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KShora (Electrical)
7 Nov 06 15:52
Hi

Can I calculate the efficiency of a 3 phase motor using the formula  W= ?3 x V x I x PF x efficiency.
I have the Input power=1600 Watts, Input Voltage=460 Volts, Rated Current= 2.6 Amps and the Power Factor=0.85. Can I use the values to calculate the efficiency?

Looking forward to the reply

Regards
KShora

 

Helpful Member!  l3city (Electrical)
7 Nov 06 16:26
E(efficiency)
Po(power out)
Pi(power in)

E=Po/Pi
thus, if you have the rating of the power delivered you just divide it and get your efficiency

OR can use the following

E=(746*HP)/[(746*HP)+(watts lost)] Where 746 is a constant and HP=horsepower

just my two humble cents, hope this helps
Regards
dpc (Electrical)
7 Nov 06 16:32
Not unless you also know the output power at the shaft.  Unless by "input" you actually mean the motor output.

Your input power of 1600 W doesn't match the current, voltage, and power factor data you gave.  That comes out to 1760 kW.  
KShora (Electrical)
7 Nov 06 16:34
Thanks for the reply

I don't have the Watt Loss data.

Does the equation I posted earlier make sense or not?

Regards
Skogsgurra (Electrical)
7 Nov 06 16:36
If you say that your output power (not input) is 1600 W, you are correct. Your efficiency will be (with your numbers) E = 1600/1761 = 0.91

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

KShora (Electrical)
7 Nov 06 16:41
skogsgurra
The power value that I have mentioned is written on the motor name plate, I suppose you are correct, it should be called as Output power.

Am I correct?

Regards
ozmosis (Electrical)
7 Nov 06 16:51
or you can get the motor efficiency from the motor manufacturers data sheets. In Europe most manufacturers have to comply with regulations according to CEMEP and in North Amercia it is EPACT, both require manfs to provide data on motor efficiency. You may even get a few who will provide part-load efficiency levels as this makes a big difference and most motors do not run at full load.
KShora (Electrical)
7 Nov 06 16:57
Yes that is by last option, however I was really hoping to be able to calculate the efficiency myself using the motor data that I had as we deal with many motor manufacturers.

Regards
Helpful Member!  CJCPE (Electrical)
7 Nov 06 18:18
KShora

In case you have been waiting for someone to state this more directly, I believe that the posts above confirm that you calculate input power using W = sqrt(3) X V X I X PF and Efficiency = output Power / input Power. We assume that the 1600 Watts marked on the nameplate is output power and efficiency = 91%.
powerjunx (Electrical)
8 Nov 06 0:20
KShora,

 You may refer to IEEE Std 112-1996 IEEE Standard Test Procedure for Polyphase Induction Motors and Generators. Or you may refer to the ieee colorbook.

 Or you may try this:

 "ProMot, a decision support tool, is being developed within the framework of a project cofinanced by SAVE, aiming to aid end-users to explore the possibility of energy savings, in motor systems of an industrial or tertiary installation: electric motors, pumps, compressed air system and chillers. The tool helps in auditing an installation and performing simple meaningful calculations of purchasing, replacing existing or retrofitting electric motor systems. These technical and economic calculations are based on equipment data retrieved from widely accepted and regularly updated European databases and methodologies."

 Download at this link,

 http://promot.cres.gr/promot_plone/downloads/tools

Regards,

bill
Helpful Member!  aolalde (Electrical)
10 Nov 06 13:52
I agree with CJCPE.

Power in = SQRT(3)*V*I*PF = 1.732*460*2.6*0.85 = 1760.8 Watts

Power out (on the shaft) = 1600 Watts (per nameplate data)

EFF = Pout/Pin = 1600/1760 = 0.9086  pu

And the total motor losses  = Pin-Pout = 160.8 Watts

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