Electrical Power Systems :
Concept 2 (Relationship between SIL and Reactive Power Compensation) :
SIL (Surge Impedance Loading) :

Due to reactive power being zero, a transmission line is completely transmitting active power. This is
measured in watts and is called SIL.

SIL of a line is the power delivered by the line to a purely resistive load equal to the surge impedance of
the line.
3-  line supplying to a resistive load Z L  Z s
Resistive load   00
P  3 VL I L cos 

P  3 VL 
VL
3 Zs

VL  linevoltage
VL2
 SIL
Zs
Z s  Surge impedance 
L = Inductance of line/phase

P
L
C
C = Capacitance of line/phase
In case of transmission line, the loading of the line may be more than SIL. Under this condition, VR  Vs
.

If the loading is less than SIL, VR  Vs . If loading = SIL then VR  Vs .
Case 1 Without Compensation : Let SIL = SIL1 ,
Z s  Z s1
where Z s1 
L
V2
and SIL1  s
C
Z s1
Case 2 With Shunt Capacitor Compensation ( Csh ) {generally employed for power factor correction}
SIL2  SIL1 1  K Csh
and
Zs2 
where K Csh is per unit Csh compensation
Z s1
1  K Csh
Conclusion : (i) Z s 2  Z s1
(ii) SIL2  SIL1
Case 3 With Shunt Inductor Compensation ( Lsh ) {generally employed to avoid Ferranti effect}
SIL3  SIL1 1  K Lsh
and
Zs3 
Z s1
1  K Lsh
where K Lsh is per unit Lsh compensation
Conclusion : (i) Z s 3  Z s1
(ii) SIL3  SIL1
Case 4 With Series Capacitor Compensation ( Cse ) {generally employed to maintain stability of the system}
SIL4 
and
SIL1
1  K Cse
where K Cse is per unit Cse compensation
Z s 4  Z s1 1  K Cse
Conclusion : (i) Z s 4  Z s1
(ii) SIL4  SIL1
Case 5 With Series Inductor Compensation ( Lse ) {generally employed to maintain ripple free current
waveform}
SIL5 
and
SIL1
1  K Lse
where K Lse is per unit Lse compensation
Z s 5  Z s1 1  K Lse
Conclusion : (i) Z s 5  Z s1
(ii) SIL5  SIL1
Case 6 With Series Capacitor and Shunt Inductor Compensation ( Cse , Lsh )
SIL6 
SIL1 1  K Lsh
1  K Cse
&
Z s 6  Z s1
1  K Cse
1  K Lsh
Conclusion : Z s 6 , SIL6 depends on values of Cse , Lsh .
Case 7 With Series and Shunt Capacitor Compensation ( Cse , Csh )
SIL7 
SIL1 1  K Csh
1  K Cse
Conclusion : (i) Z s 7  Z s1
&
Z s 7  Z s1
1  K Cse
1  K Csh
(ii) SIL7  SIL1
Example 1 :
A loss less transmission line having surge impedance loading (SIL) of 2280 MW is provided with a
uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated
transmission line will be
[GATE
2008,
IIScBangalore]
(A) 1835 MW
(B) 2280 MW
Ans.
(C)
Sol.
Question is in accordance with case 4,
SIL4 
(C) 2725 MW
(D) 3257 MW
SIL1
1  K Cse
Therefore K Cse  30 %  0.3 p.u. and SIL1  2280 MW
We get
SIL4 
2280
 2725.12 MW
1  0.3
Hence, the correct option is (C).
Ans.