Research is done... Input is Needed...

[JustAnotherAntMarching]

Hey guys so im thinkin about gettin another tent and before i order anything i just wanna put the idea out there for some final input...

Plan :
tent 4'x4'x6 1/2' high
lighting 600 watt mh hps Digital Greenhouse Ballast 95,000 lumens
reflector 4 foot SunSoaker

16 Autos in Promix-BX in 3 Gallon pots... This is basically the entire floor space...

If im doin the math right 95000/16 sq ft = 5937 lumens per sq foot... thats in the correct range right?? 5000 a foot is ideal right?

My only concern is heat...? How much of an issue is heat with a 600? Ive only used LEDS thus far in my short growin experience and heat is virtually no issue with them...

Anyone have a SunSoaker? is it worth the $? i figured it would spread the light very evenly and keep almost all the light focused towards the bottom of the tent bc its the exact same size...

As always thanks in advance...

 

[StoneyBud]

You'll need some good ventilation.

An air cooled hood would also make it much better.

Other than that, it sounds like you've got it figured out!

Nice!

 

[JustAnotherAntMarching]

Stoney do you think an air cooled hood is better then the big reflector? the tents have like 6 holes i can run hot air out of...?

 

[StoneyBud]

Stoney do you think an air cooled hood is better then the big reflector? the tents have like 6 holes i can run hot air out of...?

Amost all of your heat is produced by your light. All you need on your tent area is a 300% cubic foot per/min replacement of air which with your area, that's 4 x 4 x 6.5 x 2 = 208cu ft per/min.

That's a pretty small exhaust. Your hood exhaust will make it perfect.

Yes, the hood exhaust is much more important than a large reflector.

The inside of the tent is white, right? In that little of area, a large reflector isn't necessary.

Grow on man!

 

[JustAnotherAntMarching]

Awesome thanks for the advise... the cool tube hood is alot cheaper then the reflector anyway... and yes the tent is white mylar on the inside...

 

[StoneyBud]

"All you need on your tent area is a 300% cubic foot per/min replacement of air which with your area, that's 4 x 4 x 6.5 x 2 = 208cu ft per/min."

Hahahaha, no, I didn't smoke no weed before writing that. I gave you 300% cu ft per/min requirements and math for 200%.

Just noticed that.

Well, it's actually 4 x 4 x 6.5 x 3 = 312 cu/ft per/min needed.

You should be able to find a hood exhaust that gives that much easily.

I would suggest leaving all the tent holes open for passive fresh air intakes.

Are you going to plumb your exhaust air outside or just recirculate it inside your home via the A/C? If you're going to recirculate it, you'll need to get it into your return air system somehow.

You may even want to think about an outside air source for your air intake into the tent. That would require intake fans though, to avoid undue wear on your hood fan.

The intent is to have two criteria met:

1. The heat from the light is dealt with at the source.

2. Great CO2 and Oxygen replacement is enabled for the plants.

There are several ways to arrive there. You just have to pick the method that most fits your lifestyle and finances.

 

[JustAnotherAntMarching]

Stoney

it is actually not goin to go in the house... its gona go in a shed in the woods outside for smell reasons.... I was plannin on venting the exhaust right thru a wall directly outside...

Also i was going to use 2 inline fans that say they move 160 cfm. Will that be sufficient?

 

[StoneyBud]

Stoney

it is actually not goin to go in the house... its gona go in a shed in the woods outside for smell reasons.... I was plannin on venting the exhaust right thru a wall directly outside...

Also i was going to use 2 inline fans that say they move 160 cfm. Will that be sufficient?

Yep, you need 312 cfm and you'll have 320 cfm. Sounds perfect to me!

With that much air flow, it's doubtful that much smell will be noticeable. The Odor is particulate in nature. By having a 300% turn-over of the air in your grow area, you'll have the particulate count down to almost nothing. Vent it up, so that it will be very hard for someone to get their face in the flow.

 

[JustAnotherAntMarching]

Hey Stoney this may seem retarded but....

the 2 reflectors/hoods im looking at have an intake from outside the tent into one side of the light and then the exhaust feeds out the other side of the light and then out of the tent.... correct?

So i guess my question is: How did you come up with that calculation that i need 312 cfm of air per min?

Is there something im missing? you say i need to move 300% of the volume of air per minute but the hood is not going to draw any air from the tent...?

its just gonna pull air from the intake thru the light and out the other side... no?

Is that amount of air needed just to cool the light? or is that supposed to be the air turnover in the whole tent?

Im just confused... not high unfortunately...

 

[StoneyBud]

Hey Stoney this may seem retarded but....

the 2 reflectors/hoods im looking at have an intake from outside the tent into one side of the light and then the exhaust feeds out the other side of the light and then out of the tent.... correct?

So i guess my question is: How did you come up with that calculation that i need 312 cfm of air per min?

Is there something im missing? you say i need to move 300% of the volume of air per minute but the hood is not going to draw any air from the tent...?

its just gonna pull air from the intake thru the light and out the other side... no?

Is that amount of air needed just to cool the light? or is that supposed to be the air turnover in the whole tent?

Im just confused... not high unfortunately...

There are two things that need done.

1.The heat that is produced by the bulb needs to be drawn off and removed from your tent area.

2. The air in the tent needs to be refreshed for oxygen and CO2 rich air from outside the tent needs to be pulled into the tent.

You can do both with the single exhaust from your light by using it to pull the air from inside your tent, through the light and out of the tent.

The cubic feet of your tent are gained by multiplying width by length by height. 4 x 4 x 6.5 = 104 cu ft

By pulling 3 times that amount of cubic feet of outside air through your tent and into your light and then out of the tent again, you've replaced the air in your tent 3 times for each minute the fan is on.

That will result in your tent being the same temperature as outside the building if that fan is large enough to cool the heat created inside the tent.

Of course, if it's 90F outside the tent, it will be 90F inside the tent. You said it's 79 outside the tent, so if your fan is large enough to cool the amount of heat being created, the tent should be 79 also. If it's NOT cooling it enough, then your fan isn't large enough to deal with the created heat.

Either you have to use your hood fan to pull that air in, or use another fan to do so. Somehow, you need to replace the air in your tent for fresh air.

Let's make sure we're talking about the same thing....

 

[JustAnotherAntMarching]

Stoney thanks again for all the help... the math is the easy part...

I think that we just had our thoughts crossed...