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EET 103EET 103
Three Phase SystemThree Phase System
Chapter 3Chapter 3
((Lecture 1Lecture 1))
• Thinner conductors can be used to transmit Thinner conductors can be used to transmit the same kVA at the same voltage, which the same kVA at the same voltage, which reduces the amount of copper required reduces the amount of copper required (typically about 25% less) and turn reduces (typically about 25% less) and turn reduces construction and maintenance costs.construction and maintenance costs.
INTRODUCTION TO THREE PHASE SYSTEMINTRODUCTION TO THREE PHASE SYSTEM
In general, three phase systems are preferred In general, three phase systems are preferred over single phase systems for the transmission over single phase systems for the transmission of the power system for many reasons, including of the power system for many reasons, including the followingthe following
• The lighter lines are easier to install and the The lighter lines are easier to install and the supporting structures can be less massive and supporting structures can be less massive and farther apart.farther apart.
• In general, most larger motors are three phase In general, most larger motors are three phase because they are essentially self starting and because they are essentially self starting and do not require a special design or additional do not require a special design or additional starting circuitry.starting circuitry.
Three phase voltages Three phase voltages
A 3 phase generator basically consists of a A 3 phase generator basically consists of a rotating magnet (called the rotor) surrounded rotating magnet (called the rotor) surrounded by a stationary winding (called the stator). by a stationary winding (called the stator). Three separate windings or coils with Three separate windings or coils with terminals terminals a - a’, b - b’ a - a’, b - b’ and and c - c’ c - c’ are physically are physically placed 120placed 120oo apart around the stator. apart around the stator.
Generated VoltagesGenerated Voltages
The three phase generator can supply power to The three phase generator can supply power to both single phase and three phase loadsboth single phase and three phase loads
tVv ANmAN sin)(
)120sin()(o
BNmBN tVv
)120sin()240sin( )()(o
CNmo
CNmCN tVtVv
The sinusoidal expression for each of the phase The sinusoidal expression for each of the phase voltages voltages
7
Phase expressionPhase expression
In phase expressionIn phase expression
WhereWhere
EEM M : peak value: peak value
EEAA, E, EBB and E and ECC : rms value : rms value
02
EE M
A 1202
EE M
B 1202
EE M
C
The phasor diagram of the phase voltages The phasor diagram of the phase voltages
)AN(m)AN(m
AN V.V
V 70702
)()( 707.0
2BNm
BNmBN V
VV
)()( 707.0
2CNm
CNmCN V
VV
o)m(ANAN VV 0
o)m(BNBN VV 120
o)m(CNCN VV 120
The effective value of The effective value of each is determined each is determined by by
If the voltage sources have the same amplitude If the voltage sources have the same amplitude and frequency ω and are out of the phase with and frequency ω and are out of the phase with each other by 120each other by 120oo, the voltages are said to be , the voltages are said to be balanced. By rearranging the phasors as shown balanced. By rearranging the phasors as shown in figure below, soin figure below, so
omCN
omBN
omAN
CNBNAN
VVV
VVV
1201200 )()()(
0)866.05.0866.05.00.1( jjVm
mCNBNAN VVVV ||||||
Where Where
10
Connection in Three Phase SystemConnection in Three Phase System
A 3 phase system is equivalent to three single A 3 phase system is equivalent to three single phase circuitphase circuit
Two possible configurations in three phase Two possible configurations in three phase systemsystem
1.1. Y - connection (star connection)Y - connection (star connection)
2.2. ∆ ∆ - connection (delta connection)- connection (delta connection)
11
Three phase Voltages SourceThree phase Voltages Source
Y-connected source ∆-connected sourceY-connected source ∆-connected source
12
Three phase LoadThree phase Load
Y - connected load ∆ - connected loadY - connected load ∆ - connected load
Generator and Load ConnectionsGenerator and Load Connections
Each generator in a 3 phase system maybe either Each generator in a 3 phase system maybe either Y or Y or - - connected and loads may be mixed on a connected and loads may be mixed on a power system.power system.
