Download - Chapter 1-Three Phase
-
7/31/2019 Chapter 1-Three Phase
1/90
1
Chapter 1.
Three-Phase System
-
7/31/2019 Chapter 1-Three Phase
2/90
1.1: Review of Single-Phase System
The Sinusoidal voltage
v1(t) = Vm sin wt
i
v1
Load
ACgenerator
v2
2
-
7/31/2019 Chapter 1-Three Phase
3/90
3
1.1: Review of Single-Phase System
The Sinusoidal voltage
v(t) = Vm sin wtwhere
Vm = the amplitude of the sinusoid
w = the angular frequency in radian/s
t = time
v(t)
Vm
-Vm
wt
-
7/31/2019 Chapter 1-Three Phase
4/90
4
v(t)
Vm
-Vm
t
w 2T
T
1f
f2wThe angular frequency in radians per second
-
7/31/2019 Chapter 1-Three Phase
5/90
5
A more general expression for the sinusoid (as
shown in the figure):
v2(t) = Vm sin (wt + )where is the phase
v(t)
Vm
-Vm
wt
V1= V
msin wt
V2= V
msin wt + )
-
7/31/2019 Chapter 1-Three Phase
6/90
6
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 (t 180o) = - sin t
cos (t 180o) = - cos t
sin (t 90o
) = cos tcos (t 90o) = + sin t
-
7/31/2019 Chapter 1-Three Phase
7/90
7
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 (t + )
Time domain Phasor domain
Time domain
mVV Phasor domain
)cos( wtVm mV
)sin( wtVmo
m 90V
)cos( wtmI mI
)sin( wtmIo
m 90I
-
7/31/2019 Chapter 1-Three Phase
8/90
8
Time domain
Phasor domain
v(t)
Vm
-Vm
wt
V1= V
msin wt
V2
= Vm
sin wt + )
V1
V2
v2(t) = Vm sin (wt + )
v1(t) = Vm sinwt
m
VV2
01 mVV
or
or 01 rmsVV
rms
VV2
-
7/31/2019 Chapter 1-Three Phase
9/90
9
1.1.1: Instantaneous and Average Power
The instantaneous power is the power at any instantof 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
)cos()cos()( ivmmivmm t2IV2
1IV
2
1t wp
-
7/31/2019 Chapter 1-Three Phase
10/90
10
The average power is the average of the
instantaneous power over one period.
T
dttpT
P0
)(1
)cos( ivmmIV21 P
p(t)
t
)cos( ivmmIV2
1
mmIV2
1
-
7/31/2019 Chapter 1-Three Phase
11/90
11
The effective value is the root mean square (rms) of
the periodic signal.The average power in terms of the rms values is
Where
)cos(iv
P rm srms
IV
2
VV mrm s
2
II
m
rm s
-
7/31/2019 Chapter 1-Three Phase
12/90
12
1.1.2: Apparent Power, Reactive Power and
Power Factor
The apparent power is the product of the rms values
of voltage and current.
The reactive power is a measure of the energy
exchange between the source and the load reactive
part.
rm srm sIVS
)sin( ivQ rm srm sIV
-
7/31/2019 Chapter 1-Three Phase
13/90
13
The power factor is the cosine of the phase difference
between voltage and current.
The complex power:
)cos( ivfactorPower S
P
iv
jQP
rm srm s IV
-
7/31/2019 Chapter 1-Three Phase
14/90
14
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.
-
7/31/2019 Chapter 1-Three Phase
15/90
15
1.3: Generation of Three-Phase
Three separate windings or coils with terminals R-R,Y-Y and B-B are physically placed 120o apartaround the stator.
Y
BY
B
Stator
Rotor
Y
R
B
R
R
N
S
-
7/31/2019 Chapter 1-Three Phase
16/90
16
Vor vis generally represented a voltage, butto differentiate the emf voltage of generatorfrom voltage drop in a circuit, it is convenient touse eor Efor induced (emf) voltage.
