heat transfer of submersible pumps

[Kurt03]

if you do this please post it. im interested in how close it is.
I would think so many of the watts on the power head are going to turn it and there would be less heat. I know there would still be heat from friction and stuff but i would think it would be less then the heater all the way on.

 

[Qwiv]

You send me the killawatt and I'll do it. I don't have one. I think the experiment is close enough. The powerhead was clearly labled a higher wattage than the heater itself. About 25%. The powerhead ran 24/7, the heater was turned off some of the time on sunday due to the built in temp controller. I would stir the 1 gal tank a little to make sure it was mixed up. I don't think adding a top would make it resemble a reef tank better as most reef tanks have open or vented tops which is what I am trying to replicate.

I am not trying to disprove the second law of thermo dynamics as I know how it works and am not stupid. If I did this same experiment with insulated coolers, I bet the temps would have been much closer if not the same or the opposite as the heater was limited by a controller.

 

[BeanAnimal]

Qwiv.. most reef tanks have moving water in them Stirring your bucket and a poweheard moving water are quite different. So for the experiment to mean anything both test setups need to have similar flow.

In other words you will not find a tank with a heater and no pump. You will find tanks with pumps and no heaters though

I would love to see the experiment again with closed lids. As you state with closed lids (and/or insulation), the turnout will more than likely prove that some of the earlier brainiacs that came up with these theories were pretty good.

 

[Kurt03]

how much water did the tank with the powerhead in it evaporate? would covering both make it more fair?

 

[Scuba_Dave]

Uh, so a 1250 gph submersible pump acts as a 1250w heater?
I don't think so, or there would be a lot of melt downs

 

[BeanAnimal]

Heh?

Dave, I have no idea what kind of drinks they are serving in the lounge, but we were talking watts, not gallons per hour

If your 1250 GPH pump draws 20 AMPS, you have other problems.

 

[Kurt03]

the gph doesnt matter, its the watts that the pump uses i think is what they say = the watts of a heater.

 

[Qwiv]

Adding the lid would definalty cause the heat to be greater as you are trapping the energy in the container. This experiment has been done, see the wiki link above. That would not make it "fair" as it isn't what I am trying to prove. Evaporation is one of the things I want in my model

I was aiming to show that the statement of "adding a powerhead and a heater of the same wattage" is not the same as G1 stated. My little example did exagerate the difference as the water in the tank with the powerhead looked like a blender and the one with the heater just sat there and got hot. I was aiming to show that the kinetic energy produced by the pump does not nessesarily get converted to heat 100% inside the reeftank. I can't say for sure where the energy other than heat went but we could assume some went to noise (very little), and most went to disturbing the surface of the water. This could release the kinetic energy into the air and also increase evaporation. There might also be a potential for the magnetic field created by the pump to react with something outside the tank therefore releasing more energy not as heat. There are probably other energy exports I have not thought of. The experiment does prove that a heater and powerhead do not have the same impact to a reeftank in terms of heating it. I would also like to note that I used a low quality preassure rated powerhead that was absolute garbage and produces more heat than any good quality powerhead. When a preasure rated pump is run with 0 head, it uses more energy that it is rated for so if anything, the pump was doing a good amount of heating. For comparison, I could drop my Tunze 6100 in the bucket, but the water would be on the ceiling. I wanted to use the 50 watt heater vs. a tunze in a large container, but I didn't want people to point out a Tunze is only 45 watts and call my experiment bogus. Also, the larger water volume would take more time to show results and the difference wouldn't be as great. Maybe someone else could give it a try.

I am also not trying to bash Goby in any way. I have found this experiment fun and am looking forward to learning more about the energy exports of out tanks.

 

[BeanAnimal]

the experiment does prove that a heater and powerhead do not have the same impact to a reeftank in terms of heating it.

I agree with most of your statements and feel you have a full grasp of the concepts and science here. However, I don't agree with your conclusion. I agree that they don't have the "same" impact, but as you mentioned your example shows an extreme condition. In the real world, the evaporation level is already somewhat high. The submerged pumps are circulating water in a system that is already in a high state of agitation.

Let me put this another way. Lets take your average 100 gallon system with 3,000 GPH flow, all with external pumps (1 return and 2 closed loops). Lets assume we have to of these systems side by side. Adding a 75W heater to one system and a 75W submerged pump to the second system will both affect the tanks temperature rather equally.

