 We shall be presenting an IEC standard equipment in our sample calculation.

A 120 mm2 3core XLPE insulated cable 120m in length supplies a 110 kW induction motor that has a starting current of 6 times the full-load current of 180 amps. The starting power factor is 0.25 lagging.

If the sending end line-to-line voltage is 400 volts, the specific resistance r and reactance x for the cable are 0.197 and 0.072 ohm/km respectively at 90OC and 50 Hz.

Find the percentage volt-drop on starting the motor.

### Solution

For the 120 mm2 cable
the series impedance is:
R = rl = 0.214 x 120 / 1000 = 0.02568 ohms/phase
X = xl = 0.0808 x 120 / 1000 = 0.009696 ohms/phase

For the 110kW motor
the starting current is:
I = 6 Ã— 180.0 = 1080.0 A
The power factor is:
cosØ = 0.2500, therefore sinØ = 0.9682

Let us assume the sending voltage remains constant at 400 volts,

AB = 1080.0 x 0.02568 x 0.2500 = 6.9336 volts/phase
BE = 1080.0 x 0.02568 x 0.9682 = 26.6805 volts/phase
EF = 1080.0 x 0.009696 x 0.9682 = 10.1386 volts/phase
DF = 1080.0 x 0.009696 x 0.2500 = 2.6179 volts/phase %Vd = ( 1.732 / 400 ) x 1080.0 x ( 0.02568 x 0.25 + 0.009696 x 0.9682 ) x 100%
%Vd = 4.6764 x ( 0.00642 + 0.006738 ) x 100%
%Vd = 6.15%

Therefore,
Vr = 400 x ( 1 - 0.0615 ) / 1.732
Vr = 216.74 volts/phase

Typical minimum starting voltage for motors is 80% x rated voltage. Since the calculated volt-drop in our example is less than 20%, the motor will accelerate to full speed easily.

In practice, since the actual motor parameters are not known during the design stage, the minimum starting voltage is raised to 85% to provide a factor for safety.

For Part 3, we shall be providing more examples for voltage drop calculations.