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Geared systems
Important examples of equivalent systems are given by devices
called mechanical transformers. Such devices as levers, gears,
cams, and chains transform the motion at the input into a related
motion at the output. Their analysis is simplified by describing
the effects of their mass, elasticity, and damping relative to a
single location, such as the input location. The result is a
lumped-parameter model whose coordinate system is centered at
that location.
We can characterie a transformer by a single parameter bsuch
that in terms of the general displacement variable Q,
Q1= bQ2 !".#-$%&where the subscripts $ and " refer to the input and output states,
respectively. Since the rate variable ris related to Qby dQ/dt=r,
we also have
r1=br2 !".#-$'&
if bis constant. (n ideal transformer neither stores nor
dissipates energy. Therefore, the output power equals the input
power, or e1r1= e2r2. This shows the effort variables to be
related as
"$
$e
be = !".#-$)&
( specific example is the gear pair shown in Table ".#, with=Q , Te= , and =r . The parameter bis the gear ratioN. If
N>1, the pair is a speed reducer whose output torque T2is
greater than the input torque T1.
Suppose an element possessing inertia, elasticity, or damping
properties is connected to the output shaft of a gear pair, asshown in Table ".#. *ow can we represent the geared system by
an equivalent gearless system+ If we wish to reference all
motion to the input shaft, we can proceed as follows. We
assume that the inertia, twisting, and damping of the input shaft
are negligible. If not, they can be included in a manner similar to
the following. We derive an equivalent system by requiring its
inetic energy, potential energy, and power dissipation terms to
be identical to those of the original system. The expressions forthese quantities in the equivalent system are
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Table ".# ear pair relations
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inetic energy "
$
"
$eI
potential energy "
$
"
$ek
power dissipation "
$$$$$ &! ee ccT ==
If we neglect the gear inertias, gear tooth elasticity, and gear
friction, the corresponding expressions for the original system
are
inetic energy "
""
$I
potential energy "
""
$k
power dissipation
"
""" cT =
/quating these expressions and using the gear ratio formulas, we
obtain the parameters for the equivalent system
IN
IIe "
"
$
" $&! ==
!".#-$0&
kN
kke ""
$
" $&! ==
!".#-"1&
cN
cce "
"
$
" $&! ==
!".#-"$&
The next two examples illustrate the lumping process.
System with three inertias
2or systems with more than one inertia on a single shaft, it is
frequently possible to add the inertias to produce an equivalent
system. ( pump is presented in 2igure "."3. The pump impellerI2is driven by the shaftI3connected to an electric motor
producing a torque T1on its rotorI1. The viscous friction of the
fluid acting on the impeller is represented by the damper. (
model is desired for the behavior of the angular displacement "
due to the applied torque T1. 4iscuss how a lumped-parameter
model can be developed !2igure "."3b&.
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2igure "."3 System with three inertias. !a& 5riginal system. !b&
6umped-parameter equivalent system.
( good initial assumption might be to neglect the twist "$
between the inertiasI1andI2. This corresponds to neglecting thepotential energy in the shaft and implies that "$ = and "$ = .
/quating the inetic energy of the original and equivalent
systems gives"
$#
"
$"
"
$$
"
"
$
"
$
"
$
"
$ IIII ++=
With a coordinate reference on the motor shaft, $= , and thus,#"$ IIII ++= !".#-""&
The equivalent system model is given by
dt
dcT
dt
dI
= $"
"
or, sincedt
d $$
=
$$$
cT
dt
dI = !".#-"#&
If the damping force is large or if the torque T$is applied
suddenly, the shaft twist will become important, and a model
incorporating shaft elasticity will be needed.Geared pump-motor system
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Suppose that the pump motor drives the impeller through a gear
pair !2igure "."7&.
2igure "."7 eared system.
(ssume also that the gear and shaft inertias are significant.
8eglect the elasticity of the gear teeth and shafts, and assume
the friction and baclash in the gears are small.4evelop a lumped-inertia model of the impeller velocity as a
function of the motor torque.
The angular displacements of the gears are related through the
gear ratio 8 such thatCB N = !".#-"3&
B
C
D
DN=
8ote that a positive displacement B produces a positivedisplacement C . Thus a negative sign is not needed in !".#-"3&
even through the directions of rotation are opposite. 2rom this,
we also see that CB N = .
With negligible shaft elasticity, the twist is ero, and thus,
DC
DC
BA
BA
=
=
=
=
and
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DA N =
(lso, we can use the results from !".#-""& to obtainI1, the
lumped inertia on shaftAB, as follows9BABA
IIII ++=$
Similarly, the lumped inertia on shaft :4 isDCDC IIII ++="
2or a single-inertia model, we must reference all quantities to
one shaft. 6et us choose the motor shaftABas the reference.
Then, the inertia equivalent toI2referenced to shaftABis
"""
$I
NI
e =
The total equivalent inertiaIis the sum of all the inertias
referenced to shaftAB, or"$ eIII += !".#-"7&
The viscous damping acting on shaft :4 has an equivalent value
on shaft (; of
cN
ce "$
= !".#-"%&
Therefore, the desired model is
AeAA cT
dt
dI
= !".#-"'&
WithIand cegiven earlier.
Suppose that we had chosen shaft :4 as the reference shaft. (
good exercise is to use energy equivalence to derive the
following model.
DA
D cNTdt
dINI
=+ &! "
"
$ !".#-")&
5f course, !".#-"'& and !".#-")& are equivalent, as can be
verified substituting DA N = into !".#-"'&.
This model can be used to compute the motor torque required to
maintain the load at some desired constant speed. 2or constant
speed, !".#-"'& shows the required torque to beDeAeA NccT == !".#-"0&
Rotational-translational systems
There are many mechanisms which involve the conversion of
rotational motion to translational motion or vice versa. 2or
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example, there are rac-and-pinion, shafts with lead screws,
pulley and cable systems, etc.
To illustrate how such systems can analyed, consider a rac-
and-pinion system shown in 2igure "."%.
2igure "."%
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/liminating Toutfrom equations !".#-#1& and !".#-#"& gives
dt
dvr
r
IrcvT
in &! +=
and so
&&!! " rcvTrIr
dtdv in
+=
The result is a first-order differential equation describing how
the output is related to the input.
