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Flat Plate External Free Convection Problem

  1. Aug 3, 2014 #1
    I am trying to calculate heat transfer from an inclined flat plate under free convection conditions. I am referring to Fundamentals of heat and mass transfer (Incropera / DeWitt). The equations and a spreadsheet which can be used to find a solution can be found at:

    http://www.brighthubengineering.com...transfer-coefficient-estimation-calculations/

    "Natural Convection Heat Transfer from an Inclined Plate"

    These calculations involve calculating the Rayleigh number and then using a correlation to find the Nussult number and thus the heat transfer coefficient.

    The problem is that i don't understand what happens when the plate is cooled (at lower temperature than the surrounding air). The plate i am trying to calculate this for is an evaporator. The because the temperature between the plate and air is negative this makes the Rayleigh number negative and i don't think this is correct. The correlation is then erroneous because you need to take Ra^(1/4), which clearly is incorrect for a negative number.

    If you make the surface temperature of the plate lower than the air in the example spreadsheet reference above it goes into error so the spreadsheet has not addressed the problem.

    I am presuming that the Rayleigh number is negative because the air is being cooled rather than heated, or related to that cooling is occurring.

    I am thinking I should just times the Rayleigh number by -1 to make it positive and calculate the heat transfer co-efficient from this but i have no idea if this is valid.

    The book i am referring to has a diagram of an inclined plate being cooled in the same chapter, so it would be strange if this cannot be made to work.

    My values are as follows: Air temp = -0.4 deg C, plate temp = -50 deg C, Tf = -25.2 deg C, g = 9.81cos60 = 4.905, β=1/(-25.2+273) = 0.00404, L^3 = 4.913, v = 1.132 * 10^-5 (m^2/s), α = 15.9*10^-6 (m^2/s), pr = 0.72.

    Any ideas?
     
  2. jcsd
  3. Aug 3, 2014 #2
    If you want the long and short of it, read section 9.6.2 of Incropera and Dewitt (great textbook by the way). If you read this section, they basically say you handle either the top side of a cool surface (relative to surroundings) or the bottom side of a hot surface (relative to surroundings) using equation 9.26 and assuming gravity is equal to [itex]gcos(\theta)[/itex]. Bear in mind, that equation doesn't distinguish whether the vertical surface is warm or cold, so h should be the same either way for a given temperature difference and thermal properties (which in turn affect the Raleigh number, which will be positive as you use the magnitude of the temperature difference). There are no equations in the book to handle the lower side of a cool surface or the upper side of a hot surface for an inclined surface, at least not in the edition that I'm using (6th).

    I hope that helps somewhat. Best of luck with that problem you're working on.
     
  4. Aug 4, 2014 #3
    Hi Jlefevre, Yes if i use the magnitude of the temperature difference i get the same answer but the Raleigh number is positive. I have got a heat transfer coeff however this seems a bit low. That's good enough for now i think.
     
  5. Aug 4, 2014 #4
    Yeah, it kind of depends on what you're looking at in terms of application. Is this for a project or research? Or is this for a homework problem? If it's for homework, then this is as good as your textbook gives you in terms of options. If it's for something else, I can discuss some possibilities for approximating something closer to what your setup is (if you can describe it in detail).
     
  6. Aug 5, 2014 #5
    Yes its for a research project. We have a flat plate evaporator on a roof, it is a micro-channel style heat exchanger. This is then used as part of a heat pump for hot water generation. The heat transfer from the evaporator is affected both by radiation from the sun and convective heat transfer from the air flowing across the panel.

    We are measuring wind speed, irradance & panel temperature. But the difficulty is in estimating the different convective and radiation heat transfer elements or the proportion of which with respect to external conditions. I need to be-able to estimate these both accurately so that i can generate a simulation model for a given set of climate data.

    I'm interested in free convective conditions because often the wind speed is very low or 0 and i believe that this is driving the panel heat transfer. Other research suggests the convective heat transfer is a high proportion of the total circa 70%.

    Currently the model is grossly underestimating the reality of the situation (heat transfer).
     
  7. Aug 5, 2014 #6
    You can use the Duffie and Beckman equations to estimate a convection coefficient, that would be the easiest. One such example is h=max(5, 8.6V^0.6/L^0.4) where V is wind speed in m/s, and L is a characteristic length, the cube root of the volume that would enclose your system.

    Otherwise, it gets more complicated than I have time to address right this second (I'm responding from a tablet). I have an old method I was using to do something similar, I'll look it up and see if I can find it. Is it safe to assume there is convection off the top and bottom of the collector, or just the top?
     
  8. Aug 5, 2014 #7
    Okay, back on the laptop here and I can address this using a little bit more detail.

    So, one way to APPROXIMATE (this is not physically realistic, but it won't make you end up too far off), is to use a weighting function that interpolates between the results you get for cases you know (100% vertical or 100% horizontal). This becomes increasing complicated because the upper and lower surface behave differently in terms of free convection, so you actually have to calculate 3 Nusselt numbers (with three different Raleigh numbers because they use different characteristic lengths I think). You then use your three Nusselt numbers to get your three convection coefficients, and then can use the averaging shown in the figure.

    Again, this is likely non-physical, but you're going to have issues anyway as far as I can tell, because wind direction is going to change the boundary layer (theoretically) and therefore change your convection coefficients. Incropera and Dewitt give instructions for combining free and forced convection, but in your case they'll likely be non-physical anyway because the way the wind flows across the surface will change depending on your collector's orientation.

    Theta in the figure is the inclination angle from horizontal, in case you needed to know (0 = horizontal, 90 = vertical).
     

    Attached Files:

  9. Sep 7, 2014 #8
    I have actually had a chance to work through this now. I think I see what you mean. If you like you could perhaps have a look at the attachment and see if you think I have interpreted this correctly (cooled plate).

    This is way more complex than I expected and its easy to get this sort of thing wrong.

    I think for the purposes of my simulation this is more than enough. I think some of the detail given in the Duffie and Beckman book is also very useful and is far more relevant than what I have found in the general heat transfer books. Even if the attached calculation is inaccurate and i don't use it, I am fairly happy that I can justify with reference to the Duffie and Beckman book.

    Thanks a lot for the help here this may have saved me untold hours.
     

    Attached Files:

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