The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz, designA, Y-connected induction motor having a rated current of 28 A.DC test:Vdc = 13.6 V, Idc= 28.0ANo-load test:VT = 208V, IA = 8.12A , IB = 8.20A , IC = 8.18A , f= 60 Hz , Pin= 420WLocked-rotor test:VT = 25 V, IA = 28.IA, IB = 28.0A , Ic = 27.6A ,f= 15 Hz , Pin = 920W(a) Sketch the per-phase equivalent circuit for this motor.(b) Find the slip at the pullout torque, and find the value of the pullout torqueitself.%3D

It is given that: Pr=7.5 hpPoles=P=4 V=208 Vf=60 Hz

Answered: The following test data were taken on a…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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The following test data were taken on a 7.5 hp, 220-V, 60-Hz, four-pole, design class C, Y- connected induction motor. DC test: 10 V, 19.08 A No-load test: 220 V, 5.7 A, 380 W, 60 Hz Locked-rotor test: 26.5 V, 18.57 A, 675 W, 15 Hz a) Determine the elements of the per-phase equivalent circuit. Draw the equivalent circuit.
A four-pole, single-phase, 120 V, 60 Hz induction motor gave the following standstill impedances when tested at rated frequency:Main winding : Zm =1.5+j4 ohmsAuxiliary winding : Za =3+j6 ohmsa) Determine the value of external resistance to be inserted in series with the auxiliary winding to obtain maximum starting torque as a resistor split-phase motor.b) Determine the value of the capacitor to be inserted in series with the auxiliary winding to obtain maximum starting torque as a capacitor-start motor.c) Determine the value of the capacitor to be inserted in series with the auxiliary winding to obtain maximum starting torque per ampere of the starting current as a capacitor-start motor.d) Compare the starting torques and starting currents in parts (a), (b), and (c) expressed as per unit of the starting torque without any external element in the auxiliary circuit, when operated at 120V, 60Hz.
A 4-pole induction motor drives a load at 2500 rpm. This is to be accomplished by using an electronic converter to convert 400 Volts DC into a set of 3-phase AC voltages. Find the frequency required for the AC voltages assuming that the slip is 4 percent. The load requires 2 hp. If the DC-to-AC converter has a power efficiency of 88 percent and the motor has a power efficiency of 80 percent, estimate the current taken from the DC source.
A 3 phase, 400-V, 50 Hz 4 pole induction motor is fed from a 3-phase supply and runs of 1425 rpm. The frequency of the rotor emf is.....
The no-load speed of an induction motor is 1914 rpm, the specification shows that it is operated 3-phase, 60 hertz, 1758 rpm and supplied by 425 V. what is the percent speed regulation?
18. A single-phase, 120 V, 60 Hz, four-pole, split-phase induction motor has the following standstill impedances: Main winding: Zm =5+j6.25 Auxiliary winding: Za =8+j6 a) Determine the value of capacitance to be added in series with the auxiliary winding to obtain maximum starting torque. b) Compare the starting torques and starting current with and without the added capacitance in the auxiliary winding circuit when operated from a 120V, 60HZ supply.
Design a DC motor circuit to drive A fan-type load with a torque equation equal to: TL=1.34×10−7ω2TL=1.34×10−7ω2 N.m The fan speed range is from 500 to 1200 rpm. To achieve the required operation, the armature voltage of the motor is adjusted by a single-phase, full-wave, ac/dc converter. The input voltage and frequency to the converter are 480 V (rms) and 50 Hz. choose the motor type, show the motor rating, αα range, Supply voltage, and frequency, and any required other information Modify the design to automatically adjust the voltage drive the load within the specified speed range. show the control variable, controller type, connection, and controller weights.
4-25- six pole. Three phase. 40-hp, 60-Hz induction motor has an input when loaded of 35 kw , 51A, 440V And a speed of 1152 rpm ,when uncoupled from the load .the readn are found to be 440-v ,21.3 A ,2.3 Kw , and 1199 rpm,the resistance... Between terminal for the stator winding is 0.25 2 for a Y-connection , the stator core losses and the rotational losse are.determine: a-the power factor b-the horspower
Design a DC motor circuit to drive A fan-type load with a torque equation equal to: TL=1.34×10−7ω2TL=1.34×10−7ω2 N.m The fan speed range is from 500 to 1200 rpm. To achieve the required operation, the armature voltage of the motor is adjusted by a single-phase, full-wave, ac/dc converter. The input voltage and frequency to the converter are 480 V (rms) and 50 Hz. choose the motor type, show the motor rating, αα range, Supply voltage, and frequency, and any required other information Modify the design to automatically adjust the voltage drive the load within the specified speed range. show the control variable, controller type, connection, and controller weights. design the single-phase, full-wave, ac/ dc converter. show design specifications, the reason for DC motor selection. correct manual circuit, and correct automatic closed-loop control circuit
12. The IR-compensation in frequency control of induction motors is used for: a) compensation of the voltage drop in the active resistance of the stator winding. b) compensation of the voltage drop in the inductive reactance of the stator winding. c) keeping the motor voltage constant. d) keeping the motor speed constant U₁ U₁ Us0 0 f. f₁
18. A single-phase, 120 V, 60 Hz, four-pole, split-phase induction motor has the following standstill impedances:Main winding: Zm =5+j6.25Auxiliary winding: Za =8+j6a) Determine the value of capacitance to be added in series with the auxiliary winding to obtain maximum starting torque.b) Compare the starting torques and starting current with and without the added capacitance in the auxiliary winding circuit when operated from a 120V, 60Hz supply.
4- A 3-phase A -connected 220-V (line-to-neutral), 60-Hz 6-pole induction motor has the following parameter values in O/phase referred to the stator: R1 = 0.294, R2 = 0.144, X, = 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 5%, a. Output torque b. Stator current, c. Power factor, d. Converted power e. Efficiency.

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