LED PAR math

DoctorDelta9

DoctorDelta9

Well-Known Member
OK guys I was hoping I could get a check on my math here:
College was like 25 years ago for me...lol

CBX 3590 72V 3000k x12
1050mA
Using @SupraSPL graph: See spreadsheet post below.
https://www.rollitup.org/t/cob-efficiency-spreadsheets.865238/

Assuming 325 LER

11192.95 LM/COB x 12 = 134315.4 total LM
134315.4 / 325 =413.2781 Total PAR?

4x4=16 ft2
5x5=25 ft2
6x6=36ft2

so 413.2781 / 16 = 25.83 Par/ft2
/ 25 = 16.53 Par/ft2
/ 36 = 11.47 Par/ft2

Are these calculation correct?
Thanks and forgive my newb-ology.
 
C

churchhaze

Well-Known Member
Correct... exactly what I was going to say! (lol)

You need the power dissipated to find the luminous efficacy of source. Once you have that, divide the luminous efficacy of source by the luminous efficacy of radiation to get efficiency.

Efficiency = luminous efficacy of source / luminous efficacy of radiation.

Once you have efficiency, you can find PAR watts output.

Output power = efficiency * input power
 
Last edited:
alesh

alesh

Well-Known Member
churchhaze said:
Not even close!

You need the power dissipated to find the luminous efficacy of source. Once you have that, divide the luminous efficacy of source by the luminous efficacy of radiation to get efficiency.

Efficiency = luminous efficacy of source / luminous efficacy of radiation.

Once you have efficiency, you can find PAR watts output.

Output power = efficiency * input power
Click to expand...
Oh you! His math is actually correct. And so is yours.
 
medicinehuman

medicinehuman

Well-Known Member
I miss the par 3 golf course that was by my place. I'm working my brain as hard as possible but this calculating sucks.
 
Greengenes707

Greengenes707

Well-Known Member
The 3590 at that current dissipates 75.54w...at 45.59% efficiency...making 34.44PARw's per cobs...12X 3590's in use...=413.26PARwatts total

Each PAR watt emits 4.88µmols...
So 413.26 x 4.88µmols= 2016.73µmols PPF

EDIT...just realized you're trying for PARw/sqft...
Forget the last step and just divide the PARwatts(413.26) by the sqft....which you did, and got it correct.
 
DoctorDelta9

DoctorDelta9

Well-Known Member
Greengenes707 said:
The 3590 at that current dissipates 75.54w...at 45.59% efficiency...making 34.44PARw's per cobs...12X 3590's in use...=413.26PARwatts total

Each PAR watt emits 4.88µmols...
So 413.26 x 4.88µmols= 2016.73µmols PPF

EDIT...just realized you're trying for PARw/sqft...
Forget the last step and just divide the PARwatts(413.26) by the sqft....which you did, and got it correct.
Click to expand...
is it more accurate to use umols/PPF?
 
bicit

bicit

Well-Known Member
Greengenes707 said:
The 3590 at that current dissipates 75.54w...at 45.59% efficiency...making 34.44PARw's per cobs...12X 3590's in use...=413.26PARwatts total

Each PAR watt emits 4.88µmols...
So 413.26 x 4.88µmols= 2016.73µmols PPF

EDIT...just realized you're trying for PARw/sqft...
Forget the last step and just divide the PARwatts(413.26) by the sqft....which you did, and got it correct.
Click to expand...
Isn't the number of micro moles emitted dependant on the cct?
 
Greengenes707

Greengenes707

Well-Known Member
DoctorDelta9 said:
is it more accurate to use umols/PPF?
Click to expand...
Not more accurate, but more useful/relevant to plant response and what plant science uses. All plants use photons, and is quantifiable in how many they use. Not every PARw is the same amount of base on what spectrum that energy is being emitted from because different nm's carry different amounts of photons per the same energy. But when looked at in photons, rather than energy, it is more directly relatable to how well it will be used for photosynthesis. The ideal light needed is a very well documented to be in the 700-1000µmols PPF per m^2. So that is what should be aimed for to be supplied.
 
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