Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf
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With
a
20%
decrease,
E
A2
=
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 V
146
A2
⎞
10,
584
35.8
70
°
44
0
780
14.0
A
E
V
A
I
=
=
jX
S
8.18
j
°
=
°
With
a
25%
decrease,
E
A2
=
9,
923 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 V
A2
⎞
9,
923
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
current
below.
I
A
as a function
of
E
A
is shown
% M-file: prob5_28f.m
%
M-file to calculate and
plot the
armature current
% supplied to an infinite bus as Ea is varied.
% Define
values for this
generator
Ea
= (0.65:0.01:1.00)*13230; %
Ea
Vp
= 7044;
%
Phase
voltage
d1
= 27.9*pi/180;
%
torque angle
at
full Ea
Xs
= 8.18;
%
Xs
(ohms)
% Calculate
delta for
each Ea
d = asin( 13230 ./
Ea
.*
sin(d1));
%
Calculate Ia for each
flux
Ea
= Ea .* ( cos(d) +
j.*sin(d) );
Ia
= ( Ea -
Vp ) ./ (j*Xs);
% Plot the armature current
versus Ea
figure(1);
plot(abs(Ea)/1000,abs(Ia),'b-','LineWidth',2.0);
title ('\bfArmature
current versus
\itE_{A}\rm');
xlabel ('\bf\itE_{A}\rm\bf (kV)');
ylabel ('\bf\itI_{A}\rm\bf (A)');
grid on;
hold off;
147
The
resulting plot is
shown below:
148
Chapter 6: Synchronous Motors
6-1.
A 480-V, 60 Hz, four-pole synchronous motor draws 50 A from
the
line
at
unity
power
factor
and
full
load.
Assuming that the
motor is lossless, answer the following questions:
(a)
What
is the output
torque of this motor? Express the answer
both in newton-meters
and in pound-feet.
(b)
What
must
be done to change the power factor to 0.8 leading? Explain your answer, using phasor
diagrams.
(c)
What
will the magnitude of the line current be
if the power factor is adjusted
to 0.8
leading?
S
OLUTION
(a)
If this motor is assumed lossless,
then the input power is equal to the output power. The input
power to
this motor is
(
)(
=
=
)(
)
=
IN
3
co
T
L
s
P
V
I
⎝
The output torque would be
3
480
V
50 A
1.0
41.6 kW
⎮
OUT
P
=
=
41.6 kW
221 N
=
m
⊕
LOAD
⎤
m
In
English
units,
(
)
1800 r/min
1 min
2
rad
60 s
1 r
(
)(
)
7.04
OU
P
T
7.04
41.6
=
=
kW
163 lb
ft
=
⊕
⎮
LOAD
n
m
(
)
1800 r/min
(b)
To
change the motor’s power factor
to 0.8 leading, its
field current must be increased. Since the
power supplied to the load is independent of the field current level, an
increase
in
field
current
increases
A
E
while keeping the
distance
E
A
sin
™
constant.
This increase in
E
A
changes the angle
of
the
current
I
A
,
eventually causing it
to reach a power factor
of 0.8
leading.
P
I
A
2
I
A
1
⎞
V
jX
I
S
A
}
P
Q
I
sin
⎝
A
(c)
The
magnitude
of the line
current will be
EE
A
1
A
2
I
P
41.6 kW
=
=
62.5 A
=
L
3
PF
3
(
)
480 V
(
0.8
)
T
V
6-2.
A 480-V, 60 Hz,
400-hp
0.8-PF-leading six-pole
-connected synchronous motor
has a synchronous
reactance of 1.1
&
and
negligible armature
resistance. Ignore its friction, windage, and core losses for the
purposes of this
problem.
149
(a)
If this motor is initially supplying 400 hp
at
0.8 PF
lagging, what are
the
magnitudes
and
angles of
E
A
and
I
A
?
(b)
How
much
torque is this motor producing? What
is the torque angle
™
? How near is this value to the
maximum possible induced torque
of
the
motor
for
this field current setting?
(c)
If
E
A
is
increased by 15 percent, what is the new magnitude of the armature current? What
is the
motor’s new power
factor?
(d)
Calculate and
plot the motor’s
V-curve for this
load condition.
S
OLUTION
(a)
If losses
are
being ignored,
the
output power is equal to
the
input power, so the input
power will be
(
)(
)
IN
P
400 hp
746 W/hp
298.4 kW
=
=
This situation is
shown in the phasor diagram below:
V
⎞
I
A
The line current
flow under these circumstances is
jX
I
S
A
A
E
I
P
298.4 kW
=
=
449 A
=
L
3
PF
3
(
)
480
V
(
0.8
)
T
V
Because the motor
is
-connected,
the
corresponding phase current is
449
/
3
259 A
I
=
=
. The angle
of
A
the current
is
cos
1
(
)
0.80
36.87
=
°
, so
I
A
=
E
V
I
259
36
=
.87
A
°
.
The internal generated voltage
E
A
is
A
S
⎞
jX
A
(
)
480
0
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
Hz, so its
rotation
speed
is
m
n
= 1200
r/min. The induced
torque is
⎮
OUT
P
=
=
298.4 kW
2375 N
=
m
⊕
ind
⎤
m
(
)
1200 r/min
1 min
2
rad
60 s
1 r
The maximum possible
induced
torque for the motor at this
field setting is
3
3
(
)
480 V
(
384
V
)
⎮
⎞
A
V
E
=
=
ind,max
1 min
2
rad
4000 N
=
m
⊕
⎤
m
S
X
(
)
1200 r/min
(
)
1.1
&
60 s
1 r
(c)
If the magnitude
of
the internal generated voltage
be
found
from the fact that
E
A
sin
™
P
= constant
.
(
)
=
=
=
2
1
1.15
1.15
E
E
38
A
A
4 V
441.6 V
E
A
is increased by
15%, the new
torque
angle
can
150
sin
E
1
A
sin
sin
384 V
™
™
sin
=
=
(
)
1 1
36
.4
31.1
°
=
°
2 1
E
A2
The new
armature current
is
441.6 V
⎞
A
2
V
E
I
480
0
V
°
4
41.6
31
.1
V
°
227
24.1
A
=
=
=
°
2A
jX
S
j
1.1
&
The magnitude of the armature current is 227
A, and
the
power factor is cos (-24.1
°
)
= 0.913 lagging.
(d)
A MATLAB program to calculate
and plot
the
motor’s V-curve
is shown
below:
%
M-file: prob6_2d.m
%
M-file create a
plot of armature current
versus Ea
%
for
the synchronous
motor
of
Problem 6-2.
%
Initialize values
Ea =
(1:0.01:1.70)*384;
%
Magnitude
of
Ea
volts
Ear = 384;
%
Reference
Ea
deltar = -36.4
* pi/180;
%
Reference
torque angle
Xs =
1.1;
%
Synchronous reactance
Vp =
480;
%
Phase voltage at 0
degrees
Ear = Ear * (cos(deltar) +
j * sin(deltar));
%
Calculate delta2
delta2 = asin
( abs(Ear) ./ abs(Ea)
.*
sin(deltar) );
%
Calculate the phasor Ea
Ea =
Ea
.*
(cos(delta2)
+ j .* sin(delta2));
%
Calculate Ia
Ia =
( Vp - Ea
) / (
j * Xs);
%
Plot the
v-curve
figure(1);
plot(abs(Ea),abs(Ia),'b','Linewidth',2.0);
xlabel('\bf\itE_{A}\rm\bf (V)');
ylabel('\bf\itI_{A}\rm\bf (A)');
title ('\bfSynchronous Motor V-Curve');
grid
on;
151
