6 Symmetrical Components

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SYMMETRICAL	
  COMPONENTS	
  
and	
  
SEQUENCE	
  NETWORKS	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   1	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   2	
  
Introduc;on	
  
	
  
One	
  of	
  the	
  most	
  powerful	
  tools	
  for	
  dealing	
  with	
  unbalanced	
  
polyphase	
  circuits	
  is	
  the	
  method	
  of	
  symmetrical	
  components.	
  
	
  
Developed	
  by	
  C.	
  L.	
  Fortescue	
  in	
  1918,	
  he	
  proved	
  that	
  an	
  
unbalanced	
  system	
  of	
  n	
  related	
  phasors	
  can	
  be	
  resolved	
  into	
  n	
  
systems	
  of	
  balanced	
  phasors	
  called	
  the	
  symmetrical	
  components	
  
of	
  the	
  original	
  phasors.	
  The	
  n	
  phasors	
  of	
  each	
  set	
  of	
  components	
  
are	
  equal	
  in	
  length,	
  and	
  the	
  angles	
  between	
  adjacent	
  phasors	
  of	
  
the	
  set	
  are	
  equal.	
  	
  
	
  
Although	
  the	
  method	
  is	
  applicable	
  to	
  any	
  unbalanced	
  polyphase	
  
system,	
  we	
  consider	
  only	
  three-­‐phase	
  systems.	
  
11/20/13	
   Symmetrical	
  Components	
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  Sequence	
  Networks	
   3	
  
Introduc;on	
  
	
  
In	
  a	
  three-­‐	
  phase	
  system	
  which	
  is	
  normally	
  balanced,	
  unbalanced	
  
fault	
  condiOons	
  generally	
  cause	
  unbalanced	
  currents	
  and	
  voltages	
  
to	
  exist	
  in	
  each	
  of	
  the	
  phases.	
  	
  
	
  
If	
  the	
  currents	
  and	
  voltages	
  are	
  related	
  by	
  constant	
  impedances,	
  
the	
  system	
  is	
  linear	
  and	
  the	
  principle	
  of	
  superposiOon	
  applies.	
  The	
  
voltage	
  response	
  of	
  the	
  linear	
  system	
  to	
  the	
  unbalanced	
  currents	
  
can	
  be	
  determined	
  by	
  considering	
  the	
  separate	
  responses	
  of	
  the	
  
individual	
  elements	
  to	
  the	
  symmetrical	
  components	
  of	
  the	
  
currents.	
  	
  
	
  
The	
  system	
  elements	
  of	
  interest	
  are	
  the	
  machines,	
  transformers,	
  
transmission	
  lines,	
  and	
  loads	
  connected	
  to	
  Δ	
  or	
  Y	
  configuraOons.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   4	
  
Introduc;on	
  
	
  
We	
  will	
  study	
  symmetrical	
  components	
  and	
  show	
  that	
  the	
  
response	
  of	
  each	
  system	
  element	
  depends	
  on	
  its	
  connecOons	
  and	
  
the	
  component	
  of	
  the	
  current	
  being	
  considered.	
  	
  
	
  
Equivalent	
  circuits,	
  called	
  sequence	
  circuits,	
  will	
  be	
  developed	
  to	
  
reflect	
  the	
  separate	
  responses	
  of	
  the	
  elements	
  to	
  each	
  current	
  
component.	
  	
  
	
  
There	
  are	
  three	
  equivalent	
  circuits	
  for	
  each	
  element	
  of	
  the	
  three-­‐
phase	
  system.	
  By	
  organizing	
  the	
  individual	
  equivalent	
  circuits	
  into	
  
networks	
  according	
  to	
  the	
  interconnecOons	
  of	
  the	
  elements,	
  we	
  
arrive	
  at	
  the	
  concept	
  of	
  three	
  sequence	
  networks.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
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   5	
  
Introduc;on	
  
	
  
Solving	
  the	
  sequence	
  networks	
  for	
  the	
  fault	
  condiOons	
  gives	
  
symmetrical	
  current	
  and	
  voltage	
  components	
  which	
  can	
  be	
  
combined	
  together	
  to	
  reflect	
  the	
  effects	
  of	
  the	
  original	
  unbalanced	
  
fault	
  currents	
  on	
  the	
  overall	
  system!	
  
	
  
Analysis	
  by	
  symmetrical	
  components	
  is	
  a	
  powerful	
  tool	
  which	
  
makes	
  the	
  calculaOon	
  of	
  unsymmetrical	
  faults	
  almost	
  as	
  easy	
  as	
  
the	
  calculaOon	
  of	
  three-­‐phase	
  faults!!!	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   6	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s	
  Theorem)	
  
	
  
(Fortescue's	
  Theorem)	
  Three	
  unbalanced	
  phasors	
  of	
  a	
  three-­‐phase	
  
system	
  can	
  be	
  resolved	
  into	
  three	
  balanced	
  systems	
  of	
  phasors.	
  	
  
The	
  balanced	
  sets	
  of	
  components	
  are:	
  
1.  PosiOve-­‐sequence	
  components	
  consisOng	
  of	
  three	
  phasors	
  
equal	
  in	
  magnitude	
  ,	
  displaced	
  from	
  each	
  other	
  by	
  120o	
  in	
  
phase,	
  and	
  having	
  the	
  same	
  phase	
  sequence	
  as	
  the	
  original	
  
phasors.	
  
2.  NegaOve-­‐sequence	
  components	
  consisOng	
  of	
  three	
  phasors	
  
equal	
  in	
  magnitude	
  ,	
  displaced	
  from	
  each	
  other	
  by	
  120o	
  in	
  
phase,	
  and	
  having	
  the	
  opposite	
  phase	
  sequence	
  to	
  that	
  of	
  the	
  
original	
  phasors.	
  
3.  Zero-­‐sequence	
  components	
  consisOng	
  of	
  three	
  phasors	
  equal	
  
in	
  magnitude	
  and	
  with	
  zero	
  phase	
  displacement	
  from	
  each	
  
other.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   7	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s	
  Theorem)	
  
	
  
When	
  solving	
  a	
  problem	
  by	
  symmetrical	
  components	
  to	
  designate	
  
the	
  three	
  phases	
  of	
  the	
  system	
  as	
  a,	
  b,	
  and	
  c	
  in	
  such	
  a	
  manner	
  that	
  
the	
  phase	
  sequence	
  of	
  the	
  voltages	
  and	
  currents	
  in	
  the	
  system	
  is	
  
abc.	
  Thus,	
  the	
  phase	
  sequence	
  of	
  the	
  posiOve-­‐sequence	
  
components	
  of	
  the	
  unbalanced	
  phasors	
  is	
  abc,	
  and	
  the	
  phase	
  
sequence	
  of	
  the	
  negaOve-­‐sequence	
  components	
  is	
  acb.	
  	
  
	
  
If	
  the	
  original	
  phasors	
  are	
  voltages,	
  they	
  may	
  be	
  designated	
  Va,	
  Vb,	
  
and	
  Vc.	
  The	
  three	
  sets	
  of	
  symmetrical	
  components	
  are	
  designated	
  
by	
  the	
  addiOonal	
  superscript	
  1	
  for	
  the	
  posiOve-­‐sequence	
  
components,	
  2	
  for	
  the	
  negaOve-­‐sequence	
  components,	
  and	
  0	
  for	
  
the	
  zero-­‐sequence	
  components.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   8	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s	
  Theorem)	
  
	
  
Superscripts	
  are	
  chosen	
  so	
  as	
  not	
  to	
  confuse	
  bus	
  numbers	
  with	
  
sequence	
  indicators.	
  	
  
	
  
The	
  posiOve-­‐sequence	
  components	
  of	
  Va,	
  Vb,	
  and	
  Vc	
  are	
  V(1)a,	
  V(1)b,	
  
and	
  V(1)c,	
  respecOvely.	
  Similarly,	
  the	
  negaOve-­‐sequence	
  
components	
  are	
  V(2)a,	
  V(2)b,	
  and	
  V(2)c,	
  respecOvely,	
  and	
  the	
  zero-­‐
sequence	
  components	
  are	
  V(0)a,	
  V(0)b,	
  and	
  V(0)c,	
  ,	
  respecOvely.	
  
	
  
Phasors	
  represenOng	
  currents	
  will	
  be	
  designated	
  by	
  I	
  with	
  
superscripts	
  as	
  for	
  voltages.	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   9	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s
Theorem)	
  
 Va
0( ) Vb
0( )
 Vc
0( )
 Va
1( )
 Vb
1( )
 Vc
1( )
 Vc
2( )
 Va
2( )
 Vb
2( )
Posi;ve	
  Sequence	
  
Components	
  
Nega;ve	
  Sequence	
  
Components	
  
Zero	
  Sequence	
  
Components	
  
11/20/13	
   Symmetrical	
  Components	
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   10	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s	
  Theorem)	
  
	
  
Since	
  each	
  of	
  the	
  original	
  unbalanced	
  phasors	
  is	
  the	
  sum	
  of	
  its	
  
components,	
  the	
  original	
  phasors	
  expressed	
  in	
  terms	
  of	
  their	
  
components	
  are	
  the	
  synthesis	
  of	
  a	
  set	
  of	
  three	
  unbalanced	
  phasors	
  
from	
  the	
  three	
  sets	
  of	
  symmetrical	
  components.	
  
 
Va =Va
0( ) +Va
1( ) +Va
2( )
Vb =Vb
0( ) +Vb
1( ) +Vb
2( )
Vc =Vc
0( ) +Vc
1( ) +Vc
2( )
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   11	
  
SYNTHESIS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  FROM	
  THEIR	
  
SYMMETRICAL	
  COMPONENTS	
  (Fortescue’s	
  Theorem)	
  
	
  
The	
  many	
  advantages	
  of	
  analysis	
  of	
  power	
  systems	
  by	
  the	
  method	
  
of	
  symmetrical	
  components	
  will	
  become	
  apparent	
  as	
  we	
  apply	
  the	
  
method	
  to	
  the	
  study	
  of	
  unsymmetrical	
  faults	
  on	
  otherwise	
  
symmetrical	
  systems.	
  
