Volt Drop – ConnectUs Electrical Contractors

Volt Drop

The voltage drop in a cable is directly proportional to the circuit current and the length of the cable run. BS7671:2011 limits the voltage drop permitted between the origin of the installation and the terminals of the load to 3% of the nominal supply voltage for lighting circuits and 5% of the nominal supply voltage for other circuits.

To calculate the volt drop of a circuit, you need to know the tabulated value of voltage drop (mV/A/m) listed in appendix 4 of BS7671, the design current (Ib) and the length of the circuit.

$voltage drop = \frac{(mV/A/m) \times I_{b} \times L}{1000}$

For example, a 4.5kW 230V 50Hz single phase motor with a power factor of 0.7 and an efficiency of 75% is wired with pvc insulated single core cables in steel conduit. The length of the run is 30m, the ambient temperature is 35degC and the circuit will be protected by BS88 fuses. Calculate the size of the conductors if the voltage drop in the circuit must not exceed 5V.

Firstly we will calculate the design current (Ib),

$P=\frac{Output}{Efficiency}$

$P=\frac{4500}{0.75}$

$P=6000 W$

$I_{b} = \frac{P}{U \times \textit{p.f.}}$

$I_{b} = \frac{6000}{230 \times 0.7}$

$= 37.3 A$

Minimum BS88 fuse rating (In) will be 40A, Correction factor for ambient temperature (Ca) for 35degC will be 0.94 and the correction factor for fuses (Cr) is 1 for BS88 fuses.

Thus the minimum tabulated current rating (It) is calculated

$I_{t} = \frac{40}{0.94 \times 1}$

$= 42.55 A$

Selecting a 10mm cable from BS7671 Table 4D1A which can carry 57A and has a 4.4 mV/A/m tabulated volt drop listed in Table 4D1B, we can calculate the volt drop as follows.

$voltage drop = \frac{4.4 \times 37.3 \times 30}{1000}$

$=4.91V$

Therefore the 10mm cable will be suitable.

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