Let me take a go at this....feel free to correct me anyone...
PSCC = Eb / Rbbr , where Eb is the open circuit voltage of the batteries, and Rbbr is the total resistance of the network in ohms incl. internal resistance of battery and resistance of conductors.
So...
Lets calculate Eb first of all ;
Eb = 1.05 x Number of cells x 2V (page 272 in your Sans 10142)
= 1.05 x 100 x 2
= 210V
Next lets try to calculate Rbbr ;
Rbbr = (0,9 x Resistance of the battery) + Resistance of the battery connections + Resistance of the conductors (page 272/3 Sans 10142)
= (0,9 x 0,011 x 100 (there's 100 cells in this battery) ) + 1 + 2(x/1000 x 100m) - this is because there are 2 conductors and you need to find the Resistance value using Table E1 the 3rd column for DC circuits on page 307 of Sans 10142
= 1,99 + 1 + 2(x/10) simplify the equation in brackets - 1000 / 100 = 10
= 2,99 + 2(x/10) -- NOTE : we cannot find the resistance using table E1 as yet, as we do not know the size of the conductors -- this is what we are asked to calculate in the question
Using the information we now have, we substitute it into the original equation and solve....
PSCC = Eb / Rbbr
therefore,
95,522 = 210 / (1,99 + 2(x/10))
therfore,
1,99 + 2(x/10) = 210 / 95,522
1,99 + 2(x/10) = 2,198
2(x/10) = 2,198 - 1,99
2(x/10) = 0,208
x/10 = 0,208 / 2
x/10 = 0,104
x = 1,04 ohms
Now, using Table E1 on page 307 of Sans 10142, we look up the value of 1,04 ohms in the copper conductor column, which falls between 16mm and 25mm conductor size,
therefore,
we should choose the 25mm cable to be safe.
I hope this answers your question.
Regards
Simon
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