Question 6.9: Consider a 4 kV, 5000 hp four-pole motor, with full-load kVA......
Consider a 4 kV, 5000 hp four-pole motor, with full-load kVA rating = 4200. The following parameters in per unit on motor-base kVA are specified:
r_{1} = 0.0075
r_{2} = 0.0075
X_{1} = 0.0656
X_{2} = 0.0984
R_{m} = 100
X_{m} = 3.00
Calculate the motor short-circuit current equations for a sudden terminal fault.
The following parameters are calculated using Equations 6.128 through 6.132:
X^{\prime }=X_{1}+ \frac{X_{m}X_{2}}{X_{m} \ + \ X_{2}} (6.128)
T^{\prime }_{0}= \frac{X_{2} \ + \ X_{m}}{\omega r_{2}} (6.129)
T^{\prime }= T_{0}^{\prime} \frac{X^{\prime}}{X_{1} \ + \ X_{m}} (6.130)
T^{\prime }= \frac{X^{\prime}}{\omega r_{2}} (6.131)
T_{dc} = \frac{X^{\prime}}{\omega r_{1}} (6.132)
X^{\prime} = 0.608 pu
T^{\prime} = 0.057 s
T_{dc} = 0.057 s
T_{0} = 1.09 s
The ac component of the short-circuit current is
i_{ac} = 6.21 \ e^{-t/0.057}
At t = 0, the ac symmetrical short-circuit current is 6.21 times the full-load current.
The dc component of the short-circuit current is
i_{dc} = 8.79 e^{-t/0.057}
The nature of short-circuit currents is identical to that of synchronous machines; however, the currents decay more rapidly. The decay is a function of the motor rating, inertia, and load–torque characteristics. Typically, the effect of short-circuit currents from induction machines can be ignored after 6 cycles.