Level 3 C&G cable sizing calculations

[EFLImpudence]

So what you do is, get a figure for It, look up a cable size that has capacity of equal to or greater than that, find the voltage drop, calculate it, check against limits (<11.5V) and if bigger, go to next cable up. That's how you get the numbers.

Yes, I know but in the C&G example, the volt drop on 2.5mm² is very low so no need to have larger cable.

Regarding resistance, when you say quarter, you mean because it's Line and Neutral lengths in parallel with Line and Neutral again to form the ring?

A ring is two cables in parallel each of which is half the length of the circuit.
18 x 20 x 37 / 4 / 1000 = 3.32V

I'm not sure thats right as the length given is the length of the whole cable, not the middle of the ring.

It works out the same 2 x 9 = 17 + 1.

Table F6 gives drop for a 2.5mm2 cable of 18mV,

Yes.

so this is for the line and neutral, so 9mV drop on each per meter, per amp. 9mV drop on 2.5mm2 wire gives a resistance of 0.009 ohms. A resistivity of 2.25x10-8, pretty close to pure copper of 1.77x10-8

The resistance of 2.5mm² conductors is 7.32 mΩ/m

7.32 x 1.2 (temperature adjustment for 70º) = 8.784 x 2 = 17.568 (the 18 value) mV/A/m.

 

[richardindorset]

Yes, I know but in the C&G example, the volt drop on 2.5mm² is very low so no need to have larger cable.


A ring is two cables in parallel each of which is half the length of the circuit.
18 x 20 x 37 / 4 / 1000 = 3.32V


It works out the same 2 x 9 = 17 + 1.


Yes.


The resistance of 2.5mm² conductors is 7.32 mΩ/m

7.32 x 1.2 (temperature adjustment for 70º) = 8.784 x 2 = 17.568 (the 18 value) mV/A/m.

OK, so let me see if I have this right, per metre, per amp, there's 9mV drop in the live and 9mV drop in the neutral. For a ring, you have another pair in parallel, so the volt drop is half on each side. 4.5mV in all 4 wires, per amp, per metre. 9mV in total. Half of the 18mV.

The length of the ciruit in a ring is the whole length of the cable from the DB and back again.

I can only see the 18mV divded by 2 as the 18mV is for a pair of conductors.

 

[EFLImpudence]

It is the same reason as why R1 for a ring circuit is r1/4.

 

[flameport]

The length of the ciruit in a ring is the whole length of the cable from the DB and back again.

The conductors are that length, but the furthest point from the consumer unit is half of that length.

If the wires were 40m long, the distance to the furthest point would be 20m = ½
and there are two sets of wires = ½ again.

 

[EFLImpudence]

@richardindorset

Firstly let's make sure you understand that in the VD calculations the 'mV' value is just the same as the resistance of the conductors.
This being the case because that nice Mr.Ohm arranged it so that 1V through 1Ω is 1A - lucky that.
The near enough figure of '18' (17.568) is '9' (8.784) for each of the L & N in series. '9' mΩ being the resistance per metre of the conductors as we assume they are at 70ºC when loaded.
This is unlikely to be the case - especially for 20A on a ring circuit.

OK, so let me see if I have this right, per metre, per amp, there's 9mV drop in the live and 9mV drop in the neutral.

For a radial, yes, but it would be better if you just use the resistance until you have the final value - so 9mΩ per metre.

For a ring, you have another pair in parallel, so the volt drop is half on each side.

Yes, but they are both only half the length as well.

4.5mV in all 4 wires, per amp, per metre. 9mV in total. Half of the 18mV.

Well, no - the resistance doesn't work like that. That is why it is confusing using the 'mV' value.

The resistance of two identical conductors in parallel is half of the resistance of one of them; so one quarter. It just is.

The length of the ciruit in a ring is the whole length of the cable from the DB and back again.

Yes but that is not relevant.
It is the current path which matters; it does not 'go round' the length of the circuit and back again.

I can only see the 18mV divded by 2 as the 18mV is for a pair of conductors.

You're not looking in the right place.

I hope that helps.

 

[richardindorset]

The conductors are that length, but the furthest point from the consumer unit is half of that length.