Electromechanical systems
/lectromechanical devices, such as potentiometers, motors and
generators, transform electrical signals to rotational motion or
vice versa. ( potentiometer has an input of a rotation and an
output of a potential difference. (n electric motor has an input
of a potential difference and an output of rotation of a shaft. (
generator has an input of rotation of a shaft and an output of a
potential difference.
!otentioeter
The rotary potentiometer, 2igure "."', is a potential divider and
thus9
max
=
"
"o
where "is the potential difference across the full length of the
potentiometer trac and max is the total angle swept out by the
slider in being rotated from one end of the trac to the other.
The output is "outfor the input .
2igure "."'
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DC otor
The 4: motor is used to convert an electrical input signal into a
mechanical output signal, a current through the armature coil of
the motor resulting in a shaft being rotated and hence the load
rotated !2igure ".")&.
2igure ".") 4: motor is driving a load.
The motor basically consists of a coil, the armature coil, which
is free to rotate. This coil is located in the magnetic field
provided by a current through field coils or permanent magnet.When a current #i flows through the armature coil then, because
it is in a magnetic field, forces act on the coil and cause it to
rotate !2igure "."0&
2igure "."0 5ne wire of armature coil.
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The force$acting on a wire carrying a current i#and of length
%in a magnetic field of flux densityBat right angles to the wire
is given by$=Bi#%and withNwires is$= Nbi#%. The forces on
the armature coil wires result in a torque T, where T = $b, with
bbeing the breadth of the coil. Thus
T=Nbi#%b
The resulting torque is thus proportional toBi#, the other factors
all being constants. *ence we can write
T=k1Bi#
Since the armature is a coil rotating in a magnetic field, a
voltage will be induced in it as a consequence of
electromagnetic induction. This voltage will be in such adirection as to oppose the change producing it and is called the
bac e.m.f. This bac e.m.f. vbis proportional to the rate or
rotation of the armature and flux lined by the coil, hence the
flux densityB. ThusBkvb "= , where is the shaft angular velocity and k2a
constant.
:onsider a 4: motor which has the armature and field coilsseparately excited. With a so-called armature-controlled motor
the field current i&is held constant and the motor controlled by
ad=usting the armature voltage "#. ( constant field current
means a constant magnetic flux densityBfor the armature coil.
Thus #" kBkvb == , where k3is a constant. The armature circuit can
be considered to be a resistance'#in series with an inductance
%#!2igure ".#1&.
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2igure ".#1 4: motor circuits.If v#is the voltage applied to the armature circuit then, since
there is a bac e.m.f., we have
##
#
#b# i'dt
di%vv +=
We can thin of this equation in terms of the bloc diagram
shown in 2igure ".#$a.
2igure ".#$ 4: motors9 !a& (rmature-controlled, !b& 2ield-
controlled.
The input to the motor part of the system is v#and this is
summed with the feedbac signal of the bac e.m.f. vbto give an
error signal which is the input to the armature circuit. The above
equation thus describes the relationship between the input of the
error signal to the armature coil and the output of the armaturecurrent i#. Substituting for vb9
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##
#
## i'dt
di%kv += #
The current i#in the armature generates a torque T. Since, for the
armature-controlled motor,Bis constant we have## ikBikT 3$ ==
where kis a constant. This torque then becomes the input to the
load system. The net torque acting on the load will be
8et torque T- damping torque
The damping torque is c , where c is a constant. *ence, if any
effects due to the torsional springiness of the shaft are neglected,
8et torque cik #3
This will cause an angular acceleration ofdt
d, hence
cikdt
dI
# = 3
We thus have two equations that describe the conditions
occurring for an armature-controlled motor, namely
##
#
## i'dt
di%kv += # and
cik
dt
dI
# = 3
We can thus obtain the equation relating the output with the
input v#to the system by eliminating i#.
With a so-called field-controlled motor the armature current is
held constant and the motor controlled by varying the field
voltage. 2or the field circuit !2igure ".#1& there is essentially
=ust inductance%&in series with a resistance'&. Thus for that
circuit
dt
di%i'v
&
&&&& +=
We can thin of the field-controlled motor in terms of the bloc
diagram shown in 2igure ".#$b. The input to the system is v&.The field circuit converts this into a current i&, the relationship
between v&and if being the above equation. This current leads to
the production of a magnetic field and hence a torque on the
armature coil, as given by #BikT $= . ;ut the flux densityBis
proportional to the field current i&and i#is constant, hence ikBikT 7$ == , where kis a constant. This torque output is then
converted by the load system into an angular velocity . (s
earlier, the net torque acting on the load will be
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8et torque T- damping torque
The damping torque is c , where c is a constant. *ence, if any
effects due to the torsional springiness of the shaft are neglected,
8et torque cik & 7
This will cause an angular acceleration ofdt
d, hence
cikdt
dI
& = 7
The conditions occurring for a field-controlled motor are thus
described by the equations
dt
di%i'v
&
&&&& += and
cikdt
dI
& = 7
We can thus obtain the equation relating the output with the
input v&to the system by eliminating i&.
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2.3. Hydraulic and liquid level systems
We can model fluid behavior as incompressible, which means
that the fluid>s density remains constant despite changes in the
fluid pressure. If the density changes with pressure, the fluid is
compressible. ;ecause all real fluids are compressible to some
extent, incompressibility is an approximation. ;ut this
approximation is usually sufficiently accurate for most liquids
under typical conditions, and it results in a simpler model of the
system. *ydraulics is the study of incompressible liquids, and
hydraulic devices use an incompressible liquid, such as oil, for
their woring medium. 6iquid level systems consisting of
storage tans and connecting pipes are a class of hydraulicsystems whose driving force is due to relative differences in the
liquid heights in the tans.
*ere, we will avoid complex system models by describing only
the gross system behavior instead of the details of the fluid
motion patterns. To do this, we need only one basic physical
law9 conservation of mass. 2or incompressible fluids,
conservation of mass is equivalent to conservation of volume,
since the fluid density is constant. If we now the density andthe volume flow rate, we can compute the mass flow rate. That
is, ** = , where *and *are the mass and volume flow rates,
and is the fluid density.
Basic principles
The primary variables for hydraulic systems are pressure, mass,
and flow rate.
2luid resistance is the constitutive relation between pressure andmass flow rate.
2luid capacitance is the constitutive relation between pressure
and mass.
