Electric Machinery Fundamentals 5th Edition Chapman Solution Manual PDF

Preview

Summary

Ton of a product that I have...

Description

Solutions Manual

to accompany

Chapman

Electric Machinery Fundamentals Fifth Edition

Stephen J. Chapman BAE Systems Australia

i

Solutions Manual to accompany Electric Machinery Fundamentals, Fifth Edition Third Revision, 26 August 2011 Copyright  2012 McGraw-Hill, Inc. All rights reserved. Produced 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.

ii

TABLE OF CONTENTS

Preface

iv

1 2

Introduction to Machinery Principles Transformers

1 23

3 4 5

AC Machine Fundamentals Synchronous Generators Synchronous Motors

73 81 123

6 7

Induction Motors DC Machinery Fundamentals

152 202

8 9

DC Motors and Generators Single-Phase and Special Purpose Motors

214 276

A B C

Review of Three-Phase Circuits Coil Pitch and Distributed Windings Salient-Pole Theory of Synchronous Machines

287 295 302

S1 E

Introduction to Power Electronics Errata

308 348

iii

PREFACE TO THE INSTRUCTOR

This Instructor’s Manual is intended to accompany the fifth edition of Electric Machinery Fundamentals. 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. 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 p81_mag.dat. 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 p81_mag.dat". 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. 0 0 0.0132 6.67 0.03 13.33 0.033 16 0.067 31.30 0.1 45.46 0.133 60.26 0.167 75.06 0.2 89.74 0.233 104.4 0.267 118.86 0.3 132.86 0.333 146.46 0.367 159.78 0.4 172.18 0.433 183.98 0.467 195.04 iv

0.5 0.533 0.567 0.6 0.633 0.667 0.7 0.733 0.767 0.8 0.833 0.867 0.9 0.933 0.966 1 1.033 1.067 1.1 1.133 1.167 1.2 1.233 1.267 1.3 1.333 1.367 1.4 1.433 1.466 1.5

205.18 214.52 223.06 231.2 238 244.14 249.74 255.08 259.2 263.74 267.6 270.8 273.6 276.14 278 279.74 281.48 282.94 284.28 285.48 286.54 287.3 287.86 288.36 288.82 289.2 289.375 289.567 289.689 289.811 289.950

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 p81_mag.dat 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. 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 [email protected]. 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 http://www.mhhe.com/chapman/. Thank you.

Stephen J. Chapman Melbourne, Australia March 31, 2011 v

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 min  2 rad   60 s  1 r

  1800 r/min   1-2.

   188.5 rad/s 

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:

6 N m    t    t   5 s   7.5 rad/s J 4 kg  m 2   The speed in revolutions per minute is:  1 r  60 s  n   7.5 rad/s     71.6 r/min  2  rad  1 min 

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:

 ind  rF sin , CCW  ind   0.15 m 10 N sin 30  0.75 N  m, CCW The resulting angular acceleration is:

 1-4.

 J



0.75 N  m  0.188 rad/s 2 4 kg  m 2

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 min  2 rad  P   50 N  m 1500 r/min      7854 W  60 s  1 r   1 hp  P  7854 W   10.5 hp  746 W 

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 0.005 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. 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 l 1 = 2(27.5 cm) = 55 cm, l2 = 30 cm, and l 3 = 30 cm. The reluctances of these regions are: R1 

l l 0.55 m    72.9 kA  t/Wb 7  A r o A  800   4   10 H/m   0.05 m  0.15 m 

R2 

0.30 m l l    59.7 kA  t/Wb 7  A  r o A  800   4 10 H/m   0.05 m  0.10 m 

R3 

l

A



l

r o A



800  4   10

0.30 m

7

H/m  0.05 m  0.05 m 

 119.4 kA  t/Wb

The total reluctance is thus

RTOT  R1  R2  R3  72.9  59.7  119.4  252 kA  t/Wb and the magnetomotive force required to produce a flux of 0.005 Wb is F   R   0.005 Wb  252 kA  t/Wb   1260 A  t

and the required current is i

F 1260 A  t   2.5 A 500 t N

The flux density on the top of the core is

2

B

 A



0.005 Wb

 0.15 m 0.05 m 

 0.67 T

The flux density on the right side of the core is B

1-6.

