In handwritten form The core of a simple de motor is shown in Figure Q1. Assume that the effective crosssection area of each air gap is 18 cm² and the length of each air gap is 0.05cm. Theeffective diameter of the rotor core is 4 cm and the cross section area of the iron core is16 cm?. The relative permeability of the stator and rotor cores is 3800;i. Determine the total reluctance of the magnetic circuit.ii. Determine the value of the current that will produce a flux of 0.002 Wb.30 cmScm1000 turns4 cm:5cm30 cm ..A = 16 cm?%3DFigure Q1

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Answered: The core of a simple de motor is shown

Engineering

Electrical Engineering

The core of a simple de motor is shown

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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The armature resistance of a DC motor whose poles are made of permanent magnets, connected to a DC voltage source with a voltage of Vt= 627 volts, is 2 Ohm and drives a load of 20 Nm. In the meantime, it draws 4 A current from the source. How many rpm is the steady state speed at this load? (magnetic circuit linear- no saturation and armature reaction effects and mechanical losses will be neglected)
- A separately excited 4 - pole, 900 rpm wave wound dc generator has an induced emf of 240V at rated speed and rated field current. When connected to a load, the terminal voltage is observed to be 200V. If the armature resistance is 0.2 Ω and the flux per pole is 10 m Wb, compute the armature current and the number of conductors
A 240-V-rms 100-hp 60-Hz six-pole delta-connected synchronous motor operates with a developed power (including losses) of 100 hp and a power factor of 85 percent lagging. The synchronous reactance is Xs=0.5 Ω. The field current is If=10 A. What must the new field current be to produce 100 percent power factor? Assume that magnetic saturation does not occur so that Br is proportional to If.
The magnetic field produced in the air gap of an induction motor by the stator windings is given by B=Bm cos(ωt−2θ), in which θ is angular displacement in the counterclockwise direction as illustrated in Figure 16.4 on page 797. How many poles does this machine have? Given that the frequency of the source is 50 Hz, determine the speed of rotation of the field. Does the field rotate clockwise or counterclockwise? Repeat for a field given by B=Bm cos(ωt+3θ).
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(TRUE OR FALSE) - Energy conversion of DC generator is based on the principle ofproduction of dynamically terminal voltage. - Increasing the speed of DC motor, the back emf will decrease but theline current falls. - The current drawn by the armature of DC motor is directly proportionalto torque.
A 4 pole, 250 V, DC shunt motor has a lap connected armature with 960 conductors. The flux per pole is 2x10-2Wb. Calculate the torque developed by the armature and the useful torque in Nm when the current taken by the motor is 30 A. The armature resistance is 0.12 ohm and the field resistance is 125 ohm. The rotational losses 825 W.
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Of a shunt excited DC generator, Excitation winding internal resistance Rf = 2.5 Ω Armature winding internal resistance Ra = 0.5 Ω A rheostat has been placed in the excitation circuit to keep the machine's end voltage constant at Va = 30 V. Machine Voltage magnetization induced in the armature while operating at n = 1400 rpm at If = 7 A excitation current It is determined as Ea = 30 V from its characteristic. According to this, a) With peak voltage Va = 30 V, excitation current If = 7 A, power supplied to the load when the speed n = 1400 rpm What is the resistance of the rheostat? b) Spindle revolutions are reduced to n = 1000 rpm. In this case Va = 30 V and Ea = 37 V constant What value should the rheostat be adjusted to keep it? Calculate the power given to the load in this new case.
A separately excited D.C motor is rated for a full load torque of 100Nm. For an applied voltage of 240V and for flux (φ) per pole of 30 mWb, the no-load and full load speeds of the motor are 1595 r.p.m and 1195 r.p.m respectively. (a) Find the armature resistance (b) If the applied voltage and the flux per pole are reduced to 200V and 20mWb respectively, Find the corresponding full load speed.
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