Download topic 4: three phase circuit

ELECTRICAL TECHNOLOGY
(EET 103)
PN HAZIAH ABDUL HAMID
SYLLABUS – PART II
SYLLABUS
• TOPIC 4: THREE PHASE CIRCUIT
• TOPIC 5: TRANSFORMER
• TOPIC 6: ELECTRICAL MACHINES
TOPIC 4: THREE PHASE CIRCUIT
SINGLE PHASE TWO WIRE
V p 
SINGLE PHASE SYSTEM
• A generator connected through a pair
of wire to a load – Single Phase Two
Wire.
• Vp is the magnitude of the source
voltage, and  is the phase.
SINLGE PHASE THREE WIRE
V p 
V p 
SINGLE PHASE SYSTEM
• Most common in practice: two
identical sources connected to two
loads by two outer wires and the
neutral: Single Phase Three Wire.
• Terminal voltages have same
magnitude and the same phase.
POLYPHASE SYSTEM
• Circuit or system in which AC
sources operate at the same
frequency but different phases
are known as polyphase.
TWO PHASE SYSTEM THREE WIRE
V p 
V p   90
POLYPHASE SYSTEM
• Two Phase System:
– A generator consists of two coils placed
perpendicular to each other
– The voltage generated by one lags the
other by 90.
POLYPHASE SYSTEM
• Three Phase System:
– A generator consists of three coils placed
120 apart.
– The voltage generated are equal in
magnitude but, out of phase by 120.
• Three phase is the most economical
polyphase system.
THREE PHASE FOUR WIRE
IMPORTANCE OF THREE PHASE SYSTEM
• All electric power is generated and
distributed in three phase.
– One phase, two phase, or more than
three phase input can be taken from
three phase system rather than
generated independently.
– Melting purposes need 48 phases
supply.
IMPORTANCE OF THREE PHASE SYSTEM
• Uniform power transmission and less
vibration of three phase machines.
– The instantaneous power in a 3 system
can be constant (not pulsating).
– High power motors prefer a steady
torque especially one created by a
rotating magnetic field.
IMPORTANCE OF THREE PHASE SYSTEM
• Three phase system is more
economical than the single phase.
– The amount of wire required for a three
phase system is less than required for an
equivalent single phase system.
– Conductor: Copper, Aluminum, etc
THREE PHASE GENERATION
FARADAYS LAW
•
Three things must be present in
order to produce electrical current:
a) Magnetic field
b) Conductor
c) Relative motion
•
•
Conductor cuts lines of magnetic
flux, a voltage is induced in the
conductor
Direction and Speed are important
GENERATING A SINGLE PHASE
S
N
Motion is parallel to the flux.
No voltage is induced.
GENERATING A SINGLE PHASE
S
N
Motion is 45 to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
x
N
Motion is perpendicular to flux.
Induced voltage is maximum.
GENERATING A SINGLE PHASE
S
N
Motion is 45 to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is parallel to flux.
No voltage is induced.
GENERATING A SINGLE PHASE
S
N
Notice current in the
conductor has reversed.
Motion is 45 to flux.
Induced voltage is
0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is perpendicular to flux.
Induced voltage is maximum.
GENERATING A SINGLE PHASE
S
N
Motion is 45 to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is parallel to flux.
No voltage is induced.
Ready to produce another cycle.
THREE PHASE GENERATOR
GENERATOR WORK
• The generator consists of a rotating
magnet (rotor) surrounded by a
stationary winding (stator).
• Three separate windings or coils with
terminals a-a’, b-b’, and c-c’ are
physically placed 120 apart around
the stator.
• As the rotor rotates, its magnetic field
cuts the flux from the three coils and
induces voltages in the coils.
• The induced voltage have equal
magnitude but out of phase by 120.
GENERATION OF THREE-PHASE AC
S
x
x
N
THREE-PHASE WAVEFORM
Phase 1
120
Phase 2
Phase 3
120
120
240
Phase 2 lags phase 1 by 120.
Phase 3 lags phase 1 by 240.
Phase 2 leads phase 3 by 120.
Phase 1 lags phase 3 by 120.
GENERATION OF 3 VOLTAGES
Phase 1 Phase 2 Phase 3
The
algebraic
sum of the
instantaneous
voltages of
the three
phases equals
zero.
Phase 1 is ready to go positive.
