Heat Loss from an Enclosure using Fourier's law & m.Cp.DeltaT 1

[EngAP]

Dear all,

I have calculated the predicted heat loss from a sealed enclosure for various deltaT's (internal temperature to external temperature) using Fourier's Law and achieved reasonable accuracy, even when assuming to be through a straight wall rather than from an enclosure. The theory was confirmed by measuring average internal and external temperatures compared to the heat input.

I then ran physical tests, heating the internal air to set temperatures and measuring the rate of cooling to confirm my theory. But I am getting strange results.

Assuming 0.004 Kg air (m).
Specific heat (Cp): 1005 J/Kg.K
Delta T (initial temp at time = 0 to final temp at time = 100 seconds): 10K
t (time) = 100 secs

Therefore heat removed = Q = m.Cp.deltaT = 0.004 * 1005 * 10 = 40 Joules.

Therefore the avaerage heat flux during the 100 seconds should be Q/t = 40/100 = 0.4W. But this is a lot lower than I was expecting.

If anyone has any ideas or suggestions where I may have gone wrong (in calculation, assumptions etc.) I would love to know.

Note that an internal fan spread the heat as much as possible to achieve an even internal temperature.

Grateful for any advice.

 

[BigInch]

Was it a sealed pressure container? If not, you might have lost half the heat with the escaping air volume.

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[dcasto]

How did you keep the temperature of the cooling medium exactly the same differential? Q = U A delta T if delta changes, then Q changes.

 

[IRstuff]

What were you expecting? Just because you only need 40 J, doesn't mean that you can get the 40 J onto the right molecules.

Just consider a typical home A/C. There are losses EVERYWHERE, from the ducting, the fan, the registers, the attic insulation, the windows, the walls, etc. If Maxwell's demon were able to only cool the air molecules, your A/C would only need to be half the capacity that you have.

Another thought experiment is to consider a 1/2 W resistor running at max power suspended in still air. It'll get hot enough to blister the resistor.

TTFN

FAQ731-376

 

[prex]

0.004 kg of air makes about 4 cubic decimeters, so a cube of some 15 cm side. If you have a fan inside the enclosure, then the heat capacity of the fan itself, the heat input to the fan (100 W?), and presumably the heat capacity of the enclosure, total to something that is orders of magnitudes higher than the heat content of air. I don't think that you can obtain anything from such tests.
Only measurements in equilibrium conditions (that you tell us you have done) can give meaningful results (but don't forget the fan, if it is running during the test).

prex

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[EngAP]

Thanks for the prompt replies.

BigInch - yes - fully sealed and insulated enclosure.

prex - Thanks - this is what I was looking for, errors in my asumptions. I had not considered the heat capacity of the enclosure and am not sure how this would affect the cooling down time. The fan produces 1.8W heat.

IRStuff - steadystate experiment and theory suggested approximately 10-20W to maintain deltaT in similar region to that from which it cooled down from.

dcasto - measured the cooling air temperature during the cool down period and took this into account.

I am only looking to get an approximate answer but I would have thought that if I had made the correct assumptions/calcualtions then I would get a heat loss in the same magnitude as that predicted by theory/steadystate calcs. I am hoping to be able to graph heat loss vs. deltaT by taking small time steps and assuming constant deltaT during the small time step.