Question 6.7: Calculate the component short-circuit currents at the instan......
Calculate the component short-circuit currents at the instant of three-phase terminal short circuit of the generator (particulars as shown in Table 6.1). Assume that phase a is aligned with the field at the instant of short circuit, maximum asymmetry, i.e., δ = 0. The generator is operating at no load prior to short circuit.
| TABLE 6.1 Generator Data |
|||
| Description | Symbol | Data | |
| Generator 112.1 MVA, 2-pole, 13.8 kV, 0.85 PF, 95.285 MW, 4690 A, 0.56 SCR, 235 field V, wye connected |
|||
| Per unit reactance data, direct axis | |||
| Saturated synchronous | X_{dv} | 1.949 | |
| Unsaturated synchronous | X_{d} | 1.949 | |
| Saturated transient | X_{dv}^{\prime} | 0.207 | |
| Unsaturated transient | X_{d}^{\prime} | 0.278 | |
| Saturated subtransient | X_{dv}^{\prime \prime} | 0.164 | |
| Unsaturated subtransient | X_{d}^{\prime \prime} | 0.193 | |
| Saturated negative sequence | X_{2v} | 0.137 | |
| Unsaturated negative sequence | X_{2I} | 0.185 | |
| Saturated zero sequence | X_{0v} | 0.092 | |
| Leakage reactance, overexcited | X_{0I} | 0.111 | |
| Leakage reactance, underexcited | X_{LM,OXE} X_{LM,UXE} |
0.164 0.164 |
|
| Per unit reactance data, quadrature axis | |||
| Saturated synchronous | X_{qv} | 1.858 | |
| Unsaturated synchronous | X_{q} | 1.858 | |
| Unsaturated transient | X_{q}^{\prime} | 0.434 | |
| Saturated subtransient | X_{qv}^{\prime \prime} | 0.140 | |
| Unsaturated subtransient | X_{q}^{\prime \prime} | 0.192 | |
| Field time constant data, direct axis | |||
| Open circuit | T_{d0}^{\prime} | 5.615 | |
| Three-phase short-circuit transient | T_{d3}^{\prime} | 0.597 | |
| Line-to-line short-circuit transient | T_{d2}^{\prime} | 0.927 | |
| Line-to-neutral short-circuit transient | T_{d1}^{\prime} | 1.124 | |
| Short-circuit subtransient | T_{d}^{\prime \prime} | 0.015 | |
| Open circuit subtransient | T_{d0}^{\prime \prime} | 0.022 | |
| Field time constant data quadrature axis | |||
| Open circuit | T_{q0}^{\prime} | 0.451 | |
| Three-phase short-circuit transient | T_{q}^{\prime} | 0.451 | |
| Short-circuit subtransient | T_{q}^{\prime \prime} | 0.015 | |
| Open circuit subtransient | T_{q0}^{\prime \prime} | 0.046 | |
| Armature dc component time constant data | |||
| Three-phase short circuit | T_{a3} | 0.330 | |
| Line-to-line short circuit | T_{a2} | 0.330 | |
| Line-to-neutral short circuit | T_{a1} | 0.294 | |
The calculations are performed by substituting the required numerical data from Table 6.1 into (6.113):
i_{a} = \sqrt{2} E [(\frac{1}{X_{d}}) \sin (ωt+δ)+(\frac{1}{X^{\prime}_{d}}-\frac{1}{X_{d}})e^{-t/T^{\prime}_{d}} \sin (ωt+δ) + (\frac{1}{X^{\prime \prime}_{d}}-\frac{1}{X_{d}^{\prime}})e^{-t/T^{\prime \prime}_{d}} \sin (ωt+δ) – \frac{(X^{\prime \prime}_{d} \ + \ X^{\prime \prime}_{q})}{2X^{\prime \prime}_{d}X^{\prime \prime}_{q}}e^{-t/T_{a}} \sin δ – \frac{(X^{\prime \prime}_{d} \ – \ X^{\prime \prime}_{q})}{2X^{\prime \prime}_{d}X^{\prime \prime}_{q}}e^{-t/T_{a}} \sin (2ωt+δ)] (6.113)
Steady-state current = 2.41 kA rms
Decaying transient current = 20.24 kA rms
Decaying subtransient current = 5.95 kA rms
Decaying DC component = 43.95 kA
Decaying second-harmonic component = 2.35 kA rms
Note that the second-harmonic component is zero if the direct axis and quadrature axis subtransient reactances are equal. Also, the dc component in this case is 40.44 kA.