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POLYPHASE CIRCUITS
LEARNING GOALS
Three Phase Circuits
Advantages of polyphase circuits
Three Phase Connections
Basic configurations for three phase circuits
Source/Load Connections
Delta-Wye connections
Power Relationships
Study power delivered by three phase circuits
Power Factor Correction
Improving power factor for three phase circuits
THREE PHASE CIRCUITS
ia
ib
Vm 120 2
ic
Instantane ous Phase Voltages
van (t ) Vm cos( t )(V )
Balanced Phase Currents
ia (t ) I m cos( t )
vbn (t ) Vm cos( t 120)(V )
vc (t ) Vm cos( t 240)(V )
ib (t ) I m cos( t 120) Instantane ous power
p(t ) van (t )ia (t ) vbn (t )ib (t ) vcn (t )ic (t )
ic (t ) I m cos( t 240)
Theorem
For a balanced three phase circuit the instantaneous power is constant
p( t ) 3
Vm I m
cos (W )
2
Proof of Theorem
For a balanced three phase circuit the instantaneous power is constant
V I
p(t ) 3 m m cos (W )
2
Instantane ous power
p(t ) van (t )ia (t ) vbn (t )ib (t ) vcn (t )ic (t )
cos t cos( t )
p(t ) Vm I m cos( t 120) cos( t 120 )
cos( t 240) cos( t 240 )
1
cos cos cos( ) cos( )
2
3 cos cos(2 t )
p(t ) Vm I m cos(2 t 240 )
cos(2 t 480 )
t
cos( 240) cos( 120)
cos( 480) cos( 120)
cos(120) 0.5
Lemma
cos cos( 120) cos( 120) 0
Proof
cos cos
cos( 120) cos cos(120) sin sin( 120)
cos( 120) cos cos(120) sin sin( 120)
cos cos( 120) cos( 120) 0
THREE-PHASE CONNECTIONS
Positive sequence
a-b-c
Y-connected
loads
Delta connected loads
SOURCE/LOAD CONNECTIONS
BALANCED Y-Y CONNECTION
Line voltages
Van | V p | 0
Vab
Vca
Vbc
Vbn | V p | 120
Vcn | V p | 120
Positive sequence
phase voltages
Vab Van Vbn
| V p | 0 | V p | 120
| V p | 1 (cos120 j sin 120)
1
3
| V | p | V p | j
2
2
3 | V p | 30
Vbc 3 | V p | 90
Ia
Van
V
V
; I b bn ; I c cn
ZY
ZY
ZY
I a | I L | ; I b | I L | 120; I c | I L | 120
Vca 3 | V p | 210
VL 3 | V p | Line Voltage
I a I b I c I n 0 For this balanced circuit it is enough to analyze one phase
LEARNING EXAMPLE
For an abc sequence, balanced Y - Y three phase circuit
Vab 208 30
Determine the phase voltages
The phasor diagram could be rotated by any angle
Positive sequence
a-b-c
Balanced Y - Y
Van 120 60
Vbn 120 180
Van 12060
Van | V p | 0
Vbn | V p | 120
Vcn | V p | 120
Positive sequence
phase voltages
Vab 3 | V p | 30
Van lags Vab by 30
Vab 208 30
Van
| Vab |
(30 30)
3
Relationship between
phase and line voltages
LEARNING EXAMPLE
For an abc sequence, balanced Y - Y three phase circuit
source | V phase | 120(V )rms , Zline 1 j1, Z phase 20 j10 Because circuit is balanced
Determine line currents and load voltages
data on any one phase is
sufficient
I bB 5.06 120 27.65( A)rms
I cC 5.06120 27.65( A)rms
VAN I aA (20 j10) I aA 22.3626.57
1200
VAN 113.15 1.08(V )rms
Chosen
as reference
Van 1200
Vbn 120 120
Van 120120
Abc sequence
Van
1200
21 j11 23.7127.65
5.06 27.65( A)rms
I aA
VBN 113.15 121.08(V )rms
VCN 113.15118.92(V )rms
LEARNING EXTENSION
For an abc sequence, balanced Y - Y three phase circuit
Van 12090 (V )rms . Find the line voltages
Vab leads Van by 30
Vab 3 | V p | 30
Vab 3 120120 (V )rms
Van lags Vab by 30
Vbc 3 1200 (V )rms
Vca 3 120240 (V )rms
Relationship between
phase and line voltages
Vab 2080 (V )rms . Find the phase voltages
Van lags Vab by 30
Van
208
30 (V )rms
3
208
150 (V )rms
3
208
Vcn
90 (V )rms
3
Vbn
LEARNING EXTENSION
For an abc sequence, balanced Y - Y three phase circuit
load | V phase | 104.0226.6(V )rms , Zline 1 j1, Z phase 8 j3
Determine source phase voltages
Currents are not required. Use inverse
voltage divider
(8 j 3) (1 j1)
V AN
8 j3
9 j 4 8 j 3 84 j 5
1.153.41
8 j3 8 j3
73
Van
8
104.0226.6(V )rms
j 3
Van 12030
Vbn 120 90
Vcn 120150
Positive sequence
a-b-c
DELTA CONNECTED SOURCES
Convert to an equivalent Y connection
Vab VL0
VL
V
30
an
3
Vbc VL 120
Vca VL120 V VL 150
bn
3
VL
Vab 3 | V p | 30
V
90
cn
3
Van lags Vab by 30 Example
Relationship between
phase and line voltages
Vab 20860
Vbc 208 60
Vca 208180
Van 12030
Vbn 120 90
V 120150
cn
LEARNING EXAMPLE
Determine line currents and line voltages at the loads
Source is Delta connected.
