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MA1506

Mathematics IIChapter 2

More Applications of ODEs

1Chew T S MA1506-14 Chapter 2

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In Chapter two,

we are interested in propertiesof solutions

We have learnt how to solve those ODEsin Chapter one, so we are not interested in

how to solve those ODE here

You dont have to memorizethose complicated solutions

But you need to know how to

derive the properties from the solutions


My presentation is different from the L. N.. However

the content remains unchanged.

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In this chapter , first we study animportance system called Harmonic

Oscillatorwhich is an application of 2nd

order linear ODE

The ODE for harmonic oscillator is given

by

'' ' ( )mx bx kx F t + + =


Introduction

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Notation

In this chapter

may be denoted by or

may be denoted by or

dx

dt'x

2

2

d x

dt x

''x

4

x

Chew T S MA1506-14 Chapter 2

Introduction

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We shall consider four types of HO(1)Simpleharmonic oscillator (pp 1-11)

where m (mass) and k (spring constant)

are positivenumbers

Its motion is periodic, called simple

harmonic motion (SHM)


0mx kx+ =

Introduction

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(2)Dampedharmonic oscillator (pp 11-16)

In real oscillator, friction (damping) slows

down the motion of the system. The

frictional force is given by

m >0 b >0 k > 0

springconstant

Dampingconstant

6

bx Chew T S MA1506-14 Chapter 2

0mx bx kx+ + =

Introduction

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(3) Forced harmonic oscillator without

damping (pp17-23)

The system is a simple harmonic oscillator

driven by EXTERNALLY applied force F(t)

( )mx kx F t + =


Introduction

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m, k >0, k called spring constant

9

The equation of simple harmonic motion

is


0mx kx+ =

2.1 The simple harmonic oscillator (motion)

I have changed the subtitle The harmonic oscillator

to The simple harmonic oscillator (motion),

since it is more precise.

I shall introduce the general theory before talking aboutPendulum.

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Simple harmonic motion SHM can serve

as a mathematical model of a variety of

motions, such as a pendulum with small

amplitudes and a mass on a spring

See

http://en.wikipedia.org/wiki/Simple_harmonic_motion


2.1 Simple harmonic oscillator(cont)
http://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motionhttp://en.wikipedia.org/wiki/Simple_harmonic_motion

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For SHM equationIt is typical to define the quantity

and write this equation

as


0mx kx+ =

/k m=0mx kx+ =

2 0x x+ =

(cont) 2.1 Simple harmonic oscillator

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The general soln of the equation is

which can be written as

the phase-amplitude form(see Appendix 1)

( ) cos( )x t A t =

amplitude phase orphase angle

12

( ) cos( ) sin( )x t C t D t = +


See Chapter 1


cos( )

cos( ) cos( ) sin( ) sin( )

A t

A t A t

= +

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Note that

Hence SHM is periodic with period

and amplitude A

cos( )x A t = cos( 2 )A t = + 2

cos( ( ) )A t

= +

2 2 mk

=



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Some Technical Terms

Period T=

The time for a single oscillation (cycle)

Chew T S MA1506-14 Chapter 2 15

2 2 mk

=

Frequency f = the reciprocal of the

period T =

The number of cycles per unit time

1

2

k

m

2.1 Simple harmonic oscillator

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(cont) Some technical terms

Angular frequency=

The number of cycles per unit time

Amplitudeis the maximal

displacement from the equilibriumposition


2 kfm

=

2

=


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Why ?

17

cos( )A t

sin( )A t cos( )A t +

In fact it can also be written

as one of the following forms



sin( )A t +

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Two constants A andare determined by initial conditions

Suppose

Thenand

18

0(0)x x= 0'(0)x v=

0 cos( )x A = 0 ( )sin( )v A =

So we can find A and , for example,

22 00

vA x

= +



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An equilibrium soln (point) is said to be

stable if any soln with an initial point close

tothe equilibrium soln stays close to the

equilibrium soln (point)



A solution x(t) of ODE is said to be an

equilibrium solution (equilibrium point)if x(t) is a constant function(never move)

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2

0x x+ =Hence zerosolution of SMH

is an equilibrium solution (equilibrium point)

This zero solution is stable

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2x x=

We remark that in the above discussion of SHM

Minus is crucial

What happens if2

x x= (NOT SHM)


Zero function is a solution, but it is NOT stable

Why

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t tx Ce De

= +

(cont)

The general soln of

is

Instead of 0 = 0, 0 = 0

(which give zero solution), we assume

initial condition

(0) , (0) 0x x= =


2x x

=

where is small. Hence this new initial

condition is close to zero solution

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(cont)

