Transcript POLYPHASE CIRCUITS LEARNING GOALS Three Phase Circuits Advantages of polyphase circuits Three Phase Connections Basic configurations for three phase circuits Source/Load Connections Delta-Y connections Power Relationships Study power delivered.

POLYPHASE CIRCUITS
LEARNING GOALS
Three Phase Circuits
Advantages of polyphase circuits
Three Phase Connections
Basic configurations for three phase circuits
Source/Load Connections
Delta-Y connections
Power Relationships
Study power delivered by three phase circuits
Power Factor Correction
Improving power factor for three phase circuits
THREE PHASE CIRCUITS
Van
3 phase
voltage
Vbn
Vcn
0
120
240
Instantane ous Phase Voltages
van (t )  Vm cos( t )(V )
vbn (t )  Vm cos( t  120)(V )
vc (t )  Vm cos( t  240)(V )
Vm  120 2
a
a
+
_
V0
Wye Connected
Source
n
_
V-240
+
c
_
V-120
+
b
b
c
Delta Source
a
a
_
+
+
Delta
Source
_
Vbc = Vab  -120
b
c
_
Vab = | Vab |  0
b
+
c
Vca = Vab  -240
Wye – Wye System
a
A
Zl
ZL
n
N
ZL
c
b
B
Zl
Zl
ZL
C
Delta – Delta System
a
A
Zl
+
ZL
ZL
_
+
c
_
_
+
b
B
Zl
Zl
C
ZL
Delta – Wye System
a
A
Zl
_
ZL
+
+
_
ZL
c
_
+
b
B
Zl
Zl
ZL
C
a
a
A
I aA
I AB
+
_
I CA
V0
Z
Z
n
_
V-240
+
_
I BC
V-120
+
b
b
c
C
B
Z
c
Vcn
- Vbn
Vab
30 o
Van
Vbn
Vab = Van - Vbn
Vab = 3 Van 30o
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
Vca  3 | V p |   210
VL  3 | V p |  Line Voltage
I CA
I aA = 3 I AB -30o
I AB
I aA
I BC
- I CA
INSTANTANEOUS POWER
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 )
ib (t )  I m cos( t    120)
vc (t )  Vm cos( t  240)(V )
ic (t )  I m cos( t    240)
Instantane ous power
p(t )  van (t )ia (t )  vbn (t )ib (t )  vcn (t )ic (t )
Theorem
For a balanced three phase circuit the instantaneous power is constant
p( t )  3
Vm I m
cos (W )
2
REVIEW OF
Y
Transformations
Ra 
R1R2
R1  R2  R3
Rb 
R2 R3
R1  R2  R3
Rc 
R3 R1
R1  R2  R3
 Y
R1 
Ra Rb  Rb Rc  Rc Ra
Rb
R2 
Ra Rb  Rb Rc  Rc Ra
Rc
R3 
Ra Rb  Rb Rc  Rc Ra
Ra
Y 
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 
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
Q  Qnew  Qold
Reactive Power to be added
To use capacitors this value
should be negative
pf
pf  cos f  sin  f  1  pf 2 tan  
f
2
1

pf
Q  P tan  f
lagging  Q  0
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 BY DESIGN
Proposed new store
#4ACSR wire rated at 170 A rms
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
|Q
per capacitor
|  CV
2
876.72 10 / 3
C
2  60  13.8  10  / 3
3
3 2
 12.2  F