Volt Drop

[switched on]

Hi I have a query one of you may know the answer to. I have installed some lights in the woodlands for a customer and he is using 12watt LED lights he is only installing a few i think it was 186 watts or something. I have used SWA cable 1.5 mm as it is about 200metres and worked out the voltdrop. He now wants to extend it another 100m and add on a few more lights!! I have told the customer that it will incur too much volt drop as the resistance on the cable.

For my own sake was wandering the dangers of volt drop obviously appliances wouldnt work correctly as the volt drop and the resistance is high so surely it cant do the cable any good.

Was wandering what you guys would do and if you have any past experiences or knowledge.

Thanks

 

[telectrix]

so, from that, a ball park figure would be approx 50% of the VD if it were a single load at the end of the cable.

 

[Knobhead]

You work from the 220V source to the first fitting to give 220V-Vd1 as the voltage at the 1[SUP]st[/SUP] fitting. On to the second fitting where (220V-Vd1)-Vd2=voltage at the second fitting. At the third fitting the voltage is ((220V-Vd1)-Vd2)-Vd3 and so on. The volt drop gets exponentially smaller with each step due to the progressive reduction in voltage and current along the line. The end result is towards the end of the line voltage drop is reduced as is the current.

If any of you are in to doing excel spreadsheets it’s a fantastic maths problem!

(**** I’ve started something now, this is going to take some time)

 

[JUD]

You work from the 220V source to the first fitting to give 220V-Vd1 as the voltage at the 1[SUP]st[/SUP] fitting. On to the second fitting where (220V-Vd1)-Vd2=voltage at the second fitting. At the third fitting the voltage is ((220V-Vd1)-Vd2)-Vd3 and so on. The volt drop gets exponentially smaller with each step due to the progressive reduction in voltage and current along the line. The end result is towards the end of the line voltage drop is reduced as is the current.

If any of you are in to doing excel spreadsheets it’s a fantastic maths problem!

(**** I’ve started something now, this is going to take some time)

Already done 1 a while back

 

[JUD]

All this talk about volt drop.

Has anybody considered the circuit impedance with regards to disconnecting the circuit in the event of a short circuit.

The R1 + Rn of 200m is 4.84Ω and would be OK.

However, 300m would push it up to 7.26Ω and that is before taking into account the impedance of the supply cables and transformer winding.

As the maximum measured loop impedance (using the 80% rule of thumb) for a 6A Type B breaker should be 6.18Ω you would already be over.

You can get away with 1667Ω for the EFLI by using an RCD but that won't work for short circuit as we all know.

Just a thought.

 

[ackbarthestar]

All this talk about volt drop.

Has anybody considered the circuit impedance with regards to disconnecting the circuit in the event of a short circuit.

The R1 + Rn of 200m is 4.84Ω and would be OK.

However, 300m would push it up to 7.26Ω and that is before taking into account the impedance of the supply cables and transformer winding.

As the maximum measured loop impedance (using the 80% rule of thumb) for a 6A Type B breaker should be 6.18Ω you would already be over.

You can get away with 1667Ω for the EFLI by using an RCD but that won't work for short circuit as we all know.

Just a thought.

Very good point.
That's probably why it should be a ring or the extra 100m run as a separate circuit.
I did a basic Volt drop calculation on the same idea as yours but using 4 1A loads at 10m intervals with a theoretical volt drop of 10mV/A/m and found that the basic calc (end of line * total load) gave a VD of 1.6V and doing successive addition of volt drops with diminishing loads gave a VD of 2V .
Not that it makes much difference since you have already proved the potential addition to the circuit has flawed.

 

[pushrod]

What would the volt drop calc be if it was converted to a ring - use total load and effectively quarter the distance?

 

[Engineer54]

I notice that the OP hasn't come back since starting this thread, as so often happens here!!!

 

[ackbarthestar]

What would the volt drop calc be if it was converted to a ring - use total load and effectively quarter the distance?

Using my example you would need to increase the end load by one length so that you could get back to the supply.
The mV/m/A would now be 1/2 but the load would stay the same but with less effective outlets (paralleled) giving a total VD of 0.45V which is approximately 1/4 of the difference of 2V and 1.6V.