--- title: "Current, pf and efficiency of 100 hp motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:motor-100hp-efficiency" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-100hp-efficiency site: BCT Engineering --- # Current, pf and efficiency of 100 hp motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2082 Bhadra (new course), 3 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/motor-100hp-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A 3 phase, 100 hp, 440 V, star connected synchronous motor has a synchronous impedance per phase of 0.1+j1 Ω. The excitation and torque losses are 4 kW and may be assumed constant. Calculate the current, power factor and efficiency when operating at full load with an excitation equivalent to 400 line volts. ## Short answer Ia = 114.5 A, pf = 0.946 lagging, efficiency = 90.4%. ## Answer **Assumptions:** 1 hp = 746 W; "full load" means 100 hp shaft output; the 4 kW excitation and friction losses are supplied by the developed power. $$ \begin{aligned} V &= \frac{440}{\sqrt3} = 254.03\ \text{V},\quad E_f = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ Z_s &= 0.1 + j1 = 1.0050\angle 84.29^\circ\ \Omega \\ P_{dev} &= 74.6 + 4 = 78.6\ \text{kW} = 26.2\ \text{kW per phase} \end{aligned} $$ #### Load angle Mechanical power developed per phase: $P_{dev} = \dfrac{E_fV}{Z_s}\cos(\theta - \delta) - \dfrac{E_f^2}{Z_s}\cos\theta$ $$ \begin{aligned} 26200 &= 58376\cos(84.29^\circ - \delta) - 5281 \\ \cos(84.29^\circ - \delta) &= 0.5393 \Rightarrow 84.29^\circ - \delta = 57.37^\circ \\ \delta &= 26.92^\circ \end{aligned} $$ #### Current and power factor $$ \begin{aligned} \vec I_a &= \frac{\vec V - \vec E_f}{Z_s} = \frac{254.03 - 230.94\angle -26.92^\circ}{1.0050\angle 84.29^\circ} \\ &= 108.30 - j37.30 = 114.54\angle -19.00^\circ\ \text{A} \end{aligned} $$ Power factor $= \cos19.00^\circ = 0.9455$ **lagging** (under-excited, 400 V < 440 V). #### Efficiency $$ \begin{aligned} P_{in} &= \sqrt3 \times 440 \times 114.54 \times 0.9455 = 82.54\ \text{kW} \\ P_{cu} &= 3 \times 114.54^2 \times 0.1 = 3.94\ \text{kW} \quad (\text{check: } 82.54 - 3.94 = 78.6\ \text{kW}) \\ \eta &= \frac{74.6}{82.54} \times 100 = 90.38\ \% \end{aligned} $$ **Answer:** $I_a = 114.5$ A, pf $= 0.946$ lagging, efficiency $= 90.4\ \%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the power developed per phase obtained? A. Output only: 74.6 kW B. Output plus the 4 kW excitation and friction losses = 78.6 kW, i.e. 26.2 kW per phase C. Output minus 4 kW D. Output divided by efficiency Answer: B. The constant losses are supplied by the developed power, so Pdev = 74.6 + 4 = 78.6 kW. (Key point: Developed power) 2. Why is the power factor lagging? A. Excitation of 400 V is below the 440 V supply, so the motor is under-excited B. Because Ra is negligible C. Because efficiency is below 100 percent D. Because the motor is over-excited Answer: A. Ef < V means under-excitation and a lagging pf; the result is 0.946 lagging at Ia = 114.5 A. (Key point: Power factor) 3. How is efficiency obtained? A. Output 74.6 kW divided by input 82.54 kW = 90.4 percent B. Input divided by output C. 78.6/74.6 D. cos(phi) x 100 Answer: A. Input = sqrt3 x 440 x 114.54 x 0.9455 = 82.54 kW, consistent with 78.6 + 3.94 kW copper loss. 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