Submersible Pump - heat question

[Rekonn]

A single Maxijet 1200 powerhead is rated at 20 watts. An Eheim 1260 is rated at 65 watts. Can I assume the Eheim is going to put as much heat into my setup as about 3 Maxijet powerheads?

I'm asking because I have two MJ1200's in my tank, and I can measure the tank temp difference with them on vs off. From there I'd like to know if I can guage how much higher my tank temp will be if I add an Eheim 1200.

 

[sjm817]

I would say that is a reasonable experiment.

 

[BeanAnimal]

Rekonn, your assumptions are correct...

The added flow the 1260 creates may aide in evaporation and negate some of the effect but make no mistake that every bit of heat generated by the pump goes into the water. The funny part is the more efficient the pump, the more energy is transfered to the water instead of being lost on the neutral line

The big question is really if the Eheim is truley using 65 watts, and if the MJs are really 20 watts. The labels may not indicate what the steady state operation of the pumps are in your flow situation...

Enjoy!

 

[Rekonn]

Have you found the lables to be off before? Have you or others plugged in a meter and found the results to vary?

How about external pumps? I've read that they're better because they transfer less heat to the water. Are there any actual numbers on that? A Panworld 50PX-X is rated at 90 watts. It has a fan on it, and lets assume there's good airflow around it. What fraction of the 90 watts would be transferred to heat in water? Is there a general rule of thumb about external pumps?

Now, what if you started to insulate the external pump for the purpose of noise reduction... Without proper airflow around the pump would you wind up with more of the 90 watts transferrng as heat to water?

 

[BeanAnimal]

External pumps transfer less heat to the water, this is a simple fact of physics. An internal pump transfers ALL of it's heat to the water.

Yes, labels are off... the numbers depend on a LOT of factors. They are a good rule of thumb to go by however.

Regarding the heat transfer of different external pumps... well that depends on their design. Some pumps transfer heat directly from the motor to the water, or down the drive shaft. Others have seperate wet ends and transfer very little heat.

Insulating the external pump MAY cause it to transfer more heat to the water depending on the pump design. Locking the external pump in your closed stand with little airflow will be almost as bad as putting it under water in your sump.

 

[Rekonn]

Thanks Bean! That's exactly the info I was looking for. Right now I have an external pump locked in a closed stand. But, in an effort to make it quieter still, I made a box around it out of SoundStop board. That made it significantly quieter, and next I was thinking about using some sound barrier foam.

But now I've learned that a submersible and an insulated external pump that put out the same GPH and use the same wattage will wind up heating my tank the same. I'm coming to the conclusion that I'm better off just getting a submersible pump since being under water should be quieter than any soundproofing box I come up with, right?

 

[BeanAnimal]

Well it all depends on the actual power consumed by the pump. You may be better off with a different style of pump.

You may want to look at some of the mods that involve removing the noisy fan from the pump and using a heatsink and small fan instead.

You could add some fans to your insulated box, but you would need to build some baffles so that there was not a direct line of sight into the box (sound does not go around corners as easy).

Bean

 

[fishbaiter]

I wonder since that is a fan cooled pump and when you have t enclosed like that, doesn't it heat the pump up due to insufficient air circulation which inturn may transfer heat to your water being pumped?
Just wondering..............

 

[Rekonn]

Fishbaiter, I'm thinking the same thing, and that's why I'm starting to think I'm better off with a submerged pump. My attempts to eliminate noise by insulation are removing the main advantage of external pumps, lower heat transfer.

I'm thinking about redesigning my sump and a submerged pump would be much more convenient in that setup as well. Here's the big question - Will a submerged Eheim 1260 transfer less heat than an insulated external Panworld 50PX-X?

Panworld 50PX-X is rated at 90 watts
Eheim 1260 is rated at 65 watts

I'm also considering the Eheim 1250, which would only be 28 watts.

 

[JER-Z]

<a href=showthread.php?s=&postid=7444134#post7444134 target=_blank>Originally posted</a> by BeanAnimal
The funny part is the more efficient the pump, the more energy is transfered to the water instead of being lost on the neutral line

forgive me, as i am a little rusty in thermodynamics, but this doesn't make much sense to me..

For example, if two pumps are consuming 65w...

pump A is pumping 600 gph (more efficient)
pump B is pumping 300 gph (less efficient)

thermal energy is a byproduct of friction, etc.

in other words if the energy isn't transferred kinetically, becomes thermal, right?

wouldn't a pump that is both more efficient, and produces more heat defy the law of energy conservation?

is there something i am missing in my assumption?

i don't think Watts consumed by a submersible pump is a tell all way of knowing how much heat will transfer...Eheims are probably alot more efficient (more work, less heat) than maxi-jets, meaning that if you used an equivalent wattage, the more efficient solution would produce less heat. That's my theory anyway.

But i could be wrong.

 

[JoeESSA]

Yes, but that kinetic energy that goes into the water from the pump must also end up as heat when the water finally loses kinetic energy due to pressure drop in the plumbing and finally in the tank - at least in a steady state condition. That's the conclusion I've come to rightly or wrongly

Need someone good at physics to analyze this closed system...

