# Fault Analysis of a 3-Phase Star Connected Generator Given Data: - **Rated Power, $S_{rated}$:** 12.75 MW - **Power factor:** .85 lagging - **Line Voltage, $V_{LL}$:** 6.6 kV - **Connection:** Star (Y) - **Positive sequence reactance, $X_1$:** 58.08 Ω - **Negative sequence reactance, $X_2$:** 58.08 Ω - **Zero sequence reactance, $X_$:** .1 pu - **Neutral grounding resistance, $R_n$:** 5% of base value - **Base MVA, $S_{base}$:** 4 MVA --- ## 1. **Base Values Calculation** ### a. **Base Voltage (phase-neutral):** \[ V_{base,ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{660}{\sqrt{3}} \approx 381\, V \] ### b. **Base Current:** \[ I_{base} = \frac{S_{base}}{\sqrt{3} V_{LL}} = \frac{4 \times 10^6}{\sqrt{3} \times 660} \approx 350\, A \] ### c. **Base Impedance:** \[ Z_{base} = \frac{V_{base,ph}^2}{S_{base,ph}} \] But for per-phase calculations, \[ Z_{base,ph} = \frac{V_{base,ph}}{I_{base,ph}} \] Since $I_{base}$ calculated above is per phase, we use: \[ Z_{base,ph} = \frac{381}{350} \approx 10.89\, \Omega \] --- ## 2. **Per Unit Reactances** ### a. **Convert Given Reactances to Per Unit:** \[ X_{1,pu} = X_{2,pu} = \frac{58.08}{10.89} \approx 5.335 \] \[ X_{,pu} = .1\, (already\, per\, unit) \] --- ## 3. **Neutral Grounding Resistance (per unit)** \[ R_n = .05 \times Z_{base,ph} = .05 \times 10.89 = .5445\, \Omega \] \[ R_{n,pu} = \frac{.5445}{10.89} = .05\, pu \] --- # A. **Single Line to Ground (SLG) Fault Current** For an SLG fault at the generator terminal (assuming prefault voltage $E=1$ pu): \[ I_f = \frac{3 E}{Z_1 + Z_2 + Z_ + 3 Z_n} \] Where: - $Z_1, Z_2, Z_$ are per unit positive, negative, zero sequence impedances. - $Z_n$ is the per unit neutral grounding resistance. Substitute values: - $Z_1 = Z_2 = j5.335$ - $Z_ = j.1$ - $Z_n = .05$ So, \[ I_f = \frac{3 \times 1}{j5.335 + j5.335 + j.1 + 3 \times .05} = \frac{3}{j10.77 + .15} \] Express denominator in complex form: \[ Z_{total} = .15 + j10.77 \] \[ |Z_{total}| = \sqrt{.15^2 + 10.77^2} \approx 10.77 \] \[ \theta = \arctan\left(\frac{10.77}{.15}\right) \approx 89.2^\circ \] \[ I_f = \frac{3}{.15 + j10.77} = \frac{3(.15 - j10.77)}{(.15)^2 + (10.77)^2} \] \[ = \frac{3 \times .15 - j3 \times 10.77}{.0225 + 116.08} = \frac{.45 - j32.31}{116.1025} \] \[ = .00387 - j.2784\, pu \] \[ |I_f| = \sqrt{(.00387)^2 + (.2784)^2} \approx .2784\, pu \] ### **Actual Fault Current:** \[ I_{f,actual} = |I_f| \times I_{base} = .2784 \times 350 \approx 97.4\, A \] --- # B. **Double Line to Ground Fault (LLG):** For a double line to ground fault (say, phases B and C to ground): \[ I_B = I_C = \frac{E (Z_ + Z_2 + 3 Z_n)}{(Z_1 Z_2 + Z_2 Z_ + Z_ Z_1) + 3 Z_n (Z_1 + Z_2)} \] - $E = 1\, pu$ First, calculate numerator: \[ Z_ + Z_2 + 3Z_n = j.1 + j5.335 + .15 = .15 + j5.435 \] Now, denominator: \[ Z_1 Z_2 = (j5.335)^2 = -28.46 \] \[ Z_2 Z_ = (j5.335) \times (j.1) = -.5335 \] \[ Z_ Z_1 = (j.1) \times (j5.335) = -.5335 \] \[ \Rightarrow Z_1 Z_2 + Z_2 Z_ + Z_ Z_1 = -28.46 - .5335 - .5335 = -29.527 \] \[ 3Z_n (Z_1 + Z_2) = 3 \times .05 \times (j5.335 + j5.335) = .15 \times j10.67 = j1.6 \] \[ Denominator = -29.527 + j1.6 \] Now, plug values into expression for $I_B$ (remember, $I_B$ and $I_C$ are equal): \[ I_B = \frac{.15 + j5.435}{-29.527 + j1.6} \] Convert numerator and denominator to polar form: - Numerator: $|N| = \sqrt{.15^2 + 5.435^2} \approx 5.44$, angle $\theta_N = \arctan(\frac{5.435}{.15}) \approx 88.4^\circ$ - Denominator: $|D| = \sqrt{29.527^2 + 1.6^2} \approx 29.57$, angle $\theta_D = \arctan(\frac{1.6}{-29.527}) \approx -3.1^\circ$ So, \[ |I_B| = \frac{5.44}{29.57} \approx .184 \] \[ \text{Angle} = 88.4^\circ - (-3.1^\circ) = 91.5^\circ \] \[ I_B = .184 \angle 91.5^\circ\, pu \] ### **Actual Line Current $I_B$:** \[ I_{B,actual} = .184 \times 350 \approx 64.4\, A \] --- # **Summary Table** | Fault Type | Per Unit Fault Current | Actual Current (A) | |------------------------------------|:---------------------:|:------------------:| | Single Line to Ground (SLG) | .278 | 97.4 | | Double Line to Ground (LLG), $I_B$ | .184 | 64.4 | --- ## **Notes** - Results are rounded to 3 significant digits. - All calculations use per-phase analysis with base values derived from 4 MVA and 6.6 kV (line-to-line). - For more detail, include angles if required for phase relationships. --- ## **References** - W. H. Stevenson, **Elements of Power System Analysis** - P. Kundur, **Power System Stability and Control** --- **[Image: Sequence networks for SLG and LLG faults](https://i.imgur.com/fault_networks.png)** *Alt text: Sequence network diagrams for single line to ground and double line to ground faults*