	
  
It	
  suffices	
  to	
  say	
  that	
  the	
  method	
  consists	
  in	
  finding	
  the	
  
symmetrical	
  components	
  of	
  current	
  at	
  the	
  fault.	
  Then,	
  the	
  values	
  
of	
  current	
  and	
  voltage	
  at	
  various	
  points	
  in	
  the	
  system	
  can	
  be	
  found	
  
by	
  means	
  of	
  the	
  bus	
  impedance	
  matrix.	
  The	
  method	
  is	
  simple	
  and	
  
leads	
  to	
  accurate	
  predicOons	
  of	
  system	
  behavior.	
  
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
Is	
  this	
  a	
  balanced	
  three-­‐phase	
  voltage?	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   12	
  
 Va
 Vb
 Vc
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
But	
  look:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   13	
  
 Va
1( )
 Vb
1( )
 Vc
1( ) 
Vc
2( )
 Va
2( )
 Vb
2( ) Va
0( )
 Vb
0( )
 Vc
0( )
 Va
 Vb
 Vc
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
Is	
  this	
  balanced?	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   14	
  
 Va
1( )
 Vb
1( )
 Vc
1( )
 Vc
2( )
 Va
2( )
 Vb
2( )
 Va
0( )
 Vb
0( )
 Vc
0( )
 Va
 Vb
 Vc
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
11/20/13	
   Symmetrical	
  Components	
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   15	
  
+	
   +	
  
=	
  
Posi;ve	
  Sequence	
  
Components	
  
Nega;ve	
  Sequence	
  
Components	
  
Zero	
  Sequence	
  
Components	
  
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
How	
  do	
  we	
  do	
  this?	
  Namely,	
  how	
  do	
  we	
  resolve	
  three	
  
unsymmetrical	
  phasors	
  into	
  symmetrical	
  components?	
  
	
  
First	
  note	
  the	
  following:	
  
	
  
SelecOng	
  Va(0)	
  as	
  a	
  reference,	
  
	
  
SelecOng	
  Va(1)	
  as	
  a	
  reference,	
  
	
  
SelecOng	
  Va(2)	
  as	
  a	
  reference,	
  
11/20/13	
   Symmetrical	
  Components	
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   16	
  
 Vc
1( ) =Va
1( )e j120° Vb
1( ) =Va
1( )e j240°
 Vb
0( ) =Va
0( ) Vc
0( ) =Va
0( )
 Vb
2( ) =Va
2( )e j120° Vc
2( ) =Va
2( )e j240°
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Let:	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   17	
  
 
Va =Va
0( ) +Va
1( ) +Va
2( )
Vb =Vb
0( ) +Vb
1( ) +Vb
2( ) =Va
0( ) +Va
1( )e j240° +Va
2( )e j120°
Vc =Vc
0( ) +Vc
1( ) +Vc
2( ) =Va
0( ) +Va
1( )e j120° +Va
2( )e j240°
 a = e j120°
 
Va =Va
0( ) +Va
1( ) +Va
2( )
Vb =Va
0( ) + a2Va
1( ) + aVa
2( )
Vc =Va
0( ) + aVa
1( ) + a2Va
2( )
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Here	
  is	
  the	
  decomposiOon	
  we	
  seek:	
  
 
Va
Vb
Vc
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= Α
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
, Α =
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⇒
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
−1 Va
Vb
Vc
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= 1
3
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Va
Vb
Vc
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   18	
  
THE	
  SYMMETRICAL	
  COMPONENTS	
  OF	
  UNSYMMETRICAL	
  PHASORS	
  
	
  
Observe	
  that	
  there	
  are	
  no	
  zero-­‐sequence	
  components	
  if	
  the	
  sum	
  of	
  
the	
  unbalanced	
  phasors	
  is	
  zero.	
  	
  
	
  
	
  
	
  
Since	
  the	
  sum	
  of	
  the	
  line-­‐line	
  voltage	
  phasors	
  in	
  a	
  three-­‐phase	
  
system	
  is	
  always	
  zero,	
  zero-­‐sequence	
  components	
  are	
  never	
  
present	
  in	
  the	
  line	
  voltages	
  regardless	
  of	
  the	
  degree	
  of	
  unbalance.	
  	
  
	
  
The	
  sum	
  of	
  the	
  three	
  line-­‐to-­‐line	
  neutral	
  voltage	
  phasors	
  is	
  not	
  
necessarily	
  zero,	
  and	
  voltages	
  to	
  neutral	
  may	
  contain	
  zero-­‐sequence	
  
components.	
  
	
  
The	
  system	
  of	
  equaOons	
  derived	
  apply	
  equally	
  well	
  to	
  the	
  currents.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   19	
  
 Va
0( ) = 0⇒Va +Vb +Vc = 0
Example	
  
	
  
One	
  conductor	
  of	
  a	
  three-­‐phase	
  line	
  is	
  open.	
  The	
  current	
  flowing	
  to	
  
the	
  Δ-­‐connected	
  load	
  through	
  line	
  a	
  is	
  10	
  A.	
  With	
  the	
  current	
  in	
  line	
  
a	
  as	
  a	
  reference	
  and	
  assuming	
  that	
  line	
  c	
  is	
  open,	
  find	
  the	
  
symmetrical	
  components	
  of	
  the	
  line	
  currents.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   20	
  
Z	
  
Z	
  Z	
  
 Ia = 10∠0°
 Ic = 0
a	
  
b	
  
c	
  
 Ib = −10∠0° = 10∠180°
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   21	
  
 
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= 1
3
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= 1
3
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
10
−10
0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
0( ) = 1
3
10−10( ) = 0
Ia
1( ) = 10
3
1− a( ) = 103 1− cos 120°( )− jsin 120°( )( ) =
10
3
3
2
− j 3
2
⎛
⎝
⎜
⎞
⎠
⎟
= 5
3
3 − j( ) = 5
3
3+1∠− 30° = 10
3
∠− 30°
Ia
2( ) = 10
3
1− a2( ) = 103 1− cos 240°( )− jsin 240°( )( ) =
10
3
3
2
+ j 3
2
⎛
⎝
⎜
⎞
⎠
⎟
= 10
3
∠30°
Example	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Note	
  that	
  components	
  IC(1)	
  and	
  IC(2)	
  have	
  nonzero	
  values	
  although	
  
line	
  c	
  is	
  open	
  and	
  can	
  carry	
  no	
  net	
  current.	
  As	
  is	
  expected,	
  the	
  sum	
  
of	
  the	
  components	
  in	
  line	
  c	
  is	
  zero.	
  Of	
  course,	
  the	
  sum	
  of	
  
the	
  components	
  in	
  line	
  a	
  is	
  10	
  A,	
  and	
  the	
  sum	
  of	
  the	
  components	
  in	
  
line	
  b	
  is	
  –	
  10	
  A.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   22	
  
 
Ia
0( ) = 0, Ib
0( ) = Ia
0( ) = 0, Ic
0( ) = Ia
0( ) = 0
Ia
1( ) = 10
3
∠− 30°, Ib
1( ) = Ia
1( )e j240° = 10
3
∠−150°, Ic
1( ) = Ia
1( )e j120° = 10
3
∠90°
Ia
2( ) = 10
3
∠30°, Ib
2( ) = Ia
2( )e j120° = 10
3
∠150°, Ic
2( ) = Ia
2( )e j240° = 10
3
∠− 90°
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
In	
  three-­‐phase	
  systems	
  circuit	
  elements	
  are	
  connected	
  between	
  
lines	
  a,	
  b,	
  and	
  c	
  in	
  either	
  a	
  Y	
  or	
  Δ	
  configuraOon.	
  RelaOonships	
  
between	
  the	
  symmetrical	
  components	
  of	
  Y	
  and	
  Δ	
  currents	
  and	
  
voltages	
  can	
  be	
  established	
  by	
  referring	
  to	
  the	
  figure	
  (next	
  slide)	
  
which	
  shows	
  symmetrical	
  impedances	
  connected	
  in	
  Y	
  and	
  Δ.	
  	
  
	
  
Choose	
  the	
  reference	
  phase	
  for	
  Δ	
  quanOOes	
  as	
  branch	
  a-­‐b.	
  The	
  
choice	
  of	
  reference	
  phase	
  is	
  arbitrary	
  and	
  does	
  not	
  affect	
  the	
  
results.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   23	
  
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   24	
  
ZΔ	
  
a	
  
b	
  
c	
  
 Ia
 Ib
 Ic
ZΔ	
  
ZΔ	
  
 
+
Vab
−
 
−
Vca
+
 
+
−
Vbc
 Iab Ica
 Ibc 
Ia = Iab − Ica
Ib = Ibc − Iab
Ic = Ica − Ibc
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
For	
  a	
  balanced	
  three-­‐phase	
  system:	
  
	
  
	
  
	
  
or	
  the	
  line	
  currents	
  into	
  a	
  Δ–connected	
  circuit	
  have	
  no	
  zero-­‐
sequence	
  currents.	
  
