fundamentals of electric circuits solutions manual chapter 9
fundamentals of electric circuits
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... 190
Problems 191
Comprehensive Problems 200
Contents
xi
Chapter 2 Basic Laws 27
Chapter 3 Methods of Analysis 75
PART 1 DC CIRCUITS 1
Chapter 1 Basic Concepts 3
Chapter 4 Circuit Theorems 1 19
Chapter ... Odd-Numbered Problems 893
Selected Bibliography
92 9
Index
93 3
xiv
CONTENTS
Chapter 16 The Fourier Series 707
Chapter 17 Fourier Transform 7 59
Chapter 18 Two-Port Networks 795
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Textbook Table of Contents
Problem Solving Workbook Contents
1
DC CIRCUITS
PART 1
Chapter
1
Basic Concepts
Chapter
2
Basic Laws
Chapter
3
Methods of Analysis
Chapter
4
Circuit Theorems
Chapter
5
Operational...
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FUNDAMENTALS OF ELECTRIC DRIVERS doc
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... reserved.
Any use is subject to the Terms of Use as given at the website.
FUNDAMENTALS OF ELECTRIC SYSTEMS
FUNDAMENTALS OF ELECTRIC SYSTEMS 1 .9
FIGURE 1 .9 Iron filings around a wire carrying a ... 2.15 55,000 80 10.7
Supermendur 49% Co, V 2.4 15 ,90 0 80,000 8 26
Vanadium Permendur 49% Co, V 2.3 12,700 4 ,90 0 92 40 93 2
Hyperco 27 27% Co 2.36 70,000 2,800 198 19 925
Flake iron Carbonal power ... subject to the Terms of Use as given at the website.
FUNDAMENTALS OF ELECTRIC SYSTEMS
FUNDAMENTALS OF
ELECTRIC SYSTEMS
CAPACITORS
Figure 1.1 illustrates a capacitor. It consists of two insulated...
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Fundamentals of Corporate Finance 8th edition: Solutions Manual
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01/07/2014, 14:22
... = .5 294
Now we can use the sustainable growth rate equation to get:
Sustainable growth rate = (ROE ì b) / [1 (ROE ì b)]
Sustainable growth rate = [.1 899 (.5 294 )] / [1 – .1 899 (.5 294 )] ... $10,157 / (1 – 0.34) = $15,3 89. 39
Now, we can add interest to EBT to get EBIT as follows:
EBIT = EBT + Interest paid = $15,3 89. 39 + 3,405 = $18, 794 . 39
CHAPTER 4 B-35
10. Below is ... $41,7 69 + 85,000 = $126,7 69
Now we can calculate Tobin’s Q, which is:
Tobin’s Q = (Market value of equity + Book value of debt) / Book value of assets
Tobin’s Q = ($860,000 + 126,7 69) ...
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Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
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06/08/2014, 11:20
... limit of the capability curve would be
drawn from an origin
of
3
V
2
3
(
)
1328 V
2
Q
=
⎞
=
=
4810 kVAR
S
X
1.1
&
The radius of the ...
generate the power by means
of a
motor-generator set
consisting of a synchronous motor
driving a synchronous generator. How
many
poles
should
each
of
the
two machines have in
...
voltage
of
200
V,
and
the
maximum
I
F
is 10 A. The resistance of the field circuit is adjustable
over
the
range
from
20
to
200
&
.
The OCC of this generator...
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Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
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... speed of 1800 r/min and a
full-load speed
of 1785
r/min.
The loads supplied
by the two generators consist of 100 kW at 0.85 PF
lagging.
(a)
Calculate
the
speed droops of generator ... advantage of this
fact.
% M-file: prob5_4e.m
%
M-file to calculate and
plot the
terminal voltage
% of a synchronous
generator as a
function of load
% for
power
factors of 0.8 ... magnitude of
E
A
is 12,040 V.
(b)
The torque angle of the generator
at rated
conditions is
™
= 17.6
°
.
(c)
Ignoring
R
A
, the
maximum output
power
of the generator
...
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Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
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06/08/2014, 11:20
... of Gen
A
fnlB = 61.5;
% No-load freq of Gen
B
fnlC = 60.5;
% No-load freq of Gen
C
spA =
1.5;
% Slope of Gen A (MW/Hz)
spB =
1.676;
% Slope of Gen B (MW/Hz)
spC =
1 .96 1;
...
/Hz
(
)
sys
90 00 kW
+
0.15 MW
/Hz
sys
f
f
0.25 MW/Hz
f
sys
6050 kW
90 00 kW
=
+
100
kW
f
sys
14,
95 0 kW
59. 8 Hz
=
=
0.25 MW/Hz
...
61.5
sys
1 .96 1
60.5
f
f
sys
f
+
+
7
MW
91 .5
1.
=
5.137
f
sys
=
306.2
f
sys
=
59. 61 Hz
5
sys
103.07
1.676
sys
118.64
1 .96 1
sys
f
...
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Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
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06/08/2014, 11:20
...
P
sin
A
3
(
)
277 V
(
5 09
V
)
(
)
sin
471 kW
sin
™
™
™
=
=
=
S
X
G
0. 899
&
The power supplied
as a function of the torque angle
™
may be ...
% M-file: prob5_ 19. m
% M-file
to
plot the power output
of
a
% synchronous generator as a function of
% the
torque angle.
% Calculate
Xs
delta
= (0:1 :90 );
%
Torque
... reactance of this generator at the rated
conditions?
(b)
What is the
unsaturated synchronous reactance
of
this generator?
(c)
Plot the saturated synchronous reactance
of
this...
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Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
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06/08/2014, 11:20
...
