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Motor Contribution in Short Circuit Current

Motor Contribution in Short Circuit Current

(OP)
I have a question about the contribution of motor in short circuit current. For a simple circuit of a 1500kW 13.8kV generator, which feeds a 1000kVA 13.8kV/480V transformer, and the transformer feeds a 480V 250HP motor. We have to calculate the short circuit current at the primary terminals of the transformer. If the full load current of the motor is 270A and the impedance of the transformer is 5.75%, then which formulas should be used to calculate that how much short circuit current this motor will produce, and how much of this short circuit current (produced by motor) will pass through from the secondary to the primary of the transformer? Thanks for help!

RE: Motor Contribution in Short Circuit Current

Simply include the motor as a source in your impedance diagram. It becomes effectively a second source. You then have to reduce the network to an equivalent impedance looking in from the bus/node you are faulting. In reality, motor contribution on the primary side of the transformer will be quite small due to the transformer impedance.

RE: Motor Contribution in Short Circuit Current

You have me thinking about the motor contribution. The safe figure is to use the locked rotor current for the motor contribution.
But I am going to start another thread for a discussion on induction motor contribution to a non bolted fault.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Motor Contribution in Short Circuit Current

(OP)
Is there some document available on internet which gives example of hand calculations of reduction of network to an equivalent impedance? Thanks

RE: Motor Contribution in Short Circuit Current

I agree with dpc.
IEC 60909-0 Short-circuit currents in three-phase a.c. systems -Part O: Calculation of currents.
Chapter 3.8.2 Contribution to short-circuit currents by asynchronous motors
The impedance ZM = RM + jXM of asynchronous motors in the positive- and negative-sequence systems can be determined by:
ZM=1/(ILR/IrM)*UrM/√3/IrM=1/(ILR/IrM)*UrM^2/SrM (26)
where:
UrM is the rated voltage of the motor;
IrM is the rated current of the motor;
SrM is the rated apparent power of the motor SrM = PrM/(TM.COSTM)
TM=efficiency ; COS(TM)=power factor
If RM/XM is known, then XM shall be calculated as follows:
XM=ZM/√(1+(RM/XM)^2)
The following relations may be used with sufficient accuracy:
RM/XM = 0.10, with XM = 0.995 ZM for medium-voltage motors with powers PrM per pair of poles>= 1MW;
RM/XM = 0.15, with XM = 0.989 ZM for medium-voltage motors with powers PrM per pair of poles <1MW;
RM/XM = 0.42, with XM = 0.922 ZM for low-voltage motor groups with connection cables.
For the calculation of the initial short-circuit currents according to 4.2, asynchronous motors are substituted by their impedances ZM according to equation (26) in the positive-sequence and negative-sequence systems. The zero-sequence system impedance Z(0)M of the motor shall be given by the manufacturer, if needed (see 4.7).

RE: Motor Contribution in Short Circuit Current

Be advised that motors fed by drives won't contribute to s/c.

RE: Motor Contribution in Short Circuit Current

Recent studies revealed that the variable speed drives having four quadrant operation would contribute to the short circuit current.

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