1. **State the problem:** We have a synchronous generator and motor with given ratings and sub-transient reactances. The motor is drawing 18,000 kW at 0.85 power factor leading and terminal voltage 11.8 kV. A symmetrical three-phase fault occurs at the motor terminal. We need to find the sub-transient current flowing in both the generator and motor using the internal voltage of the machines. 2. **Given data:** - Generator rating: $S_{G} = 35,000$ kVA, $V_{G} = 13.8$ kV, $X''_{G} = 15\%$ - Motor rating: $S_{M} = 28,000$ kVA, $V_{M} = 12.8$ kV, $X''_{M} = 15\%$ - Transformers reactances: $X''_{T1} = 10\%$ on 28,000 kVA, 15/220 kV and $X''_{T2} = 12\%$ on 35,000 kVA, 230/13 kV - Motor load: $P = 18,000$ kW, power factor $pf = 0.85$ leading, terminal voltage $V_{t} = 11.8$ kV 3. **Base selection and per unit conversion:** Choose a common base for reactances, typically the motor base $S_{base} = 28,000$ kVA and voltage base $V_{base} = 12.8$ kV. 4. **Convert generator reactance to motor base:** $$X''_{G,base} = X''_{G} \times \frac{S_{base}}{S_{G}} = 0.15 \times \frac{28,000}{35,000} = 0.12$$ 5. **Transformers reactances are already given on their respective bases. Convert $X''_{T2}$ to motor base:** $$X''_{T2,base} = X''_{T2} \times \frac{S_{base}}{S_{G}} = 0.12 \times \frac{28,000}{35,000} = 0.096$$ 6. **Total reactance from generator to motor terminal:** Sum reactances in series: $$X_{total} = X''_{G,base} + X''_{T2,base} + X''_{T1} + X''_{M} = 0.12 + 0.096 + 0.10 + 0.15 = 0.466$$ 7. **Calculate motor load current:** Apparent power $S = \frac{P}{pf} = \frac{18,000}{0.85} = 21,176.47$ kVA Motor rated current: $$I_{rated} = \frac{S_{base} \times 10^3}{\sqrt{3} \times V_{base}} = \frac{28,000 \times 10^3}{\sqrt{3} \times 12,800} = 1261.5 \text{ A}$$ Load current in per unit: $$I_{load} = \frac{S}{S_{base}} = \frac{21,176.47}{28,000} = 0.756$$ 8. **Calculate internal voltage of motor $E''$ before fault:** Load angle $\theta = \cos^{-1}(0.85) = 31.79^\circ$ (leading means current leads voltage) Voltage drop across sub-transient reactance: $$V_{drop} = j X''_{M} I_{load} = j 0.15 \times 0.756 = j0.1134$$ Express current as complex number (leading power factor): $$I = I_{load} (\cos(-31.79^\circ) + j \sin(-31.79^\circ)) = 0.756 (0.85 - j0.526) = 0.6426 - j0.3977$$ Voltage drop: $$V_{drop} = j0.15 \times (0.6426 - j0.3977) = j0.0964 + 0.0597 = 0.0597 + j0.0964$$ Internal voltage: $$E'' = V_{t} + V_{drop} = 1 + (0.0597 + j0.0964) = 1.0597 + j0.0964$$ (per unit) Magnitude: $$|E''| = \sqrt{1.0597^2 + 0.0964^2} = 1.064$$ 9. **Calculate sub-transient fault current at motor terminal:** Fault voltage $V_f = 0$ (three-phase fault) Fault current: $$I_{fault} = \frac{E''}{X_{total}} = \frac{1.064}{0.466} = 2.283 \text{ p.u.}$$ Convert to amperes: $$I_{fault,A} = 2.283 \times I_{rated} = 2.283 \times 1261.5 = 2878.5 \text{ A}$$ 10. **Calculate sub-transient current in generator:** Generator base current: $$I_{G,rated} = \frac{35,000 \times 10^3}{\sqrt{3} \times 13,800} = 1463.5 \text{ A}$$ Fault current in generator base: $$I_{fault,G} = I_{fault} \times \frac{S_{base}}{S_{G}} = 2.283 \times \frac{28,000}{35,000} = 1.827 \text{ p.u.}$$ In amperes: $$I_{fault,G,A} = 1.827 \times 1463.5 = 2674.5 \text{ A}$$ **Final answers:** - Sub-transient fault current at motor terminal: $2878.5$ A - Sub-transient fault current in generator: $2674.5$ A