Phase 2 is going more negative.
Phase 3 is going less positive.
S
x
x
N
THREE PHASE QUANTITIES
BALANCED 3 VOLTAGES
• Balanced three phase voltages:
– same magnitude (VM )
– 120 phase shift
v an (t )  VM cos t 
vbn (t )  VM cos t  120
vcn (t )  VM cos t  240  VM cos t  120
BALANCED 3 CURRENTS
• Balanced three phase currents:
– same magnitude (IM )
– 120 phase shift
ia (t )  I M cos t   
ib (t )  I M cos t    120
ic (t )  I M cos t    240
PHASE SEQUENCE
van (t )  VM cos t
vbn (t )  VM cost  120
vcn (t )  VM cost  120
Van  VM 0
Van  VM 0
Vbn  VM   120
Vbn  VM   120
Vcn  VM   120
Vcn  VM   120
POSITIVE
SEQUENCE
NEGATIVE
SEQUENCE
PHASE SEQUENCE
EXAMPLE
• Determine the phase sequence of
the set voltages:
van  200 cost  10
vbn  200 cost  230
vcn  200 cost  110
Ans: Negative Sequence
BALANCED VOLTAGE AND LOAD
• Balanced Phase Voltage: all phase
voltages are equal in magnitude and
are out of phase with each other by
120.
• Balanced Load: the phase
impedances are equal in magnitude
and in phase.
THREE PHASE CIRCUIT
• POWER
– The instantaneous power is constant
p(t )  pa (t )  pb (t )  pc (t )
VM I M
3
cos 
2
 3Vrms I rms cos( )
THREE PHASE CIRCUIT
• Three Phase Power,
S T  S A  S B  S C  3 S
THREE PHASE QUANTITIES
QUANTITY
SYMBOL
Phase current
I
Line current
IL
Phase voltage
V
Line voltage
VL
PHASE VOLTAGES and LINE VOLTAGES
• Phase voltage is measured
between the neutral and any line:
line to neutral voltage
• Line voltage is measured between
any two of the three lines: line to
line voltage.
PHASE CURRENTS and LINE CURRENTS
• Line current (IL) is the current in
each line of the source or load.
• Phase current (I) is the current in
each phase of the source or load.
THREE PHASE CONNECTION
SOURCE-LOAD CONNECTION
SOURCE
LOAD
CONNECTION
Wye
Wye
Y-Y
Wye
Delta
Y-
Delta
Delta
- 
Delta
Wye
-Y
SOURCE-LOAD CONNECTION
• Common connection of source: WYE
– Delta connected sources: the
circulating current may result in the
delta mesh if the three phase voltages
are slightly unbalanced.
• Common connection of load: DELTA
– Wye connected load: neutral line may
not be accessible, load can not be
added or removed easily.
WYE CONNECTION
WYE CONNECTED GENERATOR
Ia
a
Van
Vab
n
Vbn
Ib
Vca
b
Vcn
Vbc
Ic
c
WYE CONNECTED LOAD
a
a
Y
Z
ZY
b
OR
b
Load
ZY
c
n
c
n
ZY
ZY
ZY
Load
BALANCED Y-Y CONNECTION
PHASE CURRENTS AND LINE CURRENTS
• In Y-Y system:
IL  Iφ
PHASE VOLTAGES, V
• Phase voltage is
measured between
the neutral and any
line: line to neutral
voltage
Ia
a
VVanan
n
VVbnbn
Vab
Ib
Vca
b
VVcn
cn
Vbc
Ic
c
PHASE VOLTAGES, V
Van  VM 0
volt
Vbn  VM   120 volt
Vcn  VM 120
volt
LINE VOLTAGES, VL
Ia
• Line voltage is
measured between
any two of the three
lines: line to line
voltage.
a
Van
Vab
ab
n
Vbn
Ib
b
Vcn
V
V bc
bc
Ic
c
VVcaca
LINE VOLTAGES, VL
Vab  Van  Vbn
Vbc  Vbn  Vcn
Vca  Vca  Van
Vab  3VM 30
Vbc  3VM   90
Vca  3VM 150
Van  VM 0
volt
Vbn  VM   120 volt
Vcn  VM 120 volt
LINE
VOLTAGE
(VL)
PHASE
VOLTAGE (V)
Vab  3 VM 30 volt
Vbc  3 VM   90 volt
Vca  3 VM 150 volt
PHASE DIAGRAM OF VL AND V
Vca
Vcn
Vab
30°
120°
Vbn
Vbc
-Vbn
Van
PROPERTIES OF PHASE VOLTAGE
• All phase voltages have the same
magnitude,
V  Van  Vbn  Vcn
• Out of phase with each other by 120
PROPERTIES OF LINE VOLTAGE
• All line voltages have the same
magnitude,
VL  Vab  Vbc  Vca
• Out of phase with each other by 120
RELATIONSHIP BETWEEN V and VL
1. Magnitude
VL  3 V
2. Phase
- VL LEAD their corresponding V by 30
VL  V  30
EXAMPLE
• Calculate the line currents
Single Phase Equivalent Circuit
• Phase ‘a’ equivalent circuit
5-j2
10  j8
Van
Ia 
ZY
1100
Ia 
 6.81  21.8
16.15521.8
I b  I a   120
 6.81  141.8A
I c  I a   240
 6.81  261.8  6.8198.2A
DELTA CONNECTION
DELTA CONNECTED SOURCES
DELTA CONNECTED LOAD
OR
BALANCED -  CONNECTION
PHASE VOLTAGE AND LINE VOLTAGE
• In - system, line voltages equal to
phase voltages:
VL  Vφ
PHASE CURRENTS, I
• Line voltages are equal to the
voltages across the load impedances.



