HVAC pairing with LEDs (BTU calculation)

Growmau5

Well-Known Member
Hey growers,
I am in the market for a new mini split a/c unit, and I really want to pair it properly with my lights. I am specifically looking at the Fujitsu rls3 family of units. Their 9000 btu is the most efficient highest SEER rating in the world right now. Efficiency is very important to me, but sadly the 9000 btu is not going to be enough. To my understanding, with all mini splits as you increase the cooling power in BTUs you begin to drop in SEER incrementally. So bigger is not always better, and for this application, I really want this thing to be appropriately sized to bank the kwh savings.

As a base line, I would like to know what the approximate BTU output of a (3) cxa3070 AD bin on a HLG185h-c1400 is. Then I can simply multiply this by how ever many bars of this combination I have.

I came across this article and it threw a wrench in my understanding of LED efficiency.
http://www.etcconnect.com/Support/Articles/How-To-Quickly-Calculate-BTUs-Emitted-from-a-Source-Four-For-HVAC-Purposes.aspx

Calculating Heat Load from Lighting Fixtures
Whenever doing heat-load calculations, we must always count 100% of the fixture wattage-any fixture's wattage-as heat, regardless of efficiencies at the source. In any enclosed space, virtually all of the light produced by a lighting fixture is eventually absorbed by surrounding materials (walls, ceilings, etc.). When light gets absorbed by these materials, photons are converted into heat.

New lighting technologies, such as LEDs, allow luminaires to produce light more efficiently, or with higher efficacy than incandescent lamps. This means that we get more lumens per watt; however, heat per watt is constant. What matters is that fewer watts are used to produce comparable brightness, not that these fixtures somehow produce less heat for the wattage that they consume.

1 watt of electricity used by a luminaire = 3.4128 BTUs per hour

The same math works for any lighting fixture.

For average industrial heat-pump or electric air-conditioner types of HVAC equipment, it takes approximately 0.4 watts of cooling power to offset each 1 watt of power consumed by a light source. Smaller-scale HVAC systems might require more power than this, and certain specialized systems (such as evaporative coolers) may use somewhat less, but this is a useful generalization for most applications.

If anyone can clear this up or provide me with a BTU calculation of the example posted above, it would be much appreciated.
 

Growmau5

Well-Known Member
I might have answered my own question, based on the article posted above. But please check my math.
If a (3) cxa3070 light @1400mA uses approx 175w, then (175w) x (3.4128 BTU/whr) = 597.24 BTUs of cooling to offset that heat load.

Extrapolating from that calculation, a (36) cxa 3070 @1400mA setup requires 7,167 BTUs of cooling to offset just that heat.
 

Greengenes707

Well-Known Member
It's all true...watts are watts. And it is the ability to run less wattage and get the same results is what allows LED's to save on cooling of the ENVIRONMENT. But eventually all watts are converted to heat whether by reflection, absorption, or just about anything.
Cooling of the LED ITSELF on the other hand is easier as efficiency of the source increase. This is because, as you know, higher efficiency means more light:heat at the point emission. But as we said, it all "ends up" as heat not too long after being emitted. So the junctions/case temps are easier to keep cool because more is emitted as light than heat...and then the light gets away from the source before it converts to environmental heat.
 

PSUAGRO.

Well-Known Member
I would stick with mitsu's, very solid reliability compared to the other mini split manufacturers.

Also suggest paying extra for the heat pump/inverter(if applicable for your location).....set it on auto temp and forget it all year round.
 

Growmau5

Well-Known Member
@Greengenes707 thanks for the response, super helpful as always. Would I be correct in saying that a cxa3070 z4 bin and an AD bin require the same exact amount of BTUs of cooling power to cool?

@PSUAGRO. thanks for the recommendation. I may go the mitsu route after reading horror stories that Fujitsu DOES NOT honor warranty claims on units purchased from online vendors. (such as but not limited to acwholesalers.com etc) I know Fujitsu is a quality product, but I tend to see mitsubishi "around" a lot more, and I would imagine parts are more available and technicians are more familiar with their units.
 

