Pumps and heat

[myv65]

[sarcasm mode on]
You guys really need to brush up on your thermodynamics.
[sarcasm mode off]

Some have mentioned that some of the energy gets put into the fluid as mechanical energy. So what? What do you suppose happens to that energy eventually?

There's a very basic law of thermodynamics that goes essentially like thus:

Input + produced = output + destroyed + stored

The "produced" and "destroyed" parts of the equation don't generally apply when discussing heat transfer. These apply more to availability and entropy, a couple of other thermo concepts. "Stored" doesn't apply to steady-state operation.

For a pump, it gets energy input in the form of electricity. This input gets split into a few different forms. By far the two dominant ones are mechanical energy into the water and heat. Heat gets produced directly by motor inefficiency and friction. Most electric motors that I'm familiar with run around 75-80% efficiency. Most pumps that I'm familiar with (industrial stuff in the 100-5000 gpm range) peak between 60 and 75% efficiency. As choked as we tend to run water cooling pumps, their efficiency is even lower.

So we've got some heat produced directly. As someone already noted, submerged pumps put all of this into the water. Inline pumps get rid of much of this motor inefficiency heat through convection to air. The remainder of it is due to impeller/volute inefficiency which generates a lot of water currents that aren't going where we wish them to go. These secondary flows are basically nothing more than non-productive churning of the water. A little of this can get out as convection off the pump casing, but with the water temperature so close to ambient, it isn't much.

OK, so we're left with the "useful" work produced by the pump, namely volumetric flow rate multiplied by pressure rise over the pump. This, along with resistance, is what determines our flow. Now comes the real crux of the matter. Where does the energy go that produced this flow? Hello, it can't just build up indefinitely. It has got to get out just the same as the thermal energy from our chips. Assuming a conventional system of pump, radiator, tubing, and block, then virtually all gets out through the radiator as thermal energy exchange.

You need to remember that energy isn't so different from water or any other physical entity. It can't "just disappear". If you could draw an imaginary box that totally enclosed your cooling system, then the energy entering the box must ultimately be balanced by the energy leaving the box. Energy can come and go in various forms, but what comes in must get back out eventually. Knowing this, figure out how it can get out. For all practical purposes, all of it goes in through electricity to the pump. For all practical purposes, all of it gets out as heat transfer. Some of it directly off the pump, for inline pumps, but the majority of it through the radiator.

[Sirpent]

Just one remark though.

Originally posted by myv65:

Inline pumps get rid of much of this motor inefficiency heat through convection to air.

In a typical system, this heated air goes through the radiator. Thus deltaT between air and water is lowered and some cooling efficiency is lost. I do not think there is a noticeable difference between in-line and submerged arrangements in term of overall cooling efficiency.

[Skulemate]

Ok. I understand what you said, and I guess it's my fault for any confusion. My main reason for this thread was to talk about all of the heat transferred to the water through a pump's inefficiency.

One question though. When energy is lost to friction (resistance from the pipe walls for example) is it fair to assume that the heat created is dissapated partly by the tubing, or would you expect most of that heat to be carried away by the moving fluid?

[sunblade]

Originally posted by Skulemate
One question though. When energy is lost to friction (resistance from the pipe walls for example) is it fair to assume that the heat created is dissapated partly by the tubing, or would you expect most of that heat to be carried away by the moving fluid?

I'm really just guessing here, but I would think that both the walls of the tubing and the water would absorb some of the heat caused by friction, but whichever medium can absorb heat faster will absorb more???

I would think that it would be easier for heat to be absorbed into the water, unless the tubing is copper, which would act like a radiator.

[redleader]

"ALL energy is converted to heat", wrong. a pump moves water by converting somewhere around 20% (varies greatly) of the electrical energy into fluid momentum, the rest creates heat.

This has already been addressed, so lets let it go and continue the discussion.

-------------------------------------------

Sirpent brings up a good point:

In a typical system, this heated air goes through the radiator. Thus deltaT between air and water is lowered and some cooling efficiency is lost. I do not think there is a noticeable difference between in-line and submerged arrangements in term of overall cooling efficiency.

