another 2391 question

[wrighty]

regarding volts drop

i have an equation of

mv x ib x l / 1000

so what is the equation

(r1 + rn) x cf x ib
answer x 100 / 230

will somebody please clear this up

 

[JUD]

Voltage = Current x Resistance (V = I x R)

r1 + rn is the resistance of your circuit at 20°C.

cf is the correction factor for conductor operating temperature. As resistance increases with temperature you need to adjust it for 70°C. You do this by multiplying your measured resistance by 1.2

You then multiply the new value by your design current which gives you the voltage drop on the circuit at 70°C.

 

[wrighty]

so which eqution do i use to calculate volts drop or does it depend on the question?

 

[JUD]

Both. It's up to you. They will both give you the same result.

The mV/A/m values from your first equation are calculated using the other equation.

mV/A/m = (r"1 + r"n) x 1.2 where r"1 = resistance per meter of line conductor and r"n = resistance per meter of neutral conductor.

So in the case of 2.5mm² you would get...

r"1 = 7.41mΩ per meter
r"n = 7.41mΩ per meter
r"1 + r"n = 14.82mΩ per meter
(r"1 + r"n) x 1.2 = 17.78 mΩ per meter (rounded up to 18) which is the value in BS 7671