Transcript 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”.
d 3 y dy 2
x

(
)

4
y

4
e
cos x
3
dx
dx
Ordinary differential equation -------only one independent variable involved: x
T
 2T  2T  2T Partial differential equation --------------C p
 k ( 2  2  2 ) more than one independent variable involved: x, y, z, 

x
y
z
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.
3
d y dy 2
x

(
)

4
y

4
e
cos x
3
dx
dx
3rd order O.D.E.
1st degree O.D.E.
Linear or non-linear
• Differential equations are said to be nonlinear if any products exist between the
dependent variable and its derivatives, or
between the derivatives themselves.
d 3 y dy 2
x

(
)

4
y

4
e
cos x
3
dx
dx
Product between two derivatives ---- non-linear
dy
 4 y 2  cos x
dx
Product between the dependent variable themselves ---- non-linear
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
Exact equations
• Exact?
M ( x, y )dx  N ( x, y )dy  0
F
F
dx 
dy  dF
x
y
General solution: F (x,y) = C
For example
dy
x  y sin x  (cos x  2 y )
0
dx
3
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
dy
y
 sin x
dx
Homogeneous equations
• Homogeneous/nearly homogeneous?
• A differential equation of the type,
dy
 y
 f 
dx
x
is termed a homogeneous differential equation
of the first order.
• Such an equation can be solved by making the
substitution u = y/x and thereafter integrating the
transformed 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 +HCl
C6H5Cl+Cl2  C6H4Cl2 + HCl
C6H4Cl2 + Cl2  C6H3Cl3 + HCl
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
0  k1 pq  V
dq
d
dr
d
ds
k 2 pr  k3 ps  V
d
dt
k3 ps  V
d
k1 pq  k 2 pr  V
u = r/q
dr k 2 r
 ( ) 1
dq k1 q
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”.
dy
 Py  Q
dx
where P and Q are functions of x only
Assuming the integrating factor R is a function of x only, then
dy
R  yRP  RQ
dx
dy
dR d
R y
 (Ry )
dx
dx dx

R  exp  Pdx

is the integrating factor
Example
2


dy

3
x
4
Solve xy 

 y exp 
dx
 2 
Let z = 1/y3
dz
3
 4
dy
y
dz
3 dy
 4
dx
y dx
  3x 2 
dz

3xz 
 3 exp 
dx
2


integral factor
 3x 2 
1
  3x  C
exp 
y3
2




 3x 2 

exp  3xdx  exp 
2


 3x 2  dz
 3x 2 
  exp 
  3
3xz exp 
2
dx
2





 3x 2  
 z exp 

 2    3x  C




 3x 2  
d 
 z exp 
   3



dx 
 2 
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.
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.
Area of tank (A) = ( x 1 x 2) + 2 ( 1 / 4  x 12 ) = 2.5 m2
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)
At steady state, the three rates are equal:
kA
(Tw  Ts )  h2 A(Ts  t )
l
h1 A(T  Tw ) 
Ts  0.326T  0.674t
Considering the thermal equilibrium of the liquid,
input rate - output rate = accumulation rate
t
Ts
T
Tw
0  h2 A(Ts  t )  Vs
dT
d
dT
 0.0235T  0.235  0.235 cos( / 12)
d
B.C.  = 0 , T = 95
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.
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.
Solve
Let
d2y
dy
x
 ax
2
dx
dx
dy
p
dx
and therefore
dp d 2 y

dx dx 2
dp
 xp  ax
dx
integral factor
1 
exp  x 2 
2 
1 2
1 2
x
x
dp 12 x 2
2
2
e  xpe  axe
dx
 1 
y  ax  C  exp   x 2 dx
 2 
 x 
y  ax  Aerf 
 B
 2
error function
d
( pe
dx
1 2
x
2
)  axe
1 2
x
2
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
p
dy
dx
d 2 y dp dp dy
dp
 2 

p
dx
dx dy dx
dy
Solve
Let
d2y
dy 2
y 2 1  ( )
dx
dx
dy
p
dx
and therefore
d2y
dp

p
dx 2
dy
dp
yp  1  p 2
dy
Separating the variables
p
dy
 (a 2 y 2  1)
dx
p
1
dp

dy
2
p 1
y
x
ln y  ln a 
1
ln( p 2  1)
2
dy
(a 2 y 2  1)
 
x  1 sinh 1 (ay )  b
a
Non-linear 2nd O.D.E.- Homogeneous equations
dy
 y
 f 
dx
x
• The homogeneous 1st O.D.E. was in the form:
• The corresponding dimensionless group containing
d2y
nd
the 2 differential coefficient is x 2
dx
• In general, the dimensionless
group containing the
n
n 1 d y
th
x
n coefficient is
dx n
• The second order homogenous differential equation
dy
 y

