INSTRUCTOR'S SOLUTION MANUAL - Power Unit

1 I solutions MANUAL to accompany Chapman Electric Machinery Fundamentals Fifth Edition Stephen J. Chapman BAE Systems Australia ii solutions MANUAL to accompany Electric Machinery Fundamentals, Fifth Edition Copyright 2012 McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. No part of this book may be used or reproduced in any manner whatsoever without written permission, with the following exception: homework solutions may be copied for classroom use. iii TABLE OF CONTENTS Preface iv1 Introduction to Machinery Principles 12 Transformers 233 AC Machine Fundamentals 734 Synchronous Generators 815 Synchronous Motors 1236 Induction Motors 1527 DC Machinery Fundamentals 2028 DC Motors and Generators 2149 Single-Phase and Special Purpose Motors 276A Review of Three-Phase Circuits 287B Coil Pitch and Distributed Windings 295C Salient-Pole Theory of Synchronous Machines 302S1 Introduction to Power Electronics 308E Errata 348 iv PREFACE TO THE INSTRUCTOR This Instructor s MANUAL is intended to accompany the fifth edition of Electric Machinery Fundamentals.

2 To make this MANUAL easier to use, it has been made self-contained. Both the original problem statement and the problem SOLUTION are given for each problem in the book. This structure should make it easier to copy pages from the MANUAL for posting after problems have been assigned. Many of the problems in Chapters 2, 4, 5, and 8 require that a student read one or more values from a magnetization curve. The required curves are given within the textbook, but they are shown with relatively few vertical and horizontal lines so that they will not appear too cluttered. Electronic copies of the corresponding open-circuit characteristics, short-circuit characteristics, and magnetization curves as also supplied with the book. They are supplied in as ASCII text files.

3 Students can use these files for electronic solutions to homework problems. The ASCII files can be read into MATLAB and used to interpolate points along the curve. Each curve is given in ASCII format with comments at the beginning. For example, the magnetization curve in Figure P8-1 is contained in file Its contents are shown below: % This is the magnetization curve shown in Figure % P8-1. The first column is the field current in % amps, and the second column is the internal % generated voltage in volts at a speed of 1200 r/min. % To use this file in MATLAB, type "load ". % The data will be loaded into an N x 2 array named % "p81_mag", with the first column containing If and % the second column containing the open-circuit voltage. % MATLAB function "interp1" can be used to recover % a value from this curve.

4 0 0 16 v 238 278 1 To use this curve in a MATLAB program, the user would include the following statements in the program: % Get the magnetization curve. Note that this curve is % defined for a speed of 1200 r/min. load if_values = p81_mag(:,1); ea_values = p81_mag(:,2); n_0 = 1200; The solutions in this MANUAL have been checked twice, but inevitably some errors will have slipped through.

5 If you locate errors which you would like to see corrected, please feel free to contact me at the address shown below, or at my email address I greatly appreciate your input! My physical and email addresses may change from time to time, but my contact details will always be available at the book s Web site, which is Thank you. Stephen J. Chapman Melbourne, Australia March 31, 2011 Chapter 1: Introduction to Machinery Principles 1-1. A motor s shaft is spinning at a speed of 1800 r/min. What is the shaft speed in radians per second? SOLUTION The speed in radians per second is 1 min2 rad1800 rad/s60 s1 r 1-2.

6 A flywheel with a moment of inertia of 4 kg m2 is initially at rest. If a torque of 6 N m (counterclockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after 5 s? Express that speed in both radians per second and revolutions per minute. SOLUTION The speed in radians per second is: 26 N m 5 rad/s4 kg mttJ The speed in revolutions per minute is: 1 r60 r/min2 rad 1 minn 1-3. A force of 10 N is applied to a cylinder, as shown in Figure P1-1. What are the magnitude and direction of the torque produced on the cylinder? What is the angular acceleration of the cylinder? SOLUTION The magnitude and the direction of the torque on this cylinder is: CCW ,sinind rF m 10 N sin N m, CCW The resulting angular acceleration is: N rad/s4 kg mJ 1-4.

7 A motor is supplying 50 N m of torque to its load. If the motor s shaft is turning at 1500 r/min, what is the mechanical Power supplied to the load in watts? In horsepower? SOLUTION The mechanical Power supplied to the load is 1 1 min2 rad50 N m 1500 r/min7854 W60 s1 rP 1 hp7854 hp746 WP 1-5. A ferromagnetic core is shown in Figure P1-2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 800. SOLUTION There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms a third region.

8 If we assume that the mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path lengths are = 2( cm) = 55 cm, = 30 cm, and = 30 cm. The reluctances of these regions are: 1l2l3l kA t/Wb800 410 H/m m mrollAA R kA t/Wb800 410 H/m m mrollAA R kA t/Wb800 410 H/m m mrollAA R The total reluctance is thus 252 kA t/Wb RRRR and the magnetomotive force required to produce a flux of Wb is Wb 252 kA t/Wb1260 A t FR and the required current is 1260 A A500 tiN F The flux density on the top of the core is 2 m mBA The flux density on the right side of the core is m mBA 1-6.

9 A ferromagnetic core with a relative permeability of 1500 is shown in Figure P1-3. The dimensions are as shown in the diagram, and the depth of the core is 5 cm. The air gaps on the left and right sides of the core are and cm, respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. If there are 300 turns in the coil wrapped around the center leg of the core and if the current in the coil is A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap? SOLUTION This core can be divided up into five regions. Let be the reluctance of the left-hand portion of the core, be the reluctance of the left-hand air gap, R be the reluctance of the right-hand portion of the core, be the reluctance of the right-hand air gap, and be the reluctance of the center leg of the core.

10 Then the total reluctance of the core is 1R32R4R5R 1234 TOT51234 RRRRRRRRRR m168 kA t/Wb1500 410 H/m m mrlA R m152 kA t/Wb410 H/m m m R m168 kA t/Wb1500 410 H/m m mrlA R m108 kA t/Wb410 H/m m m R 3 kA t/Wb1500 410 H/m m mrlA R The total reluctance is 1234 TOT51234168 152 168 kA t/Wb168 152 168 108 RRRRRRRRRR The total flux in the core is equal to the flux in the center leg: centerTOTTOT300 t Wb204 kA t/Wb FR The fluxes in the left and right legs can be found by the flux divider rule , which is analogous to the current divider rule. 34leftTOT1234168 Wb168 152 168 108 RRRRRR 12rightTOT1234168 Wb168 152 168 108 RRRRRR The flux density in the air gaps can be determined from the equation BA : cm cm cm cm 1-7.