How to Select a Electrical Wire or Cable
When selecting a cable for your specific application, a number of variables require attention. These are
(a) Wire Size and type of load to be supplied
(b) Permissible voltage drop
(c) Prospective fault current
(d) Circuit protection
(e) Environmental conditions of installation
strong text Load to be supplied
In order to select the appropriate cable, it is necessary to know the voltage
and the load current in amps. This information will be available either
directly in amps or as kW or kVA.
The following formulae apply:
I_FL = (KW ×1000)/(√3×V ×cos∅) Amps if we know KW, voltage, as well as power factor
I_FL = (KVA ×1000)/(√3×V) Amps if we know the KVA rating as well as the voltage
Use this value of current to determine the Electrical Wire or cable size by reference to the Relevant tables given in Section 4 (Paper insulated), Section 5 (XLPE insulated Medium voltage) or Section 6 (PVC and XLPE insulated low voltage) For Copper or Aluminum conductors.
A slightly larger conductor size may be chosen for safety aspects, and to provide for the higher than usual current which may be experienced during starting of electric motors.
Example of Cable selection for low Voltage:
Suppose it is required to supply a 3 phase, 400 volt, 100kW motor connected
in star/delta, over a distance of 50m buried direct in ground. The
motor load is known to have a power factor of 0,9 lagging. The full load line
current, IFL can be calculated as follows:
I_FL = (100×1000)/(√3×400 ×0.9) = 161 Amps
We now refer to table 6.2 on pg 38 and note that the smallest copper conductor, PVC insulated cable , that can supply a current of 161 amps in the ground, is a 50 mm² rated area cable. This cable can carry 169 amps continuously if installed under standard conditions.
Permissible Voltage Drop
Calculate the highest current drawn by the load, by multiplying the current as calculated in 1.1 by an appropriate factor. If a Star/Delta motor starter is used on a motor, this factor is 3. If the motor is started direct on line, then use a factor of 6. Where the load is resistive heating, lighting or a transformer, it is not necessary to increase the current as calculated in 1.1.
Calculate the volt drop which will be experienced at the load terminals by reference to table 6.2 or 6.3 on pg 38 or pg 39. The maximum volt drop allowed by SANS 10142-1 during full load running condition. There is no hard and fast rule as to the allowable volt drop under starting conditions. Depending on the type of load to be started, there is the possibility that the torque may be compromised during starting, if the motor is subjected to difficult starting conditions. A reasonable volt drop should be chosen in these cases
The volt drop may be calculated in two different ways:
(a) Multiplying the current by the impedance of the length of cable. Calculate the percentage volt drop by reference to the phase to earth voltage.
(b) Multiply the current by the length of cable, and then multiply the result by the volt drop per amp per metre figure as given in table 6.2 or 6.3 on page 38 or page 39, depending on the type of conductor.
Starting Current = 3 x Running Current
= 3 x 161Amps
= 483Amps
Impedance of 50m of 50mm² Cable (Table 6.2) pg 38
= 0.4718/1000×50
= 0,02359 ohms
Volt Drop = 483 x 0,02359
= 11.394/230×100/1
= 4,95% (Acceptable)
Using Method (b)
Starting Current = 3 x 161Amps
= 483Amps
Volt Drop per amp per metre = 0,817 mV/A/m (Table 6.2) pg 38
Volt Drop = 0,817 x 10 x 483 x 50
= 19,73 volts
Percentage Volt Drop = 19.73/400×100/1
= 4,93% (Acceptable)
Not:- It often happens on long runs of electric cable that a larger conductor than that calculated in 1.1 is required for volt drop reasons.
Example of Cable section for medium Voltage (11KV)
We wish to supply a 2MVA 11kV transformer from an Eskom supply point which is 3km away. We are to use an underground paper insulated, copper conductor cable. The depth of burial of the cable is 1,25m. Ground thermal resistivity is 2K.m/W. The ground temperature is 25°C and there are no other cables in the trench. Short circuit level may be assumed to be 250MVA, and the earth fault level 100 MVA, and it may be assumed that a fault will be cleared in half a second.
I_FL = 2000000/(√3×11000)
= 105 Amps before derating for non standard conditions
Derating factor for Depth of Burial at 1,25mis 0,96.
Derating factor for Soil Thermal Resistivity at 2 K.m/W is 0,84.
Derating factor for Ground Temperature of 25°C is 1,00.
Total Derating = 0,96 x 0,84 x 1,00
= 0,8064
= 105/(0.8064)
= 130Amps
Table 4.2 on pg 28 shows that a 35 mm² Copper conductor cable would be capable of carrying this load (130A). This is confirmed by reference to the Paper Insulated Cable brochure (Page 13).
The cable size required is thus 35 mm² Copper conductor, 3 Core General purpose belted cable. (Table 17 SANS 97).