If a good deal of the heat transfered to air eventually does pass through the radiator in most systems, it further bolster the theory that much if not most pump heat enters the coolant.

Skulemate:

I'm hesitant to accept real world data here, its very difficult to determine how much is actually entering the coolant and how much is being bleed off as we measure. Not to mention your setup was atypical for most people.

However your results suggust that even with truely enormous pumps (BTW what pump is that?) the influence on coolant temps is pretty minor. Another argument for maxing flow?

I'm really just guessing here, but I would think that both the walls of the tubing and the water would absorb some of the heat caused by friction, but whichever medium can absorb heat faster will absorb more???

Tube and coolant will be almost the exact same temp. Unless its copper which will (as you pointed out) function as a sort of radiator. So I don't think it really matters which absorbs the heat, both are heated to the same temp.

If you meant which will have more heat, easily the water because of its high specific heat and much larger volume.

[myv65]

Others have already stated that the tubing won't give off much heat. Sure it sees the same (roughly) fluid temperature as the radiator, but the fluid temperature is only part of the equation. Fluid to solid convection is many times more efficient that solid to air convection. This is why radiators have fins. The surface area of the fins will be many times that of the wetted solid to compensate for air's lower convection coefficient. This fact, combined with the relatively low conduction coefficient of tubing, means that the radiator really dominates the energy dissipation from our systems.

[Skulemate]

That test was hardly meant to simulate someone's setup. Rather, it was an attempt to capture all of the heat transferred from the pump to the coolant for the duration. I imagine that some small amount was lost through convection from the surface of the hoses, but I don't know how much.

Oh, and it's the same pump that Hellion_Prime uses... the Little Giant 3-MD-SC.

[bigben2k]

I was glad to see Skulemate's numbers. Can you tell us more about the pump?

myv65: I hope that you don't still believe that all the power applied to a pump turns into heat. To answer your question directly, the energy that results from the pump is by inducing movement to a mass, in this case a body of water. Your question should be: where does the energy go when the pump is turned off, and as the coolant stops circulating?

No, turbulent coolant flow does not create a significant amount of heat, nor does rubbing against the tube walls. Heat does get transfered through the tube, but since it is vinyl, which is a great isolator, the amount transfered is negligeable.

I still maintain that the heat that a pump induces in the coolant is from its own motor coils.

[Sirpent]

Skulemate, can you measure the actual power consumption of your pump?

[Skulemate]

Perhaps, but I would have to borrow some equipment to do that. I'll see if I can arrange that.

[Skulemate]

I actually have something I'd like clarified... doesn't a pump actually give the water potential energy rather than kinetic energy? The resulting motion is for the same reason that water flows down a slope...

[BrianH]

Has anyone tried to add heatsinks to the sides of an external pump. The heatsinks could be something like the Wing Fin Aluminum Heatsink, one one each side. With proper case cooling, this should perhaps remove some of the heat from a pump and channel it to the exterior of the case.

[myv65]

BigBen2k,

It is a relatively simply matter to calculate userful work of the pump. It is simply flow rate times pressure rise. Just because this is useful work to us makes no difference. It is still an energy form generated by the electrical input to the motor. Eventually all energy consumed by the pump needs to make its way out and does so as thermal energy.

If someone wishes to calculate the useful pump work, they need a means to measure suction pressure, discharge pressure, and flow rate. Water manometers can do the task, but need to be fairly tall for powerful pumps. I've personally never cared because there isn't a whole lot I can do to change the numbers.

I guess the main point I would make is that all energy consumed by the pump eventually must "leave the box" and does so as thermal energy. Even the portion that generates flow still must leave as thermal energy. Once you turn a pump off, there is some stored energy (both kinetic quickly dissipated and thermal slowly dissipated) that leaves as everything reaches an equilibrium with the ambient conditions.

