Analysis and Control of Electric Drives. Ned Mohan

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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 causes the cylinder to rotate by a differential angle dθ, the differential work done is

      (2-26)

Fig. 2-9 Torque, work, and power.

      If this differential rotation takes place in a differential time dt, the power can be expressed as

(2-27)

where ω_m = dθ/dt is the angular speed of rotation. Substituting for T from Eq. (2-21) into Eq. (2-27),

(2-28)

Integrating both sides of Eq. (2-28) with respect to time, assuming that the speed ω_m and the kinetic energy W at the time t = 0 are both zero, the kinetic energy stored in the rotating mass of inertia J is

(2-29)

      This stored kinetic energy can be recovered by making the power p(t) reverse direction, that is, by making p(t) negative.

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