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UNIT 5UNIT 5UNIT 5UNIT 5
THREE PHASETHREE PHASESYSTEMSYSTEMPREPARED BY:PREPARED BY:
NABILAH BINTI MAZALANNABILAH BINTI MAZALAN
OBJECTIVES• Understand three phase system• Explain the basic principles of a three phase
system• Understand three system configurations• Advantage and application of three phase
compare to single phasecompare to single phase• Explain three phase e.m.f generation• Identify connection and vector phase diagram
for delta and star system• Determine the phase voltage, phase current,
line voltage, line current• Define balance load in three phase system• Calculate power for three phase system
INTRODUCTION• Three-phase electricity supply is a common way of
transmission of alternating current electricity supply.• It is a multi-phase systems, and is the most
commonly used in electrical distribution grid systemin the world. It is also used as a source of power forlarge electric motors and other high load.
• In general, three-phase system is the most• In general, three-phase system is the mosteconomical system as it uses less conductormaterial to transmit electricity from the systemsingle-phase or two phases are equal at the samevoltage.
• In the three-phase system, three conductors carrythree alternating current that reached peak valuesat different times, with intervals between the phasesof 1 / 3 cycle.
THREE PHASE SYSTEM
• A three phase ac power systems consistsof three phase generators, transmissionlines and load
• A three phase generator consists of threesingle phase generators, with voltagessingle phase generators, with voltagesequal in magnitude but differing in phaseangle from the others by 120
GENERATION OF 3-PHASE
• A 3-phase generator consists of 3 single phasegenerators, with voltages equal in magnitude butdiffering in phase angle from the others by 120
• Rotating at a constant speed in a uniformmagnetic field.magnetic field.
• Each voltage source is called phase.
• Usually, three phases are labeled as– Red (R-Red)
– Yellow (Y-Yellow)
– Blue (B-Blue).
GENERATION OF 3-PHASE
Flux densityB [T]
l
d
L
A
M
N
Current induces in the coil as thecoil moves in the magnetic field
Current produced at terminal
Generator for single phase
NoteInduction motor cannot start byitself. This problem is solved byintroducing three phase system
GENERATION OF 3-PHASE
Instead of using one coil only , three coils are used arranged inone axis with orientation of 120o each other. The coils are R-R1, Y-Y1 and B-B1. The phases are measured in this sequenceR-Y-B. I.e Y lags R by 120o , B lags Y by 120o.
GENERATION OF 3-PHASE
The three winding can be represented by the above circuit. Inthis case we have six wires. The emf are represented by eR ,eY, eB.
te sinEmR
Finish R
L1eR
)120-sin(EmYte
)240-sin(EmBte
LoadFinish
Finish
start
start
start
R1
Y
Y1
B
B1
L2
L3
eY
eB
GENERATION OF 3-PHASE
vector diagram for three-phase system
GENERATION OF 3-PHASE
ADVANTAGES OF 3 PHASESYSTEM
1. 3-phase power has a constant magnitude butsingle phase power is a pulsating one
2. A 3-phase system can be set up a rotatingmagnetic field in stationary windings. This is notpossible in single phase systempossible in single phase system
3. For the same rating, 3-phase machine are smallerin size and have better operating characteristicsthan single phase machine
4. 3-phase induction motors are self startingwhereas single phase induction motors are notself starting
ADVANTAGES OF 3 PHASESYSTEM
5. 3-phase motors have better power factor andefficiency over single phase motor.
6. Generation, transmission and utilization ofpower is more economical in three phasepower is more economical in three phasesystem compared to single phase system.
7. There is saving of conductor material whenthe same power is to be transmitted over agiven distance by a 3-phase systemcompared to a single phase system
PHASE SEQUENCE
• Definition :-
The order in which the voltages in thethree phases reach their maximumvaluesvalues
PHASE & LINE ELEMENT
• The voltage between any one line and naturalis called the phase voltage (VPH)
• The current flowing in a phase is known asthe current phase (IPH)the current phase (IPH)
• The voltage between any two lines or phasesis called the line voltage (VL)
• The current flowing in a line is called the linecurrent (IL)
DELTA CONNECTION
a) Physical connection diagram
b) Conventional connectiondiagram