[Sirpent]

Skulemate,
No, it's kinetic energy. But there is no big difference. If you are pumping water from a lower tank to a higher one, the pump converts electromagnetic energy to the kinetic energy of flowing water, but then this kinetic energy (or rather the part of it which is not transfered to thermal energy by friction) ends up as potential energy. If the loop is closed, the total potential energy is constant, it does not change with time. (If you look at a small "piece" of water, its potential energy changes, but this is not important for the big picture). The total kinetic energy is also constant (at equilibrium). Thus, at equilibrium, if the pump consumes a certain amount of electromagnetic energy during some interval of time, the same amount of energy will be produced by the system as heat during that interval of time. This means that when some thermal energy is converted into kinetic energy by the pump, the same amount of kinetic energy is converted to thermal energy through various kinds of friction. Botttom line: kinetic/potential energy is not worth mentioning.

myv65,
What is your definition of "useful work"? I would think that the useful work in a closed system is zero.

[myv65]

"Useful work" is defined by the task at hand. In the case of a pump, I define it as flow rate multiplied by pressure rise. Another way to look at it is it is the energy input devoted to the desired task. If the motor and pump each worked at 100% efficiency, the motor input would equal the useful work.

I'm not so sure that skulemate's question is easily answered. Total energy of a fluid (ignoring thermal considerations for a moment) is defined by Bernoulli's equation. It basically says that energy = p + v^2 + z. I've left out a few pertinent terms like density and gravity to simplify. "p" is pressure, "v" is velocity, and "z" is elevation. "p" and "z" are what I would consider as potential energy while "v" is what I would consider kinetic energy.

Does a pump add kinetic or potential? Well, the fluid moves the same speed on each side of the pump, so one could argue that the kinetic energy is a constant. In this case, the pump couldn't add kinetic energy, right? However, without the pump there would be no flow, so the pump must add kinetic energy, right?

Like Sirpent said, it really doesn't matter. One energy form or another, it's still energy and must balance out at the end of the day.

[Skulemate]

Yeah... I am familiar with Bernoulli's one dimensional flow equation. Suppose you write the equation for a point just before the pump and another just after the pump. Assuming that the elevation head is very close to the same on both sides, what you end up with is an increase in the pressure head, since the velocity head is also going to be approximately constant. So, I guess then that I was right... that the pump does indeed add potential energy to the fluid, which immediately becomes kinetic energy. As for not mattering... well, that may be true, but I was still curious.

P.S. Useful work would not be that hard to calculate, assuming that you have a good pump curve. If that were the case, you only need to know flow rate to determine the useful work, since the head produced by the pump can be read from the graph.

[bigben2k]

I see your point, myv65. I agree that calculating the energy by using the flow rate and pressure drop would give us a proper number. If we converted that number to Watts, then we could compare it to the electrical power, and this is where I am saying that since these pumps are about 70% efficient, 70% of the electrical power is converted to a movement of the mass (body of water). Wether the coolant returns to the pump or not is irrelevant: the wheel of a car returns to its position, several times per second. A ferris wheel also returns to its position.

As for what happens to the energy (what you refer to as something has to leave the "box"), I would agree that it is friction, and to be clear, we're talking about the moment when the pump is turned off, so the water has potential energy, as it is still spinning in the loop, but it does come to a stop, as the flow restrictions cause the water to slow down, by friction (so yes, heat, but also, and mostly, pressure, aka pressure drop). The coolant comes to a full stop, and there is no more energy (other than potential, from gravity, but that's outside of what we're discussing).

Can we try to apply some numbers to Skulemate's results? Just for kicks... let's do the math.

Skulemate: the pressure is just as important as the flow rate. Flow rate alone cannot give you an idea of the energy involved. Ex: 10 gpm at 10 psi, versus 10 gpm at 100 psi. There is a difference in energy level. You otherwise have it correct.

[Skulemate]

Bigben, I think you misunderstood. I said that the only thing you need to measure is flow rate if you have a good Q-H curve for your pump. If that's the case, you can read pressure drop (in terms of psi or feet) directly from your graph for the measured flow rate, and you have all the terms required for the calculation.