	
  
SubsOtuOng	
  components	
  of	
  current	
  in	
  the	
  equaOon	
  for	
  Ia	
  gives:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   25	
  
 
Ia
0( ) = 1
3
Ia + Ib + Ic( ) = 0
 
Ia = Iab − Ica
⇒ Ia
1( ) + Ia
2( ) = Iab
0( ) + Iab
1( ) + Iab
2( )( )− Ica0( ) + Ica1( ) + Ica2( )( )
= Iab
0( ) − Ica
0( )( )
= 0
  
+ Iab
1( ) − Ica
1( )( ) + Iab2( ) − Ica2( )( )
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
If	
  a	
  nonzero	
  value	
  of	
  circulaOng	
  current	
  Iab(0)	
  exists	
  in	
  the	
  Δ	
  circuit	
  it	
  
cannot	
  be	
  determined	
  from	
  the	
  line	
  currents	
  alone.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   26	
  
 
Ia
1( ) + Ia
2( ) = Iab
0( ) − Ica
0( )( )
= 0
  
+ Iab
1( ) − Ica
1( )( ) + Iab2( ) − Ica2( )( )
ZΔ	
  
a	
  
b	
  
c	
  
 Ib
 Ic
ZΔ	
  
ZΔ	
  
 
+
Vab
−
 
−
Vca
+
 
+
−
Vbc
 Iab Ica
 Ibc
 Ia
 Iab
0( )
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
Now	
  since:	
  
	
  
SubsOtuOng:	
  
	
  
	
  
	
  
Similarly	
  for	
  phase	
  b:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   27	
  
 
Ia
1( ) + Ia
2( ) = Iab
1( ) − Ica
1( )( ) + Iab2( ) − Ica2( )( )
= 1− a( ) Iab1( ) + 1− a2( ) Iab2( )
 Ica
1( ) = aIab
1( ) , Ica
2( ) = a2Iab
2( )
 
Ib
1( ) + Ib
2( ) = 1− a( ) Ibc1( ) + 1− a2( ) Ibc2( )
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
Note:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   28	
  
 
1− a = 1− e j120° = 1− cos 120°( )− jsin 120°( )
= 1+ 1
2
− j 3
2
= 3
2
3 − j( ) = 3∠− 30°
1− a2 = 1− e j240° = 1− cos 240°( )− jsin 240°( )
= 1+ 1
2
+ j 3
2
= 3
2
3 + j( ) = 3∠30°
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
Thus:	
  
	
  
	
  
	
  
	
  
or:	
  
	
  
	
  
	
  
	
  
This,	
  for	
  the	
  Δ-­‐connected	
  load,	
  the	
  relaOonship	
  between	
  line	
  
currents	
  and	
  phase	
  currents	
  follows	
  the	
  customary	
  relaOonship.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   29	
  
 
Ia
1( ) + Ia
2( ) = 1− a( ) Iab1( ) + 1− a2( ) Iab2( )
= 3Iab
1( )∠− 30° + 3Iab
2( )∠30°
 
Ia
1( ) = 3Iab
1( )∠− 30°
Ia
2( ) = 3Iab
2( )∠30°
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   30	
  
 Ia
1( )
 Ib
1( )
 Ic
1( )
 Iab
1( ) 
Ica
1( )
 Ibc
1( )
 30°
 30°
 30°
Posi;ve	
  Sequence	
  
Components	
  
Nega;ve	
  Sequence	
  
Components	
  
 Ia
2( )
 Ib
2( )
 Ic
2( )
 Iab
2( )
 Ibc
2( )
 Ica
2( )
 30°
 30°
 30°
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
The	
  voltages	
  for	
  a	
  Y-­‐connected	
  load	
  follow	
  analogously:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   31	
  
ZY	
  
ZY	
  ZY	
  
 
−
Vca
+
 
+
Vab
−
 
+
−
Vbc
a	
  
b	
  
c	
  
 Ia
 Ib
 Ic
n	
  
 
+
Van
−
 
Vab =Van −Vbn
Vbc =Vbn −Vcn
Vca =Vcn −Van
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
Proceeding	
  in	
  a	
  similar	
  way,	
  
	
  
For	
  a	
  balanced	
  three-­‐phase	
  system:	
  
	
  
	
  
	
  
	
  
	
  
Here	
  as	
  well	
  a	
  nonzero	
  value	
  of	
  the	
  zero-­‐sequence	
  voltage	
  V(0)an	
  
cannot	
  be	
  determines	
  from	
  the	
  line-­‐to-­‐line	
  voltages	
  alone.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   32	
  
 
Vab
1( ) +Vab
2( ) = Van
0( ) − Ibn
0( )( )
= 0
  
+ Van
1( ) −Vbn
1( )( ) + Van2( ) −Vbn2( )( )
 
Vab
0( ) = 1
3
Vab +Vbc +Vca( ) = 0
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
Using	
  similar	
  calculaOons	
  as	
  for	
  the	
  current,	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   33	
  
 
Vab
1( ) = 1− a2( )Van1( ) = 3Van1( )∠30°
Vab
2( ) = 1− a( )Van2( ) = 3Van2( )∠− 30°
 Vab
1( )
 Van
1( )
 30°
Posi;ve	
  Sequence	
  
Components	
  
Nega;ve	
  Sequence	
  
Components	
  
 Vbc
1( ) Vca
1( )
 Vbn
1( )
 Vcn
1( )
 30°
 30°
 Vbc
2( )
 Vab
2( )
 Vca
2( )
 Vbn
2( )
 Vcn
2( )
 Van
2( )
 30° 30°
 30°
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
If	
  the	
  voltages	
  to	
  neutral
are	
  in	
  per	
  unit	
  referred	
  to	
  the	
  base	
  voltage	
  
to	
  neutral	
  and	
  the	
  line	
  voltages	
  are	
  in	
  per	
  unit	
  referred	
  to	
  the	
  base	
  
voltage	
  from	
  line-­‐to-­‐line,	
  the	
  √3	
  mulOpliers	
  must	
  be	
  omimed	
  the	
  
equaOons.	
  	
  
	
  
If	
  both	
  voltages	
  are	
  referred	
  to	
  the	
  same	
  base,	
  however,	
  the	
  
equaOons	
  are	
  correct	
  as	
  given	
  .	
  Similarly,	
  when	
  line	
  and	
  Δ	
  currents	
  
are	
  expressed	
  in	
  per	
  unit,	
  each	
  on	
  its	
  own	
  base,	
  the	
  √3	
  disappears	
  
since	
  the	
  two	
  bases	
  are	
  related	
  to	
  one	
  another	
  in	
  the	
  raOo	
  of	
  √3:1.	
  
	
  
When	
  the	
  currents	
  are	
  expressed	
  on	
  the	
  same	
  base,	
  the	
  equaOon	
  is	
  
correct	
  as	
  wrimen.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   34	
  
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
From	
  the	
  figure	
  for	
  the	
  Δ-­‐connected	
  load:	
  
	
  
	
  
	
  
	
  
SubsOtuOng:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   35	
  
 
Vab
Iab
= ZΔ ⇒
Vab
1( )
Iab
1( ) = ZΔ =
Vab
2( )
Iab
2( )
 
Vab
1( )
Iab
1( ) =
3Van
1( )∠30°
1
3
Ia
1( )∠30°
= ZΔ =
Vab
2( )
Iab
2( ) =
3Van
2( )∠− 30°
1
3
Ia
2( )∠− 30°
⇒
Van
1( )
Ia
1( ) =
ZΔ
3
=
Van
2( )
Ia
2( )
SYMMETRICAL	
  Y	
  AND	
  Δ	
  CIRCUITS	
  
	
  
But,	
  from	
  the	
  figure	
  for	
  the	
  Y-­‐connected	
  load:	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
As	
  far	
  as	
  posiOve-­‐	
  or	
  negaOve-­‐sequence	
  circuits	
  are	
  concerned.	
  
	
  
Of	
  course	
  this	
  is	
  merely	
  a	
  verificaOon	
  of	
  a	
  fact	
  we’ve	
  known	
  all	
  
along.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   36	
  
 
Van
1( )
Ia
1( ) =
ZΔ
3
=
Van
2( )
Ia
2( )
Van
2( )
Ia
2( ) = ZY ∴
ZΔ
3
= ZY
Example	
  
	
  
Three	
  idenOcal	
  10.58-­‐Ω,	
  Y-­‐connected	
  resistors	
  form	
  a	
  load	
  bank	
  
with	
  a	
  three-­‐phase	
  voltage	
  raOng	
  of	
  2300	
  V	
  and	
  500	
  kVA.	
  If	
  the	
  load	
  
bank	
  has	
  applied	
  voltages:	
  	
  
	
  
	
  
	
  
find	
  the	
  line	
  voltages	
  and	
  currents	
  in	
  per	
  unit	
  into	
  the	
  load.	
  Assume	
  
that	
  the	
  neutral	
  of	
  the	
  load	
  is	
  not	
  connected	
  to	
  the	
  neutral	
  of	
  the	
  
system	
  and	
  select	
  a	
  base	
  of	
  2300	
  V	
  and	
  500	
  kVA.	
  
