Centifugal Pump Flowrate
Centifugal Pump Flowrate
(OP)
Dear all,
We measured the power take up of an old cenrifugal pump.
It has 50 kW motor which is taking 35 kW electrical power.
Is there any generic method I could apply to get a very rough idea (+/- 30%) what flowrate this pump is producing?
The feed piping is 400 mm, 2 meters long , 1 meter elevation.
The upstream piping is 300 mm, 84 meters long, 6 meter elevation.
Many thanks for helping,
CARF
We measured the power take up of an old cenrifugal pump.
It has 50 kW motor which is taking 35 kW electrical power.
Is there any generic method I could apply to get a very rough idea (+/- 30%) what flowrate this pump is producing?
The feed piping is 400 mm, 2 meters long , 1 meter elevation.
The upstream piping is 300 mm, 84 meters long, 6 meter elevation.
Many thanks for helping,
CARF





RE: Centifugal Pump Flowrate
CARF
RE: Centifugal Pump Flowrate
http://www
RE: Centifugal Pump Flowrate
Brake horsepower = (flow in gpm * head in feet * specific gravity)/(3960 * efficiency)
You would have to take your electrical power and change that to brake horsepower based on the motor efficiency. Or you might have motor curves that show shaft power versus amp draw.
There should be a comparable formula out there for your metric units so you don't have to do all the unit conversions. But that is the basic methodology I would suggest.
RE: Centifugal Pump Flowrate
Power (kW) = ( flow (m3/s) x Pressure (kPa)) / Efficiency
Note that the efficiency must be expressed as a fraction less than 1.00 e.g. an efficiency of 60% is used as 0.60 in this formula.
The friction in your suction piping will be negligible. A flow of 0.26 m3/s (= 940 m3/h) will give a pressure drop of 81 kPa (made up of 49 kPa of static head and 32 kPa of friction head). If the efficiency is 60% the power is 35.1 kW and the liquid velocity in the pipe is 3.6 m/s.
All a bit rough, but you get the idea.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Centifugal Pump Flowrate
You have indicated you only want +/- 30% and even this is wishful-thinking.
you should also measure the pressure at the pump discharge - then using using:
H(Metres)x flow(litres/sec) / 102 (constant)/ efficiency (motor and pump as a decimal) = power in Kw.
Of course what you don't know the pump / motor efficiency and this makes an enormous difference to the out-come.
There are otherways of estimating flow but you will need to give a description of the installation
Naresuan University
Phitsanulok
Thailand
RE: Centifugal Pump Flowrate
RE: Centifugal Pump Flowrate
Another point to consider: if you have a three phase electric motor, find power using:
P = SQRT(3)*V*I*cos(phi), where cos(phi) is the power factor. The power factor term must be obtained from the motor faceplate. With a single phase motor, the SQRT(3) term is deleted.
Also, don't forget to apply the motor efficiency to P to get the actual power delivered to the pump.
RE: Centifugal Pump Flowrate
Thank you very much for thinking with me. Yes the power we did through the above equation. What we will do is simple but time consuming, we will disconnect the pipe at the very end and collect the water in a buffertank. Volume increase / time will hopefully give us a good value of the pumps flowrate.
Many thanks, any comments on this method?
CARF
RE: Centifugal Pump Flowrate
RE: Centifugal Pump Flowrate
Once you have measured this flowrate please come back and let us know what it is. I would be interested to know how far out my 940 m3/h "calculation" was.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Centifugal Pump Flowrate
Procedure
The simplified power law of the water pump can be,
P (kW) = Q (cu.mtr/s) x H (m) x 9.81/eff
(actually, Q should be in kgs but as specific weight of water is 1000 kg/cu.mtr, and the right side of the equation yields the units into watts, I straight away considered cu.mtr/hr)
We, now, can get one equation in terms of Q and H.
My next step was rather crude as I initially considered the head as frictional only. By Darcy-Weisbach equation, we can get another equation in terms of Q and H. Once I knew the flowrate, I fiddled with a pipeloss software to match with the total head and power drawn by the pump.
