Pennychew's 18th April competition question and answer

[Richard Burns]

Following on from the completion of Pennychew's competition I thought it may help to provide some more feedback on the question just to help clarify for those still feeling short on information.
I will try and split the question and answer over multiple posts so as to keep them separate.

So with the original competition question:

This is intended to be a fairly simple question that should be answerable by anyone at any level of training, possibly with a little help.

Do read the question carefully and answer what it asks.

This is meant to get you thinking about the requirements of the regulations and methods of providing compliant solutions in a easy manner.

A new radial circuit has been designed to add to an existing compliant installation to provide power to socket outlets in a large office adjacent to a stable complex and across a path from a farmhouse where the service head is located. There is no extraneous metalwork in the office.

The supply is a single phase TNS supply protected by a BS88-3 100A fuse, the Ze is 0.37Ω and Zdb is 0.39Ω

The final circuit runs from the consumer unit in the room and is protected by a B curve RCBO rated at 32A 30mA Ics 6kA, it is anticipated that the circuit will be used to capacity.

The circuit is run in 4.0mm² PVC insulated copper cable in plastic trunking with drops in plastic conduit to surface mounted BS1363 sockets
There are double sockets spaced around the walls of the room to create a maximum circuit length of 45m..

Is this design compliant with BS7671:2008 AMD1 2011?

If not what straightforward change(s) could be made to the design to ensure that the circuit is compliant?

The design diagram is attached along with some design notes

 

[Richard Burns]

If you are reading this for the first time and want to answer the question first, STOP reading at this point.
This is the answer

So I have already posted a brief answer on the competition thread.
This answer breaks down the whole question and details calculations that you could have done to determine compliance and references to (some of) the relevant regulations applicable.

The intent of the question was to provide the sort of information you might be given prior to installing a circuit and you would need to sort out the information to gain the relevant data and not be distracted by irrelevant information.
It should also give you some insight into how relatively minor changes can make the difference between compliance and non compliance and help to clarify circuit design.

You are being asked to say if this new circuit addition complies with the AMD2 version of BS7671. It is not asking for good practice or compliance with any other documents.
If it is not complaint how can you simply redesign the circuit to be compliant.

Now we can list out the information as provided in the question:
New Radial circuit.
No current non compliances with the rest of the installation.
For BS1363 double sockets only.
In an office on an agricultural installation.
No bonding requirements.
TNS earthing.
EFLI to which you are adding your circuit: 0.39 Ω
PD: 30 mA 32 A BSEN61009 RCBO
Design current: 32 A
Cable and installation method: 4.0 mm² singles in trunking/conduit
Layout: one branch of 45 m round three sides of room, one branch of 15 m along one side of room.


So as a starting point we can say we can consider the existing installation is suitable for our addition as there are no current problems.
The bonding for the office is suitable as none is required.

The fact that it is on agricultural premises prompts you to check special location 705:
The majority of these requirements do not apply in offices of agricultural premises but:
The circuit is RCD protected.
The circuit does not need IP44 protection.
The diagram shows no roadway and no tractors will be in a stable or office so there are no mechanical, abrasion or vibration issues.
We have a layout of the electrical equipment and a single line diagram.
There is no livestock present in the location so protection from and for livestock in terms of containment and supplementary bonding is not required.

So the agricultural area is compliant for this question.
I might have made it easier by using the house for the circuit as the first section in 705 excludes household applications entirely and may well cover offices, but not specifically.

Considering the earthing arrangement the Ze of the TNS installation is <0.8 Ω and so complies with the DNO specification (though this is not specifically required by BS7671, only the OSG)

We know that the supply to the consumer unit in the office is compliant.

So now we are down to circuit design.

You need to know the type of circuit, the maximum load and the planned installation method, then you can select appropriate cables and protective device(s) and so calculate the earth fault loop impedance and volt drop.

Regulation 411.3.3 requires sockets rated at <20 A to have additional protection by a 30 mA RCD with a trip time of <=40 ms at 5*IDn. This is in place via the RCBO.

The design current is 32 A as the circuit will be used to capacity.

As the overcurrent protective device has a rating of 32 A this is just compliant as the design current must not be greater than the rating of the protective device (if using an MCB/RCBO).

The current carrying capacity of 4.0 mm² thermoplastic insulated cables run in trunking or conduit is 32 A from table 4D1A reference method B, single phase.

(This value was not given in the question but was for you to check.
The on site guide standard circuits wired in twin and earth, only permits reference method C (clipped direct) for a 32 A RCBO protected 4.0/1.5 mm² circuit as it errs on the simplistic side, however we are concerned with BS7671 only.)

Any additional rating factors would reduce the current carrying capacity below the permitted level, however the cable is not grouped with other circuits, has no ambient temperature issues, is not buried, is not in any further insulation beyond the conduit and is not on a semi enclosed fuse so it is compliant (just, but just is enough). The rated current of the protective device does not exceed the current carrying capacity of the cables.

Now to consider permitted disconnection times:
A final circuit on a TN system at 230 V and not exceeding 32 A has a maximum disconnection time of 0.4s.
This will be achieved by the RCD portion of the RCBO with a Zs up to 1667 Ω.
This will be achieved by the overcurrent portion of the RCBO with a Zs up to 1.16 Ω at ambient temperature as given in the diagram. (this is 80% of the AMD2 Zs of 1.44 Ω, not the AZMD3 Zs of 1.37 Ω at maximum operating temperature or 1.10 Ω at ambient temperature).

