Motor Fault Contribution

[EddyWirbelstrom]

I am using PTW software for IEC60909 short-circuit calcualtions.
Motor entry data includes motor 'Locked Rotor Current' and motor R/X.
Is the motor R/X for short-circuit contribution equal to ( Stator R + standstill Rotor R ) / ( Stator X + standstill Rotor X ) ?
Is this same ratio used for motor starting power-factor ?

 

[ijl]

The IEC-standard defines the R/X ratio in the following way:

For voltages over 1000V and power per pole pair greater than or equal to 1 MW the ratio R/X = 0.10, but if the power per pole pair is less than 1 MW, then R/X = 0.15. For voltages less than or equal to 1000V the ratio R/X = 0.42.

I do not know about PTW, but my guess is that the R/X given as input has no effect in the short circuit calculation, because these standard values should be programmed in the software itself.

 

[EddyWirbelstrom]

Thankyou ijl. I use your recommended R/X values for individual medium voltage motors which I model with motor cables.
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

 

[mykh]

Check table 8C from

http://www.arcadvisor.com/pdf/GET3550Fapp4041.pdf

showing typical x/r ratios for induction motors ranging from 0.5HP up to approx 3000HP

 

[ijl]

Just some thoughts, not a definite answer.

1. Why do you need a better accuracy than that given by the standard? The accuracy of the standard is sufficient for practical purposes. That is the (main) point of the standard.

2. With a "typical" value of R/X you get a "typical" result, that is not necessarily more accurate than that from the standard.

3. Even with an accurate R/X-value, the result is not necessarily very accurate, because the standard itself describes an approximate method.

4. If you need more accurate results, you should try to simulate the transient with ATP, EMTP, or similar. But you will get a (relatively) accurate result for the simulated case only. The standard gives the maximum (worst case) values of the currents. (The standard can also give the minimum values of the currents, if so desired, of course.)

5. If you have access to a simulator, and have the necessary parameters for the motor, you might try the following:
Calculate manually the initial current Ik". Simulate the fault at the terminals of the (unloaded) motor. Deduce the R/X from the results. Use this R/X in the PTW. No guarantee for a success, but an interesting experiment, I think.

Besides, isn't it so that the peak current occurs ½ cycle after the occurrence of the fault, not ¼ cycle? (in the case of an unloaded motor) In addition, the DC-offset is due to the motor, not to the network.