synchronous machines
DESCRIPTION
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 30 o , determine the following The power The torque The pull-out torque of the motor. - PowerPoint PPT PresentationTRANSCRIPT
Synchronous Machines
Example 1
A 150 kW, 460 V, 1200 r/min, Y-connected synchronous motor has a synchronous reactance of 0.8 /phase. The internal voltage is 300 V. If the power angle is 30o, determine the following
a) The powerb) The torquec) The pull-out torque of the motor
mNT
torqueoutPullc
mNT
torqueOutputb
kWX
EVP
powerOutput
VVvoltagephaseThea
o
o
s
ft
t
.800
60
21200
90sin8.0
300266
:)
.400
60
21200
000,50
:)
5030sin8.0
300266sin
:
2663
460,)
max
Example 1 Solution
Example 2
A 2000-hp, 2300-V, unity power factor, Y-connected, 30-pole, 60-Hz synchronous motor has a synchronous reactance of 1.95 /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 phaseRated current = 497000/1328 = 374 A/phase-Y
From the phasor diagram,
The maximum power and torque,
mkNT
rad
phasethreekWP
VXIVE
s
satf
.2.12383096
sec/860230
2
309695.1
151513283
151595.13741328
max
max
2222
1328 V
1515 V
374 A
729 V
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. The apparent power supplied to the motorb. The power factorc. The reactive power absorbedd. The number of poles on the rotor
Example 3 Solution
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95720002217)
902.022172000cos)
221732040003)
22
polesofNod
kVARQc
PFb
kVASa
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. Motor speed and mechanical power outputb. Power angle, c. Position of the rotor polesd. Power factore. 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,
will increase tVSI 3
Example 5
A 4000-hp, 6.9-kV synchronous motor has a synchronous reactance of 10 /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 powerb. The line currentc. The internal voltage per phase with corresponding
phasor diagramd. The power anglee. The total reactive power supplied to the systemf. The approximate maximum power [in hp] the motor
can develop without pulling out of step
Example 5 Solution
VVEXIVE
VVc
AIb
kVAS
kWP
kWPa
tfsatf
t
a
i
58892790cos2
2789.0cos;39843
6900)
3.2899.63
3457)
345789.03076
307697.02984
2984746.04000)
0022
0
0
Vt
Ef
Ia
IaXs
270
Example 5 Solution (cont’d)
hpkWPf
kVARQe
X
EVPd
s
ftphaseper
9436703990sin10
588939843)
157830763457)
9.253
3076000sin)
0max
22
0_0
Example 6
A 1500-kW, 4600-V, 600 r/min, 60-Hz synchronous motor has a synchronous reactance of 16 /phase and a stator resistance of 0.2 /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. The power dissipated in the armature at 600 r/minb. The power dissipated in the armature at 150 r/minc. The kinetic energy at 600 r/mind. The kinetic energy at 150 r/mine. The time required for the speed to fall from 600 r/min
to 150 r/min
Example 6 Solution
460015016
,tan
1560015060
,
6006001502400
min/150,
int,)
5.132.015033
150162400
16162.0)
22
2222
s
f
aadissipated
fa
sa
X
soandspeedthetoalproportionalsoiscereacssynchronouThe
Hzf
soandspeedthetoalproportionalsoisfrequencyThe
VE
rtodropsspeedthewhenlyConsequent
speedtoalproportionisvoltageernalthefixediscurrentexcitingtheBecauseb
kWRIP
phasesthreeallindissipatedPower
AZE
I
phaseperCurrent
XRZa
IaEf Vt=0
jXs Ra+
+
Example 6 Solution (Cont’d)
kJK
isratenergykineticThed
kJJnK
isratenergykineticThec
sametheisphasesthree
indissipatedpowerunchangedremainscurrentcircuitshorttheAs
kWRIP
phasesthreeallindissipatedPower
AZE
I
phaseperCurrent
XRZ
isratphaseperimpedancenewThe
E
E
aadissipated
fa
sa
9.331502751048.5
min/150)
5.5426002751048.51048.5
min/600)
.
,
5.132.015033
1504600
442.0
min/150
232
23231
22
2222
Example 6 Solution (Cont’d)
.min
.sec7.376.5085.13
min/150min/600
.tan
6.5089.335.542
min/150min/600)
21
alsterstatortheacross
conectedwereresistorsexternalifsoonermuchstopwouldmotorThe
tttWP
isrtorfromdropto
speedthefortimeTheceresisarmaturetheinheataslostisenergyThe
kJKKW
isrtorfromondeceleratiinenergykineticinlossThee
EE