PS: I just hope to know how katmar arrived at his figure.
The equation I alluded to in my earlier post is called as Brooke's equation, given as,
Q = 1.04 x a x l
Where Q is flowrate in gpm,
a is cross sectional area of pipe in sq.in
l is length in inches from the pipe end, where the water falls down by one foot.
RE: Centifugal Pump Flowrate
The proper equation for power for liquid pumping (or gas compression) is:
P[kW] = W[kg/sec] * Hp[m.kgf/kg] * 9.81E-3[kW/(m.kgf/sec)] / effp
where W is the mass flow rate, Hp is the polytropic head, and effp is the polytropic efficiency.
I believe the units in quark's equation as posted are incorrect in that he uses:
(a) [m3/hour] for flow instead of [kg/sec],
(b) [m] for head, instead of [m.kgf/kg], and
(c) 9.81 instead of 9.81E-3 for the conversion factor from [m.kgf/sec] to [kW].
RE: Centifugal Pump Flowrate
Power = volumetric flow x pressure differential / Eff
What could be simpler? KISS!
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Centifugal Pump Flowrate
I'm with you - KISS
plus the poster was asking for +/- 30%
Naresuan University
Phitsanulok
Thailand
RE: Centifugal Pump Flowrate
Thanks for expanding the equation. If you see my post below the equation, I tried to give some justification about the simplication. Yet, I made two mistakes in it. I wrote 'Q should be in kgs', but it is a typo and should be read as kg/s. Secondly, I wrote 'specific weight of water is 1000kg/cu.mtr', but in this context I should have written as density of water is 1000kg/cu.mtr.
So, power (kW or kJ/sec) = Q(kg/s) x H (m) x 9.81 m/s2/eff
Rearranging the terms on RHS,
9.81 x Q xH(kg/s)(m/s2)(m) = 9.81 x Q xH (kg.m/s2)(m/s) = 9.81 xQ x H N(m/s) = 9.81QH (N.m/s) = 9.81QH W. = 9.81QH x 10-3kW
If I consider flowrate in volumetric terms (m3/s) then Q(kg/s) = Q (m3/s) x 1000 kg/m3)
So, 1000 in numerator (density) and 1000 in denominator (W to kW) cancel out each other and you have my equation. This is what Artisi and I considered. (Artisi's eq. 1 liter - 1x10-3 m 3, so 1000 goes to denominator and 9.81 is in the numerator. The final constant becomes 1000/9.81 (in the denominator) = 101.9367
But, I feel that this kind of step by step solution is redundant in these forums. Further, when dealing with so many unknowns, my laziness gets multipled. However, I think I should take care not to create some confusion.
RE: Centifugal Pump Flowrate
(a) There is no such thing as a "polytropic head meter". You simply measure the pressure differential across the pump and convert that into the head. This requires knowing the fluid density.
(b) Head is DEFINED simply the work done[m.kgf] per unit mass[kg] of fluid pumped/compressed, in appropriate units. The fact that most people refer to head in units of length is simply a result of the fact that the earth exerts a force of 1[kgf] on a mass of 1[kg].
In English units, however:
Power[kW] = Flow[lb/s]*Head[ft.lbf/lb]/eff/737.56
In my opinion, we would eliminate units confusion forever if we simply remember the proper definition of head. I say this only because I have seen instances of units errors causing much confusion between clients and vendors, and I'm sure I'm not alone in this. For example, tests with water were applied incorrectly to light hydrocarbons without proper accounting for density.
(c) I use the term polytropic head merely to reinforce the idea that the associated efficiency must also be polytropic. This issue is far more relevant for compressors than pumps, however.
RE: Centifugal Pump Flowrate
I feel my thread is being hijacked ; ). To give you all some more detail: Yes, we tried a Controlotron Ultra Sonic flowmeter. But I guess the pipe was rusted on the inside so I did not get any signal. With new pipes and well develloped flows it works well.