Z
Z Z Z
Z
Z
Wye Connected Generator Wye Connected Generator
gL II
NBANBNANAB VVVVV
NCBNCNBNBC VVVVV
NACNANCNCA VVVVV
Applying KVL around the indicated loop in figure Applying KVL around the indicated loop in figure above, we obtain above, we obtain
BAAB VVV
00 1200 VV
For line-to-line voltage VFor line-to-line voltage VABAB is given by is given by
VjVV
2
3
2
1
VjV2
3
2
3
2
1
2
33 jV
0303 V
Phasor DiagramPhasor Diagram
00 30330 ANABAB VVV
01503 CNCA VV
02703 BNBC VV
VVLL 3
The relationship between the magnitude of the The relationship between the magnitude of the line-to-line and line-to-neutral (phase) voltage isline-to-line and line-to-neutral (phase) voltage is
The line voltages are shifted 30The line voltages are shifted 3000 with respect to with respect to the phase voltages. Phasor diagram of the line the phase voltages. Phasor diagram of the line and phase voltage for the Y connection is shown and phase voltage for the Y connection is shown below.below.
RearrangeRearrange
VAB
VAN
VBCVBN
VCA
VCN
Line-to-line voltagesPhase voltages
Delta Connected Generator Delta Connected Generator
VVLL
CAABA III
00 2400 II
For line-to-line voltage VFor line-to-line voltage VABAB is given by is given by
IjII
2
3
2
1
IjI2
3
2
3
2
1
2
33 jI
0303 I
IIL 3
The relationship between the magnitude of the The relationship between the magnitude of the line and phase current isline and phase current is
The line currents are shifted 30The line currents are shifted 3000 relative to the relative to the corresponding phase current. Phasor diagram of corresponding phase current. Phasor diagram of the line and phase current for the Y connection is the line and phase current for the Y connection is shown below.shown below.
IA
IAB
IBC
IB
ICA
Line-to-line currentsPhase currents
IC
Phase sequencePhase sequence
The The phase sequence phase sequence is the order in which the is the order in which the voltages in the individual phases peak. voltages in the individual phases peak.
VA
VB
VC
VA
VC
VB
abc abc phase sequencephase sequence acbacb phase sequencephase sequence
21
EXAMPLE 3.1EXAMPLE 3.1
Calculate the line currents in the three-wire Y - Y Calculate the line currents in the three-wire Y - Y system as shown below.system as shown below.
22
Solution 3.1Solution 3.1
Single Phase Equivalent CircuitSingle Phase Equivalent Circuit
Phase ‘a’ equivalent circuitPhase ‘a’ equivalent circuit
23
21.86.8121.816.155
0110I
8.21155.16)810()25(Z;Z
VI
Aa
TT
ANAa jj
A2.986.811.8266.81
024II
A141.86.81
120II
AaCc
AaBb
24
EXAMPLE 3.2EXAMPLE 3.2
A balanced delta connected load having an A balanced delta connected load having an impedance 20 - j15 impedance 20 - j15 is connected to a delta is connected to a delta connected, positive sequence generator connected, positive sequence generator having Vhaving VABAB = 330 = 33000 V. Calculate the phase V. Calculate the phase
currents of the load and the line currents.currents of the load and the line currents.
25
Solution 3.2Solution 3.2
V 0330V
87.3625 j1520Z
AB
Δ
26
Phase CurrentsPhase Currents
A87.15613.2120II
A13.83-13.2120II
A36.8713.238.8725
0330
Z
VI
abca
abbc
Δ
abab
27
A 87.12686.22120II
A 13.311-86.22120II
A 87.686.22
30336.8713.2
303II
AaCc
AaBb
abAa
Line CurrentsLine Currents
28
∆ ∆ - Connected Generator with a Y - - Connected Generator with a Y - Connected LoadConnected Load
29
EXAMPLE 3.3EXAMPLE 3.3
A balanced Y - connected load with a phase A balanced Y - connected load with a phase impedance 40 + j25 impedance 40 + j25 is supplied by a balanced, is supplied by a balanced, positive-sequence positive-sequence ΔΔ-connected source with a -connected source with a line voltage of 210V. Calculate the phase line voltage of 210V. Calculate the phase currents. Use Vcurrents. Use VABAB as reference. as reference.
