# 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)

### 1 thought on “Application – Fault Sequence Network Connection”

1. Ozoguji Emeka

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