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Calculating heat loss from open tank

  1. Dec 14, 2009 #1
    I have a large circular tank made of concrete that has a volume of 9,300,000 litres. The diameter of the tank is 48.7 meters. The side wall of the tank has a height of 4.57 meters. The tank is set into the ground by 3.57 meters and 1 meter of the tank protrudes from the ground. The tanks' walls and bottom are 8" concrete with no insulation. The water in the tank has a temperature of +3degC. The tank has no top and is open to the environment, and using the weather here lately that would mean the temperature is -25degC. How would I go about calculating heat loss on this tank?

    Essentially this tank is a clarifier in a waste water treatment facility. There is 1000 cubic meters of effluent that is pumped into the bottom centerwell of this tank at a constant rate over a 24 hour period each and every day. 1000m3 leaves the tank by means of spillage over weirs at the top of the tank. The incoming effluent is 3degC. I need to be able to provide enough heat to increase the eflluent temp to 5degC and then maintain that temp.

    All help is much appreciated.
     
  2. jcsd
  3. Dec 14, 2009 #2
    The losses will be form 3 different sections -

    1) The top (calculate it using newton's law of cooling).

    2) The elevated section (from the ground) -
    You did not provide the temperature of the walls...whatever will it be, you gotta calculate the heat lost/gained from the water to the wall using the same newtons law of cooling.

    Actually, since the wall does not generate it's own heat, you can remove the temperature part of the wall...it will actually be a function of the temperature of the surroundings and the water.

    This heat transferred will go thought the wall to the surroundings (-25d)...you gotta use Fourier's law to calculate that.

    3) This section is the interaction between the wall + base and water...similar to case 2, but there instead of air, there's ground.
     
  4. Dec 14, 2009 #3
    I'm also interested in the answer and I believe only a full numerical simulation can give the result.
    I think dE_logics solution is only a correct solution if someone would constantly stirr the water in the container. With temperature gradients in the water the solution would be completely different.
     
  5. Dec 15, 2009 #4
    Yeah, I forgot about the k of water itself.

    It's not that easy...you need to involve calculus in this since water continuously enters.
     
  6. Dec 15, 2009 #5
    Tell me about it. I've been at this equation for the last four hours and am no further ahead than I was when I started.
     
  7. Dec 15, 2009 #6
    I did all this long ago...

    We have to use 2 equations basically -

    Q = -kA(dt/dx)

    Q = hA(dt)

    Here dt and dx and not the calculus ones.
     
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