Voltage Drop calculation...

[virgilns]

Just starting to plan my rewire of the house and looking splitting the house in to zones. The first one I've done in consideration to voltage drop is the lounge/hall ring. my calculation is for a 38m long 'ring'.

As it's a ring do I halve the current AND the length of the circuit for the calc?

Taking into consideration you should never protect a circuit with a device rated less than the cable capacity and I'm using a 32amp rcbo, my cable 'in use' needs to be rated at 32 amps which I presume even with diversity of use etc taken into consideration I should use 32A for the voltage drop calc?

If I'm halving the curernt AND length I get

2.5mm cable 18mv/m = 18x19x16/1000 = 5.4 volt drop

If I halve just the current (which makes no sense in my head??) I get

2.5mm cable 18mv/m - 18x38x16/1000 = 10.9 volt sdrop (too much)

In the last example, either I can shorten the circuit to 33m OR I'm forced to use 4mm cable (expensive/overkill?)

OR am I allowed to accept that I'll never realistically run 32amps on the ring, therefore the voltage drop *should* be OK?

I guess radials are easier to work out.

16m on 2.5mm cable, 20amp RCBO

18x20x16/1000 = 5.76 volt drop...perfect!

Advice gladly received...

edited tio add - all being done under LBCO oversight/control...

 

[bernardgreen]

You could run the two legs from the CU to the first sockets in 4 mm and then 2.5 mm for the rest of the ring.

 

[Scoby_Beasley]

Put all of the load at the mid point and calculate the volt drop based on half load and half length.

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

 

[virgilns]

Put all of the load at the mid point and calculate the volt drop based on half load and half length.

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

Perfect. That's what I thought, but wanted to be sure.

Thanks.

 

[ban-all-sheds]

Eric will be along soon with an explanation of the current assumed by the IET for the OSG which explain where the circuit lengths come from in there - 106m for a ring on a B32.

But if you had a copy of this:

http://www.amazon.co.uk/Electrical-Installation-Design-Guide-Calculations/dp/0863415504
or http://www.theiet.org/publishing/books/wir-reg/electrical-installation-design-guide.cfm

you would already know why...

 

[virgilns]

Eric will be along soon with an explanation of the current assumed by the IET for the OSG which explain where the circuit lengths come from in there - 106m for a ring on a B32.

But if you had a copy of this:

http://www.amazon.co.uk/Electrical-Installation-Design-Guide-Calculations/dp/0863415504
or http://www.theiet.org/publishing/books/wir-reg/electrical-installation-design-guide.cfm

you would already know why...

yeah - been trying to find a free version on line

if the design current of the ring is 20 amps, then 53m gives 9.5 volt loss...

 

[ericmark]

I attended the lecturers given by the IET when 17th first came out and I also could not get the figures to match so I asked the question in one of these meeting.

It would seem with the final ring it is assumed that 20 amp is drawn from centre of the ring and the remaining 12 amp is even distributed so we work on 26A not 32A for the calculations.

Because the cable is rated at around 20A so total around 40A the correction factors also play a large part in the calculations.

As the sad guy I am I wrote it all into excel and did some playing and yes once everything is included I did get the 106 meters voiced at the IET lecturers.

Personally I think 106 meters is silly. The main point is a short circuit must cause 160A to flow to trip the magnetic part of the B32 trip so neutral - line impedance of 1.4375 ohms. Personally again I think you should round down not to nearest so 1.43 ohms but book says 1.44 ohms.

In real terms it's the 160 amp that we look for so why measure in ohms when meter will read in amps to start with I don't know.

If you calculate the limits for volt drop and for current to activate the B32 trip are nearly the same so if you get 160A or 1.43 ohm then the volt drop will be within limits.

 

[ericmark]

As a PS when 17th came out there was no on-site guide so had to work it out.

 

[ban-all-sheds]

yeah - been trying to find a free version on line

Doubt it.

if the design current of the ring is 20 amps, then 53m gives 9.5 volt loss...

No it doesn't.

 

[JohnW2]

If you calculate the limits for volt drop and for current to activate the B32 trip are nearly the same so if you get 160A or 1.43 ohm then the volt drop will be within limits.

Whether that is true surely depends upon what you assume (or measure) to be the L-N impedance extrenal to the installation?

