Question

A 33 kV power cable 7 km long has a capacitance of 0.32 uF/km and a conductance associated with its insulation of 5 LS /km. If the cable is operating at rated voltage at a frequency of 50 Hz, calculate: (a) The value of tand associated with the cable (b) The power dissipated in the cable insulation

Answer 1

Given

Length = 7km

Capacitance = 0.32×10-6F/Km

Total capacitance (C)= 0.32×10-6×7=2.24×10-6 F

Coductance=5×10-6S/km

Total conductance (g)=5×10-6×7 = 35×10-6 S

Resistance R= 1/g= 1/[35×10-6]=0.0285714×106 ohm

a) tan$\delta$= 1/WRC

W= 2πf= 2×π×50= 314.16 rad/s

Tan$\delta$= 1/(314.16×0.0285714×106×2.24×10-6)=0.0497

$\therefore$ tan$\delta$= 0.0497

b) power dissipation = WCV​2$\delta$

$\delta$= tan$\delta$=0.0497 and V = Phase voltage = 33/√3 kV= 19.0525kv

powerdissipation = 314.26×2.24×10-6×19.05252×106×0.0497

= 12699.81 Watts

Power dissipation = 12699.81 W

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