Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
...
290 MW
34
62 .21
=
8.529
1 86. 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
MW
to ...
jX
63 51
0
(
)
0.75
(
5247
36. 87
A
)
E
A
j
=
°
+
&
°
E
A
9 264
19.9
=
V
°
The resulting
voltage regulation is
9 264...
Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
...
jX
I
7 967
0
(
)
1.5
(
418
36. 87
A
)
(
12.0
)(
418
36. 87
A
)
E
A
=
°
+
&
°
j
+
&
°
E
A
12,
040
17 .6
V
=
... the generator at this
point is
3
sin
3
(
)
7 967 V
(
119
=
=
⎞
⎝
A
Q
V
I
4
A
)
sin
(
0
40 .6
)
18
°
.6
MVA
°
R
=
The generator is act...
Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
...
5.7
A
2 86
7 .6
V
If we make
⎞
V
the reference (as we usually
do),
these voltages
and currents
become:
E
,
A
g
280
7 .6
V
=
°
V
⎞
2 86
0
V
=
...
=
3 .6 A
)
cos
(
9 .6
)
47.7 kW
°
=
⎞
A
P
V
I
⎝
and the reactive power consumed by
this motor
is
3
sin
3
(
)
480 V
(
3
=
=
3 .6 A
)
sin
(
9 ....
Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
...
66 00 V
66 00 V
3811 V
V
⎞
,base
=
=
3
=
=
,base ,base
14
A
L
I
I
04 A
=
=
(
)
=
(
)(
)
=
base
rated
S
P
3
P
T
L
V
F
I
3
66 00 ...
42
=
.6
pu
°
The base phase voltage
of this motor is
66 00 /
3
= 3810 V,
so
E
A
is
A
E
(
)
1.33
42 .6
=
(
381
°
0 V
)
5 067
=
42 .6
...
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
...
6. 667
3
0°
-
The magnitude of the phase current
flowing
in this generator is
I
E
A
1377 V
=
=
1377 V
1 86 A
=
=
A
+
+
0.15
1.1
+
6. 667
+
...
Therefore, the magnitude of the
phase
voltage
is
(
)
1 86 A
=
=
⎞
A
V
I
Z
and the terminal voltage is
(
6. 667
)
12
&
40
=
V
3
3
(
)
12...
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
...
1.5
61 .0
sys
1 .67 6
61 .5
sys
1. 961
60 .5
f
f
sys
f
+
+
7
MW
91.5
1.
=
5.137
f
sys
=
3 06. 2
f
sys
=
59 .61 Hz
5
sys
103.07
1 .67 6
...
LOAD
1.5
61 .0
sys
1 .67 6
61 .5
P
f
sys
1. 961
60 .5
f
sys
f
=
+
+
LOAD
91.5
1.5
sys
103.07
1 .6
P
f
76
sys
118 .64...
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
...
+
A
A
E
V
⎞
A
I
S
A
R
jX
I
277
0
(
)
0.
(
0 16
565 .3
31.8
A
)
(
)
0.899
(
565 .3
31.8
A
)
E
A
=
°
+
&
°
j
+
&
...
S
I
I
=
=
47
=
0 kVA
565 A
=
3
3
(
)
480
V
A L
T
V
The power factor is 0.85 lagging, so
I
A
565 .3
31.8
=
A
°
.
The rate...
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
...
0
(
0.0187
)(
)
461 9
36. 87
A
(
1.7 16
)(
)
461 9
36. 87
A
E
A
=
°
+
&
°
j
+
&
°
E
A
13,
590
27 .6
V
=
°
Therefore, ...
36. 8
=
7
A
°
.
The
phase voltage is 13.2 kV
/
3
= 762 1
V.
Therefore, the internal generated voltage
is
=
+
E
V
+
I
I
A
A
⎞
A S A...
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
...
27.9
38 .6
°
=
°
2
2
E
A
2
Therefore, the new armature
current is
9,923 V
A2
⎞
9,
923
38 .6
70
°
44
0
762
6. 6
A
E
V
...
(
1.1
)(
259
36. 87
A
)
384
36. 4
V
E
A
=
°
j
&
°
=
°
(b)
This motor
has 6 poles and
an electrical
frequency of
60...
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
...
260
Synchronous Motor V-Curve
250
240
230
220
210
200
350 400 450 500 550 60 0 65 0 700
E (V)
A
6- 3.
A 2300-V 1000-hp 0.8-PF leading 60 -Hz ...
frequency.
If the frequency is
changed from 60 Hz
to
50
Hz, the synchronous
reactance
will be
decreased
by a factor
of 5 /6.
6- 6.
A 480-V
100-kW 0.85-PF lea...
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