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General Liquid/Water Cooling Discussion For discussion about Full Cooling System kits, or general cooling topics. Keep specific cooling items like pumps, radiators, etc... in their specific forums.

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Unread 02-07-2004, 12:59 PM   #1
Reaper_Unreal
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Default More about pumps...

Well, on another unnamed web site, a large discussion has erupted about pumps. I'm probably the cause of this. My "calculations" have caused some disagreement between certain people. Now, while I realize that they're not entirely correct, I have explained my reasoning several times. So now I seek for pros to give their opinion.

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Unread 02-07-2004, 01:56 PM   #2
UberBlue
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Your calculations are out to lunch. Max flow and max head will never occur at the same time.

The formula you want to use is Watts = m(H2O)x lpm/6

Look at a pumps P/Q graph and find flow and head from there, then plug them into the formula.

As you do that with multiple flow points, you'll find the plotted points look like an upsidedown "U" with the pump being most efficient at one point. Usually that point is about halfway down the P/Q graph.

Even at the pumps point of best efficiency, centrifugal pumps tend to be horribly inefficient.

Look here for examples.
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Unread 02-07-2004, 02:35 PM   #3
lolito_fr
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"Look at a pumps P/Q graph and find flow and head from there"

Yes, just need to specify that the flow and head are found at the intersection between the pump P-Q and the system P-Q curves. The other points of the pump P-Q are pretty much irrelevent here.
Hence you need to know the system P-Q, if you want to select the best pump.

Using max flow and max head corresponds to the pumps' useful output at full flow and (ironically) zero head. (It equates to the kinetic energy of the water) So it does actually have some mathematical basis, to give credit where credit is due
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Unread 02-07-2004, 03:51 PM   #4
Les
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Quote:
Originally Posted by UberBlue
The formula you want to use is Watts = m(H2O)x lpm/6
................
Look here for examples.
It may be a thought to correct your graphs to show the values given by "Watts = m(H2O)x lpm/6".
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Unread 02-07-2004, 04:57 PM   #5
UberBlue
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Quote:
Originally Posted by Les
It may be a thought to correct your graphs to show the values given by "Watts = m(H2O)x lpm/6".
Maybee. I presented that formula as it's the popular one around here.

I'm not convinced either way as to which formula is correct. Untill I know for certain, I'm just gonna let them sit for the time being.

Last edited by UberBlue; 02-07-2004 at 05:49 PM.
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Unread 02-07-2004, 07:19 PM   #6
BillA
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popularity is not (ever) a measure of technical correctness
what is the derivation of your 'formula' ?
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Unread 02-07-2004, 08:19 PM   #7
UberBlue
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Quote:
Originally Posted by unregistered
popularity is not (ever) a measure of technical correctness
what is the derivation of your 'formula' ?
Pumps.org's (aka Hydraulic Institute) definition and formula 1.2.6.3

That formula yields results as water horsepower. From there I convert to watts. 1 water horsepower = 746.043 watts.

(metric) P = Q*H*s/366

where:
Q = flow M^3/hr
H = Head in meters
s = specific gravity

The two formulas are very similar, but something has been lost in translation for one of them.

No offence Les, but I tend to trust a formula from a respected trade group more than "beer mats and slide rules". I have no doubt you can get medieval on a math problem, but I need to know why the formula is wrong before I go fixin stuff.

I would like to know for certain which one is correct before I do anything.
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Last edited by UberBlue; 02-07-2004 at 08:36 PM.
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Unread 02-08-2004, 04:03 AM   #8
Les
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Quote:
Originally Posted by UberBlue
Pumps.org's (aka Hydraulic Institute) definition and formula 1.2.6.3

That formula yields results as water horsepower. From there I convert to watts. 1 water horsepower = 746.043 watts.

(metric) P = Q*H*s/366

where:
Q = flow M^3/hr
H = Head in meters
s = specific gravity
......
The formula
(metric) P = Q*H*s/366

is the formula for "Pump output power (P w )" in Watts
where:
Q = flow lph
H = Head in meters
s = specific gravity.

