itex]\frac{∂^2T}{∂x^2}[/itex] = 0

    There are two boundary conditions prescribed:

    T(1m) = 50°C and T(3m) = 30°C (3m was obtained as the x-position of the right face of the cone during Part 1)

    My issue begins after integrating the relevant form of the heat equation twice. Once integrated, the temperature will fit the form:

    T(x) = ax + b

    Which is not the form outlined in the statement of the problem. Why does this happen? Is it an error of my own or in the formulation of the problem statement?

    Any help anyone can provide will be greatly appreciated.
  2. jcsd
  3. Sep 26, 2013 #2
    When the problem statement says "assuming that the system is one-dimensional", what they are really saying is that the included angle of the cone is very small, so that you can assume locally that the heat conduction is taking place axially along a cylinder of constant cross section. Under these circumstances, the total rate of heat flow Q is a constant, independent of x. So, if A(x) is the cross sectional area at x, and q''(x) is the heat flux at x, how is q"(x) related to Q and A(x)? Under the 1D assumption being used, the temperature T(x) is uniform over then local cross section. What is q" in terms of k and the derivative of T with respect to x? Combine the equations, and you will have what you need.
  4. Sep 26, 2013 #3
    So, I'm not sure if I made my point exactly clear.

    If I'm reading your post correctly, you're suggesting I use Fourier's Law:

    q"(x) = [itex]\frac{Q}{A(x)}[/itex] = -k[itex]\frac{∂T}{∂x}[/itex]

    I can obtain the information to substitute for [itex]\frac{∂T}{∂x}[/itex] by differentiating T(x). My problem is not principally this though. It's that in the statement for Part 2, its been stated that T(x) will take the form:

    T(x) = a(1/x) + b

    But I found the solution to be of the form:

    T(x) = ax + b

    So my question here is really, why does my form not match the form given?
  5. Sep 27, 2013 #4
    In the equation I gave you, A is a function of x. In particular A = π(Cx)2. So when you integrate my equation with respect to x, you get a 1/x term. Do the math and see.

    As I said, in the 1D approximation, locally, you are approximating the behavior as if it were a cylinder of constant cross section. But, over larger distances, you are taking the diameter variation into account.
  6. Sep 27, 2013 #5
    Okay, I think I get it now.

    By separation of variables:

    [itex]\frac{Q}{A(x)}[/itex]dx = -k dT = -1/∏ dT

    But A(x) = ∏r2 = ∏(Cx)2

    So by integration:

    [itex]\frac{Q}{C^2x}[/itex] = T + n

    And solving for T, I get:

    T(x) = [itex]\frac{Q}{C^2x}[/itex] - n

    Which now does fit the form:

    T(x) = ax + b

    Is this correct?
  7. Sep 27, 2013 #6
    This is not quite what I got. I got:

    [tex]T=\frac{Q}{\pi C^2kx}+D[/tex]

    where D is a constant of integration.

    You're not done yet. You need to go back to the problem description (geometry) and determine the value of C. You also need to determine the values of x at the beginning and end of the cone. Then you need to substitute the temperature boundary conditions at the two values of x into the equation for T to get the values of Q and D.
  8. Sep 28, 2013 #7
    We essentially have the same solution, because in the problem statement, k is given as 1/∏, so the ∏ and k multiply to unity (although there will of course be units left over from k). My constant of integration was the n term, just because C was in use so I picked another letter.

    Part 1 of this problem involved the geometry of the problem, and from that I know that C=2, and the cone is bounded by x=1 and x=3. This leads to the conditions that T(1)=50 and T(3)=30. With these I can solve for Q and D, which are constants, and I can also find an expression for the heat flux, which will just be [itex]\frac{Q}{2∏x^2}[/itex].

    I think I'm mostly good on this problem, but my last question is, why the disparity of results between the two methods of approach; Fourier's Law and the Heat Equation produce different temperature profiles. You were saying about one being an approximation and the other taking into consideration the changing cross sectional area along the length of the cone, but I'm still not quite clear on how that works. Would you please explain it once more?

    Thanks so much for all the help you've already provided.
  9. Sep 28, 2013 #8
    In the above equation, the 2 should be a 4.

    There are some issues that I have with the problem statement. First, a value of C=2 is too large for the 1D approximation to provide an accurate answer. C should be <<1 for the 1D approximation to be accurate.

    Secondly, the problem statement should have read "locally one dimensional heat flow," rather than just "one dimensional heat flow." To people experienced in heat transfer, it would have been unnecessary to add the word "locally," but to those new to the subject, omitting it has led to confusion. As a result of all this, you immediately dropped the partials with respect to y and z from the differential equation. Because the cross section is changing with x, to be precise mathematically, one would need to first integrate the differential equation over the cross section, and properly apply the boundary condition of no normal component of heat flux on the conical surface. One then ends up with a differential equation for the average temperature (averaged over the cross section). This would turn out to be the exact same equation (for the case of C<<1) that we wrote down using the locally 1D approximation. I'm not going to go through the mathematical details of the derivation here.

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