UNIT 5
THREE PHASE
SYSTEM
PREPARED BY:
NABILAH 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 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 system
in the world. It is also used as a source of power for
large electric motors and other high load.
• In general, three-phase system is the most
economical system as it uses less conductor
material to transmit electricity from the system
single-phase or two phases are equal at the same
voltage.
• In the three-phase system, three conductors carry
three alternating current that reached peak values
at different times, with intervals between the phases
of 1 / 3 cycle.
THREE PHASE SYSTEM
• A three phase ac power systems consists
of three phase generators, transmission
lines and load
• A three phase generator consists of three
single phase generators, with voltages
equal in magnitude but differing in phase
angle from the others by 120
GENERATION OF 3-PHASE
• A 3-phase generator consists of 3 single phase
generators, with voltages equal in magnitude but
differing in phase angle from the others by 120
• Rotating at a constant speed in a uniform
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
d
Flux density
B [T]
l
 L
N
M
A

Generator for single phase
Current induces in the coil as the
coil moves in the magnetic field
Note
Induction motor cannot start by
itself. This problem is solved by
introducing three phase system
Current produced at terminal
GENERATION OF 3-PHASE
Instead of using one coil only , three coils are used arranged in
one axis with orientation of 120o each other. The coils are R-R1
, Y-Y1 and B-B1. The phases are measured in this sequence
R-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. In
this case we have six wires. The emf are represented by eR ,
eY, eB.
eR  E m sin t
eY  E m sin(t-120 )
Finish R
eR
L1
eY
L2
eB
L3
start R1
Finish Y
start Y1
Finish B
eB  Em sin(t-240 )
start
B1
Load
GENERATION OF 3-PHASE
vector diagram for three-phase system
GENERATION OF 3-PHASE
ADVANTAGES OF 3 PHASE
SYSTEM
1. 3-phase power has a constant magnitude but
single phase power is a pulsating one
2. A 3-phase system can be set up a rotating
magnetic field in stationary windings. This is not
possible in single phase system
3. For the same rating, 3-phase machine are smaller
in size and have better operating characteristics
than single phase machine
4. 3-phase induction motors are self starting
whereas single phase induction motors are not
self starting
ADVANTAGES OF 3 PHASE
SYSTEM
5. 3-phase motors have better power factor and
efficiency over single phase motor.
6. Generation, transmission and utilization of
power is more economical in three phase
system compared to single phase system.
7. There is saving of conductor material when
the same power is to be transmitted over a
given distance by a 3-phase system
compared to a single phase system
PHASE SEQUENCE
• Definition :The order in which the voltages in the
three phases reach their maximum
values
PHASE & LINE ELEMENT
• The voltage between any one line and natural
is called the phase voltage (VPH)
• The current flowing in a phase is known as
the current phase (IPH)
• The voltage between any two lines or phases
is called the line voltage (VL)
• The current flowing in a line is called the line
current (IL)
DELTA CONNECTION
b) Conventional connection
diagram
a) Physical connection diagram
DELTA CONNECTION
DELTA CONNECTION
IR
• IR, IY and IB are called line current
• I1, I2 and I3 are called phase current
I1
From Kirchoff current law we have
I3
I R  I1  I 3  I1  (  I 3 )
I Y  I 2  I1  I 2  (  I1 )
I B  I3  I 2  I 3  ( I 2 )
In phasor diagram
R
IY
VP
Y
VL
I2
IB
B
DELTA CONNECTION
Since the loads are balanced, the magnitude of currents are
equaled but 120o out of phase. i.e I1 =I2=I3 ,=IP Therefore:IR = IL30;
I1 = VP30;
IY = IL-90;
I2 = VP-90;
IB = IL150;
I3 = VP150;
Where IP is a phase current and IL is a line current
I R  2 I1 cos 30  ( 3 ) I P
I Y  2 I 2 cos 30  ( 3 ) I P
I B  2 I 3 cos 30 
Thus IR=IY=IB = IL
 3 I
P
DELTA CONNECTION
Voltage
VL  VP
Current
IL 
 3 I
P
STAR CONNECTION
b) Conventional connection
diagram
a) Physical connection diagram
STAR CONNECTION
STAR CONNECTION
•VRY, VYB and VBR are called line voltage
•VR, VY and VB are called phase voltage
R
VRY
VBR
From Kirchoff voltage law we have
VRY  VR  VY  VR  ( VY )
VYB  VY  VB  VY  ( VB )
VBR  VB  VR  VB  ( VR )
In phasor diagram
IR
VR
N
Y
VB
IY
VY
B
VYB
IB
STAR CONNECTION
For balanced load VR , VY and
VB are equaled but out of phase
VRY = VL30;
VR = VP30;
VY = VP-90;
VYB = VL-90;
VB = VP150;
VBR = VL150;
therefore
 
