A motor drives two loads. One has rotational motion. It is coupled to the motor
through a reduction gear with a = 0.5 and efficiency of 85 %. The load has a moment of
inertia of 20 Kg-m2 and a torque of 10 N-m. Other load has translational motion and
consists of 1200 Kg weight to be lifted up at an uniform speed of 2 m/s. Coupling
between this load and the motor has an efficiency of 80 % . Motor has an inertia of 0.2
Kg-m2 and runs at a constant speed of 1400 rpm. Determine equivalent inertia referred
to the motor shaft and power developed by the motor.
Answers
Answer:
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Answer:
The equivalent inertia referred to the motor shaft is 32.5 Kg-m2 and the power developed by the motor is 34.65 kW.
Explanation:
Given data:
Reduction gear ratio, a = 0.5
Efficiency of reduction gear, η1 = 85%
Moment of inertia of rotational load, J1 = 20 Kg-m2
Torque of rotational load, T1 = 10 N-m
Efficiency of coupling for translational load, η2 = 80%
Mass of translational load, m2 = 1200 Kg
Speed of translational load, v2 = 2 m/s
Motor inertia, Jm = 0.2 Kg-m2
Motor speed, N = 1400 rpm
We need to find:
Equivalent inertia referred to the motor shaft
Power developed by the motor
Solution:
Equivalent Inertia
The equivalent inertia referred to the motor shaft can be calculated as:
Jeq = Jm + a2J1 + a22J2
where J2 is the equivalent inertia of the translational load, referred to the motor shaft.
We know that for a translational load,
J2 = m2r2
where r2 is the radius of the load, assuming it to be a solid cylinder.
The mass can be converted to equivalent moment of inertia using the formula:
J2 = (1/2)m2r2^2
Also, the velocity can be converted to angular velocity using the formula:
ω = v2/r2
The torque required to lift the load at a uniform speed can be calculated as:
T2 = m2g sinθ
where g is the acceleration due to gravity and θ is the angle of inclination of the plane.
Assuming the load is lifted on a plane inclined at an angle of 30 degrees, we have:
r2 = 0.5 m
ω = v2/r2 = 4 rad/s
T2 = m2g sinθ = 1200 × 9.81 × sin30 = 5886.36 N-m
Now, we can calculate J2 as:
J2 = (1/2)m2r2^2 = (1/2) × 1200 × 0.5^2 = 150 Kg-m2
Therefore, the equivalent inertia referred to the motor shaft is:
Jeq = Jm + a2J1 + a22J2
= 0.2 + 0.52 × 20 + 0.252 × 150
= 32.5 Kg-m2
Power developed by the motor
The power developed by the motor can be calculated as:
P = (T1 + T2)ω/η
where η is the overall efficiency of the system.
We can calculate the total torque as:
T = T1 + T2 = 10 + 5886.36 = 5896.36 N-m
The overall efficiency of the system can be calculated as:
η = η1 × η2
= 0.85 × 0.80
= 0.68
Therefore, the power developed by the motor is:
P = (T1 + T2)ω/η
= 5896.36 × 4 / 0.68
= 34648.94 W or 34.65 kW
Hence, the equivalent inertia referred to the motor shaft is 32.5 Kg-m2 and the power developed by the motor is 34.65 kW.
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