--- title: "Speed, input current and output of 2/3 HP motor" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:two-thirds-hp-motor-performance" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/two-thirds-hp-motor-performance site: BCT Engineering --- # Speed, input current and output of 2/3 HP motor Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2071 Chaitra, 7 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/two-thirds-hp-motor-performance - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question A 2/3 HP, 230 V, 50 Hz, 6-pole single phase induction motor has following parameter: R1 = 3.04 ohm, X1 = 6.2 ohm, X0 = 105.6 ohms, R0 = 85 ohms R2' = 6.26 ohm, X2' = 2.12 ohm, No-load loss = 122 watts. The motor is operating at 4% slip. Determine: (i) Motor speed (ii) Input current and power factor (iii) Output power ## Short answer Speed N = 1000*(1-0.04) = 960 rpm. Using split-phase equivalent circuit, total impedance Z = 28.75+j43.45 ohm, I1 = 230/|Z| = 4.42 A, pf = cos(56.51 deg) = 0.552 lag. Forward and backward air-gap powers give developed mechanical power, subtract no-load loss yields output = 301.7 W. ## Answer Solve using the double revolving field equivalent circuit, with the magnetising and rotor branches split into forward and backward halves. **Assumption:** $X_0 = 105.6\ \Omega$ is taken as the magnetising reactance. Core, friction and windage losses are taken together as the given no-load loss of 122 W, so $R_0$ is not used again (using it would count core loss twice). **Given:** $V = 230$ V, $f = 50$ Hz, $P = 6$, $s = 0.04$, $R_1 = 3.04\ \Omega$, $X_1 = 6.2\ \Omega$, $R_2' = 6.26\ \Omega$, $X_2' = 2.12\ \Omega$. #### (i) Motor speed $$ \begin{aligned} N_s &= \frac{120f}{P} = \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ N &= N_s(1-s) = 1000 \times 0.96 = 960\ \text{rpm} \end{aligned} $$ #### (ii) Input current and power factor Half values: $0.5X_0 = 52.8\ \Omega$, $0.5X_2' = 1.06\ \Omega$, $\dfrac{0.5R_2'}{s} = \dfrac{3.13}{0.04} = 78.25\ \Omega$, $\dfrac{0.5R_2'}{2-s} = \dfrac{3.13}{1.96} = 1.597\ \Omega$. $$ \begin{aligned} Z_F &= \frac{(j52.8)(78.25 + j1.06)}{78.25 + j53.86} = 24.17 + j36.16\ \Omega \\ Z_B &= \frac{(j52.8)(1.597 + j1.06)}{1.597 + j53.86} = 1.533 + j1.085\ \Omega \\ Z &= (3.04 + j6.2) + Z_F + Z_B \\ &= 28.75 + j43.45 = 52.10\angle 56.51^\circ\ \Omega \\ I_1 &= \frac{230}{52.10\angle 56.51^\circ} = 4.415\angle -56.51^\circ\ \text{A} \\ \text{pf} &= \cos 56.51^\circ = 0.552\ \text{lagging} \end{aligned} $$ Input power: $P_{in} = 230 \times 4.415 \times 0.552 = 560.4$ W. #### (iii) Output power $$ \begin{aligned} P_{gF} &= I_1^2 R_F = 4.415^2 \times 24.17 = 471.2\ \text{W} \\ P_{gB} &= I_1^2 R_B = 4.415^2 \times 1.533 = 29.9\ \text{W} \\ P_m &= (1-s)(P_{gF} - P_{gB}) = 0.96 \times 441.3 = 423.7\ \text{W} \\ P_{out} &= P_m - P_{NL} = 423.7 - 122 = 301.7\ \text{W} \end{aligned} $$ Efficiency $= 301.7/560.4 = 53.8\%$ (low, as is common for small single-phase motors with large no-load loss). | Quantity | Value | |---|---| | Speed | 960 rpm | | Input current | 4.42 A | | Power factor | 0.552 lag | | Output power | 301.7 W (0.40 HP) | **Answer:** (i) 960 rpm; (ii) $I_1 \approx 4.42$ A at 0.552 lagging pf; (iii) $P_{out} \approx 302$ W. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 6-pole, 50 Hz motor at 4% slip, the speed is: A. 1000 rpm B. 960 rpm C. 1500 rpm D. 1440 rpm Answer: B. N_s = 120 x 50/6 = 1000 rpm; N = 1000 x 0.96 = 960 rpm. (Key point: Speed) 2. The forward rotor resistance term 0.5R2'/s with R2' = 6.26 ohm and s = 0.04 is: A. 3.13 ohm B. 78.25 ohm C. 1.597 ohm D. 156.5 ohm Answer: B. 3.13/0.04 = 78.25 ohm. (Key point: Half values) 3. Why is R_0 not used again in the solution? A. It is zero B. The 122 W no-load loss already covers core, friction and windage, so using R_0 would double count core loss C. It is in series D. Because s is small Answer: B. Core loss is already included in the given no-load loss. (Key point: Why R_0 not used) 4. Output power is found as: A. P_m = (1-s)(P_gF - P_gB), then subtract the 122 W no-load loss B. P_gF + P_gB C. Input power minus copper loss only D. P_gF - P_gB Answer: A. Mechanical power 423.7 W minus 122 W gives about 301.7 W. 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