-
7/31/2019 Chapter 1-Three Phase
17/90
1717
v(t)
wt
vR
vY
vB
The instantaneous e.m.f. generated in phase R, Y and B:
eR= EmR sin wteY= EmY sin (wt -120
o)
eB= EmB sin (wt -240o)= EmBsin (wt +120
o)
-
7/31/2019 Chapter 1-Three Phase
18/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN
18
-
7/31/2019 Chapter 1-Three Phase
19/90
The instantaneous e.m.f. generated in phase R, Y and B:
eR= EmR sin t
eY= EmY sin (t -120o)
eB= EmB sin (t -240o) = EmBsin (t +120
o)
In phasor domain:
ER= ERrms 0o
EY= EYrms -120o
EB= EBrms 120o
Phase voltage
120o
-120o
0o
ERrms= EYrms= EBrms= Ep Magnitude of phase voltage19
-
7/31/2019 Chapter 1-Three Phase
20/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN
Line voltage
ERY
ERY
= ER
- EY
-
7/31/2019 Chapter 1-Three Phase
21/90
21
Line voltage
ERY
= ER
- EY
120o
-120o
0o
-EYERY
= Ep 0o - Ep -120
o
= 1.732Ep
ERY
30o= 3 Ep
= EL
30o
30o
-
7/31/2019 Chapter 1-Three Phase
22/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN
Line voltage
EYB
EYB
= EY
- EB
-
7/31/2019 Chapter 1-Three Phase
23/90
Line voltage
EYB
= EY
- EB
120o
-120o
0o
-EB
EYB
= Ep -120o - Ep
120o
= 1.732Ep
EYB
-90o
-90o= 3 Ep
= EL
-90o
-
7/31/2019 Chapter 1-Three Phase
24/90
ER
Three-phaseLoad
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN
Line voltage
EBR
EBR = EB- ER
-
7/31/2019 Chapter 1-Three Phase
25/90
2525
Line voltage
EBR
= EB
- ER
120o
-120o
0o
-ER
EBR= Ep 120
o - Ep 0o
= 1.732Ep
EBR
150o
150o= 3 Ep
= EL
150o
For star connected supply, EL= 3 Ep
-
7/31/2019 Chapter 1-Three Phase
26/90
26
120o
-120o
0o
Phase voltages
ER= Ep 0o
EY= Ep -120o
EB= Ep 120o
Line voltages
ERY= EL 30o
EYB= EL -90o
EBR= EL 150o
It can be seen that the phasevoltage ER is reference.
-
7/31/2019 Chapter 1-Three Phase
27/90
2727
Phase voltages
ER= Ep -30o
EY= Ep -150o
EB= Ep 90o
Line voltages
ERY= EL 0o
EYB= EL -120o
EBR= EL 120o
Or we can take the line voltageERYas reference.
-
7/31/2019 Chapter 1-Three Phase
28/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
ERY
Delta connected Three-Phase supply
ERY= ER = Ep 0o
-
7/31/2019 Chapter 1-Three Phase
29/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
EYB
EBR
For delta connected supply, EL= Ep
Delta connected Three-Phase supply
-
7/31/2019 Chapter 1-Three Phase
30/90
30
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)
Star-Connected Balanced Loadsa)4-wire system b) 3-wire system
3-wire system (no neutral line ), delta connected load
Delta-Connected Balanced Loads
a) 3-wire system
-
7/31/2019 Chapter 1-Three Phase
31/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN ZN
VN
4-wire system (neutral line with impedance)
VN= INZNVoltage drop across neutralimpedance: 1.1
-
7/31/2019 Chapter 1-Three Phase
32/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN ZN
VN
4-wire system (neutral line with impedance)
IR + IY+ IB= IN
Applying KCL at star point
1.2
-
7/31/2019 Chapter 1-Three Phase
33/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN ZN
