ece330 hw5 solution
DESCRIPTION
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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).