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化化化化化化 化化化化 化化化 Lecture 4 Ordinary Differential Equations

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Page 1: Differential Equation

化工應用數學

授課教師: 郭修伯

Lecture 4 Ordinary Differential Equations

Page 2: Differential Equation

Differential equation

• An equation relating a dependent variable to one or more independent variables by means of its differential coefficients with respect to the independent variables is called a “differential equation”.

xeydx

dy

dx

yd x cos44)( 23

3

Ordinary differential equation --------only one independent variable involved: x

)(2

2

2

2

2

2

z

T

y

T

x

Tk

TC p

Partial differential equation ---------------more than one independent variable involved: x, y, z,

Page 3: Differential Equation

Order and degree

• The order of a differential equation is equal to the order of the highest differential coefficient that it contains.

• The degree of a differential equation is the highest power of the highest order differential coefficient that the equation contains after it has been rationalized.

xeydx

dy

dx

yd x cos44)( 23

3

3rd order O.D.E.

1st degree O.D.E.

Page 4: Differential Equation

Linear or non-linear

• Differential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves.

xeydx

dy

dx

yd x cos44)( 23

3

Product between two derivatives ---- non-linear

xydx

dycos4 2

Product between the dependent variable themselves ---- non-linear

Page 5: Differential Equation

First order differential equations

• No general method of solutions of 1st O.D.E.s because of their different degrees of complexity.

• Possible to classify them as:– exact equations– equations in which the variables can be separated– homogenous equations– equations solvable by an integrating factor

Page 6: Differential Equation

Exact equations

• Exact? 0),(),( dyyxNdxyxM

dFdyy

Fdx

x

F

General solution: F (x,y) = C

For example

0)2(cossin3 dx

dyyxxyx

Page 7: Differential Equation

Separable-variables equations

• In the most simple first order differential equations, the independent variable and its differential can be separated from the dependent variable and its differential by the equality sign, using nothing more than the normal processes of elementary algebra.

For example

xdx

dyy sin

Page 8: Differential Equation

Homogeneous equations

• Homogeneous/nearly homogeneous?• A differential equation of the type,

• Such an equation can be solved by making the substitution u = y/x and thereafter integrating the transformed equation.

x

yf

dx

dy

is termed a homogeneous differential equationof the first order.

Page 9: Differential Equation

Homogeneous equation example

• Liquid benzene is to be chlorinated batchwise by sparging chlorine gas into a reaction kettle containing the benzene. If the reactor contains such an efficient agitator that all the chlorine which enters the reactor undergoes chemical reaction, and only the hydrogen chloride gas liberated escapes from the vessel, estimate how much chlorine must be added to give the maximum yield of monochlorbenzene. The reaction is assumed to take place isothermally at 55 C when the ratios of the specific reaction rate constants are:

k1 = 8 k2 ; k2 = 30 k3

C6H6+Cl2 C6H5Cl +HClC6H5Cl+Cl2 C6H4Cl2 + HClC6H4Cl2 + Cl2 C6H3Cl3 + HCl

Page 10: Differential Equation

Take a basis of 1 mole of benzene fed to the reactor and introducethe following variables to represent the stage of system at time ,

p = moles of chlorine presentq = moles of benzene presentr = moles of monochlorbenzene presents = moles of dichlorbenzene presentt = moles of trichlorbenzene present

Then q + r + s + t = 1and the total amount of chlorine consumed is: y = r + 2s + 3tFrom the material balances : in - out = accumulation

d

dtVpsk

d

dsVpskprk

d

drVprkpqk

d

dqVpqk

3

32

21

10

1)(1

2 q

r

k

k

dq

dr

u = r/q

Page 11: Differential Equation

Equations solved by integrating factor

• There exists a factor by which the equation can be multiplied so that the one side becomes a complete differential equation. The factor is called “the integrating factor”.

