Garden lighting.

ulfr · Jun 15, 2014

Hello everyone,
I'm new in the industry, and soon will get 2 Level in Electrical installations.

At a class tutor gave us a homework, to make a diagram, and all necessary calculations for cable size, voltdrop, Zs, e.t.c What's needed in "real world".

I have a few questions about a schematic(?) wiring diagram, and in general. Pics are below.
For example, customer asked for 5 separate circuits(each of them will turn on individually), and 1 switch to turn all 5 circuits at the same time.
Length - 84m of wires required
Circuit 0 – all at the same time - 2140W / 230V
Circuit 1, 2 flood lights, 250W x2
Circuit 2, 2 beds 4 garden lamps, 120W x4
Circuit 3, 2 flood lights, 250W x2
Circuit 4, 4 tree lamps, 4 x 120W
Circuit 5, 4 brick lights, 4 x 40W
Questions are:
1) Is it possible to wire in that way, like in the pic2?
2) Which circuits are final here if a customer will turn on Circuit 0?
3) Customer can turn on circuits randomly e.g. 1-5, 1-2-3, 2-5-0. Then all circuits should be counted as final, and therefore 1.8 diversity factor should be applied to counting?
In that case:
Maximum demand = (2140W/230V*1.8(A1(Note 2)) * 0.66(A2) = 11.05A
Voltdrop (Reference method A)= (7.3*11)/1000*84=6.77V
Therefore, 6mm2 cable should be used. But it's quite costly, isn't it?

View attachment 25128 View attachment 25129
wires will be buried in the ground.

Thank you.

ulfr · Jun 17, 2014

Image 1

Image 2

Marvo · Jun 17, 2014

I've moved this thread to the trainee forum.

Richard Burns · Jun 18, 2014

I am afraid I cannot follow what you have drawn in pic 2 as your wiring arrangement.

Since each section of the circuit needs to be individually switched, presumably at a central location, you would have to have separate switched supplies to each section.
It does not appear that you have this in the diagram, although if a cross is a switch or a light then you may have some switches in place local to the light.

This is effectively one single final circuit with individual sections switched separately. If you think of a domestic lighting circuit this is one circuit with each room individually switched.

The 1.8 factor is only applicable to discharge lighting which I would not expect you to be using in this case unless you are using metal halide floods.
For real world calculations take the maximum potential wattage for each light fitting and use the current drawn as the design current for that fitting (unless it is discharge lighting)
If you have an arrangement like this

Then you can calculate the maximum volt drop from
the start of the circuit to the end of circuit one
the start of the circuit to the end of circuit Two
the start of the circuit to the end of circuit Three
the start of the circuit to the end of circuit Four
the start of the circuit to the end of circuit Five
The highest of these is the maximum volt drop for the circuit.

To do this
Use the length from the start of the circuit to the first (all off/on) switch and the total current (9.3 A) to calculate the initial section's volt drop.
Then for each section calculate the volt drop (in that section) using the length from the individual section switch to the end of the individual section and the current in that section (e.g 2.17 A for section one (2*250 W))
Add the initial section and the individual section's volt drop to get the volt drop for that part.
e.g. if the distance from the CU to the section one switch were 5 m and the distance from the section one switch to the end of section one were 10 m.
Then you have for say 2.5mm cable (Vd 18 mV/A/m)
Vd * Ib * L / 1000
18*9.3*5 / 1000 = 0.837 V
plus
18 * 10 * 2.17 / 1000 = 0.391 V

so 1.228 V drop for section one and so on .

Work out your volt drop in this way and you should find you need much thinner cables.