Integral causality relates mass m to mass flow rate, because
= dt* !".#-$&
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:onservation of mass can be stated as follows. 2or a container
holding a mass of fluid , the time rate of change of mass d/dt
inthe container must equal the total mass inflow rate minus the
total mass outflow rate. That is,
oi **
dt
d= !".#-"&
where *iis the inflow rate and *ois the outflow rate. The fluid
mass m is related to the container volume "by
" = !".#-#&
2or incompressible fluid, is constant, anddt
d"
dt
d= . 6et *1
and *2be the total volume inflow and outflow rates. Thus,$**i = and "**o = . Substituting these relationships into !".#-"&
gives
"$ **dt
d" =
or"$ **
dt
d"= !".#-3&
This is a statement of conservation of volume for the fluid, and
it is equivalent to conservation of mass, equation !".#-"&.
2luid capacitance and resistance
The tan shown in 2igure ".#" illustrates these concepts. 6etA
be the surface area of the tan>s bottom. If the tan>s sides arevertical, the liquid height +is related to m by
A+ = !".#-7&
In analogy with the electrical capacitance relation, v=Q/C,
pressure plays the role of voltage, and mass m plays the role of
charge. Thus, the fluid capacitance relation is
C
= !".#-%&
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where : is the fluid capacitance of the system. The hydrostatic
pressure due to the height +is,+- = !".#-'&
2igure ".#" ( liquid-level system.
where,is the acceleration due to gravity.
:omparing !".#-7&-!".#-'&, we see that
A
,- = !".#-)&
Thus, the fluid capacitance of the tan in 2igure ".#" is C=A/,.
When the container does not have vertical sides, the relations
between m and h, and between p and m are not linear. In this
case, we define the capacitance to be the slope of the m versus p
curve, and thus there is no single value for the container>scapacitance.
With reference to 2igure ".#", if atmospheric pressure pa exists
at both the liquid>s surface and the pipe outlet, the pressure
difference across the pipe is !?#& -#. The outflow rate *o
should obviously depend onsomehow. The greater the
pressure, the greater the outflow rate. 2or now, let us assume
that the relation between *oandis linear. In analogy with the
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electrical resistance relation, i=v/r, the linear fluid resistance
relation is written as
'
-*
o
=
!".#-0&where are is the fluid resistance.
Liquid heiht model
(ssume that we are given *ias a function of time and that we
want a model for the behavior of the height +for the system in
2igure ".#".
@sing !".#-"&, !".#-7&, !".#-'&, and !".#-0&, we obtain
+'
,*
dt
d+A i
= !".#-$1&
If we write this equation in terms of the volume inflow rate $* ,
the density can be divided out of the equation to obtain
+'
,*
dt
d+A = $ !".#-$$&
!nterconnected storae elements
When a hydraulic system contains more than one storageelement, apply the conservation of mass equation !".#-"& to each
element. Then use the appropriate resistance relations to couple
the resulting equations. It is necessary to assume that some
pressures are greater than others and assign the positive-flow
directions accordingly. If you are consistent, the mathematics
will handle the reversals of flow direction automatically.
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#0e 2.1
4evelop a model for the heights +1and +2in the liquid level
system shown in 2igure ".##a. The input volume rate *is given.
(ssume that laminar flow exists in the pipes. The laminar
resistances are'1and'2, and the bottom areas of the tans are
A1andA2. (lso draw the system bloc diagram.
Solution
(ssume that +1>+2so that the mass flow rate *1is positive if
going from tan $ to tan ". :onservation of mass applied to
each tan gives
2igure ".## ( system of two coupled tan elements and its
bloc diagram.
2or tan $
&! "$$
$
$$
$
++'
,*
**dt
d+A
=
=
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2or tan ",
"
"
$"
"
+
'
,*
**dt
d+A
o
o
=
=
Substituting for *1and *oand dividing by gives the desired
model.
&! "$$
$
$ ++'
,*
dt
d+A = !".#-$"&
"
"
"$
$
"
" &! +'
,++
'
,
dt
d+A = !".#-$#&
The density does not appear in the final model because of the
incompressibility assumption. The bloc diagram shown in
2igure ".##b is easily constructed from the preceding equations.
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2." Hydraulic devices
*ydraulic devices use an incompressible liquid, such as oil, for
the woring medium. They are widely used with high pressures
to obtain large forces in control systems, allowing large loads to
be moved or high accelerations to be obtained. *ere, we analye
three devices commonly found in hydraulic systems9 !$&
dampers limit the velocity of mechanical elementsA !"& hydraulic
servomotors produce motion from a pressure sourceA !#&
accumulators reduce fluctuations in pressure or flow rates.
# $luid damper
( damper or dashpot is a mechanical device that exerts a forceas a result of a velocity difference across it. 5ne common design
relying on fluid friction is shown in figure ".#3, and it is similar
to that used in automotive shoc absorbers.
2igure ".#3 *ydraulic damper
( piston of diameter 4 and thicness 6 has a cylindrical hole of
diameter d drilled through it. The housing is assumed to be
stationary here, but the results can easily be generalied to
include a movable housing. The piston rod extends out of the
housing, which is sealed and filled with a viscous
incompressible fluid, such as an oil. ( force f acting on the
piston rod creates a pressure difference !p$-p"& across the piston
such that if the acceleration is small,
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!A--A& == &! "$ !".3-$&
The net cross-sectional piston area is
"" &"
!&"
! dD
A =
2or a very viscous fluid at relatively small velocities, laminar
flow can be assumed, and the *agen-Boiseuille law can be
applied to give the volume flow rate
-C-%
d* == $
3
$")
!".3-"&
The volume flow rate *is also given by
dt
d/
A*= !".3-#&because the fluid is incompressible. Substituting !".3-$& and
!".3-"& into !".3-#& shows that
dt
d/c& = !".3-3&
where the damping coefficient is !from Tables&"
"
$
"
$)
==d
D%
C
Ac !".3-7&
2rom the principle of action and reaction,&can also be
considered to be the force exerted by the damper when moving
at a velocitydt
d/. If the housing is also in motion,
dt
d/is taen to
be the relative velocity between housing and piston.
If the relative velocity is large enough to produce turbulent flow
in the passage, the nonlinear resistance formulas should be used.
8ote that the fluid inertia, or inetic-energy storage
compartment, has been assumed to be negligible and thatpotential energy resulting from pressure difference acts against
the fluid resistance to create the system>s dynamics. ( similar
analysis can be performed for air dampers in which the woring
fluid is air. The common storm door damper is such a device.
