Confusion with calculating cable runs

[Liam151807]

Hi all, just a quick query really.

I'm and apprentice on me second year and currently doing my design project and confused about length of run per circuit.

I know you can't exceed 30m per circuit but what about lighting? Does this include separate switches? Like can I power one grid switch then have a few 20-30m runs going to different lights in order to cover a large area? Then calculate the longest "switch"?

Hope this makes sense, really grateful for any help too

 

[timhoward]

Where did 30m per circuit come from?

Circuit length increases resistance which will both reduce fault current and increase volt drop.
As long as the design considers these two things the length needn’t be limited.

Increasing conductor sizes can usually mitigate these things.

The R1+R2 needs working out and the Ze is needed to work out the loop impedance and hence whether it will trip the protective device in the required time.
Similarly volt drop needs working out.

This free volt drop calculator has the benefit that it explains how it is working

Voltage Drop Calculator

This free voltage drop calculator calculates the voltage drop based on the load, cable size, length of circuit and type of cable, single & three phase

www.procertssoftware.com

 

[Liam151807]

Where did 30m per circuit come from?

Circuit length increases resistance which will both reduce fault current and increase volt drop.
As long as the design considers these two things the length needn’t be limited.

Increasing conductor sizes can usually mitigate these things.

The R1+R2 needs working out and the Ze is needed to work out the loop impedance and hence whether it will trip the protective device in the required time.
Similarly volt drop needs working out.

This free volt drop calculator has the benefit that it explains how it is working

Voltage Drop Calculator

This free voltage drop calculator calculates the voltage drop based on the load, cable size, length of circuit and type of cable, single & three phase

www.procertssoftware.com

At college they've stipulated a circuit can only have a 30m maximum run, and lighting ends up being way over that

 

[timhoward]

At college they've stipulated a circuit can only have a 30m maximum run, and lighting ends up being way over that

If you've been told that this is a hard and fast rule for all circumstances then that is simply wrong.
I'm suspecting that their statement may have applied to a specific worked example and may be now being taken out of context.
They may have imposed an artificial limit to force a certain design, e.g. singles in conduit can sometimes require less wire.
You are going to have to clarify this with them.

As I said - what actually matters is the total resistance and will the protective device operate, and whether the load will result in excessive voltage drop. I assure you that much longer circuits are regularly installed in real life and can comply.

If you want to confirm this for yourself here's a challenge:
Assume a TNS supply with Ze of 0.7 ohms and a final circuit using 60m of 1.0 sq m T+E
Protective device is BSEN 60898 B6 MCB. Load is 1.5 amps.
Work out expected Zs, Max Zs value for that MCB, and volt drop (either in V or as percentage).

 

[MaintenanceSpark]

If you've been told that this is a hard and fast rule for all circumstances then that is simply wrong.
I'm suspecting that their statement may have applied to a specific worked example and may be now being taken out of context.
They may have imposed an artificial limit to force a certain design, e.g. singles in conduit can sometimes require less wire.
You are going to have to clarify this with them.

As I said - what actually matters is the total resistance and will the protective device operate, and whether the load will result in excessive voltage drop. I assure you that much longer circuits are regularly installed in real life and can comply.

If you want to confirm this for yourself here's a challenge:
Assume a TNS supply with Ze of 0.7 ohms and a final circuit using 60m of 1.0 sq m T+E
Protective device is BSEN 60898 B6 MCB. Load is 1.5 amps.
Work out expected Zs, Max Zs value for that MCB, and volt drop (either in V or as percentage).

Hi Tim,

As the OP hasn’t replied in a couple of months, I hope you don’t mind if I take up the challenge as I’m a trainee. Was good for me to look up the relevant tables in the regs book etc.

Expected Zs:
230 x 0.95 = 218.5v
Type B = 5 x In = 5 x 6a = 30a = 218.5 / 30 = 7.28
Expected Zs = 7.28 ohms

Maximum Zs value for a B6 MCB = 7.28 ohms - Table 41.3 page 68 of the BBB

Volt drop:
mV/A/m = 44 - Table 4D5 page 456 of the BBB
44 x 1.5a x 60m / 1000 = 3.96v
Volt Drop = 3.96v

Any guidance will be truly appreciated.

