Volt drop maths

[Dan Carroll]

Morning team
Question about volt drop. I presume I am missing something here. So firstly the facts out the OSG.
Table 7.1, page 66, lighting circuit, 6 amp, RCBO protection, 1.5mm cable. The max allowed length is stated as 106m. Using table F6 for the cable info, for 1.5mm cable the mV/A/m is 29.
The maths. VD= mV/A/m x Ib x L / 1000. So VD= 29 x 6 x 106 / 1000 = 18.4v.
That’s quite a bit over the 6.9v allowed. To get VD of 6.9 you have to reduce the length to about 39m.
So what am I doing wrong here?
Any help or abuse is welcome

 

[James]

What you have calculated is the volt drop for a single 6A load at the end of 106m cable.

In real life, the loads are likely to be spaced out over the length of the circuit. This leads to different currents passing down different parts of the cable run.
Therefore the volt drop is going to less than calculated.
The calculated volt drop will be correct if you know all the circuit loads and dimensions.

You have calculated the absolute worst case scenario
Imagine another scenario where 5.5A load is on the circuit at a distance of 2m from source and 0.5A is a further 104m away.
[automerge]1594892646[/automerge]
29 x 6 x 2 /1000=0.35v
29 x 0.5 x 104 /1000= 1.51v
Total vdrop 0.35+1.51=1.86v

 

[telectrix]

in simple terms, after the 1st load, the current reduces, then again after the 2nd load.and on and on and on. till the last load is only that of the last fitting.

 

[Dan Carroll]

Ok, that makes a bit more sense.
So for an example if I had a 100m garden (I wish) & I wanted to put LED lights outside every 2 meters with 1.5mm cable. (ignoring its not swa etc etc). And even though it's LED we still say 100W so the Ib at that point would be 0.4.
VD for the first light would be 29 x 0.4 x 2 / 1000 = 0.0232.
times that by 50 for all the lights and my volt drop would total 1.16v over 100m.
Correct or still going wrong

 

[James]

Ok, that makes a bit more sense.
So for an example if I had a 100m garden (I wish) & I wanted to put LED lights outside every 2 meters with 1.5mm cable. (ignoring its not swa etc etc). And even though it's LED we still say 100W so the Ib at that point would be 0.4.
VD for the first light would be 29 x 0.4 x 2 / 1000 = 0.0232.
times that by 50 for all the lights and my volt drop would total 1.16v over 100m.
Correct or still going wrong

Close,
First 2m of cable would have load from all 50 lights
Next 2m load from 49 lights, so on and so on until the last length of cable only has load from 1 light.

 

[Dan Carroll]

Ok I see.
So the maths can get very long winded very quickly. Is it better to just use the tables in section 7 then as general guides and only do the maths if its getting close. i.e. if i'm putting lights in the garden direct from the consumer unit i know i can get about 100m. But if the lights are coming from a distribution board at a distance from the CU, then i'ts best to go the long way and do the maths.
I can't believe that someone goes around working it out for every point. for volt drop

 

[James]

A good approximation would be to work out the volt drop assuming the whole load is applied at the furthest point, then half the calculated volt drop. This will work out close to the actual drop if the loads are roughly even spaced.
It is roughly what they have done in the onsite guide.

 

[Zerax]

Just use a spreadsheet if there's lots to calculate... for your 50 light example it would take <1 min to calculate after you've tapped in the length to each light and it's current draw etc.

 

[ElectroChem]

A good approximation would be to work out the volt drop assuming the whole load is applied at the furthest point, then half the calculated volt drop. This will work out close to the actual drop if the loads are roughly even spaced.
It is roughly what they have done in the onsite guide.

That is actually exactly what the Australian rules say we can do for loads evenly spread along a circuit, although in ours it's expressed as taking half the total load applied at the end of the cable.

 

[James]

Also @Dan Carroll
In real life, you are rarely going to have more than 1 load on a circuit you are calculating volt drop.

It is important to learn how to do the full calculations correctly because A on your exam there are bound to be more than 1 load and B sometimes you may need to calculate it out for an unusual circuit in real life.

 

[Lucien Nunes]

If all the loads are equal and evenly spaced, with or without a different length of cable from the DB to the first point, you will get exactly the correct drop with the normal formula if you insert as the length:
length from DB to first point + half the remaining length.
So if you have 15 points each drawing 0.1A, spaced 1m apart, the first of which is 5m from the board, you can use 5 + (14/2) = 12m as the effective length and 1.5A as the current. The answer will be exactly the same as if you calcluate the drops separately, e.g. 1.5A over 5m + 1.4A over 1m + 1.3A over 1m etc.

In many cases where the drop is critical, such as large warehouses and factories with high power luminaires spaced evenly, this is entirely representative. On small circuits with a mixture of fittings, e.g. domestic, the cable run is rarely long enough to merit calculating the drop precisely.

 

[Dan Carroll]

Thanks guys for your replies. As usual they have helped me massively in understanding what the course only glance over. It's one thing to know it in a book, it's another to apply it to the real world.
Can I ask a favour from one of you. I have designed a circuit for a client. My mate is the one signing it off, but rather than present him with a pile of crap, could I message one of you to have a look and critique it please.

Many thanks again

 

[James]

Thanks guys for your replies. As usual they have helped me massively in understanding what the course only glance over. It's one thing to know it in a book, it's another to apply it to the real world.
Can I ask a favour from one of you. I have designed a circuit for a client. My mate is the one signing it off, but rather than present him with a pile of crap, could I message one of you to have a look and critique it please.