 

[JoeESSA]

Basically what I'm saying is that all consumed power from a submersible pump must finally end up as heat in the water. E.g. 150W power consumption = 150W heater for submersible.

If an external pump can be cooled then x% heat end up in the water and y% ends up costing you for air conditioning. x%+y%= power consumption of pump.

 

[Rekonn]

For those of you that would like to discuss physics, I think this thread would be more up your alley.

Now, back to the point of this thread - Will a submerged Eheim 1260 transfer less heat than an insulated external Panworld 50PX-X?

 

[fishbaiter]

Rekonn, I can't see why either one of those manufacturers wouldn't be able to answer that question. I mean they must have their own statistics. Maybe ask them both to see what each says just incase one is more bias than the other...

 

[Wryknow]

Rekonn - In answer to your question: no, you cannot determine how much heat energy will be added to the system based solely on the watts of energy consumed by the pump. The pump is doing work moving water, so the amount of heat generated will be a function of how much energy is consumed - how much energy is the useful work that we bought the pump to perform (this is a broad simplification.) The watts consumed is only a useful comparison when the amount of water being moved by the different pumps is (practically) the same.

Having an external pump complicated things a little bit because some cooling will be done by the external air, depending on the design of the pump. The vast majority of the cooling is still done by the water moving through the pump though I suspect since water is a much better thermal conductor than air is.

My direct observation leads me to the conclusion that the E-heim has very little heat transfer to the tank relative to other pumps. At night, when my 215 tank is running an E-heim 1262, mag 12, a couple of seios, and a 150W MH light on the sump, it stays about 2 degrees above the ambient room temperature with no chiller. Since a considerable amount of heat is transfered from the other pumps and lights in the tank the E-heim is a very small contributor indeed. When I replaced my old mag 24 return pump (run externally) with the E-heim 1262 (run internally) my average tank temperature dropped 3.5-4 degrees.

For a 90 gallon tank I think that you will be very happy with the 1260.

 

[BeanAnimal]

Wryknow... if the pump is submerged in water, then 100% of the energy goes into the water. To argue any differently is to go against the basic laws of physics. It's that simple! If your internal mag 7 draws 60 watts, and your internal eheim draws 60 watts, they both contribute the EXACT same amount of heat into the water.... this has been hashed out a dozen times. You simply can not re-write the laws of physics. Put each pump into an insulated box and they will both produce the same amount of heat.

That said, one pump may produce more flow and that flow may aid evaporative or conductive/convective cooling, but that is beside the point. Both pumps (drawing the same amount of power) will produce the same amount of heat. Energy is energy.

The same goes for MH vs FLUORESCENT. MH may give off mor einfrared heat, and fluorescent more visible light, but close them in a box and they will produce hte same amount of heat in the box. (if you want to nit pik, radio waves may not be contained in the box, so the lamp that produce more RFI may produce less heat).

Your observations have way too many variables to be valid for anything more than anecdotal observations.

 

[BeanAnimal]

REKONN

to answer your last question... it's hard to tell. It would depend on how insulated the "external" box is in relation to the tank and how much of that heat escapes into the outside room before it is added to the tank.

That said, insulating the external pump will more than likely make it less efficient (heat causes resistance in the motor windings), which in turn creates more current draw) and shorten the life of the pump. The coolr a pump and motor run the longer the bearings and seals last

I highly doubt either company has statistics on heat transfer in different situations. Even if they did, finding a person to contact and relate the information to you in meaningful terms would be an even greater longshot.

 

[Wryknow]

I agree that my anecdotal observations are just that - that's why I am pointing out that they are just my anecdotal observations.

The "closed box" arguement is of limited value because an aquarium is not a closed box. Are you arguing that moving 900 gallons of water 5' in elevation takes no energy or that every bit of energy is ultimately converted to friction and trapped inside the "closed box" tank? Like I said, I don't know of anyway to calculate a specific pump/heat-exchange scenario without direct measurement and that only tells you the what happened in that particular system.

 

[BeanAnimal]

please see the thread in the advanced forum. I am saying that 60 watts consumed, is 60 watts consumed. The "tank" is a closed system with respect to the PUMP and therefore ALL of the energy is imparted into the water.

The "energy" can only escape the the SYSTEM via convection, conduction, evaporation or vibration. So the moving water can make sound, cause air movement or radiate heat.

The "closed box" parameter just makes this easy to illustrate or measure. Like I said, more flow may allow the water to release the energy by evaporation, convection and conduction, and vibration (sound) but these are secondary to what is happening and really only relavant when trying to compare a pump to a heater. Even if both submerged pumps don't move the same amount of water, they will be fairly close in their ability to aid energy in leaving the "system". In other words 2 times the flow is not 2 times the evaporation, vibration, or surface area turnover for conduction/convection. So the point remains the same... in general 2 fully submerged pumps of the same wattage will impart the same amount of heat into the system. Remember this is the real power usage of these devices, not what the label says.

 

[Wryknow]

But you have to agree that the pump must follow the principles of conservation of energy yes? When you add potential energy to the water by elevating it in the air that uses energy. When I say that a pump is more efficient than another pump I am trying to measure the ratio of useful work to wasted energy. If you were ambitious enough you could harness the energy of the water flowing through the return line to do something more useful than creating turbulence in the sump - including generate more electricity.