214 HANDBOOK OF ELECTRICAL ENGINEERING
is 33,700 V. The specific resistancer, inductive reactancexland capacitancecare 0.100 ohm/km,
0.110 ohm/km and 0.24 microfarad/km respectively at 90◦C and 50 Hz. Find the percentage volt-drop
at the receiving end.
The cable.
The series impedance is:-
R=r.l= 0. 100 × 25 =2.50 ohms/phase
Xl=xl.l= 0. 110 × 25 =2.75 ohms/phase
C=c.l= 0. 24 × 25 = 6. 00 μF/phase
Xc=
106
2 πf C
=
106
2 π× 50 × 6. 0
=530.52 ohms/phase
The load.
The system nominal voltageVnis 33,000 volts.
The line currentIrreceived at the load is,
Ir=
Load MVA× 106
√
3 line voltage
=
SL× 106
√
3 Vn
=
20. 0 × 106
√
3 × 33000. 0
=349.91 amps/phase
The load star connected impedanceZLis,
ZL=
Vn
√
3 Ir
=
33000
√
3 × 349. 91
=54.45 ohms/phase
The resistive componentRLis,
RL=ZLcos Ø= 54. 45 × 0. 95 =51.728 ohms/phase
The inductive componentXLis,
XL=ZLsin Ø= 54. 45 × 0. 3123 =17.002 ohms/phase.
a) The ‘Tee’ equivalent circuit.
The two series elements are,
R
2
+j
X 1
2
= 1. 25 +j1.375 ohms/phase
The single shunt element is
Xc=−j 530 .52 ohms/phase