# Moist is a horrible word

January 29, 2011 3:14 AM Subscribe

My new dehumidifier has me curious...how many gallons of water are in the air in my building at a given temp and relative humidity? Plus a couple of related questions.

I have a storage building divided into two sides, one (newly) insulated and heated, the other not. Each side has a volume of roughly 9500 cubic feet, or 270 cubic meters.

The temperature in the "cold" half is 50F (10C), and the relative humidity is 75% (ouch). If this was a completely static system, how many gallons (or litres) of water needs to be extracted from the air to bring it down to 50% humidity?

Likewise, the "warm" side is now 60F and has a relative humidity of 65%. How many gallons or litres need to be extracted to bring it down to 50%?

(I recognize that the slab and other objects in the building are now quite damp and will be giving that dampness back, plus, of course, the building isn't air-tight, so the actual amount of moisture that needs to be removed from the entire system is substantially more than the amount I'm asking about. I'm just curious at looking at a "snapshot" of conditions at any one time.)

I have a storage building divided into two sides, one (newly) insulated and heated, the other not. Each side has a volume of roughly 9500 cubic feet, or 270 cubic meters.

The temperature in the "cold" half is 50F (10C), and the relative humidity is 75% (ouch). If this was a completely static system, how many gallons (or litres) of water needs to be extracted from the air to bring it down to 50% humidity?

Likewise, the "warm" side is now 60F and has a relative humidity of 65%. How many gallons or litres need to be extracted to bring it down to 50%?

(I recognize that the slab and other objects in the building are now quite damp and will be giving that dampness back, plus, of course, the building isn't air-tight, so the actual amount of moisture that needs to be removed from the entire system is substantially more than the amount I'm asking about. I'm just curious at looking at a "snapshot" of conditions at any one time.)

Best answer: Meh, of course I forgot the question and misread some numbers:

Warm side has (total) 0.5 gallons, you want it to have (9.4*0.5*270) mL in gallons = 0.34 gallons, thus you would need to remove about 0.16 gallons.

Cold side has (total) (12.83*

I think I fixed things that time... I hope.

posted by anaelith at 4:56 AM on January 29, 2011

Warm side has (total) 0.5 gallons, you want it to have (9.4*0.5*270) mL in gallons = 0.34 gallons, thus you would need to remove about 0.16 gallons.

Cold side has (total) (12.83*

**0.6***270) mL in gallons = 0.55 gallons, you want it to have 0.46 gallons, you would need to remove about 0.09 gallons.I think I fixed things that time... I hope.

posted by anaelith at 4:56 AM on January 29, 2011

Response by poster: Thanks much!

(Fairly astonishing that the amount is so minor, guess I hadn't really thought about it--assuming a completely air-tight building of that size and temperature [and perfectly dry air to start], a spilled gallon of water would never evaporate completely.)

posted by maxwelton at 12:02 PM on January 29, 2011

(Fairly astonishing that the amount is so minor, guess I hadn't really thought about it--assuming a completely air-tight building of that size and temperature [and perfectly dry air to start], a spilled gallon of water would never evaporate completely.)

posted by maxwelton at 12:02 PM on January 29, 2011

No problem, yeah, air doesn't hold all that much water... although weight wise it makes more sense. If it were fully saturated, the air would be holding roughly one percent of its weight in water...

(One more correction, in the second post where I said warm it should be cold, and cold should read warm. Meh.)

posted by anaelith at 12:17 PM on January 29, 2011

(One more correction, in the second post where I said warm it should be cold, and cold should read warm. Meh.)

posted by anaelith at 12:17 PM on January 29, 2011

Although the static moisture in the air might be an interesting figure, it has little to do with de-humidifying in the real world. This is because, as you mentioned, buildings aren't air tight. And you might be surprised at how rapidly air turns over.

The term is air exchanges per hour which means how rapidly the total volume of air turns over in a space. For a warehouse it might be 1 or 2 times per hour. So, using anaelith's numbers you might need to remove about half a gallon an hour or more just to keep up with the new air flow.

For example if you want to know how big a drain you need for your bathtub to make sure it never overflows, it really doesn't matter whether the tub holds 20 gallons or 30 gallons. What you want to know is how fast the tap pours into the tub when the faucet is full on. If you can't keep up with the inflow, it doesn't really matter how much the tub holds. The same goes for how much moisture the air holds.

posted by JackFlash at 2:40 PM on January 29, 2011

The term is air exchanges per hour which means how rapidly the total volume of air turns over in a space. For a warehouse it might be 1 or 2 times per hour. So, using anaelith's numbers you might need to remove about half a gallon an hour or more just to keep up with the new air flow.

For example if you want to know how big a drain you need for your bathtub to make sure it never overflows, it really doesn't matter whether the tub holds 20 gallons or 30 gallons. What you want to know is how fast the tap pours into the tub when the faucet is full on. If you can't keep up with the inflow, it doesn't really matter how much the tub holds. The same goes for how much moisture the air holds.

posted by JackFlash at 2:40 PM on January 29, 2011

Response by poster: Thanks, JackFlash. I have a dehumidifier out there which can remove .25 gallon per hour at full bore. The building is pretty tight from a liquid-water-getting-inside standpoint, but I'm sure it's leaky as heck from an air standpoint. I suspect I'm going to need a bigger hammer.

posted by maxwelton at 6:51 PM on January 29, 2011

posted by maxwelton at 6:51 PM on January 29, 2011

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270 cubic m.

rel hum 75%

s. vapor density 9.4 gm/m^3

9.4*0.75 = 7.05 g/m^3

7.05*270 = 1903.5 g

1903.5 mL in gallons = 0.5 gallons

15C

270 cubic m.

rel hum 50%

s. vapor density: 12.83 gm/m^3

(12.83*0.5*270) mL in gallons = 0.46 gallons

So about half a gallon on the cold/wet side and a little less on the warm side. I used this chart and this and the fact that 1 gram of water is 1 mL, and Google calculator.

posted by anaelith at 4:43 AM on January 29, 2011