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Heat Transfer: Finding temperature at the junction

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A furnace is constructed with 0.5 m of fire brick, 0.15 m of insulating brick and 0.25 m of ordinary building brick. The inside surface-temperature is 1530K and the outside surface temperature is 525K. The thermal conductivities of the fire, insulating and building bricks are 1.4, 0.21, and 0.7 W/m-K, respectively. Calculate the following per square meter area:
    (a) Total resistance of the wall in K/W
    (b) Rate of heat flow in W
    (c) The temperatures at the junctions of the bricks.

    2. Relevant equations
    Q/T = kAΔT / L

    3. The attempt at a solution
    (a) Total resistance of the wall in K/W: answer 1.43 K/W
    (b) Rate of heat flow in W: answer 703.5 W
    (c) The temperatures at the junctions of the bricks.

    (Temperature drop over firebrick+insulating brick)/(Total temperature drop) = (0.3571+0.7143/1.43)
    Temperature drop over firebrick+insulating brick:
    (1530K-525K) ⋅ (0.3571+0.7143/1.43) = 752.9769 K
    Hence the temperature at the (firebrick & insulating)-ordinary brick interface = (1530K - 752.9769 K) = 777.0231 K

    The correct answer according to the answer sheet is 778 K, there might be a slight difference but I want to make sure that my computation is right?
     
  2. jcsd
  3. Mar 12, 2017 #2
    Looks OK.
     
  4. Jul 2, 2017 #3
    why is the solution on letter C like that? i tried equating by (1530 - X) / (0.5 / 1.4) = (X - 525) / (0.25 / 0.7) but its not showing the correct answer...
     
  5. Jul 2, 2017 #4
    My heat flow rate equations for the three layers are as follows:
    $$Q=1.4\frac{(1530-T_1)}{0.5}$$
    $$Q=0.21\frac{(T_1-T_2)}{0.15}$$
    $$Q=0.7\frac{(T_2-525)}{0.25}$$
    What is the solution to these equations for ##T_1## and ##T_2##?
     
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