actual work 380V vs 460V

[mark754]

If I have a motor running a piece of equipment in the US on 460V 60HZ and it is drawing 100amps, how does that compare to one in Asia running on 380V 50HZ drawing 100 amps?

Are they both doing the same amount of work? What is the formula to use to adjust them the amps so they are both doing the same amount of work?

Thanks

 

[busbar]

In 380V 3ø circuits, each kVA of capacity corresponds to a per-phase current 1.52 amperes. In 460V 3ø circuits, the ratio is 1.26 amperes/kVA, representing some conductor savings per unit power, but at the expense of heavier insulation.

Frequency has no direct effect on electrical-power calculations. The same ratios hold for 50- or 60-Hertz systems.

 

[jbartos]

Comment on mark754 (Electrical) Apr 5, 2004 marked ///\\If I have a motor running a piece of equipment in the US on 460V 60HZ and it is drawing 100amps, how does that compare to one in Asia running on 380V 50HZ drawing 100 amps?
///
1. 460V 60Hz motor will deliver:
HP=.746 x sqrt3 x E x I x PF x EFF/1000 = 79.7 x PF x EFF
2. 380V 50Hz motor will deliver:
HP=.746 x sqrt3 x E x I x PF x EFF/1000 = 49.1 x PF x EFF
Assuming that Efficiencies EFFs and Power Factors PFs of both motors are the same, the first motor will deliver more HPs.\\

 

[rbulsara]

In much simpler terms power includes product of voltage AND current. You need to know both the Current and voltage to determine the total power.

For 3 Phase system kVA = (sqrt 3)*V*I

kW = kVA*power factor = real 'Work"

HP = 0.746 kW
You can use the above relationship for a 3 phase motor.

What you will get by above calculation for a motor is input to the motor. The output of the motor is input*efficiency

 

[rbulsara]

correction:

kVA= sqrt (3)*kV*I or Sqrt(3)*V*I/1000

 

[jbartos]

Correction, I beg your pardon:
1. 460V 60Hz motor will deliver:
HP=(1/746) x sqrt3 x E x I x PF x EFF = 107 x PF x EFF
2. 380V 50Hz motor will deliver:
HP=(1/746) x sqrt3 x E x I x PF x EFF = 88 x PF x EFF
Assuming that Efficiencies EFFs and Power Factors PFs of both motors are the same, the first motor will deliver more HPs.

 

[aolalde]

The electric input of a circuit is; Total Power= Volts * Amps
For three phase circuits KVAin= 1.732*V*A/1000

For your post values:

KVAin1 = 1.732 *460*100/1000 = 79.674
KVAin 2 = 1.732*380*100/1000 = 65.817

If you want to know the power delivered in the motor shaft , the efficiency (EFF) and the power factor (PF) of that particular motor is needed. If we assume same figures EFF=0.90 and PF=0.85.

KWout1 = 79.674*0.90*0.85 = 60.95 (KW)= 81.7 (HP)
KWout2 = 65.817*0.90*0.85 = 50.35 (KW)= 67.49 (HP)

 

[jbartos]

Suggestion: This size of motor tend to have better efficiency and power factor than .9 and .85. Visit for example:

http://www.baldor.com

for similar-motor specs