DELTA CONNECTION
DELTA CONNECTION
• IR, IY and IB are called line current• I1, I2 and I3 are called phase current
)(III 3131R II
From Kirchoff current law we haveI3
R
Y
IR
IY
I
I1
VL
VP
)(III 1212Y II
)(III 2323B II
In phasor diagram
BIB
I2
DELTA CONNECTIONSince the loads are balanced, the magnitude of currents areequaled but 120o out of phase. i.e I1 =I2=I3 ,=IP Therefore:-
I1 = VP30;
I2 = VP-90;
I3 = VP150;
IR = IL30;
IY = IL-90;
IB = IL150;
PR III )3(30cos2 1
PY III )3(30cos2 2
PB III 330cos2 3
Thus IR=IY=IB = IL
Where IP is a phase current and IL is a line current
3 PB L
DELTA CONNECTION
Voltage
PL VV
PL II 3
Current
STAR CONNECTION
a) Physical connection diagram
b) Conventional connectiondiagram
STAR CONNECTION
)( VVVVV
STAR CONNECTION
R
Y
B
N
VRY
VYB
IR
IY
R
Y
B
N
V
VRYVBR
VYB
IR
IY
R
VYVB
•VRY, VYB and VBR are called line voltage•VR, VY and VB are called phase voltage
From Kirchoff voltage law we have
)( YRYRRY VVVVV BIB
BIB
)( BYBYYB VVVVV
)( RBRBBR VVVVV
In phasor diagram
For balanced load VR , VY andVB are equaled but out of phase
VR = VP30;
VY = VP-90;
VB = VP150;
VRY = VL30;
VYB = VL-90;
VBR = VL150;
STAR CONNECTION
Po
RRY VVV 330cos2
therefore
Po
YYB VVV 330cos2
Po
BBR VVV 330cos2
PL VV 3
Voltage
STAR CONNECTION
PL II
Current
POWER IN 3-PHASESYSTEM
• DC power system
• AC power system
POWER IN 3-PHASESYSTEM
• Power for 3-phase system (refer to phaseelement) cos3 PP IVP
• Power for 3-phase system (refer to lineelement)
cos3 LL IVP
SUMMARYCharacteristic Star Connection Delta Connection
Symbol or
Voltage
Current
BalanceBalanceCondition
Power in 1
Power in 3
- Phase element
- Line element
EXAMPLE 1
Three similar resistors are connectedin star acroos 400V, 3 phase lines. Theline current is 5A. Calculate the valueof each resistor. To what value shouldof each resistor. To what value shouldthe line voltage be changed to obtainthe same line current with theresistors are in delta connected?
SOLUTION 1
• Line voltage, VL = 400V– Phase voltage, Vph = VL / = 400 / =230.94V
• Line current, IL = 5A
3 3
• Line current, IL = 5A– Phase current Iph = 5A (in star IL = Iph)
• Zph = Rph = Vph / Iph = 230.94 / 5 =46.19
• If the same resistors are connected indelta– VL = Vph and IL = Iph
– Vph = Iph Zph = Iph Rph = IL / x Rph =
3
3
EXAMPLE 2
Three identical impedances areconnected in delta to a three phase 400Vsupply. The line current is 34.65A and thetotal power taken from the supply is14.4kW. Calculate the resistance and14.4kW. Calculate the resistance andreactance values of each impedance.
SOLUTION 2
– Line Voltage, VL = 400V
• Phase Voltage, Vph = VL = 400V
– Line Current, IL = 34.65A
• Phase Current, IPH = IL / 3 = 34.65 / 3 = • Phase Current, IPH = IL / 3 = 34.65 / 3 =20A
– Total power, P = 14.4kW
• But P = 3 Iph Rph
Rph = 14.4k / 3 Iph = 14.4k / 3(20) = 12
– Zph = Vph / Iph = 400 / 20 = 20
• XL = Zph - Rph = 20 - 12 = 16
2
2 2
2 2 2 2
EXAMPLE 3
Three similar coils are connected instar taken at a total power of 1.5kW ata p.f of 0.2 lagging from a 3-phase400V 50Hz supply. Calculate theresistance and inductance of eachresistance and inductance of eachphase.
SOLUTION 3
• P = 3 VL IL cos θ
• IL = P / 3 VL cos θ = 1.5k / ( 3 x 400x 0.2)
= 10.83A
√
√ √
• IL = Iph = 10.83A
• Vph = VL / 3 = 400 / 3 = 230.94V
• Zph = Vph / Iph = 230.94 / 10.83 =21.32Ω
• Cos θ = Rph / Zph
Rph = Zph Cos θ = 21.32 (0.2) =4.264Ω
√ √
SOLUTION 3
• XL = Zph2 - Rph2 = 21.322 – 4.2642
= 20.89Ω
XL = 2πfL
√ √
XL = 2πfL
L = XL / 2πf = 20.89 / (2π x 50) =66mH
EXAMPLE 4
Find :-
a) Phase Impedance, ZPH
b) Phase Current, IPH
c) Line Current, Ic) Line Current, IL
d) Phase Voltage, VPH
e) Single phase power, P1
f) Three phase power (phase element),P3
g) Three phase power (line element), P3
SOLUTION 4
EXAMPLE 5
Find :-Find :-
a) Phase Impedance, ZPH
b) Phase Current, IPH
c) Line Current, IL
d) Three phase power (line element), P3
SOLUTION 5
EXAMPLE 6
Three coil each having a resistance of20Ω and inductive reactance of 15Ωare connected to a 400V, 3 phase 50Hzsupply. Calculate:-
a)The line currenta)The line current
b)Power factor
c)Power drawn from the supply
SOLUTION 6a) The line current
– Zph = Rph2 + Xph2 = 202 + 152 = 25Ω
– Vph = VL / √3 = 400 / √3 = 230.94V
– Iph = Vph / Zph = 230.94 / 25 = 9.24A
– IL = Iph = 9.24A
√ √
– IL = Iph = 9.24A
b) Power factor
- cos θ = Rph / Zph = 20 / 25 = 0.8
c) Power
- P = √3 VL IL cos θ = √3(400)(9.24)(0.8)
= 5.121kW
THE END