Help with ring main calculations

[bazastan]

A client has asked us to provide cable sizing calculations including voltage drop, short circuit rating, source impedance, circuit impedance and fault current of a ring main we have installed to ensure that our 2.5mm ring is suitable for the application.

Cable size 2.5mm
System voltage 254v
Run length 30m
Circuit load 1x3000w and 1x1500w
Cable is installed on cable tray
Type B 32a Merlin mcb

We are struggling for the calculations due to it being a ring main, any help would be greatly appreciated.

 

[malcolmsanford]

I'm assuming that your 2.5mm^ ring main is in fact a ring final circuit incorporating BS 1363 sockets.

If that is the case then ring final circuits are not normally designed as you are doing. Things like voltdrop is normally calculated by the Ib of a circuit so your volt drop cal is

mV/a x Ib X length/1000 as you can't guarantee what is to be plugged into a ring final there are certain design aids we use in the OSG and appendix 15 of the regs.

Ie a A1 ring final circuitwired in 2.5mm^ protected by a 32 amp PD can serve a maximum of 100m squared, so I would imagine your 30m ring should be fine.

So going back to your OP if this is a ring final circuit where are the loadings coming from?

If this a normal distribution circuit wired into a ring with 2 fixed loads then the above cal will be dependant on what the cable is that is on that cable tray. if it were normal SWA then it would be

18x 19.56 (@ 230volt) x 30 /1000 = 10.56 volts so just accepable as 5% of 230 is 11.5 volts

Can not work out your EFLI as Zs = Ze + (R1 + R2). We can work out by calculation your R1 and R2 on design but without the Ze you can't get your Zs. if you have a 2.5 cable with a 2.5 cpc then your formula for R1 + R2 will be

14.82 (2.5/2.5 cable resitance per metre) x 1.2 ( temperature factor) X 30 (length)/1000 which is 0.53 ohms. So therefore your Zs would be Ze + 0.53.

 

[i=p/u]

couldnt of put it better myself

 

[telectrix]

malcolm, on this topic, a question for your respected knowledgeable self. if using SWA cable , whererin you have a conductor as cpc and also the armour, when measuring R1+R2, would you include the resistance of the armour in your measurement, or ignore it and just measure the R1+R2 of the cable . i ask this because if you ignore the armour, when measuring Zs, you could end up with a value for Zs<R1+R2. cheers, TEL.

 

[i=p/u]

i do think your encougaging them tho as is 1st/2nd year stuff

 

[willij]

You guys are amazing!

 

[malcolmsanford]

malcolm, on this topic, a question for your respected knowledgeable self. if using SWA cable , whererin you have a conductor as cpc and also the armour, when measuring R1+R2, would you include the resistance of the armour in your measurement, or ignore it and just measure the R1+R2 of the cable . i ask this because if you ignore the armour, when measuring Zs, you could end up with a value for Zs<R1+R2. cheers, TEL.

If your using a conductor within the SWA as your CPC then you would just use line and CPC conductor as your R1 + R2 in the design stage.

If your designing a circuit where you feel you need to combine a conductor and the SWA as a CPC, though never seen this or done this myself then you would use a combined design to get that R1+R2

As you know though if you take Zs by measurement say on a PIR or extending a circuit then you will get, or should get a really lower than calcualted Zs because of the Parallel paths, which will included your SWA .

This is why unlike the nice people at NICEIC I like to do a Zs by measurement, it's ok doing Zs by calcualtion on a design, but when there is a fault on an installation it won't happen on a piece of paper it will happen when there is bonding. parallel paths etc etc and a truer reflection of what is going to happen to that fault.

I honestly think the NICEIC have advised the calculation method becasue they feel that some, lets call them electrical operatives, are not capable of live testing, and they maybe right!!!

 

[Sintra]

Surely if it is a ring final R1 + R2 with 2.5/2.5 would be.

r1 0.00741 x 30 = 0.22
r2 0.00741 x 30 = 0.22

r1 + r2 / 4 = R1 + R2

so R1 + R2 = 0.11

 

[Sintra]

Formula for VD on a ring is.

mV/A/M x ib x L / 1000 / 4

so in your case

18 x 19.5 @ 230V x 30 / 1000 / 4 = 2.63V

 

[i=p/u]

why by 4. just curious

 

[malcolmsanford]

Surely if it is a ring final R1 + R2 with 2.5/2.5 would be.

r1 0.00741 x 30 = 0.22
r2 0.00741 x 30 = 0.22

r1 + r2 / 4 = R1 + R2

so R1 + R2 = 0.11

Formula for VD on a ring is.

mV/A/M x ib x L / 1000 / 4

so in your case

18 x 19.5 @ 230V x 30 / 1000 / 4 = 2.63V

Yes quite right 10.56/4= 2.64 and 0.53/4=0.13 Thanks Sintra. That what you get when you leave the UK and don't work on rings as much as you should.

Cheers for that my mistake so apologies to the OP

 

[Sintra]

Always divide by 4 on ring calcs due to there being 4 times less resistance than that of a radial.

 

[i=p/u]

just the tudor didnt mention that to us when we did bit of dead testing last year and working through them this year in are practicals, for volt drop that is, hes in for it

 

[i=p/u]

sorry for last reply , he didnt say that you quarter vd on a ring circuit

 

[ian.settle1]

I have to ask why you are asking this question now as surely you would have been done when doing your cable calcs during the design stage