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1D transient problem

  1. May 12, 2012 #1
    Say you have a furnace wall at a temperature at 400 degrees C and you have oven gas at 200 degrees C and you place an copper rod in to the furnace. The copper rod has an initial temperature of 25 degrees C and a length of 1 meter and a diameter of 2cm. You want to find what find the time it takes for the copper rod to reach 150 degrees C. The furnace is a sqaure with a LxWXH of 3x3x3 meters respectively.

    You can assume 1D transient problem in the radius direction and you only need to take into consideration convection of the gas and the radiation of the walls of the oven.

    The oven can be treated as a black body and the copper rod as a grey body with an emissitivity of 0.5 to 0.8.


    Should I be using heat equations to solve this problem or should I use lumped capacitance method?

    Also is heat equations always related to only conduction or can it be applied to both convection and radiation?

    Thanks in advance.
     
  2. jcsd
  3. May 12, 2012 #2
    Due to the fact the rod is copper which has a very high thermal conductivity and is thin, the easiest solution is the lumped mass solution. When this assumption is made you are assuming that there is very little temperature gradient in the object as it heats. There is a parameter called the Biot number which is

    Bi = h*d/k

    where h is an effective film coefficient, d is a characteristic length, and k is the thermal conductivity. When Bi<1, you can assume a lumped mass system. In your case, d would be the radius. Effective h can be computed by factoring the radiation condition

    q=g(t^4-tinf^4)=g(t^2+tinf^2)(t^2-tinf^2)=g(t^2+tinf^2)(t+tinf)*(t-tinf)

    Look at g(t^2+tinf^2)(t+tinf). It is an effective h for radiation since it is multiplied by (t-tinf) as would a standard convection coefficient.

    In your case you have the radiation condition plus a convection condition. So figure out the effective h by adding above to the convective coefficient due to the oven air and determine the Biot number. Then decide whether lumped mass is appropriate.
     
  4. May 13, 2012 #3
    So this means that q=g(t^4-tinf^4)+h(t-tinf) = g(t^2+tinf^2)(t+tinf)*(t-tinf)+h(t-tinf)

    But wouldn't that mean that h=g(t^2+tinf^2)(t+tinf)+h or did I do something wrong?

    Thanks
     
  5. May 13, 2012 #4
    That's right. I only factored the radiation portion because I thought it might be a little subtle. You add the two parts to get an effective H. Plug in the greatest temperature differences which are the initial rod, air, and wall temperatures so you get the greatest H possible. Then figure your Biot number based on that because it only gets less from that point on as the rod heats. g, obviously, is the product of Stephan-Boltzmann constant and shape factor which is unity. Use the radius for the Biot number equation.
     
  6. May 13, 2012 #5
    What do you mean by air and wall???

    Say:
    ε = 0.7
    σ = 5.67 x 10^-8 w/m^2K
    h = 17 w/m^2K
    t = 673K
    tinf = 298K
    k = 401w/m^2K

    H = 0.7*5.67 x 10^-8*(673^2+298^2)(673+298)+17 ≈ 31.03 w/m^2K

    Bi = HL/k = (31.03*5 x 10^-3)/401 ≈ 3.87 x 10^-4

    Is this right so far, but I didn't include the air temperature, does this matter? Or was I just meant to take the extremes for the different temperatures?
     
  7. May 13, 2012 #6
    The rod is in an oven. An object being heated in an oven gets its heat from two sources. One is radiation from the oven walls which the problem statements says are 400 C. The other source is the hot air in the oven that is at 200 C. That is where convection enters the picture either in the form of natural or forced convection.

    I did not check your multiplications but what you have done so far is correct. You have not included the film coefficient for convection yet but it won't make much difference. As figured, Bi<<1 so the lumped mass assumption is a good one.
     
  8. May 14, 2012 #7
    When finding the film coefficient for convection, do I use it to find the Bi number?

    Now assume that Bi <<1, to find the time it takes, do I just go:
    Ein - Eout = Est
    qrad - qcov = ρVc(dT/dt)

    Am I on the right track?
     
  9. May 14, 2012 #8
    Yes, you add the film coefficient to the radiation factorization to come up with an overall coefficient for heat transfer. Use that to determine the Biot number << 1. Once you confirm that, you can treat the problem as a lumped mass heat transfer problem.

    The way you solve the equation depends on what approximations you wish to make. Radiation is a function of temperature to the 4th power. That makes the problem nonlinear. You have two sources of heat at different temperatures so separating the variables and integrating is a problem.

    If I were doing it, I'd numerically integrate the following equation.
    rho*cp*V*dT/dt = -qrad - qconv

    being careful that units all match up. A simple backwards difference scheme (explicit) would suffice with sufficiently small time step.


    On the other hand, you may be able to omit the convection because much more heat is reaching the rod from radiation. You should check this out by evaluating the boundary condition and noting the strength of the heat fluxes. If you omit convection, you can solve the equation by separating variables and integrating after a variable transformation to eliminate the wall temperature.
     
  10. May 15, 2012 #9
    why is it -qrad - qconv? shouldn't it be qrad - qconv?
     
  11. May 15, 2012 #10
    rho*cp*V*dT/dt = -qrad - qconv

    I can rewrite to make it less cluttered.

    dT/dt = -qrad - qconv = -a * (T-Twall) - b * (T-Tair) where Twall and Tair are the oven wall and air temperatures, respectively.

    The initial temperature of the Cu is 25 C, the air is 200 C, and the walls are 400 C. Constants a and b are positive.

    So if you integrate and let dT = Tnew - Told and time step = dtheta

    Tnew = Told +(- a * (Told-Twall) - b * (Told-Tair) )*dtheta

    If Twall and Tair are greater than Told, Tnew should be higher after the integration time step is accomplished as above.
     
  12. May 15, 2012 #11
    that all makes sense, but instead of making dt = time step, I wanted to find a the time, could I just make it temp step (i.e. dT=temp step; e.g. dT=10deg.C) and find out how long it takes...
    So say if I wanted to iterate temperature over a give range (e.g. from 25 deg.C to 300 deg.C). How would I go about rearranging that?
     
  13. May 15, 2012 #12
    You can't solve it that way. The solution has to be marched through time to arrive at the 150 C result summing all the time steps up as the solution moves along.
     
  14. May 15, 2012 #13
    I mistyped something a few posts back. The Biot number should be less than 0.1 for lumped mass calculations.
     
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