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Time taken to freeze ice vertically from top exposed surface

  1. Jul 12, 2014 #1
    Hi guys,
    New to the forum wanted a little help in designing an insulated ice mold. Basically the objective is to force the water in the mold to freeze axially or in one direction in order to make clear ice. Ideally, the time taken for the ice cube to freeze from the top surface exposed to the ambient air should be less than the time taken for the inside surfaces of the mold to reach zero.

    Ta = -10°C
    P = 1ATM
    g = 9.8

    Ti = 20°C
    Tf = 0°C
    Cp = 4.18
    h(fg) = 355 KJ/kg (Latent Heat of Fusion)
    rho = 1000 kg/m^3

    Cube Dimension:-
    L = 0.05m
    m = 0.05^2*1000 = 0.125 kg

    Initially, I've calculated the energy required to freeze the cube as:

    Q = m[Cp*∆T - m*h(fg)]
    = 0.125[4.18(0-20) - 355]
    = -52.14375 KJ

    Then I get stuck on determining the convective coefficients. I've tried calculating the Nusselt number for free convection on a hot horizontal plate top surface (top exposed surface) but Rayleigh number is less than 10^4. And for the sides (assuming the bottom is sitting on the inside surface which is practically adiabatic) I get:

    Tf = (Ti-Ta) / 2 = 278°K
    Beta = 1/Tf = 3.597*10^-3 °K^-1

    From Linear Interpolation of Data for air @ Tf (Fundamentals of Heat and Mass Transfer, Frank P. Incropera: Table A.4)
    Nu = 13.932*10^6 m^2 / s
    Alpha = 19.596*10^6 m^2 / s
    Pr = 0.71272
    k = 24.14 *10^-3 W / m.K

    Ra = g*Beta*(Ti-Ta)*L^3 / (Alpha*Nu)
    = 4.842*10^5
    *as Ra~<10^9 use laminar equation. C=0.59, n=1/4

    Nu = 0.68 + 0.67*Ra^(n) / [1+(0.492/Pr)^(9/16)]^(4/9)
    = 14.251

    Nu = h*L/k
    h = 6.8804 W / m^2.K (Convection Coefficient)

    Then developing a thermal circuit for the side walls which is 3mm (tpe) LDPE sandwiched between two 2mm (tsi) layers of silicone rubber.

    Ksi = 0.2 W / m.K
    Kpe = 0.034 W / m.K

    The source would be the water and the sink would be the ambient freezer temp.

    1/Rt = 2*tsi/Ksi +tpe/Kpe +1/h
    Rt = 3.94 W / m^2.K

    q = Rt*A*∆T
    = 3.94*(0.05)^2*((-10)-20)
    = 0.2958 W

    I know if you divide that through Q/q = 176298.49 sec which works out to ~48 hours. That seems a little excessive to me especially considering that q is an instantaneous rate when the temperature difference is greatest and will only decrease meaning that it would take ~>50 hours for the inside surface temperature to reach 0°C (unless it is an average rate, I forget). Knowing my luck it's possible I'm going in the wrong direction for determining this altogether. Am I doing the right thing and how do I work out the time taken for the whole ice cube to freeze from the top surface?


  2. jcsd
  3. Jul 12, 2014 #2
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