Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx
Electric Machinery Fundamentals (Solutions Manual) Part 11 ppsx
...
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
™
will
be constant ... power
P
and reactive power
Q
supplied or consumed by the machine under the
conditions in
part
(a)
. Is
the
machine operating within
its ratings under these circumstance...
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
... field current in amps, and
column
2 contains short-circuit
termina
current in amps.
5 -11.
(a)
What is
the
saturated synchronous reactance of this generator at the rated
conditions?
... to read these files and
calculate and
plot
X
S
is
shown
below.
% M-file: prob5_11c.m
%
M-file to calculate and
plot the
saturated
% synchronous reactance of...
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
...
A
)
E
A
j
=
°
+
&
°
E
A
11, 120
19.9
V
=
°
The resulting
voltage regulation is
11, 120
7621
RV
100%
45.9%
=
⋅
=
7621
...
power rating) of the generator will be reduced by
a
factor
of
5/6.
5
(
)
13.2 kV
11. 0 kV
=
=
T
V
,rated
6
(b)
If armature
resistance
is
ignored,
the pow...
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
...
2
(
=
=
)
2
(
)
&
=
111
CU
3
A
A
P
I
3
R
251 A
0.15
28.4 kW
F&
P
=
W
24 kW
110
114
⎜
OUT
P
100%
=
⋅
554 ...
)
335 A
=
=
⎞
A
V
I
Z
and the terminal voltage is
(
3.333
)
11
&
17
=
V
3
3
(
)
111 7 V
1934 V
T
V
V
⎞
=
=
=
(f)
To
restore
the
te...
Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
...
119 4
40.6
A
=
=
=
°
A
1.5
12.0
+
+
&
A S
R
jX
j
The reactive
power produced by the generator at this
point is
3
sin
3
(
)
7967 V
(
119
=
... plot is
shown below:
117
system? What
is the generator’s rate of shaft rotation after paralleling occurs?
118
5-5.
Assume that the field current...
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
... will be supplying its rated
real and reactive power.
(d)
Under
the
conditions
of
part
(c)
, which
are
the rated conditions
of the
generators, the
internal
generated
...
1.
=
5.137
f
sys
=
306.2
f
sys
=
59.61 Hz
5
sys
103.07
1.676
sys
118 .64
1.961
sys
f
f
f
The power supplied
by each generator
will
be...
Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
...
5247
=
A
3
3
(
)
11. 0
kV
A L
T
V
The power factor is 0.8
lagging, so
I
A
5247
36.8
=
7
A
°
.
The
phase voltage is 11. 0 kV
/
3
= 6351
V.
...
=
jX
S
j
8.18
°
=
°
(e)
Repeating
part
(d)
:
With
a
10%
decrease,
E
A2
=
11,
907 V
, and
sin
1
1
E
1
A
sin
sin
13,...
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
...
=
°
j
&
°
=
°
(b)
This motor
has 6 poles and
an electrical
frequency of
60
Hz, so its
rotation
speed
is
m
n
= 1200
r/min. The induced
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
... maximum power that
the
motor
can produce
at rated
speed with the value of
E
A
from part
(b)
is
158
⎤
m
S
X
(
)
3600 r/min
(
)
2.8
&
60 s
... when calculating
the V-curves.
d_nl = -1.27 *
pi/180;
% delta at
no-load
d_half =
-11. 2
* pi/180;
% delta at
half-load
d_full =
-22.7
* pi/180;
% delta
at
full...
Electric Machinery Fundamentals (Solutions Manual) Part 9 ppt
...
consuming
from
or
supplying
to the electrical system? How much reactive
power
Q
is this machine
consuming from or supplying to the electrical
system?
S
OLUTION
This ...
This is
consistent with
what
we learned about reactive power sharing in Chapter 5.
6 -11.
A
480-V,
100-kW,
50-Hz,
four-pole, Y-connected synchronous motor has a rated power ......
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