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Thermo, Temperature profile of rod

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider a rod that is 1m long whose radius changes from 1cm at one end to 4cm at the other end in a uniform fashion. Assume that the end with r=1cm i kept at 0°C and that the end with r=4cm is kept at 100° C. Determine the temperature profile along the rod. One can assume that at any position along the rod the temperature is uniform across its cross section. Hint: Determine an expression for the incrimental change in temperature, dT, at any position x along the rod and integrate to get a general expression for the total temperature change between the ends of the rod. Remember that the heat flowing through any cross section of the rod is constant.


    2. Relevant equations
    H=(kaΔT)/x
    A=∏r^2

    3. The attempt at a solution
    I think what I have done so far is wrong, as I have trouble with integral problems. (My problem is with setting them up, not solving them). I tried to follow the hint given, but i am stuck. Here is what I have so far:
    (Hx)/(kA) = ΔT
    dΔT=[Hdx]/[k∏(dr)^2]
    ΔT=∫H/[k∏(dr)^2]dx (This is supposed to be aDefinite integral from 0 to 1-couldn't figure out the format).
    ΔT= H/[k∏(dr)^2] and here is where I get stuck, and think I set up the equation wrong, as 1/dr doesn't exist. Any help would be greatly appreciated. :)
     
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2
    Write an expression for the area of the rod radius as a function of its length, x. Use that expression in formulating the differential equation that equates the heat entering a slice of length dx to the heat leaving at x+dx. The derivation of the ODE is similar to the case for 'standard' 1-D conduction except the area (pi*radius^2) is within the derivative operator that you get from using the first term of a Taylor series expansion to evaluate the heat leaving at x+dx. Integrate the differential equation and evaluate the two constants of integration by the boundary conditions.
     
  4. Sep 3, 2012 #3
    Thanks!
     
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