--- title: "Open circuit emf and mechanical power of 75 kW motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-75kw-emf-mechanical-power" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-75kw-emf-mechanical-power site: BCT Engineering --- # Open circuit emf and mechanical power of 75 kW motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2082 Baisakh, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-75kw-emf-mechanical-power - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 75 kW, 400 V, 4-pole, 3-phase star connected synchronous motor has a resistance and synchronous reactance per phase of 0.04 Ω and 0.4 Ω respectively. Compute for full load 0.8 pf lead the open circuit emf per phase and mechanical power developed. Assume an efficiency of 92.5%. ## Short answer Input power = 75 kW / 0.925 = 81.08 kW. Per-phase V=230.94 V, Ia=146.29 A at 0.8 lead. Using Eb = V - Ia Zs gives Eb = 266.2 V/phase (461 V line), δ = 10.9°. Mechanical power developed = Pin - 3I²R = 78.51 kW. ## Answer **Given:** $P_{out} = 75$ kW, $V_L = 400$ V (star), $R_a = 0.04\ \Omega$, $X_s = 0.4\ \Omega$, pf = 0.8 leading, $\eta = 92.5\%$. #### Input power and current $$ \begin{aligned} P_{in} &= \frac{P_{out}}{\eta} = \frac{75}{0.925} = 81.081\ \text{kW} \\ I_a &= \frac{P_{in}}{\sqrt3\,V_L\cos\phi} = \frac{81081}{\sqrt3 \times 400 \times 0.8} = 146.29\ \text{A} \\ V &= \frac{400}{\sqrt3} = 230.94\ \text{V per phase} \end{aligned} $$ #### Open-circuit (excitation) emf For a motor $\vec V = \vec E_f + \vec I_a Z_s$, so $\vec E_f = \vec V - \vec I_a Z_s$. Take $V$ as reference; leading current is at $+36.87^\circ$. $$ \begin{aligned} Z_s &= 0.04 + j0.4 = 0.402\angle 84.29^\circ\ \Omega \\ \vec I_a &= 146.29\angle 36.87^\circ = 117.03 + j87.77\ \text{A} \\ \vec I_a Z_s &= 58.81\angle 121.16^\circ = -30.43 + j50.32\ \text{V} \\ \vec E_f &= 230.94 - (-30.43 + j50.32) \\ &= 261.37 - j50.32 \\ &= 266.17\angle -10.90^\circ\ \text{V} \end{aligned} $$ Line value: $\sqrt3 \times 266.17 = 461.0$ V. As expected for leading pf, $E_f > V$ (over-excited), and the load angle is $\delta = 10.90^\circ$. #### Mechanical power developed $$ \begin{aligned} P_{cu} &= 3I_a^2R_a = 3 \times 146.29^2 \times 0.04 = 2568\ \text{W} \\ P_{mech} &= P_{in} - P_{cu} = 81081 - 2568 = 78513\ \text{W} \end{aligned} $$ Check: $3\,\text{Re}(\vec E_f \vec I_a^{*}) = 78.51$ kW, the same value. (The difference between 78.51 kW and the 75 kW output is the friction, windage, core and stray loss.) **Answer:** $E_f = 266.2$ V per phase (461.0 V line), load angle $10.9^\circ$; mechanical power developed $= 78.51$ kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the input power found from the 75 kW output? A. Pin = 75 x 0.925 B. Pin = 75/0.925 = 81.08 kW C. Pin = 75 + 0.925 D. Pin = 75/0.8 Answer: B. Efficiency = output/input, so Pin = 75/0.925 = 81.08 kW. (Key point: Input power) 2. What is the full-load armature current? A. 146.3 A B. 117.0 A C. 84.5 A D. 75.0 A Answer: A. Ia = 81081/(sqrt3 x 400 x 0.8) = 146.29 A. (Key point: Armature current) 3. For 0.8 leading, how is Ef obtained, and what is its size relative to V? A. Ef = V + Ia Zs; Ef < V B. Ef = V - Ia Zs with Ia at +36.87 degrees; Ef = 266.2 V > V (over-excited) C. Ef = V - Ia Zs; Ef < V D. Ef = Ia Zs only Answer: B. For a motor Ef = V - Ia Zs; with leading current Ef = 266.17 V per phase, larger than V = 230.94 V. (Key point: Leading pf phasor) 4. How is the mechanical power developed obtained and what is it? A. Pin - 3 Ia^2 Ra = 78.51 kW B. Pout = 75 kW C. Pin + copper loss = 83.6 kW D. Pin x pf = 64.9 kW Answer: A. Mechanical power developed is input minus armature copper loss: 81081 - 2568 = 78513 W; friction, windage and core losses account for the rest to 75 kW. 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