Pump rated wattage vs actual power draw vs pumping power
 

Okay, this has had me puzzled for a little while now.

With these Johnson pumps, they are rated at 2.3A @ 12V, or around 27W.

Plugging one in and powering it with my lab bench power supply I find that when attached to a radiator and a waterblock (i.e. some restriction) it's only drawing around 14W of power at 12V (~1.2A). In wide-open mode (no restriction) the pump was drawing out 19W (~1.6A). At turn on it did appear to spike to just over 2A.

Some of you may remember back to some experiments I ran at OCAU to attempt to crudely measure the amount of heat that my Eheim 1250 was dumping into the water, and it worked out to around 7-8W or so when in-line and pushing through a complete system.

Now an Eheim 1250 is doing a little over 2W of real work pushing the water about by looking at the PQ curve, or about 10W in total for heat + work. Assuming that some of the heat also escapes via the pump's casing, we're probably again looking at about half of the pump's rated wattage actually being drawn by the pump.

Now clearly this is neither a comprehensive or complete analysis of the issue, but I thought it enough to kick off a thread about it.

Is a pump's rated power actually just a peak in-rush power draw that occurs at power-on, and as one loads up the pump further beyond wide-open mode, the pump actually draws significantly less power, even going down to as low as 50% of its rated power draw?

Thoughts? Experiences? Knowledge?

I would assume the watt value is of the motor itself with nothing attached to it or water runningthrough it. Try running it dry for a few seconds and see how much juice it uses. Unless of course thats what you ment by wide open mode. If so then maybe the impellor and housing are causing enough resistance to lower the wattage. I would assume the specs are from a motor with nothing attached to it. :shrug:

By "wide-open" I meant pumping water through the pump with about 30cm of tubing on the inlet and outlet to a reservoir.

Yes, you're right. If the motor was spinning in free air with absolutely nothing attached, including the impeller or anything else, it probably would draw the full 27W.

Well, you've got a few things going on here.

First, the Johnson pump is a direct drive, versus the Eheim's magnetic drive, so the Johnson is going to be more efficient (but alas, not more reliable!)

From other users of "the Johnson", the power draw does vary with the load. I believe Since87 ran some tests that way, among others. Other reported results are similar, for other pumps.

So I'd have to go with: "It's a max rating".

Yes, but both the Eheim and the Johnson are electro-magnetic motors, no?

Hum... yes, but... you'll get more torque from the Johnson pump, because of the direct drive. The Eheim has quite a gap, between the impeller's magnetic shaft, and the motor coil.

All in all, I think you'd find great variations in energy efficiency, if you compared the popular water cooling puimps, but mag drives are notoriously inefficient: I believe that a rough averaged efficiency is around 30%, from others who have studied this (myv65?).

Some info here.

I don't think that a pump is likely to draw it's spec'd current when run dry. Electric motors draw more power as the load increases, and a dry pump will have a fairly light load (at least until the bearing surfaces wear out.)

I'll have a chance to test this though. Bruce at Cooltechnica is sending me a pump to test with a "modified sinewave" as input power. After that testing is done, I can do some dry running testing as well.

Sean
Thanks for the link .
Had lost that thread

indeed, an exemplary thread
so have we lost Dave too ?
hope not, he has sure helped me

Thats what I originally thought, but then after seeing Cathars results I thought I had it backwards (again). If I am reading it right the more Cathar loaded the pump the less power it used. :confused:

This was counterintuitive to me as well, and myv65 explains it in the thread I linked above.

A centrifugal pump is actually more loaded when running wide open than when restricted, when pumping water.

However, I expect the load on the pump to be lower when 'pumping air' than when pumping water, regardless of whether the water case was wide open or deadheaded. (Till the 'bearings' wear out anyway.)

So if I understand this correct the following would be somewhat accurate:

Your peddeling a bike and it takes much more power to get it to 10mph (you have to peddle harder), but once it is at 10mph it doesn't take as much power to keep it at 10mph? because momentum and kenetic energy are working?

So if a pump is runing dry then it is in the process of peddeling harder to find it's cruising speed and being there is little resistance it has to keep peddling harder untill the motor itself is the limit?

Same would go for a car. It takes a lot more gas to get it to 55, but once there it doesn't take as much gas to keep it there.

