Chemical Engineering Mathematics

CHEMICAL ENGINEERING MATHEMATICS II

ALI ALTWAY

REFERENCESRichard G.Rice, Duong D.Do,”Applied Mathematics and Modeling for chemical Engineers”, John Wiley,New YorkMickley, Reed, Sherwood,”Applied Mathematics in Chemica Engineering”,MsGraw-HillJenson and Jeffrey,”Mathematical Methods in Chemical Engineering”,Academic Press

Course Outline1. Mathematical Formulation of Physicochemical Problems 2. Analytical Series Solution of Ordinary Differential Equation and Special Functions3. Analytical Solution of Partial Differential Equation

Course ObjectiveStudents have capability to apply mathematics to solve Physicochemical Problems.

Evaluation Assignment 20% QUIZ 40%

Exam 40% ------- 100%

MATHEMATICAL FORMULATION OF PHYSICOCHEMICAL PROBLEMS

Objective: Students have capability to develop mathematical formulation of Physicochemical problems and solve the mathematical problems using known mathematical methods.

MATHEMATICAL FORMULATION The mathematical treatment of engineering problems involves three basic steps: the expression of the problem in mathematical language, the solution of mathematical problems, and the interpretation of the results.

Physicochemical Problems

Mathematical Formulation

Solution of Mathematical Formulation

Interpretation

Assumption

Law Conservation

Equilibrium

Analytical

Numerical

FUNDAMENTAL LAWS There are three basic physical and chemical law, they are: Conservation law, Rate expression, and Equilibrium relation.

CONSERVATION LAWS

Rate of mass accumulation in system = Rate of mass in – Rate of mass out

Mass, Overall:

Mass, Component:

Rate of mass accumulation of component i in system = Rate of mass of Component i in – Rate of mass of component i out +Rate of mass generation of Component i

Rate of Energy Accumulation = Rate of Energy in – Rate of Energy out + Rate of Energy Generation

Energy

Rate of Momentum Accumulation = Rate of Momentum in – Rate of Momentum Out + Rate of Momentum Generation

Momentum

Body Force Surface Force

RATE EXPRESSION

FUNDAMENTAL LAWS

Heat Transfer

Conduction

Convection (Interface Transport)

Mass Transfer

Diffusion :

Convection (Interface Transport):

TkAq x

)TT(hAq fs

CSDN A

)CC(SkN AbASCA

Molecular (Newtonian) :

Interface Transport

Chemical Reaction

Momentum Transfer

cCbBaA BAA CkCr

EQUILIBRIUM RELATION

FUNDAMENTAL LAWS

Phase Equilibrium : Vapor-liquid ------- Raoult Law Liquid-liquid Gas/vapor-solid Liquid-solid

Chemical Equilibrium

cCbBaA

LUMP PARAMETER AND DISTRIBUTED PARAMETER MODELS

Mathematical models can be classified into two distinct types: Lump parameter and Distributed parameter models. The first type is characterized by the uniformity of the parameter value in the system such as mixed flow reactor, while the second type is characterized by the variability of the variable/parameter value in the system such as plug flow reactor. The distributed parameter model is usually called Transport Phenomena models, because it involves the phenomena of heat, mass or momentum transport. For the distributed parameter models, we have to consider more specifically on the boundary condition of the system.

GENERAL TYPE OF BOUNDRY CONDITION

1. The temperature at a surface may be specified2. The heat flux at a surface may be given, e.g. q=qo. When a surface is assumed completely isolated, then the heat flux at that surface is equal to zero. 3. At solid-fluid interface the heat flux may related to the difference between the temperature at the interface and that in the fluid, thus

fluidTThq

4. At solid-solid interface, the continuity of temperature and the normal component of the heat flux may be specified.5. At plane, axes or point of symmetry, the heat flux is equal to zero, except the symmetry is treated as the heat source.

h=heat transfer coefficient

HEAT TRANSFER

1. The concentration at a surface can be specified.2. The mass flux at a surface can be specified.3. If diffusion is occurring in a solid, it may happen that at the solid surface substance A is lost to a surrounding fluid stream according to the relation,

MASS TRANSPORT

AfAcA CCkN 00

4. The rate of chemical reaction at the surface can be specified. For example, if a substance A disappears at a surface by a first-order chemical reaction,

AA CkN "10

5. At the plane, axes, or point of symmetry the mass flux is equal to zero.

kc = mass transfer coefficient

1. At solid-fluid interfaces the fluid velocity equals the velocity with which the surface itself is moving.2. At liquid-gas interface, the momentum flux (hence the velocity gradient) is equal to zero.3. At liquid-liquid interfaces the momentum flux perpendicular to the interface, and the velocity, are continuous across the interface.4. At the plane, axes, or point of symmetry the momentum flux is equal to zero.

MOMENTUM TRANSPORT

Gas-liquid interface

Solid-Liquid interfaceliquid

GENERAL STEPS

1. Draw the sketch of the system to be modeled and label/define the various geometric, physical and chemical quantities.2. Carefully select the important variables, and list the parameters that are expected to be important3. Establish a ”control volume” for a differential or finite element of the system to be modeled.4. Write the conservation law on the control volume and use the necessary rate expression and equilibrium relation to derive equations describing the system. 5. Write boundary and initial condition6. Solve the equations7. Interpret the solution

ILLUSTRATIVE EXAMPLES (1)

Two tanks each contains 100 liters salt solution (20 gr/lit). A stream of water is fed into the first tank at a rate of 5 liters/min. The liquid flows from the tank to the second tank at a rate of 8 liters/min. The liquid flows from the second tank at a rate of 8 liters/min where part of it (3 liters/min) is directed to the first tank and the balance flows to some points out of the system. Determine the salt concentration (gr/lit) in the first and second tank as a function of time. Assume is constant in all streams.

