1) Ordinary differential equations relate a dependent variable to one or more independent variables by means of differential coefficients. They can be classified based on order, degree, whether they are linear or non-linear, and type (exact, separable variables, homogeneous).
2) First order differential equations can sometimes be solved by separation of variables, or by finding an integrating factor. Homogeneous equations can be transformed by substitution.
3) Second order linear differential equations can be reduced to a system of two first order equations. The complementary function and particular solutions combine to form the general solution. Unequal or equal roots of the characteristic equation determine the form of the complementary function.
This ppt covers the topic of B.Sc.1 Mathematics,unit - 5 , paper - 2, calculus- Introduction of Linear differential equation of second order , complete solution in terms of known integral belonging to the complementary function.
Partial Differential Equation plays an important role in our daily life.In mathematics, a partial differential equation (PDE) is a differential equation that contains unknown multivariable functions and their partial derivatives. PDEs are used to formulate problems involving functions of several variables, and are either solved by hand, or used to create a computer model. A special case is ordinary differential equations (ODEs), which deal with functions of a single variable and their derivatives.
PDEs can be used to describe a wide variety of phenomena such as sound, heat, diffusion, electrostatics, electrodynamics, fluid dynamics, elasticity, or quantum mechanics. These seemingly distinct physical phenomena can be formalised similarly in terms of PDEs. Just as ordinary differential equations often model one-dimensional dynamical systems, partial differential equations often model multidimensional systems. PDEs find their generalisation in stochastic partial differential equations.
This ppt covers the topic of B.Sc.1 Mathematics,unit - 5 , paper - 2, calculus- Introduction of Linear differential equation of second order , complete solution in terms of known integral belonging to the complementary function.
Partial Differential Equation plays an important role in our daily life.In mathematics, a partial differential equation (PDE) is a differential equation that contains unknown multivariable functions and their partial derivatives. PDEs are used to formulate problems involving functions of several variables, and are either solved by hand, or used to create a computer model. A special case is ordinary differential equations (ODEs), which deal with functions of a single variable and their derivatives.
PDEs can be used to describe a wide variety of phenomena such as sound, heat, diffusion, electrostatics, electrodynamics, fluid dynamics, elasticity, or quantum mechanics. These seemingly distinct physical phenomena can be formalised similarly in terms of PDEs. Just as ordinary differential equations often model one-dimensional dynamical systems, partial differential equations often model multidimensional systems. PDEs find their generalisation in stochastic partial differential equations.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
2. 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”.
xey
dx
dy
dx
yd x
cos44)( 2
3
3
Ordinary differential equation --------
only one independent variable involved: x
)( 2
2
2
2
2
2
z
T
y
T
x
T
k
T
Cp
Partial differential equation ---------------
more than one independent variable involved: x, y, z,
3. 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.
xey
dx
dy
dx
yd x
cos44)( 2
3
3
3rd order O.D.E.
1st degree O.D.E.
4. 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.
xey
dx
dy
dx
yd x
cos44)( 2
3
3
Product between two derivatives ---- non-linear
xy
dx
dy
cos4 2
Product between the dependent variable themselves ---- non-linear
5. 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
6. Exact equations
• Exact? 0),(),( dyyxNdxyxM
dFdy
y
F
dx
x
F
General solution: F (x,y) = C
For example
0)2(cossin3
dx
dy
yxxyx
7. 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
x
dx
dy
y sin
8. 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
y
f
dx
dy
is termed a homogeneous differential equation
of the first order.
9. 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 +HCl
C6H5Cl+Cl2 C6H4Cl2 + HCl
C6H4Cl2 + Cl2 C6H3Cl3 + HCl
10. Take a basis of 1 mole of benzene fed to the reactor and introduce
the following variables to represent the stage of system at time ,
p = moles of chlorine present
q = moles of benzene present
r = moles of monochlorbenzene present
s = moles of dichlorbenzene present
t = moles of trichlorbenzene present
Then q + r + s + t = 1
and the total amount of chlorine consumed is: y = r + 2s + 3t
From the material balances : in - out = accumulation
d
dt
Vpsk
d
ds
Vpskprk
d
dr
Vprkpqk
d
dq
Vpqk
3
32
21
10
1)(
1
2
q
r
k
k
dq
dr
u = r/q
11. 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”.
