a friend of mine that has several degrees in engineering finally laid this out in terms I can understand. hopefully this will also help others better understand and explain why LEDs are more efficient and produce less heat then an HID. ready for a nerd riot?
friend A:
If you put 100W worth of fluros and a 100W heat lamp in a sealed room, they will heat it exactly the same amount. CAS even agreed with that in his last post. Because of the laws of entropy and conservation of energy, (the 2nd and 1st laws of thermodynamics, respectively), ANY spectrum of light that is not used by the plant for photosynthesis will be turned into heat.
The only thing it seems we disagree on is how much light the plant uses that DOES NOT get converted to heat, and I'm not sure how efficient photosynthesis is, or what percentage of light that strikes a leaf surface is used for photosynthesis. If we DID know these figures, we could calculate exactly how much difference there would be between a HID and a LED light's heat input to the room.
Friend B response:
First of all, sorry everyone for the long read.
D, we are certainly on the same page with the thermo. However, that explains that the total energy in the system is not changing and that unabsorbed energy essentially "degrades" upon reflection of surfaces and "decays" to heat. If electrochemical conversion of photons was not in the equation and there was a black room (emissivity = 1 = black body radiation), they certainly would produce the same heat over a long period of time (steady-state condition) since the spectrum output wouldn't matter. As (mostly) always, energy in = energy out.
What I am trying to get across (respectfully of course) is that mush more is going on than simply the 1st (energy conservation) and 2nd (entropy) law of thermo can describe. There is also physics at play via photon absorption, converting photons into "plant energy" instead of heat. The plants are absorbing energy and converting it via photosynthesis into sugars, essentially. The spectrum, efficiency, and amount of light that plants absorb for photosynthesis is well known (not only in plant books, but also in PV text books as well). The absorption spectrum is shown below (commonly referred to as the action spectrum). See how it stops absorbing wavelengths above 700 nm. So in other words, they don't absorb ANY thing in the IR (heat). You can also see that it absorbs below 400 nm. In fact, it absorbs all the way down into the UV.
Now lets examine an HID light. For fairness, lets consider a light that a favorite and claims its success on having a wide spectrum. Below is the output emission spectrum of the EYE Hortilux Blue bulb. First notice how the emission above 700 nm is quite high, and clearly cuts off at the end of the graph (750nm = IR). In fact we know it goes rather far into the IR since you can clearly feel it when the HID light is on. If I'm not running my AC hood, my tent gets CRAZY hot. I've touched the glass on LEDs (which is like 2-3" away from the diodes) and it isn't hot at all. You can also measure these temperature differences and are written abount in many articles doing tests comparing LEDs to CFLs (for home lighting). Remember, CFLs produce even LESS in the IR than HIDs (I know this also by physically measuring the temperature as a function if difference away from the bulbs). Not only can you see that the bulb DIRECTLY emits heat but it also has a peak intensity in green. Green happens to be, as shown above, where plants absorb the least (hence why plants are green). This light, as we agree on, gets reflected and also decays to heat.
Now an LED spectrum. First, they can be tuned by picking different band gap diodes so you can pick and choose what wavelengths you want to emit light at. LEDs are the reverse of a PV solar cell. The first image below is a typical flowering LED spectrum (Fertilight). The next image is another one with a company (StealthGrow) comparing their spectrum vs. the action spectrum, chlorophyll a and b absorption, and a "typical" HID. What you can clearly see is that these LEDs don't emit ANY light in the IR (it would only do so if they specifically put an IR emitting diode in it, like what is in night vision cameras). Additionally, they don't put out any (neglegible) light in the green. This is the exact reason why a 240W LED can be recommended to replace a 400W HPS setup, they deliver only a spectrum that the plant uses, not a bunch in the IR and green.
So looking back at thermo, a 400W LED vs a 400W HID would input the same amount of energy to the system. However, for the LED, a significantly larger portion of that energy will be going to the plant and converted into electrons (via photosynthesis). For the HID, more is going IMMEDIATELY into the IR (enough to make a difference between needing cooling and not as seen by many users) along with peak emissivity in the green-greenish yellow region, which eventually gets absorbed and turned into heat. If I explained anything poorly, let me know. I'm not the best explainer.
As for calculating it, it would be a bit of a pain since you would need to integrate under the curves of these graphs and normalize per coulomb/electron.
Glad to have a fellow engineer on the board
I have my BS in Chemical Engineering (VERY thermo heavy), a MS in Materials Science, and am working on my Ph.D. Additionally I work closely with the National Renewable Energy Lab (NREL) with materials characterization (including photoemission spectroscopy) in addition to doing work with people from the OLED and Semiconductor (MOSFETs) industries. I was talking with a computational physicist (models multi-exition generation in solar cells) about this problem at a poster session today and he helped explain to me a bit better about why LEDs are more efficient and produce less heat. This stuff geeks me out and is my passion so I LOVE talking about it.
Light emitting plasmas are really cool technology also. I can go on for ages on the cool physics they use to make those lights have tuned emission AND higher brightness. If only LEDs and LEPs weren't so expensive