Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 219

And from Figure 9.1,

AB= 349. 91 × 2. 50 × 0. 9500 = 831. 04

BE= 349. 91 × 2. 50 × 0. 3123 = 273. 19

EF= 349. 91 × 2. 75 × 0. 3123 = 300. 51

DF= 349. 91 × 2. 75 × 0. 9500 = 914. 14

V

√

3 ( 831. 04 + 300. 51 ) 100

33700

= 5 .816%

Alternatively the volt-drop can be calculated by solving the circuit conditions shown in Figure 9.4,
as follows:-

By simple proportionsVrcan be found fromVsas follows,

Vr

Vs

=

ZL

R+Xl+ZL

=

51. 728 +j 17. 002

2. 5 +j 2. 75 + 51. 728 +j 17. 002

= 0. 94299 +j 0. 02995

Therefore,

Vr=( 0. 94299 +j 0. 02995 )( 19456. 7 )

= 18347. 47 +j 582. 73

And

|Vr|=18356.73 volts/phase

The total actual volt-drop

V =

|Vs|−|Vr|

|Vs|

× 100

=

( 19456. 7 − 18356. 73 )× 100

19456. 7

= 5 .653%

Figure 9.4 Equivalent simple circuit of a long cable. A long cable as a simple series circuit.