QPydx

dy where P and Q are functions of x only

Assuming the integrating factor R is a function of x only, then

)(Rydx

d

dx

dRy

dx

dyR

RQyRPdx

dyR

PdxR exp is the integrating factor

Page 12: Differential Equation

Example

Solve

2

3exp

24 x

ydx

dyxy

Let z = 1/y3

4

3

ydy

dz

dx

dy

ydx

dz4

3

2

3exp33

2x

dx

dzxz

integral factor

2

3exp3exp

2xxdx

32

3exp

2

3exp3

22

x

dx

dzxxz

32

3exp

2

xz

dx

dCx

xz

3

2

3exp

2

Cxx

y

3

2

3exp

1 2

3

Page 13: Differential Equation

Summary of 1st O.D.E.

• First order linear differential equations occasionally arise in chemical engineering problems in the field of heat transfer, momentum transfer and mass transfer.

Page 14: Differential Equation

First O.D.E. in heat transfer

An elevated horizontal cylindrical tank 1 m diameter and 2 m long is insulated withasbestos lagging of thickness l = 4 cm, and is employed as a maturing vessel for abatch chemical process. Liquid at 95 C is charged into the tank and allowed tomature over 5 days. If the data below applies, calculated the final temperature of theliquid and give a plot of the liquid temperature as a function of time.

Liquid film coefficient of heat transfer (h1) = 150 W/m2CThermal conductivity of asbestos (k) = 0.2 W/m CSurface coefficient of heat transfer by convection and radiation (h2) = 10 W/m2CDensity of liquid () = 103 kg/m3

Heat capacity of liquid (s) = 2500 J/kgCAtmospheric temperature at time of charging = 20 CAtmospheric temperature (t) t = 10 + 10 cos (/12)Time in hours ()Heat loss through supports is negligible. The thermal capacity of the lagging can be ignored.

Page 15: Differential Equation

T

Area of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2

Tw

Ts

Rate of heat loss by liquid = h1 A (T - Tw)Rate of heat loss through lagging = kA/l (Tw - Ts)Rate of heat loss from the exposed surface of the lagging = h2 A (Ts - t)

t

At steady state, the three rates are equal:

)()()( 21 tTAhTTl

kATTAh ssww tTTs 674.0326.0

Considering the thermal equilibrium of the liquid,

input rate - output rate = accumulation rate

d

dTsVtTAh s )(0 2

)12/cos(235.0235.00235.0

Td

dT

B.C. = 0 , T = 95

Page 16: Differential Equation

Second O.D.E.

• Purpose: reduce to 1st O.D.E.• Likely to be reduced equations:

– Non-linear• Equations where the dependent variable does not occur explicitly.

• Equations where the independent variable does not occur explicitly.

• Homogeneous equations.

– Linear• The coefficients in the equation are constant

• The coefficients are functions of the independent variable.

Page 17: Differential Equation

Non-linear 2nd O.D.E. - Equations where the dependent variables does not occur explicitly

• They are solved by differentiation followed by the p substitution.

• When the p substitution is made in this case, the second derivative of y is replaced by the first derivative of p thus eliminating y completely and producing a first O.D.E. in p and x.

Page 18: Differential Equation

Solve axdx

dyx

dx

yd

2

2

Let dx

dyp 2

2

dx

yd

dx

dpand therefore

axxpdx

dp

2

2

1exp xintegral factor

222

2

1

2

1

2

1xxx

axexpeedx

dp

22

2

1

2

1

)(xx

axepedx

d

Bx

Aerfaxy

dxxCaxy

2

2

1exp 2

error function

Page 19: Differential Equation

Non-linear 2nd O.D.E. - Equations where the independent variables does not occur explicitly

• They are solved by differentiation followed by the p substitution.

