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```					Chapter 1.
Three-Phase System

1
1.1: Review of Single-Phase System

The Sinusoidal voltage
v1(t) = Vm sin wt

i

v1                   v2

AC
generator
2
1.1: Review of Single-Phase System

The Sinusoidal voltage
v(t) = Vm sin wt
where
Vm = the amplitude of the sinusoid
w = the angular frequency in radian/s
t = time
v(t)
Vm

                          wt
-Vm

3
v(t)
Vm

                             t
-Vm

2               1
T              f 
w                T

w  2f
The angular frequency in radians per second
4
A more general expression for the sinusoid (as
shown in the figure):
v2(t) = Vm sin (wt + )
where  is the phase
v(t)
V1 = Vm sin wt
Vm

                                            wt
-Vm

V2 = Vm sin wt + )

5
A sinusoid can be expressed in either sine or cosine form.
When comparing two sinusoids, it is expedient to express
both as either sine or cosine with positive amplitudes.
We can transform a sinusoid from sine to cosine form or
vice versa using this relationship:
sin (ωt ± 180o) = - sin ωt
cos (ωt ± 180o) = - cos ωt
sin (ωt ± 90o) = ± cos ωt
cos (ωt ± 90o) = + sin ωt

6
Sinusoids are easily expressed in terms of phasors.
A phasor is a complex number that represents the
amplitude and phase of a sinusoid.

v(t) = Vm cos (ωt + θ)                  Time domain

V  Vm                          Phasor domain

Time domain         Phasor domain
Vm cos( wt   )     Vm 
Vm sin( wt   )    Vm   90 o
I m cos( wt   )    I m 
I m sin( wt   )    I m   90 o
7
Time domain
v(t)
V1 = Vm sin wt
v2(t) = Vm sin (wt + )
Vm
v1(t) = Vm sinwt

                                                   wt
-Vm

V2 = Vm sin wt + )

Phasor domain

V1  Vm 0         V1  Vrms 0 
V2
or
θ
V2  Vm      or   V2  Vrms 
V1
8
1.1.1: Instantaneous and Average Power

The instantaneous power is the power at any instant
of time.
p(t) = v(t) i(t)

Where          v(t) = Vm cos (wt + v)
i(t) = Im cos (wt + i)

Using the trigonometric identity, gives
1                       1
p( t )  Vm I m cos( v  i )  Vm I m cos( 2wt  v  i )
2                       2
9
The average power is the average of the
instantaneous power over one period.

1 T
P   p(t ) dt
T 0
1
P     Vm I m cos(  v   i )
2
p(t)

1
Vm I m
2

1
Vm I m cos(  v   i )
2

                                           t

10
The effective value is the root mean square (rms) of
the periodic signal.
The average power in terms of the rms values is

P  VrmsI rms cos( v  i )
Vm
Where           Vrm s 
2

Im
I rm s 
2

11
1.1.2: Apparent Power, Reactive Power and
Power Factor

The apparent power is the product of the rms values
of voltage and current.
S  VrmsI rms

The reactive power is a measure of the energy
exchange between the source and the load reactive
part.
Q  VrmsI rms sin( v  i )

12
The power factor is the cosine of the phase difference
between voltage and current.
P
Power factor   cos( v  i )
S

The complex power:  P  jQ
 Vrm s I rm s v  i

13
1.2: Three-Phase System

In a three phase system the source consists of three
sinusoidal voltages. For a balanced source, the three
sources have equal magnitudes and are phase
displaced from one another by 120 electrical degrees.

A three-phase system is superior economically and
advantage, and for an operating of view, to a single-
phase system. In a balanced three phase system the
power delivered to the load is constant at all times,
whereas in a single-phase system the power pulsates
with time.
14
1.3: Generation of Three-Phase
Three separate windings or coils with terminals R-R’,
Y-Y’ and B-B’ are physically placed 120o apart
around the stator.
R
Stator
Y
R

B        B’        Y’
N

Rotor

Y    S    B

R’

15
V or v is generally represented a voltage, but
to differentiate the emf voltage of generator
from voltage drop in a circuit, it is convenient to
use e or E for induced (emf) voltage.