VN
4-wire system (neutral line with impedance)
Applying KVL on R-phase loop
33
-
7/31/2019 Chapter 1-Three Phase
34/90
ER
Three-phase
Load
Three-phaseAC generator
IR
VR ZR
IN ZN
VN
Applying KVL on R-phase loop
ERVRVN= 0
ERIRZRVN= 0
IR=
ThusERVN
ZR
1.3
4-wire system (neutral line with impedance)
34
-
7/31/2019 Chapter 1-Three Phase
35/90
4 i ( l li i h i d )
-
7/31/2019 Chapter 1-Three Phase
36/90
36
Three-phase
Load
Three-phaseAC generator
VYEY
IY
ZY
IN ZN
VN
Applying KVL on Y-phase loop
4-wire system (neutral line with impedance)
EYVYVN= 0
EYIYZYVN= 0IY=
ThusEYVN
ZY
1.4
4 i t ( t l li ith i d )
-
7/31/2019 Chapter 1-Three Phase
37/90
Three-phase
Load
Three-phaseAC generator
EB
IB
ZB
VB
IN ZN
VN
4-wire system (neutral line with impedance)
Applying KVL on B-phase loop
EBVBVN= 0
EBIBZBVN= 0IB=
ThusEBVN
ZB
1.5
37
4 i t ( t l li ith i d )
-
7/31/2019 Chapter 1-Three Phase
38/90
38
4-wire system (neutral line with impedance)
IR + IY+ IB= IN
Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into
Eq. 1.1:
=EBVN
ZB+
EYVN
ZY
ERVN
ZR +
VN
ZN
ERVN EYVN+ + EBVN =VN
ZNZR ZR ZY ZY ZB ZB
ER
ZR+
EY
ZY+
EB
ZB=
1
ZN+
1
ZR+
1
ZY
VN +1
ZB
4 i t ( t l li ith i d )
-
7/31/2019 Chapter 1-Three Phase
39/90
39
4-wire system (neutral line with impedance)
ER
ZR+
EY
ZY+
EB
ZB=
1
ZN+
1
ZR+
1
ZY
VN +1
ZB
VN =
ERZR
+ EYZY
+ EBZB
1
ZN
+1
ZR
+1
ZY
+1
ZB
1.6
4 i t ( t l li ith i d )
-
7/31/2019 Chapter 1-Three Phase
40/90
4-wire system (neutral line with impedance)
VN =
ERZR
+ EYZY
+ EBZB
1
ZN
+1
ZR
+1
ZY
+1
ZB
1.6
VN is the voltage drop across neutral line impedance
or the potential different between load star point andsupply star point of three-phase system.
We have to determine the value of VN in order to find thevalues of currents and voltages of star connected loads of
three-phase system. 40
-
7/31/2019 Chapter 1-Three Phase
41/90
Example
ER
Three-phase
Load
ZY= 2
IR
VR
EY
EB
ZR= 5
IY
IB
ZB= 10
VB
IN ZN=10
VN
Find the line currents IR,IYand IB. Also find
the neutral current IN.
EL = 415 volt
Th h 3 wire system (no neutral line )
-
7/31/2019 Chapter 1-Three Phase
42/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
IN ZN
VN
3-wire system (no neutral line )
Th h 3 wire system (no neutral line )
-
7/31/2019 Chapter 1-Three Phase
43/90
ER
Three-phase
Load
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
VN
3-wire system (no neutral line )
No neutral line = open circuit , ZN=
43
3 wire system (no neutral line )
-
7/31/2019 Chapter 1-Three Phase
44/90
44
VN =
ER
ZR+ E
Y
ZY+ E
B
ZB
1
ZN +
1
ZR +
1
ZY +
1
ZB
1.6
3-wire system (no neutral line )
=ZN
1
= 0
3 wire system (no neutral line )
-
7/31/2019 Chapter 1-Three Phase
45/90
45
VN =
ER
ZR+ E
Y
ZY+ E
B
ZB
1
ZR +
1
ZY +
1
ZB
1.7
3-wire system (no neutral line )
-
7/31/2019 Chapter 1-Three Phase
46/90
Example
ER
Three-phase
Load
ZY= 2
IR
VR
EY
EB
ZR= 5
IY
IB
ZB= 10
VBVN
EL = 415 volt
Find the line currents IR,IYand IB. Also find
the voltages VR, VY and VB.
