Question 3: Answer the following question below:a. Consider a permanent-magnet de motor with the following parameters: R-0.4ohm, L.-1 mH,kE-0.4 V/(rad/s), kr=0.4 Nm/A, and J-0.03 kg m. The rated torque of this motor is 5 Nm. Plotthe steady state torque-speed characteristics for V.-60 Vb. The motor in (a) is driving a load whose torque requirement remains constant at 5 Nm, independentof speed. Calculate the armature voltage V. to be applied in steady state, if this load is to be drivenat 1,800 rpm.c. In an ECM drive, kg-kr-0.8 in SI units. Plot the phase currents and the induced-emf waveforms,as a function of 5, if the motor is operating at a speed of 150 rad/s and delivering a torque of 7 Nm.

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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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Example 4-13 A 5.5-kW 220-V shunt generator has a field resistance of 140 , armature resistance of 0.52, and stray power losses of 95 W at rated conditions. Find at rated load: (a) The field current (b) The load current (c) The armature current (d) The generated voltage (e) The total copper loss (f) The efficiency (g) The voltage regulation (for this, assume that the flux, hence E., remains constant)*
A DC series motor is driving a load whose torque is constant and independent of the speed. The motor speed is increased from 1000 RPM to 1200 RPM by connecting a diverter resistance across its field. Find the Armature to the field winding current. Assume negligible drop across the field winding and a linear magnetic circuit
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.
- The speed of dc shunt motor can be varied by: a) The voltage V applied to armature. b) The armature circuit resistance Ra. O c) The flux Ø. d) All the mentioned above.
A separately excited D.C. generator has armature circuit resistance of 0.15 . When running at 900 r.p.m., it delivers a current of 95 A at 200 V to a load of constant resistance. A) If the generator speed drop to 800 r.p.m., with field-current unaltered, find the current delivered to load. B)With what load resistance will the current be 95 amp, at 800 r.p.m.?
What is Equation to find Slip, Output horse power, Torque, Stator current, p.f, Efficiency, From this result and we have Frequency and poles this result from load test of an induction motor
A DC shunt motor is producing EMF of 220V running with 500rpm. And producing EMF of 200V with the speed of 1000rpm by adding external resistance with shunt field resistance. The original shunt field resistance is 250 ohm.Find the external resitance added resistance with shunt field resistance at 1000rpm. a. 300 ohm b. 250 ohm c. 500 ohm d. 550 ohm
7. The total rotational losses of an induction motor are a combination of ------- A) Friction losses B) Windage losses. D) Stray losses. C) Core losses. 8. The developed mechanical power can be calculated using the equation of- A) PAG - PRCL. B) PAG w sync C) (1-s) PAG. D) 3 E2Gc. 9. For an induction motor, if the synchronous speed is 1800 r/m, and the rotor speed 180rad/sec, then the slip will be: A) 5.5% C) 6% B) 4.5% D) 4% 10. The power losses in the stator part of the induction motor are B) Core losses D) Friction and windage losses A) Stator copper losses C) Air-gap losses
A 35.2 HP, 250 V, 100 A. 1500 RPM, series DC motor with ndmature reaction is considered in this problem. This motor has a series-field winding of 20 turns per pole. Its armature resistance in RA-0.05 £2, and its series field resistance is Rs 0.05 52. The magnetization curve at 1500 RPM is given by the following table: EA (V) F (Amp.tums) 50 95 125 150 180 200 212 229 243 200 250 253 500 755 1000 1280 1500 2000 2500 3000 3500 4000 The motor is operating at rated voltage, armature current of 100 Amperes, and with rotational losses of 1500 Watt. Find the speed of the motor. Answer: The speed is n- RPM.
A 200-V, 100 A armature full-load line current, 1800 RPM. hunt DC motor is considered in this problem. Its characteristics are given below: RA=0.20 52 RE 50 $2 NF-2000 turns This motor has compensating windings and interpoles. The magnetization curve for this motor at 1800 RPM is shown in a tabular form below If.A 0 0.1 0.2 0.4 0.5 0.6 0.8 1.0 1.2 1.6 1.8 2.0 2.2 2.4 EA-V 6.0 25 50 110 135 160 205 235 255 275 285 288 290 291 For Rad between 50 52 and 450 2. find the range of no-load speed of this motor. Answer: No-load Speed Range from RPM to RPM.
-A 250V, DC series motor has total resistance of 0.8 Q. Its full load current and speed are 10 A and 1500 r.p.m respectively. If the torque is reduced to 50% .Determine the current and speed. Assume straight line magnetization characteristics.
Two DC generators, each with armature resistance of 0.02 ohm and fieldresistance of 50 ohms run in parallel and supply a total current of 1000 amperesto the load circuit. If their emfs are 270 V and 265 V, their bus bar voltagewill be. [Ans. 267.5 V]

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