PHASE CURRENTS, I
• The phase currents are obtained:
VBC
VCA
VAB
I AB 
, I BC 
, I CA 
ZΔ
ZΔ
ZΔ
LINE CURRENTS, IL
• The line currents are obtained from the
phase currents by applying KCL at
nodes A,B, and C.



LINE CURRENTS, IL
I a  I AB  I CA
I b  I BC  I AB
I c  I CA  I BC
I a  3 I AB  30
I b  I a   120
I c  I a   120
PHASE
CURRENTS (I)
I AB
I BC
I CA
VAB

ZΔ
VBC

ZΔ
VCA

ZΔ
LINE CURRENTS (IL)
I a  3 I AB  30
I b  I a   120
I c  I a   120
PHASE DIAGRAM OF IL AND I
PROPERTIES OF PHASE CURRENT
• All phase currents have the same
magnitude,
Iφ  I AB  I BC  ICA 
Vφ
ZΔ
• Out of phase with each other by 120
PROPERTIES OF LINE CURRENT
• All line currents have the same
magnitude,
I L  Ia  I b  Ic
• Out of phase with each other by 120
RELATIONSHIP BETWEEN I and IL
1. Magnitude
IL  3 IF
2. Phase
- VL LAG their corresponding V by 30
IL  IF  30
EXAMPLE
A balanced delta connected load having
an impedance 20-j15  is connected to
a delta connected, positive sequence
generator having Vab = 3300 V.
Calculate the phase currents of the load
and the line currents.
Given Quantities
 ZΔ  20  j15   25  36.87
 Vab  3300
Phase Currents
VAB
3300
I AB 

 13.236.87A
ZΔ
25  38.87
I BC  I AB  120  13.2 - 83.13A
I CA  I AB  120  13.2156.87A
Line Currents
I a  I AB 3  30


 13.236.87 3  30 A
 22.866.87
I b  I a   120  22.86 - 113.13A
I c  I a   120  22.86126.87A
BALANCED DELTA-WYE SYSTEM
EXAMPLE
A balanced positive sequence Yconnected source with Van=10010 V
is connected to a -connected
balanced load (8+j4) per phase.
Calculate the phase and line currents.
Given Quantities
• Balanced WYE source
–Van= 10010 V
• Balanced DELTA load
–Z= 8+j4 
Phase Currents
VAB
I AB 
ZΔ
VAB= voltage across Z
= Vab= source line voltage
 VAB  3 Van 30
VAB  173.240 V
173.240
 I AB 
 19.3613.43
8  j4
Phase Currents
I AB  19.3613.43
 I BC  I AB13.43  120
I BC  19.36  106.57 A
 I CA  19.3613.43  120
I CA  19.36133.43 A
Line Currents
I a  3 I AB   30
 3 (19.36) 13.43  30
I a  33.53   16.57 A
 I b  I a   120  33.53   136.57 A
 I c  I a   120  33.53 103.43 A
DELTA TO WYE CONVERSION
Z
Z 
3
THREE PHASE POWER
MEASUREMENT
FOUR WIRE SYSTEM
• Each phase measured separately:
IA
Phase A
A
VAN
Phase B
PA
W
V
IB
A
VBN
Phase C
PB
W
V
IC
A
VCN
Neutral (N)
PC
W
V
THREE PHASE THREE WIRE SYSTEM
The three
phase power is
the sum of the
two wattmeters reading
PT  PAB  PCB
IA
PAB
A
VAB = VA - VB
Phase A
W
V
Phase B
VCB = VC - VB
IC
A
V
W
Phase C
PCB
EXAMPLE
Determine the total power (P), reactive
power (Q), and complex power (S) at the
source and at the load
Single Phase Equivalent Circuit
• Phase ‘a’ equivalent circuit
5-j2
10  j8
Known Quantities
• V =Van = 1100 V
• ZY = 10+j8 
• Zline =5-j2 
Line/Phase Currents
Van
Ia 
Zline  Z Y
1100
Ia 
 6.81  21.8
16.15521.8
Source & Load Power

φ φ
Ssource  3V I
 (2087  j834.6)VA
 Ps  2087 W, Qs  834.6 Var
2
SLoad  3 I φ Z 
 (1392  j1113)VA
 PL  1392 W, Q L  1113 Var
EXAMPLE
A three phase motor can be
regarded as a balanced Y-load. A
three phase motor draws 5.6 kW
when the line voltage is 220 V and
the line current is 18.2 A. Determine
the power factor of the motor
Known Quantities
• PLoad =5600 W
• VL = 220 V
• IL = 18.2 A
Power factor
• Power factor = cos 
|S|
Q
S  3Vφ I φ  3 VL I L
 6935.13 VA

P
P  S cos θ
P
5600
 cos θ  
 0.8
S 6935.13