Page 233 - Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 217

b) The ‘Pye’ equivalent circuit.

The single series element is,
R + jX 1 = 2.50 + j2.75 ohms/phase

The two-shunt elements are,
2X c =−j1061.04 ohms/phase

The solution sequence.
i) Calculate the total impedance seen by the sending end voltage.
ii) Calculate the total sending end current.
iii) Calculate the sending end shunt current.
iv) Calculate the receiving end voltage.

i) The parallel combination of the load impedance and the right-hand side shunt capacitive reac-
tance is,

(R L + jX L )2X c
Z 4 =
R L + jX L + 2X c
(51.728 + j17.002)(−1061.04)
=
51.728 + j17.002 − j1061.04
= 53.296 + j14.638 ohms/phase

Z 4 is connected in series with the series impedance of the cable, hence their total is,

Z 5 = Z 4 + R + jX 1
= 53.296 + j14.638 + 2.50 + j2.75
= 55.795 + j17.388 ohms/phase

Z 5 is connected in parallel with the left-hand side shunt capacitive reactance, hence this total Z 6 is,

(55.795 + j17.388)(−j1061.04)
Z 5 2X c
Z 6 = =
Z 5 + 2X c 55.795 + j17.388 − j1061.04
= 57.506 + j14.603 ohms/phase

This impedance is seen by the sending end phase voltage V s , hence the sending end current I s is,
33700.0
V s
I s = = √
Z 6 3(57.506 + j14.603)
19456.7(57.506 − j14.603)
=
3520.19
= 317.85 − j80.713 amps

|I s |= 327.93 amps