As I see it ... We know (per your post) that, using the IET's apparent loading assumptions, the maximum length of a 2.5mm² 32A ring final which satisfies VD requirements is 106m. At the 'worst-case' point (mid-point of ring), the impedance of the wiring back to the CU will be 0.477Ω. As you've said, to achieve required B32 MCB operation, one needs as least 160A, which (at 230V) translates to a L-N loop impedance of no more than 1.4375Ω - so that would only be the limiting case in terms of MCB operation/disconnection time (as well as in terms of VD) if the external part of the L-N loop impedance is about 0.96Ω (1.4375Ω - 0.477&#937. It therefore seems that if the external impedance were appreciably different from that figure, the circuit length limits would be appreciable different for VD and disconnection considerations. Am I wrong?

Kind Regards, John

 

[virgilns]

yeah - been trying to find a free version on line

Doubt it.

if the design current of the ring is 20 amps, then 53m gives 9.5 volt loss...

No it doesn't.

Eh??

It might not be how the IET get 106 max ring lenght, but why are my maths wrong...

20 amp design current = 10 amps per leg
106m ring lenght = 53m per leg
2.5mm cable vloss = 18mv

10x53x18/1000 = 9.54 volts loss

Or am I doing something really stupid??

 

[virgilns]

If you calculate the limits for volt drop and for current to activate the B32 trip are nearly the same so if you get 160A or 1.43 ohm then the volt drop will be within limits.

Whether that is true surely depends upon what you assume (or measure) to be the L-N impedance extrenal to the installation?

As I see it ... We know (per your post) that, using the IET's apparent loading assumptions, the maximum length of a 2.5mm² 32A ring final which satisfies VD requirements is 106m. At the 'worst-case' point (mid-point of ring), the impedance of the wiring back to the CU will be 0.477Ω. As you've said, to achieve required B32 MCB operation, one needs as least 160A, which (at 230V) translates to a L-N loop impedance of no more than 1.4375Ω - so that would only be the limiting case in terms of MCB operation/disconnection time (as well as in terms of VD) if the external part of the L-N loop impedance is about 0.96Ω (1.4375Ω - 0.477&#937. It therefore seems that if the external impedance were appreciably different from that figure, the circuit length limits would be appreciable different for VD and disconnection considerations. Am I wrong?

Kind Regards, John

It's only when you try to really understand something you realise how little you know...most people attempting DIY electrics know "you use 2.5mm for rings and 1.5 for lights with type B MCBs" but not many I;m guessing know why or their limitations.

Out of interest, if (the qualified of..) you were extending a ring main for a house extention, how would you calculate whether the new ring was going to be in length? Would it be done on the existing resistance of the ring then calculate the additional and then, during install, test/prove that it's still within tollerance on resistaqnce and therefore must be on length??
(I'm basing this assumption on that the important factor for current capacity and voltage drop is resistance which in itself is made up of various factors, one of which is length??)

 

[ban-all-sheds]

Or am I doing something really stupid??

Well, I wouldn't say stupid, but as the entire discussion so far has been about ring length it was pretty daft for you to write:

if the design current of the ring is 20 amps, then 53m gives 9.5 volt loss...

and expect people to realise that you were for some reason quoting half of the ring length...

 

[bernardgreen]

My opinion is that the regulations and BS7671 set a minimum standard to ensure that cost cutting exercises by profit minded tradesmen do not result in electrical installations that are dangerous..

A design based on BS7671 is likely to be safe but may not be the best design for the property.

A safe design based on what is needed will always be better for purpose than a design, often cost conscious, based on what is permissable rather than what is needed.

So if the customer is concerned about voltage drop and/or the power wasted in heating the cable ( 20 amps, 5 volt drop = 100 watts lost in the cable as heat ) then use 4 mm instead of 2.5 mm.

Ironic that we "must not" waste heat in a 100 watt incandescent lamp providing light and radiant heat to a room ( reducing the need for other heating in the winter ) but we can waste 100 watts heating the underside of floor boards.

 

[virgilns]

Or am I doing something really stupid??

Well, I wouldn't say stupid, but as the entire discussion so far has been about ring length it was pretty daft for you to write:

if the design current of the ring is 20 amps, then 53m gives 9.5 volt loss...

and expect people to realise that you were for some reason quoting half of the ring length...

AH!! makes sense. It was in reply to your post re 106m, so assumed it would be clear. I guess clarity and never making assumptions is all part of staying alive when fiddling with electrons

What's going to be of more use to me...Instllation Planning Guide or the full 7671 regs? I presume the full regs cover nuclear powerstations down to a 2 way CU in a garage?...or are they domestic only (more relevant to me)?