5 lt/min

8 lt/min

5 lt/min

SOLUTION

0835Vdt

1211 830,5 CCCVdt

121 C8C3

088Vdt

2122 C8C8CVdt

212 C8C8

dC5.12CC 2

V1= const = 100V2 = const = 100

Tank I:Consevation of mass (overall)

Conservation of mass (salt)

Tank II

Consevation of mass (overall)

Conservation of mass (salt)

Cd5.12

dC100C8C3

Cd1250

dC100 2

Cd250 2

t129.02

t031.012 eKeKC

tt eKeKdt

dC 129.02

031.01

2 129.0031.0

tt eeC 129.031.02 33.63261.26

tt eeC 129.0031.01 875.3125.16

Initial Condition: t = 0 C1=20, C2 = 20

Eq (2) is differentiated with respected to t:

Eq. (2) is substituted into Eq. (1):

m1 = - 0.031 ; m2 = - 0.129 :

Initial Condition:

t = 0 C2=20 02 dt

C2 = 20 = K1 + K2

0 = - 0.031 K1 – 0.129 K2

K1 = 26.33

K2 = - 6.33

(a) A container is maintained at a constant temperature of 800o F and is fed with a pure gas A at a steady rate of 1 lbmole/min; the gas product gas stream is withdrawn from the container at the rate necessary to keep the total pressure constant at a value of 3 atm. The container contents are vigorously agitated, and the gas mixture is always well mixed. The following irreversible second order gas phase reaction occurs in the container:

At a temperature of 800 oF, the reaction rate constant for the reaction has the numerical value of 1000 ft3 /(lbmole min ). Both A and B are perfect gases. Because of their low temperature, no reaction occurs in the lines to and from the vessel. If under steady state condition, the product stream is to contain 33 1/3 mole % B, how large ( in cubic feet ) should be the volume of the reaction container ?(b) After the steady state of a) has been attained, the valve on the exit pipe ofisothermal vessel is abruptly closed. The feed rate is controlled so that the total tank pressure is maintained at 3 atm. If the mixing is still perfect, how many minutes will it take (after the instant of closing the valve) for the tank content to be 90 mole % B. The feed rate is controlled so that the total tank pressure is maintained at 3 atm.

2A → B

P = 3 atmT = 800oC

Product

ILLUSTRATIVE EXAMPLES (2)Solution

BfAf F2F10

Mass Balance:

Bf AfBf FF 2

AfAf FF10

5.0AfF

nk5.010

VPn AA

223)3/2()1000(5.010

2Ay 8.105

)3()3/2)(1000(

)1260( )7302.0)(5.0(22

dn AB2

VPn AA

A KtV2

0345.0)1260)(7302.0(

8.105*3*1.0An

23.0 2287.0)1260)(7302.0(

8.105*3*)3/2(

K023.0

348.4t)8.105(*2

348.4726.41

348.4t726.40345.0

b) Mass Balance:

348.4K

t=?→ nA = 0.0345t=5.213

Solution

nkF AA

The apparatus shown diagrammatically in fig 1 is to be used for the continuous extraction of benzoic acid from toluene, using water as the extracting solvent the two stream are fed into a tank A where they are stirred vigorously, and the mixture is then pumped into tank B where it is allowed to settle into two layers. The upper toluene layer and the lower water layer are discharged separately and the problem is to find what proportion of the benzoic acid has passed into the solvent phase.

Toluene + benzoic acid water

Water+ benzoic acid

Toluene + benzoic acid

R m3/s TolueneC kg/m3 benzoic acid

R m3/s Toluene X kg/m3 benzoic acid

S m3/s water S m3/s water Y kg/m3 benzoic acid

The mixture is so efficient that the two stream leaving the stage are always in equilibrium with one another. This can be expressed mathematically : y = m x

Mass balance :Benzoic acid :

R . C + S . 0 = R . X + S . Y

R . C = R . X + S . m X SmR

Solution:

Proportion of benzoic acid extracted is :

Two Stages :

Stage 1 Stage 2R R R

The above eample will now be reconsidered, but two stages will be used for the extraction.

Mass balance :Stage 1 : R C + S Y2 = R X1 + S Y1

R C + S m X2 = R X1 + S m X1 ……………(1)

Stage 2 : R X1 + S 0 = R X2 + S Y2

= R X2 + S m X2 …………………(2) From Eq.(1), we obtain, X2=((R+Sm)X1-RC)/Sm And from EQ.(2), we obtain , X2=RX1/(R+Sm) From the last two equations we obtain, X1=RC(R+Sm)/(R2+RSm+S2m2) And the proportion of Benzoic acid extracted is, E=Sm(R+Sm)/(R2+RSm+S2m2)

No gram of solid material was placed in W gram of water at time to . The liquid was continuously stirred and maintained at a constant temperature. At the end of t1 second, N1 gram of solid remained undissolved. At the end of very long period of time, N2 gram of solid remained undissolved. Set up the differential equation required to determined the rate of solution of the solid in term of No, N1, N2, and to, t1 . Do not integrate your expression.Notes:It is assumed that the rate of solution is proportional to (1) the surface area of the material and (2) the concentration driving force , where the concentration is expressed as gram of solid per gram of water. The original solid consisted of S sphere each of initial diameter Do ft.

