How to Size the Motor Drive?

[Kiribanda]

Hello to all members,

I am new to Pump engineering forum.

This refers to the correct sizing of a motor drive for a pump to estimate the input electrical power (kVA) requirement.

I understand that the following equation gives the required HP for a pump to give out a given discharge at a given head.

BHP in HP = ( HEAD in Feet) x (Flow rate in gallons per
minute) x (SPGR of the fluid being pumped) /
( 3960 x Efficiency of the pump )

(Normally my observation is substituting 85% for the Efficiency of the pump term. If I am wrong please correct me)

Now based on the above,

1)Does it give the HP(kW) requirement which should be available at the pump shaft before the coupling OR at the motor shaft after the coupling?

2)If it is the required power in HP(kW) at the pump shaft, how much is the reasonable figure for typical coupling efficiency to calculate the required power at the motor shaft?

3)If it is the required power in HP(kW) at the motor shaft, where is the value for coupling efficiency in the equation?

4)If it is a compressor, how do we estimate the required power at the motor shaft? Is it different from a pump?

Your valuable inputs are greatly appreciated!!

Thanks in advance!!!

Kiribanda

 

[rmw]

Kiribanda,

I just submitted a post in this forum for twins, where your "slip" reply was immediately before mine. Good post! I may have answered some of your questions there, but I will go through it again in more detail here.

The efficiency of the pump is a function of several things shown like lines on a map on the pump curve. A centrifugal volute type pump of a given size will generally have a "family" of curves based on differing impeller diameters, for a given speed. (some synchronous speed, neglecting slip)

If you look a typical pump curve, you can see that there is a point where each flow curve intersects the highest efficiency available, or, best efficiency point (BEP). Unless there are other constraints, it is best to select the pump at or near the BEP.

There is also a family of curves for the horsepower required as a function of flow for each impeller size.

The motor, however, is normally selected so that if the flow increases such that it is way out toward the right side of the curve, (a function of, or should I say limited by, system resistance to flow, barring an abnormal situation) the motor can handle it without kicking out the overloads. The selection is then known as 'non over loading.' The horsepower determined from the curves, is at the pump shaft, and incorporates bearing and seal resistance.

I will leave coupling efficiency, etc. to others I believe it to be negligable.

rmw

 

[TD2K]

By coupling, do you mean some sort of viscous coupling like an automotive transmission or a magnetic speed adjustment coupling? A simple mechanical coupling is 100% efficient.

Your pump efficiency might be on the high side. I have seen some pumps (low head, high flow) with 90% efficiencies but more commonly I would use 70% for preliminary sizing. Of course if you have similar pumps, use their efficiencies. The lowest efficiency I believe I've seen was a low flow, high head Sundyne, it was about 15%. More of an expensive heater than a pump almost ;-).

For a compressor, the approach is similar but different equations are used. Any compressor book will give you the equations.

 

[quark]

I agree with TD2K on pump efficiency(though this is the first time I read him to be a bit pessimistic) and efficiency of mechanical coupling.

I always prefer to consider 10% higher size motor, for we are always conservative in calculating frictional losses.

Check these links for compressor power consumption.

Reciprocating Compressors

http://www.processassociates.com/process/rotating/recip_s.htm

Centrifugal Compressors

http://www.processassociates.com/process/rotating/centri_s.htm

Regards,




Eng-Tips.com : Solving your problems before you get them.

 

[smckennz]

There are a number of points that your should consider:

1) pump power is what is required at the pump shaft, and does not allow for the coupling or any ancillaries. This is standard practice.

2) If it is a mechanical coupling, you can usually ignore the coupling loss. In the rare instance that it is an eddy current or a hydraulic coupling, I would allow 5% of pump power for preliminary sizing purposes only.
Do not select motor size on duty point; if unsure add 20% and go for the next standard motor size up. Only for planning. It is quite possible that changes in duty point will require more power.

4) Compressors can be a lot different than pumps - always higher power demand. If the compressors are just 100psi air compressors, preliminary sizing on 4cfm/hp.

For gross pump efficiency, I use 50% for planning on "miscellaneous medium sized pumps (2"-4"), 35% on small pumps (under 2") and estimate larger pump efficiency according to the specific speed. The mechanical guys should be doing this for you.

Cheers

Steve

 

[vanstoja]

Pump hydraulic efficiency is best approximated from specific speed (at best efficiency flow and head)and rated flow using plots like the so-called "Worthington curves" (see Pump Handbook, 1976, Page 2-130). Your cited 85% pump hydraulic efficiency is OK for rated flows around 3000 GPM and specific speeds from about 1400 to 4000 (RPM-GPM-Ft.) Peak efficiencies for flowrates of 500, 200 and 100 GPM are about 78, 73 and 66%, at specific speeds between 1500 and 2000. A 10000 GPM rated flow peaks at about 90% efficiency around a specific speed of 2500. These are all based on presumably "good design practices". Consult manufacturers test data for more exact pump hydraulic efficiencies for specific designs.

 

[Kiribanda]

Thanks to all for your valuable inputs!!

rmw,
As per your post and others I can understand that the equation gives the required power at the PUMP Shaft.

Before we wind up one more point about a relevant standard.

Table 3.1 of API Standard 610 -Selection of Cent. Pumps-
gives some percentage values to select the motor drive. Is it correct to multiply the pump BHP - given by the above equation - by the percentage values given in the Table 3.1
to obtain the required motor output power?

Regards!

Kiribanda