Download Introduction to Distance Relaying

Introduction to Distance Relaying
Consider a balanced (transposed) transmission line (Fig 1)
 Va   Z s Z m Z m   I a 

 
  
 Vb    Z m Z s Z m   Ib 
 Vc   Z m Z m Z s   I c 
1 1 1 


Let T =  1 a 2 a  (sequence transformation matrix)

2
1 a a 
Then V abc   T  V 012 




abc
Similarly, I   T   I 012 




Applying sequence transformation matrix,
V 012   T 1  Z abc  T  I 012 



 

V 012   diag  Z0 , Z1, Z2   I 012 




Where Z0 = Zs + 2Zm ; zero sequence impedance
Z1 = Zs - Zm ; positive sequence impedance
Z2 = Zs - Zm ; negative sequence impedance
V0  Z0 I 0 ;
V1  Z1I1;
V2  Z 2 I 2 ;
Now let a 3   fault occur at the percentage (%) distance, x of the line (Fig 2)
fig2.eps
Then the fault model is given by
Va  1
 Zs Zm Zm   Ia 
  

  
Vb   1 Vn  x  Z m Z s Z m   Ib 
Vc  1
 Z m Z m Z s   I c 
For a solid 3   fault, Vn = 0
V abc   x  Z abc   I abc 





In sequence domain
V abc   x  Z abc   I abc 





012
V   x diag (Z0 , Z1 , Z 2 )  I 012 




For a 3-phase fault
Ib = a2Ia
Ic = aIa
By equation (4), we get
I0 = 0; I1 = Ia; I2 = 0;
(1)
(2)
(3)
(4)
So only positive sequence network is excited.
V1
 xZ1
(5)
I1
This way, using equation (5), we can locate the 3   fault on a transmission line.
Because, for the case of 3   fault
Va Vb Vc V1


  xZ1
(6)
I a I b I c I1
It follows that, a relay which monitor line current and phase voltages can locate 3  
fault by use of equation (6). In absence of fault Ia, Ib and Ic are much smaller in
magnitude. Thus, the apparent impedance seen by the relay is much higher. Hence, a
simple logic to locate 3   fault is provided by equation (6)
It can be easily seen that for 3   fault
Va  Vb Vb  Vc Vc  Va


 xZ1 [prove yourself]
(7)
I a  Ib Ib  I c I c  I a
Thus equation (7) provides an alternate way of locating 3   fault. Note that per unit
distance to fault is given by ration of apparent impedance seen by relay to the positive
sequence impedance of the relay.
L-L Fault:
Consider a bolted L-L fault on phase b-c of the system.
{fig3.eps}
System is considered unloaded for simplicity. Then the governing equations in 3-phase
given by
 Va 
Zs Zm Zm   0 



 

(8)
 Vb   x  Z m Z s Z m   Ib 
 Vc 
 Z m Z m Z s    Ib 
where
Va  Va  V fa
Vb  Vb  V fb
Vc  Vc  V fc
Further
V fb  V fc  V
Simplifying equation (8), we get
Va  0
Vb  x(Z s  Z m )( Ib )  xZ1Ib
Vc   x(Z s  Z m )( Ib )   xZ1Ib  xZ1I c
Subtract equation (11) from equation (10)
Vb  Vc  Vb  Vc  xZ1 ( Ib  I c )
(9)
(10)
(11)
Vb  Vc
 xZ1
(12)
Ib  I c
From equation (7) and equation (12) we conclude that a relay input configured as per
equation (7) can measure both 3-phase fault and L-L fault.
For a-c L-L fault
Vc  Va
 xZ1
Ic  I a
And for a-b L-L fault
Va  Vb
 xZ1
I a  Ib
Therefore, traditionally the distance relays are configured as per equation (7) to detect
and locate both L-L and 3-phase faults. Note that thus configuration enables to locate
fault exactly to per unit distance x , of the positive sequence impedance (Z1). This is the
fundamental principle of distance relaying.
S-L-G Fault:
We now derive the governing equation for S-L-G Fault case. Consider a-g fault on a
unloaded transmission line at a per unit distance x.
{fig4.eps}
Then 3-phase model is given by
 Va 
Zs Zm Zm   Ia 



  
 Vb   x  Z m Z s Z m  0 
 Vc 
 Z m Z m Z s  0 
Then governing equations is given by
Va  xZ s I a
(13)
V
Thus, ratio a equals xZ s and not xZ1 . Since, a distance relay inputs are configured
Ia
subjected to irrespective of fault. It always measure a fault impedance of xZ1 . It follows
that we should modify equation (13) suitably.
Va  xZ s I a
Va  x  Zs  Zm  Zm  I a
Va  x  Z1  Zm  I a
  Z 
 xZ1 1   m   I a
  Z1  
Va
 xZ1
 Zm 
Ia  
 Ia
 Z1 
Z 0  Z s  2Z m
Z1  Z s  Z m
(14)
(15)
 Z  Z1 
Zm   0
(16)

 3 
Substituting in equation (15) we get,




Va

  xZ
(17)
1

 Z 0  Z1  I a 
 I a  
  
 Z1  3 

I
Since I0 = a
3
Equation (17) can be written as
Va
 xZ1
(18)
I a  mI 0
Where m is called compensation factor for zero sequence current. Similarly, it can be
shown for b-g and c-g faults.
Vb
 xZ1
(19)
I b  mI 0
and
(20)
Vc
 xZ1
(21)
I c  mI 0
It is clear that traditionally the ground fault relays, require a different configuration. Form
phase fault (3-phase and L-L relaying)
A L-L-G can be detected by the S-L-G relay equations. There are total of 10 fault types to
be protected in the system, which are
(1) 3-phase fault
1
(2) L-L faults
3
(3) S-L-G faults
3
(4) L-L-G faults
3
The relaying units configured by equation (7) and (18,19,20) do this job satisfactorily.
Remarks: We had so far assumed bolted and unloaded faults. Therefore, there would be
errors introduced when the fault has some impedance Z f . Hence the apparent impedance
seen by the relay will not exactly lie on transmission line impedance AB, but would lie in
a region shown by trapezoid in figure as shown. Note that arcing faults are primarily
resistive in nature.