A 3-phase, 226-V, 50-Hz, 35 HP, 6-pole, Y-connected induction motor is operating with a line current of I, = 92L-48.5° A, for a per-unit of 0.12, and R, = 0.2 Q The total friction, windage, and core losses can be assumed to be constant at 3 KW. What are the Air-Gap power and the converted power? Select one: O a. None O b. PAG = 24.4 KW and Pcony = 21.4 KW c. PAG = 18.8 KW and Pcony = 16.5 KW O d. PAG = 21.16 KW and Pcony = 18.6 KW Clear my choice

A 3-phase, 226-V, 50-Hz, 35 HP, 6-pole, Y-connected induction motor is operating with a line current of I, = 92L-48.5° A, for a per-unit of 0.12, and R, = 0.2 Q The total friction, windage, and core losses can be assumed to be constant at 3 KW. What are the Air-Gap power and the converted power? Select one: O a. None O b. PAG = 24.4 KW and Pcony = 21.4 KW c. PAG = 18.8 KW and Pcony = 16.5 KW O d. PAG = 21.16 KW and Pcony = 18.6 KW Clear my choice

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Q2 Complete the answer : The rotor reactance of the three phase induction motor is varies with ................... of the rotor current.
A 3-phase, 50-Hz induction motor with its rotor star-connected gives 500-V(r.m.s.) at standstill between the slip-rings on open-circuit. Calculate the current and power factor at standstill when the rotor winding is joined to a star-connected external circuit, each phase of which has a resistance of 10 Q and an inductance of 0.04 H. The resistance per phase of the winding is 0.2 Q and its inductance is 0.04 H. Also, calculate the current and power factor when the slip-rings are short-circuited and the motor is running with a slip of 5 per cent. Assume flux remain constant.
Q2. A 230 V single-phase induction motor, 50 Hz, has the following winding impedances (at rest): Zm=28 <70° Q. Za=42 <45°Q A) Calculate the input current at start-up. B) Obtain the amount of capacitor in series with the auxiliary winding for which the torque Maximum startup. Then compare the input current, auxiliary winding current and torque all at start-up in two modes with and without starter capacitor.
What is the relationship between slip s) and load (Pa) in an induction motor? Please explain.
A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be
2) A large 3-phase, 4000 Va (phase to phase), 60 Hz squirrel-cage induction motor draws a current of 385 A/phase. The total active power at the stator input terminals is 2344 kW, when operating at full load. The corresponding speed is accurately measured and found to be 709.2 r/min. The stator is connected in wye and the resistance between two stator terminals is 0.10 Q. The total iron losses are 23.4 kW and the windage and friction losses are 12 kW. Calculate the following: c) The total 1²-R losses in the rotor d) The load mechanical power (kW), torque (kN•m), and efficiency. S = Va · V3 - I S= Vy•3·I %3D
In a 3-phase, 4-pole induction motor, The power crossing the air gap of a 60 Hz is 42.6 kW, and the power converted from electrical to mechanical form in the motor is 39.6 kW. If the total mechanical losses are 1000 W, find the load torque of this motor? Select one: O a. None O b. Tload = 208.4 N.m. O c. Tjoad = 232.6 N.m. O d. Tjoad = 220.3 N.m. Clear my choice
3. A 440V 60-hp six pole 60 hz A connected induction motor has the following parameters R1=0.048 R2= 0.035 Xm= 9.50 X1= 0.33 0 X2= 0.37 0 Pcore= 600W PF&W= 350W Use the Thevenin's Equivalent circuit concept to find: (a)Vth and Zth seen by the rotor circuit (b)The slip at which maximum torque occurs (c) the value of Max/pullout Torques (d) The starting torque.
A three-phase induction motor has: n-1500 rpm, nm-1440 rpm, the input power of the motor is Pin = 37.5 kW, if the no-load losses (i.e. stator core losses Pscl & core losses Pa) are equal 1.5 kW. Then the rotor core losses is: 1.44 [kW] 1.50 [kW] 3.56 [kW] 4.56 [kW]
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