Like I said, I agree with most of what you are saying, but don't feel that the "bucket" eperiment showed anything but how good evaporative cooling works. In other words the bucket with the pump somehwat mimiced a real world reef system, while the bucket with the heater did not mimic a real world system.

Interesting results to say the least. No matter how many times you try to tell people how efficient evaporation is, they don't believe you.

Thanks for taking the time to play with this setup... it has been fun!

 

[Qwiv]

O, I agree Bean, but in your example, the heater might produce 100% heat and the pump could produce 99.99999% heat and .00001% other. I would still be right, but by just a smaller amount.

My experiment was definatly using evaporation to its fullest, but you can think of other ways energy could be exported, like noise, increased surface movement of the water releasing kinetic energy by moving air over the tank, then there could be biochemical things I am not even considering as I know very little about that stuff. I have read that increased flow accelerates calcification. Is it possible that the kinetic energy is being used to form the actual coral? Maybe. I need someone else to speak up on that.

If my experiment only proved that the pump increased the evaporation in the water, I have shown that a pump can be used to cool the water, while a heater can only heat the water. Therefore adding a 50 watt heater is not the same as adding a 50 watt pump in terms of heating an aquarium.

 

[goby1]

It heats it the same, but does not raise the temperature the same amount. The cooling by evaporation is a secondary process, and is why the T is lower for the circulated water.

Adding thermal energy is not equivalent to raising the temperature.

I'd agree with what Bean and ChemE are saying....

Is this killawatt thing just a 0 resistance ammeter?

 

[ChemE]

The Kill-A-Watt is able to measure both VA power and real power so it will correctly read the power consumption of a device with a power factor other than 1 (which is most everything in my tank). As an example, my 20 watt MJ 1200 only draws 13 watts which is why I'm not totally convinced that the powerhead vs. heater is a valid experiment yet. Both have to be drawing the same amount of power not just be rated to draw the same amount. Powerheads are inductive and reject a lot of what they draw back to the source which means they can draw less than they are rated for. A heater is resistive and won't be rejecting much back to the source so its draw will be closer to its rating.

I'm in England for 2 weeks but when I get back I'll do my own experiment with 2 10 gallon closed tanks and a Kill-A-Watt. Even better would be using 2 Kill-A-Watts to verify that the kWh's drawn are the same over the duration of the experiment but I'm not shelling another $25 to be that precise.

 

[Mogrash]

In addition to what you suggested, perhaps even better would be to take a 50 watt heater and a 50 watt powerhead, compare that to a 100 watt powerhead with flow reduced to the same gph rate as the 50 watt powerhead (this assumes that actual total draw would be equal).

Oh and my Mag 7 heats my 44 gallon salt making brute can pretty well...

 

[ChemE]

Heh, I don't own a 50 watt PH. Biggest I've got is my MJ1200 @ 20 watts but at 4' of static head it only draws 13 watts. I love having a low power tank. I get skimming, return pumping, and main circulation done for 24 watts.

 

[BeanAnimal]

Qwiv,

Yes the movement of the coral may absorb some energy into it's growth, but again we are talking such a small amount that it is not worth considering. The same with the sound, vibration, surface agitation that in turn moves air...etc.

In the real world in a real reef tank, a submersed pump that consumes 50 watts and a heater that consume 50 watts will heat the water the same, unless of course the 50 watt pump is the only means of circulation. In other words the pump will help evaporation and the heater will not. As Goby stated, the energy input into the water is the same. However, the motion created by the pump allows some of that energy to escape.

So I think we are all in agreement of what is going on, we just seem to view the symantics of it a bit different with regards to the "real world" results and how they are interpreted with regards to putting a simple explanation to the whole mess.

 

[JerseyReef]

<a href=showthread.php?s=&postid=7285275#post7285275 target=_blank>Originally posted</a> by ChemE
I've been meaning to come back to this for a while and kept forgetting. It turns out this is very very easy to calculate.

The latent heat of vaporization of water is 2258 kJ/kg or 2258 J/g or 2258 J/mL (g=mL because we are evaporating pure water not brine).

My 55 at peak load is putting right at 200 watts into the water column or 200 J/s. So I need to evaporate (200 J/s)/(2258 J/mL) or 0.0886 mL/s to offset this heat load. On a good day, my one 80mm PC fan evaporates about a gallon of water (based on my manual kalk additions) which works out to be 0.044 mL/s. This is confirmed by the fact that my tank heats about 3-4 degrees F during the day. Although I've never calculated it until now, if I put a second fan in and put them both on a temperature controller, I could keep my tank temp rock solid stable. I just haven't gotten around to affording a temperature controller yet.

