Calculating volt drop finding ib

[Jamie']

I'm an apprentice and I was going through my college work to do with volt drop and everything I needed to know about the design current was in the question. What I'm confused about is in the real world how you would know this?
Am I right in thinking for arguments sake if I'm supplying an out house with 16mm armoured and it's going into a 50amp breaker that then becomes my design current?

Any help is appreciated!

 

[Richard Burns]

For any electrical circuit to be designed the anticipated loading of that circuit needs to be known.
If you start with the Iz>In>Ib then obviously before you start anything you need to know Ib (the design current) before you can select a protective device and you need to know both the preceding before you can select a cable.

Assess all the anticipated loads and sum them and use this as your design current. Where loads are variable such as socket circuit make a likely assessment based on the the intended use of the area, bedroom sockets would be very under utilised maybe only 5-10A max, industrial kitchen sockets would be over utilised and a design current of the breaker size would be appropriate.

As an example in your scenario of an out house the first thing you start with is what will be used in the outhouse.
Are you supplying a socket circuit and what is intended to be used on this socket circuit (in general), are you supplying lighting (and if so what type of lighting), are there any dedicated loads for say electric heating and if so what rating are the heaters.
So if you were having a socket circuit with ten double sockets to be used for general use, occasional lawnmower duty, a 500W pillar drill and a 3Kw kettle.
A lighting circuit for a 500W outside floodlight and two twin 58W fluorescent fittings inside. A dedicated circuit for two 1kW panel heaters.

So starting in the outhouse a 16A radial for the sockets is probably too low as you have 13A from the kettle to start with and anything else pushes the load up. With ten sockets there could be a lot of use and depending on the length of the circuit to the outhouse you want to minimise voltage drop so a ring final circuit at 32A may be appropriate. This can only be an estimate for sockets as loads can change dramatically. But you might expect that with an outhouse and limited use the total load may reach 20A for a short while at times. So the design current for the sockets is 20A.
Lighting is a total load of 732W just over 3 A but the fluorescent will have a high inrush current so it is recommended to apply a 1.8 factor to account for this so 1.8 *232 = 418W so give a total amps of ~4.
The heaters take 8.7A.

So 20A plus 4A plus 8.7A =32.7A as a design current for the outhouse, so a 40A breaker at the supply may be fine but if you need to consider discrimination to prevent the main breaker tripping before the outhouse breakers then different choices may be made.
Perhaps you want to take into account some future plans in place and provide another 10A of load availability so you put a 50A breaker for the main supply.

We now have a circuit for an outhouse with a 50A breaker but the design current at the moment is only 32.7A (possibly increasing to 43A later). Therefore you would size your cable based on a 43A design current when considering voltage drop.

 

[Startinez]

I'd say it depends on the load type, personally.

If you are using a fixed appliance at the other end (motor, charger, etc) then you can assume that the load is going to remain relatively stable at all points after startup/switching. So let's say a motor is drawing 13A in use but you've got a C20 MCB covering it, then that's a big difference (20A-13A) in v.d. you're accounting for when realistically it won't reach the upper limits.

However, if you take a ring, then your load will differ hour by hour, day by day, and the only real way you'll cover that is to go with the max. current which will be what your protective device will allow.

 

[Jamie']

Cheers guys!