--- title: "Excitation emf of 6600 V salient pole motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:salient-motor-6600v-excitation" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-6600v-excitation site: BCT Engineering --- # Excitation emf of 6600 V salient pole motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2075 Chaitra, 7 marks - Video: [#machines Question 24: Excitation emf in salient pole synchronous machines. #ee #gate](https://www.youtube.com/watch?v=2rRoVImgOvQ) — Electrical Questions - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-6600v-excitation - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 6600 V, 2 MW, 3 phase star connected synchronous motor has Xd = 5 ohm/phase and Xq = 3.1 ohm/phase. Neglecting all losses, calculate the excitation e.m.f. when the motor supplies rated load at 0.8 p.f. ## Short answer For 6600 V, 2 MW, 0.8 leading pf, star-connected motor, phase voltage = 3810.5 V, armature current = 218.69 A. Using two-reaction theory, excitation e.m.f. = 4542 V per phase (7.87 kV line). ## Answer **Assumption:** the power factor is not stated as lagging or leading; the usual textbook version of this problem takes **0.8 leading** (over-excited motor). The lagging result is given at the end. Star connection, all losses neglected, so input = 2 MW. $$ \begin{aligned} V &= \frac{6600}{\sqrt3} = 3810.51\ \text{V per phase} \\ I_a &= \frac{2\times10^6}{\sqrt3 \times 6600 \times 0.8} = 218.69\ \text{A} \\ \vec I_a &= 218.69\angle 36.87^\circ = 174.95 + j131.22\ \text{A} \end{aligned} $$ **Method (two-reaction theory, motor, $R_a = 0$):** $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$. 1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$. 2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$. 3. $E_f = E' + (X_d - X_q)I_d$. #### Load angle $$ \begin{aligned} jX_q\vec I_a &= j3.1(174.95 + j131.22) = -406.77 + j542.36 \\ \vec E' &= 3810.51 - (-406.77 + j542.36) = 4217.28 - j542.36 \\ &= 4252.0\angle -7.33^\circ\ \text{V} \end{aligned} $$ So $\delta = 7.33^\circ$. #### Excitation emf $$ \begin{aligned} \psi &= \delta + \phi = 7.33^\circ + 36.87^\circ = 44.20^\circ \\ I_d &= 218.69\sin44.20^\circ = 152.46\ \text{A},\quad I_q = 156.79\ \text{A} \\ E_f &= 4252.0 + (5 - 3.1)(152.46) = 4541.7\ \text{V per phase} \end{aligned} $$ Line value $= \sqrt3 \times 4541.7 = 7866$ V. Check: $3\left[\dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right)\sin2\delta\right] = 2.000$ MW. Correct. ```text Ia (leads V by 36.87 deg) / / -----+-------------------> V = 3811 V \ delta = 7.33 deg '----------> Ef = 4542 V ``` **If the pf is 0.8 lagging:** $\vec E' = 3446.7\angle -9.05^\circ$, $\psi = -27.82^\circ$, $I_d = -102.05$ A, $E_f = 3446.7 - 1.9 \times 102.05 = 3252.8$ V per phase (5634 V line). **Answer (0.8 leading):** $E_f = 4542$ V per phase $= 7.87$ kV line, load angle $\delta = 7.33^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The problem does not say lagging or leading. Which case does the main answer take? A. 0.8 lagging B. 0.8 leading C. Unity D. 0.8 lagging, then leading is ignored Answer: B. The usual textbook version uses 0.8 leading (over-excited); the lagging result is given only at the end. (Key point: Power factor assumption) 2. In the two-reaction method, what does E' = V - jXq Ia give? A. The magnitude of Ef directly B. The direction of the q-axis, i.e. the angle of Ef (-delta) C. The direct-axis current Id D. The pull-out torque Answer: B. E' lies along the q-axis, so its angle gives the load angle; Ef is then found by adding (Xd - Xq)Id. (Key point: Direction of Ef from E') 3. How is Ef obtained once E' and Id are known? A. Ef = E' - (Xd - Xq)Id B. Ef = E' + (Xd - Xq)Id C. Ef = E' + Xd Iq D. Ef = V + Xq Id Answer: B. Ef = E' + (Xd - Xq)Id, giving 4252 + 1.9 x 152.46 = 4542 V per phase. (Key point: Final formula for Ef) 4. The motor input is 2 MW at 6600 V and 0.8 pf. What are the phase current and excitation emf found? A. Ia = 218.69 A, Ef = 4542 V per phase B. Ia = 378.8 A, Ef = 4542 V per phase C. Ia = 218.69 A, Ef = 3253 V per phase D. Ia = 175 A, Ef = 7866 V per phase Answer: A. Ia = 2e6/(sqrt3 x 6600 x 0.8) = 218.69 A; for 0.8 leading Ef = 4542 V per phase (7866 V line). 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