From the point of view of the question we can therefore comply with BS7671 using the RCD portion for Zs on a quick assessment of the resistance of the conductors and need no further calculations.
From the point of view of better safety to see if we comply with the 32A type B overcurrent device, we can calculate using the given figure of 4.61 mΩ/m for a 4.0 mm² cable at 20 °C.

Maximum worst case scenario length of cable is 45 m from origin to furthest point.
We are using singles so both line and cpc are 4.0 mm².
Maximum length of line and cpc (the internal fault path) is 90 m (out and back).
4.61 mΩ/m /1000 = 0.00461 Ω/m
0.00461 Ω/m * 90 m = 0.415 Ω.
Add this to the Zdb given of 0.39 Ω to give 0.8 Ω this is less than the maximum of 1.16 Ω and so the Zs will comply for the overcurrent portion of the RCBO, good.

So the circuit is OK up to this point.

Now consider voltage drop
We are given the voltage drop at 70°C for 4.0 mm² cable as 11 mV/A/m this value covers both the line and neutral cable in a circuit.

Our worst case scenario is a 45m run carrying 32A at maximum (although in reality the 32A could not all be used at the termination of the cable in designing we are considering the worst case scenario)
So maximum volt drop in volts is = mV/A/m * A * m / 1000
i.e. 11 * 32 * 45 / 1000 = 15.84 V
Because we are considering sockets and we do not know what is going to be connected we can apply no equipment standard that may allow a lower voltage drop than the limits specified in table 4Ab in appendix 4: 5% of the nominal voltage of the installation.
For a 230V supply this is 5*230/100 = 11.5 V

Anyone who considered that the maximum current draw at the end of the cable would be limited to 26A (one double socket with two 13A plugs at full capacity) would find the volt drop to be 12.87 V

Both of these values are non compliant.

So what methods can used to reduce voltage drop?

Voltage drop is affected by the length (m), the current drawn (A) and the mV/A/m value.

We can:
Shorten the length of the cable to reduce m.
Reduce the current draw on the cable to reduce A.
Increase the cross sectional area of the cable to reduce mV/A/m.


Length
We still want to supply the sockets required by the design so we cannot just remove sockets.

However looking at the diagram the circuit has a long length of cable round three walls and a short length of cable along one wall.
If the cable on the short length were extended out to cover two walls and the long length reduced to only cover two walls the maximum circuit length would be 30m giving a complaint 10.6 V volt drop (11*32*30/1000).
This could also be split over two 20A RCBO protected radial circuits one for each side to give you some leeway to make changes later if required.
Then the current carrying capacity would be well within the limits and the volt drop also.
The maximum current in any one branch is now 20 A compared to a 32A ccc.
Volt drop is now 11 mV/A/m * 20 A * 30 m / 1000 = 6.6 V

Another solution from looking at the diagram is if you were to extend the 15 m run out to the next socket round and join it in you would have created a ring circuit.
Volt drop on this circuit with an end to end length of (45+15) =60 m
Assuming an absolute maximum load of 32 A at the midpoint, volt drop would then be
mV/A/m * A * m / 1000 / 4
(divided by 4 because you have two equal resistors in parallel i.e. each leg of the ring)
So 11 * 32 * 60 /1000 / 4 = 5.2 V, easily compliant
This is the cheapest and easiest option and you could even drop the cable size to 2.5mm² singles and your Vd, Zs and current carrying capacity would still be compliant.
Volt drop = 18 mV/A/m * 32 A * 60 m / 1000 / 4 = 8.6 V
r1 = r2 = (7.41 mΩ/m * 60 m / 1000) = 0.44 Ω
R1+R2 = (r1+r2)/4 = (0.44 Ω+ 0.44 Ω) / 4 = 0.22 Ω
Zs = 0.39 Ω + 0.22 Ω = 0.61 Ω
ccc for 2.5 mm² cable reference method B is 24 A, greater than the minimum csa and ccc required by 433.1.103.

Current
The overall demand will not change as this is the design specification so you could consider splitting the circuit up into multiple radials so dividing the current demand across circuits.

There are many ways of arranging this but the easiest is to do the same as above, have one radial for each two walls, each on a 20 A 30 mA RCBO. This would give you compliant voltage drop for a maximum 30 m circuit and a maximum 20 A current.
Volt drop = 11 * 20 * 30 / 1000 = 6.6 V
The 45m and the 15m separate circuits would also comply as the Volt drop = 11 *20 * 45 /1000 = 9.9V
For this scenario there would still be no grouping factors so you could also drop to 2.5 mm² cable. Other layouts where two circuits run in the same trunking beside each other would drop the current carrying capacity to 19.2 A for 2.5 mm² cable because of grouping factors.


Cross sectional area
If we increase the cable size to 6.0 mm² then using 7.3 mV/A/m we get a compliant voltage drop of 10.5 V, but this will increase the cost considerably. However this is still a valid choice for compliance, if not for customer satisfaction.

 

[Richard Burns]

This is a diagram of some of the possible solutions to the question as described above.

 

[Richard Burns]

I should also mention that it is probably not good planning to design a circuit that is right on the limits of the regulations as it gives no further leeway for making any changes to a circuit or for maintaining compliance as the circuit deteriorates slightly over time.
The intent is to design a circuit that should last for thirty or forty years.
However it is also important to remember that the limits are limits and can be can still be compliant even at that limit, though maybe not easily compliant.
However for the purposes of the question I was pushing to the limits.