How much does each generator supply in part (i)?

115.2 A, equal because they are identical

With 5 ohm reactors between A-B and C-D, which generators supply less current and why?

G1 and G4, because they reach the feeder through the extra 5 ohm

G2 and G3, because they have the extra 5 ohm

G1 and G4 supply more because of the reactors

2074 Asoj · 4+4 marks

Fault current: four 1600 kVA generators

Short answer

Using 11 kV phase voltage 6350.85 V and generator reactance 15.125 Ω, four parallel generators plus 10 Ω feeder give 13.781 Ω, so fault current is 460.83 A, 115.21 A per generator. With 5 Ω bus reactors between A-B and C-D, total current is 443.57 A; G1/G4 supply 95.16 A each, G2/G3 supply 126.62 A each.

Below figure shows 4 Nos of identical generators operating in parallel. Each generator is rated as 1600 kVA, 11kV, Xg = 0.2 pu. [Figure: G1, G2, G3, G4, each through Xg, connected to an 11 kV busbar at points A, B, C, D; an outgoing feeder from the busbar between B and C through a circuit breaker and a reactor X = 10 Ω, with a 3-phase to ground fault at its end] (i) Calculate fault current in outgoing feeder and fault current supplied by each generator. (ii) If reactors of 5 ohm each are connected between point A and B, point C and D respectively. Calculate fault current in outgoing feeder and fault current supplied by each generator.

Work in ohms at 11 kV. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V. Fault current $I_f = V_{ph}/X_{total}$ .

\begin{aligned} Z_{base} &= \frac{11^2}{1.6} = 75.625\ \Omega \\ X_g &= 0.2\times75.625 = 15.125\ \Omega \end{aligned}

  G1      G2       G3      G4
  |Xg     |Xg      |Xg     |Xg
  A--[5]--B----+---C--[5]--D   11 kV bus
               |       ([5] only in part ii)
              CB
               |
            X=10 ohm
               |
               F

(i) Without bus reactors

All four generators are directly in parallel:

\begin{aligned} X_{par} &= \frac{15.125}{4} = 3.781\ \Omega \\ X_{total} &= 3.781 + 10 = 13.781\ \Omega \\ I_f &= \frac{6350.85}{13.781} = 460.83\ \text{A} \end{aligned}

Each generator supplies $460.83/4 = 115.21$ A. (Fault MVA $=\sqrt3\times11\times0.46083 = 8.78$ MVA.)

(ii) With 5 Ω reactors between A–B and C–D

G1 and G4 now reach the feeder through 5 Ω; G2 and G3 connect directly.

\begin{aligned} X_{G1} = X_{G4} &= 15.125 + 5 = 20.125\ \Omega \\ \frac{1}{X_{par}} &= \frac{2}{20.125} + \frac{2}{15.125} = 0.23161 \\ X_{par} &= 4.3176\ \Omega \\ X_{total} &= 4.3176 + 10 = 14.3176\ \Omega \\ I_f &= \frac{6350.85}{14.3176} = 443.57\ \text{A} \end{aligned}

The bus voltage behind the reactors is $I_f X_{par} = 443.57\times4.3176 = 1915.2$ V, so

\begin{aligned} I_{G1} = I_{G4} &= \frac{1915.2}{20.125} = 95.16\ \text{A} \\ I_{G2} = I_{G3} &= \frac{1915.2}{15.125} = 126.62\ \text{A} \end{aligned}

Check: $2(95.16) + 2(126.62) = 443.57$ A.

Answer: (i) $I_f$ = 460.83 A, each generator 115.21 A. (ii) $I_f$ = 443.57 A; G1 and G4 supply 95.16 A each, G2 and G3 supply 126.62 A each.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Fault current: four 1600 kVA generators

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