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Power Calculation of Three Phase Motor
Calculation of power consumed by three phase induction motor

I want to caculate power consumption of three phase induction motors. I have measured following data like current, voltage & overall power factor. But when I am calculating Power consumed by motors using formula P=1.732* V*I*Power Factor, But I found that from calculation KW coming is more than specified by motor vendor. I am taking overall power factor for calculation. Can anybody help me how to calculate exact power consumed by motor?

By William Hinton Sr. Electrical Engineer @ Delphi on 17 November, 2002 - 9:53 pm

Three-Phase electric motors have different code letters, starting currents, duty cycles and some are WYE and some are DELTA. The amount of KW divided by the amount of KVA is power factor and the reactive power is in KVAR found by taking the SIN of the phase angle multiplied by the KVA. All that said, it seems your motor is working a bit harder than the manufacturer recommends. Motors are thermal devices, as they can take an overload for quite a long time depending on design.

NEMA rated electric motors, transformers, fuses, circuit breakers and so on can withstand a continuous 10% overload 24 hours per day, 7 days per week for about 35 years. NEMA motors are also rated CLASS 10 CLASS 20 and CLASS 30. Class 20 motors are standard motors that can withstand 20 seconds of locked rotor current without damage and class 30 motors can withstand 30 seconds of locked rotor current.

I'm not sure what the IEC motor ratings are but I believe they are generally CLASS 10.

There is another factor: motor efficiency. There are standard and premium effeciency motors which have significantly different starting currents. Boy, there are alot of things to know about electric motors including the fact their current, efficiency and power factor all increase as the load increases.

Dear Ramesh,
So far your motor applied voltage is correct and the motor current is within name plate the Kw rating will usually fall within the motor limits.
( for normal 3 phase induction motors)
But generally most motors are designed to operate at 80-90% of full load amps. In your case how did you read the power factor? The best power factor and the best efficiency never occur at the same load and they keep varying.The power factor is slightly lower for multi pole motors. Also substitute different measuring instruments before concluding if in doubt.
Best regards,

1 out of 1 members thought this post was helpful...

Responding to Ramesh's, Sun, Nov 17, 2:32 pm, query.

This subject was covered in great detail in an earlier thread.

Proper measurement is dependent on parameters used. For example, If you used phase-to-phase voltage and line current to provide measurement of one phase, then, the result will be in error!

If you want additional detail contact me at tal-2(AT)

Regards, Phil Corso, PE (Boca Raton, FL)

By Ramesh B. Parmar on 1 December, 2002 - 11:56 am

Dear Phils,

Thanks for your reply on Power calculaton Subject. You have mentioned that details has been discussed in thread, if you can give the details of the thread so that I can visit the same.

Also, I want to know for calculating power consumption of individiual motor is it right way to take overall system power factor for calculation? If not then please let me know how to calculate power factor of Individual motor.
Waiting for your reply.
Thanking you.

Ramesh B. Parmar

1 out of 1 members thought this post was helpful...

Replying to Ramesh Parmar's, Sun, Dec 1, 4:50 pm, query:

The thread, starting on June 02, was titled "ENGR: Power Consumption for Motor w/VFD!" It covered procedure to determine kW, kVAR, power factor, and efficiency, regardless of waveshape, harmonics, or the degree of unbalance.

Phil Corso, PE
(Boca Raton, FL)

1 out of 1 members thought this post was helpful...

Further to my earlier response to Ramesh Parmar:

In the past I was advised that instructions via E-mail were too confusing to follow. Therefore, if you send me your Fax # I will forward the procedure, complete with schematic and vector diagrams.

This offer is also made to anyone interested in this topic.

Phil Corso, PE
(Boca Raton, FL)

By Electro Mechanical Systems on 11 December, 2002 - 12:27 am


I would like to take you up on your offer... Fax# 315-894-3365

Mike Adkins
Electro Mechanical Systems, Inc.

By Quentin Guhr on 21 January, 2003 - 5:53 pm

I would like to accept your offer.

