Application – Fault Sequence Network Connection

In my previous article, I have provided the different formula for calculating fault currents in different fault scenarios. We will put that into application at this time.

A utility have given these values to our design team:

Voltage: 11kV
3-Phase Fault Current: 2424 A
Phase-ground Fault Current: 294 A

System Impedances:
Z1 = 1.06 + j2.4 ohms
Z0 = 2.55 + j16.30 ohms

From the above values, which ever values your use, current or impedance, your calculation will provide the same result. To verify this,

Three Phase Fault:
I3-phase-fault = Van / (Z1)

I3-phase-fault = (11000 / 1.732) / (1.06 + j2.4)

I3-phase-fault = (6350.85) / (2.62 < 66.17)

I3-phase-fault = 2423.988 A or 2424 A (as provided by the utility)

Phase-to-Ground Fault:
Iphase-to-ground-fault = Van / (Z1 + Z2 Z0)

Note: Since we are distant from the generator, Z1 = Z2.

I3-phase-fault = (11000 / 1.732) / (1.06 + j2.4 + 1.06 + j2.4 + 2.55 + j16.30)

I3-phase-fault = (6350.85) / (4.67 + j21.1) = 6350.85 / (21.610 < 77.52)

I3-phase-fault = 293.884 A or 294 A (as provided by the utility)

About the Author

Ver Pangonilo
A Filipino Engineer, Registered Professional Engineer of Queensland (RPEQ) - Australia and Professional Electrical Engineer (PEE) - Philippines with extensive experience in concept select, front-end engineering, HV & LV detail design, construction and commissioning of Hazardous and Non-Hazardous Area electrical installations in water and waste water pipeline and pumping facilities, offshore platforms, hydrocarbon process plants and pipelines including related facilities. Hazardous area classification and design certification (UEENEEM015B, UEENEEM016B, UEENEEM017B).
  • Ozoguji Emeka

    Sir, write a program on single line to ground fault.

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