--- title: "Excitation emf and maximum power of 4 kVA motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:salient-motor-4kva-max-power" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-4kva-max-power site: BCT Engineering --- # Excitation emf and maximum power of 4 kVA motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2076 Chaitra, 6 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-4kva-max-power - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 4 kVA, 110 V, 50 Hz, 3 phase star connected synchronous motor has Xd = 3 ohm/phase and Xq = 2 ohm/phase, when the motor is delivering full load at 0.8 pf lagging at rated voltage. Calculate the excitation emf, load angle and maximum power that motor can develop. ## Short answer For the star-connected motor, Vph=63.51 V, Ia=20.99 A at 0.8 lagging. Using two-reaction theory, E'=50.96∠-41.24° V, so δ=41.24°; ψ=4.37°, Id=1.600 A, Ef=52.56 V/phase (91.0 V line). The maximum power for this excitation is found by dP/dδ=0, giving Pmax=3.80 kW at δm=66.1°. ## Answer **Given:** 4 kVA, 110 V, star, $X_d = 3\ \Omega$, $X_q = 2\ \Omega$, full load at 0.8 lagging, $R_a$ neglected. $$ \begin{aligned} V &= \frac{110}{\sqrt3} = 63.51\ \text{V},\quad I_a = \frac{4000}{\sqrt3 \times 110} = 20.99\ \text{A} \\ \vec I_a &= 20.99\angle -36.87^\circ = 16.80 - j12.60\ \text{A} \end{aligned} $$ **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 $$ \begin{aligned} jX_q\vec I_a &= j2(16.80 - j12.60) = 25.19 + j33.59 \\ \vec E' &= 63.51 - 25.19 - j33.59 = 38.32 - j33.59 = 50.96\angle -41.24^\circ\ \text{V} \end{aligned} $$ Load angle $\delta = 41.24^\circ$. #### Excitation emf $$ \begin{aligned} \psi &= \delta - \phi = 41.24^\circ - 36.87^\circ = 4.37^\circ \\ I_d &= 20.99\sin4.37^\circ = 1.600\ \text{A},\quad I_q = 20.99\cos4.37^\circ = 20.93\ \text{A} \\ E_f &= 50.96 + (3 - 2)(1.600) = 52.56\ \text{V per phase} \end{aligned} $$ Line value $= \sqrt3 \times 52.56 = 91.03$ V. #### Maximum power for this excitation $$ P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right] = 3\left[1112.57\sin\delta + 336.11\sin2\delta\right] $$ Check at $\delta = 41.24^\circ$: $3(733.5 + 334.2) = 3200$ W $= 4000 \times 0.8$. Correct. Setting $dP/d\delta = 0$ with $a = 1112.57$, $b = V^2(1/X_q - 1/X_d) = 672.22$: $$ \begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = \frac{-1112.57 + \sqrt{1112.57^2 + 8(672.22)^2}}{4(672.22)} = 0.4055 \\ \delta_m &= 66.08^\circ \\ P_{max} &= 3[1112.57\sin66.08^\circ + 336.11\sin132.15^\circ] \\ &= 3[1017.0 + 249.2] = 3798\ \text{W} \end{aligned} $$ **Answer:** $E_f = 52.56$ V per phase (91.0 V line), load angle $= 41.24^\circ$, maximum power $\approx 3.80$ kW at $\delta = 66.1^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are phase voltage and full-load current for the 4 kVA, 110 V star motor? A. 63.51 V and 20.99 A B. 110 V and 36.4 A C. 63.51 V and 36.4 A D. 110 V and 20.99 A Answer: A. V = 110/sqrt3 = 63.51 V; Ia = 4000/(sqrt3 x 110) = 20.99 A. (Key point: Full-load current) 2. How is the load angle obtained and what is it? A. From E' = V - jXq Ia: 41.24 degrees B. From the pf alone: 36.87 degrees C. From E' = V - jXd Ia: 20 degrees D. From pf: 53.13 degrees Answer: A. E' = 38.32 - j33.59 = 50.96 at -41.24 degrees, so delta = 41.24 degrees. (Key point: Load angle) 3. What is Ef per phase? A. 52.56 V B. 50.96 V C. 91.03 V D. 63.51 V Answer: A. Ef = E' + (Xd - Xq) Id = 50.96 + 1 x 1.600 = 52.56 V; 91.03 V is the line value. (Key point: Excitation emf) 4. How is maximum power found and what is it? A. Set dP/d(delta)=0 for the salient expression: delta_m = 66.08 degrees, Pmax about 3.80 kW B. Put delta = 90 degrees in the full-load power: 4 kW C. Pmax = 3 V^2/Xd D. Pmax = full-load power 3.2 kW Answer: A. For a salient motor the maximum is at delta below 90 degrees, found by solving dP/d(delta) = 0, giving 3798 W. 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