30
Solution 3.3Solution 3.3
the load impedance, Zthe load impedance, ZYY and the source voltage, V and the source voltage, VABAB are are
V 0210V
3217.47 j2540Z
AB
Y
31
When the ∆ - connected source is transformed When the ∆ - connected source is transformed to a Y - connected source, to a Y - connected source,
V 30-121.2
3013
0210
303
VV AB
an
32
The line currents areThe line currents are
A 582.57120II
A 182-2.57120II
A 62-2.573247.17
03121.2
Z
VI
AaCc
AaBb
Y
anAa
33
Summary of Relationships in Y and Summary of Relationships in Y and ∆ - connections∆ - connections
Y-connectionY-connection ∆∆-connection-connection
Voltage Voltage magnitudesmagnitudes
Current Current magnitudesmagnitudes
Phase Phase sequencesequence
VVLL leads leads VVφφ by by
30°30°IILL lags lags IIφφ by 30° by 30°
φV3VL φVVL
φI3IL φIIL
EET 103EET 103
Three Phase SystemThree Phase System
Chapter 3Chapter 3
((Lecture 2Lecture 2))
PowerPower
Y - Connected Balanced LoadY - Connected Balanced Load
Average PowerAverage Power
The average power delivered to each phaseThe average power delivered to each phase
The total power to the balanced load isThe total power to the balanced load is
Reactive PowerReactive Power
The reactive power of each phase isThe reactive power of each phase is
The total reactive power of the load isThe total reactive power of the load is
Apparent PowerApparent Power
The apparent power of each phase is The apparent power of each phase is
The total apparent power of the load isThe total apparent power of the load is
Power FactorPower Factor
The power factor of the system isThe power factor of the system is
∆ ∆ - Connected Balanced Load- Connected Balanced Load
Average PowerAverage Power
Reactive PowerReactive Power
Apparent PowerApparent Power
Power FactorPower Factor
EXAMPLE 3.4 EXAMPLE 3.4
Determine the total power (P), reactive power Determine the total power (P), reactive power (Q) and complex power (S) at the source and at (Q) and complex power (S) at the source and at the load. the load.
Single Phase Equivalent CircuitSingle Phase Equivalent Circuit
Phase Phase ‘a’‘a’ equivalent circuit equivalent circuit
Known quantitiesKnown quantitiesVg =VAN= 1100 V
ZY = 10 + j8
Zline = 5 - j2
Solution 3.4Solution 3.4
Line / Phase CurrentsLine / Phase Currents
A 21.86.8121.816.155
0110I
ZZ
VI
A
Yline
ANA
Source & Load PowerSource & Load Power
VAR 834.6Q W,2087P
j834.6)VA(2087
I3VS
ss
φφsource
VAR 1113Q W,1392P
j1113)VA(1392
ZI3S
LL
2
φLoad
EXAMPLE 3.5EXAMPLE 3.5
A three phase motor can be regarded as a A three phase motor can be regarded as a balanced Y - load. A three phase motor balanced Y - load. A three phase motor draws 5.6 kW when the line voltage is 220 V draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the and the line current is 18.2 A. Determine the power factor of the motorpower factor of the motor
Known QuantitiesKnown Quantities
• PPLoad Load = 5600 W= 5600 W
• VVLL = 220 V = 220 V
• IILL = 18.2 A = 18.2 A
Power factor = cos Power factor = cos
VA 6935.13
IV 3
I3VS
LL
φφ
0.86935.13
5600
S
Pθ cos
θ cosSP
|S|Q
P
Solution 3.5Solution 3.5
Example 3.6Example 3.6
For the Y - connected load in FigureFor the Y - connected load in Figure
a)a) find the average power to each phase and find the average power to each phase and the total loadthe total load
b)b) determine the reactive power to each phase determine the reactive power to each phase and the total reactive power and the total reactive power
c)c) find the apparent power to each phase and find the apparent power to each phase and the total apparent power the total apparent power
d)d) find the power factor of the loadfind the power factor of the load
Figure
Solution 3.6Solution 3.6
a) The average power to each phase isa) The average power to each phase is