If the wires were 40m long, the distance to the furthest point would be 20m = ½
and there are two sets of wires = ½ again.

OK, think I have it, finally!

The ring doubles the cable length and halves the resistance so your mV/A/M drops to a quarter the book value.

Thanks for sticking with me!

 

[richardindorset]

Just to plug the numbers in:

2.5mm2 drop per the OSG is 18mV/A/M

The drop on radial of 20M with 2.5mm2 with 32A load is 11.52v

If we did a ring we'd have 40M of cable (assuming same run for the return) BUT we're measuring the drop at the halfway point AND the wires are parallel so we use a quarter of 18mV/A/M which gives 5.76v drop.

This is why I kept thinking of half and not quarter! You quarter the mV/A/M and you end up with half the drop.

(still doens't answer the original question but at least I've got it in my head now!)

 

[DetlefSchmitz]

If we did a ring we'd have 40M of cable (assuming same run for the return) BUT we're measuring the drop at the halfway point AND the wires are parallel so we use a quarter of 18mV/A/M

As you're doing exams and stuff you should probably get into the habit of using a small 'm' for metres (as it isn't some famous scientists name). 'M' has a different use.

 

[EFLImpudence]

2.5mm2 drop per the OSG is 18mV/A/M

The drop on radial of 20M with 2.5mm2 with 32A load is 11.52v

You can't have a 2.5mm² radial with 32A load.

If we did a ring we'd have 40M of cable (assuming same run for the return) BUT we're measuring the drop at the halfway point AND the wires are parallel so we use a quarter of 18mV/A/M which gives 5.76v drop.

It is - but the calculation is actually half of 18 x half of 40; i.e. 9 x 20 x 32 / 1000.

This is why I kept thinking of half and not quarter! You quarter the mV/A/M and you end up with half the drop.

but if you had halved the two amounts it would have been obvious that that was a quarter.

(still doens't answer the original question

There isn't one.

but at least I've got it in my head now!)

 

[richardindorset]

You can't have a 2.5mm² radial with 32A load.


It is - but the calculation is actually half of 18 x half of 40; i.e. 9 x 20 x 32 / 1000.


but if you had halved the two amounts it would have been obvious that that was a quarter.


There isn't one.

Ha! Yes I know you wouldn't have a 2.5mm2 radial with a 32A load but just picked some numbers, anyway back to the original question which is how does C&G calculate cable size and voltage dropfor a ring final?

In the second example I gave, from a C&G task: Ring final, design current 20A, breaker 32A type B, cable length 37M, Correction factor 0.94.

Why does C&G come to an answer of 4mm2? For some reason they take the design current and apply factors to it then look up the figure in table F6 for a cable. No idea why though, when for every other case they take the breaker rating and apply factors to that before looking up the cable. Obviously this can't work for a ring whichever method you use.

Using what I've learnt here, the voltage drop in the exam question would acutally be 1/4 x 20A x 37M x 0.018V/A/M = 3.33V which is for 2.5mm2 and is well inside the drop limit. It also follows what the Regs say: Use 2.5mm2.

 

[plugwash]

They do - but then state you cannot use 1.5mm² which has a CCC of 20A but you must use 2.5mm².

So you do the calculations based on 20A, compare them to the minimum in the regs and then pick whichever is smaller.

 

[EFLImpudence]

So you do the calculations based on 20A, compare them to the minimum in the regs and then pick whichever is smaller.

I don't understand what you mean.

 

[plugwash]

Sorry it should have said whichever was larger.

 

[EFLImpudence]

Still not with you.

The ring regulation 433.1.204 states:
"The circuit shall be wired with copper conductors having line and neutral conductors with a minimum cross-sectional area of 2.5 mm2 except for two-core mineral insulated cables..."

but then states:
"Such circuits are deemed to meet the requirements of Regulation 433.1.1 if the current-carrying capacity (Iz) of the cable is not less than 20A"

which 1.5mm² is with a 32A MCB but not a BS3036 30A fuse.

 

[plugwash]

At least my reading of that is you must meet both the current rating requirement and the size requirement.

Which one sets the size will depend on the installation method and any applicable derating factors.