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Posted By Topic: Voltage drop over distances

Kelectrical
Jan 29 2019 18:43

Hi guys,

I've always had an issue understanding fully the concept of calculating voltage drop when you go from one cable size down to another which I was hoping that you could help me to understand.

If I have a long power circuit run with 2.5mm from a 20A circuit, to compensate for the voltage drop, we run 4mm to the first outlet and then 2.5 after that.

How does the calculation for this work?

Am I right in saying that if the run is for example 40m, we run 10m of 4mm to the first power outlet and then calculate the rest of the run as 30m to ensure voltage drop is correct?

Could I do the same for a mains run?

E.g if I ran a 25mm mains cable protected by a 63Amp fuse 40m and then the remaining 20m with 16mm, is this acceptable?
   

pluto
Jan 29 2019 19:57

On the EWRB website in the sreach box write past exam questions. when the page loads then look for Electrical regulations and the exam paper ER 70 questions than look at question 9 and then at ER 70 answers for question 9.

This will give similar exm question and the model answer for the question you have asked above.
   

Kelectrical
Jan 29 2019 20:24

Thank you Pluto, I'll try to figure it out. I appreciate what you do for the community!
   

Kelectrical
Jan 29 2019 20:34

over 50m I calculate 25mm at 2.43% voltage drop (using an online calculator)

and over 20m I calculate 16mm at 1.54%

Making the total well under 5%

Could I therefore join these cables, 25mm from the pole fuse (63A) and 16mm the rest of the way to the main switch?

Just trying to get my head around this and make sure I completely understand.
   

gregmcc
Jan 29 2019 20:45

that will give you 4% volt drop on your mains, leaving 1% max volt drop for the final sub circuits, generally allow for up to 2.5% for mains and 2.5% for final sub circuits otherwise your will end up having to run much bigger cables for power and light circuits which could well cost more than increasing the size of the mains to lower it's volt drop

   

DougP
Jan 29 2019 21:25

Section 3.6, including reading exception 1. Then try some calculations.

And as gregmcc says, the 5% is from the point of supply to the end of the final subcircuit. So you first need to know the VD on the mains, before you know how much of the 5% you have left to work with.
   

AlecK
Jan 30 2019 09:16

Personally I've never bothered with trying to work in % units.
Nor with the packaged systems such as Gencalc (as was); i found those OK for a single rune, but unable to handle complexity.

I was brought up using the data that cable manufacturers provide, ie mV per amp per metre.

it's the only way I've found to deal with examples like this where there are several lengths of cable of different sizes. And as you work through you get an understanding of it - which you'll never get from just plugging numbers into a program.

Starting from the 5% max allowance, turn that into volts; or better still mV. 5% of 230 V is 11.5 / 11500 mV.

1st step; as gregmcc said, is allow for some volt drip in the mains. If you need a truly accurate answer, you need to know the cable size, route length, and current, and do a calculation that might look like 25 m cable length x 63A max load x 2.81 (for 16 mm2; from Table 42 of "3008.1.2", using the column for cable temp 75 deg C).
So 25 x 63 x 2.81 = 4425.7 mV.
Now subtract that from the max allowed : 11500 - 4426 = 7074. No need to turn that into %, just use it as the available mV at MSB.

Alternatively, many people just allocate half the 5% total allowable to mains, and start at mSB with 5750 mV. Years ago we had to use this "2.5 % from MSB" figure; and it saves a little bit of time while (usually) keeping us on the safe side - especiallt for an existing installation, where we probably don't know the cable size or the distance with any certainty. So I'll use it for the rest of this example

For the subcircuit, you just do the same thing again, for this example in two stages for the two lengths.
First stage(4 mm2): 10 m x 20A x 11.22 = 2244.
So at end of 4 mm2, there will be 5750 - 2244 = 3506 mV "available".
second stage (2.5 mm2): 30 m x 20 A x 18.02 = 10812 ; which roughly 3 x what you have available.

That's pretty much "worst case", arrived at by using max available load [option (c) of 3.6.2]. On a socket circuit, I think it's reasonable to consider that users may well want a full 10 A on the last socket, as well as another 10 A somewhere else on the circuit. By applying the full 20 A at last point; the calc is simpler, and the result errs on the safe side.
Alternatively, as DougP said,you can apply Exception 1 to 3.6.2; and use half the rating of the mcb.
So, since this first calc ended up indicating excessive VD; I'll do it again using the reduced current, which is dead simple 'cos it will be exactly half the figure for 20A. Also usingh the lower figure for mains arrived at by previous calc instead of just assuming 2.5%.

7074 mV "available" at MSB.
7074 - 1122 gives 5952 mV at step-down.
5952 - 5406 = 546 mV "spare".

I'd accept that, since the current in the circuit is stated to be OK to use; the current in the mains is worst-case,and the VD figures are for conductors at max permitted temp - which they will never reach. So any error is on the safe side. File the calc in the job file; and if there's ever any question you can prove that you complied.

That takes a long time to explain, but stuff-all time to actually do. The hardest bit is getting the figure for the cables. "3008.1.2" gives 3-phase values. But it doesn't take much effort to calculate & save / print off single-phase values for the cable types you generally use.

App C gives "simplified volt drop" in C4 & Table C7; doing the calc in %.
Even simpler, can use Table B1 in App B of 2018 edition; again in %.
But both of these provide a max distance you can run; which actually makes them difficult to apply to a mixed-cable circuit.