Total loadTotal load
W1200
13.53cos20100
cos
VIIVP
W3600W120033 PPT
b) The reactive power to each phase isb) The reactive power to each phase is
Total reactive powerTotal reactive power
VAR1600
13.53sin20100
sin
VIIVQ
VAR4800160033 QQT
c) The apparent power to each phase isc) The apparent power to each phase is
Total apparent powerTotal apparent power
VA2000
20100
IVS
VA6000200033 SST
d) The power factord) The power factor
lagging
FP
6.0VA 6000
W3600
S
P
T
T
Power relationship - Phase quantitiesPower relationship - Phase quantities
The power equations applied to Y-or The power equations applied to Y-or load in a load in a balanced 3-phase system arebalanced 3-phase system are
cosIVP 3
sinIVQ 3
IVS 3
cosZIP 23
sinZIQ 23
ZIS 23
Real powerReal power
Watts (W)Watts (W)
Apparent powerApparent power
Volt-Amps (VA)Volt-Amps (VA)
Reactive powerReactive power
Volt-Amps-Reactive (VAR)Volt-Amps-Reactive (VAR)
- angle between voltage and current in any phase of the load- angle between voltage and current in any phase of the load
Power relationship - Line quantitiesPower relationship - Line quantities
The power equations applied to Y-or The power equations applied to Y-or load in a load in a balanced 3-phase system arebalanced 3-phase system are
cosIVP LLL3
sinIVQ LLL3
LLL IVS 3
Real powerReal power
Apparent powerApparent power
Reactive powerReactive power
- angle between phase voltage and phase current - angle between phase voltage and phase current in any phase of the loadin any phase of the load
Since both the three phase source and the three Since both the three phase source and the three phase load can be either Y or phase load can be either Y or connected, we connected, we have 4 possible connectionshave 4 possible connections
1.1.Y - Y connections Y - Y connections (Y - connected source with (Y - connected source with Y - connected load)Y - connected load)
2.2.Y - Y - connection connection (Y - connected source with (Y - connected source with - connected load)- connected load)
3.3. - - connection connection (( - connected source with - connected source with - connected load)- connected load)
4.4. - Y connection - Y connection (( - connected source with Y - connected source with Y - connected load)- connected load)
1. Y connected generator / source with Y 1. Y connected generator / source with Y connected loadconnected load
LLg III
EV
VEL 3
321 ZZZ
2. Y - 2. Y - ConnectionConnection
3
ZZY
Z
Z
Z
Z/3
Z/3Z/3
must consists of three equal impedancesmust consists of three equal impedances
A balanced Y - A balanced Y - system consists of a system consists of a balanced Y - connected source feeding a balanced Y - connected source feeding a balanced balanced - connected load - connected load
3. 3. ∆ ∆ - - ∆∆ConnectionConnection
Z
Z
Z
A balanced A balanced ∆ ∆ - - system consists of a system consists of a balanced balanced ∆ ∆ - connected source feeding a - connected source feeding a balanced balanced - connected load - connected load
Z
Z
Z
4. 4. YYConnectionConnection
Z/3
Z/3Z/3
A balanced A balanced - Y system consists of a - Y system consists of a balanced balanced - connected source feeding a - connected source feeding a balanced Y - connected loadbalanced Y - connected load
Z
Z
Z
Example 3.7Example 3.7
Each transmission line of the 3 wire, three phase Each transmission line of the 3 wire, three phase system in Figure has an impedance of 15 Ω + j system in Figure has an impedance of 15 Ω + j 20 Ω. The system delivers a total power of 160 20 Ω. The system delivers a total power of 160 kW at 12,000 V to a balanced three-phase load kW at 12,000 V to a balanced three-phase load with a lagging power factor of 0.86.with a lagging power factor of 0.86.
