A three-phase, two-pole induction motor, connected at A of 5HP, 220V, 60HZ, has rotational losses of 250W, when operating with a slip of 2.5%. it is known that the motor has the following parameters per phase: r1=r2'=0.250, x1=x'2=1.2jÒ , rc=900, xp=15jQ. Consider 1HP=746W. Under these conditions, calculate:
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- Name three factors that determine the torque produced by an induction motor.A 480V, 60Hz, six-pole, Y connected cage rotor induction motor has the followingparameters in ohms per phase referred to the stator: R1=0.5ΩX1=1.1Ω R2=0.45ΩX2=0.6Ω Xm=30Ω The total rotational losses (friction, windage, and core losses) are 1100W and are assumedto be constant. For a rotor slip of 2.2% at the rated voltage and rated frequency, calculate:d) The air-gap power (Pag) and the developed mechanical power (Pconv)e) Developed electromechanical torque, and the output mechanical torquef) EfficiencyA 460V, 25hp, 60Hz, 4 pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1 = 0.641 Ω R2 = 0.332 Ω X1 = 1.106 Ω X2 = 0.436 Ω Xm = 26.3 Ω The total rotational losses are 1100W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 3.5% at the rated voltage and rated frequency, find the motor’s Speed Stator current Power factor Pconv and Pout Ʈǐnd and Ʈload Efficiency
- A 3-phase, 6 pole, 440 V, 60 Hz, Y-connected induction motor has the following impedances per phase referred to the stator circuit: R1=0.43 Ω, R2=0.54 Ω, X1=1.263 Ω, X2=0.608 Ω, XM=43Ω The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2 percent at the rated voltage and rated frequency, Find the motor speed, stator current, output powerA 440-V, 60-Hz, 2-pole, Y-connected, three-phase induction motor has the following parameters in ohms/phase: R1 = 0.3, X1 = 0.9, R2 = 0.6, X2 = 0.9, Rc = 150, Xm = 60. If the rotational loss is 4% of the powerdeveloped, determine the efficiency of the motor when it runs at 4% slip.A 480V, 60Hz, six-pole, Y connected cage rotor induction motor has the followingparameters in ohms per phase referred to the stator: R1=0.5Ω X1=1.1Ω R2=0.45ΩX2=0.6Ω Xm=30Ω The total rotational losses (friction, windage, and core losses) are 1100W and are assumedto be constant. For a rotor slip of 2.2% at the rated voltage and rated frequency, calculate:a) Rotor speedb) Stator currentc) Complex, real, and reactive input power and the power factor of the machined) The air-gap power (Pag) and the developed mechanical power (Pconv)e) Developed electromechanical torque, and the output mechanical torquef) Efficiency
- A 240 V, 60 Hz, Delta connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the following per phase parameters: R1 = 0.07, R2 = 0.05, X1 = X2 = 0.2, Xm = 6.5, Draw the approximate equivalent circuit and determine the following: I. Starting Torque II. Maximum Torque III. Rated Torque IV. Stator Current V. Power Factor VI. Input Power VII. Rotor Current VIII. Magnetizing Current IX. Stator copper loss X. Rotor copper lossA 200V, 3, star connected induction motor takes 30A at a line voltage of 35V with rotor locked. With this line voltage, power input to motor is 450W and core loss is 60W.The dc resistance between a pair of stator terminals is 0.15Ω. If the ratio of ac to dc resistance is 1.6, find the equivalent leakage reactance/phase of the motor and the stator and rotor resistance per phase.QI: The parameters of a 4 kW, 220V. 2-phase induction motor, in ohms per phase, are: r1=0.53, x1 =2.4, XM= 70, r2 = 0.96, x2 = 3.0. The motor is operated from unbalanced 2- phase source whose voltages are 230V and 210V, the smaller voltage is lagging the larger by 800. For a slip of 5% and rotational losses of 30 W, find the positive- and negative- sequence components of: (a) the applied voltages. (b) Find the values of phase currents (c) the input current. (d) Find the developed mechanical power. (e) Find the efficiency.
- A 208 V, 60 Hz, 2-pole, Y-connected induction motor is rated at 16 hp. Its equivalentcircuit components have the following values:(R1 = 0.200 ohm, X1 =0.410 ohm); (R2 = 0.120 ohm, X2 = 0.410 ohm); Xm = 15.0 ohm.Rotor slip is 4.5% at the rated voltage and frequency. Mechanical and core losses are 230Wand 200W respectively. Find (a) rotor speed (b) line current and motor p.f (c) Air-gap andconverted power (d) Induced torque.A single-phase, six-pole, capacitor start, 120V/60Hz induction motor, has the following parameters:R1=1.48Ω R2=3.2 ΩX1=2.05 Ω X2=1.6 ΩThe core losses are negligible, while the friction and windage losses are constant Pf,wi=20W. The motor is operating at the rated voltage and frequency with its starting winding open, and the motor’s slip is 4.5%. Calculate:a) Stator current and power factor of the machineb) Developed electromechanical torque and the load torquec) The motor efficiencyAn approximate per phase equivalent circuit of a 3-phase, 220-V, wye-connected, 6 pole, 60 Hz induction motor is shown in figure below for which R1 = 0.3 ohm; X1 = 0.4 ohm; R’2 = 0.4 ohm; X’2 = 0.6 ohm; Xm = 15 ohms. For the given numerical values calculate the total developed(electromagnetic) torque in (kg-cm) .The motor speed is 1080 rpm.