Download Example 5 Solution

Synchronous Machines
Example 1
A 150 kW, 460 V, 1200 r/min, Y-connected synchronous
motor has a synchronous reactance of 0.8 W/phase. The
internal voltage is 300 V. If the power angle is 30o,
determine the following
a) The power
b) The torque
c) The pull-out torque of the motor
Example 1 Solution
a ) The phase voltage, Vt  460
3
 266 V
Output power :
Vt E f
266  300
P
sin  
 sin 30 o  50 k W
Xs
0.8
b) Output torque:
50,000
T
 400 N .m
2
1200 
60
c) Pull  out torque:
266  300
 sin 90 o
0.8
Tmax 
 800 N .m
2
1200 
60
Example 2
A 2000-hp, 2300-V, unity power factor, Y-connected, 30pole, 60-Hz synchronous motor has a synchronous
reactance of 1.95 W/phase. Neglect all losses. Compute
the maximum power and torque which this motor can
deliver if it is supplied with power directly from a 60-Hz,
2300-V supply. Assume field excitation is maintained
constant at the value which will result in unity power
factor at rated load.
Example 2 Solution
Rated kVA = 2000X0.746 = 1492 kVA, three phase
= 497 kVA/phase
Rated voltage = 2300/1.732=1328 V per phase
Rated current = 497000/1328 = 374 A/phase-Y
374 A
1515 V
1328 V
729 V
From the phasor diagram,
E f  Vt 2  I a X s   1328 2  374  1.95   1515 V
2
2
The maximum power and torque,
1328  1515
 3096 kW three  phase
1.95
 2 
 s   2 60   8 rad / sec
 30 
Tmax  3096
 123 .2 kN.m
8
Pmax  3 
Example 3
A three-phase, 225 r/min synchronous motor is connected
to a 4-kV, 60-Hz line draws a current of 320 A and
absorbs 2000 kW. Calculate
a.
b.
c.
d.
The apparent power supplied to the motor
The power factor
The reactive power absorbed
The number of poles on the rotor
Example 3 Solution
a ) S  3  4000  320  2217 k VA
b) PF  cos   2000
2217
 0.902
c) Q  2217 2  2000 2  957 k VAR
d ) No. of poles  120  60 
225
 32
Example 4
A three-phase synchronous motor rated at 800-hp, 2.4-kV,
60-Hz operates at unity power factor. The line voltage
suddenly drops to 1.8 kV, but the exciting current remains
unchanged. Explain how the following quantities are
affected.
a.
b.
c.
d.
e.
Motor speed and mechanical power output
Power angle, 
Position of the rotor poles
Power factor
Stator current
Example 4 Solution
a. The speed is constant, hence the load does not know that
the line voltage has dropped. Therefore, the mechanical
power will remain unchanged
b. P=(VtEf/Xs)sin, P, Ef and Xs are the same but Vt has fallen;
consequently sin must increase, which means that 
increases
c. The poles fall slightly behind their former position,
because  increases
d. Terminal voltage is smaller than before, the motor internal
voltage is bigger than the terminal voltage and as a result,
the power factor will be less than unity and leading
e. As power factor is less than unity, apparent power S is
greater now. The terminal voltage is smaller, I  S 3Vt
will increase


Example 5
A 4000-hp, 6.9-kV synchronous motor has a synchronous
reactance of 10 W/phase. The stator is connected in wye, and
the motor operates at full-load (4000 hp) with a leading power
factor of 0.89. If the efficiency is 97%, calculate the
following:
a. The apparent power
b. The line current
c. The internal voltage per phase with corresponding
phasor diagram
d. The power angle
e. The total reactive power supplied to the system
f. The approximate maximum power [in hp] the motor
can develop without pulling out of step
Example 5 Solution
a ) P0  4000  0.746  2984 k W
Pi  2984
0.97
 3076 k W
S  3076
 3457 k VA
0.89
b) I a  3457
 289 .3 A
3  6 .9
c) Vt  6900
 3984 V ; cos   0.89    27 0
3


Ia
Vt
270
IaXs
Ef


E f  Vt 2  I a X s   2 E f Vt cos 90 0  27 0  5889 V
2
Example 5 Solution (cont’d)
d ) P0 _ per phase 
Vt E f
Xs
sin  
3076000
   25.9 0
3
e) Q  3457 2  3076 2  1578 kVAR
f ) Pmax  3 
3984  5889
 sin 90 0  7039 k W  9436 hp
10
Example 6
A 1500-kW, 4600-V, 600 r/min, 60-Hz synchronous motor has
a synchronous reactance of 16 W/phase and a stator resistance
of 0.2 W/phase. The excitation voltage is 2400 V, and the
moment of inertia of the motor and its load is 275 kg.m2. We
wish to stop the motor by short-circuiting the armature while
keeping the dc rotor current fixed. Calculate
a.
b.
c.
d.
e.
The power dissipated in the armature at 600 r/min
The power dissipated in the armature at 150 r/min
The kinetic energy at 600 r/min
The kinetic energy at 150 r/min
The time required for the speed to fall from 600 r/min
to 150 r/min
Example 6 Solution
a ) Z  Ra2  X s2  0.2 2  16 2  16 W
Current per phase
Ef
Ia 
 2400  150 A
Z
16
Power dissipated in all three phases
jXs
+
Ef
Ia
Ra
+
Vt=0
Pdissipated  3  I a2 Ra  3  150 2  0.2  13.5 kW
b) Becausethe exciting currentis fixed, the int ernal voltageis proportional to speed
Consequently , when the speed dropsto 150 r / min
E f  2400  150
 600 V
600
The frequencyis also proportional to the speed, and so
f  60  150
 15 Hz
600
The synchronous reac tan ce is also proportional to the speed, and so
X s  16  150
 4W
600






Example 6 Solution (Cont’d)
The newimpedance per phaseat 150 r / min is
Z  Ra2  X s2  0.2 2  4 2  4 W
Current per phase
Ia 
Ef
 600  150 A
Z
4
Power dissipated in all three phases
Pdissipated  3  I a2 Ra  3  150 2  0.2  13.5 k W
As the short  circuit currentremains unchanged, power dissipated in
three phasesis the same.
c) The k ineticenergyat 600 r / min is
K E1  5.48  10 3  Jn 2  5.48  10 3  275  600 2  542 .5 k J
d ) The k ineticenergyat 150 r / min is
K E 2  5.48  10 3  275  150 2  33.9 k J
Example 6 Solution (Cont’d)
e) The loss in kineticenergyin deceleration from600 r / min to 150 r / min is
W  K E1  K E 2  542 .5  33.9  508 .6 kJ
The energyis lost as heat in the armatureresis tan ce. The time for the speed
to drop from600 r / min to 150 r / min is
P  W  13.5  508 .6  t  37.7 sec .
t
t
The motor would stop much soonerif externalresistors were conected
acrossthe stator ter min als.