Can a light get too efficient

SSGrower

Well-Known Member
Since we are in this discusion id like to gauge what people think about this:


As for enviroment, chip temprature be damned, but how does the temp of the heatsink affect the enviroment?


You could for example have a fixture which dissipates say 500w of heat thru a heatsink which rise in temprature say 45 degrees over ambient.


And in an identical room and fixture with only difference being that the heat sinks are larger and the 500w only raise the temp of the heatsink by 5 degrees.



All temps at steady state operation.


Same watt in but how would enviroment react in each case? I dont think the same but would like to know what people think.


And since keeping your room hot enough is important when using leds, are there cases where you actually like your leds to run hot so you dont have to put in a 2000w radiator?
Assuming we want light that is PAR. If you have to add a heater the power it consunes must be less than the difference in PAR watts for the HID vs LED. In order for it to make fiscal sence this advantage must be well in favor of the LED due to higher initial cost in order for LED to make sence.
Since most current operations are not optimized for led use because they are currently optimized for hid its kind of like pounding a square peg in a round hole to not build a new environment for the led.
 

NanoGadget

Well-Known Member
The whole 1000 watts is a 1000 watts in a sealed room is true, but it's also semantics in this case. What we are trying to maintain here is plant health, not ambient room temps. We aren't growing warm air here. I can run my LED room 10 degrees warmer than my hid room and maintain great plant health. So not only is an LED room easier to extract if you're not running a sealed room due to the way it radiates heat, but you can run the temps higher than you can with HID in a sealed room which means real world energy savings.
 

Fubard

Well-Known Member
because thats the way thermo dynamics work. a photon contains energy. when it strikes something, part of that energy is absorbed by the object it is striking, and released as heat
So if part of the wavelength is absorbed and consumed, where is the heat created by that when absorbed and not reflected?

Heat is one part of the spectrum, where does the rest come from? Again, you're trying to create more energy than is consumed.
 

Fubard

Well-Known Member
if you give it enough time, the interior of the closets WILL be the same temperature. i promise. you can trust me, i don't know you well enough to lie to you.
No, the ambient will change to be equal to the temperature produced by the light and no higher without an external influence, and if one light produces more light energy than heat compared to the other then in a sealed room the temperature will rise to a lower maximum level as otherwise you are creating more energy than consumed.
 

NanoGadget

Well-Known Member
I am more worried about winter than summer honestly. The LED plants don't want to go under about 78 degrees. luckily it is cheaper to heat the room than cool it.
True story. I went from spending the last 20 years constantly worrying about what to do with all the heat to worrying about my ladies getting frostbite on their delicate parts when the lights are off.
 

Ryante55

Well-Known Member
What you ALL are neglecting is than none of your measurements or observations are happening in an perfectly insulated cube, which is what you would need to prove and disprove the missguided thoughts on this matter.

The laws of thermodynamics dont get broken.

You need to look at bigger picture. Its not that temperature of the room rises so much its the mass of the heatsink that heats up, this holds the energy and radiates it mack to the room over time. It is not correct to say that it does it at the same rate but more that the energy is transferred at the same rate. A bulb heats the air directly with led the heatsink heats the air, there is a lag with the heatsink having to heat up.
He will never understand :/
 

Fubard

Well-Known Member
True story. I went from spending the last 20 years constantly worrying about what to do with all the heat to worrying about my ladies getting frostbite on their delicate parts when the lights are off.
Its why I have a 1kW fan heater in my room, a HID might raise the temperatures better but the extra heat generated, and power used, is a waste outside of a couple of months.
 

SSGrower

Well-Known Member
He will never understand :/
I am trying to get it to where people dont just quote laws of physics and walk away with a partial understing. Some of the correct statments on this matter lack clarity, this causes the confusion you see here.

Edit.....
On second thought you might be right.
 

Humple

Well-Known Member
No, the ambient will change to be equal to the temperature produced by the light and no higher without an external influence, and if one light produces more light energy than heat compared to the other then in a sealed room the temperature will rise to a lower maximum level as otherwise you are creating more energy than consumed.
Energy cannot be consumed. Only transferred.
 

wietefras

Well-Known Member
I'd rather use two grow rooms and flip those instead of using a space heater. So there is always one grow with the light on and this heat can be used to at least keep the other tent above some minimum.

When the light is on you shouldn't need a space heater unless you live in a really cold area. Just make sure you utilize the heats coming off the leds properly.
 

Fubard

Well-Known Member
I'd rather use two grow rooms and flip those instead of using a space heater. So there is always one grow with the light on and this heat can be used to at least keep the other tent above some minimum.

When the light is on you shouldn't need a space heater unless you live in a really cold area. Just make sure you utilize the heats coming off the leds properly.
Been in minus Celsius for the last week here, low Celsius all year. Come next month, temperatures will start rising so less heating will be needed, that means the extra heat from a hid would need to be countered by something else for most of the year, using more energy.
 
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