[bigben2k]

Originally posted by Skulemate
Bigben, I think you misunderstood. I said that the only thing you need to measure is flow rate if you have a good Q-H curve for your pump. If that's the case, you can read pressure drop (in terms of psi or feet) directly from your graph for the measured flow rate, and you have all the terms required for the calculation.

Yes, that's correct, so you were right.

As for the energy otherwise, it appears in the system in the form of pressure (and that's the answer to myv65). If you take a balloon for example, there is work being done, in every direction, so you don't see anything move, but the work is there.

In our case, our rigs have different pressure points, and that's the "mechanical" energy that the pump provides, with about 70% of the power it is provided with. Everything else (30%) is attributed to (not necessarily in order):
1-EMI
2-Heat
3-Noise
4-Friction (of water against water, water against inner surface, impeller against bearing, impeller against water, etc...)

When the pump is turned off, the pressure drops, and that's the energy that is released.

[nemaste]

NICE myv65!!! good points, lots of technical merit (you must've not gotten get sick of excel presentations like i did in school). i knew friction will eventually take mechanical energy out of the system (turning it in to heat), but literally speaking there is in fact a small % of energy going into maintaining a steady state fluid momentum whether or not it ends up being stored as air pressure (mostly for large air volumes), elastic energy (mostly tubing stretch), or lost as heat through friction. that includes just about all the energetic arguments for an h2o system: kinetic, potential, & losses (no radioactivity in or out, no electromagnetic energy created, which are good assumptions in this context).

we all should know the basics of energy conservation so well, energy in=energy out (at least at equilibrium, right?). i'm stoked about your experiment; it was extensive enough for us to claim that ~1/4 of the power will be picked up by the water. i'm not sure how useful that data is for other pumps, since the results will change with a different pump or if the pump has some degree of aircooling.

BTW,
EMI=interference is not a dominant issue here
HEAT=the result of dissipative loss (friction/resistance)
NOISE=vibration is kinetic energy (just oscillatory)
FRICTION=loss of mechanical energy to heat

perhaps we may more appropriately describe the energy system with:
EMF=electro-motive force (E&M field energy, sorry about the old jargon)
KE=kinetic energy (vibration, water movement...)
Dissipation=friction/resistance (both in the pump inefficiency & from the water path)
PE=potential/energy storage (pressure, hieght, capacitance, structural stress...)
n=coefficient of restitution (structural dissipation)

i'm no EE major, but i'm sure there are some momentum-like effects in electronics too (eddy-currents, inductance...?).

[bigben2k]

That's all good, but what I'm seeing is a claim from myv65 that most of the power (10W for an Eheim 1048) is induced in the water in the form of heat. I don't believe that that's the case, but again, maybe I just can't picture it. I believe that most of the energy appears in the form of the circulating water, fighting its way through all the restrictions, which in turn, creates some heat, but certainly not anything near the pumps power input, or even significantly, for that matter. I don't believe that water rubbing against itself produces much heat at all, and neither does water against any other kind of surface. I believe that the energy induced by the pump, which is seen as water movement, is transformed into pressure against the innards of the cooling rig. Since this pressure is omnidirectional, there is no movement observed, but the work/energy is there.

I believe that the heat that a pump induces in the water mostly comes from the heat from the coils. In other types of pumps, the coils can heat up the driveshaft, which can transmit this heat to the water.

EMI/EMF has to be significant, since it does drive the impeller. Hopefully though, most of that energy is absorbed/converted to the motion of the impeller. It certainly explains why mag drive pumps are 70% efficient, at best.

As for noise, I also believe that it is minimal. It can be simply estimated (roughly) by calculating the amount of power required for a speaker to make the same noise/noise level.

Here's an idea: since the movement of the water is measurable, we can calculate the energy applied, using physics formulaes. What's the equation (my physics is rusty!) to calculate the energy required to move a mass (of water), over a specific amount of time?

