High Pressure Pump Electric Drive Motor Cos Phi

Power Factor for a Three-Phase Electrical Motor

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Power factor of an AC induction motor under load varies by the motor design and the amount of load. There is no way to accurately predict it in advance of the selection of a specific motor and the effect of a specific load on that specific motor. Values are typically measure after the fact, or estimated. That is why in most places, PF is required to be corrected to .95 or better, because that can usually be accomplished by estimating. I understand (from this forum) that some countries have enacted rules requiring .99 PF, which in my opinion is draconian. The ONLY way I think that can be accomplished is empirically. Measure under the exact operating conditions and determine the exact correction needed. Good luck.

The only number I have seen on motor nameplates is 0.9, which would presumably be at full load. If loaded to less than full capability, the actual power factor would be lower.

That was a bona fide and relevant answer. I wonder who thought it was OT.

Not I.

Just like any other drive-- try to make cos phi closest to 1.0- and save on energy bills.

Crabtree-- Perfect Logic!

And more so - Where is the possibility of manipulating pressures and flowrates?

Let the Power factor FIND ITS BALANCE- even if the energy bill is not minimum!

Check curves of motor characteristics(pf,efficiency vs load) and load the motor accordingly.

Thanx a lot guys all of the replies are quite helpful.

Regarding the rules in "some countries" and the value of .99 does anybody have any idea about the Middle east, particularly Oman?

I have only heard that issue mentioned in this forum with regards to some parts of India. If you can determine the name of the electric utility serving your destination, they might have information on their tariff structure on a web site.

Let's calculate the required power:

Ph = q ρ g h / (3.6* 10^6) [kW] and hwater = (p2 - p1) / ρ g where:

q = flow capacity (m3/h) ; ρ = density of fluid (kg/m3) for water =1000

g = gravity (9.81 m/s2) ; h = differential head (m) ; p[N/m^2]

Ph=q*(p2-p1)/ (3.6* 10^6)

1 bar=1.02*9.81*10^4 [N/m^2]

Ph=170*(72-1)*1.02*9.81*10^4/(3.6*10^6)=335.5 kW [450 HP]

From ABB catalogue:

For 400 V 50 Hz 355 kW General purpose steel motors

2 poles p.f.= 0.88; 4 poles p.f.=0.86; 6 poles p.f.=0.84

NEMA HV Cast iron motors 4000 V 60 Hz 450 HP

2 poles p.f.= 0.91; 4 poles p.f.=0.86; 6 poles p.f.=0.80 ;8 poles p.f.=0.76

you can estimate from the table