QPy
dx
dy
where P and Q are functions of x only
Assuming the integrating factor R is a function of x only, then
)(Ry
dx
d
dx
dR
y
dx
dy
R
RQyRP
dx
dy
R
PdxR exp is the integrating factor
12. Example
Solve
2
3
exp
2
4 x
y
dx
dy
xy
Let z = 1/y3
4
3
ydy
dz
dx
dy
ydx
dz
4
3 2
3
exp33
2
x
dx
dz
xz
integral factor
2
3
exp3exp
2
x
xdx
3
2
3
exp
2
3
exp3
22
x
dx
dzx
xz
3
2
3
exp
2
x
z
dx
d
Cx
x
z 3
2
3
exp
2
Cx
x
y
3
2
3
exp
1 2
3
13. 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.
14. First O.D.E. in heat transfer
An elevated horizontal cylindrical tank 1 m diameter and 2 m long is insulated with
asbestos lagging of thickness l = 4 cm, and is employed as a maturing vessel for a
batch chemical process. Liquid at 95 C is charged into the tank and allowed to
mature over 5 days. If the data below applies, calculated the final temperature of the
liquid and give a plot of the liquid temperature as a function of time.
Liquid film coefficient of heat transfer (h1) = 150 W/m2C
Thermal conductivity of asbestos (k) = 0.2 W/m C
Surface coefficient of heat transfer by convection and radiation (h2) = 10 W/m2C
Density of liquid ( ) = 103 kg/m3
Heat capacity of liquid (s) = 2500 J/kgC
Atmospheric temperature at time of charging = 20 C
Atmospheric 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.
15. 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 tTAhTT
l
kA
TTAh ssww tTTs 674.0326.0
Considering the thermal equilibrium of the liquid,
input rate - output rate = accumulation rate
d
dT
sVtTAh s )(0 2
)12/cos(235.0235.00235.0 T
d
dT
B.C. = 0 , T = 95
16. 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.
17. 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.
18. Solve ax
dx
dy
x
dx
yd
2
2
Let dx
dy
p 2
2
dx
yd
dx
dp
and therefore
axxp
dx
dp
2
2
1
exp xintegral factor
222
2
1
2
1
2
1
xxx
axexpee
dx
dp
22
2
1
2
1
)(
xx
axepe
dx
d
B
x
Aerfaxy
dxxCaxy
2
2
1
exp 2
error function
19. 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
Let
dx
dy
p
dy
dp
p
dx
dy
dy
dp
dx
dp
dx
yd
2
2
20. Solve 2
2
2
)(1
dx
dy
dx
yd
y
Let dx
dy
p and therefore
2
1 p
dy
dp
yp
Separating the variables
dy
dp
p
dx
yd
2
2
dy
y
dp
p
p 1
12
)1ln(
2
1
lnln 2
pay
)1( 22
ya
dx
dy
p
bay
a
x
ya
dy
x
)(sinh1
)1(
1
22
21. 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
y
f
dx
dy
2
2
dx
yd
x
n
n
n
dx
yd
x 1
x
y
f
dx
dy
dx
dy
x
y
f
dx
yd
x ,2
2 Assuming u = y/x
dx
du
xuf
dx
ud
x ,2
2
2
Assuming x = et
dt
du
uf
dt
du
dt
ud
,2
2
If in this form, called homogeneous 2nd ODE
23. A graphite electrode 15 cm in diameter passes through a furnace wall into a water
cooler which takes the form of a water sleeve. The length of the electrode between
the outside of the furnace wall and its entry into the cooling jacket is 30 cm; and as
a 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 transfer
from 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 is
1500 C, estimate the duty of the water cooler if the temperature of the electrode at
the entrance to the cooler is to be 150 C.
The following additional information is given.
Surrounding temperature = 20 C
Thermal conductivity of graphite kT = k0 - T = 152.6 - 0.056 T W/m C
The temperature of the electrode may be assumed uniform at any cross-section.
x
T
T0
24. 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
dT
Ak
dx
d
dx
dT
Ak
dx
dT
Ak TTT
where U = overall heat transfer coefficient from the electrode to the surroundings
D = electrode diameter
xTT
A
DU
x
dx
dT
k
dx
d
T )( 0
0)()( 00 TT
dx
dT
Tk
dx
d
0)()( 0
2
2
2
0 TT
dx
dT
dx
Td
Tk
dx
dT
p 2
2
dx
Td
dT
dp
p
0)()( 0
2
0 TTp
dT
dp
pTk
25. 0)()( 0
2
0 TTp
dT
dp
pTk
zp2
)( 0TTy
022])[( 0 yz
dy
dz
yTk
Integrating factor 2
00
00
2
exp yTk
yTk
dy
])(32))(([
)(
3
0
2
00
0
TTTTTkC
dTTk
x
26. 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 xyP
dx
dy
P
dx
yd
P
dx
yd
P nnn
n
n
n
where (x) is any function of x.