	
  
In	
  per	
  unit:	
  ZY	
  =	
  1.0	
  and	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   37	
  
 Vab = 1840V Vbc = 2760V Vca = 2300V
 Vab = 0.8 Vbc = 1.2 Vca = 1.0
Example	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Law	
  of	
  cosines:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   38	
  
 Vbc = 1.2
ZY	
  
ZY	
  ZY	
  
 
−
Vca
+
 
+
Vab
−
 
+
−
Vbc
a	
  
b	
  
c	
  
n	
  
 Vab +Vbc +Vca = 0
 Vca = 1.0
 Vab = 0.8
θ
 
Vab
2
= Vca
2
+ Vbc
2
− 2 Vca Vbc cosθ
Vca
2
= Vbc
2
+ Vab
2
− 2 Vab Vbc cosϕ
ϕ
Example	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   39	
  
 
Vab = 0.8 Vbc = 1.2 Vca = 1.0
cosθ =
Vca
2
+ Vbc
2
− Vab
2
2 Vca Vbc
⇒θ = 41.41°
cosϕ =
Vbc
2
+ Vab
2
− Vca
2
2 Vbc Vab
⇒ϕ = 55.77°
 Vbc = 1.2 Vca = 1.0
 Vab = 0.8
 Vca = 1.0∠82.82°
 82.82°
 82.82°
 41.41°
 55.77°
 55.77°
 Vbc = 1.2∠−55.77°
 Vab = 0.8∠180°
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   40	
  
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   41	
  
Resistors	
  are	
  1.0	
  pu	
  
Example	
  
	
  
The	
  absence	
  of	
  a	
  neutral	
  connecOon	
  means	
  that	
  zero-­‐sequence	
  
currents	
  are	
  not	
  present	
  in	
  the	
  circuit.	
  Therefore,	
  the	
  phase	
  voltages	
  
at	
  the	
  load	
  contain	
  posiOve-­‐	
  and	
  negaOve-­‐sequence	
  components	
  
only.	
  The	
  phase	
  voltages	
  are	
  found	
  from	
  	
  
	
  
	
  
But	
  the	
  √3	
  factor	
  is	
  omimed	
  since	
  the	
  line	
  voltages	
  are	
  expressed	
  in	
  
terms	
  of	
  the	
  base	
  voltage	
  from	
  line	
  to	
  line	
  and	
  the	
  phase	
  voltages	
  
are	
  desired	
  in	
  per	
  unit	
  of	
  the	
  base	
  voltage	
  to	
  neutral	
  .	
  Thus,	
  
	
  
	
  
	
  
	
  
all	
  per	
  unit.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   42	
  
 Vab
1( ) = 3Van
1( )∠30°,Vab
2( ) = 3Van
2( )∠− 30°
 
Van
1( ) =Vab
1( )∠− 30° = 0.23468∠ 42.576°− 30°( ) = 0.23468∠12.576°
Van
2( ) =Vab
2( )∠30° = 0.98568∠ −170.73° + 30°( ) = 0.98568∠−140.73°
Power	
  in	
  Terms	
  of	
  Symmetrical	
  Components	
  
	
  
If	
  the	
  symmetrical	
  components	
  of	
  current	
  and	
  voltage	
  are	
  known,	
  
the	
  power	
  expended	
  in	
  a	
  three-­‐phase	
  circuit	
  can	
  be	
  computed	
  
directly	
  from	
  the	
  components.	
  
	
  
The	
  total	
  complex	
  power	
  flowing	
  into	
  a	
  three-­‐phase	
  circuit	
  through	
  
three	
  lines	
  a,	
  b,	
  and	
  c	
  is	
  
	
  
	
  
where	
  Va,	
  Vb,	
  and	
  Vc	
  are	
  the	
  voltages	
  to	
  reference	
  at	
  the	
  terminals	
  
and	
  la,	
  lb,	
  and	
  Ic	
  are	
  the	
  currents	
  flowing	
  into	
  the	
  circuit	
  in	
  the	
  three	
  
lines.	
  A	
  neutral	
  connecOon	
  may	
  or	
  may	
  not	
  be	
  present.	
  If	
  there	
  is	
  
impedance	
  in	
  the	
  neutral	
  connecOon	
  to	
  ground	
  (see	
  next	
  slide),	
  
then	
  the	
  voltages	
  Va,	
  Vb,	
  and	
  Vc	
  must	
  be	
  interpreted	
  as	
  voltages	
  
from	
  the	
  line	
  to	
  ground	
  rather	
  than	
  to	
  neutral.	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   43	
  
 
Sthree− phase = P + jQ =Va Ia
* +VbIb
* +Vc Ic
*
Power	
  in	
  Terms	
  of	
  Symmetrical	
  Components	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   44	
  
ZY	
  
ZY	
  ZY	
  
 
−
Vca
+
 
+
Vab
−
 
+
−
Vbc
a	
  
b	
  
c	
  
n	
  
Zn	
  
Power	
  in	
  Terms	
  of	
  Symmetrical	
  Components	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   45	
  
 
Sthree− phase = V⎡⎣ ⎤⎦
T
I *⎡⎣ ⎤⎦
= ΑV012⎡⎣ ⎤⎦
T
ΑI012⎡⎣ ⎤⎦
*
= V012⎡⎣ ⎤⎦
T
ΑTΑ* I012
*⎡⎣ ⎤⎦
= V012⎡⎣ ⎤⎦
T
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
*
I012⎡⎣ ⎤⎦
*
= V012⎡⎣ ⎤⎦
T
1 1 1
1 a2 a
1 a a2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
I012⎡⎣ ⎤⎦
*
= 3 V012⎡⎣ ⎤⎦
T
1 0 0
0 1 0
0 0 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
I012⎡⎣ ⎤⎦
*
= 3 V012⎡⎣ ⎤⎦
T
I012⎡⎣ ⎤⎦
*
 
V012⎡⎣ ⎤⎦ =
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
, I012⎡⎣ ⎤⎦ =
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Power	
  in	
  Terms	
  of	
  Symmetrical	
  Components	
  
	
  
	
  
	
  
	
  
which	
  shows	
  how	
  complex	
  power	
  (in	
  voltamperes	
  )	
  can	
  be	
  
computed	
  from	
  the	
  symmetrical	
  components	
  of	
  the	
  voltages	
  to	
  
reference	
  (in	
  volts)	
  and	
  line	
  currents	
  (in	
  amperes)	
  of	
  an	
  unbalanced	
  
three-­‐phase	
  circuit.	
  	
  
	
  
Note	
  how	
  the	
  transformaOon	
  of	
  a-­‐b-­‐c	
  voltages	
  and	
  currents	
  to	
  
symmetrical	
  components	
  is	
  power-­‐invariant	
  only	
  if	
  each	
  product	
  of	
  
sequence	
  voltage	
  (in	
  volts)	
  Omes	
  the	
  complex	
  conjugate	
  of	
  the	
  
corresponding	
  sequence	
  current	
  (in	
  amperes)	
  is	
  mulOplied	
  by	
  3,	
  as	
  
shown	
  above.	
  When	
  the	
  complex	
  power	
  Sthree-­‐phase	
  is	
  expressed	
  in	
  
per	
  unit	
  of	
  a	
  three-­‐phase	
  voltampere	
  base	
  the	
  3	
  disappears.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   46	
  
 
Sthree− phase =Va Ia
* +VbIb
* +Vc Ic
* = 3 V012⎡⎣ ⎤⎦
T
I012⎡⎣ ⎤⎦
*
= 3Va
0( )Ia
0( )* + 3Va
1( )Ia
1( )* + 3Va
2( )Ia
2( )*
Power	
  in	
  Terms	
  of	
  Symmetrical	
  Components	
  
	
  
From	
  the	
  previous	
  example:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   47	
  
 
Van
0( ) = 0
Van
1( ) = 0.23468∠12.576°
Van
2( ) = 0.98568∠−140.73°
 
Ia
0( ) =
Van
0( )
1∠0°
= 0
Ia
1( ) =
Van
1( )
1∠0°
= 0.23468∠12.576°
Ia
2( ) =
Van
2( )
1∠0°
=V 0.98568∠−140.73°
 
Sthree− phase =Va
0( )Ia
0( )* +Va
1( )Ia
1( )* +Va
2( )Ia
2( )*
= 0+ 3 0.23468( )2 + 3 0.98568( )2
= 1.02664⇒1.02664×500 = 513.32 kW
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
If	
  impedance	
  Zn	
  is	
  inserted	
  between	
  the	
  neutral	
  and	
  ground	
  of	
  the	
  
Y-­‐connected	
  impedances	
  then	
  the	
  sum	
  of	
  the	
  line	
  currents	
  is	
  equal	
  
to	
  the	
  current	
  In	
  in	
  the	
  return	
  path	
  through	
  the	
  neutral:	
  
	
  
	
  
Expressing	
  the	
  unbalanced	
  line	
  currents	
  in	
  terms	
  of	
  their	
  
symmetrical	
  components	
  gives:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   48	
  
 In = Ia + Ib + Ic
 
In = Ia
0( ) + Ia
1( ) + Ia
2( )( ) + Ib0( ) + Ib1( ) + Ib2( )( ) + Ic0( ) + Ic1( ) + Ic2( )( )
= Ia
0( ) + Ib
0( ) + Ic
0( )( ) + Ia1( ) + Ib1( ) + Ic1( )( )
= 0
  
+ Ia
2( ) + Ib
2( ) + Ic
2( )( )
= 0
  
= 3Ia
0( )
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Since	
  the	
  posiOve-­‐sequence	
  and	
  negaOve-­‐sequence	
  currents	
  add	
  
separately	
  to	
  zero	
  at	
  neutral	
  point	
  n,	
  there	
  cannot	
  be	
  any	
  posiOve-­‐
sequence	
  or	
  negaOve	
  sequence	
  currents	
  in	
  the	
  connecOons	
  from	
  
neutral	
  to	
  ground	
  regardless	
  of	
  the	
  value	
  of	
  Zn.	
  
	
  
We	
  see	
  that	
  the	
  zero-­‐sequence	
  currents	
  combining	
  together	
  at	
  n	
  
become	
  3In	
  producing	
  the	
  voltage	
  drop	
  3In(0)/Zn	
  between	
  neutral	
  
and	
  ground.	
  It	
  is	
  important	
  to	
  disOnguish	
  between	
  voltages	
  to	
  
neutral	
  and	
  voltages	
  to	
  ground	
  under	
  unbalanced	
  condiOons.	
  	