The source makes the, perhaps, misleading statement that "Pump output power (P w )" is "also called water horsepower." They also use "Q = rate of flow, m 3 /h" for another formula on the same page.
These two facts could lead to your interpretation of the equation.
However the correct interpretation is as shown.

Here lolito_fr gives a derivation of (Pw) in Watts:
"Derived formula: P=0.16Q.H
P in Watts
Q in lpm
H in m"
These are (allowing for approximations) the same formula.

My usual statement of the formula is "Watts ~ m(H2O)*LPM/6". The " ~" used to imply an approximation.
My non-rigorous derivation for "Pump output power (P w )" gives:
Power(Horsepower) = 0.00365 x Head(mH2O) x Flow(m^3/h)
I stand by this as being numerically correct.

Last edited by Les; 02-08-2004 at 04:15 AM.
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Unread 02-08-2004, 04:57 AM   #9
lolito_fr
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[pumps.org] Formulas without units, for Gods sake . Give me a slide rule anyday.

edit(2): Deleted some nonsense. Les is correct, Q in lph and P in Watts.

Last edited by lolito_fr; 02-08-2004 at 06:57 AM.
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Unread 02-08-2004, 10:21 AM   #10
BillA
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thanks for the lucid explanation Les
I hit "water horsepower" and slid off the road
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Unread 02-08-2004, 10:25 AM   #11
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I'm pretty sure that using the standard mgh/t formula works fine, if I change it around a bit, I can get fgh/3600, where f is flow rate in L/h, g is 9.81m/s^2 and h is height in m. The units should cancel out properly if we take into account that 1L = 1kg for water, and that 1h = 3600s. I remember doing something similar to find the maximum height of a water jet in 1st year physics, I'm pretty sure that's right. I don't really know much about specific gravity, but I'm assuming that's also right.
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Unread 02-08-2004, 01:52 PM   #12
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Alright, well I've re-done the calculations as suggested, and I've even written a program to find all of the stats for me (I'll post it later if needed). This time, I took loads of points off of the P-Q graphs of each pump, and found the power at each point. I took the average of the powers I had found (not including 0 head and 0 flow). I then took the average of all of the heads, and the average of all of the flows, then found the power that way. I then took the average (not geometric mean) of the two power values I had found to give me my value for power. That's about as accurate as it'll get. I've got here the results for the 5 pumps I did. The program I wrote will be posted later if requested when I can find somewhere to host it.

Code:
Pump Name:      L20
Max Flow Rate:  685.16L/h
Max Head:       1.31m
Avg Flow Rate:  395.86L/h
Avg Head:       0.69m
Pump Power:     0.70W
Power Used:     11.00W
Efficiency:     6.33%

Pump Name:      L30
Max Flow Rate:  1271.90L/h
Max Head:       2.01m
Avg Flow Rate:  747.24L/h
Avg Head:       1.00m
Pump Power:     1.90W
Power Used:     23.00W
Efficiency:     8.24%

Pump Name:      Eheim 1250
Max Flow Rate:  1211.33L/h
Max Head:       2.00m
Avg Flow Rate:  638.32L/h
Avg Head:       1.18m
Pump Power:     1.93W
Power Used:     28.00W
Efficiency:     6.90%

Pump Name:      Eheim 1048
Max Flow Rate:  598.10L/h
Max Head:       1.50m
Avg Flow Rate:  299.43L/h
Avg Head:       0.93m
Pump Power:     0.76W
Power Used:     10.00W
Efficiency:     7.64%

Pump Name:      MCP600
Max Flow Rate:  651.09L/h
Max Head:       3.18m
Avg Flow Rate:  404.50L/h
Avg Head:       1.54m
Pump Power:     1.58W
Power Used:     9.00W
Efficiency:     17.51%
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Unread 02-08-2004, 05:05 PM   #13
UberBlue
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Quote:
Originally Posted by Les
The formula
(metric) P = Q*H*s/366

is the formula for "Pump output power (P w )" in Watts
where:
Q = flow lph
H = Head in meters
s = specific gravity.

