Why is the generator reactance multiplied by (6/6.6)^2?

To correct 6 kV rating to the 6.6 kV base voltage of that zone

Because the generator rating is in MVA

Because the line voltage is 132 kV

At P, the two paths to the sources are

0.2033 (G1 + T1) and 0.3468 (line + T2 + G2) in parallel

The 7.80 pu is converted to 1706 A using

I_b = 50 MVA/(sqrt(3) x 132 kV) = 218.7 A

2069 Chaitra · 6+6 marks

Fault at both ends of a 50 Ω line

Short answer

Using 50 MVA, 132 kV base, per-unit reactances: G1,G2 0.1033, T1,T2 0.10, line 0.1435. At P, Thevenin reactance is parallel combination of (0.2033 and 0.3468) = 0.1282 pu. Fault current = 1/0.1282 = 7.80 pu = 1706 A (1.71 kA). Same at Q due to symmetry.

The single line diagram of a power system is shown in figure below. Compute the fault current in p.u. and in absolute value if a 3 phase to ground fault occurs at (consider no fault impedance): i. the point P i.e. connection point between transmission line and HV side of transformer T1 ii. the point Q i.e. connection point between transmission line and HV side of transformer T2 (printed as T1). [Figure: G1 6 kV, 40 MVA, X = 10% - bus - T1 6.6/132 kV, 50 MVA, X = 10% - point P - line X = 50 Ω - point Q - T2 132/6.6 kV, 50 MVA, X = 10% - bus with load - G2 6 kV, 40 MVA, X = 10%]

For a bolted 3-phase fault, $I_f = 1/X_{th}$ pu, where $X_{th}$ is the parallel combination of the reactances from the fault point back to each generator (prefault voltage 1 pu, load current neglected).

Data and per unit values

Base: 50 MVA; 132 kV on the line, hence 6.6 kV on the generator buses (from the 6.6/132 kV transformers). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$ .

Element	Calculation	X (pu)
G1, G2 (6 kV, 40 MVA, 10%)	$0.10 \times \frac{50}{40} \times \left(\frac{6}{6.6}\right)^2$	0.1033 each
T1, T2 (50 MVA, 10%)	on own base	0.10 each
Line (50 Ω)	$50/348.48$	0.1435

Base current at 132 kV: $I_b = \dfrac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7$ A.

 G1      T1      P    Line     Q     T2     G2
(~)-0.1033-0.10--o--0.1435--o--0.10-0.1033-(~)
 |                                          |
 +============= reference (EMF 1 pu) =======+

i. Fault at point P

\begin{aligned} X_{left} &= X_{G1} + X_{T1} = 0.1033 + 0.10 = 0.2033 \\ X_{right} &= X_{line} + X_{T2} + X_{G2} = 0.1435 + 0.10 + 0.1033 = 0.3468 \\ X_{th} &= \frac{0.2033 \times 0.3468}{0.2033 + 0.3468} = 0.1282\ \text{pu} \\ I_f &= \frac{1}{0.1282} = 7.80\ \text{pu} \\ I_f &= 7.80 \times 218.7 = 1706\ \text{A} \end{aligned}

Contribution from G1 side $= 7.80 \times 0.3468/0.5501 = 4.92$ pu, from G2 side $= 2.88$ pu. Fault MVA $= 50/0.1282 = 390$ MVA.

ii. Fault at point Q

\begin{aligned} X_{left} &= X_{G1} + X_{T1} + X_{line} = 0.3468 \\ X_{right} &= X_{T2} + X_{G2} = 0.2033 \\ X_{th} &= \frac{0.3468 \times 0.2033}{0.5501} = 0.1282\ \text{pu} \\ I_f &= 7.80\ \text{pu} = 1706\ \text{A} \end{aligned}

Now G2 side supplies 4.92 pu and G1 side 2.88 pu.

The network is symmetrical about the middle of the line, so the fault current magnitude is the same at P and Q; only the sharing between the generators changes.

Answer: At P and at Q, $I_f$ = 7.80 pu ≈ 1706 A (≈ 1.71 kA at 132 kV), fault level ≈ 390 MVA.

✓ Answer checked by Claude Sonnet 5.5

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Fault at both ends of a 50 Ω line

Data and per unit values

i. Fault at point P

ii. Fault at point Q

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