Analysis and Control of Electric Drives. Ned Mohan

Fig. 2-3 Motion of a mass M due to the action of forces.

      This movement is opposed by the load, represented by a force f_L. The linear momentum associated with the mass is defined as M × u. As shown in Fig. 2-3b, in accordance with Newton’s Law of Motion, the net force f_M(=f_e − f_L) equals the rate of change of momentum, which causes the mass to accelerate:

(2-1)

where a is the acceleration in m/s², which from Eq. (2-1) is

      (2-2)

      In MKS units, a net force of 1 Newton (or 1 N), acting on a constant mass of 1 kg, results in an acceleration of 1 m/s². Integrating the acceleration with respect to time, we can calculate the speed as

      (2-3)

and, integrating the speed with respect to time, we can calculate the position as

      (2-4)

      where τ is a variable of integration.

      The differential work dW done by the mechanism supplying the force f_e is

(2-5)

Power is the time‐rate at which the work is done. Therefore, differentiating both sides of Eq. (2-5) with respect to time t, and assuming that the force f_e remains constant, the power supplied by the mechanism exerting the force f_e is

(2-6)

It takes a finite amount of energy to bring a mass to a speed from rest. Therefore, a moving mass has stored kinetic energy that can be recovered. Note that in the system of Fig. 2-3, the net force f_M(=f_e − f_L) is responsible for accelerating the mass. Therefore, assuming that f_M remains constant, the net power p_M(t) going into accelerating the mass can be calculated by replacing f_e in Eq. (2-6) with f_M:

      (2-7)

      From Eq. (2-1), substituting f_M as ,

(2-8)

The energy input, which is stored as kinetic energy in the moving mass, can be calculated by integrating both sides of Eq. (2-8) with respect to time. Assuming the initial speed u to be zero at time t = 0, the stored kinetic energy in the mass M can be calculated as

      (2-9)

      where τ is a variable of integration.

2‐3 ROTATING SYSTEMS

Most electric motors are of a rotating type. Consider a lever, pivoted and free to move as shown in Fig. 2-4a. When an external force f is applied in a perpendicular direction at a radius r from the pivot, then the torque acting on the lever is

(2-10)

      which acts in a counterclockwise direction, considered here to be positive.

      EXAMPLE 2‐1

      In Fig. 2-4a, a mass M is hung from the tip of the lever. Calculate the holding torque required to keep the lever from turning, as a function of angle θ in the range of 0–90°. Assume that M = 0.5 kg and r = 0.3 m.

       Solution

      The gravitational force on the mass is shown in Fig. 2-4b. For the lever to be stationary, the net force perpendicular to the lever must be zero, i.e. f = M g cos β where g = 9.8 m/s² is the gravitational acceleration. Note in Fig. 2-4b that β = θ. The holding torque T_h must be T_h = f r = M g r cos θ. Substituting the numerical values,

Fig. 2-4 (a) Pivoted lever and (b) holding torque for the lever.

In electric machines, the various forces shown by arrows in Fig. 2-5 are produced due to electromagnetic interactions. The definition of torque in Eq. (2-10) correctly describes the resulting electromagnetic torque T_em that causes the rotation of the motor and the mechanical load connected to it by a shaft.

Fig.

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