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Heat flow through metal cylinder?

  1. Sep 5, 2013 #1
    Suppose I have a closed end aluminum pipe of 3cm diameter with .5cm walls

    Cylinder is 10cm tall , 8mm of which is submerged in a cold bath of approximately constant temperature of -10°c

    I feel it with water at 3°c to 7cm height and stick a cork in it. How long till the water freezes? Experimentally it seems like the water freezes from outside in, so I guess its function of the thermal conductivity of the cylinder as well as the thermal conductivity of the outer layer of ice as it freezes inside?

    Is there a formula relating, I 'd guess heat transfer capacity of the cylinder and of ice, ratio of cold surface area/volume of water, initial temp of cold bath and water, total volume of water Or something like that? Sounds like there's calculus involved of some kind...

    Many thanks.
  2. jcsd
  3. Sep 5, 2013 #2
    For an approximate answer assume heat flows only through the part of the cylinder that is in the -10 degree heat bath. Then the problem simplifies to a plate of aluminum of thickness 5mm (the area of the cylinder exposed to the -10 degree heat bath) with two temperatures on either side, -10 degrees and the temp of the water? You can easily figure out how much energy must be removed from the water at 3 degrees to turn it into ice at 0 degrees.

    If you can't make this simplification this becomes a much harder problem?
  4. Sep 5, 2013 #3


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    Figuring out how much energy must be removed is trivial (assuming the OP knows enough to attempt this problem at all).

    The OP was asking about the time to freeze the water. Assuming that water and ice have large thermal conductivities compared with aluminum (i.e. ignoring the temperature gradients inside the water and ice) will give a nonsensical estimate of the time.

    This is a tough question to solve analytically, because you need to include the latent heat required to freeze the water, and because the boundary between the water and ice is moving as more water freezes (and the motion of the boundary is part of the solution, not an initial condition for the problem).

    A way to model it numerically, with a "standard" type of finite difference or finite element formulation, is to have temperature-dependent thermal properties for the water/ice. To include the latent heat, make the approximation that the freezing takes place over a small temperature range (say 0.1C or 0.01C) and the material has a high specific heat in that temperature range. You need enough grid points or elements in the model, so the freezing is localized over a small volume at any time.

    A more sophisticated approach is to write the diffusion equation in terms of enthalpy rather than temperature.

    And if you want to make the problem REALLY hard, include the convection currents in the water as well!
  5. Sep 5, 2013 #4
    I'm with AlephZero... look to numerical solvers.

    Water at 2 degrees is less dense than at 3 degrees, so as the water is cooled at the base, it will begin to rise to the top. This convective cooling will be a major )if not the dominant) method of heat transport in the system, and neglecting it could easily throw your timings way off.

    The Aluminium will also conduct far greater heat fluxes than the water/ice, so you will end up with cold walls and cold regions in your fluid near the outside. Where the freezing front develops and how it propagates through this liquid is extremely difficult to figure out.
  6. Sep 6, 2013 #5
    Wow, more complicated than I thought. I would have thought that this type of problem would have been solved somehow or other to determine the time it takes for say, stationary water to freeze in a pipe, given a sudden drop in external temperature.

    Thanks for at identifying the various factors involved.
  7. Sep 6, 2013 #6
    /Just to check that I've got a handle on the trivial --

    Volume of water (7cm height, 3cm diam) = 115.45cm3
    Weight of water: volume x .9998 = 115.43cm3

    Energy transfer out to reduce temp to 0°c:
    dQ = m ckJ dt | (115.43g x 2.05 x 3c)/1000 = .71kj

    Heat of fusion

    (115.43 x 334 ) /1000=38.55kj

    Total energy transferred = 39.26kj . Is that the correct method/result?
  8. Sep 6, 2013 #7
    I took the diameter you gave as the outside diameter so with 5mm wall thickness that leaves an inner diameter of 2cm. I got a volume of water of about 22 cm^3 or 22 grams of water.

    Using the simplification I suggested (and further simplifications) I got an initial heat flow of about 400 J/s. This suggests that the inner surface of cylinder next to the heat bath would rapidly get covered in ice and once the ice builds up the heat flow would rapidly decrease and nullify my simplification as ice conducts heat at about 1/100 the rate of aluminum and water conducts heat at about 1/400 to rate of aluminum. This is a difficult problem (not for a computer with the right software). Can you tell us what this is for. I'm curious!
  9. Sep 6, 2013 #8
    Ignoring the volume you got, you use 2.05 above, isn't the heat capacity of water closer to 4.2 J/(g)(degreeC)
  10. Sep 6, 2013 #9
    Whoops, you're right. Read the table wrong. Also got the volume wrong. Also measuring weight in cm3 (!)
    (It was very late)

    Volume of water (7cm height, 3cm inside diam) = 49.48cm3
    Weight of water: volume x .9998 = 49.47g

    Energy transfer out to reduce temp to 0°c:
    dQ = m ckJ dt | (49.47g x 4.19 x 3c)/1000 = .62kJ

    Heat of fusion

    (49.47 x 334 ) /1000=16.5kJ
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