Mass balance:

System solid : ]CC[A.k0]N[dt

NNC 2o*

A = D2 S ;

Solution:

3oo N/NDD 3 2

o2o )N/N()D)((

A = D2 S = S

NNS )N/N(DK

dN o2o3 2o

Consider the case of starting the equilibrium still. The still is initially charged with 20 lb mole of feed stock of composition xF =0.32 mole fraction of benzene. Feed is supplied at the rate of 10 lb mole/hr, and the heat input is adjusted so that the total mole of liquid in the still remain constant at 20 . It is desired to estimate the time required for the composition of overhead product YD to fall to 0.4 moles fraction of benzene. Benzene and toluene may be assumed to follow Raoult’s law, and the relative volatility may be taken as constant at an average value of 2.48.

Source of heat

Distillate

Solution:

Mass balance

Benzene

dWD10 D = 10

x.dW WW

(10)(0.32)-DYD =

PB = PB* XB

Hub antara YD dan XW

PT = PT* XT

P = PB + PT = PB* XB + PT

* (1-XB)

P)X1(X

X).1(1

W 2048.11

48.2102.3

WW 2048.11

8.2492.52.3

W dtdXX

W 032.0 1.202.3

)48.11(20W

W dXX 1.202.3

X 6.29

X 1.202.3

dX 20t

K)32.0( 1.202.3ln()1.20(

X6.29)X1.202.3ln(

for: t = 0 Xw = 0.32 K)32.0* 1.202.3ln()1.20(

32.06.29)32.0*1.202.3ln(

)32.0* 1.202.3ln(

)1.20(

32.06.29)32.0*1.202.3ln(

)32.0* 1.202.3

X 1.202.3ln(

)1.20(

32.0X6.29)

32.0*1.202.3

X1.202.3ln(

YD = 0.4 Xw = 0.21

At Xw = 0.21t = 1.58 hr

A jacketed vessel of A m2 heating surface is heated by steam condensed at TS oC. The vessel is filled with M kg liquid [heat capacity = C joule /(kg oC)] at To oC. The value of overall heat transfer coefficient is U watt/ (m2 oC). Obtain the temperature of liquid in the tank as a function of time.

Heat balance : TTUAMCTdt

dt UATT

t A UTT

UAtoSS e )TT(TT

Solution

Disk shaped catalyst is used to accelerate the following irreversible first-order reaction :

The concentration of A at the catalyst surface is CAS. The effective diffusivity A trough the catalyst is DA a) Determine concentration distribution of A in the catalyst b) Derive an equation to estimate the mass transfer rate of A into the catalyst

A → B

CASZZZAZN

Solution

ZSkCNN0 AZZZAZZZAZ

Mass Balance:a)

ZSkCZS

)NN(0 A

ZZAZZZZAZ

As Z→0

AAA eKeKC

A AA eD

ASA ee

BZAZ dZ

dCSN AD

k tanh

AAAA S

B.C.1: 0,0 dZ

B.C.2: ASA CCBZ ,

A copper fin L ft long is rectangular in cross section. It is W ft thick and B ft wide. The base of this metal is maintained at a constant temperature TB and the fin loses heat by convection to the surrounding air which is at a temperature To. The surface coefficient of heat transfer is h Btu/hr (ft2 oF).Determine the temperature distribution in the fin?Derive the formula to estimate heat losses from the fin to the surrounding air!

x qx x qx x+x

Solution:

Accumulation = Input – output

)TT(xhB2qq0 oxxxxx

)TT(xhB2

for x 0

)TT(hB2dx

x )TT(hB2dx

)TT(kW

Td0 o2

= T-To

xx KeKe 0kW

General Solution

x = 0 T = TB = TB –To TB -To = K1 + K2

1 eKeKdx

LL eKeK 210

L212 eKK L2

11oB eKKTT

oBoB2 e1

oB ee1

)xL2(xoB

0ox )TT(hBdx2dq

dx)TT(hB2qL

dxeee1

)TT(hB2

)xL2(xL2

xL2xL2

)TT(hB2

)TT(hB2 LL2L

L2LLL2

oBx Leee

)TT(hB2q

b) Heat transfer rate

SERIES SOLUTION OF ORDINARY DIFFERENTIAL EQUATION AND SPESIAL FUNCTION

Objective:1.Students have ability to get series solution of ordinary differential equation with variable coefficients2.Student has ability to identify special functions and have ability to apply them in solving some physicochemical problems.

POWER SERIES

22)10 ................)(()( 000

nn xxAxxAAxxA

Such a series is said to “converge” if it is approaches a finite value as n approaches infinity. The simplest test of convergence is the ratio test; if the absolute value of the ratio of the (n+1)st term to the nth term in any infinite series approaches a limit J as n , then the series converges for J<1, diverges for J>1 , and the test fails for J =1.

Convergence:

1lim 001

xxLJxx

1. Within the interval of convergence of the original power series, series formed by termwise differentiation and integration of the original series is convergent.2. The product of two power series converges inside the common interval of convergence of the original series.3.The ratio of two power series converges inside the common interval of convergence of the original series, provided that the devisor does not vanish in common interval.

