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

  1. dav_mt

    Thread Starter New Member

    Nov 14, 2004
    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 ?
    Thanks in advance