Frank Bertie, Chief Technical Officer at NAPIT, explains how to establish voltage drop limits within electrical installations.
BS 7671 and requirements for voltage drop
The topic of voltage drop is covered within section 525 of BS 7671 as well as in Appendix 4 section 6.4, where details are provided regarding the requirements and maximum limitations for use with electrical installations.
For low voltage installations, the voltage drop values are found in Appendix 4 of BS 7671, for those fed from a public distribution system, the values shown in Table 1 apply.
In cases where the client has their own generator or transformer, and where the electrical distribution system is not a public supply, voltage band control will be the responsibility of the client. When this happens, values of voltage drop can be increased, as seen in Table 2.
The specified lower equipment voltage limit means that the installation voltage at the terminals has to exceed that lower value in accordance with Regulation 525.1.
In the absence of a product standard for fixed current-using equipment, the installation voltage at the terminals will not impair the safe functioning of that equipment under the requirements of Regulation 525.201.
Considering voltage drop in the design involves taking into account the electrical installation and circuit type. According to Table 1, for a DNO supply, lighting circuits are at 3% of the nominal 230 V single-phase supply voltage, and the voltage drop must not exceed 6.9 V. All other circuits are at 5% and the voltage drop must not exceed 11.5 V.
The tables in Appendix 4 of BS 7671 contain a voltage drop section in which the millivolt per amp per metre (mV/A/m) of a particular cable may be obtained. For example, Table 4D5 should be used for thermoplastic insulated and sheathed flat cables with copper conductors.
Worked example
An electrical load of 6 kW for a motor circuit is to be supplied at 230 V by a PVC insulated and sheathed PVC T&E cable of 20 metres in length.
The cable is clipped on the surface through an area with an ambient temperature of 40°C and is grouped with three other cables of similar size and loading. The protection is provided by a BS 60898 Type B circuit breaker.
We need to determine the following before we can establish whether the volt drop is satisfactory:
1. The circuit design current (Ib) Design Current Ib = Watts / Voltage Ib = 6000/230 = 26 A
2. The rated current of the protective device (In) The rating of the protective device needs to be equal to or greater than the design current of 26 A. Therefore, a 32 A BS 60898 Type B circuit breaker would be selected. In = 32 A
3. The tabulated current value of the cable (It) Tabulated Current It = In / Correction Factors
The next step is to apply any correction factors. In the worked example we need to consider the ambient temperature (Ca) and grouping (Cg).
The factors are then applied to the following formula: It >In /Ca x Cg
From Table 4B1 (BS 7671) Ca = 0.87
From Table 4C1 (BS 7671) Cg = 0.75 (4 circuits, Reference Method C)
It >32/0.87×0.75
It > 49 A
It is now time to select a cable size that is capable of carrying 49 A.
From Table 4D5 (BS 7671) Reference Method C, Column 8, we can select 10 mm², which is rated at 64 A.
At this point, we can work out our volt drop and determine whether it is satisfactory. Using the same table and conductor size, we can calculate the millivolt amps per metre value.
Volt drop from Table 4D5 (BS 7671), 10 mm² Cable, Column 9 is 4.4 mV/A/m.
By using the volt drop formula below, we can now add our mV/A/m value, the design current (Ib) and the cable length (L).
These values are multiplied together and then divided by 1,000, to convert from millivolts to volts, as shown in Fig 1.
Volt drop = (mV/A/m) x Ib x L / 1000
Volt drop = 4.4 x 26 x 20 / 1000
Volt drop = 2.29 V
The maximum volt drop for this circuit is 11.5 V, as discussed above. Therefore, the actual volt drop of 2.29 V is below the permitted value and is, therefore, satisfactory.
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