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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
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
3cos   cos(2 t   ) 
V I
p(t )  m m   cos(2 t  240   ) 
2
  cos(2 t  480   ) 
  2 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 sin120) 
 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 Vphase  120(V )rms , Zline  1 j1, Z phase  20  j10
Determine line currents and load voltages
Because circuit is balanced
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
Vcn  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
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 118.65 0.71
Determine the other phases using the balance
I bB  9.38  169.14( A)rms VBC  3 118.65  120.71
I cC  9.38  71.86( A)rms VCA  3 118.65119.29
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
Polyphase