• When the p substitution is made in this case, the second derivative of y is replaced as

Letdx

dyp

dy

dpp

dx

dy

dy

dp

dx

dp

dx

yd

2

2

Page 20: Differential Equation

Solve 22

2

)(1dx

dy

dx

ydy

Let dx

dyp and therefore

21 pdy

dpyp

Separating the variables

dy

dpp

dx

yd

2

2

dyy

dpp

p 1

12

)1ln(2

1lnln 2 pay

)1( 22 yadx

dyp

bayax

ya

dyx

)(sinh1

)1(1

22

Page 21: Differential Equation

Non-linear 2nd O.D.E.- Homogeneous equations

• The homogeneous 1st O.D.E. was in the form:• The corresponding dimensionless group containing the 2nd differential

coefficient is• In general, the dimensionless group containing the nth coefficient is• The second order homogenous differential equation can be expressed in a

form analogous to , viz.

x

yf

dx

dy

2

2

dx

ydx

n

nn

dx

ydx 1

x

yf

dx

dy

dx

dy

x

yf

dx

ydx ,

2

2 Assuming u = y/x

dx

duxuf

dx

udx ,

2

22

Assuming x = et

dt

duuf

dt

du

dt

ud,

2

2

If in this form, called homogeneous 2nd ODE

Page 22: Differential Equation

Solve2

222

222

dx

dyxy

dx

ydyx

Dividing by 2xy

2

2

2

2

1

2

1

dx

dy

y

x

x

y

dx

ydx

homogeneous

2

2

2

2

dt

du

dt

udu

dx

dy

x

yf

dx

ydx ,

2

2

Let uxy 2

22

22 22

dx

dux

dx

duux

dx

udux

Let tex

dt

dup

22 pdu

dpup

2)ln( CxBxy

Axy

Singular solution

General solution

Page 23: Differential Equation

A graphite electrode 15 cm in diameter passes through a furnace wall into a watercooler which takes the form of a water sleeve. The length of the electrode betweenthe outside of the furnace wall and its entry into the cooling jacket is 30 cm; and asa safety precaution the electrode in insulated thermally and electrically in this section,so that the outside furnace temperature of the insulation does not exceed 50 C.If the lagging is of uniform thickness and the mean overall coefficient of heat transferfrom the electrode to the surrounding atmosphere is taken to be 1.7 W/C m2 of surface of electrode; and the temperature of the electrode just outside the furnace is1500 C, estimate the duty of the water cooler if the temperature of the electrode atthe entrance to the cooler is to be 150 C.The following additional information is given.

Surrounding temperature = 20 CThermal conductivity of graphite kT = k0 - T = 152.6 - 0.056 T W/m CThe temperature of the electrode may be assumed uniform at any cross-section.

x

T

T0

Page 24: Differential Equation

x

T

T0

The sectional area of the electrode A = 1/4 x 0.152 = 0.0177 m2

A heat balance over the length of electrode x at distance x from the furnace is

input - output = accumulation

0)()( 0

xTTDUx

dx

dTAk

dx

d

dx

dTAk

dx

dTAk TTT

where U = overall heat transfer coefficient from the electrode to the surroundingsD = electrode diameter

xTTA

DUx

dx

dTk

dx

dT )( 0

0)()( 00

TT

dx

dTTk

dx

d

0)()( 0

2

2

2

0

TT

dx

dT

dx

TdTk

dx

dTp

2

2

dx

Td

dT

dpp

0)()( 02

0 TTpdT

dppTk

Page 25: Differential Equation

0)()( 02

0 TTpdT

dppTk

zp 2 )( 0TTy

022])[( 0 yzdy

dzyTk

Integrating factor 200

00

2exp yTk

yTk

dy

])(32))(([

)(3

02

00

0

TTTTTkC

dTTkx

Page 26: Differential Equation

Linear differential equations

• They are frequently encountered in most chemical engineering fields of study, ranging from heat, mass, and momentum transfer to applied chemical reaction kinetics.

• The general linear differential equation of the nth order having constant coefficients may be written:

)(... 11

1

10 xyPdx

dyP

dx

ydP

dx

ydP nnn

n

n

n

where (x) is any function of x.

