Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
...
91.5
1.5
sys
1 03. 07
1.6
P
f
76
sys
118.64
1.96
f
=
5. 137
sys
31 3.2
f
P
LOAD
1
sys
f
f
sys
=
31 3.2
L
P
OAD
5. 137
A
MATLAB
... kVA)(0.85)
=
5 .31
MW.
Generator
3
will be
the
first
machine to reach that
limit. Generator 3
will supply this power at a frequency
of
(
)
(
)
=
5 .3...
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
...
3. 333
3
0
°
Ω
.
The magnitude
of
the
phase current flowing in
this generator is
E
A
137 7 V
137 7 V
33 5 A
A
I
=
=
=
=
0.15
1.1
3. 333
30
... the
phase
voltage
is
(
)
33 5 A
=
=
⎞
A
V
I
Z
and the terminal voltage is
(
3. 333
)
11
&
17
=
V
3
3
(
)
1117 V
1 934 V
T
V
V
⎞
=...
Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
...
°
j
+
&
°
E
A
139 3
11.4
=
V
°
The voltage regulation
would be:
139 3
132 8
RV
100%
4.9%
=
⋅
=
132 8
(e)
For
this
problem, ...
angle of 36 .87
°
A
L
(
3
3
230 0
)
V
L
V
=
=
. The internal generated voltage
of
the
machineThe phase voltage of this
machine
is
⎞
T
/
V
V
3...
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
...
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 rated phase voltage
is
V
⎞
= 480 V /
3
= 277 V. ...
0
(
)
0.
(
016
565 .3
31 .8
A
)
(
)
0.899
(
565 .3
31 .8
A
)
E
A
=
°
+
&
°
j
+
&
°
E
A
509
30 .5
=
V
°...
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
...
5249
=
A
3
3
(
)
13. 2
kV
A L
T
V
The power factor is 0.8
lagging, so
I
A
5249
36 .8
=
7
A
°
.
The
phase voltage is 13. 2 kV
/
3
= 7621
V.
...
134
7217
0
(
0.0187
)(
)
4619
36 .87
A
(
1.716
)(
)
4619
36 .87
A
E
A
=
°
+
&
°
j
+
&
°
E
A
13,
590
27.6...
Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
...
f
3 n
s
l3 sysP
f
f
(
)
(
)
(
)
(
)
=
+
(
)
(
)
+
290 MW
34
62.21
=
8.529
186. 63
3
f
sys
f
sys
=
59 .37 Hz
sys
34
... test
Field current, A
32 0
36 5
38 0
475
570
Line
voltage,
kV
13. 0
13. 8
14.1
15.2
16.0
Extrapolated air-gap voltage, kV
15.4
17.5
18 .3
22.8
27.4...
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
...
sin
sin
13,
230
™
™
V
sin
=
=
27.9
38 .6
°
=
°
2
2
E
A
2
Therefore, the new armature
current is
9,9 23 V
A2
⎞
9,
9 23
38 .6
70
...
0
V
(
1.1
)(
259
36 .87
A
)
38 4
36 .4
V
E
A
=
°
j
&
°
=
°
(b)
This motor
has 6 poles and
an electrical
frequ...
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
...
OUT
P
100%
=
⋅
746 k
=
W
100%
⋅
89 .3%
=
3
(
)
137 6
=
238 3 V
. This voltage
IN
P
835 kW
(c)
To
solve this
problem, we will ...
new
power
factor is
cos 3. 5
°
= 0.998 leading, and the reactive power
supplied by the motor is
3
sin
3
(
230 0
V
)(
21
=
=
4.5 A
)
sin
(
3. 5
)
52.2 kVAR
°...
Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
...
33 .6
9.6
A
=
=
=
°
A
0.4
2.0
+
+
&
A S
R
jX
j
Therefore the real
power consumed
by this
motor is
3
cos
3
(
)
480 V
(
3
=
=
3. 6 ...
V
⎞
286
0
V
=
°
E
,
A
m
32 7
18.
=
3
V
°
I
A
95.7
13. 3
=
A
°
The new
terminal voltage is
(
)
3
286 V
495
V
T
V
=
=...
Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
...
j
°
E
A
1 .33
42
=
.6
pu
°
The base phase voltage
of this motor is
6600 /
3
= 38 10 V,
so
E
A
is
A
E
(
)
1 .33
42.6
=
(
38 1
°
0 V
)
5067
=
...
1
.33
2
4.
°
9
0.6 63
20.2
pu
=
=
=
°
+
A
0.90
A S
R
jX
j
In
actual amps, this
current is
A
I
(
)
1404
A
(
0.6 63...
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