1 ( )2

t tx e e = +


The solution of 2x x=

is

(0) , (0) 0x x= =

Exponential function grows very quickly,so this solution does not stay close to zero soln

Hence zero solution is NOT stable

canbe written as=cosh()

2

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2.1.2 Pendulum with small amplitude

(small angle) (pp1-11)

rigid


0> 00, k > 0, b >0


2.2 Damped, Unforced Oscillators pp 11-16

0mx bx kx+ + =

Damping constant spring constant

kx restoring forcedamping force , e.g., friction,

air resistancebx

Again I have changed the subtitle, forced

replaced by unforced

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2.2.1 Damped, Unforced Oscillators

(2 real roots)always negative real roots

Overdamping

No oscillation42Chew T S MA1506-14 Chapter 2

General

soln

3 2 0x x x+ + = 2

1 2( )

t t

x t c e c e

= +

Goes to zero rapidlyOverdamping

2 4

2

b b mk

m

=

2

, , all positive

so bigger than

4

b m k

b

b mk

2 4 0b b mk + 0, k > 0, b >0

http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

Damped

Mass Spring

Oscillator

Textbook

p196


0mx bx kx+ + =
http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

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2.2.4 Damped pendulum Example 2

with air

resistance


0g

m mL + = 0gm S mL

+ + =

S

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2.3 Forced Oscillators ( pp17-27)

General solution of

is (See appendix 3)


where

m

0 cosmx kx F t + =

0

2 2

/( ) cos( ) cos( )

F mx t A t t

= +

km

=

0cosF t

( ) ( ) ( )h px t x t x t= +

springexternal motor

Assume

Forced without damping Oscillators ( pp17-23)

2 3 Forced Oscillators

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2.3 Forced Oscillators

get


Assume initial condition

(See appendix 4)

02 2/( ) cos( ) cos( )F mx t A t t = +

(0) 0, (0) 0x x= =

[ ]0

2 2

/

( ) cos( ) cos( )

F m

x t t t =



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Small angular frequency

when close to


[ ]02 2/( ) cos( ) cos( )F mx t t t = 0

2 2

2 /

( ) sin sin2 2

F m

x t t t

+

=

( )A t

We shall use the above form to discuss

properties of forced oscillator



2 3 Forced OscillatorsF d ith t d i O ill t ( 17 23)

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where


Red curve

Blue curve

Red curve

( ) ( ) sin

2

x t A t t +

=

0

2 2

2 /( ) sin

2

F mA t t

=

2.3 Forced OscillatorsForced without damping Oscillators ( pp17-23)

Beating 2 3 F d O ill t

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Beating

if our ear is exposed to two sounds


we only hear the above term A(t)

1st sound 2nd sound


( )sin[ ]2

A t t + =

where

[ ]02 2

/( ) cos( ) cos( )

F mx t t t

=

0

2 2

2 /( ) sin

2

F mA t t

=

( t) B ti 2 3 F d O ill t

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(cont) Beating


A fast signal

is modulated by a slower one

This behavior is called beatinginphysics

sin[( ) ]2

t +

http://www.school-for-champions.com/science/sound_beat.htm


0

2 2

2 /( ) sin

2

F mA t t

=

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L Hospital

Rule

0

sin2 / 2

lim ( ) lim

tF m

A t

=

+

0

2F t

m=

or

sin2 2

t t




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When external frequency is close

to natural frequency , A(t) tends

to a st. line

0

2

F t

m



2 3 Forced OscillatorsForced without damping Oscillators ( pp17 23)

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lim 0 sin( )2F t

x tm

=

( ) ( ) sin2

x t A t t +

=

2.3 Forced OscillatorsForced without damping Oscillators ( pp17-23)

We can understand the above result more

by looking at the following case:when

0cosmx kx F t + =

=

The particular solution of

is of the from (sin+ cos)The solution of

0cosmx kx F t + =

(0) 0, (0) 0x x= =with is 0 sin( )2

F tx t

m

=


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Oscillations go out of control whenclose to . It is called resonance


blue curve in slide 57

tends to green st. line

when tends to

Red curveGreen curve



0

2 sin()

0

2


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Resonance

If the external force has a frequencyclose to the natural frequency of the system,

the resulting amplitudes can be very large even

for small external amplitudes.

It may cause violent motions and even disasters

in bridges and buildings



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Collapse of the Tacoma Narrow Bridge.

http://www.youtube.com/watch?v=3mclp

9QmCGs

http://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=3mclp9QmCGs

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Resonance

Avoiding resonance disasters is a major concern

in every building and bridge construction project.

As a countermeasure, a tuned mass damper can

be installed to avoid disaster.