a.a. Determine the magnitude of the line voltage Determine the magnitude of the line voltage EEABAB of the generator. of the generator.
b.b. Find the power factor of the total load Find the power factor of the total load applied to the generator.applied to the generator.
c.c. What is the efficiency of the system?What is the efficiency of the system?
Figure
Solution 3.7Solution 3.7
0V V
A 94.886.042.69363
W160000
cos3
V
PI T
a.a. VVøø (load) = (load) =
PPTT (load) = 3 (load) = 3 VVøø IIøø cos cos θθ
and
Since Since θθ = cos = cos-1-1 0.86 = 30.68 0.86 = 30.68oo (lagging) (lagging)
And assigning , a lagging power And assigning , a lagging power factor results in factor results in 68.30A94.8I
V 42.69361.73
V12000
3LV
0VZIE lineAN
V 26.12358)V5.7143)(73.1(3 gAB EE
For each phase, the system will appear as shown For each phase, the system will appear as shown in figure below.in figure below.
Or
Then
0.687143.5V
85.35V7142.98V
V42.6936V35.85V56.206
0V42.693645.22V5.223
0V42.693613.532530.68-A94.8
VZIE
j
j
lineAN
b.b.
W55.596,163
W55.3596W000,160
15A94.83kW160
)(3kW1602
2
lineL
linesloadT
RI
PPP
TLLT IVP cos3
A94.8V26.358,121.73
W55.596,163
3cos
LL
TT
IV
P
856.0pF
%8.97978.0 W3596.55kW 160
kW 160
losseso
o
i
o
PP
P
P
P
And
< 0.86 of loadAnd
c.c.
Example 3.8Example 3.8
A 208V three phase power system is shown in Figure 1. It consists A 208V three phase power system is shown in Figure 1. It consists of an ideal 208V Y - connected three phase generator connected to of an ideal 208V Y - connected three phase generator connected to a three phase transmission line to a Y - connected load. The a three phase transmission line to a Y - connected load. The transmission line has an impedance of transmission line has an impedance of 0.06 + j0.120.06 + j0.12per phase and per phase and the load has an impedance of the load has an impedance of 12 + j912 + j9per phase. For this simple per phase. For this simple system, findsystem, find
(a)(a) The magnitude of the line current The magnitude of the line current IILL
(b)(b) The magnitude of the load’s line and phase voltages The magnitude of the load’s line and phase voltages VVLLLL and and VVLL
(c)(c) The real, reactive and apparent powers consumed by the loadThe real, reactive and apparent powers consumed by the load
(d)(d) The power factor of the loadThe power factor of the load
(e)(e) The real, reactive and apparent powers consumed by the The real, reactive and apparent powers consumed by the transmission linetransmission line
(f)(f) The real, reactive and apparent powers supplied by the generatorThe real, reactive and apparent powers supplied by the generator
(g)(g) The generator’s power factorThe generator’s power factor
Figure 1
Z
ZZ
Z=12+ i9
-
+
+
0.06
_
0.06
0.06
i0.12
i0.12
i0.12
V
Van=12000
Vbn=120-1200
Vcn=120-2400
208V
Solution 3.8Solution 3.8
(a)(a)The magnitude of the line current The magnitude of the line current IILL
A
j
jj
V
ZZ
VI
loadline
lineline
1.3794.7
1.3712.15
0120
12.906.12
0120
)912()12.006.0(
0120
So, the magnitude of the line current is thus 7.94 ASo, the magnitude of the line current is thus 7.94 A
(b) The magnitude of the load’s line and phase (b) The magnitude of the load’s line and phase voltages voltages VVLLLL and and VVLL
V
A
jA
ZIV LLL
2.01.119
)9.3615)(1.3794.7(
)912)(1.3794.7(
VV L 1.119
VVV LLL 3.2063
The phase voltage on the load is the voltage across one phase of the The phase voltage on the load is the voltage across one phase of the load. This voltage is the product of the phase impedance and the load. This voltage is the product of the phase impedance and the phase current of the loadphase current of the load
