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11-step procedure for a successful electrical circuit design (LV project example)

Let’s take an example of electrical circuit design.

The basic steps in electrical circuits design
The basic steps in electrical circuits design

A consumer lives in a bungalow with a detached garage and workshop, as shown in Figure 1. The building method is traditional brick and timber.

Electrical circuit design

The mains intake position is at high level and comprises an 80 A BS 1361 240 V main fuse, an 80 A rated meter and a six-way 80 A consumer unit housing BS 3036 fuses as follows:

Ring circuit30 A
Lighting circuit5 A
Immersion heater circuit15 A
Cooker circuit30 A
Shower circuit30 A
Spare circuit

The cooker is rated at 30 A, with no socket in the cooker unit. The main tails are 16 mm2 double-insulated PVC, with a 6 mm2 earthing conductor. There is no main equipotential bonding. The earthing system is TN-S, with an external loop impedance Z of 0.3 ohms.

Bungalow layout and electrical installations route
Figure 1 – Bungalow layout and electrical installations route

The prospective short circuit current (PSCC) at the origin has been measured as 800 A. The roof space is insulated to full depth of the ceiling joists and the temperature in the roof space is not expected to exceed over 40° C.

The consumer wishes to convert the workshop into a pottery room and install an 8.6 kW/230 V electric kiln. The design procedure is as follows.

  1. Assessment of general characteristics
  2. Electric Wiring: Domestic
  3. Sizing the main tails
  4. Sizing the kiln circuit cable
  5. Correction factors
  6. Tabulated current-carrying capacity of cable
  7. Cable size based on tabulated current-carrying capacity
  8. Check on voltage drop
  9. Shock risk
  10. Thermal constraints
  11. Protection

1. Assessment of general characteristics

The present maximum demand, applying diversity, is:

Ring30 A
Lighting (66% of 5 A)3.3 A
Immersion heater15 A
Cooker (10 A + 30% of 20 A)16 A
Shower30 A
TOTAL94.3 A

Reference to the current rating tables in the IEE Regulations will show that the existing main tails are too small and should be uprated. Also, the consumer unit should be capable of carrying the full load of the installation without the application of diversity.

So, the addition of another 8.6 kW of load is not possible with the present arrangement.

The current taken by the kiln is 8600/230 = 37.4 A. Therefore, the new maximum demand is 97.3 + 37.4 = 134.7 A. Supply details are:

  • Single-phase
  • 230 V, 50 Hz
  • Earthing system: TN-S
  • Prospective short circuit current (PSCC) at origin (measured): 800 A

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2. Electric Wiring: Domestic

Decisions must now be made as to the type of cable, the installation method and the type of protective device. As the existing arrangement is not satisfactory, the supply authority must be informed of the new maximum demand, as a larger main fuse and service cable may be required.

It would seem sensible then to disconnect, say, the shower circuit, and to supply it and the new kiln circuit via a new two-way consumer unit, as shown in Figure 2.

New two-way consumer unit
Figure 2 – New two-way consumer unit

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3. Sizing the main tails

  1. The new load on the existing consumer unit will be the old load less the shower load:
    94.3 – 30 = 64.3 A. From the IEE Regulations, the cable size is 16 mm2.
  2. The load on the new consumer unit will be the kiln load plus the shower load:
    37.4 + 30 = 67.4 A. From the IEE Regulations, the cable size is 16 mm2.
  3. The total load is 64.3 + 67.4 = 131.7 A. From the IEE Regulations the cable size is 35 mm2.
  4. The earthing conductor size, from the IEE Regulations, will be 16 mm2. The main equipotential bonding conductor size, from the IEE Regulations, will be 10 mm2.

For a domestic installation such as this, a PVC flat twin-cable, clipped direct through the loft space and the garage, etc., would be most appropriate.

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4. Sizing the kiln circuit cable

Design current Ib is:
Ib = P/V = 8600/230 = 37.4 A

Rating and type of protection In:
In order to show how important this choice is, it is probably best to compare the values of current-carrying capacity resulting from each type of protection.

As we have seen, the requirement for the rating In is that In > Ib. Therefore, using the tables in the IEE Regulations, In will be as follows for the various fuse types.

  • BS 88 – 40 A BS 3036 – 45 A
  • BS 1361 – 45 A MCBs – 50 A

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5. Correction factors

  • Ca – 0.87 or 0.94 if the fuse is BS 3036
  • Cg – not applicable
  • Cf – 0.725 only if the fuse is BS 3036
  • Ci – 0.5 if the cable is totally surrounded in thermal insulation

Application of correction factors //

Some or all of the onerous conditions just outlined may affect a cable along its whole length or parts of it at the same time. So, consider the following.