Properties:

POWER SERIES

Taylor Series

The power series may be put into the useful form known as “ Taylor series “

xfxfy )(

)()( 0

A function which can be represented by a Taylor series or the completely equivalent power series about xo is said to be “ regular “ at x = xo.

SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE

COEFFICIENT

dxa x y

2 1 1 0 ( ) ( )

Standard form

The behavior of the series solutions in the neighborhood of a point xo can be predicted from the behavior of the functions a1(x) and a2 (x) near xo. It is customary to classify the point xo as follows:

1. xo is termed an “ ordinary “ point of the differential equation if both a1(x) and a2 (x) can be represented by convergent power series which include x = xo in the interval of convergence, i.e., if a1(x) and a2(x) are regular at x = xo.2. xo is termed a “singular “ point of the differential equation if either a1(x) or a2(x) fails to prove regular at x = xo.3. xo is termed a “ regular singular “ point of the differential equation if 2 holds but the products (x – xo)a1(x) and (x - xo)2a2(x) both prove to be regular at x = xo4. xo is termed an “ irregular singular “ point of the differential equation if 2 holds but 3 fails.

EXAMPLE – 1 :

Locate and identify the singular points of the differential equation

x xd y

dxy2 2 2

21 2 1 0( ) ( )

Solution :

dx x xy

2 2 2 2 2 2 2

( ) ( )d y

dx x xy

2 2 2 2 2 2 2

( ) ( ) ( )

dx x x x

dx x xy

2 2 2 2 2

( )( ) ( )

)1)(1.(

222 )1(

COEFFICIENT

1. x0 = 0 : a1(x) and a2(x) : non regular

)1)(1.(

regular

x0 = 0 : regular singular point

2. x0 = 1 : a1(x) and a2(x) : non regular

2)1.(.

2)()1( 1

regular

x0 = 1 : regular singular point

3. x0 =-1 : a1(x) and a2(x) : non regular

)1)(1(

2)()1( 1

xxxxax

non regular

regular

x0 = -1 : irregular singular point

COEFFICIENT

Solution (continued):

1.If xo is an ordinary point of (2-11), two linearly independent power-series solutions which are regular at x = xo will be obtained. Each solution will be of the form.

nn xxAy

2. If xo is a regular singular point of (2-11), a power-series solution which is regular at x = x o cannot be guaranteed. However, the method of the next section will always generate at last one of the form.

s xxAxxy

In which s is a number whose value is determined in the course of the analysis

3. If xo is an irregular singular point of (2-11), a power-series solution may or may not exist.

COEFFICIENT

EXAMPLE – 2 :

Obtain the general solution of the following differential equation, valid near x = 0.

Solution : Function a1(x) = x and a2(x) = 1 are regular for xo, so point xo is ordinary, then the power series solution as

nn xAxy

dxnA xn

dxn n A xn

( ) ( )

n n A x x n A x A xnn

.( ). . . . .

[2A2 + 6A3.x + 12A4.x2 + 20A5.x3 + 30A6.x4 + 42A7.x5 + ...] + [A1.x + 2A2.x2 + 3A3.x3 + 4A4.x4 + 5A5.x5 + ...] + [A0 + A1.x + A2.x2 + A3.x3 + A4.x4 + A5.x5 + ...] = 0

COEFFICIENT

From identity :2A2 + A0 = 0 A2 = -1/2 A06A3 + 2A1 = 0 A3 = -1/3 A112A4 + 3A2 = 0 A4 = -1/4 A2 = 1/8 A020A5 + 4A3 = 0 A5 = -1/5 A3 = 1/15 A130A6 + 5A4 = 0 A6 = -1/6 A4 = -1/48 A042A7 + 6A5 = 0 A7 = -1/7 A5 = -1/105 A1

20 ...

)1()`(

xxxxAxn

COEFFICIENT

METHOD OF FROBENIUS.

The method proceeds to find solution which is valid in the region of the point x = 0, Solution valid in the region of point x = xo may be obtained by transformation of the differential equation by use of the new variable z = x – xo. In the following discussion it is assumed that such a transformation has already been accomplished. The Standard Form:

yxVxdx

Assumtions:

1.R(0)=1

2.R(x), P(x), and V(x) are regular at x = 0

kk xRxR

kk xPxP

kk xVxV

COEFFICIENT

s xAxy

R x n s n s A xkk

1( ).( ) P x n s A xkk

( ) 00

kk xAxV

0.)()1).((0 0

snknkkk xAVPsnRsnsn k + n = l

( ).( ) ( ). .l s k l s k R l s k P V Ak k k l kk

[s(s – 1 ) R o + sPo + Vo] Ao = 0

s2 + (Po – 1)s + Vo = 0

Form of Solution:

for each fixed value of l between 0 and

0l 00 tok

Indicial Equation

COEFFICIENT

AVPsss

VVsPRssA

)1()1(

)1( 11111

q s n A

f s nn

k n kk

101 tokl

ntoknl 0

kkkkk VksRPksRsq 2

002 1 VsPssf

Recurrence Formula:

SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE COEFFICIENT

Exceptional Cases

1. If s1 and s2 do not differ by zero or a real integer, two independent solution obtained.2. If s1 = s2 only one solution is obtained.3. If s1 - s2 =N, where N is a a real integer, use of the larger value of s (s1) will always given one solution .

s xAxyForm of solution:

4. In all cases in which only one solution

11 )(1

snn xuAxAy

can be found, the second independent solution is of the form

12.)ln().(.

snn xBxxuCy

Substitution into the original differential equation will determine the coefficient Bn in terms of the arbitrary constant C.

