Answer:
the reduction in the heat gain is 2.8358 kW
Explanation:
Given that;
Shaft outpower of a motor [tex]W_{shaft}[/tex] = 75 hp = ( 75 × 746 ) = 55950 W
Efficiency of motor [tex]n_{motor}[/tex] = 91.0% = 0.91
High Efficiency of the motor [tex]n_{high-eff}[/tex] = 95.4% = 0.954
now, we know that, efficiency of motor is defined as; [tex]n_{motor}[/tex] = [tex]W_{shaft}[/tex] / [tex]W_{elec}[/tex]
where [tex]W_{elec}[/tex] is the electric input given to the motor
so
[tex]W_{elec}[/tex] = [tex]W_{shaft}[/tex] / [tex]n_{motor}[/tex]
we substitute
[tex]W_{elec}[/tex] = 55950 W / 0.91
= 61483.5 W
= 61.4835 kW
now, the electric input given to the motor due to increased efficiency will be;
[tex]W_{elec-incresed}[/tex] = [tex]W_{shaft}[/tex] / [tex]n_{high-effic}[/tex]
we substitute
[tex]W_{elec-incresed}[/tex] = 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 = [tex]W_{elec}[/tex] - [tex]W_{elec-incresed}[/tex]
we substitute
Q = 61.4835 kW - 58.6477 kW
Q = 2.8358 kW
Therefore, the reduction in the heat gain is 2.8358 kW
