Let
Z3 approximately equal 0.1 Z1
Z2 approximately equal 100 Z1
Z3 approximately equal 0.1 Z1
Z2 approximately equal 100 Z1
Case 1
Normally current flow (I ) can be calculated from the relation
Normally current flow (I ) can be calculated from the relation
I = V / ( Z1 + Z2 )
= V / 101 Z1
from the above equation it seems that load current (I ) is absolutely low.
Case 2
When s1 is switched on then equivalent impedance of Z2, and Z3 is equal to Z2 parallel with Z3 approximately equal to Z3 , and in this case current (If ) can be calculated from the formula
When s1 is switched on then equivalent impedance of Z2, and Z3 is equal to Z2 parallel with Z3 approximately equal to Z3 , and in this case current (If ) can be calculated from the formula
If = V / ( Z1 + Z3 )
= V / 1.1 Z1
Which is very high compared with the current obtained from case 1 , Transmission line healthy and faulted cases is similar to the above cases.
Where :
I .. The normal (load ) current
If .. The fault current
I .. The normal (load ) current
If .. The fault current
More details are given in the next example which is shown in figure (21).
For a transmission line let the series impedance Z1 is divided into n equal segments and the switch S1 can be switched into any segment which is an indication of fault location.
For near faults
Total impedance (ZT ) can be calculated from the formula
Total impedance (ZT ) can be calculated from the formula
ZT = Z3 = 0.1 Z1
Fault in mid point of Transmission Line
ZT = Z3 + Z1 / 2 = 0.6 Z1
Fault at Line End
ZT = Z3 + Z1 = 1.1 Z1
From the previous cases, the total impedance (ZT ) is proportional to location (distance ) of fault , Which is the idea of working distance relays.