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Moisture removal after sanitation

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hssiway

Mechanical
May 7, 2010
3
Hi,

I am new to food industry. I have a typical situation where I need to calculate the time required to remove moisture from a process room after sanitation time of 60 min.

We have AHU designed for 16000 cfm supply air and equivalent flow of outside air. The intent is to put exhaust fan for sanitation period. We have 100% dehumidified outside air supplied to room during sanitation period and exhaust fan of 16000 cfm will continuously exhausting air out of room.

The supply air temperature is 55F.
 
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Sounds like a homework problem, mass flow rate type.

Assume room is full of 100% saturated air.
With the 100% dehumidified (0 moisure in the air?), at 16000 cfm, assuming the dry air absorbs all it can and leaves at 50% saturated, do the math.
 
Thanks for interest...
Calculation for Sanitation Mode: ACH = 5.00
SA cfm = 8270 San. Time: 60 min
Conditions After San. Req'd SA
DB F 80 75 56
WB F 79.9 62.54 55.9
RH % 100% 50% 100%
Sp. Vol cuft/lb 14.09 13.68 13.194
Hum. Rat GR/LB 156 66 65
Vol. of dry air cuft/hr 148868 148868
Wt. of dry air lbs 10565 10882
Moisture in air grain/hr 1648218 718222
Moisture accumulated 929996 grains/hr
Moisture removal capacity of SA : 91 grains/lbs
Amount of air required: 134691 cuft/hr
Time req'd to remove moist: 16.29 min (with SA cfm)
Is this sound correct??
 
There is no such thing as 100% dehumidified air in the real world. This is a forum for those that solve/d their own homework.
 
I have posted my calculation above. Can some verify and let me know if I am in correct direction.
 
No, not in the correct direction - define the start and end conditions to clarify what you mean by "remove moisture" and please don't post some computer printout to make us try to decipher it.

If you have 55°F supply air and you assume a 53°F dew point, it will take infinity air changes to reach a 53°F dew point in the space. You can usually get to about the 95% point with 5 air changes.
 
When you say 100% dehumidified outside air at 55 deg, are you sure you're not confusing that with saturated air (the exact opposite)? If this is outside air coming off a cooling coil, it's close to saturation.
 
Your calculations do look like you are on the right track. You have determined the grains of moisture in the space after sanitation, and your final required, and the rate of removal.
 
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