A three-phase, 400 V, 50 Hz induction motor drive is used to lift a load. The motor has a full-load efficiency of 85% and a power factor of 0.8 lagging. If the motor lifts a 2000 kg load at a constant speed of 1.2 m/s, calculate the line current drawn by the motor.

Three Phase Induction Motor Numerical Problem

Given Data

• Line Voltage (V_L) = 400 V
• Efficiency (η) = 85% = 0.85
• Power Factor (cosφ) = 0.8
• Mass (m) = 2000 kg
• Speed (v) = 1.2 m/s
• Acceleration due to gravity (g) = 9.81 m/s²

Step 1: Calculate Mechanical Power Required

Power required to lift the load:

P_out = m × g × v

P_out = 2000 × 9.81 × 1.2

P_out = 23544 W ≈ 23.54 kW

Step 2: Calculate Electrical Input Power

Efficiency formula:

η = P_out / P_in

P_in = P_out / η

P_in = 23544 / 0.85

P_in ≈ 27699 W ≈ 27.7 kW

Step 3: Use Three Phase Power Formula

P = √3 × V_L × I_L × cosφ

I_L = P / (√3 × V_L × cosφ)

Step 4: Substitute the Values

I_L = 27699 / (1.732 × 400 × 0.8)

I_L = 27699 / 554.24

I_L ≈ 49.9 A

Final Answer

Line Current Drawn by the Motor ≈ 50 A

Concept Used

When a motor lifts a load vertically, mechanical power is required to overcome the gravitational force acting on the load. The mechanical power required is given by:

P = m × g × v

where:

• m = mass of the load (kg)
• g = acceleration due to gravity (9.81 m/s²)
• v = lifting velocity (m/s)

Since the motor is not 100% efficient, the electrical input power will be higher than the mechanical output power. Efficiency is defined as:

η = P_out / P_in

For a three-phase motor, the electrical input power is calculated using the three-phase power formula:

P = √3 × V_L × I_L × cosφ

Using these relationships, we can first determine the mechanical power required, then calculate the input power, and finally determine the line current drawn by the motor.