27. 2nd order linear differential equations
The general equation can be expressed in the form
)(2
2
xRy
dx
dy
Q
dx
yd
P
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
xRv
dx
dv
Q
dx
vd
PRu
dx
du
Q
dx
ud
P
If v is a particular solution of the original differential equation
02
2
Ru
dx
du
Q
dx
ud
P
The general solution of the linear differential equation will be the sum of
a “complementary function” and a “particular solution”.
purpose
28. The complementary function
02
2
Ry
dx
dy
Q
dx
yd
P
Let the solution assumed to be:
mx
meAy mx
mmeA
dx
dy mx
m emA
dx
yd 2
2
2
0)( 2
RQmPmeA mx
m
auxiliary equation (characteristic equation)
Unequal roots
Equal roots
Real roots
Complex roots
29. 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.
xm
eAy 1
1
xm
eAy 2
2
xmxm
eAeAy 21
21
31. 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: mx
Aey
mxmx
mVe
dx
dV
e
dx
dymx
VeyLet
mxmxmx
Vem
dx
dV
me
dx
Vd
e
dx
yd 2
2
2
2
2
2
where V is a function of x 02
2
Ry
dx
dy
Q
dx
yd
P
02
2
dx
Vd
DCxV
mx
edCxy )(
34. 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
xRy
dx
dy
Q
dx
yd
P
35. 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
xRy
dx
dy
Q
dx
yd
P
RCy /
n
n xaxaxaa ...2
210
n
n xxxy ...2
210
rx
ey
36. 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
xRy
dx
dy
Q
dx
yd
P
nxMnxLy cossin
37. Solve 3
2
2
8444 xxy
dx
dy
dx
yd
32
sxrxqxpy
2
32 sxrxq
dx
dy
sxr
dx
yd
622
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 xxxyp
0442
mmmauxiliary equation
x
c eBxAy 2
)(
pcgeneral yyy
38. Method of inverse operators
• Sometimes, it is convenient to refer to the
symbol “D” as the differential operator:
n
n
n
dx
yd
yD
dx
yd
yDDyD
dx
dy
Dy
...
)( 2
2
2
But,
2
2
)(
dx
dy
Dy
y
dx
dy
dx
yd
232
2
yDDyDDyDyyD )2)(1()23(23 22
39. The differential operator D can be treated as an ordinary algebraic
quantity with certain limitations.
(1) The distribution law:
A(B+C) = AB + AC
which applies to the differential operator D
(2) The commutative law:
AB = BA
which does not in general apply to the differential operator D
Dxy 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 D
D(Dy) = (DD)y
D(xy) = (Dx)y + x(Dy)
The basic laws of algebra thus apply to the pure operators, but the
relative order of operators and variables must be maintained.
41. 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.
42. 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
43. Solve x
ey
dx
dy 2
4
differential operator
x
eyD 2
)4(
x
e
D
y 2
)4(
1
1...])
4
1
()
4
1
()
4
1
(1[
4
1 322
DDDey x
binomial expansion
=2
x
ey 2
2
1
pxpx
epfeDf )()(
x
e
D
y 2
)
4
1
1(4
1
2p
x
ey 2
)42(
1
...])
2
1
()
2
1
()
2
1
(1[
4
1 322x
ey
44. x
e
D
y 2
)4(
1 pxpx
epfeDf )()(
2p
x
ey 2
)42(
1
pxpx
e
pf
e
Df )(
1
)(
1
如果 f(p) = 0, 使用因次分析
px
n
px
e
DpD
e
Df )()(
1
)(
1
n
px
px
pDp
e
e
Df )(
1
)()(
1
pxpx
epfeDf )()(
n
px
px
Dp
e
e
Df
1
)()(
1
非0的部分
ypDfeyeDf pxpx
)()(
y = 1, p = 0, 即將 D-p換為 D
n
px
px
D
p
e
e
Df )()(
1integration
!)()(
1
n
x
p
e
e
Df
npx
px
46. Solve
23
2
2
346 xxy
dx
dy
dx
yd
differential operator
232
34)2)(3()6( xxyDDyDD
062
mm
xx
c BeAey 23
)34(
)2)(3(
1 23
xx
DD
yp
expanding each term by binomial theorem
y = yc + yp
)34(
)2(
1
)3(
1
5
1 23
xx
DD
yp
)34(...
