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POLYPHASE CIRCUITS. LEARNING GOALS. Three Phase Circuits Advantages of polyphase circuits. Three Phase Connections Basic configurations for three phase circuits. Source/Load Connections Delta-Wye connections. Theorem - PowerPoint PPT Presentation
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POLYPHASE CIRCUITSLEARNING GOALS
Three Phase CircuitsAdvantages of polyphase circuits
Three Phase ConnectionsBasic configurations for three phase circuits
Source/Load ConnectionsDelta-Wye connections
THREE PHASE CIRCUITS
))(240cos()())(120cos()(
))(cos()(
VtVtvVtVtv
VtVtv
mc
mbn
man
VoltagesPhase ousInstantane
ai
bi
ci
)240cos()()120cos()(
)cos()(
tItitItitIti
mc
mb
ma
Currents Phase Balanced
2120mV
)()()()()()()( titvtitvtitvtp ccnbbnaan power ousInstantane
TheoremFor a balanced three phase circuit the instantaneous power is constant
)(cos2
3)( WIVtp mm
Proof of TheoremFor a balanced three phase circuit the instantaneous power is constant
)(cos2
3)( WIVtp mm
)240cos()240cos()120cos()120cos(
)cos(cos)(
)()()()()()()(
tttt
ttIVtp
titvtitvtitvtp
mm
ccnbbnaan
power ousInstantane
)cos()cos(21coscos
3cos cos(2 )( ) cos(2 240 )
2cos(2 480 )
m m
tV Ip t t
t
)120sin(sin)120cos(cos)120cos()120sin(sin)120cos(cos)120cos(
coscos
0)120cos()120cos(cos
Proof
Lemma5.0)120cos(
0)120cos()120cos(cos
2cos( 240) cos( 120)cos( 480) cos( 120)
t
THREE-PHASE CONNECTIONS
Positive sequencea-b-c
Y-connectedloads Delta connected loads
SOURCE/LOAD CONNECTIONS
BALANCED Y-Y CONNECTION
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequencephase voltages
abV
bcV
caV
Line voltages
| | 0 | | 120
| | 1 (cos120 sin120)
1 3| | | |2 2
3 | | 30
ab an bn
p p
p
p p
p
V V VV V
V j
V V j
V
210||3
90||3
pca
pbc
VV
VV
VoltageLine ||3 pL VVY
cnc
Y
bnb
Y
ana Z
VIZVI
ZVI ;;
120||;120||;|| LcLbLa IIIIII
0 ncba IIII For this balanced circuit it is enough to analyze one phase
Relationship betweenphase and line voltages
LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuit 30208abV
Determine the phase voltages
Balanced Y - Y
Positive sequencea-b-c
30||3 pab VV
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequencephase voltages
)3030(3
|| ab
anVV
30by lags aban VV
30208abV
6012018012060120
an
bn
an
VVV
The phasor diagram could be rotated by any angle
LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuitsource 120( ) , 1 1 , 20 10rmsphase line phaseV V Z j Z j
Determine line currents and load voltages
Because circuit is balanceddata on any one phase issufficient
0120 Chosenas reference
120 0120 120120 120
an
bn
cn
VVV
Abc sequence
rmsAj
VI anaA
)(65.2706.565.2771.23
01201121
rmsAIrmsAI
cC
bB
)(65.2712006.5)(65.2712006.5
57.2636.22)1020( aAaAAN IjIV
rmsVVAN )(08.115.113
rmsVVrmsVV
CN
BN
)(92.11815.113)(08.12115.113
DELTA CONNECTED SOURCES
Relationship betweenphase and line voltages
30||3 pab VV
30by lags aban VV
120120
0
Lca
Lbc
Lab
VVVVVV
903
1503
303
Lcn
Lbn
Lan
VV
VV
VV
15012090120
30120
cn
bn
an
VVV
18020860208
60208
ca
bc
ab
VVVExample
Convert to an equivalent Y connection
LEARNING EXAMPLEDetermine line currents and line voltages at the loads
Source is Delta connected.Convert to equivalent Y
120120
0
Lca
Lbc
Lab
VVVVVV
903
1503
303
Lcn
Lbn
Lan
VV
VV
VV
Analyze one phase
rmsAj
IaA )(14.4938.92.41.1230)3/208(
rmsVjVAN )(71.3065.11819.4938.9)412(
Determine the other phases using the balance
rmsAIrmsAI
cC
bB
)(86.7138.9)(14.16938.9
3 118.65 0.71ABV
3 118.65 120.71
3 118.65 119.29BC
CA
V
V
DELTA-CONNECTED LOAD
Method 1: Solve directly
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequencephase voltages
210||3
90||3
30||3
pca
pbc
pab
VV
VV
VV
120||
120||
||
IZVI
IZVI
IZVI
CACA
BCBC
ABAB
currents phase Load
BCCAcC
ABBCbB
CAABaA
IIIIIIIII
currents Line
Method 2: We can also convert the delta connected load into a Y connected one. The same formulas derived for resistive circuits are applicable to impedances
3ZZY case Balanced
ZL
ABaA
LaAY
anaA Z
VIIZVI
3/||
3/||||||
ZLZZ ||
Z 30
30||3||
lineline II
Line-phase currentrelationship
30||3||
lineline II
Line-phase currentrelationship
iprelationsh voltagephase-Line
30
||3||
phase
phaseVV
LEARNING EXTENSION
currents phase the Find .4012 aAI
19093.65093.6
7093.6
CA
BC
AB
III
Y
Y baab RRR
)(|| 312 RRRRab
321
312 )(RRRRRRRR ba
321
213 )(RRRRRRRR cb
321
321 )(RRRRRRRR ac
SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra
a
b
b
a
RRRR
RR
RR 1
33
1 c
b
c
b
RRRR
RR
RR 1
21
2
REPLACE IN THE THIRD AND SOLVE FOR R1
YRRR
RRR
RRRRRR
RRRRRR
c
b
a
321
13
321
32
321
21
YR
RRRRRRR
RRRRRRRR
RRRRRRRR
a
accbba
c
accbba
b
accbba
3
2
1
tionsTransformaY
OF REVIEW
3321
RRRRRR Y
LEARNING EXAMPLE
02.3752.1254.710 jZ
Delta-connected load consists of 10-Ohm resistance in serieswith 20-mH inductance. Source is Y-connected, abc sequence,120-V rms, 60Hz. Determine all line and phase currents
rmsVVan )(30120
54.7020.0602inductanceZ
rmsAjZ
VI ABAB )(98.2260.16
54.710603120
30||3||
lineline II
Line-phase currentrelationship
iprelationsh voltagephase-Line
30
||3||
phase
phaseVV
rmsAIaA )(02.775.28
rmsAIrmsAI
CA
BC
)(98.14260.16)(02.9760.16
rmsAIrmsAI
cC
bB
)(98.11275.28)(02.12775.28
02.3717.4YZ
Alternatively, determine first the line currentsand then the delta currents
Polyphase