 A



0.005 Wb  2.0 T  0.05 m 0.05 m 

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 0.050 and 0.070 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 1.0 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 R1 be the reluctance of the left-hand portion of the core, R2 be the reluctance of the left-hand air gap, R 3 be the reluctance of the right-hand portion of the core, R4 be the reluctance of the right-hand air gap, and R5 be the reluctance of the center leg of the core. Then the total reluctance of the core is

R TOT  R 5  R1  R2 

R3  R4 

 R 1  R 2  R 3  R 4 

l1

r 0A1

R1  R 2  R 3  R 4 

1.11 m  168 kA  t/Wb 1500  4  10 H/m  0.07 m 0.05 m  7

0.0007 m l2   152 kA  t/Wb 7  0 A2 4  10 H/m  0.07 m 0.05 m 1.05 

l3

 r 0 A3 l4

 0 A4





1.11 m  168 kA  t/Wb 7  1500 4 10 H/m   0.07 m 0.05 m   

4  10

7

0.0005 m 108 kA  t/Wb H/m  0.07 m 0.05 m 1.05 

3

R5 

l5

 r 0 A5



0.37 m  56.1 kA  t/Wb 7 1500 4 10 H/m   0.07 m  0.05 m    

The total reluctance is

R TOT  R 5 

R1 R 2  R 3  R 4   56.1  168 152 168 108   204 kA  t/Wb R 1 R 2  R 3  R 4

168 152 168 108

The total flux in the core is equal to the flux in the center leg:

center   TOT 

 300 t 1.0 A 0.00147 Wb F   R TOT 204 kA t/Wb

The fluxes in the left and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule.

left 

 R3  R4  R1  R2  R3  R4

right 

 TOT 

 R1  R2  R1  R2  R3  R4

168  108  0.00147 Wb   0.00068 Wb 168 152 168 108

 TOT 

168  152 168  152  168  108

 0.00147 Wb   0.00079 Wb

The flux density in the air gaps can be determined from the equation   BA : Bleft 

Bright  1-7.

left Aeff



right Aeff

0.00068 Wb  0.185 T 0.07 cm  0.05 cm 1.05  

0.00079 Wb  0.215 T  0.07 cm 0.05 cm 1.05

A two-legged core is shown in Figure P1-4. The winding on the left leg of the core (N1) has 600 turns, and the winding on the right (N2) has 200 turns. The coils are wound in the directions shown in the figure. If the dimensions are as shown, then what flux would be produced by currents i1 = 0.5 A and i2 = 1.0 A? Assume r = 1200 and constant.

4

SOLUTION The two coils on this core are wound so that their magnetomotive forces are additive, so the total magnetomotive force on this core is FTOT  N1i1  N 2i 2  600 t 0.5 A   200 t 1.00 A  500 A  t The total reluctance in the core is 2.60 m l R TOT    76.6 kA  t/Wb 7   r 0 A 1200  4 10 H/m   0.15 m 0.15 m  and the flux in the core is:

 1-8.

FTOT 500 A  t   0.00653 Wb RTOT 76.6 kA  t/Wb

A core with three legs is shown in Figure P1-5. Its depth is 5 cm, and there are 100 turns on the leftmost leg. The relative permeability of the core can be assumed to be 2000 and constant. What flux exists in each of the three legs of the core? What is the flux density in each of the legs? Assume a 5% increase in the effective area of the air gap due to fringing effects.

5

SOLUTION This core can be divided up into four regions. Let R1 be the reluctance of the left-hand portion of the core, R 2 be the reluctance of the center leg of the core, R3 be the reluctance of the center air gap, and R4 be the reluctance of the right-hand portion of the core. Then the total reluctance of the core is the reluctance of the left-hand leg plot the parallel combination of the reluctances of the right-hand and center legs: R TOT  R1  R1  R2 

R3  R4 

l1

r 0A1 l2

 r 0 A2

 R 2  R3  R 4 R 2  R3  R4 

1.08 m  95.5 kA  t/Wb  2000 4  10 H/m  0.09 m  0.05 m 



0.34 m 18.0 kA  t/Wb  2000 4  10 H/m  0.15 m 0.05 m

7

7

0.0005 m l3   51.0 kA  t/Wb 7  0 A3  4   10 H/m   0.15 m 0.05 m 1.04  l4

 r 0 A4



1.08 m  95.5 kA  t/Wb  2000 4  10 H/m  0.09 m 0.05 m 7

The total reluctance is R TOT  R1 

 R 2  R3  R 4 R 2  R3  R 4

 95.5 

18.0  51.0  95.5 18.0  51.0  95.0

 135.5 kA  t/Wb

The total flux in the core is equal to the flux in the left leg:

left   TOT 

100 t 2.0 A  0.00148 Wb F   RTOT 135.5 kA  t/Wb

The fluxes in the center and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule.

center 

R4 95.5  TOT  0.00148 Wb   0.00086 Wb 18.0  51.0  95.5 R2  R3  R4

6

right 

R 2  R3 18.0  51.0    0.00235 Wb   0.00062 Wb R2  R3  R4 TOT 18.0  51.0  95.5

The flux density in the legs can be determined from the equation   BA : Bleft 

left A

Bcenter  Bright  1-9.



