A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its positive, negative and zero sequence impedances are 0.30, 0.40 and 0.05 pu respectively. What value of reactance must be placed in the generator neutral so that the fault current for a line to ground fault of zero fault impedance shall not exceed the rated line current. 

Problem Statement: A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its positive, negative, and zero sequence impedances are 0.30 pu, 0.40 pu, and 0.05 pu respectively. We need to determine the value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance does not exceed the rated line current.

Solution:

To solve this problem, we will calculate the required reactance step by step:

Step 1: Determine the base current

The base current for the system can be calculated using the formula:

I_base = (S_base × 10⁶) / (√3 × V_base)

Here:

• S_base = 30 MVA = 30 × 10⁶ VA
• V_base = 13.2 kV = 13.2 × 10³ V

Substituting the values:

I_base = (30 × 10⁶) / (√3 × 13.2 × 10³)

I_base = 1311.78 A

So, the base current of the system is 1311.78 A.

Step 2: Determine the per unit fault current for a line-to-ground fault

In the case of a line-to-ground fault, the equivalent impedance is given by:

Z_LG = Z₁ + Z₂ + (Z₀ || Z_N)

Where:

• Z₁ = Positive sequence impedance = 0.30 pu
• Z₂ = Negative sequence impedance = 0.40 pu
• Z₀ = Zero sequence impedance = 0.05 pu
• Z_N = Neutral reactance to be determined

The fault current is inversely proportional to the equivalent impedance. The fault current in per unit (I_f) for a line-to-ground fault can be expressed as:

I_f = 1 / Z_LG

We are tasked with limiting the fault current to the rated line current (i.e., 1 pu). Hence, the equivalent impedance must satisfy:

Z_LG = 1 pu

Substituting the values into the equation for Z_LG:

1 = 0.30 + 0.40 + (0.05 || Z_N)

Step 3: Solve for Z_N

The parallel combination of Z₀ and Z_N is given by:

(0.05 || Z_N) = (Z₀ × Z_N) / (Z₀ + Z_N)

Substituting this into the equation for Z_LG:

1 = 0.30 + 0.40 + [(0.05 × Z_N) / (0.05 + Z_N)]

Rearranging:

0.30 + 0.40 = 0.70

1 - 0.70 = (0.05 × Z_N) / (0.05 + Z_N)

0.30 = (0.05 × Z_N) / (0.05 + Z_N)

Cross-multiplying:

0.30 × (0.05 + Z_N) = 0.05 × Z_N

Expanding and simplifying:

0.015 + 0.30 × Z_N = 0.05 × Z_N

0.30 × Z_N - 0.05 × Z_N = 0.015

0.25 × Z_N = 0.015

Solving for Z_N:

Z_N = 0.015 / 0.25 = 0.06 pu

Step 4: Convert Z_N to ohms

To convert the per unit reactance into ohms, we use the base impedance:

Z_base = (V_base)² / S_base

Substituting the values:

Z_base = (13.2 × 10³)² / (30 × 10⁶)

Z_base = 5.808 ohms

Now, the neutral reactance in ohms is:

X_N = Z_N × Z_base

X_N = 0.06 × 5.808

X_N = 0.34848 ohms