3 wire system (no neutral line ) delta connected load
-
7/31/2019 Chapter 1-Three Phase
47/90
3-wire system (no neutral line ),delta connected load
ER
Three-phaseLoad
Three-phaseAC generator
VY
IR
VR
EY
EB
ZR
IY
IB
ZB ZY
VB
3 wire system (no neutral line ) delta connected load
-
7/31/2019 Chapter 1-Three Phase
48/90
3-wire system (no neutral line ),delta connected load
ER
Three-phaseLoad
Three-phaseAC generator
IR
EY
EB
IY
IB
VRY
ZRYZBR
ZYB
VYB
VBR
Ir
IbIy
3-wire system (no neutral line ) delta connected load
-
7/31/2019 Chapter 1-Three Phase
49/90
3-wire system (no neutral line ),delta connected load
ER
Three-phaseLoad
Three-phaseAC generator
IR
EY
EB
IY
IB
VRY
ZRYZBR
ZYB
VYB
VBR
Ir
IbIy
ERY=VRY
EYB =VYB
EBR =VBR
3-wire system (no neutral line ) delta connected load
-
7/31/2019 Chapter 1-Three Phase
50/90
50
3-wire system (no neutral line ),delta connected load
Phase currents
30oIr=
VRY
ZRY=
ERY
ZRY=
EL
ZRY
-90oIy=
VYB
ZYB=
EYB
ZYB=
EL
ZYB
150oIb
=VBR
ZBR=
EBR
ZBR=
EL
ZBR
3-wire system (no neutral line ) delta connected load
-
7/31/2019 Chapter 1-Three Phase
51/90
3-wire system (no neutral line ),delta connected load
ER
Three-phaseLoad
Three-phaseAC generator
IR
EY
EB
IY
IB
VRY
ZRYZBR
ZYB
VYB
VBR
Ir
IbIy
ERY=VRY
EYB =VYB
EBR =VBR
Line currents
IR= Ir Ib-
=EL
ZRY
30o-
150oEL
ZBR
IY= Iy Ir-
=EL
ZYB
-90o-
30oEL
ZRY
-
7/31/2019 Chapter 1-Three Phase
52/90
-
7/31/2019 Chapter 1-Three Phase
53/90
-
7/31/2019 Chapter 1-Three Phase
54/90
Example
ER
Three-phase
Load
ZY= 2
IR
VR
EY
EB
ZR= 5
IY
IB
ZB= 10
VBVN
Find the line currents IR,IYand IB .
EL = 415 volt
Use star-delta conversion.
Three-phase 4-wire system (neutral line without impedance)
-
7/31/2019 Chapter 1-Three Phase
55/90
ER
Three-phase
Load
Three phaseAC generator
VY
IR
VR
EY
EB
ZR
IR
IB
ZB ZY
VB
INZ
N
VN
4 wire system (neutral line without impedance)
=0
VN = INZN= IN(0)= 0 volt55
4-wire system (neutral line without impedance)
-
7/31/2019 Chapter 1-Three Phase
56/90
56
4 wire system (neutral line without impedance)
For 4-wire three-phase system, VN
is equal to0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5become,
IB=EB
ZB
1.5EBVN
IY=EY
ZY
1.4EYVN
IR=ER
ZR1.3
ERVN
-
7/31/2019 Chapter 1-Three Phase
57/90
Example
ER
Three-phase
Load
ZY= 2
IR
VR
EY
EB
ZR= 5
IY
IB
ZB= 10
VB
IN
VN
Find the line currents IR,IYand IB. Also find
the neutral current IN.
EL = 415 volt
(t)
-
7/31/2019 Chapter 1-Three Phase
58/90
58
v(t)
wt
vR
vY
vB
The instantaneous e.m.f. generated in phase R, Y and B:
eR= EmR sinw
teY= EmY sin (wt -120
o)
eB= EmB sin (wt -240o)= EmBsin (wt +120
o)
-
7/31/2019 Chapter 1-Three Phase
59/90
59
1.4: Phase sequencesRYB and RBY
120o
-120o
120o VR
VY
VB
wo
)rms(RR0VV
o
)rms(YY120VV
o
)rms(B
o
)rms(BB
120V
240VV
VR leads VY, which in turn leads VB.This sequence is produced when the rotor rotates in
the counterclockwise direction.
(a) RYB or positive sequence
-
7/31/2019 Chapter 1-Three Phase
60/90
60
(b) RBY or negative sequence
120o
-120o
120
o
VR
VB
VY
w
o
)rms(RR0VV
o
)rms(BB120VV
o
rmsY
o
rmsYY
V
V
120
240
)(
)(
V
VR leads VB, which in turn leads VY.This sequence is produced when the rotor rotates in
the clockwise direction.
