Question 6.1: Consider a 13.8 kV, 100 MVA 0.85 power factor generator. Its......
Consider a 13.8 kV, 100 MVA 0.85 power factor generator. Its rated full-load current is 4184 A. Other data are
Saturated subtransient reactance X^{\prime \prime }_{dv} = 0.15 per unit
Saturated transient reactance X^{\prime }_{dv} = 0.2 per unit
Synchronous reactance X_{d} = 2.0 per unit
Field current at rated load i_{f} = 3 per unit
Field current at no-load rated voltage i_{fg} = 1 per unit
Subtransient short circuit time constant T^{\prime \prime }_{d} = 0.012 s
Transient short circuit time constant T^{\prime }_{d} = 0.35 s
Armature short circuit time constant T_{a} = 0.15 s
Effective resistance* = 0.0012 per unit
Quadrature axis synchronous reactance* = 1.8 per unit
A three-phase short circuit occurs at the terminals of the generator, when it is operating at its rated load and power factor. It is required to construct a fault decrement curve of the generator for (1) the ac component, (2) dc component, and (3) total current. Data marked with an asterisk are intended, e.g., 6.5.
From Equation 6.15, the voltage behind subtransient reactance at the generator rated voltage, load, and power factor is
E^{\prime \prime } = V+ X^{\prime \prime }_{d} \sin \phi = 1+ (0.15)(0.527) = 1.079 \ puFrom Equation 6.14, the subtransient component of the current is
i^{\prime \prime }_{d} =\frac{E^{\prime \prime }}{X^{\prime \prime }_{dv}}=\frac{1.079}{0.15} per unit = 30.10 kA
Similarly, from Equation 6.17,
E^{\prime } = V_{a}+ X^{\prime }_{d} \sin \phi (6.17)
E^{\prime } , the voltage behind transient reactance is 1.1054 per unit and, from Equation 6.16,
i^{\prime }_{d} =\frac{E^{\prime }}{X^{\prime }_{d}} (6.16)
the transient component of the current is 23.12 kA.
From Equation 6.18,
i_{d} =\frac{V_{a}}{X_{d}}(\frac{i_{F}}{i_{Fg}}) (6.18)
current i_{d} at constant excitation is 2.09 kA rms. For a ratio of i_{f}/i_{Fg} = 3, current i_{d} = 6.28 kA rms. Therefore, the following equations can be written for the ac component of the current:
With constant excitation:
i_{ac} = 6.98^{-t/0.012} + 20.03e^{-t/0.35} + 2.09 \ kA
With full load excitation:
i_{ac} = 6.98e^{-t/0.012} + 16.84e^{-t/0.35} + 6.28 \ kA
The ac decaying component of the current can be plotted from these two equations, with the lowest value of t = 0.01–1000 s. This is shown in Figure 6.5. The dc component is given by Equation 6.19:
i_{dc}= \sqrt{2} i^{\prime \prime }_{d} e^{-t/T_{a}}=42.57 \ e^{-t/0.15} \ kA
This is also shown in Figure 6.6. At any instant, the total current is
\sqrt{i^{2}_{ac} + i^{2}_{dc}} kA rms
The fault decrement curves are shown in Figure 6.6. Short-circuit current with constant excitation is 50% of the generator full-load current. This can occur for a stuck voltage regulator condition. Though this current is lower than the generator full-load current, it cannot be allowed to be sustained. Voltage restraint or voltage-controlled overcurrent generator backup relays (ANSI=IEEE device number 51V) or distance relays (device 21) are set to pick up on this current. The generator fault decrement curve is often required for appropriate setting and coordination of these relays with the system relays.
