Fogger?

[w4tch]

My room size is 12'x12'x8'.

I'm running 3kw (5.5 ft footprints) and I'm cooling the room with ac.

My ac unit however can dehumidify up to 8 pints/hour. Is there anything that can humidify faster than the AC can dehumidify (faster than 8pints/hour)?

Right now my rh is sitting around 40%. My temps are around 75F lights on and off. I would like to get the rh up to around 60-65%.

Please Help!

Thanks.

 

[r2k]

Yow!! That's a gallon per hour. Good luck with that. I doubt your unit will be able to pull 8 pints per hour with humidity that low, but it is still tough to handle.

Condensation forms when some surface (like the condenser coils) is colder than the dew point temperature. With 75*F and 40%, the dew point temperature is 49*F. Is there any possible way to keep your AC coils higher than that? I doubt it, or you would probably lose AC efficiency.

Getting humidity up is hard because it takes so damn much energy to evaporate water. Pulling humidity out of the air like that will make your AC work MUCH harder than if it only had to cool the air because condensing water takes cooling power. Putting water back into the air is just going to fight the job your AC unit is doing to pull it out. The energy required to evaporate the water in the air is called latent heat. "Latent" is latin for "hidden" or "concealed".

Just to do a quick calculation, let's assume you want to evaporate 8 pints of water in an hour. How many watts would you need to do that?

8 pints of water = 3.78 kilograms
latent heat of vaporization for water is 2.26 x 10^6 joules/kilogram
joules = (2.26 x 10^6) x 3.78 = 8542800 joules
1 watt = 1 joule/second
1 hour = 3600 seconds

watts = 8542800 joules / 3600 seconds = 2373 watts (ouch!)

Even if you only had to replace only 3 pints of water, that is still 889 watts. This means that to convert 3 pints of liquid water into vapor without changing the temperature, you would need to provide heat energy at a rate of 889 watts. In practice, it would probably be more because that kind of heating would raise water temperature and require even more energy. Once you evaporate that water, your AC unit will happily condense it back into water again and you are back where you started.

Ideally, what you would want to do is make a MUCH larger surface area for your condenser coils, such that the condenser does not go below about 50*F and won't form condensation. I don't know how to do that. I suppose you could run the AC chiller part in a closed space such that it will only chill the same air over and over, then circulate that air through closed pipes through your grow space to warm up. It would be a closed loop system, kind of like the old method of heating a house with hot water running through radiators in each room. (Except, you would be circulating cold instead of heat.) This becomes inefficient unless you add lots of ducting, heat exchangers, and fans.

I don't suppose anybody has ever looked at something like a water based circulation system to cool their space? Think of those old soda or beer bottle chillers where the bottles sit in a bath of cold water. Instead of bottles, you would circulate that cold water around in your grow space with radiators to collect heat and warm up the water (and cool off the air). The warmed water goes back into the big tank and cools off but is never directly exposed to the air in the grow space. I think really big buildings like college campuses use this kind of approach to have one massive central chiller and then distribute the cold by pumping water around to each building. As long as the radiators in the grow space stay above the dew point, there will be no condensation problems.

I know this post is probably not much practical help to solve your problem, thanks for listening.

-r2k

 

[w4tch]

Yow!! That's a gallon per hour. Good luck with that. I doubt your unit will be able to pull 8 pints per hour with humidity that low, but it is still tough to handle.

Condensation forms when some surface (like the condenser coils) is colder than the dew point temperature. With 75*F and 40%, the dew point temperature is 49*F. Is there any possible way to keep your AC coils higher than that? I doubt it, or you would probably lose AC efficiency.

Getting humidity up is hard because it takes so damn much energy to evaporate water. Pulling humidity out of the air like that will make your AC work MUCH harder than if it only had to cool the air because condensing water takes cooling power. Putting water back into the air is just going to fight the job your AC unit is doing to pull it out. The energy required to evaporate the water in the air is called latent heat. "Latent" is latin for "hidden" or "concealed".