16
v(t)
vR    vY    vB

wt
        

The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin wt
eY = EmY sin (wt -120o)
eB = EmB sin (wt -240o) = EmBsin (wt +120o)
17
Three-phase
AC generator

IR

VR    ZR
ER
IN
EB
VB               VY
EY               ZB   ZY
IY

IB

Three-phase
18
Phase voltage

The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin ωt
eY = EmY sin (ωt -120o)
eB = EmB sin (ωt -240o) = EmBsin (ωt +120o)

In phasor domain:
120o
ER = ERrms     0o
0o
EY = EYrms     -120o
-120o
EB = EBrms     120o

ERrms = EYrms = EBrms = Ep      Magnitude of phase voltage
19
Three-phase
AC generator   Line voltage

IR

ERY             VR    ZR
ER
IN
EB
VB               VY
EY                    ZB   ZY
IY

IB

ERY = ER - EY                Three-phase
Line voltage

ERY = Ep      0o - Ep      -120o               ERY
-EY
= 1.732Ep        30o

= √3 Ep       30o              120o

= EL       30o                        0o

-120o

ERY = ER - EY                  21
Three-phase
AC generator   Line voltage

IR

VR    ZR
ER
IN
EB
VB               VY
EY                    ZB   ZY
IY

IB                  EYB

EYB = EY - EB               Three-phase
Line voltage

EYB = Ep      -120o - Ep     120o

= 1.732Ep         -90o            120o

= √3 Ep      -90o                        0o

= EL       -90o                       -120o

-EB

EYB = EY - EB                   EYB
Three-phase
AC generator   Line voltage

IR

VR    ZR
ER     EBR
IN
EB
VB               VY
EY                   ZB   ZY
IY

IB

EBR = EB - ER              Three-phase
Line voltage

EBR = Ep     120o - Ep     0o
EBR

= 1.732Ep          150o              120o

= √3 Ep        150o                        0o

= EL    150o          -ER        -120o

For star connected supply, EL= √3 Ep

EBR = EB - ER                    25
Phase voltages      It can be seen that the phase
voltage ER is reference.
ER = Ep    0o

EY = Ep   -120o

EB = Ep    120o                       120o
0o
Line voltages
-120o
ERY = EL    30o

EYB = EL   -90o

EBR = EL   150o                                    26
Phase voltages       Or we can take the line voltage
ERY as reference.
ER = Ep     -30o

EY = Ep    -150o

EB = Ep    90o

Line voltages

ERY = EL    0o

EYB = EL    -120o

EBR = EL    120o                                  27
Three-phase
AC generator
Delta connected Three-Phase supply

IR

ERY        VR    ZR
ER
EB
VB               VY
EY                          ZB   ZY
IY

IB

Three-phase
ERY = ER = Ep   0o                Load
Three-phase
AC generator
Delta connected Three-Phase supply

IR

VR    ZR
ER
EB
EBR   VB               VY
EY                            ZB   ZY
IY

IB                   EYB

Three-phase
For delta connected supply, EL= Ep                Load
Connection in Three Phase System

4-wire system (neutral line with impedance)

3-wire system (no neutral line )
4-wire system (neutral line without impedance)

3-wire system (no neutral line ), delta connected load
a) 4-wire system b) 3-wire system

a) 3-wire system                                    30
Three-phase     4-wire system (neutral line with impedance)
AC generator

IR

VR     ZR
ER
IN   ZN
EB
VN     VB               VY
EY                      ZB   ZY
IY

IB

Three-phase
Voltage drop across neutral   VN = INZN        1.1          Load
impedance:
Three-phase    4-wire system (neutral line with impedance)
AC generator
Applying KCL at star point
IR

VR     ZR
ER
IN     ZN
EB
VN    VB                     VY
EY                             ZB   ZY
IY

IB

Three-phase
IR + IY + IB= IN              1.2                Load
Three-phase    4-wire system (neutral line with impedance)
AC generator
Applying KVL on R-phase loop
IR

