3 replies to this topic

[Steve Hall]

I've been playing with calculations for heat loss from storage tanks, using inside heat transfer coefficients predicted from .25 * (Gr*Pr) ^0.25 * k/L

and outside coefficients using .14 * (Gr*Pr) ^0.33 * k/L

 

My calculations include radiation emission, loss through unwetted sidewall, and loss through the tank bottom to soil using 2 * D * ksoil * (Tinside - Tsoil) where D = tank diameter and ksoil = thermal conductivity of soil.

 

My bright idea was to test the results with a small-scale experiment that is described in the attached file. My "tank" is a soup pot that is 266 mm diameter which I filled with hot water. Then I measured the temperature as it cooled outdoors.

 

The pot cooled much more quickly than the calculations predict, thus the reason for this post. What explains the difference? The predicted heat loss is roughly 50% of the actual loss. I'm not asking for anyone to check my actual calculations, but if you have your own program to perform the calcs I wouldn't mind knowing your results. The question is whether the small scale of my experiment, low aspect ratio, etc, explain the underprediction in your opinion?

 

Thanks,

Steve

 

Attached Files

•  Heat Loss Experiment.pdf   54.11KB   425 downloads

[Bobby Strain]

Steve,

   I barely got through my heat transfer course, so I have avoided calculations. But, as I look at your results, I expect that you underestimated the heat loss through the lid where the transfer coefficient is significantly enhanced by condensation. Maybe you can make some adjustments for this.

 

Bobby

[Steve Hall]

Bobby,

Thanks. Yes, the underside of the lid was covered with condensate during my experiment.

Steve

[Steve Hall]

Of the 5.8 kg mass of water in the pot, 0.085 kg would need to evaporate in the first 70 minutes to account for the difference between the loss calculation and the measured temperature change. Thinking about the amount of condensate clinging to the lid when I opened it to take measurements, it seems plausible. It's also 50% of the energy calculated as lost through convection. Through the buoyant forces, circulation occurs in the vapor space bringing the vapors in contact with the cold lid resulting in condensation and a driving force to create more evaporation. Is this just a calculation that balances the heat transfer through the lid to cause condensation with the saturation concentration of water vapor in the space? I say "just", but it already sounds complicated!