1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat transfer by conduction in a truncated cone

  1. Jan 12, 2010 #1
    1. The problem statement
    A truncated cone 30cm high is made of Aluminum. The dia at the top is 7.5cm, and 12.5cm at the bottom. The lower surface is maintained at 93 deg C, the upper surface at 540 deg C. the other surface is insulated. Assuming 1 dimentional heat flow, calculate the rate of heat transfer in watts.

    2. Relevant equations
    Conduction equation:

    3. The attempt at a solution

    I have the k value from a textbook at 204 w/(m C)

    Made the assumptions that k is not a function of x or t and that the rest of the cone is perfectly insulated besides the top and the bottom.

    That being the case I simply plugged in the values using

    q=k*(A1*T1-A2*T2)/x where T1 is the high temperature and T2 is the low temperature. x is the thickness.

    I'm not sure if this involves some kind of integration though. I fear I may be oversimplifying the problem.

    This question was asked previously in the forum but was unanswered.

    Thanks for the help.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 12, 2010 #2
    Perform a heat balance on the cone. Since the sides of the cone are perfectly insulated, the cone does not generate energy nor does it store energy, so the amount of heat entering the cone must balance the amount of heat leaving the cone.

  4. Jan 12, 2010 #3
    So then can it be said that




    and the overall heat transfer is equal to q1/x?

    Or are you saying that q=0?

    Or do I need to use constant specific heats to solve this question to get the change in enthalpy and use that to get q?

    I'm afraid this has just confused me further.
  5. Jan 12, 2010 #4
    No. q is not equal to zero.

    This is what I am saying,

    Ein - Eout = 0 therfore Ein = Eout

    No this is not needed.

    Sorry to have confused you further. I will work on this later today.

  6. Jan 12, 2010 #5
    Let me give you a push in the right direction.

    First lets find an expession for the area of the cone as a function of x.

    let r1 = radius of the small end
    let r2 = radius of the large end
    let h = height of the cone

    Then the function will be

    r(x) = r1 + ((r2 - r1) / h)*x

    The area as a function of x is A(x) = PI*r(x)^2


    A(x) = PI*(r1 + (((r2-r1)/h)*x)^2

    Now using Fouriers law for one-dimensional heat conduction

    Q' = -A(x)*k*(dt/dx)

    Rearanging yields

    Q'*(dx/(A(x)) = -k*dt

    Now integrate the above function between 0 and h on the left hand side and integrate the right hand side between Th and To where Th = temperature on upper surface and To the temperature on the lower surface.

    Remeber Q and PI are constant so they stay outside of the integral on the left hand side.

    Think you can take it from here?

    PS - This problem is stated in a slightly misleading manner because it is really a 2-dimensional problem and to bring it into a 1-dimensional problem the equation for the area as a function of x is required.

    Last edited: Jan 12, 2010
  7. Jan 12, 2010 #6
    Thanks so much. This makes it really clear.

    Because the area is constantly changing as we move through the cone we must transform the area into terms of one of the available variables, in this case the height. Once this is done we can take the integral with respect to that variable in order to determine the values of the area at each individual point. This allows us to state the rate of heat flow through the cone as a whole instead of as a cylinder as I was doing before.

    I don't know if I've verbalized the mathematics correctly but I feel I understand it much better.


  8. Jan 12, 2010 #7
    For the most part you are on the correct track. Again, sorry to have mislead you with the heat balance method.

    I have the final answer, so when you complete the integrations and substitutions we can compare my result with yours.

  9. Jan 12, 2010 #8
    I think I must have made a mistake somewhere because I end up with a negative heat transfer which doesn't make sense.

    My integration of dx/A(x) yields (1/PI)*(h/(r2-r1))*(r1+(r2-r1)*x/h)^-1*(-1) between 0 and h. This yields (1/PI)*(h/(r2-r1))*(1/r2)*(-1).

    Multiplying this to the opposite side where the integration of -k(dT) between T1 and T2 yields k(T1-T2) gives me an answer of -1.49 kW.

    Am I missing a sign somewhere or have I made a graver mistake?
  10. Jan 12, 2010 #9
    I neglected the 0 part of the integration. I now have a value of 2.24 kW which makes more sense.
  11. Jan 12, 2010 #10
    You have the right hand side integrated correctly.

    The mistake is on the left hand side.

    You are not far off though. Here is what I have. (Do you see your mistake?)

    1/PI * (h/(r2-r1)) * (r1 + *((r2-r1)/h)*x)^-1 evaluated between 0 and h

    Think you can take it from here?

  12. Jan 12, 2010 #11
    Looks like you have corrected yourself.

    I have a final result of 2.216 kW.

    The difference is simply the value of k that I used was 202 W/(m-C) as opposed to your value of 204 W/(m-C).

    Glad to have helped you out.

  13. Jan 12, 2010 #12
    I am so thankful for your help on this. I feel like I have a much better understanding of the material now as well.

    Thanks again

  14. Oct 25, 2010 #13
    Where did you find this question? If it's possible, I'd like to know which book this comes from and the answer to it as well. Thank you!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook