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0 pointsclayrichardson10y ago0 comments

Interesting.

  With 32.2C ambient, 14% RH, 1.35C DWP, the absolute humidity is 4.78 g/m^3
  With 24.4C ambient, 14% RH, -4.80C DWP, the absolute humidity is 3.12 g/m^3

Absolute humidity here is specified as grams of water per cubic meter of air.

This is similar to the AH at BRC. Our current model assumes a DWP of -10C — 10C, so we can predict similar performance. We're not taking into account volumetric capacity and the thermal interface of the heat exchangers. If you know anyone able to help point us in the right direction with those calculations, it would be much appreciated!

Our current understanding is the enthalpy of vaporization + fusion for water is (2257 + 333.55) = 2590.55 J/g we want to pull out of the air. This is assuming we can process enough air, since the dew point is below freezing a fair amount of the time.

This is by using this equation specified in the NASA technical note [0]:

  =((0.21668*((TMPC+273)^-1))*((DWPC+273)^-4.9283)*(10^(23.5518+(-2937.4/(DWPC+273)))))*1000

[0]: http://www.nasa.gov/centers/dryden/pdf/87878main_H-937.pdf

EDIT: Formatting is hard. Added source. Clarification.

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FiatLuxDave10y ago· 2 in thread

Physicist here. I'm confused why you are using the enthalpy of fusion in your calculation, since your project does not appear to be freezing the water after condensation. Also, you seem to be using the enthalpy of vaporization from boiling at 100 C in your calculations, which is incorrect for use around 30 C.

A better way to calculate your energy needs would be to look at a steam table, such as the one on Wikipedia's water data page, then take the deltaHvap. This is the accurate enthalpy of vaporization. https://en.wikipedia.org/wiki/Water_%28data_page%29

In theory you need to add the amount of energy to cool the air/vapor mixture to the dew point. However, this amount will likely be small enough to be within your experimental error. For low humidity applications (like Burning Man), you can assume you are cooling mostly air, which is around 1.2 Joules/K per m^3. At 30 K temp differential, and humidity of 4 g/m^3, this is only 9 joules per gram water, or pretty much negligible.

So, you are looking at a deltaHvap of around 2420 to 2450 J/g in this temperature range. This is reasonably close to your original estimate of 2590 J/g, and the good news is that the error is in your favor.

clayrichardsonOP10y ago

Thank you for the feedback, this is awesome!

There is always more to read up on and calculate :)

EDIT: The current dehumidifier does accumulate ice on the evaporator coils, depending on the ambient conditions. Once enough ice has accumulated, the compressor turns off and the ice melts into the collection tray. I'm assuming we have to depose the ice out of the dry atmosphere at BRC, since the dew point is below the freezing point of water. Is this not the case?

FiatLuxDave10y ago

If you aren't trying to make ice, then the energy which goes into making ice is waste. This cuts your efficiency. So, in that case, you do need to include the enthalpy of fusion in your calculations, just be aware that portion is waste. In fact it is doubly wasteful, because the ice reduces the thermal transfer rate and having to stop the compressor means your duty factor is lowered (less water per hour for same size machine).

There are ways to prevent this, although I'm not sure if they would work for your project. Adding salt is a classic, so you can supercool your water while remaining liquid. That's probably not great for drinking water, even with RO. If you can use the melting of the ice as a source of cooling, you can recycle the enthalpy of fusion. It really is a waste and you don't need to pay it. One way to do this is to use the melting ice to lower the heat rejection temperature of your compressor.

However, it may not be worth adding complexity just to cut your energy costs by 15%.

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