Convert to equivalent Y
Vab VL0
Vbc VL 120
Vca VL120
VL
V
30
an
3
VL
150
Vbn
3
VL
V
90
cn
3
Analyze one phase
(208 / 3) 30
9.38 49.14( A)rms
12.1 j 4.2
(12 j 4) 9.38 49.19 118.65 30.71(V )rms
I aA
VAN
Vab 3 | V p | 30 VAB 3 118.650.71
Determine the other phases using the balance
I bB 9.38 169.14( A)rms VBC 3 118.65 119.29
I cC 9.38 71.86( A)rms VCA 3 118.65120.71
LEARNING EXTENSION
Compute the magnitude of the line voltage at the load
Source is Delta connected.
Convert to equivalent Y
Vab VL0
Vbc VL 120
Vca VL120
VL
V
30
an
3
VL
150
Vbn
3
VL
V
90
cn
3
Analyze one phase
j 0.1
10
V AN
10 j 4
120 30
10.1 j 4.1
Only interested in magnitudes!
10.77
118.57(V )rms
10.90
Vab 3 | V p | 30
| VAB | 205.4(V )rms
| VAN | 120
DELTA-CONNECTED LOAD
Load phase currents
I AB
V AB
| I |
Z
I BC
VBC
| I | 120
Z
30 Z
VCA
| I | 120
Z
Line currents
I CA
Z | Z L | Z
I aA I AB I CA
I bB I BC I AB
Method 1: Solve directly
Van | V p | 0
Vbn | V p | 120
Vcn | V p | 120
Positive sequence
phase voltages
Vab 3 | V p | 30
Vbc 3 | V p | 90
Vca 3 | V p | 210
| I line | 3 | I |
line 30
Line-phase current
relationship
I cC I CA I BC
Method 2: We can also convert the delta
connected load into a Y connected one.
The same formulas derived for resistive
circuits are applicable to impedances
Z
3
|V | / 3
Van
| I aA | AB
I aA
| I aA | L
| Z | / 3
ZY
L Z
Balanced case ZY
| V | 3 | V phase |
phase 30
Line - phase voltage
relationship
| I line | 3 | I |
line 30
Line-phase current
relationship
LEARNING EXTENSION
I aA 1240.