f
can be expressed in a form analogous to dx  x  , viz.
Assuming u = y/x
Assuming x = et
2
du 

d y
 y dy 
2 d u
x

f
u
,
x


x
 f ,

2
2
dx
dx 

dx
 x dx 
d 2u du
 du 


f
 u,

If in this form, called homogeneous 2nd ODE
dt 2
dt
dt


2
Solve
2
2 d y
2
2  dy 
2x y 2  y  x  
dx
 dx 
2
Dividing by 2xy
d 2 y 1 y 1 x  dy 
x 2 

 
dx
2 x 2 y  dx 
homogeneous
Let
x
2
d2y
 y dy 
 f ,

2
dx
 x dx 
y  Ax
y  ux
2
du
2 d u
2  du 
2ux
 2ux
x  
2
dx
dx
 dx 
Let
2u
y  x( B ln x  C ) 2
2
General solution
x  et
d u  du 
 
2
dt
 dt 
2
Singular solution
2
p
du
dt
dp
2up
 p2
du
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.
T0
T
x
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
dT  
dT d
dT


 (kT A )x  DU (T  T0 )x   0
  kT A
    kT A
dx  
dx dx
dx


where
U = overall heat transfer coefficient from the electrode to the surroundings
D = electrode diameter
d  dT 
DU
k

x

(T  T0 )x
 T

dx  dx 
A
d 
dT 
 (k0  T )
   (T  T0 )  0
dx 
dx 
2
d 2T
 dT 
(k0  T ) 2   
   (T  T0 )  0
dx
 dx 
T0
T
x
p
(k0  T ) p
dT
dx
p
dp d 2T

dT dx 2
dp
 p 2   (T  T0 )  0
dT
(k0  T ) p
dp
 p 2   (T  T0 )  0
dT
p2  z
[( k0  T )  y ]
y  (T  T0 )
dz
 2z  2 y  0
dy


2dy
Integrating factor exp   
  k0  T0  y 2
 k0  T0  y 
x
(k0  T )dT
[C   (k0  T )(T  T0 ) 2  2 3 (T  T0 ) 3 ]
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:
dny
d n 1 y
dy
P0 n  P1 n 1  ...  Pn 1  Pn y   ( x)
dx
dx
dx
where (x) is any function of x.
2nd order linear differential equations
The general equation can be expressed in the form
d2y
dy
P

Q
 Ry   ( x)
dx 2
dx
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
 d 2u
  d 2v

du
dv
P

Q

Ru

P

Q

Rv

 
   ( x)
2
2
dx
dx
 dx
  dx

If v is a particular solution of the original differential equation
 d 2u

du
P

Q

Ru

0
2
dx
dx


The general solution of the linear differential equation will be the sum of
a “complementary function” and a “particular solution”.
purpose
The complementary function
d2y
dy
P

Q
 Ry  0
2
dx
dx
Let the solution assumed to be: y  Am e
mx
dy
 Am memx
dx
Am e mx ( Pm 2  Qm  R)  0
auxiliary equation (characteristic equation)
Unequal roots
Equal roots
Real roots
Complex roots
d2y
2 mx

A
m
e
m
2
dx
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:
y  A1e m1x
y  A2e m2 x
• The most general solution will be
y  A1e m1x  A2e m2 x
• If m1 and m2 are complex it is customary to replace the
complex exponential functions with their equivalent
trigonometric forms.
Solve
d2y
dy

5
 6y  0
2
dx
dx
auxiliary function
m 2  5m  6  0
m1  2
m2  3
y  Ae2 x  Be 3 x
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:
y  Aemx
Let
y  Ve
mx
dy
mx dV
e
 mVemx
dx
dx
where V is a function of x
2
d2y
mx d V
mx dV
2
mx

e

2
me

m
Ve
dx 2
dx 2
dx
d2y
dy
P

Q
 Ry  0
2
dx
dx
y  (Cx  d )e mx
d 2V
0
2
dx
V  Cx  D
Solve
d2y
dy

6
 9y  0
2
dx
dx
auxiliary function
m 2  6m  9  0
m1  m2  3
y  ( A  Bx )e 3 x
Solve
d2y
dy

4
 5y  0
2
dx
dx
auxiliary function
m 2  4m  5  0
m  2i
y  Ae
( 2i ) x
 Be
( 2 i ) x
y  e ( E cos x  F sin x)
2x
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
d2y
dy
P

