Monday, 23 February 2026

📘 Power Systems – Question Bank (Page 18)

🔹 Case Study 1 – Generator + Line + Fault + Stability

Given:
Generator internal emf E = 1.2 pu
Bus voltage V = 1 pu
Reactance X = 0.6 pu
Mechanical power Pm = 1 pu

1️⃣ Find Maximum Power Transfer:

Pmax = EV/X = (1.2 × 1) / 0.6 = 2 pu

2️⃣ Find Initial Power Angle:

Pm = Pmax sinδ 1 = 2 sinδ sinδ = 0.5 δ = 30°

3️⃣ If reactance increases to 1 pu during fault, find new Pmax:

Pmax_fault = (1.2 × 1)/1 = 1.2 pu

System becomes less stable (power transfer reduces).

🔹 Case Study 2 – Multi-Source Fault Level Comparison

Two generators feed a bus: G1: X = 0.25 pu G2: X = 0.2 pu Line reactance to fault = 0.3 pu
Step 1: Parallel generator reactance

1/Xeq = 1/0.25 + 1/0.2 = 4 + 5 = 9 Xeq = 0.111 pu

Step 2: Total reactance

Ztotal = 0.111 + 0.3 = 0.411 pu

Step 3: Fault current

If prefault voltage = 1 pu I_fault = 1 / 0.411 = 2.43 pu

🔹 Case Study 3 – Economic Dispatch with Limits

C1 = 0.01P1² + 5P1 C2 = 0.02P2² + 4P2 Load = 300 MW
Incremental cost equations: dC1/dP1 = 0.02P1 + 5 dC2/dP2 = 0.04P2 + 4 Set equal: 0.02P1 + 5 = 0.04P2 + 4 P1 + P2 = 300
Solve: 0.02(300 - P2) + 5 = 0.04P2 + 4 6 - 0.02P2 + 5 = 0.04P2 + 4 11 - 0.02P2 = 0.04P2 + 4 7 = 0.06P2 P2 = 116.67 MW P1 = 183.33 MW

🔹 Case Study 4 – Load Flow Insight

2-bus system: Line reactance = j0.5 pu Load at Bus 2 = 1.5 + j0.8 pu Approximate current:

I = S*/V = (1.5 - j0.8) Magnitude ≈ 1.7 pu

🔹 Concept Summary

Stability decreases when reactance increases
Fault current inversely proportional to impedance
Dispatch based on equal incremental cost
Voltage & power flow interlinked

Page 18 – Ultra Advanced Case Study Completed

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