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G

Graham Coughlan

2 questions here

First one... I know theres a formula you can use to work out lengths of cables and the csa. I remember it being something like pl/csa but i cant remember the rest so could someone please help me remember this

Second question...

N = E x A / F x UF x MF

I know all the meanings to this formula apart from F so again someone please help with that

Cheers
 
First one

R= rho l/A
resistance(ohm) = resistivity (ohm meter) x length (metre) over cross sectional area (metre squared)

or
L = R S 27.6
Length (m) = resistance (ohm) x cross sectional area (milimetre squared) x 27.6
for copper

no idea about the second one
 
Last edited:
over cross sectional area (metre squared)

bloody big cable, that is ! LOL.
 
so far ive for N = 500 x A / F x 0.6 x 0.8 - I still dont know what F is tho... (Ive got to work out the area still)

The question ive got to answer is....

The designer has requested the sales area at a bench level to be illuminated to a value of 500 lux reapectively as fitted with T8 single flourescent luminaires.

Assuming the MF is 0.8 and the UF is 0.6 demostrate through calculation the number of fittings required.
 
Graham, this has gone beyond a joke! You’ve gone from cable to lighting calculations. For gods sake give people a clue.
UF = utilisation factor
MF = maintenance factor
Not done lighting calculations since I was an apprentice, but can still remember them.

Better still just scan the entire test and hope someone is daft enough to do it for you!
 
I think i will stick to being thick.

I have only just got my head back around the adiabatic equation after many years of not using it, and that is only cos it can crop up in the 17th edition exam, so i need to be able to explain it to the fare paying public:banghead:
 

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