Page 234 - Handbook of Electrical Engineering

218 HANDBOOK OF ELECTRICAL ENGINEERING

The charging current at the sending end I cs is,

19456.7
V s
I cs = = =+j18.337 amps
2X c −j1061.04
Deduct I cs from I s to obtain I sr ,

I sr = I s − I cs = 317.85 − j80.713 − j18.337
= 317.85 − j99.05

The volt-drop V sr in the series impedance is,

V sr = I sr (R + jX 1 ) = (317.85 − j99.05)(2.5 + j2.75)
= 794.63 + j874.09 − j247.625 + 272.39

= 1067.02 + j626.46 volts/phase

Hence the voltage received at the load is,

V r = V s − V sr
= 19456.7 − 1067.02 − j626.46
= 18389.68 − j626.46

|V r |= 18400.35 volts/phase

The total actual volt-drop

|V s |− |V r |
V = × 100
|V s |
(19456.7 − 18400.35)100
= = 5.429%
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 − 18400.35)100
= = 3.423%
19052.6
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 is I,

S L × 10 6
I = √ = 349.91 amps/phase
3V n