Electrical Equipment and Devices

WEEK 16: POWER SYSTEM PROTECTIVE DEVICES

Sections: Protection Systems | Circuit Breakers | Example Problems

EXAMPLE 1: For a 69 kV breaker with a maximum rated voltage of 72.5 kV, k = 1.21, continuous current rating is equal to 1,200 amperes, and the rated short circuit current at maximum voltage is equal to 19,000 amperes. Could the breaker sustain the rated SCC?

Solution:
The nominal voltage is 72.5 / 1.21 = 60.0 kV

The rated SCC at 60 kV = 1,900 x 1.21 = 23,000 amperes
The rated SCC at 69 kV = (72.5 / 69) x 19,000 = 20,000 amperes

Therefore, the breaker can sustain the rated SCC.

EXAMPLE 2: A 25,000 kVA, 13.8 kV generator with x"d = 15% is connected through a transformer to a bus which supplies four identical motors. The subtransient reactances x"d of each motors is 20% on the base 5,000 kVA, 6.9 kV. The three-phase rating of the transformer is 25,000 kVA, 13.8 / 6.9 kV with a leakage reaktance of 10%. The bus voltage at the motors is 6.9 kV, when a three-phase fault occurs at point P. For the fault specified, determine:
a. the subtransient current in the fault
b. the subtransitent current in breaker A
c. the symmetrical short-circuit interrupting current in the fault and in breaker A.

Solution:
a.) For a base of 25,000 kVA, 13.8 kV in the generator circuit, the base for the motors is 25,000 kVA, 6.9 kV.

Subtransient reactance of each motor: = 1.0 per cent

For a fault in P, V_f = 1.0 per unit, Z_th = j0.125 p.u. = -j 8.0 per unit

The base current in the 6.9 kV is = 2,090 Amperes

I_f = 8 x 2090A = 16,720 Amperes

b) Through breaker A comes the contribution from the genratro and 3 of the 4 motors

The generator contributes: - j4.0 per unit

Through breaker A
I" = -j 4.0 + j (-j1.0) = -j7.0 per unit I" = 7 x 2,090 = 14,630 Amperes

c) For subtransient reactanceof synchronous motors, recommended reactance is x"_d of motor multiplied by 1.5. [150% of rating]