COEFFICIENTMETHOD OF FROBENIUS.

Using method of Frobenius, obtain the general solution of the following differential equation, valid near x = 0,

2 1 2 02

dxy ( )

Solution :The differential equation is changed to the Frobenius form:

R xd y

dx xP x

dx xV x y( ). . ( ) . ( ).

dx xx y

R(x) = 1 R(0) = 1 xx

21)( xxV

EXAMPLE – 3 :

COEFFICIENT

1............)( 2210

xRxRxRxRxRn

0..........,1 210 RRR

xxPxPPxPxPk

1..........2

0............,1,2

13210 PPPP

xxVxVVxVxVk

1.............2

0..........,2

1,0 3210 VVxVV

01 002 VsPs 01

121 ssIndicial Equation:

q s n A

f s nn

k n kk

( )1Recurrsion Formula

kkkkk VksRPksRsq 2 002 1 VsPssf

1)01(01

3)01(01

]).5,2().5,2([ 02112 f

]0.2[A

)12(........531

]0.2[A

nxxxxA

Generalisasi:

COEFFICIENT

]).1([ 011 f

for s = s1 = 0 :

[ / . ]

10011111 111

211 VRPRq

1)1)(01(012122 111

211 VRPRq

]).2().2([ 02112 f

Aq A q A q A

f31 2 2 1 3 03 3 3

[ ( ). ( ). ( ). ]

[ / . ]

Generalisasi:

COEFFICIENT

BxxxAxy

5,00 !

1............

5.0021 !

)12(......531

2)()()(

Bxnxxxx

xAxyxyxy

COEFFICIENT

Using method of Frobenius obtain the general solution of the following differential equation, valid near to x = 0

EXAMPLE – 4 :

SOLUTION

R xd y

dx xP x

dx xV x y( ). . ( ) . ( ).

1............)( 2210

xRxRxRxRxRn

2..........2210

xPxPPxPxPk

............. xxVxVVxVxVk

0..........,1 210 RRR

0............,2 3210 PPPP

0..........,1,0,0 3210 VVVV

01 002 VsPs 00122 ss 1,0 21 ss

Indicial Equation:

Recurrsion Formula A

q s n A

f s nn

k n kk

kkkkk VksRPksRsq 2 002 1 VsPssf

for s = s1 = 0 : Aq A

f11 01

[ ( ). ]

Aq A q A

f21 1 2 02 2

[ ( ). ( ). ]

[ . . ]

32 00 10 0

[ . . . ]

AA A A A A A

43 2 1 0 2 00 1 0 0

20 20 120

[ . . . . ]

)1(.)(1

COEFFICIENT

)1(.)(n

xuy c u x x B xnn s

. ( ).ln( ) .

Second Solution:

y c xn

x B xn

.ln( ).( )

nx B n x

.ln( ).( ) .

( )!. .

( )!. .( ).

1 2 2 1

.ln( ).( ) . .( )

( )!. .

nx B n n x

.( ) .( )

( )!. .( ).( ).

2 11 22 2

c xn n

.ln( ).( ) . .( )

( )!. .

1 2 2 1

2 12 1

nx B n n x

.( ) .( )

( )!. .( ).( ).

2 11 22 1

nx B n x

.ln( ).( ) .

( )!. .

( )!. . .( ).

2 12 12 1

x B xn

.ln( ).( )

2 102 1

COEFFICIENT

c xn n

.ln( ).( ) .( )

( )!. .

( ) .( )

x B n n x B xn

.ln( ).( )

( )!.( ). . .

2 11 02 1

cx x x x x x

3 5. . .ln( )

. . .ln( )

c xx x x1

x x x x x x.ln( )

.ln( )

. .ln( )

.ln( )

0...][.......]4230201262[ 55

2432 xBxBxBxBxBBxBxBxBxBxBB

identity : - term : x-1 c = 0- term : x.ln(x) c{-6/3! + 1] = 0 0 = 0- term : x3.ln(x) c(20/5! - 1/3!] = 0 0 = 0so c = 0- term : x0 2B2 + B0 = 0 B2 = -B0 /2- term : x1 6B3 + B1 = 0 B3 = -B1 /6- term : x2 12B4 + B2 = 0 B4 = -B2 /12 = B0 /24- term : x3 20B5 + B3 = 0 B5 = -B3 /20 = B1 /120

COEFFICIENT

x A Bn

( )!. ( )

COEFFICIENT

BESSEL EQUATION

0)( 222

22 ypx

xJ J x

k k pp

( )( )

0p p integer xJCxJCy PP 21

Integer=n

ln ( )( )!

( ) ( ) ( )!( )!

xYCxJCy nn 21

Bessel’s function of the second kind of order n

Bessel function of the first kind of order p

=0.5772157……

Euler’s constant

COEFFICIENT

xKCxICy nn 21

0)( 222

22 ypx

0p p integer xICxICy PP 21

0p p integer

)(.)(2

1 ixYiixJixK nnn

COEFFICIENT

dxx a bx

dxc dx b a r x b x yr s r r2

22 2 22 1 0 ( ) ( ( ). . )

rxba xs

dZcexy

.. 21)/.(2/)1(

=real integer PPPP JZJZ ,

0p p integer=n

GENERALIZED FORM OF BESSEL’S EQUATION.