Page 27: Differential Equation

2nd order linear differential equationsThe general equation can be expressed in the form

)(2

2

xRydx

dyQ

dx

ydP

where P,Q, and R are constant coefficients

Let the dependent variable y be replaced by the sum of the two new variables: y = u + v

Therefore

)(2

2

2

2

xRvdx

dvQ

dx

vdPRu

dx

duQ

dx

udP

If v is a particular solution of the original differential equation

02

2

Ru

dx

duQ

dx

udP

The general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”.

purpose

Page 28: Differential Equation

The complementary function

02

2

Rydx

dyQ

dx

ydP

Let the solution assumed to be: mxmeAy mx

mmeAdx

dy mx

m emAdx

yd 22

2

0)( 2 RQmPmeA mxm

auxiliary equation (characteristic equation)

Unequal rootsEqual rootsReal rootsComplex roots

Page 29: Differential Equation

Unequal roots to auxiliary equation

• Let the roots of the auxiliary equation be distinct and of values m1 and m2. Therefore, the solutions of the auxiliary equation are:

• The most general solution will be

• If m1 and m2 are complex it is customary to replace the complex

exponential functions with their equivalent trigonometric forms.

xmeAy 11 xmeAy 2

2

xmxm eAeAy 2121

Page 30: Differential Equation

Solve 0652

2

ydx

dy

dx

yd

auxiliary function

0652 mm

3

2

2

1

m

m

xx BeAey 32

Page 31: Differential Equation

Equal roots to auxiliary equation

• Let the roots of the auxiliary equation equal and of value m1 = m2 = m. Therefore, the solution of the auxiliary equation is: mxAey

mxmx mVedx

dVe

dx

dymxVey Let

mxmxmx Vemdx

dVme

dx

Vde

dx

yd 22

2

2

2

2

where V is a function of x 02

2

Rydx

dyQ

dx

ydP

02

2

dx

VdDCxV

mxedCxy )(

Page 32: Differential Equation

Solve 0962

2

ydx

dy

dx

yd

auxiliary function

0962 mm

321 mm

xeBxAy 3)(

Page 33: Differential Equation

Solve 0542

2

ydx

dy

dx

yd

auxiliary function

0542 mm

im 2

xixi BeAey )2()2( )sincos(2 xFxEey x

Page 34: Differential Equation

Particular integrals

• Two methods will be introduced to obtain the particular solution of a second linear O.D.E.– The method of undetermined coefficients

• confined to linear equations with constant coefficients and particular form of (x)

– The method of inverse operators• general applicability

)(2

2

xRydx

dyQ

dx

ydP

Page 35: Differential Equation

Method of undetermined coefficients

• When (x) is constant, say C, a particular integral of equation is

• When (x) is a polynomial of the form where all the coefficients are constants. The form of a particular integral is

• When (x) is of the form Terx, where T and r are constants. The form of a particular integral is

)(2

2

xRydx

dyQ

dx

ydP

RCy /

nn xaxaxaa ...2

210

nn xxxy ...2

210

rxey

Page 36: Differential Equation

Method of undetermined coefficients

• When (x) is of the form G sin nx + H cos nx, where G and H are constants, the form of a particular solution is

• Modified procedure when a term in the particular integral duplicates a term in the complementary function.

)(2

2

xRydx

dyQ

dx

ydP

nxMnxLy cossin

Page 37: Differential Equation

Solve 32

2

8444 xxydx

dy

dx

yd

32 sxrxqxpy 232 sxrxq

dx

dy

sxrdx

yd62

2

2

3322 84)(4)32(4)62( xxsxrxqxpsxrxqsxr

Equating coefficients of equal powers of x

84

0124

4486

0442

s

sr

qrs

pqr

32 26107 xxxy p

0442 mmmauxiliary equation

xc eBxAy 2)(

pcgeneral yyy

Page 38: Differential Equation

Method of inverse operators

• Sometimes, it is convenient to refer to the symbol “D” as the differential operator:

n

nn

dx

ydyD

dx

ydyDDyD

dx

dyDy

...

)(2

22

But, 2

2)(

dx

dyDy

ydx

dy

dx

yd23

2

2

yDDyDDyDyyD )2)(1()23(23 22

Page 39: Differential Equation

The differential operator D can be treated as an ordinary algebraicquantity with certain limitations.

