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Well I see you boys (and girls) finally have sorted out the question on where all the enegy goes :D . But I'll post a link anyway:
http://forums.procooling.com/vbb/sho...=&threadid=533

cheers

Geesh! You could have saved us a lot of trouble, a long time ago!

Here's a quote of the relevant part:
Now let's run the numbers for a typical flow rate, say 100 gph, for the same Eheim 1250.

1250 curve here!

At 100 gph, the head is about 1.7 meters, or 2.42 psi.

E_tot=28W (total consumed power)
F=100 gph, or 0.10 L/sec
h=1.7m
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)

E_flow=F*rho*g*h=
=0.10(l/s)*1(kg/l)*9.81(N/kg)*1.7(m)=1.67(Nm/s)=
=1.67W

E_loss=E_tot-E_flow=26.33W


That doesn't feel right...

In a word, wrong. Yes, we have debated this to death, but that does not mean that everyone reached a consensus.

The referenced post makes at least two incorrect assumptions. One, that the maximum power consumed by the motor occurs at a mythical point of both maximum flow and maximum head. These two conditions are mutually exclusive of one another. The second is that the "flow energy" or whatever they heck they called it, is the only energy put into the fluid.

You've got motor inefficiency and pump inefficiency. For all practical purposes, all of the pump inefficiency goes into the water. Whether or not the motor inefficiency goes into the water depends on the particular situation. Submerged? No question it all goes into the water. Open-air? Most will convect off the motor casing into air with very little going into water.

A "typical" motor, however, will operate at >75% efficiency. Not so the typical pump. These tend to max out around 50% (among the pumps we use) or less. The "lost energy" is due to churning of the water that doesn't produce flow. It's the eddies and shearing around the impeller and between the impeller and volute.

If someone wanted to be really geeky about this, all you need to do is measure torque on the impeller shaft. Torque times rpm equals input power to the water. The fact that only a portion of this power is useful, namely flow rate times pressure rise, is irrelevant. The power still goes into the water.

I want to be geeky!!!

We're so near, closing the gap between the theoretical, and the practical...

The thing is, it would be just as easy to measure the torque of a pump impeller, as it would to measure a fan's... I have this mental picture of a belt driven pump: it would be easier to measure that, I think!

I also have another picture: a pump mod where the impeller is replaced with one that is a bit thicker, to minimize the loss of energy at that point.

myv65
Do you dissagree that it takes 6.5W of power to move 1200l/h to a height of 2m with a 100% effective pump?

What i find interesting is that my theoretical approach gives a power input of (6.5/28)=23% and skulemate got power input of 25% with his testing (see first page in this thread).

Skulemate
I'm glad to see that someone actually tried to measure the way you did. I presented an idea of measuring C/W a while ago that is not totally different than what you did.
(But it got shot down by BillA off course;) ).
Lookie here:http://forums.procooling.com/vbb/sho...W&pagenumber=4

cheers

Actually, I do disagree. Your statement would be true only if the pressure on the suction side of the pump was atmospheric. If you were running the pump, say, 2m above a reservoir, then the true pump lift would be 4m. If you had a reservoir 10m above the pump, the pump would effectively be "in the way" of natural, gravity-fed flow. Useful flow energy is flow rate multiplied by pressure rise. In a closed system, vertical lift really has no meaning and you need to look at total head loss through the system. Note that I've also left out flow loss to get it up those two meters. What if you're going through a 1/4" ID line? What if you've got elbows or other restrictions in the way? All you are considering is the gravity lift. All the pump knows is total head, a number that includes suction as well as discharge resistance. How those resistances are generated make no difference to the pump.

Even so, all one needs to do is read a pump curve (one that include pump efficiency). The folks that make most pumps have already done this work for us. The pumps we use at the flow situations we have usually run less than 50% efficient and often below 30%.

Thanks Dave.

To get back to my practical approach, could we take a minute here and run some calculations on that Eheim 1250, at 100 gph?

Like I wrote earlier, at 100 gph, the pressure drop will be 2.4 psi. The pump uses a 28 Watt source. Where do I go from here?


