ISM to Accompany Electric Machinery and Power System Fundamentals 1/e

337 Pages·2001·3.09 MB·English

by  Stephen J. Chapman

Instructor’s Manual

to accompany

Chapman

Electric Machinery and Power System Fundamentals

First Edition

Stephen J. Chapman
BAE SYSTEMS Australia

i
Instructor’s Manual to accompany Electric Machinery and Power System Fundamentals, First Edition

Copyright  2001 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.

ISBN: ???

ii
TABLE OF CONTENTS

Preface iv
1 Mechanical and Electromagnetic Fundamentals 1
2 Three-Phase Circuits 23
3 Transformers 27
4 AC Machine Fundamentals 66
5 Synchronous Machines 69
6 Parallel Operation of Synchronous Generators 103
7 Induction Motors 114
8 DC Motors 148
9 Transmission Lines 178
10 Power System Representation and Equations 193
11 Introduction to Power-Flow Studies 205
12 Symmetrical Faults 256
13 Unsymmetrical Faults 285

iii
PREFACE

TO THE INSTRUCTOR

This Instructor’s Manual is intended to accompany the third edition of Electric Machinery and Power System
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, 5, 6, and 9 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 two forms, as MATLAB MAT-files and as ASCII text files. Students can use these files for
electronic solutions to homework problems. The ASCII files are supplied so that the information can be used with
non-MATLAB software.
The solutions in this manual have been checked carefully, 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/engcs/electrical/chapman/.
Thank you.

Stephen J. Chapman
Melbourne, Australia
August 16, 2001

Stephen J. Chapman
276 Orrong Road
Caulfield North, VIC 3161
Australia
Phone [phone]
iv
Chapter 1: Mechanical and Electromagnetic Fundamentals
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2π rad
ω= 1800 r/min  =188.5 rad/s
 60 s  1 r 
1-2. A flywheel with a moment of inertia of 4 kg ⋅ m2 is initially at rest. If a torque of 5 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:
τ 5 N⋅m ( )
ω=α t =  t = 5 s =6.25 rad/s
 J  4 kg⋅m2
The speed in revolutions per minute is:
( ) 1 r  60 s 
n= 6.25 rad/s    =59.7 r/min
2π rad1 min
1-3. A force of 5 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 α of the cylinder?

SOLUTION The magnitude and the direction of the torque on this cylinder is:
τ = rF sinθ, CCW
ind ( )
( )
τ = 0.5 m 5 kg⋅m2 sin 40°=1.607 N⋅m, CCW
ind
1-4. A motor is supplying 70 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 min2π rad
P =τω= 70 N⋅m 1500 r/min   =11,000 W
 60 s  1 r 
( ) 1 hp 
P = 11,000 W  =14.7 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 0.003 Wb.
1
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 1000.

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 = 2(27.5 cm) = 55 cm, l = 30 cm, and l = 30 cm. The reluctances of these regions are:
1 2 3
l l 0.55 m
R = = = ( )( )( )( )=58.36 kA⋅t/Wb
1 µA µµA 1000 4π×10−7 H/m 0.05 m 0.15 m
r o
l l 0.30 m
R = = = ( )( )( )( )=47.75 kA⋅t/Wb
2 µA µµA 1000 4π×10−7 H/m 0.05 m 0.10 m
r o
l l 0.30 m
R = = = ( )( )( )( )=95.49 kA⋅t/Wb
3 µA µµA 1000 4π×10−7 H/m 0.05 m 0.05 m
r o
The total reluctance is thus
R =R +R +R =58.36 +47.75+95.49=201.6 kA⋅t/Wb
TOT 1 2 3
and the magnetomotive force required to produce a flux of 0.003 Wb is
F =φ R=(0.003 Wb)(201.6 kA⋅t/Wb)=605 A⋅t
and the required current is
F 605 A⋅t
i = = =1.21 A
N 500 t
The flux density on the top of the core is
φ 0.003 Wb
B = = ( )( )=0.4 T
A 0.15 m 0.05 m
The flux density on the right side of the core is
φ 0.003 Wb
B = = ( )( )=1.2 T
A 0.05 m 0.05 m
1-6. A ferromagnetic core with a relative permeability of 2000 is shown in Figure P1-3. The dimensions are
as shown in the diagram, and the depth of the core is 7 cm. The air gaps on the left and right sides of the
2
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 R be the reluctance of the left-hand portion
1
of the core, R be the reluctance of the left-hand air gap, R be the reluctance of the right-hand portion
2 3
of the core, R be the reluctance of the right-hand air gap, and R be the reluctance of the center leg of
4 5
the core. Then the total reluctance of the core is
( )( )
+ +
R R R R
R = R + 1 2 3 4
TOT 5 R +R +R +R
1 2 3 4
l 1.11 m
R = 1 = ( )( )( )( )=90.1 kA⋅t/Wb
1 µµA 2000 4π×10−7 H/m 0.07 m 0.07 m
r 0 1
l 0.0005 m
R = 2 = ( )( )( )( )= 77.3 kA⋅t/Wb
2 µA 4π×10−7 H/m 0.07 m 0.07 m 1.05
0 2
l 1.11 m
R = 3 = ( )( )( )( )=90.1 kA⋅t/Wb
3 µµA 2000 4π×10−7 H/m 0.07 m 0.07 m
r 0 3
l 0.0007 m
R = 4 = ( )( )( )( )=108.3 kA⋅t/Wb
4 µA 4π×10−7 H/m 0.07 m 0.07 m 1.05
0 4
l 0.37 m
R = 5 = ( )( )( )( )=30.0 kA⋅t/Wb
5 µµA 2000 4π×10−7 H/m 0.07 m 0.07 m
r 0 5
The total reluctance is
( )( ) ( )( )
R +R R +R 90.1+77.3 90.1+108.3
R = R + 1 2 3 4 =30.0+ =120.8 kA⋅t/Wb
TOT 5 R +R +R +R 90.1+77.3+90.1+108.3
1 2 3 4
The total flux in the core is equal to the flux in the center leg:
( )( )
F 300 t 1.0 A
φ =φ = = =0.00248 Wb
center TOT R 120.8 kA⋅t/Wb
TOT
The fluxes in the left and right legs can be found by the “flux divider rule”, which is analogous to the
current divider rule.
3
( ) ( )
R +R 90.1+108.3 ( )
φ = 3 4 φ = 0.00248 Wb =0.00135 Wb
left R +R +R +R TOT 90.1+77.3+90.1+108.3
1 2 3 4
( ) ( )
R +R 90.1+77.3 ( )
φ = 1 2 φ = 0.00248 Wb =0.00113 Wb
right R +R +R +R TOT 90.1+77.3+90.1+108.3
1 2 3 4
The flux density in the air gaps can be determined from the equation φ= BA:
φ 0.00135 Wb
B = left = ( )( )( )=0.262 T
left A 0.07 cm 0.07 cm 1.05
eff
φ
0.00113 Wb
B = right = ( )( )( )=0.220 T
right A 0.07 cm 0.07 cm 1.05
eff
1-7. A two-legged core is shown in Figure P1-4. The winding on the left leg of the core (N ) has 600 turns,
1
and the winding on the right (N ) has 200 turns. The coils are wound in the directions shown in the
2
figure. If the dimensions are as shown, then what flux would be produced by currents i = 0.5 A and i =
1 2
1.00 A? Assume µr = 1000 and constant.

