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

3222560120.)

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