I think I am on track. :shrug:

My moderator's magic 8-ball says, "no".

Hopefully he will let us know if we get too far astray.

I'm not the person to answer this question. I haven't put enough thought into understanding it myself.

I don't think that analogy is too relevant because it is concerned with non-steady state situations. I can't make a better analogy off the top of my head though.

Maybe once I get into some pump testing, I'll put some more thought into it.

If anyone has suggestions for tests to conduct WRT pump power consumption, let me know.

All I can think of, is to send you the modified Johnson, (w/ the diode, 3 caps, and coil, once I get that last one).

Otherwise, I'd suggest trying to deadhead the pump, if you have the height...

And the difference between deadheading the pump with height, rather than a plug in the outlet is...???

lol
some things never change

What was interesting me about this, and one of the comments that Dave made, was this:

"Pretty much all of the energy that a pump draws will be converted to heat"

This is something that is of interest to me on the basis of allowing us to predict what is the largest sort of pump we want to use in any particular setup.

Bill's excellent work on quantifying radiator and waterblock performance gives us 80% of the puzzle, but it's the question of pump heat that is basically unanswered.

Assuming that Dave's statement is true, then many pumps are actually putting far less heat into the water than what they are rated at. It is this that is what puzzled me, but my eyes opened a bit when I looked at the Johnson pump's power draw, which got me to thinking that this formed the basis of what didn't make sense, that being that pumps will actually draw quite a large deal less than their peak rated power consumption when loaded, and as a result add quite a deal less heat into the water than what one can predict by merely looking at the pump's power rating.

One thing that I've been considering is that we use pumps in loops.

The kinetic energy of the system is not 'lost' as it would be if the pump were pumping from one body of water to another.

Would the power consumption be much higher if the pump were sitting in a bathtub with no tubing attached?

I think it is useful to consider the aspects of 'power' separately:
the power to run the pump motor, and
the work input to the fluid

an illustration is the recirculation pump in my chiller,
a Pabst motor with a 10" shaft to the pump head in the chilled bath
- obviously no motor heat is being input into the fluid

with submerged pumps this is quite different, and to a lesser degree also with close coupled pumps

some pump specs list the 2 power values separately; power consumed by the motor (a maximum value), and work done by the pump

max flow = max power consumption (by the motor of course)
note that in a closed loop the NPSH will be rather high

Would it be possible to put a pump into a well-insulated loop with a needle valve and pressure transmitters on intake and exhaust side of pump and then just backcalculate the W the pump is putting into the water as a function of system resistance?

With my testbed it seems that pumps dump MORE heat into the loop when I throttle them during testing; pretty sure Bill sees this as well (I recall something about having to lower the set point on his chiller at low flow rates). Doesn't that sort of contradict the idea that they are pulling less power from the wall when there is more restriction? Or is the total power less but much more of it is being lost to friction?

Hey Bill,

I'm still here, though not too often anymore. I skimmed this thread purely because JD linked it over at AMDMB. I thought I had beaten this subject to death a long time ago. Much of what I'll say next (like ALL of it) I've said before, so I'll try to abbreviate. Oh, and I gotta say that you guys are really off into the Rube Goldberg region with your block hold-down devices. It's about time for someone to draw a free-body diagram of the situation and figure out that it really isn't that tough a nut to crack. If time allows I'll get back with you on that topic, just don't hold your breath.

Power to the impeller always increases with increasing flowrate.

Head from the pump always decreases with increasing flowrate.

The impact of flow is greater on shaft power than the impact of head.

Cathar didn't see max power because he is not truly running wide open. Running wide open is a sort of utopian point that doesn't exist. As folks here know, the suction conditions can easily push you over into cavitation. Well, even when there is no cavitation there are still suction (and discharge) losses. An idealized suction and discharge would generally correspond to zero-TDH and peak shaft power.

You also need to have a basic understanding of electric motors. Motors are generally rated for a specific base speed and corresponding torque. At lower loads, the speed increases ever so slightly and torque drops. Vice versa for higher loads. Provided you don't stall the motor, that is.

With no water in the pump, the rpm probably went up a dozen or so and torque dropped dramatically. Neglecting inefficiency, power is nothing more than torque * rpm, so power drops off quickly when output torque is basically zero. ie, all energy put in is essentially lost as motor inefficiency.