DixDogfight: you wrote: "Do you dissagree that it takes 6.5W of power to move 1200l/h to a height of 2m with a 100% effective pump?". I don't quite see the relation with the Eheim 1250, since it can't achieve that. It's either 1200L/h at 0 head, or 0 L/h at 2meters. What's up?


To recap what we have so far (correct me as necessary!):

-A pump uses a fix amount of energy, supplied in electrical form. (*** correction: a variable amount of energy***)

1-In a centrifugal design, less than 50% of that power is converted to create a flow (outlet to intake). This energy is converted to heat, from friction of water-on-water within the loop.

2-The remaining energy appears in the form of: churning, and energy loss. The churning again, induces heat from the water-on-water friction.

2b-The energy loss includes:
-Motor inneficiency (i.e. not all of the EM field generated by the coils moves the impeller)
-Heat from the coils (i.e. the coil wires, not being superconductors, dissipate some power)
-Friction/noise (as the impeller rattles around the housing)
-Noise (as the coils rattle, if applicable)

Ben,

The only part of yours I'd take exception to is the statement that "A pump uses a fix amount of energy, supplied in electrical form." Energy input to the motor will vary with flow and head.

Here's a "typical" graph of an industrial centrifugal pump. There's a lot more information here than you'll normally get from the Eheim's, Via's, Rio's, Little Giant's, etc. of the world.

Among the additional info is efficiency, required suction pressure, impeller diameter, motor power at the operting point, etc. Note that many industrial pumps allow you to specify the impeller diameter so that you can vary flow at a given head. No, you can't change the diameter dynamically, but you can specify it up front.

BB2k
I don't know what amount of flow a 1250 produces at a 1m lift. But using the maximum flow at the maximum lift results in the maximum kinetic energy added.
A sort of worst case scenario if you will.
It was only a theoretical exaple to figure out a rough estimate of the energy added to the water due to pumping.

I agree on the recap.

myv65
Ok so i didn't state the pressure on the suction side. I didn't state the gravity constant either. If you change the setup (like adding 2-10m of tubing below) then of course the figure/number (6,5W) is wrong.
I'm talking about figure A and you about B and C (or close to it anyway).

All-righty then. The only thing missing is the kinetic energy of getting that flow going. So you get a V^2 * rho / 2 factor that'll raise the input power slightly. I understand where you're trying to get, and practically speaking your diagram "A" matches your previous statement about a 100% efficient pump/motor.

I think more to the point is the actual motor power. Motor nameplates merely state the maximum power the motor will consume. This does not mean that they consume that amount of power irrespective of flow and/or head.

That's the problem with skulemate's data on page 1. He assumes that his motor is drawing 180 watts with ~46 watts showing up as thermal energy within the water. In the graph I posted, the operating point (as indicated by the red right-angle) calls for ~1.45 hp. At other flows/heads, it would list a different number. We use a 1.5 hp motor because it can handle all combinations of flow/head that we use. This does not mean that the motor draws 1.5 hp.

A: about 13.5 L/min, or 810 L/h.

Ok, so how do I run these numbers?

myv65... that's not a problem with the data, but rather how I used it ;)

O.K...., just jumping in here if that's ok.


Dix or myv: For any given situation a pump WILL impart a given amount of heat to the fluid just in moving it around. @ >100% efficiency. If I'm starting to get this...

Current draw can easily be measured, with the right meter.

So if I have an Eheim 1250, pushing 100 gph, which, according to the p/q curve results in a pressure drop of 1.7 meters (2.42 psi), and let's say that I (theoretically) measure the current draw at 0.20 amps, for an AC voltage of 117, which translates in a power consumption of 23.4 Watts.

Where can I go with these numbers?

Just jumping too!

The pump will induce heat regardless of its efficiency. The efficiency will however dictate how much of the total energy (supplied by the electric co) is going to actually move water. For a centrifugal pump, that should be around 50%, according to Dave (depending on the pressure drop, of course).

So, the REAL question is how to tell which pump to buy?!!!

What are the critical factors someone needs to consider when looking at manuf. specs. All that they have is head, flow, (max) input power?