The source makes the, perhaps, misleading statement that "Pump output power (P w )" is "also called water horsepower." They also use "Q = rate of flow, m 3 /h" for another formula on the same page.
These two facts could lead to your interpretation of the equation.
However the correct interpretation is as shown.

Here lolito_fr gives a derivation of (Pw) in Watts:
"Derived formula: P=0.16Q.H
P in Watts
Q in lpm
H in m"
These are (allowing for approximations) the same formula.

My usual statement of the formula is "Watts ~ m(H2O)*LPM/6". The " ~" used to imply an approximation.
My non-rigorous derivation for "Pump output power (P w )" gives:
Power(Horsepower) = 0.00365 x Head(mH2O) x Flow(m^3/h)
I stand by this as being numerically correct.
Thank you. Now I know for certain.

I'll fix my graphs accordingly in short time.


Quote:
Originally Posted by Reaper_Unreal
Alright, well I've re-done the calculations as suggested, and I've even written a program to find all of the stats for me (I'll post it later if needed). This time, I took loads of points off of the P-Q graphs of each pump, and found the power at each point. I took the average of the powers I had found (not including 0 head and 0 flow). I then took the average of all of the heads, and the average of all of the flows, then found the power that way. I then took the average (not geometric mean) of the two power values I had found to give me my value for power. That's about as accurate as it'll get. I've got here the results for the 5 pumps I did. The program I wrote will be posted later if requested when I can find somewhere to host it.

I really don't want to begin to touch this with a ten foot pole.

IMO a single point value assinged to a pump is, for all intensive purposes, useless. The properties of a pump cannot be boiled down to such simple terms.

If you insist on pursuing a single point "efficiency", at least base it of of pump input power/pump output power and not electrical input power/pump output power. Pump input power is the power imparted to the pump shaft by the pump motor. leave the motor inefficiencies that never make it to the water out of it.

Last edited by UberBlue; 02-08-2004 at 11:53 PM.
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Unread 02-08-2004, 05:09 PM   #14
Myth
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Following this discussion, I have a few questions.

The idea was to find the most efficent pump to use for the cooling setup, and i can see the the heat created by the pump will make a small difference, but what i fail to see is the connection between these calculations and a real world setup. I run a simple setup using a maze4, 12mm tubing the radiatior from a car and have tried 2 different pumps i the setup since my Hydor L30 died i switched to an Eheim 1048 and there hasn't been any noticable changes to my cpu temp..

What it all comes down to would be to get the lowest system temp, and i besides the little difference the pump created heat would make how are the above mentioned linked to the performance of the system? wouldn't messureing flow using a 10 liter bucket be the only reliable way to surely know the flow in a setup? and wouldn't a test using different pumps i the same system be the only way really to see what differences there are in performance?
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Unread 02-08-2004, 06:14 PM   #15
Cathar
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Quote:
Originally Posted by Reaper_Unreal
Alright, well I've re-done the calculations as suggested, and I've even written a program to find all of the stats for me (I'll post it later if needed). This time, I took loads of points off of the P-Q graphs of each pump, and found the power at each point. I took the average of the powers I had found (not including 0 head and 0 flow). I then took the average of all of the heads, and the average of all of the flows, then found the power that way. I then took the average (not geometric mean) of the two power values I had found to give me my value for power. That's about as accurate as it'll get. I've got here the results for the 5 pumps I did. The program I wrote will be posted later if requested when I can find somewhere to host it.
Nice work - but - pump efficiency is always expressed as a value relative to a point on the PQ curve. It is never an "average" of the points of efficiency. What you've done is expressed an overall summary of various pump's efficiency, but it says nothing at all for how efficient a particular pump is at some point of the PQ curve.
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