Answer:
Input area=0.785x10^-4m^2
Output area=0.785x10^-6m^2
P1-p2=0.49x0.99v2
V1 =0.01v2
Explanation:
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Answer:
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Explanation:
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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?
