Pump power consumption with restiction.

[Since87]

I did some pump power consumption testing last night and posted the results in a somwhat lengthy and heated thread over at oc-forums.

Anyway, I'll quickly sum up here.

The test I did, was to measure the power consumption of my Danner Mag5 pump with a Radian Research RD-21. I did this test with unrestricted flow and restricted flow. Unrestricted was a few feet of half inch ID tubing. Restricted was with the tubing pinched off as tight as I could with my fingers after taking multiple pictures with the tubing pinched in one hand and a crappy digital camera in the other. OMG I can hear the soapbox being dragged out...

I measured simultaneously:
The voltage applied to the pump.
The current through the pump.
And the power consumed by the pump.

I had previously measured the winding resistance of the pump at 14.1 Ohms.

I got the following results.

In the unrestricted case with 123 Volts connected to the pump:
The current was 0.801 Amps.
The power consumption was 35.3 Watts.

In the restricted case with 123 Volts connected to the pump:
The current was 0.757 Amps.
The power consumption was 24.6 Watts.

This surprised me, because I would have expected the pump to consume more power with its output restricted. (Although, I had seen hints at this with cruder testing earlier.)

The pump's efficiency definitely went down with restriction due to the power consumption in the winding resistance.

Unrestricted:
Presistive = I^2 * R = 0.801 * 0.801 *14.1 = 9.05 Watts.
R-Waste-Ratio = Presistive / Ptotal = 9.05 / 35.3 = 26%

Restricted:
Presistive = 8.08 Watts.
R-Waste-Ratio = 33%

I don't have a good explanation for the overall drop in power consumption with restriction though. Anyone know?

[myv65]

Bah, I was too slow. I already replied in the other thread ya went and hijacked. The easy explanation is "that's how centrifugal pumps work". The more technical answer is that a pump creates kinetic and potential energy. The kinetic is fluid motion while the potential is pressure * flow. When you monitor pressure and flow and do the calculations, you'll find that power is more sensitive to velocity than pressure. Translation: More flow at lower pressure equals more power than low flow at higher pressure.

Efficiency is another interesting aspect, but not one I wish to toss into the mix here.

[g.l.amour]

at the risk of sounding utterly stupid, does this mean that a pump puts out less heat when severely restricted?

[BillA]

for completeness copying myv65's initial post
quote
Since87,

LOLOL. Anyone, and I mean ANYONE that works with real centrifugal pumps would tell you that power goes up with flow rate. This is true despite the lower head that you get at higher flow. All ya gotta do is look at a pump curve that includes power and it'll be obvious.

What I can't quite get is safemode's insistance otherwise. I don't frequent OC that often, but generally thought the guy was OK with heat transfer and thermo. Obviously pumps aren't one of the guy's strengths.

Fact of the matter is that the lowest energy consumption you'll ever get is when you dead-head a centrifugal pump. The highest you'll ever see is when TDH is zero. Peak efficiency is the funky one, as it peaks at different flow values for different pumps. Rarely will it exceed 90% in the best industrial pump. Aquarium-style pumps would be hard pressed to exceed 50%. Given the setups people tend to favor, I figure most water cooling pumps are in the 30%-40% range.
end quote

there are several ways to run at this 'pump power' issue, and I seem to be able to make an unintelligible mess of them
that said, a brief description of my difficulty (and I think Since87, it will come round to yours)

in 'looking' at a heater/wb/coolant system - as a black box, the input power can be calculated and the output power can be calculated

a physicist friend is slapping my head saying: no, its not so simple as the pump is inputting 'power' into that black box via the friction and the pressure drop

had great difficulty with this until another engr 'splaned to me that the 'power' associated with the pressure drop could be measured using the I*E differential at the pump motor

I'm not so sure that experimentally such can be done (substitution with a straight pipe yields also a flow inequality)

separate aspect: I can speak to pump heat generation (but not quantification)
by placing an oil cooled aluminum bodied pump within a heat bath, its heat generation can be observed at different flow rates
- restricting the pump's output causes it to 'create' MUCH more heat than in a 'free flowing' condition

myv65
care to comment on the heat balance aspects of the wb pressure drop ?

[myv65]

Originally posted by unregistered
separate aspect: I can speak to pump heat generation (but not quantification)
by placing an oil cooled aluminum bodied pump within a heat bath, its heat generation can be observed at different flow rates
- restricting the pump's output causes it to 'create' MUCH more heat than in a 'free flowing' condition

myv65
care to comment on the heat balance aspects of the wb pressure drop ?