#ccumulators
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The compliance of fluid elements can have detrimental or
beneficial effects. In hydraulic control systems operating with
high force levels or at high speeds, the accuracy of the controller
can be affected by the tendency of the nominally incompressible
fluid medium to compress slightly. This compliance effect is
greatly increased by the presence of bubbles in the hydraulic
fluid, which maes the control action less positive and causes
oscillations.
5n other hand, compliance can be useful in damping
fluctuations or surges in pressure or flow rate, such as those
resulting from reciprocating pumps and the addition or removal
of other loads on common pressure source. In such cases, a
compliance element called an accumulator is added to thesystem. This general term includes such different elements as a
surge tan for liquid level system, a fluid column acting against
an elastic element for hydraulic systems, and a bellows for gas
systems. They all have the property of compliance, which
allows them to store fluid during pressure peas and release
fluid during low pressure. The result is more efficient and safe
operation by damping out pressure pulses the way a flywheel
smoothes the acceleration in mechanical drive or an electricalcapacitor smoothes or CfiltersC voltage ripples.
Two typical arrangements for hydraulic systems are shown in
2igure ".#7. 2igure ".#7a, a plate and spring oppose the pressure
of the liquid, and in 2igure ".#7b, the compliance is provided by
a gas-filled bag that compresses as the liquid pressure increases.
The simplest arrangement is a closed tube with air at the top.
The air>s compressibility acts against the liquid. This device is
the common cure for the water-hammer effect in householdplumbing.
(ll of the preceding devices use an equivalent elasticity to
oppose the pressure, and thus they can be modeled similarly. If
we tae the displacementof the interface to be proportional to
the force&due to the liquid pressure, then k/& = , where kis the
elastic constant of the device.
6etAbe the area of the interface,the liquid pressure, and *1,
*2the volume flow rates into and out of the section.
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2igure ".#7 *ydraulic accumulators. !a& Dechanical spring
principle. !b& :ompressible gas principle.
8eglect the pressure drop between the flow =unction and the
plate. 2rom mass !volume& conservation,
"$&! **dtd/AA/
dtd ==
8eglecting the inertia of the interface gives the force balancek/-A =
The model for dynamic behavior of the pressure as a function of
the flow rates is obtained by combining the last two expressions.
"$
"
**dt
d-
k
A= !".3-'&
When the resistances upstream and downstream are defined, theflow rates *1and *2can be expressed as functions of the
upstream and downstream pressures.
(ssume that the laminar resistances in the left-and right-hand
pipe are'1and'2.Then,
&!$
$
$
$ --'
* = !".3-)&
&!
$$
"" --'* = !".3-0&
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and !".3-'& becomes
&!$
&!$
"
"
$
$
"
--'
--'dt
d-
k
A= !".3-$1&
This equation can be solved foras a function of time if we are
given1and2as time functions.
%he hydraulic servomotor
( hydraulic servomotor is shown in 2igure ".#%.
2igure ".#% *ydraulic servomotor.
The pilot valve controls the flow rate of the woring medium to
the receiving unit !a piston&. The pilot valve shown is nown as
a spool-type valve because of its shape. 2luid under pressure isavailable at the supply port. 2or the pilot valve position shown
!the Cline-on-lineC position&, both cylinder ports are bloced and
no motion occurs. When the pilot valve is moved to the right,
the fluid inters the right-hand piston chamber and pushes the
piston to the left. The fluid displaced by this motion exits
through the left-hand drain port. The action is reversed for a
valve displacement to the left. ;oth drain ports are connected to
a tan !a sump& from which the pump draws fluid to deliver tothe supply port. The filters and accumulators necessary to clean
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the recirculated fluid and dampen pressure fluctuations are not
shown.
6etdenote the displacement of the pilot valve from its line-on-
line position andthe displacement of the load and piston from
their last position before the start of the motion. 8ote that a
positive value of!to the left& results from a positive value of
!to the right&. The flow through the cylinder port uncovered by
the pilot valve can be treated as flow through an orifice, and the
orifice flow relation can be applied !from tables&. 6etbe the
pressure drop across the orifice. Thus,E"-AC* od= !".3-$$&
where *is the volume flow rate through the cylinder port,Aois
the uncovered area of the port, Cdis the discharge coefficient!which usually lies between 1.%-1.) for this application&, and
is the mass density of the fluid. The areaAois equal to , where
is the port width !into the page&. If Cd, ,, and are taen
to be constant, !".3-$$& can be written as
*=C !".3-$"&
:onservation of mass requires the flow rate into the cylinder to
equal the flow rate out. Therefore,
dt
d/A* =
:ombining the last two equations gives the model for the
servomotor.
1A
C
dt
d/= !".3-$#&
2or high accelerations, this model must be modified, because it
neglects the inertia of the load and piston. The piston force was
assumed sufficient to move the load and is generated by thepressure difference across the piston.
(lthough !".3-$#& is accurate enough for many applications, we
will now develop a more detailed model to account for the
effects of load. 2igure ".#' gives the necessary details. Dachine
tools for cutting metal are one application for such a system.
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2igure ".#' *ydraulic servomotor with load.
The applied force&can be supplied by a hydraulic servomotor.
The represents the mass of a cutting tool and the power
piston, while krepresents the combined effects of the elasticity
naturally present in the structure and that introduced by the
designer to achieve proper performance. ( similar statement
applies to the tool through its prescribed motion.
The spool valve shown in 2igure ".#' has two lands. If the
width of the land is greater than the port width, the valve is said
to be overlapped !2igure ".#)a&. In this case, a dead one exists
in which a slight change in the displacement produces no
power piston motion. Such dead ones create control difficulties
and are avoided by designing the valve to be underlapped !the
land width is less than the port width-2igure ".#)b&. 2or such
valves there will be a small flow opening even when the valve is
in the neutral position !=&. This gives it a higher sensitivity
than the overlapped valve.
6et4andodenote the supply and outlet pressures,
respectively. Then the pressure drop from inlet to outlet is4-o,
and this must equal the sum of the drops across both valve
openings and the power piston. That is,---- vo4 += "
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where the drop across the valve openings iso4v ----- == $"
and the drop across the piston is$" --- =
Therefore,&!
"
$----
o4v =
2igure ".#) Spool valve construction. !a& 5verlapped.
!b& @nderlapped.
If we tae4andoto be constant, we see that v- varies only if- changes. The derivation of !".3-$#& treated v- as a constant.
