Comprehensive Study Guide for Three-Phase Induction Motor Performance and Power Flow

Performance Analysis of a 25 kW 3-Phase Induction Motor

A 25 kW, 4-pole, 3-phase, 50 Hz induction motor operates at a full load speed of $1410\,rpm$. The synchronous speed ($N_s$) is calculated as $N_s = \frac{120 \times 50}{4} = 1500\,rpm$. The slip ($s$) is determined by $s = \frac{1500 - 1410}{1500} = 0.06$. The motor supplies full load with mechanical losses ($P_{mech}$) of $850\,W$. The stator losses are specified as $1.7$ times the rotor copper losses ($P_{rcu}$). The gross mechanical power developed ($P_m$) is the sum of the net output power and mechanical losses: $P_m = 25000 + 850 = 25850\,W$.

Using the power flow relationship $P_{rcu} = \frac{s}{1-s} \times P_m$, the rotor copper losses are calculated: $P_{rcu} = \frac{0.06}{1 - 0.06} \times 25850 = \frac{0.06}{0.94} \times 25850 \approx 1650\,W$. For the rotor resistance per phase ($R_2$), given a full load rotor current ($I_{2fl}$) of $65\,A$, the total rotor copper loss is $3 \times I_{2fl}^2 \times R_2 = 1650\,W$. Solving for $R_2$: $R_2 = \frac{1650}{3 \times 65^2} \approx 0.1302\,\Omega$. For full load efficiency ($\eta$), the input power ($P_{in}$) is the sum of $P_m$, rotor copper losses, and stator losses ($P_{sl}$). $P_{sl} = 1.7 \times 1650 = 2805\,W$. Thus, $P_{in} = 25850 + 1650 + 2805 = 30305\,W$. The efficiency is $\eta = \frac{25000}{30305} \times 100 \approx 82.49\%$.

Torque and Rotor Characteristics of Star-Connected Motors

A 3-phase, 4-pole, 50 Hz, star-connected induction motor develops a useful torque ($T_{useful}$) of $300\,Nm$. The rotor EMF frequency is given as $120$ cycles per minute, which converts to a rotor frequency ($f_r$) of $f_r = \frac{120}{60} = 2\,Hz$. The slip is $s = \frac{f_r}{f} = \frac{2}{50} = 0.04$. The torque lost in friction is $50\,Nm$, making the gross electromagnetic torque ($T_g$) equal to $300 + 50 = 350\,Nm$.

The synchronous speed is $N_s = 1500\,rpm$, so the actual rotor speed is $N_r = N_s(1 - s) = 1500(1 - 0.04) = 1440\,rpm$. The net output power is $P_{out} = T_{useful} \times \omega_r = 300 \times \frac{2 \pi \times 1440}{60} \approx 45.24\,kW$. The air-gap power ($P_g$) is $T_g \times \omega_s = 350 \times \frac{2 \pi \times 1500}{60} \approx 54.98\,kW$. Total rotor copper loss is $P_{rcu} = s \times P_g = 0.04 \times 54977.8 \approx 2199.1\,W$. The rotor copper loss per phase is $\frac{2199.1}{3} \approx 733\,W$. Rotor efficiency is defined as $(1 - s) = 0.96$ or $96\%$. If the rotor current is $60\,A$, the rotor resistance per phase is $R_2 = \frac{P_{rcu}/3}{I_2^2} = \frac{733.03}{60^2} \approx 0.2036\,\Omega$.

No-Load and Full-Load Analysis for a 10 kW Delta-Connected Motor

A 10 kW, 400 V, 4-pole, 50 Hz delta-connected induction motor undergoes testing. At no load, the line current ($I_{0L}$) is $8\,A$ and input power ($P_0$) is $660\,W$. The stator phase resistance ($R_{ph}$) is $1.2\,\Omega$. Since the motor is delta-connected, the phase current at no load is $I_{0ph} = \frac{8}{\sqrt{3}} \approx 4.62\,A$. No-load stator copper losses are $3 \times I_{0ph}^2 \times R_{ph} = 3 \times 4.618^2 \times 1.2 \approx 76.8\,W$. Stator core loss is $P_{core} = P_0 - (P_{sl,no\text{-}load} + P_{friction}) = 660 - (76.8 + 420) = 163.2\,W$.

At full load, the line current is $18\,A$ and input power is $11.2\,kW$. Full load phase current is $I_{ph} = \frac{18}{\sqrt{3}} \approx 10.39\,A$. Full load stator copper loss is $3 \times 10.39^2 \times 1.2 = 388.8\,W$. Total stator losses are $388.8 + 163.2 = 552\,W$. Air-gap power ($P_g$) is $11200 - 552 = 10648\,W$. Total rotor losses equal $P_g - P_{out} = 10648 - 10000 = 648\,W$. Total rotor ohmic losses are $P_{rcu} = P_g - P_m$. Since $P_m = P_{out} + P_{friction} = 10000 + 420 = 10420\,W$, the rotor ohmic losses are $10648 - 10420 = 228\,W$. The slip is $s = \frac{P_{rcu}}{P_g} = \frac{228}{10648} \approx 0.0214$. Full load speed is $1500(1 - 0.0214) \approx 1468\,rpm$. Internal (electromagnetic) torque is $T_g = \frac{P_g}{\omega_s} = \frac{10648}{2\pi \times 1500/60} \approx 67.79\,Nm$. Shaft torque is $T_{sh} = \frac{10000}{2\pi \times 1468/60} \approx 65.05\,Nm$. Efficiency is $\frac{10}{11.2} \times 100 \approx 89.28\%$.