On your "separate aspect", what you're seeing is the difference in where the energy goes. I guess a little more definition may be required here. A classical setup is to have a pump drawing liquid from a stilled reservoir and pumping it somewhere. In such a case, the power input to the pump impeller shaft will go up with flow rate. Fortunately, a pump has no knowledge other than the conditions that exist at its inlet and outlet. For this reason, the same is true for an inline setup and its power also increases with flow rate.

What will vary as flow changes is how the energy put into the impeller shaft shows itself within the liquid. In the simplest sense, the "useful" work of the pump would be defined as the pressure difference on either side of the pump multiplied by the flow rate through the pump. True engineering types may complain that "work" has units of force*distance while flow rate multiplied by pressure has units of force*distance/time (power). Whatever. When you look at a centrifugal pump's efficiency curve, you'll find that they define efficiency as flow*delta-P/shaft power.

At zero flow, efficiency is zero. Likewise is true for zero delta-P. In the former, all energy goes into churning the water, ie directly into heat. You immediately see this in your example. As soon as you have some flow, the energy gets split between ineffectual churning of the water and producing flow. Efficiency climbs from zero and starts marching towards its peak. The peak efficiency will occur at some nominal flow that lies between zero flow/peak head and peak flow/nearly-zero-head. Even at peak efficiency, some of the shaft power doesn't generate any useful flow and produces heat directly. Point here is that the portion showing directly as heat is inversely related to the pump's efficiency.

The rub for a closed system is that all the energy put into the water eventually turns to heat. This is the result of pressure drops through the remainder of the system. In a truly open system you can have some energy storage if you indefinitely pull water from a low elevation and pump it to a high elevation. Note that our "open systems" really aren't open from an energy perspective.

Another thing that comes into play with submerged pumps like your example is the motor behavior. A split capacitor, single phase, four-pole motor will have a peak efficiency in the range of 65%. The efficiency curve looks like a parabola. So as you move away from the "full rated load", the efficiency will drop off dramatically. A dead head condition will tend to drop the motor output load, but will also bump the motor some along its efficiency curve. I have done absolutely zero work in this area to measure the effects, so I can't say positively how significant this is. All I would point out is that energy input to the motor does not drop as rapidly as the shaft output power drops due to the efficiency curve of the motor.

In your example above, this translates to lower total energy input when dead-headed, but 100% of that energy coverting immediately to heat. In a free-flow case, more energy is input to the fluid, but a smaller amount is directly converted to heat.

*Warning: Engineering geek-speak ahead*

I'm not quite sure where you are headed with your request to "comment on the heat balance aspects of the wb pressure drop". I try to break down problems into one of two conditions. Either you define a control surface or a control volume. In the former, you analyze everything that crosses a given surface. In the latter, you analyze everything that happens to a given volume. Our systems generally lend themselves to the control surface approach.

In any analysis like this, you have a fundamental equation that governs life: Input + Created = Output + Stored + Destroyed. For our simple systems when energy is the topic, there is no "Created" or "Destroyed" terms. The "Stored" term only relates to transient conditions and can be ignored when one is only interested in examining steady-state operation under extreme load.

In this respect, I would consider drawing the imaginary box around the block, cutting the inlet and outlet lines where they meet the box and cutting the interface between the bottom of the block and interface material. Energy gets into the control surface as thermal energy from the interface material and thermal, kinetic, and potential energy in the fluid stream. Energy gets out of the control surface as thermal, kinetic, and potential energy in the outlet fluid stream. There are also minor (perhaps ignorable?) effects on the remaining block surface as it may be warmer or cooler than the surrounding ambient air as well as radiation (almost certainly ignorable due to emissivity, view factors, and surrounding absolute temperatures in comparison).

The difficulty that I know you face is one of quantification. A control surface is only as good as one's ability to measure the properties at that surface. Tell you something you don't already know, eh?

*End engineering geek-speak*

OK, no guarantees it's all 100% accurate. That's what you get for writing off-the-cuff without proofing.

[bigben2k]

Originally posted by myv65

Another thing that comes into play with submerged pumps like your example is the motor behavior. A split capacitor, single phase, four-pole motor will have a peak efficiency in the range of 65%. The efficiency curve looks like a parabola. So as you move away from the "full rated load", the efficiency will drop off dramatically. A dead head condition will tend to drop the motor output load, but will also bump the motor some along its efficiency curve. I have done absolutely zero work in this area to measure the effects, so I can't say positively how significant this is. All I would point out is that energy input to the motor does not drop as rapidly as the shaft output power drops due to the efficiency curve of the motor.

Now I'm beginning to understand why pump curves aren't straight! Thanks "uncle" Dave!

