Cable Calculation Help - Adiabatic Equation

[iDragonBox]

Need help check that my cable calculations are correct as still new to doing them.

6A Lighting Circuit 60898 MCB Type B wiring in 1MM²/1MM² T+E supplying a single 10W outdoor light

Length = 3.3M

Zs@db = 0.22

R1 + R2 mΩ/M = 36.20 mΩ/M - Table lt P218 OSG

Correction Factor - 1.2 - Table L3 P220 OSG

R1 + R2 = 3.3 x 36.20 x 1.2 / 1000 = 0.14Ω

Zs = 0.22 + 014 = 0.36Ω

Fault Current = 230 / 0.36 = 638A

Disconnection Time = 0.1s

K = 115

√ I²t / k = √ (638² x 0.1) / 115 = 1.75 Minimum CPC size.

Seems like I will have to make the circuit longer to increase the Zs value which seems ridiculous or change my cable for a larger CPC which also seem ridiculous for a single 10W light.

Can someone check I am doing my calcs right before I go down the rabbit hole of adapting!

Many thanks

 

[Pretty Mouth]

Need help check that my cable calculations are correct as still new to doing them.

6A Lighting Circuit 60898 MCB Type B wiring in 1MM²/1MM² T+E supplying a single 10W outdoor light

Length = 3.3M

Zs@db = 0.22

R1 + R2 mΩ/M = 36.20 mΩ/M - Table lt P218 OSG

Correction Factor - 1.2 - Table L3 P220 OSG

R1 + R2 = 3.3 x 36.20 x 1.2 / 1000 = 0.14Ω

Zs = 0.22 + 014 = 0.36Ω

Fault Current = 230 / 0.36 = 638A

Disconnection Time = 0.1s

K = 115

√ I²t / k = √ (638² x 0.1) / 115 = 1.75 Minimum CPC size.

Seems like I will have to make the circuit longer to increase the Zs value which seems ridiculous or change my cable for a larger CPC which also seem ridiculous for a single 10W light.

Can someone check I am doing my calcs right before I go down the rabbit hole of adapting!

Many thanks

Adiabatic's quite tricky to get the head around. You're on the right lines, but as you can see, you've come up with a larger CPC than expected. What is the circuit protective device?

 

[iDragonBox]

Adiabatic's quite tricky to get the head around. You're on the right lines, but as you can see, you've come up with a larger CPC than expected. What is the circuit protective device?

6A 60898 MCB Type B

 

[Pretty Mouth]

6A 60898 MCB Type B

P147 of the regs, fig 3A4. I guess you've already seen this to come up with what you have so far, but read the note at the bottom of the chart.

What brand circuit breaker?

 

[timhoward]

P147 of the regs, fig 3A4

Try P417 !

 

[iDragonBox]

P147 of the regs, fig 3A4. I guess you've already seen this to come up with what you have so far, but read the note at the bottom of the chart.

What brand circuit breaker?

Its a theoretical circuit design so haven't chosen a brand yet.

 

[Pretty Mouth]

Its a theoretical circuit design so haven't chosen a brand yet.

Apologies, as Tim says, p417!

 

[timhoward]

Your calculations are correct and as accurate as you can achieve right now (see the note that @Pretty Mouth refers to in order to understand why)

A general comment, remember 543.1.1 - you can either calculate OR select.
In general for smaller size line conductors and shorter lengths it's better to select from table 54.7 which would give you a much more welcome answer!
In general for larger line conductors or longer cable lengths, it's better to calculate.

 

[Pretty Mouth]

It's hard to get a realistic CSA using the data from BS7671. As you've seen the time only goes down to 0.1s, but real world a fault current like that would disconnect much quicker, giving much lower let through energy (I²t), therefore a smaller CSA.

We're concerned with the worst-case situation: for a B or C type breaker, that will be right at the start of the circuit, where the fault current is greatest. If the conductor can withstand a fault here, it can withstand it anywhere downstream.

 

[timhoward]

Let's assume you went for a Wylex B6 .
Here's some data from Wylex:

You had 638A prospective fault current and a 6 amp breaker, so the multiple is x 106
That graph shows you that the disconnection time will be less than 0.01 seconds.

If you run you calculation again with 0.01 as the disconnection time you end up with a CPC of 0.55 sq mm!
You obviously need to round this up to something that actually exists, and then be mindful that it it's separate to the cable or buried there are minimum sizes that apply.
You are using T&E so that last bit isn't relevant.

 

[Pretty Mouth]

There's also table B7 on P146 of the OSG, if you don't wish to calculate.

However, manufacturer's data is best. Here is an example from Hager. Use your fault current to find the (I²t), and run it through the adiabtic. You should get a much smaller CSA.

 

[iDragonBox]

Thanks for your help guys giving me much more detail then I get in the class room!

 

[Lucien Nunes]

I think your numerical question has been well answered but just referring back to:

Seems like I will have to make the circuit longer to increase the Zs value

You can't manipulate the Zs value by altering the length of the circuit to achieve some desired result regarding fault protection of that circuit itself, since the fault might occur anywhere along the length and the OCPD must be suitable regardless. OT: About the only time one might want to make a circuit longer artificially (and I have done this) is an extra foot or two on a short distribution circuit near the origin, to lower an unusually high fault level at the sub-DB to within the Ics rating of its MCBs.

In day-to-day work, sizing a small circuit's CPC rarely involves calculation. As per @timhoward's post No.8, for typical final circuits the default option (CPC same as live conductors or whatever size comes in the appropriate Twin & Earth) works just fine and attempts to calculate it might give an impractical result as you spotted. Your original solution of 1.75mmsq can't be applied in practice (even if the 0.1s disconnect time were appropriate) because if that size CPC is needed to prevent thermal damage then the same must be true of the live conductors. Calculation begins to bear fruit in situations where the use of an unusually small CPC is desirable for some reason (e.g. existing cable) but needs validation, or some input value like Zdb is a bit marginal, or on large circuits.

Thanks for your help guys giving me much more detail then I get in the class room!

You'll get good input because you're going about it the right way. Posting your own working clearly, then asking about the bit that (quite reasonably) puzzles you. I gave it a like because it sets a good example. You'd be amazed at the nonsense some students post on here, either a jumble of word-salad that we can't decipher, or no working at all but a blunt request to answer the question from scratch.

 

[pc1966]

@Pretty Mouth has it spot-on that you really need to get the manufacturer's data for breakers (MCB, RCBO or MCCB) as the usual current-time plots do not really give you a good idea of what happens once you enter the "instant" magnetic trip region. Some give you useful curves (like the Hager example) and others just give you a maximum value to base calculations on.

There are some generic limits that can be used as well, for example the On-Site Guide in Table B7 (page 146 of current 18th AM2 version), and usually any of the "standard circuits" in Table 7.1(ii) of the OSG are fine for domestic cases where you very rarely see above a couple of kA fault currents.

As @Lucien Nunes says it can sometimes be necessary to look at how you limit the worst-case fault current for various reasons, not just cable adiabatic but also the OCPD breaking limit, etc. However, trying to increase R1+R2 is really your last resort if you find a problem, a better option where practical is to look at OCPD that can safely break the worst-case fault and/or is able to limit the I2t let-through energy to a lower value as well.

Fuses can be your friend here! Many BS88 fuses can break over 50kA PFC (compared to 6-10kA for MCB and often 18-25kA for many MCCB, though some do go higher to 70kA and above) and they do very well at limiting I2t often to orders of magnitude lower than a similar capacity breaker.

They are one-time only devices of course, and need skilled staff for safe replacement, but otherwise are cheap, reliable and effective if those two limitations are acceptable.