George2324
Well-Known Member
how man you BTU of AC is really needed per 1000w of cob when running sealed room?
BTU AC per 1000w cob?
- Thread starter George2324
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Randomblame
Well-Known Member
Depends on ambient temperatures.
Today's COB's are ~50% efficient! That is, 50% is light, the other 50% is heat, which must be dissipated via heat sinks. In the case of a 1000w lamp, ~500w of heat must be dissipated!
If you consider that you want to have temperatures of 27-28°C in your sealed room/tent, the heat is sufficient at 20°C T.ambient and you can do without additional cooling.
However, if you have >30°C T.ambient, you can not get out without AC!
At up to 25°C T.a you should have no problems keeping the chamber below 30°C, only with good ventilation. Above 25°C it is enough to dimm the light a bit down to stay below 30°C.
Like you see, it all depends on your ambient temperatures.
Today's COB's are ~50% efficient! That is, 50% is light, the other 50% is heat, which must be dissipated via heat sinks. In the case of a 1000w lamp, ~500w of heat must be dissipated!
If you consider that you want to have temperatures of 27-28°C in your sealed room/tent, the heat is sufficient at 20°C T.ambient and you can do without additional cooling.
However, if you have >30°C T.ambient, you can not get out without AC!
At up to 25°C T.a you should have no problems keeping the chamber below 30°C, only with good ventilation. Above 25°C it is enough to dimm the light a bit down to stay below 30°C.
Like you see, it all depends on your ambient temperatures.
George2324
Well-Known Member
I have a 7m x 7m room ambient can be 30c in summer and 20c in winter.
I am thinking 24000 btu AC for 6000w at 50% efficiency but not sure if this will be enough
I am thinking 24000 btu AC for 6000w at 50% efficiency but not sure if this will be enough
Randomblame
Well-Known Member
Should be more than enough IMO. 6kw is not as much in a sealed 50m² room and if you really get heat problems you can still dimm the light down to 80 or 90%.
George2324
Well-Known Member
George2324
Well-Known Member
45w per sq ft is without co2 tho right?
45w per square foot would be what ppfd?
45w per square foot would be what ppfd?
Randomblame
Well-Known Member
35-40w or 15-20PAR/w per ft² without CO² and 40-50w or 20-25PAR/w/ft² with CO².
Depends on how hard you drive the COB's and how efficient they are.
But if you have 6.000w of COB at 50% effiency you get 3.000PAR/w of light.
Your room is 7x 7m=49m² or 525sft., so 3.000PAR/w :49m² = 61,22 PAR/w/m² or :525= 5,71PAR/w/sft.
This is okay for vegging but not enough for flowering light hungry plants
You need at least three times as much watts without CO² for the whole room!
With Cobby, you have exactly found the right contact, which can provide you with a suitable equipment. Maybe one 36v COB/sft. at 50-65w(1400-1750mA).
Depends on how hard you drive the COB's and how efficient they are.
But if you have 6.000w of COB at 50% effiency you get 3.000PAR/w of light.
Your room is 7x 7m=49m² or 525sft., so 3.000PAR/w :49m² = 61,22 PAR/w/m² or :525= 5,71PAR/w/sft.
This is okay for vegging but not enough for flowering light hungry plants
You need at least three times as much watts without CO² for the whole room!
With Cobby, you have exactly found the right contact, which can provide you with a suitable equipment. Maybe one 36v COB/sft. at 50-65w(1400-1750mA).
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George2324
Well-Known Member
The room maybe that size but I only have 7m2 of canopy.
So far I have the equipment to run 27w per cob and 4 cobs per square foot already I may have to run them at less power
So far I have the equipment to run 27w per cob and 4 cobs per square foot already I may have to run them at less power
George2324
Well-Known Member
If I run them at 115w per sq ft i get 50% efficiency.
That would be 57 par watts per square ft
That would be 57 par watts per square ft
Randomblame
Well-Known Member
Effiency depends on COB's and on how hard you drive them.
At 1,4A or 50w a gen. 6 1212C4/3000°k should be around 50% efficient and at 700mA ~55%, maybe a bit more.
3.000PAR/w : 7m²= ~430PAR/w/m² or :75sft= ~32PAR/w/ft² is a bit too much IMO.
I think, you will definately need a dimmer with such an amount of light and the abillity to keep the lights at least 24" from the tipps also in flowering.
