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

1. ### lam58 Thread Starter Member

Jan 3, 2014
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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.

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:

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

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.

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

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,

where;

j_0 = current density
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.

Potential at sheath = 25.2 kV
VAr = 47.5 MVAr

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• ###### part 2b and c.png
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Last edited: Jan 4, 2015