Q
 Ry   ( x)
2
dx
dx
Method of undetermined coefficients
P
d2y
dy
Q
 Ry   ( x )
2
dx
dx
• When  (x) is constant, say C, a particular integral of
equation is
yC/R
• When  (x) is a polynomial of the form a0  a1 x  a2 x 2  ...  an x n
where all the coefficients are constants. The form of a
particular integral is
y   0  1 x   2 x 2  ...   n x n
• When  (x) is of the form Terx, where T and r are
constants. The form of a particular integral is
y  e rx
Method of undetermined coefficients
P
d2y
dy
Q
 Ry   ( x )
2
dx
dx
• 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
y  L sin nx  M cos nx
• Modified procedure when a term in the particular
integral duplicates a term in the complementary
function.
Solve
d2y
dy

4
 4 y  4 x  8x3
2
dx
dx
auxiliary equation
m 2  4m  4m  0
y  p  qx  rx 2  sx 3
d2y
 2r  6sx
dx 2
dy
 q  2rx  3sx 2
dx
(2r  6sx)  4(q  2rx  3sx 2 )  4( p  qx  rx 2  sx 3 )  4 x  8x3
Equating coefficients of equal powers of x
2r  4q  4 p  0
6s  8r  4q  4
4r  12s  0
4s  8
y p  7  10 x  6 x 2  2 x3
yc  ( A  Bx )e 2 x
y general  yc  y p
Method of inverse operators
• Sometimes, it is convenient to refer to the
symbol “D” as the differential operator:
Dy 
dy
dx
d2y
D ( Dy )  D y 
dx 2
...
2
But,
 dy 
( Dy )   
 dx 
2
2
dny
D y 
dx n
n
d2y
dy

3
 2y
2
dx
dx
D 2 y  3Dy  2 y  ( D 2  3D  2) y  ( D  1)( D  2) y
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.
Differential operator to exponentials
De px  pe px
...
( D 2  3D  2)e px  ( p 2  3 p  2)e px
D n e px  p n e px
f ( D )e px  f ( p )e px
D( ye px )  e px Dy  yDe px  e px ( D  p) y
D 2 ( ye px )  e px ( D  p) 2 y
...
D n ( ye px )  e px ( D  p) n y
f ( D)( ye )  e f ( D  p) y
px
px
More convenient!
Differential operator to
trigonometrical functions
D n (sin px)  D n Im eipx  Im D n eipx  Im( ip ) n eipx
where “Im” represents the imaginary part of the function which follows it.
eipx  cos px  i sin px
D 2 n (sin px)  ( p 2 ) n sin px
D 2 n 1 (sin px)  ( p 2 ) n p cos px
D 2 n (cos px)  ( p 2 ) n cos px
D 2 n 1 (cos px)  ( p 2 ) n p sin px
The inverse operator
The operator D signifies differentiation, i.e.


D  f ( x)dx  f ( x)

f ( x)dx  D 1 f ( x)
•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
Solve
dy
 4 y  e2 x
dx
differential operator
( D  4) y  e2 x
1
y
e2 x
( D  4)
y
f ( D)e px  f ( p)e px
p2
1
e2 x
1
4(1  D )
4
binomial expansion
y
1 2x
1
1
1
e [1  ( D )  ( D ) 2  ( D ) 3 ...]1
4
4
4
4
=2
1
1
1
1
y   e 2 x [1  ( )  ( ) 2  ( ) 3 ...]
4
2
2
2
1
y   e2 x
2
y
1
e2 x
(2  4)
Solve
d2y
dy
4x

8

16
y

6
xe
dx 2
dx
m 2  8m  16  0
differential operator
( D 2  8D  16) y  ( D  4) 2 y  6 xe4 x
yp 
6
4x
xe
( D  4) 2
yc  ( A  Bx )e 4 x
y = yc + yp
f(p) = 0
f ( D)e px  e px f ( D  p)
y p  6 xe4 x D2
integration
2
x
y p  6 xe4 x
2!
y p  3x3e4 x
Solve
d 2 y dy
3
2


6
y

4
x

3
x
dx 2 dx
m2  m  6  0
differential operator
( D 2  D  6) y  ( D  3)( D  2) y  4 x3  3x 2
yp 
yc  Ae3 x  Be 2 x
1
(4 x 3  3x 2 )
( D  3)( D  2)
1 1
1  3
2
yp   

(
4
x

3
x
)

5  (3  D) (2  D) 
y = yc + yp
expanding each term by binomial theorem
  1 D D 2 D3

1  1 D D 2 D 3
y p     

 ...    