My fan draws 1.8 watts and runs 24/7 without a controller (read cheap). I tried a 1/10th hp chiller but it couldn't keep up and sucked down much more than its rated 1/10th hp (which would be 74.6 watts). It actually drew about 200 watts. So there is one data point for you. 1.8 watts of evaporation beat out 200 watts of chilling.

ChemE - Don't take this the wrong way. I think a note on relative humidity needs to be understood for those reading this. I don't want to come across as 'picky'.

I believe you need to take into consideration the moisture already contained in the surrounding air volume for the fans ability to evaporate water and how well two fans would do the job. This evaporation rate would vary daily, depending upon several factors.

external dehumidification (use of dehumidifiers, air exchange systems)
relative indoor and outdoor humidity levels (summer and winter related from a heating/cooling aspect)

 

[BeanAnimal]

Jersey, I agree.... but I think in the worst case we could assume 50% humidity or less in a properly maintained environment. If your humidity is much more than that, you have problems.

That said, I am sure many folks do not have AC and live with the windows open, causing 70% or so humidity. Not good for the house, it's contents, or the evaporation in your system

I drop my 100 gallon system by about 5 degress with a single 10" desk fan. I keep meaning to build a more efficient blower, but never get around to it.

 

[ChemE]

<a href=showthread.php?s=&postid=7296739#post7296739 target=_blank>Originally posted</a> by JerseyReef
ChemE - Don't take this the wrong way. I think a note on relative humidity needs to be understood for those reading this. I don't want to come across as 'picky'.

I believe you need to take into consideration the moisture already contained in the surrounding air volume for the fans ability to evaporate water and how well two fans would do the job. This evaporation rate would vary daily, depending upon several factors.

external dehumidification (use of dehumidifiers, air exchange systems)
relative indoor and outdoor humidity levels (summer and winter related from a heating/cooling aspect)

Not picky at all just thorough; no worries.

Humidity is absolutely key in how rapidly tank water will evaporate. I just operate under the basis that most will be running AC and evaporating to their home. If you were trying to evaporate to a vapor barriered fish room you would evaporate virtually nothing and get no cooling.

 

[BeanAnimal]

And lots of mold

 

[goby1]

I think the last several posts explained some of my points clearly... thanks. If you do experiments with pumps and temperature though, consider that you should ideally have little to no air space above the water, as the heat added to the lid with condensation is separated from the water with the air gap... in any case, I think we agree that deviations in T would illustrate either unaccounted evaporation or limits in experimental precision. Alternatively, maybe you could start the experiment after the air gap had reached equilibrium with respect to the partial pressure of water vapor in the air...?

I think that the point of the already turbulent tank not having its evaporation increased that much (w/ say an additional powerhead) is good - hadn't thought of it that way.

Chem, I tried to implement the "near same level" tank/fuge/sump height plan you have, as I think it makes perfect sense as long as the overflows handle the the flow you want. Unfortunately, even though I have the sump in the room behind, I couldn't quite make it work for geometric reasons. I agree that it's silly to let water crash to another level, only to have to pump it back again. That's major energy consumer for sump-in-the-stand setups. I just set up a surge system for the sole flow in my tank, but have yet to consider all of the energy dynamics. The goal is to try to have the benefits of 100x type turnover sps systems with just one dart for a 180. I made a airstone skimmer, and don't think I can get away with gravity feeding it. So I'll have to waste power on another pump....

In specific reference to the large return pump - really needed ? thread, do you think that 100x, say 15% of time, comes close to 80x 100% of time? Other than coral response, the issue is having enough flow for suspending waste solids so they reach the critical region near the overflow, such that they get carried over during one surge cycle. I they don't, the detritus will remain, and if they do, the detritus might make it over, and at this point, the concern is the percentage of detritus that does make it in relation to the rate at which it is generated by lr flock/etc. Surge only is in direct conflict with small return, tunze/closed loop systems, as all flow goes through the sump. I'm wondering if the overall efficiencies might be better for these types of systems... Will start a thread sometime about all of this, but mention it here because my original thought was to save on the pge bill, yet still not compromise on sps health. Thoughts?

G1