Thus the variablesand-
determine the volume flow rate, as&,! -1&* =
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This relation is nonlinear. 2or the reference equilibrium
condition !=, - , *=&, a lineariation gives-C1C* = "$ !".3-$3&
where the variations from equilibrium are simply, - , and *.
The lineariation constants are available from theoretical and
experimental results. The constant C1is identical to Cin
!".3-$#&, and both C1and C2are positive. The effect of the
underlapping shows up in the nonero flow predicted from
!".3-$3& when=. The pressure drop - is caused by the
reaction forces of the load mass, elasticity, and damping and by
the supply pressure difference !4-o&.
#0e 2.2!a& 4erive a model of the system shown in 2igure ".#'.
Tae into account the inertia effects of the load and
assume the valve underlapped.
!b& Show that the model reduces to !".3-$#& when
c=k=and either of the following approximations is valid9
$. /A 1 !large piston force compared to the load inertia&
". C2 1 !q independent of - &
!a& (ssume an incompressible fluid. Then conservation of mass
and !".3-$3& give
-C1C*dt
d/A == "$
The force generated by the piston is -A , and from 8ewton>s
law,
-Ak/dt
d/C
dt
/d +=
"
"
whereis measured from the equilibrium point of the mass .
Substitute - from the second equation to obtain
&!"
"
"$ /A
k
dt
d/
A
c
dt
/d
A
C1C
dt
d/A ++=
or
1C/
A
kC
dt
d/A
A
cC
dt
/d
A
C$
""
"
"
" &! =+++ !".3-$7&
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This is the desired model withas the input andas the output.
!c& 2or c=k=, !".3-$7& becomes
!d&
1Cdt
d/
Adt
/d
A
C
$"
"
"=+
!".3-$%&
When either /A 1 or C2 1, we obtain
1Cdt
d/A $= !".3-$'&
This is equivalent to !".3-$#&, because C1=C.
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2.& 'neumatic elements
while hydraulic devices use an incompressible liquid, woring
medium in a pneumatic device is a compressible liquid, such as
air. Industrial control systems frequently have pneumatic
components to provide forces greater than those available from
electrical devices. (lso, they afford more safety from fire
haards. The availability of air is an advantage for these devices,
because it may be exhausted to the atmosphere at the end of the
device>s wor cycle, thus eliminating the need for return lines.
5n the other hand, the response of pneumatic systems is slower
than that of hydraulic systems because of the compressibility ofthe woring fluid.
The analysis of gas flow is more complicated than for liquid
flow. 2or compressible fluids, the mass and volume flow rates
are not readily interchangeable. Since mass is conserved,
pneumatic systems analysis uses mass flow rate, here denoted
*to distinguish from volume flow rate *. ;ecause of the
relatively low viscosity and density of gases, their flow is more
liely to be turbulent. 2inally, the possibility of supersonic!greater than the speed of sound& flow is more liely in gas
systems.
Bressure is normally the only effort variable used for such
systems. The inetic energy of a gas is usually negligible, so the
inductance relation is not employed.
%hermodynamic properties o$ ases
Temperature, pressure, volume, and mass are functionallyrelated for a gas. The model most often used to describe this
relationship is the perfect gas law, which describes all gases if
the pressure is low enough and the temperature high enough.
The law states thatT'-" ,= !".7-$&
whereis the absolute pressure of the gas with volume (, is
the mass, Tits absolute temperature, and',the gas constant that
depends on the particular type of gas.
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The prefect gas law allows us to solve for one of the variables,
", , or Tif the other three are given. We frequently do not
now the values of three of the variables and additional
information. 2or a perfect gas, this information is usually
available in the form of a pressure-volume or CprocessC relation.
The following process models are commonly used9 constant-
pressure, constant-volume, constant-temperature !isothermal&,
and reversible adiabatic !isentropic&. In the latter process, no
heat is transferred between the gas and its surroundings. These
four processes only approximate reality, but they allow some
modeling simplifications to be made. ( real process can be
more accurately described by deriving or measuring anappropriate value for the exponent n in the polytropic process
equation
.&! con4t
"- n = !".7-"&
If the mass is constant, the polytropic process describes the four
previous processes if n is chosen as 1, , $, and ,
respectively, where is the ratio c/cvof the specific heats of
the gas.
'neumatic resistance
*ere we develop methods for determining pneumatic resistances
when the flow is compressible but subsonic.
The theory of gas flow through an orifice provides the basis for
modeling other pneumatic components. We assume that perfect
gas law applies and the effects of viscosity !friction& and heattransfer are negligible, hence, an isentropic process is assumed.
:onsider an orifice with a cross-sectional areaA. The upstream
absolute pressure is1, and the absolute bac pressurebis that
pressure eventually achieved downstream from the orifice. (sbis lowered below1, the flow rate through the orifice increases.
If the difference between1andbbecomes large enough, the
gas flows through the orifice at the speed of sound. The value of
bat which this occurs is the critical pressurec. 2or air,3.$=
and the critical pressure is
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$7").1 --c = !".7-#&
The results of chief interest to us are the expressions for the
mass flow rate as functions of the pressures1andb. 8ewton>s
laws, conservation of mass, and the isentropic process formula
can be combined to show that the mass flow rate *in the
subsonic case !b>c& is
&!"
$
$
bb
,
d ---T'
AC* = !".7-3&
where Cdis an experimentally determined discharge coefficient
that accounts for viscosity effects and T1is the absolutetemperature upstream from the orifice. 2or the sonic case, *is
independent ofbas long asb5c.
( typical condition in pneumatic systems is such that a small
pressure drop exists across the component. *ence, the flow is
frequently subsonic. (lso, the pressure changes often consist of
small fluctuations about an average or steady-state constant
pressure value. @nder these circumstances, the compressibleflow resistance of pneumatic components can be modeled in the
form of turbulent resistance relation, written here as
-*' - ="
!".7-7&
where - is the pressure drop across the component and'is
the pneumatic resistance.
To show that !".7-3& reduces to !".7-7&, we assume that the inputpressure1fluctuates about a constant pressure4by a small
amounti. Thus,1=46i. Similarly, the bac pressure can be
written asb=46o, and we assume thatois also small relative
to4. 8ote that1-b=i-oand consider the following term from
!".7-3&
oi4oi
4
o
4oio4oio4bb -----
-
------------- +=+=+= &$!&&!!&! $
because $ ? poEps $ from our assumptions. Thus, !".7-3&becomes
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oi
,
4
d --T'
-AC* =
$
"!".7-%&
This has the same form as !".7-7&, where boi ----- == $ , and
the pneumatic resistance is
""
$
" AC-
T''
d4
,
- = !".7-'&
Solve !".7-7& for *assuming that 1>-
- '
-
*
= !".7-)&
where we tae the positive square root, because *>if 1>- .