(1) The distribution law:A(B+C) = AB + ACwhich applies to the differential operator D

(2) The commutative law:AB = BAwhich does not in general apply to the differential operator DDxy xDy(D+1)(D+2)y = (D+2)(D+1)y

(3) The associative law:(AB)C = A(BC)which does not in general apply to the differential operator DD(Dy) = (DD)yD(xy) = (Dx)y + x(Dy)

The basic laws of algebra thus apply to the pure operators, but therelative order of operators and variables must be maintained.

Page 40: Differential Equation

Differential operator to exponentials

pxpx

pxnpxn

pxpx

epfeDf

epeD

peDe

)()(

...

pxpx eppeDD )23()23( 22

ypDfeyeDf

ypDeyeD

ypDeyeD

ypDeyDeDyeyeD

pxpx

npxpxn

pxpx

pxpxpxpx

)())((

)()(

...

)()(

)()(22

More convenient!

Page 41: Differential Equation

Differential operator to trigonometrical functions

ipxnipxnipxnn eipeDeDpxD )Im(ImIm)(sin

pxpppxD

pxppxD

pxpppxD

pxppxD

pxipxe

nn

nn

nn

nn

ipx

sin)()(cos

cos)()(cos

cos)()(sin

sin)()(sin

sincos

212

22

212

22

where “Im” represents the imaginary part of the function which follows it.

Page 42: Differential Equation

The inverse operator

The operator D signifies differentiation, i.e.

)()( xfdxxfD )()( 1 xfDdxxf

•D-1 is the “inverse operator” and is an “intergrating” operator.•It can be treated as an algebraic quantity in exactly the same manner as D

Page 43: Differential Equation

Solve xeydx

dy 24

differential operator

xeyD 2)4(

xeD

y 2

)4(

1

1...])4

1()

4

1()

4

1(1[

4

1 322 DDDey x

binomial expansion

=2

xey 2

2

1

pxpx epfeDf )()(

xeD

y 2

)41

1(4

1

2p

xey 2

)42(

1

...])2

1()

2

1()

2

1(1[

4

1 322 xey

Page 44: Differential Equation

xeD

y 2

)4(

1

pxpx epfeDf )()(

2p

xey 2

)42(

1

pxpx epf

eDf )(

1

)(

1

如果 f(p) = 0, 使用因次分析

pxn

px eDpD

eDf )()(

1

)(

1

n

pxpx

pDp

ee

Df )(

1

)()(

1

pxpx epfeDf )()(

n

pxpx

Dp

ee

Df

1

)()(

1

非 0 的部分

ypDfeyeDf pxpx )()(

y = 1, p = 0, 即將 D-p 換為 D

npx

px Dp

ee

Df

)()(

1

integration

!)()(

1

n

x

p

ee

Df

npxpx

Page 45: Differential Equation

Solvexxey

dx

dy

dx

yd 42

2

6168

differential operator

xxeyDyDD 422 6)4()168(

01682 mm

xc eBxAy 4)(

xp xe

Dy 4

2)4(

6

246 Dxey xp

f(p) = 0

)()( pDfeeDf pxpx

integration

!26

24 x

xey xp x

p exy 433

y = yc + yp

Page 46: Differential Equation

Solve23

2

2

346 xxydx

dy

dx

yd

differential operator

232 34)2)(3()6( xxyDDyDD

062 mm

xxc BeAey 23

)34()2)(3(

1 23 xxDD

y p

expanding each term by binomial theorem

y = yc + yp)34(

)2(

1

)3(

1

5

1 23 xxDD

y p

)34(...16842

1...

812793

1

5

1 233232

xxDDDDDD

y p

...01296

2413

216

)624(7

36

612

6

34 223

xxxxx

y p

Page 47: Differential Equation

O.D.E in Chemical Engineering

• A tubular reactor of length L and 1 m2 in cross section is employed to carry out a first order chemical reaction in which a material A is converted to a product B,

• The specific reaction rate constant is k s-1. If the feed rate is u m3/s, the feed concentration of A is Co, and the diffusivity of A is assumed to be constant at D m2/s. Determine the concentration of A as a function of length along the reactor. It is assumed that there is no volume change during the reaction, and that steady state conditions are established.

