Question About Volt Drop

[JohnW2]

Indeed so - but that's unlikely to be the question, except as an educational/exam exercise.

But the first few words of the OP state that "As part of my studies, I need some help with a question " which looks like an educational exercise to me!

It does - but, as you've said, for it to be an educational exercise that could be expected to be tackled at sub-degree level, one would have to know all the loads (as well as cable lengths) and the OP gave no indication that they had been specified. As Eric as indicated, there is no single answer to the question as stated (with information provided).

Kind Regards, John

 

[JohnW2]

Don't imagine confusion. Imagine some numbers instead. One will do, actually.

Obviously very OT, but .... I've always thought that one of the reasons that so many people have a problem with imaginary numbers is the way in which they are usually introduced by teachers. Teaching usually starts off by defining 'j' as the sqaure root of -1 (which appears impossible), and then all follows (or doesn't follow, for those already confused!) from that. If it were first introduced as a vector operator representing a 90 degree rotation, followed by an explanation that applying it twice ('squaring it') had the same effect as multiplying by -1, then I think many people would get 'into it' far more easily!

Kind Regards, John

 

[Spark123]

As someone said before, just use ohms law (with the figures from the big green book), work out the current in each leg, volt drop and tot them up.

No need to use Nortons or Thevenins (wasn't that to work out equivalent circuit impedances?) or complex numbers with complex conjugates etc.

You can also have a spur off a radial circuit - a spur is to do with a reduction in CSA hence a 2.5mm leg off a 4mm radial will classed as a spur.

 

[ban-all-sheds]

Teaching usually starts off by defining 'j' as the sqaure root of -1

'i'.....

 

[JohnW2]

As someone said before, just use ohms law (with the figures from the big green book), work out the current in each leg, volt drop and tot them up. ...No need to use Nortons or Thevenins ...

Yes - but, as I wrote to stillp, you can only do that for a given set of loads (currents in each leg) - which may be fine as a classroom exercise, or even in a situation of fixed-wired loads, but is useless in practice in relation to a sockets circuit (where the loads, and distribution of loads, are not known).

The quetsion of practical importance is clear enough. If I presented you with the cable lengths and socket positions of, say, a 20A radial circuit and asked you to ascertain the maximum VD that would be experienced at any socket (in order to confirm compliance with 525 of the regs), without the entire circuit being overloaded, how would you work out the answer for me 'Just using Ohm's Law'? As I see it, you could only do that with a lot of trial and error (considering different distributions of loads) and, even then, you would not necessarily be certain that the answer you had produced was the maximum possible.

You can also have a spur off a radial circuit - a spur is to do with a reduction in CSA hence a 2.5mm leg off a 4mm radial will classed as a spur.

That is, indeed, the sense in which Appendix 15 of BGB uses the word 'spur'. However, the definition of 'spur' in Part 2 of the BGB is much more general, and potentially confusing - simply "A branch from a ring or radial final circuit".

Kind Regards, John

 

[JohnW2]

Teaching usually starts off by defining 'j' as the sqaure root of -1

'i'.....

Interesting one, that. In mathematics it is, indeed, usually 'i'. However, in engineering disciplines, 'j' is nearly always used - and since this is an electrics forum ....

Kind Regarsd, John

 

[ban-all-sheds]

You can also have a spur off a radial circuit - a spur is to do with a reduction in CSA hence a 2.5mm leg off a 4mm radial will classed as a spur.

So what's a 4mm² leg off a 2.5mm² radial? (Done either because that was all that was available, or because that run is through insulation).

What if after that section it reverts to 2.5mm² ?

What if you had a 2.5mm² radial circuit - one simple cable run with no branches, and then at some time after installation you upgraded one of the runs between 2 sockets for thermal insulation reasons? Would the first length of 2.5mm² after that be a spur? Would the rest of the circuit after that be a spur?

What if the first run of cable from the CU had to be 4mm² but it became OK to drop to 2.5mm² before the first socket? Would you have a circuit which consisted entirely of spurs?


General Principle:

If you find that a definition creates inconsistencies or nonsense then the definition is flawed.

 

[ban-all-sheds]

Interesting one, that. In mathematics it is, indeed, usually 'i'. However, in engineering disciplines, 'j' is nearly always used - and since this is an electrics forum ....

