Transcript The Y- Circuits

Chapter 12
Three-Phase Circuits
Power Factor Correction
Medium Voltage Metal
Enclosed Capacitor Banks
A balanced three-phase circuit
Three-Phase Voltages
vaa  2 V cos  t
vbb  2 V cos( t  120)
vcc  2 V cos( t  240)
(a) The three windings on a
cylindrical drum used to
obtain three-phase voltages
(b) Balanced three-phase
voltages
Three-Phase Voltages
Generator with six terminals
Three-Phase Balanced Voltages
Vaa V 0
Vbb V   120
Vcc  V   240  V   120
Phase sequence or phase rotation is abc
Positive Phase Sequence
neutral terminal
Va V 0
Vc V   120
Vb  V   240  V   120
Phase sequence or phase rotation is acb
Negative Phase Sequence
Two Common Methods of Connection
phase voltage
(a) Y-connected sources (b) -connected sources
Phase and Line Voltages
Vab  Va  Vb
 V p 0  V p  120
 V p  V p ( 0.5  j 0.866)
The line-to-line voltage Vab of
the Y-connected source
 3V p 30
Similarly
Vbc  3Vp   90
Vca  3Vp  210
The Y-to-Y Circuit
A four-wire Y-to-Y circuit
The Y-to-Y Circuit (cont.)
Four - wire
Va
Vb
Vc
I aA 
, IbB 
, and I cC 
ZA
ZB
ZC
InN  IaA  IbB  IcC
The average power delivered by the three-phase source
to the three-phase load
P  PA  PB  PC
When ZA = ZB = ZC the load is said to be balanced
Va V 0
Vb V   120
Vc V 120
I aA 

, IbB 

, and I cC 

Z A Z 
ZB
Z 
ZC
Z 
V
V
V
I aA     , IbB  (   120), and I cC  (   120)
Z
Z
Z
The Y-to-Y Circuit(cont.)
There is no current in the wire connecting the neutral node
of the source to the neutral node of the load.
InN  IaA  IbB  IcC  0
The average power delivered to the load is
P  PA  PB  PC
V
V
V
 V cos(  )  V cos(  )  V cos(  )
Z
Z
Z
V2
 3 cos( )
Z
The Y-to-Y Circuit (cont.)
A three-wire Y-to-Y circuit
The Y-to-Y Circuit (cont.)
Three - wire
We need to solve for VNn
Va  VNn Vb  VNn Vc  VNn
0


ZA
ZB
ZC
V 0  VNn V   120  VNn V 120  VNn



ZA
ZB
ZC
Solve for VNn
(V   120) Z AZC  V 120Z AZ B  V 0Z B ZC
VNn 
Z A ZC  Z A Z B  Z B ZC
Va  VNn
Vb  VNn
Vc  VNn
I aA 
, IbB 
, and I cC 
ZA
ZB
ZC
The Y-to-Y Circuit (cont.)
When the circuit is balanced i.e. ZA = ZB = ZC
(V   120)ZZ  V 120ZZ  V 0ZZ
VNn 
ZZ  ZZ  ZZ
0
The average power delivered to the load is
P  PA  PB  PC
V2
 3 cos( )
Z
The Y-to-Y Circuit (cont.)
Transmission lines
A three-wire Y-to-Y circuit with line impedances
The Y-to-Y Circuit (cont.)
The analysis of balanced Y-Y circuits is simpler than the
analysis of unbalanced Y-Y circuits.
VNn = 0. It is not necessary to solve for VNn .
The line currents have equal magnitudes and
differ in phase by 120 degree.
Equal power is absorbed by each impedance.
Per-phase equivalent circuit
Example 12.4-1 S = ?
Va  1100 Vrms
Vb  110  120 Vrms
Vc  110120 Vrms
Z A  50  j80 
Z B  j50 
ZC  100  j 25 
Unbalanced 4-wire
I aA 
Va 1100

Z A 50  j80
IbB 
Vb 110  120

 2.2150 A rms
ZB
j50
I cC
 1.16  58 A rms
Vc 110120


 1.07  106 A rms
ZC 100  j 25
Example 12.4-1 (cont.)
S A  I*aA Va  68  j109 VA
S B  I*bB Vb  j 242 VA
SC  I*cC Vc  114  j 28 VA
The total complex power delivered to the three-phase load
is
S  S A  S B  SC  182  j379 VA
Example 12.4-2 S = ? Balanced 4-wire
Va  1100 Vrms
Z A  50  j80 
Vb  110  120 Vrms
Z B  50  j80 
Vc  110120 Vrms
ZC  50  j80 
I aA 
Va 1100

