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PostPosted: Thu Mar 31, 2011 13:00 
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Pete317 wrote:
I think the cross-sectional area has nothing to do with the power loss calculation, which is a simple matter of applying Ohm's Law.
However, it probably has a lot to do with how hot the cable gets - but that's getting a bit beyond my comfort zone.
You make a good point there Pete. The cable construction is far from simple and is often/usually oil filled. So I think this is a theoretical question more than practical.

:bunker:

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PostPosted: Thu Mar 31, 2011 14:27 
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Steve wrote:
What is the recognised written short-hand notation for '(100 millimetres) squared' and '100 square millimetres' ?


100mm^2 or 100 square millimetres means a square of side 10mm. 100 millimetre square means a square of side 100mm or 10,000 square millimetres.
It can a source of confusion and the fact that the original question gets the units for resistivity wrong makes you doubt anything they say.
But there was a rule of thumb - 1.5 amps/mm^2 for ally or 2.5 amps/mm^2 for copper - which suggests that 100 mm^2 is correct.

100mm square is a big expensive cable. Another obvious rule of thumb is that the cost of the lost power should not exceed the cost of servicing the debt incurred to buy and string the cable.

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PostPosted: Thu Mar 31, 2011 15:01 
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Guy who needed help wrote:
When i tried, i also did not see the purpose of the cross sectional area, which is what threw me. This guy seems to know what he's talking about though! im pretty sure our tutor has never metioned R = pL/A, and doubling the resistance because its single phase also goes straight through me, but thanks so much for for your time and i finallly have an answer!


Thanks again everyone.

"DCB knows what he's talking about shocker". :lol: :clap:


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PostPosted: Thu Mar 31, 2011 16:20 
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Steve wrote:
Toltec wrote:
Steve wrote:
R = rho * L/A = 0.03e-6 * 1 / (0.1 * 0.1) = 3e-6 ohms per meter.
For 10km = 0.03 Ohms



Or

R = rho * L/A = 0.03e-6 * 1 / (0.1 * 0.1) = 3e-4 ohms per meter.
For 10km = 3 Ohms

:)

I think I have to disagree with you Toltec.

Not only have you multiplied the 'mantissa' (?) by 100, you also adjusted the 'exponent' by that factor too.


Oops - I blame sunlight on the phone screen...

:)

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PostPosted: Thu Mar 31, 2011 16:51 
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Toltec, who says two wrongs doesn't make a right? Your final answer (for the net resistance) was correct anyway :lol:

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PostPosted: Thu Mar 31, 2011 17:24 
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Toltec wrote:
Oops - I blame sunlight on the phone screen...

:)


You can have 100 lines too, you whippersnapper! :listenup: :whip: :stirthepot:

I knew the cable is expensive - and pikeys keep nicking it.
Better to put a genny on site! :hehe:

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PostPosted: Tue Apr 05, 2011 00:19 
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Johnnytheboy wrote:

Still don't know why we need to know the X-sectional area of the cable or that it's AC tho'


Because they want to know the student can pick out the relevant information and discard the irrelevant.... I guess (or at least that's what I always thought when I did Physics A level).


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