413
6
F
2
S
0
8
3
4
Therefore, load current IL is:
IL = (Source voltage)/(A s/s back imp Line imp B s/s impedance)
=
(150kV/ 3 )/(0.94
×
75 + 16.8
×
(0.0197
2
+ 0.2747
2
) + 0.94
×
75)
=
594.7A
3.2
Minimum fault current
The minimum fault current Ifmin on a protected transmission line is the current of the phase to
earth fault on the nearest remote terminal.
To calculate Ifmin, zero sequence earth fault current (Io), positive sequence earth fault current
(I1) and negative earth fault current (I2) are calculated as follows:
I0 = I1 = I2 = (Source voltage)/{(Back impedance of A s/s)
+ (Transmission line zero sequence impedance)
+ (Transmission line positive sequence impedance)
×
2*}
=
(150kV/ 3 )/{(0.94
×
75) + 16.8
×
(0.4970
2
+ 1.4387
2
)
+
2
×
16.8
×
(0.0197
2
+ 0.2747
2
) }
=
822.28A
So,
Ifmin= I0 + I1+ I2 = 3
×
822.28 = 2.47kA
*Note:
Assuming that positive sequence impedance = negative sequence impedance.
B s/s
Line length: 16.8km
A s/s
M
Earth fault
GRZ100
www
. ElectricalPartManuals
. com
Summary of Contents for GRZ100-211B
Page 323: ... 322 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 343: ... 342 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 383: ... 382 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 395: ... 394 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 411: ... 410 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 423: ... 422 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 443: ... 442 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 451: ... 450 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 459: ... 458 6 F 2 S 0 8 3 4 w w w E l e c t r i c a l P a r t M a n u a l s c o m ...
Page 463: ...w w w E l e c t r i c a l P a r t M a n u a l s c o m ...