Electric Machinery and Power System Fundamentals PDF

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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....

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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

1 2 3 4 5 6 7 8 9 10

Preface Mechanical and Electromagnetic Fundamentals Three-Phase Circuits Transformers

iv 1 23 27

AC Machine Fundamentals Synchronous Machines Parallel Operation of Synchronous Generators Induction Motors

66 69 103 114

DC Motors Transmission Lines

148 178

Power System Representation and Equations

11

Introduction to Power-Flow Studies

193 205

12 13

Symmetrical Faults Unsymmetrical Faults

256 285

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 +61-3-9527-9372

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    = 188.5 rad/s  60 s  1 r 

ω = (1800 r/min) 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:

5 N ⋅m τ  (5 s ) = 6 .25 rad/s t = 4 kg ⋅ m 2 J

ω =α t = 

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 ind = rF sin θ , CCW

τ ind = (0.5 m )(5 kg ⋅ m 2 ) sin 40° = 1.607 N ⋅ m, CCW 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 l1 = 2(27.5 cm) = 55 cm, l 2 = 30 cm, and l3 = 30 cm. The reluctances of these regions are:

0.55 m l l = = = 58.36 kA ⋅ t/Wb −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.15 m ) 0.30 m l l = = = 47.75 kA ⋅ t/Wb R2 = −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.10 m ) 0.30 m l l = = = 95.49 kA ⋅ t/Wb R3 = −7 µA µ r µ o A (1000)(4π × 10 H/m )(0.05 m )(0.05 m ) R1 =

The total reluctance is thus

RTOT = R1 + R2 + R3 = 58.36 + 47.75 + 95.49 = 201.6 kA ⋅ t/Wb 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

i=

F 605 A ⋅ t = = 1.21 A N 500 t

The flux density on the top of the core is

B=

φ A

=

0.003 Wb = 0.4 T (0.15 m )(0.05 m )

The flux density on the right side of the core is

B= 1-6.

φ A

=

0.003 Wb = 1.2 T (0.05 m )(0.05 m )

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 R1 be the reluctance of the left-hand portion of the core, R2 be the reluctance of the left-hand air gap, R3 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

RTOT = R5 +

(R1 + R2 )(R3 + R4 )

R1 + R2 + R3 + R4 l1 1.11 m R1 = = = 90.1 kA ⋅ t/Wb −7 µ r µ 0 A1 (2000)(4π × 10 H/m )(0.07 m )(0.07 m ) l 0.0005 m = 77.3 kA ⋅ t/Wb R2 = 2 = −7 µ 0 A2 (4π × 10 H/m )(0.07 m )(0.07 m )(1.05) l3 1.11 m = = 90.1 kA ⋅ t/Wb R3 = −7 µ r µ 0 A3 (2000)(4π × 10 H/m )(0.07 m )(0.07 m ) l 0.0007 m R4 = 4 = = 108.3 kA ⋅ t/Wb −7 µ 0 A4 (4π × 10 H/m )(0.07 m )(0.07 m )(1.05) l5 0.37 m = = 30.0 kA ⋅ t/Wb R5 = −7 µ r µ 0 A5 (2000)(4π × 10 H/m )(0.07 m )(0.07 m )

The total reluctance is

RTOT = R5 +

(R1 + R2 )(R3 + R4 ) = 30.0 + (90.1 + 77.3)(90.1 + 108.3) = 120.8 kA ⋅ t/Wb 90.1 + 77.3 + 90.1 + 108.3

R1 + R2 + R3 + R4

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

φ center = φ TOT =

(300 t )(1.0 A ) = 0.00248 Wb F = RTOT 120.8 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.

