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geoffman - You do not seem to be that confused.

Slightly over simplified explanation:

A) That referenced equation first takes the BHP (Total input power to the pump) and subtracts the WHP (The power used to lift the water to a certain head). The equation assumes that the lifted water then "falls off a cliff" and the WHP is lost and never recovered.

B) The result of this subtraction equals the *steady state* power dissipated due to all losses which include:
1) *"friction in the pump movement itself, from seals etc"*
2) Friction due to water flow restrictions/resistances
3) Losses due to pump electrical inefficiencies etc.

C) Last is the calculation of the temperature rise of the water due to this dissipated power.

To make a long story short - The water is heated by the total power into the pump minus the power used to lift the water. That calculation might be used for a large commercial non sealed system where the main purpose of the pump is to lift a lot of water to higher elevation.

But, here is the rub:
In the sealed closed loop PC cooling systems we are discussing, WHP IS ZERO
The power used to lift the water against gravity in the up path is fully recovered by gravity pulling the water down in the down path. (The "siphon effect")
ALL of the pump input power will heat the water!
In the case of an open reservoir PC system, the typical flow rates/head elevations will still only require a small amount of water "lifting power"

Summarized: Virtually all of the pump input power in PC cooling systems is required to overcome the friction of flow restrictions in the cooling system components and tubing and thus, heats the water.
And, this pump input power can easily measured (V X I)!

If you study the posts by the folks ThePimpulator recommended, you will not go wrong.
 
Thanks JPSJPS for the explanation and clarification, as you can guess I'm a bit of noob here, after I made my post and the second edit I went back and re-read the whole thread, and I think I may be starting to see some light at the end of the tunnel here, but I still have a fair bit of research to do.

I don't mind research, I think it's fun learning new stuff that I have very little knowledge or experience with!
 
Speaking like you’re scolding me isn’t going to cover up the fact that you are actually ignorant of this subject.

…takes the BHP (Total input power to the pump)
Brake Horse Power is not the total power to the pump. It is the power measured at the shaft. This measurement is so named because it was originally performed by applying a brake to a shaft and measuring the force applied by it. If you had spent one second researching the definition you would have known that.
this case is ONLY at 0 head and 0 Flow resistance.
The pump is NOT working at an "inefficient range"
The power to the pump can be EASILY calculated by measuring the DC input Volts X Amps to the pump.
I can’t imagine how you could be more wrong. These statements testify to the extent of your ignorance in this area.

First, you can’t operate at 0 head AND 0 flow. It’s either 0 head, max flow or 0 flow, max head.

Second, the efficiency of a pump is the ratio between WHP and BHP (WHP/BHP.) BEP (Best Efficiency Point) of a pump is a measure that accounts for several factors including power consumed, efficiency, even the radial pressure on the impeller is taken into consideration. Any point of operation that is not at the BEP is considered less efficient. The operating point I was using for the pump in my example was well away from its BEP; hence the pump was operating inefficiently. Once again, you’ve proven that you do no research yourself, and simply recycle what you’ve read here and spew it out without understanding.

Third, volts x amps can only tell the electrical power consumed by the pump. It has nothing to do with measuring pump efficiency.

In the sealed closed loop PC cooling systems we are discussing, WHP IS ZERO
This time you’ve really done it. You proven that you don’t even have an understanding of basic physics. Any mass in motion has an energy level associated with it. The only time WHP is zero is when the water isn’t flowing, and that occurs at max head. Also, If the power used to move the water was truly recovered, then our systems would pump as if they were operating at 0 head. Obviously, this is not the case. In truth, the pumps we use move only a fraction of the water they are capable of due to the resistance of the system. A 350 gph pump can be reduced to 100 gph or less. It does not matter whether the reduction in flow is caused by gravity or system backpressure; the result on the operation of the pump is the same.
For a submerged pump, "all the wattage of a pump goes into the water" because there is simply NO other place for this power to go. It can't get any simpler than that!
There has never been any question that all the heat generated by a submerged pump will end up in the water. That fact has been established and accepted. All discussions of pump-heat that I’ve seen in this forum have always related to inline pumps. So much so that we just assume that a pump-heat relates to inline pumps. Apparently, you are the only one that wasn’t aware of this.
If the pump is not submerged, the body of the pump will transfer a % of its internally dissipated (wasted) power into the air instead of into the water. But, because the surface area is small, pump efficiencies are fairly high, and there is no forced cooling airflow, this % power will be VERY small.
“Small surface area”? The vast majority of the surface of the pump is exposed to air! Where do you get “small surface area” from? Do you even know what a pump looks like? Also, pump efficiencies are not very high. It is an industry standard to use an efficiency level of 55% when a pump’s efficiency is not known.