16842
1
...
812793
1
5
1 23
3232
xx
DDDDDD
yp
...0
1296
2413
216
)624(7
36
612
6
34 223
xxxxx
yp
47. 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
48. u
C0
x
L
x
u
C
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 section
due to diffusion, but will not vary in the section
following the reactor. (Wehner and Wilhelm, 1956)
x x+ x
Bulk flow of A
Diffusion of A
uC
x
dx
dC
D
dx
d
dx
dC
D
dx
dC
D
x
dx
dC
uuC
Input - Output + Generation = Accumulation
0xkCx
dx
dC
D
dx
d
dx
dC
Dx
dx
dC
uuC
dx
dC
DuC
分開兩個section
Ce
50. a
D
ux
Ba
D
ux
AC 1
2
exp1
2
exp
D
ux
C exp
B. C.
0
0
dx
dC
Lx
dx
Cd
dx
dC
x
B. C.
CCx
CCx
0
0
xL
D
ua
axL
D
ua
a
D
ux
KC
C
2
exp1
2
exp1
2
exp
2
0
DuLaaDuLaaK 2/exp12/exp1
22
if diffusion is neglected (D 0)
u
kx
C
CC
exp1
0
0
51. 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 into
the base of a spray column operated at a temperature 232 C and a pressure of
4.14 MN/m2. 0.519kg/s of water at the same temperature and pressure is sprayed
into the top of the column and descends in the form of droplets through the rising fat
phase. Glycerine is generated in the fat phase by the hydrolysis reaction and is extracted
by the descending water so that 0.701 kg/s of final extract containing 12.16% glycerine
is withdrawn continuously from the column base. Simultaneously 1.121 kg/s of fatty
acid 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 glycerine
equivalent in the entering tallow 8.53% and the distribution ratio of glycerine between
the water and the fat phase at the column temperature and pressure is 10.32, estimate
the concentration of glycerine in each phase as a function of column height. Also find
out 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 is
0.0028 s-1.
Glycerin, 甘油
52. Tallow fat Hot water
G kg/s
Extract
Raffinate
L kg/s
L kg/s
xH
zH
x0
z0
y0
G kg/s
yH
h
x
z
x+ x
z+ z y+ y
y
h
x = weight fraction of glycerine in raffinate
y = weight fraction of glycerine in extract
y*= weight fraction of glycerine in extract in equilibrium with x
z = weight fraction of hydrolysable fat in raffinate
53. Consider the changes occurring in the element of column of height h:
Glycerine transferred from fat to water phase, hyyKaS )*(
S: sectional area of tower
a: interfacial area per volume of tower
K: overall mass transter coefficient
Rate of destruction of fat by hydrolysis, hSzk
A glycerine balance over the element h is:
hyyKaS
w
hSzk
h
dh
dx
xLLx *
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
Lz
Lx
L kg/s
xH
zH
x0
z0
y0
G kg/s
yH
h
x
z
x+ x
z+ z y+ y
y
h
The glycerine equilibrium between the phases is:
mxy*
hyyKaSGyh
dh
dy
yG *
in the fat phase
in the extract phase
00GyGy
in the fat phase
in the extract phase
55. w
mz
ry
r
qhBphAy 0
0
1
1
)exp()exp(
B.C.
0,
0,0
yHh
xh
We don’t really want x here!
Apply the equations two slides earlier (replace y* with mx)
0,
,0 0
yHh
yyh
dh
dy
q
r
ymx
1
We don’t know y0, either
)exp()exp(
1
qhqBphpA
q
r
ymx
0,0 yyh
pHqHqhpHpHqHpHqH
reveeveeer
w
mz
ry
r
revey )()(
1
1
)( 0
0
q
prpq
v
Substitute y0 in terms of other variables
57. 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.
58. Elimination of variable
Independent variable or dependent variables?
),(
),(
2
1
yxf
dt
dy
yxf
dt
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 and
it would be better to make use of matrices
Solving differential equations simultaneously using
matrices will be introduced later in the term
59. Elimination of dependent variables
Solve
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
60. 0)1311)(3( zDD
x
x
BeAez 311
13
0)96()6( 22
zDDyDD
x
AeyDD 11
132
2
9
11
136
11
13
)6(
= E
x
p Ee
DD
y 11
13
2
)6(
1
xx
c JeHey 32
pxpx
epfeDf )()(
11
13
p
x
p Eey 11
13
2
)6)
11
13
()
11
13
((
1
x
p Eey 11
13
700
121
y = yc + yp