POLYPHASE CIRCUITS LEARNING GOALS Three Phase Circuits
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Transcript of POLYPHASE CIRCUITS LEARNING GOALS Three Phase Circuits
POLYPHASE CIRCUITSLEARNING GOALS
Three Phase CircuitsAdvantages of polyphase circuits
Three Phase ConnectionsBasic configurations for three phase circuits
Source/Load ConnectionsDelta-Y connections
Power RelationshipsStudy power delivered by three phase circuits
Power Factor CorrectionImproving power factor for three phase circuits
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
)4802cos()2402cos()2cos(cos3
)(
tt
tIVtp mm
)120sin(sin)120cos(cos)120cos()120sin(sin)120cos(cos)120cos(
coscos
0)120cos()120cos(cos
Proof
Lemma5.0)120cos(
0)120cos()120cos(cos
)120cos()480cos()120cos()240cos(
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
30||3
23
21||||
)120sin120(cos1||
120||0||
p
pp
p
pp
bnanab
V
jVV
jV
VVVVV
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 circuit 1020,11,)(120|| jZjZVV phaselinermsphase source
Determine line currents and load voltagesBecause circuit is balanceddata on any one phase aresufficient
0120 Chosenas reference
120120120120
0120
an
bn
an
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
LEARNING EXTENSIONFor an abc sequence, balanced Y - Y three phase circuit
voltagesline the Find .)(90120 rmsVVan
Relationship betweenphase and line voltages
30||3 pab VV
30by lags aban VV
30by leads anab VV
rmsVVab )(1201203
rmsVV
rmsVV
ca
bc
)(2401203
)(01203
voltagesphase the Find.)(0208 rmsVVab
30by lags aban VV
rmsVVan )(303
208
rmsVV
rmsVV
cn
bn
)(903
208
)(1503
208
LEARNING EXTENSIONFor an abc sequence, balanced Y - Y three phase circuit 38,11,)(6.2602.104|| jZjZVV phaselinermsphase load,
Determine source phase voltages
rmsV )(6.2602.104
8
3j
Currents are not required. Use inversevoltage divider
ANan Vj
jjV38
)11()38(
41.315.1
73584
3838
3849 j
jj
jj
30120anV
Positive sequencea-b-c
15012090120
cn
bn
VV
DELTA CONNECTED SOURCES Convert to an equivalent Y connection
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
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
30||3 pab VV 71.065.1183ABV
71.12065.1183
29.11965.1183
CA
BC
V
V
LEARNING EXTENSIONCompute the magnitude of the line voltage at the load
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
301201.41.10
410jjVAN
10
1.0j
Only interested in magnitudes!
rmsVVAN )(57.11890.1077.10120||
30||3 pab VV rmsVVAB )(4.205||
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
3
ZZY case Balanced
ZL
ABaA
LaAY
anaA Z
VIIZVI
3/||
3/||||||
ZLZZ ||
Z 30
30||3||
lineline II
Line-phase currentrelationship
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
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
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
POWER RELATIONSHIPS
iprelationsh voltagephase-Line
30
||3||
phase
phaseVV
30||3||
lineline II
Line-phase currentrelationship
*3 phasephaseTotal IVS
*3 linelineTotal IVS
*3 IVS linetotal
*3 linelineTotal IVS
lineV
lineI
Power factor anglef
flinelinetotal
flinelinetotal
IVQ
IVP
sin||||3
cos||||3
- Impedance angle
LEARNING EXAMPLE
lagging 20 anglefactor power
WPrmsVV
total
line
1200)(208||
Determine the magnitude of the line currents and the value of load impedanceper phase in the delta
