Ch11 polyphase
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Transcript of Ch11 polyphase
POLYPHASE CIRCUITS
LEARNING 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()(
VtVtv
VtVtv
VtVtv
mc
mbn
man
°−=°−=
=
ωωω
VoltagesPhase ousInstantane
ai
bi
ci
)240cos()(
)120cos()(
)cos()(
°−−=°−−=
−=
θωθωθω
tIti
tIti
tIti
mc
mb
ma
Currents Phase Balanced
2120=mV
)()()()()()()( titvtitvtitvtp ccnbbnaan ++=power ousInstantane
TheoremFor a balanced three phase circuit the instantaneous power is constant
)(cos2
3)( WIV
tp mm θ=
Proof of TheoremFor a balanced three phase circuit the instantaneous power is constant
)(cos2
3)( WIV
tp mm θ=
−−−+−−−+
−=
++=
)240cos()240cos(
)120cos()120cos(
)cos(cos
)(
)()()()()()()(
θωωθωω
θωω
tt
tt
tt
IVtp
titvtitvtitvtp
mm
ccnbbnaan
power ousInstantane
[ ])cos()cos(2
1coscos βαβαβα ++−=
3cos cos(2 )
( ) cos(2 240 )2
cos(2 480 )
m m
tV I
p t t
t
θ ω θω θω θ
+ −= + − −
+ − −
)120sin(sin)120cos(cos)120cos(
)120sin(sin)120cos(cos)120cos(
coscos
0)120cos()120cos(cos
φφφφφφ
φφ
φφφ
−=++=−
=
=++−+Proof
Lemma5.0)120cos( −=
0)120cos()120cos(cos =++−+ φφφ
2
cos( 240) cos( 120)
cos( 480) cos( 120)
tφ ω θφ φφ φ
= −− = +− = −
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 V
V 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
VI
Z
VI
Z
VI === ;;
°+∠=°−∠=°∠= 120||;120||;|| θθθ LcLbLa IIIIII
0==++ ncba IIII For this balanced circuit it is enough to analyze one phase
Relationship betweenphase and line voltages
LEARNING EXAMPLE For 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
|| °−°−∠=∴ aban
VV
°30by lags aban VV
°−∠= 30208abV
°∠=°−∠=
°−∠=
60120
180120
60120
an
bn
an
V
V
V
The phasor diagram could be rotated by any angle
LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit
source 120( ) , 1 1 , 20 10rmsphase line phaseV V Z j Z j= = + Ω = + Ω
Determine line currents and load voltages
Because circuit is balanceddata on any one phase issufficient
°∠0120Chosenas reference
120 0
120 120
120 120
an
bn
cn
V
V
V
= ∠ °= ∠ − °= ∠ °
Abc sequence
rmsA
j
VI anaA
)(65.2706.5
65.2771.23
0120
1121
°−∠=°∠
°∠=+
=
rmsAI
rmsAI
cC
bB
)(65.2712006.5
)(65.2712006.5
°−∠=°−−∠=
°∠×=+×= 57.2636.22)1020( aAaAAN IjIV
rmsVVAN )(08.115.113 °−∠=
rmsVV
rmsVV
CN
BN
)(92.11815.113
)(08.12115.113
°∠=°−∠=
DELTA CONNECTED SOURCES
Relationship betweenphase and line voltages
°∠= 30||3 pab VV
°30by lags aban VV
⇒
°∠=°−∠=
°∠=
120
120
0
Lca
Lbc
Lab
VV
VV
VV
°∠=
°−∠=
°−∠=
903
1503
303
Lcn
Lbn
Lan
VV
VV
VV
°∠=°−∠=
°∠=
150120
90120
30120
cn
bn
an
V
V
V
⇒
°∠=°−∠=
°∠=
180208
60208
60208
ca
bc
ab
V
V
VExample
Convert to an equivalent Y connection
LEARNING EXAMPLE Determine line currents and line voltages at the loads
Source is Delta connected.Convert to equivalent Y
⇒
°∠=°−∠=
°∠=
120
120
0
Lca
Lbc
Lab
VV
VV
VV
°∠=
°−∠=
°−∠=
903
1503
303
Lcn
Lbn
Lan
VV
VV
VV
Analyze one phase
rmsAj
IaA )(14.4938.92.41.12
30)3/208( °−∠=+
°−∠=
rmsVjVAN )(71.3065.11819.4938.9)412( °−∠=°−∠×+=
Determine the other phases using the balance
rmsAI
rmsAI
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||
||
θ
θ
θ
IZ
VI
IZ
VI
IZ
VI
CACA
BCBC
ABAB
currents phase Load
BCCAcC
ABBCbB
CAABaA
III
III
III
−=−=−=
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
3∆= ZZY case Balanced
−=
=⇒∠== ∆
ZL
ABaA
LaAY
anaA Z
VI
IZ
VI
θθθ 3/||
3/||||
||
ZLZZ θ∠=∆ ||
Zθθ −°=∆ 30
°−==
∆
∆
30
||3||
θθ lineline II
Line-phase currentrelationship
θ
°−==
∆
∆
30
||3||
θθ lineline II
Line-phase currentrelationship
iprelationsh voltagephase-Line
°+=
=
∆
∆
30
||3||
phase
phaseVV
θθ
LEARNING EXTENSION
currents phase the Find .4012 °∠=aAI
°∠=°−∠=
°∠=
19093.6
5093.6
7093.6
CA
BC
AB
I
I
I
Y→∆
∆ → Y baab RRR +=
)(|| 312 RRRRab +=
321
312 )(
RRR
RRRRR ba ++
+=+
321
213 )(
RRR
RRRRR cb ++
+=+
321
321 )(
RRR
RRRRR ac ++
+=+
SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra
a
b
b
a
R
RRR
R
R
R
R 13
3
1 =⇒=c
b
c
b
R
RRR
R
R
R
R 12
1
2 =⇒=
REPLACE IN THE THIRD AND SOLVE FOR R1
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
→∆++
=
++=
++=
321
13
321
32
321
21
∆−
++=
++=
++=
Y
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
a
accbba
c
accbba
b
accbba
3
2
1
tionsTransformaY
OF REVIEW↔∆
3321∆
∆ =⇒=== RRRRRR Y
LEARNING EXAMPLE
°∠=Ω+=∆ 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 °∠=
Ω=××= 54.7020.0602πinductanceZ
rmsAjZ
VI ABAB )(98.2260.16
54.710
603120 °∠=+
°∠==∆
°−==
∆
∆
30
||3||
θθ lineline II
Line-phase currentrelationship
iprelationsh voltagephase-Line
°+=
=
∆
∆
30
||3||
phase
phaseVV
θθ rmsAIaA )(02.775.28 °−∠=
rmsAI
rmsAI
CA
BC
)(98.14260.16
)(02.9760.16
°∠=°−∠=
rmsAI
rmsAI
cC
bB
)(98.11275.28
)(02.12775.28
°∠=°−∠=
°∠=⇒ 02.3717.4YZ
Alternatively, determine first the line currentsand then the delta currents
Polyphase