EE130 Electromechanics 2013 J. Arthur Wagner, Ph.D. Prof. Emeritus in EE

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Presentation transcript:

EE130 Electromechanics 2013 J. Arthur Wagner, Ph.D. Prof. Emeritus in EE

Fig. 2.1 Electric Drive System (ASD) Example of load-speed requirement ASD Load

Fig. 2.2 (Linear) motion of M

Accelerati on, Power Input, Kinetic Energy Write the formula for acceleration (sum of forces / mass) (2.2) Write the formula for power (Net force times velocity) (2.6) Write the formula for kinetic energy (Recall from physics) (2.9)

Fig. 2.3 (a) Pivoted lever (b) Holding torque for the lever Torque = force x radius

Ex. 2.1 M = 0.5 kg r = 0.3 m Calculate holding torque as a function of beta How do we work this?

Ex. 2.1 torque = force x radius force perp. to radius force = Mg perp. component = Mg cos (beta) torque = 0.5 kg 9.8 m/s^2 * 0.3 m cos(beta) = 1.47 Nm In a motor, the force is produced electromagnetically and is tangent to the cylindrical rotor. The rotor radius converts this force to torque on the shaft.

Fig. 2.4 Motor torque acting on an inertia load the inertia of a cylinder = ½ * M * r1^2 (2.20) M = cylinder mass r1 = cylinder radius TL = load torque, other than due to inertia

An inertia Load Mostly inertia Mostly friction (a grinder) Mostly mechanical torque in steady state (a belt lifting gravel) All systems have some of both (an inertia plus other torque)

Inertia of a 14 in disk 14.5 oz /(16 oz/lb) / (2.2 lb / kg) =.412 kg d1 = 14 in * (.0254 m / in) =.356 m r1 =.356 / 2 =.178 m J = ½ * M * r1^2 = ½ *.412 * (.178)^2 = kg m^2

Angular acceleration = torque / moment of inertia (2.23)

Fig. 2.6 Motor and load with rigid coupling Block diagram of acceleration equation. Two integrators to go to angular velocity and angular position

Ex. 2.3 TL negligible, each cylinder has same inertia J of one cylinder =.029 kg m^2 speed goes from 0 to 1800 rpm in 5 s Calculate the required electromagnetic torque. How do we work this? First, sketch a speed profile.

Ex rpm * pi / 30 = rad/s Jeq = 2 *.029 = kg m^2 acceleration = rad/s / 5 s = 37.7 rad/s^2 electromagnetic torque = J * accel = =.058 * 37.7 = 2.19 N m

Fig Combination of rotary and linear motion

Rotary and Linear Motion T due to load only

Homework Chapter 2, Due next Tuesday Problems 2.1, 2.11, 2.12, 2.13, 2.14