If I specify that I will have 200g/h flow.

Then can I just look for head vs power @ 200g/h?

And find the highest head with lowest power?

It would help if I knew how much head I needed. I guess I'm going to have to break open the system and measure it.

I know that this is over-simplifying things a lot, but isn't that what others will really want to know. (other than those of us who enjoy the physics side of things)

Well actually calculate it.
I have the curve for my pump and I will just measure the flow and look at the chart.

Remember that there will also be a change in kinetic energy of the fluid if the suction and discharge lines are different sizes.

I assume you're quite familiar with the Bernoulli equation?

Where else can energy enter the system if there is no CPU heat load?

Alchemy

Work is the integral of a force applied over a certain distance. If there is no movement, there is no work. If I try to pick up a heavy rock but can't budge it, I'm not doing any work, no matter how much force I exert.

It is friction that *causes* heat and noise. They are not independent forms of energy loss.

Note the friction in water due to its own turbulence is basically nil unless you have supersonic flow.

Alchemy

Thanks Alchemy.

I realized that I was wrong (your first point) , where if there's no movement, then there's no work.

For your second point, Yes, ultimately it comes down to friction, except for the heat (#2), which I didn't state clearly, comes from the motor coils. The same is true for #3, but would also include "water noise".

So now you've got me wondering: if "the friction in water due to its own turbulence is basically nil unless you have supersonic flow", then where does the heat come from? Friction of water to inner tube/channel surface?

Many folks run closed systems with the same size tubing on both the suction and discharge. Depending on where you "cut the lines" to examine the system, the average fluid velocity may be the same. I'd agree that it is not uncommon to see port sizes vary with the discharge often a little smaller than the suction. And yeah, the Bernoulli equation is about as familiar as F=ma.

In the literal sense, anywhere. There's no such thing as a perfect energy barrier. In a more practical sense, the CPU and pump are responsible for virtually all energy that you have to get out of the fluid eventually. What's comical is seeing some folks use pumps that are rated as high or higher than the energy used by a CPU. At some point people need to realize that the closer you get a fluid to ambient, the more it takes to close the gap further still. It's also possible to go over the top where a pump adds so much energy that your final CPU temperatures begin to climb. Yeah, we're talking tenths of a degree here, but when the differences are that small, what is the point of using an ever larger pump?

I got a good chuckle over your "work" comments above. People outside engineering aren't held to the definitions that we construct. Work is, after all, simply another measure of energy and it is only because someone decided to call the integral of F*d "work" that we use that definition as engineers. If Ben was an engineer, I would have expected him to say "energy" rather than "work". Us engineering geeks are the only ones really concerned about the distinctions between potential energy, kinetic energy, work, etc., etc.

I am a little confused by your final comments about friction/turbulence in water. I guess I want to separate your comment from what happens within a pump. In a pump there is a lot of wasted energy because of secondary flows that don't contribute to system flow. The best of centrifugal pumps max out well less than 100% efficient with many "pond pumps" running under 50% efficient. The excess energy put into the impeller that doesn't generate flow generates heat. This heat isn't heat transfer but rather due to non-useful "churning of the water". So in this respect, it really is the internal friction of the water. It is not, however, quite the same as the turbulence in as established flow stream. I think this is what you were getting at in your post.

Well, if you draw a boundry condition around the entire piping system, the only sources of energy entering the system would be the pump and the chips being cooled. Every point in the system will be at a higher temperature than ambient, so there would be no heat transfer into the system at all.

Actually, now that I think about it, any parts inside the case might possibly be cooler than the air inside the case. So I guess you're right after all.

Regarding pump sizing, I've been trying to understand how the so-called "sweet spot" is possible for cooling systems. Convective heat transfer resistance *must* reach the minimum limit as the flow rate through the system is brought to infinity.

My best guess of why maximum heat tranfer does not coincide with minimum CPU temperatures is because at very high flow rates (and correspondingly high TDH) the pump duty becomes large enough to overtake the benefits of improved heat transfer.

