Capacitance of a High Voltage cable.

Discussion in 'Homework Help' started by lam58, Dec 29, 2014.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    I haven't done this in a while so I would be grateful if someone could check my answer. The question states a high voltage single phase cable consists of a:

    i) Central copper conductor of diameter 4.8 cm.
    ii) Insulating layer of XPLE of thickness 2.3 cm and relative permittivity k = 2.2.
    iii) Conducting lead sheath of thickness 2.9 mm.
    iv) Insulating HDPE layer of thickness 5 mm and k = 2.4.
    v) Outer coating negligible thickness in contact with soil.

    Find the capacitance of the two insulating layer per meter of cable length?

    My attempt:

    Permittivity of free space epsilon_0 = 8.8x10^-12;

    Capacitance of a cylinder:   C = \frac{2\pi k \epsilon_0}{ln(b/a)}

    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

    The reason I'm unsure is that it seems too simple for what the question asks. Have I done this right of am I missing something.
    Last edited: Dec 29, 2014