I’m tired of your attempts to pass off your misguided and misinformed mental masturbations as knowledge. I’m also tired of your personal attacks. As you clearly have no real knowledge, and nothing of any value to contribute, I will just put you on ignore and be done with you.
 
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Graystar said:
Speaking like you’re scolding me isn’t going to cover up the fact that you are actually ignorant of this subject.
Graystar - You are the new Energizer Bunny!
The original one does not have a chance!

You need to learn to read! I had a reason to preface my explanation with the following statement:
"Slightly over simplified explanation:"
This was a KISS (Keep It Simple Stupid) explanation of that equation to demonstrate that the EQUATION DOES NOT APPLY to the closed loop type of cooling systems we are discussing.
CONGRATULATIONS! - You proved that I am too dumb to make an explanation simple enough for someone like you to understand.
That puts me with the rest of the crowd here who have an engineering education and experience in this field.

RE: Your Quotes:
1st Quote) You are correct. BHP *IS* total input power at the shaft but I approximated that as total input power ( KISS ) because it does *NOT* matter for this explanation or have any effect on the outcome.

2nd Quote) You missed the final word of the first statement: *resistance*. That changes everything.

3rd Quote) That paragraph was the ONLY meaningful part of my explanation of your Googled equation:
WHP = Water Horse Power = (GPM x 8.33 x SG x Head (in feet)) / 33,000
This equation describes the power required to lift a flow of liquid to a certain height! For water, SG=1
For some reason, you did a meaningless calculation for 5 GMP and 20 feet of head height. You pulled some assumptions out of the air for use as input parameters for other calculations.
The bottom line is that you performed a calculation you Googled and did not understand. What you got was a completely meaningless answer!
Fact: In the sealed closed loop PC cooling systems we are discussing, WHP IS ZERO
THE REASON: I will carbon copy my original post:
"The power used to lift the water against gravity in the up path is fully recovered by gravity pulling the water down in the down path. (The "siphon effect")
ALL of the pump input power will heat the water!
In the case of an open reservoir PC system, the typical flow rates/head elevations will still only require a small amount of water "lifting power"

Summarized: Virtually all of the pump input power in PC cooling systems is required to overcome the friction of flow restrictions in the cooling system components and tubing and thus, heats the water.
If we know the flow resistance of all the components in the system, we can calculate an equivalent "head" of course. This is best used to estimate flow rate from the pump charts.

And, if we know the flow rate or desired flow rate and equivalent head, we can calculate an equivalent WHP from that. But, if we have those two numbers, we can select a pump from the pump flow charts vs head and be finished.
EDIT: However, this equivalent WHP *will* heat the water so (since it is due to friction)and it would not be subtracted from BHP in the referenced equation.
In fact, that equation ignored this loss completely. It only included the pump power used to pump lots of water to a high elevation and friction loss was considered insignificant.(and not included)



And of course the pump input power can easily measured (V X I)!"

4th Quote) Submerged - OK

5th Quote) Pump not submerged:
The non air cooled pump surface area cooling is small compared to the air cooled surface area area cooling of a typical radiator.

But more important, The electrical to mechanical conversion efficiency of an electric motor is much higher than the 55% number you Googled. That electrical to mechanical conversion loss is the power lost in the motor/bearings/seals and not immediately transferred to the water.
But, some of this power will even be transfered from the motor/frame through the impeller housing to the water.
Most of the inefficiency is in the mechanical rotation to water flow conversion and this power IS transferred to the water.

You did not do the two experiments I earlier proposed did you?
Or, did you just not report back?

