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EE130 Electromechanics 2013

Explore formulas for acceleration, power, and kinetic energy in electric drive systems. Learn about torque calculations and inertia loads. Solve practical problems to enhance your knowledge.

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EE130 Electromechanics 2013

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  1. EE130Electromechanics2013 J. Arthur Wagner, Ph.D. Prof. Emeritus in EE wagneretal@sbcglobal.net

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

  3. Fig. 2.2 (Linear) motion of M

  4. Acceleration, 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)

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

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

  7. 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.

  8. 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

  9. 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)

  10. 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 • = 0.0065 kg m^2

  11. Angular acceleration = torque / moment of inertia (2.23)

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

  13. 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.

  14. Ex. 2.3 • 1800 rpm * pi / 30 = 188.5 rad/s • Jeq = 2 * .029 = 0.058 kg m^2 • acceleration = 188.5 rad/s / 5 s = 37.7 rad/s^2 • electromagnetic torque = J * accel = • =.058 * 37.7 = 2.19 N m

  15. Fig. 2.13 Combination of rotary and linear motion

  16. Rotary and Linear Motion T due to load only

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

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