  
	
  
Designate	
  voltages	
  of	
  phase	
  a	
  with	
  respect	
  to	
  neutral	
  and	
  ground	
  as	
  
Van	
  and	
  Vn,	
  respecOvely.	
  Thus,	
  the	
  voltage	
  of	
  phase	
  a	
  with	
  respect	
  to	
  
ground	
  is	
  given	
  by	
  Va	
  =	
  Van	
  +	
  Vn,	
  where	
  Vn	
  =	
  3In(0)Zn	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   49	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   50	
  
ZY	
  
ZY	
  ZY	
  
 
−
Vca
+
 
+
Vab
−
 
+
−
Vbc
a	
  
b	
  
c	
  
n	
  
Zn	
  
 Ia
 Ib
 Ic
 In = 3Ia
0( )
 
+
Vn = 3Ia
0( )Zn
−
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
We	
  can	
  write	
  the	
  voltage	
  drops	
  to	
  ground	
  as:	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
The	
  a-­‐b-­‐c	
  voltages	
  can	
  be	
  replaced	
  by	
  their	
  symmetrical	
  
components…	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   51	
  
 
Va
Vb
Vc
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
Van
Vbn
Vcn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
Vn
Vn
Vn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= ZY
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+ 3Ia
0( )Zn
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
The	
  a-­‐b-­‐c	
  voltages	
  can	
  be	
  replaced	
  by	
  their	
  symmetrical	
  
components:	
  
	
  
	
  
	
  
	
  
	
  
	
  
MulOply	
  by	
  A–	
  1:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   52	
  
 
Va
Vb
Vc
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= Α
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= ZYΑ
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ 3Ia
0( )Zn
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
 
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= ZY
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ 3Ia
0( )Zn A
−1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   53	
  
 
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= ZY
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ 3Ia
0( )Zn A
−1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
A−1
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 1
3
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1
1
1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 1
3
3
1+ a + a2
1+ a2 + a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
1
0
0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
11/20/13	
   Symmetrical	
  Components	
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  Sequence	
  Networks	
   54	
  
 
Va
0( )
Va
1( )
Va
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= ZY
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ 3Ia
0( )Zn
1
0
0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Va
0( ) = ZY Ia
0( ) + 3ZnIa
0( ) = Z0Ia
0( )
Va
1( ) = ZY Ia
1( ) = Z1Ia
1( )
Va
2( ) = ZY Ia
2( ) = Z2Ia
2( )
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND
Δ	
  IMPEDANCES	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
This	
  results	
  tells	
  us	
  that	
  currents	
  of	
  one	
  sequence	
  cause	
  voltage	
  
drops	
  of	
  only	
  the	
  same	
  sequence	
  in	
  Δ	
  or	
  Y-­‐connected	
  circuits	
  with	
  
symmetrical	
  impedances	
  in	
  each	
  phase.	
  	
  
	
  
This	
  most	
  (important)	
  result	
  allows	
  us	
  to	
  draw	
  the	
  three	
  single-­‐
phase	
  sequence	
  circuits	
  shown…	
  
11/20/13	
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  Components	
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  Sequence	
  Networks	
   55	
  
 
Va
0( ) = ZY + 3Zn( ) Ia0( ) = Z0Ia0( )
Va
1( ) = ZY Ia
1( ) = Z1Ia
1( )
Va
2( ) = ZY Ia
2( ) = Z2Ia
2( )
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
These	
  three	
  circuits,	
  considered	
  simultaneously,	
  provide	
  the	
  same	
  
informaOon	
  as	
  the	
  actual	
  (unbalanced)	
  circuit	
  on	
  Slide	
  50,	
  and	
  are	
  
independent	
  of	
  one	
  another	
  because	
  the	
  above	
  equaOons	
  are	
  
decoupled	
  .	
  Circuit	
  (a)	
  shown	
  below	
  is	
  called	
  the	
  zero-­‐sequence	
  
circuit	
  because	
  it	
  relates	
  the	
  zero-­‐sequence	
  voltage	
  Va(0)	
  to	
  the	
  zero-­‐
sequence	
  current	
  Ia(0).	
  
11/20/13	
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  Sequence	
  Networks	
   56	
  
ZY	
   n	
  
3Zn	
  
Reference	
  
(a)	
  
 
+
Va
0( )
−
 Ia
0( )
Z0	
  
ZY	
   n	
  
Reference	
  
(b)	
  
 
+
Va
1( )
−
 Ia
1( )
Z1	
  
ZY	
   n	
  
Reference	
  
(c)	
  
 
+
Va
2( )
−
 Ia
2( )
Z2	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Clearly,	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Z0	
  is	
  called	
  (defined)	
  the	
  impedance	
  to	
  zero-­‐sequence	
  current.	
  
	
  
Note	
  how	
  we	
  have	
  recaptured	
  our	
  ability	
  to	
  draw	
  a	
  per-­‐phase	
  
equivalent	
  circuit	
  for	
  an	
  unbalanced	
  system,	
  only	
  now	
  we	
  need	
  
three	
  of	
  them.	
  
11/20/13	
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   57	
  
ZY	
   n	
  
3Zn	
  
Reference	
  
 
+
Va
0( )
−
 Ia
0( )
Z0	
  
 
Va
0( )
Ia
0( ) = ZY + 3Zn = Z0
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Likewise,	
  Fig.	
  (b)	
  is	
  called	
  the	
  posiOve-­‐sequence	
  circuit	
  and	
  Z1	
  is	
  
called	
  the	
  impedance	
  to	
  posiOve-­‐sequence	
  current,	
  and	
  Fig.	
  (c)	
  is	
  
the	
  negaOve	
  sequence	
  circuit	
  and	
  Z2	
  is	
  the	
  impedance	
  to	
  negaOve-­‐
sequence	
  current.	
  	
  
	
  
The	
  names	
  of	
  the	
  impedances	
  to	
  currents	
  of	
  the	
  different	
  sequences	
  
are	
  usually	
  shortened	
  to	
  the	
  less	
  descripOve	
  terms	
  zero-­‐sequence	
  
impedance	
  Z0,	
  posiOve	
  sequence	
  impedance	
  Z1,	
  and	
  negaOve-­‐
sequence	
  impedance	
  Z2.	
  
	
  
Here	
  the	
  posiOve	
  and	
  negaOve-­‐sequence	
  impedances	
  Z1	
  and	
  Z2	
  are	
  
equal	
  to	
  the	
  usual	
  per-­‐phase	
  impedance	
  ZY	
  which	
  is	
  generally	
  the	
  
case	
  for	
  staOonary	
  symmetrical	
  circuits.	
  	
  
11/20/13	
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   58	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Each	
  of	
  the	
  three	
  sequence	
  circuits	
  presents	
  one	
  phase	
  of	
  the	
  actual	
  
three-­‐phase	
  circuit	
  when	
  the	
  lamer	
  carries	
  current	
  of	
  only	
  that	
  
sequence.	
  
	
  
When	
  the	
  three	
  sequence	
  currents	
  are	
  simultaneously	
  present,	
  all	
  
three	
  sequence	
  circuits	
  are	
  needed	
  to	
  fully	
  represent	
  the	
  original	
  
circuit	
  circuit.	
  	
  
	
  
Voltages	
  in	
  the	
  posiOve-­‐sequence	
  and	
  negaOve-­‐sequence	
  circuits	
  
can	
  be	
  regarded	
  as	
  voltages	
  measured	
  with	
  respect	
  to	
  either	
  neutral	
  
or	
  ground	
  whether	
  or	
  not	
  there	
  is	
  a	
  connecOon	
  of	
  some	
  finite	
  value	
  
of	
  impedance	
  Zn	
  between	
  neutral	
  and	
  ground.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   59	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Clearly	
  in	
  the	
  posiOve-­‐sequence	
  circuit	
  there	
  is	
  no	
  difference	
  
between	
  Va(1)	
  and	
  Van(1)	
  and	
  a	
  similar	
  statement	
  applies	
  to	
  Va(2)	
  and	
  
Van(2)	
  and	
  in	
  the	
  negaOve-­‐sequence	
  circuit.	
  	
  
	
  
However,	
  a	
  voltage	
  difference	
  can	
  exist	
  between	
  the	
  neutral	
  and	
  
the	
  reference	
  of	
  the	
  zero-­‐sequence	
  circuit.	
  In	
  Fig.	
  (a)	
  the	
  current	
  Ia(0)	
  
flowing	
  through	
  impedance	
  3Zn	
  produces	
  the	
  same	
  voltage	
  drop	
  
from	
  neutral	
  to	
  ground	
  as	
  the	
  current	
  3Ia(0)	
  flowing	
  through	
  
impedance	
  Zn	
  in	
  the	
  actual	
  circuit	
  on	
  Slide	
  50.	
  
	
  
If	
  the	
  neutral	
  of	
  the	
  Y-­‐connected	
  circuit	
  is	
  grounded	
  through	
  zero	
  
impedance,	
  we	
  set	
  Zn	
  =	
  0	
  and	
  a	
  zero-­‐impedance	
  connecOon	
  then	
  
joins	
  the	
  neutral	
  point	
  to	
  the	
  reference	
  node	
  of	
  the	
  zero-­‐sequence	
  
circuit.	
  	
  
	
  
11/20/13	
   Symmetrical	
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  Sequence	
  Networks	
   60	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
If	
  there	
  is	
  no	
  connecOon	
  between	
  neutral	
  and	
  ground,	
  there	
  cannot	
  
be	
  any	
  zero-­‐sequence	
  current	
  flow,	
  for	
  then	
  Zn	
  =	
  ∞,	
  which	
  is	
  
indicated	
  by	
  the	
  open	
  circuit	
  between	
  neutral	
  and	
  the	
  reference	
  
node	
  in	
  the	
  zero-­‐sequence	
  circuit	
  below.	
  	
