# POWER CABLES Problem

Discussion in 'Homework Help' started by dav_mt, Jun 20, 2005.

1. ### dav_mt Thread Starter New Member

Nov 14, 2004
4
0
A 400KV 50Hz Cable is 6km long. If the cable capicitance is 326nF per phase per km what are;
i) the charging current in each phase
ii) the total reactive power generated by the cable
iii) by deriving the formula for the maximum current that can be transmitted at a power factor angle φ find the maximum active power that could be transmitted at a power factor of 0.85, if the rating of the cable is 1000MVA.

as for (i) i did the following
found Xc = 9764.11ohms(for 1km)
(for 6km) Xc = 58584.66ohms
Ic = 400kv/9764.11ohms = 40.966A(1km)
Ic = 400kv/(6*9764.11ohms) = 6.828A (6km)

(ii) Q = V^2/Xc = 2.731MVAR

(iii) For part 3 i derived the formula for maximum current

Imax = sin φ * Ic + (Ic^2*sin φ ^2 - Ic^2 + Irat^2)^1/2

after I used S = √ 3 * Vl * I max
getting a value for I max of 833.33A
and to find the power P = √ 3 * Vl * Imax * cos φ

However i cant understand the link, why has the formula been derived when you can use the equation of Power ?

Can someone guide me if I m working out this problem correctly ?