Let’s take a look at what are possibly the three most used calculations within the 18th Edition Wiring Regulations for electrical installations.
- Calculation Voltage Drop
- Calculating the cross-sectional area of protective conductors
- Tabulated current carrying capacity of conductors ‘It’
1. Calculating voltage drop…
You won’t find this formula anywhere in BS7671:2018 as this is another example of were they assume some prior electrical theory knowledge, but the formula is
Voltage Drop = (mV/A/m) x Ib x L divided by 1000
Key…
(mV/A/m): To find the value for (mV/A/m) we need to look in the current carry capacity and voltage drop tables within Appendix 4, and the fastest way to finding the table you need is by using Table of Tables on page 522.
Ib: is the design current which is usually just watts divided by voltage if you need to work it out.
L: is just the length of the circuit in metres.
We divide it all by 1000 to convert the figure from Millivolts into Volts
Note: By finding the value for volt-drop (mV/A/m) within Appendix 4 we can calculate the voltage drop for every type of cable we are likely to use.
It’s worth remembering though that there are a couple of things that you cannot calculate voltage drop for using your copy of BS7671 and that is power track and lighting trunking systems etc because it doesn’t give any values for voltage drop for those systems anywhere in the book and you would need to get the information from the manufacturer.
Let’s have a look at a sample question…
A multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors supplies a single-phase load of 20 amp at 230 volts over 22m. What is the voltage drop in the cable going to be?
To work this our we firstly need to put these values into the formula. We can see that the load is 20 amp which is your design current (Ib) and the length is 22 m long.
Next, we need to find the value of (mV/A/m) for a multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors. If you go to the table of tables on page 522 to start with and I’ll meet you there myself.
Within Table of tables, just towards the bottom of page 523 you’ll find current carrying capacities and voltage drop for cables and the table for we are looking for is for a multi-core 70 degrees C thermoplastic non-armoured cable is which is 4D2A with the voltage drop table for this cable being Table 4D2B on page 404 – see you there!
Table 4D2A on page 404…
It’s a 2.5 mm sq conductor and we are looking for a single-phase so have a look in column 3 and we can see that it’s 18 mv/A/m.
And now that we have all of the figures it’s time to work it out…
18 x 20 x 22 = 7920 divided by 1000 gives us 7.92V
Which means that the voltage drop for a multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors supplying a single-phase load of 20 amp at 230 volts over 22m is 7.92V
2. Cross-sectional area of earth / protective conductors…
Let’s look at the next question which concerns the cross-sectional area of earth / protective conductors. Remember that there are two ways you can have a do this. One, is by looking the size up using Table 54.7or you can use the calculation in regulation 543.1.3.
The calculation will give you a more precise earth or protective conductor size which is going to be smaller than just using Table 54.7 which can save a lot of money, especially on a large installation.
So, let’s have a look the formula. Anything to do with earthing in its going to relate to Chapter 54 and this is formula for calculating the cross sectional area (csa) is at regulation 543.1.3.
The cross-sectional area, where calculated, shall be not less than the value determined by the following formula taken from BS7671:2018 Part 5, Chapter 54, Regulation 543.1.3
S = square root of I squared multiplied by t and then divided by k
S: Cross sectional area of the conductor
I: Fault current
t: Disconnection time in seconds
k: Rating factor for protective conductors taken from Tables 54.2 to 54.6
The question…
“A 16mm multi-core cable incorporating a copper protective conductor insulated with 90° c thermosetting insulation given a fault current of 500 amps and a disconnection time of 0.4 seconds minimum acceptable CPC cross sectional area would be?”
So, it’s asking you to use the calculation and calculate the minimum size of the earth conductor.
Just like before, the first thing first thing we need to do is to throw some values, the ones that we know into the formula, and you can see we’ve got a current fault current 500 Amp and a disconnection time of 0.4 seconds giving us so far…
How do we find the value we need for k?
The two most common tables used are 54.2 and 54.3. Using the information from the question “multi-core cable incorporating a protective conductor” you can see that we are going to be using Table 54.3 values of k for protective conductor incorporated in a cable. Plus, a little more information from the question “90 degrees C thermosetting” and “copper” gives us the value for k of 143.
Now we have…
Throw those numbers into a calculator and crunch away
500 squared if using a calculator, the easy way to do it is 500 x then press equals to give you 250000 and then multiply that by 0.4 giving you 100000.
Then hit your square root √ button = 316.22776 and then divide that by 143 and you should have an answer of 2.211 and it’s millimeters squared. As we know there’s no such size as 2.21mm² so we can round it up to the next available size which in this case is 2.5 mm² and that is your answer.
Remember, what they are doing in the exam is assessing whether you can find that value for k 143 but sometimes you may get this question simply as…
t = 500
I = 0.4
k = 143
What is the minimum size of the CPC required?
What they’re doing here is assessing if you can use the formula…
3. Tabulated current carrying capacity…
The last of our three formulas is probably the most fun, because it usually forms a two-part question…
Tabulated current carrying capacity which you need to calculate first and then find an appropriate size conductor that can carry that amount of current.
So, let’s have a look at the formula taken from Appendix 4, section 5.1 BS7671:2018
“The tabulated current carrying capacity (It) should be greater than or equal to the rating of your protective device (In) divided by any relevant rating factors ( multiplied by each other on the bottom line.”
I just want to point out one thing, you’ll never get a situation where you have all those rating factors; along the bottom line. All it is showing you is what it is possible to have on the bottom line
Let’s have a look at the third type of question…
“A single-phase load of 13 amp is supplied by a with a two-core 70 degrees C thermoplastic copper cable non-armored and installed to method 4 or reference method B. The rating factor for grouping in his 0.7 and for ambient temperature 0.87. Overload protection is to be provided by a 15A BS1361 fuse. The minimum acceptable conductor size would be?”
As I said this is a two-part question first you’ll need to find out what (It) should be greater than or equal to and secondly from that information find an appropriate conductor size to use.
First:
What information do we have that can be used in the formula?
In: Rating of the protective device = 15A (don’t confuse this with design current ‘Ib’)
Cg: Rating factor for grouping = 0.7
Ca: Rating factor for ambient temperature = 0.87
Cf: ‘Look out for this one’ Cf = 0.725 if it’s a BS3036 fuse otherwise it’s Cf = 1
Tabulated current capacity (It) greater than or equal to 24.63A
Next, we need to find a conductor size which will carry that amount of current…
From our question we know it’s a two-core 70 degrees C thermoplastic copper cable non-armoured and installed to method 4 or reference method B.
Off to good old Table of tables on page 522 and we can see that the table we want is Table 4D2A on page 403. So, off we go to there…
It’s reference method B and single-phase, which is column 4
And we can see that a 2.5mm² conductor will carry up to 23A but the required current carrying capacity of our conductor (It) needs to be greater than or equal to 24.63A. This means that we are going to have to use the next available larger size which is 4.0mm² which will carry up to 30A
Therefore, the minimum acceptable conductor size is 4.0mm²
Summary:
As I said these are the types of formula questions you may get on your exam, but it’s more than likely you won’t. It’s just in case.
These questions will take a lot of time to complete so, if you come across any, best guess, flag them and come back to them later when you are happy with every other answer on your exam. They are only worth one point, the same as “What colour is a neutral conductor?”
This is Part Six of our Electrical Foundation Series
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