ee130 electromechanics 2013 j. arthur wagner, ph.d. prof. emeritus in ee [email protected]
TRANSCRIPT
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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
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Fig. 2.2 (Linear) motion of M
Mf Mef
MLf
x
; Mdx du fu adt dt M
; e Lf fdx duu adt dt M
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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)
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Fig. 2.3 (a) Pivoted lever (b) Holding torque for the lever
f
Mgtorque
r
f
o90
M
Torque = force x radius
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Ex. 2.1
• M = 0.5 kg
• r = 0.3 m
• Calculate holding torque as a function of beta
• How do we work this?
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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.
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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
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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)
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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
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Angular acceleration = torque / moment of inertia (2.23)
em L JT T T
em L JT T Tddt J J
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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
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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.
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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
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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
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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
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Homework Chapter 2, Due next Tuesday
• Problems 2.1, 2.11, 2.12, 2.13, 2.14