General Solution:

=imaginer

integer

p=0, p=integer=n

nPnP YZJZ ,

PPPP IZIZ ,

nPnP KZIZ ,

Find general solution of the following equation:

Example -5

COEFFICIENT

1 babxa r

Solution

1 222 sdcxbxrabxdcx rrs

d real

11 , YZJZ PP

25.012

25.011

xYcxJcxxYcxJcexy

Properties of Bessel Functions.

1.0 2.0 3.0

I0(x)K0(x)

COEFFICIENT

Form of Curve

ppP xp

xJ .!2

0,)!1(2

n xxY ln

ppP xp

xI .!2

0,)!1(2)( 1 nxnxK nnn xxK ln)(0

COEFFICIENT

Approximation for small x value

P 2)( x

COEFFICIENT

Approximation for very large x value

xJ sin2

)(2/1 x

xxJ cos

2)(2/1

xI sinh2

)(2/1 x

xxI cosh

2)(2/1

)()(12

)( 2/32/12/1 xJxJx

nxJ nnn

)()(12

)( 2/32/12/1 xIxIx

nxI nnn

COEFFICIENT

Bessel functions of order equal to half an odd integer

dxx Z x

x Z x Z J Y I

x Z x Z Kp

( )( ); , ,

dxx Z x

x Z x Z J Y K

x Z x Z Ip

( ); , ,

xZ x Z J Y I

xZ x Z K

( )( ) ( ); , ,

( ) ( );

xZ x Z J Y K

xZ x Z I

( )( ) ( ); , ,

( ) ( );

dxI x I x I xp p p( ) ( ) ( )

dxK x K x K xn n n( ) ( ) ( )

COEFFICIENT

Differentiation and Integration Properties of Bessel Function

pZ x Z x Z J Yp p p( ) ( ) ( ) ; , 2 1 1

pI x I xp p p( ) ( ) ( ) 2 1 1 K x

pK x K xn n n( ) ( ) ( ) 2 1 1

integeror 0 =n when

)x(K)x(K

)x(I)x(I

)x(J)1()x(J

COEFFICIENT

Recursion Formula

COEFFICIENT

Values of x for whichJ0(x) = 0

Value of x for which J1(x) = 0

2.4048 3.8317

5.5201 7.0156

8.6537 10.1735

11.7915 13.3237

14.9390 16.4706

COEFFICIENTEXAMPLE – 7

The Wedge-shaped Fin. The differential equation resulting from the analysis of heat flow through and from a wedge-shaped fin was given below.

L sec h2

where y = T - TaT = local fin temperature at xTa = temperature of surrounding airh = heat transfer coefficient , Btu/hr.ft2.oFk = thermal conductivity of fin material,

Btu/hr.ft.oFL = total length of fin, (ft) = half wedge angle of fin

a. Prove equation (1)

b. Determine the temperature distribution in the fin

c. Determine total rate of heat transfer from the fin to the surrounding air

Ta =1000 FTL =2000 F, temperature at x = LL = 1 fth = 2 Btu/(hr)(ft2 )(0F)k = 220 Btu/(hr)(ft)(0F)Sec =1W = 1/12 ft

COEFFICIENT

Solution

Heat balance on an element of Δx thickness:

0sec220 axxxxxxxconvxxxxxxx TTWxhqqqqq

axxxxxxx

0sec2 ax TTWh

dTWbkqx

aTTWhdx

)x2(Kc)x2(IcTTy 0201a

)x2(IcTT 01a

Boundary Condition:

BC1: x=0 → y=finite

BC2: x=L=1 → y = TL- Ta

=200-100=100

218.01.1.220

12.1.2.2

COEFFICIENT

Lh sec2

General Solution:

10101 230.1934.0 218.02100200 cIcIc

100934.02.81 xIT o

2.811 c

COEFFICIENT

aa TThWdATThq0 0

dx sec 2)()(

dxxIWhcqL

1 )]2([sec2

)L2(IL

L)z(zI

1dz)]z(zI

1dx)]x2(I[ 1

)2() (sec2

)1)(218.0(2)1)(218.0(

)1)(1)(2)(1)(2.81)(2()2(

) (sec211

1 ILIL

Special FunctionGamma Function

( ) ,n t e dt nn t

( ) ( )n n n 1

( ) ( )n t e dt n t n t e dt n t e dt n nn t n t n tt e

1 00 0 00

(3.5) = 2.5(2.5) = 2.5 x 1.5(1.5) = 2.5 x 1.5 x 0.5(0.5); (0.5) = = 1.76= 2.5 x 1.5 x 0.5 x 1.76 = 3.30; (0.5) = = 1.76 dan ; (1) = 1

(n) = n!

N 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

(n) 9.51

2.99 2.22

(n) = (n+1)/n (-0.5) = (0.5)/-0.5 = -3.5222.

For n = 0 or negative integer, the value of (n) equal to infinite: (n) = , where for n = negative even number, (n) = and for n = negative odd number : (n) = - .