2or !".7-)&, the slope of *is infinite at 1=- . This physically
unrealistic result is due to certain assumptions made to derive
!".7-3& that are not true as $--b . 2orbvery close to1, the
orifice flow relation resembles that of laminar flow, and will
model it as
-'
*%
= $ !".7-0&
where'%is the equivalent resistance. If we match the solution of
!".7-)& and !".7-0& at some intermediate value of - , say, -
B, then'%is related to'by-% B'' = !".7-$1&
where the value ofBmust be determined by experiment. 2or
this value of'%, the slopes of !".7-)& and !".7-0& also match at- B.
Thus, we have three models for subsonic orifice flow9 !".7-3&
for - largeA !".7-)& for - small but greater thanBA and !".7-
0& for - very small.
'neumatic capacitance
In pneumatic systems, mass is the quantity variable and pressurethe effort variable. Thus, pneumatic capacitance is the relation
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between stored mass and pressure. Specifically, we tae
pneumatic capacitance to be the system>s capacitance C!or
compliance&, defined as the ratio of the change in stored mass to
the change in pressure, or
d-
dC= !".7-$$&
2or a container of constant volume "with a gas density , this
expression may be written as
d-
d"
d-
"dC
==
&!!".7-$"&
If the gas undergoes a polytropic process,
con4t-
"-
n
n==
&! !".7-$#&
and
n-"
n-d-
d==
2or a perfect gas, this shows the capacitance of the container to
be
Tn'
"
n-"
"C
,
== !".7-$3&
8ote that the same container can have a different capacitance for
different expansion processes, temperatures, and gases, since C
depends on n, T, and',.
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)odelin pneumatic systems
;ecause fluid inertia is usually neglected in pneumatic analysis,
the simplest model of such systems is a resistance-capacitance
model for each mass storage element.
#0e 2.3
(ir passes through a valve !modeled as an orifice& into a rigid
container, as shown in 2igure ".#0. 4evelop a dynamic model of
the pressure changein the container as a function of the inlet
pressure changei. (ssume an isothermal process and thatand
iare small variations from the steady-state pressure4.
Solution9
2rom conservation of mass, the rate of mass increase in the
container equals the mass flow rate through the valve.
2igure ".#0 as flow into a rigid container.
Thus, ifi- >
dt
d-C
dt
d-
d-
d
dt
d
*dt
d
==
=
-
i
'
--*
= !".7-$7&
where Cand'are given by !".7-$3& and !".7-'&, with T = T1and n=1. The desired model is
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-
i
'
--
dt
d-C
= !".7-$%&
Ifi-5, the flow *is reversed, and the model is
-
i
'
--
dt
d-C
= !".7-$'&
Ifi-is very close to ero, the linear resistance relation !".7-0&
can be used, and !".7-$7& is replaced by
&!$
--i
'
*%
=
!".7-$)&In this case, the system model is
&!$
--'dt
d-C i
%
= !".7-$0&
and the flow reversal is accounted for automatically ifi-
changes sign.
#0e 2.( pneumatic bellows, shown in 2igure ".31, is an expandable
chamber usually made from corrugated copper because of this
metal>s good heat conduction and elastic properties. We model
the elasticity of the bellows as a spring. The spring constant k
for the bellows can be determined from vendor>s data or
standard formulas. This device, when employed with a variable
resistance'is useful in pneumatic controllers as a feedbac
element to sense and control the pressurei. The displacement
is transmitted by a linage or beam balance to a pneumatic valve
regulating the air supply. 4evelop a model for the dynamic
behavior of withia given input.
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2igure ".31 Bneumatic bellows.
We assume that the bellows expands or contracts slowly. Thus,
the product of its mass and acceleration is negligible, and a force
balance givesk/-A = !".7-"1&
where the force exerted by the internal pressure isA,Ais the
area of the bellows, andis the displacement of the right-hand
side !the left-hand side is fixed&. The volume is "=A, and thus
"k/A/A
k/-" ==
The assumption of a slow process suggests an isothermal
process. Thus, the time derivative of the perfect gas law gives
dt
dT'-"
dt
d,=&!
2rom the law of resistance !".7-0&, assuming thati-is very
small,
&!$
--'
*dt
di
%
==
Since
dt
d/k/
dt
k/d"
&! "
=
these expressions give the following model9
&!"A
k/-
'
T'
dt
d/k/ i
%
,= !".7-"$&
This is nonlinear because of the cross-product betweenand its
derivative.
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2.* %hermal systems
In thermal systems, energy is stored and transferred as heat. 2or
this reason, many energy conversion devices are examples of
thermal systems, as are chemical processes where heat must be
added or removed in order to maintain an optimal reaction
temperature. 5ther examples occur in food processing and
environmental control in buildings.
The effort variable is temperature, or more precisely,
temperature difference T . This can cause a heat transfer rate+* by one or more of the following resistance modes9
conduction, convection, or radiation. The transfer of heat causes
a change in the system>s temperature. Thermal capacitancerelates the system temperature to the amount of heat energy
stored. It is the product of the system mass and its specific heat.
8o physical elements are nown to follow a thermal inductance
causal path.
In order to have a single lumped-parameter model, we must be
able to assign a single temperature that is representative of the
system. Thermal systems analysis is sometimes complicated by
the fact that it is difficult to assign such a representativetemperature to a body or fluid because of its complex shape or
motion and resulting complex distribution of temperature
throughout the system. *ere, we develop some guidelines for
when such an assignment can be made. When it cannot, several
coupled lumped-parameter models or even a distributed-
parameter model will be required.
Heat trans$er*eat can be transferred in three ways9 by conduction !diffusion
through a substance&, convection !fluid transport&, and radiation
!mostly infrared waves&. The effort variable causing a heat flow
is a temperature difference. The constitutive relation taes a
different form for each of the three heat transfer modes. The
linear model for heat flow rate as a function of temperature
difference is given by 8ewton>s law of cooling, which is valid
for both convection and conduction.
