Chapter 2-2.doc

8/13/2019 Chapter 2-2.doc

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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 += "

108


27/54

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&* =

109


28/54

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&

110


29/54

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.

111


30/54

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.

112


31/54

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

113


32/54

$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

114


33/54

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

115


34/54

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',.

116


35/54

)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

117


36/54

-

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.

118


37/54

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.

119


38/54

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.

120


39/54

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&

121


40/54

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.

122


41/54

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.

123


42/54

#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


43/54

# 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

125


44/54

"$ ++

**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&

126


45/54

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.

127


46/54

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&

128


47/54

#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 %

=


48/54

%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.

130


49/54

#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 +=+ $

131


50/54

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.

132


51/54

".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.

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