Calculating how much moisture removal is required..

[Badbird2000]

I am working on a project for a local 4H camp. They have dormitories where the kids sleep, up 50 per dorm, with a shared shower/toilet room. 4 showers and 4-5 toilets. Currently there is no exhaust at all (not sure how they got away with that, as the windows are not operable). The shower/toilet room is supplied from a split system DX unit, but no return. I am adding exhaust fans with CFM calculated by the code, but they also want to install a small dehumidification unit in each shower/toilet room. I am trying to come up with a calculation/formula that will tell me how many pints/day i need to remove. It varies, i know as to how many kids are showering, etc, but i just need a starting point.

Any one have any pointers?

Thanks!

 

[Noway2]

This is a psychrometrics problem and you are looking at removing latent heat (the water) in addition to the sensible heat. Something like this PDF should give you an idea of the concepts involved.
I also notice you mention wanting to run a DX unit. From experience, DX units are not usually capable of sufficient de-humidification in water laden environments, such as gymnasiums and swimming pool areas and this typically requires a water based system.

 

[trashcanman]

I must take exception to the idea that DX systems cannot dehumidify a moist environment. It is done all the time. For example go to

http://www.dry-air.com/

and see for yourself. All the hotels with indoor pools have similar units.

 

[Noway2]

We should probably clarify a little bit here. For starters, the OP mentioned wanting to add a dehumidifier. Units like the ones you link to by dry-air, are specifically designed for moisture removal and are designed with functions such as reheat and variable speed compressors and fans to handle part load conditions, which make dehumidifying with a DX system notoriously difficult because of air flow requirements as well as compressor cycling (causing re-humidification).

In any case, with respect to following up on the question of sizing, the main thing that needs to be determined is the (latent) heat load of the room. Showers will act as atomizers putting a lot of water into the air at fairly warm temperatures. If you could get a set of readings, such as RH and dry bulb, or dry bulb and wet bulb temperatures that will give you a starting point for your calculation. You can then determine how much moisture you need to remove. For example, if the area is 95F dry bulb at 40% humidity, that gives you a ~100 grains of moisture per pound of air. If you are trying to condition the space down to 75F 50% humidity, you would need to reduce the moisture content to ~66 grains per pound of air. Of course if you are not cooling and just removing moisture, you need a lower relative humidity. All of this gets us into the realm of needing the size of the room, etc, to calculate the air volume, In any case, knowing the starting conditions and the target conditions, the supplier should be able to help you size a unit properly.

 

[willard3]

If you are not familiar with psychrometrics, find somebody who is. This is not a thing you can learn on the web.

 

[chicopee]

But there are several examples in the Perry Chemical Engineering handbook from which you can learn. Learning s/b a lifetime commitment.