A B

Page 48: Differential Equation

uC0

x

L

x

uC

A material balance can be taken over the element of length x at a distance x fom the inlet

The concentraion will vary in the entry sectiondue to diffusion, but will not vary in the sectionfollowing the reactor. (Wehner and Wilhelm, 1956)

x x+x

Bulk flow of A

Diffusion of A

uC

xdx

dCD

dx

d

dx

dCD

dx

dCD

xdx

dCuuC

Input - Output + Generation = Accumulation

0

xkCx

dx

dCD

dx

d

dx

dCDx

dx

dCuuC

dx

dCDuC

分開兩個 sectionCe

Page 49: Differential Equation

0

xkCx

dx

dCD

dx

d

dx

dCDx

dx

dCuuC

dx

dCDuC

dividing by x

0

kC

dx

dCD

dx

d

dx

dCu

rearranging

02

2

kCdx

dCu

dx

CdD

auxillary function

02 kumDm

a

D

uxBa

D

uxAC 1

2exp1

2exp

In the entry section0

2

2

dx

Cdu

dx

CdD

auxillary function

02 umDm

D

uxC exp

2/41 ukDa

Page 50: Differential Equation

a

D

uxBa

D

uxAC 1

2exp1

2exp

D

uxC exp

B. C.

0

0

dx

dCLx

dx

Cd

dx

dCx

B. C.

CCx

CCx

0

0

xL

D

uaaxL

D

uaa

D

ux

KC

C

2exp1

2exp1

2exp

2

0

DuLaaDuLaaK 2/exp12/exp1 22

if diffusion is neglected (D0)

u

kx

C

CCexp1

0

0

Page 51: Differential Equation

The continuous hydrolysis of tallow in a spray column 連續牛油水解

1.017 kg/s of a tallow fat mixed with 0.286 kg/s of high pressure hot water is fed intothe base of a spray column operated at a temperature 232 C and a pressure of4.14 MN/m2. 0.519kg/s of water at the same temperature and pressure is sprayedinto the top of the column and descends in the form of droplets through the rising fatphase. Glycerine is generated in the fat phase by the hydrolysis reaction and is extractedby the descending water so that 0.701 kg/s of final extract containing 12.16% glycerineis withdrawn continuously from the column base. Simultaneously 1.121 kg/s of fattyacid raffinate containing 0.24% glycerine leaves the top of the column.If the effective height of the column is 2.2 m and the diameter 0.66 m, the glycerineequivalent in the entering tallow 8.53% and the distribution ratio of glycerine betweenthe water and the fat phase at the column temperature and pressure is 10.32, estimatethe concentration of glycerine in each phase as a function of column height. Also findout what fraction of the tower height is required principally for the chemical reaction.The hydrolysis reaction is pseudo first order and the specific reaction rate constant is0.0028 s-1.

Glycerin, 甘油

Page 52: Differential Equation

Tallow fat Hot water G kg/s

Extract

RaffinateL kg/s L kg/s

xH

zH

x0

z0

y0

G kg/syH

h

xz

x+xz+z y+y

y

h

x = weight fraction of glycerine in raffinatey = weight fraction of glycerine in extracty*= weight fraction of glycerine in extract in equilibrium with xz = weight fraction of hydrolysable fat in raffinate

Page 53: Differential Equation

Consider the changes occurring in the element of column of height h:

Glycerine transferred from fat to water phase, hyyKaS )*(

S: sectional area of towera: interfacial area per volume of towerK: overall mass transter coefficient

Rate of destruction of fat by hydrolysis, hSzk

A glycerine balance over the element h is:

hyyKaSw

hSzkh

dh

dxxLLx

*

Rate of production of glycerine by hydrolysis, whSzk /k: specific reaction rate constant: mass of fat per unit volume of column (730 kg/m3)w: kg fat per kg glycetine

