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Calculation of ac motor power
I have a motor rate in lbs-ft, the power isn't on the nameplate.
By Anonymous on 24 March, 2007 - 1:46 pm

Hi!

I have a three phase 460Vac, 115A, 235lbs-ft and with a speed of 1800RPM, but the power isn't indicated on the nameplate.

If we calculate the power with the formula P=V*I*1.732, we got 91.6KVa or 122.8HP and, if we calculate with the formula P(T*N)/5250, we
got 80.5HP. So, why we don't get the same result? Is it the power factor that make the
difference?

Hi there,

I read thorough your question. I found that your equation of P=V*I*1.732 was incomplete. You should consider both power factor and motor efficiency. As you know, you should able to find the motor power factor on the motor name plate, but most kinds of motors don't indicate the motor efficiency value. Thus, let's assume your motor efficiency value to be 88% & power factor of 0.8.

P=(V*I*1.732)*0.80*0.88
= 64.4kW

This is all because the equation of P(T*N)/5250
was based on torque (motor output). In fact, it's already included the motor efficiency.

motor efficiency = (motor output/motor input)* 100%

by
KAy
Electrical engineering
DMR

Hi,

what if i just have v and i values,

then how we will find the motor power and what will be power factor for motor in India.

Yes!! Definitely it is a low PF motor!

By Ken Stewart on 3 April, 2007 - 11:00 pm

When estimating the power input to the motor, use the formula involving voltage and current. When estimating the power output from the motor, use the formula involving torque and rotational speed. The difference is wasted energy, which manifests itself as heat produced by the motor. Be careful with the nameplate ratings on the motor. They are probably maximum ratings. Be aware that motors of this type generally accept 150% overload. To get an accurate estimate you must measure the power-in versus the power-out for your given application.

Ken

AC motors have a component of the total current, called the magnetizing current. This is non-torque-producing current. Thus, the reason you cannot equate AC motor kVA to HP.

You were right, normally in AC motor will produce
30% of magnetizing current, which has the opposite direction of vector to the input current. Thus
producing heat loss through motor body frame.

KAy
Electrical Engineer
DMR

By Phil Corso, PE on 5 April, 2007 - 11:39 pm
1 out of 1 members thought this post was helpful...

Responding to Anonymous' 24-Mar-07 (12:46) query:

1) ELECTRICAL POWER.
The first P, for which you used the equation sqrt(3)*V*I/1,000 = 91.6 kVA, is not 'real' electrical power but Apparent Power. A more correct representation is 'S' and its units are VA or kVA. Then, real power, Pi, is the product of Apparent Power times Power Factor, or S*PF! In-turn, real input-power, Pi = Po/Eff'y, where Po = 0.746*Hp/Eff'y. (CAVEAT: metric Hp and English HP are not equal!)

2) MECHANICAL POWER.
The 2nd P, for which you used the equation, 5250*RPM/Torque or 5252*1800/115 =lb(f)-ft, is shaft horsepower, or Hp, a mechanical term. In addition, you used the synchronous speed to make the calculation. Instead, output Hp is more correctly rated at the motor's as is normally shown on its nameplate, not at its synchronous speed.

3) RULE-of-THUMB.
In countries that rate motor output in shaft Hp, a rule-of-thumb value often used is: input kVA is approximately equal to output Hp. Note the term approximately because the real kVA/HP ratio does vary. Also note, in countries using metric units then output shaft-power is expressed in electrical units or kW.

4) MATCHING MOTOR.
Of course, in the absence of other performance data, like pf and efficiency, then you will never know its exact rating. However, using typical manufactuter's data, then the 115-A value you mentioned would yield a 3-ph, 460-V motor of 100-Hp.

Regards, Phil Corso, PE (cepsicon@aol.com)

By electrical on 1 March, 2011 - 10:30 pm

what if we know data from nameplate

V=945V, I=39A, eff=0.76, pf=0.82, rpm 2820, 2 pole

and i want to find out how they got nameplate data with pf motor=0.82? though we know the pf from nameplate? or efficiency motor??
thanks,
alex

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

Electrical Alex... you did not mention nameplate values for output kW or Hp. Thus, based on the data you provided:

kVAi = Sqrt (3) x Vi x Ai
kWi = kVAi x Pf
kWo = Eff x kWi
Hp (Imp Units) = kWo / 0.746,
where subscripts (i) and (o) represent Input and Output parameters, respectively.