Find the phase currents
I AB 6.9370
I BC 6.93 50
ICA 6.93190
Rab R2 || ( R1 R3 )
REVIEW OF
Y
Transformations
Y
Rab Ra Rb
Y
Ra R1
Rb R1
Rb R2
Rb R1
R1R2
R
R
3
2
Ra
Rc R1
Rc
R2 ( R1 R3 ) Ra R R R Rb R3
1
2
3
Ra Rb
REPLACE IN THE THIRD AND SOLVE FOR R1
R1 R2 R3
R2 R3
Rb
R1 R2 R3
Ra Rb Rb Rc Rc Ra
R3 ( R1 R2 )
R
1
Rb Rc
Rb
R
R
3 1
R1 R2 R3 Rc
R1 R2 R3
R R Rb Rc Rc Ra
R2 a b
Rc
Y
R (R R )
Rc Ra
1
2
3
R1 R2 R3
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
R3
Ra Rb Rb Rc Rc Ra
Ra
Y
R R1 R2 R3 RY
R
3
LEARNING EXAMPLE
Delta-connected load consists of 10-Ohm resistance in series
with 20-mH inductance. Source is Y-connected, abc sequence,
120-V rms, 60Hz. Determine all line and phase currents
Van 12030(V )rms
Zinductance 2 60 0.020 7.54
Z 10 j 7.54 12.5237.02 ZY 4.1737.02
VAB 120 360
16.6022.98( A)rms
Z
10 j 7.54
16.60 97.02( A)rms
I AB
I BC
| V | 3 | V phase |
I CA 16.60142.98( A)rms
phase 30
I aA 28.75 7.02( A)rms
Line - phase voltage
I bB 28.75 127.02( A)rms
relationship
| I line | 3 | I |
line 30
Line-phase current
relationship
I cC 28.75112.98( A)rms
Alternatively, determine first the line currents
and then the delta currents
POWER RELATIONSHIPS
| V | 3 | V phase |
- Impedance angle
phase 30
Line - phase voltage
relationship
STotal 3 V phase
*
STotal 3Vline I line
| I line | 3 | I |
line 30
Line-phase current
relationship
Vline
I *phase
Stotal 3Vline I *
*
STotal 3Vline I line
f
Power factor angle
I line
Ptotal 3 |Vline || I line | cos f
Qtotal 3 |Vline || I line | sin f
LEARNING EXAMPLE
| Vline | 208(V )rms
Ptotal 1200W
power factor angle 20 lagging
Vline
Determine the magnitude of the line
currents and the value of load impedance
per phase in the delta
- Impedance angle
Ptotal 3 |Vline || I line | cos f
Qtotal 3 |Vline || I line | sin f
Z 101.4620
Vline
Ptotal | Vline || I line |
cos f | I line | 3.54( A)rms
3
3
| I line | 3 | I |
line 30
f
Power factor angle
| Vline |
|
Z
|
101.46
| I | 2.05( A)rms
|
I
|
Line-phase current
relationship
I line
LEARNING EXAMPLE
For an abc sequence, balanced Y - Y three phase circuit
source | V phase | 120(V )rms , Zline 1 j1, Z phase 20 j10
Determine real and reactive power per phase at the load and total real, reactive and
complex power at the source
VAN I aA (20 j10) I aA 22.3626.57
VAN 113.15 1.08(V )rms
*
S phase V AN I aA
113.15 1.08 5.0627.65
S phase 572.5426.57 512 j 256.09(VA)rms
Pper phase
1200
Chosen
as reference
*
Ssource phase Van IaA
1200 5.0627.65
Van 1200
Vbn 120 120 Because circuit is balanced
Van 120120
data on any one phase is
sufficient
Abc sequence
Van
1200
21 j11 23.7127.65
5.06 27.65( A)rms
I aA
Qper phase
Ssource phase 607.227.65
537.86 j 281.78VA
Ptotal source 3 537.86(W )
Qtotal source 3 281.78 (VA)
Stotal source Ptotal source Qtotal source
1613.6 j845.2(VA)
| Stotal source | 1821.6(VA)
LEARNING EXAMPLE
Determine the line currents and the combined power factor
Circuit is balanced
Load 1 : 24kW at pf 0.6 lagging
Load 2 : 10kW at pf 1
Vline 208(V )rms
Load 3 : 12kVA at pf 0.8 leading
inductive
S P jQ
P | S | cos f
P1 24kW
| S1 | 40kVA
pf 0.6 lagging
| Q1 | | S1 |2 | P1 |2 32kVA
Q | S | sin f
pf cos f
Stotal S1 S2 S3
f
lagging inductive S1 24 j 32 kVA
capacitive
Load 2
STOTAL S1 S2 S3 43.6 j 24.8kVA 50.16029.63kVA
P2 10kW
S2 10 j 0 kVA Ptotal 3 |Vline || I line | cos f
| Stotal | 3 | Vline | | I line |
pf 1
Q
3
|
V
||
I
|
sin
total
line
line
f
Load 3
f 29.63
| S3 | 12kVA P3 9.6kW
pf 0.8
| Q3 | 7.2kVA
leading pf capacitive S3 9.6 j 7.2kVA
pf 0.869 lagging
| I line | 139.23( A)rms
Continued ...
LEARNING EXAMPLE
continued ….