-
7/31/2019 Chapter 1-Three Phase
61/90
-
7/31/2019 Chapter 1-Three Phase
62/90
62
Star Connectionb) Four wire system
VRN
VBN
VYN
ZR
ZY ZB
R
B
NY
-
7/31/2019 Chapter 1-Three Phase
63/90
63
Wye connection of Load
Z1
Z3
Z2
R
B
Y
NLoad
Z3
Z1
Z2
R
Y
B
Load
N
-
7/31/2019 Chapter 1-Three Phase
64/90
64
1.5.2: Delta Connection
R
Y
B
Y
B
R
-
7/31/2019 Chapter 1-Three Phase
65/90
65
Delta connection of load
Zc
Za
Zb
R
B
Y
Load
Zc
Zb
Za
R
Y
B
Load
-
7/31/2019 Chapter 1-Three Phase
66/90
Wye-Connected Balanced Loads
-
7/31/2019 Chapter 1-Three Phase
67/90
67
Example
ER
Three-phase
Load
ZY= 20
IR
VR
EY
EB
ZR= 20
IY
IB
ZB= 20
VBVN
EL = 415 volt
Find the line currents IR,IYand IB. Also find
the voltages VR, VY and VB.
Wye Connected Balanced Loadsb) Three wire system
-
7/31/2019 Chapter 1-Three Phase
68/90
68
Wye-Connected Balanced Loadsb) Three wire system
VN = = 0 volt
VR= ER
VY= EY
VB= EB
1.6.1: Wye-Connected Balanced Loads
) i
-
7/31/2019 Chapter 1-Three Phase
69/90
69
Example
ER
Three-phase
Load
ZY= 20
IR
VR
EY
EB
ZR= 20
IY
IB
ZB= 20
VB
IN
VN
Find the line currents IR,IYand IB. Also find
the neutral current IN.
EL = 415 volt
a)Four wire system
1.6.1: Wye-Connected Balanced Loads
-
7/31/2019 Chapter 1-Three Phase
70/90
70
VRN
VBN
Z1
Z2 Z3
R
B
N
Y
VYN
IR
IY
IB
IN
BYRNIIII
For balanced load system,
IN = 0 and Z1 = Z2 = Z3
3
o
BN
B
2
o
YN
Y
1
o
RN
R
Z
120VI
Z
120V
I
Z
0VI
BNYNRNphasa
phasaBN
phasaYN
phasaRN
VVVVwhere
120VV
120VV
0VV
a)Four wire system
-
7/31/2019 Chapter 1-Three Phase
71/90
71
Wye-Connected Balanced Loadsb) Three wire system
R
Y
B
Z1
Z 2 Z3
IR
IY
IB
VRY
VYB
VBR S
0IIIBYR
3
o
BS
B
2
o
YS
Y
1
o
RS
R
Z
120VI
Z
120VI
Z
0VI
BSYSRSphasa
phasaBS
phasaYS
phasaRS
VVVVwhere
240VV
120VV
0VV
-
7/31/2019 Chapter 1-Three Phase
72/90
72
1.6.2: Delta-Connected Balanced Loads
ZZ
Z
R
Y
B
VRY
VYB
VBR
IR
IRY
IBR
IYB
IB
IY
Phase currents:
3
o
BR
BR
2
o
YB
YB
1
o
RY
RY
Z
120VI
Z
120VI
Z
0VI
Line currents:
YBBRB
RYYBY
BRRYR
III
III
III
lineBYR
phasaBRYBRY
IIIIand
IIIIwhere
-
7/31/2019 Chapter 1-Three Phase
73/90
-
7/31/2019 Chapter 1-Three Phase
74/90
74
1.7.1: Wye-Connected Unbalanced LoadsFour wire system
VRN
VBN
Z1
Z2 Z3
R
B
N
Y
VYN
IR
IY
IB
IN
BYRNIIII
For unbalanced load system,
IN
0 and Z1
Z2
Z3
3
o
BN
B
2
o
YN
Y
1
o
RN
R
Z
120VI
Z
120V
I
Z
0VI
120VV
120VV
0VV
phasaBN
phasaYN
phasaRN
-
7/31/2019 Chapter 1-Three Phase
75/90
75
1.7.2: Delta-Connected Unbalanced Loads
Z
Z
Z
R
Y
B
VRY
VYB
VBR
IR
IRY
IBR
IYB
IB
IY
Phase currents:
3
o
BR
BR
2
o
YB
YB
1
o
RY
RY
Z
120VI
Z
120VI
Z
0VI
Line currents:
YBBRB
RYYBY
BRRYR
III
III
III
120VV
120VV
0VV
phasaBN
phasaYN
phasaRN
-
7/31/2019 Chapter 1-Three Phase
76/90
76
1.8 Power in a Three Phase
System
-
7/31/2019 Chapter 1-Three Phase
77/90
77
