Formula for Cable Cross-Section Calculation: Emergency Lighting, Fireproof Cables, 437W Power

Hello everyone,
I was looking carefully at all the topics related to the selection of cabling for the installation. However, I cannot translate this into the case that I have to calculate.

- lighting installation emergency, fireproof cables
- the necessary formula to calculate the min. cable cross-section S
- I have the following data: power of all receivers in the circuit P = 437W, Un = 230V, In = 1.88A, circuit length L = 130m, copper -> conductivity 58 S * m / mm2, max. permissible voltage drop in the circuit dU = 3%

I will have a few such calculations to do, it is mainly about the formula and a short comment on why the cable cross-section is selected in such a way. The case is divided into sections of 1.5-2.5 mm2. However, I have to justify it and present the relevant calculations.

One of the forum members advised me to calculate it from the formula below, but I don't know where I'm making a mistake

R = [ro] xl / S
[ro] - by this I mean the resistivity for copper 1.7 ? 10-8

after conversion:
S = ([ro] xl) / R

R = U / I -> R = 230 / 1.88 = 122.34

so: S = (1.7 ? 10-8 x 130) / 122.34 = 1.80 x 10-8 mm2 ????? that is nonsense; /


Colleagues, I am asking for help and possibly some materials.

Once again:
- cable length 130 x 2 = 260 m
with the formula R = U / I you calculated the resistance for which the current 1.88A will flow, i.e. what resistance should you put so that the current 1.88A will flow -> here is the basic error.
It is necessary to answer the question what can be the maximum resistance of the cable so that there is a decrease of no more than 3% on it, i.e. 230/100 * 3 = 6.9V. This is how the calculations should be. Good luck

Try to calculate again Link

I'm gonna be a total fool, but I just don't get it

kreslarz wrote:

Once again:
- cable length 130 x 2 = 260 m

why times 2 ???


buddy "sq9jjh" I got to this article, the formula R = [ro] xl / S coincides with the information on that page. but I still don't understand it, sorry; /

hakerit1 wrote:

I'm gonna be a total fool, but I just don't get it

kreslarz wrote:

Once again:
- cable length 130 x 2 = 260 m

why times 2 ??? /

Probably because the current must flow back and forth, unfortunately.

Literature: Wiatr, Boczkowski, Orzechowski ...... "Protection against electric shock and the selection of cables and their protection in low voltage electrical installations" 2010 states:
p. 162 - when Scu

The current must "flow" to the receiver and "return" to the socket, it travels the path in the cable 2 times, therefore the length times two ...

Miwhoo wrote:

The current must "flow" to the receiver and "return" to the socket, it travels the path in the cable 2 times, therefore the length times two ...

That's right, buddy. This "two" is probably already included in the formula ...

Will it be the correct way if I transform this formula and on its basis the calculation of S. For the calculations I will take my maximal value, ie 3%, as dU, and then the cross-section mm2 will come out, which will not cause voltage drops greater than 3%. The cables are always selected with a greater weight, so dU will always be below 3%.

For example:

S = 2 x In x L x 100 / gamma x dU x Un

S = 2 x 1.88A x 130m x 100/58 (Sxm / mm2) x 3 x 230V = 1.22 mm2

And now ... 1.5 mm2 is enough or 2.5 mm2 immediately?
But let's assume 1.5mm2.
So, at 1.5mm2 the actual voltage drop will be:

dU = 2 x In x L x 100 / gamma x S x Un
dU = 2 x 1.88 x 130 x 100/58 x 1.5 x 230 = 2.44%

Is this approach correct ???? If so, please explain to me what "safety" factor to choose when determining this cross-section ... when it comes out 1.22 is 1.5, should I load bigger ???

Here you have the table and formula.

My friend, from what I remember, the above formula with reactance is used for calculations on a larger cross-section (16 mm? and more) and higher power.

S = 2 x In x L x 100 / ?x ?U x Un is the correct formula

You remember well.