ME242 – MECHANICAL ENGINEERING SYSTEMS
LECTURE 20:
• Ideal Machines: Transformers and Gyrators 2.4
328
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES
e1 Ideal
q&1 Machine
e2
q&2
An ideal machine is a two port device that transmits
work from one port to the other
• No energy is stored, generated or dissipated
• Entropy is not generated
• Can be run in either direction
ME242 - Spring 2006 - Eugenio Schuster
329
1
IDEAL MACHINES
e1 Ideal
q&1 Machine
e2
q&2
Power Conservation
e1q&1 = e2 q&2
330
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES
Physical or Mechanical systems modeled as ideal machines
•
•
•
•
•
Levers
Gears
Electric motors
Piston pumps
Electric Transformers
More accurate (and more complex) models of these devices
might include other elements. Example: Real Electric Motor
ME242 - Spring 2006 - Eugenio Schuster
331
2
IDEAL MACHINES
Two special cases:
Two-port devices
T
Transformers
T
G
Gyrators
G
332
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES - TRANSFORMER
Defining Condition:
e1
e2
q&1 T q&2
q&2 = Tq&1
or
e2
T
q&2
e1
q&1
Transformer Modulus (constant)
The modulus of the Transformer, T, is defined as the ratio of the
generalized velocity or flow on the bond with the outward power
arrow to the generalized velocity or flow on the bond with the
inward power convention arrow
ME242 - Spring 2006 - Eugenio Schuster
333
3
IDEAL MACHINES - TRANSFORMER
Combining the Ideal Machine condition:
e1q&1 = e2 q&2
with the Transformer condition:
q&2 = Tq&1
yields an additional condition:
e1 = Te2
q&2 e1
=
T=
q&1 e2
The ratio of the generalized forces of an ideal tranformer equals
the inverse of the ratio of the respective generalized velocities
334
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Friction (shear forces) Drives:
Pulley Drives
Rolling Contact Drives
φ&
2
as x& = r1φ&1 = r2φ&2
φ&
1
r2
r1
ME242 - Spring 2006 - Eugenio Schuster
x&
φ&2 r1
T = & = =T
φ1 r2
335
4
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Positive Action (normal forces) Drives:
Toothed Drives
ME242 - Spring 2006 - Eugenio Schuster
336
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Electric Transformer:
ME242 - Spring 2006 - Eugenio Schuster
337
5
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Positive-displacement mechanical to fluid transducer:
piston-and-cylinder / ram
Conservation of Energy
(or Power Balance)
Conservation of Mass
(or Kinematic Constraint )
Cons. of Momentum
(or Force Equilibrium)
Fc x& = PQ
Q = Ax&
Fc = PA
338
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Pumps and Actuators:
Pump
Actuator
ME242 - Spring 2006 - Eugenio Schuster
339
6
IDEAL MACHINES – TRANSFORMER - EXAMPLES
Pump:
Conservation of Energy
(or Power Balance)
M φ& = ∆PQ
Radian Displacement
Conservation of Mass
(or Kinematic Constraint)
Cons. of Momentum
(or Force Equilibrium)
Q = Dφ&
M = D∆P
340
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES - GYRATORS
Defining Condition:
e1
e2
q&1 G q&2
e2 = Gq&1
or
e2
G
q&2
e1
q&1
Gyrator Modulus (constant)
The modulus of the Gyrator, G, is defined as the ratio of the
effort on one of the bonds – either one – to the flow on the other
bond
ME242 - Spring 2006 - Eugenio Schuster
341
7
IDEAL MACHINES - GYRATOR
Combining the Ideal Machine condition:
e1q&1 = e2 q&2
with the Gyrator condition:
e2 = Gq&1
e2 e1
G= =
q&1 q&2
yields an additional condition:
e1 = Gq&2
The ratio of the generalized forces of an ideal tranformer equals
the inverse of the ratio of the respective generalized velocities
342
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES – GYRATOR - EXAMPLES
Spinning Top (A type of Gyroscope):
Top Spinning with angular velocity ω
Has angular momentum H = Iω
H
ω
r
ME242 - Spring 2006 - Eugenio Schuster
343
8
IDEAL MACHINES – GYRATOR - EXAMPLES
Apply a perpendicular impulsive force F∆t
Causes an impulsive moment perpendicular to F,
M∆t = r × F∆t
The moment equals the change in angular momentum
M∆t = ∆H
∆H = M∆t
H
ω
r
F∆t
