How many LED's can I put on a 24x3 heat exchanger?

How many LED's can I put on a 24x3 heat exchanger? Also, would I need optics if the led's will be 4 inches from the water? The led's will be housed in a single fluorescent light housing.

Similar to this one:

 

what are the dimensions of the tank? The depth of the tank will determine what optics you will need.

 

It is a standard 29 gallon. I believe the dimensions are 30x12x18.

 

What are you planning to keep? If you want to keep anything, anywhere, you'll probably want 60deg optics. You may be able to get away with 80, but 60 would punch down better. 20-24 would probably let you keep just about anything. It depends what you plan to keep though. You wouldn't need as many 80deg, as they provide better coverage, but the light will be less intense, especially at the bottom. So, again, you might be more limited in what you can keep.

 

At 4" from the water optics will look very strange, unless you go really wide. Build and test it without optics first. You can experiment with that later. You will be good with 2 rows of 12 on that heatsink or even more. Depends on what drivers you use of course, but most fixtures are multiples of 12 LEDs. You can go as high as 14 per string as well. 24 3W XP-G/XP-E should be more than sufficient for that tank.

 

i have an 18' deep tank, i have 80* optics, mine is also 4 inches from the water.
60* will give spotlighting, and the shimmer is sometimes blue and sometimes white base on the direction of the wave/ surface agitation. my par on the sandbed is in the 250'ish, however i have 48LED's.

24x3'' heatsink will yield you 2 rows of 12, 24 LED's is plenty of light for your tnaak dimension.

 

60 deg may give some spotlighting, but will give you more PAR with less leds. You can model how much coverage you will have though. Ideally, it should be TAN(lens angle / 2) * distance * 2. However, most leds are not even coverage, but brighter in the center, so, some people reduce this by 75%, which gives TAN(lens angle / 2) * distance * 1.5.

So, if you use the latter formula and have the lights 4" above the tank we get the following coverage area, for each individual led, for 60 vs 80deg optics. So, you can use this to sketch out what your coverage would be at different depths. Each LED will produce a circle of coverage, approx this diameter. Where they overlap will give more PAR.

60 degree optics
Water Surface
4" below surface - 3.46"
6" below surface - 6.93"
12" below surface - 13.86"
16" below surface - 17.32"

80 degree optics
Water Surface
4" below surface - 5.03"
6" below surface - 10.07"
12" below surface - 20.14"
16" below surface - 25.17"

No optics 120deg

4" below surface - 10.39"
6" below surface - 20.78"
12" below surface - 41.57"
16" below surface - 51.96"


But remember, the wider the coverage (i.e. wider the lens angle) the more leds you need to get the same PAR. Also, with 120 deg optics, you probably won't get much PAR at all on the bottom regardless. Also, 51.96" is much bigger than the tank, so, much light would be wasted.

 

Thanks for the help. The led's will be recessed around 3 inches into the fixture. If I lined the inside with aluminum, would this be enough to direct the light down into the tank? It might be more trouble than it is worth.

 

No; you need optics on each diode to achieve that. They are a point-source of light, not a continuous bulb (like a T5); therefore they require different types of "reflectors".