# Example: Single Line-to-Ground Fault Calculation

**The single line-to-ground fault on a transmission line happens when one conductor accidentally to the ground or comes in some cases in contact with the neutral conductor. For this reason, the protective devices of a power system needs to be accurately responsive to avoid severe damage in the system.**

See Unbalanced Fault Analysis: Single Line to Ground Fault

**Indeed, it is very important to know the procedure of single line to ground fault calculation.**

Related articles:

- Double Line to Ground Fault
- Line to line fault
- Symmetrical three phase fault

**Example:**

Assumed that the generator below is solidly grounded and neglect the fault impedance. Determine the phase currents and phase voltage at faulted location.

Generator |

__Solution:__

**Compute the base value:**

Choose: Sb = 20 MVA and kVb = 13.8 kV

then,

Ea = 20 MVA/ 20 MVA = 1 (angle 0) per unit.

Ibase = 20 MVA / (1.73 x 13.8 kV)

Ibase = 0.837 kA

__Develop positive sequence network:__

System Positive- Sequence Network Equivalent |

where,

- Ea = the generator EMF, which has a value of 1 (angle 0) per unit.
- Xd" = the positive sequence impedance of transformer (in subtransient state)
- If-1 = positive sequence current.

__Develop negative sequence network:__

System Negative- Sequence Network Equivalent |

where,

- If-2 = negative sequence current
- X2 = negative sequence reactance

__Develop zero sequence network:__

System Zero-Sequence Network Equivalent |

- If-0 = zero sequence current
- X2 = zero sequence reactance

then, If-1 = 1 (angle 0)/ (Z0 + Z1 + Z2) =

**-j 2.22 pu or 2.22 (angle -90)**

also it means that, If-2 = If-0 =

**- j 2.22 pu**

**Multiplying the base value**

**Ibase = 0.837 kA**

**then**, If-1 = 2.22 (angle -90) * (0.837 kA) = 1.86

**kA (angle -90)**--> actual value

Since all sequence currents are equal, thus:

- If-1 = 1.86 kA (angle -90) --> + S
- If-2 = 1.86 kA (angle -90) --> -S
- If-0 = 1.86 kA (angle -90) --> 0S

Applying the formula of sequence to phase values matrix we can get:

- Fault current at A = 5.58 kA (angle -90 degrees)
- Fault current at B = 0
- Fault current at C = 0

The zero values of phases B and C confirmed that no fault current flows from it during fault condition.

Looking at this diagram, the voltage values of the un-faulted phases (phase B and C) are the only one that has a value while phase A (the faulted phase), its voltage is equal to zero (neglecting impedance).

Thus,

__Computing Voltage at the faulted point,__

Looking at this diagram, the voltage values of the un-faulted phases (phase B and C) are the only one that has a value while phase A (the faulted phase), its voltage is equal to zero (neglecting impedance).

Thus,

- Voltage at phase A = 0
- Voltage at phase B = Vb
- Voltage at phase C = Vc.

Analyzing positive sequence network equivalent,

- Vf-1 = 1 (angle 0) - (If-1) * (Z1) = 1 - (-j 2.22) (j 0.25)
**= 0.445 (angle 0) (+ sequence voltage)**

Analyzing negative sequence network equivalent,

- Vf-2 = 0 - (If2) * (Z2) = 0- (-j2.22) * (j0.1)
**= 0.222 (angle 180) ( neg. sequence voltage).**

Analyzing zero sequence network equivalent,

- Vf-0 = 0 - (If-0)* (Z0) = 0- (-j 2.22) * (j0.1)
**= 0.222 (angle 180) (zero sequence voltage).**

Applying the formula of sequence to phase values matrix we can get:

- Voltage at phase A = 0
- Voltage at phase B = 0.667 (angle -120) pu
- Voltage at phase C = 0.667 (angle 120) pu

Applying base value,

Vbase = 13.8 kV/ 1.73

Therefore, the actual values of phase voltages are,

- Voltage at phase A = 0
- Voltage at phase B = 5.31 kV (angle -120)
- Voltage at phase C = 5.31 kV (angle 120)