How is the line's per-unit impedance found?

Divide the ohmic value by 33^2/30 = 36.3 ohm

Divide the ohmic value by 11^2/30 ohm

Multiply the ohmic value by 36.3

How are the elements combined to get Z_th seen from the fault F?

G1 in parallel with G2, then in series with the transformer and line

G1 and G2 in series, then in parallel with the line

Only the line and transformer are used; generators are ignored

2078 Bhadra · 8 marks

Fault at far end of 33 kV line

Short answer

On 30 MVA, 33 kV base with pre-fault voltage 1 pu and neglecting load current, the three-phase fault current at the far end is 1605 A, and fault level is 91.8 MVA.

For the system shown in figure below, determine the fault current, fault level if three phase balanced short circuit fault occurs at the far end F. [Figure: G1 11 kV, 20 MVA, 15% and G2 11 kV, 10 MVA, 10% on a common 11 kV bus; transformer 30 MVA, 5%, ratio 1:3 (to 33 kV); line of (3 + j5) Ω at 33 kV to the fault point F at the far end]

Base: 30 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side (ratio 1:3). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit impedances

Element	Calculation	pu
G1	$0.15 \times 30/20$	0.225
G2	$0.10 \times 30/10$	0.300
Transformer	5% on 30 MVA	0.05
Line	$(3 + j5)/Z_{base}$	0.0826 + j0.1377

$Z_{base}$ at 33 kV $= 33^2/30 = 36.3\ \Omega$ .

 E=1--jX_G1--+
             +--j0.05--(3+j5 ohm)--F
 E=1--jX_G2--+

Total impedance up to F

\begin{aligned} X_G &= \frac{0.225 \times 0.300}{0.225 + 0.300} = 0.1286\ \text{pu} \\ Z_{th} &= jX_G + j0.05 + Z_L = j0.1286 + j0.05 + (0.0826 + j0.1377) \\ &= 0.0826 + j0.3163 = 0.3269\angle 75.36^\circ\ \text{pu} \end{aligned}

Fault current and fault level

\begin{aligned} I_f &= \frac{1}{|Z_{th}|} = \frac{1}{0.3269} = 3.0588\ \text{pu} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 524.86\ \text{A} \\ I_f &= 3.0588 \times 524.86 = 1605.4\ \text{A} \\ \text{Fault level} &= \frac{S_{base}}{|Z_{th}|} = \frac{30}{0.3269} = 91.76\ \text{MVA} \end{aligned}

Answer: fault current $\approx 1605$ A at 33 kV, fault level $\approx 91.8$ MVA.

✓ Answer checked by Claude Sonnet 5.5

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Fault at far end of 33 kV line

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