Answer :
Answer:
74.55Hp
Explanation:
Hello! To solve this problem we must do the following 3 steps.
1. Find the ideal power using the equation to find the power W = mgV
W=mgV
where
M=mass of elevator=1700kg
g=gravity=9.81m/S^2
V=speed=3m/S
W=(1700kg)(9.81m/S^2)(3m/s)=50031W
2.
Now as the problem indicates that 10% of energy is lost in the mechanical system, we divide the power found in the previous step by 90%
[tex]W=\frac{50031W}{0.9} =55590W[/tex]
3.
finally we convert units from W to hp.
[tex]W=55590W\frac{1HP}{745.7w} =74.55Hp[/tex]
the minimum output power rating in HP of the motor that meets the requirement is 74.55Hp
Output power rating of motor is equal to the 90% of elevator system power. Power rating of motor that meets the requirement is 74.55 horse power.
What is mechanical power?
Mechanical power is the power or energy required to perform a particular work. The mechanic power is product of force and the displacement.
Given information-
The elevator (with a full load) weights 1700 kg and is required to move upward 3 m/sec at constant speed.
The efficiency of the lifting mechanism is 90 percent.
The work done by the elevator is,
[tex]W_e=1700\times9.81\times3\\W_e=50031 \rm W[/tex]
Thus the work done by the elevator is 50031 Watt.
As the efficiency of the lifting mechanism is 90 percent and the wastage of power is 10 percent. Thus the work done by the motor is,
[tex]W_m=\dfrac{50031}{0.9}\\W_m=55590\rm W[/tex]
Thus the work done by the motor is 55590 Watt.
As the 1 horse power is equal to the 745.7 watts. Thus the work done by the motor in horse power is,
[tex]W_m=\dfrac{55590}{745.7}\\W_m=74.55\rm HP[/tex]
Hence, the output power rating in HP of the motor that meets the requirement is 74.55 horse power.
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