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voltage drop calculation

This article is the second part on the discussion about voltage drop calculation. We shall be providing real world example for you to be able to appreciate it better as it may have already been a part of what you have already done or currently doing in your design.

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 × 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

%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.