If the line impedances are Z line 0.05 j 0.02
determine line voltages and power factor at the source
inductive
f
capacitive
| I line | 139.23( A)rms
*
Sline 3 ( Zline I line ) I line
3 Zline | I line |
2
Sline 2908 j1163(VA)
Sload total 43.6 j 24.8kVA 50.16029.63kVA
Ssource total 46.508 j 25.963 53.26429.17kVA
| Stotal | 3 | Vline | | I line |
f 29.17
Vline
53,264
220.87(V )rms
3 139.13
pf cos f cos(29.17) 0.873 lagging
A Y -Y balanced three-phase circuit has a line voltage of
208-Vrms. The total real power absorbed by the load is 12kW
at pf=0.8 lagging. Determine the per-phase impedance
of the load
LEARNING EXTENSION
| V | 3 | V phase |
Impedance angle
phase 30
Line - phase voltage
relationship
Vline
STotal 3 V phase I *phase
S P jQ
P | S | cos f
208
| V phase |
120(V )rms
3
I line
Q | S | sin f
pf cos f
*
V phase
| V phase |2
3
Stotal 3V phase
*
Z
Z
phase
phase
f
Power factor angle
| Z phase|
3 | V phase |2
| Stotal |
2.88
pf 0.8 cos f f 36.87
| S total |
Ptotal
15kVA
pf
Z pahse 2.8836.87
LEARNING EXTENSION
Determine real, reactive and complex power at both load
and source
Source is Delta connected.
Convert to equivalent Y
Analyze one phase
j 0.1
10
*
Sload 3 VAN I aA
*
S source 3 Van I aA
3
| Van |2
*
Z total
phase
10 j 4
120 30
10.1 j 4.1
10.77
| 120
118.57(V )rms
10.90
V AN
| VAN
| VAN |2 3 | 118.57 |2 3 | 118.57 |2 (10 j 4)
3 *
10
j
4
102 42
Z phase
3 | 120 |2 3 | 120 |2 (10.1 j 4.1)
10.1 j 4.1
(10.1)2 (4.1)2
Sload 3 (1,212.0 j 484.8)
S source 3 (1224.1 j 496)
LEARNING EXTENSION
A 480-V rms line feeds two balanced 3-phase loads.
The loads are rated
Load 1: 5kVA at 0.8 pf lagging
Load 2: 10kVA at 0.9 pf lagging.
Determine the magnitude of the line current from the 408-V rms source
| S1 | 5kVA
S P jQ
P | S | cos f
P
P1 4kW
0.8
Q1 | S1 |2 P12 3.0kVA
Q | S | sin f
pf lagging S1 4 j 3kVA
| S2 | 10kVA
Q2 | S2 |
2
P
P 9kW
0.9
P22
4.36kVA
S2 9 j 4.36kVA
Stotal 13 j 7.36kVA
| I lineq |
pf cos f
Stotal S1 S2
Ptotal 3 |Vline || I line | cos f
Qtotal 3 |Vline || I line | sin f
| Stotal | 3 | Vline || I line |
| Stotal |
14,939
21.14( A)rms
3 | Vline | 706.68
POWER FACTOR CORRECTION
Similar to single phase case.
Use capacitors to increase the
power factor
Balanced
load
Low pf
lagging
Keep clear about total/phase
power, line/phase voltages
Sold
Qold
pf old
Q Qnew Qold
Reactive Power to be added
Pold
Qnew To use capacitors this value
pf new
should be negative
S P jQ
pf
pf cos f sin f 1 pf 2 tan
f
2
The voltage depends on how P | S | cos f
1
pf
Q P tan f
Q
|
S
|
sin
the capacitors are connected
f
lagging Q 0
pf cos f
Qper capacitor CV 2
LEARNING EXAMPLE
f 60 Hz , | Vline | 34.5kV rms . Required : pf 0.94 leading
Pold 18.72 MW
S P jQ
P | S | cos f
Q | S | sin f
pf cos f
Q P tan f
tan f
pf
1 pf 2
lagging Qold 0
pf cos f sin f 1 pf 2 0.626
| Qold | 15.02 MVA
Pold 18.72 MW
Qnew 6.8 MVA
pf new 0.94 leading
Q 6.8 15.02 21.82 MVA
Qper capacitor 7.273MVA
Y connection Vcapacitor
34.5
kV rms
3
34.5 103
6
7.273 10 2 60 C
3
C 48.6 F
2
LEARNING EXAMPLE
f 60 Hz , | Vline | 34.5kV rms . Required : pf 0.90 lagging
Pold 18.72 MW
S P jQ
P | S | cos f
Q | S | sin f
pf cos f
Q P tan f
tan f
pf
1 pf 2
lagging Qold 0
pf cos f sin f 1 pf 2 0.626
| Qold | 15.02 MVA
Pold 18.72 MW
Qnew 9.067 MVA
pf new 0.90 lagging
Q 9.067 15.02 5.953MVA
Qper capacitor 1.984 MVA
Y connection Vcapacitor
34.5
kV rms
3
34.5 103
6
1.984 10 2 60 C
3
C 13.26 F
2
LEARNING EXAMPLE
MEASURING POWER FLOW Which circuit is the source and
what is the average power supplied?
