14/15 Semester 5
Computational Method in
Chemical Engineering
(TKK-2109)
Instructor: Rama Oktavian
Email: rama.oktavian@ub.ac.id
Office Hr.: M.13-15, T. 13-15, W. 13-15, F. 13-15
Ordinary 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
Ordinary Differential Equation
Differential equation
Ordinary differential equations are classified in terms
of order and degree
Order of an ordinary differential equation is the same
as the highest order derivative
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.
Ordinary Differential Equation
Differential equation
Ordinary differential equations are classified in terms
of order and degree
3rd order O.D.E.
1st degree O.D.E.
1st order O.D.E.
2nd degree O.D.E.
Ordinary Differential Equation
Linear or non-linear
 Differential equations are said to be
non-linear if any products exist between
the dependent variable and its
derivatives, or between the derivatives
themselves.
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
Ordinary Differential Equation
Swinging pendulum
d 2 g
 sin   0
2
dt
l
A second-order
nonlinear ODE.
Falling parachutist problem
Ordinary Differential Equation
Differential equation
Ordinary differential equations are classified in terms
of order and degree
3rd order O.D.E.
1st degree O.D.E.
1st order O.D.E.
2nd degree O.D.E.
Ordinary Differential Equation
ODE in chemical engineering
Illustrative Example: A Blending Process
An unsteady-state mass balance for the blending system:
rate of accumulation   rate of   rate of 




 of mass in the tank  mass in  mass out 
(2-1)
Ordinary Differential Equation
ODE in chemical engineering
d Vρ 
or
 w1  w2  w
dt
(2-2)
where w1, w2, and w are mass flow rates.
 The unsteady-state component balance is:
d Vρx 
dt
 w1x1  w2 x2  wx
(2-3)
For constant  , Eqs. 2-2 and 2-3 become:

dV
 w1  w2  w
dt
 d Vx 
dt
 w1x1  w2 x2  wx
(2-12)
(2-13)
Ordinary Differential Equation
ODE in chemical engineering
Equation 2-13 can be simplified by expanding the accumulation term using the “chain
rule” for differentiation of a product:

d Vx 
dt
dx
dV
 V
 x
dt
dt
(2-14)
Substitution of (2-14) into (2-13) gives:
dx
dV
V   x
 w1x1  w2 x2  wx
(2-15)
dt
dt
Substitution of the mass balance in (2-12) for  dV/dt in (2-15) gives:
dx
V  x  w1  w2  w   w1x1  w2 x2  wx
dt
(2-16)
Ordinary Differential Equation
ODE in chemical engineering
After canceling common terms and rearranging (2-12) and (2-16), a more convenient
model form is obtained:
dV 1
  w1  w2  w 
dt 
w2
dx w1

x

x

1 
 x2  x 
dt V 
V
(2-17)
(2-18)
Ordinary Differential Equation
ODE in chemical engineering
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
y
dy
 f x, y , y 0  y 0
dx
Slope

