Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Problem with Heat Transfer

  1. Jul 8, 2010 #1
    I am trying to understand how heat is lost from a glass aquarium tank maintained at some temperature above ambient.

    If we imagine a direct interface between the body of water and the surrounding air, I think heat transfer would be entirely due to radiation and convection.

    When a glass pane is placed in-between the two, heat transfer is due (entirely?) to conduction.

    What I am having trouble understanding, is how the heat lost by conduction through the glass can be so significant but as soon as that glass is removed the heat loss by radiation and convection is so trivial. It seems that the presence of the glass pane encourages heat-loss?

    Here is what I am considering:

    A body of water with surface area 1m x 0.25m , or 0.25 m^2 at temperature 29C.
    The surrounding environment temperature of 25C.
    A glass pane of thickness 0.01 m.

    Heat loss by conduction:
    Watts = (k * A * (Th - Tl)) / d , where k for glass is 1.1 Watts/m*K

    = (1.1 * 0.25 * (29 - 25)) / 0.01
    = 110 Watts of heat flow.

    When the pane is removed:

    Heat loss by radiation:
    Watts = (em * sbc * A * (Th^4 - Tl^4)), where em for water is 0.95 and sbc = 5.6703x10^(-8) W/m^2*K^4

    = 0.95 * 0.000000056703 * 0.25 * ( (29^4) - (25^4) )
    = 0.0043 Watts

    Heat loss by convection:
    Watts = (h * A * (Th - Tl)), where h for water is taken to be 20 W/m^2*K

    = 100 * 0.25 * (29 - 25)
    = 20 Watts

    Here it appears that when a glass pane is present 110 watts of heat are lost, but when it is removed only 20.0043 watts are lost.

    Can anyone explain this to me?

    Thank you for your time.
     
  2. jcsd
  3. Jul 8, 2010 #2
    It may not be clear from the above, but I am considering heat loss from each side of a rectangular aquarium turn-by-turn. In the example I gave, I am only considering heat loss from one side.

    Also, is this perhaps the wrong sub-forum for this kind of question?
     
  4. Jul 9, 2010 #3
    In your calculation, you assumed that the temperature difference in case 1 is directly 29-25 = 4 degrees. However, temperature varies from the water surface, whose Th=29C, to the outside air, whose Tl=25C. So the temperature difference between the 2 faces of the pane is not necessarily 4 degrees, but smaller. Denote T the temperature of the face which faces the air, we have:
    P = (k_glass / d) * A * (Th-T)
    P = h_air * A * (T - Tl)
    We have (k_glass / d) = 110; h_air = 20. Thus: T = 28.4C, P = 16.5W!
    Now you may rethink about your conclusion :smile:
     
  5. Jul 9, 2010 #4
    hikaru1221, thank you for your reply.

    I now see that the temperature will not be a full four degree difference as you pointed out, but I am not quite clear how you settled on T = 28.4C ? Will this value be a function of the thickness and material of the pane and the overall difference in temperature?

    Also, in the convection equation:
    P = h_air * A * (T - Tl)
    Shouldn't this remain as:
    P = h_air * A * (Th - Tl), unless we are convecting heat off of the panel (rather than directly from the surface of the water)?

    And just one last question, I noticed that you used h_air for the convection equation whereas I used h_water - was I mistaken in doing so?

    Thank you for your time.
     
  6. Jul 9, 2010 #5
    Yes. Plug in the values of those two and you will get T=28.4C.

    I was considering the 1st case, where the pane is present. Convection doesn't happen inside the pane. Heat flows from the water through the pane in conduction, and then through the air in convection.

    First, have a look at the last 2 lines :smile:
    http://en.wikipedia.org/wiki/Heat_transfer_coefficient#Overall_heat_transfer_coefficient

    Again, I was considering the 1st case (sorry for not specifying this). The temperature at water surface is 29C, which is my assumption. In your solution, in case 1, you only consider the heat loss due to the pane only, which means you ignored the convection in the water (or assumed that the temperature was uniform in the water) and this is my reason for the assumption. So convection we should consider here is due to the air.

    Actually convection in the water does happen. You may recalculate heat loss rate by including it besides conduction in the pane and convection in the air, just as I did.
     
  7. Jul 9, 2010 #6
    I see, so the heat transferred by conduction through the pane must be equal to the heat transferred by convection from the pane into the air? I did not realize you were stating a system of equations, I thought they were two independent equations.

    I see that 0.1m of glass doesn't offer the water much insulation from heat loss - 16.5Watts versus 20Watts. Glass really is not much of an insulator.

    Thank you so much for your help here hikaru1221. Do you mind if I ask you another question? Does my estimation for heat loss by radiation look reasonable? Just 0.0043 watts?
     