120kV rms
125kV rms
Phase differences
determine direction
of power flow!
Determine the current
flowing. Convert line
voltages to phase voltages
*
SY 3V AN I aA
S X 3Van ( I aA )*
Equivalent 1-phase circuit
V VAN
I aA an
1 j2
12000
12000
30
25
3
3
1 j2
I aA 270.30 180.93( A) rms
Ploss ( PX PY )
SY 3 12 0.2703(25 180.93) MVA
S X 3 12 0.2703(30 180.93) MVA
PY 5.13MW
S P jQ
PX 4.91MW
System Y is the source
P | S | cos f
LEARNING EXAMPLE
INCREMENTAL COST OF POWER FACTOR CORRECTION
How much capacitance is required to improve the power factor by a fixed amount (say
Desired:
Supplied by
capacitors
• least expensive at 0.8
• very expensive as
pf
OLD
1
CAPACITOR SPECIFICATIONS
Capacitors for power factor correction are normally specified in VARs
| Qper capacitor | CV 2
The voltage and frequency must be given in order to know the capacitanc e
Assume 60 Hz unless other value is given.
LEARNING EXAMPLE
For pf 0.94 leading one needs
C 48.6 F
Choices available
Capacity
1
2
3
Rated Voltage (kV)
10
50
20
Rated Q (Mvar)
4
25
7.5
4 106
C1
106.1 F
3 2
2 60 (10 10 )
25 106
C2
26.53 F
3 2
2 60 (50 10 )
Vline 34.5kV V phase 19.9kV
Capacitor 1 is not rated at high enough
voltage!
7.5 106
C3
49.7 F
3 2
2 60 (20 10 )
Capacitor 3 is the best alternative
LEARNING BY DESIGN
Proposed new store
#4ACSR wire rated at 170 A rms
1. Is the wire suitable?
2. What capacitance
would be required to
have a composite
pf =0.92 lagging
Capacitors are to be
Y - connected
S1 70036.9
S2 100060kVA
S3 80025.8kVA
560 j 420 kVA 500 j866 kVA
720 j 349 kVA
Stotal 1780 j1635 kVA 241742.57 kVA
| Stotal |
2.417 106
| I line |
101.1A rms
3
3 Vline
3 13.8 10
Wire is OK
Pold
Qnew P tan f ( new) 758.28kVA
pf new
Q Qnew Qold 876.72kVA
P | S | cos f
Q | S | sin f
pf cos f
STotal 3 V phase I *phase
*
3 Vline I line
| Qper capacitor | CV 2
876.72 103 / 3
12.2 F
C
3 2
13.8kV
2 60 13.8 10 / 3
V | V phase |
3
Stotal S1 S2 S3
S P jQ
LEARNING EXAMPLE
DESIGN OF A 3-PHASE EMULATOR
Magnitude Adjustor
Design equations for
adjustor
Powered by 120V(RMS) – standard wall outlet
using a transformer with turns ratio
Potentiometers are more restricted
Select R 10k
P
V V 120
3-PHASE EMULATOR - CONTINUED
CN
O
1
SOLUTION:
• Use RC network for -60 deg (lag)
• Use inverter for -180 phase shift
1
1
j C
V
V
V
1
1 j R C
R
j C
2
1
1
3
3
buffer
V tan
2
1
R C 60
o
3
RC tan60 1.73 RC 4.6 10
3
3-PHASE EMULATOR - CONTINUED
V V V 0
AN
USE 10k RESISTORS FOR
UNIFORMITY
BN
CN
3-PHASE EMULATOR – PROPOSED SOLUTION
Polyphase