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T'*+ = !".%-$&
The thermal resistance for convection is
+A'
$= !".%-"&
where +is the film coefficient of the fluid-solid interface. 2or
conduction, we will see that
kA
d'= !".%-#&
where kis the thermal conductivity of the material,Ais the
surface area, and dthe material thicness.
:onvective heat transfer is divided into two categories9 forced
convection-due, for example, to fluid pumped past a surface-and
free or natural convection, due to motion produced by density
difference in the fluid. The heat transfer coefficient for
convection is a complicated function of the fluid flowcharacteristics especially.
Significant heat transfer can occur by radiation, the most notable
example is solar energy. Thermal radiation produces heat when
it stries a surface capable of absorbing it. It can also be
reflected or refracted, and all three mechanisms can occur at a
single surface. When two bodies are in visual contact, a mutual
exchange of energy occurs by emission and absorption. The net
transfer of heat occurs from the warmer to the colder body. Thisrate depends on material properties, geometric factors affecting
the portion of radiation emitted by one body and striing the
other, and the amount of surface area involved. The net heat
transfer rate can be shown to depend on the body temperatures
raised to the fourth power !a consequence of the Stefan-
;oltmann law&.
&! 3"3
$ TT*+ = !".%-3&
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The absolute body temperatures are T1and T2, and is a factor
incorporating the other effects. This factor is usually very small.
Thus, the effect of radiation heat transfer is usually negligible
compared to conduction and convection effects, unless the
temperature of one body is much greater than that of the other.
Some simpli$yin appro+imations
The thermal resistance for conduction through a plane wall
given by !".%-#& is an approximation that is valid only under
certain conditions. 6et us tae a closer loo at the problem. If
we consider the wall to extend to infinity in both directions, then
the heat flow is one dimensional. If the wall material is
homogeneous, the temperature gradient through the wall isconstant under steady-state conditions !2igure ".3$a&.
2igure ".3$ 5ne-dimensional heat transfer through a wall. !a&Steady-state temperature gradient. !b& 6umped approximation.
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2ourier>s law of heat conduction states that the heat transfer rate
per unit area within a homogeneous substance is directly
proportional to the negative temperature gradient. The
proportionality constant is the thermal conductivity . 2or the
case shown in 2igure ".3$a, the gradient is (T2-T1)/d, and the
heat transfer rate is thus
d
TTkA*+
&! $" = !".%-0&
where ( is the area in question. :omparing this with !".%-$&
shows that the thermal resistance is given by !".%-#&, with
"$ TTT = .
In transient conditions where temperatures change with time, the
temperature distribution is no longer a straight line. If the
internal temperature gradients in the body are small, as would be
the case for a large value of k, it is possible to treat the body as
being at one uniform, average temperature and thus obtain a
lumped-instead of distributed-parameter model. 2or solid bodies
immersed in a fluid, a useful criterion for determining thevalidity of the uniform-temperature assumption is based on the
;iot number, defined as
k
+%NB= !".%-%&
where%is the ratio of the volume to surface area of the body
and h is the film coefficient. If the shape of the body resembles a
plate, or sphere, it may be considered to have a uniform
temperature ifNBis small. IfNB5.1, the temperature is usually
taen to be uniform, and the accuracy of this approximation
improves if the inputs vary slowly.
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#0e 2.
:onsider a copper sphere $in in diameter with k"$" ;tuEhr-ft-o2 at 7'1o2 and immersed in a fluid such that +7 ;tuEhr-ft"-o2.
Show that its temperature can be considered uniform, and
develop a model of the sphere>s temperature as a function of the
temperature Toof the surrounding fluid.
Solution9
The surface area and volume of the sphere are""
$1$)."&"3
$!3 == /A
3# $11#.#&"3
$!&
#
3! == /"
Thus,% = "/A 1.1$#0, andNB #.")x$1-3, which is much less
than 1.$.
(ccording to the ;iot criterion, we may treat the sphere as a
lumped-parameter system with a single uniform temperature T.
The amount of heat energy in the sphere is CT, where Cis its
thermal capacitance. If the outside temperature Tois greater than
T, heat is transferred into the sphere at the rate *+given by
!".%-$&, where TTT o= . Thus, from the conservation ofenergy,
+*CT
dt
d=&!
@sing !".%-$& and noting that Cis constant, we obtain
&!$
TT'dt
dTC o= !".%-'&
2or the copper sphere, !".%-"& gives
C =c "c $'.#!#.1#x$1-3&!".0#& 1.1$73
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# wall model
:onductive and convective resistances to heat transfer
frequently occur when one or more layers of solid materials are
surrounded on both sides by fluid, such as shown in 2igure
".3$b. We will use the figure to show how thermal resistance is
computed when both conduction and convection are present.
(ssume that the wall>s material is homogeneous, and that it is to
be treated with a lumped approximation. The wall temperature is
considered to be constant throughout and is denoted by T. The
temperatures T1and T2are the temperatures in the surrounding
fluid =ust outside the wall surface. There is a temperature
gradient on each side of the wall across a thin film of fluid
adhering to the wall. This film is nown as the thermal boundarylayer. The film coefficient h is a measure of the conductivity of
this layer. The temperatures outside of these layers are denoted
by Tiand To.
The thermal capacitance Cis the product of the wall mass times
its specific heat. The capacitance of the boundary layer is
generally negligible because of its small fluid mass. Thus, since
no heat is stored in the layer, the heat flow rate through the layer
must equal the heat flow rate from the surface to the surface tothe mass at temperature Tconsidered to be located at the center
of the wall. The length of this latter path is d/2.
2or the left-hand side, this gives
&!"E
&! $$$$ i+ TTd
kATTA+* == !".%-)&
This allows us to solve for T1as a function of Tiand T.
$
$$
"
"
d+k
Td+kTT i
+
+= !".%-0&
Similar equations can be developed for the right-hand flow *+2
and temperature T2.
(n energy balance for the wall mass gives
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"$ ++
**dt
dTC = !".%-$1&
(fter T1and T2have been eliminated, this gives an equation for
Twith Tiand Toas inputs. 4efine
",$,"
"=
+= 7
d+k
Ak+#
7
7
7 !".%-$$&
and !".%-$1& becomes
&!&! "$ oi TT#TT#
dt
dTC = !".%-$"&
In !".0-$"&, #1is the reciprocal of the resistance for the path
from Tito T. We can write this resistance as
kA
d
A+Ak+
d+k
#'
"
$
"
"$
$$
$
$
$ +=+
== !".0-$#&
:omparing this with !".0-"& and !".0-#& shows that'1is the sum
of the thermal boundary layer resistance and the wall>s
conductive resistance along the path of length d/2.
