Question

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

Answer 1

Answer:

the reduction in the heat gain is 2.8358 kW

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

Explanation:  i hope this answer your question if this is wrong or correct please let me know.also no trying to be rude but can you sent me like a thanks?