Rise
Run

y1  y0
x1  x0
True value
Φ
x0,y0
y1, Predicted
value
Step size, h
 f  x0 , y 0 
y1  y0  f x0 , y0 x1  x0 
 y0  f x0 , y0 h
x
Figure 1 Graphical interpretation of the first step of Euler’s method
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
y
yi 1  yi  f xi , yi h
True Value
h  xi 1  xi
yi+1, Predicted value
Φ
yi
h
Step size
xi
xi+1
x
Figure 2. General graphical interpretation of Euler’s method
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
How does one write a first order differential equation in the form of
dy
 f  x, y 
dx
Example
dy
 2 y  1.3e  x , y 0   5
dx
is rewritten as
dy
 1.3e  x  2 y, y 0   5
dx
In this case
f x, y   1.3e  x  2 y
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
Example
The concentration of salt, x in a home made soap maker
is given as a function of time by
dx
 37.5  3.5 x
dt
At the initial time, t = 0, the salt concentration in the tank is
50g/L. Using Euler’s method and a step size of h = 1.5 min,
what is the salt concentration after 3 minutes.
dx
 37.5  3.5 x
dt
f t , x  37.5  3.5x
xi 1  xi  f t i , xi h
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
For i  0, t0  0, x0  50
Step 1:
Example
x1  x0  f t0 , x0 h
 50  f 0,501.5
 50  37.5  3.5501.5
 50   137.501.5
 156.25 g / L
x1 is the approximate concentration of salt at
t  t1  t0  h  0  1.5  1.5 min
x1.5  x1  156.25g / L
Ordinary Differential Equation
Numerical method for solving ODE
Euler’s method
Example
For i  1, t1  1.5, x1  156.25
Step 2:
x2  x1  f t1 , x1 h
 156.25  f 1.5,156.251.5
 156.25  37.5  3.5 156.251.5
 156.25  584.381.5
 720.31g / L
x 2 is the approximate concentration of salt at
t  t2  t1  h  15.  1.5  3 min
x3  x2  720.31g / L
Ordinary Differential Equation
Numerical method for solving ODE
The exact solution of the ordinary differential equation is
given by
xt   10.714  39.286e 3.5 x
The solution to this nonlinear equation at t=3 minutes is
x3  10.715 g/L
Ordinary Differential Equation
Numerical method for solving ODE
Figure 3. Comparing exact and Euler’s method
Ordinary Differential Equation
Numerical method for solving ODE
Table 1. Concentration of salt at 3 minutes as a
function of step size, h
h
Step
3
1.5
0.75
0.375
0.1875
x3
Et
|t | %
−362.50
720.31
284.65
10.718
10.714
373.22
−709.60
−273.93
−0.0024912
0.0010803
3483.0
6622.2
2556.5
0.023249
0.010082
Ordinary Differential Equation
Numerical method for solving ODE
Figure 4. Comparison of Euler’s method with
exact solution for different step sizes
Ordinary Differential Equation
Numerical method for solving ODE
Runge Kutta 2nd order method
Taylor’s expansion
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Runge Kutta 2nd order method
Taylor’s expansion
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Runge Kutta 2nd order method
However, it is relatively difficult to find second derivative of ODE
For
dy
 f ( x, y ), y (0)  y0
dx
Runge Kutta 2nd order method is given by
yi 1  yi  a1k1  a2 k2 h
where
k1  f xi , yi 
k2  f xi  p1h, yi  q11k1h 
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Heun’s method
Slope  f xi  h, yi  k1h 
y
Here a2=1/2 is chosen
1
a1 
2
p1  1
Slope  f xi , yi 
q11  1
resulting in
1 
1
yi 1  yi   k1  k2 h
2 
2
where
k1  f xi , yi 
yi+1, predicted
Average Slope 
yi
xi
1
 f xi  h, yi  k1h   f xi , yi 
2
xi+1
x
Figure 1 Runge-Kutta 2nd order method (Heun’s method)
k 2  f xi  h, yi  k1h 
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Midpoint Method
Here a2  1 is chosen, giving
a1  0
p1 
1
2
q11 
1
2
resulting in
yi 1  yi  k2h
where
k1  f xi , yi 
1
1


k 2  f  xi  h, yi  k1h 
2
2


http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Ralston’s Method
Here a 2  2 is chosen, giving
3
1
3
3
p1 
4
3
q11 
4
resulting in
a1 
2 
1
yi 1  yi   k1  k 2 h
3 
3
where
k1  f xi , yi 
3
3


k 2  f  xi  h, yi  k1h 
4
4


http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Example
The concentration of salt, x in a home made soap maker is
given as a function of time by
dx
 37.5  3.5 x
dt
At the initial time, t = 0, the salt concentration in the tank is
50g/L. Using Euler’s method and a step size of h=1.5 min,
what is the salt concentration after 3 minutes.
dx
 37.5  3.5 x
dt
f t , x  37.5  3.5x
1 
1
xi 1  xi   k1  k 2 h
2 
2
http://numericalmethods.eng.usf.edu
Solution
Step 1: i  0, t0  0, x0  50
k1  f t0 , xo   f 0,50  37.5  3.550  137.50
k 2  f t0  h, x0  k1h   f 0  1.5,50   137.501.5  f 1.5,156.25
 37.5  3.5 156.25  584.38
1 
1
x1  x0   k1  k 2 h
2 
2
1
1

 50    137.50   584.381.5
2
2

 50  223.44 1.5
 385.16 g / L
x1 is the approximate concentration of
salt at t  t  t  h  0  1.5  1.5 min
1
0
x1.5  x1  385.16 g/L
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Solution
Step 2: i  1, t1  t 0  h  0  1.5  1.5, x1  385.16 g / L
k1  f t1 , x1   f 1.5,385.16  37.5  3.5385.16  1310.5
k 2 f t1  h, x1  k1h   f 1.5  1.5,385.16   1310.51.5  f 3,1580.6
 37.5  3.5 1580.6  5569.8
1 
1
x2  x1   k1  k 2 h
2 
2
1
1