Thank you.

 

[timhoward]

As the OP hasn’t replied in a couple of months, I hope you don’t mind if I take up the challenge as I’m a trainee.

No problem at all.
Volt drop - you are spot on, and as a percentage 3.96/230 = 1.72% so well within the 3% for lighting.

You have correctly calculated the max Zs for the breaker in question. The value is in the OSG as you also said, so this isn't normally a required step. But the question is whether the circuit I described has a lower or higher Zs than this value, and therefore whether the breaker will work quickly enough under fault conditions.

Maybe have another go at calculating what the Zs will be on a circuit that is 60m long, wired in 1.0 T&E, with an external loop impedance (Ze) of 0.70 ohms? If you need a clue to get started, say so!

 

[pc1966]

It is essential to know how to do the calculations, even if a lot of domestic work is solved by a quick look in the OSG "standard circuits".

One aspect of Zs limit that is often confusing is the BBB (and calculation from MCB specifications) gives to the working resistance limit, whereas the OSG table of Zs values gives you the cold measured limit on the assumption of a higher operating temperature commensurate with the CCC limit.

 

[MaintenanceSpark]

No problem at all.
Volt drop - you are spot on, and as a percentage 3.96/230 = 1.72% so well within the 3% for lighting.

You have correctly calculated the max Zs for the breaker in question. The value is in the OSG as you also said, so this isn't normally a required step. But the question is whether the circuit I described has a lower or higher Zs than this value, and therefore whether the breaker will work quickly enough under fault conditions.

Maybe have another go at calculating what the Zs will be on a circuit that is 60m long, wired in 1.0 T&E, with an external loop impedance (Ze) of 0.70 ohms? If you need a clue to get started, say so!

Thanks very much for the feed back.

A clue to get started would be great, if you don’t mind. Thanks.

 

[mainline]

Edit: Previous posts didn't show until after i posted weird.

 

[timhoward]

A clue to get started would be great, if you don’t mind. Thanks.

Short hint - you need to use Zs=Ze+R1+R2

Longer hint and background theory:
What we are trying to work out, is
a) will it turn off if there's a fault
b) if it does, will it do it quickly enough to meet the disconnection times required

There are time/current graphs in the regs, we want fig 3A4 on page 417.
You'll see that the speed a device operate depends on the current that is flowing through it.
We know the nominal voltage. If we can find the resistance all the way back to the supplier transformer, then we can use Ohms law to find the current. This resistance is split into two parts, the supplier Ze, and the path from the consumer unit to the fault and back along the CPC (R1+R2)

A further hint is that I've given you the Ze and I've specified the length of the cable and the type of the cable.

 

[MaintenanceSpark]

Short hint - you need to use Zs=Ze+R1+R2

Longer hint and background theory:
What we are trying to work out, is
a) will it turn off if there's a fault
b) if it does, will it do it quickly enough to meet the disconnection times required

There are time/current graphs in the regs, we want fig 3A4 on page 417.
You'll see that the speed a device operate depends on the current that is flowing through it.
We know the nominal voltage. If we can find the resistance all the way back to the supplier transformer, then we can use Ohms law to find the current. This resistance is split into two parts, the supplier Ze, and the path from the consumer unit to the fault and back along the CPC (R1+R2)

A further hint is that I've given you the Ze and I've specified the length of the cable and the type of the cable.

I think I’m there.

Zs = Ze + (R1 + R2)
R1 + R2 = mΩ / M x L x Temp factor / 1000

Temp factor = 1.20 - Appendix I, table I3, page 220 of the OSG
R1 + R2 = 36.20 - Appendix I, table II, page 218 of the OSG
R1 + R2 = 36.20 x 60 x 1.20 / 1000 = 2.61Ω

Zs = 0.7Ω + 2.61Ω = 3.31Ω

Calculated Zs = 3.31Ω
Maximum Zs = 7.28Ω

Thanks for the clue to get me started mate.