  8. Jul 9, 2010 #7
    It's 0.01m, or 1mm, too thin to insulate anything. If you consider another more practical case, a glass pot of thickness 5.5mm containing boiling water (at 100C)when the surrounding is at 20C, you will see that T = 60C! They don't produce glass pot for no reason :smile:

    Well I was kind of surprised when seeing that em of water = 0.95. Is water that close to black body? I did a little search and found that em of water at 38C = 0.67, so you should check your source.
    Anyway your calculation is fine. Even if em=1, it doesn't matter much.
     
    Last edited: Jul 9, 2010
  9. Jul 10, 2010 #8
  10. Jul 10, 2010 #9
    Wow, I again did a search then found, just like you said in most sites, it's around 0.95 at 300K or 27C. I didn't expect the emissivity coefficient of water to vary that much from 0.95 at 27C to 0.67 at 38C. I had better checked my source :biggrin:
    Anyway even if emissivity coefficient = 1, the heat loss rate due to radiation is still infinitesimal, right?

    I don't know much about this topic, so I cannot tell if it's right or not. The formula is empirically derived, and as we can see, h only depends on v. I think it's much more complicated, depending on the conditions we conduct the experiment. For example, in the extreme case where the density of the air is infinitesimal (i.e. ~ vacuum), convection hardly happens. However it does agree with Wikipedia about the range h should be under normal conditions, i.e. from around 10 to 100.

    I guess you want to do some exact calculations for some practical purpose. However this is the best I can do - to give qualitative opinions :smile:

    EDIT:
    I'm sorry, I made a big mistake. Your calculation for heat loss rate due to radiation is wrong. You must use the absolute temperature instead. The result is about 6W, not so small at all!
     
    Last edited: Jul 11, 2010
  11. Jul 11, 2010 #10
    This is quite embarrassing, I can see my mistake now.

    Actually the first time I used the equation, I thought it was (Th - Tl)^4 and so I thought to myself celsius/kelvin wouldn't matter. I forgot to re-consider that when I corrected it to (Th^4 - Tl^4). Thanks for the correction.
     
  12. Jul 11, 2010 #11
    So since the heat loss by radiation is non-trivial, should we revise the system of equations to:

    P = (k_glass / d) * A * (Th-T)
    P = ( h_air * A * (T - Tl) ) + (em * sbc * A * (T^4 - Tl^4))

    and use this to solve for P?
     
  13. Jul 11, 2010 #12
    Actually there is problem with the heat loss rate by radiation. The exact equation is:

    P = P(29) - Ps = em(29)*sbc*A*Th^4 - Ps(25/29)

    Ps(25/29) is the power of radiation that is emitted by the surrounding at 25C and absorbed by the water at 29C. However I don't think we can find the exact formula for Ps(25/29) for the following reason. The power of the radiation coming in the area A, emitted from the surrounding may be written as:

    P' = e(surrounding) * sbc * A * Tl^4

    However, since water doesn't absorb all the radiations, Ps(25/29) < P', or Ps(25/29)=kP', where k<1. We may luckily find e(surrounding) on some website, but not k as k depends on the material (water), the wavelength of the radiation (we don't know at which range of wavelength the air radiates), the dimensions of the tank and water's temperature. It's too specific and thus, most likely unavailable on the internet.

    At first, I thought of some way to go around. If the water is at 25C, then it's in thermal equilibrium with the surrounding, which means:

    the radiation power it emits P(25) = the one it absorbs from the surrounding = em(25)*sbc*A*Tl^4.

    P(25) = the power the water absorbs when it's at 25C. It should be different from the power it absorbs when it's at 29C, though the same surrounding's temperature. However as emissivity = 0.95, very close to 1, we may expect that it changes slightly from 25C to 29C, i.e. em(25)=em(29). The same idea arose in my mind. We may expect that the ability to absorb radiation of water (or its absorptivity) changes slightly from 25C to 29C. Thus: Ps(25/29) = P(25), and that leads to your equation.

    I'm not really sure about my reasoning. I'll think of some other way or search on the internet; hopefully I can find something. Any other opinion would be appreciated.
     
  14. Jul 12, 2010 #13
    hikaru, I don't completely understand the issue.

    The equation I am using for heat-loss by an object via radiation to it's surroundings is:

    q = ε σ (Th^4 - Tc^4) A

    from: http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

    They do not indicate on that page anything about wavelengths? Is this a simplified form the of the equation, some sort of approximation?
     
  15. Jul 12, 2010 #14
    I'm not really sure about the issue. Both my thought and the website lead to the same equation, i.e. your equation. However I still think it's just an approximation. Anyway, since I'm not quite sure and the website seems to be reliable, you may apply the equation :smile:
     
  16. Jul 12, 2010 #15
    Alright then :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic Problem with Heat Transfer
  1. Heat transfer problem (Replies: 2)

  2. Heat Transfer Problem (Replies: 3)

Loading...