HV bus fault and limiting reactor

Base: 30 MVA, 11 kV / 132 kV. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). The load on the HV bus is neglected.

Per-unit reactances (30 MVA base)

 G1--j0.75--T1 j0.075--+
                       +-- HV bus (F)
 G2--j0.50--T2 j0.05--[X]--+

(i) Fault at the HV bus

$$\begin{aligned} X_1 &= 0.75 + 0.075 = 0.825, \quad X_2 = 0.50 + 0.05 = 0.55 \\ X_{th} &= \frac{0.825 \times 0.55}{0.825 + 0.55} = 0.33\ \text{pu} \\ \text{SC MVA} &= \frac{30}{0.33} = 90.91\ \text{MVA} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 131.22\ \text{A} \\ I_f &= \frac{1}{0.33} \times 131.22 = 3.0303 \times 131.22 = 397.6\ \text{A} \end{aligned}$$

(ii) Reactor to reduce the fault level by 25%

"Limit the fault level by 25%" is read as reducing it by 25%, i.e. to 75% of the value above. (Reducing it to 25% is impossible here, because the G1 branch alone gives $30/0.825 = 36.4$ MVA, which is 40% of the present level.)

$$\begin{aligned} \text{New fault level} &= 0.75 \times 90.91 = 68.18\ \text{MVA} \\ X_{th,new} &= \frac{30}{68.18} = 0.44\ \text{pu} \end{aligned}$$

With reactor $X$ (pu) in series with T2 on its 132 kV (secondary) side:

$$\begin{aligned} \frac{1}{0.44} &= \frac{1}{0.825} + \frac{1}{0.55 + X} \\ \frac{1}{0.55 + X} &= 2.2727 - 1.2121 = 1.0606 \\ 0.55 + X &= 0.9429 \\ X &= 0.3929\ \text{pu} \end{aligned}$$

In ohms on the 132 kV side, $Z_{base} = 132^2/30 = 580.8\ \Omega$:

$$X = 0.3929 \times 580.8 = 228.2\ \Omega$$

Answer: (i) SC MVA $= 90.91$ MVA, fault current $\approx 397.6$ A at 132 kV. (ii) A reactor of about 228 Ω per phase (0.393 pu on 30 MVA) on the 132 kV side of T2 reduces the fault level to 68.2 MVA.