  
11/20/13	
   Symmetrical	
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  Sequence	
  Networks	
   61	
  
ZY	
  
ZY	
  ZY	
  
n	
  
ZY	
   n	
  
Reference	
  
 
+
Va
0( )
−
 Ia
0( )
a	
  
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Obviously,	
  a	
  Δ-­‐connected	
  circuit	
  cannot	
  provide	
  a	
  path	
  through	
  
neutral,	
  and	
  so	
  line	
  currents	
  flowing	
  into	
  a	
  Δ-­‐connected	
  load	
  or	
  its	
  
equivalent	
  Y	
  circuit	
  cannot	
  contain	
  any	
  zero-­‐sequence	
  components.	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
To	
  see	
  this…	
  
11/20/13	
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   62	
  
ZΔ	
  
a	
  
b	
  
c	
  
ZΔ	
  
ZΔ	
  
ZΔ	
  
Reference	
  
 
+
Va
0( )
−
a	
  
 Ia
0( )
 Iab
0( ) = 0
 −Vab
0( ) +
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
For	
  the	
  Δ-­‐connected	
  circuit:	
  
	
  
	
  
	
  
Recall	
  Slide	
  19:	
  
	
  
	
  
Thus	
  
	
  
and	
  since	
  the	
  sum	
  of	
  the	
  line-­‐to-­‐line	
  voltages	
  is	
  always	
  zero,	
  
11/20/13	
   Symmetrical	
  Components	
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  Sequence	
  Networks
63	
  
 Vab = ZΔ Iab Vbc = ZΔ Ibc Vca = ZΔ Ica
 
Vab
0( ) = 1
3
Vab +Vbc +Vca( )
 Vab +Vbc +Vca = 3Vab
0( ) = 3ZΔ Iab
0( )
 Vab
0( ) = 0, Iab
0( ) = 0
SEQUENCE	
  CIRCUITS	
  OF	
  Y	
  AND	
  Δ	
  IMPEDANCES	
  
	
  
Thus,	
  in	
  Δ-­‐connected	
  circuits	
  with	
  impedances	
  only	
  and	
  no	
  sources	
  
or	
  mutual	
  coupling	
  there	
  cannot	
  be	
  any	
  circulaOng	
  currents.	
  	
  
	
  
SomeOmes	
  single-­‐phase	
  circulaOng	
  currents	
  can	
  be	
  produced	
  in	
  the	
  
Δ	
  circuits	
  of	
  transformers	
  and	
  generators	
  by	
  either	
  inducOon	
  or	
  
zero-­‐sequence	
  generated	
  voltages.	
  	
  
	
  
A	
  Δ	
  circuit	
  and	
  its	
  zero-­‐sequence	
  circuit	
  were	
  shown	
  on	
  the	
  previous	
  
slide.	
  Note,	
  however,	
  that	
  even	
  if	
  zero-­‐sequence	
  voltages	
  were	
  
generated	
  in	
  the	
  phases	
  of	
  the	
  Δ,	
  no	
  zero-­‐sequence	
  voltage	
  could	
  
exist	
  between	
  the	
  Δ	
  terminals,	
  for	
  the	
  rise	
  in	
  voltage	
  in	
  each	
  phase	
  
would	
  then	
  be	
  matched	
  by	
  the	
  voltage	
  drop	
  in	
  the	
  zero-­‐sequence	
  
impedance	
  of	
  each	
  phase.	
  Consider	
  an	
  example.	
  
11/20/13	
   Symmetrical	
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   64	
  
Example	
  
	
  
Three	
  equal	
  impedances	
  of	
  j21	
  Ω	
  are	
  Δ-­‐connected.	
  Determine	
  the	
  
sequence	
  impedances	
  and	
  circuits	
  of	
  the	
  combinaOon.	
  Repeat	
  for	
  
the	
  case	
  where	
  a	
  mutual	
  impedance	
  of	
  j6	
  Ω	
  exists	
  between	
  each	
  
pair	
  of	
  adjacent	
  branches	
  in	
  the	
  Δ.	
  
11/20/13	
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   65	
  
a	
  
b	
  
c	
  
 j21 j21
 j21
a	
  
b	
  
c	
  
 j21 j21
 j21
 j6 j6
 j6
Example	
  
	
  
The	
  line-­‐to-­‐line	
  voltages	
  are	
  related	
  to	
  the	
  phase	
  currents	
  by	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   66	
  
 
Vab
Vbc
Vca
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
j21 0 0
0 j21 0
0 0 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Iab
Ibc
Ica
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Example	
  
	
  
Transforming	
  to	
  symmetrical	
  components:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   67	
  
 
Α
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j21 0 0
0 j21 0
0 0 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⇒
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= j21Α−1Α
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j21 0 0
0 j21 0
0 0 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   68	
  
 
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j21 0 0
0 j21 0
0 0 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Zero-­‐sequence	
  
ZY	
  
Reference	
  
Posi;ve-­‐sequence	
  
 
+
Va
1( )
−
 Ia
1( )
ZY	
  
Reference	
  
Nega;ve	
  sequence	
  
 
+
Va
2( )
−
 Ia
2( )
ZΔ	
  
Reference	
  
 
+
Va
0( )
−
a	
  
 Ia
0( ) 
ZΔ
3
= ZY
 j7 j7 j21
Example	
  
	
  
For	
  the	
  second	
  part:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   69	
  
 
Α
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j21 j6 j6
j6 j21 j6
j6 j6 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
 
j21 j6 j6
j6 j21 j6
j6 j6 j21
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= j15
1 0 0
0 1 0
0 0 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
+ j6
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Remember	
  this	
  trick!	
  
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   70	
  
 
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= Α−1 j15
1 0 0
0 1 0
0 0 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
+ j6
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎧
⎨
⎪
⎩
⎪
⎫
⎬
⎪
⎭
⎪
Α
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⇒
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= j15+ j6Α−1
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
⎧
⎨
⎪
⎩
⎪
⎫
⎬
⎪
⎭
⎪
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= j15+ j6
3 0 0
0 0 0
0 0 0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎧
⎨
⎪
⎩
⎪
⎫
⎬
⎪
⎭
⎪
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j33 0 0
0 j15 0
0 0 j15
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   71	
  
 
Vab
0( )
Vab
1( )
Vab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
j33 0 0
0 j15 0
0 0 j15
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Iab
0( )
Iab
1( )
Iab
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Zero-­‐sequence	
  
ZY	
  
Reference	
  
Posi;ve-­‐sequence	
  
 
+
Va
1( )
−
 Ia
1( )
ZY	
  
Reference	
  
Nega;ve	
  sequence	
  
 
+
Va
2( )
−
 Ia
2( )
ZΔ	
  
Reference	
  
 
+
Va
0( )
−
a	
  
 Ia
0( )
 j5 j5 j33
 
ZΔ
3
= ZY
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Our	
  concern	
  is	
  mostly	
  with	
  systems	
  that	
  are	
  essenOally	
  balanced	
  but	
  
which	
  become	
  unbalanced	
  upon	
  the	
  occurrence	
  of	
  an	
  
unsymmetrical	
  fault.	
  	
  
	
  
PracOcally	
  perfect	
  symmetry	
  is	
  an	
  idealizaOon,	
  but	
  	
  since	
  the	
  effect	
  
of	
  the	
  departure	
  from	
  symmetry	
  is	
  usually	
  small,	
  perfect	
  balance	
  
between	
  phases	
  is	
  ouen	
  assumed,	
  especially	
  if	
  the	
  lines	
  are	
  
transposed.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   72	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Consider	
  a	
  three-­‐phase	
  line	
  with	
  a	
  neutral	
  conductor.	
  The	
  self-­‐	
  
impedance	
  Zaa	
  is	
  the	
  same	
  for	
  each	
  phase	
  conductor,	
  and	
  the	
  
neutral	
  conductor	
  has	
  self-­‐impedance	
  Znn.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   73
Zaa
 Zaa
 Zaa
 Znn 
+
Van
_ 
+
Vbn
_
 
+
−
Vcn
 Ia
 Ib
 Ic
 In
 
+
V ′a ′n
_ 
+
V ′b ′n
_
 
V ′c ′n
+
−
a’	
  
b’	
  
c’	
  
n’	
  
a	
  
b	
  
c	
  
n	
  
 Zab
 Zab
 Zab
 Zan
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
When	
  currents	
  in	
  the	
  line	
  are	
  unbalanced,	
  then	
  the	
  neutral	
  
conductor	
  serves	
  as	
  a	
  return	
  path.	
  	
  
	
  
All	
  the	
  currents	
  are	
  assumed	
  posiOve	
  in	
  the	
  direcOons	
  shown	
  
even	
  though	
  some	
  of	
  their	
  numerical	
  values	
  may	
  be	
  negaOve	
  under	
  
unbalanced	
  condiOons	
  caused	
  by	
  faults.	
  	
  
	
  
Because	
  of	
  mutual	
  coupling,	
  current	
  flow	
  in	
  any	
  one	
  of	
  the	
  phases	
  
induces	
  voltages	
  in	
  each	
  of	
  the	
  other	
  adjacent	
  phases	
  and	
  in	
  the	
  
neutral	
  conductor.	
  	
  
	
  
Similarly,	
  current	
  in	
  the	
  neutral	
  conductor	
  induces	
  voltages	
  in	
  
each	
  of	
  the	
  phases.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   74	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
The	
  coupling	
  between	
  all	
  three	
  phase	
  conductors	
  is	
  regarded	
  as	
  
being	
  symmetrical	
  and	
  mutual	
  impedance	
  Zab	
  is	
  assumed	
  between	
  
each	
  pair.	
  The	
  mutual	
  impedance	
  between	
  the	
  neutral	
  conductor	
  
and	
  each	
  of	
  the	
  phases	
  is	
  taken	
  to	
  be	
  Zan.	
  
