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 the application at this time.

A utility has given these values to our design team:

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

System Impedances:
$Z_{1} = {1.06} + j{2.4}~Omega$
$Z_{0} = {2.55} + j{16.30}~Omega$

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

Three Phase Fault:
$I_{3-phase-fault} = {V_{an}} / {Z_{1}}$

$I_{3-phase-fault} = {lbrace 11000/sqrt{3} rbrace} / {1.06 + j{2.4}}$

$I_{3-phase-fault} = {6350.85} / {2.62< 66.17^o}$

$I_{3-phase-fault} = 2423.988~A~ or~ 2424~ A~ (as~ provided~ by~ the~ utility)$

Phase-to-Ground Fault:
$I_{phase-to-ground-fault} = {V_{an}} / {lbrace{Z_{1}} + { Z_{2} + Z_{0} }rbrace}$

Note: Since we are distant from the generator, $Z_{1} = Z_{2}$ .

$I_{phase-to-ground-fault} = { lbrace 11000 / sqrt{3} rbrace } / { lbrace (1.06 + j{2.4}) + (1.06 + j{2.4}) + (2.55 + j{16.30}) rbrace }$

$I_{phase-to-ground-fault} = { 6350.85 } / { (4.67 + j{21.1}) } = 6350.85 / {(21.610 < 77.52^o)}$

$I_{phase-to-ground-fault} = 293.884~A~ or~ 294~ A~ (as~ provided~ by~ the~ utility)$

1 thought on “Application – Fault Sequence Network Connection”