A 75-hp (shaft output) motor that has an efficiency of 91.0% is worn out and is replaced by a high-efficiency 75-hp motor that has an efficiency of 95.4%. Determine the reduction in the heat gain of the room due to higher efficiency under full-load conditions.

the reduction in the heat gain is 2.8358 kW

Explanation:

Given that;

Shaft outpower of a motor = 75 hp = ( 75 × 746 ) = 55950 W

Efficiency of motor = 91.0% = 0.91

High Efficiency of the motor = 95.4% = 0.954

now, we know that, efficiency of motor is defined as; = /

where  is the electric input given to the motor

so

= /

we substitute

= 55950 W / 0.91

= 61483.5 W

= 61.4835 kW

now, the electric input given to the motor due to increased efficiency will be;

= /

we substitute

= 55950 W / 0.954

= 58647.79 W

= 58.6477 kW

so the reduction of the heat gain of the room due to higher efficiency will be;

Q = -

we substitute

Q = 61.4835 kW - 58.6477 kW

Q = 2.8358 kW

Therefore, the reduction in the heat gain is 2.8358 kW

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