# LED BTU impact/calculation in indoor application?



## svjung (Dec 23, 2014)

Hi all,

I have a project that has a lot of LED that would be installed indoor, basically a big LED screen in a lobby.

The client-site HVAC engineer is concerning about the BTU impact to their system, and hence ask us to provide a estimated BTU from our product.

I haven't done anything like this before and I didn't find some related thread about this topic, so I tried a somehow logic( at least for myself ) approach, and I really some advise form you guys.

Here is my approach:
Assume the driver is 100W, and has 85% efficiency. From what I found from some literature, the LED efficiency (not efficacy) is about 30%, please correct me if I was wrong here.

So for a single fixture, the loss heat is:
Driver Power Heat loss: 100W * 10% = 10W
LED Heat Loss: Power transferred from Driver * (1-efficiency)
(100W-10W)* (1-30%)= 90W*70%=63W

W->BTU conversion
63W* 3.412=215BTU

End of calculation.


First of all, 63W out of 100W kind doesn't make sense to me, 63% of the power goes to heat?? Please correct me if I was wrong, sometimes math proved itself so good that you cannot argue with...

Under my calculation, The BTU of our project for the big screen is gigantic number that I am not comfortable with. Or it tends to be this way? more than half of the power would somehow goes to heat for LED system?


===

In addition to the somewhat academic kind approach, I found something online called BTU solar meter, 
video link: https://www.youtube.com/watch?v=c0Zgh0obdgM
That meter can get a reading of BTU/ft2 or W/m2


I am thinking maybe buy one and point that at the screen, get the reading, times the square foot I have, and get the total BTU for the system?
Is that a reasonable approach? Or I completely misunderstood the use of BTU meter?


Thank all you guys' help in advance and Happy Holiday!

Jung


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## Steve K (Dec 23, 2014)

Does the light from the LEDs stay in the room? If so, then all of the power delivered to the LEDs is eventually turned into heat in that room, regardless of the efficacy of the LEDs when turning some electrical power into light. If that light stays in the room, it is eventually turned into heat. Therefore, 100 watts delivered to the driver is 100 watts of heat in the room.


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## Anders Hoveland (Dec 23, 2014)

To be fair, some of the energy is lost as radio-frequency energy in the driver. This might not all necessarily translate into heat.
But yes, I agree that a 100 Watt device is generally going to drop 100 Watts of heat energy into a room. That's like leaving a light bulb on (the REAL ones we used to have).


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## svjung (Dec 23, 2014)

Anders, Steve,

1. radio frequency is minor, for simple estimate I ignore that part first.

2. Form my understanding, light and heat are different energy transfer..., 100W of energy going to LED, some portion goes to light (30% I assume) and some portion (70%) to light.
So if we just consider HVAC issue, we take consideration the heat part only. I still think this should be the case, however the portion might not be correct.

Thanks for replying.

Jung


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## DIWdiver (Dec 23, 2014)

Steve was saying that if the light doesn't leave the room (through a window or door), it will get absorbed by some material in the room, and converted to heat at that point. Even if it gets reflected 3 or 4 times first, if it doesn't leave, it eventually gets turned to heat in the room. 

Oh, and he's right.


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## BurmaJones (Dec 25, 2014)

svjung said:


> . . .
> W->BTU conversion
> 63W* 3.412=215BTU
> 
> ...



Hmmm. You have to begin before you can end. Probably you should review the difference between energy and power.
Also, use input apparent power, not real power.


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## Epsilon (Dec 26, 2014)

That's not his fault, it's the industries fault. BTU/hour is the unit which he needs, but BTU is the unit which is used by the industry and public to size an AC. Though the op should use the correct units, but I think he understands the difference. Ppl who doesn't can't calculate the thermal impact

That said, a 60w heat source is less than a human in the room I think and far less than the accurate tuning capability of the HVAC guy . It's good that he asks, because if it was 500w of heat, then he should take measures.


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## brickbat (Dec 26, 2014)

svjung said:


> Here is my approach:
> Assume the driver is 100W, and has 85% efficiency.



Assuming most of the light from the LEDs stays in the room, and also assuming you are specifying the LED driver OUTPUT power, the calculation is lead-pipe simple:

100W/0.85 = 118W
118W*3.4 BTU/hr/W = 400BTU/hr.


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## brickbat (Dec 26, 2014)

BurmaJones said:


> Also, use input apparent power, not real power.



Incorrect.

Use real power.


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## BurmaJones (Dec 26, 2014)

brickbat said:


> Incorrect.
> 
> Use real power.



Are you saying that reactive currents don't release heat? Please advise.


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## brickbat (Dec 26, 2014)

Not exactly. You're muddying the water with imprecise terminology. I'm saying that, when calculating the heat generated by an electrical load in this case, all one needs to consider is the real power.

'Real' power is measured in Watts, and is a measure of energy per unit time.
'Apparent Power' is measured in Volt Amps, and is not a measure of energy per unit time. 



Read this:

http://www.allaboutcircuits.com/vol_2/chpt_11/2.html


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## mattheww50 (Dec 26, 2014)

The pure reactance component does NOT release heat, however because the circuit is not a super conductor, there will always be a real component even in what would seem to be pure reactive load. Power is always E*I. Energy is the integral of power over time, and the reactive component will always end up as zero power over time since it is at quadrature. So while E*I in this case seems to be 118 watts, The real component is 100 watts. Anytime there is a reactive component in the load, voltage and current will NOT be precisely in phase, so while E*I may be 118 (actually it is 118VoltAmps), the integral over time would in fact be equivalent to a pure resistive 100 watt load. 

Exercise for those who don't believe me:
Compare the Integral over time of E*sin(wt)*I*sin(wt+theta) with E*sin(wt)*I*sin(wt). Where theta is the phase angle. The difference between the two integrals will be the reactive component.


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## brickbat (Dec 26, 2014)

It appears you may have misconstrued the 0.85 number in my post #8 above. It was NOT meant to be the power factor of the LED driver. Rather it represents the efficiency of the supply, which was stated to be 85%. This 100W watt power supply (operating at full load) will consume a total of 118 W, inclusive of its internal losses. Apparent power, power factor, reactive loads, and phase angle (not to mention RF energy) are all red herrings, and as such simply not relevant to this problem.

(I note that an AC LED driver operating with a low power factor will cause increased losses in the AC mains wiring supplying it, but as a practical matter, I'd challenge anyone to show that it is significant in this case, where the object is to approximate the heat load of a system.)


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## onetrickpony (Dec 26, 2014)

Correct me if I'm wrong here, but I think everyone is over thinking this. 

Tell the hvac guy that the screen consumes 100w and to assume 0% efficiency. Or just tell him it's 341 btu/hr. It's almost negligable unless he ignored all the other heat sources in the building like overhead lighting or something like a stove/oven in a commercial kitchen that runs all day.


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## brickbat (Dec 26, 2014)

If it uses a 100W driver that's 85% efficient, it consumes 118W. The 118W number should be all the HVAC guy needs to know.


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