Ned Mohan - Analysis and Control of Electric Drives

In electric drives, the power flow may be in either direction. For example, in an electric vehicle in driving mode, power flows from the electric source, a battery, to the electric motor, a mechanical system, through a power electronic converter. On the other hand, while slowing down the vehicle, the roles are reversed. The kinetic energy of the moving vehicle energy is extracted (called regenerative braking), and power flows from the electric motor to the battery, again through a power electronic converter.

Electric drives must satisfy the requirements of torque and speed imposed by mechanical loads connected to them. The load in Fig. 2-2, for example, may require a trapezoidal profile for the angular speed, as a function of time. In this chapter, we will briefly review the basic principles of mechanics for understanding the requirements imposed by mechanical systems on electric drives. This understanding is necessary for selecting an appropriate electric drive for a given application.

Fig. 2-2 (a) Electric drive system and (b) example of a load‐speed profile requirement.

This analysis equally applies when the load becomes the source of power, as in a wind turbine, and the electric drive generates and transfers power to the utility grid, an electric system.

We will begin by applying physical laws of motion in their simplest form, starting with linear systems. In Fig. 2-3a, a load of a constant mass M is acted upon by an external force f ethat causes it to move in the linear direction x at speed u = dx / dt .

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/s 2, 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/s 2. 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 eis

(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 eremains constant, the power supplied by the mechanism exerting the force f eis

(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 Mremains constant, the net power p M( t ) going into accelerating the mass can be calculated by replacing f ein Eq. (2-6)with f M:

(2-7)

From Eq. (2-1), substituting f Mas ,

(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.

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.

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.

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/s 2is the gravitational acceleration. Note in Fig. 2-4b that β = θ . The holding torque T hmust be T h= f r = M g r cos θ . Substituting the numerical values,