If you pick the most important part of the system and consider the change in fluid temperature across the radiator as Q=UA(deltaT), UA is increasing very slightly as flow rate increases, but if you have to increase Q more than that to match it, you're going to end up with an even larger temperature gradient. Not good.

What do you think?

I'm still a bit wary of calling pressure change in the system "potential energy" because it really isn't, but I'm going with it here b/c I think it's too pedantic and too complicated to call it anything else, or explain why it can be expressed in terms of height.

I assume from your posts you're a mechanical engineer?

Well, perhaps I should have said "due *entirely* to its own turbulence," that is, not turbulence created in the pump.

If you think about it, when you dead-head a centrifugal pump so that the impeller does no shaft work, you're still doing work on the fluid. It's essentially the same setup Joule used to prove the first law of thermodynamics.

On that note:

Yep. Exactly. And since most plastic tubing is very smooth, the vast majority of the heat is being created at the fittings.

Alchemy

Interesting.

Since you brought up fittings, maybe you can answer something that's been bugging me for the longest time:

We all know that a 90 deg elbow is very restrictive. What we end up doing, as much as possible, is taking the 90 degree bend over a larger radius, by bending a tube.

My question: how much of a benefit is this, really?

I mean, the mass of water will still be re-directed 90 degrees, so the same amount of work is applied. I do understand that because of the sharp bend, the flow will tend to "keep going straight", which is restrictive, but isn't the same being done over a much larger radius?

Centripedal acceleration?

The "sweet spot" is, IMHO, largely imaginary unless one uses a ridiculously powerful pump. Within reason, more flow will yield temperatures that continue to approach ambient. As you surmise, once the power of the pump becomes a dominant source of the total energy, it'll cause the chip temperatures to begin climbing (unless you also have a ridiculously large radiator). With a very large pump, you may be better off throttling flow as pump power will drop with flow in spite of the higher TDH. I can see it now, someone is going to begin selling the "Spaceballs" system with ludicrous pump, radiator, and water block. LOL.

Yes. And unlike most folks that get the degree, I have chosen not to use it as a stepping stone to other areas. The common joke at most companies that I know (at least the ones that employ engineers) is that "you must check your calculator at the door" when you transfer to any other department.

This alone does not make me any brighter than any other engineer. In fact some would argue I'm thick in the head for staying on the engineering side indefinitely. It has allowed me to continue in my field, learning new things and keeping my hands directly involved with engineering issues on a daily basis. I now manage the mechanical engineering department for a small OEM company and have ~15 people reporting to me. Fortunately, they fly pretty well on their own and I get to spend a decent part of my time thinking about designing and technical issues and not too much on "managing".

I gather you've already got an engineering degree of some fashion or another, too. I recall you said something about a "licensing exam" when you first began posting here.

I can't find any numbers on that, despite checking three of my texts. Sorry. The best I can tell you is that Perry's Chemical Engineers Handbook says it is "significant."

I *think* the issue is that the fitting serves to convert some of the fluid's energy into creating a radial angular acceleration. The force required to change the fluid's direction would be equal to the velocity of the fluid squared over the radius of the bend. I *believe* that that would be integrated over the angle of 90 degrees, so the work would be less when the radius is larger.

Just a guess, though.

Alchemy

No, just a senior in college. The licensing exam was the "Fundamentals of Engineering," the first half of qualification as a registered professional engineer. The second part I can't do until I get out in the real world and gain this strange item called "experience."

I'm looking at grad schools right now. It would probably be easy for me to get a management position once I get out, but since I'm not terribly interested in that, I'd rather get a doctorate and do consulting work.

If I could make a living hanging around boards like these and doing serious engineering consulting, I would.

Alchemy

I hear ya. Consulting is nice work when you can find it. You can also work for a consulting engineering firm, but then the "middle men" take a good portion of what your work is worth. I've done some independent consulting based upon my prior career. The pay was phenomenal, but required travel. That was OK prior to kids, but not my bag any longer.

I forgot about Ben's elbow question above. The gist of it is a question of momentum change vs time. A shorter radius equals a quicker change. Rapidly changing momentum in a fluid tears up energy. Same thing you said above in a slightly modified wording.