📘 Power Systems – Question Bank (Page 13)
🔹 Section A: Multi-Bus Fault Case Study
Q1. A 3-bus system has following reactances (pu on 100 MVA base):- Generator at Bus 1: X = j0.2 pu
- Line 1–2: j0.3 pu
- Line 2–3: j0.25 pu
- Load Bus 3
Total impedance to Bus 3: Ztotal = 0.2 + 0.3 + 0.25 = 0.75 pu Fault current = 1 / 0.75 = 1.33 pu
🔹 Section B: Stability + Fault Combined
Q2. A generator has E = 1.1 pu, V = 1 pu, X = 0.5 pu. Find maximum power transfer and power at δ = 45°.Pmax = EV/X = (1.1 × 1)/0.5 = 2.2 pu At 45°: Pe = 2.2 × sin45° = 2.2 × 0.707 = 1.555 pu
🔹 Section C: Economic Dispatch with Limits
Q3. Two generators:C1 = 0.02P1² + 4P1 C2 = 0.01P2² + 5P2 Load = 250 MW Limits: 50 ≤ P1 ≤ 200 50 ≤ P2 ≤ 200
Incremental cost: dC1/dP1 = 0.04P1 + 4 dC2/dP2 = 0.02P2 + 5 Set equal: 0.04P1 + 4 = 0.02P2 + 5 P1 + P2 = 250 Solve: Multiply by 100: 4P1 - 2P2 = 100 P1 = 250 - P2 Substitute: 4(250 - P2) - 2P2 = 100 1000 - 4P2 - 2P2 = 100 1000 - 6P2 = 100 6P2 = 900 P2 = 150 MW P1 = 100 MW Check limits → Within limits ✔
🔹 Section D: Load Flow Quick Numerical
Q4. A 2-bus system has line reactance j0.4 pu. Load at Bus 2 = 1.2 + j0.6 pu. Estimate approximate current.I = S*/V = (1.2 - j0.6)/1 = 1.2 - j0.6 pu
🔹 Section E: Protection + Fault Level
Q5. A 220 kV bus has fault level 5000 MVA. Find short circuit current.I = MVA / (√3 × V) = 5000 / (1.732 × 220) = 13.12 kA (approx)
🔹 Final Revision Strategy
- Combine per-unit + fault + protection
- Link stability with power angle
- Understand dispatch with constraints
- Practice multi-step numerical
Page 13 – Advanced Multi-Concept Practice Completed
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