--- title: "Two machines 11∠0° and 11∠40° kV via j15 Ω" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:two-machines-11kv-j15" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-11kv-j15 site: BCT Engineering --- # Two machines 11∠0° and 11∠40° kV via j15 Ω Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2078 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-11kv-j15 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Two three-phase machines have generated EMF of 11∠0° kV and 11∠40° kV. They are connected through a 3-phase line having an impedance of 0 + j15 Ω per phase. Find: (i) whether each machine is acting as generator or motor and the real power generated or consumed by them; (ii) whether each machine is delivering or consuming reactive power and the amount of reactive power. ## Short answer With per-phase EMFs 6.351∠0° kV and 6.351∠40° kV, line current I = (E1-E2)/j15 = 289.6∠-160° A. Machine 1 consumes 5.185 MW (motor) and supplies 1.887 Mvar; machine 2 generates 5.185 MW (generator) and also supplies 1.887 Mvar. Total line reactive absorption is 3.774 Mvar. ## Answer Assumption: 11 kV is the line-to-line EMF. The per-phase EMFs are therefore $E_1 = 6.351\angle0^\circ$ kV and $E_2 = 6.351\angle40^\circ$ kV. The current $I$ is taken as flowing from machine 1 to machine 2 through $Z = j15\ \Omega$. #### Current $$ \begin{aligned} I &= \frac{E_1 - E_2}{jX} = \frac{6351\angle0^\circ - 6351\angle40^\circ}{j15}\\ &= 289.6\angle-160^\circ\ \text{A} \end{aligned} $$ #### Complex power (three-phase) Machine 1, with power taken out of its terminal: $$ S_1 = 3E_1I^* = -5.185 + j1.887\ \text{MVA} $$ Machine 2, with power taken into it (since $I$ enters it): $$ S_2 = 3E_2I^* = -5.185 - j1.887\ \text{MVA} $$ Check with the standard formulas: $$ \begin{aligned} P &= \frac{E_1E_2\sin\delta}{X} = \frac{11\times11\times\sin40^\circ}{15} = 5.185\ \text{MW}\\ Q &= \frac{E^2(1-\cos\delta)}{X} = \frac{121(1-\cos40^\circ)}{15} = 1.887\ \text{Mvar} \end{aligned} $$ #### (i) Real power - Machine 1: $P_1 = -5.185$ MW out of it, so it **consumes 5.185 MW** and acts as a **motor**. - Machine 2: $P_{2,in} = -5.185$ MW, so it **generates 5.185 MW** and acts as a **generator**. Machine 2's EMF leads by 40°, so real power flows from 2 to 1. The line is lossless ($R = 0$), so the two powers are equal. #### (ii) Reactive power - Machine 1: $Q_1 = +1.887$ Mvar out of it, so it **delivers 1.887 Mvar**. - Machine 2: $Q_{2,in} = -1.887$ Mvar, so it also **delivers 1.887 Mvar**. Both machines supply reactive power to the line, which absorbs: $$ 3|I|^2X = 3\times289.6^2\times15 = 3.774\ \text{Mvar} $$ **Answer:** Machine 1 is a motor consuming 5.185 MW, and machine 2 is a generator producing 5.185 MW. Each machine supplies 1.887 Mvar, a total of 3.774 Mvar absorbed by the line reactance. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The given 11 kV EMFs are line-to-line. What per-phase EMF is used in the calculation? A. 11 kV B. 11 x sqrt(3) kV C. 11/sqrt(3) = 6.351 kV D. 11/3 kV Answer: C. Phase value = line value / sqrt(3), so 11/sqrt(3) = 6.351 kV. (Key point: Per-phase EMF) 2. Machine 2's EMF leads machine 1's by 40°. Which machine is the generator? A. Machine 1, because it has the reference angle B. Machine 2, because its leading EMF sends real power out C. Both act as generators D. Neither; both are motors since R = 0 Answer: B. Real power flows from the machine with the leading angle to the lagging one, so machine 2 generates and machine 1 is a motor. (Key point: Direction of real power) 3. The EMF magnitudes are equal. How do the machines share reactive power? A. Machine 1 supplies it and machine 2 absorbs it B. Machine 2 supplies it and machine 1 absorbs it C. Both absorb reactive power from the line D. Each supplies 1.887 Mvar, which the line reactance absorbs Answer: D. With equal magnitudes and an angle between them, both supply Q = E^2(1-cos δ)/X = 1.887 Mvar and the line absorbs 3.774 Mvar. (Key point: Reactive power) 4. Which formula gives the real power transferred over the lossless line? A. P = E1E2 sinδ / X B. P = E1E2 cosδ / X C. P = E^2 (1 - cosδ) / X D. P = 3|I|^2 X Answer: A. For a purely reactive line P = E1E2 sinδ/X = 11x11xsin40°/15 = 5.185 MW. 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