A glycerine balance between the element and the base of the tower is:

00

w

Lz

w

LzLx

L kg/sxH

zH

x0

z0

y0

G kg/syH

h

xz

x+xz+z y+y

y

h

The glycerine equilibrium between the phases is:

mxy *

hyyKaSGyhdh

dyyG *

in the fat phase

in the extract phase

00 GyGy

in the fat phase

in the extract phase

Page 54: Differential Equation

02

2

00

2

dh

dy

L

KaSm

dh

yd

dh

dy

G

KaS

dh

dyy

G

KaS

L

Skyy

L

mG

w

mz

LG

KaSk

L

mGr

L

Skp

)1( rG

KaSq

w

mzry

r

pqpqy

dh

dyqp

dh

yd 002

2

1)( 2nd O.D.E. with constant coefficients

Complementary functionParticular solution

0)(2 pqmqpm Constant at the right hand side, yp = C/R

)exp()exp( qhBphAyc pqw

mzry

r

pqy p /

10

0

Page 55: Differential Equation

w

mzry

rqhBphAy 0

01

1)exp()exp(

B.C.

0,

0,0

yHh

xhWe don’t really want x here!

Apply the equations two slides earlier (replace y* with mx)

0,

,0 0

yHh

yyh

dh

dy

q

rymx

1

We don’t know y0, either

)exp()exp(1

qhqBphpAq

rymx

0,0 yyh

pHqHqhpHpHqHpHqH reveeveeerw

mzry

rrevey

)()(

1

1)( 0

0

q

prpqv

Substitute y0 in terms of other variables

Page 56: Differential Equation

qH

pHqHqH

qH

pHpH

er

revee

er

vee

vrw

mzy

)(0

730

66.04

0028.0

32.10

165.0286.0701.0

276.01216.0701.0~

403.02

519.0286.0~

069.12

121.1017.1~

2

0

S

k

m

y

G

L

Page 57: Differential Equation

Simultaneous differential equations

• These are groups of differential equations containing more than one dependent variable but only one independent variable.

• In these equations, all the derivatives of the different dependent variables are with respect to the one independent variable.

Our purpose: Use algebraic elemination of the variables until only one differential equation relating two of the variables remains.

Page 58: Differential Equation

Elimination of variableIndependent variable or dependent variables?

),(

),(

2

1

yxfdt

dy

yxfdt

dx

Elimination of independent variable

),(

),(

2

1

yxf

yxf

dy

dx

較少用

Elimination of one or more dependent variables

It involves with equations of high order andit would be better to make use of matrices

Solving differential equations simultaneously usingmatrices will be introduced later in the term

Page 59: Differential Equation

Elimination of dependent variablesSolve

0)23()103(

0)96()6(

22

22

zDDyDD

and

zDDyDD

0)1)(2()5)(2(

0)3()2)(3( 2

zDDyDD

and

zDyDD)5( D

)3( D 0)1)(2)(3()5)(2)(3(

0)3)(5()5)(2)(3( 2

zDDDyDDD

and

zDDyDDD

0)23()158()3(

0)1)(2)(3()3)(5(

22

2

zDDDDD

zDDDzDD

0)1311)(3( zDD

Page 60: Differential Equation

0)1311)(3( zDD

xxBeAez 311

13

0)96()6( 22 zDDyDD

xAeyDD 11

1322 9

11

136

11

13)6(

= E

x

p EeDD

y 11

13

2 )6(

1

xxc JeHey 32

pxpx epfeDf )()( 11

13p

x

p Eey 11

13

2 )6)1113

()1113

((

1

x

p Eey 11

13

700

121 y = yc + yp

Page 61: Differential Equation

1.25 kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be cooled in a two-stage counter-current cooler of the following type. Hot acid at 174 C is fed to a tank where it is well stirred incontact with cooling coils. The continuous discharge from this tank at 88 C flows to a secondstirred tank and leaves at 45C. Cooling water at 20 C flows into the coil of the second tank andthence to the coil of the first tank. The water is at 80 C as it leaves the coil of the hot acid tank.To what temperatures would the contents of each tank rise if due to trouble in the supply, the cooling water suddenly stopped for 1h?