Advanced Problems in Induction Machine Power Flow

For a 50 hp, 6-pole, 50 Hz motor running at $960\,rpm$, the synchronous speed is $1000\,rpm$, resulting in a slip of $s = 0.04$. The output power is $50 \times 746 = 37300\,W$. Mechanical losses are $1200\,W$, so gross mechanical power is $38500\,W$. Rotor copper loss ($P_{rcu}$) follows the ratio $P_{rcu} = \frac{s}{1-s} P_m = \frac{0.04}{0.96} \times 38500 \approx 1604.17\,W$. Subtracting the $300\,W$ copper loss in the short-circuiting gear gives the winding loss as $1304.17\,W$. With a rotor current of $40\,A$, resistance $R_2 = \frac{1304.17}{3 \times 40^2} \approx 0.2717\,\Omega$.

In a GATE 2008 problem, a $400\,V$, $50\,Hz$ motor draws $50\,A$ at $0.8\,pf$. Input power is $P_{in} = \sqrt{3} \times 400 \times 50 \times 0.8 = 27712.8\,W$. Stator copper loss is $1.5\,kW$ and core loss is $1.2\,kW$. The air-gap power ($P_g$) is $27.71 - 1.5 - 1.2 = 25.01\,kW$.

Regarding slip dependency (GATE 2012), the slip normally does not depend on the core-loss component. In power ratios (GATE 2005), the ratio of gross power output ($P_m$) to air-gap power ($P_g$) is equal to $(1 - s)$.

Motor-Generator Coupled Systems and Magnetic Field Speeds

A 3-phase, 4-pole motor is fed from the rotor side. The motor runs at $1410\,rpm$. The synchronous speed of the field relative to the rotor structure is the speed of the stator magnetic field. The motor is coupled to a DC generator with a $10\,\Omega$ load. Input power is measured by two wattmeters: $W_1 = 1800\,W$, $W_2 = -200\,W$, giving a total input of $1600\,W$.

The speed of rotation of the stator magnetic field with respect to the rotor structure is calculated by the difference between synchronous speed and rotor speed: $1500 - 1410 = 90\,rpm$. Since the supply is to the rotor, the field rotates at synchronous speed relative to the rotor. The relative speed of the stator field to the rotor structure is $90\,rpm$. Neglecting losses, the DC generator output power matches the mechanical power. If input power is $1600\,W$ and slip is $0.06$, $P_m = P_{in} (1 - s) = 1600(0.94) = 1504\,W$. The current through the $10\,\Omega$ resistance is $I = \sqrt{\frac{P}{R}} = \sqrt{\frac{1504}{10}} \approx 12.26\,A$.

Efficiency Calculations in Exam Contexts

For a 400 V, 15 kW, 4-pole, 50 Hz motor with $4\%$ slip (GATE 2004), the rotor speed is $1500(0.96) = 1440\,rpm$. Output torque is $T = \frac{15000}{2\pi \times 1440/60} \approx 99.47\,Nm$.

In a GATE 2000/ESE 2009 problem, a 40 kW input motor with 1 kW stator losses and 2 kW friction/windage losses runs at 975 rpm ($N_s = 1000$, $s = 0.025$). $P_g = 40 - 1 = 39\,kW$. $P_{rcu} = 0.025 \times 39 = 0.975\,kW$. $P_m = 39 - 0.975 = 38.025\,kW$. Net output is $38.025 - 2 = 36.025\,kW$. Efficiency is $\frac{36.025}{40} \approx 90\%$.

For an ESE 2016 problem, a motor with $5\%$ slip develops $20\,kW$ rotor power output ($P_m$). Rotor copper loss is $P_{rcu} = \frac{s}{1-s} P_m = \frac{0.05}{0.95} \times 20000 \approx 1052.6\,W$, roughly $1050\,W$. In another ESE 2016 task, a motor with 2 kW stator loss, $4\%$ slip, and 90 kW input power has $P_g = 90 - 2 = 88\,kW$. The mechanical power developed is $P_m = P_g(1 - s) = 88(0.96) = 84.48\,kW$.

For GATE 2021, a 500 V motor with 40 kW input, 1 kW stator losses, and 2.025 kW friction/windage losses runs at 975 rpm ($s = 0.025$). $P_g = 39\,kW$, $P_{rcu} = 0.975\,kW$, $P_m = 38.025\,kW$. Output power is $38.025 - 2.025 = 36\,kW$. Efficiency is $\frac{36}{40} = 90\%$. Finally, for GATE 2024, a 6-pole, 50 Hz motor at 960 rpm ($N_s = 1000$, $s = 0.04$) with negligible rotational and stator losses has an efficiency equal to the rotor efficiency: $(1 - s) \times 100 = 96\%$.