The Taipei 101 building relies

on a 730-ton pendulum

a tuned mass damper to avoid resonance.



(cont)

Taipei 101 Tuned Mass Damper

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Taipei 101 Tuned Mass Damper

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A tuned mass damper is a device mounted in

structures to prevent damage caused by

vibration


http://en.wikipedia.org/wiki/File:Tuned_mass_damper.gif
http://en.wikipedia.org/wiki/File:Tuned_mass_damper.gifhttp://en.wikipedia.org/wiki/File:Tuned_mass_damper.gif

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Liquid tuned mass damper


2.3 Forced OscillatorsForced without damping Oscillators ( pp17 23)

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has two important phenomena

beating, resonance



Forced , NO damped, Oscillator

0cosmx kx F t + =

We have learnt



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Forced Damped Oscillators (pp 24-27)


Hence a particular solution is of the form

0 cos( )mx bx kx F t + + =

(see Appendix 5)

sin cospx B t C t = +

2

0 0

2 2 2 2

( )cos( ) sin( )

( ) ( )p

F k m t F b t

x t k m b

+

= +


sint or cost never appear in see Section 2.2

We can find A and B, so

2.3 Forced OscillatorsForced Damped Oscillators (pp 24-27)

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+ Gen Sol of


General soln is

The 2ndpart (damped oscillation)

tends to zero rapidly, see Section 2.2

Hence 2ndpart called transient soln

2

0 0

2 2 2 2

( )cos( ) sin( )( )

( )

F k m t F b t x t

k m b

+= +

0mx bx kx+ + =

o ced a ped Osc a o s (pp )


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http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html


So when t big enough, the general soln

becomes

called steady-state soln (response)

2

0 0

2 2 2 2

( )cos( ) sin( )( )

( )

F k m t F b t x t

k m b

+=

+

p (pp )

http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html

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Oscillation at

angular

frequency

http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html


X(t) can be written as

0

22 2 2 2

2

( / )cos( )

( )( )

F m t

x t b

m

=

+

/k m=where

when t big enough

p (pp )

http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.htmlhttp://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html

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Although the steady-state oscillation has

the same frequency as the external force

but it is NOT in phase with the external

Force. (compare with )

The amplitudes of the steady-state

soln and the external force are also

different


cos t cos( )t

p (pp )


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0

2

2 2 2 2

2

( / )cos( )( )

( )

F m tx tb

m

= +

( )cos( )A t =

0

2

2 2 2 2

2

( / )( )

( )

F mAb

m

=

+

where =amplitude

p (pp )

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2.4 Conservation pp28-29


We shall prove it in next slide

21( )2

dx xdx

=

We need the following formula in this section

dxx

dt=Recall

2.4 Conservation

( t)

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Chain rule


2

12

d dxdt dt

=

2

2

1 22

dx d x dt dt dt dx

=

2

2d x xdt

= =

2

21 1( )2 2

d d dxxdx dx dt

=

(cont)

dtdx

2.4 Conservation

In this section we shall look at

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21

2 mx

21

2

kx

Consider SHM

We shall show that E is constant

Kinetic energy

+ potential energy

mx kx= Let E=

In this section, we shall look at

conservative systems

(cont)2.4 Conservation

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simple harmonic motion

0mx kx+ =

2

2

1( )

2

1( )

2

dmx m x

dx

dm x

dx

=

=

First

21 1 (2 )2 2

d kx k x kxdx

= =

(cont)

from slide 80

Next

( t)2.4 Conservation

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2 21 1 02 2

d mx kxdx + =

0mx kx+ =From

get

(cont)

and previous slide

(cont)2.4 Conservation

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2 21 1

( )2 2m x kx E + =

Hence

Hence the kinetic energy + potential energy

of the system remains constant

This system is called conservative system

(cont)

kinetic energy potential energy

where E

is aconstant

2.5 EULERs equation (Cantilevered Beams) pp30-37

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2.5 EULER s equation (Cantilevered Beams) pp30 37


Beamlong, thin object

Cantilevered Beam (supported at only one end)

beam must bend

x

y

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3.7 Cantilevered Beams

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In this section, we will discuss bending of

a cantilevered beam

Why bending?

It is due to the weight of the beam and

other forces acting on the beam, called

load W(x)=force per unit length at point x

OUR Convention is that W(x) is positive in

UPWARD direction


y

xW(x)


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Load=weight of

springboard Load=weight of bamboo

stick+clothes

Load=weight of

springboard +swimmers


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Bending depends on W(x), stiffness, and

shape of the cross-section.

The stiffnessis measured by a constant E

called Youngs modulus.

