Handbook of Electrical Engineering

218 HANDBOOK OF ELECTRICAL ENGINEERING

The charging current at the sending endIcsis,

Ics=

Vs

2 Xc

=

19456. 7

−j 1061. 04

=+j 18 .337 amps

DeductIcsfromIsto obtainIsr,

Isr=Is−Ics= 317. 85 −j 80. 713 −j 18. 337

= 317. 85 −j 99. 05

The volt-dropVsrin the series impedance is,

Vsr=Isr(R+jX 1 )=( 317. 85 −j 99. 05 )( 2. 5 +j 2. 75 )

= 794. 63 +j 874. 09 −j 247. 625 + 272. 39

= 1067. 02 +j626.46 volts/phase

Hence the voltage received at the load is,

Vr=Vs−Vsr

= 19456. 7 − 1067. 02 −j 626. 46

= 18389. 68 −j 626. 46

|Vr|=18400.35 volts/phase

The total actual volt-drop

V=

|Vs|−|Vr|

|Vs|

× 100

=

( 19456. 7 − 18400. 35 ) 100

19456. 7

= 5 .429%

The receiving end volt-drop with respect to the nominal system voltage isVn,

Vn=

|Vn|−|Vr|

|Vn|

× 100

=

( 19052. 6 − 18400. 35 ) 100

19052. 6

= 3 .423%

c) Neglecting the shunt capacitive reactance.

The method of 9.4.3.1 can be used for a long cable to compare the results and accuracy
obtained. The current in the load based on the nominal system voltage isI,

I=

SL× 106

√

3 Vn

=349.91 amps/phase