Z A 50  j80
 1.16  58 A rms
S A  I*aA Va  68  j109 VA
The total complex power delivered to the three-phase load
is
S  3S A  204  j326 VA
Also
IbB  1.16  177 Arms , IcC  1.1662 Arms
S B  68  j109 VA  SC
Example 12.4-3 S = ?
Va  1100 Vrms
Vb  110  120 Vrms
Vc  110120 Vrms
Z A  50  j80 
Z B  j50 
ZC  100  j 25 
Unbalanced 3-wire
Determine VNn
(110  120) Z AZC  110120Z AZ B  1100Z B ZC
VNn 
Z AZC  Z AZ B  Z B ZC
 56  151 Vrms
Va  VNn
Vb  VNn
Vc  VNn
I aA 
, IbB 
, and I cC 
ZA
ZB
ZC
Example 12.4-3 (cont.)
IaA  1.71  48, IbB  2.453, and IcC  1.1979
S A  I*aA Va  I*aA (I aAZ A )  146  j 234 VA
S B  I*bB Vb  I*bB (I bB Z B )  j 94 VA
SC  I*cC Vc  I*cC (I cC ZC )  141  j 35 VA
The total complex power delivered to the three-phase load
is
S  S A  S B  SC  287  j364 VA
Example 12.4-4 S = ? Balanced 3-wire
Va  1100 Vrms
Z A  50  j80 
Vb  110  120 Vrms
Z B  50  j80 
Vc  110120 Vrms
ZC  50  j80 
I aA 
Va 1100

Z A 50  j80
 1.16  58 A rms
S A  I*aA Va  68  j109 VA
The total complex power delivered to the three-phase load
is
S  3S A  204  j326 VA
Example 12.4-5 PLoad = ? PLine = ? PSource = ?
Balanced 3-wire
Va
1000
I aA ( ) 

Z A 50  j (377)(0.045)
per-phase
equivalent circuit
 1.894  18.7 A
The phase voltage at the load is
VAN ( )  (40  j(377)(0.04))IaA ( )  812 V
Example 12.4-5 (cont.)
The power delivered by the source is
Vm I m
Pa 
cos(V   I )
2
(100)(1.894)

cos(18.7)  89.7 W
2
The power delivered to the load is
I m2
(1.894)2
PA  Re( Z A ) 
40  71.7 W
2
2
The power lost in the line is
I m2
(1.894)2
PaA  Re( Z Line ) 
10  17.9 W
2
2
Line loss  20 %
The -Connected Source and Load
Vab  1200 Vrms
Vbc  120.1  121 Vrms
Vca  120.2121 Vrms
Circulating
current
( Vab  Vbc  Vca )
I
 3.75 A
1
Unacceptable
Total resistance around the loop
Therefore the  sources connection is seldom used
in practice.
The -Y and Y-  Transformation
Z1Z3
ZA 
Z1  Z 2  Z3
ZA
ZB
ZC
Z1Z 2
ZC 
Z1  Z 2  Z3
Z AZ B  Z B Z C  Z AZ C
Z1 
ZB
Z3
Z1
Z 2 Z3
ZB 
Z1  Z 2  Z3
Z2
Z AZ B  Z B ZC  Z AZC
Z2 
ZA
Z AZ B  Z B ZC  Z AZC
Z3 
ZC
Example
The Y-  Circuits
I aA  I AB  ICA
I bB  I BC  I AB
I cC  ICA  I BC
where
I AB
VAB

Z3
I BC
VBC

Z1
ICA 
VCA
Z2
The Y-  Circuits (cont.)
I aA  I AB  ICA
 I cos   j sin   I cos(  120)  j sin(  120)
 3I (  30)
or
IaA  3 I
 I L  3I p
Example 12.6-1 IP = ? IL = ?
220
  30 Vrms
3
220
Vb 
  150 Vrms
3
220
Vc 
90 Vrms
3
Va 
The -connected load is balanced with Z  1050
VAB  Va  Vb  2200 Vrms
VBC  Vb  Vc  220  120 Vrms
VCA  Vc  Va  220  240 Vrms
The line currents are
I AB
I
BC
ICA
VAB

