A motor drives two loads. One has rotational motion. It is coupled to the motor through a reduction gear with a = 0.1 and efficiency of 90%. The load has a moment of inertia of 10.kg- m ^ 2 and a torque of 10 N-m. Other load has translational motion and consists of 1000 kg weight to be lifted up at an uniform speed of 1.5 m/s. Coupling between this load and the motor has an efficiency of 85%. Motor has an inertia of 0.2kg – m ^ 2 and runs at a constant speed of 1420 rpm. Determine equivalent inertia referred to the shaft and power developed by the motor.

• 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.