--- title: "Performance of 7.5 kW induction motor from equivalent circuit" subject: "Electrical Machines I" chapter: 5 key: "semester_4/electrical_machines_i:im-400v-7p5kw-equivalent-circuit-slip2" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-400v-7p5kw-equivalent-circuit-slip2 site: BCT Engineering --- # Performance of 7.5 kW induction motor from equivalent circuit Electrical Machines I · Chapter 5: Three-Phase Induction Machines - Asked in: 2080 Chaitra, 2+2+2+2 marks - Video: [(English)EM || Example 7.3 (4ed) || Equivalent Circuit of Induction Motor's || The Ultimate Guide](https://www.youtube.com/watch?v=VZdshRtBWXE) — Electrical Engineering Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/im-400v-7p5kw-equivalent-circuit-slip2 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-5 ## Question A three phase Y-connected 400 V, 7.5 kW, 50 Hz, 6 pole induction motor has following parameter value in Ω/phase referred to the stator: R1 = 0.29 Ω, R2 = 0.14 Ω, X1 = 0.503 Ω, X2 = 0.209 Ω, Xm = 13.25 Ω. The total friction, windage and core losses may be assumed to be constant at 405 W, independent of load. For a slip of 2 %, calculate the speed, output torque and power, stator current, power factor and efficiency when the motor is operated at rated voltage and frequency. ## Short answer Using the exact per‑phase equivalent circuit with R2/s = 7 Ω, stator current is 34.88 A at 0.850 lagging pf. Output power is 18.69 kW, torque 182.1 N·m at 980 rpm, and efficiency 90.98%. ## Answer Use the exact per-phase equivalent circuit (magnetizing reactance in parallel with the rotor branch; core loss is included in the 405 W constant loss). ```text I1 R1=0.29 jX1=0.503 jX2=0.209 o--[====]---[~~~~~]---+-----[~~~~~]----+ | | V1=230.94 V jXm=13.25 R2/s = 7.0 | | o---------------------+----------------+ ``` **Given:** $V_L = 400$ V (Y), $f = 50$ Hz, $P = 6$, $s = 0.02$. #### Speed $$ \begin{aligned} N_s &= \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ N &= (1 - s)N_s = 0.98 \times 1000 = 980\ \text{rpm} \end{aligned} $$ #### Impedances $$ \begin{aligned} V_1 &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ Z_2 &= \frac{R_2}{s} + jX_2 = \frac{0.14}{0.02} + j0.209 = 7 + j0.209\ \Omega \\ Z_p &= \frac{jX_m\, Z_2}{jX_m + Z_2} = \frac{j13.25 (7 + j0.209)}{7 + j13.459} = 5.340 + j2.983\ \Omega \\ Z_{in} &= (0.29 + j0.503) + Z_p = 5.630 + j3.486 = 6.622\angle 31.77^\circ\ \Omega \end{aligned} $$ #### Stator current and power factor $$ \begin{aligned} I_1 &= \frac{V_1}{Z_{in}} = \frac{230.94}{6.622} = 34.88\ \text{A}\ (\angle -31.77^\circ) \\ \text{pf} &= \cos 31.77^\circ = 0.850\ \text{lagging} \end{aligned} $$ #### Powers Air-gap power (all power into $Z_p$ goes to $R_2/s$): $$ \begin{aligned} P_{ag} &= 3 I_1^2 R_p = 3 \times 34.876^2 \times 5.340 = 19\,485\ \text{W} \\ P_{m} &= (1 - s)P_{ag} = 0.98 \times 19\,485 = 19\,096\ \text{W} \\ P_{out} &= P_m - 405 = 18\,691\ \text{W} \approx 18.69\ \text{kW} \end{aligned} $$ #### Output torque $$ \begin{aligned} \omega &= \frac{2\pi \times 980}{60} = 102.63\ \text{rad/s} \\ T_{out} &= \frac{P_{out}}{\omega} = \frac{18\,691}{102.63} = 182.1\ \text{N·m} \end{aligned} $$ #### Efficiency $$ \begin{aligned} P_{in} &= 3V_1 I_1\cos\phi = 3 \times 230.94 \times 34.876 \times 0.8502 = 20\,543\ \text{W} \\ \eta &= \frac{18\,691}{20\,543} = 0.9098 = 90.98\% \end{aligned} $$ Check of losses: stator Cu $= 3 \times 34.876^2 \times 0.29 = 1058$ W; rotor Cu $= sP_{ag} = 390$ W; fixed $= 405$ W; total $= 1853$ W $= 20\,543 - 18\,691$. (checks) | Quantity | Value | |---|---| | Speed | 980 rpm | | Output power | 18.69 kW | | Output torque | 182.1 N·m | | Stator current | 34.88 A | | Power factor | 0.850 lag | | Efficiency | 90.98% | Note: at 400 V the computed output at 2% slip exceeds the 7.5 kW rating; these are the values the given circuit data produce. **Answer: N = 980 rpm, $T_{out}$ = 182.1 N·m, $P_{out}$ = 18.69 kW, $I_1$ = 34.88 A, pf = 0.850 lagging, η = 90.98%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the exact per-phase equivalent circuit used here, how is the magnetizing reactance Xm connected? A. In parallel with the rotor branch (R2/s + jX2), after the stator impedance B. In series with the stator impedance R1 + jX1 C. In parallel with the whole circuit including R1 and X1 D. In series with the rotor resistance R2/s Answer: A. Xm sits as a shunt branch after R1 and X1, in parallel with the rotor branch, giving Zp = jXm(Z2)/(jXm + Z2). (Key point: Equivalent circuit model) 2. At slip s = 0.02 with R2 = 0.14 Ω, what is the rotor branch resistance used in the circuit? A. 0.14 Ω B. 7 Ω (R2/s) C. 0.0028 Ω (R2 x s) D. 6.86 Ω (R2(1-s)/s only) Answer: B. The rotor branch uses R2/s = 0.14/0.02 = 7 Ω, which includes both rotor copper loss and mechanical power. (Key point: Rotor branch resistance) 3. After finding the air-gap power Pag, how is the output power obtained? A. Pout = Pag - stator copper loss B. Pout = s x Pag C. Pout = (1 - s)Pag minus the constant 405 W loss D. Pout = Pag minus rotor copper loss only Answer: C. Pm = (1 - s)Pag is the mechanical power developed; subtracting the 405 W friction, windage and core loss gives the output. (Key point: Air-gap power to output) 4. How are the input power and power factor found from the circuit? A. pf = cos of the angle of Zp alone; Pin = Pag B. pf = cos of the angle of Zin (31.77°) = 0.850 lagging; Pin = 3V1 I1 cos(phi) C. pf = sin of the angle of Zin; Pin = 3V1 I1 D. pf = 1 because the motor is resistive at 2% slip Answer: B. I1 = V1/Zin lags V1 by the angle of Zin, so pf = cos 31.77° = 0.850 and Pin = 3V1I1cos(phi). 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