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Voltage-monitored overcurrent relay timing

Short answer

For a fault just beyond R2, fault current 0.5405 pu reduces terminal voltage to 0.054 pu, activating the voltage monitor. R2 sees 354.6 A: PSM 3.55, time 1.64 s. R1, with plug setting reduced to 40%, has PSM 1.33 and time 2.45 s; it backs up R2.

Figure shows a portion of power system in a single line diagram. Find out the time of operation of relays R1 and R2 for a fault immediately after relaying point R2. Relay R1 is voltage monitored over current relay, the plug setting (PS) of which reduces to 40% of the set value if voltage collapses below 70% of rated voltage. (Assume suitable data if necessary.) [Figure: Generator (Xs = 175%, Xd = 30%, 11 kV, 120 MW, 0.8 pf) with relay R1 (PS = 100%, TDS = 0.1) at its terminals, feeding a step-up transformer (X = 10%, 11/132 kV, 150 MVA) through a breaker to a 132 kV bus. Three outgoing feeders leave the bus through breakers; on one of them relay R2 with CT 200/1 (PS = 50%, TDS = 0.3) and a fault just beyond R2.]

PSM235101520
Time (sec)1064.132.52.2

A voltage-monitored (voltage-controlled) overcurrent relay is used on generators because the sustained fault current of a generator (limited by synchronous reactance) can be less than full-load current, so a plain overcurrent relay would never operate. When the terminal voltage collapses during a fault, its plug setting is reduced so that it can pick up.

Assumptions

  • Base 150 MVA (generator 120 MW / 0.8 = 150 MVA); prefault voltage 1.0 pu; three-phase fault just beyond R2; only this generator feeds the fault.
  • Relay times are of the order of seconds, so the sustained current with X_s = 175% is used (AVR action neglected).
  • R1 CT ratio 8000/1 A (generator rated current 7873 A); R2 CT 200/1 A as given; relay rated current 1 A.
  • Table interpolated linearly; for PSM below 2 the IEC normal inverse curve t = 0.14/(PSM^0.02 − 1), which matches the table (gives 10.0 s at PSM 2 and 2.97 s at PSM 10), is used.

Fault current

X_total = X_s + X_T = 1.75 + 0.10 = 1.85 pu
I_f = 1 / 1.85 = 0.5405 pu
I_base(11 kV)  = 150×10⁶/(√3×11×10³)  = 7872.96 A
I_base(132 kV) = 150×10⁶/(√3×132×10³) = 656.08 A
I_f at 11 kV  = 0.5405 × 7872.96 = 4255.7 A
I_f at 132 kV = 0.5405 × 656.08  = 354.6 A
Generator terminal voltage = I_f × X_T = 0.5405 × 0.10
                           = 0.054 pu (< 0.7 pu)

Relay R2 (feeder)

Relay current = 354.6 / 200 = 1.773 A
Pick-up = 50% × 1 = 0.5 A
PSM = 1.773 / 0.5 = 3.546
t (TDS 1) = 6 − (3.546 − 3)/(5 − 3) × (6 − 4.1) = 5.481 s
t_R2 = 0.3 × 5.481 = 1.644 s

Relay R1 (generator, voltage monitored)

Relay current = 4255.7 / 8000 = 0.532 A
At PS 100%: PSM = 0.532/1 = 0.53 < 1
            → would never operate
Voltage 0.054 pu < 0.7 pu
→ PS reduced to 40% of 100% = 40%
Pick-up = 0.4 A
PSM = 0.532 / 0.4 = 1.33
t (TDS 1) = 0.14 / (1.33^0.02 − 1) = 24.48 s
t_R1 = 0.1 × 24.48 = 2.45 s

Result

RelayCurrentPSPSMTDSTime
R2354.6 A (132 kV)50%3.550.31.64 s
R14255.7 A (11 kV)100% → 40%1.330.12.45 s

Answer: R2 operates in about 1.64 s and R1 in about 2.45 s, so R1 correctly backs up R2 with about 0.8 s margin. Without voltage monitoring, R1 (PSM 0.53) would not operate at all. (Times depend on the assumed R1 CT ratio.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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