Centifugal Pump Flowrate 2

[CARF]

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

 

[CARF]

Pumping cooling water @ 14 degC,

CARF

 

[bulkhandling]

There was a discussion lately. See link:

http://www.eng-tips.com/viewthread.cfm?qid=148464&page=5

 

[JJPellin]

I am sorry, but I don't have anything in metric units. But, we regulary estimate brake horsepower if we know the flow, head and pump efficiency. If you know the brake horsepower and head and can estimate the pump efficiency, you can calculate the flow. The formula that I use is

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.

 

[katmar]

The method proposed by JJPellin is the only way you will get an answer. In SI units the formula is

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

 

[Artisi]

If you cannot measure the flow by a direct method - ie, flow meter - pitot tube - orifice plate etc you are only guessing without having an actual performance curve of the pump and a lot of other acurate information.
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

 

[quark]

If you are free discharging the water horizontally, then you can approximate calculate the flow by checking the drop in elevation of the water jet. Otherwise, by above methods, if you know head and power.

 

[UmeshMathur]

I agree with Artisi, but I would state that he likely is recommending use of the DIFFERENTIAL head across the pump (based on the differential pressure).

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.

 

[CARF]

Dear all,

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

 

[Zapster]

Have you looked into a portable strap on flow meter? Perhaps you could rent one if you have no further need for one.

 

[katmar]

CARF,

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

 

[quark]

I agree with Katmar on the calculated value if 1 meter in the suction piping complements the head of the pump. If it is a negative suction, then the elevation can be 7 meters or 68 kPa.

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.

 

[UmeshMathur]

At the risk of sounding pedantic, I would like to address the issue of units and dimensions for the power calculations implicit in quark's post of 15 Apr 06 1:46.

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].

 

[katmar]

Umesh, your equation may be correct, but it is not very practical and it is more complicated than necessary. None of my pumps have polytropic head meters on them. The meters all measure kPa. All my pump curves are in volumetric terms. That is why (see above) I stick to the basic formula of

Power = volumetric flow x pressure differential / Eff

What could be simpler? KISS!


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Artisi]

Katmar
I'm with you - KISS
plus the poster was asking for +/- 30%

Naresuan University
Phitsanulok
Thailand

 

[quark]

UmeshMathur,

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.

 

[UmeshMathur]

Continuing the pedantic discussion, the following notes are in order:

(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.

 

[CARF]

Dear all,

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

 

[freefallingbody]

Carf,

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.

 

[quark]

0.2 bara approximately adding 8 meters of head,and so with a total static head of (4.5-1+8) 11.5 meters, with the roughest pipe existing (e/D of 0.01) and two globe valves (worst case), I got about 550 cu.mtr/hr with a resistance of 1.4329 bar at an overall efficiency of 60%.