Growmau5

Well-Known Member
I wanted to post this here even tho it is a little outdated. Thread and spread sheets were originally posted by Stonefree69 on ICMAG. I mainly want to have it here for easy reference for myself and others.mini split 1.jpg mini split 2.jpg
 

Greengenes707

Well-Known Member
@Greengenes707 thanks for the response, super helpful as always. Would I be correct in saying that a cxa3070 z4 bin and an AD bin require the same exact amount of BTUs of cooling power to cool?
They would both contribute the same amount of heat to the environment that would need to be cooled by your AC or extraction fans.

The AD would require less heatsink to maintain the same Tc. Very minimal in practice..but for the theory of a more efficient source, that's what it would mean.
 

PSUAGRO.

Well-Known Member
Just make sure to install/or get installed a surge protector by the power disconnect, it's a $30 part that will save your outdoor unit in case of a surge after an outage/storms/etc.

They are delicate in this sense and learned it the hard way when the board on the outdoor unit fried......complete swap was cheaper than fixing it($3500 with install)........ go figure ;-)

It's a must on these mini's IMO

Good luck
 

Merkin Donor

Well-Known Member
So a question for the lighting pro's then. If I have two 4x4x8 rooms, both are closed with no ventilation - one has a HPS assembly that draws 600 watts at the wall and the other a 12 cob Super Panel w/cxb3070's drawing 600 watts at the wall- when turned on they should both raise the ambient temp 5 degrees in the exact same time?
 

alesh

Well-Known Member
So a question for the lighting pro's then. If I have two 4x4x8 rooms, both are closed with no ventilation - one has a HPS assembly that draws 600 watts at the wall and the other a 12 cob Super Panel w/cxb3070's drawing 600 watts at the wall- when turned on they should both raise the ambient temp 5 degrees in the exact same time?
Yes the ambient temperature should be almost the same. Surface temperature of anything in the room, however, will be significantly higher in the case of the HPS due to its IR radiation (and this might affect measurement of the temperature).
Another thing is that you need less draw with the COBs [CXB3070] to produce the same number of photons as the HPS, therefore you need less power to achieve similar results which leads to lower temperature provided that you want to keep the same intensity as the HPS.

It would be appropriate to add that I'm no expert and I might be wrong. But this seems logical to me.

e: The temperatures will definitely not be the same (but very very close) as CXBs provide more photons of which more will be stored in chemical bonds in the plants. I'm not sure how much energy is not dissipated as heat but it's not much for sure so I'd say that this fact is (almost) negligible.

Everyone is welcome to correct me if needed.
 
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Greengenes707

Well-Known Member
@alesh
It doesn't make perfect sense to me or any in depth thinker understanding how complicated all the interactions can be.
But that is what a professional HVAC crew goes by and is educated on in school. And for all intents and purposes despite what would seem like an all to simple calculation...it does work out.
You are correct on IR in the light beam itself heating surfaces, special canopy temps.
 

salmonetin

Well-Known Member
I might have answered my own question, based on the article posted above. But please check my math.
If a (3) cxa3070 light @1400mA uses approx 175w, then (175w) x (3.4128 BTU/whr) = 597.24 BTUs of cooling to offset that heat load.

Extrapolating from that calculation, a (36) cxa 3070 @1400mA setup requires 7,167 BTUs of cooling to offset just that heat.
(3) cxa3070 light @1400mA uses approx 175w, then (175w) x (3.4128 BTU/whr) = 597.24 BTUs

my note,.. i assume 175w measured on wall through a true rms power meter or maybe kilawatt or similar... with driver fans etc... maybe im wrong too...

...my maths... ...yeah only the leds... i know driver and fans count too... but i miss the hardware... then miss this values too...

(3) cxa3070 light @1400 mA

Vf 50ºC = 37,30 V ...If = 1400 mA ....Diss W 50ºC = 52,22 W * 3 = 156,66 Diss W

...% 50ºC = 43,31 % ...PAR W... 156,66 * 0,4331 = 67,85 PAR W ... 156,66 - 67,85 = 88, 81 W heat???...

