Ned Mohan - Analysis and Control of Electric Drives

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

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

Fig. 2-5 Torque in an electric motor.

In a rotational system, the angular acceleration, due to the net torque acting on it, is determined by its moment‐of‐inertia J . The example below shows how to calculate the moment‐of‐inertia J of a rotating solid cylindrical mass.

1 Calculate the moment‐of‐inertia J of a solid cylinder that is free to rotate about its axis, as shown in Fig. 2-6a, in terms of its mass M and the radius r1.

2 Given that a solid steel cylinder has a radius r1 = 6 cm, length ℓ = 18 cm, and material density ρ = 7.85 × 103 kg/m3, calculate its moment‐of‐inertia J.

1 From Newton’s Law of Motion, in Fig. 2-6a, to accelerate a differential mass dM at a radius r, the net differential force df required in a perpendicular (tangential) direction, from Eq. (2-1), is (2-11)

where the linear speed u in terms of the angular speed ω m(in rad/s) is

(2-12)

Multiplying both sides of Eq. (2-11) by the radius r , recognizing that ( r df ) equals the net differential torque dT and using Eq. (2-12),

(2-13)

The same angular acceleration is experienced by all elements of the cylinder. With the help of Fig. 2-6b, the differential mass dM in Eq. (2-13)can be expressed as

(2-14)

where ρ is the material density in kg/m 3. Substituting dM from Eq. (2-14)into Eq. (2-13),

(2-15)

The net torque acting on the cylinder can be obtained by integrating over all differential elements in terms of r , θ , and ℓ as

(2-16)

Carrying out the triple integration yields

(2-17)

(2-18)

where the quantity within the brackets in Eq. (2-17)is called the moment‐of‐inertia J , which for a solid cylinder is

(2-19)

Since the mass of the cylinder in Fig. 2-6a is , the moment‐of‐inertia in Eq. (2-19)can be written as

(2-20)

1 Substituting r1 = 6 cm, length ℓ = 18 cm, and ρ = 7.85 × 103 kg/m3 in Eq. (2-19), the moment‐of‐inertia Jcyl of the cylinder in Fig. 2-5a is

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

The net torque T Jacting 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/s 2is

(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 ⋅ m 2results in an angular acceleration of 1 rad/s 2.

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 Lopposing the rotation. In most systems, we can assume that the rotating part of the motor with inertia J Mis 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 Land the equivalent combined inertia is J eq= J M+ J L.

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

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

The angular acceleration is

Therefore, from Eq. (2-23),