The cross-sectionis measured by a

constant called the second moment of

area.Chew T S MA1506-14 Chapter 2 89

I


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2 2

2 2 ( )

d d yEI W x

dx dx

=

Recall the directionof load W(x)

The equation of the beam is given by 4th

order ODE, Eulers equation, (proof omitted)

y

xW(x)

4

4

( )d y W x

EIdx

=


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y

x

4

4

( )d y W x

EIdx =

( )W x

Find max deflection 3.7 Cantilevered Beams

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Find max deflection


L

Assume: uniform mass

( )W x =

4

4

( )d y W x

EIdx=

Then

( )y L =

y

x

Recall W(x)=force per unit length at point x

for all x


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To find i.e. to find ( )y L

we need to solve4

4( )d y W x

EIdx=

This is 4th order ODE,

we need FOUR conditions

(0) 0y =

00

x

dy

dx = =

3

3 0x L

d y

dx=

=

2

2 0x L

d y

dx=

=

Here are the conditions

EI=


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L


y

x(0) 0y =

00

x

dy

dx = = No moment at L

No shear force at L

2

20

x L

d y

dx ==

3

3 0

x L

d y

dx=

=

Proof omitted

Proof omitted


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integrate

both sides get

4

4

d y

EIdx

=3

3

d y xA

EIdx

= +

LA

EI

=3

3 0

x L

d y

dx=

=By get

Now solve this 4thorder ODE with four conditions


3d y x L

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3

d y x L

EI EIdx

= +

2 2

2 2

d y x LxB

EI EIdx

= + +

2 2 2

2 2

L L LB

EI EI EI

= =

2

2 0

x L

d y

dx=

=By

get

integrate

both sides get


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0

0x

dy

dx ==

Byget

2 2 2

2

2 2

d y x Lx L

EI EI EIdx

= +

3 2 2

6 2 2

dy x Lx L xC

dx EI EI EI

= + +

0C=

integrate

both sides get


3 2 2

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3 2 2

6 2 2

dy x Lx L x

dx EI EI EI

= +

By (0) 0y =get

4 3 2 2

24 6 4x Lx L xy DEI EI EI

= + +

0D=

Integrate both sides get


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Beam equation

y

x

4 3 24 1 1 1

2 12 3 2

L x x xy EI L L L

= +


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Cantilever deflection

formula


4 3 24 1 1 1

2 12 3 2

L x x xy

EI L L L

= +

4 1 1 1

( )

2 12 3 2

Ly L

EI

= = +

4

8

L

EI

=

R k

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Remark:

In the above example, we assume that

( )W x = for all x

In tutorial and past year exam questions,we will consider the following cases:

( ) 2 1 x

W x

L

=

( ) 2 x

W x

L

=

( ) cos2

xW x

L

=

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2.6 Plug flow reactor (PFR) model pp38-48

The plug flow reactor(PFR) model is used todescribe chemical reactions in continuous,

flowing systems.

PFR is like a long tube into which you pushsome mixture of chemicals which move through

the tube while they react with each other.


3.6 PFR model

A

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Assume

Velocity uof flow is constant

Cross-sectional areaA is constant

Temp constant

No mixing upstream or downstream----everything that happens in a small

region of the PER is controlled by

chemical reactions IN that region,and by mixture of chemicals following

in and out ( see next two slides)Chew T S MA1506-14 Chapter 2 103

Now we shall use the following example

to illustrate the idea of PFR model

3.6 PFR model

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to illustrate the idea of PFR model


x

H2O

O2

H2

Assumption: A lot of Oxygen pumped in at velocity u

Hydrogen pumped in at velocity u

Keep pumping

H2

O2

H2O

H2

O2

H2O

H2

O2

3.6 PFR modelQ:At point x, how many molecules of 2H

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Let C(x) be concentration of at point x.


2

H

i.e., C(x)= # of molecules /cubic meter at x2

H

are passing by second?

Then in a time t Let N be the number of 2H

that pass by

Since ( )

volume

volume

number ofnumber of

moleculesmolecules

per

=

we have ( ) ( ( ))N C x A u t = So ( )dN

C x A udt

=

3.6 PFR model

Now consider a small piece of tube called plug

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2H

2H

Now consider a small piece of tube, called plug,

Let the length of plug is x

x

flowing in at x,

at a rate

( )

dN

C x A udt =

flowing out at x+x,

at a rate

( )

dN

C x x Audt = +

2H molecules are being destroyed inside the plug

at a rate ( 2)( )( )r A x

( 2)( )( )r A xWhy What is r ?