There's an easy way to hear this with fans, which follow the same basic laws as pumps. LEGAL WARNING TO BONEHEADS: Do not stick your fingers in a fan doing this! Hold a 60 or 80mm fan in your hand and plug it into a molex. It'll have a distinct pitch. Cover the inlet with your other hand. The pitch will increase, yet no air gets pumped. The speed goes up because the shaft torque goes down. Delta_speed is a heckuva lot smaller than delta_torque, so net power goes down a lot. Cover the outlet and the result is the same. Wanna be really *cool*? Do this in your BIOS screen with a three plug fan or with some fan monitoring software running for a (somewhat inaccurate) measure of rpm.

Phaestus,

I haven't done what you suggest, but it's pretty routine to keep pumps running 24/7 in the systems we used in the paper industry. The pump energy is enough to keep the oil warm during machine outages. A fun one was our hot oil systems. At room temperature the oil is pretty thick and our flow control valves would be wide open yet system flow well below setpoint. Pump energy would get you up to ~140°F and almost bring the valve off its wide-open position. By the time the burners took it to 550°F and the oil's viscosity dropped below water the flow control valve would darn near close to keep from overshooting the desired loop flow.

Not how I interpret your results.
My interpretation is an increase in Heat imput as flow rate increases.
http://forums.procooling.com/vbb/sho...5&pagenumber=7


Billa's problem was discussed here:
http://forum.oc-forums.com/vb/showth...0&pagenumber=1
Bill may have a better recollection of the resolution

Oh, and I haven't followed Phaestus' data, but there is a distinction to be made. Heat per volume of water does indeed increase with decreasing flow. Total heat drops with decreasing flow, however. It's a minor point, but an easy source of confusion.

The easy example is to dead head a pump. The water in the pump casing will get hot in a hurry, but it really doesn't take much energy to heat that little volume. Let the baby run wide open with a reservoir and you'll put more energy in, but a lower amount per unit mass vs time.

Les:

Was referring to testing with straight water and a radiator and not chiller. Chiller data has more than enough other complications to make my head hurt. Still not even sure what I am going to say in that review :)

deleted...

Damn it takes me WAY to long to answer a post. Someone else almost always beats me to it.

And does a better job...

Back to just reading all of this.

Rated power is sort of "cover your ass number" as i understand it. Sort of like the 8 amp rating on 250w power supplies :)

myv65 has answered this realy thoroughly, but the I'll just add a little in case some people aren't buying it. The job of the pump is to add kinetic energy the water. This is represented by mass times velocity squared. The higher the velocity of the water, the harder the pump works.

The fan effect myv65 spoke of has been experienced by all of us when vacuum cleaning. When something gets clogged in the vacuum cleaner, the motor pitch increases. That is because the motor is no longer sucking a flow. While the vacuum is blocked the flow there is no flow so the motor doesn't realy do any work, and runs free up to it's max rpm.

Also with regards to water heating from pumps. All the energy imparted by the pump to the water gets converted to heat. How efficient a pump is, has got very little to do with it, it will simply spend less overall energy in giving the water it's energy.

Example; a system loop is sized so that it will have a flow of 3L/min. You have the choice of either using a low head/high flow pump rated at 80W, or a high head /low flow pump rated at 20W. You'd naturally pick the later, thinking that the first will impart much more heat to the water. However the first pump will be working so far back on it's pump curve that it will be drawing nowhere near 80W. It could well be drawing 20W, just like the second pump.
Since the flow of the water is the same for both pumps, it is safe to assume that frictional losses between the water and the piping will be exactly that same. Therefore the heating of the water should be exctly the same.
I.E. it's the flow thorugh the tubing that's causing the heat, and not the motor itself. A certain flow through a system will always result in much the same heating, regardless of what pump you use.
The pump will of course have some effect, but that effect will be limited to how much thrashing and friction there is within the pump. A pump which is far from it's optimum working range will impart more heat since it will be thrashing and casuing more friction with the water, than a pump that's designed to work at the flowrate in question.

Be sure to correct me if I'm wrong about this.

<Off topic>
Who's black liquor recovery boiler do you have??