#1 – If a cable ran for the whole of its length, grouped with others of the same size in a high ambient temperature, and was totally surrounded with thermal insulation, it would seem logical to apply all the CFs, as they all affect the whole cable run.

Certainly, the factors for the BS 3036 fuse, grouping and thermal insulation should be used.

However, it is doubtful if the ambient temperature will have any effect on the cable, as the thermal insulation, if it is efficient, will prevent heat reaching the cable. Hence, apply Ca, Cg and Cf.

#2 – If, however, the cable first runs grouped, then leaves the group and runs in high ambient temperature and finally is enclosed in thermal insulation, there will be three different conditions, each affecting the cable in different areas.

The BS 3036 fuse affects the whole cable run and, therefore, Cf must be used, but there is no need to apply all of the remaining factors as the worst one will automatically compensate for the others.

Having chosen the relevant correction factors, we now apply them as divisors to the rating of the protective device In in order to calculate the tabulated current-carrying capacity It of the cable to be used.

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6. Tabulated current-carrying capacity of cable

For each of the different types of protection, the current-carrying capacity will be as shown in Table 1(a).

Table 1
Table 1 (click to enlarge table)

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7. Cable size based on tabulated current-carrying capacity

Table 1(b) shows the sizes of cable for each type of protection (from the IEE Regulations). Clearly the BS 88 fuse gives the smallest cable size if the cable is kept clear of thermal insulation, i.e. 6.0 mm2.

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8. Check on voltage drop

The actual voltage drop is given by:

The actual voltage drop

This voltage drop, whilst not causing the kiln to work unsafely, may mean inefficiency, and it is perhaps better to use a 10 mm2 cable. This also gives a wider choice of protection type, except BS 3036 rewirable. This decision can be left until later.

For a 10 mm2 cable, the voltage drop is checked as:

The actual voltage drop

So, at this point, we have selected a 10 mm2 twin cable and we have at our disposal a range of protection types, the choice of which will be influenced by the loop impedance.

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9. Shock risk

The circuit protective conductor (CPC) associated with a 10 mm2 twin 6242 Y cable is 4 mm2. Hence, the total loop impedance will be:

Total loop impedance

Note //

6.44 is the tabulated (R1 + R2) value and the multiplier 1.2 takes account of the conductor resistance at its operating temperature. This means that all protective devices except 50 A types 3, C and D MCBs could be used (by comparison with Zs values in the IEE Regulations).

As only BS EN 60898 types B, C and D are now available, any CB used must be a type B.

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10. Thermal constraints

We still need to check that the 4 mm2 circuit protective conductor (CPC) is large enough to withstand damage under earth fault conditions. So, the fault current would be

I = Uoc / Zs = 240 / 0.489 = 490 A

The disconnection time t for this current for each type of protection (from the relevant curves in the IEE Regulations) is as follows:

  • 40 A BS 88 – 0.05 s
  • 45 A BS 1361 – 0.18 s
  • 50 A CB type B – 0.01 s

From the regulations, the factor for k = 115. We can now apply the adiabatic equation:

Adiabatic equation

Therefore, for each type of protection we have the following minimum sizes of circuit protective conductor (CPC).

  • 40 A BS 88 – 0.9 mm2
  • 45 A BS 1361- 1.7 mm2
  • 50 A CB type B – 0.466 mm2

Hence, our 4 mm2 circuit protective conductor (CPC) is of adequate size.

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11. Protection

It simply remains to decide which of the types of protection to use. A type B CB is probably the most economical. However, if this is chosen, a check should be made on the shower circuit to ensure that this type of protection is also suitable.

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Reference // Electric Wiring Domestic by Scaddan

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About Author //

author-pic

Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on

6 Comments


  1. Wilbert
    Jul 25, 2017

    helpful as always. Thanks!


  2. Muluken
    Jul 20, 2017

    Hi Edvard, you are so generous, THANK you very much.


  3. Anilkumar
    Jul 19, 2017

    Hello Edvard

    We engineering community is thankful to you for your invaluable contribution. This type of project examples is a great help.
    Could you also post if possible a typical medium voltage distribution for a reference from transformer sizing like 72 kv/4160v then medium voltage switch gear, relay selection etc
    Thanks

    • Edvard
      Edvard
      Jul 19, 2017

      Good suggestion Anilkumar, I’ll take a look for it. Thank you!


  4. Nagarajan Ravi Shankar
    Jul 19, 2017

    Excellent Article …Kudos

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