Reactor to protect circuit breaker of small station

The C.B. must not see a fault level above its rating (150,000 kVA). G3 is connected to the bus through reactor $X$. For a fault on the outgoing circuit just beyond the C.B., all three machines feed the fault.

Base: 10,000 kVA, 3.3 kV; $Z_{base} = \dfrac{3.3^2}{10} = 1.089\ \Omega$. Percentage reactance on new base: $\%X_{new} = \%X\times\dfrac{\text{kVA}_b}{\text{kVA}}$.

   A 2500kVA   B 500kVA    C 10000kVA,7.5%
     8%          6%          |
     |           |          [X]
 ====+===========+===========+==== 3.3 kV
           |
          CB 150,000 kVA
           |--- F

Reactances on 10,000 kVA base

$A\parallel B$: $\dfrac{32\times120}{152} = 25.263\ \%$, giving fault kVA $= 10000\times100/25.263 = 39{,}583$ kVA (safe).

If C were connected directly: $25.263\parallel7.5 = 5.783\ \%$, fault kVA $= 172{,}917$ kVA > 150,000 kVA. So a reactor is needed.

Required reactor

Allowed total reactance:

$$\%X_{eq} = \frac{10000\times100}{150000} = 6.667\ \%$$ $$\begin{aligned} \frac{1}{6.667} &= \frac{1}{25.263} + \frac{1}{7.5+X} \\ 0.15 - 0.039583 &= 0.110417 = \frac{1}{7.5+X} \\ 7.5 + X &= 9.0566 \\ X &= 1.5566\ \% \end{aligned}$$

In ohms:

$$X_\Omega = \frac{1.5566}{100}\times1.089 = 0.01695\ \Omega$$

Answer: Reactor $X$ ≈ 1.557 % on 10,000 kVA base ≈ 0.017 Ω in series with the new generator keeps the fault level at 150,000 kVA.