205.MET November 2022 Q.7(b)

Determine the line current taken by a 440V, three-phase, star-connected motor having an output of 45kW at 0.88(lagging) power factor and an efficiency of 93 per cent

 Apparent Power (S): Apparent power is the product of voltage and current in amperes reactive (A). It represents the total electrical energy flowing in a circuit.

S = P × (1 + (PF)^2)

Where: P = Real Power (kW) = 45,000 W PF = Power Factor = 0.88 (lagging)

S = 45,000 W × (1 + (0.88)^2) = 63,761.2 W

Real Power (P): Real power is the actual useful power being consumed by the motor. It is equal to the product of apparent power and power factor.

P = S × PF P = 63,761.2 W × 0.88 = 54,937.2 W or 54.9 kW

Reactive Power (Q): Reactive power represents the electrical energy that flows back and forth between the supply source and load due to capacitive or inductive effects in the circuit. It can be calculated as follows:

Q = S × (1 - PF) Q = 63,761.2 W × (1 - 0.88) = 7,824.2 W or 7.8 kVar

Line Current: Line current can be calculated using the following formula: Iline = S / VL^2 or Iline = P / (VL^2 × PF) where VL is line voltage in volts: VL for this problem is given as 440V.**

Iline_star_connected = S / VL^2 or Iline_star_connected = P / (VL^2 × PF) Iline_star_connected = 63,761.2 W / (440V)^2 or Iline_star_connected = 54,937.2 W / (440V)^2×(0.88) Iline_star_connected≈195 A or≈195 Amperes for star-connected motors.