[000]

well I've done quite a few tests with a danner 700 gph magdrive pump which is rated at 60 watts. It transmits about 43% of that to water when inline and about 53% submerged ( 26 & 32 watts).

for more detail we have a thread going on about this exact subject over at the overclockers forum here:
http://forum.oc-forums.com/vb/showth...0&pagenumber=1

Sorry if I'm not supposed to provide links to other forums, let me know and I'll remove it.

-Sidney

[bigben2k]

Nice link, and pump...

I was particularly interested by this statement:
Water pumps serve many different functions in the aquarium. Submersible pumps such as the Rio and Supreme series are designed to run under water. This helps eliminate the noise, but puts more heat into the water. External pumps are designed to deliver more velocity at higher pressure.
(quote from the Aquadirect website)

What I couldn't find however, is the website to that pumps manufacturer (Hi Tech). I'd like to see the chart for it. 750 GPH@0 is not enough information for me, to consider purchasing one, but I am interested!

[myv65]

BigBen2k,

I think you're confusing my intent a little. It isn't so much that a given portion of the energy input to the pump immediately shows itself as heat. The point I am making is that all the energy consumed by the pump must at some point escape from the system. It does this as thermal energy.

As to how much does immediately show itself as thermal energy, one must either measure this directly or believe the data provided by the pump manufacturer (assuming they provide it). The best industrial centrifugal pumps peak around 80% efficient. These pumps have very closely held tolerances while dealing with flows in excess of 1000 gpm. These comparatively dinky pumps we use have tolerances on the same order of magnitude, but pump less than 1000 gph. This means even the best of our pumps probably does no better than 70% efficiency.

The efficiency peaks at only one specific flow/discharge head. As flow nears zero, efficiency nears zero. As discharge head nears zero, efficiency also nears zero. Ya gotta remember that pump work is delta-P * flow rate. If either of these terms approaches zero, so does work hence efficiency.

So that I am clear, I'll rehash the various energies and efficiencies involved here. A pump's work output is delta-P * flow rate. Energy required to generate this flow is work divided by pump efficiency. Assuming a "best case" scenario, the shaft power must be about work divided by 0.7 (more realistically, probably 0.5 to 0.6). Shaft power is the electric motor's output power. The motor's input power is work divided by motor efficiency. Motor efficiency is probably between 80 and 85%.

So, line power to the motor should roughly equal delta-P * flow rate / ~0.7 (pump efficiency) / ~ 0.8 (motor efficiency). I'll leave it to anyone interested to apply the proper unit conversions.

The motor's inefficiency mainly goes off as convection for an inline pump. It obviously goes into the water for a submerged pump. The pump's inefficiency goes into thermal energy in the water. The pump's useful energy goes into circulating flow. All of the energy put into the pump's motor must eventually leave the box. Under steady-state conditions, energy leaving equals energy entering. For a while after starting up, energy leaving is less than entering with the excess showing up as thermal energy in the fluid (and kinetic energy of the moving fluid). After powering off, flow will slow to a halt and the fluid will cool to ambient. This is the release of stored energy and results in an energy output with no energy input.

[bigben2k]

I can start seeing it... kinda like a light at the end of the tunnel.

From my previous statements, I said that most of the energy is used to maintain the coolant flow. That's obviously incorrect. Since the coolant flow is steady, the energy still has to go somewhere, so it goes against what makes the water slow down, i.e. the restrictions, ergo friction.

I also said that the energy appears in the form of pressure, but again, once the tubing has expanded (for example), the energy still has to go somewhere. I just don't know where.

So I can see that there is friction, between water against water, and water against all inner surfaces.

What I don't see is how the pressure against the inner surfaces is accounted for.

(continuing ramblings)
The pressure is trying to burst through everything (tube, waterblock), but it's resisted by the "solid" state of the materials (rubber, copper). So how do these materials react when subjected to pressure? Do they emit heat, like when you compress vapor to make water?