Analysis and Control of Electric Drives. Ned Mohan

Fig. 2-6 Calculation of the inertia, J_cyl, of a solid cylinder.

The net torque T_J acting on the rotating body of inertia J causes it to accelerate. Similar to systems with linear motion, where f_M = M a Newton’s Law in rotational systems becomes

(2-21)

      where the angular acceleration α(=dω/dt) in rad/s² is

(2-22)

which is similar to Eq. (2-18) in the previous example. In MKS units, a torque of 1 Nm acting on the inertia of 1 kg ⋅ m² results in an angular acceleration of 1 rad/s².

In systems such as the one shown in Fig. 2-7a, the motor produces an electromagnetic torque T_em. The bearing friction and wind resistance (drag) can be combined with the load torque T_L opposing the rotation. In most systems, we can assume that the rotating part of the motor with inertia J_M is rigidly coupled (without flexing) to the load inertia J_L. The net torque, which is the difference between the electromagnetic torque developed by the motor and the load torque opposing it, causes the combined inertias of the motor and the load to accelerate in accordance with Eq. (2-22):

(2-23)

      where the net torque T_J = T_em − T_L and the equivalent combined inertia is J_eq = J_M + J_L.

EXAMPLE 2‐3

In Fig. 2-7a, each structure has the same inertia as the cylinder in Example 2-2. The load torque T_L is negligible. Calculate the required electromagnetic torque, if the speed is to increase linearly from rest to 1800 rpm in 5 s.

       Solution

      Using the results of Example 2-2, the combined inertia of the system is

      The angular acceleration is

Therefore, from Eq. (2-23),

Fig. 2-7 Motor and load torque interaction with a rigid coupling.

      Equation (2-23) shows that the net torque is the quantity that causes acceleration, which in turn leads to changes in speed and position. Integrating the acceleration α(t) with respect to time,

(2-24)

where ω_m(0) is the speed at t = 0 and τ is a variable of integration. Further integrating ω_m(t) in Eq. (2-24) with respect to time yields

(2-25)

where θ(0) is the position at t = 0 and τ is again a variable of integration. Equations (2-23) through (2-25) indicate that torque is the fundamental variable for controlling speed and position. Equations (2-23) through (2-25) can be represented in a block‐diagram form, as shown in Fig. 2-6b.

      EXAMPLE 2‐4

      Consider that the rotating system shown in Fig. 2-7a, with the combined inertia J_eq = 2 × 0.029 = 0.058 kg ⋅ m², is required to have the angular speed profile shown in Fig. 2-1b. The load torque is zero. Calculate and plot, as functions of time, the electromagnetic torque required from the motor, and the change in position.

       Solution

      In the plot of Fig. 2-2b, the magnitude of the acceleration and the deceleration is 100 rad/s². During the intervals of acceleration and deceleration, since T_L = 0,

as shown in Fig. 2-8.

      During intervals with a constant speed, no torque is required. Since the position θ is the time‐integral of speed, the resulting change of position (assuming that the initial position is zero) is also plotted in Fig. 2-8.

Fig. 2-8 Speed, torque, and angle variations with time.

In a rotational system shown in Fig. 2-9, if a net torque T