Power Systems
Power Systems asks you to model a network on a common base and then reason about faults, load flow, protection, and stability. Per-unit fluency is the single most important skill here.
Per unit
- Z_pu = Z_actual/Z_base with Z_base = kV²/MVA.
- Base change: Z_new = Z_old × (MVA_new/MVA_old) × (kV_old/kV_new)².
Y-bus and load flow
- Diagonal entry = sum of admittances connected to that bus; off-diagonal entry = negative of the admittance between the two buses.
- Gauss–Seidel is simple but slow; Newton–Raphson converges quadratically and is the standard.
Faults and stability
- Symmetrical fault current in per unit equals 1/X_th.
- Symmetrical components: I_a0 = (I_a + I_b + I_c)/3.
- Swing equation M d²δ/dt² = P_m − P_e; the equal-area criterion gives transient stability limit.
Worked example: a 50 MVA generator with X″ = 0.2 pu is re-based to 100 MVA, giving X = 0.2 × (100/50) = 0.4 pu. For a bolted three-phase fault at its terminals with X_th = 0.4 pu, I_f = 1/0.4 = 2.5 pu. On an 11 kV base, I_base = 100×10⁶/(√3 × 11×10³) = 5249 A, so the fault current is 2.5 × 5249 ≈ 13.1 kA. The per-unit fault current is simply the reciprocal of the per-unit Thevenin reactance.
Traps that cost marks
- Inverting the base-change voltage ratio.
- Mixing line-to-line and phase quantities in three-phase formulas.
- Ignoring transformer and line impedance, which reduces the fault current substantially.