VR     ZR
ER
IN   ZN
EB
VN     VB                  VY
EY                         ZB   ZY
IY

IB

Three-phase
33
Three-phase     4-wire system (neutral line with impedance)
AC generator
Applying KVL on R-phase loop
IR

VR    ZR
ER
IN   ZN

VN
ER – VR – VN = 0

ER – IRZR – VN = 0

Thus            ER – VN
IR =                            1.3           Three-phase
ZR                                           34
Three-phase    4-wire system (neutral line with impedance)
AC generator
Applying KVL on Y-phase loop
IR

VR     ZR
ER
IN   ZN
EB
VN     VB                  VY
EY                         ZB   ZY
IY

IB

Three-phase
35
Three-phase     4-wire system (neutral line with impedance)
AC generator
Applying KVL on Y-phase loop

EY – VY – VN = 0             Thus               EY – VN
IY =                    1.4
EY – IYZY – VN = 0                               ZY

IN    ZN

VN                          VY
EY
ZY
IY

Three-phase
36
Three-phase     4-wire system (neutral line with impedance)
AC generator
Applying KVL on B-phase loop

EB – VB – VN = 0            Thus                EB – VN
IB =                         1.5
EB – IBZB – VN = 0                                  ZB

IN   ZN
EB
VN          VB
ZB

IB

Three-phase
37
4-wire system (neutral line with impedance)

Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into
Eq. 1.1:
IR + IY + IB= IN

ER – VN             EY – VN         EB – VN               VN
+                 +                   =
ZR              ZY              ZB                   ZN

ER – VN + EY – VN +               EB – VN             VN
=
ZR ZR     ZY ZY                   ZB ZB               ZN

ER         EY          EB             1            1        1       1
+         +          = VN           +           +        +
ZR         ZY          ZB            ZN        ZR          ZY       ZB   38
4-wire system (neutral line with impedance)

ER            EY               EB
+             +
ZR            ZY               ZB                                    1.6
VN   =
1        1             1               1
+          +               +
ZN           ZR        ZY              ZB

ER        EY               EB                     1               1       1        1
+            +            = VN                      +           +        +
ZR            ZY           ZB                 ZN              ZR          ZY       ZB     39
4-wire system (neutral line with impedance)

VN is the voltage drop across neutral line impedance
or the potential different between load star point and
supply star point of three-phase system.

ER        EY               EB
+             +
ZR        ZY               ZB              1.6
VN   =
1       1             1        1
+         +            +
ZN       ZR        ZY           ZB

We have to determine the value of VN in order to find the
values of currents and voltages of star connected loads of
three-phase system.                                       40
Example

IR

EL = 415 volt      VR
ER                                     ZR = 5 Ω
IN ZN =10 Ω
EB
VN      VB                ZY= 2 Ω
EY
ZB = 10 Ω
IY

IB
Find the line currents IR ,IY and IB. Also find
Three-phase
Three-phase    3-wire system (no neutral line )
AC generator

IR

VR        ZR
ER
IN   ZN
EB
VN     VB                    VY
EY                          ZB   ZY
IY

IB

Three-phase
Three-phase     3-wire system (no neutral line )
AC generator

IR

VR        ZR
ER
VN
EB
VB                    VY
EY                         ZB   ZY
IY

IB

Three-phase
No neutral line = open circuit ,    ZN = ∞                 43
3-wire system (no neutral line )

ER        EY               EB
+             +
ZR        ZY               ZB
VN =                                             1.6
1       1             1        1
+         +            +
ZN       ZR        ZY           ZB

ZN = ∞
1
=   0
∞
44
3-wire system (no neutral line )

ER        EY               EB
+             +
ZR        ZY               ZB
VN =                                         1.7
1             1        1
+            +
ZR        ZY           ZB

