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# The max current of this motor?

GOOD DAY PPL, can anyone help me with the following datas? i would like to know what would be the max current of this motor?W = 75KW, V = 380V, A = 152.5,F = 50Hz,HP = 100HP. kindly help me how to calculate, that's the motor used to lift granite blocks. usually i know its 10 times the normal Ampere rating. still can anyone help me with the formulas and get me the nearby approx current rating of this motor?

#1

09-05-2014 05:27 AM

Top #2

The motor nameplate should list the locked-rotor amps or locked-rotor kVA/hp or NEMA kVA Code Letter: A, B, C, ... up to V see the following link...

http://www.engineeringtoolbox.com/locked-rotor-code-d_917.html.

The actual locked-rotor current will depend on the motor design. Depending on what you use the locked-rotor amp calculation for will also dictate how accurate the number needs to be. For fault studies that include motor contribution, using a value of 4-6X times full-load current is a good estimate for a typical induction motor with 4X being used for larger motors and 6X for smaller motors.

http://www.engineeringtoolbox.com/locked-rotor-code-d_917.html.

The actual locked-rotor current will depend on the motor design. Depending on what you use the locked-rotor amp calculation for will also dictate how accurate the number needs to be. For fault studies that include motor contribution, using a value of 4-6X times full-load current is a good estimate for a typical induction motor with 4X being used for larger motors and 6X for smaller motors.

09-05-2014 07:43 AM

Top #3

Also, according to your data, the full-load current is listed as 152.5 A. Based on 100 hp at the shaft, this can be estimated...

FLA = (100 hp)(746 W/hp)/sqrt(3)/V/eff/pf,

where,

eff = efficiency,

pf = power factor at full-load,

V = line voltage,

Example:

For eff = 0.93, pf = 0.8, and V = 380 volt, FLA = 152 amp.

FLA = (100 hp)(746 W/hp)/sqrt(3)/V/eff/pf,

where,

eff = efficiency,

pf = power factor at full-load,

V = line voltage,

Example:

For eff = 0.93, pf = 0.8, and V = 380 volt, FLA = 152 amp.

09-05-2014 10:23 AM

Top #4

If the LRC (Locked Rotor Current) is not shown on your nameplate, you should be able to find that on the manufacturers datasheet.

If you only need the value to design starting equipment, you can generally use a standard multiplier between your full load current and the LRC, for a similar motor type (typically 6-8 times).

It is not possible to set up any general formula, because this ratio is highly dependent on the internal design of the motor. Things like wiring layout, air gap distance, iron cross section etc. play a big role. In addition to this, comes variations caused by number of poles and so on.

Newer motors tend to have relatively high LRC, because the trend toward more efficient motors means reduced winding resistances, to reduce stator losses, reduced rotor resistance, to reduce the slip, these factors will both mean an increase in the LRC factor, if all else is equal, and the list can go on.

If you only need the value to design starting equipment, you can generally use a standard multiplier between your full load current and the LRC, for a similar motor type (typically 6-8 times).

It is not possible to set up any general formula, because this ratio is highly dependent on the internal design of the motor. Things like wiring layout, air gap distance, iron cross section etc. play a big role. In addition to this, comes variations caused by number of poles and so on.

Newer motors tend to have relatively high LRC, because the trend toward more efficient motors means reduced winding resistances, to reduce stator losses, reduced rotor resistance, to reduce the slip, these factors will both mean an increase in the LRC factor, if all else is equal, and the list can go on.

09-05-2014 01:08 PM

Top #5

Apart from design of starting equipment, you could need to know this LRC to determine if your cable impedance is sufficiently low to allow the required starting torque for the motor, and on the other side, if the starting current would cause too much voltage drop at the switchboard, so other loads may be affected.

When you calculate the actual starting current, you must of course also take into account the voltage drop along the supply cable, which will reduce the starting current somewhat.

When you calculate the actual starting current, you must of course also take into account the voltage drop along the supply cable, which will reduce the starting current somewhat.

09-05-2014 03:09 PM

Top #6

Agrees with you Jim, however for information motors are no longer made with letter codes above M. Letters codes above "M" are not deemed efficient.Some of these motors will still found in the system.

Davidson J, assuming 10 times starting current is very risky, the only method is to identify the letter code. Please provide us wit the same.

Regards

Davidson J, assuming 10 times starting current is very risky, the only method is to identify the letter code. Please provide us wit the same.

Regards

09-05-2014 06:02 PM

Top #7

at starting this current will take 400 amp........

09-05-2014 08:09 PM

Top #8

Ashvini Bhardwaj how did you come to that conclusion?