Beta Function

Special Function

11 )1(, dxxxnm nm

)().(,

dyyayanma

nmnm 0

111 )(,

dyyynm nmn

1 )1(,

dnm nm 2/

1212 sin.cos2,

Bentuk bentuk Fungsi Betta:

y= a x

Error Function

Special Function

n dnexerf0

X Er. f(x) X Er. f(x)

0.0 0.0000 1.6 0.9763

0.2 0.2227 1.8 0.9891

0.4 0.4284 2.0 0.9953

0.6 0.6039 2.2 0.9981

0.8 0.7421 2.4 0.9993

1.0 0.8427 2.6 0.9998

1.2 0.9103 2.8 0.9999

1.4 0.9523

32222 )2(

x > 2.8,

PARTIAL DIFEERENTIAL EQUATION

yFu f x y

22 ( , )

Partial differential equation is a differential equation containing an unknown function ( or dependent variable) and several independent variables. So, the partial differential equation involves partial derivatives

The “order” of a partial differential equation is the order of the highest derivative appearing in the equation. A partial differential equation is said to be linear if no powers or product of the dependent variable or its partial derivatives are present.

If all terms in the partial differential equation have the same order than the equation is said to be homogeneous

Linear second order partial differential equation with constant coefficients:

Homogeneous linear second order partial differential equation with constant coefficients:

The second order partial differential equation can be classified as :- Elliptic, if B2-AC < 0

- Hyperbolic, if B2 - AC > 0 - Parabolic, if B2 - AC = 0

Two dimensional Laplace equation or Potential equation is Elliptic

A = 1, B = 0, and C = 1, therefore B2 - AC = 0 - 1.1 = -1 < 0.

One dimensional Heat equation is Parabolic

*A = c2 , B = 0, and C = 0, therefore B2 - AC = 0 – c2 .0 = 0.

One dimensional Wave equation is Hyperbolic

A = C2, B = 0, and C = -1, therefore B2 - AC = 0 – c2 (-1) = c2 > 0.

CLASSIFICATION OF PARTIAL DIFFERENSIAL EQUATION

The solution of a partial differential equation in a region R of its independent variables is a function where all the partial derivatives are exist in this region and the function satisfies the equation at all points in this region. The function has to be continue at boundary of R. Such functions are very numerous. A unique solution is limited by boundary condition and initial condition.

Theorem 1 : If u1 , u2 , ..., uk are the solution of Eq.(2) then u = c1u1 + c2u2 + ... + ck uk , where c1 , c2 ,..., ck are constants , is also the solution.

Theorem 2 : If u1 , u2 , ..., un , ... are the solution of Eq.(2) then

nn uCu is also the solution.

SOLUTION OF PARTIAL DIFFERENTIAL EQUATION

There two kinds of problems: Initial value problems and Boundary value problems. For the Boundary value problem, the region is finite and the boundary condition is set at all part of region boundary, while for initial value problem, the region is not finite.

There are some methods to solve the partial differential equation. The following are some of the methods: Laplace Transform. Separation of Variables. Combination of Variables.

TYPES OF PROBLEMS AND METHODS OF SOLUTION

Laplace Transform

Generally Laplace Transform method is used to solve Initial value Problems.Steps of solution are:1) Operate Laplace Transform on P.D.E and the boundary conditions using the initial conditions. This operation results an ordinary differential equation of

dependent variable in Laplace domain.2) Solve the equation in (1) to obtain an expression of dependent variable in

Laplace domain.3) Operate inverse transform on the expression of dependent variable obtained in

Example 3.1:Heat transfer in semi infinite wall.

A very thick slab (slab with infinite thick) initially at temperature of T0 through-out. Suddenly one of the slab surface is contacted with hot liquid at temperature Ts. Determine the temperature distribution in the slab.

PARTIAL DIFEERENTIAL EQUATION Solution :The heat transfer phenomena in the slab is described by the following partial differential equation (Prove it!):

Initial condition: T(x,0) = T0 (E1-2)Boundary conditions: 1. T(0,t) = Ts (E1-3)

2. T(,t) = T0 (E1-4)

Step 1 :Operate Laplace trasform on Equation(E1-1):

(E1-1)

2. s T T x

dx. ( , ) 0 2

Operate Laplace transform on The boundary conditions ( Eq.(E1-3) and Eq.(E1-4) 1. L{T(0,t)} = L{Ts}

ss( , )0

2. L{T(,t)} = L{T0} T sT

s( , ) 0

Step 2 :The general solution of Eq.(E1-5) is:

T K e K eT

1 20. .

From B.C.2 (Eq.(E1-7)) and Eq.(E1-8)) we obtain K1 = 0. Therefore Eq.(E1-8) become:

T K eT

From BC.1 (Eq.(E1-6)) and Eq.(E1-9) the following equation is obtained:

Substitute Eq.(E1-10) into Eq.(E1-9), TT T

Step3 : ),(),( 1 sxTLtxT

xErfTTs

using boundary condition x = 0 erf(0) = 0 and x = erf() = 1 Eq.(E1-12) become:

xErfTTtxT S

PARTIAL DIFEERENTIAL EQUATION SEPARATION OF VARIABLES METHOD

For the application of this method the following conditions have to be satisfied:1. The differential equation is homogeneous.2. The boundary condition is homogeneous.

General steps:1)Perform variable separation to obtain two ordinary differential equations.2)Solve these two equations satisfying boundary conditions to obtain partial solutions.3)Obtain total solution satisfying initial condition.