45
Example

IR

EL = 415 volt       VR
ER                                     ZR = 5 Ω

EB
VN      VB                ZY= 2 Ω
EY
ZB = 10 Ω
IY

IB

Find the line currents IR ,IY and IB . Also find   Three-phase
the voltages VR, VY and VB.                               Load
3-wire system (no neutral line ),delta connected load
Three-phase
AC generator
IR

VR      ZR
ER

EB
VB                    VY
EY                       ZB    ZY
IY

IB

Three-phase
3-wire system (no neutral line ),delta connected load
Three-phase
AC generator
IR
Ir

ER                                             VRY
VBR
ZBR ZRY
EB
Ib        ZYB
EY                                   Iy
IY
VYB
IB

Three-phase
3-wire system (no neutral line ),delta connected load
Three-phase
AC generator
IR
Ir

ER             ERY =VRY                        VRY
VBR
EBR =VBR                    ZBR ZRY
EB
EY                   Ib       ZYB       Iy
IY
VYB
IB                    EYB =VYB

Three-phase
3-wire system (no neutral line ),delta connected load

Phase currents

VRY       ERY       EL    30o
Ir =         =         =
ZRY       ZRY       ZRY
VYB       EYB       EL    -90o
Iy =         =         =
ZYB       ZYB       ZYB

VBR       EBR       EL    150o
Ib =         =         =
ZBR       ZBR       ZBR

50
3-wire system (no neutral line ),delta connected load
Three-phase
AC generator
IR
Ir
Line currents

IR = Ir - Ib      ER            ERY =VRY                          VRY
VBR
EL     30o    EBR =V150o
E                            ZBR ZRY
EB
=                 - L    BR
ZRY          EZBR
Y                  Ib          ZYB       Iy
IY = Iy - Ir                    IY
VYB
EL   -90o EL 30oYB =VYB
E                      IB
=             -
ZYB        ZRY
Three-phase
3-wire system (no neutral line ),delta connected load
Three-phase
AC generator
IR
Ir
Line currents

IB =     Ib - Iy   ER               ERY =VRY                          VRY
VBR
EL     150o   EBRL =V-90o
E                            ZBR ZRY
EB
=                   -      BR
ZBR            EZYB
Y                  Ib          ZYB       Iy
IY
VYB
EYB =VYB          IB

Three-phase
Star to delta conversion

ZRZY + ZYZB + ZBZR
ZRY =
ZB                           ZR
ZBR                    ZRY

ZYB = ZRZY
+ ZYZB + ZBZR
ZY
ZR
ZB

ZBR = ZRZY
+ ZYZB + ZBZR
ZYB
ZY
Example               Use star-delta conversion.

IR

EL = 415 volt       VR
ER                                       ZR = 5 Ω

EB
VN         VB               ZY= 2 Ω
EY
ZB = 10 Ω
IY

IB

Find the line currents IR ,IY and IB .               Three-phase
Three-phase    4-wire system (neutral line without impedance)
AC generator

IR

VR     ZR
ER
IN   ZN = 0 Ω
EB
VN     VB                  VY
EY                        ZB   ZY
IR

IB

Three-phase
VN = INZN = IN(0) = 0 volt                55
4-wire system (neutral line without impedance)

For 4-wire three-phase system, VN is equal to
0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5
become,
ER – VN
IR =                       1.3
ZR

EY – VN
IY =                      1.4
ZY

EB – VN
IB =                      1.5
ZB
56
Example

IR

EL = 415 volt       VR
ER                                     ZR = 5 Ω
IN
EB
VN     VB                ZY= 2 Ω
EY
ZB = 10 Ω
IY

IB

Find the line currents IR ,IY and IB . Also find   Three-phase
v(t)
vR    vY    vB

wt
        

The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin wt
eY = EmY sin (wt -120o)
eB = EmB sin (wt -240o) = EmBsin (wt +120o)
58
1.4: Phase sequences
RYB and RBY
(a) RYB or positive sequence
VB
w                 VR  VR ( rms) 0o
120o
VY  VY ( rms)   120   o