	
  
The	
  voltages	
  induced	
  in	
  phase	
  a,	
  for	
  example,	
  by	
  currents	
  in	
  the	
  
other	
  two	
  phases	
  and	
  the	
  neutral	
  conductor	
  can	
  be	
  treated	
  as	
  
sources	
  as	
  follows…	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   75	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Applying	
  Kirchhoff's	
  voltage	
  law	
  around	
  the	
  loop	
  circuit	
  gives…	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   76	
  
 
+
Van
_
 Ia
 In
 
+
V ′a ′n
_
a’	
  
n’	
  
a	
  
n	
  
 Zaa
 Znn
 
ZabIb
+ − 
ZabIc
+ − 
ZanIn
+ −
 
ZanIb
+ − 
ZanIc
+ − 
ZanIc
+ −
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Applying	
  Kirchhoff's	
  voltage	
  law	
  around	
  the	
  loop	
  circuit	
  gives:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   77	
  
 
Van = Zaa Ia + ZabIb + ZabIc + ZanIn   +V ′a ′n − ZnnIn + ZanIc + ZanIb + ZanIa( )  
 Zaa
 Zaa
 Zaa
 Znn 
+
Van
_ 
+
Vbn
_
 
+
−
Vcn
 Ia
 Ib
 Ic
 In
 
+
V ′a ′n
_ 
+
V ′b ′n
_
 
V ′c ′n
+
−
a’	
  
b’	
  
c’	
  
n’	
  
a	
  
b	
  
c	
  
n	
  
 Zab
 Zab
 Zab
 Zan
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Applying	
  Kirchhoff's	
  voltage	
  law	
  around	
  the	
  loop	
  circuit	
  gives:	
  
	
  
	
  
	
  
	
  
The	
  voltage	
  drop	
  across	
  the	
  line	
  is	
  
	
  
	
  
	
  
Similarly:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   78	
  
 
Van = Zaa Ia + ZabIb + ZabIc + ZanIn +V ′a ′n
− ZnnIn + ZanIc + ZanIb + ZanIa( )
 Van −V ′a ′n = Zaa − Zan( ) Ia + Zab − Zan( ) Ib + Ic( ) + Zan − Znn( ) In
 
Vbn −V ′b ′n = Zaa − Zan( ) Ib + Zab − Zan( ) Ia + Ic( ) + Zan − Znn( ) In
Vcn −V ′c ′n = Zaa − Zan( ) Ic + Zab − Zan( ) Ia + Ib( ) + Zan − Znn( ) In
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
When	
  the	
  line	
  currents	
  return	
  in	
  the	
  neutral	
  conductor:	
  
	
  
	
  
SubsOtuOng:	
  
	
  
	
  
	
  
	
  
	
  
	
  
Algebra…	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   79	
  
 In = − Ia + Ib + Ic( )
 
Van −V ′a ′n = Zaa − Zan( ) Ia + Zab − Zan( ) Ib + Ic( )− Zan − Znn( ) Ia + Ib + Ic( )
Vbn −V ′b ′n = Zaa − Zan( ) Ib + Zab − Zan( ) Ia + Ic( )− Zan − Znn( ) Ia + Ib + Ic( )
Vcn −V ′c ′n = Zaa − Zan( ) Ic + Zab − Zan( ) Ia + Ib( )− Zan − Znn( ) Ia + Ib + Ic( )
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Grouping	
  terms,	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
We	
  see	
  that	
  the	
  presence	
  of	
  the	
  neutral	
  conductor	
  changes	
  the	
  self-­‐	
  
and	
  mutual	
  impedances	
  of	
  the	
  phase	
  conductors	
  to	
  the	
  following	
  
effecOve	
  values:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   80	
  
 
Van −V ′a ′n = Zaa − 2Zan + Znn( ) Ia + Zab − 2Zan + Znn( ) Ib + Ic( )
Vbn −V ′b ′n = Zaa − 2Zan + Znn( ) Ib + Zab − 2Zan + Znn( ) Ia + Ic( )
Vcn −V ′c ′n = Zaa − 2Zan + Znn( ) Ic + Zab − 2Zan + Znn( ) Ia + Ib( )
 
Zs = Zaa − 2Zan + Znn
Zm = Zab − 2Zan + Znn
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Since	
  this	
  does	
  not	
  explicitly	
  include	
  the	
  neutral	
  conductor,	
  Zs	
  and	
  
Zm	
  can	
  be	
  regarded	
  as	
  parameters	
  of	
  the	
  phase	
  conductors	
  alone,	
  
without	
  any	
  self	
  or	
  mutual	
  inductance	
  being	
  associated	
  with	
  the	
  
return	
  path.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   81	
  
 
Van −V ′a ′n = ZsIa + Zm Ib + Ic( )
Vbn −V ′b ′n = ZsIb + Zm Ia + Ic( )
Vcn −V ′c ′n = ZsIc + Zm Ia + Ib( )
 
Va ′a
Vb ′b
Vc ′c
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
Van −V ′a ′n
Vbn −V ′b ′n
Vcn −V ′c ′n
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
Zs Zm Zm
Zm Zs Zm
Zm Zm Zs
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Now	
  express	
  the	
  a-­‐b-­‐c	
  voltage	
  drops	
  and	
  the	
  currents	
  on	
  the	
  line	
  in	
  
terms	
  of	
  their	
  symmetrical	
  components	
  with	
  phase	
  a	
  as	
  the	
  
reference.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   82	
  
 
Zs Zm Zm
Zm Zs Zm
Zm Zm Zs
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
Zs − Zm 0 0
0 Zs − Zm 0
0 0 Zs − Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
Zm Zm Zm
Zm Zm Zm
Zm Zm Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Remember	
  the	
  trick!	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   83	
  
 
Va ′a
0( )
Va ′a
1( )
Va ′a
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= Α
Va ′a
Vb ′b
Vc ′c
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
=
Zs − Zm 0 0
0 Zs − Zm 0
0 0 Zs − Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
Zm Zm Zm
Zm Zm Zm
Zm Zm Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎧
⎨
⎪⎪
⎩
⎪
⎪
⎫
⎬
⎪⎪
⎭
⎪
⎪
Α
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Va ′a
0( )
Va ′a
1( )
Va ′a
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= Zs − Zm( )Α−1
1 0 0
0 1 0
0 0 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α+ ZmΑ
−1
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
⎧
⎨
⎪
⎩
⎪
⎫
⎬
⎪
⎭
⎪
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
Va ′a
0( )
Va ′a
1( )
Va ′a
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
Zs + 2Zm 0 0
0 Zs − Zm 0
0 0 Zs − Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Now	
  define	
  zero-­‐,	
  posiOve-­‐,	
  and	
  negaOve-­‐sequence	
  impedances	
  as:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   84	
  
 
Va ′a
0( )
Va ′a
1( )
Va ′a
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
Zs + 2Zm 0 0
0 Zs − Zm 0
0 0 Zs − Zm
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
 
Z0 = Zs + 2Zm
= Zaa + 2Zab + 3Znn − 6Zan
Z1 = Z2
= Zs − Zm = Zaa − Zab
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   85	
  
 
Va ′a
0( )
Va ′a
1( )
Va ′a
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
Van
0( ) −V ′a ′n
0( )
Van
1( ) −V ′a ′n
1( )
Van
2( ) −V ′a ′n
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
=
Z0 0 0
0 Z1 0
0 0 Z2
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   86	
  
 Z0
 
+
Van
0( )
_ 
+
V ′a ′n
0( )
_
a’	
  
n’	
  
a	
  
n	
  
 Ia
0( )
 Z1
a’	
  
n’	
  
a	
  
n	
  
 Z2
a’	
  
n’	
  
a	
  
n	
  
 
+
Van
1( )
_
 
+
Van
2( )
_
 
+
V ′a ′n
1( )
_
 
+
V ′a ′n
2( )
_
 Ia
1( )
 Ia
2( )
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Because	
  of	
  the	
  assumed	
  symmetry	
  of	
  the	
  circuit	
  of	
  the	
  line,	
  we	
  see	
  
that	
  the	
  zero-­‐,	
  posiOve-­‐,	
  and	
  negaOve-­‐sequence	
  equaOons	
  decouple	
  
from	
  one	
  another,	
  and	
  corresponding	
  uncoupled	
  zero-­‐,	
  posiOve-­‐,	
  
and	
  negaOve-­‐sequence	
  circuits	
  can	
  be	
  drawn.	
  	
  
	
  
Despite	
  the	
  simplicity	
  of	
  the	
  line	
  model,	
  the	
  model	
  has	
  incorporates	
  
important	
  characterisOcs	
  of	
  the	
  sequence	
  impedances	
  which	
  apply	
  
to	
  more	
  elaborate	
  and	
  pracOcal	
  line	
  models.	
  We	
  note	
  that	
  the	
  
posiOve-­‐	
  and	
  negaOve-­‐sequence	
  impedances	
  are	
  equal	
  and	
  do	
  not	
  
include	
  the	
  neutral	
  –conductor	
  impedances	
  Znn	
  and	
  Zan	
  which	
  enter	
  
into	
  the	
  calculaOon	
  of	
  only	
  the	
  zero-­‐sequence	
  impedance	
  Zo.	
  	
  
	
  
Impedance	
  parameters	
  of	
  the	
  return-­‐path	
  conductors	
  enter	
  into	
  the	
  
values	
  of	
  the	
  zero-­‐sequence	
  impedances,	
  but	
  they	
  do	
  not	
  affect	
  the	
  
posiOve	
  or	
  negaOve	
  –sequence	
  impedance.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   87	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
Most	
  aerial	
  transmission	
  lines	
  have	
  at	
  least	
  two	
  overhead	
  
conductors	
  called	
  ground	
  wires	
  ,	
  which	
  are	
  grounded	
  at	
  uniform	
  
intervals	
  along	
  the	
  length	
  of	
  the	
  line.	
  	
  
	
  
The	
  ground	
  wires	
  combine	
  with	
  the	
  earth	
  return	
  path	
  to	
  consOtute	
  
an	
  effecOve	
  neutral	
  conductor	
  with	
  impedance	
  parameters	
  much	
  
like	
  Znn	
  and	
  Zan	
  but	
  which	
  depend	
  on	
  the	
  resisOvity	
  of	
  the	
  earth.	
  	