...other way...



= Photometric Flux = ...here we can use Lm 50ºC = 7350???.... or Lm / Watt 50ºC = 140,75....????... ...what are the correct value?...

LER = 325

7350 / 325 = 22,615 ....then i assume Lm 50ºC... more sense to me...

140,75 / 325 = 0,433 ... i dont think so...

Pheat = 156,66 Diss W – 22,615 = 134,045 Heat W....



= Radiant Flux

...then
= 22,615 Radiant Flux....;)

...curious LER calc...



.....but im too :joint::bigjoint: today and maybe my math are wrong...sorry... and its only my inexpert pov too...

pd... upsss...here i saw the technics using a BTU meter too... named solar power meter or similar ... i dont remember well...;-) the hvac its only a dream for me... too expensive too ...electricity here its expensive too...

:peace:

Saludos
 
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alesh

Well-Known Member
(3) cxa3070 light @1400mA uses approx 175w, then (175w) x (3.4128 BTU/whr) = 597.24 BTUs

(3) cxa3070 light @1400 mA

Vf 50ºC = 37,30 V ...If = 1400 mA ....Diss W 50ºC = 52,22 W * 3 = 156,66 W

...% 50ºC = 43,31 % ...PAR W... 156,66 * 0,4331 = 67,85 W ... 156,66 - 67,85 = 88, 81 W

...other way...


= Photometric Flux = ...here we can use Lm 50ºC = 7350???.... or Lm / Watt 50ºC = 140,75....????...

LER = 325

.....but im too stoned to follow the rest...sorry... and its only my inexpert pov too...

pd... upsss...here i saw the technics using a BTU meter too... named solar power meter or similar ... i dont remember well...;-)

:peace:

Saludos
That's the heat created in (on?) the COB itself. The rest of the energy is radiated but ends up being converted to heat anyway (most of it).
 

ttystikk

Well-Known Member
Does someone know how much energy does NOT ultimately get converted into heat (ie is stored in chemical bonds inside plants)?
I am curious about this, and the much larger amount of heat that's used in transpiration.
 

salmonetin

Well-Known Member
....the measurements on the wall ... a true rms power meter give us the best values off course... on the wall count driver fan etc... off course...;)

...the meditions on the DC side give us efficiency values too... power input - power output...

...but im too stooned for tech talks today too... sorry

saludos
 
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ttystikk

Well-Known Member
....the measurements on the wall ... a true rms power meter give us the best values off course... on the wall count driver fan etc... off course...;)

...the meditions on the DC side give us efficiency values too... power input - power output...

...but im too stooned for tech talks today too... sorry

PD...maybe the heat used in traspiration its a bit offtopic here....my pov...

but maybe some pics and links (only for examples or ideas) help a bit...



http://elhocino-adra.blogspot.com.es/2012/07/cultivando-bajo-invernadero-en-verano-1.html

http://elhocino-adra.blogspot.com.es/2012/07/cultivando-bajo-invernadero-en-verano-2.html

sorry for my off topic MD...

...partial e function activated... sorry .... ...fast...

saludos
I'd like English versions of those charts. Are they available?

The second one is VPD, and I can say from experience that it's effective.
 

ttystikk

Well-Known Member
Regarding transpiration; it's not off topic or beating a dead horse when discussing the movement of heat through a growroom, because this process is so clearly a part of it.

As part of my ongoing upgrade project, I noticed two things; first, that an empty growroom needs A LOT more cooling than one full of plants and second, that the more plant material is in the room and being productive, the easier it is to control temperature and the harder it is to control RH.

I'm stuck ventilating the excess RH because my cooling/dehuey system simply can't keep up with the massive amount of moisture being added to the air by my plants. On the other hand, I'm well aware that ventilating warm, moist air is the same as removing a lot of excess heat.

In my case, the room can draw outside air in that's WARMER than my growroom and by the time the plants have added their moisture load, it's become a net cooling effect.
 
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