In many applications, the wall>s thermal capacitance is small
compared to the capacitance of the mass of fluid on either side
of the wall. In this case, we model the wall as a pure resistance,
and the total wall resistance is the sum of the path resistances.
/quation !".%-$"& can be used to show this. If either the rate of
change of Tor the wall capacitance Cis very small, the right-
hand side of !".%-$"& can be taen to be ero and thus *+1equals
*+2. This relation can be manipulated to show that the walltemperature is
"$
"$
##
T#T#T oi
+
+= !".%-$3&
and that the heat flow rate is
&!
"$
"$
oi+ TT
##
##*
+= !".%-$7&
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Thus, the total resistance between Tiand To is !#16#2&E#1#2.
With !".%-$$& and some algebra, the total resistance can be
shown to be
A+k
d
k
d
+'
$&
$
""
$!
"$
+++= !".%-$%&
When it is not possible to identify one representative
temperature for a system, several temperatures can be chosen,
one for each distinct mass. The resistance paths between the
masses are then identified, and conservation of heat energy is
applied to each mass. This results in a model whose order equals
the number of representative temperatures.#0e 2.8
( simplified representation of temperature dynamics of a room
is shown in 2igure ".3".
2igure ".3" ( model of room temperature.
The room is perfectly insulated on all sides except one, which
has a wall with a thermal capacitance C2. The inner wall
resistance is'1, as given by !".%-$#&. The outer resistance is'2
and is found by similar expression. The representative
temperature of the room>s air is designated Ti, that of the wall is
T, and the outside air temperature is To. The room aircapacitance is C1. 4evelop a model of the behavior of Ti.
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Solution
(ssume that To>T>Ti. Then the heat flow is from Toto Tto Ti.
:onservation of energy applied to the capacitances C1and C2
gives
&!$
$
$ ii TT
'dt
dTC = !".%-$'&
&!$
&!$
$"
" io
TT'
TT'dt
dTC = !".%-$)&
This is the desired model. If the heat is not in the assumed
direction, the mathematics taes care of the reversalautomatically.
If the wall capacitance is negligible, then C2=, and we can
substitute Tfrom !".%-$)& into !".%-$'&. 5r, equivalently, we
can use the principle of !".%-$%& to find the total wall resistance
'. This is
' = '16 '2 !".%-$0&
The heat flow rate through the wall is (To-Ti)/', and the model
becomes
&!$
$ ioi TT
'dt
dTC = !".%-"1&
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#0e 2.9
;uild up a model for the thermal system shown in 2igure ".3#.
2igure ".3# ( thermometer-liquid thermal system.
:onsider a thermometer at temperature Twhich has =ust been
inserted into a liquid at temperature T%. If the thermal resistance
to heat flow from the liquid to the thermometer is', then,
'
TT* %
=
where *is the net rate of heat flow from liquid to thermometer.
The thermal capacitance Cof the thermometer is given by the
equation
dt
dTC** = "$
Since there is only a net flow of heat from the liquid to thethermometer, *1=*and *2=. Thus
dt
dTC*=
Substituting this value of q in the earlier equation gives
'
TT
dt
dTC %
=
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%TT
dt
dT'C =+
This equation, a first-order differential equation, describes howthe temperature indicated by the thermometer Twill vary with
time when the thermometer is inserted into a hot liquid.
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#0e 2.:
2igure ".33 shows a thermal system consisting of an electric fire
in a room. This fire emits heat at the rate *1and the room loses
heat at the rate *2. (ssuming that the air in the room is at a
uniform temperature Tand that there is no heat storage in the
walls of the room, derive an equation describing how the room
temperature will change with time.
2igure ".33 /lectric fire-room thermal system.
If the air in the room has a thermal capacity C, then
dt
dTC** = "$
If the temperature inside the room is Tand that outside the room
To, then
'TT* o="
where'is the resistivity of the walls. Substituting for *2gives
dt
dTC
'
TT* o =
$
*ence
oT'*T
dt
dT'C +=+ $
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Broblems
".$ 4erive the mathematical model for the following system.
The input is the force$, and the output is the displacement.
"." 4erive an equation relating the input angular displacement
iwith the output angular displacement o
".# 4erive the relationship between the output, the potential
difference across the resistor'of v', and the input vfor the
series%C' circuit shown below.
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".3 4erive the relationship between the height +2and time for
the hydraulic system shown below. 8eglect inertance.
".7 The following figure shows a thermal system involving
two compartments, with one containing a heater. If the
temperature of the compartment containing the heater is T1, the
temperature of the other compartment T2and the temperature
surrounding the compartments T3, develop equations describing
how the temperature T1and T2will vary with time. (ll the walls
of the containers have the same resistance and negligible
capacity. The two containers have the same capacity C.
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".% 4erive the differential equation for a motor driving a load
through a gear system, as shown in the following figure, which
relates the angular displacement of the load with time.
".' 4erive the mathematical model for a 4: generator. The
generator may be assumed to have a constant magnetic field.The armature circuit has the armature coil, having both
resistance and inductance, in series with the load. (ssume that
the load has both resistance and inductance.
".) 4erive the mathematical model for a permanent magnet
4: motor.
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".0 4evelop the mathematical model for the field-controlled
dc motor shown in the following figure with e&as input and
as output.
".$1 The following system might represent a machine tool slide
driven through a rac and pinion by a motor, or an
electrohydraulic actuator in which the mass 1represents the
spool valve.
The gear shaft is supported by frictionless bearings and thus can
only rotate. (ssume that the only significant masses in thesystem are the motor with inertiaIand the slide with mass 1.
5btain the system>s differential equations with the motor torque
Tas the input.
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".$$ ( load inertiaI2is driven through gears by a motor with
inertiaI1. The shaft inertias are I#and I3A the gear inertias areI
andI8. The gear ratio is 79$ !the motor shaft has the greater
speed&. The motor torque is T1, and the viscous damping
coefficient is c=1.2 0b-&t-4ec/r#d. 8eglect elasticity in the
system, and use the following inertia values !in4ec2-&t-0b/r#d&9
I1=.1 I 2=.
I3=.1 I =.
I=.2 I 8=.3
!a& 4erive the model for the motor shaft speed 1with T1as
input.
!b& 4erive the model for the load shaft speed 2with T1as
input.