Page 232 - Handbook of Electrical Engineering

216 HANDBOOK OF ELECTRICAL ENGINEERING

iii) Hence the voltage across the capacitance is,

V c = V s − V sc
= 19456.7 + j0.0 − 508.35 − j336.28
= 18948.35 − j336.28 volts/phase

The charging current I c for the capacitance is,

18949.35 − j336.28
V c
I c = =
X c −j530.52
= 0.634 + j35.716 amps
Deduct I c from I s to find I r ,

I r = I s − I c = 317.92 − j80.691 − 0.634 − j35.716

= 317.286 − j116.41 amps

The volt-drop in the right-hand side of the cable is V cr ,
R X 1

V cr = I r + j = (317.286 − j116.41)(1.25 + j1.375)
2 2
= 556.671 + j290.756 volts/phase

iv) Hence the voltage received at the load is,

V r = V c − V cr
= 18948.35 − j336.28 − 556.671 − j290.756
= 18391.68 − j627.04 volts/phase
|V r |= 18402.36 volts/phase

The total actual volt-drop

|V s |− |V r |
V = × 100
|V s |
(19456.7 − 18402.36)100
= = 5.419%
19456.7
The receiving end volt-drop with respect to the nominal system voltage is V n ,

|V n |− |V r |
V n = × 100
|V n |
(19052.6 − 18402.36)
= 100 = 3.413%
19052.6