Please Fax to: 563-285-1198
attn: Quentin

Quentin Guhr
Systems Engineering

By Mike Adams on 19 March, 2003 - 4:55 pm

I am also interested in this information. I have someone trying to tell me they measured the voltage and current seperately in the three hot wires going to the 3-phase motor, and then used the average volts and average amps in the calculation hp=E*I/746 to get horsepower. This didn't seem right to me, because it gave the answer that a 20 hp motor was operating at 7 hp under a condition where it should have been nearly fully loaded (i.e. driving a centrifugal pump working against a near shut-off condition). Any info you have regarding this will be very appreciated. Email me at, or fax to 216-221-8126, hit *51 then the start button on your fax machine. I am a mechanical engineer so I am somewhat unfamiliar with the electrical properties of motors.

Mike Adams

By Steve Myres on 20 March, 2003 - 10:21 am

Hi Mike,

If you examine the pump curves, you'll probably find that your pump loads most heavily at open or near open discharge. If you multiply the discharge head in feet times the mass flow in lbs/min, you will get a product in ft- lbs/min, which is power. So at max head zero flow, there is actually zero fluid horsepower output. (There will still be some shaft horsepower draw due to mechanical inefficiency in the pump head). At max flow, on the other hand, there is little static head but substantial velocity head. Apparently, the presence of the velocity head, along with the high mass flow creates more power consumption than the high head, low mass flow product.


Hi Mike,

Agree about the maths of the method - but a centrifugal pump operating against a closed is not "fully loaded" - it is common practice to start a centrifugal pump against a closed valve, then open it after about 5 seconds to ease the starting problem of the motor.


1 out of 1 members thought this post was helpful...

Three comments:

1) The correct equation for input power is:
Pin = Sqrt(3) x Vpp x I x PF, where Vpp = ph-to-ph volts.
I = Line current.
PF = Power Factor.

2) The equivalent current is not Iavg, but instead, it should be obtained from the square root of the (sum of the squares of line current) / 3.

3) Output power is determined by one of three methods, all of which are explained in the List archives:
a) Measure power input via two-wattmeter method if system is 3-phase, 3-wire, or by the single- or three-wattmeter method if the system is 3-phase, 4-wire. Then, deduct estimated losses provided by the motor manufacturer.
b) Determine the relative output by ratioing the measured temperature-rise above ambient, to the rated temperature-rise at full load.
c) Determine the relative output by ratioing the measured slip, to the rated slip at full-load.

Phil Corso, PE
Boca Raton, FL
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1 out of 1 members thought this post was helpful...

Further to my earlier reply:

A pump operating near shutoff is not operating at full-load. Lookup the "fluid" Hp formula!

Phil Corso, PE
Boca Raton, FL
[] ( {}

Does it matter whether the motor is delta or wye connected when using this formula?

Don Gall

By Jose Lourenco on 1 August, 2004 - 12:41 pm

Dear Mike,

Your interpretation of pump requirement is not correct. A centrifugal pump, working with a near shut-off valve, wil draw very little power. Power for a centrifugal pump is proportional to head, and flow-rate.

So the 7hp could easily be right.



> In the past I was advised that instructions via E-mail were too confusing to follow. Therefore, if you send me your Fax # I will forward the procedure, complete with schematic and vector diagrams. <

Could you kindly fax me to +65 6368 9281?

1 out of 1 members thought this post was helpful...

Win... repeating an earlier Caveat, "I no longer accept anonymous requests for information or help... something to do with the fact that anonymity breeds boorishness"

Therefore, anyone requiring help must provide a name, affiliation with a company (or school) and a location! In return, I will send a resume of my qualifications and experience... an abridged or complete version!

Regards, Phil Corso (cepsicon [at] aol [dot] com)

By Anonymous on 13 May, 2004 - 11:07 pm

P=I x V x pf for single phase
P= 1.732 X I x V X pf /1000 for 3 phase

so I= P x 1000/1.732 X 415 X pf(.85)

exp: p=2hp

I= 2 x 1000/610
= ? amp

By Anonymous on 23 May, 2006 - 12:57 am

dear friend,

the power that is on the data of motor is the output power but the power that u get it and it was more than the datapower is the input power.

so the difference input power-output power = loss power