flinelinetotal
flinelinetotal
IVQ
IVP
sin||||3
cos||||3
flinelinetotal IVP cos
3||||
3 rmsAI line )(54.3||
30||3||
lineline II
Line-phase currentrelationship
rmsAI )(05.2||
46.101
||||||
IVZ line
2046.101Z
lineV
lineI
Power factor anglef
- Impedance angle
lineV
LEARNING EXAMPLEFor an abc sequence, balanced Y - Y three phase circuit 1020,11,)(120|| jZjZVV phaselinermsphase source
Because circuit is balanceddata on any one phase aresufficient
0120 Chosenas reference
120120120120
0120
an
bn
an
VVV
Abc sequence
rmsAj
VI anaA
)(65.2706.565.2771.23
01201121
Determine real and reactive power per phase at the load and total real, reactive andcomplex power at the source
57.2636.22)1020( aAaAAN IjIV
rmsVVAN )(08.115.113
65.2706.508.115.113*aAANphase IVS
rmsVAjS phase )(09.25651257.2654.572
phase per P phase per Q 65.2706.50120*
aAan IVS phase source
VAj
S
78.28186.537
65.272.607
phase source
)(78.2813)(86.5373
VAQWP
source total
source total
)(6.1821||)(2.8456.1613
VASVAj
QPS
source total
source totalsource totalsource total
LEARNING EXAMPLE
rmsVVline )(208 leading 0.8pfat 12kVA :3 Load
1 pfat 10kW :2 Loadlagging 0.6pfat 24kW :1 Load
balanced isCircuit
Determine the line currents and the combined power factor
kVASlaggingpf
kWP40||
6.024
11
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
kVAPSQ 32|||||| 21
211
kVAjSinductivelagging 32241
kVAjSpf
kWP010
110
22
2 Load
kVAQkWP
pfkVAS
2.7||6.9
8.012||
3
33
3 Load
kVAjScapacitivepfleading 2.76.93
kVAkVAjSSSSTOTAL 63.29160.508.246.43321
flinelinetotal
flinelinetotal
IVQ
IVP
sin||||3
cos||||3
63.29
||||3||
f
linelinetotal IVS
laggingpf 869.0
inductive
capacitive
f
rmsAI line )(23.139||
321 SSSStotal
Continued ...
source theat factor power and voltagesline determine are impedances line the If 02.005.0 jZ lineLEARNING EXAMPLE
continued ….
rmsAI line )(23.139||
2* ||3)(3 linelinelinelinelineline IZIIZS
)(11632908 VAjSline
kVAkVAjS 63.29160.508.246.43total load
kVAjS 17.29264.53963.25508.46total source
17.29||||3||
f
linelinetotal IVS
inductive
capacitive
f
rmsVVline )(87.22013.1393
264,53
laggingpf f 873.0)17.29cos(cos
LEARNING EXTENSIONA Y -Y balanced three-phase circuit has a line voltage of208-Vrms. The total real power absorbed by the load is 12kWat pf=0.8 lagging. Determine the per-phase impedanceof the load
iprelationsh voltagephase-Line
30
||3||
phase
phaseVV
*3 phasephaseTotal IVS
rmsVVphase )(1203
208||
*
2*||
33phase
phase
phase
phasephasetotal Z
VZV
VS
lineV
lineI
Power factor anglef
Impedance angle
87.36cos8.0 ffpf
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
kVApfPS total
total 15||
88.2||
||3||
2
total
phasephase S
VZ
87.3688.2pahseZ
301201.41.10
410jjVAN
Source is Delta connected.Convert to equivalent Y
Analyze one phase
10
1.0j
rmsVVAN )(57.11890.1077.10120||
LEARNING EXTENSIONDetermine real, reactive and complex power at both loadand source
*
2* ||33
phase
ANaAANload Z
VIVS
*
2* ||33
phasetotal
anaAansource Z
VIVS
22
22
410)410(|57.118|3
410|57.118|3
j
j
22
22
)1.4()1.10()1.41.10(|120|3
1.41.10|120|3
jj
)4961.1224(3)8.4840.212,1(3
jSjS
source
load
LEARNING EXTENSIONA 480-V rms line feeds two balanced 3-phase loads. The loads are ratedLoad 1: 5kVA at 0.8 pf laggingLoad 2: 10kVA at 0.9 pf lagging.
