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Analysis and Control of Electric Drives. Ned Mohan
Читать онлайн.Название Analysis and Control of Electric Drives
Год выпуска 0
isbn 9781119584551
Автор произведения Ned Mohan
Жанр Физика
Издательство John Wiley & Sons Limited
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 Th must be Th = 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 Tem that 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.
Solution
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 ωm (in rad/s) is
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),
The same angular acceleration
where ρ is the material density in kg/m3. 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
where the quantity within the brackets in Eq. (2-17) is called the moment‐of‐inertia J, which for a solid cylinder is
Since the mass of the cylinder in Fig. 2-6a is
(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