Induction Motor Short Circuit Current

[Vegemite]

Given the following resistance and reactance values how can we work out the magnitude and power factor of the three phase short circuit current for a fault near to the motor terminals.
The givens are per unit values of:-
-stator (R1, X1)
-rotor running values at rated speed (R2, X2 at slip ~ very small)
-rotor stalled values (R2, X2 at slip = 1).

The induction motors concerned are 3 phase squirrel cage with 415 V voltage rating.

 

[mpparent]

Sounds like a homework problem...

Mike

 

[Vegemite]

Dear readers

This is a genuine enquiry. Although it is well known that an induction motor's short circuit current's magnitude is equal to locked rotor current what I am unsure about is the power factor or X/R ratio of the short circuit current.

Is it the same as the motor starting power factor?

 

[EddyWirbelstrom]

My enquiry is the similar to John Archer’s.
I use the IEC60909 recommended R/X values when entering asynchronous motor data into the PTW software program for short-circuit calculations to IEC60909.
The starting power factor supplied by motor manufacturers varies from 0.1 for large MV motors to 0.4 for small LV motors.

However the IEC60909 R/X value for LV motors assumes all motors at a bus are lumped into one equivalent motor.
The R/X value of 0.42 includes motor cables.
I wish to model LV motors individually, with their cables.
Therefore the motor R/X would be different than 0.42
I need the R/X value to determine the motor peak short-circuit current contribution.

The peak short-circuit current (ip) specified in IEC60909 is the peak or crest value of the current wave 1/4 of a cycle after fault inception assuming the fault occured at a voltage zero crossing. This would result in a theoretical maximum d.c. offset of 2*sqrt2*I"k due to the inductive reactance of the network for a fault which is assumed 100% inductive.
However the network resistance results in a d.c. decrement of this d.c. offset current :-
iDC = sqrt(2)*I"k*e^-L/R = sqrt(2)*I"k*e^-2pi*f*t*R/X

The value for the peak short-circuit current 1/4 of a cycle after fault inception is :-
ip = sqrt(2)*I"k ( 1.02 + 0.98*e^-3* R/X ).

In IEC60909 the value of ip for each source - including asynchronous induction motors - is calculated and added to give the total ip at the fault location.

For LV (415V) motors IEC60909 allows the simplified lumping of all motors on one bus into one equivalent lumped motor. ( To simplify hand calculations )
All the motors various cable lengths and sizes are accommodated by modelling thie lumped motor with :-

- Motor initial symmetrical rms fault contribution = 5 x motor rated full load current.
- Motor R/X = 0.42

Software packages such as PTW allow the modelling of each motor and associated cables indiviually. This would give a more accurate motor short-circuit contribution than the simplified lumped motor.

Therefore I need to know the R/X value for typical 415V, 50Hz asynchronous motors.
I use an R/X of 0.2

Your comments and suggestions on how to determine motor R/X for motor peak short-circuit contribution would be most welcome.

Murray Newman