 ...  (4 x 3  3x 2 )
5  3 9 27 81
8 16
 2 4

4 x3  3x 2 12 x 2  6 x 7(24 x  6) 13  24
yp  



 0...
6
36
216
1296
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,
A  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.
L
u
C0
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)
u
C
Ce
x
x
x
Bulk flow of A
uC
Diffusion of A
dC
D
dx
x+x
dC
x
dx
dC d 
dC 
D
  D
x
dx dx 
dx 
uC  u
A material balance can be taken over the element of length x at a distance x fom the inlet
Input - Output + Generation = Accumulation

dC  
dC  
dC d 
dC  

uC     D dx    uC  u dx x     D dx  dx   D dx x   kCx  0

 
 

 


dC  
dC  
dC d 
dC  



uC


D

uC

u

x



D


D

 
 

x   kCx  0

dx  
dx  
dx dx 
dx  


dividing by x
 dC   d 
dC  
  u
     D
   kC  0
dx
dx
dx
  
 

rearranging
d 2C
dC
D 2 u
 kC  0
dx
dx
In the entry section
auxillary function


d 2C
dC
D 2 u
0
dx
dx
auxillary function
Dm 2  um  k  0
Dm 2  um  0
 ux
1  a   B exp  ux 1  a 
C  A exp 
 2D

 2D


 ux 
C     exp  
D
a
1  4kD / u 
2

 ux 
C     exp  
D
 ux
1  a   B exp  ux 1  a 
C  A exp 
 2D

 2D

B. C.
B. C.

dC dC
x0

dx dx
dC
xL
0
dx

x   C  C0

x0 C C
C
2
 ux 
 ua
L  x   a  1exp  ua L  x  
 exp 
a  1 exp 
C0 K
 2 D 
 2D

 2D

K  a  1 exp uLa / 2 D   a  1 exp  uLa / 2 D 
2
if diffusion is neglected (D0)
2
C0  C
  kx 
 1  exp 

C0
u


Raffinate
L kg/s
L kg/s
xH
zH
x+x
z+z
h
Tallow fat
x
z
G kg/s
yH
y+y
h
y
Hot water
G kg/s
Extract
x0
z0
y0
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
Consider the changes occurring in the element of column of height h:
Glycerine transferred from fat to water phase, KaS ( y *  y )h
kSzh
Rate of destruction of fat by hydrolysis,
Rate of production of glycerine by hydrolysis, kSzh / w
S: sectional area of tower
k: specific reaction rate constant
a: interfacial area per volume of tower : mass of fat per unit volume of column (730 kg/m3)
K: overall mass transter coefficient
w: kg fat per kg glycetine
A glycerine balance over the element h is:
dx  kSzh 

Lx  L x 
h  
 KaS  y *  y h

dh

in the fat phase
w
dy 

G y 
h   Gy  KaS  y  y *h
dh 

L kg/s
xH
zH
G kg/s
yH
in the extract phase
A glycerine balance between the element and the base of the tower is:
Lz Lz0 

in the fat phase

 Lx 
0
w
w


in the extract phase
h
Gy  Gy   0
x+x
z+z
y+y
h
x
z
y
0
The glycerine equilibrium between the phases is:
y*  mx
x0
z0
y0
kS 2 Ka  mz0 mG
dy   KaS dy d 2 y  KaSm dy
 kS  KaS

 y  y0  

y

0


LG  w
L
L  G
dh   G dh dh 2 
L dh

p
kS
L
r
mG
L
q
KaS
(r  1)
G
mz0 
d2y
dy
pq 

(
p

q
)

pqy

ry

 0

dh 2
dh
r 1 
w 
Complementary function
m2  ( p  q)m  pq  0
yc  A exp(  ph)  B exp( qh)
2nd O.D.E. with constant coefficients
Particular solution
Constant at the right hand side, yp = C/R
yp 
mz0 
pq 
 ry0 
 / pq
r 1 
w 
y  A exp(  ph)  B exp(  qh) 
mz0 
1 
 ry0 

r 1 
w 
B.C.
We don’t know y0, either
h  0, x  0
h  H, y  0
We don’t really want x here!
h  0, y  y0
h  H, y  0
Apply the equations two slides earlier (replace y* with mx)
mx  y 
mx  y 
r 1
 pA exp(  ph)  qB exp( qh)
q
y (veqH  re  pH ) 
q  rp  p
v
q
r  1 dy
q dh

mz 
1 
 qH
 pH
 (e  pH  v)e  qh  ve qH  re  pH
 ry0  0  (r  e )e
r 1 
w 
Substitute y0 in terms of other variables
h  0, y  y0

mz0   pH  e  pH  v   qH  ve qH  re  pH
e  
y
 
e
 qH 
 qH
w(r  v) 
r

e
r

e



1.017  1.121
 1.069
2
0.286  0.519
G~
 0.403
2
0.701 0.1216  0.276
y0 ~
 0.165
0.701  0.286
m  10.32
L~
k  0.0028
S