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( )( )( )

r is rate per volume at which the chemical reaction

2H2+ O2= 2H2O

happens in each unit of volume

Why minus 2 ? Because we are losing 2H

and the 2 because each reaction costs 2 molecules

Hence ( )( )r A x is the # of reactions happensin the plug per second

being destroyed inside the plug is2

HSo in each second, the # of

( 2)( )( )r A xmolecules

x

A Volume ofplug isA x

fl i i flowing out

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( )C x A u ( )C x x Au+( 2)( )( )r A x

flowing in

at a rate

flowing out

at a rate2H 2H

2H destroyed

at rate

Hence ( ) 2 ( )C x Au rA x C x x Au = +So ( ) ( ) 2

C x x Au C x AurA

x

+ =

Hence ( ) ( ) 2C x x u C x u rx

+ =

( )2

dC xu r

dx =

3.6 PFR model

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It is known that the rate of chemical reaction

r depends on concentration C(x) of andtemperature


2H

We assume temperature is constant

Hence r=k C(x) where k is a constant

( )implies 2 ( )

dC xu kC x

dx

= ( )

2dC x

u r

dx

= So

3.6 PFR model

( ) 2dC x k

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Solve the above ODE, get

How to find B? Let x=0, we get

2

( )

kx

u

C x Be

=

2 0

(0)k

uC Be B

= =

( ) 2( )

dC x k C x

dx u =

Therefore2

( ) (0)

kx

C C

3.6 PFR model

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Therefore


( ) (0) uC x C e=

END

( ) 2 ( )C x Au rA x C x x Au = +

Remark: In the above, we assumearea A of cross section is constant

and we have

If A is not constant, then we will have

( ) ( ) 2 ( ) ( ) ( )C x A x u rA x x C x x A x x u = + +

In this case, following the same idea as above, we will have

( ) ( ) 2 ( )d C x A xu rA xdx

= ( ) ( )implies 2 ( ) ( )d C x A xu kC x A xdx

=

Appendix1

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pp


Hence

cos( )

cos( ) cos( ) sin( ) sin( )

A t

A t A t

= +

By the above formula, get

cos( )

sin( )

C A

D A

=

= where

cos( )A t

cos( ) cos cos sin sin + = +

cos( ) sin( )C t D t = +

Appendix 2 (Defintions of cosh and sinh)

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Appendix 2 (Defintions of cosh and sinh)


Formulae

cosh2

x x

e ex

+= sinh2

x x

e ex

=

2 2(cosh ) (sinh ) 1x x =

sinhcosh

d xxdx =

cosh

sinh

d x

xdx =

Appendix 2 (cont) Graphs of sinhx, coshx, tanhx

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sinhxcoshxtanhx

http://www.graphmatica.com

sinhx

tanhx

In slide 36, we compute t using cosh function as follows:

Appendix 2 (cont)
http://www.graphmatica.com/http://www.graphmatica.com/

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Suppose now

1/ cosh (2)t L g =

p g

( ) 2t =

( )( ) 2 cosh /t g L t = = Hence

Find time t such that

2/ 100 / secg L=

9.8L= centimeters

Hence

11 cosh (2) 0.132sec100

t =

Appendix 2 (cont)

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Suppose we use the following exp. function instead of cosh

( / ) ( / )

2g L t g L t e e = +

Then

Let We get 4 =+ 1

Solve the above eq , get the value x,

and then get the value t (choose positive t).

Hence using cosh is easier.

2 3x=

( ) ( )/ /2

2

g L t g L t e e

= +

( )/g L tx e=

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Wsin cosx B t C t = +

(cont)

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We guess

Then subst. into

0cosmx kx F t + =

get

sin cosp

x B t C t = +

0

20,

FB C

k m= =

0

2 2

/F m

=

0

2 2/cos( ) cosF mx A t t

= +

,p px x

Assume

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Appendix 4 (cont)0cosmx kx F t + =

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By the initial conditions in previous slide,

get


0

2 2

/F m

A = 0=

0

So0

2 2

/

[cos cos ]

F m

x t t =

Appendix 5

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0 cos( )mx bx kx F t + + = p

xWe shall find a particular soln

sin cospx B t C t = +We guess

Then subst.

into 0 cos( )mx bx kx F t + + =

, ,p p p

x x x

Appendix 5 (cont.)

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pp ( )


0

2 2 2 2( )

F bB

k m b

=

+

2

0

2 2 2 2

( )

( )

F k mC

k m b

=

+

get

Appendix 5 (cont)

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pp ( )

so2

0 0

2 2 2 2 2 2 2 2

( )sin cos

( ) ( )p

F b F k mx t t

k m b k m b

= +

+ +

END

Chapter 2

chewma1506-14 ch2

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