)
46 19
36.87
A
(
1.716
)(
)
46 19
36.87
A
E
A
=
°
+
&
°
j
+
&
°
E
A
13,
590
27.6
V
=
°
Therefore, the magnitude of ...
I
A
A
⎞
A S A
R
jX
7621
0
(
)
0 .9
(
52 49
36.87
A
)
E
A
j
=
°
+
&
°
E
A
11,120
19. 9
V
=
°
The resulting
voltage regulation ...
1 39
When
this program is
executed,
the
plot
of reactive
power versus flux is
(d)
The program in part
(c)
of this program calculated
I
A
as
a function of
...
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Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
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06/08/2014, 11:20
...
j
=
°
+
&
°
E
A
92 64
19. 9
=
V
°
The resulting
voltage regulation is
92 64
6351
RV
100%
45 .9%
=
⋅
=
6351
Because
voltage,
... MW/Hz
f
nl1
=
61 .91 Hz
f
nl1
60
Hz
(d)
If the swing generator trips off the
line, the other three
generators would have to supply all 290 MW
of
the load.
Therefore, ...
290 MW
34
62.21
=
8.5 29
186.63
3
f
sys
f
sys
=
59. 37 Hz
sys
34
62.21
sys
34
62.21
f
f
sys
f
Each generator will
supply
96 .7
...
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Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
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06/08/2014, 11:20
...
9,
92 3 V
, and
sin
1
1
E
1
A
sin
sin
13,
230
™
™
V
sin
=
=
27 .9
38.6
°
=
°
2
2
E
A
2
Therefore, the new armature
current is
9, 923 ...
298 .4 kW
=
=
4 49 A
=
L
3
PF
3
(
)
480
V
(
0.8
)
T
V
Because the motor
is
-connected,
the
corresponding phase current is
4 49
/
3
2 59 A
I
...
⎞
9,
92 3
38.6
70
°
44
0
762
6.6
A
E
V
A
I
=
=
jX
S
j
8.18
°
=
°
(f)
A
MATLAB program to plot the magnitude of the
armature
...
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Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
Ngày tải lên :
06/08/2014, 11:20
... change?
(Hint:
Think about the derivation
of
X
S
.)
157
198
&
0
A
°
14 39
=
22.
7
V
°
These
values
of
E
A
and
™
at
unity
...
(
141.5
31.8
A
)
E
A
=
°
j
&
°
E
A
4 29
24
=
.9
V
°
So
E
A
= 4 29 V at rated conditions.
The
resulting plot is
shown below
260 ... voltage of
would require a field current of 4.6 A.
(b)
The motor’s efficiency at full load
and unity
power factor is
⎜
OUT
P
100%
=
⋅
746 k
=
W
100%
⋅
89. 3%
=
...
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Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
Ngày tải lên :
06/08/2014, 11:20
...
°
E
A
124
14 .9
=
V
°
The torque angle
™
of
this machine is –14 .9
°
.
(b)
A
phasor
diagram of the motor operating at a power factor of 0.78 leading is shown
...
generated
voltage of the motor is
161
=
E
V
I
,m
⎞
,m
jX
,mA S A
(
)(
)
E
,m
A
E
,mA
277
0
V
=
°
j
1.1
297
20.
=
9
V
°
96
&
... the magnitudes and
angles of
E
A
for
both machines.
(b)
If the
flux of the motor is increased by 10 percent, what happens to the terminal voltage of the power
system? What
is...
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Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
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06/08/2014, 11:20
...
are
168
reactance
of 0 .90 and
a
per-unit resistance of 0.02.
(a)
What
is the rated input power of this motor?
(b)
What
is the magnitude of
E
A
at
rated conditions?
...
What
is the speed of
rotation of this motor?
(b)
What
is the output
torque of this motor at the rated conditions?
(c)
What
is the internal generated voltage
of
this motor ...
hp
(
)
1200 r/min
91 ,
91 0 lb
ft
=
=
=
⊕
m
n
6-13.
A
440-V
three-phase
Y-connected synchronous motor has a synchronous reactance of 1.5
&
per phase.
...
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Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx
Ngày tải lên :
06/08/2014, 11:20
...
3
A
A
P
I
3
R
79. 4 A
0.22
4.16 kW
(f)
co
P
nv
at rated conditions is
=
=
=
conv IN C
P
P
U
83
P
.8 kW
4.16 kW
79. 6 kW
(g)
If
E
A ...
If
E
A
is
decreased by 10%, the
new
value if
E
A
= (0 .9) (603 V)
= 543
V.
To
simplify
this part of the
problem,
we will ignore
R
A
. Then the quantity
E
A
sin
...
E
A
changes. Therefore,
sin
E
1
A
sin
sin
603 V
™
™
sin
(
)
1 1
=
=
19. 5
21.8
°
=
°
2 1
Therefore,
I
E
A2
⎞
A
V
...
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fundamentals of heat and mass transfer solutions manual phần 1 docx
Ngày tải lên :
08/08/2014, 17:20
... differences of 50 to 100°C.
Coefficients (W/m
2
⋅K)
T
s
(°C) ε
h
r
h
r,a
h
35 0.05 0.32 0.32 2.1
0 .9 5.7 5.7
135 0.05 0.51 0.50 4.7
0 .9 9.2 9. 0
100 300 500 700 90 0
Surroundings ... 308 298 2 K 0.32W m K
=ì × × ⋅ + = ⋅
<
The free convection coefficient with T
s
= 35°C and
T
∞
= T
sur
= 25°C, find that
() ( )
1/3 1/3
1/3 2
s
h 0 .98 T 0 .98 T T 0 .98 308 298 2.1W ... 1 .9
KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that
through a composite wall of prescribed thermal conductivity and thickness.
FIND: Thickness of...
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- 4
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