--- title: "Power transfer and pf of motor on infinite bus" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-infinite-bus-power-transfer" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-infinite-bus-power-transfer site: BCT Engineering --- # Power transfer and pf of motor on infinite bus Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2071 Shrawan, 6 marks - Video: [EM || Example 6.3 (4th ed) || Synchronous Motors || The infinite bus](https://www.youtube.com/watch?v=F7bq8JR0zVk) — Electrical Engineering Academy - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-infinite-bus-power-transfer - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 3 phase, 10 MVA, 2300 V, 60 Hz synchronous motor has Xs = 0.9 pu and negligible stator resistance. The motor is connected to infinite bus. If terminal voltage, Vt = 2300∠0 V and excitation emf Ef = 3450∠120 V [printed "3450V<120"]. Determine power transfer and power factor of machine. Also draw phasor diagram. ## Short answer With Ef = 1.5∠120° pu, power transfer = 1.443 pu = 14.43 MW delivered to bus (generating); power factor = 0.596 leading. If Ef angle is -120°, motor absorbs 14.43 MW at 0.596 lagging. ## Answer **Reading of data:** $V_t = 2300$ V and $E_f = 3450$ V are taken as line values (the rating is 2300 V line), with $E_f$ at $+120^\circ$ as printed. Work in per unit on 10 MVA, 2300 V. $$ \begin{aligned} V_t &= \frac{2300}{2300} = 1\angle0^\circ\ \text{pu},\quad E_f = \frac{3450}{2300}\angle120^\circ = 1.5\angle120^\circ = -0.75 + j1.299\ \text{pu} \\ Z_{base} &= \frac{2300^2}{10\times10^6} = 0.529\ \Omega,\quad X_s = 0.9 \times 0.529 = 0.4761\ \Omega \\ I_{base} &= \frac{10\times10^6}{\sqrt3 \times 2300} = 2510.2\ \text{A} \end{aligned} $$ #### Power transfer $$ P = \frac{V_tE_f}{X_s}\sin\delta = \frac{1 \times 1.5}{0.9}\sin120^\circ = 1.4434\ \text{pu} = 14.43\ \text{MW} $$ #### Current and power factor Using the motor equation $\vec V_t = \vec E_f + jX_s\vec I_a$: $$ \begin{aligned} \vec I_a &= \frac{\vec V_t - \vec E_f}{jX_s} = \frac{1.75 - j1.299}{j0.9} = -1.4434 - j1.9444 \\ &= 2.4216\angle -126.59^\circ\ \text{pu} = 6079\ \text{A} \\ \vec S &= \vec V_t\vec I_a^{*} = -1.4434 + j1.9444\ \text{pu} \end{aligned} $$ The real part is **negative**: since $E_f$ **leads** $V_t$, the machine is not taking power; it is **delivering 14.43 MW** to the bus, i.e. acting as a generator. Taking the generated current $-\vec I_a = 2.4216\angle 53.41^\circ$ pu, it leads $V_t$ by $53.41^\circ$: $$ \text{pf} = \cos53.41^\circ = 0.596\ \text{leading} $$ ```text Ef (1.5 pu) \ \ at 120 deg from Vt \ O-----------> Vt (1.0 pu) \ \ Ia (motor convention) v at -126.6 deg jXs.Ia = Vt - Ef joins tip of Ef to tip of Vt Ia(gen) = -Ia leads Vt by 53.4 deg ``` **Note:** with $\delta = 120^\circ$ the machine is beyond the steady-state limit ($90^\circ$) and the current (2.42 pu) is far above rating, so this point is not a stable working point. If the angle was meant as $-120^\circ$ (motor action, $E_f$ lagging), the same magnitudes result: the motor **absorbs 14.43 MW** at a pf of **0.596 lagging**. **Answer:** Power transferred $= 1.443$ pu $= 14.43$ MW; power factor $= 0.596$ (leading as printed, i.e. generating; lagging if $E_f$ is at $-120^\circ$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the power transferred here? A. P = Vt Ef sin(delta)/Xs = 1.5/0.9 x sin 120 = 1.443 pu B. P = Vt Ef cos(delta)/Xs C. P = Vt^2/Xs D. P = 3 Vt Ia Answer: A. P = (1 x 1.5/0.9) sin120 degrees = 1.443 pu = 14.43 MW. (Key point: Power transfer formula) 2. Because Ef leads Vt (delta = +120 degrees), what is the machine doing? A. Absorbing 14.43 MW as a motor B. Delivering power to the bus, acting as a generator C. Running as a synchronous condenser with zero power D. Supplying no reactive power Answer: B. The real part of S = Vt Ia* is negative with the motor convention, so power flows to the bus. (Key point: Direction of power flow) 3. Why is this not a stable working point? A. Because Vt is too large B. Because delta = 120 degrees exceeds the 90 degree steady-state limit and the current (2.42 pu) is far above rating C. Because Xs is zero D. Because the frequency is 60 Hz Answer: B. Beyond 90 degrees the steady-state stability limit is passed. 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