In part (i), what is the total reactance to the fault and the fault current?

18.15 + 20 = 38.15 ohm, 166.47 A

Why does each alternator supply 83.24 A in part (i)?

The alternators are identical so they share the fault current equally

Each supplies the full 166.47 A

G1 supplies more because it is nearer the feeder

Each supplies the square root of the total

With a 10 ohm reactor between A and B, which result is correct?

G1 supplies less current than G2 because its path has higher reactance

G1 supplies more current than G2 because its path has higher reactance

The feeder fault current rises above 166.47 A

2073 Chaitra · 6 marks

Fault current: two 500 kVA alternators

Short answer

Using ohmic values, each generator reactance is 36.3 Ω; with 20 Ω feeder, total is 38.15 Ω, giving 166.47 A fault current, 83.24 A per alternator. With 10 Ω between A and B, total is 40.347 Ω, giving 157.40 A; G1 supplies 69.17 A and G2 supplies 88.23 A.

Below figure shows two Alternators operating in parallel. [Figure: G1 500 kVA, X1 = 15%, connected through a breaker to point A; G2 500 kVA, X2 = 15%, connected through a breaker to point C; points A, B, C lie on the 11 kV busbar; the feeder leaves from point B through a breaker and a reactor X = 20 ohms, with a 3 phase to ground fault on the feeder] (i) Calculate the fault current in outgoing feeder and fault current supplied by each alternator. (ii) If a reactor of 10 ohm is connected between point A and point B, calculate new fault current in outgoing feeder and fault current supplied by each alternator.

Work in ohms. Phase voltage $V_{ph} = 11000/\sqrt3 = 6350.85$ V, and $I_f = V_{ph}/X_{total}$ .

\begin{aligned} Z_{base} &= \frac{11^2}{0.5} = 242\ \Omega \\ X_{G1} = X_{G2} &= 0.15\times242 = 36.3\ \Omega \end{aligned}

   G1 500kVA        G2 500kVA
    15%              15%
     |                |
    CB               CB
     |                |
  ===A==[10]===B======C=== 11 kV
              |   ([10] only in part ii)
             CB
              |
            X=20 ohm
              |
              F

(i) Without reactor between A and B

\begin{aligned} X_{par} &= \frac{36.3}{2} = 18.15\ \Omega \\ X_{total} &= 18.15 + 20 = 38.15\ \Omega \\ I_f &= \frac{6350.85}{38.15} = 166.47\ \text{A} \end{aligned}

Each alternator supplies $166.47/2 = 83.24$ A.

(ii) With 10 Ω between A and B

\begin{aligned} X_{G1,path} &= 36.3 + 10 = 46.3\ \Omega \\ X_{par} &= \frac{46.3\times36.3}{46.3+36.3} = 20.347\ \Omega \\ X_{total} &= 20.347 + 20 = 40.347\ \Omega \\ I_f &= \frac{6350.85}{40.347} = 157.40\ \text{A} \end{aligned}

Sharing (inversely as path reactance):

\begin{aligned} I_{G1} &= 157.40\times\frac{36.3}{82.6} = 69.17\ \text{A} \\ I_{G2} &= 157.40\times\frac{46.3}{82.6} = 88.23\ \text{A} \end{aligned}

Answer: (i) $I_f$ = 166.47 A, each alternator 83.24 A. (ii) $I_f$ = 157.40 A; G1 supplies 69.17 A, G2 supplies 88.23 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Fault current: two 500 kVA alternators

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