4000w should be more than enough, also with CO², but you could easily expand to 10m² later to use it's full potential.
At 1,4A or 50w a gen. 6 1212C4/3000°k should be around 50% efficient and at 700mA ~55%, maybe a bit more.
3.000PAR/w : 7m²= ~430PAR/w/m² or :75sft= ~32PAR/w/ft² is a bit too much IMO.
I think, you will definately need a dimmer with such an amount of light and the abillity to keep the lights at least 24" from the tipps also in flowering.
4000w should be more than enough, also with CO², but you could easily expand to 10m² later to use it's full potential.
REALSTYLES
Well-Known Member
@George2324 what cobs are you running because 27w has a higher efficiency if you are running Cree CXB3590. Funny thing is no one asked what you are using.
VegasWinner
Well-Known Member
Here is a BTU calculator for HID lights. I think you can interpolate from HID to COB 1000w of heat is 1000w of heat.
https://www.sunlightsupply.com/tools/btu-calculator
https://www.sunlightsupply.com/tools/btu-calculator
PSUAGRO.
Well-Known Member
I don't recommend that, 4000btu per thouie is absolute bare minimum, compressor generally will never cycle off in the summer........op I would be around 6000+btu per 1000w for no issues/strain on controls
he told us his total light draw...........which specific cob doesn't matter for btu demand
REALSTYLES
Well-Known Member
Well funny thing is I don't use AC and running 1800w @56% efficiency my room isn't sealed but have done sealed rooms. All I can say is look at my grow that's going right now no AC and no synthetic nutes
Randomblame
Well-Known Member
To get theoretical PPF and PPFD values you need at least two values.
LER which is usually given in lm/w and
QER which is given in μMol/j.
In short, LER is the theoretical maximum at 100% effiency and QER is the effiency of the light spectrum.
For instance, full deep red light/660-670nm is ~5,5μMol/j, white COB's are between 4,5 and 5μMol/j.
For Citizen COB's in 3000°k/CRI 80 the QER should be 4,86μMol/j and LER should be 328lm/w if we use the values from the sheet below.
Now I will try to show how you to use this values.
If we use your 6000w expample, we need at first the lm/w values for the given current from the datasheet.
In this case at 27w one 1212C4 would run at ~800mA, which results in 160lm/w.
LER(theoretical max.) is 328lm/w but the real output is 160lm/w at 800mA(from datasheet), so the effiency is 49%.
You have 6000w and your light is 49% efficient so 6000w x 0,49%= 2.940PAR/w and 3.060w heat.
Now you need the QER which is 4,86 and multiply this with PAR/w to get the PPF, so 2.940PAR/w x 4,86μMol/j= 14.288,4μMol/s PPF, thats the theroretical maximum photon flux of the whole lamp.
Divided by 7m2 gives you 2.041,2μMol/s/m².
But the real intensity is ~10-20% less because of wall loses, driver effiency, reflectors or not so you should have at least 1600-1800μMol/s/m² PPFD.
Thats the whole story, nothing special, no hidden secrets...
The screenie below is from an old thread about Citizen COB effiency here in the LED section, but don't ask me where.
My old brain can not remember anything like that, but you're lucky that I still had the screenie on my andorid tablet.
LER which is usually given in lm/w and
QER which is given in μMol/j.
In short, LER is the theoretical maximum at 100% effiency and QER is the effiency of the light spectrum.
For instance, full deep red light/660-670nm is ~5,5μMol/j, white COB's are between 4,5 and 5μMol/j.
For Citizen COB's in 3000°k/CRI 80 the QER should be 4,86μMol/j and LER should be 328lm/w if we use the values from the sheet below.
Now I will try to show how you to use this values.
If we use your 6000w expample, we need at first the lm/w values for the given current from the datasheet.
In this case at 27w one 1212C4 would run at ~800mA, which results in 160lm/w.
LER(theoretical max.) is 328lm/w but the real output is 160lm/w at 800mA(from datasheet), so the effiency is 49%.
You have 6000w and your light is 49% efficient so 6000w x 0,49%= 2.940PAR/w and 3.060w heat.
Now you need the QER which is 4,86 and multiply this with PAR/w to get the PPF, so 2.940PAR/w x 4,86μMol/j= 14.288,4μMol/s PPF, thats the theroretical maximum photon flux of the whole lamp.
Divided by 7m2 gives you 2.041,2μMol/s/m².