Determine the magnitude of the line current from the 408-V rms source
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
21 SSStotal
kVAPSQ
kWPPkVAS
0.3||
48.0
5||
21
211
11
kVAjSlaggingpf 341
kVAjSkVAPSQ
kWPPkVAS
36.4936.4||
99.0
10||
2
22
222
2
kVAjStotal 36.713 ||||3||
sin||||3
cos||||3
linelinetotal
flinelinetotal
flinelinetotal
IVS
IVQ
IVP
rmsAV
SIline
totallineq )(14.21
68.706939,14
||3||||
POWER FACTOR CORRECTION
Balanced loadLow pflagging
Similar to single phase case.Use capacitors to increase thepower factor
Keep clear about total/phasepower, line/phase voltages
oldold
old QpfS
newnew
old QpfP
added be toPower Reactive
oldnew QQQ
To use capacitors this valueshould be negative
connected are capacitors thehow on depends voltageThe
capacitorper 2CVQ
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
21sincos pfpf ff
21tan
pfpf
f
0 QlaggingfPQ tan
LEARNING EXAMPLE leading pf rms kV V Hz fline94 . 0 . 5. 34 | | , 60 : Required
626.01sincos 2 pfpf ff
MWPMVAQ
old
old
72.1802.15||
21tan
pfpf
f
0 oldQlagging
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
fPQ tan
MVAQleadingpfMWP
newnew
old 8.694.0
72.18
MVAQMVAQ
273.782.2102.158.6
capacitorper
rmskVVY35.34
capacitorconnection23
6
3105.3460210273.7
C
FC 6.48
LEARNING EXAMPLE lagging pf rms kV V Hz fline90 . 0 . 5. 34 | | , 60 : Required
626.01sincos 2 pfpf ff
MWPMVAQ
old
old
72.1802.15||
21tan
pfpf
f
0 oldQlagging
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
fPQ tan
MVAQlaggingpfMWP
newnew
old 067.990.0
72.18
MVAQMVAQ
984.1953.502.15067.9
capacitorper
rmskVVY35.34
capacitorconnection23
6
3105.3460210984.1
C
FC 26.13
Determine the currentflowing. Convert linevoltages to phase voltages
LEARNING EXAMPLEMEASURING POWER FLOW
rmskV 012 rmskV 512
Which circuit is the source and what is the average power supplied?
Equivalent 1-phase circuit
21
253
12000303
12000
21 jjVVI ANan
aA
rmsAIaA )(93.18030.270
*
*
)(3
3
aAanX
aAANY
IVS
IVS
MVASY )93.18025(2703.0123
MWPMWP
X
Y
91.413.5
fSPjQPScos||
System Y is the source
MVASX )93.18030(2703.0123
)( YXloss PPP
Phase differencesdetermine directionof power flow!
CAPACITOR SPECIFICATIONS
Capacitors for power factor correction are normally specified in VARs
given. is eother valu unless Assumeecapacitanc the know toorder in given bemust frequency and voltageThe
capacitorper
Hz
CVQ
60
|| 2
FC
53.26)1050(602
102523
6
2
FC
7.49)1020(602
105.723
6
3
FC
1.106)1010(602
10423
6
1
kVVkVV phaseline 9.195.34
LEARNING EXAMPLE
FC 6.48 needs one leading 0.94pfFor Capacity Rated Voltage (kV) Rated Q (Mvar)
1 10 42 50 253 20 7.5
Choices available
Capacitor 1 is not rated at high enoughvoltage!
Capacitor 3 is the best alternative
LEARNING BY DESIGNProposed new storerms A 170at rated wire #4ACSR
1. Is the wire suitable?
2. What capacitancewould be required tohave a composite pf =0.92 laggingCapacitors are to beY - connected
f
f
f
pf
SQ
SPjQPS
cos
sin||
cos||
*
*
3
3
lineline
phasephaseTotal
IV
IVS
321 SSSStotal
kVAjS
4205609.367001
kVAjkVAS
8665006010002
kVAjkVAS
3497208.258003
kVAkVAjStotal 57.42241716351780
rmsAV
SIline
totalline 1.101
108.13310417.2
3|||| 3
6
Wire is OK
newnew
old QpfP
oldnew QQQ
)(tan newfP kVA28.758
kVA72.8762|| CVQ capacitorper
38.13|| kVVV phase
3/108.13602
3/1072.87623
3
C F2.12
Polyphase