Transcript No Slide Title

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  12060
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  5.06120  27.65( A)rms
VAN  I aA  (20  j10)  I aA  22.3626.57
1200
VAN  113.15  1.08(V )rms
Chosen
as reference
Van  1200
Vbn  120  120
Van  120120
Abc sequence
Van
1200

21  j11 23.7127.65
 5.06  27.65( A)rms
I aA 
VBN  113.15  121.08(V )rms
VCN  113.15118.92(V )rms
LEARNING EXTENSION
For an abc sequence, balanced Y - Y three phase circuit
Van  12090 (V )rms . Find the line voltages
Vab leads Van by 30
Vab  3 | V p | 30
Vab  3 120120 (V )rms
Van lags Vab by 30
Vbc  3 1200 (V )rms
Vca  3 120240 (V )rms
Relationship between
phase and line voltages
Vab  2080 (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 | 104.0226.6(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


 1.153.41
8  j3 8  j3
73
Van 
8
 104.0226.6(V )rms
j 3
Van  12030
Vbn  120  90
Vcn  120150
Positive sequence
a-b-c
DELTA CONNECTED SOURCES
Convert to an equivalent Y connection
Vab  VL0
VL
 
V

  30
  an
3
Vbc  VL  120 

Vca  VL120  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  20860 

Vbc  208  60 
Vca  208180 
Van  12030

Vbn  120  90
V  120150
 cn
LEARNING EXAMPLE
Determine line currents and line voltages at the loads
Source is Delta connected.
Convert to equivalent Y
Vab  VL0


Vbc  VL  120 
Vca  VL120 
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 118.650.71
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  118.65120.71
LEARNING EXTENSION
Compute the magnitude of the line voltage at the load
Source is Delta connected.
Convert to equivalent Y
Vab  VL0


Vbc  VL  120 
Vca  VL120 
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  1240.
Find the phase currents
I AB  6.9370
I BC  6.93  50
ICA  6.93190
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  12030(V )rms
Zinductance  2  60  0.020  7.54
Z   10  j 7.54  12.5237.02  ZY  4.1737.02
VAB 120 360

 16.6022.98( A)rms
Z
10  j 7.54
 16.60  97.02( A)rms
I AB 
I BC
| V | 3 | V phase |
I CA  16.60142.98( 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  28.75112.98( 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   101.4620
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  22.3626.57
VAN  113.15  1.08(V )rms
*
S phase  V AN I aA
 113.15  1.08  5.0627.65
S phase  572.5426.57  512  j 256.09(VA)rms
Pper phase
1200
Chosen
as reference
*
Ssource phase  Van  IaA
 1200  5.0627.65
Van  1200
Vbn  120  120 Because circuit is balanced
Van  120120
data on any one phase is
sufficient
Abc sequence
Van
1200

21  j11 23.7127.65
 5.06  27.65( A)rms
I aA 
Qper phase
Ssource phase  607.227.65
 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  50.16029.63kVA
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  50.16029.63kVA
Ssource total  46.508  j 25.963  53.26429.17kVA
| 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  2.8836.87
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?
 120kV rms
 125kV 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  70036.9
S2  100060kVA
S3  80025.8kVA
 560  j 420 kVA  500  j866 kVA
 720  j 349 kVA
Stotal  1780  j1635 kVA  241742.57 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