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Heat equation with radiation effect

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Let's say you have a 3m long copper pipe, 3mm in thickness with a diameter of 170mm. You fix one end at 1K and insulate it to prevent conduction or convection between the air and the pipe itself. There is still radiation. Assume that the inside of the pipe has no effect (radiative, conductive or otherwise). emissivity is 0.1, find the steady state temperature at the other end of the pipe.

    The atmosphere is at 300K. We can assume the problem is 1 dimensional and approximate that the temperature will be constant across the 3 mm thickness.

    2. Relevant equations
    [tex]\alpha*d^2 u/dx^2 = du/dt = 0 [/tex](steady state)

    [tex]Q = mCdT[/tex]
    [tex]P = \epsilon \sigma A*T^4[/tex]

    3. The attempt at a solution

    The problem I'm having with solving this scenario is that the differential equation becomes something I've never seen before. There is heat emitted through radiation at every dx due to the temperature at that x coordinate, which, coupled with the conduction across the pipe will then be equal to the heat gained from the input radiation.

    The differential equation turns out to look like:

    [tex]d^2 T(x)/dx^2 = C - \epsilon \sigma A*T(x)^4[/tex]

    where [tex]C = \epsilon \sigma A*T(outer)^4 [/tex]

    I don't have a second boundary condition, and this turns into a DE that I wouldn't wish on my worst enemies. I've considered approximating it by saying that T_copper << T_atm so T^4 effect from copper is insignificant, which makes the equation a much more manageable:

    [tex]d^2 T(x)/dx^2 = C[/tex]

    But I would prefer not to make this assumption. Does anyone have any suggestions? I realize if the pipe was long enough I would just be able to say that the other end is fixed at 300K equal to the surroundings, but I would need to show that 3m is enough distance from the infinite 1K source to use this assumption.
  2. jcsd
  3. Nov 12, 2009 #2


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    I don't agree with your governing equation. Right away, we can see that the units don't match up. The pipe thermal conductivity and the inner diameter don't appear, but should.
  4. Nov 13, 2009 #3
    you're right, there should be a constant in front of the second derivative of temperature wrt x:

    \alpha d^2 T(x)/dx^2 = C - \epsilon \sigma A*T(x)^4

    However, I'm changing the problem into a 1D heat conduction problem because I want to solve the simpler version first. In that case, I'm only using the surface area to determine the magnitude of radiation on each dx strip. If i do this then I won't need inner diameter. I'm also assuming that the temperature is constant along depth (even though it is not).
  5. Nov 13, 2009 #4


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    So what is [itex]\alpha[/itex] here? A 1-D conduction + radiation equation necessarily includes the inner diameter; after all, heat is being conducted through the annulus down the length of the pipe. If you do a proper energy balance, you'll see that the solution can be simplified considerably.
  6. Nov 13, 2009 #5
    Perhaps I'm not approaching the problem the same way you are:

    [tex]\alpha = \kappa / c_p \rho [/tex]

    [tex] C = \epsilon \sigma A * T(outer)^4 / (m*c_p) [/tex]

    --> this is the delta T as contributed from external radiation, adjusted for the mass (which will have the inner and outer diameters in the calculation from area * density) as well as the heat capacity of the material

    Were you thinking that I should include an inner radiation term? Where radiation is emitted towards the inside of the pipe using the inner surface area, and another term for the amount absorbed from the inside?

    If that is the case, then I need to clarify that this is only part of the problem, the pipe is actually used to carry liquid helium, and my task is to investigate the effect on pipe temperature of the radiation shining on the pipe given that the liquid helium is treated as an infinite heat sink at 1K on the left side of the pipe (yes, this means superfluid state and all kinds of cool thermal properties).
  7. Nov 13, 2009 #6


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    I'm not sure where you're getting these terms; can you give a reference? The units still don't match up, and [itex]c_p[/itex] and [itex]\rho[/itex] don't belong in a steady state equation.

    The inner diameter isn't related to inner radiation; it's related to heat conduction down the length of the tube, which should be proportional to [itex]d_o^2-d_i^2[/itex].
  8. Nov 13, 2009 #7


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    Do you mean

    [tex]\frac{\kappa}{\rho c_p}\left(\frac{d^2T(x)}{dx^2}\right)=\frac{\epsilon \sigma A [T(x)^4-T^4_\infty]}{mc_p}[/tex]

    where [itex]m[/itex] is the total mass and [itex]A[/itex] is the total surface area? If so, cancel the [itex]c_p[/itex], [itex]\rho[/itex], and [itex]L[/itex] out of there. Assume [itex]T(x)<<T_\infty[/itex] to get something that can be easily integrated. Find the solution by applying the boundary conditions, then go back and check your assumption.

    EDIT: Fixed equation typo.
    Last edited: Nov 13, 2009
  9. Nov 13, 2009 #8
    Yes, that is what I meant, except that I used the second derivative of temperature. I know there's a major difference, is your first derivative just a typo or is there something I missed?

    By assuming that T(x) << T(outer) I can get a temperature of approximately 175K. Thanks for all of your help!

    In terms of estimating the error, I can get T(x) but integrating to get the power emitted is still a hassle since it's a quadratic taken to the fourth power. Do you have any suggestions on simplifying this?
  10. Nov 13, 2009 #9


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    Yep, that was a typo, sorry.

    I get

    [tex]T(x)=(2xL-x^2)\frac{r_0\sigma \epsilon T_\infty^4}{k(r_0^2-r_i^2)}+1[/tex]

    which satisfies [itex]T=1\,\mathrm{K}[/itex] at [itex]x=0[/itex] and [itex]dT/dx=0[/itex] at [itex]x=L[/itex], and which gives me [itex]T=37\,\mathrm{K}[/itex] at [itex]x=L[/itex]. I used 1000 W m-1 K-1 for copper's thermal conductivity, which might account for the difference between our answers. (Copper's thermal conductivity increases considerably at low temperatures.)

    To get the power, how about looking at the conduction heat flux [itex]-\pi(r_0^2-r_i^2)\kappa \,(dT/dx)[/itex] at [itex]x=0[/itex]?
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