Just to do a quick calculation, let's assume you want to evaporate 8 pints of water in an hour. How many watts would you need to do that?

8 pints of water = 3.78 kilograms
latent heat of vaporization for water is 2.26 x 10^6 joules/kilogram
joules = (2.26 x 10^6) x 3.78 = 8542800 joules
1 watt = 1 joule/second
1 hour = 3600 seconds

watts = 8542800 joules / 3600 seconds = 2373 watts (ouch!)

Even if you only had to replace only 3 pints of water, that is still 889 watts. This means that to convert 3 pints of liquid water into vapor without changing the temperature, you would need to provide heat energy at a rate of 889 watts. In practice, it would probably be more because that kind of heating would raise water temperature and require even more energy. Once you evaporate that water, your AC unit will happily condense it back into water again and you are back where you started.

Ideally, what you would want to do is make a MUCH larger surface area for your condenser coils, such that the condenser does not go below about 50*F and won't form condensation. I don't know how to do that. I suppose you could run the AC chiller part in a closed space such that it will only chill the same air over and over, then circulate that air through closed pipes through your grow space to warm up. It would be a closed loop system, kind of like the old method of heating a house with hot water running through radiators in each room. (Except, you would be circulating cold instead of heat.) This becomes inefficient unless you add lots of ducting, heat exchangers, and fans.

I don't suppose anybody has ever looked at something like a water based circulation system to cool their space? Think of those old soda or beer bottle chillers where the bottles sit in a bath of cold water. Instead of bottles, you would circulate that cold water around in your grow space with radiators to collect heat and warm up the water (and cool off the air). The warmed water goes back into the big tank and cools off but is never directly exposed to the air in the grow space. I think really big buildings like college campuses use this kind of approach to have one massive central chiller and then distribute the cold by pumping water around to each building. As long as the radiators in the grow space stay above the dew point, there will be no condensation problems.

I know this post is probably not much practical help to solve your problem, thanks for listening.

-r2k

No, you did solve my problem! Thanks. I know know i need to gut my bitch and find a way to make this shit run how i want it too.

Thank you. Seriously!

 

[growshopfrank]

A small "swamp cooler" would do it

Try looking up DIY swamp cooler or swamp cooler attachment for fan for inspiration

 

[r2k]

If you were really clever (and I mean, REALLY clever), you could kill two birds with one stone. You have 3.3 KW of lights, so that's free heat. It's already in the room. If you could use it to evaporate water instead of discarding it with the AC, you could:

1) Cut down the total amount of heat the AC would need to move. This cuts AC load and reduces your power bill.

2) Increase humidity as desired.

Again, I'm not sure how to do that safely, but the math works out. I think you would need to warm the water with heat from the light (not sure how) and then cause it to evaporate, maybe with one of those cool mist humidifiers that uses warm water. This gets to be quite a challenge in a confined space if you assume fire and shock hazards are to be avoided.

Like I said, REALLY clever. Let us know how it turns out.


-r2k

 

[w4tch]

A small "swamp cooler" would do it

Try looking up DIY swamp cooler or swamp cooler attachment for fan for inspiration

Thank you very much! I'll check those out!

If you were really clever (and I mean, REALLY clever), you could kill two birds with one stone. You have 3.3 KW of lights, so that's free heat. It's already in the room. If you could use it to evaporate water instead of discarding it with the AC, you could:

1) Cut down the total amount of heat the AC would need to move. This cuts AC load and reduces your power bill.

2) Increase humidity as desired.

Again, I'm not sure how to do that safely, but the math works out. I think you would need to warm the water with heat from the light (not sure how) and then cause it to evaporate, maybe with one of those cool mist humidifiers that uses warm water. This gets to be quite a challenge in a confined space if you assume fire and shock hazards are to be avoided.

Like I said, REALLY clever. Let us know how it turns out.


-r2k

Thanks!

I've heard about soaking towels and hanging them. I've also heard of blowing a fan on a surface of water. I'm gunna give everything a try. Guess you dont really know until you try!