  
	
  
More	
  advanced	
  literature	
  shows	
  that	
  the	
  model	
  developed	
  here	
  is	
  
sOll	
  valid	
  with	
  the	
  numerical	
  parameters	
  adjusted	
  accordingly.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   88	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
In	
  deriving	
  the	
  equaOons	
  for	
  inductance	
  and	
  capacitance	
  of	
  
transposed	
  lines,	
  we	
  assumed	
  balanced	
  three-­‐phase	
  currents	
  and	
  
did	
  not	
  specify	
  phase	
  order.	
  Those	
  results	
  are	
  therefore	
  thus	
  valid	
  
for	
  both	
  posiOve-­‐	
  and	
  negaOve-­‐sequence	
  impedances.	
  	
  
	
  
When	
  only	
  zero-­‐sequence	
  current	
  flows	
  in	
  a	
  transmission	
  line,	
  the	
  
current	
  in	
  each	
  phase	
  is	
  idenOcal	
  .	
  The	
  current	
  returns	
  through	
  the	
  
ground,	
  through	
  over	
  head	
  ground	
  wires,	
  or	
  through	
  both.	
  
	
  	
  
Because	
  zero-­‐sequence	
  current	
  is	
  idenOcal	
  in	
  each	
  phase	
  conductor	
  
(rather	
  than	
  equal	
  only	
  in	
  magnitude	
  and	
  displaced	
  in	
  phase	
  by	
  120o	
  
from	
  other	
  phase	
  currents),	
  the	
  magneOc	
  field	
  due	
  to	
  zero-­‐
sequence	
  current	
  is	
  very	
  different	
  from	
  the	
  magneOc	
  field	
  caused	
  by	
  
either	
  posiOve-­‐	
  or	
  negaOve-­‐sequence	
  current.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   89	
  
SEQUENCE	
  CIRCUITS	
  OF	
  A	
  SYMMERTICAL	
  TRANSMISSION	
  LINE	
  
	
  
The	
  difference	
  in	
  magneOc	
  field	
  results	
  in	
  the	
  zero-­‐sequence	
  
inducOve	
  reactance	
  of	
  overhead	
  transmission	
  lines	
  being	
  2	
  to	
  3.5	
  
Omes	
  as	
  large	
  as	
  the	
  posiOve-­‐sequence	
  reactance.	
  	
  
	
  
The	
  raOo	
  is	
  toward	
  the	
  higher	
  porOon	
  of	
  the	
  specified	
  range	
  for	
  
double-­‐circuit	
  lines	
  and	
  lines	
  without	
  ground	
  wires.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   90	
  
Example	
  
	
  
The	
  terminal	
  voltages	
  at	
  the	
  leu-­‐hand	
  and	
  right-­‐hand	
  ends	
  of	
  a	
  line	
  
are	
  given	
  by:	
  
	
  
	
  
	
  
	
  
	
  
The	
  line	
  impedances	
  in	
  ohms	
  are:	
  
	
  
	
  
	
  
Determine	
  the	
  line	
  currents	
  Ia,	
  Ib,	
  and	
  Ic	
  using	
  both	
  symmetrical	
  
components	
  and	
  without	
  symmetrical	
  components.	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   91	
  
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence
Networks	
   92	
  
Example	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
	
  
Since	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   93	
  
 Ia
1( ) = Ia
2( ) = 0⇒ Ia = Ib = Ic = 262.5− j175
Example	
  
	
  
Without	
  using	
  symmetrical	
  components,	
  find	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   94	
  
Example	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   95	
  
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
Consider	
  a	
  synchronous	
  generator	
  (next	
  slide),	
  grounded	
  through	
  a	
  
reactor.	
  
	
  
When	
  a	
  fault	
  (not	
  shown	
  in	
  the	
  figure)	
  occurs	
  at	
  the	
  terminals	
  of	
  the	
  
generator,	
  currents	
  la,	
  lb,	
  and	
  Ic	
  flow	
  in	
  the	
  lines.	
  	
  
	
  
If	
  the	
  fault	
  involves	
  ground,	
  the	
  current	
  flowing	
  into	
  the	
  neutral	
  of	
  
the	
  generator	
  is	
  designated	
  In	
  and	
  the	
  line	
  currents	
  can	
  be	
  resolved	
  
into	
  their	
  symmetrical	
  components	
  regardless	
  of	
  how	
  unbalanced	
  
they	
  may	
  be.	
  
	
  
The	
  equaOons	
  developed	
  earlier	
  this	
  semester	
  for	
  the	
  idealized	
  
synchronous	
  machine	
  are	
  all	
  based	
  on	
  the	
  assumpOon	
  of	
  balanced	
  
instantaneous	
  armature	
  currents.	
  We	
  assumed	
  that	
  ia	
  +	
  ib	
  +	
  ic	
  =	
  0.	
  	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   96	
  
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
Consider	
  again	
  a	
  synchronous	
  generator	
  grounded	
  through	
  a	
  
reactor:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   97	
  
+ 
_ 
~ 
 Ean Zn
 Ib
 Ia
 Ic
 Ecn
 Ebn
 In
a	
  
b	
  c	
  
n	
  
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
When	
  a	
  fault	
  (not	
  indicated	
  in	
  the	
  figure)	
  occurs	
  at	
  the	
  terminals	
  o	
  f	
  
the	
  generator,	
  currents	
  la,	
  lb,	
  and	
  Ic	
  flow	
  in	
  the	
  lines.	
  	
  
	
  
If	
  the	
  fault	
  involves	
  ground,	
  the	
  current	
  flowing	
  into	
  the	
  neutral	
  of	
  
the	
  generator	
  is	
  designated	
  In	
  and	
  the	
  line	
  currents	
  can	
  be	
  resolved	
  
into	
  their	
  symmetrical	
  components	
  regardless	
  of	
  how	
  unbalanced	
  
they	
  may	
  be.	
  
	
  
Recall	
  the	
  equaOons	
  developed	
  for	
  the	
  idealized	
  synchronous	
  
machine	
  are	
  all	
  based	
  on	
  the	
  assumpOon	
  of	
  balanced	
  instantaneous	
  
armature	
  currents.	
  We	
  assumed	
  that	
  ia	
  +	
  ib	
  +	
  ic	
  =	
  0,	
  and	
  then	
  set	
  ia	
  =	
  
	
  –	
  (ib	
  +	
  ic)	
  in	
  order	
  to	
  arrive	
  at	
  the	
  terminal	
  voltage	
  of	
  phase	
  a	
  in	
  the	
  
form:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   98	
  
 
van = −Ria − Ls + Ms( ) diadt + ean
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
In	
  sinusoidal	
  steady-­‐state:	
  
	
  
Had	
  we	
  NOT	
  then	
  set	
  ia	
  =	
  –	
  (ib	
  +	
  ic)	
  then	
  we	
  would	
  have	
  found:	
  
	
  
	
  
	
  
In	
  steady-­‐state:	
  
	
  
	
  
Similarly:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   99	
  
 Van = −RIa − jω Ls + Ms( ) Ia + Ean
 
van = −Ria − Ls
dia
dt
+ Ms
d
dt
ib + ic( ) + ean
 Van = −RIa − jωLsIa + jωMs Ib + Ic( ) + Ean
 
Vbn = −RIb − jωLsIb + jωMs Ia + Ic( ) + Ebn
Vcn = −RIc − jωLsIc + jωMs Ia + Ib( ) + Ecn
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
In	
  matrix	
  form:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   100	
  
 
Van
Vbn
Vcn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= − R + jωLs( )
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+ jω Ms
0 1 1
1 0 1
1 1 0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
Ean
Ebn
Ecn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= − R + jω Ls + Ms( )⎡⎣ ⎤⎦
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+ jω Ms
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+
Ean
Ebn
Ecn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
Following	
  the	
  usual	
  procedure	
  we	
  express	
  the	
  a-­‐b-­‐c	
  quanOOes	
  of	
  
the	
  machine	
  in	
  terms	
  of	
  symmetrical	
  components	
  of	
  phase	
  a	
  of	
  the	
  
armature:	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   101	
  
 
Α−1
Van
Vbn
Vcn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= − R + jω Ls + Ms( )⎡⎣ ⎤⎦Α−1
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+ jω MsΑ
−1
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
Ib
Ic
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
+Α−1
Ean
Ebn
Ecn
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Van
0( )
Van
1( )
Van
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= − R + jω Ls + Ms( )⎡⎣ ⎤⎦
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ jω MsΑ
−1
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+Α−1
Ean
a2Ean
aEan
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
Α =
1 1 1
1 a a2
1 a2 a
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
1 1 1
1 e j240° e j120°
1 e j120° e j240°
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Since	
  the	
  synchronous	
  generator	
  is	
  
designed	
  to	
  supply	
  balanced	
  three-­‐phase	
  
voltages,	
  we	
  have	
  shown	
  the	
  generated	
  
voltages	
  Ean,	
  Ebn,	
  and	
  Ecn	
  as	
  a	
  
posi;ve-­‐	
  sequence	
  set	
  of	
  phasors.	
  
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine	
  
	
  
11/20/13	
   Symmetrical	
  Components	
  &	
  Sequence	
  Networks	
   102	
  
 
Van
0( )
Van
1( )
Van
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
= − R + jω Ls + Ms( )⎡⎣ ⎤⎦
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ jω MsΑ
−1
1 1 1
1 1 1
1 1 1
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Α
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+Α−1
Ean
a2Ean
aEan
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= − R + jω Ls + Ms( )⎡⎣ ⎤⎦
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+ jω Ms
3 0 0
0 0 0
0 0 0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ia
0( )
Ia
1( )
Ia
2( )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
+
0
1
0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
Ean
=
− R + jω Ls + Ms( )⎡⎣ ⎤⎦ Ia0( ) + j3ω MsIa0( )
− R + jω Ls + Ms( )⎡⎣ ⎤⎦ Ia1( ) + Ean
− R + jω Ls + Ms( )⎡⎣ ⎤⎦ Ia2( )
⎧
⎨
⎪
⎪
⎩
⎪
⎪
Sequence	
  Circuits	
  of	
  the	
  Synchronous	
  Machine