120o                  VR
VB  VB ( rms)   240o
-120o
 VB ( rms) 120o
VY
This sequence is produced when the rotor rotates in
the counterclockwise direction.
59
(b) RBY or negative sequence

VY                          VR  VR ( rms) 0o
w
120o               VB  VB( rms)   120o
120o                 VR
VY  VY ( rms )   240 o
-120o
 VY ( rms ) 120 o
VB

This sequence is produced when the rotor rotates in
the clockwise direction.
60
1.5: Connection in Three Phase System

1.5.1: Star Connection
a) Three wire system
R

ZR

ZY        Z
B
Y

B
61
Star Connection
b) Four wire system

R

VRN                         ZR

V BN         V YN

ZY         Z
B
Y
N
B
62

R                                 R
Z1
Y                   Z1
2
Z

Y
Z2
Z3
B                                 B
Z3

N                                 N

63
1.5.2: Delta Connection

R
R

Y

Y

B
B

64

Zc
Zb

Y
c
Z

Zb

Y                              Za
Za
B

65
3-Phase System

66
Example
b) Three wire system
IR

EL = 415 volt      VR
ER                                    ZR = 20 Ω

EB
VN     VB                ZY= 20 Ω
EY
ZB = 20 Ω
IY

IB

Find the line currents IR ,IY and IB . Also find   Three-phase
67
the voltages VR, VY and VB.                               Load
b) Three wire system

VN = = 0 volt

VR = ER

VY = EY

VB = EB

68
Example                a) Four wire system
IR

EL = 415 volt       VR
ER                                    ZR = 20 Ω
IN
EB
VB                ZY= 20 Ω
VN
EY
ZB = 20 Ω
IY

IB

Find the line currents IR ,IY and IB . Also find   Three-phase
69
a) Four wire system
IR
R
IN  IR  IY  IB
Z1
VRN
IN
V YN
Z2        Z   IN = 0 and Z1 = Z2 = Z3
V BN             IY                      3

Y
VRN 0 o
B                                              IR 
IB
Z1
VRN  Vphasa 0                               VYN   120o
IY 
VYN  Vphasa   120                               Z2
VBN  Vphasa 120
VBN 120o
where Vphasa  VRN  VYN  VBN            IB 
Z3
70
b) Three wire system
IR
R
IR  IY  IB  0
Z1
VBR VRY
S
IY
VRS 0 o
Y
Z2            Z
3
IR 
VYB                                  Z1
B
VYS   120o
IB
IY 
Z2
VRS  Vphasa 0
VBS 120o
VYS  Vphasa   120               IB 
Z3
VBS  Vphasa   240
where Vphasa  VRS  VYS  VBS
71
IR
R
Phase currents:
VRY 0 o
I RY                          I BR         I RY   
VRY                                                           Z1

Z

Z
VYB   120o


VBR
I YB   
IY                                                        Z2
Y                            Z                                  VBR 120o
I YB                                   I BR   
VYB                                                             Z3
B
IB              Line currents:

where I RY  I YB  I BR  I phasa                      I R  I RY  I BR
and         I R  I Y  I B  I line                    I Y  I YB  I RY
I B  I BR  I YB        72

73
Four wire system
IR
R                                        IN  IR  IY  IB
Z1
VRN
IN
N                                       IN  0 and Z1  Z2  Z3
V YN                       Z
IY        Z2        3
V BN
Y                                                VRN 0 o
IR 
B                                                        Z1
IB                                   VYN   120o
IY 
VRN  Vphasa 0                                            Z2
VBN 120o
VYN  Vphasa   120                             IB 
Z3
VBN  Vphasa 120
74
IR
Phase currents:
VRY 0 o
R
I RY   
I RY                          I BR                      Z1
VRY

Z

Z                                         VYB   120o


VBR
I YB   
IY                                                         Z2
VBR 120o
Y                         Z
VYB
I YB
I BR   
B
Z3
IB                Line currents:

VRN  Vphasa 0                                             I R  I RY  I BR
VYN  Vphasa   120                                        I Y  I YB  I RY
VBN  Vphasa 120                                           I B  I BR  I YB 75
1.8 Power in a Three Phase
System

