So summer is coming and I'm already having problems with heat, last summer my room got as high as the mid 90's and I need a way to cool it off, I have an AC unit thAt doesn't work for shit and I'm sure I just completely wasted a bill on last summer when I didn't know anything about growing, and it barely cools my room at all. So what I'm trying to cool is a 5 by 10 tent in my garage so it's kinda difficult to cool. I heard a 5000 btu AC would cool it down enough, but others have said there is no way it would cool down my tent in my garage. So does anyone have any advice on cooling a room with high temps like mine? I do t wanna use co2 cause Its not an airtight room so I just have a couple exhale bags in there.
Will this be too hot?
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fir3dragon
Well-Known Member
Bigger ac only solution I think
The only thing I can think of is to insulate your tent. You could buy some rigid foam panals at Home Depot and construct a box around your tent. That should help to contain the cool air being pumped in by your ac unit, but I'm no expert on hvac. Good luck and happy growing.
shadedform1213
Member
Intermitten misting systems are often used by commercial nurseries to combat overheating issues also what is the circulation of the room like and the ventilation. Misting systems can be pieced together to lower the overall temperature however this would also require carefull calculation and an exhaust to compliment the total cfms of the intake fan idk if this helps but it could be a start on something. Let me know what you think
shadedform1213
Member
Ok heres something to think about say you have a water pressure of 45psi attached to a fine mister yeilding approximately 1629.8 grams of water an hour (.37gph) at about so 27.16grams of water is expended over the course of 60 seconds. The heat absorption rate of water is
4.18j/gram*°c meaning for each 4.18j a gram of water is raised by 1°c. And. 37gph*24 would equals 8.88 gallons of water a day... just wait for it. 347,487.7 grams of water per day so 347,487.7g*4.18j=1,452,498.5j of heat absorbed every 24 hour period. In other words that amount of heat energy is transferred into water with an abdorption rate of 4.18j per grm of x°C of water... ok now broken down further ventilation and all if done correctly should be roughly anywhere between an 8-15°f difference. There are other variables that i would need to know to give you a better idea. Please keep in mind ventilatiin will be needed to remove a certain amount of cfms to maintain a balanced air circulation rate ideally replacing the air every 5-8 minutes to reduce the effect of bacteria within the room. The convection current caused by the acr will also lower the overall temperature of the tent by pushing cooler air upward and forcing warm air lower and out of the tenr via exhaust.
4.18j/gram*°c meaning for each 4.18j a gram of water is raised by 1°c. And. 37gph*24 would equals 8.88 gallons of water a day... just wait for it. 347,487.7 grams of water per day so 347,487.7g*4.18j=1,452,498.5j of heat absorbed every 24 hour period. In other words that amount of heat energy is transferred into water with an abdorption rate of 4.18j per grm of x°C of water... ok now broken down further ventilation and all if done correctly should be roughly anywhere between an 8-15°f difference. There are other variables that i would need to know to give you a better idea. Please keep in mind ventilatiin will be needed to remove a certain amount of cfms to maintain a balanced air circulation rate ideally replacing the air every 5-8 minutes to reduce the effect of bacteria within the room. The convection current caused by the acr will also lower the overall temperature of the tent by pushing cooler air upward and forcing warm air lower and out of the tenr via exhaust.
shadedform1213
Member
Im assuming you have an intake fan and that its pointed upwards in some fashion inside ur tent. And that you have an exhaust at the top of ur tent pushing air either out of the tent but prefferably outside ur garage. If not then i hate to tell you you r pretty much roasting yourself.
The air you are exhausting from your tent must be directed outside ir to another room... while pulling fresh air or carbonated air from outside ideally. However the misting refers to a mister or set of misters placed at a point where the mists can be circulated to the rest of the confined space. Generally cold water for heat reduction applications but misters are set in the pathway of incoming air usually close to the intake or an intake booster fan (depends on the space) which the mist coming from the misters is the driven upward toward the warmer air fircing a rapid heat transfer and cooling of warmer air, air that doesnt cool is the exauhsted via exhaust fan and exoelled from the tent. Keeo in mind this scenerio will need about even airfliw both in and out of the cyclic repitition or airfliw to reach peak efficiency.
cheechZS
Active Member
Most garages r not insulated to begin with so its gonna b a battle. Your gonna have to ventilate the garage along with adequate tent ventilation to remove the removed heat. I would jump up to a 12000 ac(that provides 1 ton of conditioned air,which will cool aprox 400 sq ft). I know its more than needed but depending on where u live and lights used it may b the only way to combat the heat load. Best way would b to insulate garage and install mini split ac.......hvac expert 10 yrs experience
dtowndabber
Well-Known Member
The only reason the ac wouldn't work is because you have to vent the exhaust out of the garage. Other then that I don't see why it wouldn't work. But I would think more like 7500 if you're cooling the whole garage.
hello sir seens you have lots of know how please bare with me my floweringroom is 11 x 11 with 1000 w hps im Exhusting my room to my basement pushing the hot air down a 10x10 inch vent chase i built in my downstairs closet so it dumps into my basement can i still use my 300 sq.ft cfm fan to push the hot air down the total push down is about 15 feet ? please help