Many thanks again

Either post it here in the thread, or if you don’t want to post it publicly, feel free to send it to me.

 

[Mark Wright]

Can you take a guess at the way the OSG length was calculated and how the loads were spaced? knowing the OSG each light was probably rated 100W/0.44A.
Don't worry if this is too much work just curious.

 

[Dan Carroll]

So the scenario. I’ve been asked to design this for a friend, who is the one signing it off. I know its on his call. But also I know that bad habits creep into peoples work. So I would appreciate what you guys say as well.
Here’s what I am doing anyway.

1. Update home consumer unit and move to garage

1. Install DB in spa and associated cabling & garden lighting

1. Install but not connect cable for future annex at bottom of garden

1) Update home consumer unit.
Currently a 6 way re-wireable fuse board with 3 x 30 amp powering 2 rings (1 with an unknown spur) and presumably the cooker, and 2 x 5 amp seemingly powering 3 lighting circuits.
Board is located in the meter box.
New CU to be placed on inside of wall and across into garage about 2m from current position
Henley block tails coming from meter and run extension tails to new board.
The new CU will be an 8 way board for spare capacity. Circuits replaced respectively with RCBO’s (3 x 32 amp, 1 x 20amp & 2 x 6 amp) And 63amp MCB type B for distribution to DBs.
2) Install Spa DB and cabling. Spa wood built
SPA power requirements:
Estimate 40 Amps
Hot tub. Standard requirement 32 amp radial
Sockets. radial 16 amp protection.
Lights. A few LED internal & external lights. 10 max @ 50W = 2.2amp. 6 amp protection
Garden lights. 20 LED all around the garden at regular intervals. @ 50W = 4.3 amp. 6 amp protection
Spa DB to be 4 way board with RCBO’s. DP 60 amp isolator
Cable to run from 63amp MCB at CU, underground to Henley block on the spa building. (Henley block on spa for future power to lower annex.
Hot tub will run from RCBO TYPE C to 45 amp isolator, to hot tub position
Garden lights will run around whole garden in SWA with lights approx. every 2 meters. The SWA will come up to a wiska box and the light will take power from that
3) Annex. Brick built with wiring in walls
Estimate 20 amps
Annex power requirement:
Potential for shower/water heater max 5kw
Sockets 16 amp
Lighting, A few LED lights both internal and external.
Building will not be built until next year.

Supply cable for distribution.
Run underground to spa into Henley block, then (disconnected for time being) underground to annex.
House is currently PME. No extraneous parts in either building so no need for TT
Volt drop calculated by taking mV/A/m x Ib x length / 1000 then divide by 2 as load isn’t at the end.
Cable spec:
Distribution,
16mm SWA, 25m to spa, 25m to annex. 50m total. Capacity= 91amp Volt drop = 1.6V at halfway, 3.2V to annex
Garden lights
1.5mm SWA, 65 m around garden from spa. Capacity= 25 amps Volt drop = 4V
Other cable as standard

 

[Dan Carroll]

Is the silence a good thing or a bad thing

 

[James]

What does your friend who is signing the work off think of it?

There are many knowledgeable people here who will gladly give there time up to assist in your learning.

This may not be the case in this instance but,
Quite a few diy, trainee, or chancers of some sort post things like this with the hope that they will get an endorsement from some qualified sparks.
Most have a “friend” or other contact who will be signing it off.

The reality is that this imaginary signature giver should be involved from the beginning to design, specify and oversee the work from start to finish.

It is not possible to have someone come in after it is installed to “check it over and write the certificate “
When you are signing an installation certificate, you are confirming that you are responsible for the design, installation and testing of it.

That, I am afraid is why the thread has gone a bit quiet.

 

[Dan Carroll]

OK well I appreciate what you are saying.
Allow me to say.
I completed my 2365 level 3, 18th and part p course last year. I am qualified enough now to actually become a certified domestic installer. I haven't because I want to get the NVQ and thought that I would be working for someone by now and therfore it would be wasted money for something in not using. I might revisit that option.
I come from a background of 16 years as a mechanical helicopter engineer, where safety is paramount. We were trained to never be afraid to ask a question. Even when we were trained to do all the jobs, there is a vast difference between a book and real world answers.
One person doesn't know everything and never be afraid to challenge those above you. They may have been making the same mistake with out realising it for years.

I do fully understand what you are saying, and I imagine there are people out there who would use this as authorisation to do the work.

My friend is happy with the design.
With the training I've had, I'd be happy to sign it off.

Please don't take this message as a rant or moan.
All I was after was some guidance that I'm going down the right track and my friend is correct as well. I'm not after someone to do the work for me.

 

[EricMark]

I did a quick calculation and get around 44.85 meters, correction factor Ct of 0.8854166666666666 yes I used java script which is why such a long number, but without the correction factor you will not get a true reading.

The formula for the correction factor is not simple, and I realised too easy to make an error, so by writing a program using java script it allows one to enter know figures, we know a ring final allows 106 meters and uses 26 amp as the design current Ib. So since the program gets that right, there is a good chance I have got the maths correct.

 

[Mark Wright]

I did a quick calculation and get around 44.85 meters, correction factor Ct of 0.8854166666666666 yes I used java script which is why such a long number, but without the correction factor you will not get a true reading.

The formula for the correction factor is not simple, and I realised too easy to make an error, so by writing a program using java script it allows one to enter know figures, we know a ring final allows 106 meters and uses 26 amp as the design current Ib. So since the program gets that right, there is a good chance I have got the maths correct.

Can you upload the source?