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Time to reach thermal equilibrium and/or a steady state

  1. Jul 20, 2011 #1
    I want to calculate the time it takes for a spherical sample (radius, r) to reach thermal equilibrium in an oven where the temperature is held constant.

    I don’t need a super accurate answer, within 5-10 seconds is sufficient. Is it reasonable to simply say that when the Fourier Number (Fo) is 1 that the system is almost in a steady state and almost in equilibrium? Or does Fo have to be higher to make this assumption? I read somewhere that Fo = 1 equates to an energy transfer of ~97%, is this true? (There was no reference given).

  2. jcsd
  3. Jul 20, 2011 #2
    First try the lumped capacity method. See Equation 6.9 in:

    http://ec.pathways-news.com/Text-PDF/Part B-6.pdf

    Check your Biot number is within limits (Equation 6.15), or your result will not be very accurate.

    If your Biot number is out of range, let me know - there are other (more complex) models to treat this case.
  4. Jul 21, 2011 #3
    Thanks for the reply. Is there another link to the PDF? The link you posted doesn’t work for me, but I am in living China so many it is not too much of a surprise.

    I know very little about thermodynamics and my hope of a simple solution may not be so simple. For the lumped capacity method is the solution what is on the wiki page? http://en.wikipedia.org/wiki/Lumped-capacitance_model#Solution_in_terms_of_object_heat_capacity

    For this and the Biot number I need h, the heat transfer coefficient. On wiki this is describes in terms of convective heat transfer, so should I use the value for the media? In this case it is air (10-100 W m^-2 K^-1).

    Thanks again for the help
  5. Jul 21, 2011 #4
    For your heat transfer coefficient h, use the T. Yuge correlation for external flow on a sphere in the wiki link:

    NuD = 2 + 0.43 RaD^1/2

    Your Rayleigh number is a function (amoung other things) of the temperature difference between the surface and external. So this also varies with time.

    To get started and learn the method, assume RaD = 0, and thus NuD = 2. This will overestimate the heatup time. When you get comfortable with this, you can include the Rayleigh number in the treatment.
    Last edited: Jul 21, 2011
  6. Jul 25, 2011 #5
    Thanks for the info. I haven't had a chance to work on this yet, but I'll let you know how it goes (hopefully by the end of the week). Rayleigh and Nusselt numbers are ringing old bells (I did an undergrad degree in Geophysics where they are used in core and mantle dynamics).

  7. Jul 27, 2011 #6
    OK, so I have found some time to crunch through this in more detail.

    The furnace cavity is cylindrical with diameter, D = 0.1 m, and length > D (more than 10 times D). The furnace is held at constant temperature, [itex]T_{env}[/itex]= 873 K. A spherical sample of radius = 0.0125 m at temperature, [itex]T_{samp}(t=0)[/itex]= 293 K is "instantaneous" introduced into the furnace with no heat loss.

    So the Rayleigh Number, [itex]\mathrm{Ra}[/itex], (when [itex]\mathrm{Pr}=1[/itex]) is given by:

    [itex]\mathrm{Ra} = \mathrm{Gr}= \frac{g \beta} {\nu^2} (T_{env} - T_{samp}) D^3[/itex]

    To first order I assume that [itex]\beta[/itex] and [itex]\nu[/itex] are temperature independent and values for air at 873 K are used (1.17e-3 [itex]K^{-1}[/itex] and 98.65e-6 [itex]m^2s^{-1}[/itex], respectively)

    Following the equation for [itex]\mathrm{Nu}[/itex], the heat transfer coefficient, [itex]h[/itex] can be determined by:


    where k is the thermal conductivity of the sample (taken to be ~2.5 [itex]W.m^{-1}.K^{-1}[/itex] for a basalt sample). Should this be for air?
    L is the characteristic length scale, here I use the ratio of the sample volume/area [itex]\left(L\approx{\frac{V}{A}}\right)[/itex].

    The solution to the lumped capacity method is given by:

    [itex]T_{samp}(t) = T_{env} + (T_{samp}(0) - T_{env}) \ e^{-r t}[/itex],

    where [itex]r=\frac{hA}{C_p\rho{}V}[/itex].
    My sample is a basalt so...
    The mass specific heat capacitance, [itex]C_p=1200 J. K^{-1}. kg^{-1}[/itex]
    and the density, [itex]\rho{}=2800 kg. m^{-1}[/itex]
    Again, these are assumed to be invariant of temperature.

    Still following me? I'll admit it's interesting learning this stuff! :smile:

    Using the above I calculate Ra(t=0) = 685730 and h (t=0) = 35.9.
    By t = 1000 s, Ra(t=1000) = 114221, h (t=1000) = 24.8, and [itex]T_{samp}[/itex](t=1000) = 776.5 K.
    And by t = 2000 s Ra(t=2000) = 38227, h (t=2000) = 20.0, and [itex]T_{samp}[/itex](t=2000) = 840.8 K.

    Since Ra, and hence [itex]r[/itex], depend on [itex]T_{samp}[/itex] I evaluate [itex]T_{samp}(t_n)[/itex] using [itex]r(t_{n-1})[/itex] and use 50 s time steps. This will introduce an error, which will be reduced as the time steps decrease, correct?

    Are these reasonable values? They seem OK to me. But am I using the appropriate values? By that I mean, am I making the correct distinction between fluid (air) and sample (rock) properties in the equations?

    I know I skipped ahead of the simple example you suggested, but I find it easier to learn by applying methods to more realistic situations, assuming of course I have done it correctly.

    If I am on the correct track then I will build in the temperature variation of the "constants" ([itex]\beta[/itex] and [itex]\nu[/itex]). Should these be evaluated at


    Also, in the strictest sense, the furnace cavity is cylindrical (so is the sample) and [itex]\mathrm{Pr}\neq1[/itex]. Would it then be more appropriate to calculate h using:

    [itex]h = \frac{k} {D}\left({0.6 + \frac{0.387 \mathrm{Ra}^{1/6}}{\left[1 + (0.559/\mathrm{Pr})^{9/16} \, \right]^{8/27} \,}}\right)^2[/itex]

    Cheers for all the help so far!
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