0.66 2
4
  730



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.
Elimination of variable
Independent variable or dependent variables?
dx
 f1 ( x, y )
dt
dy
 f 2 ( x, y )
dt
Elimination of independent variable
dx f1 ( x, y )

dy f 2 ( x, y )
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
Elimination of dependent variables
Solve
( D 2  D  6) y  ( D 2  6 D  9) z  0
and
( D 2  3D  10) y  ( D 2  3D  2) z  0
( D  3)( D  2) y  ( D  3) 2 z  0
and
( D  2)( D  5) y  ( D  2)( D  1) z  0
 ( D  5)
 ( D  3)
( D  3)( D  2)( D  5) y  ( D  5)( D  3) 2 z  0
and
( D  3)( D  2)( D  5) y  ( D  3)( D  2)( D  1) z  0
( D  5)( D  3) 2 z  ( D  3)( D  2)( D  1) z  0

( D  3)(11D  13) z  0


( D  3) ( D 2  8D  15)  ( D 2  3D  2) z  0
( D  3)(11D  13) z  0
z  Ae

13
x
11
 Be 3 x
( D 2  D  6) y  ( D 2  6D  9) z  0
 13  2  6 13   13 x
( D  D  6) y      
  9 Ae 11
 11   11  
13
 x
1
yp  2
Ee 11
( D  D  6)
2
=E
yc  He  Je
2x
f ( D)e px  f ( p)e px
yp 
3 x
y = yc + yp
p
13
11
1
13
13
((  ) 2  ( )  6)
11
11
x
121 13
yp  
Ee 11
700
Ee

13
x
11
Example of simultaneous O.D.E.s
1.25 kg/s of sulphuric acid (heat capacity 1500 J/kg C) is to be cooled in a two-stage countercurrent cooler of the following type. Hot acid at 174 C is fed to a tank where it is well stirred in
contact with cooling coils. The continuous discharge from this tank at 88 C flows to a second
stirred tank and leaves at 45C. Cooling water at 20 C flows into the coil of the second tank and
thence 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. Calculate
the 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 tank
750 W/m2 C. These constants may be assumed constant.
Tank 1
Tank 2
80 C
40 C
20 C
0.96 kg/s
88 C
1.25 kg/s
45 C
174 C
Heat transfer area A1
Heat transfer area A2
Steady state calculation:
Heat capacity of water 4200 J/kg C
1.25 1500  (174  45)  Fwater  4200  (80  20)
Fwater  0.96kg / s
1.25 1500  (88  45)  0.96  4200  (Tmiddle  20)
Tmiddle  40 C
1.25  1500  (174  88)  1150  A1  T
and
T 
(88  80)  (88  40)
 22.32
 (88  80) 
ln 
 (88  40) 



T 
(174  80)  (88  40)
 68.44
 (174  80) 
ln 
 (88  40) 



When water fails for 1 hour, heat balance for tank 1 and tank 2:
Tank 1
Tank 2
dT1
dt
dT
MCT1  MCT2  VC 2
dt
MCT0  MCT1  VC
dT1
dt
dT2

dt
T0  T1 
T1  T2
B.C.
t = 0, T1 = 88
M: mass flow rate of acid
C: heat capacity of acid
V: mass capacity of tank
Ti: temperature of tank i
T1  174  86e t
t = 1, T1=142.4 C
dT2 integral factor, et
T2  174  (86t  129)e t
174  86e  T2 
dt
t
B.C.
t = 0, T2 = 45
t = 1, T2 = 94.9 C
When water supply restores after 1 hour, heat balance for tank 1 and tank 2:
Tank 1
Tank 2
WC wt2  MCT0   WC wt1  MCT1   VC dT1
dt
WCwt3  MCT1   WCwt2  MCT2   VC dT2
dt
t1
T0
t2
1
T1
t3
2
T2
W: mass flow rate of water
Cw: heat capacity of water
t1: temperature of water leaving tank 1
t2: temperature of water leaving tank 2
t3: temperature of water entering tank 2
Heat transfer rate equations for the two tanks:
 (T1  t1 )  (T1  t2 ) 
WCw (t1  t2 )  U1 A1 

 ln( T1  t1 )  ln( T1  t2 ) 
 (T2  t2 )  (T2  t3 ) 
WCw (t2  t3 )  U 2 A2 

 ln( T2  t2 )  ln( T2  t3 ) 
4 equations have to be
solved simultaneously