But the real intensity is ~10-20% less because of wall loses, driver effiency, reflectors or not so you should have at least 1600-1800μMol/s/m² PPFD.
Thats the whole story, nothing special, no hidden secrets...
The screenie below is from an old thread about Citizen COB effiency here in the LED section, but don't ask me where.
My old brain can not remember anything like that, but you're lucky that I still had the screenie on my andorid tablet.
Attachments
George2324
Well-Known Member
I am using citizen 1212 gen 5 cobs .
4 cobs per square foot.
On hlg-600h drivers I have 20 cobs per driver.
I have all this already setup I'm just in the process of finishing the room. Things such as AC etc.
If I run the drivers at max I'll have around 27w per cob at 50% efficiency aprox and 115w per square foot.
I was aiming for 1500 ppfd and 1500 ppm co2 when I designed this thinking I'd have little heat.
Now I'm thinking maybe I would be able to have them running that high due to heat issues.
The max AC I could possibly do where I am is get. 3 double hose portable AC units at 12000 btu each and vent them out the chiney stack.
Getting a mini split installed here without raising suspicions would be next to impossible even with it being legal I don't want people knowing about my grow.
I may have to just buy the AC units and start running and keep dimming the lights if I get issues.
I was hoping people on here might have ppfd figures from how many watts of cobs
4 cobs per square foot.
On hlg-600h drivers I have 20 cobs per driver.
I have all this already setup I'm just in the process of finishing the room. Things such as AC etc.
If I run the drivers at max I'll have around 27w per cob at 50% efficiency aprox and 115w per square foot.
I was aiming for 1500 ppfd and 1500 ppm co2 when I designed this thinking I'd have little heat.
Now I'm thinking maybe I would be able to have them running that high due to heat issues.
The max AC I could possibly do where I am is get. 3 double hose portable AC units at 12000 btu each and vent them out the chiney stack.
Getting a mini split installed here without raising suspicions would be next to impossible even with it being legal I don't want people knowing about my grow.
I may have to just buy the AC units and start running and keep dimming the lights if I get issues.
I was hoping people on here might have ppfd figures from how many watts of cobs
George2324
Well-Known Member
If I dim my cobs down to 6000w they would be running at 18.75w each 4 cobs per square foot(ish)To get theoretical PPF and PPFD values you need at least two values.
LER which is usually given in lm/w and
QER which is given in μMol/j.
In short, LER is the theoretical maximum at 100% effiency and QER is the effiency of the light spectrum.
For instance, full deep red light/660-670nm is ~5,5μMol/j, white COB's are between 4,5 and 5μMol/j.
For Citizen COB's in 3000°k/CRI 80 the QER should be 4,86μMol/j and LER should be 328lm/w if we use the values from the sheet below.
Now I will try to show how you to use this values.
If we use your 6000w expample, we need at first the lm/w values for the given current from the datasheet.
In this case at 27w one 1212C4 would run at ~800mA, which results in 160lm/w.
LER(theoretical max.) is 328lm/w but the real output is 160lm/w at 800mA(from datasheet), so the effiency is 49%.
You have 6000w and your light is 49% efficient so 6000w x 0,49%= 2.940PAR/w and 3.060w heat.
Now you need the QER which is 4,86 and multiply this with PAR/w to get the PPF, so 2.940PAR/w x 4,86μMol/j= 14.288,4μMol/s PPF, thats the theroretical maximum photon flux of the whole lamp.
Divided by 7m2 gives you 2.041,2μMol/s/m².
But the real intensity is ~10-20% less because of wall loses, driver effiency, reflectors or not so you should have at least 1600-1800μMol/s/m² PPFD.
Thats the whole story, nothing special, no hidden secrets...
The screenie below is from an old thread about Citizen COB effiency here in the LED section, but don't ask me where.
My old brain can not remember anything like that, but you're lucky that I still had the screenie on my andorid tablet.
320 cobs over 7.2 sqm
How much more efficient would running them at this wattage make it?
George2324
Well-Known Member
I have large heat sinks with 4 cobs on them 2 of them are 3000k 90 Cri and 2 of them are 3000k 80 Cri per heat sink.
I'm aiming for 1500 ppfd maybe I won't need to run them anywhere near 115w per square foot
I'm aiming for 1500 ppfd maybe I won't need to run them anywhere near 115w per square foot