 385.16    1310.5  5569.81.5
2
2

 385.16  2129.6 1.5
x1 is the approximate concentration of
 3579.7 g / L
salt at
t  t2  t1  h  1.5  1.5  3 min
x3  x1  3579.7 g/L
http://numericalmethods.eng.usf.edu
Solution
Table 1. Effect of step size for Heun’s method
Step size, h
3
1.5
0.75
0.375
0.1875
x3
|t | %
Et
1803.1
−1792.4
16727
3579.6
−3568.9
33306
442.05
−431.34
4025.4
11.038 −0.32231
3.0079
10.718 −0.0024979 0.023311
x(3)  10.715 (exact)
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Ordinary Differential Equation
Numerical method for solving ODE
Runge Kutta 4th order method
Taylor’s expansion
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Numerical method for solving ODE
Runge Kutta 4th order method
However, it is relatively difficult to find second and third derivative of ODE
For
dy
 f ( x, y ), y (0)  y0
dx
Runge Kutta 4th order method is given by
http://numericalmethods.eng.usf.edu
Ordinary Differential Equation
Example
The concentration of salt, x in a home made soap maker is
given as a function of time by
dx
 37.5  3.5 x
dt
At the initial time, t = 0, the salt concentration in the tank is
50g/L. Using Euler’s method and a step size of h=1.5 min,
what is the salt concentration after 3 minutes.
dx
 37.5  3.5 x
dt
f t , x  37.5  3.5x
http://numericalmethods.eng.usf.edu
Solution
Step 1: i  0, t0  0, x0  50 g / L
k1  f t0 , x0   f 0, 50  37.5  3.550  137.5
1
1
1
1




k 2  f  t0  h, x0  k1h   f  0  1.5, 50   137.51.5 
2
2
2
2




 f 0.75,  53.125  37.5  3.5 53.125  223.44
1
1
1




k3  f  t0  k 2 h   f  0  1.5, 50  223.44 1.5 
2
2
2




 f 0.75, 217.58  37.5  3.5217.58  724.02
k 4  t0  h, x0  k3h   f 0  1.5, 50   724.031.5
 f 1.5,  1036.0  37.5  3.51036.0  3663.6
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Solution
1
k1  2k2  2k3  k4 h
6
1
 50   137.5  2223.44  2 724.02  3663.6 1.5
6
1
 50  2525.01.5
6
 681.24 g/L
x1  x0 
x1
is the approximate concentration of salt at
t  t1  t0  h  0  1.5  1.5
x1.5  x1  681.24 g / L
http://numericalmethods.eng.usf.edu
Solution
Step 2:
i  1, t1  1.5, x1  681.24 g / L
k1  f t1 , x1   f 1.5, 681.24  37.5  3.5681.24  2346.8
1
1
1
1




k 2  f  t1  h, x1  k1h   f 1.5  1.5,681.24   2346.81.5 
2
2
2
2




 f 2.25,1078.9   37.5  3.5 1078.9  38.13.6
1
1
1
1




k3  f  t1  h, x1  k 2 h   f 1.5  1.5, 681.24  3813.6 1.5 
2
2
2
2




 f 2.25, 3541.4   37.5  3.53541.4   12358
k 4  f t1  h, x1  k3h   f 1.5  1.5, 681.24   123581.5
 f 3,17855  37.5  3.5 17855  62530
http://numericalmethods.eng.usf.edu
Solution
1
x2  x1  k1  2k 2  2k3  k 4 h
6
1
 681.24   2346.8  23813.6  2 12358  625301.5
6
1
 681.24  430961.5
6
 11455 g/L
x 2 is the approximate concentration of salt at
t2  t1  h  1.5  1.5  3 min .
x3  x2  11455g / L
http://numericalmethods.eng.usf.edu
Solution
Table 1 Value of concentration of salt at 3 minutes for
different step sizes
Step size, h
x3
Et
|t | %
3
1.5
0.75
0.375
0.1875
14120
11455
25.559
10.717
10.715
−14109
−11444
−14.843
−0.0014969
−0.00031657
131680
106800
138.53
0.013969
0.0029544
x(3)  10.715
(exact)
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