Semiconductive Cable Screen End to End Resistance, Calculation Assistance

Seems to me that you could treat the problem as a wire having the same cross-section and resistivity as the sheath and go from there.

unless:

1. It is in contact with seawater,

2. It is carrying an AC signal, in which case skin-effect will drive up the impedance as a function of frequency and geometry.

Is this how you modeled it? As a wire?

I think i've treated as a wire using the data on good on wikipedia (http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity) . And converting the 25Ω/cm to the appropriate value for the x-sectional area.

AC is involved but at present a simple DC equiv value will suffice and then look into the skin effect from there.

My process is what is troubling me and the final value result (from my calc) while tangible just 'looks wrong'.

Been far too many years since I had to do anything close to this type of calc hence my trepidation with my results. Unfortunately despite searching fairly well I can't find a 'simple' description of how to perform this calc.

OK just reworked this as I found a units error. Would I be correct with the following based on the above data.

Final value = 9.218 MΩ (end to end resistance, not allowing any skin effects)

Just a note: resistivity is in units of Ω cm, not Ω/cm.

Look at the definition of resistivity:

Now do a dimensional analysis (use cm for linear measurements to jive with given units):

ρ = Ω * cm² / cm = ohm * cm * cm / cm = Ω cm

Treating the screen as an annular cylinder, where

t = thickness

r_o = OD/2 [outer radius]

r_i = ID/2 = r_o - t [inner radius]

A = πr_o² - πr_i² = π(r_o² - r_i²) [Area of annulus]

------

t = 0.7 mm

r_o = 14.66/2 mm = 7.33 mm

r_i = 7.33 - 0.7 = 6.63 mm

A = π * (7.33² - 6.63²) = 30.7 mm²

Converting to cm²,

30.7 mm² / 100 mm²/cm² = 0.307 cm²

----

Converting cable length, L, to cm,

L = 5800 m * 100 cm/m = 580000 cm

----

R = ρ * L / A

where

A = 0.307 cm²

ρ = 2.5 Ω-cm

L = 580,000 cm

and so

R = 2.5 Ω-cm * 580,000 cm / 0.307 cm²

R = 4,723,127 Ω or about 4.7 MΩ

Correction, ρ is not 2.5 Ω cm, but 25 Ω cm, so scale result by x10:

47 MΩ

Sorry 'bout that. My mistake.

Does this look reasonable to you?

Looks perfectly reasonable, thanks for clarifiying my math and the principle.

You did highlight another error I had made with respect to the area of the cross section. As I started with diameter I used Half D to give me the radius, but when allowing for the thickness I had halved this also, incorrectly.

Hence my 9.2MΩ aas opposed to your 47 MΩ which now I have amended my figures I concur with.

Thanks all for the input on this.

You're welcome! If you've any other questions, feel free to post them.

e2

Why do you want end to end values for a starter/ There is a simple calculation to measure and obtain the value of any semi con in XLPE. (Semi con screen over the XLPE insulation and for the inner core).. Simply refer to the IEC standards 60502, test method, and you will have the value for each core over the length of the cable.

IQ, this is a custom made cable and not sure it complies to the standard stated. It is for special use subsea and has gas / water blocking qualities but I will look at the standard and quiz the manufacturer accordingly.

End to end value is critical to the operation of our Line insulation monitor device. Also of concern to us is EM blocking which this will not provide as the cable also carries data.

We are just trying to understand the implications of the construction and relative values accordingly. Unfortuantely I have several 100's of km already manufactured and we need to input the effects into our system models.