A 3-phase 15 kW motor has the name plate details as 415 V, 18.2 amps and 0.85 PF. Actual input measurement shows 415 V, 10 A and 0.75 PF which was measured with power analyzer during motor running. Find out the motor loading and actual input power of the motor.

Solution:

To find out the motor loading and actual input power of the motor, we can use the following formulas:

1. Apparent Power (S) = Voltage (V) × Current (I)
2. Real Power (P) = Apparent Power (S) × Power Factor (PF)

Given:

• Nameplate details: Voltage (V) = 415 V, Current (I) = 18.2 A, Power Factor (PF) = 0.85
• Actual input measurement: Voltage (V) = 415 V, Current (I) = 10 A, Power Factor (PF) = 0.75

Let's calculate:

1. Nameplate Apparent Power (S1): S1 = V × I = 415 V × 18.2 A = 7543 VA

2. Actual Apparent Power (S2): S2 = V × I = 415 V × 10 A = 4150 VA

3. Actual Real Power (P2): P2 = S2 × PF2 = 4150 VA × 0.75 = 3112.5 W

Now, let's find out the motor loading:

Motor loading is typically expressed as the ratio of actual power to rated power.

4. Rated Real Power (P1): P1 = S1 × PF1 = 7543 VA × 0.85 = 6406.55 W

5. Motor Loading: Motor Loading = P2 / P1 = 3112.5 W / 6406.55 W ≈ 0.485 (or 48.5%)

So, the motor loading is approximately 48.5%, and the actual input power of the motor is approximately 3112.5 watts.