Plot the V curves (no load, partial loads and full load) for a 7.2 kW, 208 V, three phase, star connected synchronous motor by varying the power factor from 0.5 lagging to 0.5 leading. The synchronous reactance of the motor is 4Ω/phase and the winding resistance is negligible. Assume Ea= 100 Ifwhere Ifis the field winding current
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Plot the V curves (no load, partial loads and full load) for a 7.2 kW, 208 V, three phase, star connected synchronous motor by varying the power factor from 0.5 lagging to 0.5 leading. The synchronous reactance of the motor is 4Ω/phase and the winding resistance is negligible. Assume Ea= 100 Ifwhere Ifis the field winding current
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- What are the advantages and disadvantages of using the following types of motors? shaded-pole motor PSC motor split-phase motor capacitor-start–capacitor-run motorPlot the V curves (no load, partial loads and full load) for a 7.2 kW, 208 V, three phase, star connected synchronous motor by varying the power factor from 0.5 lagging to 0.5 leading. The synchronous reactance of the motor is 4Ω/phase and the winding resistance is negligible. Assume Ea= 100If where Ifis the field winding currentReduced parameters of 480V, 60 Hz, 6 pole, 3 phase, delta connected induction motor to stator are Rs= 0.461 ohm, Rr'=0.258 ohm, Xs=0.507 ohm, Xr'= 0.309 ohm, Rc=100 ohm and Xm=30.74 ohm. given. The motor rotates at 1170 rpm. According to the given values; a) Find the synchronous speed (Ns) in rpm. b) Find the slip (s). c) Find the stator phase current (Is). d) Find the input power (Pin). e) Find the output power (Pout).
- Calculate and plot the torque-speed characteristics for a four-pole induction motor. Themotor parameters are rs = 24.5 Ω, rr =23 Ω, Xs = 10 Ω, Xr = 40 Ω, and Xm = 25 Ω. Therated voltage is 110 V. The maximum frequency of the supplied phase voltages is fmax =60 Hz. Assume the phase voltages are supplied with frequencies 20, 40 and 60 Hz. ...A 3,3 kV, 50 HZ, three-phase, star connected induction motor, with an output of 100 kW, has a synchronous speed of 500 r/min. The full-load slip is 1,8 % and the full load power factor is 0,85 lagging. The stator copper loss is 2,44 kW, the stator iron loss is 3,5 kW and the rotational losses are 1,2 kW. Calculate: a) the rotor copper loss (PCuR), b) the line current and c) the full load efficiencyA three-phase Y-connected 230-V (line-to-line) 7.5-kW 60-Hz six-pole induction motor has the following parameter values in Ω/phase referred to the stator: R1 = 0.294, R2 = 0.144 X1 = 0.503, X2 = 0.209, Xm = 13.25 The total friction, windage, and core losses may be assumed to be constant at 403 W, independent of load. For a slip of 4 percent, determine: (a) stator current, power factor, (b) output torque and power,(c) efficiency when the motor is operated at rated voltage and frequency, (d) the load component 12 of the stator current, the induced torque Tind, and the converted power Pconv for a slip s= 0.04,(e) the maximum torque and the corresponding speed; and (f) the starting torque Tst and the corresponding stator load current 12 start
- A 110-V, 50-Hz single phase induction motor (split -phase) has the following constants for the main and auxiliary windings: Main winding, Zm (1.2 +j 25) ohm and auxiliary winding, Z,= (12+j 5) ohm. Determine (i) total current and power factor currents at time of starting, and (ii) phase angle between the main winding current and starting winding current.The stator resistance of a 1-phase induction motor is 2.5 ohm and its leakage reactance is 2.0 ohm. Onno-load, the motor takes 4 A at 96 V and at 0.25 lagging power factor. The no-load friction andwindage loss is negligible. Under the blocked-rotor condition, the input power is 130 W at 6 A and42 V. obtain the equivalent circuit parameters.A single-phase, 120 V, 60 Hz, 1/2 hp, 1740 rpm split-phase fan motor gave the following testresults:V I PNo load test 120 4 110Standstill test (main winding) 41 5.8 115Standstill test (auxiliary winding) 38 7 200Main winding resistance = 1.85 Ω.a) Obtain an equivalent circuit for the motor for running conditions.b) Determine the no-load rotational loss.c) Determine the efficiency at the rated speed.
- A 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.A three-phase, two-pole, 35 hp, 480 V, 60 Hz, Y-connected induction motor has the following constants in ohms per phase referred to the stator: Rl = 0.322Ω R2 = 0.196 Ω X1 = 0.675Ω X2 = 0.510Ω Xm = 12.5Ω The total rotational losses are 1850 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 3% at the rated voltage and rated frequency, determine the following: (a) The speed in rpm and in rad/s (b) The stator current (c) The power factor (d) The developed power and output power (e) The developed torque and output torque (f) The efficiency.A 209-V, three-phase, six-pole, Y-connected induction motor has the following parameters: R1 = 0.128 Ω, R’2 = 0.0935 Ω, Xeq =0.49Ω. The motor slip at full load is 2%. Assume that the motor load is a fan-type. If an external resistance equal to the rotor resistance is added to the rotor circuit, calculate the following: a. Motor speed, Starting torque, Starting current, and Motor efficiency (ignore rotational and core losses) b. Calculate the value of the resistance that should be added to the rotor circuit to reduce the speed at full load by 20%. Find the motor efficiency. c. If the voltage is reduced by 20%, calculate the Motor speed, Starting torque, Starting current, and Motor efficiency (ignore rotational and core losses) d. If the supply frequency is reduced by 20%, calculate the Motor speed, Starting torque, Starting current, and Motor efficiency (ignore rotational and core losses) ..