If the boundary conditions and differential equation are not homogeneous then before the application of this method it is necessary to perform variable transformation so that the boundary condition and the differential equation become homogeneous.For non-homogeneous boundary conditions for instance U(0,t) = U0 and U(L,t) = UL perform the following variable substitution (in case of Cartesian coordinate):

V = U + a + bxV(0,t) = U(0,t) + a

0 = U0 + a a = -U0 ,V(L,t) = U(L,t) + a + bL

0 = UL + a + bL

UUbLbUU L

UUUUV L

UUUVU L

(See Example 3.5 for the explanation of this)

For non-homogeneous differential equation, dependent variable is expressed as the summation of steady state solution ( function of only spatial coordinate) and deviation variable ( function of time and spatial coordinate ).For example consider the following partial differential equation:

tg x t 2

2( , )

with initial condition : T(x,0) = f(x) and boundary condition: T(0,t) = T(L,t) = 0. First, perform the following substitution: T(x,t) = V(x,t) - W(x,t). Then, the differential equation becomes the following differential equations:

where V(x,0) = f(x) and V(0,t) = V(L,t) = 0

tg x t 2

2( , )

where W(x,0) = 0 and W(0,t) = W(L,t) = 0.

Example 3.2 :A rod with isolated side surfaces has initial temperature distribution: T(x,0) = f(x). Suddenly (at t = 0), the two ends of the rod are contacted with iced water ( the temperature is maintained at 0 oC ).

0 oC 0 oC

Determine the temperature of the rod as function of x and t .

Solution:Heat transfer phenomena in the rod can be described as :

Initial condition: T(x,0) = f(x) (E2-2)Boundary condition: T(0,t) = 0 dan T(L,t) = 0 (E2-3)Eq.(E2-1) is homogeneous partial differential equation and the boundary conditions are also homogeneous. Then the method of separation variable can be applied.

Step 1 :T(x,t) = F(x).G(t) (E2-4)

Substitute Eq.(E2-4) into Eq.(E2-1)

F G F G. ' ". 2G

•It can be concluded that both sides of Eq.(E2-5) are constant:

Eq.(E2-6) can be decomposed into 2 equations GCG 2'

0" FCF

(E2-5)

•(E2-6)

•(E2-7)

•(E2-8)

•The constant C may have the following values:

a. C > 0 :The solution of Eq.(E2-8) is: F x K e K eC x C x( ) . .. .

•From boundary conditions: 0 = K1 + K2

LCLC eKeK .21 ..0

•then the values of K1 and K2 are: K1 = 0 and K2 = 0, there we get trivial solution.

b. C = 0 :The solution of Eq.(E2-8) is : F x K K x( ) . 1 2

•From the boundary conditions: 0 = K1

• 0 = K1 + K2.L

•and the value of integration constants: K1 = 0 and K2 = 0, again we obtain trivial

solution.

c. C < 0 :The solution of Eq.(E2-8) is: F x K C x K C x( ) .cos( . ) .sin( . ) 1 2

From the boundary conditions: •0 = K1

)sin(.).cos(.0 21 LCKLCK

•Then the value of integration constants: K1 = 0 dan K2 0. In this case we obtain non-trivial solution, the solution we expected.

•Therefore the value of C must be negative,for instance C = -p2.Then the Eq.(E2-8) become:

0" 2 FpF GpG 22'

Step 2 :The solution of Eq.(E2-9) :The general solution of Eq.(E2-9) is: F(x) = K1.cos (px) + K2.sin (px)From boundary condition: F(0) = 0 , we obtain: 0 = K1 + 0 or K1 = 0 Then the general solution becomes:F(x) = K2.sin (px), and using the other boundary condition: 0 = K2.sin (pL), then sin (pL) = 0. so, pL = n. or p = n./L, where n = 1,2,3, ... .. Finally, the solution of Eq.(E2-9) becomes:

F x Kn x

Ln n( ) .sin. .

*The solution of Eq.(E2-10):

G e p t 2 2. . G t en

Therefore:

)().(,,n

n tGxFtxTtxT

..sin.),(

xnKtxT

..sin.

Step 3 :From initial condition:

..sin.)()0,(

xnAxfxT n

f xn x

( ).sin. .

It can be seen that f(x) is Sinus Fourier series where:

Therefore the overall solution is Eq.(E2-11) with constants An obtained from Eq. (E2-12).

Kombinasi Variabel.

An illustrative example of this method is heat transfer problem in semi infinite (very thick) wall. We define new variable:

Partial differential equation describing this heat transfer problem is:

Initial Condition: t=0 0 < x < ~ T = T0

Boundary Condition: t > 0 x = 0 T = Ts

t > 0 x = ~ T = T0

Solution:

. . ..

4. . ..

. .. .

Substitution :

T R then:

RR 2. .

R 2. .

R K e 1

T K e 1

. T K e K 1

Kondisi-kondisi batas dalam variabel adalah :

= 0 : T = Ts (8) = : T = T0 (9)

dari kondisi batas pers. (8) dan pers. (7) :

Ts = 0 + K2 K2 = Ts, sehingga : T K e Ts 1

T K Erf Ts 1 2. ( )

dari kondisi batas pers. (9) dan pers. (11) :

T K Ts0 1 21 . .( )

K T Ts1 02

Dengan kondisi batas x = 0 erf(0) = 0 dan x = erf() = 1, sehingga pers. (11) menjadi :

xErfTTtxT S

Chemical Engineering Mathematics

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