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AquaticShrimpNoob

Heat Waste from Electrical Equipment

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Hi Everyone,

 

I recently did some calculations on heat waste produce by my light and water pump and I am not impressed. See the calculations below:

 

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1402664711_ScreenShot2018-12-20at9_53_18PM.thumb.png.9e5b9e6ccfb824359ee3187a5c5ed687.png

 

I did lots of assumption here. I also don't take the possibility that I have made some mistakes. But I think this is sufficient enough to convince me to do some modification. Clearly, my LED light and water pump add heat to the water, which increases my water temperature. In addition, there is no such thing as perfect insulator. That said, there is still heat transfer from surrounding to my tank. 

 

I am thinking of adding a small fan to my water pump or submerge the pump under water and let evaporative cooling cool the pump. I am also thinking of modifying my canopy so that the back surface of my LED light is exposed to the outside for convection. I may add some sort of metal plate or heat sink just to enhance the heat transfer.

 

Anyways, I just want to report this just in case some one might ask. 

 

Sincerely,

AquaticShrimpNoob

 

 

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Hey, those 2 elements alone provide more than enough heat for my indoor tanks.  None have a heater & my shrimp appear to do better with that.  In my setups, the magnetic impeller driven water filter generates the heat waste...

 

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It seems like I was able to remove my water pump as possible source of heat waste. I just submerge the pump in a container with water and my aquarium water barely change temperature through the course of the day. We will see how effective this is in the summer. For now, I think my biggest challenge is my freaking LED light. I need to engineer it to enhance heat transfer. BTW, yes, I was able to observe the LED light effect on my aquarium temperature.

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I’m confused by your LED heat waste calculation. I’m probably misunderstanding but are you assuming that 20% of the LEDs power draw is converted to heat in the water? Why?

 

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@aotf

 

Sorry for late response. Yes, the assumption is that the LED light is only 80% efficient, which means that 20% of the energy (i.e. electricity) is being converted to heat. I am not really sure if this is true for my LED. Keep in mind that not all LED light have the same quality. And not all companies provide some information about efficiency of their product.

 

I kind of see where the confusion is coming from. If you have an LED light, you have probably felt the backside to be a bit warm because of the heatsink. We are dealing with three different type of heat transfer: convection, conduction, and radiation.

 

Indirectly: LED light release heat through their heatsink (i.e. conduction) and transfer to the air (i.e. convection). Ideally, you want this heat to dissipate through the air. But life is not that simple. If you don't have good air flow in that room or if your LED is inside a canopy, you can imagine that the energy from the stagnant air will transfer to the water (BTW, water has high heat capacity). This is an engineering problem that you can solve and optimize.

 

Directly: light energy transforms to heat energy (i.e. radiation) as light wave excites matter, which is pretty much everything under the light source.  I won't explain the science behind this here. You can look this up. You can't really do anything about this.

 

I hope that helps.

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Got it.

 

What confused me (and still does) is why you assume that all of the 20% of heat waste generated by the LEDs get transferred to the water. As you say, it's an engineering problem and I won't pretend to be able to solve it, but I would imagine that the amount of heat transferred from the LEDs to a water column 4-12" below (depending on the setup) through convection is relatively low, not 100% of the heat generated on the light fixture (air is a bad conductor, there's a lot of air in the room, and hot air tends to rise anyway so the convective transfer seems limited). Not saying it's negligible, just that it would be a fraction of the 20% you use in the calculated rate of temp increase.

Radiation, on the other hand, seems like it could account for a lot. Soils absorb roughly 80% of the light that hits them, so if you had an amazonia bottomed tank with no plants, you'd be looking at a decent amount of heat generated by the LEDs.

 

(37.533(0.8*0.8))/(0.999*62.31*2.94)= 0.13 F/hr

dT = 1.05 F in 8 hours.

...so not as much as the pump, but still something! One easy way to get around it is to lower the light output (dimmer), which would also decrease the waste heat/convective transfer or go with a higher albedo bottom in the tank.

 

Fun stuff!

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