On restoration of the water supply, water is put on the system at the rate of 1.25 kg/s. Calculatethe acid discharge temperature after 1 h. The capacity of each tank is 4500 kg of acid and the overall coefficient of heat transfer in the hot tank is 1150 W/m2 C and in the colder tank750 W/m2 C. These constants may be assumed constant.

Example of simultaneous O.D.E.s

Page 62: Differential Equation

1.25 kg/s

0.96 kg/s

88 C

45 C

174 C

20 C40 C80 C

Steady state calculation: Heat capacity of water 4200 J/kg C

)2080(4200)45174(150025.1 waterF skgFwater /96.0

)20(420096.0)4588(150025.1 middleT CTmiddle40

Heat transfer area A1 Heat transfer area A2

32.22

)4088()8088(

ln

)4088()8088(

1150)88174(150025.1 1

T

and

TA

Tank 1 Tank 2

Note: 和單操課本不同

44.68

)4088()80174(

ln

)4088()80174(

T

Page 63: Differential Equation

When water fails for 1 hour, heat balance for tank 1 and tank 2:

dt

dTVCMCTMCT 1

10

dt

dTVCMCTMCT 2

21

Tank 1

Tank 2

M: mass flow rate of acidC: heat capacity of acidV: mass capacity of tankTi: temperature of tank i

dt

dTTT

dt

dTTT

221

110

B.C.t = 0, T1 = 88 teT 861741

t = 1, T1=142.4 C

dt

dTTe t 2

286174 integral factor, et tetT )12986(1742

t = 1, T2 = 94.9 C

B.C.t = 0, T2 = 45

Page 64: Differential Equation

When water supply restores after 1 hour, heat balance for tank 1 and tank 2:

dt

dTVCMCTtWCMCTtWC ww

22213

Tank 1

Tank 2

W: mass flow rate of waterCw: heat capacity of watert1: temperature of water leaving tank 1t2: temperature of water leaving tank 2t3: temperature of water entering tank 2

dt

dTVCMCTtWCMCTtWC ww

11102

1 2T0 T1 T2

t3t2t1

Heat transfer rate equations for the two tanks:

)ln()ln(

)()()(

2111

21111121 tTtT

tTtTAUttWCw

)ln()ln(

)()()(

3222

32222232 tTtT

tTtTAUttWCw

4 equations have to besolved simultaneously

Page 65: Differential Equation

dt

dTVCMCTtWCMCTtWC ww

22213

dt

dTVCMCTtWCMCTtWC ww

11102

)ln()ln(

)()()(

2111

21111121 tTtT

tTtTAUttWCw

)ln()ln(

)()()(

3222

32222232 tTtT

tTtTAUttWCw

觀察各 dependent variable 出現次數 , 發現 t1 出現次數最少,先消去! (i.e. t1= *** 代入 )

211 )1( ttT

322 )1( ttT

wWC

AU

e11

wWC

AU

e22

再由出現次數次少的 t2 消去

…..

30975.708.6 22

22

2

Tdt

dT

dt

Td

代入各數值

…..

B.C. t = 0, T2 = 94.9 C 同時整理 T1

Page 66: Differential Equation

基本上, 1st order O.D.E. 應該都解的出來,方法不外乎:

Check exactSeparate variableshomogenous equations, u = y/xequations solvable by an integrating factor

2nd order 以上的 O.D.E Non-linear O.D.E.

Linear O.D.E. 缺 x 的 O.D.E., reduced to 1st O.D.E.缺 y 的 O.D.E., reduced to 1st O.D.E. homogeneous 的 O.D.E., u = y/x

我們會解的部分

General solution = complementary solution + particular solution

我們會解的部分

constant coefficientsThe method of undetermined coefficientsThe method of inverse operators

尋找特殊解的方法variable coefficient?用數列解, next course