Why are three different base impedances (1.452, 145.2, 0.48387 ohm) used?

Each voltage zone has its own base voltage, so Z_base = kV^2/MVA differs per zone

The generators have different MVA ratings

Because the loads have different power factors

How is X_th at load B (3.81 kV bus) formed?

(G1||G2) + Tr1 + line + Tr2 = 0.6206, in parallel with G3 (0.3001)

All six reactances added in series

Which base current converts the pu fault current to amperes?

The 3.81 kV base current (4546 A), since the fault is on that bus

2075 Asoj · 10 marks

Short circuit at load B (ohmic data)

Short answer

On 30 MVA, 3.81 kV base, the short-circuit MVA is 148.3 MVA and the short-circuit current is 22.48 kA at the load B bus.

Calculate short circuit MVA and short circuit current when three phase fault occur at load B of the following single line diagram. [Figure: G1 20 MVA, 6.6 kV, X = 0.655 ohm and G2 10 MVA, 6.6 kV, X = 1.31 ohm on a common 6.6 kV bus which also supplies Load A (15 MVA, 6.6 kV, pf = 0.9 lag); transformer Tr-1 30 MVA, 6.6/66 kV, X = 14.52 ohm (referred to HV side); transmission line x = 17.4 ohm; transformer Tr-2 30 MVA, 66/3.81 kV, X = 14.52 ohm (referred to HV side); 3.81 kV bus with generator G3 (30 MVA, 3.81 kV, X = 0.1452 ohm) and Load B (30 MW, 3.81 kV, pf = 0.9 lag)]

Base: 30 MVA, with base voltages 6.6 kV, 66 kV and 3.81 kV in the three zones. Loads are neglected and the pre-fault voltage is 1.0 pu (usual for fault level), so all generator EMFs are 1 pu. The fault is on the 3.81 kV bus of load B.

Base impedances

\begin{aligned} Z_{b,6.6} &= \frac{6.6^2}{30} = 1.452\ \Omega \\ Z_{b,66} &= \frac{66^2}{30} = 145.2\ \Omega \\ Z_{b,3.81} &= \frac{3.81^2}{30} = 0.48387\ \Omega \end{aligned}

Per-unit reactances

Element	Ohms	Zone base (Ω)	pu
G1	0.655	1.452	0.4511
G2	1.31	1.452	0.9022
Tr-1 (HV)	14.52	145.2	0.1000
Line	17.4	145.2	0.1198
Tr-2 (HV)	14.52	145.2	0.1000
G3	0.1452	0.48387	0.3001

 G1--+
     +--Tr1--line--Tr2--+-- 3.81 kV bus (F, load B)
 G2--+                  |
                        G3

Thevenin reactance at load B

\begin{aligned} X_{G1\|G2} &= \frac{0.4511 \times 0.9022}{0.4511 + 0.9022} = 0.3007 \\ X_{left} &= 0.3007 + 0.1000 + 0.1198 + 0.1000 = 0.6206 \\ X_{th} &= \frac{0.6206 \times 0.3001}{0.6206 + 0.3001} = 0.2023\ \text{pu} \end{aligned}

Short-circuit MVA and current

\begin{aligned} \text{SC MVA} &= \frac{30}{0.2023} = 148.32\ \text{MVA} \\ I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 3.81 \times 10^3} = 4546.1\ \text{A} \\ I_f &= \frac{1}{0.2023} \times 4546.1 = 4.9439 \times 4546.1 = 22475\ \text{A} \end{aligned}

Answer: short-circuit MVA $\approx 148.3$ MVA, short-circuit current $\approx 22.48$ kA at 3.81 kV.

✓ Answer checked by Claude Sonnet 5.5

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Short circuit at load B (ohmic data)

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