SOLUTION The two coils on this core are would so that their magnetomotive forces are additive, so the
total magnetomotive force on this core is
( )( ) ( )( )
F = N i +N i = 600 t 0.5 A + 200 t 1.0 A =500 A⋅t
TOT 11 2 2
The total reluctance in the core is
l 2.60 m
R = = ( )( )( )( )=92.0 kA⋅t/Wb
TOT µµA 1000 4π×10−7 H/m 0.15 m 0.15 m
r 0
and the flux in the core is:
F 500 A⋅t
φ= TOT = =0.0054 Wb
R 92.0 kA⋅t/Wb
TOT
4
1-8. A core with three legs is shown in Figure P1-5. Its depth is 5 cm, and there are 200 turns on the leftmost
leg. The relative permeability of the core can be assumed to be 1500 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 4% increase in
the effective area of the air gap due to fringing effects.

SOLUTION This core can be divided up into four regions. Let R be the reluctance of the left-hand
1
portion of the core, R be the reluctance of the center leg of the core, R be the reluctance of the center
2 3
air gap, and R be the reluctance of the right-hand portion of the core. Then the total reluctance of the
4
core is
( )
+
R R R
R =R + 2 3 4
TOT 1 R +R +R
2 3 4
l 1.08 m
R = 1 = ( )( )( )( )=127.3 kA⋅t/Wb
1 µµA 1500 4π×10−7 H/m 0.09 m 0.05 m
r 0 1
l 0.34 m
R = 2 = ( )( )( )( )=24.0 kA⋅t/Wb
2 µµA 1500 4π×10−7 H/m 0.15 m 0.05 m
r 0 2
l 0.0004 m
R = 3 = ( )( )( )( )= 40.8 kA⋅t/Wb
3 µA 4π×10−7 H/m 0.15 m 0.05 m 1.04
0 3
l 1.08 m
R = 4 = ( )( )( )( )=127.3 kA⋅t/Wb
4 µµA 1500 4π×10−7 H/m 0.09 m 0.05 m
r 0 4
The total reluctance is
( ) ( )
R +R R 24.0+40.8127.3
R =R + 2 3 4 =127.3+ =170.2 kA⋅t/Wb
TOT 1 R +R +R 24.0+40.8+127.3
2 3 4
The total flux in the core is equal to the flux in the left leg:
( )( )
F 200 t 2.0 A
φ =φ = = =0.00235 Wb
left TOT R 170.2 kA⋅t/Wb
TOT
The fluxes in the center and right legs can be found by the “flux divider rule”, which is analogous to the
current divider rule.
R 127.3 ( )
φ = 4 φ = 0.00235 Wb =0.00156 Wb
center R +R +R TOT 24.0+40.8+127.3
2 3 4
R +R 24.0+40.8 ( )
φ = 2 3 φ = 0.00235 Wb =0.00079 Wb
right R +R +R TOT 24.0+40.8+127.3
2 3 4
5
The flux density in the legs can be determined from the equation φ= BA:
φ 0.00235 Wb
B = left = ( )( )=0.522 T
left A 0.09 cm 0.05 cm
φ 0.00156 Wb
B = center = ( )( )=0.208 T
center A 0.15 cm 0.05 cm
φ 0.00079 Wb
B = left = ( )( )=0.176 T
right A 0.09 cm 0.05 cm
1-9. 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 0.35 T =0.7 N, into the page
1-10. The wire 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.
e =(v×B)⋅l =vBl cos 45°=(6 m/s)(0.2 T)(0.75 m) cos 45°=0.636 V, positive down
ind
6