Substituting the values given yields:
kVAi = 63.8
kWi = 52.3
kWo = 39.8.
Hp (Imperial Units) = 53.4
Hp (In Metric Units = 54.1

Let me know if I misunderstood your question. Regards, Phil Corso (epsicon [at] aol [dot] com)

By curt wuollet on 2 March, 2011 - 5:30 pm

I'm fairly sure that they loaded the motor to the rated HP and took readings. Phase angle gives you PF, mechanical power /electrical power gives you efficiency, etc. Easy except for the mechanical power which requires a torque sensor, a brake of some sort, a tachometer and math.

Regards
cww

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

Hi,

Many motor nameplate does not indicate motor Eff, or PF.
So based on motor voltage and current listed on name plate, how can match the power calculated with the power listed on nameplate?

Regards
Wael

By Bob Peterson on 27 February, 2012 - 8:15 am

You cannot calculate the PF of a motor from its nameplate. The PF varies with the actual load on the motor. The more energy the motor is actually delivering to the load, the higher the PF will be. A lightly loaded motor will have a low PF.

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

Wael… if the motor's nameplate gives its rated temperature-rise there is way to obtain the information you seek!

Determine its temperature-rise over ambient, then, Hp(out), kW(in), kVA(in), PF, and Efficiency, can be derived, with reasonable accuracy!

Please note, this was the procedure used way before the introduction of Clamp-On Watt meters!

I suggest you search the Control.Com Archives for similar topics. If your search proves inadequate for your needs, then contact me.

Regards, Phil

What does 1.732 represent?

By Tomy Zacharia on 17 April, 2007 - 10:50 pm

Square root(3)

Tomy Zacharia

By Teddy nugroho on 24 June, 2007 - 10:52 am

if we measure electric power in various load from induction motor. How we determine pf and eff? I think this is the biggest problem when we try to find the input power.

>if we measure electric power in various load from induction motor. How we determine pf and eff? I think this is the biggest problem when we try to find the input power. <

hi there,

By Jun Pascasio on 26 March, 2014 - 4:11 am

I secured an Induction motor properties chart from Teco and compared it with other motor properties chart. The I've got provides a graph for the torque and current plotted against a load/torque current for vertical values and % synchronous speed for horizontal axes.

Dividing the heights of the full load current and full load torque into 100 segments and manipulating these in auto CAD, I was able to get values for the RPM, Torque, and with some math, the efficiency and the power factor.

By Neville Ellens on 19 August, 2007 - 7:11 pm

1.732 is Root 3 of the voltage in a 3 phase power supply.

Formula

Reactive Power = VxIx1.1732X PF

I think the formula gives true power. If PF is cosine theta as it was when I went to school.

Bill Code

By Neville Ellens on 5 May, 2010 - 7:31 am

Formula

True Power = VxIx1.732X PF

>Formula
>
>True Power = VxIx1.732X PF

I have question regarding the power shown on motor nameplate, is it the the output power n the shaft or the input power???

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

Nicolas... Depending on country of manufacture, OUTPUT-shaft power is given in Hp or the equivalent kW!

Regards, Phil Corso

> Nicolas... Depending on country of manufacture, OUTPUT-shaft power is given
> in Hp or the equivalent kW!

I am wondering if I need to size a motor, what %age losses should be considered. I need guaranteed 772Kw shaft power output from the motor. What will be appropriate size of motor to deliver 772kw shaft power? Am I too conservative if I consider 20% losses?

Regards,
Baig

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

Baig... if you are only concerned about the motor loss, and not the driven-machine loss, then 20% is way too high.

If you are interested in a more realistic value, then please provide the motor's nameplate Hp or kW, its rated voltage, and its rated speed or number of poles!

Regards, Phil Corso