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Bus reactor between 100 MVA and 200 MVA generators

Short answer

On 100 MVA base, generator reactances are 0.1 pu and 0.04 pu. To limit fault level to 1500 MVA, total equivalent reactance must be 0.0667 pu; solving the parallel paths yields a bus-bar reactor of 0.16 pu.

A 100 MVA generator with 10% reactance and a 200 MVA generator with 8% reactance are connected to a common bus. The fault level on bus 1 is to be restricted to 1500 MVA. If a reactor is added in between these two generators in the bus bar, then on 100 MVA base Calculate the value of reactance added.

The bus-bar is split into two sections joined by a reactor XX: G1 on bus 1 and G2 on bus 2. For a fault on bus 1, G1 feeds directly and G2 feeds through XX. Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}.

Base: 100 MVA.

XG1=0.10 pu,XG2=0.08×100200=0.04 puX_{G1} = 0.10\ \text{pu},\qquad X_{G2} = 0.08\times\frac{100}{200} = 0.04\ \text{pu} 
  G1 100MVA,10%        G2 200MVA,8%
       |                     |
  ===[Bus 1]=====[ X ]=====[Bus 2]===
       |
       F (fault on bus 1)

Without reactor: Xeq=0.1∥0.04=0.02857X_{eq} = 0.1\parallel0.04 = 0.02857 pu, fault MVA =100/0.02857=3500= 100/0.02857 = 3500 MVA (too high).

With reactor: required

Xeq=1001500=0.06667 puX_{eq} = \frac{100}{1500} = 0.06667\ \text{pu}

The two paths are in parallel:

1Xeq=1XG1+1XG2+X15=10+10.04+X0.04+X=15=0.2X=0.16 pu\begin{aligned} \frac{1}{X_{eq}} &= \frac{1}{X_{G1}} + \frac{1}{X_{G2}+X} \\ 15 &= 10 + \frac{1}{0.04+X} \\ 0.04 + X &= \frac{1}{5} = 0.2 \\ X &= 0.16\ \text{pu} \end{aligned}

Answer: Bus-bar reactor XX = 0.16 pu (16 %) on 100 MVA base. (If the bus voltage VV kV is known, XΩ=0.16 V2/100X_\Omega = 0.16\,V^2/100; e.g. at 11 kV, 0.194 Ω.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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