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Heat Loss From a Closed Box - Experimental Help

  1. Jun 15, 2011 #1
    Hi everyone. I am currently carrying out an experiment whereby I have a closed tank of water submerged in a larger tank. The water in the smaller box is heated and I am interested in the heat loss from the tank. I have done a rough theoretical calculation based on U values, where by I obtain:

    Q (heat loss) = A*U*DeltaT

    and end up with a value of 10W. I have calculated the Wattage of the heat source I am supplying to the box as 1KW so in theory this should mean the heat loss will be negligible, correct?

    Obviously this heat loss calculation is only rough, I had to calculate U values from calculated resistance values of the materials (as the walls of the tank are of composite construction) by the relation R=d/k where d is the thickness of the material and k is the thermal conductivity of the material. Then calculated the U value as U=1/(R1+R2+R3). Providing this is even correct the calculation will still be rough as I am not taking into account any thermal bridging effects of the joins in the tank etc.

    I want to check this calculation experimentally, to do this I will fill the tank with heated water, seal it and then submerge it in the larger tank of cooler water and monitor the temperature drop within the small tank. I am then just a little puzzled as to how to relate this to my theoretical calculation and obtain Q (the heat loss) from the experimental data. Will I just use Newton's law of cooling and the relation: T=(Tint - Toutside)*exp(-A*t/m*C*R) + Tout to calculate R from the experimental data and then plug this back in, in the form U=1/R to my initial calculation for Q above and compare?

    Any help/advice would be much appreciated.
     
  2. jcsd
  3. Jun 15, 2011 #2
    First, I would mention that natural convection resistance terms in the 'U' variable of

    Q (heat loss) = A*U*DeltaT

    are likely much larger then the conduction resistance. I can refer you to some natural convection correlations for enclosures if you need.

    The temperature of the small tank will increase enough to transfer whatever heat source Q you put in the small tank.

    You can apply the lumped capacity model you mentioned if the Biot number is less than 0.1. If not, there are other models that can be used. Let me know what Biot number you figure.
     
  4. Jun 16, 2011 #3
    When calculating the Biot number:

    Bi=(h*V)/(A*k)

    Am I supposed to take the h value of the water in my smaller tank and the k value of the shell? i.e. ktotal=k1+k2+k3 where k1, k2, k3 are the conductivities of each part of the composite construction?

    Thanks for taking the time to reply.

    As an aside:

    I should have made myself more clear originally in regards to when I said the heat loss will be negligible, the system heats itself to a steady temperature and ends up in a steady state, so the temperature difference between the inside and outside of the tank is constant. In this scenario there is then a constant heat loss from the tank but also a constant supply in terms of the heat source, my meaning of negligible is that the steady state temperature won't be far off what I predict in assuming the tank is perfectly insulated as the heat loss is only 1% of the heat I am pumping in, is my understanding of this fair?
     
  5. Jun 16, 2011 #4
    For the Biot number in relation to the lumped capacity method, h will be the convection heat transfer coefficient for the surface of the "solid" to the fluid around it. As you can see, the lumped capacity method will be an approximation to your system.

    For h in the Biot number, I would apply a natural circulation correlation for a vertical plate. I think the best approximation for k in the Biot number will be for the water inside your tank.

    I did not quite understand your second question. From your equation in your first post:

    Q (heat input to small box) = A*U* ( T-small box - T-large box)

    and solve for T-small box. This assumes the large box is large enough that its temperature does not change much.
     
  6. Jun 16, 2011 #5
    I see, ok thanks for your help with the Biot number stuff. So is this the only way of getting a handle on the U value or just the heat loss in general from the small tank experimentally? In my naivety I assumed it would be an easier thing to do that it seems to be...I also assumed that it would give a more accurate idea of the heat loss in the tank as I would have a plot of the tank actually 'losing' heat.

    Sorry my second question wasn't exactly clear, that isn't such an important issue, was just trying to think out loud and check I was understanding something.

    Thanks for your ongoing help.

    Jamie.
     
  7. Jun 16, 2011 #6
    Plug away at it and you will get the hang of it. Try to set up the calculations in MathCAD or Excel or similar package you like. After some practice, it will be second nature to you.
     
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