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Heat transfer in series and parallel

  1. May 6, 2017 #1

    knc

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

    2. Relevant equations
    [tex] P = k A \frac{dT}{dx}[/tex]

    3. The attempt at a solution
    a)
    Assuming steady state transfer, energy transfers through rods at the same rate everywhere.
    Letting T be the temperature at the point of welding.

    [tex] P_1 = k_1 A \frac{T_h-T}{L} \\ P_2 = k_2 A \frac{T - T_c}{L} [/tex]
    Setting rates equal to represent steady-state system:
    [tex] k_1 A \frac{T_h-T}{L} = k_2 A \frac{T - T_c}{L} [/tex]
    Solving for T:
    [tex] T = \frac{k_1 T_h + k_2 T_c}{k_1 + k_2} [/tex]
    Plugging this into either equation for individual rod:
    [tex] P = \frac{T_h - T_c}{L} \frac{k_1k_2}{k_1 + k_2} A [/tex]

    a)
    I don't know where to begin with part B. I would imagine I would try to find a ratio between the conductors in series and parallel.

    Thanks in advance
     
  2. jcsd
  3. May 7, 2017 #2

    TSny

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    Setting up the ratio of the powers for the two cases would be a reasonable approach. See if you can show that part (b) does not have a definite answer. So, the question is not well posed. [EDIT: Consider the ratio of the powers for the case where k1 and k2 are equal and the case where k1 is much smaller than k2 (i.e., k1 is now a good insulator.)]
     
    Last edited: May 7, 2017
  4. May 7, 2017 #3

    TSny

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    You will need to give the answer in terms of k1 and k2 rather than a numerical answer.
     
  5. May 7, 2017 #4
    In b, what is the rate of heat flow through rod A? What is the rate of heat flow through rod B. What is the total rate of heat flow through the two rods in parallel.
     
  6. May 7, 2017 #5

    knc

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    I can easily find the heat flow through each individual rod, I am just not sure if I can add them together to represent the overall rate of heat...
     
  7. May 7, 2017 #6

    TSny

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    Yes, for the parallel rods, the total rate of heat flow is the sum of the individual rates. What is your reservation about this?
     
  8. May 7, 2017 #7

    knc

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    Finding the total heat flow in configuration b)
    [tex] P = P_1 + P_2 [/tex]
    [tex] P_1 = k_1 A \frac{T_h - T_c}{L} \\ P_2 = k_2 A \frac{T_h - T_c}{L} [/tex]
    [tex] P = (k_1 + k_2) A \frac{T_h - T_c}{L} [/tex]

    We are trying to find time taken to conduct 10J, difference in temperature is 100C.

    [tex] P = \frac{\Delta Q}{\Delta t} \\ \frac{10}{t} = (k_1+k_2)A \frac{100}{L} [/tex]
    Rearranging:
    [tex] t = \frac{L}{10 A (k_1 + k_2)} [/tex]

    I don't see how the first bit of information regarding the heat flow rate of the first configuration is relevant.
     
  9. May 7, 2017 #8

    TSny

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    Your expression for t is OK. You could use the first bit to eliminate L/A so that the answer for t is expressed in terms of just k1 and k2.
    I found the question to be a little odd. When I first read it, I thought that the answer to part (b) should be a numerical answer. But, the best you can do is express the answer in terms of k1 and k2. Actually, you can express it in terms of just the ratio k1/k2. But you can't get a numerical answer.
     
  10. May 7, 2017 #9

    knc

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    Here is what I got:
    [tex] P_a = \frac{10}{120} = \frac{100}{L}\frac{k_1 k_2}{k_1 + k_2} A [/tex]
    [tex] \frac{L}{A} = 1200 \frac{k_1 k_2}{k_1 + k_2} [/tex]
    Plugging into t:
    [tex] t = \frac{120 (k_1 k_2)}{(k_1 + k_2)^2} [/tex]
     
  11. May 7, 2017 #10

    TSny

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    That looks correct.
     
  12. May 7, 2017 #11
    One thing we can conclude is that, irrespective of k1 and k2, the time for (b) is less than 30 seconds.
     
  13. May 7, 2017 #12

    knc

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    Thanks for the help.
     
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