New Zealand Volt drop cont.

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New Zealand Volt drop cont.

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Monkeyblaine

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Ok I have another VD question. I'm installing 9 LED floods around my friends garden which totals at just over 100m in length. Obviously for each branch the VD varies so for argument sake I worked It out as if all the cable would be taking all the load (rather than dropping relevant current after each light, worse case scenario) the load is minimal being LEDS but to be safe I'm putting 2.5mm 3c hightuf in to cope with the distance and to help for the zs reading. Now I am also using a wise box control box (wireless switching) now am I right in thinking I can supply this box from the CU with a length of 1mm t+e as this is a very short distance and is my thinking correct that as long as no voltage has been dropped through the 1mm to the wise box then I have a full 230v (give or take) for the garden circuit and as its run in 2.5mm it will stop any VD issues. Basically what I'm asking is, whatever voltage you have at a certain point regardless of cable size, then that's what you have from that point onwards. I.e if I lost 3volts through initial 1mm cable then I'm down to 227v at wise box (for example)
 
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It's one reason why branched radials are that bit more tricky to work with, and why many folk avoid installing them, even if we do have the IET's blessing to install them. Personally, I have a branched 4mm 32amp radial supplying my kitchen sockets, pretty much as Richard has described. I wouldn't have installed it this way if rewiring from scratch, but as I added a bank of sockets to one side of the kitchen and left the existing sockets in place on the other side, it was the only way to do it without making a mess. Richard has described things pretty much bang on, the 'main' supply cable will be subject to all the voltage drop from both branches but as the circuit has two 'ends' I needed to calculate it twice. Actually it wasn't much of an issue in my example, cos it was an extremely short cable run, but you get my gist! :)
 
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Engineer54 said:
I'm not sure i'm missing something here or not, but ....
I'm not quite following what your saying here?? What has the voltage at the origin (apart from complying with said norms, eg, 240 true/230 make believe) got to do with calculating the volt drop of a branch circuit over given distances and known loads?? We have formula's for calculating VD regardless of what the voltage present at the origin is, ....that voltage will typically remain constant, it's the design of the circuit and the cable/conductors sizes selected, that will determine if a VD anywhere on that branch circuit complies or doesn't comply....
Click to expand...

You have missed it completely. Monkeyblaine is discussing calculating the volt drop on different parts of a circuit (lighting) and then applying these values to those parts of the circuit and sizing the cable accordingly, a clear nonsense because the volt drop must be calculated across the length of the circuit as the maximum VD will be on the leg from the protective device to the first accessory/equipment.

I don't get what you're not getting?
 
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Richard Burns said:
Monkeyblaine I think your method of explaining can leave others wondering what you mean (and possibly me as well!).
However I believe that what you are saying is that if you have a branched radial circuit then the volt drop of one branch is not added to other branches.
In this you would be correct.
If you assume you have a circuit with a supply cable going to a junction box and then splitting to two cables (one going left, one going right) each of which has two luminaires, each at a different point along the cable.
The volt crop calculation would include the cable from the CU to the point of the first branch for the full load of the whole circuit; added to that would be the volt drop to the first luminaire on the left hand branch using the current load for two luminaires; added to that would be the volt drop to the last luminiare on the left hand branch using the current for the last luminaire. This would give the max volt drop for the left hand branch.

Then you would calculate in the same way for the supply cable, the cable to the first luminaire on the right hand branch and the cable to the last luminaire of the right hand branch. This would give you the maximum volt drop on the right hand branch.

The highest of the left hand and right hand calculations would be the maximum volt drop for the whole circuit.
Click to expand...

Richard, in the example you have given you are wrong i am afraid to say.

Voltage drop can only be calculated along the entire length of a circuit for design purposes. In your example the Voltage drop on the supply to the JB is ZERO because to measure this properly you need to disconnect from the branch to avoid loading by the installed fittings, and as there would be no load then there would be no volt drop.

This section of the circuit will suffer the maximum voltage drop on the circuit, the voltage drop on the branch sections is irrelevant in this example because the circuit should be designed to take into account all the volt drop along it's length.

This is why we use the equation VD=Mv/A/m
 
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Haha once again totally confused. Oustpoken in your example is that the same if you take a feed in and out numerous times across the length of a circuit in different jbs. And at each JB A few lights branched out. Surely at each JB. You start with a different Voltage as volts have been lost over previous joints. So the last JB to the last light, Adding all those volt drops from origin to each JB (which will be different to each point) then that total is the
Max VD for that circuit. Surely that's right??
 
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Monkeyblaine said:
Haha once again totally confused. Oustpoken in your example is that the same if you take a feed in and out numerous times across the length of a circuit in different jbs. And at each JB A few lights branched out. Surely at each JB. You start with a different Voltage as volts have been lost over previous joints. So the last JB to the last light, Adding all those volt drops from origin to each JB (which will be different to each point) then that total is the
Max VD for that circuit. Surely that's right??
Click to expand...

