• If the motor is rated at 19kW, is the motor sufficient to drive the two loads?
• The translational motion load now has to lift a weight of 1200 kg at the same speed of 1.5m/s. Is the motor still capable to drive both loads at the same motor speed of 1420 rpm?
Answer
Given data:
For the rotational load:
Moment of inertia, I1 = 10 kg-m^2
Torque, T1 = 10 N-m
Gear ratio, a = 0.1
Efficiency of the gear, η1 = 90% = 0.9
For the translational load:
Mass, m2 = 1000 kg
Speed, v2 = 1.5 m/s
Efficiency of the coupling, η2 = 85% = 0.85
Motor inertia, Im = 0.2 kg-m^2
Motor speed, N = 1420 rpm
Motor power, P = 19 kW
1. Equivalent inertia referred to the shaft:
The equivalent inertia referred to the shaft can be calculated as:
Ieq = I1 + 1/a21/𝑎2 * I2 + Im
Where I2 is the equivalent inertia of the translational load.
Given that the load has to lift a weight of 1000 kg at 1.5 m/s, the equivalent inertia for the translational load is:
I2 = m2 * v2𝑣2^2 = 1000 * 1.51.5^2 = 2250 kg-m^2
Therefore,
Ieq = 10 + 1/0.121/0.12 * 2250 + 0.2 = 10 + 22500 + 0.2 = 22710.2 kg-m^2
2. Power developed by the motor:
The power developed by the motor can be calculated using the formula:
P = 2 * π * N * T / 60
Where T is the torque developed by the motor.
For the rotational load, the torque developed by the motor is:
T1 = T / a = 10 / 0.1 = 100 N-m
For the translational load, the force required to lift the weight is:
F = m2 * g = 1000 * 9.81 = 9810 N
The power developed by the motor for the translational load is:
P2 = F * v2 = 9810 * 1.5 = 14715 W = 14.715 kW
Therefore, the total power developed by the motor is:
Ptotal = η1 * P1 + η2 * P2
P1 = 2 * π * N * T1 / 60 = 2 * π * 1420 * 100 / 60 = 1489.77 W = 1.48977 kW
Ptotal = 0.9 * 1.48977 + 0.85 * 14.715 = 1.340793 + 12.50775 = 13.848543 kW
3. Motor sufficiency:
The motor is rated at 19 kW, and the total power developed by the motor is 13.848543 kW.
Since the total power developed is less than the rated power of the motor, the motor is sufficient to drive both loads at 1420 rpm.
4. Increased load scenario:
If the load has to lift a weight of 1200 kg at the same speed of 1.5 m/s, the force required is:
F’ = 1200 * 9.81 = 11772 N
The power developed by the motor for the increased load is:
P2′ = F’ * v2 = 11772 * 1.5 = 17658 W = 17.658 kW
The total power developed by the motor is now:
Ptotal’ = 0.9 * 1.48977 + 0.85 * 17.658 = 1.340793 + 15.0093 = 16.350093 kW
Since the total power developed is still less than the rated power of the motor, the motor is still capable of driving both loads at 1420 rpm.