Protection, Stability and Economic Dispatch (GATE EE)

Overcurrent, distance and differential relays, plug setting multiplier, circuit breakers, swing equation, equal area criterion, critical clearing angle, and economic load dispatch with equal incremental cost.

This lesson covers how a power system is protected against faults, how it stays in step after disturbances, and how generation is shared at least cost.

Protection

Requirements: selectivity, speed, sensitivity, reliability, and back-up.

Overcurrent relays: plug setting multiplier PSM = fault current/(relay setting current × CT ratio). The time multiplier setting (TMS) scales the operating time from the relay curve (IDMT: inverse definite minimum time).
Distance (impedance) relays measure V/I and trip if it is below a setting: impedance, reactance (good for arc faults on short lines), mho (directional, long lines). Zones 1, 2, 3 cover about 80%, 120% and further.
Differential relays compare currents entering and leaving equipment (transformers, generators, buses); percentage (biased) differential relays avoid false trips from CT mismatch. Transformers also need harmonic restraint against magnetising inrush.
Buchholz relay: gas-operated, for incipient faults in oil-filled transformers.

Circuit breakers interrupt the arc at current zero: oil, air-blast, SF₆ and vacuum breakers. Restriking voltage and the rate of rise of restriking voltage (RRRV) decide whether the arc reignites.

Stability

Swing equation (per unit): (2H/ω_s) d²δ/dt² = P_m − P_e, with H = stored kinetic energy/MVA rating (MJ/MVA).

Steady-state stability: small changes; limit P_max = EV/X at δ = 90°.
Transient stability: large disturbances (faults).

Equal area criterion (single machine, infinite bus): the system is stable if the accelerating area (P_m > P_e during the fault) can be matched by an equal decelerating area after clearing. The critical clearing angle is where the two areas are just equal. Faster breakers and auto-reclosing improve stability.

For a three-phase fault where P_e = 0 during the fault and the post-fault curve equals the pre-fault curve: cos δ_cr = (π − 2δ₀) sin δ₀ − cos δ₀.

Economic dispatch

Each generator has a cost curve; the incremental cost IC = dC/dP. Without losses, the cheapest sharing of a load has all units at equal incremental cost λ (within their limits).

With losses: IC_i × penalty factor L_i = λ, with L_i = 1/(1 − ∂P_L/∂P_i).

Worked example

Two units: IC₁ = 0.02P₁ + 10 and IC₂ = 0.04P₂ + 8 (₹/MWh), total load 300 MW.

0.02P₁ + 10 = 0.04(300 − P₁) + 8 → 0.06P₁ = 10 → P₁ = 166.7 MW, P₂ = 133.3 MW, λ = ₹13.33/MWh.

Practice

GATE-styleNAT · numerical answer

A relay has a 5 A setting with a 400/5 CT. A fault current of 4000 A flows. What is the PSM?

Your answer

Show answer and explanation

Answer: 10

Secondary current = 4000/80 = 50 A; PSM = 50/5 = 10.

GATE-styleNAT · numerical answer

In the worked example, what is the system incremental cost λ in ₹/MWh? (Give 2 decimal places.)

Your answer

Show answer and explanation

Answer: 13.33

λ = 0.02 × 166.67 + 10 = 13.33 ₹/MWh.

GATE-styleNAT · numerical answer

For a fault with P_e = 0 during the fault and the same pre- and post-fault curve, the initial angle is 30°. What is the critical clearing angle in degrees? (Give 1 decimal place.)

Your answer

Show answer and explanation

Answer: 79.6

cos δ_cr = (π − π/3) × sin 30° − cos 30° = 1.0472 − 0.8660 = 0.1812, so δ_cr = 79.6°.

GATE-styleMCQ · one correct answer

For economic dispatch neglecting losses, generators should share the load so that:

Their incremental costs are equal
Their outputs are equal
Their efficiencies are equal
The cheapest unit is fully loaded first, always

Show answer and explanation

Answer: A

Equal incremental cost minimises total cost (subject to generator limits).