ME242 - Spring 2006 - Eugenio Schuster
344
IDEAL MACHINES – GYRATOR - EXAMPLES
To get a ∆H requires a ∆ω
The ∆ω is perpendicular to both ω and F
∆ω
∆H
H
ω
r
F ∆t
If F∆t is in direction 1 and r is indirection 3
then ∆ω is indirection 2
ME242 - Spring 2006 - Eugenio Schuster
345
9
IDEAL MACHINES – GYRATOR - EXAMPLES
Gyroscope:
346
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES – GYRATOR - EXAMPLES
Gyroscope:
top view :
φ&2
M2 = 0
M1
φ&
1
L
φ&2 is the precession rate
side view :
φ&1 = 0
ω
M1
ME242 - Spring 2006 - Eugenio Schuster
mg
M2
φ&2
G = Iω = mrg2ω
M 1 = Lmg
M 2 = Gφ&1 = 0
φ&2 =
support
G
M 1 Lmg
Lg
=
=
G mrg2ω rg2ω
347
10
IDEAL MACHINES – GYRATOR - EXAMPLES
Electric Motor / Generator (Tachometer)
348
ME242 - Spring 2006 - Eugenio Schuster
IDEAL MACHINES – GYRATOR - EXAMPLES
DC motor with N coils of radius r rotating in magnetic
field B:
F = 2π rNBi
M = G2i
F = G1i
G1 = 2π rNB
M = rF
T1 = 1/ r
e
i
G1
F
x&
ME242 - Spring 2006 - Eugenio Schuster
M
T1
φ&
G2 = 2π r 2 NB
e = G φ&
2
or
e
i
G2
M
φ&
349
11
SYSTEMS WITH TRANSFORMER AND GYRATORS
Objective 1: Examine implication of two or more
cascaded ideal machines (two-port elements)
T1
T2
T1
G2
G1
G1
T2
G2
350
ME242 - Spring 2006 - Eugenio Schuster
SYSTEMS WITH TRANSFORMER AND GYRATORS
Objective 2: Examine implication of an ideal machine
(two-port element) in sequence with a source or a
resistance (a one-port element)
G
R
R
S
T
S
G
T
ME242 - Spring 2006 - Eugenio Schuster
351
12
SYSTEMS WITH TRANSFORMER AND GYRATORS
We are interested in analyzing the implications
associated with:
• Effective modulus
• Effective Load Characteristic
• Effective Source Characteristic
• Source-load matching / power transfered
ME242 - Spring 2006 - Eugenio Schuster
352
CASCADED TRANSFORMER
ME242 - Spring 2006 - Eugenio Schuster
353
13
e1
q&1
CASCADED TRANSFORMER
T1
e2
q&2
e3
T2
q&3
The definition of transformer requires that
q&3 = T2 q&2 = T2T1q&1
e1 = T1e2 = T1T2 e3
Then,
e1
q&1
T
e3
q&3
T = T1T2
ME242 - Spring 2006 - Eugenio Schuster
354
CASCADED GYRATORS
ME242 - Spring 2006 - Eugenio Schuster
355
14
CASCADED GYRATORS
e1
q&1
G1
e2
e3
G2
q&3
q&2
The definition of transformer requires that
q&3 =
Then,
e2 G1
G
=
q&1 e1 = G1q&2 = 1 e3
G2
G2 G2
e1
q&1
e3
q&3
T
T=
G1
G2
356
ME242 - Spring 2006 - Eugenio Schuster
TRANSFORMER – GYRATORS PAIRS
e1
f1
Gyrator: e3
T
= Gf 2
Transformer: e1
e2
f2
G
Transformer:
= Te2
Gyrator:
e3
f3
f 2 = Tf1 ⇒ e3 = GTf1
e2 = Gf 3 ⇒ e1 = TGf3
Then,
e1
f1
ME242 - Spring 2006 - Eugenio Schuster
G
'
e3
f3
with G ' = TG
357
15
SOURCE - TRANSFORMER – RESISTOR
S
T
R
Two equivalent
views:
R′
S
S′
R
358
ME242 - Spring 2006 - Eugenio Schuster
EFFECTIVE RESISTANCE (AS SEEN BY SOURCE)
S
e1
q&1
T
e2
q&2
R
e2
(example; T=2)
e2T R ′
R
⇒
q&2
T
q&2
S
ME242 - Spring 2006 - Eugenio Schuster
R′
359
16
EFFECTIVE SOURCE (AS SEEN BY RESISTANCE)
S
e1
e1
q&1
T
e2
q&2
S
R
(example; T=2)
e1
T
⇒
q&1
S′
q&1T
S′
ME242 - Spring 2006 - Eugenio Schuster
R
360
SOURCE - TRANSFORMER – RESISTOR
CASE STUDY: Automobile Drive Train
ME242 - Spring 2006 - Eugenio Schuster
361
17
SOURCE - TRANSFORMER – RESISTOR
M 72 ft lb 1 ft
x&
123 ft / s
1 ft
=
=
=
T=
T= &=
F
φ 386rad / s π rad
226 lb π rad
ME242 - Spring 2006 - Eugenio Schuster
Either T or desired
equilibrium given.
362
SOURCE - TRANSFORMER – RESISTOR
M 111.5 ft lb
ft
x& 73.5 ft / s
ft
T=
=
= .242
T= =
= .242
&
461 lb
F
rad
rad
φ 304rad / s
different
gear ratios
required
for max.
power
From last slide.
ME242 - Spring 2006 - Eugenio Schuster
363
18
SOURCE - GYRATOR – RESISTOR
G
S
R
Two equivalent
views:
R′
S
S′
R
364
ME242 - Spring 2006 - Eugenio Schuster
EFFECTIVE RESISTANCE (AS SEEN BY SOURCE)
S
e1
q&1
G
e2
q&2
R
e2
(example; G=.5)
q&2G
R
⇒
e2
G
q&2
S
ME242 - Spring 2006 - Eugenio Schuster
R′
R′
365
19
EFFECTIVE SOURCE (AS SEEN BY RESISTANCE)
S
e1
e1
q&1
G
e2
q&2
R
(example; G=2)
q&1G
S
S′
⇒
e1
G
q&1
S′
ME242 - Spring 2006 - Eugenio Schuster
R
366
SOURCE - GYRATOR – RESISTOR
CASE STUDY: DC Motor
ME242 - Spring 2006 - Eugenio Schuster
367
20