Motor Data (equivalent wye connected circuit parameters):
500hp; 460V; 4 pole; R1=0.262??; X1=1.206??; XM=54.02 ?; X2’=1.206??; R2’=0.187??
2250hp; 2300V; 4 pole; R1=0.029??; X1=0.226??; XM=13.04 ?; X2’=0.226??; R2’=0.022??
Assumptions:
Source is “ideal” with zero impedance and frequency of 60 Hz
Mechanical losses are neglected so that Pdev=Pmech.
The 2250hp motor is used as an induction generator. When connected to an ideal three-phase 2300V power system (the “source”),
Determine the real and reactive power output and power factor of the generator when it is turning at a speed of 1820 rpm.
Determine the speed at which the generator must run so that the developed power is equal to the 2250hp rating. (Hint—you can do this using a plot of the developed power vs. speed or it can be done analytically.) (speed = 1813 rpm)
What is the size of the capacitor bank, expressed in three-phase kvar @ 2300 V, that must be connected in parallel with the generator so that the combined (capacitor & generator) output power factor is 0.95 lagging when the induction generator is turning at 1813 rpm? (Capacitor size = 108 kvar @ 2300 V)
By what amount must the speed change in order to result in an developed power change of 50% (from rated)? Based on this, would you say the output power of an induction generator is very sensitive (or not) to shaft speed? (speed change of 7 rpm, about 0.4%; very sensitive—a speed change of 0.4% results in a power change of about 840 kW—enough to power several hundred homes!)
The 500hp machine is operated as a motor at 1750 rpm with an unbalanced supply consisting of Van=277/5° V, Vbn=286/-110° V, and Vcn=270/127° V. Determine the following:
The zero, positive, and negative sequence voltages at the motor terminals.
The zero, positive, and negative sequence currents at the motor terminals.
The zero, positive, and negative sequence developed torques.
The zero, positive, and negative sequence electrical losses.
The developed torque and…