Power Systems

GATE EE syllabus for power systems, with per-unit quantities, transmission line models and performance, power transfer, surge impedance loading, economic dispatch, load flow, symmetrical components and fault analysis, protection and stability, with solved examples.

Syllabus

Power generation concepts; AC and DC transmission; models and performance of transmission lines and cables; economic load dispatch (with and without losses); series and shunt compensation; electric field distribution and insulators; distribution systems; per-unit quantities; bus admittance matrix; Gauss-Seidel and Newton-Raphson load flow; voltage and frequency control; power factor correction; symmetrical components; symmetrical and unsymmetrical faults; over-current, differential, directional and distance protection; circuit breakers; stability and the equal area criterion.

Key concepts and formulas

Per-unit system. Zbase = Vbase²/Sbase, and Zpu = Z/Zbase. Changing base: Zpu,new = Zpu,old (Snew/Sold)(Vold/Vnew)². A transformer has the same per-unit impedance on both sides.

Transmission lines.

  • Short line (< 80 km): series impedance only; A = D = 1, B = Z, C = 0.
  • Medium line: nominal π or T models. Long line: distributed parameters, characteristic impedance Zc = √(z/y) ≈ √(L/C).
  • Surge impedance loading SIL = VL²/Zc. Above SIL a line absorbs reactive power; below it, a lightly loaded long line shows a voltage rise at the receiving end (Ferranti effect).
  • Voltage regulation = (VR,no load − VR,full load)/VR,full load.

Power transfer. Between two buses connected by a reactance X: P = (V₁V₂/X) sin δ, maximum V₁V₂/X at δ = 90°. Reactive power flows from the higher voltage to the lower voltage bus.

Economic dispatch. Without losses, generators run at equal incremental cost: dC₁/dP₁ = dC₂/dP₂ = λ, subject to ΣP = load.

Load flow. Bus types: slack (V, δ given), PV or generator (P, |V| given), PQ or load (P, Q given). Ybus: diagonal Yii = sum of admittances connected to bus i; off-diagonal Yij = −(admittance between i and j). Newton-Raphson converges quadratically and needs few iterations; Gauss-Seidel is simpler but slower.

Faults. Symmetrical components: a = 1∠120°. For a fault at a point with Thevenin pre-fault voltage E = 1 pu:

FaultFault current
Three-phaseE/Z₁
Line to ground (LG)3E/(Z₁ + Z₂ + Z₀)
Line to line (LL)√3E/(Z₁ + Z₂)
Double line to ground (LLG)Sequence networks: positive in series with (negative ∥ zero)

Three-phase fault MVA = base MVA/Zpu.

Stability. Swing equation (2H/ωs) d²δ/dt² = Pm − Pe. Equal area criterion: the system stays stable if the accelerating area can be matched by an equal decelerating area. Inertia constant H = stored kinetic energy/machine rating (MJ/MVA).

Protection. Differential relays protect transformers and generators; distance (impedance) relays protect lines; over-current relays with time grading protect feeders. Plug setting multiplier = fault current/(relay setting × CT ratio).

Worked example

A generator rated 50 MVA, 11 kV has X = 0.2 pu. Express it on a 100 MVA, 11 kV base, and find the three-phase fault MVA and current at its terminals. Then find the SIL of a 400 kV line with Zc = 400 Ω, and dispatch 300 MW between two units with incremental costs 20 + 0.1P₁ and 16 + 0.08P₂ (₹/MWh, losses ignored).

import math

Xpu = 0.2 * 100 / 50
print(f"X on 100 MVA base = {Xpu:.2f} pu, fault MVA = {100 / Xpu:.0f} MVA")
Ibase = 100e6 / (math.sqrt(3) * 11e3)
print(f"fault current = {Ibase / Xpu / 1000:.3f} kA")
print(f"SIL = {400e3 ** 2 / 400 / 1e6:.0f} MW")
# 20 + 0.1 P1 = 16 + 0.08 P2 = lambda, P1 + P2 = 300
P1 = (16 - 20 + 0.08 * 300) / (0.1 + 0.08)
P2 = 300 - P1
print(f"P1 = {P1:.2f} MW, P2 = {P2:.2f} MW, lambda = {20 + 0.1 * P1:.3f} per MWh")
Output
X on 100 MVA base = 0.40 pu, fault MVA = 250 MVA
fault current = 13.122 kA
SIL = 400 MW
P1 = 111.11 MW, P2 = 188.89 MW, lambda = 31.111 per MWh

Practice questions

GATE-styleNAT · numerical answer

A machine has X = 0.25 pu on its own base of 20 MVA. What is its reactance on a 100 MVA base (same voltage)?

Show answer and explanationAnswer: 1.25

Znew = 0.25 × (100/20) = 1.25 pu.

GATE-styleNAT · numerical answer

A line has a characteristic impedance of 250 Ω and operates at 220 kV. What is its surge impedance loading in MW?

Show answer and explanationAnswer: 193.6

SIL = V²/Zc = (220 × 10³)²/250 = 193.6 MW.

GATE-styleNAT · numerical answer

At a fault point, Z₁ = Z₂ = 0.2 pu and Z₀ = 0.1 pu, and the pre-fault voltage is 1 pu. What is the line-to-ground fault current in pu?

Show answer and explanationAnswer: 6

If = 3E/(Z₁ + Z₂ + Z₀) = 3/0.5 = 6 pu.

GATE-styleNAT · numerical answer

Two buses at 1.0 pu are connected by a reactance of 0.5 pu. What is the maximum power (pu) that can be transferred?

Show answer and explanationAnswer: 2

Pmax = V₁V₂/X = 1/0.5 = 2 pu, at δ = 90°.

GATE-styleMCQ · one correct answer

Which bus type has its real power and voltage magnitude specified in a load flow study?

Show answer and explanationAnswer: B

At a PV bus, P and |V| are specified, and Q and δ are found. The slack bus has V and δ specified; a PQ bus has P and Q specified.