3

φ left =

(R3 + R4 ) R1 + R2 + R3 + R4

φ TOT =

(R1 + R2 )

φ right =

R1 + R2 + R3 + R4

(90.1 + 108.3) 90.1 + 77.3 + 90.1 + 108.3

φ TOT =

(0.00248 Wb) = 0.00135 Wb

(90.1 + 77.3) 90.1 + 77.3 + 90.1 + 108.3

(0.00248 Wb) = 0.00113 Wb

The flux density in the air gaps can be determined from the equation φ = BA :

Bleft =

φ left Aeff

Bright = 1-7.

=

φ right Aeff

0.00135 Wb = 0.262 T (0.07 cm )(0.07 cm )(1.05)

=

0.00113 Wb = 0.220 T (0.07 cm )(0.07 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 = 1.00 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

FTOT = N 1i1 + N 2 i2 = (600 t )(0.5 A ) + (200 t )(1.0 A ) = 500 A ⋅ t The total reluctance in the core is

RTOT =

l

µ r µ0 A

=

2.60 m = 92.0 kA ⋅ t/Wb (1000)(4π × 10 H/m )(0.15 m )(0.15 m ) −7

and the flux in the core is:

φ=

FTOT 500 A ⋅ t = 0.0054 Wb = RTOT 92.0 kA ⋅ t/Wb 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 R1 be the reluctance of the left-hand portion of the core, R2 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

RTOT = R1 +

l1

(R2 + R3 )R4 R 2 + R3 + R 4

1.08 m = 127.3 kA ⋅ t/Wb µ r µ 0 A1 (1500)(4π × 10 H/m )(0.09 m )(0.05 m ) l2 0.34 m = = 24.0 kA ⋅ t/Wb R2 = −7 µ r µ 0 A2 (1500)(4π × 10 H/m )(0.15 m )(0.05 m ) l 0.0004 m = 40.8 kA ⋅ t/Wb R3 = 3 = −7 µ 0 A3 (4π × 10 H/m )(0.15 m )(0.05 m )(1.04 ) l4 1.08 m R4 = = = 127.3 kA ⋅ t/Wb −7 µ r µ 0 A4 (1500)(4π × 10 H/m )(0.09 m )(0.05 m )

R1 =

=

−7

The total reluctance is

RTOT = R1 +

(R2 + R3 )R4 R 2 + R3 + R 4

= 127.3 +

(24.0 + 40.8)127.3 24.0 + 40.8 + 127.3

= 170.2 kA ⋅ t/Wb

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

φ left = φ TOT =

(200 t )(2.0 A ) = 0.00235 Wb F = RTOT 170.2 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.

φ center = φ right =

R4 127.3 (0.00235 Wb) = 0.00156 Wb φ TOT = 24.0 + 40.8 + 127.3 R2 + R3 + R4 R2 + R3 24.0 + 40.8 (0.00235 Wb) = 0.00079 Wb φ TOT = R 2 + R3 + R 4 24.0 + 40.8 + 127.3 5

The flux density in the legs can be determined from the equation φ = BA :

Bleft =

φ left A

Bcenter = Bright = 1-9.

=

0.00235 Wb = 0.522 T (0.09 cm )(0.05 cm )

φ center A

φ left A

=

=

0.00156 Wb = 0.208 T (0.15 cm )(0.05 cm )

0.00079 Wb = 0.176 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 )(0.35 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.

eind = ( v × B ) ⋅ l = vBl cos 45° = ( 6 m/s )( 0.2 T )( 0.75 m ) cos 45° = 0.636 V, positive down

6

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 1000 a good assumption for these conditions? Is it a good assumption in general?

7

SOLUTION The magnetization curve for this core is shown below:

0.16

193

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 = N 1i1 + N 2 i2 = (600 t )(0.5 A ) + (200 t )(1.0 A ) = 500 A ⋅ t Therefore, the magnetizing intensity H is

H=

F 500 A ⋅ t = = 193 A ⋅ t/m lc 2.60 m

From the magnetization curve,

B = 0.16 T and the total flux in the core is

φ TOT = BA = (0.16 T )(0.15 m )(0.15 m ) = 0.0036 Wb The relative permeability of the core can be found from the reluctance as follows:

R=

FTOT

φ TOT

=

l

µ r µ0 A

Solving for µr yields

µr =

φ TOT l (0.0036 Wb)(2.6 m ) = = 662 FTOT µ 0 A (500 A ⋅ t )(4π × 10 -7 H/m )(0.15 m )(0.15 m )

The assumption that µ r = 1000 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 8 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? 8

(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?