Therefore, the magnitude of the load’s phase voltage isTherefore, the magnitude of the load’s phase voltage is
and the magnitude of the load’s line voltage isand the magnitude of the load’s line voltage is
W
AV
IVPLoad
2270
9.36cos)94.7)(1.119(3
cos3
var1702
9.36sin)94.7)(1.119(3
sin3
AV
IVQLoad
VA
AV
IVSLoad
2839
)94.7)(1.119(3
3
(c) The real power consumed by the load is(c) The real power consumed by the load is
The reactive power consumed by the load isThe reactive power consumed by the load is
The apparent power consumed by the load isThe apparent power consumed by the load is
(d) The load power factor is (d) The load power factor is
lagging
PFLoad
8.0
9.36cos
cos
A 1.3794.7 )12.006.0( j
4.63134.0
W
A
ZIPLine
3.11
4.63cos)134.0()94.7(3
cos32
2
var7.22
4.63sin)134.0()94.7(3
sin32
2
A
ZIQLine
VA
A
ZISLine
3.25
)134.0()94.7(3
32
2
(e) The current in the (e) The current in the transmission line is transmission line is
TThe impedance he impedance per phase of the line is of the line is or
Therefore, the real, reactive and apparent powers consumed in the line are
W
WW
PPP loadlinegen
2281
22703.11
var1725
var1702var7.22
loadlinegen QQQ
VA
QPS gengengen
2860
22
(f) The real and reactive powers supplied by the (f) The real and reactive powers supplied by the generator are the sum of the powers generator are the sum of the powers consumed by the line and the loadconsumed by the line and the load
The apparent power of the generator is the square root The apparent power of the generator is the square root of the sum of the squares of the real and reactive of the sum of the squares of the real and reactive powerspowers
1.372281
1725tan
tan
1
1
W
VAR
P
Q
gen
gengen
laggingPFgen 798.01.37cos
(g) From the power triangle, the power factor (g) From the power triangle, the power factor angle angle is is
Therefore, the generator’s power factor isTherefore, the generator’s power factor is
Assignment 3.1Assignment 3.1
A 208V three phase power system is shown in Figure 2. It consists of A 208V three phase power system is shown in Figure 2. It consists of an ideal 208V Y - connected three phase generator connected to a an ideal 208V Y - connected three phase generator connected to a three phase transmission line to a three phase transmission line to a - connected load. The - connected load. The transmission line has an impedance of transmission line has an impedance of 0.06 + j0.120.06 + j0.12per phase and per phase and the load has an impedance of the load has an impedance of 12 + j912 + j9per phase. For this simple per phase. For this simple system, findsystem, find
a.a. The magnitude of the line current The magnitude of the line current IILL
b.b. The magnitude of the load’s line and phase voltages The magnitude of the load’s line and phase voltages VVLLLL and and VVLL
c.c. The real, reactive and apparent powers consumed by the loadThe real, reactive and apparent powers consumed by the load
d.d. The power factor of the loadThe power factor of the load
e.e. The real, reactive and apparent powers consumed by the The real, reactive and apparent powers consumed by the transmission linetransmission line
f.f. The real, reactive and apparent powers supplied by the The real, reactive and apparent powers supplied by the generatorgenerator
g.g. The generator’s power factorThe generator’s power factor
Figure 2