o
VRYV  2V2VR cos
30
cos 30 o 
  33VVP
BR
B
VYB  2VY cos 30 o 
 3 V
P
P
STAR CONNECTION
Voltage
 
VL  3 VP
Current
IL  IP
POWER IN 3-PHASE
SYSTEM
• DC power system
• AC power system
POWER IN 3-PHASE
SYSTEM
• Power for 3-phase system (refer to phase
element) P  3V P I P cos 
 3V L system
I L cos (refer to line
• Power forP 3-phase
element)
SUMMARY
Characteristic
Symbol
Voltage
Current
Balance
Condition
Power in 1
Power in 3
- Phase element
- Line element
Star Connection
or
Delta Connection
EXAMPLE 1
Three similar resistors are connected
in star acroos 400V, 3 phase lines. The
line current is 5A. Calculate the value
of each resistor. To what value should
the line voltage be changed to obtain
the same line current with the
resistors are in delta connected?
SOLUTION 1
• Line voltage, VL = 400V
– Phase voltage, Vph = VL3/
230.94V
= 400
3 /
=
• Line current, IL = 5A
– Phase current Iph = 5A (in star IL = Iph)
• Zph = Rph = Vph / Iph = 230.94 / 5 =
46.19
3
• If the same resistors are3connected in
delta
– VL = Vph and IL = Iph
– Vph = Iph Zph = Iph Rph = IL /
x Rph =
EXAMPLE 2
Three identical impedances are
connected in delta to a three phase 400V
supply. The line current is 34.65A and the
total power taken from the supply is
14.4kW. Calculate the resistance and
reactance values of each impedance.
SOLUTION 2
– Line Voltage, VL = 400V
• Phase Voltage, Vph = VL = 400V
– Line Current, IL = 34.65A
• Phase Current, IPH = IL / 3 = 34.65
/ 3 =

20A
– Total power,2 P = 14.4kW
2
• But P = 3 Iph Rph 2
Rph = 14.4k / 3 Iph = 14.4k / 3(20) = 12
2 / 20
– Zph = Vph /2 Iph =2 400
2 = 20

• XL = Zph - Rph = 20 - 12 = 16
EXAMPLE 3
Three similar coils are connected in
star taken at a total power of 1.5kW at
a p.f of 0.2 lagging from a 3-phase
400V 50Hz supply. Calculate the
resistance and inductance of each
phase.
SOLUTION 3
• P =√ 3 VL IL cos θ
• IL = P √/ 3 VL cos θ = 1.5k√ / ( 3 x 400
x 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
L = XL / 2πf = 20.89 / (2π x 50) =
66mH
EXAMPLE 4
Find :a) Phase Impedance, ZPH
b) Phase Current, IPH
c) 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 :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 of
20Ω and inductive reactance of 15Ω
are connected to a 400V, 3 phase 50Hz
supply. Calculate:a)The line current
b)Power factor
c)Power drawn from the supply
SOLUTION 6
a) 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
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