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WTP dehumidification

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m2e

Mechanical
Jun 28, 2006
92
Hi

I'm designing a ventilation system on a water treatment plant. We're not using dehumidifier or AHU, we're only using fan for basic ventilation. There is going to be open water tank inside, and the latent load is going to be high.

The ambient condition: DB=68F, WB=60F

I'm trying to use the Psy chart to find the answer, but the explaination regarding this scenario is very limited, so I couldn't find the answer on my on even thought I've spent days researching on it. Is it possible to use fan only and exhaust away the moisture so that there's little or no condensation? If so, how do u do the calculation?

Thanks.
 
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Thank you so much for your link. Is it possible to demonstrate a sample calculation because I couldn't figure out how to do the psychometric exercise... thanks.
 
I'd use a more simple pool calculation to get a rough idea for evaporation rate.

Evap Rate = 0.1 * Area * (Vw - Vp)

Vw is the vapour pressure of the water, which is dependant on the water temperature.

Vp is the vapour pressure (vapor pressure in the US I guess) of the air, which is dependant on air temp and RH.

Once you have your evaporation rate you can determine what the change in humidity would be given your air.

The next step would be to determine the maximum desired RH within the space (say 60%) and then you would calculate the required outdoor air rate (at design wetbulb conditions) to dehum the space to your desired maximum.

The main issue I imagine is that, in conditions of high outdoor wetbulb you will little to no ability to dehumidify. If that is a concern, you'll have to look at some form of mechanical cooling, or dessicant dehumidication.
 
If condensation (and is assume mould or corrosion) is your concern, convert your resulting temperature and humidity into dewpoint (psychro chart or find the formula) and check against the inside wall temperature of your building, based on outdoor conditions and insulation rating.
 
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