Is this a ruddy wind up or do you simply not see it...Calculating the individual legs of a circuit will give an inaccurate final figure, you need to calculate for the whole circuit...see this I have just knocked up for you..

Volt drop.jpg
 
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Outspoken said:
Is this a ruddy wind up or do you simply not see it...Calculating the individual legs of a circuit will give an inaccurate final figure, you need to calculate for the whole circuit...see this I have just knocked up for you..

View attachment 19414
Click to expand...


what software do you use for these drawings? i've got visio and am struugling with it. it's too complex for this type of drawing work.
 
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Tele, I use AutoCad for proper stuff, but this and the others you have seen me post on here are all done in Visio chap. I create them as a drawing, save as a drawing, then save it a second time as a jpg for posting here.

Patience my friend...:D
 
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I was taught that way at college. The reason I'm struggling with this is because not the entire length of the circuit is taking all the load stated. So therefore they must be dropping different voltages across certain lengths. That's why I don't get it. Clearly I've been taught wrong. Sorry to frustrate you.
 
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liken it to a bus load of 60 passengers. bus drops off 10 at the first load. so it's now got 50. at the 2nd load it drops off 5 more. so now its lost another 5 on top of the 10 already lost and now has 45. . and so on.
 
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That is exactly what I'm saying In my earlier post. With a circuit with say 5 joints down its length with lights coming off joints it's dropping volts at every joint. So the VD will be at its worse at the last joint. It's what I said and outspoken said that's not right???
 
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the lowest voltage will befoundat the kast load, but the biggest vd is on the 1st section, as that section from the ocpd will be the only part of the circuit carrying the full load.
 
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no. max. vd for the circuit is from ocpd to last load. the biggest vd is from ocpd to 1st load.
 
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There's conflicting information going on here. Outspoken info understand what your saying and just because at the start of the circuit is the max VD as it has the most load upon it. Doesn't mean it's not going to drop more by the time it reaches furthest point and got operation of things working, that's what im interested in. So an accumulation of volts dropped by the time it reaches last point. Which is surely what the regs require us to obtain.
 
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Monkeyblaine said:
There's conflicting information going on here. Outspoken info understand what your saying and just because at the start of the circuit is the max VD as it has the most load upon it. Doesn't mean it's not going to drop more by the time it reaches furthest point and got operation of things working, that's what im interested in. So an accumulation of volts dropped by the time it reaches last point. Which is surely what the regs require us to obtain.
Click to expand...

Monkey, FFS, the calculation of the TOTAL voltage drop across the entire circuit tells you the voltage at the END of the circuit FFS....

If the Volt drop is calculated at 4V (Thus dropping to 226V) the the voltage at the end of the circuit is ...Yep, you guessed it 226V ...it is not Rocket Science!
 
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i'm bowing out before i suggest that you got 10V to play with anyway as we have 240V at source, but calculate VD from 230V. courtesy of the meddling euro rats in brussels.
 
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I'm not stupid and I do get what you are saying. I know how to calculate the entire load of a circuit then use VD calcs to obtain total VD. I was always taught that you had to work out each leg individually. As the way your explaining it is assuming the total load is at the end of the circuit rather than it branching off at various points. Do you not see my point. But I will go with everything you say as you clearly have worlds more knowledge/experience. And if in your examples is correct way to
Determine total VD for a branch circuit, then I will leave it as that. As its a simpler way to calculate.
 
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I had considered pointing out that if you measure voltages in the UK the averages for single and three phase are 236.7V and 415.8V respectively...but realised he was confused enough as it is!! :)
 
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Monkeyblaine said:
I'm not stupid and I do get what you are saying. I know how to calculate the entire load of a circuit then use VD calcs to obtain total VD. I was always taught that you had to work out each leg individually. As the way your explaining it is assuming the total load is at the end of the circuit rather than it branching off at various points. Do you not see my point. But I will go with everything you say as you clearly have worlds more knowledge/experience. And if in your examples is correct way to
Determine total VD for a branch circuit, then I will leave it as that. As its a simpler way to calculate.
Click to expand...

Monkey, FFS, whoever taught you that is a frigging idiot as you will always end up with incorrect figures, that is why i gave you the example I did, to PROVE you MUST calculate the total maximum demand on the maximum length of the circuit to establish the true maximum volt drop you may experience. Obviously in the example I gave this will be variable as lights are turned on and off on various branches, thus you need to calculate the worst case scenario, if this is within the permitted levels with BS7671 then everything else is a pointless waste of your time!
 
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Monkey, FFS, the calculation of the TOTAL voltage drop across the entire circuit tells you the voltage at the END of the circuit FFS....

If the Volt drop is calculated at 4V (Thus dropping to 226V) the the voltage at the end of the circuit is ...Yep, you guessed it 226V ...it is not Rocket Science!

Well put and better than a thousand diagrams :)
 
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