76
Power Calculation

The three phase power is equal the sum of
the phase powers

P = PR + PY + PB

P = 3 Pphase = 3 Vphase Iphase cos θ

77
1.8.1: Wye connection system:
I phase = I L and
VLL  3 Vphase
Real Power, P = 3 Vphase Iphase cos θ

 3 VLL I L cos Watt
Reactive power,
Q = 3 Vphase Iphase sin θ

 3 VLL I L sin  VAR
Apparent power,
S = 3 Vphase Iphase

 3 VLL I L VA
or S = P + jQ
78
1.8.2: Delta connection system

IL  3 Iphase
VLL= Vphase

P = 3 Vphase Iphase cos θ

 3 VLL I L cos  Watt

79
1.9: Three phase power
measurement

80
Power measurement

In a four-wire system (3 phases and a
neutral) the real power is measured using
three single-phase watt-meters.

81
Three Phase Circuit
Four wire system,
Each phase measured separately

IA            PA
Phase A
A         W

VAN   V        IB            PB
Phase B
A         W

VBN   V
IC           PC
Phase C
A        W

VCN      V
Neutral (N)

82
watt-meter connection

W

Current coil (low impedance)

voltage coil (high impedance)
83
a) Four wire system
Example
WR
IR

VR       ZR = 5         30o Ω
ER    EL = 415 volt
IN
EB
VB                   ZY = 10     90o
EY        VN                                              Ω
WY
ZB = 20   45o Ω
IY
IB
WB
Three-phase
Find the three-phase total power, PT.                              Load
b) Three wire system
Example
WR
IR

VR       ZR = 5         30o Ω
ER    EL = 415 volt

EB
ZY = 10     90o
EY        VN                                         Ω
WY
ZB = 20   45o Ω
IY
IB
WB
Three-phase
Find the three-phase total power, PT.                         Load
b) Three wire system
Example
WR
IR

VR         ZR = 5         30o Ω
ER    EL = 415 volt

EB                                     VB               ZY = 10     90o
EY        VN                                          Ω
WY
ZB = 20   45o Ω
IY
IB
WB
Three-phase
Find the three-phase total power, PT.                          Load
Three Phase Circuit
Three wire system,

The three phase power is the sum of the two watt-
P         AB
Phase A
A            W

VAB = VA - VB   V
Phase B

VCB = VC - VB
V

PT  PAB  PCB
Phase C
A            W
IC                     PCB

87
Proving:                                PT  PAB  PCB

The three phase power (3-wire system) is the
sum of the two watt-meters reading
IA            PA
Phase A
A         W
Instantaneous power:
VAN   V        IB            PB
pA = v A i A                        Phase B
A         W

VBN
pB = v B i B                                                           V
IC                PC
Phase C
A             W
pC = v C i C
VCN      V
Neutral (N)

pT = pA + pB + pC = vA iA + vB iB +vC iC

= vA iA + vB iB +vC iC = vA iA + vB (-iA -iC) +vCiC                                        88
Proving:

The three phase power (3-wire system) is the
sum of the two watt-meters reading
IA                PA
Phase A                                                 PAB
A IA          W
Instantaneous power:                                            A                      W
Phase A

pT = vA iA + vB (-iA –iC) +vCiC                               V
VAN
IB                PB
Phase B        VAB = VA - VB           V
A             W
Phase B
= (vA – vB )iA + (vC – vB )iC                                        VBN       V
IC               PC
Phase C
VCB = VC - VB                            A            W
V
= vAB iA + vCBiC                                                                          VCN     V Phase C
A                      W
Neutral (N)        IC                                    PCB

pT = pAB + pCB
PT  PAB  PCB                                                      89
Power measurement

In a four-wire system (3 phases and a
neutral) the real power is measured using
three single-phase watt-meters.

In a three-wire system (three phases
without neutral) the power is measured
using only two single phase watt-meters.

The watt-meters are supplied by the line
current and the line-to-line voltage.
90

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