[BillA]

"The rub for a closed system is that all the energy put into the water eventually turns to heat. This is the result of pressure drops through the remainder of the system."

which is precisely the point being pressed upon me
but (being an experimentalist) why can I not measure such ?

take the case of a high flow resistance wb at a high flow rate
the energy represented by the pressure drop (~3psi @ 3gpm) should, if being converted to heat, be more than sufficient to measure
(isolate and insulate the wb, measure temp diff across inlet and outlet)

any ideas why I see NO change ?
- got me stumped
aahhh . . . .
and having written it out, I believe I 'see' the problem
at 40W input from the die, the coolant temp rise is 0.05°C
if the pressure drop energy is nominally 5W, then I'm trying to measure ~0.0125°C; way too small for my instrumentation and setup to consistently and reliably detect

(but such will, or should, enter into the 'efficiency' determination of the heat die)

quantification indeed !

Ben
well yes, and also no
consider also a positive displacement pump

[myv65]

You've hit it exactly. Anyone familiar with pressure relief valves on high pressure hydraulic systems knows about this subject well. If you have to blow 30 gpm at 2000 psig across a relief valve, you'd better believe that valve will get hot in a hurry. As you've noted, a few gpm at < 10 psig pales in comparison.

[BaconGrease]

sweet mother of god!
This post is so over my head!

[Since87]

Ok, I hope you guys will bear with me. I'm a mere EE. In my work, motion is something to be avoided. It is much more often an annoying aspect of the product's environment, than a desired goal.

I guess the main thing that was throwing me, was that in the case of a heavily loaded or stalled electric motor, the power consumption is high compared to an unloaded motor. My initial assumption was that restricting a centrifigual pump's output would have an effect on the impeller equivalent to loading the shaft of an electrical motor. That's obviously wrong as indicated by my test results.

myv65 a lot of what you wrote is over my head, but thinking about what you wrote, I've developed the following mental model/paraphrase.

I'm picturing the power delivered by the impeller being used to do two distinct things.

1. The impeller is maintaining the rotation of a cylinder of water the shape of the impeller chamber. Sort of a flywheel, albeit one with fairly lousy bearings, so it takes a fair amount of power to maintain its rotation. (And that power is dissipated in the lousy bearings.) This "water flywheel" component is capable of storing potential energy and that stored energy is effectively the same as pressure.

2. The impeller is also tossing water out the pump outlet. Obviously kinetic energy.

At zero flow, all the energy imparted by the impeller is being used to overcome the "bearing friction" of this "water flywheel". This is the state of lowest power consumption once this flywheel is up to speed. It is also the state in which the impeller chamber is heated the most, because all of the power dissipation is occuring inside the impeller chamber, and most of the heat is staying there.

At max flow, none of the impeller power is going into maintaining the rotation of the water-flywheel. All of the power is going into tossing water out the outlet. However, even though no energy is being stored, the "bearing friction" associated with the "water flywheel" must still be overcome, because the bearing friction is a function of the the impeller and the impeller chamber chacteristics. The power required to toss water out the outlet is added to the power required to overcome the "bearing friction". This is the state of highest power consumption, but most of the heat is moving out of the impeller chamber with the water.

This is undoubtedly a gross oversimplification, but it explains the data as I understand it.

Are there major factors I'm missing here?

[BillA]

Since87
where on earth did you get that RD-12 ? awfully nice piece of gear
(WAY beyond my budget !)

I think your model is a bit 'loose'

the pump shaft (or rotating magnet if mag drive) is connected to an impeller having curved vanes rotating in an enclosure
- at no flow there is no rotating 'disk' of water, but rather a violently churned mass in which the vanes can get no clean 'bite'; they have zero efficiency albiet max pressure - due to the cavitating vanes the motor is relatively unloaded
- as the flow increases, the effectiveness of the vanes does so as well, increasing the load on the motor and slowing it slightly
- maximum efficiency relates to the specific pump design and motor combination and is decsribed in Mgal / kWh (units probably wrong - all this was years ago)
- as the (back) pressure continues to decrease the flow will increase but as the vane configuration is no longer ideal the specific efficiency will start decreasing as well, though the total work performed will still be increasing

jeez, I'm sorry I started this; despite tweaking its not all that good a model either

I guess you don't have access to a pump handbook ?
even Marks describes this as I recall

well, leave it for Dave

[Since87]

I got the RD-21 from the cal lab downstairs. It is a marketing demo unit. I designed and/or had a hand in designing a fair amount of the analog and digital electronics in it. I'm currently working on final polish of the digital logic for its big brother - a three phase version.