Following is another VERY GOOD experiment to perform:

Step 1) Run a test with everything off except the pump and radiator fan.
After the water temperature stabilizes, measure that water temperature. Then, measure the temperature of the incoming air to the radiator/fan. Record the difference which is the temperature rise of the water above ambient temperature due to the pump power only.
(Move the same temp sensor from place to place to eliminate calibration errors that you could get with two sensors.)

Step 2) Turn on the computer.
After the water temperature stabilizes, Take the same two measurements as Step 1 and record the difference. This is the temperature rise of the water above ambient temperature due to the combined CPU power and pump power.

Compare the results of Steps 1 & 2.
What does this tell you?

Instead of using Googled calculations that you do not understand, you can measure pump power very easily. Buy a cheap digital voltmeter with a manual and you are all set up.

Here is a good $90 pump to try if you need more water flow ;-)
http://www.harborfreight.com/cpi/ctaf/displayitem.taf?Itemnumber=02725

Darn, I forgot - You have me on ignore.
Ya Right!!
 
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There is an elementary equation from basic thermodynamics that states that the rate of heat transfer (Q) equals the mass flow rate (M) times a constant (the specific heat of water) times the delta T (fluid temp out minus fluid temp in).

Q=M x c x Delta T

In other words, the rate of heat transfer is directly proportional to mass flow rate. You increase the flow rate, you will then increase the rate of heat transfer. Since you cannot mess with mother nature it is very naive to think it works any other way.

Assume the CPU inserts a constant rate of energy (Q) into the cooling system. Then, from the relationship above, increasing the mass flow rate must result in a smaller delta T because Q remains constant. This smaller Delta T (fluid out - fluid in) also means that the average fluid temperature in the water block is somewhat lower even though the rate of heat transfer has not changed.

Now lets look at the heat transfer from the CPU to the water. The rate of heat transfer between two points is proportional to the temperature difference between those points. In our case this Delta T (not to be confused with the one above) is the temperature of the CPU minus the average water temperature in the water block. Lowering the average water temperature, as we did above by increasing the flow rate, means we have a little better heat transfer from the CPU to the now somewhat cooler water. The result is that the CPU runs a little cooler.

This all says that if you increase the flow rate, and everything else remains constant, you will decrease the CPU temperature. However, everything else will not remain constant if you increase the flow rate by using a larger pump. The pump uses some amount of electrical energy. This energy must end up somewhere. A relatively small amount of it is dissapated as heat from the motor. The overwhelming majority of it is converted from electrical energy to mechanical energy in the form of a rotating shaft that does real work on the water. This energy ends up in the water by increasing its temperature. It is called "pump heat" and can be very significant. An Eheim 1048 is rated at 10 watts, almost all of which ends up in the water. I understand a very overclocked CPU is good for upwards of 75 watts. As you can see a smaller pump like the 1048 contributes about 13% to the total heat load on a system with an energy hungry CPU. With other more common CPUs running at 25 to 50 watts, this percentage is much higher and is therfore much more significant.

As an interesting aside for those non-believers, this is also why excessive use of a blender to mix up frozen orange juice results in the juice not being as cold as expected. Also, nuclear power plants use primarily pump heat (from three or four 6,000 HP pumps) to heat up almost 75,000 gallons of water from 200 degrees F to about 550 degrees in about six hours or less.

The point here is that there is a trade off in how big a pump to use to increase the flow rate. More flow is beneficial. It is best to achieve the desired flow with a small a pump as possible and flow paths with minimum flow resistance. The bigger the pump, the more heat is added to the system. Eheim makes a 50 watts unit that I see talked about every now and then. This guy is probably a bigger heat load on the cooling system than the CPU itself.

Bottom Line: If you increase flow rate with the same pump your temperatures will trend in the direction of goodness. If you increase flow rate by going to a bigger pump you will reach a trade off somewhwere where the pump starts putting too much energy into the system and temperatures will start increasing.

I did not intend this to be so long but I do hope this helps remove some of the confusion from this issue.

Exactly
It would need more heat discipation to run at a higher flow rate,which would probably drop the temperatures.
 
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