But I was taught maths at school, not engineering

 

[JohnW2]

You can also have a spur off a radial circuit - a spur is to do with a reduction in CSA hence a 2.5mm leg off a 4mm radial will classed as a spur.

So what's a 4mm² leg off a 2.5mm² radial? ... What if ... What if ... What if ....
General Principle: If you find that a definition creates inconsistencies or nonsense then the definition is flawed.

I agree - and, as I've pointed out, the BGB is actually internally inconsistent in this respect.

Kind Regards, John

 

[JohnW2]

Interesting one, that. In mathematics it is, indeed, usually 'i'. However, in engineering disciplines, 'j' is nearly always used - and since this is an electrics forum ....

But I was taught maths at school, not engineering

Do I take it that you didn't do A-Level Physics which (at least in my day) qualified in this respect as 'an engineering discipline' (i.e. it used 'j')?

Kind Regards, John

 

[ban-all-sheds]

I did do A-level physics, and A-level maths. I guess my school was internally consistent.

 

[JohnW2]

I did do A-level physics, and A-level maths. I guess my school was internally consistent.

Ah - mine certainly wasn't. I'm not sure it actually had that much choice (without risking even more confusion) since, IIRC, the textbooks of Maths and Physics differed in this respect!

Kind Regards, John

 

[bernardgreen]

It is not that difficult.

Break the network into sections of cable joining nodes. A node is formed at every junction and every change in type or size of cable.

Work out the current in each section, use Ohms law, the current and the resistance of the cable to calculate the voltage drop for each section. Then add the voltage drops along the path of interest.

In the original posted question there are three sections of voltage drop between CU and no 4 that need to be calculated and added

 

[JohnW2]

One may try guessing/deducing what might be the 'worst case situation' but (without thinking too deeply) I somewhat doubt that would always necessarily give the right answer (I'll think a bit more about that!) - which, as far as I can see, leaves one stuck with either trial and error or clever maths.

OK, I’ve been thinking about this, and have got to ‘first base’ ....

I think that one can easily deduce the arrangement of loadings which will result in the maximum voltage drop at the socket which has the longest path back to the CU. Assuming, for simplicity, all cables of same CSA, that situation will, I think, always arise when that ‘furthest socket’ is maximally loaded (** see below) and the remainder of the available current (i.e. circuit ‘design current’ / In of the OPD) being drawn by socket(s) which have the greatest length of common cable path with the furthest socket.

[ ** If one is talking about a 20A radial and the ‘furthest socket’ is a double one, then, if one worked on the basis that a double socket can supply 20A (although some would say 13A), the problem obviously becomes trivial , because the greatest VD see-able on the circuit is that which arises when there is a 20A load on that furthest socket. To get into ‘more interesting’ scenarios, one therefore has to consider single sockets and/or a 32A radial.]

What I have yet to convince myself is that the situation which results in maximum VD at the furthest socket is necessarily the situation which results in the maximum VD at any socket. ‘First impression intuition’ suggests to be that such probably is the case, but I am not yet totally convinced. Watch this space!

Kind Regards, John

 

[JohnW2]

Something totally bizarre is happening. My penultimate post seems to have vanished, but when I click to 'edit' my latest one, I see the penultimate one in the edit window!. Just to be sure, the message which seems to have disappeared was:

It is not that difficult. Break the network into sections of cable joining nodes. A node is formed at every junction and every change in type or size of cable. Work out the current in each section, use Ohms law, the current and the resistance of the cable to calculate the voltage drop for each section. Then add the voltage drops along the path of interest. In the original posted question there are three sections of voltage drop between CU and no 4 that need to be calculated and added

Yes, but as I keep saying, that's no use in the practical (as opposed to classroom exercise) situation (of wanting to know the maximum possible VD at a given socket), because one doesn't know the various loads (hence 'the current in each section') which will be operative in the situation which gives the maximum possible voltage drop at a particular socket. One may try guessing/deducing what might be the 'worst case situation' but (without thinking too deeply) I somewhat doubt that would always necessarily give the right answer (I'll think a bit more about that!) - which, as far as I can see, leaves one stuck with either trial and error or clever maths.

Kind Regards, John