 2250 A rms
Z
VBC

 22  70 A rms
Z
VCA

 22  190 A rms
Z
IaA  I AB  ICA  22 320, IbB  22 3  100, IcC  22 3  220
The Balanced Three-Phase Circuits
Y-to- circuit
equivalent Y-to-Y circuit
Z
ZY 
3
per-phase equivalent circuit
Example 12.7-1 IP = ?
Va  1100 Vrms
Vb  110  120 Vrms
Vc  110120 Vrms
Z L  10  j5 
Z   75  j 225 
Z
ZY 
 25  j 75 
3

Va
I aA 
 1.26  66 A rms
Z L  ZY

Example 12.7-1 (cont.)
IbB  1.26  186 Arms and IcC  1.26  54 Arms
The voltages in the per-phase equivalent circuit are
VAN  I aAZY  99.65 Vrms
VBN  99.6  115 Vrms
VCN  99.6125 Vrms
The line-to-line voltages are
VAB  VAN  VBN  17235 Vrms
VBC  VBN  VCN  172  85 Vrms

I AB 
VAB
 0.727  36 A rms
Z
I BC 
VBC
 0.727  156 A rms
Z
VCA  VCN  VAN  172155 Vrms
ICA
VCA

 0.72784 A rms
Z
Instantaneous and Average Power in BTP Circuits
One advantage of three-phase power is the smooth flow of
energy to the load.
vab  V cos  t , vbc  V cos( t  120),
The instantaneous power
and vca  V cos( t  240)
2
vab
vbc2 vca2
(1  cos 2 t )
2
p (t ) 


cos  t 
R
R
R
2
V2

1  cos 2 t  1  cos 2( t  120)  1  cos 2( t  240) 
2R
3V 2 V 2


cos 2 t  cos(2 t  240) cos(2 t  480)
2R 2R
0
3V 2

2R
Instantaneous and Average Power in BTP Circuits
The total average power delivered to the balanced
Y-connected load is
IaA  I L AI , VAN  VP AV
Phase A
 3VL I L cos( )
3
3
VL
I L cos( )
 3VP I L cos( )
PY  3PA  3VP I L cos( AV   AI )
Instantaneous and Average Power in BTP Circuits
The total average power delivered to the balanced
-connected load is
 3VP I L cos( )
3
 3( 3VP ) L cos( )
I
P  3PAB  3VAB I AB cos( )
Example 12.8-1 P = ?
Va  1100 Vrms
Vb  110  120 Vrms
Vc  110120 Vrms
Z L  10  j5 
Z   75  j 225 
Va
I aA 
 1.26  66 A rms
Z L  ZY
VAN  IaAZY  99.65 Vrms
P  3(99.6)(1.26) cos(5  ( 66))  122.6 W
Two-Wattmeter Power Measurement
cc = current coil
vc = voltage coil
W1 read
P1  VAB I A cos1
W2 read
P2  VCB I C cos 2
For balanced load with abc phase sequence
1   a  30 and  2   a  30
 a is the angle between phase current and phase voltage of phase a
Two-Wattmeter Power Measurement(cont.)
P  P1  P2
 2VL I L cos cos30
 3VL I L cos
To determine the power factor angle
P1  P2  VL I L 2cos cos30
P1  P2  VL I L (2sin sin30)
P1  P2
VL I L 2 cos  cos 30
 3


P1  P2 VL I L ( 2sin  sin 30) tan 
P1  P2
 tan   3
P1  P2

P1  P2 
or   tan  3

P1  P2 

1
Example 12.9-1 P = ?
Z  1045
line-to-line voltage = 220Vrms
The phase voltage
220
VA 
  30
3
The line current
VA 220  30
IA 

 12.7  75 and I B  12.7  195
Z 10 345
P1  VAC I A cos1  2698 W
P2  VBC I B cos 2  723

W
P  P1  P2  3421 W
Electrodynamic Wattmeter
Digital Power Meter
pf Meter
VAR Meter
Summary
Three-Phase voltages
The Y-to-Y Circuits
The -Connected Source and Load
The Y-to-  Circuits
Balanced Three-Phase Circuits
Instantaneous and Average Power in Bal. 3 Load
Two-Wattmeter Power Measurement