Solution
For the 120 mm2 cable
the series impedance is:
R = r*l = {0.214 * 120} / 1000 = 0.02568 Omega/phase
X = x*l = {0.0808 * 120} / 1000 = 0.009696 Omega/phase

For the 110kW motor
the starting current is:
I = 6 * 180.0 = 1080.0 A
The power factor is:
cos phi = 0.2500, therefore sin phi = 0.9682

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

AB = 1080.0 * 0.02568 * 0.2500 = 6.9336 volts/phase
BE = 1080.0 * 0.02568 * 0.9682 = 26.6805 volts/phase
EF = 1080.0 * 0.009696 * 0.9682 = 10.1386 volts/phase
DF = 1080.0 * 0.009696 * 0.2500 = 2.6179 volts/phase

%V_d = ( 1.732 / 400 ) * 1080.0 * ( 0.02568 * 0.25 + 0.009696 * 0.9682 ) * 100%
%V_d = 4.6764 * ( 0.00642 + 0.006738 ) * 100%
%V_d = 6.15%

Therefore,
V_r = 400 * ( 1 - 0.0615 ) / 1.732
V_r = 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.

On the next part, we shall be providing more examples for voltage drop calculations.

Pages: 1 2 3 4 5