EE130Electromechanics

2013

J. Arthur Wagner, Ph.D.

Prof. Emeritus in EE

wagneretal@sbcglobal.net

Fig. 2.1 Electric Drive

System (ASD)

Example of load-speed requirement

ASD Load

L

0 1 2 3 4 5 6 7

100

sect

/ secL raddesired speed profile

Fig. 2.2 (Linear) motion of M

Mf Mef

MLf

x

; Mdx du fu adt dt M

; e Lf fdx duu adt dt M

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)

Fig. 2.3 (a) Pivoted lever (b) Holding torque for the lever

f

Mgtorque

r

f

o90

M

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

emT LTMotorLoad

the inertia of a cylinder = ½ * M * r1^2 (2.20)M = cylinder massr1 = cylinder radiusTL = 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

• = 0.0065 kg m^2

Angular acceleration = torque / moment of inertia (2.23)

em L JT T T

em L JT T Tddt J J

Fig. 2.6 Motor and load with rigid coupling

MotorLoad

m

emT

emT

LT

m JT

eq

1

J

LT

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. 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

Fig. 2.13 Combination of rotary and linear motion

Lfu

Motor

Jm

Tem

M

r

Jm motor inertia

M = mass of load

r = pulley radius

Rotary and Linear Motion

required to accelerate due to load motor

2mem m L

d dT J r M r f

dt dt

L

m2 m

L

d uf M f

dtu r

dT r f r M rf

dt

T due to load only

Homework Chapter 2, Due next Tuesday

• Problems 2.1, 2.11, 2.12, 2.13, 2.14