Page 231 - Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 215

The solution sequence.

i) Calculate the total impedance seen by the sending end voltage.

ii) Calculate the total sending current.
iii) Calculate the voltage at the centre of the cable, which supplies the shunt capacitance.
iv) Calculate the voltage at the load.

i) The impedance Z 1 to the right-hand side of the shunt reactance is,

R X 1
Z 1 = + j + R L + jX L
2 2
= 1.25 + j1.375 + 51.728 + j17.002 ohms/phase
= 52.978 + j18.377 ohms/phase

Z 1 is connected in parallel with X c and so their total impedance is Z 2 ,which is,

(52.978 + j18.377)(j530.52)
Z 1 .X c
Z 2 = =
Z 1 + X c 52.978 + j18.377 − j530.52
= 56.246 + j13.218 ohms/phase
Z 2 is connected in series with the left-hand side series impedance; hence their total is,

R X 1
Z 3 = Z 2 + + j
2 2
= 56.246 + j13.218 + 1.25 + j1.375
= 57.496 + j14.593 ohms/phase

ii) 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 3 3(57.496 + j14.593)
19456.7(57.496 − j14.593)
=
3518.75
= 317.92 − j80.691 amps

|I s |= 328.00 amps

The volt-drop in the left-hand side of the cable is V sc ,

R X 1

V sc = I s + j = (317.92 − j80.691)(1.25 + j1.375)
2 2
= 508.35 + j336.28 volts/phase