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A 40 HP motor with a load factor of 75% and an efficiency of 89.3% will be replaced with a 30 HP motor with a load factor of 100% and an efficiency of 93.6%. How many kW of power savings will be obtained from this project?​

User Esel
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1 Answer

7 votes

First, we need to calculate the actual power consumption of the 40 HP motor with a load factor of 75% and an efficiency of 89.3%.

Actual power consumption = Rated power x Load factor / Efficiency

= 40 HP x 0.75 / 0.893

= 33.6 kW

Next, we need to calculate the actual power consumption of the 30 HP motor with a load factor of 100% and an efficiency of 93.6%.

Actual power consumption = Rated power x Load factor / Efficiency

= 30 HP x 1 / 0.936

= 31.9 kW

The power savings from this project can be calculated as the difference between the actual power consumption of the 40 HP motor and the actual power consumption of the 30 HP motor.

Power savings = Actual power consumption of 40 HP motor - Actual power consumption of 30 HP motor

= 33.6 kW - 31.9 kW

= 1.7 kW

Therefore, the project will result in a power savings of 1.7 kW.