Power Calculation
The three phase power is equal the sum ofthe phase powers
P = PR+ PY+ PB
If the load is balanced:
P = 3 Pphase = 3 Vphase Iphase cos
1.8.1: Wye connection system:
-
7/31/2019 Chapter 1-Three Phase
78/90
78
I phase= I L and
Real Power, P = 3 Vphase Iphase cos
Reactive power,
Q = 3 Vphase
Iphase
sin
Apparent power,
S = 3 Vphase Iphase
or S = P + jQ
phaseLL VV 3
WattIV LLL cos3
VARIV3 LLL sin
VAIV3 LLL
-
7/31/2019 Chapter 1-Three Phase
79/90
79
1.8.2: Delta connection system
VLL= Vphase
P = 3 Vphase Iphase cos
phaseL I3I
WattIV LLL cos3
-
7/31/2019 Chapter 1-Three Phase
80/90
80
1.9: Three phase power
measurement
-
7/31/2019 Chapter 1-Three Phase
81/90
81
Power measurement
In a four-wire system (3 phases and a
neutral) the real power is measured using
three single-phase watt-meters.
Three Phase Circuit
-
7/31/2019 Chapter 1-Three Phase
82/90
82
Three Phase Circuit
Four wire system,
Each phase measured separately
A
A
V
W
W
Phase A
Phase B
Phase C
VAN
IA
IC
V
A
V
W
IB
VBN
VCN
Neutral (N)
PA
PB
PC
watt-meter connection
-
7/31/2019 Chapter 1-Three Phase
83/90
8383
Current coil (low impedance)
voltage coil (high impedance)
W
Examplea)Four wire system
-
7/31/2019 Chapter 1-Three Phase
84/90
Example
ER
Three-phase
Load
IR
VR
EY
EB
ZR= 5
IYIB
ZB= 20
VB
IN
VN
Find the three-phase total power, PT.
EL = 415 volt30o
ZY= 10 90o
45o
WR
WB
WY
-
7/31/2019 Chapter 1-Three Phase
85/90
Exampleb)Three wire system
-
7/31/2019 Chapter 1-Three Phase
86/90
Example
ER
Three-phase
Load
IR
VR
EY
EB
ZR= 5
IYIB
ZB= 20
VN
Find the three-phase total power, PT.
EL = 415 volt30o
ZY= 10 90o
45o
WR
WB
WY
VB
Th Ph Ci it
-
7/31/2019 Chapter 1-Three Phase
87/90
87
Three Phase Circuit
Three wire system,
The three phase power is the sum of the two watt-
meters readingA
A
V
V
W
W
Phase A
Phase B
Phase C
VAB
= VA
- VB
VCB
= VC
- VB
IA
IC
PAB
PCBCBABT PPP
CBABT PPP Proving:
-
7/31/2019 Chapter 1-Three Phase
88/90
8888
The three phase power (3-wire system) is the
sum of the two watt-meters reading
Instantaneous power:
pA = vA iA
pB= vBiB
pC= vCiC
A
A
V
W
W
Phase A
Phase B
Phase C
VAN
IA
IC
V
A
V
W
IB
VBN
VCN
Neutral (N)
PA
PB
PC
pT= pA + pB+ pC= vA iA + vBiB+vCiC
= vA iA + vBiB+vCiC= vA iA + vB(-iA -iC)+vCiC
Proving:
-
7/31/2019 Chapter 1-Three Phase
89/90
8989
The three phase power (3-wire system) is the
sum of the two watt-meters reading
CBABT PPP
Instantaneous power:A
A
V
W
W
Phase A
Phase B
Phase C
VAN
IA
IC
V
A
V
W
IB
VBN
VCN
Neutral (N)
PA
PB
PC
pT= pAB+ pCB
pT= vA iA + vB(-iAiC) +vCiC
= (vA vB)iA + (vC vB)iC
= vABiA + vCBiC
A
A
V
V
W
W
Phase A
Phase B
Phase C
VAB
= VA
- VB
VCB
= VC
- VB
IA
IC
PAB
PCB
-
7/31/2019 Chapter 1-Three Phase
90/90
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.