Reactive Power on a Cable, please help.

Discussion in 'Homework Help' started by lam58, Jan 4, 2015.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    So I'm stuck on part c of this question (attached in pics below), and to be honest I'm not entirely sure if I've even the first two parts right. Could someone please have a look at what I've done and/or help me with part c.

    My answer to part a:

    Using permittivity of free space epsilon_0 = 8.8x10^-12;

    Capacitance of XPLE layer:

     C = \frac{2\pi * 2.2 * 8.8x10^{-12}}{ln(0.071/0.048)} = 310.73x10^{-12} F/m

    Capacitance of HDPE:

     C = \frac{2\pi * 2.4 * 8.8x10^{-12}}{ln(0.0789/0.0739)} = 2.03x10^{-9} F/m

    Answer to part b:

    To find the potential I assumed the charge between the outer layer of the cable is equal to negative the charge on the inner surface of the sheath i.e. +'ve Q (cable) = -'ve Q (sheath). Thus that implies if the sheath is ungrounded the charge on the outer layer of the sheath should be equal to +'ve Q. Hence I can find the potential on the sheath.

     Q = CV = 310.73x10^{-12} * \frac{220}{\sqrt{3}} = 3.95x10^{-5} C

    Then to find potential I use:

     v = \frac{Q}{2\pi\epsilon_0\epsilon_r}.ln(b/a)

    \Rightarrow v = \frac{310.73x10^{-12}}{2\pi * 8.8x10^{-12} * 2.4}.ln(0.0789/0.0739) = 19.5 kV

    Which implies the potential on the lead sheath = \frac{-220}{\sqrt{3}} kV + 19.5kV = -107.5kV

    For part c I'm lost. I guess perhaps I could find the total current flow as described by current density i.e:

     I = j_0 d 2 \pi r = I j_0,<br />
<br />
where;<br />
<br />
j_0 = current density<br />
d = skin depth

    Then I could find the rms power loss using:

     P_{rms} = j_0^2 \frac{2 \pi r d}{2 \sqrt{2} \sigma} = P j_0^2

    This way I can find the apparent a.c resistance using:

     \frac{P_{rms}}{I^2} = R_{ac} ohms/m

    At this point I'm not sure what to do. Could I find the inductance here? Am I even on the right track?

    EDIT/UPDATE: I solved it, I was doing it completely wrong but I've found the right way to do it now.
    My new answers

    Potential at sheath = 25.2 kV
    VAr = 47.5 MVAr
    Last edited: Jan 4, 2015