To be more precisely:
The feed piping is 400 mm, 2 meters long , 1 meter positive elevation, ABOVE OF THIS A DEEP VACUUM 0.2 BARA (It is a barometric condenser). (BARA = BAR Absolute pressure)
The upstream piping is 300 mm, 84 meters long, 4.5 meters positive elevation (so not 6!). The pipe contains 6 x 90deg bends and 2 valves and is rusty on the inside.
Normally the pump is rated for 500 m3/hr. But since it is old and has been cavitating alot (deep vacuum!) I say the pump cannot do more than 330 m3/hr.
Next week I will give you the answer, after the open tank trials. So if you want to give a calculated guess, guess lower then 500 m3/hr (940 m3/hr is definitively too much).
Many thanks for helping, I enjoy the discussion and sharing of knowledge,
CARF
RE: Centifugal Pump Flowrate
I also faced similar problems related to guessing pump flow. The pump curves are useful for giving you within +/-30% if pumps are new. For that, you have to measure differential head across the pump under operation and find out the corresponding flow. If pump is old, I will also measure the shut off power and shut off pressure by closing the discharge valve fully with the pump in operation.
Looking at the pump curve, see whether what you have measured(shut off pressure and shut off power) coincides with what is given in the curve. If it is more or less OK, I will proceed with checking the flow from the pump curves corresponding to the measured head as well as measured power, independantly. I will then compute the average of the flows.
I have "measured" flow like this in 100s of pumping systems as part of energy conservation studies. I don't know how accurate is this very crude method. This was the only way I could think of. Yeah,... I know, I may be absolutely wrong. But clients who implemented energy saving measures based on this guessed flow method never complained, as they got power saving invariably!
In Umeshmathur's post dated 13th April, the following sentence is found.
" P = SQRT(3)*V*I*cos(phi), where cos(phi) is the power factor. The power factor term must be obtained from the motor faceplate."
I do not think that guessing electrical power input from measured values of current, voltage and rated p.f on the motor name plate is correct. If a 50 kW motor is operatiung at 35 kW input power, the shaft loading may be 70 to 75%. P.f will be less than rated p.f.
So, measure the power input directly than using equations.
RE: Centifugal Pump Flowrate
RE: Centifugal Pump Flowrate
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Centifugal Pump Flowrate
I worked with a very knowledgeable enginner who would use the "Shut-in" method and measure the watts at no flow. He would measure the watts at full flow and take the difference as an indication of work done.
You need a wattmeter.
Re; Power factor. the power factor on the nameplate of a motor is the power factor at 100% load and rated voltage only. At any other loading or applied voltage the power factor will be different.
respectfully
RE: Centifugal Pump Flowrate
We MEASURED current at full load (water pumping):
Current was 62 Ampere.
So the power take up was SRT(3) * 380 * 62 * 0.85 = 34.7 kW
So say 35 kW.
Indeed we could do the same at zero load (no water, pump and motor running)only meachanical load. Will the pump survive a small run without water? Mechanical seals.
Thanks all, I will let you know the results of the tank trial real soon.
CARF
RE: Centifugal Pump Flowrate
With the pump pumping and the meter in place, close the discharge valve for just long enough to read the meter and then open the valve again.
You should try to aquire a wattmeter for the test. The power factor will be enough different at part load to seriously affect your results.
If you have access to power factor capacitors, try connecting different values until you get minimum current. At that point the power factor wil;l be close to 100% and When the power factor is corrected to 100% it will stay at 100% through out the load range from light loads to heavy loads. The ammeter will then give quite an accurate reflection of the watts. (After you have added capacitors for minimum current use 100% for the power factor in your equation.)
respectfully
RE: Centifugal Pump Flowrate
We will do a Waros said and let you know the results.
We will install a pressure gauge at the discharge of the pump. So we will measure pump power take up and pressure at several valve settings.
Best regards and thanks for helping,
CARF