ECE330 – Spring 2014 6.2 Total per-phase complex power is

( ) 10050033001500 jjSTp +=+= kVA

Assuming 00∠== aLpaMpaLp VVV , the per-phase load complex power is

79.18779.1874558.2654520003

2300 000* jIVS aLpaLpLp +=∠=∠×∠=×= kVA

The per-phase complex power consumed by the motor is 071.1532.32479.8721.31279.18779.187100500 −∠=−=−−+=−= jjjSSS LpTpMp kVA

a) Line current taken by the motor

0

0

0** 71.1523.244

032300

71.15324320 −∠=∠

−∠===aMp

MpaMpaMl V

SII A

Hence, 071.1523.244 ∠=aMlI A

b) Power factor of the motor

( ) 9627.071.15cos 0 =−=PF leading (since negative phase difference) c) Ear of the motor

( )( ) 000 85.1715342.4702.146071.1523.244203

2300 −∠=−=∠−∠=−= jjIjxVE aMpsaMpar V

Torque angle is 085.17−=δ

6.8

a) Generator speed

30012

606060 =×==p

fns rpm

Torque of electric origin

25464812602

1108000

2

3

=×

××===ππω pf

PPT T

m

Te N.m

b) Phase current of the generator

( ) 5.135833400

8000000=== alap II A

Ear

Vap

jxsIap

Iap δ

005.1358 ∠=apI A (due to unity PF).

Assuming phase voltage is 00 0196303

3400 ∠=∠=apV V, Ear can be expressed as

( ) ( )( ) 000 69.342.238705.1358101963 ∠=∠+∠=+= jIxjVE apsapar V

Torque angle is 069.34=δ

c)

Assuming the magnetic core is not saturated, the new RMS value of arE is

7.28642.23872.1_ =×=newarE V

Due to constant input power, the new torque angle newδ must satisfy

( ) ( )newnewarar EE δδ sinsin _=

Therefore

( ) ( ) ( ) 4743.069.34sin2.1

1sinsin 0

_

=== δδnewar

arnew E

E => 031.28=newδ

The new stator current is

000

__ 37.221.1469

2.1

0196331.287.2864 −∠=∠−∠=−

=jjx

VEI

s

apnewarnewap A

The reactive power output: ( ) 32933806.01.146919633sin3 =×××=×××= θapap IVQ kVAR

6.10 a) Phase voltage is

5.3173

550 ==apV V

The torque angle can be determined from ( )

3

sin T

s

aar P

x

VE=

δ, which leads to

( ) 6847.05.3174603

2.1250000

3sin =

×××==

aar

sT

VE

xPδ => 021.43=δ

b) The new stator current is

Ear

Vap

jxsIap

Iap

δ

θ

000

23.388.2622.1

05.31721.43460 −∠=∠−∠=−

=jjx

VEI

s

aparap A

Power factor ( ) 9984.0cos == θPF The generated reactive power is

( ) ( ) 105.1405633.088.2625.3173sin3 =×××=×××= θapap IVQ kVAR

c) The new generator current is

( ) 08.3288.05.3173

250000

cos3_ =××

==newap

Tnewap V

PI

ϕ A

The new excitation voltage can be obtained by ( ) 010

_ 63.2901.6378.0cos08.3282.105.317 ∠=−∠×+∠=+= −jIjxVE newapsapar

The new torque angle is 063.29=newδ

6.13 a) The reactive power required by the three-phase load is

( )( ) ( )( ) 5.228.0costan300008.0costan 11 =×== −−LL PQ kVAR

b) Since the PF of the motor is 0 (zero real power), the magnitude of the motor’s current is

( ) ( ) 48.562303

22500

3===

Ml

MMl

V

QI A

c) Since the load has a lagging PF, the motor must have a leading PF. Assuming a wye-connected stator, the per-phase internal voltage is

75.24548.5623

230 =×+=+= MpsMpar IxVE V

d) The new reactive power required by the combination of three-phase load and motor is

( )( ) ( )( ) 861.995.0costan3000095.0costan 11 =×== −−Lnew PQ kVAR

Hence, the reactive supplied by the motor is 639.125.22861.9_ −=−=−= LnewnewM QQQ kVAR

The magnitude of the motor’s current is

( ) ( ) 73.312303

12639

3

_ ===Ml

newMMl

V

QI A

Since the load has a lagging PF, the motor must have a leading PF. Assuming a wye-connected stator, the new per-phase internal voltage is

Vap

Iap

-jxsIap

Earp

25.19673.3123

230_ =×+=+= MpsMpnewar IxVE V

e) Assuming the magnetic core is not saturated, the new machine’s field current as a percentage of the old field current can be obtained by

7986.075.245

25.196_

_

_ ===ar

newar

oldf

newf

E

E

I

I

This means the field current is reduced by 20% from the value in c).