Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
Electric Machinery Fundamentals (Solutions Manual) Part 10 ppt
... efficiency of this
motor
at
full load is
(
)(
)
⎜
OUT
P
100 %
=
⋅
2100
=
0 hp
746 W/hp
100 %
⋅
97.6%
=
IN
P
16.05 MW
(f)
The output
torque in SI ... supplied to the motor
is
10
MW
/
16.05
MW = 0.623.
The maximum
E
A
is 5067 V
or
1.33 pu,
so
with
E
A
set to maximum
and the motor
consuming 10 MW, the to...
Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt
...
OUT
P
P
CU
P
F&W
P
core
P
stray
P
870.4 kW
η
OUT
P
100 %
800 kW
100 %
91.9%
=
⋅
=
⋅
=
IN
P
870.4 kW
(b)
If the generator
is ...
The generator is actually consuming
reactive power at this time.
5-8.
A 480-V,
100 -kW, two-pole, three-phase, 60-Hz
synchronous generator’s
prime
mover
has
a
no-load...
Electric Machinery Fundamentals (Solutions Manual) Part 6 ppt
... you were an engineer planning a new electric co-generation facility for a
plant
with
excess
process steam.
You
have a
choice
of
either
two 10 MW turbine-generators or a single ... and
more efficient
than
two
10
MW
generators, but if the 20 MW generator goes down all 20 MW of generation would be lost at
once.
If two
10
MW
generators
are
chosen,...
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 ... 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 its new
value?...
Electric Machinery Fundamentals (Solutions Manual) Part 1 pdf
...
P
1
5
10
Therefore, a 10- pole synchronous
motor
must
be
coupled to a 12-pole
synchronous generator
to accomplish
this frequency
conversion.
5-2.
A
2300-V
100 0-kVA ... synchronous generator.
% Calculate
the waveforms for times from 0 to 1/30 s
Q = -4 810;
DE =
5567;
S = 100 0;
% Get
points for stator current limit
theta
= -95:1:95;...
Electric Machinery Fundamentals (Solutions Manual) Part 3 docx
...
nl
fl
100 %
=
⋅
3630 r/m
=
in
3570 r/min
100 %
1.
⋅
68%
=
n
fl
3570 r/min
The speed droop of generator 2 is given by
n
n
DS
2
nl
fl
100 %
18
=
...
)
(
)
(
)
(
)
=
+
100 kW
0.1 MW/Hz
60.5
Hz
sys
0.15 MW/Hz
60.0 Hz
sys
f
f
(
)
(
)
100 kW
6050 kW
0 .10 MW
=
/Hz
(
)
sy...
Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx
...
at
SCC
is 532 A.
I
F
= 0.50
A, the air-gap
line voltage is 104 0
V, and
the
X
Su
(
)
104 0 V
/
3
1.13
=
=
&
532 A
(c)
This
task
can ...
880.400
129
generators
would have
more margin than
this.
131
E
A
2825
10. 9
=
V
°
Problems 5-11
to
5-21
refer to a two-pole Y-connected synchronous gene...
Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx
...
75%
to 100 %
of the flux
at rated conditions.
(c)
Plot the reactive power supplied by this generator as a
function of the flux
⎞
as the flux
is
varied
from
75% to
100 % of ...
19.9
V
=
°
The resulting
voltage regulation is
11,120
7621
RV
100 %
45.9%
=
⋅
=
7621
(b)
If the generator is
to be operated at 50 Hz with t...
Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
...
10, 584 V
, and
sin
1
1
E
1
A
sin
sin
13,
230
™
™
V
sin
=
=
27.9
35.8
°
=
°
2
2
E
A
2
Therefore, the new armature
current is
10, 584 ...
148
A2
⎞
10,
584
35.8
70
°
44
0
780
14.0
A
E
V
A
I
=
=
jX
S
8....
Electric Machinery Fundamentals (Solutions Manual) Part 8 pps
...
(
)
6
e
f
m
n
P
=
=
120
120
15 Hz
=
A speed of 100 0 r/min
corresponds
to a frequency
of
(
)
100 0 r/min
e
f
m
n
P
=
=
120
120
(
)
6
50 Hz
=
... V-Curve
250
240
230
220
210
200
350 400 450 500 550 600 650 700
E (V)
A
6-3.
A 2300-V 100 0-hp 0.8-PF leading 60-Hz two-pole
Y-connec...
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