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Rate of heat loss with two thermal conductivity values

  1. Mar 16, 2008 #1
    Strangely, this is for a Mat Sci Class. It is probably not as complicated as I am making it out to be, but the book has no suggestions on how to handle this problem.

    1. The problem statement, all variables and given/known data
    Calculate rate of heat loss through a fireclay refractory wall of wall with thickness .1m. Furnace operates at 1000C and outside is at 100C. Thermal conductivity of the clay at 1000C is 1.5, and 1.1 at 100C

    2. Relevant equations
    Rate of heat loss per square meter = q = -kdT/dx
    dT = 900C
    dx = .1m
    k(1000) = 1.5
    k(100) = 1.1
    (q is what I am trying to find)

    3. The attempt at a solution

    After realizing that there are two thermal conductivity values, so I could not just plug in the numbers...I figured I would assume a linear relationship between K and Temperature, so that k = 1.1 + .4T/900 - 100(.4/900). But I'm not sure if that gets me anywhere as for solving the problem. Or...could I just take the average k value to be (1.1+1.5)/2 and use that? I'm not sure what the appropriate way to go about this problem is...
  2. jcsd
  3. Mar 17, 2008 #2


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    I agree that a good first approximation would use the average thermal conductivity to give


    A better approximation can be found by using the exact governing equation. You can find this by doing an energy balance on a control volume, if you're familiar with that. You'd have


    going in and


    coming out. (The second term is the first expression in a Taylor series expansion of the first term.) An energy balance gives


    which is the governing equation. You would likely have to solve this numerically. (If k is varies linearly with T, there might be an exact solution; I just don't know it offhand.)
  4. Mar 17, 2008 #3
    Thanks! I have a feeling that the professor is looking for the simple average solution, since this was part of a question from a Mat Sci test, on a chapter that just briefly touched on thermal properties of materials.

    But it is nice to know some of the uses for these other things I am learning in some of my classes (like Taylor series and math processing software)!
  5. Apr 15, 2008 #4


    Mapes, I would like to continue in developing a discrete equation to solve numerically. What is the next step and what would be the solution of this equation.


    After reading around, one have to discretized the above equation, and identify the initial and limits conditions. Your help would be appreciated.

    Thank you.
  6. Apr 15, 2008 #5


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    Exactly, you'll discretize as (if I did this right)

    [tex]\frac{dk(T)}{dT}\left(\frac{T_\mathrm{R}-T_\mathrm{L}}{2\Delta x}\right)^2+k(T)\frac{T_\mathrm{R}+T_\mathrm{L}-2T}{(\Delta x)^2}=0[/tex]

    [tex]T=\frac{T_\mathrm{R}+T_\mathrm{L}}{2}+\frac{dk(T)}{dT}\frac{(T_\mathrm{R}-T_\mathrm{L})^2}{8k(T)} [/tex]

    Solve for T at each node and iterate until the solution converges.

    I just did this for the case of k(T) varying linearly with T and interestingly enough, the exact solution (11,700 W) is the same as if you use the average value of k = 1.3. So flying fish really had the right idea earlier.
  7. Apr 16, 2008 #6

    Thank you for your inputs. I don't really understand what you mean by solving for T at each node and iterate until the solution converges.

    Please tell me if I am wrong:

    solving for T means,

    T - \frac{T_\mathrm{R}+T_\mathrm{L}}{2}+\frac{dk(T)} {dT}\frac{(T_\mathrm{R}-T_\mathrm{L})^2}{8k(T)} = 0

    at each node means,
    discretize the domain which is the wall thickness.
    Let say
    Tl | .T1 .T2 .T3 .T4 | Tr
    Tl and Tr are known.

    Here is my logic:
    I can find an equation for k(T) varying linearly with T,

    k(T) = 0.0004 T + 1.0556


    \frac{dk(T)}{dT} = 00004

    I don't know what to do next to solve (using MATLAB),


    it look straight forward but I am stock.

    Thank you if you can help me more.
  8. Apr 17, 2008 #7


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    You're totally on the right track. You have several simultaneous equations (one for each node):

    [tex]T_1 - \frac{T_\mathrm{2}+1000^\circ\mathrm{C}}{2}+\frac{dk} {dT}\frac{(T_\mathrm{2}-1000^\circ\mathrm{C})^2}{8k(T_1)} = 0[/tex]

    [tex]T_2 - \frac{T_\mathrm{3}+T_\mathrm{1}}{2}+\frac{dk} {dT}\frac{(T_\mathrm{3}-T_\mathrm{1})^2}{8k(T_2)} = 0[/tex]

    [tex]T_3 - \frac{T_\mathrm{4}+T_\mathrm{2}}{2}+\frac{dk} {dT}\frac{(T_\mathrm{4}-T_\mathrm{2})^2}{8k(T_3)} = 0[/tex]

    [tex]T_4 - \frac{100^\circ\mathrm{C}+T_\mathrm{3}}{2}+\frac{dk} {dT}\frac{(100^\circ\mathrm{C}-T_\mathrm{3})^2}{8k(T_4)} = 0[/tex]

    So you see, the left and right temperatures are relative to the node, not absolute. You can have matlab solve these equations by making a matrix of the coefficients and inverting it, but you'll need to iterate because k is a function of T.

    Your equation for Q is exactly right, and you need to discretize it too. You should get the same Q between any two nodes (this is a good check on your results).
  9. Apr 21, 2008 #8

    It is clearer for me. Thank you for your precise and valuable inputs.

    I solved the equations using MATLAB (for fun) and I found:

    1 node, q = 11,700 W (2 iterations)
    11 nodes, q = 10,200 W (115 iterations)
    101 nodes, q = 9,935 W (6124 iterations)
    1001 nodes, q = 9,901 W (125713 iterations)
    10001 nodes, q = 9,899 W (too much iterations)
    100001 nodes, q = 9,898 W (way too much interations even with less precision, residue = 0.01)
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