0.00148 Wb  0.329 T 0.09 m  0.05 m  

center A

left A





0.00086 Wb  0.115 T 0.15 m  0.05 m  

0.00062 Wb  0.138 T 0.09 cm 0.05 cm 

A wire is shown in Figure P1-6 which is carrying 2.0 A in the presence of a magnetic field. Calculate the magnitude and direction of the force induced on the wire.

SOLUTION The force on this wire can be calculated from the equation

F  i l  B   ilB  2 A 1 m 0.5 T  1.00 N, into the page 1-10.

The wire is shown in Figure P1-7 is moving in the presence of a magnetic field. With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire.

SOLUTION The induced voltage on this wire can be calculated from the equation shown below. The voltage on the wire is positive downward because the vector quantity v  B points downward. 7

eind  v  B  l  vBl cos 45   10 m/s  0.2 T  0.25 m  cos 45  0.354 V, positive down 1-11.

Repeat Problem 1-10 for the wire in Figure P1-8.

SOLUTION The induced voltage on this wire can be calculated from the equation shown below. The total voltage is zero, because the vector quantity v B points into the page, while the wire runs in the plane of the page. eind   v  B   l  vBl cos 90  1 m/s 0.5 T  0.5 m cos 90  0 V 1-12.

The core shown in Figure P1-4 is made of a steel whose magnetization curve is shown in Figure P1-9. Repeat Problem 1-7, but this time do not assume a constant value of µr. How much flux is produced in the core by the currents specified? What is the relative permeability of this core under these conditions? Was the assumption in Problem 1-7 that the relative permeability was equal to 1200 a good assumption for these conditions? Is it a good assumption in general?

8

SOLUTION The magnetization curve for this core is shown below:

192

9

The two coils on this core are wound so that their magnetomotive forces are additive, so the total magnetomotive force on this core is FTOT  N1i1  N 2i 2  600 t 0.5 A   200 t 1.00 A  500 A  t

Therefore, the magnetizing intensity H is H

F 500 A  t   192 A  t/m lc 2.60 m

From the magnetization curve, B  0.17 T and the total flux in the core is

TOT  BA   0.17 T  0.15 m  0.15 m   0.00383 Wb The relative permeability of the core can be found from the reluctance as follows:

R

FTOT

 TOT



l

 r 0 A

Solving for µr yields

r 

TOT l 0.00383 Wb  2.6 m    704 FTOT  0 A  500 A  t  4  10-7 H/m  0.15 m 0.15 m 

The assumption that  r = 1200 is not very good here. It is not very good in general. 1-13.

A core with three legs is shown in Figure P1-10. Its depth is 5 cm, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure 1-10c. Answer the following questions about this core: (a) What current is required to produce a flux density of 0.5 T in the central leg of the core? (b) What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part (a)? (c) What are the reluctances of the central and right legs of the core under the conditions in part (a)? (d) What are the reluctances of the central and right legs of the core under the conditions in part (b)? (e) What conclusion can you make about reluctances in real magnetic cores?

10

SOLUTION The magnetization curve for this core is shown below:

(a)

A flux density of 0.5 T in the central core corresponds to a total flux of

TOT  BA   0.5 T  0.05 m  0.05 m   0.00125 Wb By symmetry, the flux in each of the two outer legs must be 1   2  0.000625 Wb , and the flux density in the other legs must be B1  B2 

0.000625 Wb  0.25 T  0.05 m 0.05 m

The magnetizing intensity H required to produce a flux density of 0.25 T can be found from Figure 1-10c. It is 50 A·t/m. Similarly, the magnetizing intensity H required to produce a flux density of 0.50 T is 75 A·t/m. The mean length of the center leg is 21 cm and the mean length of each outer leg is 63 dm, so the total MMF needed is FTOT  H center lcenter  H outer l outer

FTOT  75 A  t/m 0.21 m   50 A  t/m 0.63 m   47.3 A  t and the required current is

i (b)

FTOT 47.3 A  t   0.12 A N 400 t

A flux density of 1.0 T in the central core corresponds to a total flux of

TOT  BA  1.0 T  0.05 m 0.05 m   0.0025 Wb By symmetry, the flux in each of the two outer legs must be 1  2  0.00125 Wb , and the flux density in the other legs must be

B1  B2 

0.00125 Wb  0.50 T  0.05 m 0.05 m 11

The magnetizing intensity H required to produce a flux density of 0.50 T can be found from Figure 1-10c. It is 75 A·t/m. Similarly, the magnetizing intensity H required to produce a flux density of 1.00 T is about 160 A·t/m. Therefore, the total MMF needed is FTOT  Hcenter Icenter  Houter Iouter F TOT  160 A  t/m 0.21 m  75 A  t/m 0.63 m  80.8 A  t and the required current is i

TOT N



80.8 A  t  0.202 A 400 t

This current is not twice the current in part (a).

(c)

The reluctance of the central ...