Cable Size Calculation Formula

If you don't divide by that value, you will need a really humongous cable.

If you did not get this formula from your region's electrical code book then you probably have the wrong formula. I don't like the mixture of units nor having constant integers that are multiples of each other in the numerator and denominator of this equation. However, a formula like this might be a simplified method to generate a look-up table.

If this formula and your mathematics is correct, your conclusion is dangerously wrong. The code book and any formula inside it produces the minimum safe wiring for your application. You calculated 36.17 so 35 mm^2 is undersized wiring for this application. You should find the next larger available wire.

When evaluating any formula, one must understand all the units of measure used in the formula, and the answer must be in consistent units. In your formula the only units you give are amps and meters (PF is a unitless ratio). How do you get mm2 from these two units? The answer to this should explain where the other numbers come from, and will help you to understand the formula.

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Good Answer John! It never ceases to amaze me how many people seem to think that all you need to do is to memorize a "magical" list of formulas.

These guys end up doing serious damage as managers, more often than not. They are always looking for the quick answer without having to understand where it came from.

Knowledge is handy. Understanding is better. Knowing how to apply it, is best.

Interested to know who thought it was an off-topic answer (not that I give a shit. Let the OP fry themself if that's the consensus).

Maybe OP has an idea who?

Be pleasantly surprised if OP has an idea .

Touche′.

This appears to be a formula for single-phase voltage drop, with some of the variables already plugged in. The 2 in the numerator would account for conductor length both ways. Wild guess, the 4 in the denominator could be the chosen allowable percentage of voltage drop.

For this length of run, the voltage drop calculation leads to a larger wire size than ampacity alone would require.

If 4 is 4 mm^2 [the conductor cross sectional area] then this formula could be the voltage drop in a case of a single phase [line and neutral] supply cable.

Voltage drop=2*I*[R*cos(fi)+X*sin(fi)]-simplified formula- and neglecting X it is Voltage drop= 2*I*R*cos(fi). R=ro*L/mm^2 ro=1/58 ohm. mm^2/m [copper 20 dgr.C] and for 90 dgr.C =1/58*(234.5+90)/(234.5+20)=0.022=1/45.5. (neglecting skin and proximity effect-for 4 mm^2 it can be neglected indeed). I is in [ A] and L in [m].

Then if I=22.3 A cosfi=0.9 L=164 m voltage drop=40.2 V

For 230 V rated voltage 40/230=17.4% [usually maximum permissible for steady state could be 5% then a 16 mm^2 will be required].

Of course, as JohnDG explained it very well, it is not the single criterion-ampacity is the first and short-circuit current withstand could be another.

Volts drop in a cable is I x R, and power is I² x R. Power factor is always 1.

Sorry, Horace40, if it is a joke IMO it is not funny!

Hello 7anoter4, for the benefit of others, perhaps you would be good enough to explain the unfunny bit.

The OP is speaking about "pf" [power factor=cosine of angle between Voltage and Current in the same circuit] .This cannot be something else but "alternative current" [a.c.] and "pf" it may be any value from 0 to 1 [lagging or leading].

A cable supplied at one end with a voltage lets through it a current as required by supplied installation at the other end and the cosine is determined by the receiver mainly.

In a.c. a cable has resistance and reactance always.

See [ for instance ]:

http://okonite.com/engineering/voltage-regulation.html

Thanks 7anoter4 for the explanation and the link.

The current in my formula is I, resulting from the combined effect of all the resistances and reactances in the whole circuit, and the voltages and frequency causing the current to flow.

R is the resulting impedance of a copper wire where capacitance and inductance are usually marginal at mains frequency and likely to cancel out, just leaving the pure resistance component.

The instantaneous volt drop is I x R, where there is no cosΦ component.

That's my simplistic view. Simple Ohm's law is a powerful tool to have when the going gets complicated.

You really need to get a look at the IEE regs for cable sizing. There are many things to take into account such as where the cable is running (under insulation or under the ground) is it a single cable or is it run with a group of many other possibly hot cables. Will your cable run with other cables that might only be rated to 70deg C. Spare a few thoughts on the protection available. Again this will add to the cable size. God forbid but Volt drop .....% ?? Leave Horace with his DC - he seems happy!! Over and out.

By-the-way, the NEC 310.15(B)(16)- any running system but 30 dgr.C ambient-ampacity of 6 Awg (90dgr.ins.) is 55A [6 Awg is 13.3 mm^2].IEC 60364-5-52 installation group A2" in a thermally insulated wall" for 16 mm^2 30 dgr.C ambient recommends 73 A. From BS7671 the minimum-minimorum will be 68 A permissible for 16 mm^2.

Usually in low voltage system of 400/230 V the maximum short-circuit current could be 5 kA [rms].Neither NEC nor BS7671 recommends any appreciation about conductor short-circuit withstanding.IEC 60364-5-52 I=K*mm^2/sqrt(time) K=143 -XLPE ins.Following this 16 mm^2 withstand only 4.57 kA within 0.25 sec fault clearing time.

This really is time to pay an engineer to do this calc. It will save you a lot of money and won't waste the fire brigades time.

You don't say what voltage you are operating at. In your last post you give us the load as 31.7kVA. At 230V its a 137A but at 110V its 288A. Who knows it might even be 400V TP&N.

Strangely it is now many metres further to your apartment. Have you stopped off at Mr Slacks for some tea. Incidentally Mr Slacks copy of the IEE regulations will show you how to make this calculation.

Regards

Silverfox

British Standard 7671.