I don't really know what we charge for them, but I can't afford one either.

I imagine my model is a lot loose.

Getting from my knowledge about this stuff to Dave's is like crossing the Grand Canyon on foot. I'm just reporting my view from the rim of the abyss at this point. It's unlikely I'll actually decide to go through with the trip.

I do best thinking about this stuff with some sort of picture in my head. Even a very inaccurate picture is a better starting place for me than none at all. Presenting my picture for dissection is only intended as a way for me to get from here to there. I wouldn't necessrily expect anyone else to find it useful.

Of course fluid flow has got to be one of the most difficult things to picture there is. So, i realize my crude picture is only going to get me so far.

I'm sure I could borrow a copy of a pump handbook from the engineering library at Purdue U. I don't have enough hardcore interest at the moment to make me want to go that far. I don't expect to ever be more than a dabbler in this stuff.

[myv65]

Since87,

I hear you on the EE bit. I'm an ME, and I prefer it 'cause I can SEE what's happening (at least usually). Darn double-Es deal with all the crap that can't be DIRECTLY observed (and I'm not talking about observing a meter, LOL).

As Bill pointed out, your "model" is a little hazy, but if it works for you I suppose that's fine. Ya got the gist of the overall picture as far as I can tell.

Bill,

You mentioned that you should be able to measure the effect on the waterblock. You already are. You're trying to detect it with a thermometer rather than relying on your pressure and flow readings. If you really wish to convince yourself and measure it thermally, you're going to need a lot more pressure. Get yourself a positive displacement pump with some balls, insulate the bejeezus out of a valve, then monitor temperature and pressure on each side of the valve (along with flow rate). Once you convince yourself of the reality, you can forget about trying to measure such minute temperature changes and rely on your pressure/flow readings to determine the energy.

[myv65]

Originally posted by Since87
Of course fluid flow has got to be one of the most difficult things to picture there is. So, i realize my crude picture is only going to get me so far.

It's all a matter of perspective, and I guarantee you that an ME would put something else in place of "fluid flow" in your sentence. I like to delude myself into believing that ALL physical systems have the same basic set of rules and that I have a pretty good handle on them. Yeah, I sound like a conceited fathead when I say that, but it's quite true to a degree. That degree being my ability or lack thereof to actually pull it off.

I've had the great fortune of working in an environment where I got to deal with hydraulics, pneumatics, rotating machinery, automated controls (PLCs, Servos, etc.), heat transfer equipment (electric, natural gas, fuel oil, etc), and even a little refrigeration on rare occasions. The company I used to work for even developed their own CAD system beginning in the late 70s and had ANSYS before I arrived (on old Tectronix green-screens with thermal paper print-outs, woohoo!). There is no substitute for a wide range of experiences.

Now when I look at pumps, fans, hydraulics, RLC circuits, beams, etc., etc., I *see* pretty much the same thing. I won't kid you and say that all of it makes sense all of the time, but it sure seems a whole lot more straightforward and common sense than it did in college.

[BillA]

no, no
I quite understand (now), and quite agree
I have a Blackmer vane pump with a 1hp motor - but need no proof
I unna stan, plez

I'm now trying to sort out the calcs to quantify this very small amount, that I cannot measure, to add to the correction factor for the heat die's efficiency determination
- but that 3psi @ 3gpm represents the worst/maximum case
methinks I'm picking the pepper out of the flysh*t

any simple way to translate the pressure drop into an energy term ?
(am I overlooking the obvious again ?)

Thanks, Bill

[myv65]

Hey Bill,

I thought from your prior post that you had already done that. Flow rate (L^3/T) multiplied by pressure drop (F/L^2) = units of F*L/T (power). It's as simple as that. Measure this stuff in metric and the unit conversions are a little easier, but it's not so bad in imperial units. I'm sure you're up to the task. (3 gpm at 3 psi = 3.91 watts, hint, hint)

It gets back to the "balance" equation I posted earlier. Your total energy change in the fluid is m_dot * c_sub_p * delta_T + flow * delta_P. You know the power you're putting in with the heater (within reason) and can accurately measure both flow rate and pressure drop. What's tough is trying to reconcile your temperature measurement accuracy against your ability to ensure that all the energy from the heater actually enters the block (and that no heat escapes from the block but by the fluid). Come on, it doesn't *sound* all that hard.

[BillA]

obviously there are several steps
- identifying all of the influencing factors
- quantifying those constituent 'bits'
-> then the kicker: trying to understand if the results 'make sense'

not so easy at all
Thanks for the input, ck your PM on a different topic - Bill