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

φ TOT = BA = (0.5 T )(0.08 m )(0.08 m ) = 0.0032 Wb By symmetry, the flux in each of the two outer legs must be φ1 = φ 2 = 0.0016 Wb , and the flux density in the other legs must be

B1 = B2 =

0.0016 Wb = 0.25 T (0.08 m )(0.08 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 70 A·t/m. Therefore, the total MMF needed is

FTOT = H center lcenter + H outer l outer 9

FTOT = (70 A ⋅ t/m )(0.24 m ) + (50 A ⋅ t/m )(0.72 m ) = 52.8 A ⋅ t and the required current is

i= (b)

FTOT 52.8 A ⋅ t = = 0.13 A 400 t N

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

φ TOT = BA = (1.0 T )(0.08 m )(0.08 m ) = 0.0064 Wb By symmetry, the flux in each of the two outer legs must be φ1 = φ 2 = 0.0032 Wb , and the flux density in the other legs must be

B1 = B2 =

0.0032 Wb = 0.50 T (0.08 m )(0.08 m )

The magnetizing intensity H required to produce a flux density of 0.50 T can be found from Figure 1-10c. It is 70 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 = H center I center + H outer I outer FTOT = (160 A ⋅ t/m )(0.24 m ) + (70 A ⋅ t/m )(0.72 m ) = 88.8 A ⋅ t and the required current is

i= (c)

φ TOT N

=

88.8 A ⋅ t = 0.22 A 400 t

The reluctance of the central leg of the core under the conditions of part (a) is:

Rcent =

FTOT

φ TOT

=

(70 A ⋅ t/m )(0.24 m ) = 5.25 kA ⋅ t/Wb 0.0032 Wb

The reluctance of the right leg of the core under the conditions of part (a) is:

Rright = (d)

FTOT

φ TOT

=

(50 A ⋅ t/m )(0.72 m ) = 22.5 kA ⋅ t/Wb 0.0016 Wb

The reluctance of the central leg of the core under the conditions of part (b) is:

Rcent =

FTOT

φ TOT

=

(160 A ⋅ t/m )(0.24 m ) = 6.0 kA ⋅ t/Wb 0.0064 Wb

The reluctance of the right leg of the core under the conditions of part (b) is:

Rright = (e) 1-14.

FTOT

φ TOT

=

(70 A ⋅ t/m)(0.72 m ) = 15.75 kA ⋅ t/Wb 0.0032 Wb

The reluctances in real magnetic cores are not constant.

A two-legged magnetic core with an air gap is shown in Figure P1-11. The depth of the core is 5 cm, the length of the air gap in the core is 0.07 cm, and the number of turns on the coil is 500. The magnetization curve of the core material is shown in Figure P1-9. Assume a 5 percent increase in effective air-gap area to account for fringing. How much current is required to produce an air-gap flux density of 0.5 T? What are the flux densities of the four sides of the core at that current? What is the total flux present in the air gap? 10

SOLUTION The magnetization curve for this core is shown below:

An air-gap flux density of 0.5 T requires a total flux of

φ = BAeff = (0.5 T )(0.05 m )(0.05 m )(1.05) = 0.00131 Wb This flux requires a flux density in the right-hand leg of

Bright =

φ A

=

0.00131 Wb = 0.524 T (0.05 m )(0.05 m )

The flux density in the other three legs of the core is

Btop = Bleft = Bbottom =

φ A

=

0.00131 Wb = 0.262 T (0.10 m )(0.05 m )

The magnetizing intensity required to produce a flux density o...